This research is supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIP) (No. 2021R1A4A1027378).
Research supported in part by a grant from
the Simons Foundation (#960480, Renming Song)
Research supported in part by the Croatian Science Foundation under the project 4197.
(Zoran Vondraček)
1. Introduction and main results
In this paper, we study both conservative and non-conservative purely discontinuous (self-similar) Markov processes in
the upper half-space of ℝ d {\mathbb{R}}^{d}
with jump kernels of the form J ( x , y ) = | x − y | − d − α ℬ ( x , y ) J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y) ,
α ∈ ( 0 , 2 ) \alpha\in(0,2) . The function ℬ ( x , y ) {\cal B}(x,y) may tend to 0 when x x or y y tends to boundary of the half-space, and so the jump kernel may be degenerate.
The main focus of the paper is to show the existence and continuity of
the transition densities of the processes
(or the heat kernels of the corresponding non-local operators),
and to establish their sharp two-sided estimates.
Heat kernel estimates for non-local operators have been the subject of many papers in the last twenty years,
see
[2 , 4 , 5 , 10 , 12 , 13 , 15 , 16 , 17 , 19 , 24 , 26 ]
and the references therein.
In all of the papers mentioned above,
the function ℬ {\cal B} is assumed to be bounded between two positive constants, which can be viewed as a uniform ellipticity condition for non-local operators. To the best of our knowledge, the current paper is the first one to study
sharp two-sided heat kernel estimates when the jump kernel is degenerate.
Boundary Harnack principle and sharp two-sided Green function estimates for
purely discontinuous Markov processes with degenerate jump kernels, which can be viewed as the elliptic counterpart of results of the current paper, have been recently obtained in [30 , 31 , 32 ] .
The first type of processes we look at are conservative jump processes on
ℝ ¯ + d := { x = ( x ~ , x d ) : x ~ ∈ ℝ d − 1 , x d ≥ 0 } \overline{{\mathbb{R}}}^{d}_{+}:=\{x=(\widetilde{x},x_{d}):\widetilde{x}\in{\mathbb{R}}^{d-1},x_{d}\geq 0\} with jump kernel
J ( x , y ) = | x − y | − d − α ℬ ( x , y ) J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y) , where the function ℬ ( x , y ) {\cal B}(x,y)
is symmetric, homogeneous, horizontally translation invariant, and is allowed to approach 0 at the boundary at arbitrary fixed polynomial rate in terms of
some non-negative parameters β 1 , β 2 , β 3 , β 4 \beta_{1},\beta_{2},\beta_{3},\beta_{4} . The generator of such a process is the non-local operator
L α ℬ f ( x ) = p . v . ∫ ℝ + d ( f ( y ) − f ( x ) ) | x − y | − d − α ℬ ( x , y ) 𝑑 y . L^{{\cal B}}_{\alpha}f(x)=\mathrm{p.v.}\int_{{{\mathbb{R}}}^{d}_{+}}(f(y)-f(x))|x-y|^{-d-\alpha}{\cal B}(x,y)\,dy.
When ℬ ( x , y ) {\cal B}(x,y) is bounded between two positive constants,
the heat kernel estimates for such
processes are of the form min { t − d / α , t J ( x , y ) } \min\{t^{-d/\alpha},tJ(x,y)\} . This was first established in the pioneering work [15 ] , even for the case of metric measure spaces.
This form of the estimates reflects the fact that the main contribution to the heat kernel comes from one (big) jump from x x to y y .
This feature has been observed in all subsequent studies, see
[2 , 16 , 10 , 17 , 19 , 24 ] and the references therein.
In our setting of
jump kernel degenerate at the boundary,
there are two novel features in the heat kernel estimates. The first one appears in the case when the involved parameters satisfy β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} , in which case the heat kernel is comparable to min { t − d / α , t J ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) } \min\{t^{-d/\alpha},tJ(x+t^{1/\alpha}\mathbf{e}_{d},y+t^{1/\alpha}\mathbf{e}_{d})\} , where 𝐞 d = ( 0 ~ , 1 ) \mathbf{e}_{d}=(\widetilde{0},1) . In words, the form of the heat kernel estimates
shows that the main contribution to the heat kernel at time t t comes from
one jump from the point t 1 / α t^{1/\alpha} units above x x to the point t 1 / α t^{1/\alpha} units above y y .
Due to the fact the jump kernel vanishes at the boundary, it is very unlikely that the process will make one (big) jump from (or to) a point very close to the boundary.
The second feature is more striking and indicates a sort of a phase-transition
at the level β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} :
When the parameters satisfy β 2 ≥ α + β 1 \beta_{2}\geq\alpha+\beta_{1} ,
in addition to the already mentioned part min { t − d / α , t J ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) } \min\{t^{-d/\alpha},tJ(x+t^{1/\alpha}\mathbf{e}_{d},y+t^{1/\alpha}\mathbf{e}_{d})\} , the sharp heat kernel estimates include a part
which reflects a significant contribution to the heat kernel coming from two jumps connecting x x and y y .
The precise description is given in Theorem 1.1 .
The second type of processes we look at are the ones described in the previous paragraph but killed either by a critical potential
κ x d − α \kappa x_{d}^{-\alpha} or upon hitting the boundary of
ℝ + d := { x = ( x ~ , x d ) : x ~ ∈ ℝ d − 1 , x d > 0 } {\mathbb{R}}^{d}_{+}:=\{x=(\widetilde{x},x_{d}):\widetilde{x}\in{\mathbb{R}}^{d-1},x_{d}>0\} (the latter happens only when α ∈ ( 1 , 2 ) \alpha\in(1,2) ).
The generator of such a process is the non-local operator
L ℬ f ( x ) = L α ℬ f ( x ) − κ x d − α f ( x ) L^{{\cal B}}f(x)=L^{{\cal B}}_{\alpha}f(x)-\kappa x_{d}^{-\alpha}f(x) .
We study the effect of such killings on the heat kernel.
When ℬ ( x , y ) {\cal B}(x,y) is bounded between two positive constants,
the effect of killing on the heat kernel of the process
is, after intensive research during the last fifteen years, fairly well understood.
In most cases, the heat kernel of the killed process has the so called approximate factorization property:
It is comparable to the product of the heat kernel of non-killed (original) process and
the survival probabilities starting from points x x and y y .
In case of smooth open sets, the survival probability can be expressed in terms of the distance between the point and the boundary, see
[4 , 5 , 8 , 12 , 13 , 21 , 27 ] and the references therein.
In our setting of jump kernel degenerate at the boundary, we establish the same property: The heat kernel of the killed process enjoys the approximate factorization property with
survival probabilities decaying as the q q -th power
of the distance to the boundary, where q q is in one-to-one correspondence with the constant κ ≥ 0 \kappa\geq 0 , see Theorem 1.2 .
When κ = 0 \kappa=0 , this theorem generalizes [14 ] (for half spaces) where the factorization property for censored stable process is established. Due to the quite complicated form of the heat kernel estimates in Theorem 1.1 , obtaining the factorization property in Theorem 1.2 is a formidable task.
We now introduce the precise setup and state
the main results of this paper.
This setup was first introduced in [30 ] and was motivated by the results of [28 , 29 ] on subordinate killed Lévy processes.
In fact, subordinate killed Lévy processes, whose analytical counterparts are fractional powers of Dirichlet fractional Laplacians,
are the main natural examples of Markov processes with jump kernels satisfying the assumptions below.
Let d ≥ 1 d\geq 1 and 0 < α < 2 0<\alpha<2 . Recall that
ℝ + d = { ( x ~ , x d ) ∈ ℝ d : x d > 0 } {\mathbb{R}}^{d}_{+}=\{(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:x_{d}>0\} and ℝ ¯ + d = { ( x ~ , x d ) ∈ ℝ d : x d ≥ 0 } \overline{\mathbb{R}}^{d}_{+}=\{(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:x_{d}\geq 0\} .
Here and below, a ∧ b := min { a , b } a\wedge b:=\min\{a,b\} , a ∨ b := max { a , b } a\vee b:=\max\{a,b\} , and a ≍ b a\asymp b means that c ≤ b / a ≤ c − 1 c\leq b/a\leq c^{-1} for some c ∈ ( 0 , 1 ) c\in(0,1) .
For b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 , let
B b 1 , b 2 , b 3 , b 4 ( x , y ) \displaystyle B_{b_{1},b_{2},b_{3},b_{4}}(x,y)
:= ( x d ∧ y d | x − y | ∧ 1 ) b 1 ( x d ∨ y d | x − y | ∧ 1 ) b 2 \displaystyle:=\left(\frac{x_{d}\wedge y_{d}}{|x-y|}\wedge 1\right)^{b_{1}}\left(\frac{x_{d}\vee y_{d}}{|x-y|}\wedge 1\right)^{b_{2}}
× log b 3 ( e + ( x d ∨ y d ) ∧ | x − y | x d ∧ y d ∧ | x − y | ) log b 4 ( e + | x − y | ( x d ∨ y d ) ∧ | x − y | ) . \displaystyle\qquad\times\log^{b_{3}}\left(e+\frac{(x_{d}\vee y_{d})\wedge|x-y|}{x_{d}\wedge y_{d}\wedge|x-y|}\right)\,\log^{b_{4}}\left(e+\frac{|x-y|}{(x_{d}\vee y_{d})\wedge|x-y|}\right).
(1.1)
Let J ( x , y ) = | x − y | − d − α ℬ ( x , y ) J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y) , where
the function ℬ : ℝ ¯ + d × ℝ ¯ + d → [ 0 , ∞ ) {\cal B}:\overline{{\mathbb{R}}}^{d}_{+}\times\overline{{\mathbb{R}}}^{d}_{+}\to[0,\infty)
will be
assumed to satisfy some or all of the following hypotheses:
(A1) ℬ ( x , y ) = ℬ ( y , x ) {\cal B}(x,y)={\cal B}(y,x) for all x , y ∈ ℝ ¯ + d x,y\in\overline{{\mathbb{R}}}^{d}_{+} .
(A2) If α ≥ 1 \alpha\geq 1 , then there exist θ > α − 1 \theta>\alpha-1 and
C 1 > 0 C_{1}>0 such that
| ℬ ( x , x ) − ℬ ( x , y ) | ≤ C 1 ( | x − y | x d ∧ y d ) θ , x , y ∈ ℝ + d . |{\cal B}(x,x)-{\cal B}(x,y)|\leq C_{1}\left(\frac{|x-y|}{x_{d}\wedge y_{d}}\right)^{\theta},\quad x,y\in{\mathbb{R}}^{d}_{+}.
(A3) (I)
There exist C 2 ≥ 1 C_{2}\geq 1
and β 1 , β 2 , β 3 , β 4 ≥ 0 \beta_{1},\beta_{2},\beta_{3},\beta_{4}\geq 0 , with β 1 > 0 \beta_{1}>0 if β 3 > 0 \beta_{3}>0 , and β 2 > 0 \beta_{2}>0 if β 4 > 0 \beta_{4}>0 , such that
C 2 − 1 B β 1 , β 2 , β 3 , β 4 ( x , y ) ≤ ℬ ( x , y ) ≤ C 2 , x , y ∈ ℝ ¯ + d . C_{2}^{-1}B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x,y)\leq{\cal B}(x,y)\leq C_{2},\quad x,y\in\overline{{\mathbb{R}}}^{d}_{+}.
(II)
There exists C 3 > 0 C_{3}>0 such that
ℬ ( x , y ) ≤ C 3 B β 1 , β 2 , β 3 , β 4 ( x , y ) , x , y ∈ ℝ ¯ + d , {\cal B}(x,y)\leq C_{3}B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x,y),\quad x,y\in\overline{{\mathbb{R}}}^{d}_{+},
where β 1 , β 2 , β 3 , β 4 \beta_{1},\beta_{2},\beta_{3},\beta_{4} are the same constants as in (I).
(A4) For all x , y ∈ ℝ ¯ + d x,y\in\overline{{\mathbb{R}}}^{d}_{+} and a > 0 a>0 , ℬ ( a x , a y ) = ℬ ( x , y ) {\cal B}(ax,ay)={\cal B}(x,y) . In case d ≥ 2 d\geq 2 , for all x , y ∈ ℝ ¯ + d x,y\in\overline{{\mathbb{R}}}^{d}_{+} and z ~ ∈ ℝ d − 1 \widetilde{z}\in{\mathbb{R}}^{d-1} , ℬ ( x + ( z ~ , 0 ) , y + ( z ~ , 0 ) ) = ℬ ( x , y ) {\cal B}(x+(\widetilde{z},0),y+(\widetilde{z},0))={\cal B}(x,y) .
These four hypotheses were introduced in [30 ] , and with the same notation as above repeated in [31 , 32 ] . Condition (A2) is not needed in Theorems 1.1 and 1.3 , while in Theorems 1.2 and 1.4 it is used
through several results from [30 , 31 , 32 ] .
Throughout this paper, we always assume that ℬ ( x , y ) {\cal B}(x,y) satisfies (A1) , (A3) (I) and (A4) .
Consider the following symmetric form
ℰ 0 ( u , v ) := 1 2 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) J ( x , y ) 𝑑 y 𝑑 x . {\cal E}^{0}(u,v):=\frac{1}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))(v(x)-v(y))J(x,y)dydx.
Since ℬ ( x , y ) {\cal B}(x,y) is bounded, C c ∞ ( ℝ ¯ + d ) C_{c}^{\infty}({\overline{\mathbb{R}}}^{d}_{+}) is closable in L 2 ( ℝ ¯ + d , d x ) = L 2 ( ℝ + d , d x ) L^{2}({\overline{\mathbb{R}}}^{d}_{+},dx)=L^{2}({\mathbb{R}}^{d}_{+},dx) by Fatou’s lemma. Let ℱ ¯ {\overline{\cal F}} be the closure of C c ∞ ( ℝ ¯ + d ) C_{c}^{\infty}({\overline{\mathbb{R}}}^{d}_{+}) in L 2 ( ℝ + d , d x ) L^{2}({\mathbb{R}}^{d}_{+},dx) under the norm ℰ 1 0 := ℰ 0 + ( ⋅ , ⋅ ) L 2 ( ℝ + d , d x ) {\cal E}^{0}_{1}:={\cal E}^{0}+(\cdot,\cdot)_{L^{2}({\mathbb{R}}^{d}_{+},dx)} . Then ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},{\overline{\cal F}}) is a regular Dirichlet form on L 2 ( ℝ + d , d x ) L^{2}({\mathbb{R}}^{d}_{+},dx) . Let Y ¯ = ( Y ¯ t , t ≥ 0 ; ℙ x , x ∈ ℝ ¯ + d ∖ 𝒩 ′ ) \overline{Y}=(\overline{Y}_{t},t\geq 0;{\mathbb{P}}_{x},x\in\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}^{\prime}) be the Hunt
process associated with ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},{\overline{\cal F}}) , where 𝒩 ′ {\cal N}^{\prime} is an exceptional set.
Here is our first main result. The heat kernel estimates are expressed in different but equivalent forms,
each providing a different viewpoint.
Recall that 𝐞 d = ( 0 ~ , 1 ) ∈ ℝ d {\mathbf{e}}_{d}=(\widetilde{0},1)\in{\mathbb{R}}^{d} .
Theorem 1.1 .
Suppose that (A1) , (A3) and (A4) hold. Then the process Y ¯ \overline{Y} can be refined to be a
conservative Feller process with strong Feller property starting from every point in ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+} and has a jointly continuous heat kernel p ¯ : ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d → ( 0 , ∞ ) \overline{p}:(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+}\to(0,\infty) . Moreover, the heat kernel p ¯ \overline{p} has the following estimates: For all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d (t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} ,
p ¯ ( t , x , y ) ≍ t − d / α ∧ [ t J ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) \displaystyle\overline{p}(t,x,y)\asymp t^{-d/\alpha}\wedge\bigg{[}tJ(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})
+ 𝟏 { β 2 ≥ α + β 1 } t 2 ∫ ( x d ∨ y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 J ( x + t 1 / α 𝐞 d , x + r 𝐞 d ) J ( x + r 𝐞 d , y + t 1 / α 𝐞 d ) r d − 1 d r ] . \displaystyle\,+{\bf 1}_{\{\beta_{2}\geq\alpha+\beta_{1}\}}t^{2}\int_{(x_{d}\vee y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\!\!\!\!\!\!\!\!\!\!\!J(x+t^{1/\alpha}{\mathbf{e}}_{d},x+r{\mathbf{e}}_{d})\,J(x+r{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})r^{d-1}dr\bigg{]}.
(1.2)
Furthermore the heat kernel estimates in (1.1 ) can be rewritten in terms of B b 1 , b 2 , b 3 , b 4 B_{b_{1},b_{2},b_{3},b_{4}}
explicitly by considering three cases separately:
(i) If β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} , then for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d (t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} ,
p ¯ ( t , x , y ) \displaystyle\overline{p}(t,x,y)
≍ ( t − d / α ∧ t B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) | x − y | d + α ) \displaystyle\asymp\bigg{(}t^{-d/\alpha}\wedge\frac{tB_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})}{|x-y|^{d+\alpha}}\bigg{)}
≍ ( t − d / α ∧ t | x − y | d + α ) B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) . \displaystyle\asymp\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d}).
(1.3)
(ii) If β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} , then for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d (t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} ,
p ¯ ( t , x , y ) ≍ ( t − d / α ∧ t | x − y | d + α ) [ B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) \displaystyle\overline{p}(t,x,y)\asymp\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})
+ ( 1 ∧ t | x − y | α ) B β 1 , β 1 , 0 , β 3 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) + t 1 / α ) ∧ | x − y | ) ] . \displaystyle+\!\bigg{(}\!1\wedge\frac{t}{|x-y|^{\alpha}}\!\bigg{)}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\!\bigg{(}\!e\!+\!\frac{|x-y|}{((x_{d}\wedge y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}\!\bigg{]}.
(1.4)
(iii) If β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} , then for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d (t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} ,
p ¯ ( t , x , y ) ≍ ( t − d / α ∧ t | x − y | d + α ) [ B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) \displaystyle\overline{p}(t,x,y)\asymp\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})
+ ( 1 ∧ t | x − y | α ) B β 1 , β 1 , 0 , β 3 + β 4 + 1 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) + t 1 / α ) ∧ | x − y | ) ] . \displaystyle\!+\!\bigg{(}\!1\wedge\frac{t}{|x-y|^{\alpha}}\!\bigg{)}B_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\!\bigg{(}\!e\!+\!\frac{|x-y|}{((x_{d}\wedge y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}\bigg{]}.
(1.5)
Note that if x x or y y
is close to the boundary, then J ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) J(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d}) is not comparable to J ( x , y ) J(x,y) in our setting.
Thus, even in case β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} , the form of the heat kernel is different from the usual form.
The appearance of the two terms in the brackets on the right-hand sides of (1.4 )–(1.1 )
reflects the fact that the dominant contribution to the heat kernel may come from either one jump or two jumps. Moreover, it
is easy to see that neither of these two terms dominates the other one for all ( t , x , y ) (t,x,y) . Below we illustrate this feature when β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} .
Let β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} . When | x − y | > 6 t 1 / α |x-y|>6t^{1/\alpha} , the second term in the brackets in (1.4 ), i.e.,
( 1 ∧ t | x − y | α ) B β 1 , β 1 , 0 , β 3 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) + t 1 / α ) ∧ | x − y | ) \displaystyle\bigg{(}\!1\wedge\frac{t}{|x-y|^{\alpha}}\!\bigg{)}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\!\bigg{(}\!e\!+\!\frac{|x-y|}{((x_{d}\wedge y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}
(1.6)
is comparable to
t | x − y | d + α ∫ B ( x + 2 − 1 | x − y | 𝐞 d , 4 − 1 | x − y | ) J ( x + t 1 / α 𝐞 d , z ) J ( z , y + t 1 / α 𝐞 d ) 𝑑 z , \displaystyle t|x-y|^{d+\alpha}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}J(x+t^{1/\alpha}{\mathbf{e}}_{d},z)\,J(z,y+t^{1/\alpha}{\mathbf{e}}_{d})dz,
(1.7)
see Remark 6.5 below.
In the special case when β 3 = β 4 = 0 \beta_{3}=\beta_{4}=0 (and β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} ), Figure 1(b) ,
where the comparability of (1.6 ) and (1.7 ) is used,
illustrates the regions where one jump and two jumps dominate.
The different forms of the heat kernel estimates in Theorem 1.1 are consequences of the relationship among β 1 \beta_{1} , β 2 \beta_{2} and α \alpha only, the values of β 3 \beta_{3} and β 4 \beta_{4}
do not play any role in this.
To reduce technicalities, the reader, on a first reading, may assume β 3 = β 4 = 0 \beta_{3}=\beta_{4}=0 , without loosing
the essential features of this paper. Note that, even in this special case, a logarithmic term appears in the estimates when β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} . We need the logarithmic terms in the assumptions to cover the important examples in [21 , 29 ] .
Figure 1. Dominant path from
x x to y y (with x d ≤ y d x_{d}\leq y_{d} ) at time
t < ( | x − y | / 6 ) α t<(|x-y|/6)^{\alpha}
when β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} , β 3 = β 4 = 0 \beta_{3}=\beta_{4}=0
Let ℱ 0 {\cal F}^{0} be the closure of C c ∞ ( ℝ + d ) C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) in L 2 ( ℝ + d , d x ) L^{2}({\mathbb{R}}^{d}_{+},dx) under the norm ℰ 1 0 {\cal E}^{0}_{1} . Then ( ℰ 0 , ℱ 0 ) ({\cal E}^{0},{\cal F}^{0}) is also a regular Dirichlet form on L 2 ( ℝ + d , d x ) L^{2}({\mathbb{R}}^{d}_{+},dx) and it is the part form of ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},{\overline{\cal F}}) on ℝ + d {\mathbb{R}}^{d}_{+} .
Let Y 0 = ( Y t 0 , t ≥ 0 ; ℙ x , x ∈ ℝ + d ∖ 𝒩 0 ) Y^{0}=(Y^{0}_{t},t\geq 0;{\mathbb{P}}_{x},x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}_{0}) be the Hunt
process associated with ( ℰ 0 , ℱ 0 ) ({\cal E}^{0},{\cal F}^{0}) where 𝒩 0 {\cal N}_{0} is an exceptional set.
Y 0 Y^{0} can be regarded as the part process of Y ¯ \overline{Y} killed upon exiting ℝ + d {\mathbb{R}}^{d}_{+} .
For κ ∈ ( 0 , ∞ ) \kappa\in(0,\infty) , define
ℰ κ ( u , v ) \displaystyle{\cal E}^{\kappa}(u,v)
:= ℰ 0 ( u , v ) + ∫ ℝ + d u ( x ) v ( x ) κ x d − α 𝑑 x , \displaystyle:={\cal E}^{0}(u,v)+\int_{{\mathbb{R}}^{d}_{+}}u(x)v(x)\kappa x_{d}^{-\alpha}dx,
ℱ κ \displaystyle{\cal F}^{\kappa}
:= ℱ ~ 0 ∩ L 2 ( ℝ + d , κ x d − α d x ) , \displaystyle:=\widetilde{\cal F}^{0}\cap L^{2}({\mathbb{R}}^{d}_{+},\kappa x_{d}^{-\alpha}dx),
where ℱ ~ 0 \widetilde{\cal F}^{0} is the family of all ℰ 1 0 {\cal E}^{0}_{1} -quasi-continuous functions in ℱ 0 {\cal F}^{0} . Then ( ℰ κ , ℱ κ ) ({\cal E}^{\kappa},{\cal F}^{\kappa}) is a regular Dirichlet form on L 2 ( ℝ + d , d x ) L^{2}({\mathbb{R}}^{d}_{+},dx) with C c ∞ ( ℝ + d ) C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) as a special standard core, see [23 , Theorems 6.1.1 and 6.1.2] .
Let Y κ = ( Y t κ , t ≥ 0 ; ℙ x , x ∈ ℝ + d ∖ 𝒩 κ ) Y^{\kappa}=(Y^{\kappa}_{t},t\geq 0;{\mathbb{P}}_{x},x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}_{\kappa}) be the Hunt process associated with
( ℰ κ , ℱ κ ) ({\cal E}^{\kappa},{\cal F}^{\kappa}) where 𝒩 κ {\cal N}_{\kappa} is an exceptional set. For κ ∈ [ 0 , ∞ ) \kappa\in[0,\infty) , we denote by ζ κ \zeta^{\kappa} the lifetime of Y κ Y^{\kappa} . Define Y t κ = ∂ Y^{\kappa}_{t}=\partial for t ≥ ζ κ t\geq\zeta^{\kappa} , where ∂ \partial is a cemetery point added to the state space ℝ + d {\mathbb{R}}^{d}_{+} .
We now associate with the constant κ \kappa a positive parameter q q which plays an important role in the paper.
For q ∈ ( − 1 , α + β 1 ) q\in(-1,\alpha+\beta_{1}) , set
C ( α , q , ℬ ) = { ∫ ℝ d − 1 1 ( | u ~ | 2 + 1 ) ( d + α ) / 2 ∫ 0 1 ( s q − 1 ) ( 1 − s α − q − 1 ) ( 1 − s ) 1 + α ℬ ( ( ( 1 − s ) u ~ , 1 ) , s 𝐞 d ) 𝑑 s 𝑑 u ~ if d ≥ 2 , ∫ 0 1 ( s q − 1 ) ( 1 − s α − q − 1 ) ( 1 − s ) 1 + α ℬ ( 1 , s ) 𝑑 s if d = 1 . \displaystyle C(\alpha,q,{\cal B})\!=\!\begin{cases}\displaystyle\int_{{\mathbb{R}}^{d-1}}\!\frac{1}{(|\widetilde{u}|^{2}+1)^{(d+\alpha)/2}}\!\int_{0}^{1}\frac{(s^{q}-1)(1-s^{\alpha-q-1})}{(1-s)^{1+\alpha}}{\cal B}\big{(}((1-s)\widetilde{u},1),s{\mathbf{e}}_{d}\big{)}ds\,d\widetilde{u}\!&\mbox{if }d\geq 2,\\[14.22636pt]
\displaystyle\int_{0}^{1}\frac{(s^{q}-1)(1-s^{\alpha-q-1})}{(1-s)^{1+\alpha}}{\cal B}\big{(}1,s\big{)}ds&\mbox{if }d=1.\end{cases}
If we additionally assume that (A3) (II) holds, then the constant C ( α , q , ℬ ) C(\alpha,q,{\cal B}) is
well defined and finite
for every q ∈ ( − 1 , α + β 1 ) q\in(-1,\alpha+\beta_{1}) , C ( α , q , ℬ ) = 0 C(\alpha,q,{\cal B})=0 if and only if q ∈ { 0 , α − 1 } q\in\{0,\alpha-1\} , and lim q → − 1 C ( α , q , ℬ ) = lim q → α + β 1 C ( α , q , ℬ ) = ∞ \lim_{q\to-1}C(\alpha,q,{\cal B})=\lim_{q\to\alpha+\beta_{1}}C(\alpha,q,{\cal B})=\infty (see [30 , Lemma 5.4 and Remark 5.5] ). Note that for every s ∈ ( 0 , 1 ) s\in(0,1) , q ↦ ( s q − 1 ) ( 1 − s α − q − 1 ) q\mapsto(s^{q}-1)(1-s^{\alpha-q-1}) is strictly decreasing on ( − 1 , ( α − 1 ) / 2 ) (-1,(\alpha-1)/2) and strictly increasing on ( ( α − 1 ) / 2 , α + β 1 ) ((\alpha-1)/2,\alpha+\beta_{1}) . Thus, the shape of the map q ↦ C ( α , q , ℬ ) q\mapsto C(\alpha,q,{\cal B}) is given as follows.
Consequently, for every κ ≥ 0 \kappa\geq 0 , there exists a unique
q κ ∈ [ ( α − 1 ) + , α + β 1 ) q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1})
such that
κ = C ( α , q κ , ℬ ) . \kappa=C(\alpha,q_{\kappa},{\cal B}).
(1.8)
Theorem 1.2 .
Suppose that (A1) – (A4) and (1.8 ) hold with q κ ∈ [ ( α − 1 ) + , α + β 1 ) q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) . Then the process Y κ Y^{\kappa}
can be refined to start from every point in ℝ + d {\mathbb{R}}^{d}_{+} and
has a jointly continuous heat kernel p κ : ( 0 , ∞ ) × ℝ + d × ℝ + d → ( 0 , ∞ ) p^{\kappa}:(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}\to(0,\infty) . Moreover, the following approximate factorization holds for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ + d × ℝ + d (t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} :
p κ ( t , x , y ) ≍ ( 1 ∧ x d t 1 / α ) q κ ( 1 ∧ y d t 1 / α ) q κ p ¯ ( t , x , y ) ≍ ℙ x ( ζ κ > t ) ℙ y ( ζ κ > t ) p ¯ ( t , x , y ) , \displaystyle p^{\kappa}(t,x,y)\asymp\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\overline{p}(t,x,y)\asymp{\mathbb{P}}_{x}(\zeta^{\kappa}>t){\mathbb{P}}_{y}(\zeta^{\kappa}>t)\,\overline{p}(t,x,y),
(1.9)
where p ¯ ( t , x , y ) \overline{p}(t,x,y) is the heat kernel of Y ¯ \overline{Y} .
As a consequence of Lemma 2.3 below,
we will see that, when α ≤ 1 \alpha\leq 1 , the process Y ¯ \overline{Y} started from ℝ + d {\mathbb{R}}^{d}_{+} will never hit ∂ ℝ + d \partial{\mathbb{R}}^{d}_{+} and is equal to Y 0 Y^{0} . Thus for x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} we have that p 0 ( t , x , y ) = p ¯ ( t , x , y ) p^{0}(t,x,y)=\overline{p}(t,x,y) , implying that the non-trivial content of
Theorem 1.2 is for κ > 0 \kappa>0 when α ≤ 1 \alpha\leq 1 .
Let
G ¯ ( x , y ) = ∫ 0 ∞ p ¯ ( t , x , y ) 𝑑 t and G κ ( x , y ) = ∫ 0 ∞ p κ ( t , x , y ) 𝑑 t . \overline{G}(x,y)=\int_{0}^{\infty}\overline{p}(t,x,y)dt\quad\text{and}\quad G^{\kappa}(x,y)=\int_{0}^{\infty}p^{\kappa}(t,x,y)dt.
(1.10)
When G ¯ ( ⋅ , ⋅ ) \overline{G}(\cdot,\cdot) is not identically infinite, it is called the Green function of Y ¯ \overline{Y} ,
and when G κ ( ⋅ , ⋅ ) G^{\kappa}(\cdot,\cdot) is not identically infinite, it is called the Green function of Y κ Y^{\kappa} .
As a consequence of
the heat kernel estimates, we get the Green function estimates.
The following theorem says that
the Green function of Y ¯ \overline{Y} is comparable to that of the isotropic α \alpha -stable process in ℝ d {\mathbb{R}}^{d} even
though the jump kernel of Y ¯ \overline{Y} may be degenerate.
Theorem 1.3 .
Suppose that (A1) , (A3) (I) and (A4) hold. If d > α d>\alpha , then
G ¯ ( x , y ) ≍ 1 | x − y | d − α , x , y ∈ ℝ ¯ + d . \overline{G}(x,y)\asymp\frac{1}{|x-y|^{d-\alpha}},\quad x,y\in{\overline{\mathbb{R}}}_{+}^{d}.
(1.11)
If d ≤ α d\leq\alpha , then G ¯ ( x , y ) = ∞ \overline{G}(x,y)=\infty for all x , y ∈ ℝ ¯ + d x,y\in{\overline{\mathbb{R}}}_{+}^{d} .
When d > ( α + β 1 + β 2 ) ∧ 2 d>(\alpha+\beta_{1}+\beta_{2})\wedge 2 , sharp two-sided estimates on G κ ( x , y ) G^{\kappa}(x,y) were obtained in [31 , 32 ] . In the following theorem, we extend those results by removing the restriction on d d and give another proof
using the heat kernel estimates.
The advantage of the new proof is that it explains the reason for the phase transition in the
Green function estimates for d ≥ 2 d\geq 2 .
Let
H q ( x , y ) = { 1 if q < α + 1 2 ( β 1 + β 2 ) , log β 4 + 1 ( e + | x − y | ( x d ∨ y d ) ∧ | x − y | ) if q = α + 1 2 ( β 1 + β 2 ) , ( x d ∨ y d | x − y | ∧ 1 ) 2 α + β 1 + β 2 − 2 q log β 4 ( e + | x − y | ( x d ∨ y d ) ∧ | x − y | ) if q > α + 1 2 ( β 1 + β 2 ) . \displaystyle H_{q}(x,y)=\!\begin{cases}\displaystyle 1&\mbox{if }q<\alpha+\frac{1}{2}(\beta_{1}+\beta_{2}),\\[6.0pt]
\displaystyle\log^{\beta_{4}+1}\left(e+\frac{|x-y|}{(x_{d}\vee y_{d})\wedge|x-y|}\right)&\mbox{if }q=\alpha+\frac{1}{2}(\beta_{1}+\beta_{2}),\\[9.0pt]
\displaystyle\left(\frac{x_{d}\vee y_{d}}{|x-y|}\wedge 1\right)^{2\alpha+\beta_{1}+\beta_{2}-2q}\log^{\beta_{4}}\left(e+\frac{|x-y|}{(x_{d}\vee y_{d})\wedge|x-y|}\right)&\mbox{if }q>\alpha+\frac{1}{2}(\beta_{1}+\beta_{2}).\end{cases}
Theorem 1.4 .
Suppose that (A1) – (A4) and (1.8 ) hold with q κ ∈ [ ( α − 1 ) + , α + β 1 ) q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) . When α ≤ 1 \alpha\leq 1 , suppose also that q κ > 0 q_{\kappa}>0 (or, equivalently, κ > 0 \kappa>0 ). Then G κ G^{\kappa} has the following estimates:
(i) If d ≥ 2 d\geq 2 , then for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
G κ ( x , y ) \displaystyle G^{\kappa}(x,y)
≍ H q κ ( x , y ) | x − y | d − α ( x d ∧ y d | x − y | ∧ 1 ) q κ ( x d ∨ y d | x − y | ∧ 1 ) q κ . \displaystyle\asymp\frac{H_{q_{\kappa}}(x,y)}{|x-y|^{d-\alpha}}\left(\frac{x_{d}\wedge y_{d}}{|x-y|}\wedge 1\right)^{q_{\kappa}}\left(\frac{x_{d}\vee y_{d}}{|x-y|}\wedge 1\right)^{q_{\kappa}}.
(ii) If d = 1 d=1 , then for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
G κ ( x , y ) ≍ { 1 | x − y | 1 − α ( x ∧ y | x − y | ∧ 1 ) q κ if α < 1 , ( x ∧ y | x − y | ∧ 1 ) q κ log ( e + ( x ∧ y ) ∨ | x − y | | x − y | ) if α = 1 , ( x ∧ y ) α − 1 ( x ∧ y | x − y | ∧ 1 ) q κ − α + 1 if α > 1 . \displaystyle G^{\kappa}(x,y)\asymp\begin{cases}\displaystyle\frac{1}{|x-y|^{1-\alpha}}\left(\frac{x\wedge y}{|x-y|}\wedge 1\right)^{q_{\kappa}}&\mbox{ if }\alpha<1,\\[9.0pt]
\displaystyle\left(\frac{x\wedge y}{|x-y|}\wedge 1\right)^{q_{\kappa}}\log\left(e+\frac{(x\wedge y)\vee|x-y|}{|x-y|}\right)&\mbox{ if }\alpha=1,\\[9.0pt]
\displaystyle(x\wedge y)^{\alpha-1}\left(\frac{x\wedge y}{|x-y|}\wedge 1\right)^{q_{\kappa}-\alpha+1}&\mbox{ if }\alpha>1.\end{cases}
Note that when d ≥ 2 d\geq 2 , at the threshold q κ = α + 1 2 ( β 1 + β 2 ) q_{\kappa}=\alpha+\frac{1}{2}(\beta_{1}+\beta_{2}) , there is a transition
from the usual behavior of the Green function estimates to anomalous behavior.
We describe now the strategy for proving our main results and the organization of the paper.
It is well known that an appropriate Nash-type inequality implies the existence of the heat kernel (outside an exceptional set) and its α \alpha -stable-type upper bound. So we start in Section 2 with establishing
a Nash-type inequality, see Proposition 2.6 .
To this end, we consider a certain Feller process in ℝ + d {\mathbb{R}}^{d}_{+} with continuous paths, subordinate it by an independent α / 2 \alpha/2 -stable subordinator, and show a Nash-type inequality for the Dirichlet form of the subordinate
process. In case α < 1 \alpha<1 , one can estimate
the Dirichlet form of the subordinate process from above by ℰ 0 {\cal E}^{0} and thus prove a Nash-type
inequality for ℰ 0 {\cal E}^{0} . In case α ≥ 1 \alpha\geq 1 , we use the scaling property of the process and a truncation of the jump kernel together with
the already obtained inequality for α < 1 \alpha<1 .
In Section 3 , we
prove parabolic Hölder regularity for both Y ¯ \overline{Y} and Y κ Y^{\kappa} and use this to remove the exceptional sets and extend p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y)
continuously
to ( 0 , ∞ ) × ℝ ¯ + d × ℝ ¯ + d (0,\infty)\times\overline{{\mathbb{R}}}^{d}_{+}\times\overline{{\mathbb{R}}}^{d}_{+} , respectively ( 0 , ∞ ) × ℝ + d × ℝ + d (0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} .
To prove the parabolic Hölder regularity, we need an interior lower bound on the heat kernels. The proof of this lower bound is based on an argument that has already appeared in [30 , Section 3] and is given here in Proposition 3.1 .
Next, we prove several
lower bounds on the heat kernels, mean exit times and exit distributions
of the underlying process and its killed version that allow us to apply the standard arguments for establishing the parabolic Hölder regularity.
We end Section 3 with the elementary Lemma 3.15 which will be used repeatedly
in deriving the heat kernel upper bounds.
Following the arguments from [9 , 21 ] , in Section 4 we establish the parabolic Harnack inequality, see Theorem 4.3 , and use it to establish the important preliminary off-diagonal lower bound p κ ( t , x , y ) ≥ c t J ( x , y ) p^{\kappa}(t,x,y)\geq ctJ(x,y) for t t small compared to | x − y | |x-y| , x d x_{d} and y d y_{d} , see Proposition 4.5 .
In Section 5 we prove the following preliminary upper bound:
p κ ( t , x , y ) ≤ C ( 1 ∧ x d t 1 / α ) q κ ( 1 ∧ y d t 1 / α ) q κ ( t − d / α ∧ t | x − y | d + α ) , t > 0 , x , y ∈ ℝ + d , p^{\kappa}(t,x,y)\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),\quad t>0,\ x,y\in{\mathbb{R}}^{d}_{+},
for all q κ ∈ [ ( α − 1 ) + , α + β 1 ) q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) , see Proposition 5.1 .
Proposition 5.1 is proved in two steps:
Using Lemma 3.15 and several
results from [30 , 31 , 32 ]
(see Lemmas 5.10 -5.11 )
we first prove Proposition 5.1 in case q κ < α q_{\kappa}<\alpha .
The real challenge is to extend it to the full range of q κ q_{\kappa} by covering the
case q κ ≥ α q_{\kappa}\geq\alpha .
For this we use the upper bound on Green potentials of powers of the distance to the boundary. In case d ≥ 2 d\geq 2 , such bound was proved in [31 , 32 ] , and here we prove the corresponding result for d = 1 d=1 . The restriction q κ < α q_{\kappa}<\alpha is removed in Lemma 5.13 by using a bootstrap (induction) argument.
Section 6 is devoted to proving sharp heat kernel lower bounds.
The estimates are given in terms of a function A b 1 , b 2 , b 3 , b 4 ( t , x , y ) A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y) , a space-time version of the function B b 1 , b 2 , b 3 , b 4 ( x , y ) B_{b_{1},b_{2},b_{3},b_{4}}(x,y) .
We first prove a preliminary lower bound, Proposition 6.2 , by using the semigroup property and some interior lower bound on the heat kernel. This bound turns out to be sufficient in case β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} . For the case β 2 ≥ α + β 1 \beta_{2}\geq\alpha+\beta_{1} we need a sharper lower bound obtained in the key Lemma 6.4 . There we apply the semigroup property
(and the preliminary lower bound)
on a carefully chosen interior set on which we have good control of
the terms appearing in the preliminary lower bound.
By combining Proposition 6.2 and Lemma 6.4 , we obtain in Proposition 6.6 the sharp lower bound in cases β 2 ≠ α + β 1 \beta_{2}\neq\alpha+\beta_{1} . The remaining case β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} is more delicate and is covered in Proposition 6.9 .
Sharp heat kernel upper bound is more difficult to establish and
Section 7 is devoted to this task. The first step is to establish in Lemma 7.2 an upper bound which includes the function A β 1 , 0 , β 3 , 0 A_{\beta_{1},0,\beta_{3},0} . The proof uses Lemma 3.15 and an induction argument which consecutively increases the first parameter of the function A A until reaching β 1 \beta_{1} , thus making the decay successively sharper. The sharp upper bound in case β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} , respectively β 2 ≥ α + β 1 \beta_{2}\geq\alpha+\beta_{1} , is given in Theorem 7.5 and Corollary 7.9 , respectively Theorem 7.10 .
The proofs are based on Lemma 3.15 , Lemma 7.2 ,
and a number of delicate technical lemmas involving multiple space-time integrals of
the preliminary heat kernel estimates.
In Section 8 ,
we combine the upper bounds obtained
in Section 7
with the lower bounds from Section 6 and give the proofs of Theorems 1.1 –1.2 .
It is well known that the Green function is the integral over time of the heat kernel.
In Section 9 we first use this observation together with the estimates of
p ¯ ( t , x , y ) \overline{p}(t,x,y) obtained in Proposition 2.7 and Lemma 3.7 (see also Remark 3.12 ) to prove Theorem 1.3 . By using the same method of integrating
the heat kernel estimates of p κ ( t , x , y ) p^{\kappa}(t,x,y)
over time we establish Theorem 1.4 , thus reproving and extending the main result of [31 ] . This new proof sheds more light on
the anomalous behavior of these Green function estimates.
Lemma 9.1 clearly shows that they are
consequences of the different forms of the small time heat kernel estimates.
The paper ends with an appendix which contains a number of technical results not depending on
the preliminary estimates of the heat kernel.
Throughout this paper, the constants β 1 \beta_{1} , β 2 \beta_{2} , β 3 \beta_{3} , β 4 \beta_{4}
will remain the same, and κ \kappa always stands for a non-negative number.
The notation C = C ( a , b , … ) C=C(a,b,\ldots) indicates that the constant C C
depends on a , b , … a,b,\ldots . The dependence on κ , d \kappa,d and α \alpha may not be mentioned explicitly.
Lower case letters
c i , i = 1 , 2 , … c_{i},i=1,2,\dots are used to denote the constants in the proofs
and the labeling of these constants starts anew in each proof.
We denote by m d m_{d} the Lebesgue measure on ℝ d {\mathbb{R}}^{d} .
For Borel subset D ⊂ ℝ d D\subset{\mathbb{R}}^{d} , δ D ( x ) \delta_{D}(x) denotes the distance
between x x and the boundary ∂ D \partial D .
2. Nash inequality and the existence of the heat kernel
The goal of this section is (i) to prove a Nash type inequality
(Proposition 2.6 ); (ii) to deduce the existence of the heat kernels of Y ¯ \overline{Y} and Y κ Y^{\kappa} ; and (iii) to establish their preliminary upper bounds (Proposition 2.7 ).
We begin the section
by introducing the notation for the relevant semigroups and establishing their scale invariance and horizontal translation invariance.
For κ ≥ 0 \kappa\geq 0 ,
we denote by ( P ¯ t ) t ≥ 0 (\overline{P}_{t})_{t\geq 0} and ( P t κ ) t ≥ 0 (P_{t}^{\kappa})_{t\geq 0} the semigroups corresponding to ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},\overline{\cal F}) and ( ℰ κ , ℱ κ ) ({\cal E}^{\kappa},{\cal F}^{\kappa}) respectively. ( P ¯ t ) t ≥ 0 (\overline{P}_{t})_{t\geq 0} and ( P t κ ) t ≥ 0 (P_{t}^{\kappa})_{t\geq 0} define
contraction semigroups
on
L p ( ℝ ¯ + d , d x ) = L p ( ℝ + d , d x ) L^{p}({\overline{\mathbb{R}}}^{d}_{+},dx)=L^{p}({\mathbb{R}}^{d}_{+},dx) for every p ∈ [ 1 , ∞ ] p\in[1,\infty] , and when p ∈ [ 1 , ∞ ) p\in[1,\infty) , these semigroups are strongly continuous.
For t > 0 t>0 and p , q ∈ [ 1 , ∞ ] p,q\in[1,\infty] , define
∥ P ¯ t ∥ p → q = sup { ∥ P ¯ t f ∥ L q ( ℝ + d , d x ) : f ∈ L p ( ℝ + d , d x ) , ∥ f ∥ L p ( ℝ + d , d x ) ≤ 1 } . \displaystyle\lVert\overline{P}_{t}\rVert_{p\to q}=\sup\left\{\lVert\overline{P}_{t}f\rVert_{L^{q}({\mathbb{R}}^{d}_{+},dx)}:f\in L^{p}({\mathbb{R}}^{d}_{+},dx),\,\lVert f\rVert_{L^{p}({\mathbb{R}}^{d}_{+},dx)}\leq 1\right\}.
For f : ℝ d → ℝ f:{\mathbb{R}}^{d}\to{\mathbb{R}} and r > 0 r>0 , define f ( r ) ( x ) = f ( r x ) f^{(r)}(x)=f(rx) . The following scaling property of ( P t κ ) t ≥ 0 (P_{t}^{\kappa})_{t\geq 0} comes from [30 , Lemma 5.1] and [32 , Lemma 2.1] . By the same proof, ( P ¯ t ) t ≥ 0 (\overline{P}_{t})_{t\geq 0} also has the
same scaling property. We give the proof for the reader’s convenience.
Lemma 2.1 .
Let p ∈ [ 1 , ∞ ] p\in[1,\infty] and κ ≥ 0 \kappa\geq 0 . For any f ∈ L p ( ℝ + d , d x ) f\in L^{p}({\mathbb{R}}^{d}_{+},dx) , t > 0 t>0 and r > 0 r>0 , we have
P ¯ t f ( x ) = P ¯ r − α t f ( r ) ( x / r ) and P t κ f ( x ) = P r − α t κ f ( r ) ( x / r ) in L p ( ℝ + d , d x ) . \overline{P}_{t}f(x)=\overline{P}_{r^{-\alpha}t}f^{(r)}(x/r)\quad\text{and}\quad P^{\kappa}_{t}f(x)=P^{\kappa}_{r^{-\alpha}t}f^{(r)}(x/r)\quad\text{in }\,L^{p}({\mathbb{R}}^{d}_{+},dx).
In particular, we have
∥ P ¯ t ∥ 1 → ∞ = t − d / α ∥ P ¯ 1 ∥ 1 → ∞ for all t > 0 . \displaystyle\lVert\overline{P}_{t}\rVert_{1\to\infty}=t^{-d/\alpha}\lVert\overline{P}_{1}\rVert_{1\to\infty}\quad\text{for all }\,t>0.
(2.1)
Proof. Using (A4) , changes of the variables and the correspondence between
Dirichlet forms and semigroups,
we see that for any nice function f f ,
lim t ↓ 0 1 t ∫ ℝ + d ( f ( r ) ( z / r ) − P r − α t κ f ( r ) ( z / r ) ) f ( r ) ( z / r ) 𝑑 z \displaystyle\lim_{t\downarrow 0}\frac{1}{t}\int_{{\mathbb{R}}^{d}_{+}}\big{(}f^{(r)}(z/r)-P_{r^{-\alpha}t}^{\kappa}f^{(r)}(z/r)\big{)}f^{(r)}(z/r)dz
= r d − α lim t ↓ 0 1 r − α t ∫ ℝ + d ( f ( r ) ( x ) − P r − α t κ f ( r ) ( x ) ) f ( r ) ( x ) 𝑑 x \displaystyle=r^{d-\alpha}\lim_{t\downarrow 0}\frac{1}{r^{-\alpha}t}\int_{{\mathbb{R}}^{d}_{+}}\big{(}f^{(r)}(x)-P_{r^{-\alpha}t}^{\kappa}f^{(r)}(x)\big{)}f^{(r)}(x)dx
= r d − α 2 ∫ ℝ + d ∫ ℝ + d ( f ( r ) ( x ) − f ( r ) ( y ) ) 2 ℬ ( x , y ) | x − y | d + α 𝑑 x 𝑑 y + κ r d − α ∫ ℝ + d f ( r ) ( x ) 2 x d − α 𝑑 x \displaystyle=\frac{r^{d-\alpha}}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(f^{(r)}(x)-f^{(r)}(y))^{2}\frac{{\cal B}(x,y)}{|x-y|^{d+\alpha}}dxdy+\kappa r^{d-\alpha}\int_{{\mathbb{R}}^{d}_{+}}f^{(r)}(x)^{2}x_{d}^{-\alpha}dx
= r 2 d 2 ∫ ℝ + d ∫ ℝ + d ( f ( r x ) − f ( r y ) ) 2 ℬ ( r x , r y ) | r x − r y | d + α 𝑑 x 𝑑 y + κ r d ∫ ℝ + d f ( r x ) 2 ( r x d ) − α 𝑑 x \displaystyle=\frac{r^{2d}}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(f(rx)-f(ry))^{2}\frac{{\cal B}(rx,ry)}{|rx-ry|^{d+\alpha}}dxdy+\kappa r^{d}\int_{{\mathbb{R}}^{d}_{+}}f(rx)^{2}(rx_{d})^{-\alpha}dx
= 1 2 ∫ ℝ + d ∫ ℝ + d ( f ( x ) − f ( y ) ) 2 ℬ ( x , y ) | x − y | d + α 𝑑 x 𝑑 y + ∫ ℝ + d f ( x ) 2 κ x d − α 𝑑 x = ℰ κ ( f , f ) . \displaystyle=\frac{1}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(f(x)-f(y))^{2}\frac{{\cal B}(x,y)}{|x-y|^{d+\alpha}}dxdy+\int_{{\mathbb{R}}^{d}_{+}}f(x)^{2}\kappa x_{d}^{-\alpha}dx={\cal E}^{\kappa}(f,f).
Thus, by the uniqueness of the corresponding semigroup, we deduce the scaling property of ( P t κ ) t ≥ 0 (P^{\kappa}_{t})_{t\geq 0} . By a similar proof, we also get the scaling property of ( P ¯ t ) t ≥ 0 (\overline{P}_{t})_{t\geq 0} .
Now, we have that, for every t > 0 t>0 ,
∥ P ¯ t ∥ 1 → ∞ = t − d / α sup { ∥ P ¯ t f ∥ L ∞ ( ℝ + d , d x ) : ∥ f ∥ L 1 ( ℝ + d , d x ) ≤ t d / α } \displaystyle\lVert\overline{P}_{t}\rVert_{1\to\infty}=t^{-d/\alpha}\sup\left\{\lVert\overline{P}_{t}f\rVert_{L^{\infty}({\mathbb{R}}^{d}_{+},dx)}:\lVert f\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq t^{d/\alpha}\right\}
= t − d / α sup { ∥ P ¯ 1 f ( t 1 / α ) ∥ L ∞ ( ℝ + d , d x ) : ∥ f ∥ L 1 ( ℝ + d , d x ) ≤ 1 } = t − d / α ∥ P ¯ 1 ∥ 1 → ∞ , \displaystyle=t^{-d/\alpha}\sup\left\{\lVert\overline{P}_{1}f^{(t^{1/\alpha})}\rVert_{L^{\infty}({\mathbb{R}}^{d}_{+},dx)}:\lVert f\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1\right\}=t^{-d/\alpha}\lVert\overline{P}_{1}\rVert_{1\to\infty},
which proves (2.1 ). The proof is complete. □ \Box
Let d ≥ 2 d\geq 2 . For f : ℝ d → ℝ f:{\mathbb{R}}^{d}\to{\mathbb{R}} and z ~ ∈ ℝ d − 1 \widetilde{z}\in{\mathbb{R}}^{d-1} , define f z ( x ) = f ( x + ( z ~ , 0 ) ) f_{z}(x)=f(x+(\widetilde{z},0)) .
From (A4) , we also get the following horizontal translation invariance property of the semigroups ( P ¯ t ) t ≥ 0 (\overline{P}_{t})_{t\geq 0} and ( P t κ ) t ≥ 0 (P_{t}^{\kappa})_{t\geq 0} .
Lemma 2.2 .
Let d ≥ 2 d\geq 2 , p ∈ [ 1 , ∞ ] p\in[1,\infty] and κ ≥ 0 \kappa\geq 0 . For any f ∈ L p ( ℝ + d , d x ) f\in L^{p}({\mathbb{R}}^{d}_{+},dx) , t > 0 t>0 and z ~ ∈ ℝ d − 1 \widetilde{z}\in{\mathbb{R}}^{d-1} , we have
P ¯ t f ( x ) = P ¯ t f z ( x − ( z ~ , 0 ) ) and P t κ f ( x ) = P t κ f z ( x − ( z ~ , 0 ) ) in L p ( ℝ + d , d x ) . \overline{P}_{t}f(x)=\overline{P}_{t}f_{z}(x-(\widetilde{z},0))\quad\text{and}\quad P^{\kappa}_{t}f(x)=P^{\kappa}_{t}f_{z}(x-(\widetilde{z},0))\quad\text{in }\,L^{p}({\mathbb{R}}^{d}_{+},dx).
A consequence of the next lemma is that, in case when α ≤ 1 \alpha\leq 1 , the process Y ¯ \overline{Y} started from ℝ + d {\mathbb{R}}^{d}_{+} will never hit ∂ ℝ + d \partial{\mathbb{R}}^{d}_{+} and is equal to Y 0 Y^{0} .
Lemma 2.3 .
If α ≤ 1 \alpha\leq 1 , then
ℱ 0 = ℱ ¯ {\cal F}^{0}={\overline{\cal F}} .
Proof. Define
𝒞 ~ ( u , v ) \displaystyle\widetilde{\cal C}(u,v)
:= 1 2 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) | x − y | d + α 𝑑 y 𝑑 x , \displaystyle:=\frac{1}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{d+\alpha}}dydx,
𝒟 ( 𝒞 ~ ) \displaystyle{\cal D}(\widetilde{\cal C})
:= closure of C c ∞ ( ℝ ¯ + d ) in L 2 ( ℝ + d , d x ) under 𝒞 ~ + ( ⋅ , ⋅ ) L 2 ( ℝ + d , d x ) . \displaystyle:=\text{closure of $C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+})$ in $L^{2}({\mathbb{R}}^{d}_{+},dx)$ under $\widetilde{\cal C}+(\cdot,\cdot)_{L^{2}({\mathbb{R}}^{d}_{+},dx)}$}.
Then ( 𝒞 ~ , 𝒟 ( 𝒞 ~ ) ) (\widetilde{\cal C},{\cal D}(\widetilde{\cal C})) is a regular Dirichlet form associated with the reflected α \alpha -stable process in ℝ ¯ + d \overline{\mathbb{R}}^{d}_{+} in the sense of [6 ] .
By (A3) (I), ℰ 0 ( u , u ) ≤ C 2 𝒞 ~ ( u , u ) {\cal E}^{0}(u,u)\leq C_{2}\widetilde{\cal C}(u,u) for all u ∈ C c ∞ ( ℝ ¯ + d ) u\in C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+})
and hence 𝒟 ( 𝒞 ~ ) ⊂ ℱ ¯ {\cal D}(\widetilde{\cal C})\subset{\overline{\cal F}} . By [6 , Theorem 2.5(i) and Remark 2.2(1)] , since α ≤ 1 \alpha\leq 1 , ℝ ¯ + d ∖ ℝ + d \overline{\mathbb{R}}^{d}_{+}\setminus{\mathbb{R}}^{d}_{+} is
( 𝒞 ~ , 𝒟 ( 𝒞 ~ ) ) (\widetilde{\cal C},{\cal D}(\widetilde{\cal C})) -polar
and hence is also ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},{\overline{\cal F}}) -polar.
Therefore, when starting from ℝ + d {\mathbb{R}}^{d}_{+} , Y ¯ \overline{Y} will never exit ℝ + d {\mathbb{R}}^{d}_{+} . Hence
Y ¯ \overline{Y} and Y 0 Y^{0} are the same when they
start from x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and thus ℱ 0 = ℱ ¯ {\cal F}^{0}={\overline{\cal F}} .
□ \Box
In order to prove the Nash type inequality, we first consider a Brownian motion on ℝ + d {\mathbb{R}}^{d}_{+}
killed with a critical potential
and a subordinate process obtained by time changing this killed Brownian motion with an independent α / 2 \alpha/2 -stable subordinator. Then using results
from [21 ] , the Hardy inequality in
[32 , Proposition 3.2] and comparing the Dirichlet form corresponding to the subordinate process with ( ℰ 0 , ℱ ¯ ) ({\cal E}^{0},{\overline{\cal F}}) , we deduce the desired result.
For any γ ≥ 0 \gamma\geq 0 , denote by I γ I_{\gamma} the modified Bessel function of the first kind which is defined by
I γ ( r ) = ∑ m = 0 ∞ 1 m ! Γ ( γ + 1 + m ) ( r 2 ) 2 m + γ , I_{\gamma}(r)=\sum_{m=0}^{\infty}\frac{1}{m!\,\Gamma(\gamma+1+m)}\left(\frac{r}{2}\right)^{2m+\gamma},
where Γ ( r ) := ∫ 0 ∞ u r − 1 e − u 𝑑 u \Gamma(r):=\int_{0}^{\infty}u^{r-1}e^{-u}du is the Gamma function. It is known that (see, e.g. [1 , (9.6.7) and (9.7.1)] )
I γ ( r ) ≍ ( 1 ∧ r ) γ + 1 / 2 r − 1 / 2 e r , r > 0 . I_{\gamma}(r)\asymp(1\wedge r)^{\gamma+1/2}r^{-1/2}e^{r},\quad r>0.
(2.2)
Define for t > 0 t>0 and
x = ( x 1 , … , x d ) , y = ( y 1 , … , y d ) ∈ ℝ + d x=(x_{1},\dots,x_{d}),y=(y_{1},\dots,y_{d})\in{\mathbb{R}}^{d}_{+} ,
q γ ( t , x , y ) = x d y d 2 t I γ ( x d y d 2 t ) exp ( − x d 2 + y d 2 4 t ) ∏ i = 1 d − 1 [ 1 4 π t exp ( − | x i − y i | 2 4 t ) ] . q^{\gamma}(t,x,y)=\frac{\sqrt{x_{d}y_{d}}}{2t}I_{\gamma}\left(\frac{x_{d}y_{d}}{2t}\right)\exp\left(-\frac{x_{d}^{2}+y_{d}^{2}}{4t}\right)\prod_{i=1}^{d-1}\left[\frac{1}{\sqrt{4\pi t}}\exp\left(-\frac{|x_{i}-y_{i}|^{2}}{4t}\right)\right].
Note that by (2.2 ),
q γ ( t , x , y ) ≍ ( 1 ∧ x d y d t ) γ + 1 / 2 t − d / 2 exp ( − | x − y | 2 4 t ) . q^{\gamma}(t,x,y)\asymp\left(1\wedge\frac{x_{d}y_{d}}{t}\right)^{\gamma+1/2}t^{-d/2}\exp\left(-\frac{|x-y|^{2}}{4t}\right).
(2.3)
By [33 , Lemma 4.1 and Theorem 4.9] , q γ ( t , x , y ) q^{\gamma}(t,x,y) is
the transition density of the Feller process
(killed Brownian motion with critical potential)
W γ = ( W t γ ) t ≥ 0 W^{\gamma}=(W^{\gamma}_{t})_{t\geq 0}
on ℝ + d {\mathbb{R}}^{d}_{+} associated with the following regular Dirichlet form ( 𝒬 γ , 𝒟 ( 𝒬 γ ) ) ({\cal Q}^{\gamma},{\cal D}({\cal Q}^{\gamma})) :
𝒬 γ ( u , v ) \displaystyle{\cal Q}^{\gamma}(u,v)
:= ∫ ℝ + d ∇ u ( x ) ⋅ ∇ v ( x ) 𝑑 x + ( γ 2 − 1 4 ) ∫ ℝ + d u ( x ) v ( x ) x d − 2 𝑑 x , \displaystyle:=\int_{{\mathbb{R}}^{d}_{+}}\nabla u(x)\cdot\nabla v(x)dx+\Big{(}\gamma^{2}-\frac{1}{4}\Big{)}\int_{{\mathbb{R}}^{d}_{+}}u(x)v(x)x_{d}^{-2}dx,
𝒟 ( 𝒬 γ ) \displaystyle{\cal D}({\cal Q}^{\gamma})
:= closure of C c ∞ ( ℝ + d ) in L 2 ( ℝ + d , d x ) under 𝒬 1 γ = 𝒬 γ + ( ⋅ , ⋅ ) L 2 ( ℝ + d , d x ) . \displaystyle:=\text{closure of $C_{c}^{\infty}({\mathbb{R}}^{d}_{+})$ in $L^{2}({\mathbb{R}}^{d}_{+},dx)$ under ${\cal Q}^{\gamma}_{1}={\cal Q}^{\gamma}+(\cdot,\cdot)_{L^{2}({\mathbb{R}}^{d}_{+},dx)}$}.
Let S = ( S t ) t ≥ 0 S=(S_{t})_{t\geq 0} be an α / 2 \alpha/2 -stable subordinator independent of W γ W^{\gamma} , and let X γ = ( X t γ ) t ≥ 0 X^{\gamma}=(X^{\gamma}_{t})_{t\geq 0} be the subordinate process X t γ := W S t γ X^{\gamma}_{t}:=W^{\gamma}_{S_{t}} . Then X γ X^{\gamma} is a Hunt process with no diffusion part. The transition density p γ ( t , x , y ) p^{\gamma}(t,x,y) of X t γ X^{\gamma}_{t} exists and is given by
p γ ( t , x , y ) = ∫ 0 ∞ q γ ( s , x , y ) d d s ℙ ( S t ≤ s ) . p^{\gamma}(t,x,y)=\int_{0}^{\infty}q^{\gamma}(s,x,y)\frac{d}{ds}{\mathbb{P}}(S_{t}\leq s).
Also, the jump kernel J γ ( d x , d y ) J^{\gamma}(dx,dy) and the killing measure κ γ ( d x ) \kappa^{\gamma}(dx) of X γ X^{\gamma} have densities J γ ( x , y ) J^{\gamma}(x,y) and κ γ ( x ) \kappa^{\gamma}(x) that are given by the following formulas
(see, for instance, [34 , (2.8)–(2.9)] ):
J γ ( x , y ) \displaystyle J^{\gamma}(x,y)
= ∫ 0 ∞ q γ ( t , x , y ) ν α / 2 ( t ) 𝑑 t , κ γ ( x ) = ∫ 0 ∞ ( 1 − ∫ ℝ + d q γ ( t , x , y ) 𝑑 y ) ν α / 2 ( t ) 𝑑 t , \displaystyle=\int_{0}^{\infty}q^{\gamma}(t,x,y)\,\nu_{\alpha/2}(t)dt,\qquad\kappa^{\gamma}(x)=\int_{0}^{\infty}\bigg{(}1-\int_{{\mathbb{R}}^{d}_{+}}q^{\gamma}(t,x,y)dy\bigg{)}\nu_{\alpha/2}(t)dt,
where ν α / 2 ( t ) = α / 2 Γ ( 1 − α / 2 ) t − 1 − α / 2 \nu_{\alpha/2}(t)=\frac{\alpha/2}{\Gamma(1-\alpha/2)}t^{-1-\alpha/2} is the Lévy density of the α / 2 \alpha/2 -stable subordinator S S .
Lemma 2.4 .
(i) There exists a constant c γ , α > 0 c_{\gamma,\alpha}>0 such that κ γ ( x ) = c γ , α x d − α \kappa^{\gamma}(x)=c_{\gamma,\alpha}x_{d}^{-\alpha} for every x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} .
(ii) It holds that for any x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
J γ ( x , y ) \displaystyle J^{\gamma}(x,y)
≍ ( 1 ∧ x d | x − y | ) γ + 1 / 2 ( 1 ∧ y d | x − y | ) γ + 1 / 2 1 | x − y | d + α = B γ + 1 / 2 , γ + 1 / 2 , 0 , 0 ( x , y ) | x − y | d + α . \displaystyle\asymp\left(1\wedge\frac{x_{d}}{|x-y|}\right)^{\gamma+1/2}\left(1\wedge\frac{y_{d}}{|x-y|}\right)^{\gamma+1/2}\frac{1}{|x-y|^{d+\alpha}}=\frac{B_{\gamma+1/2,\gamma+1/2,0,0}(x,y)}{|x-y|^{d+\alpha}}.
(iii) There exists a constant C > 0 C>0 such that
p γ ( t , x , y ) ≤ C t − d / α , t > 0 , x , y ∈ ℝ + d . p^{\gamma}(t,x,y)\leq Ct^{-d/\alpha},\quad\,t>0,\;x,y\in{\mathbb{R}}^{d}_{+}.
Proof. (i) Observe that the process W γ W^{\gamma} satisfies the scaling property and horizontal translation invariance, namely, for any t > 0 t>0 , x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} , λ > 0 \lambda>0 and z ~ ∈ ℝ d − 1 \widetilde{z}\in{\mathbb{R}}^{d-1} ,
q γ ( λ 2 t , λ x , λ y ) = λ − d q γ ( t , x , y ) and q γ ( t , x + ( z ~ , 0 ) , y + ( z ~ , 0 ) ) = q γ ( t , x , y ) . q^{\gamma}(\lambda^{2}t,\lambda x,\lambda y)=\lambda^{-d}q^{\gamma}(t,x,y)\quad\text{ and }\quad q^{\gamma}(t,x+(\widetilde{z},0),y+(\widetilde{z},0))=q^{\gamma}(t,x,y).
Therefore, for every x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} , using the change of variables z = ( y − ( x ~ , 0 ) ) / x d z=(y-(\widetilde{x},0))/x_{d} in the second equality below and s = x d − 2 t s=x_{d}^{-2}t in the third, we get that
κ γ ( x ) \displaystyle\kappa^{\gamma}(x)
= α / 2 Γ ( 1 − α / 2 ) ∫ 0 ∞ ( 1 − ∫ ℝ + d q γ ( t , ( 0 ~ , x d ) , ( y ~ − x ~ , y d ) ) 𝑑 y ) t − α / 2 − 1 𝑑 t \displaystyle=\frac{\alpha/2}{\Gamma(1-\alpha/2)}\int_{0}^{\infty}\bigg{(}1-\int_{{\mathbb{R}}^{d}_{+}}q^{\gamma}(t,(\widetilde{0},x_{d}),(\widetilde{y}-\widetilde{x},y_{d}))dy\bigg{)}t^{-\alpha/2-1}dt
= α / 2 Γ ( 1 − α / 2 ) ∫ 0 ∞ ( 1 − x d d ∫ ℝ + d q γ ( t , ( 0 ~ , x d ) , x d z ) 𝑑 z ) t − α / 2 − 1 𝑑 t \displaystyle=\frac{\alpha/2}{\Gamma(1-\alpha/2)}\int_{0}^{\infty}\bigg{(}1-x_{d}^{d}\int_{{\mathbb{R}}^{d}_{+}}q^{\gamma}(t,(\widetilde{0},x_{d}),x_{d}z)dz\bigg{)}t^{-\alpha/2-1}dt
= α / 2 Γ ( 1 − α / 2 ) ∫ 0 ∞ ( 1 − ∫ ℝ + d q γ ( x d − 2 t , ( 0 ~ , 1 ) , z ) 𝑑 z ) t − α / 2 − 1 𝑑 t \displaystyle=\frac{\alpha/2}{\Gamma(1-\alpha/2)}\int_{0}^{\infty}\bigg{(}1-\int_{{\mathbb{R}}^{d}_{+}}q^{\gamma}(x_{d}^{-2}t,(\widetilde{0},1),z)dz\bigg{)}t^{-\alpha/2-1}dt
= α / 2 Γ ( 1 − α / 2 ) x d − α ∫ 0 ∞ ( 1 − ∫ ℝ + d q γ ( s , ( 0 ~ , 1 ) , z ) 𝑑 z ) s − α / 2 − 1 𝑑 s = κ γ ( ( 0 ~ , 1 ) ) x d − α . \displaystyle=\frac{\alpha/2}{\Gamma(1-\alpha/2)}x_{d}^{-\alpha}\int_{0}^{\infty}\bigg{(}1-\int_{{\mathbb{R}}^{d}_{+}}q^{\gamma}(s,(\widetilde{0},1),z)dz\bigg{)}s^{-\alpha/2-1}ds=\kappa^{\gamma}((\widetilde{0},1))x_{d}^{-\alpha}.
(ii) In view of (2.3 ), the condition 𝐇𝐊 𝐔 𝐡 \mathbf{HK_{U}^{h}} in [21 ] holds with C 0 = 0 C_{0}=0 , Φ ( r ) = r 2 \Phi(r)=r^{2} and
h ( t , x , y ) = ( 1 ∧ ( x d y d / t ) ) γ + 1 / 2 h(t,x,y)=(1\wedge(x_{d}y_{d}/t))^{\gamma+1/2} .
Note that the tail of the Lévy measure of S S is given by ∫ r ∞ ν α / 2 ( u ) 𝑑 u = r − α / 2 / Γ ( 1 − α / 2 ) \int_{r}^{\infty}\nu_{\alpha/2}(u)du=r^{-\alpha/2}/\Gamma(1-\alpha/2) . Thus, the condition (Poly-∞ \infty ) in [21 ] also holds and we obtain from [21 , Theorem 4.1] that for x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
J γ ( x , y ) \displaystyle J^{\gamma}(x,y)
≍ ( 1 ∧ x d y d | x − y | 2 ) γ + 1 / 2 1 | x − y | d + α ≍ ( 1 ∧ x d | x − y | ) γ + 1 / 2 ( 1 ∧ y d | x − y | ) γ + 1 / 2 1 | x − y | d + α . \displaystyle\asymp\left(1\wedge\frac{x_{d}y_{d}}{|x-y|^{2}}\right)^{\gamma+1/2}\!\frac{1}{|x-y|^{d+\alpha}}\asymp\left(1\wedge\frac{x_{d}}{|x-y|}\right)^{\gamma+1/2}\!\left(1\wedge\frac{y_{d}}{|x-y|}\right)^{\gamma+1/2}\!\frac{1}{|x-y|^{d+\alpha}}.
In the second comparison, we used the fact that ( 1 ∧ x d | x − y | ) ( 1 ∧ y d | x − y | ) ≤ ( 1 ∧ x d y d | x − y | 2 ) ≤ 2 ( 1 ∧ x d | x − y | ) ( 1 ∧ y d | x − y | ) (1\wedge\frac{x_{d}}{|x-y|})(1\wedge\frac{y_{d}}{|x-y|})\leq(1\wedge\frac{x_{d}y_{d}}{|x-y|^{2}})\leq 2(1\wedge\frac{x_{d}}{|x-y|})(1\wedge\frac{y_{d}}{|x-y|}) for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} , which can be proved elementarily by using y d ≤ x d + | x − y | y_{d}\leq x_{d}+|x-y| .
(iii) Since the conditions (Poly-∞ \infty ) and 𝐇𝐊 𝐔 𝐡 \mathbf{HK_{U}^{h}} in [21 ] hold, the result follows from [21 , Proposition 4.5(ii)] . □ \Box
Denote by ( 𝒞 γ , 𝒟 ( 𝒞 γ ) ) ({\cal C}^{\gamma},{\cal D}({\cal C}^{\gamma})) the regular Dirichlet form associated with the subordinate process X γ X^{\gamma} . Then since X γ X^{\gamma} has no diffusion part, we get from Lemma 2.4 (i) that
𝒞 γ ( u , v ) = ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) J γ ( x , y ) 𝑑 y 𝑑 x + c γ , α ∫ ℝ + d u ( x ) v ( x ) x d − α 𝑑 x . {\cal C}^{\gamma}(u,v)=\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))(v(x)-v(y))J^{\gamma}(x,y)dydx+c_{\gamma,\alpha}\int_{{\mathbb{R}}^{d}_{+}}u(x)v(x)x_{d}^{-\alpha}dx.
Also, we have C c ∞ ( ℝ + d ) ⊂ 𝒟 ( 𝒞 γ ) C_{c}^{\infty}({\mathbb{R}}^{d}_{+})\subset{\cal D}({\cal C}^{\gamma}) since 𝒟 ( 𝒬 γ ) ⊂ 𝒟 ( 𝒞 γ ) {\cal D}({\cal Q}^{\gamma})\subset{\cal D}({\cal C}^{\gamma}) , see [34 ] .
Lemma 2.5 .
There exists a constant C > 0 C>0 such that
∥ u ∥ L 2 ( ℝ + d , d x ) 2 ( 1 + α / d ) ≤ C 𝒞 γ ( u , u ) for every u ∈ C c ∞ ( ℝ + d ) with ∥ u ∥ L 1 ( ℝ + d , d x ) ≤ 1 . \lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+\alpha/d)}\leq C{\cal C}^{\gamma}(u,u)\quad\text{for every }\;u\in C_{c}^{\infty}({\mathbb{R}}^{d}_{+})\text{ with }\lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1.
Proof. By [7 , Theorem 2.1] (see also [11 , Theorem 3.4] and [22 , Theorem II.5] ), the result follows from Lemma 2.4 (iii). □ \Box
Proposition 2.6 .
There exists a constant C > 0 C>0 such that
∥ u ∥ L 2 ( ℝ + d , d x ) 2 ( 1 + α / d ) ≤ C ℰ 0 ( u , u ) for every u ∈ ℱ ¯ with ∥ u ∥ L 1 ( ℝ + d , d x ) ≤ 1 . \lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+\alpha/d)}\leq C{\cal E}^{0}(u,u)\quad\text{for every }\;u\in{\overline{\cal F}}\text{ with }\lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1.
(2.4)
Proof. We first assume that α < 1 \alpha<1 .
Let γ = β 1 ∨ β 2 \gamma=\beta_{1}\vee\beta_{2} .
Using Lemmas 2.5 and
2.4 (i)–(ii), the Hardy inequality in [32 , Proposition 3.2]
and (A3) (I),
we get that for any u ∈ C c ∞ ( ℝ + d ) u\in C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) with ∥ u ∥ L 1 ( ℝ + d , d x ) ≤ 1 \lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1 ,
∥ u ∥ L 2 ( ℝ + d , d x ) 2 ( 1 + α / d ) \displaystyle\lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+\alpha/d)}
≤ c 1 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) 2 B γ + 1 / 2 , γ + 1 / 2 , 0 , 0 ( x , y ) | x − y | d + α 𝑑 y 𝑑 x + c γ , α ∫ ℝ + d u ( x ) 2 x d − α 𝑑 x \displaystyle\leq c_{1}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))^{2}\frac{B_{\gamma+1/2,\gamma+1/2,0,0}(x,y)}{|x-y|^{d+\alpha}}dydx+c_{\gamma,\alpha}\int_{{\mathbb{R}}^{d}_{+}}u(x)^{2}x_{d}^{-\alpha}dx
≤ c 2 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) 2 B γ + 1 / 2 , γ + 1 / 2 , 0 , 0 ( x , y ) | x − y | d + α 𝑑 y 𝑑 x ≤ c 3 ℰ 0 ( u , u ) , \displaystyle\leq c_{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))^{2}\frac{B_{\gamma+1/2,\gamma+1/2,0,0}(x,y)}{|x-y|^{d+\alpha}}dydx\leq c_{3}{\cal E}^{0}(u,u),
where B γ + 1 / 2 , γ + 1 / 2 , 0 , 0 B_{\gamma+1/2,\gamma+1/2,0,0} is defined in (1 ). By Lemma 2.3 , ℱ ¯ {\overline{\cal F}} is the closure of C c ∞ ( ℝ + d ) C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) under ℰ 1 0 {\cal E}^{0}_{1} . Therefore, we conclude that (2.4 ) is true when α < 1 \alpha<1 .
Now, we assume that α ≥ 1 \alpha\geq 1 . Since (2.4 ) is valid when α < 1 \alpha<1 and C c ∞ ( ℝ ¯ + d ) ⊂ ℱ ¯ C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+})\subset{\overline{\cal F}} ,
we get that for every u ∈ C c ∞ ( ℝ ¯ + d ) u\in C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+}) with ∥ u ∥ L 1 ( ℝ + d , d x ) ≤ 1 \lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1 ,
ℰ 0 ( u , u ) \displaystyle{\cal E}^{0}(u,u)
≥ 1 2 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) 2 ℬ ( x , y ) | x − y | d + 1 / 2 𝟏 { | x − y | ≤ 1 } 𝑑 y 𝑑 x \displaystyle\geq\frac{1}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))^{2}\frac{{\cal B}(x,y)}{|x-y|^{d+1/2}}{\bf 1}_{\{|x-y|\leq 1\}}dydx
= 1 2 ∫ ℝ + d ∫ ℝ + d ( u ( x ) − u ( y ) ) 2 ℬ ( x , y ) | x − y | d + 1 / 2 ( 1 − 𝟏 { | x − y | > 1 } ) 𝑑 y 𝑑 x \displaystyle=\frac{1}{2}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}(u(x)-u(y))^{2}\frac{{\cal B}(x,y)}{|x-y|^{d+1/2}}(1-{\bf 1}_{\{|x-y|>1\}})dydx
≥ c 4 ∥ u ∥ L 2 ( ℝ + d , d x ) 2 ( 1 + 1 / ( 2 d ) ) − 2 C 2 ∥ u ∥ L 2 ( ℝ + d , d x ) 2 sup x ∈ ℝ + d ∫ ℝ + d , | x − y | > 1 d y | x − y | d + 1 / 2 \displaystyle\geq c_{4}\lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+1/(2d))}-2C_{2}\lVert u\rVert^{2}_{L^{2}({\mathbb{R}}^{d}_{+},dx)}\sup_{x\in{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+},|x-y|>1}\frac{dy}{|x-y|^{d+1/2}}
≥ c 4 ∥ u ∥ L 2 ( ℝ + d , d x ) 2 ( 1 + 1 / ( 2 d ) ) − c 5 ∥ u ∥ L 2 ( ℝ + d , d x ) 2 . \displaystyle\geq c_{4}\lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+1/(2d))}-c_{5}\lVert u\rVert^{2}_{L^{2}({\mathbb{R}}^{d}_{+},dx)}.
Since ℱ ¯ {\overline{\cal F}} is the closure of C c ∞ ( ℝ ¯ + d ) C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+}) under ℰ 1 0 {\cal E}^{0}_{1} , [7 , Theorem 2.1] yields that
∥ P ¯ t ∥ 1 → ∞ ≤ c 6 t − 2 d e c 5 t \lVert\overline{P}_{t}\rVert_{1\to\infty}\leq c_{6}t^{-2d}e^{c_{5}t} for all t > 0 t>0 .
By (2.1 ), it follows that
∥ P ¯ t ∥ 1 → ∞ = t − d / α ∥ P ¯ 1 ∥ 1 → ∞ ≤ c 6 e c 5 t − d / α \lVert\overline{P}_{t}\rVert_{1\to\infty}=t^{-d/\alpha}\lVert\overline{P}_{1}\rVert_{1\to\infty}\leq c_{6}e^{c_{5}}t^{-d/\alpha} for all t > 0 t>0 . Using [7 , Theorem 2.1] again, we conclude that (2.4 ) holds for α ≥ 1 \alpha\geq 1 and finish the proof. □ \Box
As a consequence of the Nash-type inequality (2.4 ), we get the existence and a priori upper bounds of the heat kernels p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y) of Y ¯ \overline{Y} and Y κ Y^{\kappa} respectively.
Proposition 2.7 .
Let κ ≥ 0 \kappa\geq 0 . The processes Y ¯ \overline{Y} and Y κ Y^{\kappa} have heat kernels p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y) defined on ( 0 , ∞ ) × ( ℝ ¯ + d ∖ 𝒩 ) × ( ℝ ¯ + d ∖ 𝒩 ) (0,\infty)\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})
and ( 0 , ∞ ) × ( ℝ + d ∖ 𝒩 ) × ( ℝ + d ∖ 𝒩 ) (0,\infty)\times({\mathbb{R}}^{d}_{+}\setminus{\cal N})\times({\mathbb{R}}^{d}_{+}\setminus{\cal N})
respectively, where 𝒩 ⊂ ℝ ¯ + d {\cal N}\subset\overline{\mathbb{R}}^{d}_{+} is a properly exceptional set for
Y ¯ \overline{Y} .
Moreover, there exists a constant C > 0 C>0 such that
p κ ( t , x , y ) ≤ p ¯ ( t , x , y ) , x , y ∈ ℝ + d ∖ 𝒩 p^{\kappa}(t,x,y)\leq\overline{p}(t,x,y),\quad x,y\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}
(2.5)
and
p ¯ ( t , x , y ) ≤ C ( t − d / α ∧ t | x − y | d + α ) , t > 0 , x , y ∈ ℝ ¯ + d ∖ 𝒩 . \overline{p}(t,x,y)\leq C\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),\quad t>0,\;x,y\in\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}.
(2.6)
Proof. By our Nash-type inequality (2.4 ) and [3 ] ,
Y ¯ \overline{Y} has a heat kernel p ¯ ( t , x , y ) \overline{p}(t,x,y) on ( 0 , ∞ ) × ( ℝ ¯ + d ∖ 𝒩 ) × ( ℝ ¯ + d ∖ 𝒩 ) (0,\infty)\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}) for a properly exceptional set 𝒩 {\cal N} and
p ¯ ( t , x , y ) ≤ c t − d / α , t > 0 , x , y ∈ ℝ ¯ + d ∖ 𝒩 . \overline{p}(t,x,y)\leq ct^{-d/\alpha},\quad t>0,\;x,y\in\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}.
(2.7)
Since ℬ ( x , y ) {\cal B}(x,y) is bounded above, using (2.7 ), we can follow the arguments given in [11 , Example 5.5] line by line and conclude that
(2.6 ) holds.
According to [32 ] (see the discussion before Lemma 2.1 there),
Y κ Y^{\kappa} can be realized as a subprocess of Y ¯ \overline{Y} .
Thus,
Y κ Y^{\kappa} has
a heat kernel p κ ( t , x , y ) p^{\kappa}(t,x,y) defined on
( 0 , ∞ ) × ( ℝ + d ∖ 𝒩 ) × ( ℝ + d ∖ 𝒩 ) (0,\infty)\times({\mathbb{R}}^{d}_{+}\setminus{\cal N})\times({\mathbb{R}}^{d}_{+}\setminus{\cal N}) and
we also obtain
(2.5 ).
□ \Box
For notational convenience, we extend
the domain of p κ ( t , x , y ) p^{\kappa}(t,x,y)
to ( 0 , ∞ ) × ( ℝ ¯ + d ∖ 𝒩 ) × ( ℝ ¯ + d ∖ 𝒩 ) (0,\infty)\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}) by letting p κ ( t , x , y ) = 0 p^{\kappa}(t,x,y)=0 if
x ∈ ∂ ℝ + d ∖ 𝒩 x\in\partial{\mathbb{R}}^{d}_{+}\setminus{\cal N} or y ∈ ∂ ℝ + d ∖ 𝒩 y\in\partial{\mathbb{R}}^{d}_{+}\setminus{\cal N} .
3. Parabolic Hölder regularity and consequences
For κ ≥ 0 \kappa\geq 0 and an open set D ⊂ ℝ ¯ + d D\subset{\overline{\mathbb{R}}}^{d}_{+} relative to the topology on ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+} ,
we denote by Y ¯ D \overline{Y}^{D} and P ¯ t D \overline{P}^{D}_{t} the part of the process Y ¯ \overline{Y} killed upon exiting D D and its semigroup, and by Y κ , D Y^{\kappa,D} and P t κ , D P_{t}^{\kappa,D} the part of the process Y κ Y^{\kappa} killed upon exiting D ∩ ℝ + d D\cap{\mathbb{R}}^{d}_{+} and its semigroup, respectively.
The Dirichlet forms of Y ¯ D \overline{Y}^{D} and Y κ , D Y^{\kappa,D} are ( ℰ 0 , ℱ ¯ D ) ({\cal E}^{0},\overline{{\cal F}}^{D}) and ( ℰ κ , ℱ κ , D ) ({\cal E}^{\kappa},{\cal F}^{\kappa,D}) , where
ℱ ¯ D = { u ∈ ℱ ¯ : u = 0 q.e. on ℝ ¯ + d ∖ D } \overline{{\cal F}}^{D}=\{u\in\overline{{\cal F}}:u=0\mbox{ q.\/e. on }{\overline{\mathbb{R}}}^{d}_{+}\setminus D\} and ℱ κ , D = { u ∈ ℱ κ : u = 0 q.e. on ℝ + d ∖ D } {\cal F}^{\kappa,D}=\{u\in{\cal F}^{\kappa}:u=0\mbox{ q.\/e. on }{\mathbb{R}}^{d}_{+}\setminus D\} respectively.
For u , v ∈ ℱ κ , D u,v\in{\cal F}^{\kappa,D} ,
ℰ κ ( u , v ) = 1 2 ∫ D ∫ D ( ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) J ( x , y ) d y d x + ∫ D u ( x ) v ( x ) κ D ( x ) d x , {\cal E}^{\kappa}(u,v)=\frac{1}{2}\int_{D}\int_{D}((u(x)-u(y))(v(x)-v(y))J(x,y)dydx+\int_{D}u(x)v(x)\kappa_{D}(x)dx,
(3.1)
where
κ D ( x ) = ∫ ℝ + d ∖ D J ( x , y ) 𝑑 y + κ x d − α . \kappa_{D}(x)=\int_{{\mathbb{R}}^{d}_{+}\setminus D}J(x,y)dy+\kappa x_{d}^{-\alpha}.
(3.2)
Let τ ¯ D = inf { t > 0 : Y ¯ t ∉ D } \overline{\tau}_{D}=\inf\{t>0:\overline{Y}_{t}\notin D\} and τ D κ = inf { t > 0 : Y t κ ∉ D ∩ ℝ + d } \tau^{\kappa}_{D}=\inf\{t>0:Y^{\kappa}_{t}\notin D\cap{\mathbb{R}}^{d}_{+}\} .
For x , y ∉ 𝒩 x,y\notin\mathcal{N} , let
p ¯ D ( t , x , y ) \displaystyle\overline{p}^{D}(t,x,y)
= p ¯ ( t , x , y ) − 𝔼 x [ p ¯ ( t − τ ¯ D , Y ¯ τ ¯ D , y ) ; τ ¯ D < t ] , \displaystyle=\overline{p}(t,x,y)-{\mathbb{E}}_{x}\big{[}\overline{p}(t-\overline{\tau}_{D},\overline{Y}_{\overline{\tau}_{D}},y);\,\overline{\tau}_{D}<t\big{]},
p κ , D ( t , x , y ) \displaystyle p^{\kappa,D}(t,x,y)
= p κ ( t , x , y ) − 𝔼 x [ p κ ( t − τ D κ , Y τ D κ κ , y ) ; τ D κ < t ] . \displaystyle=p^{\kappa}(t,x,y)-{\mathbb{E}}_{x}\big{[}p^{\kappa}(t-\tau_{D}^{\kappa},Y^{\kappa}_{\tau^{\kappa}_{D}},y);\,\tau_{D}^{\kappa}<t\big{]}.
(3.3)
By the strong Markov property, p ¯ D ( t , x , y ) \overline{p}^{D}(t,x,y) and p κ , D ( t , x , y ) p^{\kappa,D}(t,x,y) are the transition densities of Y ¯ D \overline{Y}^{D} and Y κ , D Y^{\kappa,D} respectively.
In case when D D is a relatively compact open subset of ℝ + d {\mathbb{R}}^{d}_{+} , one can show that the process Y κ , D Y^{\kappa,D} can start from every point in D D and provide an interior lower bound for its transition density. We accomplish this by identifying the semigroup P t κ , D P^{\kappa,D}_{t} with the Feynman-Kac semigroup of the part process on D D of an auxiliary process.
This idea has already been used in [30 , Subsection 3.1] . For the benefit of the reader, we repeat some of the details here.
Note that, unlike [30 , Subsection 3.1] , (A3) (II) is not assumed.
Recall that we denote by m d m_{d} the Lebesgue measure on ℝ d {\mathbb{R}}^{d} .
Proposition 3.1 .
If D D is a relatively compact open subset of ℝ + d {\mathbb{R}}^{d}_{+} , then Y κ , D Y^{\kappa,D}
has a transition density
p κ , D ( t , x , y ) p^{\kappa,D}(t,x,y) defined for any ( t , x , y ) ∈ ( 0 , ∞ ) × D × D (t,x,y)\in(0,\infty)\times D\times D .
Furthermore, for any T > 0 T>0 and b ∈ ( 0 , 1 ] b\in(0,1] , there exists
a constant C = C ( b , T , D ) > 0 C=C(b,T,D)>0 such that
p κ , D ( t , x , y ) ≥ C ( t − d / α ∧ t | x − y | d + α ) t ∈ ( 0 , T ) , x , y ∈ D such that δ D ( x ) ∧ δ D ( y ) > b t 1 / α . p^{\kappa,D}(t,x,y)\geq C\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\quad t\in(0,T),\;x,y\in D\textrm{ such that }\delta_{D}(x)\wedge\delta_{D}(y)>bt^{1/\alpha}.
Proof. For γ > 0 \gamma>0 let J γ : ℝ d × ℝ d → [ 0 , ∞ ] J_{\gamma}:{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}\to[0,\infty] be defined by J γ ( x , y ) = J ( x , y ) J_{\gamma}(x,y)=J(x,y) if x , y ∈ D x,y\in D , and J γ ( x , y ) = γ | x − y | − α − d J_{\gamma}(x,y)=\gamma|x-y|^{-\alpha-d} otherwise. It follows from (A3) (I) and the relative compactness of D D that J γ ( x , y ) ≍ | x − y | − α − d J_{\gamma}(x,y)\asymp|x-y|^{-\alpha-d} . Hence by [16 , Theorem 1.2] , there exists a Feller and strongly Feller process Z Z (that can start from every point in ℝ d ) {\mathbb{R}}^{d}) with a continuous transition density q ~ ( t , x , y ) \widetilde{q}(t,x,y) on ( 0 , ∞ ) × ℝ d × ℝ d (0,\infty)\times{\mathbb{R}}^{d}\times{\mathbb{R}}^{d} satisfying
c 1 − 1 ( t − d / α ∧ t | x − y | d + α ) ≤ q ~ ( t , x , y ) ≤ c 1 ( t − d / α ∧ t | x − y | d + α ) , t > 0 , x , y ∈ ℝ d , c_{1}^{-1}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\leq\widetilde{q}(t,x,y)\leq c_{1}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),\quad t>0,\;x,y\in{\mathbb{R}}^{d},
(3.4)
for some constant c 1 ≥ 1 c_{1}\geq 1 .
Denote the part of the process Z Z killed upon exiting D D by Z D Z^{D} . The Dirichlet form of Z D Z^{D} is ( 𝒞 , 𝒟 D ( 𝒞 ) ) ({\cal C},\mathcal{D}_{D}({\cal C})) , where for u , v ∈ 𝒟 D ( 𝒞 ) u,v\in{\cal D}_{D}({\cal C}) ,
𝒞 ( u , v ) \displaystyle{\cal C}(u,v)
= \displaystyle=
1 2 ∫ D ∫ D ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) J γ ( x , y ) 𝑑 y 𝑑 x + ∫ D u ( x ) v ( x ) κ D Z ( x ) 𝑑 x \displaystyle\frac{1}{2}\int_{D}\int_{D}(u(x)-u(y))(v(x)-v(y))J_{\gamma}(x,y)\,dy\,dx+\int_{D}u(x)v(x)\kappa^{Z}_{D}(x)\,dx
= \displaystyle=
1 2 ∫ D ∫ D ( u ( x ) − u ( y ) ) ( v ( x ) − v ( y ) ) J ( x , y ) 𝑑 y 𝑑 x + ∫ D u ( x ) v ( x ) κ D Z ( x ) 𝑑 x \displaystyle\frac{1}{2}\int_{D}\int_{D}(u(x)-u(y))(v(x)-v(y))J(x,y)\,dy\,dx+\int_{D}u(x)v(x)\kappa^{Z}_{D}(x)\,dx
with
κ D Z ( x ) = ∫ ℝ d ∖ D J γ ( x , y ) 𝑑 y = γ ∫ ℝ d ∖ D | x − y | − d − α 𝑑 y , x ∈ D , \kappa^{Z}_{D}(x)=\int_{{\mathbb{R}}^{d}\setminus D}J_{\gamma}(x,y)\,dy=\gamma\int_{{\mathbb{R}}^{d}\setminus D}|x-y|^{-d-\alpha}\,dy\,,\quad x\in D\,,
(3.5)
and 𝒟 D ( 𝒞 ) = { u ∈ 𝒟 ( 𝒞 ) : u = 0 q.e. on ℝ d ∖ D } \mathcal{D}_{D}({\cal C})=\{u\in\mathcal{D}({\cal C}):\,u=0\textrm{ q.e.~{}on }{\mathbb{R}}^{d}\setminus D\} .
Let δ D = dist ( D , ∂ ℝ + d ) \delta_{D}=\mathrm{dist}(D,\partial{\mathbb{R}}^{d}_{+}) and let V V be the δ D / 2 \delta_{D}/2 -neighborhood of D D , that is V := { x ∈ ℝ + d : dist ( x , D ) < δ D / 2 } V:=\{x\in{\mathbb{R}}^{d}_{+}:\,\mathrm{dist}(x,D)<\delta_{D}/2\} . Then
κ D ( x ) = ∫ ℝ + d ∖ V J ( x , y ) 𝑑 y + ∫ V ∖ D J ( x , y ) 𝑑 y + κ x d − α . \kappa_{D}(x)=\int_{{\mathbb{R}}^{d}_{+}\setminus V}J(x,y)\,dy+\int_{V\setminus D}J(x,y)\,dy+\kappa x_{d}^{-\alpha}\,.
It follows from (A3) (I) and the relative compactness of D D that c 2 | x − y | − d − α ≤ J ( x , y ) ≤ C 2 | x − y | − d − α c_{2}|x-y|^{-d-\alpha}\leq J(x,y)\leq C_{2}|x-y|^{-d-\alpha} for all x , y ∈ V x,y\in V with c 2 := c 2 ( D ) > 0 c_{2}:=c_{2}(D)>0 .
It is easy to see that
sup x ∈ D ∫ ℝ + d ∖ V J ( x , y ) d y = : c 3 < ∞ \sup_{x\in D}\int_{{\mathbb{R}}^{d}_{+}\setminus V}J(x,y)\,dy=:c_{3}\ <\infty . Therefore
c 2 ∫ V ∖ D | x − y | − d − α 𝑑 y ≤ κ D ( x ) ≤ c 3 + C 2 ∫ V ∖ D | x − y | − d − α 𝑑 y + c 4 , x ∈ U , c_{2}\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\leq\kappa_{D}(x)\leq c_{3}+C_{2}\int_{V\setminus D}|x-y|^{-d-\alpha}dy+c_{4}\,,\quad x\in U\,,
where c 4 := κ sup x ∈ D x d − α c_{4}:=\kappa\sup_{x\in D}x_{d}^{-\alpha} . Since
inf x ∈ D ∫ V ∖ D | x − y | − d − α d y ≥ m d ( V ∖ D ) diam ( V ) − d − α = : c 5 > 0 , \inf_{x\in D}\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\geq m_{d}(V\setminus D)\,\mathrm{diam}(V)^{-d-\alpha}=:c_{5}>0\,,
we conclude that
c 2 ∫ V ∖ D | x − y | − d − α 𝑑 y ≤ κ D ( x ) ≤ c 6 ∫ V ∖ D | x − y | − d − α 𝑑 y , x ∈ D . c_{2}\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\leq\kappa_{D}(x)\leq c_{6}\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\,,\quad x\in D\,.
Further, since
κ D Z ( x ) = γ ( ∫ ℝ d ∖ V | x − y | − d − α 𝑑 y + ∫ V ∖ D | x − y | − d − α 𝑑 y ) , x ∈ D \kappa_{D}^{Z}(x)=\gamma\bigg{(}\int_{{\mathbb{R}}^{d}\setminus V}|x-y|^{-d-\alpha}\,dy+\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\bigg{)}\,,\quad x\in D\,
and sup x ∈ D ∫ ℝ d ∖ V | x − y | − d − α d y = : c 7 < ∞ \sup_{x\in D}\int_{{\mathbb{R}}^{d}\setminus V}|x-y|^{-d-\alpha}\,dy=:c_{7}<\infty , we see that there is a constant c 8 > 0 c_{8}>0 such
∫ V ∖ D | x − y | − d − α 𝑑 y ≤ γ − 1 κ D Z ( x ) ≤ c 8 ∫ V ∖ D | x − y | − d − α 𝑑 y , x ∈ D , . \int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\leq\gamma^{-1}\kappa_{D}^{Z}(x)\leq c_{8}\int_{V\setminus D}|x-y|^{-d-\alpha}\,dy\,,\quad x\in D,.
It follows that
c 6 − 1 κ D ( x ) ≤ γ − 1 κ D Z ( x ) ≤ c 8 c 2 − 1 κ D ( x ) c_{6}^{-1}\kappa_{D}(x)\leq\gamma^{-1}\kappa_{D}^{Z}(x)\leq c_{8}c_{2}^{-1}\kappa_{D}(x)
for all x ∈ D x\in D
with positive constants c 2 , c 6 , c 8 c_{2},c_{6},c_{8} not depending on γ \gamma . Now we choose γ > 0 \gamma>0 so small that γ c 8 c 2 − 1 ≤ 1 \gamma c_{8}c_{2}^{-1}\leq 1 . With this choice we get that κ D Z ( x ) ≤ κ D ( x ) \kappa_{D}^{Z}(x)\leq\kappa_{D}(x) for all x ∈ D x\in D . In particular, with c 9 := γ c 6 − 1 c_{9}:=\gamma c_{6}^{-1} we see that
c 9 κ D ( x ) ≤ κ D Z ( x ) ≤ κ D ( x ) , x ∈ D . c_{9}\kappa_{D}(x)\leq\kappa_{D}^{Z}(x)\leq\kappa_{D}(x)\,,\qquad x\in D\,.
(3.6)
It follows that for u ∈ C c ∞ ( D ) u\in C_{c}^{\infty}(D) ,
ℰ 1 κ , D ( u , u ) \displaystyle{\cal E}^{\kappa,D}_{1}(u,u)
= \displaystyle=
ℰ κ , D ( u , u ) + ∫ U u ( x ) 2 𝑑 x \displaystyle{\cal E}^{\kappa,D}(u,u)+\int_{U}u(x)^{2}\,dx
= \displaystyle=
1 2 ∫ D ∫ D ( u ( x ) − u ( y ) ) 2 J ( x , y ) 𝑑 y 𝑑 x + ∫ D u ( x ) 2 κ D ( x ) 𝑑 x + ∫ U u ( x ) 2 𝑑 x \displaystyle\frac{1}{2}\int_{D}\int_{D}(u(x)-u(y))^{2}J(x,y)\,dy\,dx+\int_{D}u(x)^{2}\kappa_{D}(x)\,dx+\int_{U}u(x)^{2}dx
≍ \displaystyle\asymp
1 2 ∫ D ∫ D ( u ( x ) − u ( y ) ) 2 J γ ( x , y ) 𝑑 y 𝑑 x + ∫ D u ( x ) 2 κ D Z ( x ) 𝑑 x + ∫ D u ( x ) 2 𝑑 x \displaystyle\frac{1}{2}\int_{D}\int_{D}(u(x)-u(y))^{2}J_{\gamma}(x,y)\,dy\,dx+\int_{D}u(x)^{2}\kappa_{D}^{Z}(x)\,dx+\int_{D}u(x)^{2}dx
= \displaystyle=
𝒞 ( u , u ) + ∫ D u ( x ) 2 𝑑 x = 𝒞 1 ( u , u ) . \displaystyle{\cal C}(u,u)+\int_{D}u(x)^{2}dx={\cal C}_{1}(u,u)\,.
Since C c ∞ ( D ) C_{c}^{\infty}(D) is a core of both ( ℰ κ , ℱ κ , D ) ({\cal E}^{\kappa},{\cal F}^{\kappa,D}) and ( 𝒞 , 𝒟 D ( 𝒞 ) ) ({\cal C},\mathcal{D}_{D}({\cal C})) , we conclude that ℱ κ , D = 𝒟 D ( 𝒞 ) {\cal F}^{\kappa,D}=\mathcal{D}_{D}({\cal C}) .
Define κ ~ : D → ℝ \widetilde{\kappa}:D\to{\mathbb{R}} by
κ ~ ( x ) := κ D ( x ) − κ D Z ( x ) \widetilde{\kappa}(x):=\kappa_{D}(x)-\kappa_{D}^{Z}(x) , x ∈ D . x\in D.
By the choice of γ \gamma we have that κ ~ ≥ 0 \widetilde{\kappa}\geq 0 . On the other hand,
it follows from (3.5 ) and (3.6 ) that there is a constant c 10 > 0 c_{10}>0 such that
κ ~ ( x ) ≤ κ D ( x ) ≤ c 10 δ D ( x ) − α , x ∈ D . \widetilde{\kappa}(x)\leq\kappa_{D}(x)\leq c_{10}\delta_{D}(x)^{-\alpha}\,,\quad x\in D.
(3.7)
Let μ ( d x ) = κ ~ ( x ) d x \mu(dx)=\widetilde{\kappa}(x)\,dx be a measure on D D . Using (3.4 ) and (3.7 ) one can check that
μ ∈ 𝐊 1 ( D ) \mu\in{\bf K}_{1}(D) , where the class 𝐊 1 ( D ) {\bf K}_{1}(D) is defined in [20 , Definition
2.12] .
For any Borel function f : D → [ 0 , ∞ ) f:D\to[0,\infty) let
T t D , κ ~ f ( x ) = 𝔼 x [ exp ( − ∫ 0 t κ ~ ( Z s D ) 𝑑 s ) f ( Z t D ) ] , t > 0 , x ∈ D . T^{D,\widetilde{\kappa}}_{t}f(x)={\mathbb{E}}_{x}\left[\exp\left(-\int_{0}^{t}\widetilde{\kappa}(Z^{D}_{s})ds\right)f(Z^{D}_{t})\right],\quad t>0,\;x\in D.
(3.8)
be the Feynman-Kac transform of the semigroup corresponding to the killed process Z D Z^{D} .
By [20 , Theorem 2.15] , the semigroup ( T t D , κ ~ ) t ≥ 0 (T^{D,\widetilde{\kappa}}_{t})_{t\geq 0} has a transition density p Z , D ( t , x , y ) p^{Z,D}(t,x,y) (with respect to the Lebesgue measure) such that for every T > 0 T>0 and b ∈ ( 0 , 1 ] b\in(0,1] there exists a constant c 2 = c 2 ( b , T , D ) > 0 c_{2}=c_{2}(b,T,D)>0 such that
p Z , D ( t , x , y ) ≥ c 2 q ~ ( t , x , y ) , t ∈ ( 0 , T ) , x , y ∈ D such that δ D ( x ) ∧ δ D ( y ) > b t 1 / α . p^{Z,D}(t,x,y)\geq c_{2}\widetilde{q}(t,x,y),\quad t\in(0,T),\;x,y\in D\textrm{ such that }\delta_{D}(x)\wedge\delta_{D}(y)>bt^{1/\alpha}.
(3.9)
Finally, by computing the Dirichlet forms of Y κ , D Y^{\kappa,D} and Z D Z^{D} (for the latter use [23 , Theorem 6.1.2] ), we conclude that they coincide. This implies that P t κ . D = T t D , κ ~ P^{\kappa.D}_{t}=T^{D,\widetilde{\kappa}}_{t} .
Combining the lower bound in (3.4 ) with (3.9 ), the proof is complete.
□ \Box
Recall that 𝐞 d = ( 0 ~ , 1 ) {\mathbf{e}}_{d}=(\widetilde{0},1) .
Define for a ∈ ( 0 , 1 ] a\in(0,1] ,
S ( a ) = { ( z ~ , z d ) ∈ ℝ + d : | z ~ | < 2 / a , a / ( 2 a + 2 ) < z d < ( 2 a + 2 ) / a } . S(a)=\{(\widetilde{z},z_{d})\in{\mathbb{R}}^{d}_{+}:|\widetilde{z}|<2/a,\,a/(2a+2)<z_{d}<(2a+2)/a\}.
Note that B ( 𝐞 d , 1 / 8 ) ⊂ S ( a ) B({\mathbf{e}}_{d},1/8)\subset S(a) for all a ∈ ( 0 , 1 ] a\in(0,1] . Moreover, for any a ∈ ( 0 , 1 ] a\in(0,1] , x ∈ B ( 𝐞 d , 1 / 8 ) x\in B({\mathbf{e}}_{d},1/8) and y = ( y ~ , y d ) ∈ ℝ + d y=(\widetilde{y},y_{d})\in{\mathbb{R}}^{d}_{+} with x d ∧ y d > a | x − y | x_{d}\wedge y_{d}>a|x-y| , we have | y ~ | ≤ | x ~ | + | y − x | < 1 / 8 + x d / a < 2 / a |\widetilde{y}|\leq|\widetilde{x}|+|y-x|<1/8+x_{d}/a<2/a , y d + y d / a ≥ x d − | x − y | + | x − y | ≥ 7 / 8 y_{d}+y_{d}/a\geq x_{d}-|x-y|+|x-y|\geq 7/8 and y d ≤ x d + | x − y | ≤ x d + x d / a < ( 2 a + 2 ) / a y_{d}\leq x_{d}+|x-y|\leq x_{d}+x_{d}/a<(2a+2)/a . Thus,
{ ( t , ( x ~ , x d ) , ( y ~ , y d ) ) ∈ ℝ + 1 × B ( 𝐞 d , 1 / 8 ) × ℝ + d : x d ∧ y d > a ( t 1 / α ∨ | x − y | ) } \displaystyle\big{\{}(t,(\widetilde{x},x_{d}),(\widetilde{y},y_{d}))\in{\mathbb{R}}^{1}_{+}\times B({\mathbf{e}}_{d},1/8)\times{\mathbb{R}}^{d}_{+}:x_{d}\wedge y_{d}>a(t^{1/\alpha}\vee|x-y|)\big{\}}
⊂ ( 0 , 2 / a α ) × S ( a ) × S ( a ) for every a ∈ ( 0 , 1 ] . \displaystyle\subset(0,2/a^{\alpha})\times S(a)\times S(a)\qquad\text{for every}\;\,a\in(0,1].
(3.10)
Proposition 3.2 .
For any a ∈ ( 0 , 1 ] a\in(0,1] , there exists a constant C = C ( a ) > 0 C=C(a)>0
such that the following estimates hold: For any t > 0 t>0 and x ∈ ℝ + d ∖ 𝒩 x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} , there is a measurable set N t , x ⊂ ℝ + d N_{t,x}\subset{\mathbb{R}}^{d}_{+}
of zero Lebesgue measure
such that for all y ∈ ℝ + d ∖ N t , x y\in{\mathbb{R}}^{d}_{+}\setminus N_{t,x} with x d ∧ y d > a ( t 1 / α ∨ | x − y | ) x_{d}\wedge y_{d}>a(t^{1/\alpha}\vee|x-y|) ,
p ¯ ( t , x , y ) ≥ p κ ( t , x , y ) ≥ C ( t − d / α ∧ t | x − y | d + α ) . \overline{p}(t,x,y)\geq p^{\kappa}(t,x,y)\geq C\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
(3.11)
Proof. By (2.5 ),
it suffices to prove the second inequality in (3.11 ).
Since 𝒩 {\cal N} has
zero Lebesgue measure,
there exist r > 0 r>0 and z ~ ∈ ℝ d − 1 \widetilde{z}\in{\mathbb{R}}^{d-1} such that r − 1 ( x − ( z ~ , 0 ) ) ∈ B ( 𝐞 d , 1 / 8 ) ∖ 𝒩 r^{-1}(x-(\widetilde{z},0))\in B({\mathbf{e}}_{d},1/8)\setminus{\cal N} . By Lemmas 2.1 and 2.2 , we have
p κ ( t , x , y ) = r − d p κ ( r − α t , r − 1 ( x − ( z ~ , 0 ) ) , r − 1 ( y − ( z ~ , 0 ) ) ) for a.e. y ∈ ℝ + d . \displaystyle p^{\kappa}(t,x,y)=r^{-d}p^{\kappa}(r^{-\alpha}t,r^{-1}(x-(\widetilde{z},0)),r^{-1}(y-(\widetilde{z},0)))\qquad\text{for a.e.}\;\,y\in{\mathbb{R}}^{d}_{+}.
(3.12)
Therefore, by (3 ), it is enough to prove that there exists a constant c = c ( a ) > 0 c=c(a)>0 such that for all 0 < t < 2 / a α 0<t<2/a^{\alpha} ,
x ∈ B ( 𝐞 d , 1 / 8 ) ∖ 𝒩 x\in B({\mathbf{e}}_{d},1/8)\setminus\mathcal{N}
and a.e. y ∈ S ( a ) ∖ 𝒩 y\in S(a)\setminus{\cal N} ,
p κ , S ( a / 2 ) ( t , x , y ) ≥ c ( t − d / α ∧ t | x − y | d + α ) , \displaystyle p^{\,\kappa,S(a/2)}(t,x,y)\geq c\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),
(3.13)
which implies
the second inequality in (3.11 )
by (3.3 ) and (3.12 ).
Note that δ S ( a / 2 ) ( y ) ≥ a / ( a 2 + 3 a + 2 ) \delta_{S(a/2)}(y)\geq a/(a^{2}+3a+2) for every y ∈ S ( a ) y\in S(a) . Hence for all 0 < t < 2 / a α 0<t<2/a^{\alpha} and z , y ∈ S ( a / 2 ) z,y\in S(a/2) it holds that δ S ( a / 2 ) ( z ) ∧ δ S ( a / 2 ) ( y ) ≥ ( a 2 / ( 2 a 2 + 6 a + 4 ) ) t 1 / α \delta_{S(a/2)}(z)\wedge\delta_{S(a/2)}(y)\geq(a^{2}/(2a^{2}+6a+4))t^{1/\alpha} .
Now it follows from Proposition 3.1 (with b = ( a 2 / ( 2 a 2 + 6 a + 4 ) ) b=(a^{2}/(2a^{2}+6a+4)) ) that (3.13 ) holds.
□ \Box
As a direct consequence of Proposition 3.2 , for any a ∈ ( 0 , 1 ] a\in(0,1] , there exists a constant C > 0 C>0 such that
p ¯ ( t , x , y ) ≥ p κ ( t , x , y ) ≥ C t − d / α \overline{p}(t,x,y)\geq p^{\kappa}(t,x,y)\geq Ct^{-d/\alpha}
(3.14)
for all t > 0 t>0 , x ∈ ℝ + d ∖ 𝒩 x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} and a.e. y ∈ ℝ + d ∖ 𝒩 y\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} with x d ∧ y d > a t 1 / α x_{d}\wedge y_{d}>at^{1/\alpha} and | x − y | ≤ t 1 / α |x-y|\leq t^{1/\alpha} .
By repeating the proofs of [21 , Lemmas 6.1 and 6.3] , we obtain the following two results from (3.3 ), Proposition 2.7 and (3.14 ).
Lemma 3.3 .
There exist constants C > 0 C>0 and η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) such that
for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} , r ∈ ( 0 , x d ) r\in(0,x_{d}) , t ∈ ( 0 , ( η r ) α ] t\in(0,(\eta r)^{\alpha}] and z ∈ B ( x , η t 1 / α ) ∖ 𝒩 z\in B(x,\eta t^{1/\alpha})\setminus{\cal N} ,
p ¯ B ( x , r ) ( t , z , y ) ≥ p κ , B ( x , r ) ( t , z , y ) ≥ C t − d / α for a.e. y ∈ B ( x , η t 1 / α ) . \overline{p}^{B(x,r)}(t,z,y)\geq p^{\kappa,B(x,r)}(t,z,y)\geq Ct^{-d/\alpha}\quad\text{for a.e. }y\in B(x,\eta t^{1/\alpha}).
Lemma 3.4 .
There exists a constant C > 1 C>1 such that for all x ∈ ℝ + d ∖ 𝒩 x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} and r ∈ ( 0 , x d ) r\in(0,x_{d}) ,
C − 1 r α ≤ 𝔼 x [ τ B ( x , r ) κ ] ≤ sup z ∈ B ( x , r ) ∖ 𝒩 𝔼 z [ τ ¯ B ( x , r ) ] ≤ C r α . C^{-1}r^{\alpha}\leq{\mathbb{E}}_{x}[\tau^{\kappa}_{B(x,r)}]\leq\sup_{z\in B(x,r)\setminus{\cal N}}{\mathbb{E}}_{z}[\overline{\tau}_{B(x,r)}]\leq Cr^{\alpha}.
(3.15)
The Lévy system formula (see [23 , Theorem 5.3.1] and the arguments in [15 , p.40] )
state
that for any
non-negative Borel function F F on ℝ + d × ℝ + d {\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} vanishing on the diagonal and any stopping time T T for Y κ Y^{\kappa} , it holds that
𝔼 x ∑ s ≤ T F ( Y s − κ , Y s κ ) = 𝔼 x ∫ 0 T ∫ ℝ + d F ( Y s κ , y ) J ( Y s κ , y ) 𝑑 y 𝑑 s , x ∈ ℝ + d ∖ 𝒩 . \displaystyle{\mathbb{E}}_{x}\sum_{s\leq T}F(Y^{\kappa}_{s-},Y^{\kappa}_{s})={\mathbb{E}}_{x}\int^{T}_{0}\int_{{\mathbb{R}}^{d}_{+}}F(Y^{\kappa}_{s},y)J(Y^{\kappa}_{s},y)dyds,\quad x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}.
(3.16)
Here Y s − κ = lim t ↑ s Y t κ Y^{\kappa}_{s-}=\lim_{t\uparrow s}Y^{\kappa}_{t} denotes the left limit of the process Y Y at time s > 0 s>0 .
Similarly, for any
non-negative Borel function F F on ℝ ¯ + d × ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} vanishing on the diagonal and any stopping time T T for Y ¯ \overline{Y} , it holds that
𝔼 x ∑ s ≤ T F ( Y ¯ s − , Y ¯ s ) = 𝔼 x ∫ 0 T ∫ ℝ + d F ( Y ¯ s , y ) J ( Y ¯ s , y ) 𝑑 y 𝑑 s , x ∈ ℝ ¯ + d ∖ 𝒩 . \displaystyle{\mathbb{E}}_{x}\sum_{s\leq T}F(\overline{Y}_{s-},\overline{Y}_{s})={\mathbb{E}}_{x}\int^{T}_{0}\int_{{\mathbb{R}}^{d}_{+}}F(\overline{Y}_{s},y)J(\overline{Y}_{s},y)dyds,\quad x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N}.
(3.17)
See [31 , (3.3) and (3.4)] for a simper form following from (3.16 ), which will be used in this paper too.
For x = ( x ~ , x d ) ∈ ℝ ¯ + d x=(\widetilde{x},x_{d})\in{\overline{\mathbb{R}}}^{d}_{+} and t > 0 t>0 , we define
V x ( t ) \displaystyle V_{x}(t)
= { z = ( z ~ , z d ) ∈ ℝ d : | z ~ − x ~ | < 2 t 1 / α , z d ∈ [ 0 , 2 t 1 / α ) } , \displaystyle=\{z=(\widetilde{z},z_{d})\in{\mathbb{R}}^{d}:\,|\widetilde{z}-\widetilde{x}|<2t^{1/\alpha},\,z_{d}\in[0,2t^{1/\alpha})\},
W x ( t ) \displaystyle W_{x}(t)
= { z = ( z ~ , z d ) ∈ ℝ d : | z ~ − x ~ | < 2 t 1 / α , z d ∈ [ x d + 5 t 1 / α , x d + 8 t 1 / α ) } . \displaystyle=\{z=(\widetilde{z},z_{d})\in{\mathbb{R}}^{d}:\,|\widetilde{z}-\widetilde{x}|<2t^{1/\alpha},\,z_{d}\in[x_{d}+5t^{1/\alpha},x_{d}+8t^{1/\alpha})\}.
(3.18)
In dimension 1, we abuse notation and use V x ( t ) = [ 0 , 2 t 1 / α ) V_{x}(t)=[0,2t^{1/\alpha}) and
W x ( t ) = [ x + 5 t 1 / α , x + 8 t 1 / α ) W_{x}(t)=[x+5t^{1/\alpha},x+8t^{1/\alpha}) .
Lemma 3.5 .
(i)
There exists C > 0 C>0 such that for all t > 0 t>0 and x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} ,
ℙ x ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) ≥ C . {\mathbb{P}}_{x}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\big{)}\geq C.
(ii) There exists a constant C > 0 C>0 such that for all n ≥ 1 n\geq 1 , t > 0 t>0 and x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} ,
ℙ x ( τ ¯ V x ( t ) > n t ) ≤ 2 e − C n . {\mathbb{P}}_{x}(\overline{\tau}_{V_{x}(t)}>nt)\leq 2e^{-Cn}.
Proof. (i) Define V x ( t , r ) = { z ∈ V x ( t ) : δ V x ( t ) ( z ) > r t 1 / α } V_{x}(t,r)=\{z\in V_{x}(t):\delta_{V_{x}(t)}(z)>rt^{1/\alpha}\} for r > 0 r>0 . For any
r ∈ ( 0 , 1 ) r\in(0,1) , z ∈ V x ( t , r ) z\in V_{x}(t,r) , u ∈ B ( z , r t 1 / α / 2 ) u\in B(z,rt^{1/\alpha}/2) and w ∈ W x ( t ) w\in W_{x}(t) , we have
r t 1 / α / 2 ≤ u d ≤ 5 t 1 / α ≤ w d and | u − w | ≤ | u − z | + | z − x | + | x − w | < 16 t 1 / α . \displaystyle rt^{1/\alpha}/2\leq u_{d}\leq 5t^{1/\alpha}\leq w_{d}\quad\text{and}\quad|u-w|\leq|u-z|+|z-x|+|x-w|<16t^{1/\alpha}.
(3.19)
Using the Lévy system formula in (3.17 ), (A3) (I), (3.19 ) and Lemma 3.4 , we get that for any r ∈ ( 0 , 1 ) r\in(0,1) and z ∈ V x ( t , r ) ∖ 𝒩 z\in V_{x}(t,r)\setminus{\cal N} ,
ℙ z ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) \displaystyle{\mathbb{P}}_{z}(\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t))
≥ ℙ z ( Y ¯ τ ¯ B ( z , r t 1 / α / 2 ) ∈ W x ( t ) ) = 𝔼 z [ ∫ 0 τ ¯ B ( z , r t 1 / α / 2 ) ∫ W x ( t ) J ( Y ¯ s , w ) 𝑑 w 𝑑 s ] \displaystyle\geq{\mathbb{P}}_{z}(\overline{Y}_{\overline{\tau}_{B(z,rt^{1/\alpha}/2)}}\in W_{x}(t))={\mathbb{E}}_{z}\bigg{[}\int_{0}^{\overline{\tau}_{B(z,rt^{1/\alpha}/2)}}\int_{W_{x}(t)}J(\overline{Y}_{s},w)dwds\bigg{]}
≥ c 1 𝔼 z [ ∫ 0 τ ¯ B ( z , r t 1 / α / 2 ) ∫ W x ( t ) ( r / 2 16 ) β 1 ( 5 16 ) β 2 ( 16 t 1 / α ) − d − α 𝑑 w 𝑑 s ] \displaystyle\geq c_{1}{\mathbb{E}}_{z}\bigg{[}\int_{0}^{\overline{\tau}_{B(z,rt^{1/\alpha}/2)}}\int_{W_{x}(t)}\left(\frac{r/2}{16}\right)^{\beta_{1}}\left(\frac{5}{16}\right)^{\beta_{2}}(16t^{1/\alpha})^{-d-\alpha}dwds\bigg{]}
≥ c 2 r β 1 t − 1 − d / α 𝔼 z [ τ ¯ B ( z , r t 1 / α / 2 ) ] ∫ W x ( t ) 𝑑 w ≥ c 3 r β 1 + α . \displaystyle\geq c_{2}r^{\beta_{1}}t^{-1-d/\alpha}{\mathbb{E}}_{z}\big{[}\overline{\tau}_{B(z,rt^{1/\alpha}/2)}\big{]}\int_{W_{x}(t)}dw\geq c_{3}r^{\beta_{1}+\alpha}.
(3.20)
By Proposition 2.7 and [25 , Remark 3.3] , the condition (i) in [25 , Theorem 3.1] holds true with ρ ( s ) = s 1 / α \rho(s)=s^{1/\alpha} . Then, since Y ¯ \overline{Y} is conservative, by the implication (i) ⇒ \Rightarrow (ii) of [25 , Theorem 3.1] , there exists a constant ε 0 > 0 {\varepsilon}_{0}>0 independent of t t and x x such that
ℙ x ( τ ¯ V x ( t ) > ε 0 t ) ≥ 1 / 2 . \displaystyle{\mathbb{P}}_{x}(\overline{\tau}_{V_{x}}(t)>{\varepsilon}_{0}t)\geq 1/2.
(3.21)
On the other hand, for all r > 0 r>0 , we get from Proposition 2.7 that
ℙ x ( τ ¯ V x ( t ) ∖ V x ( t , r ) > ε 0 t ) ≤ ℙ x ( Y ¯ ε 0 t ∈ V x ( t ) ∖ V x ( t , r ) ) \displaystyle{\mathbb{P}}_{x}(\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r)}>{\varepsilon}_{0}t)\leq{\mathbb{P}}_{x}(\overline{Y}_{{\varepsilon}_{0}t}\in V_{x}(t)\setminus V_{x}(t,r))
≤ ∫ V x ( t ) ∖ V x ( t , r ) p ( ε 0 t , x , y ) 𝑑 y ≤ c 4 t − d / α ∫ V x ( t ) ∖ V x ( t , r ) 𝑑 y ≤ c 5 r . \displaystyle\leq\int_{V_{x}(t)\setminus V_{x}(t,r)}p({\varepsilon}_{0}t,x,y)dy\leq c_{4}t^{-d/\alpha}\int_{V_{x}(t)\setminus V_{x}(t,r)}dy\leq c_{5}r.
(3.22)
Set r 0 := 1 / ( 4 c 5 + 1 ) r_{0}:=1/(4c_{5}+1) . Using the strong Markov property and (3 )–(3 ), we obtain
ℙ x ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) ≥ ℙ x ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) , τ ¯ V x ( t ) ∖ V x ( t , r 0 ) ≤ ε 0 t < τ ¯ V x ( t ) ) \displaystyle{\mathbb{P}}_{x}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\big{)}\geq{\mathbb{P}}_{x}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t),\,\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r_{0})}\leq{\varepsilon}_{0}t<\overline{\tau}_{V_{x}(t)}\big{)}
≥ ℙ x ( ℙ Y ¯ τ ¯ V x ( t ) ∖ V x ( t , r 0 ) ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) : τ ¯ V x ( t ) ∖ V x ( t , r 0 ) ≤ ε 0 t < τ ¯ V x ( t ) ) \displaystyle\geq{\mathbb{P}}_{x}\Big{(}{\mathbb{P}}_{\overline{Y}_{\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r_{0})}}}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\big{)}\,:\,\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r_{0})}\leq{\varepsilon}_{0}t<\overline{\tau}_{V_{x}(t)}\Big{)}
≥ c 3 r 0 β 1 + α ℙ x ( τ ¯ V x ( t ) ∖ V x ( t , r 0 ) ≤ ε 0 t < τ ¯ V x ( t ) ) \displaystyle\geq c_{3}r_{0}^{\beta_{1}+\alpha}{\mathbb{P}}_{x}\left(\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r_{0})}\leq{\varepsilon}_{0}t<\overline{\tau}_{V_{x}(t)}\right)
≥ c 3 r 0 β 1 + α ( ℙ x ( τ ¯ V x ( t ) > ε 0 t ) − ℙ x ( τ ¯ V x ( t ) ∖ V x ( t , r 0 ) > ε 0 t ) ) ≥ c 3 r 0 β 1 + α / 4 . \displaystyle\geq c_{3}r_{0}^{\beta_{1}+\alpha}\left({\mathbb{P}}_{x}\left(\overline{\tau}_{V_{x}(t)}>{\varepsilon}_{0}t\right)-{\mathbb{P}}_{x}\left(\overline{\tau}_{V_{x}(t)\setminus V_{x}(t,r_{0})}>{\varepsilon}_{0}t\right)\right)\geq c_{3}r_{0}^{\beta_{1}+\alpha}/4.
(ii) By Proposition 2.7 , there exists a constant k 0 > 0 k_{0}>0 independent of t t and x x such that for any z ∈ V x ( t ) ∖ 𝒩 z\in V_{x}(t)\setminus{\cal N} ,
ℙ z ( τ ¯ V x ( t ) > k 0 t ) ≤ ℙ z ( Y ¯ k 0 t ∈ V x ( t ) ) ≤ ∫ V x ( t ) p ¯ ( k 0 t , z , y ) 𝑑 y ≤ c 1 t d / α ( k 0 t ) − d / α ≤ 1 2 . \displaystyle{\mathbb{P}}_{z}\big{(}\overline{\tau}_{V_{x}(t)}>k_{0}t\big{)}\leq{\mathbb{P}}_{z}\big{(}\overline{Y}_{k_{0}t}\in V_{x}(t)\big{)}\leq\int_{V_{x}(t)}\overline{p}(k_{0}t,z,y)dy\leq c_{1}t^{d/\alpha}(k_{0}t)^{-d/\alpha}\leq\frac{1}{2}.
(3.23)
For r > 0 r>0 , let ⌊ r ⌋ := sup { m ∈ ℕ : m ≤ r } \lfloor r\rfloor:=\sup\{m\in{\mathbb{N}}:m\leq r\} . Now for any n ≥ 1 n\geq 1 , using the Markov property and (3.23 ), we get that
ℙ x ( τ ¯ V x ( t ) > n t ) ≤ ℙ x ( τ ¯ V x ( t ) > ⌊ n / k 0 ⌋ k 0 t ) \displaystyle{\mathbb{P}}_{x}(\overline{\tau}_{V_{x}(t)}>nt)\leq{\mathbb{P}}_{x}(\overline{\tau}_{V_{x}(t)}>\lfloor n/k_{0}\rfloor k_{0}t)
≤ ℙ x ( ℙ X ( ⌊ n / k 0 ⌋ − 1 ) k 0 t ( τ ¯ V x ( t ) > k 0 t ) : τ ¯ V x ( t ) > ( ⌊ n / k 0 ⌋ − 1 ) k 0 t ) \displaystyle\leq{\mathbb{P}}_{x}\big{(}{\mathbb{P}}_{X_{(\lfloor n/k_{0}\rfloor-1)k_{0}t}}(\overline{\tau}_{V_{x}(t)}>k_{0}t):\overline{\tau}_{V_{x}(t)}>(\lfloor n/k_{0}\rfloor-1)k_{0}t\big{)}
≤ 2 − 1 ℙ x ( τ ¯ V x ( t ) > ( ⌊ n / k 0 ⌋ − 1 ) k 0 t ) ≤ ⋯ ≤ 2 − ⌊ n / k 0 ⌋ ≤ 2 e − ( log 2 ) n / k 0 . \displaystyle\leq 2^{-1}{\mathbb{P}}_{x}\big{(}\overline{\tau}_{V_{x}(t)}>(\lfloor n/k_{0}\rfloor-1)k_{0}t\big{)}\leq\cdots\leq 2^{-\lfloor n/k_{0}\rfloor}\leq 2e^{-(\log 2)n/k_{0}}.
□ \Box
Lemma 3.6 .
There exist constants M > 1 M>1 and C > 0 C>0 such that
for all t > 0 t>0 and x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} ,
ess inf z ∈ W x ( t ) p ¯ ( M t , x , z ) ≥ C t − d / α . \operatorname*{ess\,inf}_{z\in W_{x}(t)}\,\overline{p}(Mt,x,z)\geq Ct^{-d/\alpha}.
Proof. Suppose t > 0 t>0 and x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} .
For all w , z ∈ W x ( t ) w,z\in W_{x}(t) , we have w d ∧ z d ≥ 5 t 1 / α w_{d}\wedge z_{d}\geq 5t^{1/\alpha} and | w − z | < 7 t 1 / α |w-z|<7t^{1/\alpha} . Therefore, by Proposition 3.2 , for any M > 1 M>1 , there is c 1 = c 1 ( M ) > 0 c_{1}=c_{1}(M)>0 independent of t t and x x such that
for all w ∈ W x ( t ) ∖ 𝒩 w\in W_{x}(t)\setminus{\cal N} , z ∈ W x ( t ) z\in W_{x}(t) and ε ∈ ( 0 , t 1 / α ) {\varepsilon}\in(0,t^{1/\alpha}) ,
inf t ≤ s ≤ M t ℙ w ( Y ¯ s ∈ B ( z , ε ) ) = inf t ≤ s ≤ M t ∫ B ( z , ε ) p ¯ ( s , w , y ) 𝑑 y ≥ c 1 t − d / α m d ( B ( z , ε ) ) . \displaystyle\inf_{t\leq s\leq Mt}\,{\mathbb{P}}_{w}\big{(}\overline{Y}_{s}\in B(z,{\varepsilon})\big{)}=\inf_{t\leq s\leq Mt}\int_{B(z,{\varepsilon})}\overline{p}(s,w,y)dy\geq c_{1}t^{-d/\alpha}m_{d}(B(z,{\varepsilon})).
(3.24)
By the strong Markov property and (3.24 ), for all M > 1 M>1 , z ∈ W x ( t ) z\in W_{x}(t) and ε ∈ ( 0 , t 1 / α ) {\varepsilon}\in(0,t^{1/\alpha}) ,
ℙ x ( Y ¯ M t ∈ B ( z , ε ) ) ≥ 𝔼 x [ ℙ Y ¯ τ ¯ V x ( t ) ( Y ¯ M t − τ ¯ V x ( t ) ∈ B ( z , ε ) ) : τ ¯ V x ( t ) ≤ ( M − 1 ) t , Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ] \displaystyle{\mathbb{P}}_{x}\left(\overline{Y}_{Mt}\in B(z,{\varepsilon})\right)\geq{\mathbb{E}}_{x}\left[{\mathbb{P}}_{\overline{Y}_{\overline{\tau}_{V_{x}(t)}}}\big{(}\overline{Y}_{Mt-\overline{\tau}_{V_{x}(t)}}\in B(z,{\varepsilon})\big{)}:\overline{\tau}_{V_{x}(t)}\leq(M-1)t,\,\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\right]
≥ ( inf t ≤ s ≤ M t inf w ∈ W x ( t ) ∖ 𝒩 ℙ w ( Y ¯ s ∈ B ( z , ε ) ) ) ℙ x ( τ ¯ V x ( t ) ≤ ( M − 1 ) t , Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) \displaystyle\geq\Big{(}\inf_{t\leq s\leq Mt}\,\inf_{w\in W_{x}(t)\setminus{\cal N}}{\mathbb{P}}_{w}\big{(}\overline{Y}_{s}\in B(z,{\varepsilon})\big{)}\Big{)}\,{\mathbb{P}}_{x}\left(\overline{\tau}_{V_{x}(t)}\leq(M-1)t,\,\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\right)
≥ c 1 t − d / α m d ( B ( z , ε ) ) ( ℙ x ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) − ℙ x ( τ ¯ V x ( t ) > ( M − 1 ) t ) ) . \displaystyle\geq c_{1}t^{-d/\alpha}m_{d}(B(z,{\varepsilon}))\left({\mathbb{P}}_{x}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\big{)}-{\mathbb{P}}_{x}\big{(}\overline{\tau}_{V_{x}(t)}>(M-1)t\big{)}\right).
(3.25)
By Lemma 3.5 (i)-(ii), there are constants c 2 , c 3 , c 4 > 0 c_{2},c_{3},c_{4}>0 such that for all M > 1 M>1 ,
ℙ x ( Y ¯ τ ¯ V x ( t ) ∈ W x ( t ) ) − ℙ x ( τ ¯ V x ( t ) > ( M − 1 ) t ) ≥ c 2 − c 3 e − c 4 M . \displaystyle{\mathbb{P}}_{x}\big{(}\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)\big{)}-{\mathbb{P}}_{x}\big{(}\overline{\tau}_{V_{x}(t)}>(M-1)t\big{)}\geq c_{2}-c_{3}e^{-c_{4}M}.
Choosing M = c 4 − 1 log ( 2 c 3 / c 2 ) + 1 M=c_{4}^{-1}\log(2c_{3}/c_{2})+1 , we arrive at the result by (3 )
and the Lebesgue differentiation theorem. □ \Box
Lemma 3.7 .
There exists C > 0 C>0 such that for all t > 0 t>0
and x , y ∈ ℝ ¯ + d ∖ 𝒩 x,y\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} with | x − y | ≤ t 1 / α |x-y|\leq t^{1/\alpha} ,
p ¯ ( t , x , y ) ≥ C t − d / α . \displaystyle\overline{p}(t,x,y)\geq Ct^{-d/\alpha}.
Proof. Let M > 1 M>1 be the constant in
Lemma 3.6 .
Note that for all ( z , w ) ∈ W x ( t / M ) × W y ( t / M ) (z,w)\in W_{x}(t/M)\times W_{y}(t/M) , we have z d ∧ w d ≥ 5 ( t / M ) 1 / α z_{d}\wedge w_{d}\geq 5(t/M)^{1/\alpha} and | z − w | ≤ | z − x | + | x − y | + | y − w | ≤ 21 ( t / M ) 1 / α |z-w|\leq|z-x|+|x-y|+|y-w|\leq 21(t/M)^{1/\alpha} .
Therefore, by Proposition 3.2 , there is c 1 > 0 c_{1}>0 independent of t , x , y t,x,y such that
ess inf z ∈ W x ( t / ( 3 M ) ) ess inf w ∈ W y ( t / ( 3 M ) ) p ¯ ( t / 3 , z , w ) ≥ c 1 t − d / α . \displaystyle\operatorname*{ess\,inf}_{z\in W_{x}(t/(3M))}\operatorname*{ess\,inf}_{w\in W_{y}(t/(3M))}\overline{p}(t/3,z,w)\geq c_{1}t^{-d/\alpha}.
(3.26)
By the semigroup property, (3.26 ) and Lemma 3.6 ,
p ¯ ( t , x , y ) \displaystyle\overline{p}(t,x,y)
≥ ∫ W x ( t / ( 3 M ) ) ∫ W y ( t / ( 3 M ) ) p ¯ ( t / 3 , x , z ) p ¯ ( t / 3 , z , w ) p ¯ ( t / 3 , w , y ) 𝑑 z 𝑑 w \displaystyle\geq\int_{W_{x}(t/(3M))}\int_{W_{y}(t/(3M))}\overline{p}(t/3,x,z)\overline{p}(t/3,z,w)\overline{p}(t/3,w,y)dzdw
≥ ( ess inf z ∈ W x ( t / ( 3 M ) ) p ¯ ( t / 3 , x , z ) ) ( ess inf w ∈ W y ( t / ( 3 M ) ) p ¯ ( t / 3 , y , w ) ) \displaystyle\geq\left(\operatorname*{ess\,inf}_{z\in W_{x}(t/(3M))}\overline{p}(t/3,x,z)\right)\left(\operatorname*{ess\,inf}_{w\in W_{y}(t/(3M))}\overline{p}(t/3,y,w)\right)
× ( ess inf z ∈ W x ( t / ( 3 M ) ) ess inf w ∈ W y ( t / ( 3 M ) ) p ¯ ( t / 3 , z , w ) ) ∫ W x ( t / ( 3 M ) ) ∫ W y ( t / ( 3 M ) ) 𝑑 z 𝑑 w \displaystyle\qquad\times\left(\operatorname*{ess\,inf}_{z\in W_{x}(t/(3M))}\operatorname*{ess\,inf}_{w\in W_{y}(t/(3M))}\overline{p}(t/3,z,w)\right)\int_{W_{x}(t/(3M))}\int_{W_{y}(t/(3M))}dzdw
≥ c 2 t − 3 d / α + 2 d / α = c 2 t − d / α . \displaystyle\geq c_{2}t^{-3d/\alpha+2d/\alpha}=c_{2}t^{-d/\alpha}.
The proof is complete.
□ \Box
For x ∈ ℝ ¯ + d x\in{\overline{\mathbb{R}}}^{d}_{+} and r > 0 r>0 , we denote B + ( x , r ) := B ( x , r ) ∩ ℝ ¯ + d {B_{+}}(x,r):=B(x,r)\cap{\overline{\mathbb{R}}}^{d}_{+} .
We observe that
m d ( B + ( x , r ) ) ≍ r d for all x ∈ ℝ ¯ + d , r > 0 . \displaystyle m_{d}({B_{+}}(x,r))\asymp r^{d}\qquad\text{for all}\;\,x\in{\overline{\mathbb{R}}}^{d}_{+},\,r>0.
(3.27)
Now, using (2.6 ) and Lemma 3.7 (instead of (3.14 )), we extend the results in Lemmas 3.3 and 3.4 to Y ¯ \overline{Y} removing the restrictions on x x and r r .
Lemma 3.8 .
There exist constants C > 0 C>0 and η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) such that for all x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} , r > 0 r>0 , t ∈ ( 0 , ( η r ) α ] t\in(0,(\eta r)^{\alpha}] and z ∈ B + ( x , η t 1 / α ) ∖ 𝒩 z\in{B_{+}}(x,\eta t^{1/\alpha})\setminus{\cal N} ,
p ¯ B + ( x , r ) ( t , z , y ) ≥ C t − d / α for a.e. y ∈ B + ( x , η t 1 / α ) ∖ 𝒩 . \overline{p}^{{B_{+}}(x,r)}(t,z,y)\geq Ct^{-d/\alpha}\quad\text{for a.e. }y\in{B_{+}}(x,\eta t^{1/\alpha})\setminus{\cal N}.
Lemma 3.9 .
There exists a constant C > 1 C>1 such that for all
x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N}
and r > 0 r>0 ,
C − 1 r α ≤ 𝔼 x [ τ ¯ B + ( x , r ) ] ≤ sup z ∈ B + ( x , r ) ∖ 𝒩 𝔼 z [ τ ¯ B + ( x , r ) ] ≤ C r α . C^{-1}r^{\alpha}\leq{\mathbb{E}}_{x}\left[\overline{\tau}_{{B_{+}}(x,r)}\right]\leq\sup_{z\in{B_{+}}(x,r)\setminus{\cal N}}{\mathbb{E}}_{z}\left[\overline{\tau}_{{B_{+}}(x,r)}\right]\leq Cr^{\alpha}.
(3.28)
Let X ¯ := ( T s , Y ¯ s ) s ≥ 0 \overline{X}:=(T_{s},\overline{Y}_{s})_{s\geq 0} and X κ := ( T s , Y s κ ) s ≥ 0 X^{\kappa}:=(T_{s},Y^{\kappa}_{s})_{s\geq 0} be time-space processes where T s = T 0 − s T_{s}=T_{0}-s .
The law of the time-space process s ↦ X ¯ s s\mapsto\overline{X}_{s} or s ↦ X s κ s\mapsto X^{\kappa}_{s} starting from ( t , x ) (t,x) will be denoted by
ℙ ( t , x ) {\mathbb{P}}_{(t,x)} .
For
every open subset U U of [ 0 , ∞ ) × ℝ d [0,\infty)\times{\mathbb{R}}^{d} , define τ ¯ U = inf { s > 0 : X ¯ s ∉ U } \overline{\tau}_{U}=\inf\{s>0:\,\overline{X}_{s}\notin U\} and τ U κ = inf { s > 0 : X s κ ∉ U } \tau^{\kappa}_{U}=\inf\{s>0:\,X_{s}^{\kappa}\notin U\} .
We also define σ ¯ U = inf { s > 0 : X ¯ s ∈ U } \overline{\sigma}_{U}=\inf\{s>0:\,\overline{X}_{s}\in U\} and σ U κ = inf { s > 0 : X s κ ∈ U } \sigma^{\kappa}_{U}=\inf\{s>0:\,X_{s}^{\kappa}\in U\} .
A Borel function u : [ 0 , ∞ ) × ℝ ¯ + d → ℝ u:[0,\infty)\times\overline{\mathbb{R}}^{d}_{+}\to{\mathbb{R}} is
said to be parabolic in ( a , b ] × B + ( x , r ) ⊂ ( 0 , ∞ ) × ℝ ¯ + d (a,b]\times{B_{+}}(x,r)\subset(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}
with respect to Y ¯ \overline{Y} if for every relatively compact open set U ⊂ ( a , b ] × B + ( x , r ) U\subset(a,b]\times{B_{+}}(x,r) with respect to the topology on [ 0 , ∞ ) × ℝ ¯ + d [0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+} ,
it holds that
u ( t , z ) = 𝔼 ( t , z ) u ( X ¯ τ ¯ U ) u(t,z)={\mathbb{E}}_{(t,z)}u(\overline{X}_{\overline{\tau}_{U}}) for all
( t , z ) ∈ U (t,z)\in U with z ∉ 𝒩 z\notin{\cal N} .
Similarly, a Borel function u : [ 0 , ∞ ) × ℝ + d → ℝ u:[0,\infty)\times{\mathbb{R}}^{d}_{+}\to{\mathbb{R}} is
said to be parabolic in ( a , b ] × B ( x , r ) ⊂ ( 0 , ∞ ) × ℝ + d (a,b]\times B(x,r)\subset(0,\infty)\times{\mathbb{R}}^{d}_{+}
with respect to Y κ Y^{\kappa} if for every relatively compact open set U ⊂ ( a , b ] × B ( x , r ) U\subset(a,b]\times B(x,r) , it holds that
u ( t , z ) = 𝔼 ( t , z ) u ( X τ U κ κ ) u(t,z)={\mathbb{E}}_{(t,z)}u(X^{\kappa}_{\tau^{\kappa}_{U}}) for all
( t , z ) ∈ U (t,z)\in U with z ∉ 𝒩 z\notin{\cal N} .
Lemma 3.10 .
(i) Let η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) be the constant from Lemma 3.3 .
For every δ ∈ ( 0 , η ] \delta\in(0,\eta] , there exists a constant C > 0 C>0 such that for all x ∈ ℝ + d ∖ 𝒩 x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} , r ∈ ( 0 , x d ) r\in(0,x_{d}) ,
t ≥ δ r α t\geq\delta r^{\alpha} , and any compact set A ⊂ [ t − δ r α , t − δ r α / 2 ] × B ( x , ( η δ / 2 ) 1 / α r ) A\subset[t-\delta r^{\alpha},t-\delta r^{\alpha}/2]\times B(x,(\eta\delta/2)^{1/\alpha}r) ,
ℙ ( t , x ) ( σ A κ < τ [ t − δ r α , t ] × B ( x , r ) κ ) ≥ C m d + 1 ( A ) r d + α . {\mathbb{P}}_{(t,x)}\big{(}\sigma^{\kappa}_{A}<\tau^{\kappa}_{[t-\delta r^{\alpha},t]\times B(x,r)}\big{)}\geq C\frac{m_{d+1}(A)}{r^{d+\alpha}}.
(ii) Let η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) be the constant from Lemma 3.8 .
For every δ ∈ ( 0 , η ] \delta\in(0,\eta] , there exists a constant C > 0 C>0 such that for all x ∈ ℝ ¯ + d ∖ 𝒩 x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} , r > 0 r>0 ,
t ≥ δ r α t\geq\delta r^{\alpha} , and any compact set A ⊂ [ t − δ r α , t − δ r α / 2 ] × B + ( x , ( η δ / 2 ) 1 / α r ) A\subset[t-\delta r^{\alpha},t-\delta r^{\alpha}/2]\times{B_{+}}(x,(\eta\delta/2)^{1/\alpha}r) ,
ℙ ( t , x ) ( σ ¯ A < τ ¯ [ t − δ r α , t ] × B + ( x , r ) ) ≥ C m d + 1 ( A ) r d + α . {\mathbb{P}}_{(t,x)}\left(\overline{\sigma}_{A}<\overline{\tau}_{[t-\delta r^{\alpha},t]\times{B_{+}}(x,r)}\right)\geq C\frac{m_{d+1}(A)}{r^{d+\alpha}}.
Proof. By repeating the proofs of [21 , Lemma 6.5]
(using the Lévy system formulas in (3.16 ) and (3.17 )),
we deduce the results from Lemmas 3.3 and 3.8 respectively.
□ \Box
Theorem 3.11 .
(i) For any δ ∈ ( 0 , 1 ) \delta\in(0,1) ,
there exist λ ∈ ( 0 , 1 ] \lambda\in(0,1] and C > 0 C>0 such that
for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} ,
r ∈ ( 0 , x d ) r\in(0,x_{d}) , t 0 ≥ 0 t_{0}\geq 0 , and any function u u on ( 0 , ∞ ) × ℝ + d (0,\infty)\times{\mathbb{R}}^{d}_{+} which is parabolic in ( t 0 , t 0 + r α ] × B ( x , r ) (t_{0},t_{0}+r^{\alpha}]\times B(x,r) with respect to Y κ Y^{\kappa} and bounded in ( t 0 , t 0 + r α ] × ℝ + d (t_{0},t_{0}+r^{\alpha}]\times{\mathbb{R}}^{d}_{+} , we have
| u ( s , y ) − u ( t , z ) | ≤ C ( | s − t | 1 / α + | y − z | r ) λ ess sup [ t 0 , t 0 + r α ] × ℝ + d | u | , |u(s,y)-u(t,z)|\leq C\bigg{(}\frac{|s-t|^{1/\alpha}+|y-z|}{r}\bigg{)}^{\lambda}\operatorname*{ess\,sup}_{[t_{0},t_{0}+r^{\alpha}]\times{\mathbb{R}}^{d}_{+}}|u|,
(3.29)
for every s , t ∈ ( t 0 + ( 1 − δ α ) r α , t 0 + r α ] s,t\in(t_{0}+(1-\delta^{\alpha})r^{\alpha},t_{0}+r^{\alpha}] and y , z ∈ B ( x , δ r ) ∖ 𝒩 y,z\in B(x,\delta r)\setminus{\cal N} .
(ii) For any δ ∈ ( 0 , 1 ) \delta\in(0,1) ,
there exist λ ∈ ( 0 , 1 ] \lambda\in(0,1] and C > 0 C>0 such that
for all
x ∈ ℝ ¯ + d x\in{\overline{\mathbb{R}}}^{d}_{+} , r > 0 r>0 , t 0 ≥ 0 t_{0}\geq 0 , and any function u u on ( 0 , ∞ ) × ℝ ¯ + d (0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+} which is parabolic in ( t 0 , t 0 + r α ] × B + ( x , r ) (t_{0},t_{0}+r^{\alpha}]\times{B_{+}}(x,r) with respect to Y ¯ \overline{Y} and bounded in ( t 0 , t 0 + r α ] × ℝ ¯ + d (t_{0},t_{0}+r^{\alpha}]\times\overline{\mathbb{R}}^{d}_{+} , ( 3.29 ) holds true
for every s , t ∈ ( t 0 + ( 1 − δ α ) r α , t 0 + r α ] s,t\in(t_{0}+(1-\delta^{\alpha})r^{\alpha},t_{0}+r^{\alpha}] and
y , z ∈ B + ( x , δ r ) ∖ 𝒩 y,z\in{B_{+}}(x,\delta r)\setminus{\cal N} .
Proof. Using Lemmas
3.3 , 3.4 and 3.10 (i) for (i), and (3.27 ) and Lemmas
3.8 , 3.9 and 3.10 (ii) for (ii), we get the desired results
using the same argument as in the proof of [15 , Theorem 4.14] (see also the proof of [18 , Proposition 3.8] ). We omit details here.
□ \Box
From Proposition 2.7 , since the exceptional set 𝒩 {\cal N} is removed, we obtain
Corollary 3.13 .
If d > α d>\alpha ,
G κ ( x , y ) ≤ G ¯ ( x , y ) ≤ c | x − y | d − α , x , y ∈ ℝ ¯ + d . G^{\kappa}(x,y)\leq\overline{G}(x,y)\leq\frac{c}{|x-y|^{d-\alpha}},\quad x,y\in{\overline{\mathbb{R}}}_{+}^{d}.
Since the heat kernels p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y) are jointly continuous, we get the next lemma from the strong Markov properties of Y ¯ \overline{Y} and Y κ Y^{\kappa} .
This lemma is a refined version of
[13 , Lemma 4.2] which was inspired by [35 ] .
In this paper, this lemma will play an important role in the bootstrap method to prove sharp upper estimates on the heat kernels.
Although the proof of next lemma is standard, we give it for the reader’s convenience.
Lemma 3.15 .
(i) Let V 1 V_{1} and V 3 V_{3} be open subsets of ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+} with dist ( V 1 , V 3 ) > 0 {\rm dist}(V_{1},V_{3})>0 . Set V 2 := ℝ ¯ + d ∖ ( V 1 ∪ V 3 ) V_{2}:={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}) . For any x ∈ V 1 x\in V_{1} , y ∈ V 3 y\in V_{3} and t > 0 t>0 , it holds that
p ¯ ( t , x , y ) \displaystyle\overline{p}(t,x,y)
≤ ℙ x ( τ ¯ V 1 < t ) sup s ≤ t , z ∈ V 2 p ¯ ( s , z , y ) \displaystyle\leq{\mathbb{P}}_{x}(\overline{\tau}_{V_{1}}<t)\sup_{s\leq t,\,z\in V_{2}}\overline{p}(s,z,y)
+ dist ( V 1 , V 3 ) − d − α ∫ 0 t ∫ V 3 ∫ V 1 p ¯ V 1 ( t − s , x , u ) ℬ ( u , w ) p ¯ ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s . \displaystyle\quad+{\rm dist}(V_{1},V_{3})^{-d-\alpha}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}\overline{p}^{V_{1}}(t-s,x,u){\cal B}(u,w)\overline{p}(s,y,w)dudwds.
(ii) Let V 1 V_{1} and V 3 V_{3} be open subsets of ℝ + d {\mathbb{R}}^{d}_{+} with dist ( V 1 , V 3 ) > 0 {\rm dist}(V_{1},V_{3})>0 . Set V 2 := ℝ + d ∖ ( V 1 ∪ V 3 ) V_{2}:={\mathbb{R}}^{d}_{+}\setminus(V_{1}\cup V_{3}) . For any x ∈ V 1 x\in V_{1} , y ∈ V 3 y\in V_{3} and t > 0 t>0 , it holds that
p κ ( t , x , y ) \displaystyle p^{\kappa}(t,x,y)
≤ ℙ x ( τ V 1 κ < t < ζ κ ) sup s ≤ t , z ∈ V 2 p κ ( s , z , y ) \displaystyle\leq{\mathbb{P}}_{x}(\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa})\sup_{s\leq t,\,z\in V_{2}}p^{\kappa}(s,z,y)
+ dist ( V 1 , V 3 ) − d − α ∫ 0 t ∫ V 3 ∫ V 1 p κ , V 1 ( t − s , x , u ) ℬ ( u , w ) p κ ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s . \displaystyle\quad+{\rm dist}(V_{1},V_{3})^{-d-\alpha}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{\kappa,V_{1}}(t-s,x,u){\cal B}(u,w)p^{\kappa}(s,y,w)dudwds.
Proof. Since the proofs are the same, we only give the proof of (ii).
By the strong Markov property, the Lévy system formula in (3.16 ) and symmetry, we get that for any x ∈ V 1 x\in V_{1} and y ∈ V 3 y\in V_{3} ,
p κ ( t , x , y ) \displaystyle p^{\kappa}(t,x,y)
= 𝔼 x [ p κ ( t − τ V 1 κ , Y τ V 1 κ κ , y ) : τ V 1 κ < t < ζ κ ] \displaystyle={\mathbb{E}}_{x}\left[p^{\kappa}(t-\tau^{\kappa}_{V_{1}},Y^{\kappa}_{\tau^{\kappa}_{V_{1}}},y):\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa}\right]
= 𝔼 x [ p κ ( t − τ V 1 κ , Y τ V 1 κ κ , y ) : τ V 1 κ < t < ζ κ , Y τ V 1 κ κ ∈ V 2 ] \displaystyle={\mathbb{E}}_{x}\left[p^{\kappa}(t-\tau^{\kappa}_{V_{1}},Y^{\kappa}_{\tau^{\kappa}_{V_{1}}},y):\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa},Y^{\kappa}_{\tau^{\kappa}_{V_{1}}}\in V_{2}\right]
+ 𝔼 x [ p κ ( t − τ V 1 κ , Y τ V 1 κ κ , y ) : τ V 1 κ < t < ζ κ , Y τ V 1 κ κ ∈ V 3 ] \displaystyle\quad+{\mathbb{E}}_{x}\left[p^{\kappa}(t-\tau^{\kappa}_{V_{1}},Y^{\kappa}_{\tau^{\kappa}_{V_{1}}},y):\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa},Y^{\kappa}_{\tau^{\kappa}_{V_{1}}}\in V_{3}\right]
≤ ℙ x ( τ V 1 κ < t < ζ κ ) sup s ≤ t , z ∈ V 2 p κ ( s , z , y ) \displaystyle\leq{\mathbb{P}}_{x}(\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa})\sup_{s\leq t,\,z\in V_{2}}p^{\kappa}(s,z,y)
+ ∫ 0 t ∫ V 3 ∫ V 1 p κ , V 1 ( t − s , x , u ) ℬ ( u , w ) | u − w | d + α p κ ( s , w , y ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\quad+\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{\kappa,V_{1}}(t-s,x,u)\frac{{\cal B}(u,w)}{|u-w|^{d+\alpha}}p^{\kappa}(s,w,y)dudwds
≤ ℙ x ( τ V 1 κ < t < ζ κ ) sup s ≤ t , z ∈ V 2 p κ ( s , z , y ) \displaystyle\leq{\mathbb{P}}_{x}(\tau^{\kappa}_{V_{1}}<t<\zeta^{\kappa})\sup_{s\leq t,\,z\in V_{2}}p^{\kappa}(s,z,y)
+ dist ( V 1 , V 3 ) − d − α ∫ 0 t ∫ V 3 ∫ V 1 p κ , V 1 ( t − s , x , u ) ℬ ( u , w ) p κ ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s . \displaystyle\quad+{\rm dist}(V_{1},V_{3})^{-d-\alpha}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{\kappa,V_{1}}(t-s,x,u){\cal B}(u,w)p^{\kappa}(s,y,w)dudwds.
□ \Box
4. Parabolic Harnack inequality and preliminary lower bound of heat kernels
In this section, we prove that the parabolic Harnack inequality holds for
Y ¯ \overline{Y} and Y κ Y^{\kappa} , and get some preliminary lower bounds for
the heat kernels p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y) .
Recall that B + ( x , r ) = B ( x , r ) ∩ ℝ ¯ + d {B_{+}}(x,r)=B(x,r)\cap{\overline{\mathbb{R}}}^{d}_{+} .
Lemma 4.1 .
(i) Let η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) be the constant from Lemma 3.3 and let δ ∈ ( 0 , 2 − α − 2 η ) \delta\in(0,2^{-\alpha-2}\eta) . There exists a constant C > 0 C>0 such that for all y ∈ ℝ + d y\in{\mathbb{R}}^{d}_{+} , R ∈ ( 0 , y d ) R\in(0,y_{d}) , r ∈ ( 0 , ( η δ / 2 ) 1 / α R / 2 ] r\in(0,(\eta\delta/2)^{1/\alpha}R/2] ,
δ R α / 2 ≤ t − s ≤ 4 δ ( 2 R ) α \delta R^{\alpha}/2\leq t-s\leq 4\delta(2R)^{\alpha} ,
x ∈ B ( y , ( η δ / 2 ) 1 / α R / 2 ) x\in B(y,(\eta\delta/2)^{1/\alpha}R/2) and z ∈ B ( x 0 , ( η δ / 2 ) 1 / α R ) z\in B(x_{0},(\eta\delta/2)^{1/\alpha}R) ,
ℙ ( t , z ) ( σ { s } × B ( x , r ) κ ≤ τ [ s , t ] × B ( y , R ) κ ) ≥ C ( r / R ) d . {\mathbb{P}}_{(t,z)}\big{(}\sigma^{\kappa}_{\{s\}\times B(x,r)}\leq\tau^{\kappa}_{[s,t]\times B(y,R)}\big{)}\geq C(r/R)^{d}.
(ii) Let η ∈ ( 0 , 1 / 4 ) \eta\in(0,1/4) be the constant from Lemma 3.8 and let δ ∈ ( 0 , 2 − α − 2 η ) \delta\in(0,2^{-\alpha-2}\eta) . There exists a constant C > 0 C>0 such that for all y ∈ ℝ ¯ + d y\in{\overline{\mathbb{R}}}^{d}_{+} , R > 0 R>0 , r ∈ ( 0 , ( η δ / 2 ) 1 / α R / 2 ] r\in(0,(\eta\delta/2)^{1/\alpha}R/2] ,
δ R α / 2 ≤ t − s ≤ 4 δ ( 2 R ) α \delta R^{\alpha}/2\leq t-s\leq 4\delta(2R)^{\alpha} ,
x ∈ B + ( y , ( η δ / 2 ) 1 / α R / 2 ) x\in{B_{+}}(y,(\eta\delta/2)^{1/\alpha}R/2) and z ∈ B + ( x 0 , ( η δ / 2 ) 1 / α R ) z\in{B_{+}}(x_{0},(\eta\delta/2)^{1/\alpha}R) ,
ℙ ( t , z ) ( σ ¯ { s } × B + ( x , r ) ≤ τ ¯ [ s , t ] × B + ( y , R ) ) ≥ C ( r / R ) d . {\mathbb{P}}_{(t,z)}\big{(}\overline{\sigma}_{\{s\}\times{B_{+}}(x,r)}\leq\overline{\tau}_{[s,t]\times{B_{+}}(y,R)}\big{)}\geq C(r/R)^{d}.
Proof. Using Lemmas 3.3 and 3.8 , and (3.27 ), the result can be proved by the same argument as that of [21 , Lemma 6.7] . We omit details here. □ \Box
In order to obtain
the parabolic Harnack inequality for Y ¯ \overline{Y} and Y κ Y^{\kappa} , we introduce two conditions:
(UBS) There exists a constant C > 0 C>0 such that for all x , y ∈ ℝ ¯ + d x,y\in{\overline{\mathbb{R}}}_{+}^{d} and 0 < r ≤ | x − y | / 2 0<r\leq|x-y|/2 ,
ℬ ( x , y ) ≤ C r d ∫ B + ( x , r ) ℬ ( z , y ) 𝑑 z . \displaystyle{\cal B}(x,y)\leq\frac{C}{r^{d}}\int_{{B_{+}}(x,r)}{\cal B}(z,y)dz.
(4.1)
(IUBS)
There exists a constant C > 0 C>0 such that (4.1 ) holds for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}_{+}^{d} and 0 < r ≤ ( | x − y | ∧ x d ) / 2 0<r\leq(|x-y|\wedge x_{d})/2 . (Note that, B + ( x , r ) = B ( x , r ) {B_{+}}(x,r)=B(x,r) for this range of r r .)
Lemma 4.2 .
If (A3) (II)
also holds,
then (UBS) is satisfied. In particular, (IUBS) is satisfied.
Proof. Let x , y ∈ ℝ ¯ + d x,y\in{\overline{\mathbb{R}}}_{+}^{d} and 0 < r ≤ | x − y | / 2 0<r\leq|x-y|/2 . Note that for all z ∈ B + ( x , r ) z\in{B_{+}}(x,r) , | x − y | / 2 ≤ | z − y | ≤ 2 | x − y | |x-y|/2\leq|z-y|\leq 2|x-y| by the triangle inequality. Thus, by (A3) , there is c 1 > 0 c_{1}>0 independent of x , y x,y and r r such that for all z ∈ B + ( x , r ) z\in{B_{+}}(x,r) with z d ≥ x d z_{d}\geq x_{d} , ℬ ( x , y ) ≤ c 1 ℬ ( z , y ) {\cal B}(x,y)\leq c_{1}{\cal B}(z,y) . Using this, we get
1 r d ∫ B + ( x , r ) ℬ ( z , y ) 𝑑 z ≥ ℬ ( x , y ) c 1 r d ∫ B ( x , r ) : z d ≥ x d 𝑑 z ≥ c 2 ℬ ( x , y ) . \displaystyle\frac{1}{r^{d}}\int_{{B_{+}}(x,r)}{\cal B}(z,y)dz\geq\frac{{\cal B}(x,y)}{c_{1}r^{d}}\int_{B(x,r):z_{d}\geq x_{d}}dz\geq c_{2}{\cal B}(x,y).
□ \Box
We now show that the following parabolic Harnack inequalities hold.
Theorem 4.3 .
(i) Suppose that ℬ ( x , y ) {\cal B}(x,y)
satisfies (IUBS) .
Then there exist constants δ > 0 \delta>0 and C , M ≥ 1 C,M\geq 1 such that for all t 0 ≥ 0 t_{0}\geq 0 , x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and R ∈ ( 0 , x d ) R\in(0,x_{d}) , and any non-negative function u u on ( 0 , ∞ ) × ℝ + d (0,\infty)\times{\mathbb{R}}^{d}_{+} which is parabolic on
Q := ( t 0 , t 0 + 4 δ R α ] × B ( x , R ) Q:=(t_{0},t_{0}+4\delta R^{\alpha}]\times B(x,R)
with respect to Y ¯ \overline{Y} or Y κ Y^{\kappa} , we have
sup ( t 1 , y 1 ) ∈ Q − u ( t 1 , y 1 ) ≤ C inf ( t 2 , y 2 ) ∈ Q + u ( t 2 , y 2 ) , \sup_{(t_{1},y_{1})\in Q_{-}}u(t_{1},y_{1})\leq C\inf_{(t_{2},y_{2})\in Q_{+}}u(t_{2},y_{2}),
(4.2)
where Q − = [ t 0 + δ R α , t 0 + 2 δ R α ] × B ( x , R / M ) Q_{-}=[t_{0}+\delta R^{\alpha},t_{0}+2\delta R^{\alpha}]\times B(x,R/M) and Q + = [ t 0 + 3 δ R α , t 0 + 4 δ R α ] × B ( x , R / M ) Q_{+}=[t_{0}+3\delta R^{\alpha},t_{0}+4\delta R^{\alpha}]\times B(x,R/M) .
(ii) Suppose that ℬ ( x , y ) {\cal B}(x,y) satisfies
(UBS) . Then there exist constants δ > 0 \delta>0 and C , M ≥ 1 C,M\geq 1 such that for all t 0 ≥ 0 t_{0}\geq 0 , x ∈ ℝ ¯ + d x\in{\overline{\mathbb{R}}}^{d}_{+} and R > 0 R>0 , and any non-negative function u u on ( 0 , ∞ ) × ℝ ¯ + d (0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+} which is parabolic on
Q 0 := ( t 0 , t 0 + 4 δ R α ] × B + ( x , R ) Q^{0}:=(t_{0},t_{0}+4\delta R^{\alpha}]\times{B_{+}}(x,R) with respect to Y ¯ \overline{Y} , we have
sup ( t 1 , y 1 ) ∈ Q − 0 u ( t 1 , y 1 ) ≤ C inf ( t 2 , y 2 ) ∈ Q + 0 u ( t 2 , y 2 ) , \sup_{(t_{1},y_{1})\in Q^{0}_{-}}u(t_{1},y_{1})\leq C\inf_{(t_{2},y_{2})\in Q^{0}_{+}}u(t_{2},y_{2}),
(4.3)
where Q − 0 = [ t 0 + δ R α , t 0 + 2 δ R α ] × B + ( x , R / M ) Q^{0}_{-}=[t_{0}+\delta R^{\alpha},t_{0}+2\delta R^{\alpha}]\times{B_{+}}(x,R/M) and Q + 0 = [ t 0 + 3 δ R α , t 0 + 4 δ R α ] × B + ( x , R / M ) Q^{0}_{+}=[t_{0}+3\delta R^{\alpha},t_{0}+4\delta R^{\alpha}]\times{B_{+}}(x,R/M) .
Proof. (i) By (IUBS) ,
there is a constant C > 0 C>0 such that for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} and 0 < r ≤ ( | x − y | ∧ x d ) / 2 0<r\leq(|x-y|\wedge x_{d})/2 ,
J ( x , y ) = ℬ ( x , y ) | x − y | d + α ≤ C r d ∫ B + ( x , r ) ℬ ( z , y ) | z − y | d + α 𝑑 z = C r d ∫ B + ( x , r ) J ( z , y ) 𝑑 z . \displaystyle J(x,y)=\frac{{\cal B}(x,y)}{|x-y|^{d+\alpha}}\leq\frac{C}{r^{d}}\int_{{B_{+}}(x,r)}\frac{{\cal B}(z,y)}{|z-y|^{d+\alpha}}dz=\frac{C}{r^{d}}\int_{{B_{+}}(x,r)}J(z,y)dz.
(4.4)
Using Proposition 2.7 and (4.4 ), one can follow the proof of [21 , Lemma 6.10] and see that
[21 , Lemma 6.10] is also
valid for our case. (Note that, in the proof of [21 , Lemma 6.10] , a pointwise comparison for the jump kernel from [21 , Proposition 6.8] was used to bound the term I 2 I_{2} therein which can be replaced by (4.4 ).)
Using this and
Lemmas 3.3 , 3.4 , 3.10 , 4.1 , the result can be proved
using the same argument as in the proof of [9 , Lemma 5.3] (see also the proof of [18 , Lemma 4.1] ). We omit details here.
(ii) Since (UBS) implies that (4.4 ) is satisfied for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} and 0 < r ≤ | x − y | / 2 0<r\leq|x-y|/2 , using Proposition 2.7 and (3.27 ), one can also deduce that
[21 , Lemma 6.10] is
valid for this case with B + ( x 0 , ⋅ ) {B_{+}}(x_{0},\cdot) instead of B ( x 0 , ⋅ ) B(x_{0},\cdot) in the definitions of Q i Q_{i} , 1 ≤ i ≤ 4 1\leq i\leq 4 therein. Then using Lemmas 3.8 , 3.9 , 3.10 (ii), 4.1 (ii), one can follow the arguments in the proof of [9 , Lemma 5.3] and conclude the result. □ \Box
Using Lemma 3.3 , and Theorem 4.3 (i), we obtain the following lemma.
Lemma 4.4 .
Suppose that ℬ ( x , y ) {\cal B}(x,y)
satisfies (IUBS) . For any positive
constants a , b a,b , there exists a constant
C = C ( a , b , κ ) > 0 C=C(a,b,\kappa)>0 such that for all z ∈ ℝ + d z\in{\mathbb{R}}_{+}^{d} and r > 0 r>0
with B ( z , 2 b r ) ⊂ ℝ + d B(z,2br)\subset{\mathbb{R}}^{d}_{+} ,
inf y ∈ B ( z , b r / 2 ) ℙ y ( τ B ( z , b r ) κ > a r α ) ≥ C . \inf_{y\in B(z,br/2)}{\mathbb{P}}_{y}\big{(}\tau^{\kappa}_{B(z,br)}>ar^{\alpha}\big{)}\,\geq\,C.
Proof. By Lemma 3.3 , there exist constants
c 1 , c 2 , ε 1 > 0 c_{1},c_{2},{\varepsilon}_{1}>0 such that for all z ∈ ℝ + d z\in{\mathbb{R}}^{d}_{+} and r > 0 r>0 with B ( z , 2 b r ) ⊂ ℝ + d B(z,2br)\subset{\mathbb{R}}^{d}_{+} ,
ℙ z ( τ B ( z , b r / 2 ) κ > ε 1 r α ) = ∫ B ( z , b r / 2 ) p κ , B ( z , b r / 2 ) ( ε 1 r α , z , w ) 𝑑 w \displaystyle{\mathbb{P}}_{z}\big{(}\tau^{\kappa}_{B(z,br/2)}>{\varepsilon}_{1}r^{\alpha}\big{)}=\int_{B(z,br/2)}p^{\kappa,B(z,br/2)}({\varepsilon}_{1}r^{\alpha},z,w)dw
≥ ∫ B ( z , c 1 r ) p κ , B ( z , b r / 2 ) ( ε 1 r α , z , w ) 𝑑 w ≥ c 2 . \displaystyle\geq\int_{B(z,c_{1}r)}p^{\kappa,B(z,br/2)}({\varepsilon}_{1}r^{\alpha},z,w)dw\geq c_{2}.
(4.5)
Thus it suffices
to prove the lemma for a > ε 1 a>{\varepsilon}_{1} . Applying the parabolic Harnack inequality (Theorem 4.3 ) repeatedly,
we conclude that there
exists c 3 > 0 c_{3}>0 such that for any
w , y ∈ B ( z , b r / 2 ) w,y\in B(z,br/2) ,
p κ , B ( z , b r ) ( a r α , y , w ) ≥ c 3 p κ , B ( z , b r ) ( ε 1 r α , z , w ) . p^{\kappa,B(z,br)}(ar^{\alpha},y,w)\geq c_{3}\,p^{\kappa,B(z,br)}({\varepsilon}_{1}r^{\alpha},z,w).
Thus, using (4 ), we deduce that for any y ∈ B ( z , b r / 2 ) y\in B(z,br/2) ,
ℙ y ( τ B ( z , b r ) κ > a r α ) = ∫ B ( z , b r ) p κ , B ( z , b r ) ( a r α , y , w ) 𝑑 w \displaystyle{\mathbb{P}}_{y}\big{(}\tau^{\kappa}_{B(z,br)}>ar^{\alpha}\big{)}=\int_{B(z,br)}p^{\kappa,B(z,br)}(ar^{\alpha},y,w)dw
≥ c 3 ∫ B ( z , b r / 2 ) p κ , B ( z , b r ) ( ε 1 r α , z , w ) 𝑑 w ≥ c 3 ℙ z ( τ B ( z , b r / 2 ) κ > ε 1 r α ) ≥ c 2 c 3 . \displaystyle\geq c_{3}\int_{B(z,br/2)}p^{\kappa,B(z,br)}({\varepsilon}_{1}r^{\alpha},z,w)dw\geq c_{3}{\mathbb{P}}_{z}\big{(}\tau^{\kappa}_{B(z,br/2)}>{\varepsilon}_{1}r^{\alpha}\big{)}\geq c_{2}c_{3}.
This proves the lemma.
□ \Box
Now, we follow the proof of [8 , Proposition 3.5] to get the following preliminary lower bound.
Proposition 4.5 .
Suppose that ℬ ( x , y ) {\cal B}(x,y)
satisfies (IUBS) . For any a > 0 a>0 ,
there exists a constant C = C ( a , κ ) > 0 C=C(a,\kappa)>0 such
that for any ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ + d × ℝ + d (t,x,y)\in(0,\infty)\times{\mathbb{R}}_{+}^{d}\times{\mathbb{R}}_{+}^{d} with a t 1 / α ≤ ( 4 | x − y | ) ∧ x d ∧ y d at^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d} ,
p ¯ ( t , x , y ) ≥ p κ ( t , x , y ) ≥ C t J ( x , y ) . \overline{p}(t,x,y)\geq p^{\kappa}(t,x,y)\geq CtJ(x,y).
Proof. The first inequality holds true by (2.5 ) and Remark 3.12 .
By Lemma 4.4 , starting at z ∈ B ( y , ( 12 ) − 1 a t 1 / α ) z\in B(y,\,(12)^{-1}at^{1/\alpha}) , with probability at least c 1 = c 1 ( a ) > 0 c_{1}=c_{1}(a)>0 , the process Y κ Y^{\kappa} does not move more than ( 18 ) − 1 a t 1 / α (18)^{-1}at^{1/\alpha} by time t t .
Thus, using the strong Markov property and the Lévy system formula in (3.16 ), we obtain
that for
a ( t / 2 ) 1 / α ≤ ( 4 | x − y | ) ∧ x d ∧ y d a(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d} ,
ℙ x ( Y t κ ∈ B ( y , 6 − 1 a t 1 / α ) ) \displaystyle{\mathbb{P}}_{x}\left(Y^{\kappa}_{t}\in B\big{(}y,\,6^{-1}at^{1/\alpha}\big{)}\right)
≥ c 1 ℙ x ( Y t ∧ τ B ( x , ( 18 ) − 1 a t 1 / α ) κ κ ∈ B ( y , ( 12 ) − 1 a t 1 / α ) and t ∧ τ B ( x , ( 18 ) − 1 a t 1 / α ) κ is a jumping time ) \displaystyle\geq c_{1}{\mathbb{P}}_{x}\big{(}\,Y^{\kappa}_{t\wedge\tau^{\kappa}_{B(x,(18)^{-1}at^{1/\alpha})}}\in B(y,\,(12)^{-1}at^{1/\alpha})\hbox{ and }t\wedge\tau^{\kappa}_{B(x,(18)^{-1}at^{1/\alpha})}\hbox{ is a jumping time\,}\big{)}
= c 1 𝔼 x [ ∫ 0 t ∧ τ B ( x , ( 18 ) − 1 a t 1 / α ) κ ∫ B ( y , ( 12 ) − 1 a t 1 / α ) J ( Y s κ , u ) 𝑑 u 𝑑 s ] . \displaystyle=c_{1}{\mathbb{E}}_{x}\bigg{[}\int_{0}^{t\wedge\tau^{\kappa}_{B(x,(18)^{-1}at^{1/\alpha})}}\int_{B(y,\,(12)^{-1}at^{1/\alpha})}J(Y^{\kappa}_{s},u)duds\bigg{]}.
(4.6)
By (4.4 ), we obtain that for
a ( t / 2 ) 1 / α ≤ ( 4 | x − y | ) ∧ x d ∧ y d a(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d} ,
𝔼 x [ ∫ 0 t ∧ τ B ( x , ( 18 ) − 1 a t 1 / α ) κ ∫ B ( y , ( 12 ) − 1 a t 1 / α ) J ( Y s κ , u ) 𝑑 u 𝑑 s ] \displaystyle{\mathbb{E}}_{x}\bigg{[}\int_{0}^{t\wedge\tau^{\kappa}_{B(x,(18)^{-1}at^{1/\alpha})}}\int_{B(y,\,(12)^{-1}at^{1/\alpha})}J(Y^{\kappa}_{s},u)duds\bigg{]}
= 𝔼 x [ ∫ 0 t ∫ B ( y , ( 12 ) − 1 a t 1 / α ) J ( Y s κ , B ( x , ( 18 ) − 1 a t 1 / α ) , u ) 𝑑 u 𝑑 s ] \displaystyle={\mathbb{E}}_{x}\bigg{[}\int_{0}^{t}\int_{B(y,\,(12)^{-1}at^{1/\alpha})}J(Y^{\kappa,B(x,(18)^{-1}at^{1/\alpha})}_{s},u)duds\bigg{]}
≥ c 2 t d / α ∫ 0 t 𝔼 x [ J ( Y s κ , B ( x , ( 18 ) − 1 a t 1 / α ) , y ) ] 𝑑 s \displaystyle\geq c_{2}t^{d/\alpha}\int_{0}^{t}{\mathbb{E}}_{x}\left[J(Y^{\kappa,B(x,(18)^{-1}at^{1/\alpha})}_{s},y)\right]ds
≥ c 2 t d / α ∫ t / 2 t ∫ B ( x , ( 72 ) − 1 a ( t / 2 ) 1 / α ) J ( w , y ) p κ , B ( x , ( 18 ) − 1 a t 1 / α ) ( s , x , w ) 𝑑 w 𝑑 s . \displaystyle\geq c_{2}t^{d/\alpha}\int_{t/2}^{t}\int_{B(x,(72)^{-1}a(t/2)^{1/\alpha})}J(w,y)p^{\kappa,B(x,(18)^{-1}at^{1/\alpha})}(s,x,w)dwds.
(4.7)
Note that, for t / 2 < s < t t/2<s<t and w ∈ B ( x , ( 72 ) − 1 a ( t / 2 ) 1 / α ) w\in B(x,(72)^{-1}a(t/2)^{1/\alpha}) , we have
δ B ( x , ( 18 ) − 1 a t 1 / α ) ( w ) ≥ ( 18 ) − 1 a t 1 / α − ( 72 ) − 1 a ( t / 2 ) 1 / α ≥ ( 36 ) − 1 a s − 1 / α \delta_{B(x,(18)^{-1}at^{1/\alpha})}(w)\geq(18)^{-1}at^{1/\alpha}-(72)^{-1}a(t/2)^{1/\alpha}\geq(36)^{-1}as^{-1/\alpha}
and
| x − w | < ( 72 ) − 1 a ( t / 2 ) 1 / α ≤ 4 − 1 ( 18 ) − 1 a s 1 / α . |x-w|<(72)^{-1}a(t/2)^{1/\alpha}\leq 4^{-1}(18)^{-1}as^{1/\alpha}.
Thus by Lemma 3.3 and the parabolic Harnack inequality (Theorem 4.3 )
we see that for
t / 2 < s < t t/2<s<t and w ∈ B ( x , ( 72 ) − 1 a ( t / 2 ) 1 / α ) w\in B(x,(72)^{-1}a(t/2)^{1/\alpha}) ,
p κ , B ( x , ( 18 ) − 1 a t 1 / α ) ( s , x , w ) ≥ c 3 t − d / α . \displaystyle p^{\kappa,B(x,(18)^{-1}at^{1/\alpha})}(s,x,w)\geq c_{3}\,t^{-d/\alpha}.
(4.8)
Combining (4.6 ), (4.7 ) with (4.8 ) and applying (4.4 ) again, we get that for
a ( t / 2 ) 1 / α ≤ ( 4 | x − y | ) ∧ x d ∧ y d a(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d} ,
ℙ x ( Y t κ ∈ B ( y , 6 − 1 a t 1 / α ) ) ≥ c 4 t ∫ B ( x , ( 72 ) − 1 a ( t / 2 ) 1 / α ) J ( w , y ) 𝑑 w ≥ c 5 t 1 + d / α J ( x , y ) . \displaystyle{\mathbb{P}}_{x}\left(Y^{\kappa}_{t}\in B\big{(}y,\,6^{-1}at^{1/\alpha}\big{)}\right)\geq c_{4}t\int_{B(x,(72)^{-1}a(t/2)^{1/\alpha})}J(w,y)dw\geq c_{5}t^{1+d/\alpha}J(x,y).
(4.9)
Note that for
y d ≥ a t 1 / α y_{d}\geq at^{1/\alpha} and z ∈ B ( y , a ( t / 2 ) 1 / α / 6 ) z\in B(y,a(t/2)^{1/\alpha}/6) ,
z d ≥ y d − | z − y | ≥ a ( 1 − ( 1 / 2 ) 1 / α / 6 ) t 1 / α ≥ ( 5 a / 6 ) t 1 / α \displaystyle z_{d}\geq y_{d}-|z-y|\geq a(1-(1/2)^{1/\alpha}/6)t^{1/\alpha}\geq(5a/6)t^{1/\alpha}
(4.10)
The proposition now follows from the Chapman-Kolmogorov equation along with (4.9 ) and Proposition 3.2 (using (4.10 )): for
a t 1 / α ≤ ( 4 | x − y | ) ∧ x d ∧ y d at^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d} ,
p κ ( t , x , y ) \displaystyle p^{\kappa}(t,x,y)
= ∫ D p κ ( t / 2 , x , z ) p κ ( t / 2 , z , y ) 𝑑 z ≥ ∫ B ( y , a ( t / 2 ) 1 / α / 6 ) p κ ( t / 2 , x , z ) p κ ( t / 2 , z , y ) 𝑑 z \displaystyle=\int_{D}p^{\kappa}(t/2,x,z)p^{\kappa}(t/2,z,y)dz\geq\int_{B(y,\,a(t/2)^{1/\alpha}/6)}p^{\kappa}(t/2,x,z)p^{\kappa}(t/2,z,y)dz
≥ c 6 t − d / α ℙ x ( Y t / 2 κ ∈ B ( y , a ( t / 2 ) 1 / α / 6 ) ) ≥ c 7 t J ( x , y ) . \displaystyle\geq c_{6}t^{-d/\alpha}\,{\mathbb{P}}_{x}\left(Y^{\kappa}_{t/2}\in B(y,a(t/2)^{1/\alpha}/6)\right)\geq c_{7}\,{t}J(x,y).
□ \Box
5. Preliminary upper bounds of heat kernels
The goal of this section is to prove the following proposition.
Proposition 5.1 .
Suppose that (A1) – (A4) and (1.8 ) hold true. Then there exists a constant C > 0 C>0 such that
p κ ( t , x , y ) ≤ C ( 1 ∧ x d t 1 / α ) q κ ( 1 ∧ y d t 1 / α ) q κ ( t − d / α ∧ t | x − y | d + α ) , t > 0 , x , y ∈ ℝ + d , p^{\kappa}(t,x,y)\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),\quad t>0,\,x,y\in{\mathbb{R}}^{d}_{+},
(5.1)
where q κ ∈ [ ( α − 1 ) + , α + β 1 ) q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) is the constant from (1.8 ).
Note that, since t − d / α ∧ t | x − y | d + α t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}} is comparable to the transition density of
the isotropic α \alpha -stable process in ℝ d {\mathbb{R}}^{d} ,
there exists a constant C > 0 C>0 such that for all t , s > 0 t,s>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
∫ ℝ + d ( t − d / α ∧ t | x − z | d + α ) 𝑑 z ≤ C , \displaystyle\int_{{\mathbb{R}}^{d}_{+}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-z|^{d+\alpha}}\right)dz\leq C,
(5.2)
and
∫ ℝ + d ( t − d / α ∧ t | x − z | d + α ) ( s − d / α ∧ s | y − z | d + α ) 𝑑 z ≤ C ( ( t + s ) − d / α ∧ t + s | x − y | d + α ) , \displaystyle\int_{{\mathbb{R}}^{d}_{+}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-z|^{d+\alpha}}\right)\left(s^{-d/\alpha}\wedge\frac{s}{|y-z|^{d+\alpha}}\right)dz\leq C\left((t+s)^{-d/\alpha}\wedge\frac{t+s}{|x-y|^{d+\alpha}}\right),
(5.3)
where in (5.3 ) we used the semigroup property.
Before giving the proof of Proposition 5.1 , we record its simple consequence.
Corollary 5.2 .
There exists a constant C > 0 C>0 such that
ℙ x ( ζ κ > t ) ≤ C ( 1 ∧ x d t 1 / α ) q κ , t > 0 , x ∈ ℝ + d . {\mathbb{P}}_{x}(\zeta^{\kappa}>t)\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}},\quad t>0,\;x\in{\mathbb{R}}^{d}_{+}.
Proof. By Proposition 5.1 and (5.2 ),
ℙ x ( ζ κ > t ) = ∫ ℝ + d p κ ( t , x , y ) 𝑑 y \displaystyle{\mathbb{P}}_{x}(\zeta^{\kappa}>t)=\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t,x,y)dy
≤ c 1 ( 1 ∧ x d t 1 / α ) q κ ∫ ℝ + d ( t − d / α ∧ t | x − y | d + α ) 𝑑 y ≤ c 2 ( 1 ∧ x d t 1 / α ) q κ . \displaystyle\leq c_{1}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}\int_{{\mathbb{R}}^{d}_{+}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)dy\leq c_{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q_{\kappa}}.
□ \Box
When q κ = 0 q_{\kappa}=0 , Proposition 5.1 follows from Proposition 2.7 . Hence,
in the remainder of this section,
we assume that (A1) –(A4) hold, fix
κ ∈ [ 0 , ∞ ) \kappa\in[0,\infty) such that q κ > 0 q_{\kappa}>0 and denote by q q the constant q κ q_{\kappa} in (1.8 ).
For the proof we will need several results from [30 , 31 , 32 ] that we now recall for the convenience of the reader.
For r > 0 r>0 , we define a subset U ( r ) U(r) of ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+} , d ≥ 2 d\geq 2 , by
U ( r ) := { x = ( x ~ , x d ) ∈ ℝ d : | x ~ | < r / 2 , 0 ≤ x d < r / 2 } . U(r):=\{x=(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:\,|\widetilde{x}|<r/2,\,0\leq x_{d}<r/2\}.
When d = 1 d=1 , we abuse notation and use U ( r ) = [ 0 , r / 2 ) U(r)=[0,r/2) . For t > 0 t>0 and an open set V ⊂ ℝ + d V\subset{\mathbb{R}}^{d}_{+} , denote by Y t κ , d Y_{t}^{\kappa,d} and Y t κ , V , d Y_{t}^{\kappa,V,d} the last coordinates of Y t κ Y_{t}^{\kappa} and Y t κ , V Y_{t}^{\kappa,V} respectively.
Lemma 5.3 .
There exists a constant C > 0 C>0 such that for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} satisfying | x − y | ≥ x d |x-y|\geq x_{d} , it holds that
ℬ ( x , y ) ≤ C x d β 1 ( | log x d | β 3 ∨ 1 ) ( 1 + 𝟏 | y | ≥ 1 ( log | y | ) β 3 ) | x − y | − β 1 . \displaystyle{\cal B}(x,y)\leq Cx_{d}^{\beta_{1}}(|\log x_{d}|^{\beta_{3}}\vee 1)\big{(}1+{\bf 1}_{|y|\geq 1}(\log|y|)^{\beta_{3}}\big{)}|x-y|^{-\beta_{1}}.
Proof. See [30 , Lemma 5.2(a)] . □ \Box
Lemma 5.4 .
For any R > 0 R>0 , there exists a constant C > 0 C>0 such that for all r ∈ ( 0 , R ] r\in(0,R] and x ∈ U ( 2 − 4 r ) ∩ ℝ + d x\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+} ,
𝔼 x ∫ 0 τ U ( r ) κ ( Y t κ , d ) β 1 | log Y t κ , d | β 3 𝑑 t ≤ C x d q . \displaystyle{\mathbb{E}}_{x}\int_{0}^{\tau^{\kappa}_{U(r)}}(Y^{\kappa,d}_{t})^{\beta_{1}}|\log Y^{\kappa,d}_{t}|^{\beta_{3}}dt\leq Cx_{d}^{q}.
Proof. The result follows from scaling (Lemma 2.1 ), and [30 , Lemma 5.7(a)] if κ > 0 \kappa>0 and [32 , Lemma 5.3] if κ = 0 \kappa=0 . □ \Box
Lemma 5.5 .
There exists a constant C > 0 C>0 such that for all r > 0 r>0 and x ∈ U ( 2 − 4 r ) ∩ ℝ + d x\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+} ,
ℙ x ( τ U ( r ) κ < ζ κ ) = ℙ x ( Y τ U ( r ) κ κ ∈ ℝ + d ) ≤ C ( x d r ) q . {\mathbb{P}}_{x}\big{(}\tau^{\kappa}_{U(r)}<\zeta^{\kappa}\big{)}={\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(r)}}\in{\mathbb{R}}^{d}_{+}\big{)}\leq C\left(\frac{x_{d}}{r}\right)^{q}.
Proof. When κ > 0 \kappa>0 , the result follows from [31 , Lemma 3.4] . When κ = 0 \kappa=0 , using [32 , Lemma 5.5] and Lemma 5.4 , one can follow the proof of [31 , Lemma 3.4] and deduce the result. □ \Box
Lemma 5.6 .
There exists a constant C > 0 C>0 such that for all r > 0 r>0 and x ∈ U ( 2 − 4 r ) ∩ ℝ + d x\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+} ,
𝔼 x [ τ U ( r ) κ ] ≤ C ( x d r ) q . {\mathbb{E}}_{x}\big{[}\tau^{\kappa}_{U(r)}\big{]}\leq C\left(\frac{x_{d}}{r}\right)^{q}.
Proof. The result follows from scaling (Lemma 2.1 ), and [30 , Lemma 5.13] if κ > 0 \kappa>0 and [32 , Lemma 4.5] if κ = 0 \kappa=0 . □ \Box
Proposition 5.7 .
There exists a constant C > 0 C>0 such that for any w ∈ ∂ ℝ + d w\in\partial{\mathbb{R}}^{d}_{+} , r > 0 r>0 and any non-negative function f f in ℝ + d {\mathbb{R}}^{d}_{+} that is harmonic in ℝ + d ∩ B ( w , r ) {\mathbb{R}}^{d}_{+}\cap B(w,r) with respect to Y κ Y^{\kappa} and vanishes continuously on ∂ ℝ + d ∩ B ( w , r ) \partial{\mathbb{R}}^{d}_{+}\cap B(w,r) , we have
f ( x ) ≤ C f ( x ^ ) for all x ∈ ℝ + d ∩ B ( w , r / 2 ) , \displaystyle f(x)\leq Cf(\widehat{x})\quad\text{for all}\;\,x\in{\mathbb{R}}^{d}_{+}\cap B(w,r/2),
where x ^ ∈ ℝ + d ∩ B ( w , r ) \widehat{x}\in{\mathbb{R}}^{d}_{+}\cap B(w,r) with x ^ d ≥ r / 4 \widehat{x}_{d}\geq r/4 .
Proof. The result follows from [30 , Theorem 1.2] if κ > 0 \kappa>0 and [32 , Theorem 5.6] if κ = 0 \kappa=0 . □ \Box
After recalling the known results above, we now continue with several auxiliary lemmas leading to the proof of Proposition 5.1 .
Lemma 5.8 .
For all γ ≥ 0 \gamma\geq 0 , t > 0 t>0 and x ∈ U ( 1 ) ∩ ℝ + d x\in U(1)\cap{\mathbb{R}}^{d}_{+} , it holds that
∫ ℝ + d p κ ( t , x , z ) ( 1 ∧ z d ) γ d z ≤ 𝔼 x [ ( 1 ∧ Y t κ , U ( 1 ) , d ) γ : τ U ( 1 ) κ > t ] + ℙ x ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) . \int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t,x,z)(1\wedge z_{d})^{\gamma}dz\leq{\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,U(1),d}_{t})^{\gamma}:\tau^{\kappa}_{U(1)}>t\right]+{\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+}\big{)}.
(5.4)
In particular, it holds that
ℙ x ( ζ κ > t ) ≤ t − 1 𝔼 x [ τ U ( 1 ) κ ] + ℙ x ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) . {\mathbb{P}}_{x}(\zeta^{\kappa}>t)\leq t^{-1}{\mathbb{E}}_{x}\big{[}\tau^{\kappa}_{U(1)}\big{]}+{\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+}\big{)}.
(5.5)
Proof. Since Y t κ , U ( 1 ) = Y t κ Y^{\kappa,U(1)}_{t}=Y^{\kappa}_{t} for t < τ U ( 1 ) κ t<\tau^{\kappa}_{U(1)} , we have
∫ ℝ + d p κ ( t , x , z ) ( 1 ∧ z d ) γ d z = 𝔼 x [ ( 1 ∧ Y t κ , d ) γ : t < ζ κ ] \displaystyle\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t,x,z)(1\wedge z_{d})^{\gamma}dz={\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,d}_{t})^{\gamma}:t<\zeta^{\kappa}\right]
= 𝔼 x [ ( 1 ∧ Y t κ , d ) γ : τ U ( 1 ) κ > t ] + 𝔼 x [ ( 1 ∧ Y t κ , d ) γ : τ U ( 1 ) κ ≤ t < ζ κ ] \displaystyle={\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,d}_{t})^{\gamma}:\tau^{\kappa}_{U(1)}>t\right]+{\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,d}_{t})^{\gamma}:\tau^{\kappa}_{U(1)}\leq t<\zeta^{\kappa}\right]
≤ 𝔼 x [ ( 1 ∧ Y t κ , U ( 1 ) , d ) γ : τ U ( 1 ) κ > t ] + 𝔼 x [ 1 : τ U ( 1 ) κ < ζ κ ] \displaystyle\leq{\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,U(1),d}_{t})^{\gamma}:\tau^{\kappa}_{U(1)}>t\right]+{\mathbb{E}}_{x}\big{[}1:\tau^{\kappa}_{U(1)}<\zeta^{\kappa}\big{]}
= 𝔼 x [ ( 1 ∧ Y t κ , U ( 1 ) , d ) γ : τ U ( 1 ) κ > t ] + ℙ x ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) . \displaystyle={\mathbb{E}}_{x}\left[(1\wedge Y^{\kappa,U(1),d}_{t})^{\gamma}:\tau^{\kappa}_{U(1)}>t\right]+{\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+}\big{)}.
By taking γ = 0 \gamma=0 in (5.4 ) and using Markov’s inequality, we get
ℙ x ( ζ κ > t ) ≤ ℙ x ( τ U ( 1 ) κ > t ) + ℙ x ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) ≤ t − 1 𝔼 x [ τ U ( 1 ) κ ] + ℙ x ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) . {\mathbb{P}}_{x}(\zeta^{\kappa}>t)\leq{\mathbb{P}}_{x}(\tau^{\kappa}_{U(1)}>t)+{\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+}\big{)}\leq t^{-1}{\mathbb{E}}_{x}\big{[}\tau^{\kappa}_{U(1)}\big{]}+{\mathbb{P}}_{x}\big{(}Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+}\big{)}.
□ \Box
Lemma 5.9 .
There exists C > 0 C>0 such that
p κ ( t , x , y ) ≤ C t − d / α ℙ x ( ζ κ > t / 3 ) ℙ y ( ζ κ > t / 3 ) , t > 0 , x , y ∈ ℝ + d . p^{\kappa}(t,x,y)\leq Ct^{-d/\alpha}{\mathbb{P}}_{x}(\zeta^{\kappa}>t/3){\mathbb{P}}_{y}(\zeta^{\kappa}>t/3),\quad t>0,\,x,y\in{\mathbb{R}}^{d}_{+}.
Proof. By the semigroup property, the symmetry of p κ ( t , ⋅ , ⋅ ) p^{\kappa}(t,\cdot,\cdot) and Proposition 2.7 , we obtain
p κ ( t , x , y ) = ∫ ℝ + d ∫ ℝ + d p κ ( t / 3 , x , z ) p κ ( t / 3 , z , w ) p κ ( t / 3 , y , w ) 𝑑 z 𝑑 w \displaystyle p^{\kappa}(t,x,y)=\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t/3,x,z)p^{\kappa}(t/3,z,w)p^{\kappa}(t/3,y,w)dzdw
≤ c 1 t − d / α ∫ ℝ + d p κ ( t / 3 , x , z ) 𝑑 z ∫ ℝ + d p κ ( t / 3 , y , w ) 𝑑 w = c 1 t − d / α ℙ x ( ζ κ > t / 3 ) ℙ y ( ζ κ > t / 3 ) . \displaystyle\leq c_{1}t^{-d/\alpha}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t/3,x,z)dz\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t/3,y,w)dw=c_{1}t^{-d/\alpha}{\mathbb{P}}_{x}(\zeta^{\kappa}>t/3){\mathbb{P}}_{y}(\zeta^{\kappa}>t/3).
□ \Box
The next lemma shows that (5.1 )
(hence Proposition 5.1 ) is a consequence of the following, seemingly weaker, inequality: There exists C > 0 C>0 such that
p κ ( t , x , y ) ≤ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q t − d / α , t > 0 , x , y ∈ ℝ + d . p^{\kappa}(t,x,y)\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}t^{-d/\alpha},\quad t>0,\,x,y\in{\mathbb{R}}^{d}_{+}.
(5.6)
Lemma 5.10 .
If (5.6 ) holds true, then (5.1 ) also holds.
Proof. We claim that there exists a constant c 1 > 0 c_{1}>0 such that
p κ ( t , x , y ) ≤ c 1 ( 1 ∧ x d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) . p^{\kappa}(t,x,y)\leq c_{1}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
(5.7)
By (3.30 ),
we can assume x ~ = 0 ~ \widetilde{x}=\widetilde{0} and t = t 0 = ( 1 / 2 ) α t=t_{0}=(1/2)^{\alpha} . If x d ≥ 2 − 4 t 0 1 / α x_{d}\geq 2^{-4}t_{0}^{1/\alpha} or | x − y | ≤ 4 t 0 1 / α |x-y|\leq 4t_{0}^{1/\alpha} , then (5.7 ) follows from
Proposition 2.7 or the assumption (5.6 ) respectively.
Hence, we assume x d < 2 − 4 t 0 1 / α x_{d}<2^{-4}t_{0}^{1/\alpha} and | x − y | > 4 t 0 1 / α |x-y|>4t_{0}^{1/\alpha} , and will show that
p κ ( t 0 , x , y ) ≤ c 1 ( t 0 ) x d q | x − y | d + α . p^{\kappa}(t_{0},x,y)\leq c_{1}(t_{0})\frac{x_{d}^{q}}{|x-y|^{d+\alpha}}.
(5.8)
Let V 1 = U ( t 0 1 / α ) V_{1}=U(t_{0}^{1/\alpha}) , V 3 = { w ∈ ℝ + d : | w − y | < | x − y | / 2 } V_{3}=\{w\in{\mathbb{R}}^{d}_{+}:|w-y|<|x-y|/2\} and V 2 = ℝ + d ∖ ( V 1 ∪ V 3 ) V_{2}={\mathbb{R}}^{d}_{+}\setminus(V_{1}\cup V_{3}) . By Lemma 5.5 , we have
ℙ x ( τ V 1 κ < t 0 < ζ κ ) ≤ ℙ x ( Y τ V 1 κ κ ∈ ℝ + d ) ≤ c 2 ( t 0 − 1 / α x d ) q . \displaystyle{\mathbb{P}}_{x}(\tau^{\kappa}_{V_{1}}<t_{0}<\zeta^{\kappa})\leq{\mathbb{P}}_{x}(Y^{\kappa}_{\tau^{\kappa}_{V_{1}}}\in{\mathbb{R}}^{d}_{+})\leq c_{2}(t_{0}^{-1/\alpha}x_{d})^{q}.
(5.9)
Also, we get from Proposition 2.7 that
sup s ≤ t 0 , z ∈ V 2 p κ ( s , z , y ) ≤ c 3 sup s ≤ t 0 , z ∈ ℝ + d , | z − y | > | x − y | / 2 s | z − y | d + α = 2 d + α c 3 t 0 | x − y | d + α . \sup_{s\leq t_{0},\,z\in V_{2}}p^{\kappa}(s,z,y)\leq c_{3}\sup_{s\leq t_{0},\,z\in{\mathbb{R}}^{d}_{+},|z-y|>|x-y|/2}\frac{s}{|z-y|^{d+\alpha}}=2^{d+\alpha}c_{3}\frac{t_{0}}{|x-y|^{d+\alpha}}.
(5.10)
Next, we note that by the triangle inequality, for any u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} ,
| u − w | ≥ | x − y | − | x − u | − | y − w | ≥ | x − y | − t 0 1 / α − | x − y | 2 ≥ | x − y | 4 ≥ t 0 1 / α ≥ u d . |u-w|\geq|x-y|-|x-u|-|y-w|\geq|x-y|-t_{0}^{1/\alpha}-\frac{|x-y|}{2}\geq\frac{|x-y|}{4}\geq t_{0}^{1/\alpha}\geq u_{d}.
(5.11)
In particular, recalling that β 1 > 0 \beta_{1}>0 if β 3 > 0 \beta_{3}>0 ,
we see that ( 1 + 𝟏 | w | ≥ 1 ( log | w | ) β 3 ) | u − w | − β 1 ≤ c 4 \big{(}1+{\bf 1}_{|w|\geq 1}(\log|w|)^{\beta_{3}}\big{)}|u-w|^{-\beta_{1}}\leq c_{4} for u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} , so by Lemma 5.3 , we have that for any u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} ,
ℬ ( u , w ) ≤ c 5 u d β 1 ( | log u d | β 3 ∨ 1 ) ( 1 + 𝟏 | w | ≥ 1 ( log | w | ) β 3 ) | u − w | − β 1 ≤ c 6 u d β 1 | log u d | β 3 . {\cal B}(u,w)\leq c_{5}u_{d}^{\beta_{1}}(|\log u_{d}|^{\beta_{3}}\vee 1)\big{(}1+{\bf 1}_{|w|\geq 1}(\log|w|)^{\beta_{3}}\big{)}|u-w|^{-\beta_{1}}\leq c_{6}u_{d}^{\beta_{1}}|\log u_{d}|^{\beta_{3}}.
(5.12)
Thus, by (A3) (II), Lemma 5.4 , (5.11 ) and (5.12 ) we get that
∫ 0 t 0 ∫ V 3 ∫ V 1 p κ , V 1 ( t − s , x , u ) ℬ ( u , w ) p κ ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\int_{0}^{t_{0}}\int_{V_{3}}\int_{V_{1}}p^{\kappa,V_{1}}(t-s,x,u){\cal B}(u,w)p^{\kappa}(s,y,w)dudwds
≤ c 7 ∫ 0 t 0 ( ∫ V 1 p κ , V 1 ( t − s , x , u ) u d β 1 | log u d | β 3 𝑑 u ) ( ∫ V 3 p κ ( s , y , w ) 𝑑 w ) 𝑑 s \displaystyle\leq c_{7}\int_{0}^{t_{0}}\left(\int_{V_{1}}p^{\kappa,V_{1}}(t-s,x,u)u_{d}^{\beta_{1}}|\log u_{d}|^{\beta_{3}}du\right)\left(\int_{V_{3}}p^{\kappa}(s,y,w)dw\right)ds
≤ c 7 ∫ 0 ∞ ( ∫ V 1 p κ , V 1 ( s , x , u ) u d β 1 | log u d | β 3 𝑑 u ) 𝑑 s \displaystyle\leq c_{7}\int_{0}^{\infty}\left(\int_{V_{1}}p^{\kappa,V_{1}}(s,x,u)u_{d}^{\beta_{1}}|\log u_{d}|^{\beta_{3}}du\right)ds
= c 7 𝔼 x ∫ 0 τ V 1 κ ( Y s κ , d ) β 1 | log Y s κ , d | β 3 𝑑 s ≤ c 8 x d q . \displaystyle=c_{7}{\mathbb{E}}_{x}\int_{0}^{\tau^{\kappa}_{V_{1}}}(Y_{s}^{\kappa,d})^{\beta_{1}}|\log Y_{s}^{\kappa,d}|^{\beta_{3}}ds\leq c_{8}x_{d}^{q}.
(5.13)
Now (5.8 ) (and so (5.7 )) follows
from (5.9 )–(5.11 ), (5.13 ) and Lemma 3.15 .
Finally, by the semigroup property, symmetry, (5.3 ) and (5.7 ),
p κ ( t , x , y ) = ∫ ℝ + d p κ ( t / 2 , x , z ) p κ ( t / 2 , y , z ) 𝑑 z \displaystyle p^{\kappa}(t,x,y)=\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t/2,x,z)p^{\kappa}(t/2,y,z)dz
≤ c 1 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ∫ ℝ + d ( ( t / 2 ) − d / α ∧ t / 2 | x − z | d + α ) ( ( t / 2 ) − d / α ∧ t / 2 | y − z | d + α ) 𝑑 z \displaystyle\leq c_{1}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\int_{{\mathbb{R}}^{d}_{+}}\left((t/2)^{-d/\alpha}\wedge\frac{t/2}{|x-z|^{d+\alpha}}\right)\left((t/2)^{-d/\alpha}\wedge\frac{t/2}{|y-z|^{d+\alpha}}\right)dz
≤ c 9 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) . \displaystyle\leq c_{9}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
□ \Box
Now we prove (5.6 ) holds. We first consider the case q < α q<\alpha .
Lemma 5.11 .
If q < α q<\alpha , then (5.6 ) holds true.
Proof. By (3.30 ), it suffices to prove (5.6 ) when t = 1 t=1 .
By Lemma 5.9 , it suffices to prove that there exists a constant c 1 > 0 c_{1}>0 such that ℙ x ( ζ κ > 1 / 3 ) ≤ c 1 ( 1 ∧ x d ) q {\mathbb{P}}_{x}(\zeta^{\kappa}>1/3)\leq c_{1}(1\wedge x_{d})^{q} for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} .
By Lemma 2.2 and the fact that ℙ x ( ζ > 1 / 3 ) ≤ 1 {\mathbb{P}}_{x}(\zeta>1/3)\leq 1 ,
without loss of generality, we can assume x ~ = 0 \widetilde{x}=0 and x d < 2 − 5 x_{d}<2^{-5} .
Then, since q < α q<\alpha , by (5.5 ) and Lemmas 5.5 –5.6 (with r = 1 r=1 ), we get ℙ x ( ζ κ > 1 / 3 ) ≤ c 2 x d q {\mathbb{P}}_{x}(\zeta^{\kappa}>1/3)\leq c_{2}x_{d}^{q} .
□ \Box
To remove the assumption q < α q<\alpha in Lemma 5.11 , we will make use of
a result from [31 , 32 ] . Recall from Remark 3.14 that all results in [31 , 32 ] are valid when d > α d>\alpha .
Now we state, and will prove later, that the desired result still holds true for d = 1 ≤ α d=1\leq\alpha (thus for all d d ).
Lemma 5.12 .
Let γ > q − α \gamma>q-\alpha . There exists C > 0 C>0 such that for any R > 0 R>0 , U ( R ) ⊂ D ⊂ U ( 2 R ) U(R)\subset D\subset U(2R) and any x = ( 0 ~ , x d ) ∈ ℝ + d x=(\widetilde{0},x_{d})\in{\mathbb{R}}^{d}_{+} with x d ≤ R / 10 x_{d}\leq R/10 , it holds that
𝔼 x ∫ 0 τ D κ ( Y s κ , D , d ) γ 𝑑 s = ∫ 0 ∞ ∫ D p κ , D ( t , x , z ) z d γ 𝑑 z 𝑑 t ≤ C R γ + α − q x d q . {\mathbb{E}}_{x}\int_{0}^{\tau^{\kappa}_{D}}(Y^{\kappa,D,d}_{s})^{\gamma}ds=\int_{0}^{\infty}\int_{D}p^{\kappa,D}(t,x,z)z_{d}^{\gamma}dzdt\leq CR^{\gamma+\alpha-q}x_{d}^{q}.
Proof. When d > α d>\alpha , by Remark 3.14 , the result follows from [31 , Proposition 6.10] if κ > 0 \kappa>0 , and from [32 , Proposition 6.8] if κ = 0 \kappa=0 . The proof of the case d = 1 ≤ α d=1\leq\alpha is postponed to the end of this section.
□ \Box
Lemma 5.13 .
Inequality (5.6 ) holds true.
Proof. Again using (3.30 ), it suffices to prove (5.6 ) when t = 1 t=1 .
Inequality (5.6 ) holds when q < α q<\alpha by Lemma 5.11 .
Now,
assume that (5.6 ) holds for all q < k α q<k\alpha
for some k ∈ ℕ k\in\mathbb{N} . We now show (5.6 ) also holds for q ∈ [ k α , ( k + 1 ) α ) q\in[k\alpha,(k+1)\alpha) and hence (5.6 ) always holds by induction.
Fix ε ∈ ( 0 , α ) {\varepsilon}\in(0,\alpha) such that q − α + ε < k α q-\alpha+{\varepsilon}<k\alpha . Then q − α + ε < q q-\alpha+{\varepsilon}<q and
C ( α , q − α + ε , ℬ ) < C ( α , q , ℬ ) = κ C(\alpha,q-\alpha+{\varepsilon},{\cal B})<C(\alpha,q,{\cal B})=\kappa .
By (3.30 ) and the induction hypothesis, it holds that for any s , u ∈ [ 0 , 1 / 4 ] s,u\in[0,1/4] and z , w ∈ ℝ + d z,w\in{\mathbb{R}}^{d}_{+} ,
p κ ( 1 − s − u , z , w ) \displaystyle p^{\kappa}(1-s-u,z,w)
= ( 1 − s − u ) − d / α p κ ( 1 , ( 1 − s − u ) − 1 / α z , ( 1 − s − u ) − 1 / α w ) \displaystyle=(1-s-u)^{-d/\alpha}p^{\kappa}(1,(1-s-u)^{-1/\alpha}z,(1-s-u)^{-1/\alpha}w)
≤ 2 d / α p C ( α , q − α + ε , ℬ ) ( 1 , ( 1 − s − u ) − 1 / α z , ( 1 − s − u ) − 1 / α w ) \displaystyle\leq 2^{d/\alpha}p^{C(\alpha,q-\alpha+{\varepsilon},{\cal B})}(1,(1-s-u)^{-1/\alpha}z,(1-s-u)^{-1/\alpha}w)
≤ c 3 ( 1 ∧ z d ) q − α + ε ( 1 ∧ w d ) q − α + ε . \displaystyle\leq c_{3}(1\wedge z_{d})^{q-\alpha+{\varepsilon}}(1\wedge w_{d})^{q-\alpha+{\varepsilon}}.
Here in the first inequality above we used the
fact that p κ ( 1 , x , y ) ≤ p C ( α , q − α + ε , ℬ ) ( 1 , x , y ) p^{\kappa}(1,x,y)\leq p^{C(\alpha,q-\alpha+{\varepsilon},{\cal B})}(1,x,y) , which is a consequence of C ( α , q , ℬ ) > C ( α , q − α + ε , ℬ ) C(\alpha,q,{\cal B})>C(\alpha,q-\alpha+{\varepsilon},{\cal B}) .
Therefore, by the semigroup property and symmetry, we get
p κ ( 1 , x , y ) = 16 ∫ 0 1 / 4 ∫ 0 1 / 4 p κ ( 1 , x , y ) 𝑑 s 𝑑 u \displaystyle p^{\kappa}(1,x,y)=16\int_{0}^{1/4}\int_{0}^{1/4}p^{\kappa}(1,x,y)dsdu
= 16 ∫ 0 1 / 4 ∫ 0 1 / 4 ∫ ℝ + d ∫ ℝ + d p κ ( s , x , z ) p κ ( 1 − s − u , z , w ) p κ ( u , y , w ) 𝑑 z 𝑑 w 𝑑 s 𝑑 u \displaystyle=16\int_{0}^{1/4}\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,x,z)p^{\kappa}(1-s-u,z,w)p^{\kappa}(u,y,w)dzdwdsdu
≤ 16 c 3 ( ∫ 0 1 / 4 ∫ ℝ + d p κ ( s , x , z ) ( 1 ∧ z d ) q − α + ε 𝑑 z 𝑑 s ) ( ∫ 0 1 / 4 ∫ ℝ + d p κ ( u , y , w ) ( 1 ∧ w d ) q − α + ε 𝑑 w 𝑑 u ) . \displaystyle\leq 16c_{3}\bigg{(}\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,x,z)(1\wedge z_{d})^{q-\alpha+{\varepsilon}}dzds\bigg{)}\bigg{(}\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(u,y,w)(1\wedge w_{d})^{q-\alpha+{\varepsilon}}dwdu\bigg{)}.
Thus, to conclude (5.6 ) by induction, it suffices to show that there exists a constant
c 4 > 0 c_{4}>0 such that
∫ 0 1 / 4 ∫ ℝ + d p κ ( s , v , z ) ( 1 ∧ z d ) q − α + ε 𝑑 z 𝑑 s ≤ c 4 ( 1 ∧ v d ) q , v ∈ ℝ + d . \int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,v,z)(1\wedge z_{d})^{q-\alpha+{\varepsilon}}dzds\leq c_{4}(1\wedge v_{d})^{q},\quad v\in{\mathbb{R}}^{d}_{+}.
(5.14)
By Lemma 2.2 , we can assume v ~ = 0 \widetilde{v}=0 . If v ∉ U ( 2 − 4 ) v\notin U(2^{-4}) , then we get
( 1 ∧ v d ) q ≥ 2 − 5 q ≥ 2 − 5 q ∫ 0 1 / 4 ∫ ℝ + d p κ ( s , v , z ) 𝑑 z 𝑑 s ≥ 2 − 5 q ∫ 0 1 / 4 ∫ ℝ + d p κ ( s , v , z ) ( 1 ∧ z d ) q − α + ε 𝑑 z 𝑑 s . (1\wedge v_{d})^{q}\geq 2^{-5q}\geq 2^{-5q}\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,v,z)dzds\geq 2^{-5q}\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,v,z)(1\wedge z_{d})^{q-\alpha+{\varepsilon}}dzds.
Otherwise, if v ∈ U ( 2 − 4 ) v\in U(2^{-4}) , then by (5.4 ), Fubini’s theorem, Lemmas 5.5 and 5.12 ,
∫ 0 1 / 4 ∫ ℝ + d p κ ( s , v , z ) ( 1 ∧ z d ) q − α + ε 𝑑 z 𝑑 s \displaystyle\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,v,z)(1\wedge z_{d})^{q-\alpha+{\varepsilon}}dzds
≤ ∫ 0 1 / 4 𝔼 v [ ( 1 ∧ Y s κ , U ( 1 ) , d ) q − α + ε : τ U ( 1 ) κ > s ] d s + ∫ 0 1 / 4 ℙ v ( Y τ U ( 1 ) κ κ ∈ ℝ + d ) d s \displaystyle\leq\int_{0}^{1/4}{\mathbb{E}}_{v}\left[(1\wedge Y^{\kappa,U(1),d}_{s})^{q-\alpha+{\varepsilon}}:\tau^{\kappa}_{U(1)}>s\right]ds+\int_{0}^{1/4}{\mathbb{P}}_{v}(Y^{\kappa}_{\tau^{\kappa}_{U(1)}}\in{\mathbb{R}}^{d}_{+})ds
≤ 𝔼 v ∫ 0 τ U ( 1 ) κ ( 1 ∧ Y s κ , U ( 1 ) , d ) q − α + ε 𝑑 s + 1 4 ℙ v ( Y τ U ( 1 ) κ ∈ ℝ + d ) ≤ c 5 v d q . \displaystyle\leq{\mathbb{E}}_{v}\int_{0}^{\tau^{\kappa}_{U(1)}}(1\wedge Y^{\kappa,U(1),d}_{s})^{q-\alpha+{\varepsilon}}ds+\frac{1}{4}{\mathbb{P}}_{v}(Y^{\kappa}_{\tau_{U(1)}}\in{\mathbb{R}}^{d}_{+})\leq c_{5}v_{d}^{q}.
This completes the proof. □ \Box
Proof of Proposition 5.1 :
The assertion is a direct consequence of Lemmas 5.13 and 5.10 .
□ \Box
In the remainder of this section
we complete the proof of Lemma 5.12 .
Lemma 5.14 .
If d = 1 ≤ α d=1\leq\alpha , then there exists C > 0 C>0 such that for all x , y ∈ ( 0 , ∞ ) x,y\in(0,\infty) ,
G κ ( x , y ) ≤ C ( 1 ∧ x ∧ y | x − y | ) q ∧ ( α − 1 2 ) ( x ∨ y ) α − 1 log ( e + x ∨ y | x − y | ) . G^{\kappa}(x,y)\leq C\left(1\wedge\frac{x\wedge y}{|x-y|}\right)^{q\wedge(\alpha-\frac{1}{2})}(x\vee y)^{\alpha-1}\log\left(e+\frac{x\vee y}{|x-y|}\right).
(5.15)
Proof. Let q ~ := q ∧ ( α − 1 2 ) ∈ ( 0 , q ] ∩ [ α − 1 , α ) \widetilde{q}:=q\wedge(\alpha-\frac{1}{2})\in(0,q]\cap[\alpha-1,\alpha) . By symmetry and scaling
(3.30 ), we can assume | x − y | = 1 |x-y|=1 and x ≤ y x\leq y without loss of generality. Then y = x + 1 > x ∨ 1 y=x+1>x\vee 1 . It suffices to show that
G κ ( x , y ) ≤ C x q ~ y α − 1 log ( e + y ) . \displaystyle G^{\kappa}(x,y)\leq Cx^{\widetilde{q}}y^{\alpha-1}\log(e+y).
Note that 1 + 2 q ~ > 1 + q ~ ≥ α 1+2\widetilde{q}>1+\widetilde{q}\geq\alpha . Since C ( α , q ~ , ℬ ) ≤ C ( α , q , ℬ ) = κ C(\alpha,\widetilde{q},{\cal B})\leq C(\alpha,q,{\cal B})=\kappa implies that
p κ ( t , x , y ) ≤ p C ( α , q ~ , ℬ ) ( t , x , y ) p^{\kappa}(t,x,y)\leq p^{C(\alpha,\widetilde{q},{\cal B})}(t,x,y) ,
using the fact y > x ∨ 1 y>x\vee 1 and
Lemmas 5.11 and 5.10 ,
we get
G κ ( x , y ) ≤ ∫ 0 ∞ p C ( α , q ~ , ℬ ) ( t , x , y ) 𝑑 t \displaystyle G^{\kappa}(x,y)\leq\int_{0}^{\infty}p^{C(\alpha,\widetilde{q},{\cal B})}(t,x,y)dt
≤ c 1 ( x q ~ ∫ 0 1 t ( α − q ~ ) / α 𝑑 t + x q ~ ∫ 1 y α t − ( 1 + q ~ ) / α 𝑑 t + x q ~ y q ~ ∫ y α ∞ t − ( 1 + 2 q ~ ) / α 𝑑 t ) \displaystyle\leq c_{1}\left(x^{\widetilde{q}}\int_{0}^{1}t^{(\alpha-\widetilde{q})/\alpha}dt+x^{\widetilde{q}}\int_{1}^{y^{\alpha}}t^{-(1+\widetilde{q})/\alpha}dt+x^{\widetilde{q}}y^{\widetilde{q}}\int_{y^{\alpha}}^{\infty}t^{-(1+2\widetilde{q})/\alpha}dt\right)
≤ c 1 x q ~ ( 1 + log ( y α ) + α 1 + 2 q ~ − α y α − q ~ − 1 ) ≤ c 2 x q ~ y α − 1 log ( e + y ) . \displaystyle\leq c_{1}x^{\widetilde{q}}\left(1+\log(y^{\alpha})+\frac{\alpha}{1+2\widetilde{q}-\alpha}y^{\alpha-\widetilde{q}-1}\right)\leq c_{2}x^{\widetilde{q}}y^{\alpha-1}\log(e+y).
The proof is complete.
□ \Box
We now improve (5.15 ) by removing the term α − 1 2 \alpha-\frac{1}{2} from the power of the first factor.
Lemma 5.15 .
If d = 1 ≤ α d=1\leq\alpha , then there exists C > 0 C>0 such that for all x , y ∈ ( 0 , ∞ ) x,y\in(0,\infty) ,
G κ ( x , y ) ≤ C ( 1 ∧ x ∧ y | x − y | ) q ( x ∨ y ) α − 1 log ( e + x ∨ y | x − y | ) . \displaystyle G^{\kappa}(x,y)\leq C\left(1\wedge\frac{x\wedge y}{|x-y|}\right)^{q}(x\vee y)^{\alpha-1}\log\left(e+\frac{x\vee y}{|x-y|}\right).
Proof. Let r = 2 − 6 r=2^{-6} .
By symmetry, (3.30 ) and Lemma 5.14 , without loss of generality, we can assume | x − y | = 1 |x-y|=1 and x < y ∧ ( r / 2 ) x<y\wedge(r/2) . Note that y = x + 1 ∈ ( 1 , 1 + r / 2 ) y=x+1\in(1,1+r/2) . It suffices to show that
G κ ( x , y ) ≤ C x q . \displaystyle G^{\kappa}(x,y)\leq Cx^{q}.
For z ∈ ( 0 , r / 2 ) z\in(0,r/2) and w ∈ ( r , ∞ ) w\in(r,\infty) , we have z < w / 2 ≤ w − z < w z<w/2\leq w-z<w and w ∨ y ≍ w w\vee y\asymp w . Thus, for any z ∈ ( 0 , r / 2 ) z\in(0,r/2) , using
Lemmas 5.3 and 5.14 , we obtain
∫ r ∞ G κ ( w , y ) ℬ ( z , w ) ( w − z ) − 1 − α 𝑑 w \displaystyle\int_{r}^{\infty}G^{\kappa}(w,y){\cal B}(z,w)(w-z)^{-1-\alpha}dw
≤ c 1 z β 1 ( | log z | β 3 ∨ 1 ) ∫ r ∞ G κ ( w , y ) w 1 + α + β 1 ( 1 + 𝟏 { w ≥ 1 } ( log w ) β 3 ) 𝑑 w \displaystyle\leq c_{1}z^{\beta_{1}}(|\log z|^{\beta_{3}}\vee 1)\int_{r}^{\infty}\frac{G^{\kappa}(w,y)}{w^{1+\alpha+\beta_{1}}}\big{(}1+{\bf 1}_{\{w\geq 1\}}(\log w)^{\beta_{3}}\big{)}dw
≤ c 2 z β 1 | log z | β 3 ∫ r ∞ log ( e + w | w − y | ) ( 1 + 𝟏 { w ≥ 1 } ( log w ) β 3 ) w 2 + β 1 𝑑 w . \displaystyle\leq c_{2}z^{\beta_{1}}|\log z|^{\beta_{3}}\int_{r}^{\infty}\log\bigg{(}e+\frac{w}{|w-y|}\bigg{)}\frac{\big{(}1+{\bf 1}_{\{w\geq 1\}}(\log w)^{\beta_{3}}\big{)}}{w^{2+\beta_{1}}}dw.
Hence, by the Lévy system formula and Lemma 5.4 , we get
𝔼 x [ G κ ( Y τ ( 0 , r / 2 ) κ κ , y ) ; Y τ ( 0 , r / 2 ) κ κ ∉ ( 0 , r ) ] \displaystyle{\mathbb{E}}_{x}\left[G^{\kappa}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}},y);Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\notin(0,r)\right]
≤ c 2 𝔼 x ∫ 0 τ ( 0 , r / 2 ) κ ( Y t κ , d ) β 1 | log ( Y t κ , d ) | β 3 𝑑 t ∫ r ∞ log ( e + w | w − y | ) ( 1 + 𝟏 { w ≥ 1 } ( log w ) β 3 ) w 2 + β 1 𝑑 w \displaystyle\leq c_{2}{\mathbb{E}}_{x}\int^{\tau^{\kappa}_{(0,r/2)}}_{0}(Y^{\kappa,d}_{t})^{\beta_{1}}|\log(Y^{\kappa,d}_{t})|^{\beta_{3}}dt\int_{r}^{\infty}\log\bigg{(}e+\frac{w}{|w-y|}\bigg{)}\frac{\big{(}1+{\bf 1}_{\{w\geq 1\}}(\log w)^{\beta_{3}}\big{)}}{w^{2+\beta_{1}}}dw
≤ c 3 x q ∫ r ∞ log ( e + w | w − y | ) ( 1 + 𝟏 { w ≥ 1 } ( log w ) β 3 ) w 2 + β 1 𝑑 w . \displaystyle\leq c_{3}x^{q}\int_{r}^{\infty}\log\bigg{(}e+\frac{w}{|w-y|}\bigg{)}\frac{\big{(}1+{\bf 1}_{\{w\geq 1\}}(\log w)^{\beta_{3}}\big{)}}{w^{2+\beta_{1}}}dw.
(5.16)
Since w − y ≥ w / 2 w-y\geq w/2 for w ∈ ( 2 y , ∞ ) w\in(2y,\infty) and y ∈ ( 1 , 1 + r / 2 ) y\in(1,1+r/2) , using a change of the variables, we obtain
∫ r ∞ log ( e + w | w − y | ) ( 1 + 𝟏 { w ≥ 1 } ( log w ) β 3 ) w 2 + β 1 𝑑 w \displaystyle\int_{r}^{\infty}\log\bigg{(}e+\frac{w}{|w-y|}\bigg{)}\frac{\big{(}1+{\bf 1}_{\{w\geq 1\}}(\log w)^{\beta_{3}}\big{)}}{w^{2+\beta_{1}}}dw
≤ 1 + ( log ( 2 + r ) ) β 3 r 2 + β 1 ∫ r 2 y log ( e + 2 + r | w − y | ) 𝑑 w + log ( e + 2 ) ∫ 2 y ∞ 1 + ( log w ) β 3 w 2 + β 1 𝑑 w \displaystyle\leq\frac{1+(\log(2+r))^{\beta_{3}}}{r^{2+\beta_{1}}}\int_{r}^{2y}\log\bigg{(}e+\frac{2+r}{|w-y|}\bigg{)}dw+\log(e+2)\int_{2y}^{\infty}\frac{1+(\log w)^{\beta_{3}}}{w^{2+\beta_{1}}}dw
≤ 2 ( 1 + ( log ( 2 + r ) ) β 3 ) r 2 + β 1 ∫ 0 1 + r / 2 log ( e + 2 + r v ) 𝑑 v + c 4 ≤ c 5 . \displaystyle\leq\frac{2(1+(\log(2+r))^{\beta_{3}})}{r^{2+\beta_{1}}}\int_{0}^{1+r/2}\log\bigg{(}e+\frac{2+r}{v}\bigg{)}dv+c_{4}\leq c_{5}.
Thus, by (5 ), it holds that
𝔼 x [ G κ ( Y τ ( 0 , r / 2 ) κ κ , y ) ; Y τ ( 0 , r / 2 ) κ κ ∉ ( 0 , r ) ] ≤ c 3 c 5 x q . \displaystyle{\mathbb{E}}_{x}\left[G^{\kappa}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}},y);Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\notin(0,r)\right]\leq c_{3}c_{5}x^{q}.
(5.17)
Since z ↦ G κ ( z , y ) z\mapsto G^{\kappa}(z,y) is harmonic in ( 0 , 2 r ) (0,2r) with respect to Y κ Y^{\kappa} and vanishes continuously as z → 0 z\to 0 by Lemma 5.14 , we get from Proposition 5.7 that G κ ( z , y ) ≤ c 6 G κ ( r , y ) G^{\kappa}(z,y)\leq c_{6}G^{\kappa}(r,y) for all z ∈ ( 0 , r ) z\in(0,r) . Therefore,
using Lemmas 5.5 and 5.14 , since y ∈ ( 1 , 1 + r / 2 ) y\in(1,1+r/2) , we obtain
𝔼 x [ G κ ( Y τ ( 0 , r / 2 ) κ κ , y ) ; Y τ ( 0 , r / 2 ) κ κ ∈ ( 0 , r ) ] ≤ c 6 G κ ( r , y ) ℙ x ( Y τ ( 0 , r / 2 ) κ κ ∈ ( 0 , r ) ) \displaystyle{\mathbb{E}}_{x}\left[G^{\kappa}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}},y);Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\in(0,r)\right]\leq c_{6}G^{\kappa}(r,y){\mathbb{P}}_{x}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\in(0,r))
≤ c 7 ( x r ) q ( r y − r ) q ∧ ( α − 1 2 ) y α − 1 log ( e + y y − r ) ≤ c 8 x q . \displaystyle\leq c_{7}\bigg{(}\frac{x}{r}\bigg{)}^{q}\bigg{(}\frac{r}{y-r}\bigg{)}^{q\wedge(\alpha-\frac{1}{2})}y^{\alpha-1}\log\bigg{(}e+\frac{y}{y-r}\bigg{)}\leq c_{8}x^{q}.
(5.18)
Combining (5.17 ) and (5 ) and using the harmonicity of z ↦ G κ ( z , y ) z\mapsto G^{\kappa}(z,y) in ( 0 , 2 r ) (0,2r) ,
we arrive at
G κ ( x , y ) = 𝔼 x [ G κ ( Y τ ( 0 , r / 2 ) κ κ , y ) ; Y τ ( 0 , r / 2 ) κ κ ∉ ( 0 , r ) ] + 𝔼 x [ G κ ( Y τ ( 0 , r / 2 ) κ κ , y ) ; Y τ ( 0 , r / 2 ) κ κ ∈ ( 0 , r ) ] ≤ c 9 x q . \displaystyle G^{\kappa}(x,y)={\mathbb{E}}_{x}\left[G^{\kappa}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}},y);Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\notin(0,r)\right]+{\mathbb{E}}_{x}\left[G^{\kappa}(Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}},y);Y^{\kappa}_{\tau^{\kappa}_{(0,r/2)}}\in(0,r)\right]\leq c_{9}x^{q}.
The proof is complete.
□ \Box
Proof of Lemma 5.12 for d = 1 ≤ α d=1\leq\alpha :
Assume that d = 1 ≤ α d=1\leq\alpha . Since
D ⊂ U ( 2 R ) = [ 0 , R ) D\subset U(2R)=[0,R) ,
using Fubini’s theorem and Lemma 5.15 , we have
∫ 0 ∞ ∫ D p κ , D ( t , x , z ) z γ 𝑑 z 𝑑 t ≤ ∫ 0 R ∫ 0 ∞ p κ , D ( t , x , z ) 𝑑 t z γ 𝑑 z ≤ ∫ 0 R G κ ( x , z ) z γ 𝑑 z \displaystyle\int_{0}^{\infty}\int_{D}p^{\kappa,D}(t,x,z)z^{\gamma}dzdt\leq\int_{0}^{R}\int_{0}^{\infty}p^{\kappa,D}(t,x,z)dt\,z^{\gamma}dz\leq\int_{0}^{R}G^{\kappa}(x,z)z^{\gamma}dz
≤ c 1 ∫ 0 x / 2 z q + γ | x − z | q x α − 1 log ( e + x | x − z | ) 𝑑 z + c 1 ∫ x / 2 2 x z γ ( 2 x ) α − 1 log ( e + 2 x | x − z | ) 𝑑 z \displaystyle\leq c_{1}\int_{0}^{x/2}\frac{z^{q+\gamma}}{|x-z|^{q}}x^{\alpha-1}\log\bigg{(}e+\frac{x}{|x-z|}\bigg{)}dz+c_{1}\int_{x/2}^{2x}z^{\gamma}(2x)^{\alpha-1}\log\bigg{(}e+\frac{2x}{|x-z|}\bigg{)}dz
+ c 1 ∫ 2 x R x q | x − z | q z γ + α − 1 log ( e + z | x − z | ) 𝑑 z \displaystyle\quad+c_{1}\int_{2x}^{R}\frac{x^{q}}{|x-z|^{q}}z^{\gamma+\alpha-1}\log\bigg{(}e+\frac{z}{|x-z|}\bigg{)}dz
= : c 1 ( I + I I + I I I ) . \displaystyle=:c_{1}(I+II+III).
Since γ + α − q > 0 \gamma+\alpha-q>0 and x ≤ R / 10 x\leq R/10 , we see that
I ≤ x q + γ + α − 1 ∫ 0 x / 2 1 ( x / 2 ) q log ( e + x ( x / 2 ) ) 𝑑 z ≤ 2 q − 1 log ( e + 2 ) x γ + α ≤ c 2 R γ + α − q x q . \displaystyle I\leq x^{q+\gamma+\alpha-1}\int_{0}^{x/2}\frac{1}{(x/2)^{q}}\log\bigg{(}e+\frac{x}{(x/2)}\bigg{)}dz\leq 2^{q-1}\log(e+2)x^{\gamma+\alpha}\leq c_{2}R^{\gamma+\alpha-q}x^{q}.
Next, using the change of the variables z = x y z=xy , we also get
I I \displaystyle II
≤ ( 2 x ) γ + α − 1 ∫ x / 2 2 x log ( e + 2 x | x − z | ) 𝑑 z = 2 γ + α − 1 x γ + α ∫ 1 / 2 2 log ( e + 2 | 1 − y | ) 𝑑 y \displaystyle\leq(2x)^{\gamma+\alpha-1}\int_{x/2}^{2x}\log\bigg{(}e+\frac{2x}{|x-z|}\bigg{)}dz=2^{\gamma+\alpha-1}x^{\gamma+\alpha}\int_{1/2}^{2}\log\bigg{(}e+\frac{2}{|1-y|}\bigg{)}dy
≤ c 3 x γ + α ≤ c 3 R γ + α − q x q . \displaystyle\leq c_{3}x^{\gamma+\alpha}\leq c_{3}R^{\gamma+\alpha-q}x^{q}.
Lastly, since | x − z | ≥ z / 2 |x-z|\geq z/2 for all z ≥ 2 x z\geq 2x , we obtain
I I I ≤ 2 q log ( e + 2 ) x q ∫ 2 x R z γ + α − q − 1 𝑑 z ≤ c 4 R γ + α − q x q . \displaystyle III\leq 2^{q}\log(e+2)x^{q}\int_{2x}^{R}z^{\gamma+\alpha-q-1}dz\leq c_{4}R^{\gamma+\alpha-q}x^{q}.
The proof is complete.
□ \Box
6. Sharp lower bounds of heat kernels
In this and the next section,
whenever we consider Y ¯ \overline{Y} , we assume that (A1) , (A3) and (A4) hold, and whenever we consider Y κ Y^{\kappa} , we assume that all (A1) – (A4) hold. Note that (IUBS) holds under this setting by Lemma 4.2 .
We fix κ ∈ [ 0 , ∞ ) \kappa\in[0,\infty) . The following notational convenience will be used throughout this and the next section.
When we consider Y κ Y^{\kappa} , we assume that κ > 0 \kappa>0 if α ∈ ( 0 , 1 ] \alpha\in(0,1] (see Lemma 2.3 ), and denote by q q the strictly positive constant q k ∈ ( ( α − 1 ) + , α + β 1 ) q_{k}\in((\alpha-1)_{+},\alpha+\beta_{1}) from (1.8 ). Additionally, we write Y Y , p ( t , x , y ) p(t,x,y) , p D ( t , x , y ) p^{D}(t,x,y) , τ D \tau_{D} and ζ \zeta instead of Y κ Y^{\kappa} , p κ ( t , x , y ) p^{\kappa}(t,x,y) , p κ , D ( t , x , y ) p^{\kappa,D}(t,x,y) , τ D κ \tau_{D}^{\kappa} and ζ κ \zeta^{\kappa} . When we consider
Y ¯ \overline{Y} , the letter q q denotes 0, and we write Y Y , p ( t , x , y ) p(t,x,y) , p D ( t , x , y ) p^{D}(t,x,y) , τ D \tau_{D} and ζ \zeta instead of Y ¯ \overline{Y} ,
p ¯ ( t , x . y ) \overline{p}(t,x.y) , p ¯ D ( t , x , y ) \overline{p}^{D}(t,x,y) , τ ¯ D \overline{\tau}_{D} and ∞ \infty .
Recall the definitions of V x ( t ) V_{x}(t) and W x ( t ) W_{x}(t) from (3 ). We let V x := V x ( 1 ) V_{x}:=V_{x}(1) and W x := W x ( 1 ) W_{x}:=W_{x}(1) .
Lemma 6.1 .
There exist constants M > 1 M>1 and C > 0 C>0 such that
for all t > 0 t>0 and x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} ,
inf z ∈ W x ( t ) p ( M t , x , z ) ≥ C ( 1 ∧ x d t 1 / α ) q t − d / α . \inf_{z\in W_{x}(t)}p(Mt,x,z)\geq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}t^{-d/\alpha}.
Proof. When q = 0 q=0 , the result
is given in Lemma 3.6 .
Suppose q > 0 q>0 .
By the scaling property (3.30 ),
it suffices to prove the lemma for t = 1 t=1 . If x d ≥ 1 x_{d}\geq 1 , then the result follows from Proposition 3.2 .
Assume x d < 1 x_{d}<1 . By Proposition 3.2 , for any M > 1 M>1 , there is c 1 = c 1 ( M ) > 0 c_{1}=c_{1}(M)>0 such that
inf w , z ∈ W x , 1 ≤ s ≤ M p ( s , w , z ) ≥ c 1 for all x ∈ ℝ + d . \displaystyle\inf_{w,z\in W_{x},\,1\leq s\leq M}p(s,w,z)\geq c_{1}\quad\text{for all $x\in{\mathbb{R}}^{d}_{+}$}.
(6.1)
Using the strong Markov property and (6.1 ),
we see that for all M > 1 M>1 and z ∈ W x z\in W_{x} ,
p ( M , x , z ) \displaystyle p(M,x,z)
≥ 𝔼 x [ p ( M − τ V x , Y τ V x , z ) : τ V x ≤ M − 1 , Y τ V x ∈ W x ] \displaystyle\geq{\mathbb{E}}_{x}\left[\,p(M-\tau_{V_{x}},Y_{\tau_{V_{x}}},z):\tau_{V_{x}}\leq M-1,\,Y_{\tau_{V_{x}}}\in W_{x}\right]
≥ ( inf w ∈ W x , 1 ≤ s ≤ M p ( s , w , z ) ) ℙ x ( τ V x ≤ M − 1 , Y τ V x ∈ W x ) \displaystyle\geq\bigg{(}\inf_{w\in W_{x},\,1\leq s\leq M}p(s,w,z)\bigg{)}{\mathbb{P}}_{x}\left(\tau_{V_{x}}\leq M-1,\,Y_{\tau_{V_{x}}}\in W_{x}\right)
≥ c 1 ( ℙ x ( Y τ V x ∈ W x ) − ℙ x ( τ V x ( t ) > M − 1 ) ) . \displaystyle\geq c_{1}\left({\mathbb{P}}_{x}(Y_{\tau_{V_{x}}}\in W_{x})-{\mathbb{P}}_{x}(\tau_{V_{x}(t)}>M-1)\right).
Note that, by [30 , Lemma 5.10] for κ > 0 \kappa>0 and [32 , Theorem 1.1] for κ = 0 \kappa=0 , we have
ℙ x ( Y τ V x ∈ W x ) ≥ 2 c 2 x d q {\mathbb{P}}_{x}(Y_{\tau_{V_{x}}}\in W_{x})\geq 2c_{2}x_{d}^{q}
and, by Corollary 5.2 , we also have
ℙ x ( τ V x > M − 1 ) ≤ ℙ x ( ζ > M − 1 ) ≤ c 3 ( x d / ( M − 1 ) 1 / α ) q . {\mathbb{P}}_{x}(\tau_{V_{x}}>M-1)\leq{\mathbb{P}}_{x}(\zeta>M-1)\leq c_{3}(x_{d}/(M-1)^{1/\alpha})^{q}.
Thus, we can choose M = 1 + ( c 3 / c 2 ) α / q M=1+(c_{3}/c_{2})^{\alpha/q} so that
2 c 2 − c 3 ( M − 1 ) − q / α = c 2 2c_{2}-c_{3}(M-1)^{-q/\alpha}=c_{2} , which implies
p ( M , x , z ) ≥ c 2 x d q p(M,x,z)\geq c_{2}x_{d}^{q} . □ \Box
For any b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 , we define for t ≥ 0 t\geq 0 and x , y ∈ ℝ ¯ + d x,y\in{\overline{\mathbb{R}}}^{d}_{+} ,
A b 1 , b 2 , b 3 , b 4 ( t , x , y ) \displaystyle A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)
:= ( ( x d ∧ y d ) ∨ t 1 / α | x − y | ∧ 1 ) b 1 ( ( x d ∨ y d ) ∨ t 1 / α | x − y | ∧ 1 ) b 2 \displaystyle:=\bigg{(}\frac{(x_{d}\wedge y_{d})\vee t^{1/\alpha}}{|x-y|}\wedge 1\bigg{)}^{b_{1}}\bigg{(}\frac{(x_{d}\vee y_{d})\vee t^{1/\alpha}}{|x-y|}\wedge 1\bigg{)}^{b_{2}}
× log b 3 ( e + ( ( x d ∨ y d ) ∨ t 1 / α ) ∧ | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) log b 4 ( e + | x − y | ( ( x d ∨ y d ) ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\times\log^{b_{3}}\bigg{(}e+\frac{((x_{d}\vee y_{d})\vee t^{1/\alpha})\wedge|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\,\log^{b_{4}}\bigg{(}e+\frac{|x-y|}{((x_{d}\vee y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
(6.2)
We first note that
A b 1 , b 2 , b 3 , b 4 ( 0 , x , y ) = B b 1 , b 2 , b 3 , b 4 ( x , y ) A_{b_{1},b_{2},b_{3},b_{4}}(0,x,y)=B_{b_{1},b_{2},b_{3},b_{4}}(x,y)
for x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} and
A b 1 , b 2 , b 3 , b 4 ( t , x , y ) ≍ B b 1 , b 2 , b 3 , b 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) for t ≥ 0 , x , y ∈ ℝ + d , \displaystyle A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\asymp B_{b_{1},b_{2},b_{3},b_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\quad\text{ for }t\geq 0,x,y\in{\mathbb{R}}^{d}_{+},
(6.3)
since a ∨ b ≍ a + b a\vee b\asymp a+b for all a , b ≥ 0 a,b\geq 0 .
Note also that for any a > 0 a>0 ,
there exists c > 0 c>0 such that
A b 1 , b 2 , b 3 , b 4 ( t , x , y ) ≥ c ( a ∧ 1 ) b 1 + b 2 , A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\geq c(a\wedge 1)^{b_{1}+b_{2}},
(6.4)
for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} with ( x d ∧ y d ) + t 1 / α ≥ a | x − y | (x_{d}\wedge y_{d})+t^{1/\alpha}\geq a|x-y| .
Proposition 6.2 .
There exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) ≥ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q A β 1 , β 2 , β 3 , β 4 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) . \displaystyle p(t,x,y)\geq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
Proof. Without loss of generality, we assume x d ≤ y d x_{d}\leq y_{d} .
Let M > 1 M>1 be the constant in Lemma 6.1 .
By the semigroup property and Lemma 6.1 ,
p ( t , x , y ) \displaystyle p(t,x,y)
≥ ∫ W x ( t / ( 3 M ) ) ∫ W y ( t / ( 3 M ) ) p ( t / 3 , x , z ) p ( t / 3 , z , w ) p ( t / 3 , w , y ) 𝑑 z 𝑑 w \displaystyle\geq\int_{W_{x}(t/(3M))}\int_{W_{y}(t/(3M))}p(t/3,x,z)\,p(t/3,z,w)\,p(t/3,w,y)dzdw
≥ c 1 t 2 d / α ( inf z ∈ W x ( t / ( 3 M ) ) p ( t / 3 , x , z ) ) ( inf w ∈ W y ( t / ( 3 M ) ) p ( t / 3 , y , w ) ) \displaystyle\geq c_{1}t^{2d/\alpha}\left(\inf_{z\in W_{x}(t/(3M))}p(t/3,x,z)\right)\left(\inf_{w\in W_{y}(t/(3M))}p(t/3,y,w)\right)
× ( inf ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) p ( t / 3 , z , w ) ) \displaystyle\quad\times\left(\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}p(t/3,z,w)\right)
≥ c 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q inf ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) p ( t / 3 , z , w ) . \displaystyle\geq c_{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}p(t/3,z,w).
(6.5)
For all ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) (z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M)) , we have z d ≥ x d + 5 ( t / ( 3 M ) ) 1 / α z_{d}\geq x_{d}+5(t/(3M))^{1/\alpha} , w d ≥ y d + 5 ( t / ( 3 M ) ) 1 / α w_{d}\geq y_{d}+5(t/(3M))^{1/\alpha} ,
| z − w | ≤ | z − x | + | x − y | + | y − w | ≤ | x − y | + 20 ( t / ( 3 M ) ) 1 / α |z-w|\leq|z-x|+|x-y|+|y-w|\leq|x-y|+20(t/(3M))^{1/\alpha} and | z − w | ≥ | x − y | − | z − x | − | y − w | ≥ | x − y | − 20 ( t / ( 3 M ) ) 1 / α |z-w|\geq|x-y|-|z-x|-|y-w|\geq|x-y|-20(t/(3M))^{1/\alpha} .
Hence, if | x − y | ≤ 21 ( t / ( 3 M ) ) 1 / α |x-y|\leq 21(t/(3M))^{1/\alpha} , then
z d ∧ w d ≥ 5 ( t 3 M ) 1 / α ≥ 5 41 ( ( t 3 M ) 1 / α ∨ | z − w | ) , ( z , w ) ∈ W x ( t 3 M ) × W y ( t 3 M ) . \displaystyle z_{d}\wedge w_{d}\geq 5(\frac{t}{3M})^{1/\alpha}\geq\frac{5}{41}\big{(}(\frac{t}{3M})^{1/\alpha}\vee|z-w|\big{)},\quad(z,w)\in W_{x}(\frac{t}{3M})\times W_{y}(\frac{t}{3M}).
(6.6)
If | x − y | > 21 ( t / ( 3 M ) ) 1 / α |x-y|>21(t/(3M))^{1/\alpha} , then using (A3) (I) in the first inequality below,
| z − w | ≤ 2 | x − y | |z-w|\leq 2|x-y| and Lemma 10.9 in the second and (6.3 ) in the third, we get
inf ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) J ( z , w ) ≥ C 2 − 1 inf ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) B β 1 , β 2 , β 3 , β 4 ( z , w ) | z − w | d + α \displaystyle\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}J(z,w)\geq C_{2}^{-1}\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}\frac{B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(z,w)}{|z-w|^{d+\alpha}}
≥ c 3 B β 1 , β 2 , β 3 , β 4 ( x + 5 ( t / ( 3 M ) ) 1 / α 𝐞 d , y + 5 ( t / ( 3 M ) ) 1 / α 𝐞 d ) | x − y | d + α ≥ c 4 A β 1 , β 2 , β 3 , β 4 ( t , x , y ) | x − y | d + α . \displaystyle\geq c_{3}\frac{B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+5(t/(3M))^{1/\alpha}{\mathbf{e}}_{d},y+5(t/(3M))^{1/\alpha}{\mathbf{e}}_{d})}{|x-y|^{d+\alpha}}\geq c_{4}\frac{A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)}{|x-y|^{d+\alpha}}.
(6.7)
Now, by Propositions 3.2 and 4.5 , (6.6 ), (6 ) and (6.4 ), we have
inf ( z , w ) ∈ W x ( t / ( 3 M ) ) × W y ( t / ( 3 M ) ) p ( t / 3 , z , w ) \displaystyle\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}p(t/3,z,w)
≥ c 5 { t − d / α if | x − y | ≤ 21 ( t / ( 3 M ) ) 1 / α , t A β 1 , β 2 , β 3 , β 4 ( t , x , y ) | x − y | − d − α if | x − y | > 21 ( t / ( 3 M ) ) 1 / α \displaystyle\geq c_{5}\begin{cases}t^{-d/\alpha}&\text{ if }|x-y|\leq 21(t/(3M))^{1/\alpha},\\[6.0pt]
tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)|x-y|^{-d-\alpha}&\text{ if }|x-y|>21(t/(3M))^{1/\alpha}\end{cases}
≥ c 6 A β 1 , β 2 , β 3 , β 4 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) . \displaystyle\geq c_{6}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
(6.8)
Combining (6.5 ) and (6 ),
we get the desired result.
□ \Box
Corollary 6.3 .
It holds that for any t > 0 t>0 and x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} ,
ℙ x ( ζ > t ) ≍ ( 1 ∧ x d t 1 / α ) q . \displaystyle{\mathbb{P}}_{x}(\zeta>t)\asymp\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}.
Proof. Using Proposition 6.2 and (6.4 ), we get
ℙ x ( ζ > t ) \displaystyle{\mathbb{P}}_{x}(\zeta>t)
≥ ∫ B ( x , 2 t 1 / α ) : y d ≥ t 1 / α p ( t , x , y ) 𝑑 y ≥ c 1 ( 1 ∧ x d t 1 / α ) q t − d / α ∫ B ( x , 2 t 1 / α ) : y d ≥ t 1 / α 𝑑 y \displaystyle\geq\int_{B(x,2t^{1/\alpha}):y_{d}\geq t^{1/\alpha}}p(t,x,y)dy\geq c_{1}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}t^{-d/\alpha}\int_{B(x,2t^{1/\alpha}):y_{d}\geq t^{1/\alpha}}dy
≥ c 2 ( 1 ∧ x d t 1 / α ) q . \displaystyle\geq c_{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}.
Combining the above with Corollary 5.2 , we arrive at the result.
□ \Box
Lemma 6.4 .
There exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) ≥ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) \displaystyle p(t,x,y)\geq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)
× ( | x − y | α ∧ t ) ∫ ( y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 . \displaystyle\quad\times\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{(y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}.
Proof. Case (i): ( x d ∧ y d ) ∨ t 1 / α ≥ | x − y | / 4 (x_{d}\wedge y_{d})\vee t^{1/\alpha}\geq|x-y|/4 . In this case, we
see from (6.4 ) that A β 1 , β 2 , β 3 , β 4 ( t , x , y ) A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y) is bounded below by a positive constant. Moreover, since sup s > 0 , z , w ∈ ℝ + d A β 1 , β 2 , β 3 , β 4 ( s , z , w ) ≤ c 1 < ∞ \sup_{s>0,\,z,w\in{\mathbb{R}}^{d}_{+}}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(s,z,w)\leq c_{1}<\infty , we get
| x − y | α ∫ ( y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle|x-y|^{\alpha}\int_{(y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≤ c 1 2 | x − y | α ∫ | x − y | / 4 ∞ d r r α + 1 = 4 α c 1 2 / α . \displaystyle\leq c_{1}^{2}|x-y|^{\alpha}\int_{|x-y|/4}^{\infty}\frac{dr}{r^{\alpha+1}}=4^{\alpha}c_{1}^{2}/\alpha.
Therefore, we get the result from Proposition 6.2 in this case.
Case (ii): ( x d ∧ y d ) ∨ t 1 / α < | x − y | / 4 (x_{d}\wedge y_{d})\vee t^{1/\alpha}<|x-y|/4 .
For r > 0 r>0 , set x ( r ) := x + r 𝐞 d x(r):=x+r{\mathbf{e}}_{d} and
K ( r ) := { z = ( z ~ , z d ) ∈ ℝ + d : | z ~ − x ~ | < r 2 , z d = x d + r } . K(r):=\left\{z=(\widetilde{z},z_{d})\in{\mathbb{R}}^{d}_{+}:|\widetilde{z}-\widetilde{x}|<\frac{r}{2},\,z_{d}=x_{d}+r\right\}.
Let
K := { z ∈ ℝ + d : z ∈ K ( r ) for some ( y d ∨ t 1 / α ) ∧ | x − y | 4 < r < | x − y | 2 } . K:=\left\{z\in{\mathbb{R}}^{d}_{+}:z\in K(r)\text{ for some }(y_{d}\vee t^{1/\alpha})\wedge\frac{|x-y|}{4}<r<\frac{|x-y|}{2}\right\}.
For any z = ( z ~ , x d + r ) ∈ K z=(\widetilde{z},x_{d}+r)\in K , since
r ≤ | x − z | ≤ 5 2 r ≤ 5 4 | x − y | , ( 1 − 5 4 ) | x − y | ≤ | y − z | ≤ ( 1 + 5 4 ) | x − y | , r\leq|x-z|\leq\frac{\sqrt{5}}{2}r\leq\frac{\sqrt{5}}{4}|x-y|,\qquad(1-\frac{\sqrt{5}}{4})|x-y|\leq|y-z|\leq(1+\frac{\sqrt{5}}{4})|x-y|,
(6.9)
we see that
A β 1 , β 2 , β 3 , β 4 ( t , x , z ) \displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)
≍ ( 1 ∧ x d ∨ t 1 / α r ) β 1 ( 1 ∧ ( x d + r ) ∨ t 1 / α r ) β 2 \displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{(x_{d}+r)\vee t^{1/\alpha}}{r}\bigg{)}^{\beta_{2}}
× log β 3 ( e + ( ( x d + r ) ∨ t 1 / α ) ∧ r ( x d ∨ t 1 / α ) ∧ r ) log β 4 ( e + r ( ( x d + r ) ∨ t 1 / α ) ∧ r ) \displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{((x_{d}+r)\vee t^{1/\alpha})\wedge r}{(x_{d}\vee t^{1/\alpha})\wedge r}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{r}{((x_{d}+r)\vee t^{1/\alpha})\wedge r}\bigg{)}
≍ ( 1 ∧ x d ∨ t 1 / α r ) β 1 log β 3 ( e + r ( x d ∨ t 1 / α ) ∧ r ) \displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{r}{(x_{d}\vee t^{1/\alpha})\wedge r}\bigg{)}
(6.10)
≍ A β 1 , β 2 , β 3 , β 4 ( t , x , x ( r ) ) \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))
and
A β 1 , β 2 , β 3 , β 4 ( t , z , y ) \displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,y)
≍ ( 1 ∧ y d ∨ t 1 / α | x − y | ) β 1 ( 1 ∧ x d + r | x − y | ) β 2 log β 3 ( e + x d + r y d ∨ t 1 / α ) log β 4 ( e + | x − y | x d + r ) \displaystyle\asymp\bigg{(}1\wedge\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}+r}{|x-y|}\bigg{)}^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}+r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{x_{d}+r}\bigg{)}
(6.11)
≍ A β 1 , β 2 , β 3 , β 4 ( t , x ( r ) , y ) . \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y).
Thus, using the semigroup property, Proposition 6.2 and (6.9 ), we get
p ( t , x , y ) ≥ ∫ K p ( t / 2 , x , z ) p ( t / 2 , z , y ) 𝑑 z \displaystyle p(t,x,y)\geq\int_{K}p(t/2,x,z)p(t/2,z,y)dz
≥ c 2 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ∫ K t A β 1 , β 2 , β 3 , β 4 ( t , x , z ) | x − z | d + α t A β 1 , β 2 , β 3 , β 4 ( t , z , y ) | y − z | d + α 𝑑 z \displaystyle\geq c_{2}^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\int_{K}\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)}{|x-z|^{d+\alpha}}\,\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,y)}{|y-z|^{d+\alpha}}dz
≥ c 3 t 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q \displaystyle\geq c_{3}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}
× ∫ ( y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x ( r ) ) r d + α A β 1 , β 2 , β 3 , β 4 ( t , x ( r ) , y ) | x − y | d + α ∫ K ( r ) d z ~ d r \displaystyle\quad\times\int_{(y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\frac{A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))}{r^{d+\alpha}}\frac{A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y)}{|x-y|^{d+\alpha}}\int_{K(r)}d\widetilde{z}dr
≥ c 4 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q t | x − y | d + α \displaystyle\geq c_{4}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\frac{t}{|x-y|^{d+\alpha}}
× t ∫ ( y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x ( r ) ) A β 1 , β 2 , β 3 , β 4 ( t , x ( r ) , y ) d r r α + 1 . \displaystyle\quad\times t\int_{(y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x(r),y)\frac{dr}{r^{\alpha+1}}.
□ \Box
Note that, for any x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} and | x − y | / 4 ≤ r ≤ | x − y | / 2 |x-y|/4\leq r\leq|x-y|/2 ,
A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) \displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})
≍ ( 1 ∧ x d ∨ t 1 / α | x − y | ) β 1 log β 3 ( e + | x − y | ( x d ∨ t 1 / α ) ∧ | x − y | ) \displaystyle\asymp\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(x_{d}\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}
≍ A β 1 , β 2 , β 3 , β 4 ( t , x , x + | x − y | 2 𝐞 d ) \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})
(6.12)
and, since y d ≤ x d + 4 r y_{d}\leq x_{d}+4r ,
A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) \displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)
≍ ( 1 ∧ y d ∨ t 1 / α | x − y | ) β 1 log β 3 ( e + | x − y | ( y d ∨ t 1 / α ) ∧ | x − y | ) \displaystyle\asymp\bigg{(}1\wedge\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(y_{d}\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}
≍ A β 1 , β 2 , β 3 , β 4 ( t , x + | x − y | 2 𝐞 d , y ) \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)
≍ A β 1 , β 2 , β 3 , β 4 ( t , y , y + | x − y | 2 𝐞 d ) . \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,y,y+\frac{|x-y|}{2}{\mathbf{e}}_{d}).
(6.13)
In particular, we have
A β 1 , β 2 , β 3 , β 4 ( t , x , x + | x − y | 2 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + | x − y | 2 𝐞 d , y ) \displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)
≍ A β 1 , β 2 , β 3 , β 4 ( t , x , x + | x − y | 2 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , y , y + | x − y | 2 𝐞 d ) \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,y,y+\frac{|x-y|}{2}{\mathbf{e}}_{d})
≍ A β 1 , β 1 , 0 , β 3 ( t , x , y ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\asymp A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
(6.14)
By (6 )–(6 ), we have that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
∫ | x − y | / 4 | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle\int_{|x-y|/4}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≍ A β 1 , β 2 , β 3 , β 4 ( t , x , x + | x − y | 2 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + | x − y | 2 𝐞 d , y ) ∫ | x − y | / 4 | x − y | / 2 d r r α + 1 \displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)\int_{|x-y|/4}^{|x-y|/2}\frac{dr}{r^{\alpha+1}}
≍ | x − y | − α A β 1 , β 1 , 0 , β 3 ( t , x , y ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\asymp|x-y|^{-\alpha}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
(6.15)
Combining (6 ), Proposition 6.2 and Lemma 6.4 , we get the following result.
Proposition 6.6 .
There exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) \displaystyle p(t,x,y)
≥ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) [ A β 1 , β 2 , β 3 , β 4 ( t , x , y ) \displaystyle\geq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)
+ ( 1 ∧ t | x − y | α ) A β 1 , β 2 , β 3 , β 4 ( t , x , x + | x − y | 2 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + | x − y | 2 𝐞 d , y ) ] \displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+\frac{|x-y|}{2}{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)\bigg{]}
≍ ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) [ A β 1 , β 2 , β 3 , β 4 ( t , x , y ) \displaystyle\asymp\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)
+ ( 1 ∧ t | x − y | α ) A β 1 , β 1 , 0 , β 3 ( t , x , y ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) ] . \displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\bigg{]}.
Lemma 6.8 .
Suppose that β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} . Then for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
( | x − y | α ∧ t ) ∫ ( x d ∨ y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{(x_{d}\vee y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\!\!\!\!\!\!\!\!\!\!\!\!A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≍ ( 1 ∧ t | x − y | α ) A β 1 , β 1 , 0 , β 3 + β 4 + 1 ( t , x , y ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
Proof. Without loss of generality, we assume x d ≤ y d x_{d}\leq y_{d} .
Assume first that y d ∨ t 1 / α < | x − y | / 4 y_{d}\vee t^{1/\alpha}<|x-y|/4 .
Using β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} and (6 )–(6 ),
since x d ≤ y d ∨ t 1 / α x_{d}\leq y_{d}\vee t^{1/\alpha} , we get
∫ y d ∨ t 1 / α | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle\int_{y_{d}\vee t^{1/\alpha}}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≍ ∫ y d ∨ t 1 / α | x − y | / 2 ( x d ∨ t 1 / α r ) β 1 ( y d ∨ t 1 / α | x − y | ) β 1 ( r | x − y | ) α + β 1 \displaystyle\asymp\int_{y_{d}\vee t^{1/\alpha}}^{|x-y|/2}\bigg{(}\frac{x_{d}\vee t^{1/\alpha}}{r}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{r}{|x-y|}\bigg{)}^{\alpha+\beta_{1}}
× log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + | x − y | r ) d r r α + 1 \displaystyle\qquad\times\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r^{\alpha+1}}
≍ ( x d ∨ t 1 / α | x − y | ) β 1 ( y d ∨ t 1 / α | x − y | ) β 1 | x − y | − α \displaystyle\asymp\bigg{(}\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}|x-y|^{-\alpha}
× ∫ y d ∨ t 1 / α | x − y | / 2 log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + | x − y | r ) d r r . \displaystyle\qquad\times\int_{y_{d}\vee t^{1/\alpha}}^{|x-y|/2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r}.
(6.20)
By a change of the variables and Lemma 10.11 , we see that
∫ y d ∨ t 1 / α | x − y | / 2 log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + | x − y | r ) d r r \displaystyle\int_{y_{d}\vee t^{1/\alpha}}^{|x-y|/2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{|x-y|}{r}\bigg{)}\frac{dr}{r}
= ∫ 4 ( y d ∨ t 1 / α ) / | x − y | 2 log β 3 ( e + | x − y | s 4 ( x d ∨ t 1 / α ) ) log β 3 ( e + | x − y | s 4 ( y d ∨ t 1 / α ) ) log β 4 ( e + 4 s ) d s s \displaystyle=\int_{4(y_{d}\vee t^{1/\alpha})/|x-y|}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|s}{4(x_{d}\vee t^{1/\alpha})}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|s}{4(y_{d}\vee t^{1/\alpha})}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{4}{s}\bigg{)}\frac{ds}{s}
≍ log β 3 ( e + | x − y | 4 ( x d ∨ t 1 / α ) ) log β 3 + β 4 + 1 ( e + | x − y | 4 ( y d ∨ t 1 / α ) ) . \displaystyle\asymp\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{4(x_{d}\vee t^{1/\alpha})}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{|x-y|}{4(y_{d}\vee t^{1/\alpha})}\bigg{)}.
(6.21)
By (6 )–(6 ),
since y d ∨ t 1 / α < | x − y | / 4 y_{d}\vee t^{1/\alpha}<|x-y|/4 , we obtain
( | x − y | α ∧ t ) ∫ y d ∨ t 1 / α | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{y_{d}\vee t^{1/\alpha}}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≍ ( 1 ∧ t | x − y | α ) ( x d ∨ t 1 / α | x − y | ) β 1 ( y d ∨ t 1 / α | x − y | ) β 1 \displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\bigg{(}\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}
× log β 3 ( e + | x − y | 4 ( x d ∨ t 1 / α ) ) log β 3 + β 4 + 1 ( e + | x − y | 4 ( y d ∨ t 1 / α ) ) . \displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{4(x_{d}\vee t^{1/\alpha})}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{|x-y|}{4(y_{d}\vee t^{1/\alpha})}\bigg{)}.
(6.22)
If y d ∨ t 1 / α ≥ | x − y | / 4 y_{d}\vee t^{1/\alpha}\geq|x-y|/4 , then
log ( e + | x − y | ( y d ∨ t 1 / α ) ∧ | x − y | ) ≍ 1 \log\big{(}e+\frac{|x-y|}{(y_{d}\vee t^{1/\alpha})\wedge|x-y|}\big{)}\asymp 1 .
Thus, by (6 ),
( | x − y | α ∧ t ) ∫ | x − y | / 4 | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 \displaystyle\big{(}|x-y|^{\alpha}\wedge t\big{)}\int_{|x-y|/4}^{|x-y|/2}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}
≍ ( 1 ∧ t | x − y | α ) A β 1 , β 1 , 0 , β 3 + β 4 + 1 ( t , x , y ) log β 3 ( e + | x − y | ( x d ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\asymp\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{(x_{d}\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
(6.23)
The proof is now complete.
□ \Box
Combining Proposition 6.2 , Lemma 6.4 and Lemma 6.8 .
we get the following
Proposition 6.9 .
Suppose that β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} . There exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) \displaystyle p(t,x,y)
≥ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) [ A β 1 , β 2 , β 3 , β 4 ( t , x , y ) \displaystyle\geq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)
+ ( 1 ∧ t | x − y | α ) A β 1 , β 1 , 0 , β 3 + β 4 + 1 ( t , x , y ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) ] . \displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}\bigg{]}.
7. Sharp upper bounds of heat kernels
In this section we prove the sharp upper bounds of heat kernels. The key results are Theorem 7.5 and its Corollary 7.9 which deals with the case β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} , and Theorem 7.10 which deals with the case β 2 ≥ α + β 1 \beta_{2}\geq\alpha+\beta_{1} .
Recall that U ( r ) = { x = ( x ~ , x d ) ∈ ℝ d : | x ~ | < r / 2 , 0 ≤ x d < r / 2 } U(r)=\{x=(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:\,|\widetilde{x}|<r/2,\,0\leq x_{d}<r/2\}
for r > 0 r>0 , d ≥ 2 d\geq 2 and U ( r ) = [ 0 , r / 2 ) U(r)=[0,r/2) for d = 1 d=1 , and that the function A β 1 , β 2 , β 3 , β 4 ( t , x , y ) A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y) is defined by (6 ).
We also recall that we use
p ( t , x , y ) p(t,x,y) for both p ¯ ( t , x , y ) \overline{p}(t,x,y) and p κ ( t , x , y ) p^{\kappa}(t,x,y) .
We remind the readers that,
for an open set D ⊂ ℝ ¯ + d D\subset{\overline{\mathbb{R}}}^{d}_{+} relative to the topology on ℝ ¯ + d {\overline{\mathbb{R}}}^{d}_{+} ,
τ D = τ ¯ D = inf { t > 0 : Y ¯ t ∉ D } \tau_{D}=\overline{\tau}_{D}=\inf\{t>0:\overline{Y}_{t}\notin D\} when we consider Y ¯ \overline{Y} , and
τ D = τ D κ = inf { t > 0 : Y t κ ∉ D ∩ ℝ + d } \tau_{D}=\tau^{\kappa}_{D}=\inf\{t>0:Y^{\kappa}_{t}\notin D\cap{\mathbb{R}}^{d}_{+}\} when we consider Y κ Y^{\kappa} .
Lemma 7.1 .
Let b 1 , b 3 ≥ 0 b_{1},b_{3}\geq 0 be constants with b 1 > 0 b_{1}>0 if b 3 > 0 b_{3}>0 .
Suppose that there exists a constant C 0 > 0 C_{0}>0 such that for all t > 0 t>0 and z , y ∈ ℝ + d z,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , z , y ) ≤ C 0 ( 1 ∧ z d t 1 / α ) q A b 1 , 0 , b 3 , 0 ( t , z , y ) ( t − d / α ∧ t | z − y | d + α ) . p(t,z,y)\leq C_{0}\left(1\wedge\frac{z_{d}}{t^{1/\alpha}}\right)^{q}A_{b_{1},0,b_{3},0}(t,z,y)\left(t^{-d/\alpha}\wedge\frac{t}{|z-y|^{d+\alpha}}\right).
(7.1)
Then there exists a constant
C = C ( C 0 ) > 0 C=C(C_{0})>0 such that for all t > 0 t>0 and x = ( 0 ~ , x d ) ∈ ℝ + d x=(\widetilde{0},x_{d})\in{\mathbb{R}}^{d}_{+} with x d ≤ 2 − 5 x_{d}\leq 2^{-5} ,
ℙ x ( τ U ( 1 ) < t < ζ ) ≤ C t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) b 1 log b 3 ( e + 1 x d ∨ t 1 / α ) . {\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\leq Ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.2)
Proof. We first note that (7.1 ) implies that for all t > 0 t>0 and y ∈ ℝ + d y\in{\mathbb{R}}^{d}_{+} ,
ℙ y ( | Y t − y | > 2 − 3 , t < ζ ) = ∫ z ∈ ℝ + d , | z − y | > 2 − 3 p ( t , y , z ) 𝑑 z \displaystyle{\mathbb{P}}_{y}(|Y_{t}-y|>2^{-3},\,t<\zeta)=\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}p(t,y,z)dz
≤ c 1 t ( 1 ∧ y d t 1 / α ) q ∫ z ∈ ℝ + d , | z − y | > 2 − 3 ( y d ∨ t 1 / α | z − y | ∧ 1 ) b 1 log b 3 ( e + | z − y | y d ∨ t 1 / α ) d z | z − y | d + α \displaystyle\leq c_{1}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}\bigg{(}\frac{y_{d}\vee t^{1/\alpha}}{|z-y|}\wedge 1\bigg{)}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{|z-y|}{y_{d}\vee t^{1/\alpha}}\bigg{)}\frac{dz}{|z-y|^{d+\alpha}}
≤ c 2 t ( 1 ∧ y d t 1 / α ) q ( y d ∨ t 1 / α ) b 1 log b 3 ( e + 2 − 3 y d ∨ t 1 / α ) ∫ z ∈ ℝ + d , | z − y | > 2 − 3 d z | z − y | d + α / 2 + b 1 \displaystyle\leq c_{2}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{2^{-3}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|>2^{-3}}\frac{dz}{|z-y|^{d+\alpha/2+b_{1}}}
≤ c 3 t ( 1 ∧ y d t 1 / α ) q ( y d ∨ t 1 / α ) b 1 log b 3 ( e + 1 y d ∨ t 1 / α ) , \displaystyle\leq c_{3}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)},
(7.3)
where in the first inequality above we used (7.1 ) and Lemma 10.2 ,
and in the second we used (10.2 ).
By Proposition 2.7 (see also Remark 3.12 ), we have
sup s ≤ t , y ∈ ℝ + d ℙ y ( | Y s − y | ≥ 2 − 2 , s < ζ ) ≤ c 1 sup s ≤ t , y ∈ ℝ + d ∫ z ∈ ℝ + d , | z − y | ≥ 2 − 2 s | z − y | d + α 𝑑 z ≤ c 2 t . \sup_{s\leq t,\,y\in{\mathbb{R}}^{d}_{+}}{\mathbb{P}}_{y}\big{(}|Y_{s}-y|\geq 2^{-2},\,s<\zeta\big{)}\leq c_{1}\sup_{s\leq t,\,y\in{\mathbb{R}}^{d}_{+}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|z-y|\geq 2^{-2}}\frac{s}{|z-y|^{d+\alpha}}dz\leq c_{2}t.
(7.4)
If t ≥ 1 / ( 2 c 2 ) t\geq 1/(2c_{2}) ,
then (7.2 ) follows from Corollary 5.2 .
Let t < 1 / ( 2 c 2 ) t<1/(2c_{2}) .
By the strong Markov property and (7.4 ), we have
ℙ x ( τ U ( 1 ) < t < ζ , | Y t − Y τ U ( 1 ) | ≥ 2 − 2 ) = 𝔼 x [ ℙ Y τ U ( 1 ) ( | Y t − τ U ( 1 ) − Y 0 | ≥ 2 − 2 ) : τ U ( 1 ) < t < ζ ] \displaystyle{\mathbb{P}}_{x}\big{(}\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|\geq 2^{-2}\big{)}={\mathbb{E}}_{x}\left[{\mathbb{P}}_{Y_{\tau_{U(1)}}}\left(|Y_{t-\tau_{U(1)}}-Y_{0}|\geq 2^{-2}\right):\tau_{U(1)}<t<\zeta\right]
≤ ℙ x ( τ U ( 1 ) < t < ζ ) sup s ≤ t , y ∈ ℝ + d ℙ y ( | Y s − y | ≥ 2 − 2 , s < ζ ) ≤ 2 − 1 ℙ x ( τ U ( 1 ) < t < ζ ) . \displaystyle\leq{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\sup_{s\leq t,\,y\in{\mathbb{R}}^{d}_{+}}{\mathbb{P}}_{y}(|Y_{s}-y|\geq 2^{-2},\,s<\zeta)\leq 2^{-1}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta).
Thus,
ℙ x ( τ U ( 1 ) < t < ζ ) \displaystyle{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)
= 2 ( ℙ x ( τ U ( 1 ) < t < ζ ) − 2 − 1 ℙ x ( τ U ( 1 ) < t < ζ ) ) \displaystyle=2\big{(}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)-2^{-1}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\big{)}
≤ 2 ( ℙ x ( τ U ( 1 ) < t < ζ ) − ℙ x ( τ U ( 1 ) < t < ζ , | Y t − Y τ U ( 1 ) | ≥ 2 − 2 ) ) \displaystyle\leq 2\big{(}{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)-{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|\geq 2^{-2})\big{)}
= 2 ℙ x ( τ U ( 1 ) < t < ζ , | Y t − Y τ U ( 1 ) | < 2 − 2 ) . \displaystyle=2{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|<2^{-2}).
(7.5)
Note that by the triangle inequality, for any y ∈ ℝ + d ∖ U ( 1 ) y\in{\mathbb{R}}^{d}_{+}\setminus U(1) and
z ∈ B ( y , 2 − 2 ) z\in B(y,2^{-2}) , we have | z − x | ≥ | y − x | − | y − z | > 15 / 32 − 1 / 4 > 2 − 3 |z-x|\geq|y-x|-|y-z|>15/32-1/4>2^{-3} .
Therefore using (7 ) and (7 ), we have
ℙ x ( τ U ( 1 ) < t < ζ ) ≤ 2 ℙ x ( τ U ( 1 ) < t < ζ , | Y t − Y τ U ( 1 ) | < 2 − 2 ) \displaystyle{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta)\leq 2{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta,\,|Y_{t}-Y_{\tau_{U(1)}}|<2^{-2})
≤ 2 ℙ x ( | Y t − x | > 2 − 3 , t < ζ ) ≤ c 4 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) b 1 log b 3 ( e + 1 x d ∨ t 1 / α ) . \displaystyle\leq 2{\mathbb{P}}_{x}(|Y_{t}-x|>2^{-3},\,t<\zeta)\leq c_{4}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
The proof is complete. □ \Box
Note that for any t , k , r > 0 t,k,r>0 and a ≥ 0 a\geq 0 ,
( 1 ∧ r t 1 / α ) a ( r ∨ t 1 / α ) k = r k ( 1 ∧ r t 1 / α ) a − k . \bigg{(}1\wedge\frac{r}{t^{1/\alpha}}\bigg{)}^{a}(r\vee t^{1/\alpha})^{k}=r^{k}\bigg{(}1\wedge\frac{r}{t^{1/\alpha}}\bigg{)}^{a-k}.
(7.6)
Lemma 7.2 .
There exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) ≤ C ( 1 ∧ x d ∧ y d t 1 / α ) q A β 1 , 0 , β 3 , 0 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) . p(t,x,y)\leq C\left(1\wedge\frac{x_{d}\wedge y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
Proof. By Proposition 5.1 , the lemma holds for β 1 = 0 \beta_{1}=0 .
We assume β 1 > 0 \beta_{1}>0 and set a n = β 1 ∧ n α 2 a_{n}=\beta_{1}\wedge\frac{n\alpha}{2} for n ≥ 0 n\geq 0 .
Below, we prove by induction that for any n ≥ 0 n\geq 0 , there exists a constant C > 0 C>0 such that for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
p ( t , x , y ) ≤ C ( 1 ∧ x d ∧ y d t 1 / α ) q A a n , 0 , β 3 , 0 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) . \displaystyle p(t,x,y)\leq C\left(1\wedge\frac{x_{d}\wedge y_{d}}{t^{1/\alpha}}\right)^{q}A_{a_{n},0,\beta_{3},0}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right).
(7.7)
The lemma is a direct consequence of (7.7 ).
By Proposition 5.1 and the fact that the logarithmic term in A β 1 , 0 , β 3 , 0 ( t , x , y ) A_{\beta_{1},0,\beta_{3},0}(t,x,y) is always larger than 1, (7.7 ) holds for n = 0 n=0 .
Suppose (7.7 ) holds for n − 1 n-1 . By symmetry and (3.30 ), we can assume without loss of generality that x d ≤ y d x_{d}\leq y_{d} , x ~ = 0 \widetilde{x}=0 and | x − y | = 5 |x-y|=5 . If t > 1 t>1 or x d > 2 − 5 x_{d}>2^{-5} , then (7.7 ) follows from Proposition 5.1 and (6.4 ).
Let t ≤ 1 t\leq 1 and x d ≤ 2 − 5 x_{d}\leq 2^{-5} . Then y d ≤ x d + | x − y | ≤ 4 + 2 − 5 y_{d}\leq x_{d}+|x-y|\leq 4+2^{-5} by the triangle inequality. Our goal is to show that there exists a constant c 1 > 0 c_{1}>0 independent of t , x , y t,x,y such that
p ( t , x , y ) ≤ c 1 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) a n log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) . p(t,x,y)\leq c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.8)
Set V 1 = U ( 1 ) V_{1}=U(1) , V 3 = B ( y , 2 ) ∩ ℝ ¯ + d V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V 2 = ℝ ¯ + d ∖ ( V 1 ∪ V 3 ) V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}) . Similarly to (5.10 ) and (5.11 ), we get from Proposition 2.7 and the triangle inequality that
sup s ≤ t , z ∈ V 2 p ( s , z , y ) ≤ c 2 sup s ≤ t , z ∈ ℝ + d , | z − y | ≥ 2 s | z − y | d + α ≤ 2 − d − α c 2 t \sup_{s\leq t,\,z\in V_{2}}p(s,z,y)\leq c_{2}\sup_{s\leq t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq 2}\frac{s}{|z-y|^{d+\alpha}}\leq 2^{-d-\alpha}c_{2}t
(7.9)
and
dist ( V 1 , V 3 ) ≥ sup u ∈ V 1 , w ∈ V 3 ( 4 − | x − u | − | y − w | ) ≥ 1 . {\rm dist}(V_{1},V_{3})\geq\sup_{u\in V_{1},\,w\in V_{3}}(4-|x-u|-|y-w|)\geq 1.
(7.10)
By the induction hypothesis,
condition (7.1 ) in Lemma 7.1 holds with b 1 = a n − 1 b_{1}=a_{n-1} and b 3 = β 3 b_{3}=\beta_{3} .
Thus, since a n − a n − 1 ≤ α / 2 a_{n}-a_{n-1}\leq\alpha/2 , we get from Lemma 7.1 and (7.9 ) that
ℙ x ( τ V 1 < t < ζ ) sup s ≤ t , z ∈ V 2 p ( s , z , y ) \displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
≤ c 3 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) a n − 1 ( t 1 / α ) α / 2 t 1 / 2 log β 3 ( e + 1 t 1 / α ) \displaystyle\leq c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{a_{n-1}}(t^{1/\alpha})^{\alpha/2}t^{1/2}\log^{\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}\,
≤ c 3 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) a n − 1 ( t 1 / α ) a n − a n − 1 ( sup s ≤ 1 s 1 / 2 log β 3 ( e + 1 s 1 / α ) ) \displaystyle\leq c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{a_{n-1}}(t^{1/\alpha})^{a_{n}-a_{n-1}}\,\left(\sup_{s\leq 1}s^{1/2}\log^{\beta_{3}}\bigg{(}e+\frac{1}{s^{1/\alpha}}\bigg{)}\right)
≤ c 4 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) a n . \displaystyle\leq c_{4}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{a_{n}}.
(7.11)
In order to apply Lemma 3.15 and get the desired result, it remains to bound ∫ 0 t ∫ V 3 ∫ V 1 p V 1 ( t − s , x , u ) ℬ ( u , w ) p ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s \int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal B}(u,w)p(s,y,w)dudwds . We consider the following two cases separately.
(Case 1) q ≥ α + a n q\geq\alpha+a_{n} and 10 x d < t 1 / α 10x_{d}<t^{1/\alpha} .
Pick ε ∈ ( 0 , β 1 ) {\varepsilon}\in(0,\beta_{1}) such that q < α + β 1 − ε q<\alpha+\beta_{1}-{\varepsilon} . Using (A3) (II),
Lemmas 10.1 (i)–(ii), 10.2 (see Remark 10.3 ), and
(7.10 ), we see that for all u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} ,
ℬ ( u , w ) ≤ c 5 B β 1 − ε , 0 , 0 , 0 ( u , w ) ≤ c 6 ( u d | u − w | ) β 1 − ε ≤ c 6 u d β 1 − ε . \displaystyle{\cal B}(u,w)\leq c_{5}B_{\beta_{1}-{\varepsilon},0,0,0}(u,w)\leq c_{6}\left(\frac{u_{d}}{|u-w|}\right)^{\beta_{1}-{\varepsilon}}\leq c_{6}u_{d}^{\beta_{1}-{\varepsilon}}.
(7.12)
By (7.12 ) and Lemma 5.12 , we have
∫ 0 t ∫ V 3 ∫ V 1 p V 1 ( t − s , x , u ) ℬ ( u , w ) p ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal B}(u,w)p(s,y,w)dudwds
≤ c 6 ∫ 0 t ∫ V 1 p V 1 ( t − s , x , u ) u d β 1 − ε 𝑑 u ∫ V 3 p ( s , y , w ) 𝑑 w 𝑑 s ≤ c 6 ∫ 0 ∞ ∫ V 1 p V 1 ( s , x , u ) u d β 1 − ε 𝑑 u 𝑑 s \displaystyle\leq c_{6}\int_{0}^{t}\int_{V_{1}}p^{V_{1}}(t-s,x,u)u_{d}^{\beta_{1}-{\varepsilon}}du\int_{V_{3}}p(s,y,w)dwds\leq c_{6}\int_{0}^{\infty}\int_{V_{1}}p^{V_{1}}(s,x,u)u_{d}^{\beta_{1}-{\varepsilon}}duds
≤ c 7 x d q = c 7 t ( x d t 1 / α ) q ( t 1 / α ) q − α ≤ c 8 t ( 1 ∧ x d t 1 / α ) q ( x d ∨ t 1 / α ) a n . \displaystyle\leq c_{7}x_{d}^{q}=c_{7}t\left(\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(t^{1/\alpha})^{q-\alpha}\leq c_{8}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{a_{n}}.
(7.13)
In the last inequality above, we used the facts that q − α ≥ a n q-\alpha\geq a_{n} , 10 x d < t 1 / α 10x_{d}<t^{1/\alpha} and x d ∨ t 1 / α ≤ 1 x_{d}\vee t^{1/\alpha}\leq 1 .
Now, using Lemma 3.15 , (7 ) and (7 ), we get (7.8 ) in this case.
(Case 2) q < α + a n q<\alpha+a_{n} or 10 x d ≥ t 1 / α 10x_{d}\geq t^{1/\alpha} .
By (A3) (II),
(7.10 ), Lemma 10.1 (ii) and Lemma 10.2
(see Remark 10.3 ),
it holds that for all u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} ,
ℬ ( u , w ) \displaystyle{\cal B}(u,w)
≤ c 9 B β 1 , 0 , β 3 , 0 ( u , w ) ≤ c 10 ( u d | u − w | ) β 1 log β 3 ( e + w d ∧ | u − w | u d ∧ | u − w | ) \displaystyle\leq c_{9}B_{\beta_{1},0,\beta_{3},0}(u,w)\leq c_{10}\left(\frac{u_{d}}{|u-w|}\right)^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}\wedge|u-w|}{u_{d}\wedge|u-w|}\bigg{)}
≤ c 10 u d β 1 log β 3 ( e + w d u d ) . \displaystyle\leq c_{10}u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}.
(7.14)
By (10.2 ) (or using that u ↦ u β 1 log β 3 ( e + t / u ) u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+t/u) is almost increasing)
and Corollary 5.2 , since β 1 > 0 \beta_{1}>0 and a n ≤ β 1 a_{n}\leq\beta_{1} , we get that for any 0 < s < t 0<s<t and w ∈ V 3 w\in V_{3} ,
∫ u ∈ V 1 : u d < x d p ( s , x , u ) u d β 1 log β 3 ( e + w d u d ) 𝑑 u ≤ ∫ u ∈ V 1 : u d < x d ∨ s 1 / α p ( s , x , u ) u d β 1 log β 3 ( e + w d u d ) 𝑑 u \displaystyle\int_{u\in V_{1}:u_{d}<x_{d}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du\leq\int_{u\in V_{1}:u_{d}<x_{d}\vee s^{1/\alpha}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du
≤ c 11 ( x d ∨ s 1 / α ) β 1 log β 3 ( e + w d x d ∨ s 1 / α ) ∫ u ∈ V 1 : u d < x d ∨ s 1 / α p ( s , x , u ) 𝑑 u \displaystyle\leq c_{11}(x_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee s^{1/\alpha}}\bigg{)}\int_{u\in V_{1}:u_{d}<x_{d}\vee s^{1/\alpha}}p(s,x,u)du
≤ c 11 ℙ x ( ζ > s ) ( x d ∨ s 1 / α ) a n log β 3 ( e + w d x d ∨ s 1 / α ) \displaystyle\leq c_{11}{\mathbb{P}}_{x}(\zeta>s)(x_{d}\vee s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee s^{1/\alpha}}\bigg{)}
≤ c 12 ( 1 ∧ x d s 1 / α ) q ( x d ∨ s 1 / α ) a n log β 3 ( e + w d x d ∨ s 1 / α ) . \displaystyle\leq c_{12}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee s^{1/\alpha}}\bigg{)}.
(7.15)
Next, using the induction hypothesis and Lemma 10.10 ,
since a n ≤ β 1 a_{n}\leq\beta_{1} and a n < α + a n − 1 a_{n}<\alpha+a_{n-1} , we get that for any 0 < s < t 0<s<t and w ∈ V 3 w\in V_{3} ,
∫ u ∈ V 1 : u d ≥ x d p ( s , x , u ) u d β 1 log β 3 ( e + w d u d ) 𝑑 u \displaystyle\int_{u\in V_{1}:u_{d}\geq x_{d}}p(s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du
≤ c 13 ( 1 ∧ x d s 1 / α ) q ∫ u ∈ V 1 : u d ≥ x d ( x d ∨ s 1 / α | x − u | ∧ 1 ) a n − 1 log β 3 ( e + | x − u | ( x d ∨ s 1 / α ) ∧ | x − u | ) \displaystyle\leq c_{13}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}\int_{u\in V_{1}:u_{d}\geq x_{d}}\bigg{(}\frac{x_{d}\vee s^{1/\alpha}}{|x-u|}\wedge 1\bigg{)}^{a_{n-1}}\log^{\beta_{3}}\bigg{(}e+\frac{|x-u|}{(x_{d}\vee s^{1/\alpha})\wedge|x-u|}\bigg{)}
× ( s − d / α ∧ s | x − u | d + α ) u d a n log β 3 ( e + w d u d ) d u \displaystyle\hskip 156.49014pt\times\left(s^{-d/\alpha}\wedge\frac{s}{|x-u|^{d+\alpha}}\right)u_{d}^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du
≤ c 14 ( 1 ∧ x d s 1 / α ) q ( x d ∨ s 1 / α ) a n log β 3 ( e + w d x d ∨ s 1 / α ) . \displaystyle\leq c_{14}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee s^{1/\alpha})^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee s^{1/\alpha}}\bigg{)}.
(7.16)
Similarly, again splitting
the integration into two parts w d < y d w_{d}<y_{d} and w d ≥ y d w_{d}\geq y_{d} , and using the induction hypothesis and
Lemma 10.10 again, we also get that for any 0 < s < t 0<s<t ,
∫ V 3 p ( s , y , w ) log β 3 ( e + w d x d ∨ ( t − s ) 1 / α ) 𝑑 w \displaystyle\int_{V_{3}}p(s,y,w)\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}dw
≤ c 15 ( 1 ∧ y d s 1 / α ) q log β 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) \displaystyle\leq c_{15}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}
≤ c 15 log β 3 ( e + y d ∨ t 1 / α x d ∨ ( t − s ) 1 / α ) . \displaystyle\leq c_{15}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}.
(7.17)
By (7 )–(7 ) and (7.6 ), we have
∫ 0 t ∫ V 3 ∫ V 1 p V 1 ( t − s , x , u ) ℬ ( u , w ) p ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal B}(u,w)p(s,y,w)dudwds
≤ c 10 ∫ 0 t ∫ V 3 p ( s , y , w ) ∫ V 1 p ( t − s , x , u ) u d β 1 log β 3 ( e + w d u d ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\leq c_{10}\int_{0}^{t}\int_{V_{3}}p(s,y,w)\int_{V_{1}}p(t-s,x,u)u_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}dudwds
≤ c 16 ∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q ( x d ∨ ( t − s ) 1 / α ) a n ∫ V 3 p ( s , y , w ) log β 3 ( e + w d x d ∨ ( t − s ) 1 / α ) 𝑑 w 𝑑 s \displaystyle\leq c_{16}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q}(x_{d}\vee(t-s)^{1/\alpha})^{a_{n}}\int_{V_{3}}p(s,y,w)\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}dwds
≤ c 17 x d a n ∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q − a n log β 3 ( e + y d ∨ t 1 / α x d ∨ ( t − s ) 1 / α ) d s = : I . \displaystyle\leq c_{17}x_{d}^{a_{n}}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds=:I.
(7.18)
When q < α + a n q<\alpha+a_{n} , we get from Lemma 10.4 that
I ≤ c 18 t x d a n ( 1 ∧ x d t 1 / α ) q − a n log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) . \displaystyle I\leq c_{18}tx_{d}^{a_{n}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.19)
When 10 x d ≥ t 1 / α 10x_{d}\geq t^{1/\alpha} , we also get from Lemma 10.4 that
I \displaystyle I
≤ c 17 x d a n ∫ 0 t log β 3 ( e + y d ∨ t 1 / α x d ∨ ( t − s ) 1 / α ) 𝑑 s ≤ c 19 t x d a n log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) \displaystyle\leq c_{17}x_{d}^{a_{n}}\int_{0}^{t}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds\leq c_{19}tx_{d}^{a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}
≤ 10 q − a n c 19 t x d a n ( 1 ∧ x d t 1 / α ) q − a n log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) . \displaystyle\leq 10^{q-a_{n}}c_{19}tx_{d}^{a_{n}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-a_{n}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.20)
Now, (7.8 ) follows from (7.6 ), (7 ), (7 ), (7 ) and Lemma 3.15 . The proof is complete.
□ \Box
Now we use Lemma 7.2 to
improve the bound in (7 ).
Lemma 7.3 .
Let 0 < t ≤ 1 0<t\leq 1 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} be such that x ~ = 0 ~ \widetilde{x}=\widetilde{0} , x d ≤ 2 − 5 x_{d}\leq 2^{-5}
and | x − y | = 5 |x-y|=5 . Set V 1 = U ( 1 ) V_{1}=U(1) , V 3 = B ( y , 2 ) ∩ ℝ ¯ + d V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V 2 = ℝ ¯ + d ∖ ( V 1 ∪ V 3 ) V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}) . There exists a constant C > 0 C>0 independent of t , x t,x and y y such that
ℙ x ( τ V 1 < t < ζ ) sup s ≤ t , z ∈ V 2 p ( s , z , y ) \displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
≤ C t 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 1 \displaystyle\leq Ct^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}
× log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
Proof. Note that, since y d ≤ x d + | x − y | ≤ 5 + 2 − 5 < 6 y_{d}\leq x_{d}+|x-y|\leq 5+2^{-5}<6 , for any 0 < s ≤ 1 0<s\leq 1 and z ∈ ℝ + d z\in{\mathbb{R}}^{d}_{+} with | z − y | ≥ 2 |z-y|\geq 2 ,
( y d ∨ s 1 / α ) ∧ | z − y | ≥ 1 3 ( y d ∨ s 1 / α ) . \displaystyle(y_{d}\vee s^{1/\alpha})\wedge|z-y|\geq\frac{1}{3}(y_{d}\vee s^{1/\alpha}).
(7.21)
By Lemmas 7.2 and 10.2 and applying (7.21 ),
sup s ≤ t , z ∈ V 2 p ( s , z , y ) ≤ c 1 sup s ≤ t , z ∈ ℝ + d , | z − y | ≥ 2 ( 1 ∧ z d ∧ y d s 1 / α ) q A β 1 , 0 , β 3 , 0 ( s , z , y ) s | z − y | d + α \displaystyle\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)\leq c_{1}\sup_{s\leq t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq 2}\left(1\wedge\frac{z_{d}\wedge y_{d}}{s^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(s,z,y)\frac{s}{|z-y|^{d+\alpha}}
≤ c 1 sup s ≤ t , z ∈ ℝ + d , | z − y | ≥ 2 ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α | z − y | ) β 1 log β 3 ( e + 3 | z − y | y d ∨ s 1 / α ) s | z − y | d + α . \displaystyle\leq c_{1}\sup_{s\leq t,\,z\in{\mathbb{R}}^{d}_{+},|z-y|\geq 2}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}\frac{y_{d}\vee s^{1/\alpha}}{|z-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{3|z-y|}{y_{d}\vee s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}.
(7.22)
Using that u ↦ u β 1 log β 3 ( e + 1 / u ) u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+1/u) is almost increasing, we have that for any 0 < s ≤ t ≤ 1 0<s\leq t\leq 1 and z ∈ ℝ + d z\in{\mathbb{R}}^{d}_{+} with | z − y | ≥ 2 |z-y|\geq 2 ,
( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α | z − y | ) β 1 log β 3 ( e + 3 | z − y | y d ∨ s 1 / α ) s | z − y | d + α \displaystyle\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}\frac{y_{d}\vee s^{1/\alpha}}{|z-y|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{3|z-y|}{y_{d}\vee s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}
≤ c 2 ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α ) β 1 log β 3 ( e + 1 y d ∨ s 1 / α ) s | z − y | d + α \displaystyle\leq c_{2}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}\frac{s}{|z-y|^{d+\alpha}}
≤ 2 − d − α c 2 s ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α ) β 1 log β 3 ( e + 1 y d ∨ s 1 / α ) . \displaystyle\leq 2^{-d-\alpha}c_{2}s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}.
(7.23)
Let ε > 0 {\varepsilon}>0 be such that q < α + β 1 − ε q<\alpha+\beta_{1}-{\varepsilon} .
Using (7.6 ), we see that
s ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α ) β 1 log β 3 ( e + 1 y d ∨ s 1 / α ) \displaystyle s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}
= s y d β 1 − ε ( 1 ∧ y d s 1 / α ) q − β 1 + ε ( y d ∨ s 1 / α ) ε log β 3 ( e + 1 y d ∨ s 1 / α ) \displaystyle=sy_{d}^{\beta_{1}-{\varepsilon}}\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q-\beta_{1}+{\varepsilon}}(y_{d}\vee s^{1/\alpha})^{{\varepsilon}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}
= s ( α + β 1 − q − ε ) / α y d β 1 − ε ( y d ∧ s 1 / α ) q − β 1 + ε ( y d ∨ s 1 / α ) ε log β 3 ( e + 1 y d ∨ s 1 / α ) . \displaystyle=s^{(\alpha+\beta_{1}-q-{\varepsilon})/\alpha}y_{d}^{\beta_{1}-{\varepsilon}}(y_{d}\wedge s^{1/\alpha})^{q-\beta_{1}+{\varepsilon}}(y_{d}\vee s^{1/\alpha})^{{\varepsilon}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}.
Thus, the map s ↦ s ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α ) β 1 log β 3 ( e + 1 y d ∨ s 1 / α ) s\mapsto s\big{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\big{)}^{q}(y_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\big{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\big{)} is almost increasing on ( 0 , t ] (0,t]
by (10.2 ). Using this and (7 )–(7 ), we get that
sup s ≤ t , z ∈ V 2 p ( s , z , y ) \displaystyle\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
≤ c 3 sup s ≤ t s ( 1 ∧ y d s 1 / α ) q ( y d ∨ s 1 / α ) β 1 log β 3 ( e + 1 y d ∨ s 1 / α ) \displaystyle\leq c_{3}\sup_{s\leq t}s\left(1\wedge\frac{y_{d}}{s^{1/\alpha}}\right)^{q}(y_{d}\vee s^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}
≤ c 4 t ( 1 ∧ y d t 1 / α ) q ( y d ∨ t 1 / α ) β 1 log β 3 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{4}t\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.24)
Note that (7.1 ) is satisfied with a 1 = β 1 a_{1}=\beta_{1} and
a 3 = β 3 a_{3}=\beta_{3} by Lemma 7.2 .
Thus, by Lemma 7.1 and (7 ) we obtain
ℙ x ( τ V 1 < t < ζ ) sup s ≤ t , z ∈ V 2 p ( s , z , y ) \displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
≤ c 4 t 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 1 \displaystyle\leq c_{4}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}
× log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
□ \Box
Lemma 7.4 .
Let η 1 , η 2 , γ ≥ 0 \eta_{1},\eta_{2},\gamma\geq 0 .
There exists a constant C > 0 C>0 such that for any x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and any s , k , l > 0 s,k,l>0 ,
∫ B + ( x , 2 ) p ( s , x , z ) z d γ log η 1 ( e + k z d ) log η 2 ( e + z d l ) 𝑑 z \displaystyle\int_{B_{+}(x,2)}p(s,x,z)z_{d}^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{z_{d}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{z_{d}}{l}\bigg{)}dz
≤ C x d γ ( 1 ∧ x d s 1 / α ) q − γ log η 1 ( e + k x d ∨ s 1 / α ) log η 2 ( e + x d ∨ s 1 / α l ) \displaystyle\leq Cx_{d}^{\gamma}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{x_{d}\vee s^{1/\alpha}}{l}\bigg{)}
+ C 𝟏 { γ > α + β 1 } s x d β 1 ( 1 ∧ x d s 1 / α ) q − β 1 log β 3 ( e + 2 x d ∨ s 1 / α ) log η 1 ( e + k ) log η 2 ( e + 1 l ) \displaystyle\quad+C{\bf 1}_{\{\gamma>\alpha+\beta_{1}\}}sx_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{2}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}(e+k)\log^{\eta_{2}}\bigg{(}e+\frac{1}{l}\bigg{)}
+ C 𝟏 { γ = α + β 1 , x d ∨ s 1 / α < 2 } s x d β 1 ( 1 ∧ x d s 1 / α ) q − β 1 \displaystyle\quad+C{\bf 1}_{\{\gamma=\alpha+\beta_{1},\,x_{d}\vee s^{1/\alpha}<2\}}sx_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}
× ∫ x d ∨ s 1 / α 2 log β 3 ( e + r x d ∨ s 1 / α ) log η 1 ( e + k r ) log η 2 ( e + r l ) d r r . \displaystyle\qquad\times\int_{x_{d}\vee s^{1/\alpha}}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}.
Proof. For any x , z ∈ ℝ + d x,z\in{\mathbb{R}}^{d}_{+} and s > 0 s>0 , by Lemmas 7.2 and 10.2 ,
p ( s , x , z ) ≤ c 1 ( 1 ∧ x d s 1 / α ) q A β 1 , 0 , β 3 , 0 ( s , x , z ) ( s − d / α ∧ s | x − z | d + α ) \displaystyle p(s,x,z)\leq c_{1}\left(1\wedge\frac{x_{d}}{s^{1/\alpha}}\right)^{q}A_{\beta_{1},0,\beta_{3},0}(s,x,z)\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right)
≤ c 2 ( 1 ∧ x d s 1 / α ) q ( 1 ∧ x d ∨ s 1 / α | x − z | ) β 1 log β 3 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) ( s − d / α ∧ s | x − z | d + α ) . \displaystyle\leq c_{2}\left(1\wedge\frac{x_{d}}{s^{1/\alpha}}\right)^{q}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{\beta_{1}}\log^{\beta_{3}}\left(e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\right)\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right).
Now combining (7.6 ) and Lemma 10.10 , we get the desired result.
□ \Box
We now state the first main result of this section.
Theorem 7.5 .
For any ε ∈ ( 0 , α / 2 ] {\varepsilon}\in(0,\alpha/2] , there exists a constant C > 0 C>0 such that
p ( t , x , y ) \displaystyle p(t,x,y)
≤ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q A β 1 , β 2 ∧ ( α + β 1 − ε ) , β 3 , β 4 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) , \displaystyle\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},\beta_{2}\wedge(\alpha+\beta_{1}-{\varepsilon}),\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),
for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} .
This theorem will proved by using several lemmas.
We first introduce some additional notation:
For b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 , t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} , define
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})
:= ∫ 0 t / 2 ∫ B + ( y , 2 ) ∫ B + ( x , 2 ) 𝟏 { u d ≤ w d } p ( t − s , x , u ) p ( s , y , w ) \displaystyle:=\int_{0}^{t/2}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}{\bf 1}_{\{u_{d}\leq w_{d}\}}p(t-s,x,u)p(s,y,w)
× u d b 1 w d b 2 log b 3 ( e + w d u d ) log b 4 ( e + 1 w d ) d u d w d s , \displaystyle\hskip 56.9055pt\times u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw\,ds,
(7.25)
ℐ t , 2 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
:= ∫ 0 t / 2 ∫ B + ( y , 2 ) ∫ B + ( x , 2 ) 𝟏 { w d ≤ u d } p ( t − s , x , u ) p ( s , y , w ) \displaystyle:=\int_{0}^{t/2}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}{\bf 1}_{\{w_{d}\leq u_{d}\}}p(t-s,x,u)p(s,y,w)
× u d b 2 w d b 1 log b 3 ( e + u d w d ) log b 4 ( e + 1 u d ) d u d w d s . \displaystyle\hskip 56.9055pt\times u_{d}^{b_{2}}w_{d}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{u_{d}}{w_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{u_{d}}\bigg{)}\,du\,dw\,ds.
(7.26)
Lemma 7.6 .
Let b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 .
Let b 1 ′ ∈ [ 0 , b 1 ] b_{1}^{\prime}\in[0,b_{1}] and b 2 ′ := b 1 + b 2 − b 1 ′ b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime} . Then
ℐ t , i ( x , y ; b 1 , b 2 , b 3 , b 4 ) ≤ ℐ t , i ( x , y ; b 1 ′ , b 2 ′ , b 3 , b 4 ) , i = 1 , 2 . {\cal I}_{t,i}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal I}_{t,i}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4}),\quad i=1,2.
Proof. If u d ≤ w d u_{d}\leq w_{d} , then u d b 1 w d b 2 ≤ u d b 1 ′ w d b 2 ′ u_{d}^{b_{1}}w_{d}^{b_{2}}\leq u_{d}^{b_{1}^{\prime}}w_{d}^{b_{2}^{\prime}} , implying that
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) ≤ ℐ t , 1 ( x , y ; b 1 ′ , b 2 ′ , b 3 , b 4 ) . {\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal I}_{t,1}(x,y;b_{1}^{\prime},b_{2}^{\prime},\linebreak b_{3},b_{4}).
The other inequality is analogous. □ \Box
Lemma 7.7 .
Let b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 with b 1 ∨ b 2 < α + β 1 b_{1}\vee b_{2}<\alpha+\beta_{1} , x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} with | x − y | = 5 |x-y|=5 , and t ∈ ( 0 , 1 ] t\in(0,1] .
(i) If b 2 > q − α b_{2}>q-\alpha or y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} , then there exists a constant C > 0 C>0 independent of t , x , y t,x,y such that
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ C t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(ii) If b 1 > q − α b_{1}>q-\alpha or y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} , then there exists a constant C > 0 C>0 independent of t , x , y t,x,y such that
ℐ t , 2 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ C t x d b 2 y d b 1 ( 1 ∧ x d t 1 / α ) q − b 2 ( 1 ∧ y d t 1 / α ) q − b 1 log b 3 ( e + x d ∨ t 1 / α y d ∨ t 1 / α ) log b 4 ( e + 1 x d ∨ t 1 / α ) . \displaystyle\leq Ctx_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
Proof. We give the proof for (i). (ii) can be proved similarly.
By using Lemma 7.4 together with the fact that
t − s ≍ t t-s\asymp t if 0 ≤ s ≤ t / 2 0\leq s\leq t/2 , we see that for 0 ≤ s ≤ t / 2 0\leq s\leq t/2 ,
∫ B + ( x , 2 ) p ( t − s , x , u ) u d b 1 log b 3 ( e + w d u d ) 𝑑 u ≤ c 1 x d b 1 ( 1 ∧ x d t 1 / α ) q − b 1 log b 3 ( e + w d x d ∨ t 1 / α ) . \displaystyle\int_{B_{+}(x,2)}p(t-s,x,u)u_{d}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}du\leq c_{1}x_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
Thus, using Lemma 7.4 again, we get that for 0 ≤ s ≤ t / 2 0\leq s\leq t/2 ,
∫ B + ( y , 2 ) ∫ B + ( x , 2 ) p ( t − s , x , u ) p ( s , y , w ) u d b 1 w d b 2 log b 3 ( e + w d u d ) log b 4 ( e + 1 w d ) 𝑑 u 𝑑 w \displaystyle\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}p(t-s,x,u)p(s,y,w)u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw
≤ c 1 x d b 1 ( 1 ∧ x d t 1 / α ) q − b 1 ∫ B + ( y , 2 ) p ( s , y , w ) w d b 2 log b 3 ( e + w d x d ∨ t 1 / α ) log b 4 ( e + 1 w d ) 𝑑 w \displaystyle\leq c_{1}x_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{B_{+}(y,2)}p(s,y,w)w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}dw
≤ c 2 x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) \displaystyle\leq c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}
From this and Lemma 10.5 , we get that
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ c 2 x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ∫ 0 t 2 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\leq c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
≤ c 2 x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ∫ 0 t ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\leq c_{2}x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\int_{0}^{t}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
≤ c 3 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{3}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
Lemma 7.8 .
Let b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 be such that b 1 > q − α b_{1}>q-\alpha and b 1 ∨ b 2 < α + β 1 b_{1}\vee b_{2}<\alpha+\beta_{1} . Assume that b 2 > 0 b_{2}>0 if b 4 > 0 b_{4}>0 . Then, there exists a constant C > 0 C>0 such that for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} with x d ≤ y d x_{d}\leq y_{d} and | x − y | = 5 |x-y|=5 , and all t ∈ ( 0 , 1 ] t\in(0,1] ,
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) + ℐ t , 2 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})+{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ C t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
(7.27)
and
ℐ t , 1 ( y , x ; b 1 , b 2 , b 3 , b 4 ) + ℐ t , 2 ( y , x ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(y,x;b_{1},b_{2},b_{3},b_{4})+{\cal I}_{t,2}(y,x;b_{1},b_{2},b_{3},b_{4})
≤ C t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.28)
Proof. Let δ ∈ ( 0 , 1 ) \delta\in(0,1) be such that
b 1 ∨ b 2 + ( 1 − δ ) ( b 1 ∧ b 2 ) < α + β 1 , \displaystyle b_{1}\vee b_{2}+(1-\delta)(b_{1}\wedge b_{2})<\alpha+\beta_{1},
(7.29)
and let b 1 ′ := δ ( b 1 ∧ b 2 ) b_{1}^{\prime}:=\delta(b_{1}\wedge b_{2}) and b 2 ′ := b 1 + b 2 − b 1 ′ b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime} . Then we see that b 1 ′ ∈ [ 0 , b 1 ∧ b 2 ] b_{1}^{\prime}\in[0,b_{1}\wedge b_{2}] , b 2 ′ ≥ b 1 > q − α b_{2}^{\prime}\geq b_{1}>q-\alpha and
b 1 ′ ≤ b 2 ′ = b 1 ∨ b 2 + ( 1 − δ ) ( b 1 ∧ b 2 ) < α + β 1 \displaystyle b_{1}^{\prime}\leq b_{2}^{\prime}=b_{1}\vee b_{2}+(1-\delta)(b_{1}\wedge b_{2})<\alpha+\beta_{1}
(7.30)
by (7.29 ). Moreover, since x d ≤ y d x_{d}\leq y_{d} and b 2 ′ − b 1 = b 2 − b 1 ′ b_{2}^{\prime}-b_{1}=b_{2}-b_{1}^{\prime} , we see that
( x d ∨ t 1 / α ) b 2 ′ − b 1 ( y d ∨ t 1 / α ) b 1 ′ − b 2 log b 4 ( e + 1 x d ∨ t 1 / α ) log − b 4 ( e + 1 y d ∨ t 1 / α ) ≤ c 1 . \displaystyle(x_{d}\vee t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}\leq c_{1}.
(7.31)
Indeed, when b 4 = 0 b_{4}=0 , (7.31 ) clearly holds with c 1 = 1 c_{1}=1 . If b 4 > 0 b_{4}>0 , then b 2 > 0 b_{2}>0 so that b 2 > δ b 2 ≥ b 1 ′ b_{2}>\delta b_{2}\geq b_{1}^{\prime} . Hence, we get (7.31 ) from (10.2 ).
(i) We first prove (7.8 ). For this, we distinguish between two cases: y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} and y d < t 1 / α y_{d}<t^{1/\alpha} .
Assume first that y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} . The desired bound for ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) {\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4}) follows from Lemma 7.7 (i). On the other hand, by using Lemma
7.6
in the first inequality, Lemma 7.7 (ii) in the second inequality (which uses (7.30 ) and y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} ), (7.6 ) in the equality, and (7.31 ) in the last inequality, we get that
ℐ t , 2 ( x , y ; b 1 , b 2 , b 3 , b 4 ) ≤ ℐ t , 2 ( x , y ; b 1 ′ , b 2 ′ , b 3 , b 4 ) \displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal I}_{t,2}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})
≤ c 2 t x d b 2 ′ y d b 1 ′ ( 1 ∧ x d t 1 / α ) q − b 2 ′ ( 1 ∧ y d t 1 / α ) q − b 1 ′ log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 x d ∨ t 1 / α ) \displaystyle\leq c_{2}tx_{d}^{b_{2}^{\prime}}y_{d}^{b_{1}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}
= c 2 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle=c_{2}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
× ( x d ∨ t 1 / α ) b 2 ′ − b 1 ( y d ∨ t 1 / α ) b 1 ′ − b 2 log b 4 ( e + 1 x d ∨ t 1 / α ) log − b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\quad\;\times(x_{d}\vee t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
≤ c 1 c 2 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{1}c_{2}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
Assume now that y d < t 1 / α y_{d}<t^{1/\alpha} . Using Lemma
7.6 and Lemma 7.7 (i) (which uses b 2 ′ > q − α b_{2}^{\prime}>q-\alpha ), we get
ℐ t , 1 ( x , y ; b 1 , b 2 , b 3 , b 4 ) ≤ ℐ t , 1 ( x , y ; b 1 ′ , b 2 ′ , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})\leq{\cal I}_{t,1}(x,y;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})
≤ c 3 t x d b 1 ′ y d b 2 ′ ( 1 ∧ x d t 1 / α ) q − b 1 ′ ( 1 ∧ y d t 1 / α ) q − b 2 ′ log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{3}tx_{d}^{b_{1}^{\prime}}y_{d}^{b_{2}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.32)
Also, since b 1 > q − α b_{1}>q-\alpha , we get from Lemma 7.7 (ii) that
ℐ t , 2 ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ c 4 t x d b 2 y d b 1 ( 1 ∧ x d t 1 / α ) q − b 2 ( 1 ∧ y d t 1 / α ) q − b 1 log b 3 ( e + x d ∨ t 1 / α y d ∨ t 1 / α ) log b 4 ( e + 1 x d ∨ t 1 / α ) . \displaystyle\leq c_{4}tx_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\!\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.33)
Since x d ≤ y d < t 1 / α x_{d}\leq y_{d}<t^{1/\alpha} and b 1 ′ + b 2 ′ = b 1 + b 2 b_{1}^{\prime}+b_{2}^{\prime}=b_{1}+b_{2} , it holds that x d ∨ t 1 / α = y d ∨ t 1 / α = t 1 / α x_{d}\vee t^{1/\alpha}=y_{d}\vee t^{1/\alpha}=t^{1/\alpha} and
x d b 1 ′ y d b 2 ′ ( 1 ∧ x d t 1 / α ) q − b 1 ′ ( 1 ∧ y d t 1 / α ) q − b 2 ′ = x d b 2 y d b 1 ( 1 ∧ x d t 1 / α ) q − b 2 ( 1 ∧ y d t 1 / α ) q − b 1 \displaystyle x_{d}^{b_{1}^{\prime}}y_{d}^{b_{2}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}=x_{d}^{b_{2}}y_{d}^{b_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}
= x d q y d q t ( 2 q − b 1 − b 2 ) / α = x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 . \displaystyle=\frac{x_{d}^{q}y_{d}^{q}}{t^{(2q-b_{1}-b_{2})/\alpha}}=x_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}.
Thus, (7.8 ) follows from (7 ) and (7 ).
(ii) Now, we prove (7.8 ).
By using Lemma
7.6 in the first inequality, Lemma 7.7 (i) in the second inequality (which uses (7.30 ) and b 2 ′ > q − α b_{2}^{\prime}>q-\alpha ), (7.6 ) together with the fact that x d ≤ y d x_{d}\leq y_{d} in the third inequality, (7.31 ) in the last inequality, we obtain
ℐ t , 1 ( y , x ; b 1 , b 2 , b 3 , b 4 ) ≤ ℐ t , 1 ( y , x ; b 1 ′ , b 2 ′ , b 3 , b 4 ) \displaystyle{\cal I}_{t,1}(y,x;b_{1},b_{2},b_{3},b_{4})\leq{\cal I}_{t,1}(y,x;b_{1}^{\prime},b_{2}^{\prime},b_{3},b_{4})
≤ c 5 t x d b 2 ′ y d b 1 ′ ( 1 ∧ x d t 1 / α ) q − b 2 ′ ( 1 ∧ y d t 1 / α ) q − b 1 ′ log b 3 ( e + x d ∨ t 1 / α y d ∨ t 1 / α ) log b 4 ( e + 1 x d ∨ t 1 / α ) \displaystyle\leq c_{5}tx_{d}^{b_{2}^{\prime}}y_{d}^{b_{1}^{\prime}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}^{\prime}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}^{\prime}}\log^{b_{3}}\bigg{(}e+\frac{x_{d}\vee t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}
≤ c 5 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq c_{5}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
× ( x d ∨ t 1 / α ) b 2 ′ − b 1 ( y d ∨ t 1 / α ) b 1 ′ − b 2 log b 4 ( e + 1 x d ∨ t 1 / α ) log − b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\quad\times(x_{d}\vee t^{1/\alpha})^{b_{2}^{\prime}-b_{1}}(y_{d}\vee t^{1/\alpha})^{b_{1}^{\prime}-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{-b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
≤ c 1 c 5 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{1}c_{5}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
On the other hand, by Lemma 7.7 (ii), it holds that
ℐ t , 2 ( y , x ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t,2}(y,x;b_{1},b_{2},b_{3},b_{4})
≤ c 3 t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{3}tx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
The proof is complete. □ \Box
Proof of Theorem 7.5 .
Set β ^ 2 := β 2 ∧ ( α + β 1 − ε ) \widehat{\beta}_{2}:=\beta_{2}\wedge(\alpha+\beta_{1}-{\varepsilon}) . As in the proof of Lemma 7.2 , by symmetry, Proposition 5.1 , (3.30 ),
and (6.4 ),
we can assume without loss of generality that x d ≤ y d ∧ 2 − 5 x_{d}\leq y_{d}\wedge 2^{-5} , x ~ = 0 \widetilde{x}=0 and | x − y | = 5 |x-y|=5 , and then it is enough to show that there exists a constant c 1 > 0 c_{1}>0 independent of x x and y y such that for any t ≤ 1 t\leq 1 ,
p ( t , x , y ) \displaystyle p(t,x,y)
≤ c 1 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β ^ 2 \displaystyle\leq c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\widehat{\beta}_{2}}
× log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.34)
Let t ≤ 1 t\leq 1 . Set V 1 = U ( 1 ) V_{1}=U(1) , V 3 = B ( y , 2 ) ∩ ℝ ¯ + d V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V 2 = ℝ ¯ + d ∖ ( V 1 ∪ V 3 ) V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}) .
By Lemma 7.3 ,
ℙ x ( τ V 1 < t < ζ ) sup s ≤ t , z ∈ V 2 p ( s , z , y ) \displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
≤ c 2 t 2 ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 1 log 2 β 3 ( e + 1 t 1 / α ) \displaystyle\leq c_{2}t^{2}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}
= c 2 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) α + β 1 − ε ( y d ∨ t 1 / α ) − α + ε t log 2 β 3 ( e + 1 t 1 / α ) \displaystyle=c_{2}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\alpha+\beta_{1}-{\varepsilon}}(y_{d}\vee t^{1/\alpha})^{-\alpha+{\varepsilon}}t\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}
≤ c 2 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β ^ 2 t ε / α log 2 β 3 ( e + 1 t 1 / α ) \displaystyle\leq c_{2}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\widehat{\beta}_{2}}t^{{\varepsilon}/\alpha}\log^{2\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}
≤ c 3 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β ^ 2 , \displaystyle\leq c_{3}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\widehat{\beta}_{2}},
(7.35)
where in the last inequality above we used (10.1 ).
Next, we show that there exists a constant C ′ > 0 C^{\prime}>0 such that
I \displaystyle I
:= ∫ 0 t ∫ V 3 ∫ V 1 p V 1 ( t − s , x , u ) ℬ ( u , w ) p ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s \displaystyle:=\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal B}(u,w)p(s,y,w)du\,dw\,ds
≤ C ′ t x d β 1 y d β ^ 2 ( 1 ∧ x d t 1 / α ) q − β 1 ( 1 ∧ y d t 1 / α ) q − β ^ 2 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq C^{\prime}tx_{d}^{\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q-\beta_{1}}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q-\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.36)
Once we get (7 ), by (7.6 ) and (7 ), we can apply Lemma 3.15 to get (7 ) and finish the proof.
By (A3) (II), since | u − w | ≍ 1 |u-w|\asymp 1
and u d ∨ w d ≤ y d + 2 ≤ 7 + 2 − 5 u_{d}\vee w_{d}\leq y_{d}+2\leq 7+2^{-5} for u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} , we have
I \displaystyle I
≤ c 4 ∫ 0 t ∫ V 3 ∫ V 1 𝟏 { u d ≤ w d } p ( t − s , x , u ) p ( s , y , w ) u d β 1 w d β ^ 2 log β 3 ( e + w d u d ) log β 4 ( e + 1 w d ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\leq c_{4}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}{\bf 1}_{\{u_{d}\leq w_{d}\}}p(t-s,x,u)p(s,y,w)u_{d}^{\beta_{1}}w_{d}^{\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}du\,dw\,ds
+ c 4 ∫ 0 t ∫ V 3 ∫ V 1 𝟏 { w d ≤ u d } p ( t − s , x , u ) p ( s , y , w ) u d β ^ 2 w d β 1 log β 3 ( e + u d w d ) log β 4 ( e + 1 u d ) 𝑑 u 𝑑 w 𝑑 s \displaystyle\;+c_{4}\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}{\bf 1}_{\{w_{d}\leq u_{d}\}}p(t-s,x,u)p(s,y,w)u_{d}^{\widehat{\beta}_{2}}w_{d}^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{u_{d}}{w_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{u_{d}}\bigg{)}du\,dw\,ds
= c 4 ( ( ∫ 0 t / 2 + ∫ t / 2 t ) ∫ V 3 ∫ V 1 𝟏 { u d ≤ w d } ⋯ + ( ∫ 0 t / 2 + ∫ t / 2 t ) ∫ V 3 ∫ V 1 𝟏 { w d ≤ u d } … ) . \displaystyle=c_{4}\bigg{(}\big{(}\int_{0}^{t/2}+\int_{t/2}^{t}\big{)}\int_{V_{3}}\int_{V_{1}}{\bf 1}_{\{u_{d}\leq w_{d}\}}\dots+\big{(}\int_{0}^{t/2}+\int_{t/2}^{t}\big{)}\int_{V_{3}}\int_{V_{1}}{\bf 1}_{\{w_{d}\leq u_{d}\}}\dots\bigg{)}.
Thus, by the change of variable s ~ = t − s \widetilde{s}=t-s in integrals ∫ t / 2 t \int_{t/2}^{t} ,
I ≤ c 4 ∑ i = 1 2 ℐ t , i ( x , y ; β 1 , β ^ 2 , β 3 , β 4 ) + c 4 ∑ i = 1 2 ℐ t , i ( y , x ; β 1 , β ^ 2 , β 3 , β 4 ) , \displaystyle I\leq c_{4}\sum_{i=1}^{2}{\cal I}_{t,i}(x,y;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4})+c_{4}\sum_{i=1}^{2}{\cal I}_{t,i}(y,x;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4}),
where the functions ℐ t , i ( x , y ; β 1 , β ^ 2 , β 3 , β 4 ) {\cal I}_{t,i}(x,y;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4}) , 1 ≤ i ≤ 2 1\leq i\leq 2 , are defined in (7.25 )–(7.26 ). Then by using Lemma 7.6 (i) and Lemma 7.8 , we conclude that (7 ) holds. The proof is complete. □ \Box
As an immediate consequence of Theorem 7.5 , we obtain the
following sharp upper bound of the heat kernel for β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} .
Corollary 7.9 .
Suppose that β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} . There exists a constant C > 0 C>0 such that
p ( t , x , y ) \displaystyle p(t,x,y)
≤ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q A β 1 , β 2 , β 3 , β 4 ( t , x , y ) ( t − d / α ∧ t | x − y | d + α ) , \displaystyle\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right),
for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} .
Here is the second main result of the section.
Theorem 7.10 .
Suppose that β 2 ≥ α + β 1 \beta_{2}\geq\alpha+\beta_{1} . There exists a constant C > 0 C>0 such that
p ( t , x , y ) \displaystyle p(t,x,y)
≤ C ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( t − d / α ∧ t | x − y | d + α ) [ A β 1 , β 2 , β 3 , β 4 ( t , x , y ) \displaystyle\leq C\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)\bigg{[}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y)
+ ( 1 ∧ t | x − y | α ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) + t 1 / α ) ∧ | x − y | ) \displaystyle\quad+\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})+t^{1/\alpha})\wedge|x-y|}\bigg{)}
× ( 𝟏 { β 2 > α + β 1 } A β 1 , β 1 , 0 , β 3 ( t , x , y ) + 𝟏 { β 2 = α + β 1 } A β 1 , β 1 , 0 , β 3 + β 4 + 1 ( t , x , y ) ) ] \displaystyle\quad\quad\times\bigg{(}{\bf 1}_{\{\beta_{2}>\alpha+\beta_{1}\}}A_{\beta_{1},\beta_{1},0,\beta_{3}}(t,x,y)+{\bf 1}_{\{\beta_{2}=\alpha+\beta_{1}\}}A_{\beta_{1},\beta_{1},0,\beta_{3}+\beta_{4}+1}(t,x,y)\bigg{)}\bigg{]}
for all t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} .
Again, we first introduce additional notation and prove a lemma.
For b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 , t > 0 t>0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} , define
ℐ t ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})
:= ∫ 0 t ∫ B + ( y , 2 ) ∫ B + ( x , 2 ) p ( t − s , x , u ) p ( s , y , w ) \displaystyle:=\int_{0}^{t}\int_{B_{+}(y,2)}\int_{B_{+}(x,2)}p(t-s,x,u)p(s,y,w)
× u d b 1 w d b 2 log b 3 ( e + w d u d ) log b 4 ( e + 1 w d ) d u d w d s . \displaystyle\qquad\times u_{d}^{b_{1}}w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{u_{d}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}\,du\,dw\,ds.
(7.37)
Lemma 7.11 .
Let b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 be such that b 1 ∧ b 2 > q − α b_{1}\wedge b_{2}>q-\alpha and b 1 < α + β 1 b_{1}<\alpha+\beta_{1} . There exists a constant C > 0 C>0 such that for all x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} with | x − y | = 5 |x-y|=5 , and all t ∈ ( 0 , 1 ] t\in(0,1] ,
ℐ t ( x , y ; b 1 , b 2 , b 3 , b 4 ) \displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})
≤ C t x d b 1 y d b 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq Ctx_{d}^{b_{1}}y_{d}^{b_{2}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e\!+\!\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\!\log^{b_{4}}\bigg{(}e\!+\!\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
+ 𝟏 { b 2 > α + β 1 } C t 2 x d b 1 y d β 1 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − β 1 log b 3 ( e + 1 x d ∨ t 1 / α ) log b 3 ( e + 1 y d ∨ t 1 / α ) \displaystyle+{\bf 1}_{\{b_{2}>\alpha+\beta_{1}\}}Ct^{2}x_{d}^{b_{1}}y_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\!\log^{b_{3}}\bigg{(}e\!+\!\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\!\log^{b_{3}}\bigg{(}e\!+\!\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
+ 𝟏 { b 2 = α + β 1 , y d < 2 } C t x d b 1 y d β 1 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − β 1 \displaystyle+{\bf 1}_{\{b_{2}=\alpha+\beta_{1},y_{d}<2\}}Ctx_{d}^{b_{1}}y_{d}^{\beta_{1}}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\!\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}
× ∫ y d 2 ( r α ∧ t ) log b 3 ( e + r x d ∨ t 1 / α ) log b 3 ( e + r y d ∨ t 1 / α ) log b 4 ( e + 1 r ) d r r . \displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.
Proof. By using Lemma 7.4 in the first inequality (integration with respect to u u ; note that
b 1 < α + β 1 b_{1}<\alpha+\beta_{1} ) and the second inequality (integration with respect to w w ),
we get
ℐ t ( x , y ; b 1 , b 2 , b 3 , b 4 ) ≤ c 1 x d b 1 ∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 \displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})\leq c_{1}x_{d}^{b_{1}}\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}
× ∫ B + ( y , 2 ) p ( s , y , w ) w d b 2 log b 3 ( e + w d x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 w d ) d w d s \displaystyle\hskip 71.13188pt\times\int_{B_{+}(y,2)}p(s,y,w)w_{d}^{b_{2}}\log^{b_{3}}\bigg{(}e+\frac{w_{d}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{w_{d}}\bigg{)}dwds
≤ c 2 ( x d b 1 y d b 2 I 1 + 𝟏 { b 2 > α + β 1 } x d b 1 y d β 1 I 2 + 𝟏 { b 2 = α + β 1 , y d < 2 } x d b 1 y d β 1 I 3 ) , \displaystyle\leq c_{2}\left(x_{d}^{b_{1}}y_{d}^{b_{2}}I_{1}+{\bf 1}_{\{b_{2}>\alpha+\beta_{1}\}}x_{d}^{b_{1}}y_{d}^{\beta_{1}}I_{2}+{\bf 1}_{\{b_{2}=\alpha+\beta_{1},y_{d}<2\}}x_{d}^{b_{1}}y_{d}^{\beta_{1}}I_{3}\right),
where
I 1 := ∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s , I_{1}:=\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\!\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds,
I 2 := ∫ 0 t s ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − β 1 log b 3 ( e + 2 y d ∨ s 1 / α ) log b 3 ( e + 1 x d ∨ ( t − s ) 1 / α ) 𝑑 s , I_{2}:=\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\!\log^{b_{3}}\bigg{(}e+\frac{2}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}ds,
and
I 3 := \displaystyle I_{3}:=
∫ 0 t s ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − β 1 \displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-\beta_{1}}
× ∫ y d ∨ s 1 / α 2 log b 3 ( e + r y d ∨ s 1 / α ) log b 3 ( e + r x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 r ) d r r d s . \displaystyle\hskip 36.98866pt\times\int_{y_{d}\vee s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds.
Applying
Lemma 10.5 to I 1 I_{1} and I 2 I_{2} ,
Lemma 10.6 to I 3 I_{3} ,
and (10.2 ), wee see that
I 1 ≤ c 3 t ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) , I_{1}\leq c_{3}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)},
I 2 ≤ c 3 t 2 ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − β 1 log b 3 ( e + 1 x d ∨ t 1 / α ) log b 3 ( e + 1 y d ∨ t 1 / α ) , I_{2}\leq c_{3}t^{2}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{b_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)},
and
I 2 ≤ \displaystyle I_{2}\leq
c 3 t ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − β 1 \displaystyle c_{3}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}
× ∫ y d 2 ( r α ∧ t ) log b 3 ( e + r x d ∨ t 1 / α ) log b 3 ( e + r y d ∨ t 1 / α ) log b 4 ( e + 1 r ) d r r . \displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.
This proves the lemma. □ \Box
Proof of Theorem 7.10 . As in the proof of Lemma 7.2 , by symmetry, (6 ),
Proposition 5.1 , (3.30 ) and (6.4 ),
we can assume without loss of generality that x d ≤ y d ∧ 2 − 5 x_{d}\leq y_{d}\wedge 2^{-5} , x ~ = 0 \widetilde{x}=0 and | x − y | = 5 |x-y|=5 ,
and then it is enough to show that there exists a constant c 1 > 0 c_{1}>0 independent of x x and y y such that for any t ≤ 1 t\leq 1 ,
p ( t , x , y ) ≤ c 1 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q ( x d ∨ t 1 / α ) β 1 \displaystyle p(t,x,y)\leq c_{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}
× [ ( y d ∨ t 1 / α ) β 2 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\qquad\times\bigg{[}(y_{d}\vee t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
(7.38)
+ 𝟏 { β 2 > α + β 1 } t ( y d ∨ t 1 / α ) β 1 log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 ( e + 1 y d ∨ t 1 / α ) \displaystyle\quad\qquad+{\bf 1}_{\{\beta_{2}>\alpha+\beta_{1}\}}t(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
+ 𝟏 { β 2 = α + β 1 } t ( y d ∨ t 1 / α ) β 1 log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 + β 4 + 1 ( e + 1 y d ∨ t 1 / α ) ] . \displaystyle\quad\qquad+{\bf 1}_{\{\beta_{2}=\alpha+\beta_{1}\}}t(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}\bigg{]}.
(7.39)
Let t ≤ 1 t\leq 1 . Set V 1 = U ( 1 ) V_{1}=U(1) , V 3 = B ( y , 2 ) ∩ ℝ ¯ + d V_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V 2 = ℝ ¯ + d ∖ ( V 1 ∪ V 3 ) V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}) .
By Lemmas 3.15 and 7.3
it remains to prove that I := ∫ 0 t ∫ V 3 ∫ V 1 p V 1 ( t − s , x , u ) ℬ ( u , w ) p ( s , y , w ) 𝑑 u 𝑑 w 𝑑 s I:=\int_{0}^{t}\int_{V_{3}}\int_{V_{1}}p^{V_{1}}(t-s,x,u){\cal B}(u,w)p(s,y,w)dudwds is bounded above by the right-hand side of (7.39 ).
By (A3) (II), since | u − w | ≍ 1 |u-w|\asymp 1 for u ∈ V 1 u\in V_{1} and w ∈ V 3 w\in V_{3} , using the change of variables s ~ = t − s \widetilde{s}=t-s we have
I ≤ c 2 ( ℐ t ( x , y ; β 1 , β 2 , β 3 , β 4 ) + ℐ t ( y , x ; β 1 , β 2 , β 3 , β 4 ) ) , \displaystyle I\leq c_{2}\left({\cal I}_{t}(x,y;\beta_{1},\beta_{2},\beta_{3},\beta_{4})+{\cal I}_{t}(y,x;\beta_{1},\beta_{2},\beta_{3},\beta_{4})\right),
(7.40)
where the functions ℐ t ( x , y ; β 1 , β 2 , β 3 , β 4 ) {\cal I}_{t}(x,y;\beta_{1},\beta_{2},\beta_{3},\beta_{4}) is defined in (7.37 ).
By Lemma 7.11 and (7.6 ), the right hand side of (7.40 ) is less than or equal to
c 3 t ( 1 ∧ x d t 1 / α ) q ( 1 ∧ y d t 1 / α ) q c_{3}t(1\wedge\frac{x_{d}}{t^{1/\alpha}})^{q}(1\wedge\frac{y_{d}}{t^{1/\alpha}})^{q} times
( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 2 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\,\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
(7.41)
+ ( x d ∨ t 1 / α ) β 2 ( y d ∨ t 1 / α ) β 1 log β 3 ( e + x d ∨ t 1 / α y d ∨ t 1 / α ) log β 4 ( e + 1 x d ∨ t 1 / α ) \displaystyle\quad+(x_{d}\vee t^{1/\alpha})^{\beta_{2}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}\vee t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}
(7.42)
+ 𝟏 { β 2 > α + β 1 } t ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 1 log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 ( e + 1 y d ∨ t 1 / α ) \displaystyle\quad+{\bf 1}_{\{\beta_{2}>\alpha+\beta_{1}\}}t(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
(7.43)
+ 𝟏 { β 2 = α + β 1 } ( x d ∨ t 1 / α ) β 1 ( y d ∨ t 1 / α ) β 1 \displaystyle\quad+{\bf 1}_{\{\beta_{2}=\alpha+\beta_{1}\}}(x_{d}\vee t^{1/\alpha})^{\beta_{1}}(y_{d}\vee t^{1/\alpha})^{\beta_{1}}
× ∫ x d 2 ( r α ∧ t ) log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + 1 r ) d r r . \displaystyle\qquad\;\times\int_{x_{d}}^{2}(r^{\alpha}\wedge t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.
(7.44)
In the last line we used x d ≤ y d ∧ 2 x_{d}\leq y_{d}\wedge 2 to get that 𝟏 { y d < 2 } ∫ y d 2 ≤ ∫ x d 2 {\bf 1}_{\{y_{d}<2\}}\int_{y_{d}}^{2}\leq\int_{x_{d}}^{2} .
Since β 2 > β 1 \beta_{2}>\beta_{1} and x d ≤ y d x_{d}\leq y_{d} , clearly
( x d ∨ t 1 / α ) β 2 log β 3 ( e + x d ∨ t 1 / α y d ∨ t 1 / α ) ≤ ( x d ∨ t 1 / α ) β 1 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) , \displaystyle(x_{d}\vee t^{1/\alpha})^{\beta_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{x_{d}\vee t^{1/\alpha}}{y_{d}\vee t^{1/\alpha}}\bigg{)}\leq(x_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)},
(7.45)
and by using
(10.2 ) (with ε = ( β 2 − β 1 ) / β 4 ) \varepsilon=(\beta_{2}-\beta_{1})/\beta_{4}) ), we get
( y d ∨ t 1 / α ) β 1 log β 4 ( e + 1 x d ∨ t 1 / α ) ≤ c 4 ( y d ∨ t 1 / α ) β 2 log β 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle(y_{d}\vee t^{1/\alpha})^{\beta_{1}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\leq c_{4}(y_{d}\vee t^{1/\alpha})^{\beta_{2}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.46)
Applying (7.45 ) and (7.46 ) to (7.42 )
and combining it with (7.41 ) and
(7.43 ), we arrive at the result in case β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} .
Assume now that β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} . From the above argument,
to prove the result, in view of (7.38 ), (7.39 ) and (7.44 ), it suffices to show that
∫ x d 2 ( r α ∧ t ) log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + 1 r ) d r r \displaystyle\int_{x_{d}}^{2}(r^{\alpha}\wedge t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≤ c 5 t log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 + β 4 + 1 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq c_{5}t\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
+ c 5 ( y d ∨ t 1 / α ) β 2 − β 1 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\quad+c_{5}(y_{d}\vee t^{1/\alpha})^{\beta_{2}-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.47)
By Lemma 10.11 (with b 1 = β 3 b_{1}=\beta_{3} , b 2 = β 4 b_{2}=\beta_{4} , k = x d ∨ t 1 / α k=x_{d}\vee t^{1/\alpha} and l = y d ∨ t 1 / α l=y_{d}\vee t^{1/\alpha} ), it holds that
∫ y d ∨ t 1 / α 2 ( r α ∧ t ) log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + 1 r ) d r r \displaystyle\int_{y_{d}\vee t^{1/\alpha}}^{2}(r^{\alpha}\wedge t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≤ t ∫ y d ∨ t 1 / α 2 log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + 1 r ) d r r \displaystyle\leq t\int_{y_{d}\vee t^{1/\alpha}}^{2}\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≤ c 6 t log β 3 ( e + 1 x d ∨ t 1 / α ) log β 3 + β 4 + 1 ( e + 1 y d ∨ t 1 / α ) . \displaystyle\leq c_{6}t\log^{\beta_{3}}\bigg{(}e+\frac{1}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}+\beta_{4}+1}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.48)
On the other hand, using (10.2 ), we see that
∫ x d y d ∨ t 1 / α ( r α ∧ t ) log β 3 ( e + r x d ∨ t 1 / α ) log β 3 ( e + r y d ∨ t 1 / α ) log β 4 ( e + 1 r ) d r r \displaystyle\int_{x_{d}}^{y_{d}\vee t^{1/\alpha}}(r^{\alpha}\wedge t)\log^{\beta_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≤ log β 3 ( e + 1 ) log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) ∫ x d y d ∨ t 1 / α log β 4 ( e + 1 r ) d r r 1 − α \displaystyle\leq\log^{\beta_{3}}(e+1)\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\int_{x_{d}}^{y_{d}\vee t^{1/\alpha}}\log^{\beta_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r^{1-\alpha}}
≤ c 7 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) ∫ x d y d ∨ t 1 / α ( y d ∨ t 1 / α r ) α / 2 d r r 1 − α \displaystyle\leq c_{7}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}\int_{x_{d}}^{y_{d}\vee t^{1/\alpha}}\bigg{(}\frac{y_{d}\vee t^{1/\alpha}}{r}\bigg{)}^{\alpha/2}\frac{dr}{r^{1-\alpha}}
≤ c 8 ( y d ∨ t 1 / α ) α log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq c_{8}(y_{d}\vee t^{1/\alpha})^{\alpha}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
= c 8 ( y d ∨ t 1 / α ) β 2 − β 1 log β 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log β 4 ( e + 1 y d ∨ t 1 / α ) . \displaystyle=c_{8}(y_{d}\vee t^{1/\alpha})^{\beta_{2}-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}.
(7.49)
Combining (7 )–(7 ), we
show that (7 ) holds true.
The proof is complete. □ \Box
8. Proofs of Theorems 1.1 and 1.2
Proof of Theorem 1.1 . Since p ¯ ( t , x , y ) \overline{p}(t,x,y) is jointly continuous
(see Remark 3.12 ),
it suffices to prove that
(1.1 )–(1.1 ) hold for ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ + d × ℝ + d (t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} .
We first note that by (A3) and (6.4 ),
t − d / α ∧ ( t J ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) ) ≍ t − d / α ∧ t B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) | x − y | d + α \displaystyle t^{-d/\alpha}\wedge\big{(}tJ(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\big{)}\asymp t^{-d/\alpha}\wedge\frac{tB_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})}{|x-y|^{d+\alpha}}
≍ ( t − d / α ∧ t | x − y | d + α ) B β 1 , β 2 , β 3 , β 4 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) , \displaystyle\asymp\left(t^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}}\right)B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d}),
(8.1)
which implies the second comparison in (1.1 ).
(i)
Using (6.3 ), we get the lower heat kernel estimate in the first comparison in (1.1 ) from Proposition 6.2 and the upper heat kernel estimate from Corollary 7.9 .
(ii) For (1.4 ), using (6.3 ), we get the lower heat kernel estimate from Proposition 6.6 (see Remark 6.7 ) and the upper heat kernel estimate from Theorem 7.10 .
(iii) Using (6.3 ), we get the lower heat kernel estimate in (1.1 ) from Proposition 6.9 .
The upper heat kernel estimate in (1.1 ) follows from Theorem 7.10 .
From the comparisons in (i)-(iii) and
(6.4 ), we have that p ¯ ( t , x , y ) ≍ t − d / α \overline{p}(t,x,y)\asymp t^{-d/\alpha} when t 1 / α ≥ | x − y | / 8 t^{1/\alpha}\geq|x-y|/8 and
this implies that (1.1 ) holds for t 1 / α ≥ | x − y | / 8 t^{1/\alpha}\geq|x-y|/8 .
Moreover, (1.1 ) for β 2 < α + β 1 \beta_{2}<\alpha+\beta_{1} follows from (1.1 ), (6.3 ) and (8 ).
By (A3) and (6.3 ), we have that when t 1 / α < | x − y | / 8 t^{1/\alpha}<|x-y|/8 ,
t ∫ ( x d ∨ y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 J ( x + t 1 / α 𝐞 d , x + r 𝐞 d ) J ( x + r 𝐞 d , y + t 1 / α 𝐞 d ) r d − 1 𝑑 r \displaystyle t\int_{(x_{d}\vee y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}J(x+t^{1/\alpha}{\mathbf{e}}_{d},x+r{\mathbf{e}}_{d})J(x+r{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})r^{d-1}dr
≍ t | x − y | d + α ∫ ( x d ∨ y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 A β 1 , β 2 , β 3 , β 4 ( t , x , x + r 𝐞 d ) A β 1 , β 2 , β 3 , β 4 ( t , x + r 𝐞 d , y ) d r r α + 1 . \displaystyle\asymp\frac{t}{|x-y|^{d+\alpha}}\int_{(x_{d}\vee y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}\!\!\!\!\!\!\!\!\!\!\!A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d})\,A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y)\frac{dr}{r^{\alpha+1}}.
(8.2)
Thus, we see
that, for β 2 = α + β 1 \beta_{2}=\alpha+\beta_{1} and t 1 / α < | x − y | / 8 t^{1/\alpha}<|x-y|/8 , (1.1 )
follows from (1.1 ), (8 ) and
Lemma 6.8 , and
that, for β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} and t 1 / α < | x − y | / 8 t^{1/\alpha}<|x-y|/8 , the lower bound in (1.1 )
follows from (8 ) and
Proposition 6.2 and Lemma 6.4 .
We have from (8 ), (6 ) and (6.3 ) that, when t 1 / α < | x − y | / 8 t^{1/\alpha}<|x-y|/8 ,
t ∫ ( x d ∨ y d ∨ t 1 / α ) ∧ ( | x − y | / 4 ) | x − y | / 2 J ( x + t 1 / α 𝐞 d , x + r 𝐞 d ) J ( x + r 𝐞 d , y + t 1 / α 𝐞 d ) r d − 1 𝑑 r \displaystyle t\int_{(x_{d}\vee y_{d}\vee t^{1/\alpha})\wedge(|x-y|/4)}^{|x-y|/2}J(x+t^{1/\alpha}{\mathbf{e}}_{d},x+r{\mathbf{e}}_{d})J(x+r{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})r^{d-1}dr
≥ c 1 t | x − y | d + 2 α B β 1 , β 1 , 0 , β 3 ( x + t 1 / α 𝐞 d , y + t 1 / α 𝐞 d ) log β 3 ( e + | x − y | ( ( x d ∧ y d ) ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\geq\frac{c_{1}t}{|x-y|^{d+2\alpha}}B_{\beta_{1},\beta_{1},0,\beta_{3}}(x+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})\log^{\beta_{3}}\bigg{(}e+\frac{|x-y|}{((x_{d}\wedge y_{d})\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
(8.3)
Now,
for β 2 > α + β 1 \beta_{2}>\alpha+\beta_{1} and t 1 / α < | x − y | / 8 t^{1/\alpha}<|x-y|/8 , the upper bound in (1.1 ) follows from (8 ),
(8 )
and the upper bound in (1.4 ).
Finally, from the joint continuity of p ¯ ( t , x , y ) \overline{p}(t,x,y) and upper heat kernel estimates, we deduce that Y ¯ \overline{Y} is a Feller process and finish the proof by Remark 3.12 .
□ \Box
Proof of Theorem 1.2 . The second comparison in (1.9 ) follows from Corollary 6.3 . By (6.3 ), Theorem 1.1 and Propositions 6.2 , 6.6 and 6.9 , p κ ( t , x , y ) ≥ c 1 ( 1 ∧ x d t 1 / α ) q κ ( 1 ∧ y d t 1 / α ) q κ p ¯ ( t , x , y ) p^{\kappa}(t,x,y)\geq c_{1}\big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\overline{p}(t,x,y)
for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ + d × ℝ + d (t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} . On the other hand, by (6.3 ), Theorem 1.1 , Corollary 7.9 , and Theorem 7.10 ,
p κ ( t , x , y ) ≤ c 2 ( 1 ∧ x d t 1 / α ) q κ ( 1 ∧ y d t 1 / α ) q κ p ¯ ( t , x , y ) p^{\kappa}(t,x,y)\leq c_{2}\big{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\big{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\big{)}^{q_{\kappa}}\overline{p}(t,x,y)
for all ( t , x , y ) ∈ ( 0 , ∞ ) × ℝ + d × ℝ + d (t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} and hence (1.9 ) holds true.
Note that for each ( t , y ) ∈ ( 0 , ∞ ) × ℝ + d (t,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+} , the map x ↦ p κ ( t , x , y ) x\mapsto p^{\kappa}(t,x,y) vanishes continuously on ∂ ℝ + d \partial{\mathbb{R}}^{d}_{+} . Hence, using the joint continuity of p κ ( t , x , y ) p^{\kappa}(t,x,y) and upper heat kernel estimates, we deduce that Y κ Y^{\kappa} is a Feller process. By Remark 3.12 , the proof is complete. □ \Box
10. Appendix
Note that for any ε > 0 {\varepsilon}>0 ,
log ( e + r ) < ( 2 + ε − 1 ) r ε for all r ≥ 1 , \log(e+r)<(2+{\varepsilon}^{-1})r^{\varepsilon}\quad\text{for all}\;r\geq 1,
(10.1)
log ( e + a r ) log ( e + r ) < ( 1 + ε − 1 ) a ε for all a ≥ 1 and r > 0 . \frac{\log(e+ar)}{\log(e+r)}<(1+{\varepsilon}^{-1})a^{{\varepsilon}}\quad\quad\text{for all }a\geq 1\text{ and }r>0.
(10.2)
Recall the definition of A b 1 , b 2 , b 3 , b 4 ( t , x , y ) A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y) from (6 ).
Lemma 10.1 .
Let b 1 , b 2 , b 3 , b 4 ≥ 0 b_{1},b_{2},b_{3},b_{4}\geq 0 .
(i) If b 1 > 0 b_{1}>0 , then for any ε ∈ ( 0 , b 1 ] {\varepsilon}\in(0,b_{1}] ,
there exists c 1 > 0 c_{1}>0 such that
A b 1 , b 2 , b 3 , b 4 ( t , x , y ) ≤ c 1 A b 1 − ε , b 2 , 0 , b 4 ( t , x , y ) for all t ≥ 0 , x , y ∈ ℝ + d . A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\leq c_{1}A_{b_{1}-{\varepsilon},b_{2},0,b_{4}}(t,x,y)\quad\text{ for all }t\geq 0,\,x,y\in{\mathbb{R}}^{d}_{+}.
(ii) If b 2 > 0 b_{2}>0 , then for any ε ∈ ( 0 , b 2 ] {\varepsilon}\in(0,b_{2}] ,
there exists c 2 > 0 c_{2}>0 such that
A b 1 , b 2 , b 3 , b 4 ( t , x , y ) ≤ c 2 A b 1 , b 2 − ε , b 3 , 0 ( t , x , y ) for all t ≥ 0 , x , y ∈ ℝ + d . A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y)\leq c_{2}A_{b_{1},b_{2}-{\varepsilon},b_{3},0}(t,x,y)\quad\text{ for all }t\geq 0,\,x,y\in{\mathbb{R}}^{d}_{+}.
Proof. The results follow from (10.1 ). □ \Box
Lemma 10.2 .
Let b 1 , b 3 ≥ 0 b_{1},b_{3}\geq 0 . Suppose that b 1 > 0 b_{1}>0 if b 3 > 0 b_{3}>0 . Then there exists a constant C > 0 C>0 such that for all t ≥ 0 t\geq 0 and x , y ∈ ℝ + d x,y\in{\mathbb{R}}^{d}_{+} ,
A b 1 , 0 , b 3 , 0 ( t , x , y ) \displaystyle A_{b_{1},0,b_{3},0}(t,x,y)
≤ C ( x d ∨ t 1 / α | x − y | ∧ 1 ) b 1 log b 3 ( e + ( y d ∨ t 1 / α ) ∧ | x − y | ( x d ∨ t 1 / α ) ∧ | x − y | ) . \displaystyle\leq C\,\bigg{(}\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1\bigg{)}^{b_{1}}\log^{b_{3}}\bigg{(}e+\frac{(y_{d}\vee t^{1/\alpha})\wedge|x-y|}{(x_{d}\vee t^{1/\alpha})\wedge|x-y|}\bigg{)}.
Proof. When x d ≤ y d x_{d}\leq y_{d} , the result is clear. Assume that x d > y d x_{d}>y_{d} .
Set r = x d ∨ t 1 / α | x − y | ∧ 1 r=\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1 and s = y d ∨ t 1 / α | x − y | ∧ 1 s=\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1 .
Then 0 < s ≤ r ≤ 1 0<s\leq r\leq 1 and the desired inequality is equivalent to
log b 3 ( e + r / s ) log − b 3 ( e + s / r ) ≤ C ( r / s ) b 1 . \log^{b_{3}}(e+r/s)\log^{-b_{3}}(e+s/r)\leq C(r/s)^{b_{1}}.
(10.3)
If b 3 = 0 b_{3}=0 , then (10.3 ) clearly holds with C = 1 C=1 . If b 3 > 0 b_{3}>0 and b 1 > 0 b_{1}>0 , then we get from (10.1 ) that
log b 3 ( e + r / s ) log − b 3 ( e + s / r ) ≤ log b 3 ( e + r / s ) ≤ c 1 ( r / s ) b 1 . \log^{b_{3}}(e+r/s)\log^{-b_{3}}(e+s/r)\leq\log^{b_{3}}(e+r/s)\leq c_{1}(r/s)^{b_{1}}.
This proves the lemma. □ \Box
Lemma 10.4 .
Let γ ∈ ℝ \gamma\in{\mathbb{R}} , b ≥ 0 b\geq 0 and t , k , l > 0 t,k,l>0 . Suppose that either γ < α \gamma<\alpha or k ≥ t 1 / α k\geq t^{1/\alpha} . Then we have
∫ 0 t ( 1 ∧ k s 1 / α ) γ log b ( e + l k ∨ s 1 / α ) 𝑑 s ≤ C t ( 1 ∧ k t 1 / α ) γ log b ( e + l k ∨ t 1 / α ) , \int_{0}^{t}\bigg{(}1\wedge\frac{k}{s^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{k\vee s^{1/\alpha}}\bigg{)}ds\leq Ct\bigg{(}1\wedge\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{k\vee t^{1/\alpha}}\bigg{)},
where C > 0 C>0 is a constant which depends only on γ \gamma and b b .
Proof. If k ≥ t 1 / α k\geq t^{1/\alpha} , then the desired inequality holds since the left hand side is t log b ( e + l / k ) t\log^{b}(e+l/k) .
Suppose that k < t 1 / α k<t^{1/\alpha} and γ < α \gamma<\alpha .
Let ε > 0 {\varepsilon}>0 be such that γ + b ε < α \gamma+b{\varepsilon}<\alpha . Then the left hand side is
log b ( e + l k ) ∫ 0 k α 𝑑 s + k γ ∫ k α t 1 s γ / α log b ( e + l s 1 / α ) 𝑑 s \displaystyle\log^{b}\bigg{(}e+\frac{l}{k}\bigg{)}\int_{0}^{k^{\alpha}}ds+k^{\gamma}\int_{k^{\alpha}}^{t}\frac{1}{s^{\gamma/\alpha}}\log^{b}\bigg{(}e+\frac{l}{s^{1/\alpha}}\bigg{)}ds
≤ k α log b ( e + l k ) + c 1 k γ log b ( e + l t 1 / α ) ∫ k α t t b ε / α s γ / α + b ε / α 𝑑 s \displaystyle\leq k^{\alpha}\log^{b}\bigg{(}e+\frac{l}{k}\bigg{)}+c_{1}k^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}\int_{k^{\alpha}}^{t}\frac{t^{b{\varepsilon}/\alpha}}{s^{\gamma/\alpha+b{\varepsilon}/\alpha}}ds
≤ c 2 k α ( t 1 / α k ) α − γ log b ( e + l t 1 / α ) + c 2 k γ t 1 − γ / α log b ( e + l t 1 / α ) \displaystyle\leq c_{2}k^{\alpha}\bigg{(}\frac{t^{1/\alpha}}{k}\bigg{)}^{\alpha-\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}+c_{2}k^{\gamma}t^{1-\gamma/\alpha}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}
= 2 c 2 t ( k t 1 / α ) γ log b ( e + l t 1 / α ) = 2 c 2 t ( 1 ∧ k t 1 / α ) γ log b ( e + l t 1 / α ) . \displaystyle=2c_{2}t\bigg{(}\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}=2c_{2}t\bigg{(}1\wedge\frac{k}{t^{1/\alpha}}\bigg{)}^{\gamma}\log^{b}\bigg{(}e+\frac{l}{t^{1/\alpha}}\bigg{)}.
We used (10.2 ) in both inequalities above. □ \Box
Lemma 10.5 .
Let b 1 , b 2 ∈ ℝ b_{1},b_{2}\in{\mathbb{R}} , b 3 , b 4 ≥ 0 b_{3},b_{4}\geq 0 and t , x d , y d > 0 t,x_{d},y_{d}>0 .
Suppose that (1) either b 1 > q − α b_{1}>q-\alpha or x d ≥ t 1 / α x_{d}\geq t^{1/\alpha} , and (2) either b 2 > q − α b_{2}>q-\alpha or y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} . Then we have
∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
≤ C t ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) log b 4 ( e + 1 y d ∨ t 1 / α ) \displaystyle\leq Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}
where C > 0 C>0 is a constant which depends only on b 1 , b 2 , b 3 b_{1},b_{2},b_{3} and b 4 b_{4} .
Proof. Observe that
∫ 0 t 2 ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
≤ c 1 ( 1 ∧ x d t 1 / α ) q − b 1 log b 3 ( e + y d ∨ t 1 / α x d ∨ t 1 / α ) ∫ 0 t 2 ( 1 ∧ y d s 1 / α ) q − b 2 log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\leq c_{1}\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee t^{1/\alpha}}\bigg{)}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
and
∫ t 2 t ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 log b 3 ( e + y d ∨ s 1 / α x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 y d ∨ s 1 / α ) 𝑑 s \displaystyle\int_{\frac{t}{2}}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee s^{1/\alpha}}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee s^{1/\alpha}}\bigg{)}ds
≤ c 2 ( 1 ∧ y d t 1 / α ) q − b 2 log b 4 ( e + 1 y d ∨ t 1 / α ) ∫ 0 t 2 ( 1 ∧ x d s 1 / α ) q − b 1 log b 3 ( e + y d ∨ t 1 / α x d ∨ s 1 / α ) 𝑑 s . \displaystyle\leq c_{2}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}\log^{b_{4}}\bigg{(}e+\frac{1}{y_{d}\vee t^{1/\alpha}}\bigg{)}\int_{0}^{\frac{t}{2}}\bigg{(}1\wedge\frac{x_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{1}}\log^{b_{3}}\bigg{(}e+\frac{y_{d}\vee t^{1/\alpha}}{x_{d}\vee s^{1/\alpha}}\bigg{)}ds.
Using Lemma 10.4 twice, we arrive at the result. □ \Box
Lemma 10.6 .
Let b 1 , b 2 ∈ ℝ b_{1},b_{2}\in{\mathbb{R}} , b 3 , b 4 ≥ 0 b_{3},b_{4}\geq 0 and t , x d , y d > 0 t,x_{d},y_{d}>0 . Suppose that (1) either b 1 > q − α b_{1}>q-\alpha or x d ≥ t 1 / α x_{d}\geq t^{1/\alpha} , and (2) either b 2 > q − α b_{2}>q-\alpha or y d ≥ t 1 / α y_{d}\geq t^{1/\alpha} . Then we have
that for y d ∨ t 1 / α < 2 y_{d}\vee t^{1/\alpha}<2 ,
∫ 0 t s ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 \displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}
× ∫ y d ∨ s 1 / α 2 log b 3 ( e + r y d ∨ s 1 / α ) log b 3 ( e + r x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 r ) d r r d s \displaystyle\quad\times\int_{y_{d}\vee s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds
≤ C t ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 \displaystyle\leq Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}
× ∫ y d 2 ( r α ∧ t ) log b 3 ( e + r x d ∨ t 1 / α ) log b 3 ( e + r y d ∨ t 1 / α ) log b 4 ( e + 1 r ) d r r , \displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r},
where C > 0 C>0 is a constant which depends only on b 1 , b 2 , b 3 b_{1},b_{2},b_{3} and b 4 b_{4} .
Proof. Using Fubini’s theorem and Lemma 10.4 twice as in the proof of Lemma 10.5 , we get
∫ 0 t s ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 \displaystyle\int_{0}^{t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}
× ∫ y d ∨ s 1 / α 2 log b 3 ( e + r y d ∨ s 1 / α ) log b 3 ( e + r x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 r ) d r r d s \displaystyle\quad\times\int_{y_{d}\vee s^{1/\alpha}}^{2}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\,ds
= \displaystyle=
∫ y d 2 ∫ 0 r α ∧ t s ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 \displaystyle\int_{y_{d}}^{2}\int_{0}^{r^{\alpha}\wedge t}s\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}
× log b 3 ( e + r y d ∨ s 1 / α ) log b 3 ( e + r x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 r ) d s d r r \displaystyle\quad\times\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\,ds\frac{dr}{r}
≤ \displaystyle\leq
∫ y d 2 ( r α ∧ t ) ∫ 0 t ( 1 ∧ x d ( t − s ) 1 / α ) q − b 1 ( 1 ∧ y d s 1 / α ) q − b 2 \displaystyle\int_{y_{d}}^{2}(r^{\alpha}\wedge t)\int_{0}^{t}\bigg{(}1\wedge\frac{x_{d}}{(t-s)^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{s^{1/\alpha}}\bigg{)}^{q-b_{2}}
× log b 3 ( e + r y d ∨ s 1 / α ) log b 3 ( e + r x d ∨ ( t − s ) 1 / α ) log b 4 ( e + 1 r ) d s d r r \displaystyle\quad\times\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee s^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee(t-s)^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\,ds\frac{dr}{r}
≤ \displaystyle\leq
C t ( 1 ∧ x d t 1 / α ) q − b 1 ( 1 ∧ y d t 1 / α ) q − b 2 \displaystyle\,Ct\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{1}}\bigg{(}1\wedge\frac{y_{d}}{t^{1/\alpha}}\bigg{)}^{q-b_{2}}
× ∫ y d 2 ( r α ∧ t ) log b 3 ( e + r x d ∨ t 1 / α ) log b 3 ( e + r y d ∨ t 1 / α ) log b 4 ( e + 1 r ) d r r . \displaystyle\quad\times\int_{y_{d}}^{2}(r^{\alpha}\wedge t)\log^{b_{3}}\bigg{(}e+\frac{r}{x_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{3}}\bigg{(}e+\frac{r}{y_{d}\vee t^{1/\alpha}}\bigg{)}\log^{b_{4}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}.
□ \Box
Lemma 10.7 .
There is a constant C > 0 C>0 such that for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and A > 0 A>0 ,
∫ B + ( x , A ) z d − 1 / 2 𝑑 z ≤ C A d ( x d ∨ A ) − 1 / 2 . \displaystyle\int_{B_{+}(x,A)}z_{d}^{-1/2}dz\leq CA^{d}(x_{d}\vee A)^{-1/2}.
Proof. We have
∫ B + ( x , A ) z d − 1 / 2 𝑑 z ≤ ∫ z ~ ∈ ℝ d − 1 , | x ~ − z ~ | < A 𝑑 z ~ ∫ ( x d − A ) ∨ 0 x d + A z d − 1 / 2 𝑑 z d ≤ c 1 A d − 1 ∫ ( x d − A ) ∨ 0 x d + A z d − 1 / 2 𝑑 z d . \displaystyle\int_{B_{+}(x,A)}z_{d}^{-1/2}dz\leq\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{x}-\widetilde{z}|<A}d\widetilde{z}\int_{(x_{d}-A)\vee 0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq c_{1}A^{d-1}\int_{(x_{d}-A)\vee 0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}.
(10.4)
If x d ≥ 2 A x_{d}\geq 2A , then
∫ ( x d − A ) ∨ 0 x d + A z d − 1 / 2 𝑑 z d ≤ 1 ( x d / 2 ) 1 / 2 ∫ x d − A x d + A 𝑑 z d = 2 3 / 2 A x d − 1 / 2 . \displaystyle\int_{(x_{d}-A)\vee 0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq\frac{1}{(x_{d}/2)^{1/2}}\int_{x_{d}-A}^{x_{d}+A}dz_{d}=2^{3/2}Ax_{d}^{-1/2}.
(10.5)
If x d < 2 A x_{d}<2A , then
∫ ( x d − A ) ∨ 0 x d + A z d − 1 / 2 𝑑 z d ≤ ∫ 0 4 A z d − 1 / 2 𝑑 z d = 4 A 1 / 2 . \displaystyle\int_{(x_{d}-A)\vee 0}^{x_{d}+A}z_{d}^{-1/2}dz_{d}\leq\int_{0}^{4A}z_{d}^{-1/2}dz_{d}=4A^{1/2}.
(10.6)
Combining (10.4 ) with (10.5 )–(10.6 ), we arrive at the result. □ \Box
Lemma 10.8 .
(i) There is a constant C > 0 C>0 such that for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and 0 < A < x d 0<A<x_{d} ,
∫ z ∈ ℝ + d , x d ≥ | x − z | > A d z z d 1 / 2 | x − z | d + α ≤ C x d − 1 / 2 A − α . \displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}\leq Cx_{d}^{-1/2}A^{-\alpha}.
(ii) Let ε ∈ ( 0 , 1 ) {\varepsilon}\in(0,1) and δ > 0 \delta>0 . There is a constant C ′ > 0 C^{\prime}>0 such that for all x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and A ≥ x d A\geq x_{d} ,
∫ z ∈ ℝ + d , | x − z | > A d z z d ε | x − z | d + δ ≤ C ′ A − ε − δ . \displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{z_{d}^{{\varepsilon}}\,|x-z|^{d+\delta}}\leq C^{\prime}A^{-{\varepsilon}-\delta}.
Proof. Without loss of generality, we assume x = ( 0 ~ , x d ) x=(\widetilde{0},x_{d}) .
(i) Note that
∫ z ∈ ℝ + d , x d ≥ | x − z | > A d z z d 1 / 2 | x − z | d + α \displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}
= ∫ z ∈ ℝ + d , x d ≥ | x − z | > A , | z ~ | ≤ | x d − z d | d z z d 1 / 2 | x − z | d + α + ∫ z ∈ ℝ + d , x d ≥ | x − z | > A , | z ~ | > | x d − z d | d z z d 1 / 2 | x − z | d + α \displaystyle=\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A,\,|\widetilde{z}|\leq|x_{d}-z_{d}|}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}+\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>A,\,|\widetilde{z}|>|x_{d}-z_{d}|}\frac{dz}{z_{d}^{1/2}|x-z|^{d+\alpha}}
= : I + I I . \displaystyle=:I+II.
First, we have
I \displaystyle I
≤ ∫ x d ≥ | x d − z d | > A 2 1 z d 1 / 2 | x d − z d | d + α ∫ z ~ ∈ ℝ d − 1 , | z ~ | ≤ | x d − z d | 𝑑 z ~ 𝑑 z d \displaystyle\leq\int_{x_{d}\geq|x_{d}-z_{d}|>\frac{A}{2}}\frac{1}{z_{d}^{1/2}|x_{d}-z_{d}|^{d+\alpha}}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\leq|x_{d}-z_{d}|}d\widetilde{z}dz_{d}
≤ c 1 ( ∫ 0 x d 2 d z d z d 1 / 2 | x d − z d | 1 + α + ∫ x d 2 x d − A 2 d z d z d 1 / 2 | x d − z d | 1 + α + ∫ x d + A 2 2 x d d z d z d 1 / 2 | x d − z d | 1 + α ) \displaystyle\leq c_{1}\bigg{(}\int_{0}^{\frac{x_{d}}{2}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}+\int_{\frac{x_{d}}{2}}^{x_{d}-\frac{A}{2}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}+\int_{x_{d}+\frac{A}{2}}^{2x_{d}}\frac{dz_{d}}{z_{d}^{1/2}|x_{d}-z_{d}|^{1+\alpha}}\bigg{)}
≤ c 1 ( 2 1 + α x d 1 + α ∫ 0 x d 2 z d − 1 / 2 𝑑 z d + 1 ( x d / 2 ) 1 / 2 ∫ x d 2 x d − A 2 d z d | x d − z d | 1 + α + x d − 1 / 2 ∫ x d + A 2 2 x d d z d | x d − z d | 1 + α ) \displaystyle\leq c_{1}\bigg{(}\frac{2^{1+\alpha}}{x_{d}^{1+\alpha}}\int_{0}^{\frac{x_{d}}{2}}{z_{d}^{-1/2}}dz_{d}+\frac{1}{(x_{d}/2)^{1/2}}\int_{\frac{x_{d}}{2}}^{x_{d}-\frac{A}{2}}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\alpha}}+{x_{d}^{-1/2}}\int_{x_{d}+\frac{A}{2}}^{2x_{d}}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\alpha}}\bigg{)}
≤ c 2 ( x d − 1 / 2 − α + x d − 1 / 2 A − α + x d − 1 / 2 A − α ) ≤ c 3 x d − 1 / 2 A − α . \displaystyle\leq c_{2}\left(x_{d}^{-1/2-\alpha}+x_{d}^{-1/2}A^{-\alpha}+x_{d}^{-1/2}A^{-\alpha}\right)\leq c_{3}x_{d}^{-1/2}A^{-\alpha}.
On the other hand, we see that for any z ∈ ℝ + d z\in{\mathbb{R}}^{d}_{+} with x d ≥ | x − z | > A x_{d}\geq|x-z|>A and | z ~ | > | x d − z d | |\widetilde{z}|>|x_{d}-z_{d}| ,
z d ≥ x d − | x d − z d | ≥ x d − 1 2 ( | z ~ | + | x d − z d | ) ≥ x d − 1 2 | x − z | ≥ ( 1 − 1 2 ) x d . \displaystyle z_{d}\geq x_{d}-|x_{d}-z_{d}|\geq x_{d}-\frac{1}{2}(|\widetilde{z}|+|x_{d}-z_{d}|)\geq x_{d}-\frac{1}{\sqrt{2}}|x-z|\geq\left(1-\frac{1}{\sqrt{2}}\right)x_{d}.
Hence, we also have that
I I \displaystyle II
≤ c 4 x d − 1 / 2 ∫ z ∈ ℝ + d , | x − z | > A d z | x − z | d + α ≤ c 5 x d − 1 / 2 A − α . \displaystyle\leq c_{4}x_{d}^{-1/2}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{|x-z|^{d+\alpha}}\leq c_{5}x_{d}^{-1/2}A^{-\alpha}.
(ii) Observe that
∫ z ∈ ℝ + d , | x − z | > A d z z d ε | x − z | d + δ \displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{z_{d}^{{\varepsilon}}\,|x-z|^{d+\delta}}
≤ ∫ z ∈ ℝ + d , | x d − z d | ≥ | z ~ | ∨ A 2 d z ~ d z d z d ε | x d − z d | d + δ + ∫ z ∈ ℝ + d , | z ~ | ≥ | x d − z d | ∨ A 2 d z ~ d z d z d ε | z ~ | d + δ = : I + I I . \displaystyle\leq\int_{z\in{\mathbb{R}}^{d}_{+},\,|x_{d}-z_{d}|\geq|\widetilde{z}|\vee\frac{A}{2}}\frac{d\widetilde{z}\,dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{d+\delta}}+\int_{z\in{\mathbb{R}}^{d}_{+},\,|\widetilde{z}|\geq|x_{d}-z_{d}|\vee\frac{A}{2}}\frac{d\widetilde{z}\,dz_{d}}{z_{d}^{{\varepsilon}}\,|\widetilde{z}|^{d+\delta}}=:I+II.
Using Fubini’s theorem, since A ≥ x d A\geq x_{d} , we see that
I I \displaystyle II
≤ ∫ z ~ ∈ ℝ d − 1 , | z ~ | ≥ A 2 1 | z ~ | d + δ ∫ 0 | z ~ | + x d z d − ε 𝑑 z d 𝑑 z ~ ≤ c 1 ∫ z ~ ∈ ℝ d − 1 , | z ~ | ≥ A 2 ( | z ~ | + A ) 1 − ε | z ~ | d + δ 𝑑 z ~ \displaystyle\leq\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{1}{|\widetilde{z}|^{d+\delta}}\int_{0}^{|\widetilde{z}|+x_{d}}z_{d}^{-{\varepsilon}}dz_{d}\,d\widetilde{z}\leq c_{1}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{(|\widetilde{z}|+A)^{1-{\varepsilon}}}{|\widetilde{z}|^{d+\delta}}d\widetilde{z}
≤ c 2 ∫ z ~ ∈ ℝ d − 1 , | z ~ | ≥ A 2 d z ~ | z ~ | d − 1 + ε + δ ≤ c 3 A − ε − δ . \displaystyle\leq c_{2}\int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\geq\frac{A}{2}}\frac{d\widetilde{z}}{|\widetilde{z}|^{d-1+{\varepsilon}+\delta}}\leq c_{3}A^{-{\varepsilon}-\delta}.
On the other hand, using the fact that ∫ z ~ ∈ ℝ d − 1 , | z ~ | ≤ | x d − z d | 𝑑 z ~ ≤ c 4 | x d − z d | d − 1 \int_{\widetilde{z}\in{\mathbb{R}}^{d-1},\,|\widetilde{z}|\leq|x_{d}-z_{d}|}d\widetilde{z}\leq c_{4}|x_{d}-z_{d}|^{d-1} , we also see that
I \displaystyle I
≤ c 5 ( ∫ 0 ( x d − A 2 ) ∨ 0 d z d z d ε | x d − z d | 1 + δ + ∫ x d + A 2 ∞ d z d z d ε | x d − z d | 1 + δ ) \displaystyle\leq c_{5}\bigg{(}\int_{0}^{(x_{d}-\frac{A}{2})\vee 0}\frac{dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{1+\delta}}+\int_{x_{d}+\frac{A}{2}}^{\infty}\frac{dz_{d}}{z_{d}^{{\varepsilon}}\,|x_{d}-z_{d}|^{1+\delta}}\bigg{)}
≤ c 6 ( 1 A 1 + δ ∫ 0 ( x d − A 2 ) ∨ 0 z d − ε 𝑑 z d + 1 A ε ∫ x d + A 2 ∞ d z d | x d − z d | 1 + δ ) \displaystyle\leq c_{6}\bigg{(}\frac{1}{A^{1+\delta}}\int_{0}^{(x_{d}-\frac{A}{2})\vee 0}z_{d}^{-{\varepsilon}}dz_{d}+\frac{1}{A^{{\varepsilon}}}\int_{x_{d}+\frac{A}{2}}^{\infty}\frac{dz_{d}}{|x_{d}-z_{d}|^{1+\delta}}\bigg{)}
≤ c 7 ( 1 A 1 + δ ( ( x d − A 2 ) ∨ 0 ) 1 − ε + 1 A ε + δ ) ≤ c 6 A ε + δ , \displaystyle\leq c_{7}\bigg{(}\frac{1}{A^{1+\delta}}\Big{(}\big{(}x_{d}-\frac{A}{2}\big{)}\vee 0\Big{)}^{1-{\varepsilon}}+\frac{1}{A^{{\varepsilon}+\delta}}\bigg{)}\leq\frac{c_{6}}{A^{{\varepsilon}+\delta}},
where we used the fact that A ≥ x d A\geq x_{d} in the last inequality. The proof is complete. □ \Box
For γ , η 1 , η 2 ≥ 0 \gamma,\eta_{1},\eta_{2}\geq 0
and k , l > 0 k,l>0 , define
f γ , η 1 , η 2 , k , l ( r ) := r γ log η 1 ( e + k r ) log η 2 ( e + r l ) . f_{\gamma,\eta_{1},\eta_{2},k,l}(r):=r^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}.
Lemma 10.9 .
Let γ , η 1 , η 2 ≥ 0 \gamma,\eta_{1},\eta_{2}\geq 0 .
(i) For any ε > 0 {\varepsilon}>0 , there exist constants C , C ′ > 0 C,C^{\prime}>0 such that for any k , l , r > 0 k,l,r>0 and any a ≥ 1 a\geq 1 ,
C a γ − ε ≤ f γ , η 1 , η 2 , k , l ( a r ) f γ , η 1 , η 2 , k , l ( r ) ≤ C ′ a γ + ε . Ca^{\gamma-{\varepsilon}}\leq\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\leq C^{\prime}a^{\gamma+{\varepsilon}}.
(10.7)
(ii) Assume that γ > 0 \gamma>0 . Then there exists a constant C > 0 C>0 such that for any k , l , r > 0 k,l,r>0 and any a ≥ 1 a\geq 1 ,
f γ , η 1 , η 2 , k , l ( a r ) f γ , η 1 , η 2 , k , l ( r ) ≥ C . \frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\geq C.
Proof. (i)
If η 2 = 0 \eta_{2}=0 , the second inequality in (10.7 ) is true with any ε ≥ 0 {\varepsilon}\geq 0 . In case η 2 > 0 \eta_{2}>0 ,
for any given ε > 0 {\varepsilon}>0 , let ε ′ := ε / η 2 {\varepsilon}^{\prime}:={\varepsilon}/\eta_{2} . We get from (10.2 ) that for all a ≥ 1 a\geq 1 and r > 0 r>0 ,
f γ , η 1 , η 2 , k , l ( a r ) f γ , η 1 , η 2 , k , l ( r ) ≤ a γ ( log ( e + a r / l ) log ( e + r / l ) ) η 2 ≤ a γ ( ( 1 + 1 / ε ′ ) a ε ′ ) η 2 = c ( ε , η 2 ) a γ + ε . \displaystyle\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(ar)}{f_{\gamma,\eta_{1},\eta_{2},k,l}(r)}\leq a^{\gamma}\left(\frac{\log(e+ar/l)}{\log(e+r/l)}\right)^{\eta_{2}}\leq a^{\gamma}\big{(}(1+1/{\varepsilon}^{\prime})a^{{\varepsilon}^{\prime}}\big{)}^{\eta_{2}}=c({\varepsilon},\eta_{2})a^{\gamma+{\varepsilon}}.
The first inequality can be proved by a similar argument.
(ii)
The desired result follows from the first inequality in (10.7 ) with ε = γ {\varepsilon}=\gamma .
□ \Box
Lemma 10.10 .
Let b 1 , b 2 , η 1 , η 2 , γ ≥ 0 b_{1},b_{2},\eta_{1},\eta_{2},\gamma\geq 0 .
There exists a constant C > 0 C>0 such that for any
x ∈ ℝ + d x\in{\mathbb{R}}^{d}_{+} and s , k , l > 0 s,k,l>0 ,
∫ B + ( x , 2 ) ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) \displaystyle\int_{B_{+}(x,2)}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}
× ( s − d / α ∧ s | x − z | d + α ) z d γ log η 1 ( e + k z d ) log η 2 ( e + z d l ) d z \displaystyle\hskip 51.21504pt\times\left(s^{-d/\alpha}\wedge\frac{s}{|x-z|^{d+\alpha}}\right)z_{d}^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{z_{d}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{z_{d}}{l}\bigg{)}dz
≤ C ( x d ∨ s 1 / α ) γ log η 1 ( e + k x d ∨ s 1 / α ) log η 2 ( e + x d ∨ s 1 / α l ) \displaystyle\leq C(x_{d}\vee s^{1/\alpha})^{\gamma}\log^{\eta_{1}}\bigg{(}e+\frac{k}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{x_{d}\vee s^{1/\alpha}}{l}\bigg{)}
+ C 𝟏 { x d ∨ s 1 / α < 2 } s ( x d ∨ s 1 / α ) b 1 [ 𝟏 { γ > α + b 1 } log b 2 ( e + 2 x d ∨ s 1 / α ) log η 1 ( e + k ) log η 2 ( e + 1 l ) \displaystyle\quad+C{\bf 1}_{\{x_{d}\vee s^{1/\alpha}<2\}}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\bigg{[}{\bf 1}_{\{\gamma>\alpha+b_{1}\}}\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}(e+k)\log^{\eta_{2}}\bigg{(}e+\frac{1}{l}\bigg{)}
+ 𝟏 { γ = α + b 1 } ∫ x d ∨ s 1 / α 2 log b 2 ( e + r x d ∨ s 1 / α ) log η 1 ( e + k r ) log η 2 ( e + r l ) d r r ] . \displaystyle\hskip 88.2037pt+{\bf 1}_{\{\gamma=\alpha+b_{1}\}}\int_{x_{d}\vee s^{1/\alpha}}^{2}\log^{b_{2}}\bigg{(}e+\frac{r}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}\bigg{]}.
Proof. Using the triangle inequality, we see that for any z ∈ ℝ + d z\in{\mathbb{R}}^{d}_{+} ,
z d ≤ x d + | x − z | ≤ 2 ( x d ∨ | x − z | ) . z_{d}\leq x_{d}+|x-z|\leq 2(x_{d}\vee|x-z|).
(10.8)
Therefore, using Lemma 10.9 (i)-(ii) and Lemma 10.7 , we get that
∫ B + ( x , s 1 / α ) ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) s − d / α f γ , η 1 , η 2 , k , l ( z d ) 𝑑 z \displaystyle\int_{B_{+}(x,s^{1/\alpha})}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}\,s^{-d/\alpha}f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})dz
= ( log b 2 ( e + 1 ) ) s − d / α ∫ B + ( x , s 1 / α ) z d − 1 / 2 f γ + 1 2 , η 1 , η 2 , k , l ( z d ) 𝑑 z \displaystyle=(\log^{b_{2}}(e+1))s^{-d/\alpha}\int_{B_{+}(x,s^{1/\alpha})}z_{d}^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(z_{d})dz
≤ c 1 s − d / α f γ + 1 2 , η 1 , η 2 , k , l ( 2 ( x d ∨ s 1 / α ) ) ∫ B + ( x , s 1 / α ) z d − 1 / 2 𝑑 z \displaystyle\leq c_{1}s^{-d/\alpha}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(2(x_{d}\vee s^{1/\alpha}))\int_{B_{+}(x,s^{1/\alpha})}z_{d}^{-1/2}dz
≤ c 2 ( x d ∨ s 1 / α ) − 1 / 2 f γ + 1 2 , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) = c 2 f γ , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) . \displaystyle\leq c_{2}(x_{d}\vee s^{1/\alpha})^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha})=c_{2}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha}).
When x d > s 1 / α x_{d}>s^{1/\alpha} , we get from (10.8 ), Lemma 10.9 (i)-(ii) and Lemma 10.8 (i) that
s ∫ z ∈ ℝ + d , x d ≥ | x − z | > s 1 / α ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) f γ , η 1 , η 2 , k , l ( z d ) | x − z | d + α 𝑑 z \displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz
= ( log b 2 ( e + 1 ) ) s ∫ z ∈ ℝ + d , x d ≥ | x − z | > s 1 / α f γ + 1 2 , η 1 , η 2 , k , l ( z d ) z d 1 / 2 | x − z | d + α 𝑑 z \displaystyle=(\log^{b_{2}}(e+1))s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\frac{f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(z_{d})}{z_{d}^{1/2}|x-z|^{d+\alpha}}dz
≤ c 3 s f γ + 1 2 , η 1 , η 2 , k , l ( 2 x d ) ∫ z ∈ ℝ + d , x d ≥ | x − z | > s 1 / α 1 z d 1 / 2 | x − z | d + α 𝑑 z \displaystyle\leq c_{3}sf_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(2x_{d})\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\geq|x-z|>s^{1/\alpha}}\frac{1}{z_{d}^{1/2}|x-z|^{d+\alpha}}dz
≤ c 4 x d − 1 / 2 f γ + 1 2 , η 1 , η 2 , k , l ( x d ) = c 4 f γ , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) . \displaystyle\leq c_{4}x_{d}^{-1/2}f_{\gamma+\frac{1}{2},\eta_{1},\eta_{2},k,l}(x_{d})=c_{4}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha}).
It remains to bound the integral over { z ∈ ℝ + d : x d ∨ s 1 / α < | x − z | < 2 } \{z\in{\mathbb{R}}^{d}_{+}:x_{d}\vee s^{1/\alpha}<|x-z|<2\}
under the assumption x d ∨ s 1 / α < 2 x_{d}\vee s^{1/\alpha}<2 .
For this, we consider the following three cases separately.
(i) Case γ < α + b 1 \gamma<\alpha+b_{1} : Fix ε ∈ ( 0 , 1 ) {\varepsilon}\in(0,1) such that γ + 3 ε < α + b 1 \gamma+3{\varepsilon}<\alpha+b_{1} . Using (10.8 ), (10.1 ), Lemma 10.9 (i)-(ii) and Lemma 10.8 (ii), we get
s ∫ z ∈ ℝ + d , | x − z | > x d ∨ s 1 / α ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) f γ , η 1 , η 2 , k , l ( z d ) | x − z | d + α 𝑑 z \displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee s^{1/\alpha}}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz
= s ( x d ∨ s 1 / α ) b 1 ∫ z ∈ ℝ + d , | x − z | > x d ∨ s 1 / α log b 2 ( e + | x − z | x d ∨ s 1 / α ) f γ + ε , η 1 , η 2 , k , l ( z d ) z d ε | x − z | d + α + b 1 𝑑 z \displaystyle=s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee s^{1/\alpha}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(z_{d})}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz
≤ c 5 s ( x d ∨ s 1 / α ) b 1 ∫ z ∈ ℝ + d , | x − z | > x d ∨ s 1 / α log b 2 ( e + | x − z | x d ∨ s 1 / α ) f γ + ε , η 1 , η 2 , k , l ( 2 | x − z | ) z d ε | x − z | d + α + b 1 𝑑 z \displaystyle\leq c_{5}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee s^{1/\alpha}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(2|x-z|)}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz
≤ c 6 s ( x d ∨ s 1 / α ) b 1 ∫ z ∈ ℝ + d , | x − z | > x d ∨ s 1 / α ( | x − z | x d ∨ s 1 / α ) ε + γ + 2 ε f γ + ε , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) z d ε | x − z | d + α + b 1 𝑑 z \displaystyle\leq c_{6}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee s^{1/\alpha}}\bigg{(}\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}^{{\varepsilon}+\gamma+2{\varepsilon}}\frac{f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha})}{z_{d}^{{\varepsilon}}|x-z|^{d+\alpha+b_{1}}}dz
= c 6 s ( x d ∨ s 1 / α ) b 1 − γ − 3 ε f γ + ε , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) ∫ z ∈ ℝ + d , | x − z | > x d ∨ s 1 / α 1 z d ε | x − z | d + α + b 1 − γ − 3 ε 𝑑 z \displaystyle=c_{6}s(x_{d}\vee s^{1/\alpha})^{b_{1}-\gamma-3{\varepsilon}}f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha})\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>x_{d}\vee s^{1/\alpha}}\frac{1}{z_{d}^{\varepsilon}|x-z|^{d+\alpha+b_{1}-\gamma-3{\varepsilon}}}dz
≤ c 7 s ( x d ∨ s 1 / α ) − α ( x d ∨ s 1 / α ) − ε f γ + ε , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) ≤ c 7 f γ , η 1 , η 2 , k , l ( x d ∨ s 1 / α ) . \displaystyle\leq c_{7}s(x_{d}\vee s^{1/\alpha})^{-\alpha}(x_{d}\vee s^{1/\alpha})^{-{\varepsilon}}f_{\gamma+{\varepsilon},\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha})\leq c_{7}f_{\gamma,\eta_{1},\eta_{2},k,l}(x_{d}\vee s^{1/\alpha}).
(ii) Case γ > α + b 1 \gamma>\alpha+b_{1} : Fix ε > 0 {\varepsilon}>0 such that γ − ε > α + b 1 \gamma-{\varepsilon}>\alpha+b_{1} . Using (10.8 ), Lemma 10.9 (i)-(ii) and (10.2 ), we get
s ∫ z ∈ ℝ + d , x d ∨ s 1 / α < | x − z | < 2 ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) f γ , η 1 , η 2 , k , l ( z d ) | x − z | d + α 𝑑 z \displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee s^{1/\alpha}<|x-z|<2}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz
≤ c 8 s ( x d ∨ s 1 / α ) b 1 ∫ z ∈ ℝ + d , x d ∨ s 1 / α < | x − z | < 2 log b 2 ( e + | x − z | x d ∨ s 1 / α ) f γ , η 1 , η 2 , k , l ( 2 | x − z | ) | x − z | d + α + b 1 𝑑 z \displaystyle\leq c_{8}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(2|x-z|)}{|x-z|^{d+\alpha+b_{1}}}dz
≤ c 9 s ( x d ∨ s 1 / α ) b 1 f γ , η 1 , η 2 , k , l ( 4 ) log b 2 ( e + 2 x d ∨ s 1 / α ) ∫ z ∈ ℝ + d , | x − z | < 2 d z | x − z | d + α + b 1 − γ + ε \displaystyle\leq c_{9}s(x_{d}\vee s^{1/\alpha})^{b_{1}}f_{\gamma,\eta_{1},\eta_{2},k,l}(4)\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee s^{1/\alpha}}\bigg{)}\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|<2}\frac{dz}{|x-z|^{d+\alpha+b_{1}-\gamma+{\varepsilon}}}
≤ c 10 s ( x d ∨ s 1 / α ) b 1 f γ , η 1 , η 2 , k , l ( 1 ) log b 2 ( e + 2 x d ∨ s 1 / α ) . \displaystyle\leq c_{10}s(x_{d}\vee s^{1/\alpha})^{b_{1}}f_{\gamma,\eta_{1},\eta_{2},k,l}(1)\log^{b_{2}}\bigg{(}e+\frac{2}{x_{d}\vee s^{1/\alpha}}\bigg{)}.
(iii) Case γ = α + b 1 \gamma=\alpha+b_{1} : In this case, we see that
s ∫ z ∈ ℝ + d , x d ∨ s 1 / α < | x − z | < 2 ( 1 ∧ x d ∨ s 1 / α | x − z | ) b 1 log b 2 ( e + | x − z | ( x d ∨ s 1 / α ) ∧ | x − z | ) f γ , η 1 , η 2 , k , l ( z d ) | x − z | d + α 𝑑 z \displaystyle s\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee s^{1/\alpha}<|x-z|<2}\bigg{(}1\wedge\frac{x_{d}\vee s^{1/\alpha}}{|x-z|}\bigg{)}^{b_{1}}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{(x_{d}\vee s^{1/\alpha})\wedge|x-z|}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(z_{d})}{|x-z|^{d+\alpha}}dz
≤ c 11 s ( x d ∨ s 1 / α ) b 1 ∫ z ∈ ℝ + d , x d ∨ s 1 / α < | x − z | < 2 log b 2 ( e + | x − z | x d ∨ s 1 / α ) f γ , η 1 , η 2 , k , l ( | x − z | ) | x − z | d + α + b 1 𝑑 z \displaystyle\leq c_{11}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}\frac{f_{\gamma,\eta_{1},\eta_{2},k,l}(|x-z|)}{|x-z|^{d+\alpha+b_{1}}}dz
= c 11 s ( x d ∨ s 1 / α ) b 1 \displaystyle=c_{11}s(x_{d}\vee s^{1/\alpha})^{b_{1}}
× ∫ z ∈ ℝ + d , x d ∨ s 1 / α < | x − z | < 2 log b 2 ( e + | x − z | x d ∨ s 1 / α ) log η 1 ( e + k | x − z | ) log η 2 ( e + | x − z | l ) d z | x − z | d \displaystyle\times\int_{z\in{\mathbb{R}}^{d}_{+},\,x_{d}\vee s^{1/\alpha}<|x-z|<2}\log^{b_{2}}\bigg{(}e+\frac{|x-z|}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{|x-z|}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{|x-z|}{l}\bigg{)}\frac{dz}{|x-z|^{d}}
≤ c 12 s ( x d ∨ s 1 / α ) b 1 ∫ x d ∨ s 1 / α 2 log b 2 ( e + r x d ∨ s 1 / α ) log η 1 ( e + k r ) log η 2 ( e + r l ) d r r . \displaystyle\leq c_{12}s(x_{d}\vee s^{1/\alpha})^{b_{1}}\int_{x_{d}\vee s^{1/\alpha}}^{2}\log^{b_{2}}\bigg{(}e+\frac{r}{x_{d}\vee s^{1/\alpha}}\bigg{)}\log^{\eta_{1}}\bigg{(}e+\frac{k}{r}\bigg{)}\log^{\eta_{2}}\bigg{(}e+\frac{r}{l}\bigg{)}\frac{dr}{r}.
The proof is complete.
□ \Box
Lemma 10.11 .
Let b 1 , b 2 ≥ 0 b_{1},b_{2}\geq 0 . For any 0 < k ≤ l < 1 0<k\leq l<1 ,
∫ l 2 log b 1 ( e + r k ) log b 1 ( e + r l ) log b 2 ( e + 1 r ) d r r ≍ log b 1 ( e + 1 k ) log b 1 + b 2 + 1 ( e + 1 l ) , \int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}\asymp\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)},
with comparison constants independent of k k and l l .
Proof. Note that
log ( e + 1 r ) ≍ log ( 2 e r ) ≍ log ( e + 1 r ) , 0 < r < 2 . \log\bigg{(}e+\frac{1}{r}\bigg{)}\asymp\log\bigg{(}\frac{2e}{r}\bigg{)}\asymp\log\bigg{(}e+\frac{1}{\sqrt{r}}\bigg{)},\quad 0<r<2.
(10.9)
Hence, we get
∫ l 2 log b 1 ( e + r k ) log b 1 ( e + r l ) log b 2 ( e + 1 r ) d r r \displaystyle\int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≥ c 1 ∫ l 2 log b 1 ( 2 e r k ) log b 1 ( 2 e r l ) log b 2 ( 2 e r ) d r r \displaystyle\geq c_{1}\int_{\sqrt{l}}^{2}\log^{b_{1}}\bigg{(}\frac{2er}{k}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2er}{l}\bigg{)}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}
≥ c 1 log b 1 ( 2 e k ) log b 1 ( 2 e l ) ∫ l 2 log b 2 ( 2 e r ) d r r \displaystyle\geq c_{1}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{k}}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}\int_{\sqrt{l}}^{2}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}
= c 1 b 2 + 1 log b 1 ( 2 e k ) log b 1 ( 2 e l ) ( log b 2 + 1 ( 2 e l ) − 1 ) \displaystyle=\frac{c_{1}}{b_{2}+1}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{k}}\bigg{)}\log^{b_{1}}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}\bigg{(}\log^{b_{2}+1}\bigg{(}\frac{2e}{\sqrt{l}}\bigg{)}-1\bigg{)}
≥ c 2 log b 1 ( e + 1 k ) log b 1 + b 2 + 1 ( e + 1 l ) . \displaystyle\geq c_{2}\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)}.
On the other hand, using (10.9 ) and (10.2 ), we get
∫ l 2 log b 1 ( e + r k ) log b 1 ( e + r l ) log b 2 ( e + 1 r ) d r r \displaystyle\int_{l}^{2}\log^{b_{1}}\bigg{(}e+\frac{r}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{r}{l}\bigg{)}\log^{b_{2}}\bigg{(}e+\frac{1}{r}\bigg{)}\frac{dr}{r}
≤ c 3 log b 1 ( e + 2 k ) log b 1 ( e + 2 l ) ∫ l 2 log b 2 ( 2 e r ) d r r ≤ c 4 log b 1 ( e + 1 k ) log b 1 + b 2 + 1 ( e + 1 l ) . \displaystyle\leq c_{3}\log^{b_{1}}\bigg{(}e+\frac{2}{k}\bigg{)}\log^{b_{1}}\bigg{(}e+\frac{2}{l}\bigg{)}\int_{l}^{2}\log^{b_{2}}\bigg{(}\frac{2e}{r}\bigg{)}\frac{dr}{r}\leq c_{4}\log^{b_{1}}\bigg{(}e+\frac{1}{k}\bigg{)}\log^{b_{1}+b_{2}+1}\bigg{(}e+\frac{1}{l}\bigg{)}.
□ \Box
Lemma 10.12 .
Let γ > 1 \gamma>1 and b 1 , b 2 ≥ 0 b_{1},b_{2}\geq 0 . For any a , k , l > 0 a,k,l>0 ,
∫ a ∞ t − γ ( 1 ∧ k t ) b 1 ( 1 ∧ l t ) b 2 𝑑 t ≍ a 1 − γ ( 1 ∧ k a ) b 1 ( 1 ∧ l a ) b 2 , \int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt\asymp a^{1-\gamma}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}},
with comparison constants independent of a , k a,k and l l .
Proof. We have
∫ a ∞ t − γ ( 1 ∧ k t ) b 1 ( 1 ∧ l t ) b 2 𝑑 t ≤ ( 1 ∧ k a ) b 1 ( 1 ∧ l a ) b 2 ∫ a ∞ t − γ 𝑑 t ≤ a 1 − γ γ − 1 ( 1 ∧ k a ) b 1 ( 1 ∧ l a ) b 2 \displaystyle\int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt\leq\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}\int_{a}^{\infty}t^{-\gamma}dt\leq\frac{a^{1-\gamma}}{\gamma-1}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}
and
∫ a ∞ t − γ ( 1 ∧ k t ) b 1 ( 1 ∧ l t ) b 2 𝑑 t \displaystyle\int_{a}^{\infty}t^{-\gamma}\bigg{(}1\wedge\frac{k}{t}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{t}\bigg{)}^{b_{2}}dt
≥ ( 1 ∧ k 2 a ) b 1 ( 1 ∧ l 2 a ) b 2 ∫ a 2 a t − γ 𝑑 t \displaystyle\geq\bigg{(}1\wedge\frac{k}{2a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{2a}\bigg{)}^{b_{2}}\int_{a}^{2a}t^{-\gamma}dt
≥ ( 1 − 2 1 − γ ) a 1 − γ 2 b 1 + b 2 ( γ − 1 ) ( 1 ∧ k a ) b 1 ( 1 ∧ l a ) b 2 . \displaystyle\geq\frac{(1-2^{1-\gamma})a^{1-\gamma}}{2^{b_{1}+b_{2}}(\gamma-1)}\bigg{(}1\wedge\frac{k}{a}\bigg{)}^{b_{1}}\bigg{(}1\wedge\frac{l}{a}\bigg{)}^{b_{2}}.
□ \Box