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Heat kernel estimates for Dirichlet forms degenerate at the boundary

Soobin Cho,  Panki Kim,  Renming Song  and  Zoran Vondraček Department of Mathematical Sciences, Seoul National University, Seoul 08826, Republic of Korea soobin15@snu.ac.kr Department of Mathematical Sciences and Research Institute of Mathematics, Seoul National University, Seoul 08826, Republic of Korea pkim@snu.ac.kr Department of Mathematics, University of Illinois, Urbana, IL 61801, USA rsong@illinois.edu Department of Mathematics, Faculty of Science, University of Zagreb, Zagreb, Croatia, and Department of Mathematical Sciences, Seoul National University, Seoul, Republic of Korea vondra@math.hr
Abstract.

The goal of this paper is to prove the existence of heat kernels for two types of purely discontinuous symmetric Markov processes in the upper half-space of d{\mathbb{R}}^{d} with jump kernels degenerate at the boundary, and to establish sharp two-sided estimates on the heat kernels. The jump kernels are of the form J(x,y)=(x,y)|xy|αdJ(x,y)={\cal B}(x,y)|x-y|^{-\alpha-d}, α(0,2)\alpha\in(0,2), where the function {\cal B} depends on four parameters and may vanish at the boundary. Our results are the first sharp two-sided estimates for the heat kernels of non-local operators with jump kernels degenerate at the boundary.

The first type of processes we consider are conservative Markov processes on the closed half-space with jump kernel J(x,y)J(x,y). Depending on the regions where the parameters belong, the heat kernels estimates have three different forms, two of them are qualitatively different from all previously known heat kernel estimates.

The second type of processes we consider are the processes above killed either by a critical potential or upon hitting the boundary of the half-space. We establish that their heat kernel estimates have the approximate factorization property with survival probabilities decaying as a power of the distance to the boundary, where the power depends on the constant in the critical potential.

Finally, by using the heat kernel estimates, we obtain sharp two-sided Green function estimates that cover the main result of [31]. This alternative proof provides an explanation of the anomalous form of the estimates.

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)

AMS 2020 Mathematics Subject Classification: Primary 60J35, 60J45; Secondary 60J50, 60J76, 31C25, 35K08


Keywords and phrases: Markov processes, Dirichlet forms, jump kernel, killing potential, heat kernel, heat kernel estimates, Green function


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 d{\mathbb{R}}^{d} with jump kernels of the form J(x,y)=|xy|dα(x,y)J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y), α(0,2)\alpha\in(0,2). The function (x,y){\cal B}(x,y) may tend to 0 when xx or yy 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 {\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 ¯+d:={x=(x~,xd):x~d1,xd0}\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)=|xy|dα(x,y)J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y), where the function (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 β1,β2,β3,β4\beta_{1},\beta_{2},\beta_{3},\beta_{4}. The generator of such a process is the non-local operator

Lαf(x)=p.v.+d(f(y)f(x))|xy|dα(x,y)𝑑y.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 (x,y){\cal B}(x,y) is bounded between two positive constants, the heat kernel estimates for such processes are of the form min{td/α,tJ(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 xx to yy. 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 β2<α+β1\beta_{2}<\alpha+\beta_{1}, in which case the heat kernel is comparable to min{td/α,tJ(x+t1/α𝐞d,y+t1/α𝐞d)}\min\{t^{-d/\alpha},tJ(x+t^{1/\alpha}\mathbf{e}_{d},y+t^{1/\alpha}\mathbf{e}_{d})\}, where 𝐞d=(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 tt comes from one jump from the point t1/αt^{1/\alpha} units above xx to the point t1/αt^{1/\alpha} units above yy. 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 β2=α+β1\beta_{2}=\alpha+\beta_{1}: When the parameters satisfy β2α+β1\beta_{2}\geq\alpha+\beta_{1}, in addition to the already mentioned part min{td/α,tJ(x+t1/α𝐞d,y+t1/α𝐞d)}\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 xx and yy. 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 κxdα\kappa x_{d}^{-\alpha} or upon hitting the boundary of +d:={x=(x~,xd):x~d1,xd>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 Lf(x)=Lαf(x)κxdαf(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 (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 xx and yy. 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 qq-th power of the distance to the boundary, where qq is in one-to-one correspondence with the constant κ0\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 d1d\geq 1 and 0<α<20<\alpha<2. Recall that +d={(x~,xd)d:xd>0}{\mathbb{R}}^{d}_{+}=\{(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:x_{d}>0\} and ¯+d={(x~,xd)d:xd0}\overline{\mathbb{R}}^{d}_{+}=\{(\widetilde{x},x_{d})\in{\mathbb{R}}^{d}:x_{d}\geq 0\}. Here and below, ab:=min{a,b}a\wedge b:=\min\{a,b\}, ab:=max{a,b}a\vee b:=\max\{a,b\}, and aba\asymp b means that cb/ac1c\leq b/a\leq c^{-1} for some c(0,1)c\in(0,1). For b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0, let

Bb1,b2,b3,b4(x,y)\displaystyle B_{b_{1},b_{2},b_{3},b_{4}}(x,y) :=(xdyd|xy|1)b1(xdyd|xy|1)b2\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}}
×logb3(e+(xdyd)|xy|xdyd|xy|)logb4(e+|xy|(xdyd)|xy|).\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)=|xy|dα(x,y)J(x,y)=|x-y|^{-d-\alpha}{\cal B}(x,y), where the function :¯+dׯ+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) (x,y)=(y,x){\cal B}(x,y)={\cal B}(y,x) for all x,y¯+dx,y\in\overline{{\mathbb{R}}}^{d}_{+}.

(A2) If α1\alpha\geq 1, then there exist θ>α1\theta>\alpha-1 and C1>0C_{1}>0 such that

|(x,x)(x,y)|C1(|xy|xdyd)θ,x,y+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 C21C_{2}\geq 1 and β1,β2,β3,β40\beta_{1},\beta_{2},\beta_{3},\beta_{4}\geq 0, with β1>0\beta_{1}>0 if β3>0\beta_{3}>0, and β2>0\beta_{2}>0 if β4>0\beta_{4}>0, such that

C21Bβ1,β2,β3,β4(x,y)(x,y)C2,x,y¯+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 C3>0C_{3}>0 such that

(x,y)C3Bβ1,β2,β3,β4(x,y),x,y¯+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 β1,β2,β3,β4\beta_{1},\beta_{2},\beta_{3},\beta_{4} are the same constants as in (I).

(A4) For all x,y¯+dx,y\in\overline{{\mathbb{R}}}^{d}_{+} and a>0a>0, (ax,ay)=(x,y){\cal B}(ax,ay)={\cal B}(x,y). In case d2d\geq 2, for all x,y¯+dx,y\in\overline{{\mathbb{R}}}^{d}_{+} and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1}, (x+(z~,0),y+(z~,0))=(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 (x,y){\cal B}(x,y) satisfies (A1), (A3)(I) and (A4).

Consider the following symmetric form

0(u,v):=12+d+d(u(x)u(y))(v(x)v(y))J(x,y)𝑑y𝑑x.{\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 (x,y){\cal B}(x,y) is bounded, Cc(¯+d)C_{c}^{\infty}({\overline{\mathbb{R}}}^{d}_{+}) is closable in L2(¯+d,dx)=L2(+d,dx)L^{2}({\overline{\mathbb{R}}}^{d}_{+},dx)=L^{2}({\mathbb{R}}^{d}_{+},dx) by Fatou’s lemma. Let ¯{\overline{\cal F}} be the closure of Cc(¯+d)C_{c}^{\infty}({\overline{\mathbb{R}}}^{d}_{+}) in L2(+d,dx)L^{2}({\mathbb{R}}^{d}_{+},dx) under the norm 10:=0+(,)L2(+d,dx){\cal E}^{0}_{1}:={\cal E}^{0}+(\cdot,\cdot)_{L^{2}({\mathbb{R}}^{d}_{+},dx)}. Then (0,¯)({\cal E}^{0},{\overline{\cal F}}) is a regular Dirichlet form on L2(+d,dx)L^{2}({\mathbb{R}}^{d}_{+},dx). Let Y¯=(Y¯t,t0;x,x¯+d𝒩)\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 (0,¯)({\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 𝐞d=(0~,1)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 ¯+d{\overline{\mathbb{R}}}^{d}_{+} and has a jointly continuous heat kernel p¯:(0,)ׯ+dׯ+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,)ׯ+dׯ+d(t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+},

p¯(t,x,y)td/α[tJ(x+t1/α𝐞d,y+t1/α𝐞d)\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{β2α+β1}t2(xdydt1/α)(|xy|/4)|xy|/2J(x+t1/α𝐞d,x+r𝐞d)J(x+r𝐞d,y+t1/α𝐞d)rd1dr].\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 Bb1,b2,b3,b4B_{b_{1},b_{2},b_{3},b_{4}} explicitly by considering three cases separately:

(i) If β2<α+β1\beta_{2}<\alpha+\beta_{1}, then for all (t,x,y)(0,)ׯ+dׯ+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) (td/αtBβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d)|xy|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{)}
(td/αt|xy|d+α)Bβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d).\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 β2>α+β1\beta_{2}>\alpha+\beta_{1}, then for all (t,x,y)(0,)ׯ+dׯ+d(t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+},

p¯(t,x,y)(td/αt|xy|d+α)[Bβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d)\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})
+(1t|xy|α)Bβ1,β1,0,β3(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)+t1/α)|xy|)].\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 β2=α+β1\beta_{2}=\alpha+\beta_{1}, then for all (t,x,y)(0,)ׯ+dׯ+d(t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+},

p¯(t,x,y)(td/αt|xy|d+α)[Bβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d)\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})
+(1t|xy|α)Bβ1,β1,0,β3+β4+1(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)+t1/α)|xy|)].\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 xx or yy is close to the boundary, then J(x+t1/α𝐞d,y+t1/α𝐞d)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 β2<α+β1\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 β2>α+β1\beta_{2}>\alpha+\beta_{1}.

Let β2>α+β1\beta_{2}>\alpha+\beta_{1}. When |xy|>6t1/α|x-y|>6t^{1/\alpha}, the second term in the brackets in (1.4), i.e.,

(1t|xy|α)Bβ1,β1,0,β3(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)+t1/α)|xy|)\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|xy|d+αB(x+21|xy|𝐞d, 41|xy|)J(x+t1/α𝐞d,z)J(z,y+t1/α𝐞d)𝑑z,\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 β3=β4=0\beta_{3}=\beta_{4}=0 (and β2>α+β1\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 β1\beta_{1}, β2\beta_{2} and α\alpha only, the values of β3\beta_{3} and β4\beta_{4} do not play any role in this. To reduce technicalities, the reader, on a first reading, may assume β3=β4=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 β2=α+β1\beta_{2}=\alpha+\beta_{1}. We need the logarithmic terms in the assumptions to cover the important examples in [21, 29].

Refer to caption
(a) One jump regime
Refer to caption
(b) Two jumps regime
Figure 1. Dominant path from xx to yy (with xdydx_{d}\leq y_{d}) at time t<(|xy|/6)αt<(|x-y|/6)^{\alpha} when β2>α+β1\beta_{2}>\alpha+\beta_{1}, β3=β4=0\beta_{3}=\beta_{4}=0

.

Let 0{\cal F}^{0} be the closure of Cc(+d)C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) in L2(+d,dx)L^{2}({\mathbb{R}}^{d}_{+},dx) under the norm 10{\cal E}^{0}_{1}. Then (0,0)({\cal E}^{0},{\cal F}^{0}) is also a regular Dirichlet form on L2(+d,dx)L^{2}({\mathbb{R}}^{d}_{+},dx) and it is the part form of (0,¯)({\cal E}^{0},{\overline{\cal F}}) on +d{\mathbb{R}}^{d}_{+}. Let Y0=(Yt0,t0;x,x+d𝒩0)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 (0,0)({\cal E}^{0},{\cal F}^{0}) where 𝒩0{\cal N}_{0} is an exceptional set. Y0Y^{0} can be regarded as the part process of Y¯\overline{Y} killed upon exiting +d{\mathbb{R}}^{d}_{+}.

For κ(0,)\kappa\in(0,\infty), define

κ(u,v)\displaystyle{\cal E}^{\kappa}(u,v) :=0(u,v)++du(x)v(x)κxdα𝑑x,\displaystyle:={\cal E}^{0}(u,v)+\int_{{\mathbb{R}}^{d}_{+}}u(x)v(x)\kappa x_{d}^{-\alpha}dx,
κ\displaystyle{\cal F}^{\kappa} :=~0L2(+d,κxdαdx),\displaystyle:=\widetilde{\cal F}^{0}\cap L^{2}({\mathbb{R}}^{d}_{+},\kappa x_{d}^{-\alpha}dx),

where ~0\widetilde{\cal F}^{0} is the family of all 10{\cal E}^{0}_{1}-quasi-continuous functions in 0{\cal F}^{0}. Then (κ,κ)({\cal E}^{\kappa},{\cal F}^{\kappa}) is a regular Dirichlet form on L2(+d,dx)L^{2}({\mathbb{R}}^{d}_{+},dx) with Cc(+d)C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) as a special standard core, see [23, Theorems 6.1.1 and 6.1.2]. Let Yκ=(Ytκ,t0;x,x+d𝒩κ)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 (κ,κ)({\cal E}^{\kappa},{\cal F}^{\kappa}) where 𝒩κ{\cal N}_{\kappa} is an exceptional set. For κ[0,)\kappa\in[0,\infty), we denote by ζκ\zeta^{\kappa} the lifetime of YκY^{\kappa}. Define Ytκ=Y^{\kappa}_{t}=\partial for tζκt\geq\zeta^{\kappa}, where \partial is a cemetery point added to the state space +d{\mathbb{R}}^{d}_{+}.

We now associate with the constant κ\kappa a positive parameter qq which plays an important role in the paper. For q(1,α+β1)q\in(-1,\alpha+\beta_{1}), set

C(α,q,)={d11(|u~|2+1)(d+α)/201(sq1)(1sαq1)(1s)1+α(((1s)u~,1),s𝐞d)𝑑s𝑑u~if d2,01(sq1)(1sαq1)(1s)1+α(1,s)𝑑sif 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,)C(\alpha,q,{\cal B}) is well defined and finite for every q(1,α+β1)q\in(-1,\alpha+\beta_{1}), C(α,q,)=0C(\alpha,q,{\cal B})=0 if and only if q{0,α1}q\in\{0,\alpha-1\}, and limq1C(α,q,)=limqα+β1C(α,q,)=\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(sq1)(1sαq1)q\mapsto(s^{q}-1)(1-s^{\alpha-q-1}) is strictly decreasing on (1,(α1)/2)(-1,(\alpha-1)/2) and strictly increasing on ((α1)/2,α+β1)((\alpha-1)/2,\alpha+\beta_{1}). Thus, the shape of the map qC(α,q,)q\mapsto C(\alpha,q,{\cal B}) is given as follows.

qq 1-1 \cdots (α1)0(\alpha-1)\wedge 0 \cdots 12(α1)\frac{1}{2}(\alpha-1) \cdots (α1)0(\alpha-1)\vee 0 \cdots α+β1\alpha+\beta_{1}
C(α,q,)C(\alpha,q,{\cal B}) \infty \searrow 0 \searrow minimum0\text{minimum}\leq 0 \nearrow 0 \nearrow \infty

Consequently, for every κ0\kappa\geq 0, there exists a unique qκ[(α1)+,α+β1)q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) such that

κ=C(α,qκ,).\kappa=C(\alpha,q_{\kappa},{\cal B}). (1.8)
Theorem 1.2.

Suppose that (A1)(A4) and (1.8) hold with qκ[(α1)+,α+β1)q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}). Then the process YκY^{\kappa} can be refined to start from every point in +d{\mathbb{R}}^{d}_{+} and has a jointly continuous heat kernel pκ:(0,)×+d×+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,)×+d×+d(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}:

pκ(t,x,y)(1xdt1/α)qκ(1ydt1/α)qκp¯(t,x,y)x(ζκ>t)y(ζκ>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 α1\alpha\leq 1, the process Y¯\overline{Y} started from +d{\mathbb{R}}^{d}_{+} will never hit +d\partial{\mathbb{R}}^{d}_{+} and is equal to Y0Y^{0}. Thus for x,y+dx,y\in{\mathbb{R}}^{d}_{+} we have that p0(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 α1\alpha\leq 1.

Let

G¯(x,y)=0p¯(t,x,y)𝑑tandGκ(x,y)=0pκ(t,x,y)𝑑t.\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 Gκ(,)G^{\kappa}(\cdot,\cdot) is not identically infinite, it is called the Green function of YκY^{\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 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>αd>\alpha, then

G¯(x,y)1|xy|dα,x,y¯+d.\overline{G}(x,y)\asymp\frac{1}{|x-y|^{d-\alpha}},\quad x,y\in{\overline{\mathbb{R}}}_{+}^{d}. (1.11)

If dαd\leq\alpha, then G¯(x,y)=\overline{G}(x,y)=\infty for all x,y¯+dx,y\in{\overline{\mathbb{R}}}_{+}^{d}.

When d>(α+β1+β2)2d>(\alpha+\beta_{1}+\beta_{2})\wedge 2, sharp two-sided estimates on Gκ(x,y)G^{\kappa}(x,y) were obtained in [31, 32]. In the following theorem, we extend those results by removing the restriction on dd 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 d2d\geq 2.

Let

Hq(x,y)={1if q<α+12(β1+β2),logβ4+1(e+|xy|(xdyd)|xy|)if q=α+12(β1+β2),(xdyd|xy|1)2α+β1+β22qlogβ4(e+|xy|(xdyd)|xy|)if q>α+12(β1+β2).\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 qκ[(α1)+,α+β1)q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}). When α1\alpha\leq 1, suppose also that qκ>0q_{\kappa}>0 (or, equivalently, κ>0\kappa>0). Then GκG^{\kappa} has the following estimates:

(i) If d2d\geq 2, then for all x,y+dx,y\in{\mathbb{R}}^{d}_{+},

Gκ(x,y)\displaystyle G^{\kappa}(x,y) Hqκ(x,y)|xy|dα(xdyd|xy|1)qκ(xdyd|xy|1)qκ.\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=1d=1, then for all x,y+dx,y\in{\mathbb{R}}^{d}_{+},

Gκ(x,y){1|xy|1α(xy|xy|1)qκ if α<1,(xy|xy|1)qκlog(e+(xy)|xy||xy|) if α=1,(xy)α1(xy|xy|1)qκα+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 d2d\geq 2, at the threshold qκ=α+12(β1+β2)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 +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 0{\cal E}^{0} and thus prove a Nash-type inequality for 0{\cal E}^{0}. In case α1\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 YκY^{\kappa} and use this to remove the exceptional sets and extend p¯(t,x,y)\overline{p}(t,x,y) and pκ(t,x,y)p^{\kappa}(t,x,y) continuously to (0,)ׯ+dׯ+d(0,\infty)\times\overline{{\mathbb{R}}}^{d}_{+}\times\overline{{\mathbb{R}}}^{d}_{+}, respectively (0,)×+d×+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 pκ(t,x,y)ctJ(x,y)p^{\kappa}(t,x,y)\geq ctJ(x,y) for tt small compared to |xy||x-y|, xdx_{d} and ydy_{d}, see Proposition 4.5.

In Section 5 we prove the following preliminary upper bound:

pκ(t,x,y)C(1xdt1/α)qκ(1ydt1/α)qκ(td/αt|xy|d+α),t>0,x,y+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 qκ[(α1)+,α+β1)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 qκ<αq_{\kappa}<\alpha. The real challenge is to extend it to the full range of qκq_{\kappa} by covering the case qκαq_{\kappa}\geq\alpha. For this we use the upper bound on Green potentials of powers of the distance to the boundary. In case d2d\geq 2, such bound was proved in [31, 32], and here we prove the corresponding result for d=1d=1. The restriction qκ<αq_{\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 Ab1,b2,b3,b4(t,x,y)A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y), a space-time version of the function Bb1,b2,b3,b4(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 β2<α+β1\beta_{2}<\alpha+\beta_{1}. For the case β2α+β1\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 β2α+β1\beta_{2}\neq\alpha+\beta_{1}. The remaining case β2=α+β1\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 Aβ1,0,β3,0A_{\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 AA until reaching β1\beta_{1}, thus making the decay successively sharper. The sharp upper bound in case β2<α+β1\beta_{2}<\alpha+\beta_{1}, respectively β2α+β1\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 1.11.2.

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 pκ(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 β1\beta_{1}, β2\beta_{2}, β3\beta_{3}, β4\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 CC 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 ci,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 mdm_{d} the Lebesgue measure on d{\mathbb{R}}^{d}. For Borel subset DdD\subset{\mathbb{R}}^{d}, δD(x)\delta_{D}(x) denotes the distance between xx 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 YκY^{\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 κ0\kappa\geq 0, we denote by (P¯t)t0(\overline{P}_{t})_{t\geq 0} and (Ptκ)t0(P_{t}^{\kappa})_{t\geq 0} the semigroups corresponding to (0,¯)({\cal E}^{0},\overline{\cal F}) and (κ,κ)({\cal E}^{\kappa},{\cal F}^{\kappa}) respectively. (P¯t)t0(\overline{P}_{t})_{t\geq 0} and (Ptκ)t0(P_{t}^{\kappa})_{t\geq 0} define contraction semigroups on Lp(¯+d,dx)=Lp(+d,dx)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>0t>0 and p,q[1,]p,q\in[1,\infty], define

P¯tpq=sup{P¯tfLq(+d,dx):fLp(+d,dx),fLp(+d,dx)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:df:{\mathbb{R}}^{d}\to{\mathbb{R}} and r>0r>0, define f(r)(x)=f(rx)f^{(r)}(x)=f(rx). The following scaling property of (Ptκ)t0(P_{t}^{\kappa})_{t\geq 0} comes from [30, Lemma 5.1] and [32, Lemma 2.1]. By the same proof, (P¯t)t0(\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 κ0\kappa\geq 0. For any fLp(+d,dx)f\in L^{p}({\mathbb{R}}^{d}_{+},dx), t>0t>0 and r>0r>0, we have

P¯tf(x)=P¯rαtf(r)(x/r)andPtκf(x)=Prαtκf(r)(x/r)in Lp(+d,dx).\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¯t1=td/αP¯11for 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 ff,

limt01t+d(f(r)(z/r)Prαtκf(r)(z/r))f(r)(z/r)𝑑z\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
=rdαlimt01rαt+d(f(r)(x)Prαtκf(r)(x))f(r)(x)𝑑x\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
=rdα2+d+d(f(r)(x)f(r)(y))2(x,y)|xy|d+α𝑑x𝑑y+κrdα+df(r)(x)2xdα𝑑x\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
=r2d2+d+d(f(rx)f(ry))2(rx,ry)|rxry|d+α𝑑x𝑑y+κrd+df(rx)2(rxd)α𝑑x\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
=12+d+d(f(x)f(y))2(x,y)|xy|d+α𝑑x𝑑y++df(x)2κxdα𝑑x=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 (Ptκ)t0(P^{\kappa}_{t})_{t\geq 0}. By a similar proof, we also get the scaling property of (P¯t)t0(\overline{P}_{t})_{t\geq 0}.

Now, we have that, for every t>0t>0,

P¯t1=td/αsup{P¯tfL(+d,dx):fL1(+d,dx)td/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\}
=td/αsup{P¯1f(t1/α)L(+d,dx):fL1(+d,dx)1}=td/αP¯11,\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 d2d\geq 2. For f:df:{\mathbb{R}}^{d}\to{\mathbb{R}} and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1}, define fz(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)t0(\overline{P}_{t})_{t\geq 0} and (Ptκ)t0(P_{t}^{\kappa})_{t\geq 0}.

Lemma 2.2.

Let d2d\geq 2, p[1,]p\in[1,\infty] and κ0\kappa\geq 0. For any fLp(+d,dx)f\in L^{p}({\mathbb{R}}^{d}_{+},dx), t>0t>0 and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1}, we have

P¯tf(x)=P¯tfz(x(z~,0))andPtκf(x)=Ptκfz(x(z~,0))in Lp(+d,dx).\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 α1\alpha\leq 1, the process Y¯\overline{Y} started from +d{\mathbb{R}}^{d}_{+} will never hit +d\partial{\mathbb{R}}^{d}_{+} and is equal to Y0Y^{0}.

Lemma 2.3.

If α1\alpha\leq 1, then 0=¯{\cal F}^{0}={\overline{\cal F}}.

Proof. Define

𝒞~(u,v)\displaystyle\widetilde{\cal C}(u,v) :=12+d+d(u(x)u(y))(v(x)v(y))|xy|d+α𝑑y𝑑x,\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 Cc(¯+d) in L2(+d,dx) under 𝒞~+(,)L2(+d,dx).\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 ¯+d\overline{\mathbb{R}}^{d}_{+} in the sense of [6]. By (A3)(I), 0(u,u)C2𝒞~(u,u){\cal E}^{0}(u,u)\leq C_{2}\widetilde{\cal C}(u,u) for all uCc(¯+d)u\in C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+}) and hence 𝒟(𝒞~)¯{\cal D}(\widetilde{\cal C})\subset{\overline{\cal F}}. By [6, Theorem 2.5(i) and Remark 2.2(1)], since α1\alpha\leq 1, ¯+d+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 (0,¯)({\cal E}^{0},{\overline{\cal F}})-polar. Therefore, when starting from +d{\mathbb{R}}^{d}_{+}, Y¯\overline{Y} will never exit +d{\mathbb{R}}^{d}_{+}. Hence Y¯\overline{Y} and Y0Y^{0} are the same when they start from x+dx\in{\mathbb{R}}^{d}_{+} and thus 0=¯{\cal F}^{0}={\overline{\cal F}}. \Box

In order to prove the Nash type inequality, we first consider a Brownian motion on +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 (0,¯)({\cal E}^{0},{\overline{\cal F}}), we deduce the desired result.

For any γ0\gamma\geq 0, denote by IγI_{\gamma} the modified Bessel function of the first kind which is defined by

Iγ(r)=m=01m!Γ(γ+1+m)(r2)2m+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):=0ur1eu𝑑u\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)])

Iγ(r)(1r)γ+1/2r1/2er,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>0t>0 and x=(x1,,xd),y=(y1,,yd)+dx=(x_{1},\dots,x_{d}),y=(y_{1},\dots,y_{d})\in{\mathbb{R}}^{d}_{+},

qγ(t,x,y)=xdyd2tIγ(xdyd2t)exp(xd2+yd24t)i=1d1[14πtexp(|xiyi|24t)].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),

qγ(t,x,y)(1xdydt)γ+1/2td/2exp(|xy|24t).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], qγ(t,x,y)q^{\gamma}(t,x,y) is the transition density of the Feller process (killed Brownian motion with critical potential) Wγ=(Wtγ)t0W^{\gamma}=(W^{\gamma}_{t})_{t\geq 0} on +d{\mathbb{R}}^{d}_{+} associated with the following regular Dirichlet form (𝒬γ,𝒟(𝒬γ))({\cal Q}^{\gamma},{\cal D}({\cal Q}^{\gamma})):

𝒬γ(u,v)\displaystyle{\cal Q}^{\gamma}(u,v) :=+du(x)v(x)𝑑x+(γ214)+du(x)v(x)xd2𝑑x,\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(𝒬γ)\displaystyle{\cal D}({\cal Q}^{\gamma}) :=closure of Cc(+d) in L2(+d,dx) under 𝒬1γ=𝒬γ+(,)L2(+d,dx).\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=(St)t0S=(S_{t})_{t\geq 0} be an α/2\alpha/2-stable subordinator independent of WγW^{\gamma}, and let Xγ=(Xtγ)t0X^{\gamma}=(X^{\gamma}_{t})_{t\geq 0} be the subordinate process Xtγ:=WStγX^{\gamma}_{t}:=W^{\gamma}_{S_{t}}. Then XγX^{\gamma} is a Hunt process with no diffusion part. The transition density pγ(t,x,y)p^{\gamma}(t,x,y) of XtγX^{\gamma}_{t} exists and is given by

pγ(t,x,y)=0qγ(s,x,y)dds(Sts).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 Jγ(dx,dy)J^{\gamma}(dx,dy) and the killing measure κγ(dx)\kappa^{\gamma}(dx) of XγX^{\gamma} have densities Jγ(x,y)J^{\gamma}(x,y) and κγ(x)\kappa^{\gamma}(x) that are given by the following formulas (see, for instance, [34, (2.8)–(2.9)]):

Jγ(x,y)\displaystyle J^{\gamma}(x,y) =0qγ(t,x,y)να/2(t)𝑑t,κγ(x)=0(1+dqγ(t,x,y)𝑑y)να/2(t)𝑑t,\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 να/2(t)=α/2Γ(1α/2)t1α/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 SS.

Lemma 2.4.

(i) There exists a constant cγ,α>0c_{\gamma,\alpha}>0 such that κγ(x)=cγ,αxdα\kappa^{\gamma}(x)=c_{\gamma,\alpha}x_{d}^{-\alpha} for every x+dx\in{\mathbb{R}}^{d}_{+}.

(ii) It holds that for any x,y+dx,y\in{\mathbb{R}}^{d}_{+},

Jγ(x,y)\displaystyle J^{\gamma}(x,y) (1xd|xy|)γ+1/2(1yd|xy|)γ+1/21|xy|d+α=Bγ+1/2,γ+1/2,0,0(x,y)|xy|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>0C>0 such that

pγ(t,x,y)Ctd/α,t>0,x,y+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 WγW^{\gamma} satisfies the scaling property and horizontal translation invariance, namely, for any t>0t>0, x,y+dx,y\in{\mathbb{R}}^{d}_{+}, λ>0\lambda>0 and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1},

qγ(λ2t,λx,λy)=λdqγ(t,x,y) and qγ(t,x+(z~,0),y+(z~,0))=qγ(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+dx\in{\mathbb{R}}^{d}_{+}, using the change of variables z=(y(x~,0))/xdz=(y-(\widetilde{x},0))/x_{d} in the second equality below and s=xd2ts=x_{d}^{-2}t in the third, we get that

κγ(x)\displaystyle\kappa^{\gamma}(x) =α/2Γ(1α/2)0(1+dqγ(t,(0~,xd),(y~x~,yd))𝑑y)tα/21𝑑t\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/2Γ(1α/2)0(1xdd+dqγ(t,(0~,xd),xdz)𝑑z)tα/21𝑑t\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/2Γ(1α/2)0(1+dqγ(xd2t,(0~,1),z)𝑑z)tα/21𝑑t\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/2Γ(1α/2)xdα0(1+dqγ(s,(0~,1),z)𝑑z)sα/21𝑑s=κγ((0~,1))xdα.\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 𝐇𝐊𝐔𝐡\mathbf{HK_{U}^{h}} in [21] holds with C0=0C_{0}=0, Φ(r)=r2\Phi(r)=r^{2} and h(t,x,y)=(1(xdyd/t))γ+1/2h(t,x,y)=(1\wedge(x_{d}y_{d}/t))^{\gamma+1/2}. Note that the tail of the Lévy measure of SS is given by rνα/2(u)𝑑u=rα/2/Γ(1α/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+dx,y\in{\mathbb{R}}^{d}_{+},

Jγ(x,y)\displaystyle J^{\gamma}(x,y) (1xdyd|xy|2)γ+1/21|xy|d+α(1xd|xy|)γ+1/2(1yd|xy|)γ+1/21|xy|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 (1xd|xy|)(1yd|xy|)(1xdyd|xy|2)2(1xd|xy|)(1yd|xy|)(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+dx,y\in{\mathbb{R}}^{d}_{+}, which can be proved elementarily by using ydxd+|xy|y_{d}\leq x_{d}+|x-y|.

(iii) Since the conditions (Poly-\infty) and 𝐇𝐊𝐔𝐡\mathbf{HK_{U}^{h}} in [21] hold, the result follows from [21, Proposition 4.5(ii)]. \Box

Denote by (𝒞γ,𝒟(𝒞γ))({\cal C}^{\gamma},{\cal D}({\cal C}^{\gamma})) the regular Dirichlet form associated with the subordinate process XγX^{\gamma}. Then since XγX^{\gamma} has no diffusion part, we get from Lemma 2.4(i) that

𝒞γ(u,v)=+d+d(u(x)u(y))(v(x)v(y))Jγ(x,y)𝑑y𝑑x+cγ,α+du(x)v(x)xdα𝑑x.{\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 Cc(+d)𝒟(𝒞γ)C_{c}^{\infty}({\mathbb{R}}^{d}_{+})\subset{\cal D}({\cal C}^{\gamma}) since 𝒟(𝒬γ)𝒟(𝒞γ){\cal D}({\cal Q}^{\gamma})\subset{\cal D}({\cal C}^{\gamma}), see [34].

Lemma 2.5.

There exists a constant C>0C>0 such that

uL2(+d,dx)2(1+α/d)C𝒞γ(u,u)for every uCc(+d) with uL1(+d,dx)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>0C>0 such that

uL2(+d,dx)2(1+α/d)C0(u,u)for every u¯ with uL1(+d,dx)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=β1β2\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 uCc(+d)u\in C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) with uL1(+d,dx)1\lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1,

uL2(+d,dx)2(1+α/d)\displaystyle\lVert u\rVert_{L^{2}({\mathbb{R}}^{d}_{+},dx)}^{2(1+\alpha/d)} c1+d+d(u(x)u(y))2Bγ+1/2,γ+1/2,0,0(x,y)|xy|d+α𝑑y𝑑x+cγ,α+du(x)2xdα𝑑x\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
c2+d+d(u(x)u(y))2Bγ+1/2,γ+1/2,0,0(x,y)|xy|d+α𝑑y𝑑xc30(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 Bγ+1/2,γ+1/2,0,0B_{\gamma+1/2,\gamma+1/2,0,0} is defined in (1). By Lemma 2.3, ¯{\overline{\cal F}} is the closure of Cc(+d)C_{c}^{\infty}({\mathbb{R}}^{d}_{+}) under 10{\cal E}^{0}_{1}. Therefore, we conclude that (2.4) is true when α<1\alpha<1.

Now, we assume that α1\alpha\geq 1. Since (2.4) is valid when α<1\alpha<1 and Cc(¯+d)¯C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+})\subset{\overline{\cal F}}, we get that for every uCc(¯+d)u\in C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+}) with uL1(+d,dx)1\lVert u\rVert_{L^{1}({\mathbb{R}}^{d}_{+},dx)}\leq 1,

0(u,u)\displaystyle{\cal E}^{0}(u,u) 12+d+d(u(x)u(y))2(x,y)|xy|d+1/2𝟏{|xy|1}𝑑y𝑑x\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
=12+d+d(u(x)u(y))2(x,y)|xy|d+1/2(1𝟏{|xy|>1})𝑑y𝑑x\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
c4uL2(+d,dx)2(1+1/(2d))2C2uL2(+d,dx)2supx+d+d,|xy|>1dy|xy|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}}
c4uL2(+d,dx)2(1+1/(2d))c5uL2(+d,dx)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 ¯{\overline{\cal F}} is the closure of Cc(¯+d)C_{c}^{\infty}(\overline{\mathbb{R}}^{d}_{+}) under 10{\cal E}^{0}_{1}, [7, Theorem 2.1] yields that P¯t1c6t2dec5t\lVert\overline{P}_{t}\rVert_{1\to\infty}\leq c_{6}t^{-2d}e^{c_{5}t} for all t>0t>0. By (2.1), it follows that P¯t1=td/αP¯11c6ec5td/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>0t>0. Using [7, Theorem 2.1] again, we conclude that (2.4) holds for α1\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 pκ(t,x,y)p^{\kappa}(t,x,y) of Y¯\overline{Y} and YκY^{\kappa} respectively.

Proposition 2.7.

Let κ0\kappa\geq 0. The processes Y¯\overline{Y} and YκY^{\kappa} have heat kernels p¯(t,x,y)\overline{p}(t,x,y) and pκ(t,x,y)p^{\kappa}(t,x,y) defined on (0,)×(¯+d𝒩)×(¯+d𝒩)(0,\infty)\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}) and (0,)×(+d𝒩)×(+d𝒩)(0,\infty)\times({\mathbb{R}}^{d}_{+}\setminus{\cal N})\times({\mathbb{R}}^{d}_{+}\setminus{\cal N}) respectively, where 𝒩¯+d{\cal N}\subset\overline{\mathbb{R}}^{d}_{+} is a properly exceptional set for Y¯\overline{Y}. Moreover, there exists a constant C>0C>0 such that

pκ(t,x,y)p¯(t,x,y),x,y+d𝒩p^{\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(td/αt|xy|d+α),t>0,x,y¯+d𝒩.\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,)×(¯+d𝒩)×(¯+d𝒩)(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)ctd/α,t>0,x,y¯+d𝒩.\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 (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), YκY^{\kappa} can be realized as a subprocess of Y¯\overline{Y}. Thus, YκY^{\kappa} has a heat kernel pκ(t,x,y)p^{\kappa}(t,x,y) defined on (0,)×(+d𝒩)×(+d𝒩)(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 pκ(t,x,y)p^{\kappa}(t,x,y) to (0,)×(¯+d𝒩)×(¯+d𝒩)(0,\infty)\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N})\times(\overline{\mathbb{R}}^{d}_{+}\setminus{\cal N}) by letting pκ(t,x,y)=0p^{\kappa}(t,x,y)=0 if x+d𝒩x\in\partial{\mathbb{R}}^{d}_{+}\setminus{\cal N} or y+d𝒩y\in\partial{\mathbb{R}}^{d}_{+}\setminus{\cal N}.

3. Parabolic Hölder regularity and consequences

For κ0\kappa\geq 0 and an open set D¯+dD\subset{\overline{\mathbb{R}}}^{d}_{+} relative to the topology on ¯+d{\overline{\mathbb{R}}}^{d}_{+}, we denote by Y¯D\overline{Y}^{D} and P¯tD\overline{P}^{D}_{t} the part of the process Y¯\overline{Y} killed upon exiting DD and its semigroup, and by Yκ,DY^{\kappa,D} and Ptκ,DP_{t}^{\kappa,D} the part of the process YκY^{\kappa} killed upon exiting D+dD\cap{\mathbb{R}}^{d}_{+} and its semigroup, respectively. The Dirichlet forms of Y¯D\overline{Y}^{D} and Yκ,DY^{\kappa,D} are (0,¯D)({\cal E}^{0},\overline{{\cal F}}^{D}) and (κ,κ,D)({\cal E}^{\kappa},{\cal F}^{\kappa,D}), where ¯D={u¯:u=0 q.e. on ¯+dD}\overline{{\cal F}}^{D}=\{u\in\overline{{\cal F}}:u=0\mbox{ q.\/e. on }{\overline{\mathbb{R}}}^{d}_{+}\setminus D\} and κ,D={uκ:u=0 q.e. on +dD}{\cal F}^{\kappa,D}=\{u\in{\cal F}^{\kappa}:u=0\mbox{ q.\/e. on }{\mathbb{R}}^{d}_{+}\setminus D\} respectively. For u,vκ,Du,v\in{\cal F}^{\kappa,D},

κ(u,v)=12DD((u(x)u(y))(v(x)v(y))J(x,y)dydx+Du(x)v(x)κD(x)dx,{\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

κD(x)=+dDJ(x,y)𝑑y+κxdα.\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¯tD}\overline{\tau}_{D}=\inf\{t>0:\overline{Y}_{t}\notin D\} and τDκ=inf{t>0:YtκD+d}\tau^{\kappa}_{D}=\inf\{t>0:Y^{\kappa}_{t}\notin D\cap{\mathbb{R}}^{d}_{+}\}. For x,y𝒩x,y\notin\mathcal{N}, let

p¯D(t,x,y)\displaystyle\overline{p}^{D}(t,x,y) =p¯(t,x,y)𝔼x[p¯(tτ¯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{]},
pκ,D(t,x,y)\displaystyle p^{\kappa,D}(t,x,y) =pκ(t,x,y)𝔼x[pκ(tτDκ,YτDκκ,y);τDκ<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 pκ,D(t,x,y)p^{\kappa,D}(t,x,y) are the transition densities of Y¯D\overline{Y}^{D} and Yκ,DY^{\kappa,D} respectively.

In case when DD is a relatively compact open subset of +d{\mathbb{R}}^{d}_{+}, one can show that the process Yκ,DY^{\kappa,D} can start from every point in DD and provide an interior lower bound for its transition density. We accomplish this by identifying the semigroup Ptκ,DP^{\kappa,D}_{t} with the Feynman-Kac semigroup of the part process on DD 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 mdm_{d} the Lebesgue measure on d{\mathbb{R}}^{d}.

Proposition 3.1.

If DD is a relatively compact open subset of +d{\mathbb{R}}^{d}_{+}, then Yκ,DY^{\kappa,D} has a transition density pκ,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>0T>0 and b(0,1]b\in(0,1], there exists a constant C=C(b,T,D)>0C=C(b,T,D)>0 such that

pκ,D(t,x,y)C(td/αt|xy|d+α)t(0,T),x,yD such that δD(x)δD(y)>bt1/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 Jγ:d×d[0,]J_{\gamma}:{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}\to[0,\infty] be defined by Jγ(x,y)=J(x,y)J_{\gamma}(x,y)=J(x,y) if x,yDx,y\in D, and Jγ(x,y)=γ|xy|αdJ_{\gamma}(x,y)=\gamma|x-y|^{-\alpha-d} otherwise. It follows from (A3)(I) and the relative compactness of DD that Jγ(x,y)|xy|αdJ_{\gamma}(x,y)\asymp|x-y|^{-\alpha-d}. Hence by [16, Theorem 1.2], there exists a Feller and strongly Feller process ZZ (that can start from every point in d){\mathbb{R}}^{d}) with a continuous transition density q~(t,x,y)\widetilde{q}(t,x,y) on (0,)×d×d(0,\infty)\times{\mathbb{R}}^{d}\times{\mathbb{R}}^{d} satisfying

c11(td/αt|xy|d+α)q~(t,x,y)c1(td/αt|xy|d+α),t>0,x,yd,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 c11c_{1}\geq 1.

Denote the part of the process ZZ killed upon exiting DD by ZDZ^{D}. The Dirichlet form of ZDZ^{D} is (𝒞,𝒟D(𝒞))({\cal C},\mathcal{D}_{D}({\cal C})), where for u,v𝒟D(𝒞)u,v\in{\cal D}_{D}({\cal C}),

𝒞(u,v)\displaystyle{\cal C}(u,v) =\displaystyle= 12DD(u(x)u(y))(v(x)v(y))Jγ(x,y)𝑑y𝑑x+Du(x)v(x)κDZ(x)𝑑x\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= 12DD(u(x)u(y))(v(x)v(y))J(x,y)𝑑y𝑑x+Du(x)v(x)κDZ(x)𝑑x\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

κDZ(x)=dDJγ(x,y)𝑑y=γdD|xy|dα𝑑y,xD,\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 𝒟D(𝒞)={u𝒟(𝒞):u=0 q.e. on dD}\mathcal{D}_{D}({\cal C})=\{u\in\mathcal{D}({\cal C}):\,u=0\textrm{ q.e.~{}on }{\mathbb{R}}^{d}\setminus D\}.

Let δD=dist(D,+d)\delta_{D}=\mathrm{dist}(D,\partial{\mathbb{R}}^{d}_{+}) and let VV be the δD/2\delta_{D}/2-neighborhood of DD, that is V:={x+d:dist(x,D)<δD/2}V:=\{x\in{\mathbb{R}}^{d}_{+}:\,\mathrm{dist}(x,D)<\delta_{D}/2\}. Then

κD(x)=+dVJ(x,y)𝑑y+VDJ(x,y)𝑑y+κxdα.\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 DD that c2|xy|dαJ(x,y)C2|xy|dαc_{2}|x-y|^{-d-\alpha}\leq J(x,y)\leq C_{2}|x-y|^{-d-\alpha} for all x,yVx,y\in V with c2:=c2(D)>0c_{2}:=c_{2}(D)>0. It is easy to see that supxD+dVJ(x,y)dy=:c3<\sup_{x\in D}\int_{{\mathbb{R}}^{d}_{+}\setminus V}J(x,y)\,dy=:c_{3}\ <\infty. Therefore

c2VD|xy|dα𝑑yκD(x)c3+C2VD|xy|dα𝑑y+c4,xU,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 c4:=κsupxDxdαc_{4}:=\kappa\sup_{x\in D}x_{d}^{-\alpha}. Since

infxDVD|xy|dαdymd(VD)diam(V)dα=:c5>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

c2VD|xy|dα𝑑yκD(x)c6VD|xy|dα𝑑y,xD.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

κDZ(x)=γ(dV|xy|dα𝑑y+VD|xy|dα𝑑y),xD\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 supxDdV|xy|dαdy=:c7<\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 c8>0c_{8}>0 such

VD|xy|dα𝑑yγ1κDZ(x)c8VD|xy|dα𝑑y,xD,.\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 c61κD(x)γ1κDZ(x)c8c21κD(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 xDx\in D with positive constants c2,c6,c8c_{2},c_{6},c_{8} not depending on γ\gamma. Now we choose γ>0\gamma>0 so small that γc8c211\gamma c_{8}c_{2}^{-1}\leq 1. With this choice we get that κDZ(x)κD(x)\kappa_{D}^{Z}(x)\leq\kappa_{D}(x) for all xDx\in D. In particular, with c9:=γc61c_{9}:=\gamma c_{6}^{-1} we see that

c9κD(x)κDZ(x)κD(x),xD.c_{9}\kappa_{D}(x)\leq\kappa_{D}^{Z}(x)\leq\kappa_{D}(x)\,,\qquad x\in D\,. (3.6)

It follows that for uCc(D)u\in C_{c}^{\infty}(D),

1κ,D(u,u)\displaystyle{\cal E}^{\kappa,D}_{1}(u,u) =\displaystyle= κ,D(u,u)+Uu(x)2𝑑x\displaystyle{\cal E}^{\kappa,D}(u,u)+\int_{U}u(x)^{2}\,dx
=\displaystyle= 12DD(u(x)u(y))2J(x,y)𝑑y𝑑x+Du(x)2κD(x)𝑑x+Uu(x)2𝑑x\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 12DD(u(x)u(y))2Jγ(x,y)𝑑y𝑑x+Du(x)2κDZ(x)𝑑x+Du(x)2𝑑x\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)+Du(x)2𝑑x=𝒞1(u,u).\displaystyle{\cal C}(u,u)+\int_{D}u(x)^{2}dx={\cal C}_{1}(u,u)\,.

Since Cc(D)C_{c}^{\infty}(D) is a core of both (κ,κ,D)({\cal E}^{\kappa},{\cal F}^{\kappa,D}) and (𝒞,𝒟D(𝒞))({\cal C},\mathcal{D}_{D}({\cal C})), we conclude that κ,D=𝒟D(𝒞){\cal F}^{\kappa,D}=\mathcal{D}_{D}({\cal C}).

Define κ~:D\widetilde{\kappa}:D\to{\mathbb{R}} by κ~(x):=κD(x)κDZ(x)\widetilde{\kappa}(x):=\kappa_{D}(x)-\kappa_{D}^{Z}(x), xD.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 c10>0c_{10}>0 such that

κ~(x)κD(x)c10δD(x)α,xD.\widetilde{\kappa}(x)\leq\kappa_{D}(x)\leq c_{10}\delta_{D}(x)^{-\alpha}\,,\quad x\in D. (3.7)

Let μ(dx)=κ~(x)dx\mu(dx)=\widetilde{\kappa}(x)\,dx be a measure on DD. Using (3.4) and (3.7) one can check that μ𝐊1(D)\mu\in{\bf K}_{1}(D), where the class 𝐊1(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

TtD,κ~f(x)=𝔼x[exp(0tκ~(ZsD)𝑑s)f(ZtD)],t>0,xD.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 ZDZ^{D}. By [20, Theorem 2.15], the semigroup (TtD,κ~)t0(T^{D,\widetilde{\kappa}}_{t})_{t\geq 0} has a transition density pZ,D(t,x,y)p^{Z,D}(t,x,y) (with respect to the Lebesgue measure) such that for every T>0T>0 and b(0,1]b\in(0,1] there exists a constant c2=c2(b,T,D)>0c_{2}=c_{2}(b,T,D)>0 such that

pZ,D(t,x,y)c2q~(t,x,y),t(0,T),x,yD such that δD(x)δD(y)>bt1/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 Yκ,DY^{\kappa,D} and ZDZ^{D} (for the latter use [23, Theorem 6.1.2]), we conclude that they coincide. This implies that Ptκ.D=TtD,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 𝐞d=(0~,1){\mathbf{e}}_{d}=(\widetilde{0},1). Define for a(0,1]a\in(0,1],

S(a)={(z~,zd)+d:|z~|<2/a,a/(2a+2)<zd<(2a+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(𝐞d,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], xB(𝐞d,1/8)x\in B({\mathbf{e}}_{d},1/8) and y=(y~,yd)+dy=(\widetilde{y},y_{d})\in{\mathbb{R}}^{d}_{+} with xdyd>a|xy|x_{d}\wedge y_{d}>a|x-y|, we have |y~||x~|+|yx|<1/8+xd/a<2/a|\widetilde{y}|\leq|\widetilde{x}|+|y-x|<1/8+x_{d}/a<2/a, yd+yd/axd|xy|+|xy|7/8y_{d}+y_{d}/a\geq x_{d}-|x-y|+|x-y|\geq 7/8 and ydxd+|xy|xd+xd/a<(2a+2)/ay_{d}\leq x_{d}+|x-y|\leq x_{d}+x_{d}/a<(2a+2)/a. Thus,

{(t,(x~,xd),(y~,yd))+1×B(𝐞d,1/8)×+d:xdyd>a(t1/α|xy|)}\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/aα)×S(a)×S(a)for everya(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)>0C=C(a)>0 such that the following estimates hold: For any t>0t>0 and x+d𝒩x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}, there is a measurable set Nt,x+dN_{t,x}\subset{\mathbb{R}}^{d}_{+} of zero Lebesgue measure such that for all y+dNt,xy\in{\mathbb{R}}^{d}_{+}\setminus N_{t,x} with xdyd>a(t1/α|xy|)x_{d}\wedge y_{d}>a(t^{1/\alpha}\vee|x-y|),

p¯(t,x,y)pκ(t,x,y)C(td/αt|xy|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>0r>0 and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1} such that r1(x(z~,0))B(𝐞d,1/8)𝒩r^{-1}(x-(\widetilde{z},0))\in B({\mathbf{e}}_{d},1/8)\setminus{\cal N}. By Lemmas 2.1 and 2.2, we have

pκ(t,x,y)=rdpκ(rαt,r1(x(z~,0)),r1(y(z~,0)))for a.e.y+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)>0c=c(a)>0 such that for all 0<t<2/aα0<t<2/a^{\alpha}, xB(𝐞d,1/8)𝒩x\in B({\mathbf{e}}_{d},1/8)\setminus\mathcal{N} and a.e. yS(a)𝒩y\in S(a)\setminus{\cal N},

pκ,S(a/2)(t,x,y)c(td/αt|xy|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 δS(a/2)(y)a/(a2+3a+2)\delta_{S(a/2)}(y)\geq a/(a^{2}+3a+2) for every yS(a)y\in S(a). Hence for all 0<t<2/aα0<t<2/a^{\alpha} and z,yS(a/2)z,y\in S(a/2) it holds that δS(a/2)(z)δS(a/2)(y)(a2/(2a2+6a+4))t1/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=(a2/(2a2+6a+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>0C>0 such that

p¯(t,x,y)pκ(t,x,y)Ctd/α\overline{p}(t,x,y)\geq p^{\kappa}(t,x,y)\geq Ct^{-d/\alpha} (3.14)

for all t>0t>0, x+d𝒩x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} and a.e. y+d𝒩y\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} with xdyd>at1/αx_{d}\wedge y_{d}>at^{1/\alpha} and |xy|t1/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>0C>0 and η(0,1/4)\eta\in(0,1/4) such that for all x+dx\in{\mathbb{R}}^{d}_{+}, r(0,xd)r\in(0,x_{d}), t(0,(ηr)α]t\in(0,(\eta r)^{\alpha}] and zB(x,ηt1/α)𝒩z\in B(x,\eta t^{1/\alpha})\setminus{\cal N},

p¯B(x,r)(t,z,y)pκ,B(x,r)(t,z,y)Ctd/αfor a.e. yB(x,ηt1/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>1C>1 such that for all x+d𝒩x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N} and r(0,xd)r\in(0,x_{d}),

C1rα𝔼x[τB(x,r)κ]supzB(x,r)𝒩𝔼z[τ¯B(x,r)]Crα.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 FF on +d×+d{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} vanishing on the diagonal and any stopping time TT for YκY^{\kappa}, it holds that

𝔼xsTF(Ysκ,Ysκ)=𝔼x0T+dF(Ysκ,y)J(Ysκ,y)𝑑y𝑑s,x+d𝒩.\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 Ysκ=limtsYtκY^{\kappa}_{s-}=\lim_{t\uparrow s}Y^{\kappa}_{t} denotes the left limit of the process YY at time s>0s>0. Similarly, for any non-negative Borel function FF on ¯+dׯ+d{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} vanishing on the diagonal and any stopping time TT for Y¯\overline{Y}, it holds that

𝔼xsTF(Y¯s,Y¯s)=𝔼x0T+dF(Y¯s,y)J(Y¯s,y)𝑑y𝑑s,x¯+d𝒩.\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~,xd)¯+dx=(\widetilde{x},x_{d})\in{\overline{\mathbb{R}}}^{d}_{+} and t>0t>0, we define

Vx(t)\displaystyle V_{x}(t) ={z=(z~,zd)d:|z~x~|<2t1/α,zd[0,2t1/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})\},
Wx(t)\displaystyle W_{x}(t) ={z=(z~,zd)d:|z~x~|<2t1/α,zd[xd+5t1/α,xd+8t1/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 Vx(t)=[0,2t1/α)V_{x}(t)=[0,2t^{1/\alpha}) and Wx(t)=[x+5t1/α,x+8t1/α)W_{x}(t)=[x+5t^{1/\alpha},x+8t^{1/\alpha}).

Lemma 3.5.

(i) There exists C>0C>0 such that for all t>0t>0 and x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N},

x(Y¯τ¯Vx(t)Wx(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>0C>0 such that for all n1n\geq 1, t>0t>0 and x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N},

x(τ¯Vx(t)>nt)2eCn.{\mathbb{P}}_{x}(\overline{\tau}_{V_{x}(t)}>nt)\leq 2e^{-Cn}.

Proof. (i) Define Vx(t,r)={zVx(t):δVx(t)(z)>rt1/α}V_{x}(t,r)=\{z\in V_{x}(t):\delta_{V_{x}(t)}(z)>rt^{1/\alpha}\} for r>0r>0. For any r(0,1)r\in(0,1), zVx(t,r)z\in V_{x}(t,r), uB(z,rt1/α/2)u\in B(z,rt^{1/\alpha}/2) and wWx(t)w\in W_{x}(t), we have

rt1/α/2ud5t1/αwdand|uw||uz|+|zx|+|xw|<16t1/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 zVx(t,r)𝒩z\in V_{x}(t,r)\setminus{\cal N},

z(Y¯τ¯Vx(t)Wx(t))\displaystyle{\mathbb{P}}_{z}(\overline{Y}_{\overline{\tau}_{V_{x}(t)}}\in W_{x}(t)) z(Y¯τ¯B(z,rt1/α/2)Wx(t))=𝔼z[0τ¯B(z,rt1/α/2)Wx(t)J(Y¯s,w)𝑑w𝑑s]\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{]}
c1𝔼z[0τ¯B(z,rt1/α/2)Wx(t)(r/216)β1(516)β2(16t1/α)dα𝑑w𝑑s]\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{]}
c2rβ1t1d/α𝔼z[τ¯B(z,rt1/α/2)]Wx(t)𝑑wc3rβ1+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)=s1/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 ε0>0{\varepsilon}_{0}>0 independent of tt and xx such that

x(τ¯Vx(t)>ε0t)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>0r>0, we get from Proposition 2.7 that

x(τ¯Vx(t)Vx(t,r)>ε0t)x(Y¯ε0tVx(t)Vx(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))
Vx(t)Vx(t,r)p(ε0t,x,y)𝑑yc4td/αVx(t)Vx(t,r)𝑑yc5r.\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 r0:=1/(4c5+1)r_{0}:=1/(4c_{5}+1). Using the strong Markov property and (3)–(3), we obtain

x(Y¯τ¯Vx(t)Wx(t))x(Y¯τ¯Vx(t)Wx(t),τ¯Vx(t)Vx(t,r0)ε0t<τ¯Vx(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{)}
x(Y¯τ¯Vx(t)Vx(t,r0)(Y¯τ¯Vx(t)Wx(t)):τ¯Vx(t)Vx(t,r0)ε0t<τ¯Vx(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{)}
c3r0β1+αx(τ¯Vx(t)Vx(t,r0)ε0t<τ¯Vx(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)
c3r0β1+α(x(τ¯Vx(t)>ε0t)x(τ¯Vx(t)Vx(t,r0)>ε0t))c3r0β1+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 k0>0k_{0}>0 independent of tt and xx such that for any zVx(t)𝒩z\in V_{x}(t)\setminus{\cal N},

z(τ¯Vx(t)>k0t)z(Y¯k0tVx(t))Vx(t)p¯(k0t,z,y)𝑑yc1td/α(k0t)d/α12.\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>0r>0, let r:=sup{m:mr}\lfloor r\rfloor:=\sup\{m\in{\mathbb{N}}:m\leq r\}. Now for any n1n\geq 1, using the Markov property and (3.23), we get that

x(τ¯Vx(t)>nt)x(τ¯Vx(t)>n/k0k0t)\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)
x(X(n/k01)k0t(τ¯Vx(t)>k0t):τ¯Vx(t)>(n/k01)k0t)\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{)}
21x(τ¯Vx(t)>(n/k01)k0t)2n/k02e(log2)n/k0.\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>1M>1 and C>0C>0 such that for all t>0t>0 and x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N},

essinfzWx(t)p¯(Mt,x,z)Ctd/α.\operatorname*{ess\,inf}_{z\in W_{x}(t)}\,\overline{p}(Mt,x,z)\geq Ct^{-d/\alpha}.

Proof. Suppose t>0t>0 and x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N}. For all w,zWx(t)w,z\in W_{x}(t), we have wdzd5t1/αw_{d}\wedge z_{d}\geq 5t^{1/\alpha} and |wz|<7t1/α|w-z|<7t^{1/\alpha}. Therefore, by Proposition 3.2, for any M>1M>1, there is c1=c1(M)>0c_{1}=c_{1}(M)>0 independent of tt and xx such that for all wWx(t)𝒩w\in W_{x}(t)\setminus{\cal N}, zWx(t)z\in W_{x}(t) and ε(0,t1/α){\varepsilon}\in(0,t^{1/\alpha}),

inftsMtw(Y¯sB(z,ε))=inftsMtB(z,ε)p¯(s,w,y)𝑑yc1td/αmd(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>1M>1, zWx(t)z\in W_{x}(t) and ε(0,t1/α){\varepsilon}\in(0,t^{1/\alpha}),

x(Y¯MtB(z,ε))𝔼x[Y¯τ¯Vx(t)(Y¯Mtτ¯Vx(t)B(z,ε)):τ¯Vx(t)(M1)t,Y¯τ¯Vx(t)Wx(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]
(inftsMtinfwWx(t)𝒩w(Y¯sB(z,ε)))x(τ¯Vx(t)(M1)t,Y¯τ¯Vx(t)Wx(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)
c1td/αmd(B(z,ε))(x(Y¯τ¯Vx(t)Wx(t))x(τ¯Vx(t)>(M1)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 c2,c3,c4>0c_{2},c_{3},c_{4}>0 such that for all M>1M>1,

x(Y¯τ¯Vx(t)Wx(t))x(τ¯Vx(t)>(M1)t)c2c3ec4M.\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=c41log(2c3/c2)+1M=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>0C>0 such that for all t>0t>0 and x,y¯+d𝒩x,y\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} with |xy|t1/α|x-y|\leq t^{1/\alpha},

p¯(t,x,y)Ctd/α.\displaystyle\overline{p}(t,x,y)\geq Ct^{-d/\alpha}.

Proof. Let M>1M>1 be the constant in Lemma 3.6. Note that for all (z,w)Wx(t/M)×Wy(t/M)(z,w)\in W_{x}(t/M)\times W_{y}(t/M), we have zdwd5(t/M)1/αz_{d}\wedge w_{d}\geq 5(t/M)^{1/\alpha} and |zw||zx|+|xy|+|yw|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 c1>0c_{1}>0 independent of t,x,yt,x,y such that

essinfzWx(t/(3M))essinfwWy(t/(3M))p¯(t/3,z,w)c1td/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) Wx(t/(3M))Wy(t/(3M))p¯(t/3,x,z)p¯(t/3,z,w)p¯(t/3,w,y)𝑑z𝑑w\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
(essinfzWx(t/(3M))p¯(t/3,x,z))(essinfwWy(t/(3M))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)
×(essinfzWx(t/(3M))essinfwWy(t/(3M))p¯(t/3,z,w))Wx(t/(3M))Wy(t/(3M))𝑑z𝑑w\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
c2t3d/α+2d/α=c2td/α.\displaystyle\geq c_{2}t^{-3d/\alpha+2d/\alpha}=c_{2}t^{-d/\alpha}.

The proof is complete. \Box

For x¯+dx\in{\overline{\mathbb{R}}}^{d}_{+} and r>0r>0, we denote B+(x,r):=B(x,r)¯+d{B_{+}}(x,r):=B(x,r)\cap{\overline{\mathbb{R}}}^{d}_{+}. We observe that

md(B+(x,r))rdfor allx¯+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 xx and rr.

Lemma 3.8.

There exist constants C>0C>0 and η(0,1/4)\eta\in(0,1/4) such that for all x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N}, r>0r>0, t(0,(ηr)α]t\in(0,(\eta r)^{\alpha}] and zB+(x,ηt1/α)𝒩z\in{B_{+}}(x,\eta t^{1/\alpha})\setminus{\cal N},

p¯B+(x,r)(t,z,y)Ctd/αfor a.e. yB+(x,ηt1/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>1C>1 such that for all x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N} and r>0r>0,

C1rα𝔼x[τ¯B+(x,r)]supzB+(x,r)𝒩𝔼z[τ¯B+(x,r)]Crα.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¯:=(Ts,Y¯s)s0\overline{X}:=(T_{s},\overline{Y}_{s})_{s\geq 0} and Xκ:=(Ts,Ysκ)s0X^{\kappa}:=(T_{s},Y^{\kappa}_{s})_{s\geq 0} be time-space processes where Ts=T0sT_{s}=T_{0}-s. The law of the time-space process sX¯ss\mapsto\overline{X}_{s} or sXsκs\mapsto X^{\kappa}_{s} starting from (t,x)(t,x) will be denoted by (t,x){\mathbb{P}}_{(t,x)}. For every open subset UU of [0,)×d[0,\infty)\times{\mathbb{R}}^{d}, define τ¯U=inf{s>0:X¯sU}\overline{\tau}_{U}=\inf\{s>0:\,\overline{X}_{s}\notin U\} and τUκ=inf{s>0:XsκU}\tau^{\kappa}_{U}=\inf\{s>0:\,X_{s}^{\kappa}\notin U\}. We also define σ¯U=inf{s>0:X¯sU}\overline{\sigma}_{U}=\inf\{s>0:\,\overline{X}_{s}\in U\} and σUκ=inf{s>0:XsκU}\sigma^{\kappa}_{U}=\inf\{s>0:\,X_{s}^{\kappa}\in U\}.

A Borel function u:[0,)ׯ+du:[0,\infty)\times\overline{\mathbb{R}}^{d}_{+}\to{\mathbb{R}} is said to be parabolic in (a,b]×B+(x,r)(0,)ׯ+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,)ׯ+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 z𝒩z\notin{\cal N}. Similarly, a Borel function u:[0,)×+du:[0,\infty)\times{\mathbb{R}}^{d}_{+}\to{\mathbb{R}} is said to be parabolic in (a,b]×B(x,r)(0,)×+d(a,b]\times B(x,r)\subset(0,\infty)\times{\mathbb{R}}^{d}_{+} with respect to YκY^{\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(XτUκκ)u(t,z)={\mathbb{E}}_{(t,z)}u(X^{\kappa}_{\tau^{\kappa}_{U}}) for all (t,z)U(t,z)\in U with z𝒩z\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>0C>0 such that for all x+d𝒩x\in{\mathbb{R}}^{d}_{+}\setminus{\cal N}, r(0,xd)r\in(0,x_{d}), tδrαt\geq\delta r^{\alpha}, and any compact set A[tδrα,tδrα/2]×B(x,(ηδ/2)1/αr)A\subset[t-\delta r^{\alpha},t-\delta r^{\alpha}/2]\times B(x,(\eta\delta/2)^{1/\alpha}r),

(t,x)(σAκ<τ[tδrα,t]×B(x,r)κ)Cmd+1(A)rd+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>0C>0 such that for all x¯+d𝒩x\in{\overline{\mathbb{R}}}^{d}_{+}\setminus{\cal N}, r>0r>0, tδrαt\geq\delta r^{\alpha}, and any compact set A[tδrα,tδrα/2]×B+(x,(ηδ/2)1/αr)A\subset[t-\delta r^{\alpha},t-\delta r^{\alpha}/2]\times{B_{+}}(x,(\eta\delta/2)^{1/\alpha}r),

(t,x)(σ¯A<τ¯[tδrα,t]×B+(x,r))Cmd+1(A)rd+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>0C>0 such that for all x+dx\in{\mathbb{R}}^{d}_{+}, r(0,xd)r\in(0,x_{d}), t00t_{0}\geq 0, and any function uu on (0,)×+d(0,\infty)\times{\mathbb{R}}^{d}_{+} which is parabolic in (t0,t0+rα]×B(x,r)(t_{0},t_{0}+r^{\alpha}]\times B(x,r) with respect to YκY^{\kappa} and bounded in (t0,t0+rα]×+d(t_{0},t_{0}+r^{\alpha}]\times{\mathbb{R}}^{d}_{+}, we have

|u(s,y)u(t,z)|C(|st|1/α+|yz|r)λesssup[t0,t0+rα]×+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(t0+(1δα)rα,t0+rα]s,t\in(t_{0}+(1-\delta^{\alpha})r^{\alpha},t_{0}+r^{\alpha}] and y,zB(x,δr)𝒩y,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>0C>0 such that for all x¯+dx\in{\overline{\mathbb{R}}}^{d}_{+}, r>0r>0, t00t_{0}\geq 0, and any function uu on (0,)ׯ+d(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+} which is parabolic in (t0,t0+rα]×B+(x,r)(t_{0},t_{0}+r^{\alpha}]\times{B_{+}}(x,r) with respect to Y¯\overline{Y} and bounded in (t0,t0+rα]ׯ+d(t_{0},t_{0}+r^{\alpha}]\times\overline{\mathbb{R}}^{d}_{+}, (3.29) holds true for every s,t(t0+(1δα)rα,t0+rα]s,t\in(t_{0}+(1-\delta^{\alpha})r^{\alpha},t_{0}+r^{\alpha}] and y,zB+(x,δr)𝒩y,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

Remark 3.12.

By Theorem 3.11, since the heat kernels p¯(t,x,y)\overline{p}(t,x,y) and pκ(t,x,y)p^{\kappa}(t,x,y) are parabolic with respect to Y¯\overline{Y} and YκY^{\kappa} respectively, they can be extended continuously to (0,)ׯ+dׯ+d(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} and (0,)×+d×+d(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+} respectively. As consequences, by Proposition 2.7, Y¯\overline{Y} and YκY^{\kappa} can be refined to be a strongly Feller processes starting from every point in ¯+d{\overline{\mathbb{R}}}^{d}_{+} and +d{\mathbb{R}}^{d}_{+} respectively, and the exceptional set 𝒩{\cal N} in Proposition 2.7 can be taken to be the empty set.

In the remainder of this paper, we take the jointly continuous version of p¯:(0,)ׯ+dׯ+d[0,)\overline{p}:(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+}\to[0,\infty) and pκ:(0,)×+d×+d[0,)p^{\kappa}:(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}\to[0,\infty), take the exceptional set 𝒩{\cal N} in Proposition 2.7 to be empty set, and replace the essinf{\rm essinf} in Lemma 3.6 by inf\inf. Again, for notational convenience, we extend the domain of pκ(t,x,y)p^{\kappa}(t,x,y) to (0,)×(0,)ׯ+dׯ+d(0,\infty)\times(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+} by letting pκ(t,x,y)=0p^{\kappa}(t,x,y)=0 if x+dx\in\partial{\mathbb{R}}^{d}_{+} or y+dy\in\partial{\mathbb{R}}^{d}_{+}.

The following scaling and horizontal translation invariance properties of the heat kernels (the latter for d2d\geq 2), which come from Lemmas 2.1 and 2.2, will be used throughout the paper: For any (t,x,y)(0,)ׯ+dׯ+d(t,x,y)\in(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+}\times{\overline{\mathbb{R}}}^{d}_{+}, r>0r>0 and z~d1\widetilde{z}\in{\mathbb{R}}^{d-1},

p¯(t,x,y)=rdp¯(t/rα,x/r,y/r)=p¯(t,x+(z~,0),y+(z~,0)),\overline{p}(t,x,y)=r^{-d}\overline{p}(t/r^{\alpha},x/r,y/r)=\overline{p}(t,x+(\widetilde{z},0),y+(\widetilde{z},0)),
pκ(t,x,y)=rdpκ(t/rα,x/r,y/r)=pκ(t,x+(z~,0),y+(z~,0)).p^{\kappa}(t,x,y)=r^{-d}p^{\kappa}(t/r^{\alpha},x/r,y/r)=p^{\kappa}(t,x+(\widetilde{z},0),y+(\widetilde{z},0)). (3.30)

From Proposition 2.7, since the exceptional set 𝒩{\cal N} is removed, we obtain

Corollary 3.13.

If d>αd>\alpha,

Gκ(x,y)G¯(x,y)c|xy|dα,x,y¯+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}.
Remark 3.14.

The assumption d>(α+β1+β2)2d>(\alpha+\beta_{1}+\beta_{2})\wedge 2 in [31, 32] is only used to show Gκ(x,y)c|xy|d+αG^{\kappa}(x,y)\leq c|x-y|^{-d+\alpha}. Thus, by Corollary 3.13, all results in [31, 32] with the assumption d>(α+β1+β2)2d>(\alpha+\beta_{1}+\beta_{2})\wedge 2 hold under the weaker assumption d>αd>\alpha.

Since the heat kernels p¯(t,x,y)\overline{p}(t,x,y) and pκ(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 YκY^{\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 V1V_{1} and V3V_{3} be open subsets of ¯+d{\overline{\mathbb{R}}}^{d}_{+} with dist(V1,V3)>0{\rm dist}(V_{1},V_{3})>0. Set V2:=¯+d(V1V3)V_{2}:={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}). For any xV1x\in V_{1}, yV3y\in V_{3} and t>0t>0, it holds that

p¯(t,x,y)\displaystyle\overline{p}(t,x,y) x(τ¯V1<t)supst,zV2p¯(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(V1,V3)dα0tV3V1p¯V1(ts,x,u)(u,w)p¯(s,y,w)𝑑u𝑑w𝑑s.\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 V1V_{1} and V3V_{3} be open subsets of +d{\mathbb{R}}^{d}_{+} with dist(V1,V3)>0{\rm dist}(V_{1},V_{3})>0. Set V2:=+d(V1V3)V_{2}:={\mathbb{R}}^{d}_{+}\setminus(V_{1}\cup V_{3}). For any xV1x\in V_{1}, yV3y\in V_{3} and t>0t>0, it holds that

pκ(t,x,y)\displaystyle p^{\kappa}(t,x,y) x(τV1κ<t<ζκ)supst,zV2pκ(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(V1,V3)dα0tV3V1pκ,V1(ts,x,u)(u,w)pκ(s,y,w)𝑑u𝑑w𝑑s.\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 xV1x\in V_{1} and yV3y\in V_{3},

pκ(t,x,y)\displaystyle p^{\kappa}(t,x,y) =𝔼x[pκ(tτV1κ,YτV1κκ,y):τV1κ<t<ζκ]\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]
=𝔼x[pκ(tτV1κ,YτV1κκ,y):τV1κ<t<ζκ,YτV1κκV2]\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]
+𝔼x[pκ(tτV1κ,YτV1κκ,y):τV1κ<t<ζκ,YτV1κκV3]\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]
x(τV1κ<t<ζκ)supst,zV2pκ(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)
+0tV3V1pκ,V1(ts,x,u)(u,w)|uw|d+αpκ(s,w,y)𝑑u𝑑w𝑑s\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
x(τV1κ<t<ζκ)supst,zV2pκ(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(V1,V3)dα0tV3V1pκ,V1(ts,x,u)(u,w)pκ(s,y,w)𝑑u𝑑w𝑑s.\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 YκY^{\kappa}, and get some preliminary lower bounds for the heat kernels p¯(t,x,y)\overline{p}(t,x,y) and pκ(t,x,y)p^{\kappa}(t,x,y).

Recall that B+(x,r)=B(x,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(0,2α2η)\delta\in(0,2^{-\alpha-2}\eta). There exists a constant C>0C>0 such that for all y+dy\in{\mathbb{R}}^{d}_{+}, R(0,yd)R\in(0,y_{d}), r(0,(ηδ/2)1/αR/2]r\in(0,(\eta\delta/2)^{1/\alpha}R/2], δRα/2ts4δ(2R)α\delta R^{\alpha}/2\leq t-s\leq 4\delta(2R)^{\alpha}, xB(y,(ηδ/2)1/αR/2)x\in B(y,(\eta\delta/2)^{1/\alpha}R/2) and zB(x0,(ηδ/2)1/αR)z\in B(x_{0},(\eta\delta/2)^{1/\alpha}R),

(t,z)(σ{s}×B(x,r)κτ[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(0,2α2η)\delta\in(0,2^{-\alpha-2}\eta). There exists a constant C>0C>0 such that for all y¯+dy\in{\overline{\mathbb{R}}}^{d}_{+}, R>0R>0, r(0,(ηδ/2)1/αR/2]r\in(0,(\eta\delta/2)^{1/\alpha}R/2], δRα/2ts4δ(2R)α\delta R^{\alpha}/2\leq t-s\leq 4\delta(2R)^{\alpha}, xB+(y,(ηδ/2)1/αR/2)x\in{B_{+}}(y,(\eta\delta/2)^{1/\alpha}R/2) and zB+(x0,(ηδ/2)1/αR)z\in{B_{+}}(x_{0},(\eta\delta/2)^{1/\alpha}R),

(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 YκY^{\kappa}, we introduce two conditions:

(UBS) There exists a constant C>0C>0 such that for all x,y¯+dx,y\in{\overline{\mathbb{R}}}_{+}^{d} and 0<r|xy|/20<r\leq|x-y|/2,

(x,y)CrdB+(x,r)(z,y)𝑑z.\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>0C>0 such that (4.1) holds for all x,y+dx,y\in{\mathbb{R}}_{+}^{d} and 0<r(|xy|xd)/20<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 rr.)

Lemma 4.2.

If (A3)(II) also holds, then (UBS) is satisfied. In particular, (IUBS) is satisfied.

Proof. Let x,y¯+dx,y\in{\overline{\mathbb{R}}}_{+}^{d} and 0<r|xy|/20<r\leq|x-y|/2. Note that for all zB+(x,r)z\in{B_{+}}(x,r), |xy|/2|zy|2|xy||x-y|/2\leq|z-y|\leq 2|x-y| by the triangle inequality. Thus, by (A3), there is c1>0c_{1}>0 independent of x,yx,y and rr such that for all zB+(x,r)z\in{B_{+}}(x,r) with zdxdz_{d}\geq x_{d}, (x,y)c1(z,y){\cal B}(x,y)\leq c_{1}{\cal B}(z,y). Using this, we get

1rdB+(x,r)(z,y)𝑑z(x,y)c1rdB(x,r):zdxd𝑑zc2(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 (x,y){\cal B}(x,y) satisfies (IUBS). Then there exist constants δ>0\delta>0 and C,M1C,M\geq 1 such that for all t00t_{0}\geq 0, x+dx\in{\mathbb{R}}^{d}_{+} and R(0,xd)R\in(0,x_{d}), and any non-negative function uu on (0,)×+d(0,\infty)\times{\mathbb{R}}^{d}_{+} which is parabolic on Q:=(t0,t0+4δRα]×B(x,R)Q:=(t_{0},t_{0}+4\delta R^{\alpha}]\times B(x,R) with respect to Y¯\overline{Y} or YκY^{\kappa}, we have

sup(t1,y1)Qu(t1,y1)Cinf(t2,y2)Q+u(t2,y2),\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=[t0+δRα,t0+2δRα]×B(x,R/M)Q_{-}=[t_{0}+\delta R^{\alpha},t_{0}+2\delta R^{\alpha}]\times B(x,R/M) and Q+=[t0+3δRα,t0+4δRα]×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 (x,y){\cal B}(x,y) satisfies (UBS). Then there exist constants δ>0\delta>0 and C,M1C,M\geq 1 such that for all t00t_{0}\geq 0, x¯+dx\in{\overline{\mathbb{R}}}^{d}_{+} and R>0R>0, and any non-negative function uu on (0,)ׯ+d(0,\infty)\times{\overline{\mathbb{R}}}^{d}_{+} which is parabolic on Q0:=(t0,t0+4δRα]×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(t1,y1)Q0u(t1,y1)Cinf(t2,y2)Q+0u(t2,y2),\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 Q0=[t0+δRα,t0+2δRα]×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=[t0+3δRα,t0+4δRα]×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>0C>0 such that for all x,y+dx,y\in{\mathbb{R}}^{d}_{+} and 0<r(|xy|xd)/20<r\leq(|x-y|\wedge x_{d})/2,

J(x,y)=(x,y)|xy|d+αCrdB+(x,r)(z,y)|zy|d+α𝑑z=CrdB+(x,r)J(z,y)𝑑z.\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 I2I_{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+dx,y\in{\mathbb{R}}^{d}_{+} and 0<r|xy|/20<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+(x0,){B_{+}}(x_{0},\cdot) instead of B(x0,)B(x_{0},\cdot) in the definitions of QiQ_{i}, 1i41\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 (x,y){\cal B}(x,y) satisfies (IUBS). For any positive constants a,ba,b, there exists a constant C=C(a,b,κ)>0C=C(a,b,\kappa)>0 such that for all z+dz\in{\mathbb{R}}_{+}^{d} and r>0r>0 with B(z,2br)+dB(z,2br)\subset{\mathbb{R}}^{d}_{+},

infyB(z,br/2)y(τB(z,br)κ>arα)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 c1,c2,ε1>0c_{1},c_{2},{\varepsilon}_{1}>0 such that for all z+dz\in{\mathbb{R}}^{d}_{+} and r>0r>0 with B(z,2br)+dB(z,2br)\subset{\mathbb{R}}^{d}_{+},

z(τB(z,br/2)κ>ε1rα)=B(z,br/2)pκ,B(z,br/2)(ε1rα,z,w)𝑑w\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,c1r)pκ,B(z,br/2)(ε1rα,z,w)𝑑wc2.\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>ε1a>{\varepsilon}_{1}. Applying the parabolic Harnack inequality (Theorem 4.3) repeatedly, we conclude that there exists c3>0c_{3}>0 such that for any w,yB(z,br/2)w,y\in B(z,br/2),

pκ,B(z,br)(arα,y,w)c3pκ,B(z,br)(ε1rα,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 yB(z,br/2)y\in B(z,br/2),

y(τB(z,br)κ>arα)=B(z,br)pκ,B(z,br)(arα,y,w)𝑑w\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
c3B(z,br/2)pκ,B(z,br)(ε1rα,z,w)𝑑wc3z(τB(z,br/2)κ>ε1rα)c2c3.\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 (x,y){\cal B}(x,y) satisfies (IUBS). For any a>0a>0, there exists a constant C=C(a,κ)>0C=C(a,\kappa)>0 such that for any (t,x,y)(0,)×+d×+d(t,x,y)\in(0,\infty)\times{\mathbb{R}}_{+}^{d}\times{\mathbb{R}}_{+}^{d} with at1/α(4|xy|)xdydat^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d},

p¯(t,x,y)pκ(t,x,y)CtJ(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 zB(y,(12)1at1/α)z\in B(y,\,(12)^{-1}at^{1/\alpha}), with probability at least c1=c1(a)>0c_{1}=c_{1}(a)>0, the process YκY^{\kappa} does not move more than (18)1at1/α(18)^{-1}at^{1/\alpha} by time tt. Thus, using the strong Markov property and the Lévy system formula in (3.16), we obtain that for a(t/2)1/α(4|xy|)xdyda(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d},

x(YtκB(y, 61at1/α))\displaystyle{\mathbb{P}}_{x}\left(Y^{\kappa}_{t}\in B\big{(}y,\,6^{-1}at^{1/\alpha}\big{)}\right)
c1x(YtτB(x,(18)1at1/α)κκB(y,(12)1at1/α) and tτB(x,(18)1at1/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{)}
=c1𝔼x[0tτB(x,(18)1at1/α)κB(y,(12)1at1/α)J(Ysκ,u)𝑑u𝑑s].\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|xy|)xdyda(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d},

𝔼x[0tτB(x,(18)1at1/α)κB(y,(12)1at1/α)J(Ysκ,u)𝑑u𝑑s]\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{]}
=𝔼x[0tB(y,(12)1at1/α)J(Ysκ,B(x,(18)1at1/α),u)𝑑u𝑑s]\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{]}
c2td/α0t𝔼x[J(Ysκ,B(x,(18)1at1/α),y)]𝑑s\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
c2td/αt/2tB(x,(72)1a(t/2)1/α)J(w,y)pκ,B(x,(18)1at1/α)(s,x,w)𝑑w𝑑s.\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<tt/2<s<t and wB(x,(72)1a(t/2)1/α)w\in B(x,(72)^{-1}a(t/2)^{1/\alpha}), we have

δB(x,(18)1at1/α)(w)(18)1at1/α(72)1a(t/2)1/α(36)1as1/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

|xw|<(72)1a(t/2)1/α41(18)1as1/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<tt/2<s<t and wB(x,(72)1a(t/2)1/α)w\in B(x,(72)^{-1}a(t/2)^{1/\alpha}),

pκ,B(x,(18)1at1/α)(s,x,w)c3td/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|xy|)xdyda(t/2)^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d},

x(YtκB(y, 61at1/α))c4tB(x,(72)1a(t/2)1/α)J(w,y)𝑑wc5t1+d/αJ(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 ydat1/αy_{d}\geq at^{1/\alpha} and zB(y,a(t/2)1/α/6)z\in B(y,a(t/2)^{1/\alpha}/6),

zdyd|zy|a(1(1/2)1/α/6)t1/α(5a/6)t1/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 at1/α(4|xy|)xdydat^{1/\alpha}\leq(4|x-y|)\wedge x_{d}\wedge y_{d},

pκ(t,x,y)\displaystyle p^{\kappa}(t,x,y) =Dpκ(t/2,x,z)pκ(t/2,z,y)𝑑zB(y,a(t/2)1/α/6)pκ(t/2,x,z)pκ(t/2,z,y)𝑑z\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
c6td/αx(Yt/2κB(y,a(t/2)1/α/6))c7tJ(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>0C>0 such that

pκ(t,x,y)C(1xdt1/α)qκ(1ydt1/α)qκ(td/αt|xy|d+α),t>0,x,y+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 qκ[(α1)+,α+β1)q_{\kappa}\in[(\alpha-1)_{+},\alpha+\beta_{1}) is the constant from (1.8).

Note that, since td/αt|xy|d+αt^{-d/\alpha}\wedge\frac{t}{|x-y|^{d+\alpha}} is comparable to the transition density of the isotropic α\alpha-stable process in d{\mathbb{R}}^{d}, there exists a constant C>0C>0 such that for all t,s>0t,s>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

+d(td/αt|xz|d+α)𝑑zC,\displaystyle\int_{{\mathbb{R}}^{d}_{+}}\left(t^{-d/\alpha}\wedge\frac{t}{|x-z|^{d+\alpha}}\right)dz\leq C, (5.2)

and

+d(td/αt|xz|d+α)(sd/αs|yz|d+α)𝑑zC((t+s)d/αt+s|xy|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>0C>0 such that

x(ζκ>t)C(1xdt1/α)qκ,t>0,x+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),

x(ζκ>t)=+dpκ(t,x,y)𝑑y\displaystyle{\mathbb{P}}_{x}(\zeta^{\kappa}>t)=\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t,x,y)dy
c1(1xdt1/α)qκ+d(td/αt|xy|d+α)𝑑yc2(1xdt1/α)qκ.\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 qκ=0q_{\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 qκ>0q_{\kappa}>0 and denote by qq the constant qκq_{\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>0r>0, we define a subset U(r)U(r) of ¯+d{\overline{\mathbb{R}}}^{d}_{+}, d2d\geq 2, by

U(r):={x=(x~,xd)d:|x~|<r/2, 0xd<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=1d=1, we abuse notation and use U(r)=[0,r/2)U(r)=[0,r/2). For t>0t>0 and an open set V+dV\subset{\mathbb{R}}^{d}_{+}, denote by Ytκ,dY_{t}^{\kappa,d} and Ytκ,V,dY_{t}^{\kappa,V,d} the last coordinates of YtκY_{t}^{\kappa} and Ytκ,VY_{t}^{\kappa,V} respectively.

Lemma 5.3.

There exists a constant C>0C>0 such that for all x,y+dx,y\in{\mathbb{R}}^{d}_{+} satisfying |xy|xd|x-y|\geq x_{d}, it holds that

(x,y)Cxdβ1(|logxd|β31)(1+𝟏|y|1(log|y|)β3)|xy|β1.\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>0R>0, there exists a constant C>0C>0 such that for all r(0,R]r\in(0,R] and xU(24r)+dx\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+},

𝔼x0τU(r)κ(Ytκ,d)β1|logYtκ,d|β3𝑑tCxdq.\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>0C>0 such that for all r>0r>0 and xU(24r)+dx\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+},

x(τU(r)κ<ζκ)=x(YτU(r)κκ+d)C(xdr)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>0C>0 such that for all r>0r>0 and xU(24r)+dx\in U(2^{-4}r)\cap{\mathbb{R}}^{d}_{+},

𝔼x[τU(r)κ]C(xdr)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>0C>0 such that for any w+dw\in\partial{\mathbb{R}}^{d}_{+}, r>0r>0 and any non-negative function ff in +d{\mathbb{R}}^{d}_{+} that is harmonic in +dB(w,r){\mathbb{R}}^{d}_{+}\cap B(w,r) with respect to YκY^{\kappa} and vanishes continuously on +dB(w,r)\partial{\mathbb{R}}^{d}_{+}\cap B(w,r), we have

f(x)Cf(x^)for allx+dB(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^+dB(w,r)\widehat{x}\in{\mathbb{R}}^{d}_{+}\cap B(w,r) with x^dr/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 γ0\gamma\geq 0, t>0t>0 and xU(1)+dx\in U(1)\cap{\mathbb{R}}^{d}_{+}, it holds that

+dpκ(t,x,z)(1zd)γdz𝔼x[(1Ytκ,U(1),d)γ:τU(1)κ>t]+x(YτU(1)κκ+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

x(ζκ>t)t1𝔼x[τU(1)κ]+x(YτU(1)κκ+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 Ytκ,U(1)=YtκY^{\kappa,U(1)}_{t}=Y^{\kappa}_{t} for t<τU(1)κt<\tau^{\kappa}_{U(1)}, we have

+dpκ(t,x,z)(1zd)γdz=𝔼x[(1Ytκ,d)γ:t<ζκ]\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]
=𝔼x[(1Ytκ,d)γ:τU(1)κ>t]+𝔼x[(1Ytκ,d)γ:τU(1)κt<ζκ]\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]
𝔼x[(1Ytκ,U(1),d)γ:τU(1)κ>t]+𝔼x[1:τU(1)κ<ζκ]\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{]}
=𝔼x[(1Ytκ,U(1),d)γ:τU(1)κ>t]+x(YτU(1)κκ+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

x(ζκ>t)x(τU(1)κ>t)+x(YτU(1)κκ+d)t1𝔼x[τU(1)κ]+x(YτU(1)κκ+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>0C>0 such that

pκ(t,x,y)Ctd/αx(ζκ>t/3)y(ζκ>t/3),t>0,x,y+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 pκ(t,,)p^{\kappa}(t,\cdot,\cdot) and Proposition 2.7, we obtain

pκ(t,x,y)=+d+dpκ(t/3,x,z)pκ(t/3,z,w)pκ(t/3,y,w)𝑑z𝑑w\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
c1td/α+dpκ(t/3,x,z)𝑑z+dpκ(t/3,y,w)𝑑w=c1td/αx(ζκ>t/3)y(ζκ>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>0C>0 such that

pκ(t,x,y)C(1xdt1/α)q(1ydt1/α)qtd/α,t>0,x,y+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 c1>0c_{1}>0 such that

pκ(t,x,y)c1(1xdt1/α)q(td/αt|xy|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=t0=(1/2)αt=t_{0}=(1/2)^{\alpha}. If xd24t01/αx_{d}\geq 2^{-4}t_{0}^{1/\alpha} or |xy|4t01/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 xd<24t01/αx_{d}<2^{-4}t_{0}^{1/\alpha} and |xy|>4t01/α|x-y|>4t_{0}^{1/\alpha}, and will show that

pκ(t0,x,y)c1(t0)xdq|xy|d+α.p^{\kappa}(t_{0},x,y)\leq c_{1}(t_{0})\frac{x_{d}^{q}}{|x-y|^{d+\alpha}}. (5.8)

Let V1=U(t01/α)V_{1}=U(t_{0}^{1/\alpha}), V3={w+d:|wy|<|xy|/2}V_{3}=\{w\in{\mathbb{R}}^{d}_{+}:|w-y|<|x-y|/2\} and V2=+d(V1V3)V_{2}={\mathbb{R}}^{d}_{+}\setminus(V_{1}\cup V_{3}). By Lemma 5.5, we have

x(τV1κ<t0<ζκ)x(YτV1κκ+d)c2(t01/αxd)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

supst0,zV2pκ(s,z,y)c3supst0,z+d,|zy|>|xy|/2s|zy|d+α=2d+αc3t0|xy|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 uV1u\in V_{1} and wV3w\in V_{3},

|uw||xy||xu||yw||xy|t01/α|xy|2|xy|4t01/αud.|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 β1>0\beta_{1}>0 if β3>0\beta_{3}>0, we see that (1+𝟏|w|1(log|w|)β3)|uw|β1c4\big{(}1+{\bf 1}_{|w|\geq 1}(\log|w|)^{\beta_{3}}\big{)}|u-w|^{-\beta_{1}}\leq c_{4} for uV1u\in V_{1} and wV3w\in V_{3}, so by Lemma 5.3, we have that for any uV1u\in V_{1} and wV3w\in V_{3},

(u,w)c5udβ1(|logud|β31)(1+𝟏|w|1(log|w|)β3)|uw|β1c6udβ1|logud|β3.{\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

0t0V3V1pκ,V1(ts,x,u)(u,w)pκ(s,y,w)𝑑u𝑑w𝑑s\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
c70t0(V1pκ,V1(ts,x,u)udβ1|logud|β3𝑑u)(V3pκ(s,y,w)𝑑w)𝑑s\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
c70(V1pκ,V1(s,x,u)udβ1|logud|β3𝑑u)𝑑s\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
=c7𝔼x0τV1κ(Ysκ,d)β1|logYsκ,d|β3𝑑sc8xdq.\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),

pκ(t,x,y)=+dpκ(t/2,x,z)pκ(t/2,y,z)𝑑z\displaystyle p^{\kappa}(t,x,y)=\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(t/2,x,z)p^{\kappa}(t/2,y,z)dz
c12(1xdt1/α)q(1ydt1/α)q+d((t/2)d/αt/2|xz|d+α)((t/2)d/αt/2|yz|d+α)𝑑z\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
c9(1xdt1/α)q(1ydt1/α)q(td/αt|xy|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<αq<\alpha.

Lemma 5.11.

If q<αq<\alpha, then (5.6) holds true.

Proof. By (3.30), it suffices to prove (5.6) when t=1t=1. By Lemma 5.9, it suffices to prove that there exists a constant c1>0c_{1}>0 such that x(ζκ>1/3)c1(1xd)q{\mathbb{P}}_{x}(\zeta^{\kappa}>1/3)\leq c_{1}(1\wedge x_{d})^{q} for all x+dx\in{\mathbb{R}}^{d}_{+}. By Lemma 2.2 and the fact that 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 xd<25x_{d}<2^{-5}. Then, since q<αq<\alpha, by (5.5) and Lemmas 5.55.6 (with r=1r=1), we get x(ζκ>1/3)c2xdq{\mathbb{P}}_{x}(\zeta^{\kappa}>1/3)\leq c_{2}x_{d}^{q}. \Box

To remove the assumption q<αq<\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>αd>\alpha. Now we state, and will prove later, that the desired result still holds true for d=1αd=1\leq\alpha (thus for all dd).

Lemma 5.12.

Let γ>qα\gamma>q-\alpha. There exists C>0C>0 such that for any R>0R>0, U(R)DU(2R)U(R)\subset D\subset U(2R) and any x=(0~,xd)+dx=(\widetilde{0},x_{d})\in{\mathbb{R}}^{d}_{+} with xdR/10x_{d}\leq R/10, it holds that

𝔼x0τDκ(Ysκ,D,d)γ𝑑s=0Dpκ,D(t,x,z)zdγ𝑑z𝑑tCRγ+αqxdq.{\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>αd>\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=1αd=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=1t=1.

Inequality (5.6) holds when q<αq<\alpha by Lemma 5.11. Now, assume that (5.6) holds for all q<kαq<k\alpha for some kk\in\mathbb{N}. We now show (5.6) also holds for q[kα,(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 qα+ε<kαq-\alpha+{\varepsilon}<k\alpha. Then qα+ε<qq-\alpha+{\varepsilon}<q and C(α,qα+ε,)<C(α,q,)=κC(\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+dz,w\in{\mathbb{R}}^{d}_{+},

pκ(1su,z,w)\displaystyle p^{\kappa}(1-s-u,z,w) =(1su)d/αpκ(1,(1su)1/αz,(1su)1/αw)\displaystyle=(1-s-u)^{-d/\alpha}p^{\kappa}(1,(1-s-u)^{-1/\alpha}z,(1-s-u)^{-1/\alpha}w)
2d/αpC(α,qα+ε,)(1,(1su)1/αz,(1su)1/αw)\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)
c3(1zd)qα+ε(1wd)qα+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 pκ(1,x,y)pC(α,qα+ε,)(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,)>C(α,qα+ε,)C(\alpha,q,{\cal B})>C(\alpha,q-\alpha+{\varepsilon},{\cal B}). Therefore, by the semigroup property and symmetry, we get

pκ(1,x,y)=1601/401/4pκ(1,x,y)𝑑s𝑑u\displaystyle p^{\kappa}(1,x,y)=16\int_{0}^{1/4}\int_{0}^{1/4}p^{\kappa}(1,x,y)dsdu
=1601/401/4+d+dpκ(s,x,z)pκ(1su,z,w)pκ(u,y,w)𝑑z𝑑w𝑑s𝑑u\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
16c3(01/4+dpκ(s,x,z)(1zd)qα+ε𝑑z𝑑s)(01/4+dpκ(u,y,w)(1wd)qα+ε𝑑w𝑑u).\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 c4>0c_{4}>0 such that

01/4+dpκ(s,v,z)(1zd)qα+ε𝑑z𝑑sc4(1vd)q,v+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 vU(24)v\notin U(2^{-4}), then we get

(1vd)q25q25q01/4+dpκ(s,v,z)𝑑z𝑑s25q01/4+dpκ(s,v,z)(1zd)qα+ε𝑑z𝑑s.(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 vU(24)v\in U(2^{-4}), then by (5.4), Fubini’s theorem, Lemmas 5.5 and 5.12,

01/4+dpκ(s,v,z)(1zd)qα+ε𝑑z𝑑s\displaystyle\int_{0}^{1/4}\int_{{\mathbb{R}}^{d}_{+}}p^{\kappa}(s,v,z)(1\wedge z_{d})^{q-\alpha+{\varepsilon}}dzds
01/4𝔼v[(1Ysκ,U(1),d)qα+ε:τU(1)κ>s]ds+01/4v(YτU(1)κκ+d)ds\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
𝔼v0τU(1)κ(1Ysκ,U(1),d)qα+ε𝑑s+14v(YτU(1)κ+d)c5vdq.\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=1αd=1\leq\alpha, then there exists C>0C>0 such that for all x,y(0,)x,y\in(0,\infty),

Gκ(x,y)C(1xy|xy|)q(α12)(xy)α1log(e+xy|xy|).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(α12)(0,q][α1,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 |xy|=1|x-y|=1 and xyx\leq y without loss of generality. Then y=x+1>x1y=x+1>x\vee 1. It suffices to show that

Gκ(x,y)Cxq~yα1log(e+y).\displaystyle G^{\kappa}(x,y)\leq Cx^{\widetilde{q}}y^{\alpha-1}\log(e+y).

Note that 1+2q~>1+q~α1+2\widetilde{q}>1+\widetilde{q}\geq\alpha. Since C(α,q~,)C(α,q,)=κC(\alpha,\widetilde{q},{\cal B})\leq C(\alpha,q,{\cal B})=\kappa implies that pκ(t,x,y)pC(α,q~,)(t,x,y)p^{\kappa}(t,x,y)\leq p^{C(\alpha,\widetilde{q},{\cal B})}(t,x,y), using the fact y>x1y>x\vee 1 and Lemmas 5.11 and 5.10, we get

Gκ(x,y)0pC(α,q~,)(t,x,y)𝑑t\displaystyle G^{\kappa}(x,y)\leq\int_{0}^{\infty}p^{C(\alpha,\widetilde{q},{\cal B})}(t,x,y)dt
c1(xq~01t(αq~)/α𝑑t+xq~1yαt(1+q~)/α𝑑t+xq~yq~yαt(1+2q~)/α𝑑t)\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)
c1xq~(1+log(yα)+α1+2q~αyαq~1)c2xq~yα1log(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 α12\alpha-\frac{1}{2} from the power of the first factor.

Lemma 5.15.

If d=1αd=1\leq\alpha, then there exists C>0C>0 such that for all x,y(0,)x,y\in(0,\infty),

Gκ(x,y)C(1xy|xy|)q(xy)α1log(e+xy|xy|).\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=26r=2^{-6}. By symmetry, (3.30) and Lemma 5.14, without loss of generality, we can assume |xy|=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

Gκ(x,y)Cxq.\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/2wz<wz<w/2\leq w-z<w and wyww\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

rGκ(w,y)(z,w)(wz)1α𝑑w\displaystyle\int_{r}^{\infty}G^{\kappa}(w,y){\cal B}(z,w)(w-z)^{-1-\alpha}dw
c1zβ1(|logz|β31)rGκ(w,y)w1+α+β1(1+𝟏{w1}(logw)β3)𝑑w\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
c2zβ1|logz|β3rlog(e+w|wy|)(1+𝟏{w1}(logw)β3)w2+β1𝑑w.\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

𝔼x[Gκ(Yτ(0,r/2)κκ,y);Yτ(0,r/2)κκ(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]
c2𝔼x0τ(0,r/2)κ(Ytκ,d)β1|log(Ytκ,d)|β3𝑑trlog(e+w|wy|)(1+𝟏{w1}(logw)β3)w2+β1𝑑w\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
c3xqrlog(e+w|wy|)(1+𝟏{w1}(logw)β3)w2+β1𝑑w.\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 wyw/2w-y\geq w/2 for w(2y,)w\in(2y,\infty) and y(1,1+r/2)y\in(1,1+r/2), using a change of the variables, we obtain

rlog(e+w|wy|)(1+𝟏{w1}(logw)β3)w2+β1𝑑w\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))β3r2+β1r2ylog(e+2+r|wy|)𝑑w+log(e+2)2y1+(logw)β3w2+β1𝑑w\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))β3)r2+β101+r/2log(e+2+rv)𝑑v+c4c5.\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

𝔼x[Gκ(Yτ(0,r/2)κκ,y);Yτ(0,r/2)κκ(0,r)]c3c5xq.\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 zGκ(z,y)z\mapsto G^{\kappa}(z,y) is harmonic in (0,2r)(0,2r) with respect to YκY^{\kappa} and vanishes continuously as z0z\to 0 by Lemma 5.14, we get from Proposition 5.7 that Gκ(z,y)c6Gκ(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

𝔼x[Gκ(Yτ(0,r/2)κκ,y);Yτ(0,r/2)κκ(0,r)]c6Gκ(r,y)x(Yτ(0,r/2)κκ(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))
c7(xr)q(ryr)q(α12)yα1log(e+yyr)c8xq.\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 zGκ(z,y)z\mapsto G^{\kappa}(z,y) in (0,2r)(0,2r), we arrive at

Gκ(x,y)=𝔼x[Gκ(Yτ(0,r/2)κκ,y);Yτ(0,r/2)κκ(0,r)]+𝔼x[Gκ(Yτ(0,r/2)κκ,y);Yτ(0,r/2)κκ(0,r)]c9xq.\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=1αd=1\leq\alpha: Assume that d=1αd=1\leq\alpha. Since DU(2R)=[0,R)D\subset U(2R)=[0,R), using Fubini’s theorem and Lemma 5.15, we have

0Dpκ,D(t,x,z)zγ𝑑z𝑑t0R0pκ,D(t,x,z)𝑑tzγ𝑑z0RGκ(x,z)zγ𝑑z\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
c10x/2zq+γ|xz|qxα1log(e+x|xz|)𝑑z+c1x/22xzγ(2x)α1log(e+2x|xz|)𝑑z\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
+c12xRxq|xz|qzγ+α1log(e+z|xz|)𝑑z\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
=:c1(I+II+III).\displaystyle=:c_{1}(I+II+III).

Since γ+αq>0\gamma+\alpha-q>0 and xR/10x\leq R/10, we see that

Ixq+γ+α10x/21(x/2)qlog(e+x(x/2))𝑑z2q1log(e+2)xγ+αc2Rγ+αqxq.\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=xyz=xy, we also get

II\displaystyle II (2x)γ+α1x/22xlog(e+2x|xz|)𝑑z=2γ+α1xγ+α1/22log(e+2|1y|)𝑑y\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
c3xγ+αc3Rγ+αqxq.\displaystyle\leq c_{3}x^{\gamma+\alpha}\leq c_{3}R^{\gamma+\alpha-q}x^{q}.

Lastly, since |xz|z/2|x-z|\geq z/2 for all z2xz\geq 2x, we obtain

III2qlog(e+2)xq2xRzγ+αq1𝑑zc4Rγ+αqxq.\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 YκY^{\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 YκY^{\kappa}, we assume that κ>0\kappa>0 if α(0,1]\alpha\in(0,1] (see Lemma 2.3), and denote by qq the strictly positive constant qk((α1)+,α+β1)q_{k}\in((\alpha-1)_{+},\alpha+\beta_{1}) from (1.8). Additionally, we write YY, p(t,x,y)p(t,x,y), pD(t,x,y)p^{D}(t,x,y), τD\tau_{D} and ζ\zeta instead of YκY^{\kappa}, pκ(t,x,y)p^{\kappa}(t,x,y), pκ,D(t,x,y)p^{\kappa,D}(t,x,y), τDκ\tau_{D}^{\kappa} and ζκ\zeta^{\kappa}. When we consider Y¯\overline{Y}, the letter qq denotes 0, and we write YY, p(t,x,y)p(t,x,y), pD(t,x,y)p^{D}(t,x,y), τD\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 Vx(t)V_{x}(t) and Wx(t)W_{x}(t) from (3). We let Vx:=Vx(1)V_{x}:=V_{x}(1) and Wx:=Wx(1)W_{x}:=W_{x}(1).

Lemma 6.1.

There exist constants M>1M>1 and C>0C>0 such that for all t>0t>0 and x+dx\in{\mathbb{R}}^{d}_{+},

infzWx(t)p(Mt,x,z)C(1xdt1/α)qtd/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=0q=0, the result is given in Lemma 3.6.

Suppose q>0q>0. By the scaling property (3.30), it suffices to prove the lemma for t=1t=1. If xd1x_{d}\geq 1, then the result follows from Proposition 3.2. Assume xd<1x_{d}<1. By Proposition 3.2, for any M>1M>1, there is c1=c1(M)>0c_{1}=c_{1}(M)>0 such that

infw,zWx, 1sMp(s,w,z)c1for all x+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>1M>1 and zWxz\in W_{x},

p(M,x,z)\displaystyle p(M,x,z) 𝔼x[p(MτVx,YτVx,z):τVxM1,YτVxWx]\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]
(infwWx, 1sMp(s,w,z))x(τVxM1,YτVxWx)\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)
c1(x(YτVxWx)x(τVx(t)>M1)).\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

x(YτVxWx)2c2xdq{\mathbb{P}}_{x}(Y_{\tau_{V_{x}}}\in W_{x})\geq 2c_{2}x_{d}^{q}

and, by Corollary 5.2, we also have

x(τVx>M1)x(ζ>M1)c3(xd/(M1)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+(c3/c2)α/qM=1+(c_{3}/c_{2})^{\alpha/q} so that 2c2c3(M1)q/α=c22c_{2}-c_{3}(M-1)^{-q/\alpha}=c_{2}, which implies p(M,x,z)c2xdqp(M,x,z)\geq c_{2}x_{d}^{q}. \Box

For any b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0, we define for t0t\geq 0 and x,y¯+dx,y\in{\overline{\mathbb{R}}}^{d}_{+},

Ab1,b2,b3,b4(t,x,y)\displaystyle A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y) :=((xdyd)t1/α|xy|1)b1((xdyd)t1/α|xy|1)b2\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}}
×logb3(e+((xdyd)t1/α)|xy|((xdyd)t1/α)|xy|)logb4(e+|xy|((xdyd)t1/α)|xy|).\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 Ab1,b2,b3,b4(0,x,y)=Bb1,b2,b3,b4(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+dx,y\in{\mathbb{R}}^{d}_{+} and

Ab1,b2,b3,b4(t,x,y)Bb1,b2,b3,b4(x+t1/α𝐞d,y+t1/α𝐞d) for t0,x,y+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 aba+ba\vee b\asymp a+b for all a,b0a,b\geq 0. Note also that for any a>0a>0, there exists c>0c>0 such that

Ab1,b2,b3,b4(t,x,y)c(a1)b1+b2,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>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+} with (xdyd)+t1/αa|xy|(x_{d}\wedge y_{d})+t^{1/\alpha}\geq a|x-y|.

Proposition 6.2.

There exists a constant C>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)C(1xdt1/α)q(1ydt1/α)qAβ1,β2,β3,β4(t,x,y)(td/αt|xy|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 xdydx_{d}\leq y_{d}. Let M>1M>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) Wx(t/(3M))Wy(t/(3M))p(t/3,x,z)p(t/3,z,w)p(t/3,w,y)𝑑z𝑑w\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
c1t2d/α(infzWx(t/(3M))p(t/3,x,z))(infwWy(t/(3M))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)Wx(t/(3M))×Wy(t/(3M))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)
c2(1xdt1/α)q(1ydt1/α)qinf(z,w)Wx(t/(3M))×Wy(t/(3M))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)Wx(t/(3M))×Wy(t/(3M))(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M)), we have zdxd+5(t/(3M))1/αz_{d}\geq x_{d}+5(t/(3M))^{1/\alpha}, wdyd+5(t/(3M))1/αw_{d}\geq y_{d}+5(t/(3M))^{1/\alpha}, |zw||zx|+|xy|+|yw||xy|+20(t/(3M))1/α|z-w|\leq|z-x|+|x-y|+|y-w|\leq|x-y|+20(t/(3M))^{1/\alpha} and |zw||xy||zx||yw||xy|20(t/(3M))1/α|z-w|\geq|x-y|-|z-x|-|y-w|\geq|x-y|-20(t/(3M))^{1/\alpha}. Hence, if |xy|21(t/(3M))1/α|x-y|\leq 21(t/(3M))^{1/\alpha}, then

zdwd5(t3M)1/α541((t3M)1/α|zw|),(z,w)Wx(t3M)×Wy(t3M).\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 |xy|>21(t/(3M))1/α|x-y|>21(t/(3M))^{1/\alpha}, then using (A3)(I) in the first inequality below, |zw|2|xy||z-w|\leq 2|x-y| and Lemma 10.9 in the second and (6.3) in the third, we get

inf(z,w)Wx(t/(3M))×Wy(t/(3M))J(z,w)C21inf(z,w)Wx(t/(3M))×Wy(t/(3M))Bβ1,β2,β3,β4(z,w)|zw|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}}
c3Bβ1,β2,β3,β4(x+5(t/(3M))1/α𝐞d,y+5(t/(3M))1/α𝐞d)|xy|d+αc4Aβ1,β2,β3,β4(t,x,y)|xy|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)Wx(t/(3M))×Wy(t/(3M))p(t/3,z,w)\displaystyle\inf_{(z,w)\in W_{x}(t/(3M))\times W_{y}(t/(3M))}p(t/3,z,w)
c5{td/α if |xy|21(t/(3M))1/α,tAβ1,β2,β3,β4(t,x,y)|xy|dα if |xy|>21(t/(3M))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}
c6Aβ1,β2,β3,β4(t,x,y)(td/αt|xy|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>0t>0 and x+dx\in{\mathbb{R}}^{d}_{+},

x(ζ>t)(1xdt1/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

x(ζ>t)\displaystyle{\mathbb{P}}_{x}(\zeta>t) B(x,2t1/α):ydt1/αp(t,x,y)𝑑yc1(1xdt1/α)qtd/αB(x,2t1/α):ydt1/α𝑑y\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
c2(1xdt1/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>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)C(1xdt1/α)q(1ydt1/α)q(td/αt|xy|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)
×(|xy|αt)(ydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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): (xdyd)t1/α|xy|/4(x_{d}\wedge y_{d})\vee t^{1/\alpha}\geq|x-y|/4. In this case, we see from (6.4) that Aβ1,β2,β3,β4(t,x,y)A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y) is bounded below by a positive constant. Moreover, since sups>0,z,w+dAβ1,β2,β3,β4(s,z,w)c1<\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

|xy|α(ydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
c12|xy|α|xy|/4drrα+1=4αc12/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): (xdyd)t1/α<|xy|/4(x_{d}\wedge y_{d})\vee t^{1/\alpha}<|x-y|/4. For r>0r>0, set x(r):=x+r𝐞dx(r):=x+r{\mathbf{e}}_{d} and

K(r):={z=(z~,zd)+d:|z~x~|<r2,zd=xd+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+d:zK(r) for some (ydt1/α)|xy|4<r<|xy|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~,xd+r)Kz=(\widetilde{z},x_{d}+r)\in K, since

r|xz|52r54|xy|,(154)|xy||yz|(1+54)|xy|,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

Aβ1,β2,β3,β4(t,x,z)\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z) (1xdt1/αr)β1(1(xd+r)t1/αr)β2\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}}
×logβ3(e+((xd+r)t1/α)r(xdt1/α)r)logβ4(e+r((xd+r)t1/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{)}
(1xdt1/αr)β1logβ3(e+r(xdt1/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)
Aβ1,β2,β3,β4(t,x,x(r))\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x(r))

and

Aβ1,β2,β3,β4(t,z,y)\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,y)
(1ydt1/α|xy|)β1(1xd+r|xy|)β2logβ3(e+xd+rydt1/α)logβ4(e+|xy|xd+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)
Aβ1,β2,β3,β4(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)Kp(t/2,x,z)p(t/2,z,y)𝑑z\displaystyle p(t,x,y)\geq\int_{K}p(t/2,x,z)p(t/2,z,y)dz
c22(1xdt1/α)q(1ydt1/α)qKtAβ1,β2,β3,β4(t,x,z)|xz|d+αtAβ1,β2,β3,β4(t,z,y)|yz|d+α𝑑z\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
c3t2(1xdt1/α)q(1ydt1/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}
×(ydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x(r))rd+αAβ1,β2,β3,β4(t,x(r),y)|xy|d+αK(r)dz~dr\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
c4(1xdt1/α)q(1ydt1/α)qt|xy|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(ydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x(r))Aβ1,β2,β3,β4(t,x(r),y)drrα+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+dx,y\in{\mathbb{R}}^{d}_{+} and |xy|/4r|xy|/2|x-y|/4\leq r\leq|x-y|/2,

Aβ1,β2,β3,β4(t,x,x+r𝐞d)\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+r{\mathbf{e}}_{d}) (1xdt1/α|xy|)β1logβ3(e+|xy|(xdt1/α)|xy|)\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{)}
Aβ1,β2,β3,β4(t,x,x+|xy|2𝐞d)\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 ydxd+4ry_{d}\leq x_{d}+4r,

Aβ1,β2,β3,β4(t,x+r𝐞d,y)\displaystyle A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+r{\mathbf{e}}_{d},y) (1ydt1/α|xy|)β1logβ3(e+|xy|(ydt1/α)|xy|)\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{)}
Aβ1,β2,β3,β4(t,x+|xy|2𝐞d,y)\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+\frac{|x-y|}{2}{\mathbf{e}}_{d},y)
Aβ1,β2,β3,β4(t,y,y+|xy|2𝐞d).\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

Aβ1,β2,β3,β4(t,x,x+|xy|2𝐞d)Aβ1,β2,β3,β4(t,x+|xy|2𝐞d,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)
Aβ1,β2,β3,β4(t,x,x+|xy|2𝐞d)Aβ1,β2,β3,β4(t,y,y+|xy|2𝐞d)\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})
Aβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|).\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>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

|xy|/4|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
Aβ1,β2,β3,β4(t,x,x+|xy|2𝐞d)Aβ1,β2,β3,β4(t,x+|xy|2𝐞d,y)|xy|/4|xy|/2drrα+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}}
|xy|αAβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|).\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)
Remark 6.5.

Here we give a proof of the comparability of (1.6) and (1.7). Let t>0t>0 and x,y¯+dx,y\in{\overline{\mathbb{R}}}^{d}_{+} be such that |xy|>6t1/α|x-y|>6t^{1/\alpha}. For any zB(x+21|xy|𝐞d,41|xy|)z\in B(x+2^{-1}|x-y|{\mathbf{e}}_{d},4^{-1}|x-y|), by using the triangle inequality several times, we have

zdt1/αzdxdyd+|xy|xdyd+|xy|\displaystyle z_{d}-t^{1/\alpha}\asymp z_{d}\asymp x_{d}\wedge y_{d}+|x-y|\asymp x_{d}\vee y_{d}+|x-y| (6.16)

and

|x+t1/α𝐞dz||y+t1/α𝐞dz||xz||yz||xy|.\displaystyle|x+t^{1/\alpha}{\mathbf{e}}_{d}-z|\asymp|y+t^{1/\alpha}{\mathbf{e}}_{d}-z|\asymp|x-z|\asymp|y-z|\asymp|x-y|. (6.17)

For any t>0t>0 and x,y¯+dx,y\in{\overline{\mathbb{R}}}^{d}_{+} with |xy|>6t1/α|x-y|>6t^{1/\alpha}, if xdyd|xy|/4x_{d}\wedge y_{d}\geq|x-y|/4, then by (A3), (6.3), (6.4), (6.16) and (6.17),

t|xy|d+αB(x+21|xy|𝐞d, 41|xy|)J(x+t1/α𝐞d,z)J(z,y+t1/α𝐞d)𝑑z\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
t|xy|d+αB(x+21|xy|𝐞d, 41|xy|)𝑑zt|xy|α\displaystyle\asymp\frac{t}{|x-y|^{d+\alpha}}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}dz\asymp\frac{t}{|x-y|^{\alpha}}
(1t|xy|α)Bβ1,β1,0,β3(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)+t1/α)|xy|),\displaystyle\asymp\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{)},

Otherwise, if xdyd<|xy|/4x_{d}\wedge y_{d}<|x-y|/4, then xdydxdyd+|xy|<5|xy|/4x_{d}\vee y_{d}\leq x_{d}\wedge y_{d}+|x-y|<5|x-y|/4 so that

zdt1/αzd|xy|for zB(x+21|xy|𝐞d, 41|xy|)z_{d}-t^{1/\alpha}\asymp z_{d}\asymp|x-y|\quad\text{for $z\in B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)$}

by (6.16). Using this, (6.17) and (6.3), we get that for zB(x+21|xy|𝐞d, 41|xy|)z\in B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|),

Bβ1,β2,β3,β4(x+t1/α𝐞d,z)Bβ1,β2,β3,β4(x+t1/α𝐞d,z+t1/α𝐞d)\displaystyle B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},z)\asymp B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(x+t^{1/\alpha}{\mathbf{e}}_{d},z+t^{1/\alpha}{\mathbf{e}}_{d})
Aβ1,β2,β3,β4(t,x,z)Aβ1,β2,β3,β4(t,x,x+21|xy|𝐞d)\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,z)\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+2^{-1}|x-y|{\mathbf{e}}_{d}) (6.18)

and

Bβ1,β2,β3,β4(z,y+t1/α𝐞d)Bβ1,β2,β3,β4(z+t1/α𝐞d,y+t1/α𝐞d)\displaystyle B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(z,y+t^{1/\alpha}{\mathbf{e}}_{d})\asymp B_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(z+t^{1/\alpha}{\mathbf{e}}_{d},y+t^{1/\alpha}{\mathbf{e}}_{d})
Aβ1,β2,β3,β4(t,z,z)Aβ1,β2,β3,β4(t,x+21|xy|𝐞d,y).\displaystyle\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,z,z)\asymp A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+2^{-1}|x-y|{\mathbf{e}}_{d},y). (6.19)

By (A3), (6.17), (6.5), (6.5) and (6), we arrive at

t|xy|d+αB(x+21|xy|𝐞d, 41|xy|)J(x+t1/α𝐞d,z)J(z,y+t1/α𝐞d)𝑑z\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
tAβ1,β2,β3,β4(t,x,x+21|xy|𝐞d)Aβ1,β2,β3,β4(t,x+21|xy|𝐞d,y)|xy|d+αB(x+21|xy|𝐞d, 41|xy|)𝑑z\displaystyle\asymp\frac{tA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,x+2^{-1}|x-y|{\mathbf{e}}_{d})A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x+2^{-1}|x-y|{\mathbf{e}}_{d},y)}{|x-y|^{d+\alpha}}\int_{B(x+2^{-1}|x-y|{\mathbf{e}}_{d},\,4^{-1}|x-y|)}\!\!\!\!\!\!\!\!dz
t|xy|αAβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|)\displaystyle\asymp\frac{t}{|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{)}
(1t|xy|α)Bβ1,β1,0,β3(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)+t1/α)|xy|).\displaystyle\asymp\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{)}.

Hence, (1.6) and (1.7) are comparable.

Combining (6), Proposition 6.2 and Lemma 6.4, we get the following result.

Proposition 6.6.

There exists a constant C>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)\displaystyle p(t,x,y) C(1xdt1/α)q(1ydt1/α)q(td/αt|xy|d+α)[Aβ1,β2,β3,β4(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)
+(1t|xy|α)Aβ1,β2,β3,β4(t,x,x+|xy|2𝐞d)Aβ1,β2,β3,β4(t,x+|xy|2𝐞d,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{]}
(1xdt1/α)q(1ydt1/α)q(td/αt|xy|d+α)[Aβ1,β2,β3,β4(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)
+(1t|xy|α)Aβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|)].\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{]}.
Remark 6.7.

If β2<α+β1\beta_{2}<\alpha+\beta_{1}, then there is a constant C>0C>0 such that for all t>0t>0 and x,ydx,y\in{\mathbb{R}}^{d},

(1t|xy|α)Aβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|)CAβ1,β2,β3,β4(t,x,y).\displaystyle\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{)}\leq CA_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y).

Indeed, for ε:=(α+β1β2)/2>0{\varepsilon}:=(\alpha+\beta_{1}-\beta_{2})/2>0, using (10.1), we get that for all t>0t>0 and x,ydx,y\in{\mathbb{R}}^{d},

(1t|xy|α)Aβ1,β1,0,β3(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|)\displaystyle\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{)}
c1(1t|xy|α)(1xdt1/α|xy|)β1ε(1ydt1/α|xy|)β1ε\displaystyle\leq c_{1}\bigg{(}1\wedge\frac{t}{|x-y|^{\alpha}}\bigg{)}\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}\bigg{(}1\wedge\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}
=c1(1t1/α|xy|)ε(1xdt1/α|xy|)β1ε(1t1/α|xy|)β1+β2+ε(1ydt1/α|xy|)β1ε\displaystyle=c_{1}\bigg{(}1\wedge\frac{t^{1/\alpha}}{|x-y|}\bigg{)}^{{\varepsilon}}\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}\bigg{(}1\wedge\frac{t^{1/\alpha}}{|x-y|}\bigg{)}^{-\beta_{1}+\beta_{2}+{\varepsilon}}\bigg{(}1\wedge\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}-{\varepsilon}}
c1(1xdt1/α|xy|)β1(1ydt1/α|xy|)β2c1Aβ1,β2,β3,β4(t,x,y).\displaystyle\leq c_{1}\bigg{(}1\wedge\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{1}}\bigg{(}1\wedge\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\bigg{)}^{\beta_{2}}\leq c_{1}A_{\beta_{1},\beta_{2},\beta_{3},\beta_{4}}(t,x,y).

Therefore, in view of Proposition 6.2, Proposition 6.6 is relevant only if β2α+β1\beta_{2}\geq\alpha+\beta_{1}.

Lemma 6.8.

Suppose that β2=α+β1\beta_{2}=\alpha+\beta_{1}. Then for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

(|xy|αt)(xdydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
(1t|xy|α)Aβ1,β1,0,β3+β4+1(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|).\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 xdydx_{d}\leq y_{d}.

Assume first that ydt1/α<|xy|/4y_{d}\vee t^{1/\alpha}<|x-y|/4. Using β2=α+β1\beta_{2}=\alpha+\beta_{1} and (6)–(6), since xdydt1/αx_{d}\leq y_{d}\vee t^{1/\alpha}, we get

ydt1/α|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
ydt1/α|xy|/2(xdt1/αr)β1(ydt1/α|xy|)β1(r|xy|)α+β1\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}}
×logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+|xy|r)drrα+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}}
(xdt1/α|xy|)β1(ydt1/α|xy|)β1|xy|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}
×ydt1/α|xy|/2logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+|xy|r)drr.\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

ydt1/α|xy|/2logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+|xy|r)drr\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(ydt1/α)/|xy|2logβ3(e+|xy|s4(xdt1/α))logβ3(e+|xy|s4(ydt1/α))logβ4(e+4s)dss\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}
logβ3(e+|xy|4(xdt1/α))logβ3+β4+1(e+|xy|4(ydt1/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 ydt1/α<|xy|/4y_{d}\vee t^{1/\alpha}<|x-y|/4, we obtain

(|xy|αt)ydt1/α|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
(1t|xy|α)(xdt1/α|xy|)β1(ydt1/α|xy|)β1\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}}
×logβ3(e+|xy|4(xdt1/α))logβ3+β4+1(e+|xy|4(ydt1/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 ydt1/α|xy|/4y_{d}\vee t^{1/\alpha}\geq|x-y|/4, then log(e+|xy|(ydt1/α)|xy|)1\log\big{(}e+\frac{|x-y|}{(y_{d}\vee t^{1/\alpha})\wedge|x-y|}\big{)}\asymp 1. Thus, by (6),

(|xy|αt)|xy|/4|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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}}
(1t|xy|α)Aβ1,β1,0,β3+β4+1(t,x,y)logβ3(e+|xy|(xdt1/α)|xy|).\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 β2=α+β1\beta_{2}=\alpha+\beta_{1}. There exists a constant C>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)\displaystyle p(t,x,y) C(1xdt1/α)q(1ydt1/α)q(td/αt|xy|d+α)[Aβ1,β2,β3,β4(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)
+(1t|xy|α)Aβ1,β1,0,β3+β4+1(t,x,y)logβ3(e+|xy|((xdyd)t1/α)|xy|)].\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{]}.
Remark 6.10.

In this section, for Y¯\overline{Y}, (A3)(II) is only used to get (IUBS). Therefore, the results of Propositions 6.6 and 6.9 hold under weaker assumptions (A1), (A3)(I), (A4) and (IUBS).

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 β2<α+β1\beta_{2}<\alpha+\beta_{1}, and Theorem 7.10 which deals with the case β2α+β1\beta_{2}\geq\alpha+\beta_{1}.

Recall that U(r)={x=(x~,xd)d:|x~|<r/2, 0xd<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>0r>0, d2d\geq 2 and U(r)=[0,r/2)U(r)=[0,r/2) for d=1d=1, and that the function Aβ1,β2,β3,β4(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 pκ(t,x,y)p^{\kappa}(t,x,y). We remind the readers that, for an open set D¯+dD\subset{\overline{\mathbb{R}}}^{d}_{+} relative to the topology on ¯+d{\overline{\mathbb{R}}}^{d}_{+}, τD=τ¯D=inf{t>0:Y¯tD}\tau_{D}=\overline{\tau}_{D}=\inf\{t>0:\overline{Y}_{t}\notin D\} when we consider Y¯\overline{Y}, and τD=τDκ=inf{t>0:YtκD+d}\tau_{D}=\tau^{\kappa}_{D}=\inf\{t>0:Y^{\kappa}_{t}\notin D\cap{\mathbb{R}}^{d}_{+}\} when we consider YκY^{\kappa}.

Lemma 7.1.

Let b1,b30b_{1},b_{3}\geq 0 be constants with b1>0b_{1}>0 if b3>0b_{3}>0. Suppose that there exists a constant C0>0C_{0}>0 such that for all t>0t>0 and z,y+dz,y\in{\mathbb{R}}^{d}_{+},

p(t,z,y)C0(1zdt1/α)qAb1,0,b3,0(t,z,y)(td/αt|zy|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(C0)>0C=C(C_{0})>0 such that for all t>0t>0 and x=(0~,xd)+dx=(\widetilde{0},x_{d})\in{\mathbb{R}}^{d}_{+} with xd25x_{d}\leq 2^{-5},

x(τU(1)<t<ζ)Ct(1xdt1/α)q(xdt1/α)b1logb3(e+1xdt1/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>0t>0 and y+dy\in{\mathbb{R}}^{d}_{+},

y(|Yty|>23,t<ζ)=z+d,|zy|>23p(t,y,z)𝑑z\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
c1t(1ydt1/α)qz+d,|zy|>23(ydt1/α|zy|1)b1logb3(e+|zy|ydt1/α)dz|zy|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}}
c2t(1ydt1/α)q(ydt1/α)b1logb3(e+23ydt1/α)z+d,|zy|>23dz|zy|d+α/2+b1\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}}}
c3t(1ydt1/α)q(ydt1/α)b1logb3(e+1ydt1/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

supst,y+dy(|Ysy|22,s<ζ)c1supst,y+dz+d,|zy|22s|zy|d+α𝑑zc2t.\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 t1/(2c2)t\geq 1/(2c_{2}), then (7.2) follows from Corollary 5.2.

Let t<1/(2c2)t<1/(2c_{2}). By the strong Markov property and (7.4), we have

x(τU(1)<t<ζ,|YtYτU(1)|22)=𝔼x[YτU(1)(|YtτU(1)Y0|22):τU(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]
x(τU(1)<t<ζ)supst,y+dy(|Ysy|22,s<ζ)21x(τU(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,

x(τU(1)<t<ζ)\displaystyle{\mathbb{P}}_{x}(\tau_{U(1)}<t<\zeta) =2(x(τU(1)<t<ζ)21x(τU(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(x(τU(1)<t<ζ)x(τU(1)<t<ζ,|YtYτU(1)|22))\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{)}
=2x(τU(1)<t<ζ,|YtYτU(1)|<22).\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+dU(1)y\in{\mathbb{R}}^{d}_{+}\setminus U(1) and zB(y,22)z\in B(y,2^{-2}), we have |zx||yx||yz|>15/321/4>23|z-x|\geq|y-x|-|y-z|>15/32-1/4>2^{-3}. Therefore using (7) and (7), we have

x(τU(1)<t<ζ)2x(τU(1)<t<ζ,|YtYτU(1)|<22)\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})
2x(|Ytx|>23,t<ζ)c4t(1xdt1/α)q(xdt1/α)b1logb3(e+1xdt1/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>0t,k,r>0 and a0a\geq 0,

(1rt1/α)a(rt1/α)k=rk(1rt1/α)ak.\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>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)C(1xdydt1/α)qAβ1,0,β3,0(t,x,y)(td/αt|xy|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 β1=0\beta_{1}=0.

We assume β1>0\beta_{1}>0 and set an=β1nα2a_{n}=\beta_{1}\wedge\frac{n\alpha}{2} for n0n\geq 0. Below, we prove by induction that for any n0n\geq 0, there exists a constant C>0C>0 such that for all t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

p(t,x,y)C(1xdydt1/α)qAan,0,β3,0(t,x,y)(td/αt|xy|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 Aβ1,0,β3,0(t,x,y)A_{\beta_{1},0,\beta_{3},0}(t,x,y) is always larger than 1, (7.7) holds for n=0n=0. Suppose (7.7) holds for n1n-1. By symmetry and (3.30), we can assume without loss of generality that xdydx_{d}\leq y_{d}, x~=0\widetilde{x}=0 and |xy|=5|x-y|=5. If t>1t>1 or xd>25x_{d}>2^{-5}, then (7.7) follows from Proposition 5.1 and (6.4).

Let t1t\leq 1 and xd25x_{d}\leq 2^{-5}. Then ydxd+|xy|4+25y_{d}\leq x_{d}+|x-y|\leq 4+2^{-5} by the triangle inequality. Our goal is to show that there exists a constant c1>0c_{1}>0 independent of t,x,yt,x,y such that

p(t,x,y)c1t(1xdt1/α)q(xdt1/α)anlogβ3(e+ydt1/αxdt1/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 V1=U(1)V_{1}=U(1), V3=B(y,2)¯+dV_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V2=¯+d(V1V3)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

supst,zV2p(s,z,y)c2supst,z+d,|zy|2s|zy|d+α2dαc2t\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(V1,V3)supuV1,wV3(4|xu||yw|)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 b1=an1b_{1}=a_{n-1} and b3=β3b_{3}=\beta_{3}. Thus, since anan1α/2a_{n}-a_{n-1}\leq\alpha/2, we get from Lemma 7.1 and (7.9) that

x(τV1<t<ζ)supst,zV2p(s,z,y)\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
c3t(1xdt1/α)q(xdt1/α)an1(t1/α)α/2t1/2logβ3(e+1t1/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{)}\,
c3t(1xdt1/α)q(xdt1/α)an1(t1/α)anan1(sups1s1/2logβ3(e+1s1/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)
c4t(1xdt1/α)q(xdt1/α)an.\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 0tV3V1pV1(ts,x,u)(u,w)p(s,y,w)𝑑u𝑑w𝑑s\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) qα+anq\geq\alpha+a_{n} and 10xd<t1/α10x_{d}<t^{1/\alpha}.

Pick ε(0,β1){\varepsilon}\in(0,\beta_{1}) such that q<α+β1εq<\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 uV1u\in V_{1} and wV3w\in V_{3},

(u,w)c5Bβ1ε,0,0,0(u,w)c6(ud|uw|)β1εc6udβ1ε.\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

0tV3V1pV1(ts,x,u)(u,w)p(s,y,w)𝑑u𝑑w𝑑s\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
c60tV1pV1(ts,x,u)udβ1ε𝑑uV3p(s,y,w)𝑑w𝑑sc60V1pV1(s,x,u)udβ1ε𝑑u𝑑s\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
c7xdq=c7t(xdt1/α)q(t1/α)qαc8t(1xdt1/α)q(xdt1/α)an.\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 qαanq-\alpha\geq a_{n}, 10xd<t1/α10x_{d}<t^{1/\alpha} and xdt1/α1x_{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+anq<\alpha+a_{n} or 10xdt1/α10x_{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 uV1u\in V_{1} and wV3w\in V_{3},

(u,w)\displaystyle{\cal B}(u,w) c9Bβ1,0,β3,0(u,w)c10(ud|uw|)β1logβ3(e+wd|uw|ud|uw|)\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{)}
c10udβ1logβ3(e+wdud).\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 uuβ1logβ3(e+t/u)u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+t/u) is almost increasing) and Corollary 5.2, since β1>0\beta_{1}>0 and anβ1a_{n}\leq\beta_{1}, we get that for any 0<s<t0<s<t and wV3w\in V_{3},

uV1:ud<xdp(s,x,u)udβ1logβ3(e+wdud)𝑑uuV1:ud<xds1/αp(s,x,u)udβ1logβ3(e+wdud)𝑑u\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
c11(xds1/α)β1logβ3(e+wdxds1/α)uV1:ud<xds1/αp(s,x,u)𝑑u\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
c11x(ζ>s)(xds1/α)anlogβ3(e+wdxds1/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{)}
c12(1xds1/α)q(xds1/α)anlogβ3(e+wdxds1/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 anβ1a_{n}\leq\beta_{1} and an<α+an1a_{n}<\alpha+a_{n-1}, we get that for any 0<s<t0<s<t and wV3w\in V_{3},

uV1:udxdp(s,x,u)udβ1logβ3(e+wdud)𝑑u\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
c13(1xds1/α)quV1:udxd(xds1/α|xu|1)an1logβ3(e+|xu|(xds1/α)|xu|)\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{)}
×(sd/αs|xu|d+α)udanlogβ3(e+wdud)du\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
c14(1xds1/α)q(xds1/α)anlogβ3(e+wdxds1/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 wd<ydw_{d}<y_{d} and wdydw_{d}\geq y_{d}, and using the induction hypothesis and Lemma 10.10 again, we also get that for any 0<s<t0<s<t,

V3p(s,y,w)logβ3(e+wdxd(ts)1/α)𝑑w\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 c15(1yds1/α)qlogβ3(e+yds1/αxd(ts)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{)}
c15logβ3(e+ydt1/αxd(ts)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

0tV3V1pV1(ts,x,u)(u,w)p(s,y,w)𝑑u𝑑w𝑑s\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
c100tV3p(s,y,w)V1p(ts,x,u)udβ1logβ3(e+wdud)𝑑u𝑑w𝑑s\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
c160t(1xd(ts)1/α)q(xd(ts)1/α)anV3p(s,y,w)logβ3(e+wdxd(ts)1/α)𝑑w𝑑s\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
c17xdan0t(1xd(ts)1/α)qanlogβ3(e+ydt1/αxd(ts)1/α)ds=: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+anq<\alpha+a_{n}, we get from Lemma 10.4 that

Ic18txdan(1xdt1/α)qanlogβ3(e+ydt1/αxdt1/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 10xdt1/α10x_{d}\geq t^{1/\alpha}, we also get from Lemma 10.4 that

I\displaystyle I c17xdan0tlogβ3(e+ydt1/αxd(ts)1/α)𝑑sc19txdanlogβ3(e+ydt1/αxdt1/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{)}
10qanc19txdan(1xdt1/α)qanlogβ3(e+ydt1/αxdt1/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<t10<t\leq 1 and x,y+dx,y\in{\mathbb{R}}^{d}_{+} be such that x~=0~\widetilde{x}=\widetilde{0}, xd25x_{d}\leq 2^{-5} and |xy|=5|x-y|=5. Set V1=U(1)V_{1}=U(1), V3=B(y,2)¯+dV_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V2=¯+d(V1V3)V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}). There exists a constant C>0C>0 independent of t,xt,x and yy such that

x(τV1<t<ζ)supst,zV2p(s,z,y)\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y) Ct2(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/α)β1\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}}
×logβ3(e+1xdt1/α)logβ3(e+1ydt1/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 ydxd+|xy|5+25<6y_{d}\leq x_{d}+|x-y|\leq 5+2^{-5}<6, for any 0<s10<s\leq 1 and z+dz\in{\mathbb{R}}^{d}_{+} with |zy|2|z-y|\geq 2,

(yds1/α)|zy|13(yds1/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),

supst,zV2p(s,z,y)c1supst,z+d,|zy|2(1zdyds1/α)qAβ1,0,β3,0(s,z,y)s|zy|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}}
c1supst,z+d,|zy|2(1yds1/α)q(yds1/α|zy|)β1logβ3(e+3|zy|yds1/α)s|zy|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 uuβ1logβ3(e+1/u)u\mapsto u^{\beta_{1}}\log^{\beta_{3}}(e+1/u) is almost increasing, we have that for any 0<st10<s\leq t\leq 1 and z+dz\in{\mathbb{R}}^{d}_{+} with |zy|2|z-y|\geq 2,

(1yds1/α)q(yds1/α|zy|)β1logβ3(e+3|zy|yds1/α)s|zy|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}}
c2(1yds1/α)q(yds1/α)β1logβ3(e+1yds1/α)s|zy|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}}
2dαc2s(1yds1/α)q(yds1/α)β1logβ3(e+1yds1/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+β1εq<\alpha+\beta_{1}-{\varepsilon}. Using (7.6), we see that

s(1yds1/α)q(yds1/α)β1logβ3(e+1yds1/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{)}
=sydβ1ε(1yds1/α)qβ1+ε(yds1/α)εlogβ3(e+1yds1/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+β1qε)/αydβ1ε(yds1/α)qβ1+ε(yds1/α)εlogβ3(e+1yds1/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 ss(1yds1/α)q(yds1/α)β1logβ3(e+1yds1/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

supst,zV2p(s,z,y)\displaystyle\sup_{s\leq t,\,z\in V_{2}}p(s,z,y) c3supsts(1yds1/α)q(yds1/α)β1logβ3(e+1yds1/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{)}
c4t(1ydt1/α)q(ydt1/α)β1logβ3(e+1ydt1/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 a1=β1a_{1}=\beta_{1} and a3=β3a_{3}=\beta_{3} by Lemma 7.2. Thus, by Lemma 7.1 and (7) we obtain

x(τV1<t<ζ)supst,zV2p(s,z,y)\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y) c4t2(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/α)β1\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}}
×logβ3(e+1xdt1/α)logβ3(e+1ydt1/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 η1,η2,γ0\eta_{1},\eta_{2},\gamma\geq 0. There exists a constant C>0C>0 such that for any x+dx\in{\mathbb{R}}^{d}_{+} and any s,k,l>0s,k,l>0,

B+(x,2)p(s,x,z)zdγlogη1(e+kzd)logη2(e+zdl)𝑑z\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
Cxdγ(1xds1/α)qγlogη1(e+kxds1/α)logη2(e+xds1/αl)\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{)}
+C𝟏{γ>α+β1}sxdβ1(1xds1/α)qβ1logβ3(e+2xds1/α)logη1(e+k)logη2(e+1l)\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{)}
+C𝟏{γ=α+β1,xds1/α<2}sxdβ1(1xds1/α)qβ1\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}}
×xds1/α2logβ3(e+rxds1/α)logη1(e+kr)logη2(e+rl)drr.\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+dx,z\in{\mathbb{R}}^{d}_{+} and s>0s>0, by Lemmas 7.2 and 10.2,

p(s,x,z)c1(1xds1/α)qAβ1,0,β3,0(s,x,z)(sd/αs|xz|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)
c2(1xds1/α)q(1xds1/α|xz|)β1logβ3(e+|xz|(xds1/α)|xz|)(sd/αs|xz|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>0C>0 such that

p(t,x,y)\displaystyle p(t,x,y) C(1xdt1/α)q(1ydt1/α)qAβ1,β2(α+β1ε),β3,β4(t,x,y)(td/αt|xy|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>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+}.

This theorem will proved by using several lemmas. We first introduce some additional notation: For b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0, t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+}, define

t,1(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4}) :=0t/2B+(y,2)B+(x,2)𝟏{udwd}p(ts,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)
×udb1wdb2logb3(e+wdud)logb4(e+1wd)dudwds,\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)
t,2(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4}) :=0t/2B+(y,2)B+(x,2)𝟏{wdud}p(ts,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)
×udb2wdb1logb3(e+udwd)logb4(e+1ud)dudwds.\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 b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0. Let b1[0,b1]b_{1}^{\prime}\in[0,b_{1}] and b2:=b1+b2b1b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime}. Then

t,i(x,y;b1,b2,b3,b4)t,i(x,y;b1,b2,b3,b4),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 udwdu_{d}\leq w_{d}, then udb1wdb2udb1wdb2u_{d}^{b_{1}}w_{d}^{b_{2}}\leq u_{d}^{b_{1}^{\prime}}w_{d}^{b_{2}^{\prime}}, implying that t,1(x,y;b1,b2,b3,b4)t,1(x,y;b1,b2,b3,b4).{\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 b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0 with b1b2<α+β1b_{1}\vee b_{2}<\alpha+\beta_{1}, x,y+dx,y\in{\mathbb{R}}^{d}_{+} with |xy|=5|x-y|=5, and t(0,1]t\in(0,1].

(i) If b2>qαb_{2}>q-\alpha or ydt1/αy_{d}\geq t^{1/\alpha}, then there exists a constant C>0C>0 independent of t,x,yt,x,y such that

t,1(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})
Ctxdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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 b1>qαb_{1}>q-\alpha or ydt1/αy_{d}\geq t^{1/\alpha}, then there exists a constant C>0C>0 independent of t,x,yt,x,y such that

t,2(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
Ctxdb2ydb1(1xdt1/α)qb2(1ydt1/α)qb1logb3(e+xdt1/αydt1/α)logb4(e+1xdt1/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 tstt-s\asymp t if 0st/20\leq s\leq t/2, we see that for 0st/20\leq s\leq t/2,

B+(x,2)p(ts,x,u)udb1logb3(e+wdud)𝑑uc1xdb1(1xdt1/α)qb1logb3(e+wdxdt1/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 0st/20\leq s\leq t/2,

B+(y,2)B+(x,2)p(ts,x,u)p(s,y,w)udb1wdb2logb3(e+wdud)logb4(e+1wd)𝑑u𝑑w\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
c1xdb1(1xdt1/α)qb1B+(y,2)p(s,y,w)wdb2logb3(e+wdxdt1/α)logb4(e+1wd)𝑑w\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
c2xdb1ydb2(1xdt1/α)qb1(1yds1/α)qb2logb3(e+yds1/αxdt1/α)logb4(e+1yds1/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

t,1(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,1}(x,y;b_{1},b_{2},b_{3},b_{4})
c2xdb1ydb2(1xdt1/α)qb10t2(1yds1/α)qb2logb3(e+yds1/αxdt1/α)logb4(e+1yds1/α)𝑑s\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
c2xdb1ydb2(1xdt1/α)qb10t(1yds1/α)qb2logb3(e+yds1/αxd(ts)1/α)logb4(e+1yds1/α)𝑑s\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
c3txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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{)}.

\Box

Lemma 7.8.

Let b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0 be such that b1>qαb_{1}>q-\alpha and b1b2<α+β1b_{1}\vee b_{2}<\alpha+\beta_{1}. Assume that b2>0b_{2}>0 if b4>0b_{4}>0. Then, there exists a constant C>0C>0 such that for all x,y+dx,y\in{\mathbb{R}}^{d}_{+} with xdydx_{d}\leq y_{d} and |xy|=5|x-y|=5, and all t(0,1]t\in(0,1],

t,1(x,y;b1,b2,b3,b4)+t,2(x,y;b1,b2,b3,b4)\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})
Ctxdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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

t,1(y,x;b1,b2,b3,b4)+t,2(y,x;b1,b2,b3,b4)\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})
Ctxdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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

b1b2+(1δ)(b1b2)<α+β1,\displaystyle b_{1}\vee b_{2}+(1-\delta)(b_{1}\wedge b_{2})<\alpha+\beta_{1}, (7.29)

and let b1:=δ(b1b2)b_{1}^{\prime}:=\delta(b_{1}\wedge b_{2}) and b2:=b1+b2b1b_{2}^{\prime}:=b_{1}+b_{2}-b_{1}^{\prime}. Then we see that b1[0,b1b2]b_{1}^{\prime}\in[0,b_{1}\wedge b_{2}], b2b1>qαb_{2}^{\prime}\geq b_{1}>q-\alpha and

b1b2=b1b2+(1δ)(b1b2)<α+β1\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 xdydx_{d}\leq y_{d} and b2b1=b2b1b_{2}^{\prime}-b_{1}=b_{2}-b_{1}^{\prime}, we see that

(xdt1/α)b2b1(ydt1/α)b1b2logb4(e+1xdt1/α)logb4(e+1ydt1/α)c1.\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 b4=0b_{4}=0, (7.31) clearly holds with c1=1c_{1}=1. If b4>0b_{4}>0, then b2>0b_{2}>0 so that b2>δb2b1b_{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: ydt1/αy_{d}\geq t^{1/\alpha} and yd<t1/αy_{d}<t^{1/\alpha}.

Assume first that ydt1/αy_{d}\geq t^{1/\alpha}. The desired bound for t,1(x,y;b1,b2,b3,b4){\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 ydt1/αy_{d}\geq t^{1/\alpha}), (7.6) in the equality, and (7.31) in the last inequality, we get that

t,2(x,y;b1,b2,b3,b4)t,2(x,y;b1,b2,b3,b4)\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})
c2txdb2ydb1(1xdt1/α)qb2(1ydt1/α)qb1logb3(e+ydt1/αxdt1/α)logb4(e+1xdt1/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{)}
=c2txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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{)}
×(xdt1/α)b2b1(ydt1/α)b1b2logb4(e+1xdt1/α)logb4(e+1ydt1/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{)}
c1c2txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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 yd<t1/αy_{d}<t^{1/\alpha}. Using Lemma 7.6 and Lemma 7.7(i) (which uses b2>qαb_{2}^{\prime}>q-\alpha), we get

t,1(x,y;b1,b2,b3,b4)t,1(x,y;b1,b2,b3,b4)\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})
c3txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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 b1>qαb_{1}>q-\alpha, we get from Lemma 7.7(ii) that

t,2(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t,2}(x,y;b_{1},b_{2},b_{3},b_{4})
c4txdb2ydb1(1xdt1/α)qb2(1ydt1/α)qb1logb3(e+xdt1/αydt1/α)logb4(e+1xdt1/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 xdyd<t1/αx_{d}\leq y_{d}<t^{1/\alpha} and b1+b2=b1+b2b_{1}^{\prime}+b_{2}^{\prime}=b_{1}+b_{2}, it holds that xdt1/α=ydt1/α=t1/αx_{d}\vee t^{1/\alpha}=y_{d}\vee t^{1/\alpha}=t^{1/\alpha} and

xdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2=xdb2ydb1(1xdt1/α)qb2(1ydt1/α)qb1\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}}
=xdqydqt(2qb1b2)/α=xdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2.\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 b2>qαb_{2}^{\prime}>q-\alpha), (7.6) together with the fact that xdydx_{d}\leq y_{d} in the third inequality, (7.31) in the last inequality, we obtain

t,1(y,x;b1,b2,b3,b4)t,1(y,x;b1,b2,b3,b4)\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})
c5txdb2ydb1(1xdt1/α)qb2(1ydt1/α)qb1logb3(e+xdt1/αydt1/α)logb4(e+1xdt1/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{)}
c5txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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{)}
×(xdt1/α)b2b1(ydt1/α)b1b2logb4(e+1xdt1/α)logb4(e+1ydt1/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{)}
c1c5txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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

t,2(y,x;b1,b2,b3,b4)\displaystyle{\cal I}_{t,2}(y,x;b_{1},b_{2},b_{3},b_{4})
c3txdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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:=β2(α+β1ε)\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 xdyd25x_{d}\leq y_{d}\wedge 2^{-5}, x~=0\widetilde{x}=0 and |xy|=5|x-y|=5, and then it is enough to show that there exists a constant c1>0c_{1}>0 independent of xx and yy such that for any t1t\leq 1,

p(t,x,y)\displaystyle p(t,x,y) c1t(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/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}}
×logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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 t1t\leq 1. Set V1=U(1)V_{1}=U(1), V3=B(y,2)¯+dV_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V2=¯+d(V1V3)V_{2}={\overline{\mathbb{R}}}^{d}_{+}\setminus(V_{1}\cup V_{3}). By Lemma 7.3,

x(τV1<t<ζ)supst,zV2p(s,z,y)\displaystyle{\mathbb{P}}_{x}(\tau_{V_{1}}<t<\zeta)\sup_{s\leq t,\,z\in V_{2}}p(s,z,y)
c2t2(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/α)β1log2β3(e+1t1/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{)}
=c2t(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/α)α+β1ε(ydt1/α)α+εtlog2β3(e+1t1/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{)}
c2t(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/α)β^2tε/αlog2β3(e+1t1/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{)}
c3t(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1(ydt1/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>0C^{\prime}>0 such that

I\displaystyle I :=0tV3V1pV1(ts,x,u)(u,w)p(s,y,w)𝑑u𝑑w𝑑s\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
Ctxdβ1ydβ^2(1xdt1/α)qβ1(1ydt1/α)qβ^2logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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 |uw|1|u-w|\asymp 1 and udwdyd+27+25u_{d}\vee w_{d}\leq y_{d}+2\leq 7+2^{-5} for uV1u\in V_{1} and wV3w\in V_{3}, we have

I\displaystyle I c40tV3V1𝟏{udwd}p(ts,x,u)p(s,y,w)udβ1wdβ^2logβ3(e+wdud)logβ4(e+1wd)𝑑u𝑑w𝑑s\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
+c40tV3V1𝟏{wdud}p(ts,x,u)p(s,y,w)udβ^2wdβ1logβ3(e+udwd)logβ4(e+1ud)𝑑u𝑑w𝑑s\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
=c4((0t/2+t/2t)V3V1𝟏{udwd}+(0t/2+t/2t)V3V1𝟏{wdud}).\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~=ts\widetilde{s}=t-s in integrals t/2t\int_{t/2}^{t},

Ic4i=12t,i(x,y;β1,β^2,β3,β4)+c4i=12t,i(y,x;β1,β^2,β3,β4),\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 t,i(x,y;β1,β^2,β3,β4){\cal I}_{t,i}(x,y;\beta_{1},\widehat{\beta}_{2},\beta_{3},\beta_{4}), 1i21\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 β2<α+β1\beta_{2}<\alpha+\beta_{1}.

Corollary 7.9.

Suppose that β2<α+β1\beta_{2}<\alpha+\beta_{1}. There exists a constant C>0C>0 such that

p(t,x,y)\displaystyle p(t,x,y) C(1xdt1/α)q(1ydt1/α)qAβ1,β2,β3,β4(t,x,y)(td/αt|xy|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>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+}.

Here is the second main result of the section.

Theorem 7.10.

Suppose that β2α+β1\beta_{2}\geq\alpha+\beta_{1}. There exists a constant C>0C>0 such that

p(t,x,y)\displaystyle p(t,x,y) C(1xdt1/α)q(1ydt1/α)q(td/αt|xy|d+α)[Aβ1,β2,β3,β4(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)
+(1t|xy|α)logβ3(e+|xy|((xdyd)+t1/α)|xy|)\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{β2>α+β1}Aβ1,β1,0,β3(t,x,y)+𝟏{β2=α+β1}Aβ1,β1,0,β3+β4+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>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+}.

Again, we first introduce additional notation and prove a lemma. For b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0, t>0t>0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+}, define

t(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4}) :=0tB+(y,2)B+(x,2)p(ts,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)
×udb1wdb2logb3(e+wdud)logb4(e+1wd)dudwds.\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 b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0 be such that b1b2>qαb_{1}\wedge b_{2}>q-\alpha and b1<α+β1b_{1}<\alpha+\beta_{1}. There exists a constant C>0C>0 such that for all x,y+dx,y\in{\mathbb{R}}^{d}_{+} with |xy|=5|x-y|=5, and all t(0,1]t\in(0,1],

t(x,y;b1,b2,b3,b4)\displaystyle{\cal I}_{t}(x,y;b_{1},b_{2},b_{3},b_{4})
Ctxdb1ydb2(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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{b2>α+β1}Ct2xdb1ydβ1(1xdt1/α)qb1(1ydt1/α)qβ1logb3(e+1xdt1/α)logb3(e+1ydt1/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{b2=α+β1,yd<2}Ctxdb1ydβ1(1xdt1/α)qb1(1ydt1/α)qβ1\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}}
×yd2(rαt)logb3(e+rxdt1/α)logb3(e+rydt1/α)logb4(e+1r)drr.\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 uu; note that b1<α+β1b_{1}<\alpha+\beta_{1}) and the second inequality (integration with respect to ww), we get

t(x,y;b1,b2,b3,b4)c1xdb10t(1xd(ts)1/α)qb1\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)wdb2logb3(e+wdxd(ts)1/α)logb4(e+1wd)dwds\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
c2(xdb1ydb2I1+𝟏{b2>α+β1}xdb1ydβ1I2+𝟏{b2=α+β1,yd<2}xdb1ydβ1I3),\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

I1:=0t(1xd(ts)1/α)qb1(1yds1/α)qb2logb3(e+yds1/αxd(ts)1/α)logb4(e+1yds1/α)𝑑s,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,
I2:=0ts(1xd(ts)1/α)qb1(1yds1/α)qβ1logb3(e+2yds1/α)logb3(e+1xd(ts)1/α)𝑑s,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

I3:=\displaystyle I_{3}:= 0ts(1xd(ts)1/α)qb1(1yds1/α)qβ1\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}}
×yds1/α2logb3(e+ryds1/α)logb3(e+rxd(ts)1/α)logb4(e+1r)drrds.\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 I1I_{1} and I2I_{2}, Lemma 10.6 to I3I_{3}, and (10.2), wee see that

I1c3t(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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{)},
I2c3t2(1xdt1/α)qb1(1ydt1/α)qβ1logb3(e+1xdt1/α)logb3(e+1ydt1/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

I2\displaystyle I_{2}\leq c3t(1xdt1/α)qb1(1ydt1/α)qβ1\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}}
×yd2(rαt)logb3(e+rxdt1/α)logb3(e+rydt1/α)logb4(e+1r)drr.\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 xdyd25x_{d}\leq y_{d}\wedge 2^{-5}, x~=0\widetilde{x}=0 and |xy|=5|x-y|=5, and then it is enough to show that there exists a constant c1>0c_{1}>0 independent of xx and yy such that for any t1t\leq 1,

p(t,x,y)c1t(1xdt1/α)q(1ydt1/α)q(xdt1/α)β1\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}}
×[(ydt1/α)β2logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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{β2>α+β1}t(ydt1/α)β1logβ3(e+1xdt1/α)logβ3(e+1ydt1/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{β2=α+β1}t(ydt1/α)β1logβ3(e+1xdt1/α)logβ3+β4+1(e+1ydt1/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 t1t\leq 1. Set V1=U(1)V_{1}=U(1), V3=B(y,2)¯+dV_{3}=B(y,2)\cap{\overline{\mathbb{R}}}^{d}_{+} and V2=¯+d(V1V3)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:=0tV3V1pV1(ts,x,u)(u,w)p(s,y,w)𝑑u𝑑w𝑑sI:=\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 |uw|1|u-w|\asymp 1 for uV1u\in V_{1} and wV3w\in V_{3}, using the change of variables s~=ts\widetilde{s}=t-s we have

Ic2(t(x,y;β1,β2,β3,β4)+t(y,x;β1,β2,β3,β4)),\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 t(x,y;β1,β2,β3,β4){\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 c3t(1xdt1/α)q(1ydt1/α)qc_{3}t(1\wedge\frac{x_{d}}{t^{1/\alpha}})^{q}(1\wedge\frac{y_{d}}{t^{1/\alpha}})^{q} times

(xdt1/α)β1(ydt1/α)β2logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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)
+(xdt1/α)β2(ydt1/α)β1logβ3(e+xdt1/αydt1/α)logβ4(e+1xdt1/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{β2>α+β1}t(xdt1/α)β1(ydt1/α)β1logβ3(e+1xdt1/α)logβ3(e+1ydt1/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{β2=α+β1}(xdt1/α)β1(ydt1/α)β1\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}}
×xd2(rαt)logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+1r)drr.\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 xdyd2x_{d}\leq y_{d}\wedge 2 to get that 𝟏{yd<2}yd2xd2{\bf 1}_{\{y_{d}<2\}}\int_{y_{d}}^{2}\leq\int_{x_{d}}^{2}.

Since β2>β1\beta_{2}>\beta_{1} and xdydx_{d}\leq y_{d}, clearly

(xdt1/α)β2logβ3(e+xdt1/αydt1/α)(xdt1/α)β1logβ3(e+ydt1/αxdt1/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=(β2β1)/β4)\varepsilon=(\beta_{2}-\beta_{1})/\beta_{4})), we get

(ydt1/α)β1logβ4(e+1xdt1/α)c4(ydt1/α)β2logβ4(e+1ydt1/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 β2>α+β1\beta_{2}>\alpha+\beta_{1}.

Assume now that β2=α+β1\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

xd2(rαt)logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+1r)drr\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}
c5tlogβ3(e+1xdt1/α)logβ3+β4+1(e+1ydt1/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{)}
+c5(ydt1/α)β2β1logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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 b1=β3b_{1}=\beta_{3}, b2=β4b_{2}=\beta_{4}, k=xdt1/αk=x_{d}\vee t^{1/\alpha} and l=ydt1/αl=y_{d}\vee t^{1/\alpha}), it holds that

ydt1/α2(rαt)logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+1r)drr\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}
tydt1/α2logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+1r)drr\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}
c6tlogβ3(e+1xdt1/α)logβ3+β4+1(e+1ydt1/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

xdydt1/α(rαt)logβ3(e+rxdt1/α)logβ3(e+rydt1/α)logβ4(e+1r)drr\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}
logβ3(e+1)logβ3(e+ydt1/αxdt1/α)xdydt1/αlogβ4(e+1r)drr1α\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}}
c7logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/α)xdydt1/α(ydt1/αr)α/2drr1α\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}}
c8(ydt1/α)αlogβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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{)}
=c8(ydt1/α)β2β1logβ3(e+ydt1/αxdt1/α)logβ4(e+1ydt1/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,)×+d×+d(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}\times{\mathbb{R}}^{d}_{+}.

We first note that by (A3) and (6.4),

td/α(tJ(x+t1/α𝐞d,y+t1/α𝐞d))td/αtBβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d)|xy|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}}
(td/αt|xy|d+α)Bβ1,β2,β3,β4(x+t1/α𝐞d,y+t1/α𝐞d),\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)td/α\overline{p}(t,x,y)\asymp t^{-d/\alpha} when t1/α|xy|/8t^{1/\alpha}\geq|x-y|/8 and this implies that (1.1) holds for t1/α|xy|/8t^{1/\alpha}\geq|x-y|/8. Moreover, (1.1) for β2<α+β1\beta_{2}<\alpha+\beta_{1} follows from (1.1), (6.3) and (8).

By (A3) and (6.3), we have that when t1/α<|xy|/8t^{1/\alpha}<|x-y|/8,

t(xdydt1/α)(|xy|/4)|xy|/2J(x+t1/α𝐞d,x+r𝐞d)J(x+r𝐞d,y+t1/α𝐞d)rd1𝑑r\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|xy|d+α(xdydt1/α)(|xy|/4)|xy|/2Aβ1,β2,β3,β4(t,x,x+r𝐞d)Aβ1,β2,β3,β4(t,x+r𝐞d,y)drrα+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 β2=α+β1\beta_{2}=\alpha+\beta_{1} and t1/α<|xy|/8t^{1/\alpha}<|x-y|/8, (1.1) follows from (1.1), (8) and Lemma 6.8, and that, for β2>α+β1\beta_{2}>\alpha+\beta_{1} and t1/α<|xy|/8t^{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 t1/α<|xy|/8t^{1/\alpha}<|x-y|/8,

t(xdydt1/α)(|xy|/4)|xy|/2J(x+t1/α𝐞d,x+r𝐞d)J(x+r𝐞d,y+t1/α𝐞d)rd1𝑑r\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
c1t|xy|d+2αBβ1,β1,0,β3(x+t1/α𝐞d,y+t1/α𝐞d)logβ3(e+|xy|((xdyd)t1/α)|xy|).\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 β2>α+β1\beta_{2}>\alpha+\beta_{1} and t1/α<|xy|/8t^{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, pκ(t,x,y)c1(1xdt1/α)qκ(1ydt1/α)qκp¯(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,)×+d×+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, pκ(t,x,y)c2(1xdt1/α)qκ(1ydt1/α)qκp¯(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,)×+d×+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,)×+d(t,y)\in(0,\infty)\times{\mathbb{R}}^{d}_{+}, the map xpκ(t,x,y)x\mapsto p^{\kappa}(t,x,y) vanishes continuously on +d\partial{\mathbb{R}}^{d}_{+}. Hence, using the joint continuity of pκ(t,x,y)p^{\kappa}(t,x,y) and upper heat kernel estimates, we deduce that YκY^{\kappa} is a Feller process. By Remark 3.12, the proof is complete. \Box

9. Green function estimates

In this section, we give proofs of Theorems 1.3 and 1.4.

Proof of Theorem 1.3. When d>αd>\alpha, we get the upper bound of (1.11) from Corollary 3.13. On the other hand, by Lemma 3.7 and Remark 3.12, we have

G¯(x,y)|xy|αp¯(t,x,y)𝑑tc1|xy|αtd/α𝑑t={c1αdα|xy|d+αif d>α;if dα.\displaystyle\overline{G}(x,y)\geq\int_{|x-y|^{\alpha}}^{\infty}\overline{p}(t,x,y)dt\geq c_{1}\int_{|x-y|^{\alpha}}^{\infty}t^{-d/\alpha}dt=\begin{cases}\frac{c_{1}\alpha}{d-\alpha}|x-y|^{-d+\alpha}\;\;&\mbox{if }d>\alpha;\\[4.0pt] \infty\;\;&\mbox{if }d\leq\alpha.\end{cases}

The proof is complete. \Box

In the remainder of this section, we assume the setting of Theorem 1.4 holds and denote by qq the constant qκq_{\kappa} in (1.8), which is strictly positive.

Let x,y+dx,y\in{\mathbb{R}}^{d}_{+} be such that xdydx_{d}\leq y_{d} and |xy|=1|x-y|=1. From Theorem 7.5, Proposition 6.2 and (6.4), we have for t1t\leq 1,

pκ(t,x,y)\displaystyle p^{\kappa}(t,x,y) Ct(1xdt1/α)q(1ydt1/α)q((xdt1/α)1)β1((ydt1/α)1)β2(α/2+β1)\displaystyle\leq Ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\big{(}(x_{d}\vee t^{1/\alpha})\wedge 1\big{)}^{\beta_{1}}\big{(}(y_{d}\vee t^{1/\alpha}\big{)}\wedge 1\big{)}^{\beta_{2}\wedge(\alpha/2+\beta_{1})}
×logβ3(e+(ydt1/α)1(xdt1/α)1)logβ4(e+1(ydt1/α)1),\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{(y_{d}\vee t^{1/\alpha})\wedge 1}{(x_{d}\vee t^{1/\alpha})\wedge 1}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{(y_{d}\vee t^{1/\alpha})\wedge 1}\bigg{)}, (9.1)
pκ(t,x,y)\displaystyle p^{\kappa}(t,x,y) ct(1xdt1/α)q(1ydt1/α)q((xdt1/α)1)β1((ydt1/α)1)β2\displaystyle\geq ct\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}\big{(}(x_{d}\vee t^{1/\alpha})\wedge 1\big{)}^{\beta_{1}}\big{(}(y_{d}\vee t^{1/\alpha}\big{)}\wedge 1\big{)}^{\beta_{2}}
×logβ3(e+(ydt1/α)1(xdt1/α)1)logβ4(e+1(ydt1/α)1),\displaystyle\quad\times\log^{\beta_{3}}\bigg{(}e+\frac{(y_{d}\vee t^{1/\alpha})\wedge 1}{(x_{d}\vee t^{1/\alpha})\wedge 1}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{(y_{d}\vee t^{1/\alpha})\wedge 1}\bigg{)}, (9.2)

and for t>1t>1,

pκ(t,x,y)td/α(1xdt1/α)q(1ydt1/α)q.\displaystyle p^{\kappa}(t,x,y)\asymp t^{-d/\alpha}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}. (9.3)
Lemma 9.1.

Let x,y+dx,y\in{\mathbb{R}}^{d}_{+} be such that xdydx_{d}\leq y_{d} and |xy|=1|x-y|=1. Set q^:=2α+β1+β2q\widehat{q}:=2\alpha+\beta_{1}+\beta_{2}-q. Then we have

01pκ(t,x,y)𝑑t{(xd1)q(yd1)q if q<q^,(xd1)q(yd1)qlogβ4+1(e+1yd1) if q=q^,(xd1)q(yd1)q^logβ4(e+1yd1) if q>q^,\int_{0}^{1}p^{\kappa}(t,x,y)dt\asymp\begin{cases}(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}&\mbox{ if }q<\widehat{q},\\[5.0pt] \displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}\log^{\beta_{4}+1}\left(e+\frac{1}{y_{d}\wedge 1}\right)&\mbox{ if }q=\widehat{q},\\[8.0pt] \displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{\widehat{q}}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}\wedge 1}\right)&\mbox{ if }q>\widehat{q},\end{cases}

where the comparison constant is independent of x,yx,y.

Proof. Set G1:=01pκ(t,x,y)𝑑tG_{1}:=\int_{0}^{1}p^{\kappa}(t,x,y)dt and β^2:=β2(α/2+β1)\widehat{\beta}_{2}:=\beta_{2}\wedge(\alpha/2+\beta_{1}).

(Case 1) xd1x_{d}\geq 1: We have from (9) and (9) that G101t𝑑t1.G_{1}\asymp\int_{0}^{1}tdt\asymp 1.

(Case 2) yd1>xdy_{d}\geq 1>x_{d}: By (9) and (7.6),

G1\displaystyle G_{1} c1xdβ11/21t(1xdt1/α)qβ1𝑑t21c1xdβ11/21(1xd)qβ1𝑑t=22c1xdq.\displaystyle\geq c_{1}x_{d}^{\beta_{1}}\int_{1/2}^{1}t\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q-\beta_{1}}dt\geq 2^{-1}c_{1}x_{d}^{\beta_{1}}\int_{1/2}^{1}(1\wedge x_{d})^{q-\beta_{1}}dt=2^{-2}c_{1}x_{d}^{q}.

Besides, since α+β1q>0\alpha+\beta_{1}-q>0, we get from (9) and (7.6) that

G1\displaystyle G_{1} c2xdβ101t(1xdt1/α)qβ1logβ3(e+1t1/α)𝑑t\displaystyle\leq c_{2}x_{d}^{\beta_{1}}\int_{0}^{1}t\bigg{(}1\wedge\frac{x_{d}}{t^{1/\alpha}}\bigg{)}^{q-\beta_{1}}\log^{\beta_{3}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt
c3xdq01t(α+β1q)/αlogβ3(e+1t1/α)𝑑tc4xdq.\displaystyle\leq c_{3}x_{d}^{q}\int_{0}^{1}t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt\leq c_{4}x_{d}^{q}.

(Case 3) 1>ydxd1>y_{d}\geq x_{d}: Note that

ydα1t1+(q^q)/αlogβ4(e+1t1/α)𝑑t{1 if q<q^;logβ4+1(e+1/yd) if q=q^;ydq^qlogβ4(e+1/yd) if q>q^.\displaystyle\int_{y_{d}^{\alpha}}^{1}t^{-1+(\widehat{q}-q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt\asymp\begin{cases}1&\mbox{ if }\,q<\widehat{q};\\[2.0pt] \displaystyle\log^{\beta_{4}+1}(e+1/y_{d})&\mbox{ if }\,q=\widehat{q};\\[2.0pt] \displaystyle y_{d}^{\widehat{q}-q}\log^{\beta_{4}}(e+1/y_{d})&\mbox{ if }\,q>\widehat{q}.\end{cases} (9.4)

For the lower bound, we get from (9) and (9.4) that

G1\displaystyle G_{1} c5xdqydqydα1t(α+β1+β22q)/αlogβ4(e+1t1/α)𝑑t\displaystyle\geq c_{5}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{(\alpha+\beta_{1}+\beta_{2}-2q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt
=c5xdqydqydα1t1+(q^q)/αlogβ4(e+1t1/α)𝑑t\displaystyle=c_{5}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{-1+(\widehat{q}-q)/\alpha}\log^{\beta_{4}}\bigg{(}e+\frac{1}{t^{1/\alpha}}\bigg{)}dt
xdqydq×{1 if q<q^;logβ4+1(e+1/yd) if q=q^;ydq^qlogβ4(e+1/yd) if q>q^.\displaystyle\asymp x_{d}^{q}y_{d}^{q}\times\begin{cases}1&\mbox{ if }\,q<\widehat{q};\\[2.0pt] \displaystyle\log^{\beta_{4}+1}(e+1/y_{d})&\mbox{ if }\,q=\widehat{q};\\[2.0pt] \displaystyle y_{d}^{\widehat{q}-q}\log^{\beta_{4}}(e+1/y_{d})&\mbox{ if }\,q>\widehat{q}.\end{cases}

For the upper bound, we see from (9) that

G1\displaystyle G_{1} c6xdβ1ydβ^2logβ3(e+ydxd)logβ4(e+1yd)0xdαt𝑑t\displaystyle\leq c_{6}x_{d}^{\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}}{x_{d}}\bigg{)}\log^{\beta_{4}}\bigg{(}e+\frac{1}{y_{d}}\bigg{)}\int_{0}^{x_{d}^{\alpha}}tdt
+c6xdqydβ^2logβ4(e+1yd)xdαydαt(α+β1q)/αlogβ3(e+ydt1/α)𝑑t\displaystyle\quad+c_{6}x_{d}^{q}y_{d}^{\widehat{\beta}_{2}}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\int_{x_{d}^{\alpha}}^{y_{d}^{\alpha}}t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}\bigg{(}e+\frac{y_{d}}{t^{1/\alpha}}\bigg{)}dt
+c6xdqydqydα1t(α+β1+β^22q)/αlogβ4(e+1t1/α)𝑑t\displaystyle\quad+c_{6}x_{d}^{q}y_{d}^{q}\int_{y_{d}^{\alpha}}^{1}t^{(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha}\log^{\beta_{4}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt
=:c6(I+II+III).\displaystyle=:c_{6}(I+II+III).

Using (10.1), we obtain

Ic7xd2α+β1ydβ^2(ydxd)2α+β1qlogβ4(e+1yd)=c7xdqyd2α+β1+β^2qlogβ4(e+1yd).\displaystyle I\leq c_{7}x_{d}^{2\alpha+\beta_{1}}y_{d}^{\widehat{\beta}_{2}}\bigg{(}\frac{y_{d}}{x_{d}}\bigg{)}^{2\alpha+\beta_{1}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)=c_{7}x_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right). (9.5)

Since α+β1>q\alpha+\beta_{1}>q, the map tt(α+β1q)/αlogβ3(e+yd/t1/α)t\mapsto t^{(\alpha+\beta_{1}-q)/\alpha}\log^{\beta_{3}}(e+y_{d}/t^{1/\alpha}) is almost increasing. Therefore,

IIc8xdqydα+β1+β^2qlogβ4(e+1yd)xdαydα𝑑tc8xdqyd2α+β1+β^2qlogβ4(e+1yd).\displaystyle II\leq c_{8}x_{d}^{q}y_{d}^{\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\int_{x_{d}^{\alpha}}^{y_{d}^{\alpha}}dt\leq c_{8}x_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right). (9.6)

We consider the cases q<q^q<\widehat{q} and qq^q\geq\widehat{q} separately. First, suppose that q<q^q<\widehat{q}, which is equivalent to q<α+(β1+β2)/2q<\alpha+(\beta_{1}+\beta_{2})/2. Since q<α+β1q<\alpha+\beta_{1}, it follows that q<α+(β1+β^2)/2q<\alpha+(\beta_{1}+\widehat{\beta}_{2})/2. Using (10.1), since yd<1y_{d}<1, we see from (9.5)–(9.6) that

I+IIc9xdqydqyd2α+β1+β^22qlogβ4(e+1yd)c10xdqydq.\displaystyle I+II\leq c_{9}x_{d}^{q}y_{d}^{q}y_{d}^{2\alpha+\beta_{1}+\widehat{\beta}_{2}-2q}\log^{\beta_{4}}\left(e+\frac{1}{y_{d}}\right)\leq c_{10}x_{d}^{q}y_{d}^{q}.

Moreover, since (α+β1+β^22q)/α>1(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha>-1, we get

IIIc7xdqydq01t(α+β1+β^22q)/αlogβ4(e+1t1/α)𝑑tc11xdqydq.\displaystyle III\leq c_{7}x_{d}^{q}y_{d}^{q}\int_{0}^{1}t^{(\alpha+\beta_{1}+\widehat{\beta}_{2}-2q)/\alpha}\log^{\beta_{4}}\left(e+\frac{1}{t^{1/\alpha}}\right)dt\leq c_{11}x_{d}^{q}y_{d}^{q}.

Therefore, we arrive at the result in this case.

Suppose that qq^q\geq\widehat{q}. In this case, we have 2α+β1+β2=q+q^2q<2α+2β12\alpha+\beta_{1}+\beta_{2}=q+\widehat{q}\leq 2q<2\alpha+2\beta_{1}. Thus, β2<β1\beta_{2}<\beta_{1} and β^2=β2\widehat{\beta}_{2}=\beta_{2}. Then we deduce the desired upper bound from (9.4)–(9.6). The proof is complete. \Box

Lemma 9.2.

Let x,y+dx,y\in{\mathbb{R}}^{d}_{+} be such that xdydx_{d}\leq y_{d} and |xy|=1|x-y|=1. Then we have

1pκ(t,x,y)𝑑t{(xd1)q(yd1)q if d>α,(xd1)q(yd1)qlog(e+(xd1)) if d=1=α,(xd1)q(yd1)q(xd1)α1 if d=1<α,\int_{1}^{\infty}p^{\kappa}(t,x,y)dt\asymp\begin{cases}(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}&\mbox{ if }d>\alpha,\\[5.0pt] (x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}\log\big{(}e+(x_{d}\vee 1)\big{)}&\mbox{ if }d=1=\alpha,\\[5.0pt] \displaystyle(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}(x_{d}\vee 1)^{\alpha-1}&\mbox{ if }d=1<\alpha,\end{cases}

where the comparison constant is independent of x,yx,y.

Proof. By (9.3), we have

1pκ(t,x,y)dt1td/α(1xdt1/α)q(1ydt1/α)qdt=:G2.\int_{1}^{\infty}p^{\kappa}(t,x,y)dt\asymp\int_{1}^{\infty}t^{-d/\alpha}\left(1\wedge\frac{x_{d}}{t^{1/\alpha}}\right)^{q}\left(1\wedge\frac{y_{d}}{t^{1/\alpha}}\right)^{q}dt=:G_{2}.

If d>αd>\alpha, then by Lemma 10.12, G2(xd1)q(yd1)qG_{2}\asymp(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}.

If d=1=αd=1=\alpha, then using Lemma 10.12, we get

G2\displaystyle G_{2} 1xd1t1𝑑t+xdqxd1t1q(1ydt)q𝑑t\displaystyle\asymp\int_{1}^{x_{d}\vee 1}t^{-1}dt+x_{d}^{q}\int_{x_{d}\vee 1}^{\infty}t^{-1-q}\left(1\wedge\frac{y_{d}}{t}\right)^{q}dt
𝟏{xd>1}log(xd)+xdq(xd1)q(1ydxd1)q\displaystyle\asymp{\bf 1}_{\{x_{d}>1\}}\log(x_{d})+x_{d}^{q}(x_{d}\vee 1)^{-q}\left(1\wedge\frac{y_{d}}{x_{d}\vee 1}\right)^{q}
{log(e+xd) if xd>1xdq(yd1)q if xd1(xd1)q(yd1)qlog(e+(xd1)).\displaystyle\asymp\begin{cases}\log(e+x_{d})&\text{ if }x_{d}>1\\ x_{d}^{q}(y_{d}\wedge 1)^{q}&\text{ if }x_{d}\leq 1\end{cases}\asymp(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}\log(e+(x_{d}\vee 1)).

If d=1<αd=1<\alpha, then since yd>2y_{d}>2 implies xdyd|xy|yd/2>1x_{d}\geq y_{d}-|x-y|\geq y_{d}/2>1, and αq10\alpha-q-1\leq 0, we get

G2\displaystyle G_{2} 𝟏{yd>2}(1xdαt1/α𝑑t+xdqxdα(2yd)αt(q+1)/α𝑑t+xdqydq(2yd)αt(2q+1)/α𝑑t)\displaystyle\asymp{\bf 1}_{\{y_{d}>2\}}\bigg{(}\int_{1}^{x_{d}^{\alpha}}t^{-1/\alpha}dt+x_{d}^{q}\int_{x_{d}^{\alpha}}^{(2y_{d})^{\alpha}}t^{-(q+1)/\alpha}dt+x_{d}^{q}y_{d}^{q}\int_{(2y_{d})^{\alpha}}^{\infty}t^{-(2q+1)/\alpha}dt\bigg{)}
+𝟏{yd2}xdqydq1t(2q+1)/α𝑑t\displaystyle\quad+{\bf 1}_{\{y_{d}\leq 2\}}x_{d}^{q}y_{d}^{q}\int_{1}^{\infty}t^{-(2q+1)/\alpha}dt
𝟏{yd>2}(xdα1+xdq+α(q+1)+xdqydαq1)+𝟏{yd2}xdqydq\displaystyle\asymp{\bf 1}_{\{y_{d}>2\}}\big{(}x_{d}^{\alpha-1}+x_{d}^{q+\alpha-(q+1)}+x_{d}^{q}y_{d}^{\alpha-q-1}\big{)}+{\bf 1}_{\{y_{d}\leq 2\}}x_{d}^{q}y_{d}^{q}
𝟏{yd>2}xdα1+𝟏{yd2}xdqydq(xd1)q(yd1)q(xd1)α1.\displaystyle\asymp{\bf 1}_{\{y_{d}>2\}}x_{d}^{\alpha-1}+{\bf 1}_{\{y_{d}\leq 2\}}x_{d}^{q}y_{d}^{q}\asymp(x_{d}\wedge 1)^{q}(y_{d}\wedge 1)^{q}(x_{d}\vee 1)^{\alpha-1}.

The proof is complete. \Box

Proof of Theorem 1.4. From scaling property (3.30), we obtain

Gκ(x,y)=|xy|d+αGκ(x/|xy|,y/|xy|),x,y+d.\displaystyle G^{\kappa}(x,y)=|x-y|^{-d+\alpha}\,G^{\kappa}(x/|x-y|,y/|x-y|),\quad x,y\in{\mathbb{R}}^{d}_{+}. (9.7)

Using (9.7), symmetry and Lemmas 9.1 and 9.2, we deduce the desired result. \Box

10. Appendix

Note that for any ε>0{\varepsilon}>0,

log(e+r)<(2+ε1)rεfor allr1,\log(e+r)<(2+{\varepsilon}^{-1})r^{\varepsilon}\quad\text{for all}\;r\geq 1, (10.1)
log(e+ar)log(e+r)<(1+ε1)aεfor all a1 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 Ab1,b2,b3,b4(t,x,y)A_{b_{1},b_{2},b_{3},b_{4}}(t,x,y) from (6).

Lemma 10.1.

Let b1,b2,b3,b40b_{1},b_{2},b_{3},b_{4}\geq 0.

(i) If b1>0b_{1}>0, then for any ε(0,b1]{\varepsilon}\in(0,b_{1}], there exists c1>0c_{1}>0 such that

Ab1,b2,b3,b4(t,x,y)c1Ab1ε,b2,0,b4(t,x,y) for all t0,x,y+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 b2>0b_{2}>0, then for any ε(0,b2]{\varepsilon}\in(0,b_{2}], there exists c2>0c_{2}>0 such that

Ab1,b2,b3,b4(t,x,y)c2Ab1,b2ε,b3,0(t,x,y) for all t0,x,y+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 b1,b30b_{1},b_{3}\geq 0. Suppose that b1>0b_{1}>0 if b3>0b_{3}>0. Then there exists a constant C>0C>0 such that for all t0t\geq 0 and x,y+dx,y\in{\mathbb{R}}^{d}_{+},

Ab1,0,b3,0(t,x,y)\displaystyle A_{b_{1},0,b_{3},0}(t,x,y) C(xdt1/α|xy|1)b1logb3(e+(ydt1/α)|xy|(xdt1/α)|xy|).\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 xdydx_{d}\leq y_{d}, the result is clear. Assume that xd>ydx_{d}>y_{d}. Set r=xdt1/α|xy|1r=\frac{x_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1 and s=ydt1/α|xy|1s=\frac{y_{d}\vee t^{1/\alpha}}{|x-y|}\wedge 1. Then 0<sr10<s\leq r\leq 1 and the desired inequality is equivalent to

logb3(e+r/s)logb3(e+s/r)C(r/s)b1.\log^{b_{3}}(e+r/s)\log^{-b_{3}}(e+s/r)\leq C(r/s)^{b_{1}}. (10.3)

If b3=0b_{3}=0, then (10.3) clearly holds with C=1C=1. If b3>0b_{3}>0 and b1>0b_{1}>0, then we get from (10.1) that

logb3(e+r/s)logb3(e+s/r)logb3(e+r/s)c1(r/s)b1.\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

Remark 10.3.

Recall that Bb1,b2,b3,b4(x,y)=Ab1,b2,b3,b4(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) for x,y+dx,y\in{\mathbb{R}}^{d}_{+}. Therefore, the results of Lemmas 10.1 and 10.2 hold true with Bb1,0,b3,0(x,y)B_{b_{1},0,b_{3},0}(x,y) instead of Ab1,0,b3,0(t,x,y)A_{b_{1},0,b_{3},0}(t,x,y).

Lemma 10.4.

Let γ\gamma\in{\mathbb{R}}, b0b\geq 0 and t,k,l>0t,k,l>0. Suppose that either γ<α\gamma<\alpha or kt1/αk\geq t^{1/\alpha}. Then we have

0t(1ks1/α)γlogb(e+lks1/α)𝑑sCt(1kt1/α)γlogb(e+lkt1/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>0C>0 is a constant which depends only on γ\gamma and bb.

Proof. If kt1/αk\geq t^{1/\alpha}, then the desired inequality holds since the left hand side is tlogb(e+l/k)t\log^{b}(e+l/k).

Suppose that k<t1/αk<t^{1/\alpha} and γ<α\gamma<\alpha. Let ε>0{\varepsilon}>0 be such that γ+bε<α\gamma+b{\varepsilon}<\alpha. Then the left hand side is

logb(e+lk)0kα𝑑s+kγkαt1sγ/αlogb(e+ls1/α)𝑑s\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
kαlogb(e+lk)+c1kγlogb(e+lt1/α)kαttbε/αsγ/α+bε/α𝑑s\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
c2kα(t1/αk)αγlogb(e+lt1/α)+c2kγt1γ/αlogb(e+lt1/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{)}
=2c2t(kt1/α)γlogb(e+lt1/α)=2c2t(1kt1/α)γlogb(e+lt1/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 b1,b2b_{1},b_{2}\in{\mathbb{R}}, b3,b40b_{3},b_{4}\geq 0 and t,xd,yd>0t,x_{d},y_{d}>0. Suppose that (1) either b1>qαb_{1}>q-\alpha or xdt1/αx_{d}\geq t^{1/\alpha}, and (2) either b2>qαb_{2}>q-\alpha or ydt1/αy_{d}\geq t^{1/\alpha}. Then we have

0t(1xd(ts)1/α)qb1(1yds1/α)qb2logb3(e+yds1/αxd(ts)1/α)logb4(e+1yds1/α)𝑑s\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
Ct(1xdt1/α)qb1(1ydt1/α)qb2logb3(e+ydt1/αxdt1/α)logb4(e+1ydt1/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>0C>0 is a constant which depends only on b1,b2,b3b_{1},b_{2},b_{3} and b4b_{4}.

Proof. Observe that

0t2(1xd(ts)1/α)qb1(1yds1/α)qb2logb3(e+yds1/αxd(ts)1/α)logb4(e+1yds1/α)𝑑s\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
c1(1xdt1/α)qb1logb3(e+ydt1/αxdt1/α)0t2(1yds1/α)qb2logb4(e+1yds1/α)𝑑s\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

t2t(1xd(ts)1/α)qb1(1yds1/α)qb2logb3(e+yds1/αxd(ts)1/α)logb4(e+1yds1/α)𝑑s\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
c2(1ydt1/α)qb2logb4(e+1ydt1/α)0t2(1xds1/α)qb1logb3(e+ydt1/αxds1/α)𝑑s.\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 b1,b2b_{1},b_{2}\in{\mathbb{R}}, b3,b40b_{3},b_{4}\geq 0 and t,xd,yd>0t,x_{d},y_{d}>0. Suppose that (1) either b1>qαb_{1}>q-\alpha or xdt1/αx_{d}\geq t^{1/\alpha}, and (2) either b2>qαb_{2}>q-\alpha or ydt1/αy_{d}\geq t^{1/\alpha}. Then we have that for ydt1/α<2y_{d}\vee t^{1/\alpha}<2,

0ts(1xd(ts)1/α)qb1(1yds1/α)qb2\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}}
×yds1/α2logb3(e+ryds1/α)logb3(e+rxd(ts)1/α)logb4(e+1r)drrds\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
Ct(1xdt1/α)qb1(1ydt1/α)qb2\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}}
×yd2(rαt)logb3(e+rxdt1/α)logb3(e+rydt1/α)logb4(e+1r)drr,\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>0C>0 is a constant which depends only on b1,b2,b3b_{1},b_{2},b_{3} and b4b_{4}.

Proof. Using Fubini’s theorem and Lemma 10.4 twice as in the proof of Lemma 10.5, we get

0ts(1xd(ts)1/α)qb1(1yds1/α)qb2\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}}
×yds1/α2logb3(e+ryds1/α)logb3(e+rxd(ts)1/α)logb4(e+1r)drrds\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= yd20rαts(1xd(ts)1/α)qb1(1yds1/α)qb2\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}}
×logb3(e+ryds1/α)logb3(e+rxd(ts)1/α)logb4(e+1r)dsdrr\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 yd2(rαt)0t(1xd(ts)1/α)qb1(1yds1/α)qb2\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}}
×logb3(e+ryds1/α)logb3(e+rxd(ts)1/α)logb4(e+1r)dsdrr\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 Ct(1xdt1/α)qb1(1ydt1/α)qb2\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}}
×yd2(rαt)logb3(e+rxdt1/α)logb3(e+rydt1/α)logb4(e+1r)drr.\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>0C>0 such that for all x+dx\in{\mathbb{R}}^{d}_{+} and A>0A>0,

B+(x,A)zd1/2𝑑zCAd(xdA)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)zd1/2𝑑zz~d1,|x~z~|<A𝑑z~(xdA)0xd+Azd1/2𝑑zdc1Ad1(xdA)0xd+Azd1/2𝑑zd.\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 xd2Ax_{d}\geq 2A, then

(xdA)0xd+Azd1/2𝑑zd1(xd/2)1/2xdAxd+A𝑑zd=23/2Axd1/2.\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 xd<2Ax_{d}<2A, then

(xdA)0xd+Azd1/2𝑑zd04Azd1/2𝑑zd=4A1/2.\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>0C>0 such that for all x+dx\in{\mathbb{R}}^{d}_{+} and 0<A<xd0<A<x_{d},

z+d,xd|xz|>Adzzd1/2|xz|d+αCxd1/2Aα.\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>0C^{\prime}>0 such that for all x+dx\in{\mathbb{R}}^{d}_{+} and AxdA\geq x_{d},

z+d,|xz|>Adzzdε|xz|d+δCAεδ.\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~,xd)x=(\widetilde{0},x_{d}).

(i) Note that

z+d,xd|xz|>Adzzd1/2|xz|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+d,xd|xz|>A,|z~||xdzd|dzzd1/2|xz|d+α+z+d,xd|xz|>A,|z~|>|xdzd|dzzd1/2|xz|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+II.\displaystyle=:I+II.

First, we have

I\displaystyle I xd|xdzd|>A21zd1/2|xdzd|d+αz~d1,|z~||xdzd|𝑑z~𝑑zd\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}
c1(0xd2dzdzd1/2|xdzd|1+α+xd2xdA2dzdzd1/2|xdzd|1+α+xd+A22xddzdzd1/2|xdzd|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{)}
c1(21+αxd1+α0xd2zd1/2𝑑zd+1(xd/2)1/2xd2xdA2dzd|xdzd|1+α+xd1/2xd+A22xddzd|xdzd|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{)}
c2(xd1/2α+xd1/2Aα+xd1/2Aα)c3xd1/2Aα.\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+dz\in{\mathbb{R}}^{d}_{+} with xd|xz|>Ax_{d}\geq|x-z|>A and |z~|>|xdzd||\widetilde{z}|>|x_{d}-z_{d}|,

zdxd|xdzd|xd12(|z~|+|xdzd|)xd12|xz|(112)xd.\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

II\displaystyle II c4xd1/2z+d,|xz|>Adz|xz|d+αc5xd1/2Aα.\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+d,|xz|>Adzzdε|xz|d+δ\displaystyle\int_{z\in{\mathbb{R}}^{d}_{+},\,|x-z|>A}\frac{dz}{z_{d}^{{\varepsilon}}\,|x-z|^{d+\delta}}
z+d,|xdzd||z~|A2dz~dzdzdε|xdzd|d+δ+z+d,|z~||xdzd|A2dz~dzdzdε|z~|d+δ=:I+II.\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 AxdA\geq x_{d}, we see that

II\displaystyle II z~d1,|z~|A21|z~|d+δ0|z~|+xdzdε𝑑zd𝑑z~c1z~d1,|z~|A2(|z~|+A)1ε|z~|d+δ𝑑z~\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}
c2z~d1,|z~|A2dz~|z~|d1+ε+δc3Aεδ.\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~d1,|z~||xdzd|𝑑z~c4|xdzd|d1\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 c5(0(xdA2)0dzdzdε|xdzd|1+δ+xd+A2dzdzdε|xdzd|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{)}
c6(1A1+δ0(xdA2)0zdε𝑑zd+1Aεxd+A2dzd|xdzd|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{)}
c7(1A1+δ((xdA2)0)1ε+1Aε+δ)c6Aε+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 AxdA\geq x_{d} in the last inequality. The proof is complete. \Box

For γ,η1,η20\gamma,\eta_{1},\eta_{2}\geq 0 and k,l>0k,l>0, define

fγ,η1,η2,k,l(r):=rγlogη1(e+kr)logη2(e+rl).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,η1,η20\gamma,\eta_{1},\eta_{2}\geq 0.

(i) For any ε>0{\varepsilon}>0, there exist constants C,C>0C,C^{\prime}>0 such that for any k,l,r>0k,l,r>0 and any a1a\geq 1,

Caγεfγ,η1,η2,k,l(ar)fγ,η1,η2,k,l(r)Caγ+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>0C>0 such that for any k,l,r>0k,l,r>0 and any a1a\geq 1,

fγ,η1,η2,k,l(ar)fγ,η1,η2,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 η2=0\eta_{2}=0, the second inequality in (10.7) is true with any ε0{\varepsilon}\geq 0. In case η2>0\eta_{2}>0, for any given ε>0{\varepsilon}>0, let ε:=ε/η2{\varepsilon}^{\prime}:={\varepsilon}/\eta_{2}. We get from (10.2) that for all a1a\geq 1 and r>0r>0,

fγ,η1,η2,k,l(ar)fγ,η1,η2,k,l(r)aγ(log(e+ar/l)log(e+r/l))η2aγ((1+1/ε)aε)η2=c(ε,η2)aγ+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 b1,b2,η1,η2,γ0b_{1},b_{2},\eta_{1},\eta_{2},\gamma\geq 0. There exists a constant C>0C>0 such that for any x+dx\in{\mathbb{R}}^{d}_{+} and s,k,l>0s,k,l>0,

B+(x,2)(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)\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{)}
×(sd/αs|xz|d+α)zdγlogη1(e+kzd)logη2(e+zdl)dz\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(xds1/α)γlogη1(e+kxds1/α)logη2(e+xds1/αl)\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{)}
+C𝟏{xds1/α<2}s(xds1/α)b1[𝟏{γ>α+b1}logb2(e+2xds1/α)logη1(e+k)logη2(e+1l)\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+b1}xds1/α2logb2(e+rxds1/α)logη1(e+kr)logη2(e+rl)drr].\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+dz\in{\mathbb{R}}^{d}_{+},

zdxd+|xz|2(xd|xz|).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,s1/α)(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)sd/αfγ,η1,η2,k,l(zd)𝑑z\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
=(logb2(e+1))sd/αB+(x,s1/α)zd1/2fγ+12,η1,η2,k,l(zd)𝑑z\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
c1sd/αfγ+12,η1,η2,k,l(2(xds1/α))B+(x,s1/α)zd1/2𝑑z\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
c2(xds1/α)1/2fγ+12,η1,η2,k,l(xds1/α)=c2fγ,η1,η2,k,l(xds1/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 xd>s1/αx_{d}>s^{1/\alpha}, we get from (10.8), Lemma 10.9(i)-(ii) and Lemma 10.8(i) that

sz+d,xd|xz|>s1/α(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)fγ,η1,η2,k,l(zd)|xz|d+α𝑑z\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
=(logb2(e+1))sz+d,xd|xz|>s1/αfγ+12,η1,η2,k,l(zd)zd1/2|xz|d+α𝑑z\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
c3sfγ+12,η1,η2,k,l(2xd)z+d,xd|xz|>s1/α1zd1/2|xz|d+α𝑑z\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
c4xd1/2fγ+12,η1,η2,k,l(xd)=c4fγ,η1,η2,k,l(xds1/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+d:xds1/α<|xz|<2}\{z\in{\mathbb{R}}^{d}_{+}:x_{d}\vee s^{1/\alpha}<|x-z|<2\} under the assumption xds1/α<2x_{d}\vee s^{1/\alpha}<2. For this, we consider the following three cases separately.

(i) Case γ<α+b1\gamma<\alpha+b_{1}: Fix ε(0,1){\varepsilon}\in(0,1) such that γ+3ε<α+b1\gamma+3{\varepsilon}<\alpha+b_{1}. Using (10.8), (10.1), Lemma 10.9(i)-(ii) and Lemma 10.8(ii), we get

sz+d,|xz|>xds1/α(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)fγ,η1,η2,k,l(zd)|xz|d+α𝑑z\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(xds1/α)b1z+d,|xz|>xds1/αlogb2(e+|xz|xds1/α)fγ+ε,η1,η2,k,l(zd)zdε|xz|d+α+b1𝑑z\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
c5s(xds1/α)b1z+d,|xz|>xds1/αlogb2(e+|xz|xds1/α)fγ+ε,η1,η2,k,l(2|xz|)zdε|xz|d+α+b1𝑑z\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
c6s(xds1/α)b1z+d,|xz|>xds1/α(|xz|xds1/α)ε+γ+2εfγ+ε,η1,η2,k,l(xds1/α)zdε|xz|d+α+b1𝑑z\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
=c6s(xds1/α)b1γ3εfγ+ε,η1,η2,k,l(xds1/α)z+d,|xz|>xds1/α1zdε|xz|d+α+b1γ3ε𝑑z\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
c7s(xds1/α)α(xds1/α)εfγ+ε,η1,η2,k,l(xds1/α)c7fγ,η1,η2,k,l(xds1/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+b1\gamma>\alpha+b_{1}: Fix ε>0{\varepsilon}>0 such that γε>α+b1\gamma-{\varepsilon}>\alpha+b_{1}. Using (10.8), Lemma 10.9(i)-(ii) and (10.2), we get

sz+d,xds1/α<|xz|<2(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)fγ,η1,η2,k,l(zd)|xz|d+α𝑑z\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
c8s(xds1/α)b1z+d,xds1/α<|xz|<2logb2(e+|xz|xds1/α)fγ,η1,η2,k,l(2|xz|)|xz|d+α+b1𝑑z\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
c9s(xds1/α)b1fγ,η1,η2,k,l(4)logb2(e+2xds1/α)z+d,|xz|<2dz|xz|d+α+b1γ+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}}}
c10s(xds1/α)b1fγ,η1,η2,k,l(1)logb2(e+2xds1/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+b1\gamma=\alpha+b_{1}: In this case, we see that

sz+d,xds1/α<|xz|<2(1xds1/α|xz|)b1logb2(e+|xz|(xds1/α)|xz|)fγ,η1,η2,k,l(zd)|xz|d+α𝑑z\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
c11s(xds1/α)b1z+d,xds1/α<|xz|<2logb2(e+|xz|xds1/α)fγ,η1,η2,k,l(|xz|)|xz|d+α+b1𝑑z\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
=c11s(xds1/α)b1\displaystyle=c_{11}s(x_{d}\vee s^{1/\alpha})^{b_{1}}
×z+d,xds1/α<|xz|<2logb2(e+|xz|xds1/α)logη1(e+k|xz|)logη2(e+|xz|l)dz|xz|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}}
c12s(xds1/α)b1xds1/α2logb2(e+rxds1/α)logη1(e+kr)logη2(e+rl)drr.\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 b1,b20b_{1},b_{2}\geq 0. For any 0<kl<10<k\leq l<1,

l2logb1(e+rk)logb1(e+rl)logb2(e+1r)drrlogb1(e+1k)logb1+b2+1(e+1l),\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 kk and ll.

Proof. Note that

log(e+1r)log(2er)log(e+1r),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

l2logb1(e+rk)logb1(e+rl)logb2(e+1r)drr\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}
c1l2logb1(2erk)logb1(2erl)logb2(2er)drr\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}
c1logb1(2ek)logb1(2el)l2logb2(2er)drr\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}
=c1b2+1logb1(2ek)logb1(2el)(logb2+1(2el)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{)}
c2logb1(e+1k)logb1+b2+1(e+1l).\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

l2logb1(e+rk)logb1(e+rl)logb2(e+1r)drr\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}
c3logb1(e+2k)logb1(e+2l)l2logb2(2er)drrc4logb1(e+1k)logb1+b2+1(e+1l).\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 b1,b20b_{1},b_{2}\geq 0. For any a,k,l>0a,k,l>0,

atγ(1kt)b1(1lt)b2𝑑ta1γ(1ka)b1(1la)b2,\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,ka,k and ll.

Proof. We have

atγ(1kt)b1(1lt)b2𝑑t(1ka)b1(1la)b2atγ𝑑ta1γγ1(1ka)b1(1la)b2\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

atγ(1kt)b1(1lt)b2𝑑t\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 (1k2a)b1(1l2a)b2a2atγ𝑑t\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
(121γ)a1γ2b1+b2(γ1)(1ka)b1(1la)b2.\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

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