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Energy measures of harmonic functions on the Sierpiński Gasket

Renee Bell1 Renee Bell
Department of Mathematics
University of California, Berkeley
Evans Hall
Berkeley, CA 94720-3840
U.S.A
rbell916@berkeley.edu
Ching-Wei Ho Ching-Wei Ho
Department of Mathematics
Lady Shaw Building
The Chinese University of Hong Kong
Shatin, Hong Kong
s0962318@cuhk.edu.hk
 and  Robert S. Strichartz2 Robert S. Strichartz
Department of Mathematics
Malott Hall
Cornell University
Ithaca, NY 14853
U.S.A
str@math.cornell.edu
Abstract.

We study energy measures on SG based on harmonic functions. We characterize the positive energy measures through studying the bounds of Radon-Nikodym derivatives with respect to the Kusuoka measure. We prove a limited continuity of the derivative on the graph VV_{*} and express the average value of the derivative on a whole cell as a weighted average of the values on the boundary vertices. We also prove some characterizations and properties of the weights.

Key words and phrases:
Sierpiński gasket, energy measures, Kusuoka measure, energy Laplacian, Radon-Nikodym derivatives
2010 Mathematics Subject Classification:
28A80
1Research supported by the National Science Foundation through the Research Experiences for Undergraduates(REU) Program at Cornell.
2Research supported by the National Science Foundation grant DMS-1162045

1. introduction

In the development of analysis on fractals, there are three basic concepts: energy, measure and Laplacian. For SGSG, which may be regarded as the “poster child” for the theory, Kigami has developed an elegant construction of a self-similar Laplacian based on a self-similar energy \mathcal{E} and self-similar measure μ\mu via an analog of integration by parts. This theory is briefly sketched below, and is described in complete detail in the book [8]. This standard Laplacian Δ\Delta is not the only Laplacian on SGSG that has been widely studied. A competing energy Laplacian Δν\Delta_{\nu}, based on the same energy but using the Kusuoka measure ν\nu in place of μ\mu, has received a lot of attention in recent years([4, 5, 6, 10, 11]). It has two striking advantages over Δ\Delta, namely the existence of a Leibniz-type product formula and the fact that the associated heat kernel satisfies Gaussian type estimates. The main drawback so far has been that the Kusuoka measure is not algorithmically transparent. The purpose of this paper is to eliminate this drawback. We will provide algorithms that describe the nature of this measure and more generally the family of energy measures νh\nu_{h} based on harmonic functions hh, and the Radon-Nikodym derivatives dνhdν\frac{d\nu_{h}}{d\nu}. Some of the results of this paper, especially Theorem 5.5, may also be derived from [6].

The Sierpiński gasket, or simply SG, is the unique nonempty compact set satisfying

SG=i=02FiSGSG=\bigcup_{i=0}^{2}F_{i}SG

where Fi=12(x+qi)F_{i}=\frac{1}{2}(x+q_{i}) and {qi}i=02\{q_{i}\}_{i=0}^{2} are vertices of an equilateral triangle in counter-clockwise direction. {qi}i=12\{q_{i}\}_{i=1}^{2} are called the boundary points of SG. If w=(w1,,wn)w=(w_{1},\cdots,w_{n}) is a finite word, we define the mapping Fw=Fw1FwnF_{w}=F_{w_{1}}\circ\cdots\circ F_{w_{n}} where |w||w| is defined to be the length of the word. Define V0={qi}V_{0}=\{q_{i}\}, Vn=Fi(Vn1)V_{n}=\bigcup F_{i}\left(V_{n-1}\right) and V=n=1VnV_{*}=\bigcup_{n=1}^{\infty}V_{n}. For the boundary of a cell C=FwSGC=F_{w}SG , denoted by C\partial C, we mean the three vertices {Fw(qi)}i=02\{F_{w}(q_{i})\}_{i=0}^{2} of the cell.

For functions uu, vv defined on SG, we define the energy on level mm by

(1.1) m(u,v)=(53)m|w|=mi<j(uFw(qi)uFw(qj))(vFw(qi)vFw(qj)).\ \mathcal{E}_{m}(u,v)=\left(\frac{5}{3}\right)^{m}\sum_{|w|=m}\sum_{i<j}(u\circ F_{w}(q_{i})-u\circ F_{w}(q_{j}))(v\circ F_{w}(q_{i})-v\circ F_{w}(q_{j})).

and energy (u,v)=limmm(u,v)\mathcal{E}(u,v)=\lim\limits_{m\to\infty}\mathcal{E}_{m}(u,v) provided that (u,u)\mathcal{E}(u,u) and (v,v)\mathcal{E}(v,v) are finite. For each cell FwSGF_{w}SG, we define the energy of uu, vv on this cell by

FwSG(u,v)=(53)|w|(uFw,vFw).\mathcal{E}_{F_{w}SG}(u,v)=\left(\frac{5}{3}\right)^{|w|}\mathcal{E}(u\circ F_{w},v\circ F_{w}).

Observe that the energy is non-negative if u=vu=v, but in general it is not. In the case u=vu=v, we write (u)\mathcal{E}(u) instead of (u,u)\mathcal{E}(u,u). We define a function hh to be harmonic if it minimizes the energy from level mm to level m+1m+1, as defined in (1.1). A simple computation shows that this minimization occurs when m(h)=m+1(h)\mathcal{E}_{m}(h)=\mathcal{E}_{m+1}(h) at which to each cell FwSGF_{w}SG, h(Fw0(q1))=25h(Fw(q0))+25h(Fw(q1))+15h(Fw(q2))h(F_{w0}(q_{1}))=\frac{2}{5}h(F_{w}(q_{0}))+\frac{2}{5}h(F_{w}(q_{1}))+\frac{1}{5}h(F_{w}(q_{2})). Equivalently, a function hh on SG is harmonic if the sequence, taking u=h=vu=h=v in (1.1), is a constant sequence, i.e. constantly i<j(h(qi)h(qj))2\sum_{i<j}(h(q_{i})-h(q_{j}))^{2}. Given three values on the boundary, in order to have a constant sequence in (1.1), all the other values are completely determined. Consequently, all the harmonic functions form a three-dimensional space and hence any values on the boundary of a cell can extend to a unique harmonic function on SG. A harmonic function hh is said to be symmetric(resp. skew-symmetric) if h(qi)=h(qi+1)h(q_{i})=h(q_{i+1})(resp. h(qi)=h(qi+1)h(q_{i})=-h(q_{i+1})) for some ii. Harmonic functions characterized by hi(qj)=δijh_{i}(q_{j})=\delta_{ij} are three symmetric functions and the notation hih_{i} will be used throughout this paper. On the space \mathcal{H} of harmonic functions modulo constants, (,(\mathcal{H},\mathcal{E}) is a two-dimensional inner product space. A harmonic function hh is said to be symmetric (resp. skew-symmetric) in \mathcal{H} if there exists a constant cc and a symmetric (resp. skew-symmetric) harmonic function ff such that h=c+fh=c+f. Explicitly, for a symmetric harmonic function hh, we mean h=ahi+ch=ah_{i}+c for some a,ca,c\in\mathbb{R} and some hih_{i}; and for a skew-symmetric harmonic function hh, we mean a(hihj)+ca(h_{i}-h_{j})+c for some a>0a>0 and iji\not=j. To each harmonic hh, we write hh^{\perp} the harmonic function having the same energy as hh and orthogonal to hh unless we specify hh^{\perp} “orthonormal to hh” in which case we mean having energy 11 and orthogonal to hh.

The standard measure μ\mu is a self-similar measure characterized by μ(FwFiSG)=13μ(FwSG)\mu(F_{w}F_{i}SG)=\frac{1}{3}\mu(F_{w}SG) for every cell CC. The standard Laplacian Δu\Delta u of uu is a continuous function satisfying

(1.2) (u,v)=Δuv𝑑μ-\mathcal{E}(u,v)=\int\Delta uvd\mu

for every vv vanishing at the boundary with (v)<\mathcal{E}(v)<\infty. Harmonic functions on SG are precisely those whose Laplacian is zero. However, for a function uu whenever Δu\Delta u is defined as a function, Δ(u2)\Delta(u^{2}) is not defined as a function. We can view Δu\Delta u as a measure m=Δudμm=\Delta ud\mu and Δu\Delta u exists as a function if and only if this measure is absolutely continuous with respect to the standard measure. The carré du champs formula given by, for all f,u,vf,u,v of finite energy, SGf𝑑νu,v=12(fu,v)+12(u,fv)12(f,uv)\int_{SG}fd\nu_{u,v}=\frac{1}{2}\mathcal{E}(fu,v)+\frac{1}{2}\mathcal{E}(u,fv)-\frac{1}{2}\mathcal{E}(f,uv) shows that

(u2,v)=2(uΔu)v𝑑μ2v𝑑νu\mathcal{E}(u^{2},v)=-2\int(u\Delta u)vd\mu-2\int vd\nu_{u}

for all u,vu,v with Δu\Delta u as a function. Thus if Δu\Delta u exists as a function and if we first view Δ(u2)\Delta(u^{2}) as a measure,

Δ(u2)=2uΔudμ+2dνu\Delta(u^{2})=2u\Delta ud\mu+2d\nu_{u}

then Δ(u2)\Delta(u^{2}) would exist as a function if this measure were absolutely continuous with respect to the standard measure μ\mu, but this is almost never the case [2].

Suppose that uu, vv are functions of finite energy. We can define a signed measure νu,v\nu_{u,v} by, on each cell,

νu,v(C)=C(u,v).\nu_{u,v}(C)=\mathcal{E}_{C}(u,v).

As usual, if it happens that u=vu=v, we write νu\nu_{u} instead. In this paper, we will restrict the attention to both uu, vv being harmonic. The energy measures of harmonic functions form a three-dimensional space. The Kusuoka measure ν\nu is defined to be ν0+ν1+ν2\nu_{0}+\nu_{1}+\nu_{2}. An easy computation shows that for any harmonic function hh, we have νh+νh=cν\nu_{h}+\nu_{h^{\perp}}=c\nu where c=13(h)c=\frac{1}{3}\mathcal{E}(h). Now, we define the energy Laplacian Δνu\Delta_{\nu}u of uu by, for every finite energy vv vanishing on the boundary,

(1.3) (u,v)=(Δνu)v𝑑ν.-\mathcal{E}(u,v)=\int(\Delta_{\nu}u)vd\nu.

By Theorem 5.3.1 of [8], every energy measure of harmonic functions is absolutely continuous with respect to the Kusuoka measure ν\nu and it makes sense to consider the Radon-Nikodym derivatives. For the energy measures of the symmetric harmonic functions, we will denote these by νi\nu_{i} instead of νhi\nu_{h_{i}}. {νi}\{\nu_{i}\} forms a basis of the space of all energy measures of harmonic functions. Since h0+h1+h2=1h_{0}+h_{1}+h_{2}=1, we have νh0,hj+νh1,hj+νh2,hj=ν1,hj=0\nu_{h_{0},h_{j}}+\nu_{h_{1},h_{j}}+\nu_{h_{2},h_{j}}=\nu_{1,h_{j}}=0 for j=0,1,2j=0,1,2. Since νf,g=νg,f\nu_{f,g}=\nu_{g,f} for all ff, gg, this gives us three equations that we solve to obtain

(1.4) νh0,h1=12(ν0ν1+ν2);νh0,h2=12(ν0+ν1ν2);νh1,h2=12(ν0ν1ν2).\begin{split}\nu_{h_{0},h_{1}}&=\frac{1}{2}(-\nu_{0}-\nu_{1}+\nu_{2});\\ \nu_{h_{0},h_{2}}&=\frac{1}{2}(-\nu_{0}+\nu_{1}-\nu_{2});\\ \nu_{h_{1},h_{2}}&=\frac{1}{2}(\hskip 8.0pt\nu_{0}-\nu_{1}-\nu_{2}).\end{split}

We will keep using the notation νf,g\nu_{f,g} for any energy measure of harmonic functions without mentioning ff and gg.

By [1], we have that there exist matrices EiE_{i} such that for every cell CC,

(νf,g(Fi0C)νf,g(Fi1C)νf,g(Fi2C))=Ei(νf,g(F0C)νf,g(F1C)νf,g(F2C)).\left(\begin{array}[]{c}\nu_{f,g}(F_{i0}C)\\ \nu_{f,g}(F_{i1}C)\\ \nu_{f,g}(F_{i2}C)\end{array}\right)=E_{i}\left(\begin{array}[]{c}\nu_{f,g}(F_{0}C)\\ \nu_{f,g}(F_{1}C)\\ \nu_{f,g}(F_{2}C)\end{array}\right).

The EiE_{i}’s are all diagonalizable, so it is easy to calculate EimE_{i}^{m}. Explicitly, after diagonalization, we have

(1.5) E0=(1730111111)(1150003500015)(7202140214072014014001212);E1=(1111730111)(1150003500015)(2140720214014072014012012);E2=(1111111730)(1150003500015)(2140214072014014072012120).\begin{split}E_{0}&=\left(\begin{array}[]{ccc}\frac{1}{7}&3&0\\[1.0pt] 1&1&-1\\[1.0pt] 1&1&1\end{array}\right)\left(\begin{array}[]{ccc}\frac{1}{15}&0&0\\[1.0pt] 0&\frac{3}{5}&0\\[1.0pt] 0&0&\frac{1}{5}\end{array}\right)\left(\begin{array}[]{ccc}\frac{-7}{20}&\frac{21}{40}&\frac{21}{40}\\[1.0pt] \frac{7}{20}&\frac{-1}{40}&\frac{-1}{40}\\[1.0pt] 0&-\frac{1}{2}&\frac{1}{2}\end{array}\right);\\ E_{1}&=\left(\begin{array}[]{ccc}1&1&-1\\[1.0pt] \frac{1}{7}&3&0\\[1.0pt] 1&1&1\end{array}\right)\left(\begin{array}[]{ccc}\frac{1}{15}&0&0\\[1.0pt] 0&\frac{3}{5}&0\\[1.0pt] 0&0&\frac{1}{5}\end{array}\right)\left(\begin{array}[]{ccc}\frac{21}{40}&\frac{-7}{20}&\frac{21}{40}\\[1.0pt] \frac{-1}{40}&\frac{7}{20}&\frac{-1}{40}\\[1.0pt] -\frac{1}{2}&0&\frac{1}{2}\end{array}\right);\\ E_{2}&=\left(\begin{array}[]{ccc}1&1&-1\\[1.0pt] 1&1&1\\[1.0pt] \frac{1}{7}&3&0\end{array}\right)\left(\begin{array}[]{ccc}\frac{1}{15}&0&0\\[1.0pt] 0&\frac{3}{5}&0\\[1.0pt] 0&0&\frac{1}{5}\end{array}\right)\left(\begin{array}[]{ccc}\frac{21}{40}&\frac{21}{40}&\frac{-7}{20}\\[1.0pt] \frac{-1}{40}&\frac{-1}{40}&\frac{7}{20}\\[1.0pt] -\frac{1}{2}&\frac{1}{2}&0\end{array}\right).\\ \end{split}

It is easy to calculate that the limits of quotients which are used to compute derivatives are as follows:

limm(111)E0mE1(xyz)(111)E0mE1(αβγ)=limm(111)E1mE0(xyz)(111)E1mE0(αβγ)\lim_{m}\frac{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}^{m}E_{1}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}^{m}E_{1}\left(\begin{array}[]{c}\alpha\\ \beta\\ \gamma\end{array}\right)}=\lim_{m}\frac{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{1}^{m}E_{0}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{1}^{m}E_{0}\left(\begin{array}[]{c}\alpha\\ \beta\\ \gamma\end{array}\right)}

where

(xyz)=(νf,g(F0C)νf,g(F1C)νf,g(F2C)),(αβγ)=(ν(F0C)ν(F1C)ν(F2C)).\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}[]{c}\nu_{f,g}(F_{0}C)\\ \nu_{f,g}(F_{1}C)\\ \nu_{f,g}(F_{2}C)\end{array}\right),\left(\begin{array}[]{c}\alpha\\ \beta\\ \gamma\end{array}\right)=\left(\begin{array}[]{c}\nu(F_{0}C)\\ \nu(F_{1}C)\\ \nu(F_{2}C)\end{array}\right).

For each infinite word ww, let wmw_{m} be the word consisting of the first mm letters of ww; then limmEwm=0\lim\limits_{m\to\infty}E_{w_{m}}=0. It follows that every energy measure is continuous.

It is natural to ask whether the energy Laplacian behaves in a similar manner to the standard Laplacian; in Section 2, we express the “self-similarity” of the energy Laplacian. In Section 3, we will study the decay rates of energy measure from a cell to its subcells. Through studying this, we give sharp bounds for the Radon-Nikodym derivatives for each νh\nu_{h}. Being in general a signed measure, it is natural to ask when it is a (positive) measure; in Section 4, we will characterize all the positive energy measures. It is well-known that the derivative is not continuous; nevertheless, in Section 5, we show that a certain restriction of the derivative is continuous. In Section 6, we will express, on each cell, the average value of derivatives by a weighted average of values on vertices of the cell. How the energy distributes is mysterious; in Section 7, we will provide some graphs and hypothesis about this question.

2. Self-similarity of energy Laplacian

In this section we discuss the self-similarity identities for energy measures and the energy Laplacian. We will see that the Radon-Nikodym derivatives Ri=dνidνR_{i}=\frac{d\nu_{i}}{d\nu} play a crucial role in these identities. Individually, none of the νi\nu_{i} is self-similar; nevertheless, together they form a self-similar family, in the sense of Mauldin-Williams [7]. We begin with considering the symmetric function h0h_{0}. We have the following equalities,

(2.1) h0F0=h0+25h1+25h2=25+35h0;h0F1=25h0+15h2;h0F2=25h0+15h1.\begin{split}&h_{0}\circ F_{0}=h_{0}+\frac{2}{5}h_{1}+\frac{2}{5}h_{2}=\frac{2}{5}+\frac{3}{5}h_{0};\\ &h_{0}\circ F_{1}=\frac{2}{5}h_{0}+\frac{1}{5}h_{2};\\ &h_{0}\circ F_{2}=\frac{2}{5}h_{0}+\frac{1}{5}h_{1}.\end{split}

Next we establish the relation of νi\nu_{i}’s from cells to subcells.

Theorem 2.1.

Let

M0=115(900221212),M1=115(221090122),M2=115(212122009),M_{0}=\frac{1}{15}\left(\begin{array}[]{ccc}9&0&0\\ 2&2&-1\\ 2&-1&2\end{array}\right),M_{1}=\frac{1}{15}\left(\begin{array}[]{ccc}2&2&-1\\ 0&9&0\\ -1&2&2\end{array}\right),M_{2}=\frac{1}{15}\left(\begin{array}[]{ccc}2&-1&2\\ -1&2&2\\ 0&0&9\end{array}\right),

then

(ν0(FiC)ν1(FiC)ν2(FiC))=Mi(ν0(C)ν1(C)ν2(C))\left(\begin{array}[]{c}\nu_{0}(F_{i}C)\\ \nu_{1}(F_{i}C)\\ \nu_{2}(F_{i}C)\end{array}\right)=M_{i}\left(\begin{array}[]{c}\nu_{0}(C)\\ \nu_{1}(C)\\ \nu_{2}(C)\end{array}\right)

for every cell C. In other words,

(ν0ν1ν2)=M0(ν0ν1ν2)F01+M1(ν0ν1ν2)F11+M2(ν0ν1ν2)F21.\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)=M_{0}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{0}^{-1}+M_{1}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{1}^{-1}+M_{2}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{2}^{-1}.
Proof.

Let ff be a contintuous function on SG, by (1.4), (2.1),

F0SGf𝑑ν0=53SGfF0𝑑ν25+35h0=53(35)2SGfF0𝑑ν0=35SGfF0𝑑ν0.\begin{split}\int_{F_{0}SG}{fd\nu_{0}}=&\frac{5}{3}\int_{SG}f\circ F_{0}\,d\nu_{\frac{2}{5}+\frac{3}{5}h_{0}}\\ =&\frac{5}{3}\left(\frac{3}{5}\right)^{2}\int_{SG}f\circ F_{0}\,d\nu_{0}\\ =&\frac{3}{5}\int_{SG}f\circ F_{0}\,d\nu_{0}.\end{split}

On the other hand,

F1SGf𝑑ν0=53SGfF1𝑑ν25h0+15h2=215SGfF1𝑑ν0+215SGfF1𝑑ν1115SGfF1𝑑ν2\begin{split}&\int_{F_{1}SG}f\,d\nu_{0}\\ =&\frac{5}{3}\int_{SG}f\circ F_{1}\,d\nu_{\frac{2}{5}h_{0}+\frac{1}{5}h_{2}}\\ =&\frac{2}{15}\int_{SG}{f\circ F_{1}\,d\nu_{0}}+\frac{2}{15}\int_{SG}{f\circ F_{1}\,d\nu_{1}}-\frac{1}{15}\int_{SG}{f\circ F_{1}\,d\nu_{2}}\end{split}

using (1.4) again. Similarly,

F2SGf𝑑ν0=53SGfF1𝑑ν25h0+15h2=215SGfF2𝑑ν0115SGfF2𝑑ν1+215SGfF2𝑑ν2.\begin{split}&\int_{F_{2}SG}f\,d\nu_{0}\\ =&\frac{5}{3}\int_{SG}f\circ F_{1}\,d\nu_{\frac{2}{5}h_{0}+\frac{1}{5}h_{2}}\\ =&\frac{2}{15}\int_{SG}{f\circ F_{2}\,d\nu_{0}}-\frac{1}{15}\int_{SG}{f\circ F_{2}\,d\nu_{1}}+\frac{2}{15}\int_{SG}{f\circ F_{2}\,d\nu_{2}}.\\ \end{split}

Since this is true for arbitrary continuous function ff, we have the following identity

ν0=915ν0F01+215ν0F11+215ν1F11115ν2F11+215ν0F21115ν1F21+215ν2F21.\begin{split}\nu_{0}=&\hskip 12.0pt\frac{9}{15}\nu_{0}\circ F_{0}^{-1}\\ &+\frac{2}{15}\nu_{0}\circ F_{1}^{-1}+\frac{2}{15}\nu_{1}\circ F_{1}^{-1}-\frac{1}{15}\nu_{2}\circ F_{1}^{-1}\\ &+\frac{2}{15}\nu_{0}\circ F_{2}^{-1}-\frac{1}{15}\nu_{1}\circ F_{2}^{-1}+\frac{2}{15}\nu_{2}\circ F_{2}^{-1}.\\ \end{split}

By symmetry, we get similar expressions for ν1\nu_{1} and ν2\nu_{2}, and we have the relation

(ν0ν1ν2)=M0(ν0ν1ν2)F01+M1(ν0ν1ν2)F11+M2(ν0ν1ν2)F21.\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)=M_{0}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{0}^{-1}+M_{1}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{1}^{-1}+M_{2}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{2}^{-1}.

Corollary 2.2.

The Kusuoka measure satisfies the variable weight self-similar identity

(2.2) ν=i=02((115+1215Ri)ν)Fi1.\nu=\sum_{i=0}^{2}\left(\left(\frac{1}{15}+\frac{12}{15}R_{i}\right)\nu\right)\circ F_{i}^{-1}.
Proof.

If we consider ν=(111)(ν0ν1ν2)\nu=\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right), where Rj=dνjdνR_{j}=\frac{d\nu_{j}}{d\nu}, the Radon-Nikodym Derivative of νj\nu_{j} w.r.t. ν\nu, we get

(2.3) ν=1215ν0F01+115νF01+1215ν1F11+115νF11+1215ν2F21+115νF21=i=02((115+1215Ri)ν)Fi1.\begin{split}\nu=&\frac{12}{15}\nu_{0}\circ F_{0}^{-1}+\frac{1}{15}\nu\circ F_{0}^{-1}+\frac{12}{15}\nu_{1}\circ F_{1}^{-1}+\frac{1}{15}\nu\circ F_{1}^{-1}+\frac{12}{15}\nu_{2}\circ F_{2}^{-1}+\frac{1}{15}\nu\circ F_{2}^{-1}\\ =&\sum_{i=0}^{2}\left(\left(\frac{1}{15}+\frac{12}{15}R_{i}\right)\nu\right)\circ F_{i}^{-1}.\end{split}

We now give the “self-similarity” of the energy Laplacian Δν\Delta_{\nu}.

Theorem 2.3.

Let Qj=115+1225RjQ_{j}=\frac{1}{15}+\frac{12}{25}R_{j}. Then the energy Laplacian Δν\Delta_{\nu} satisfies

Δν(uFj)=Qj(Δνu)Fj\Delta_{\nu}(u\circ F_{j})=Q_{j}(\Delta_{\nu}u)\circ F_{j}
Proof.

By the definition of the energy Laplacian and by (2.2),

(u,v)=SG(Δνu)v𝑑ν=j=02SG(115+1215Rj)(Δνu)FjvFj𝑑ν.\begin{split}-\mathcal{E}(u,v)&=\int_{SG}{(\Delta_{\nu}u)vd\nu}\\ &=\sum_{j=0}^{2}{\int_{SG}{(\frac{1}{15}+\frac{12}{15}R_{j})(\Delta_{\nu}u)\circ F_{j}\hskip 6.0ptv\circ F_{j}}\hskip 6.0ptd\nu}.\end{split}

On the other hand, the self-similarity of \mathcal{E} also gives

(u,v)=53j=02(uFj,vFj).-\mathcal{E}(u,v)=-\frac{5}{3}\sum_{j=0}^{2}{\mathcal{E}(u\circ F_{j},v\circ F_{j})}.

Together we have

SGj=02(115+215Rj)(Δνu)FjvFjdν=53j=02SGΔν(uFj)vFj𝑑ν.\int_{SG}{\sum_{j=0}^{2}{(\frac{1}{15}+\frac{2}{15}R_{j})(\Delta_{\nu}u)\circ F_{j}\hskip 6.0ptv\circ F_{j}\hskip 6.0ptd\nu}}=\frac{5}{3}\sum_{j=0}^{2}{\int_{SG}{\Delta_{\nu}(u\circ F_{j})\hskip 6.0ptv\circ F_{j}\hskip 6.0ptd\nu}}.

Since vv is arbitary, and so vFjv\circ F_{j} is arbitary, we must have the self-similarity of Δν\Delta_{\nu}

35Qj(Δνu)Fj=Δν(uFj).\frac{3}{5}Q_{j}(\Delta_{\nu}u)\circ F_{j}=\Delta_{\nu}(u\circ F_{j}).

Corollary 2.4.

Let w=(w1,,wm)w=(w_{1},...,w_{m}) be a finite word of length mm and Fw=Fw1.FwmF_{w}=F_{w_{1}}\circ....\circ F_{w_{m}}. Define

Qw=Qwm(Qwm1Fwm)(Qwm2Fwm1Fwm)(Qw1Fw2Fwm)Q_{w}=Q_{w_{m}}\cdot(Q_{w_{m-1}}\circ F_{w_{m}})\cdot(Q_{w_{m-2}}\circ F_{w_{m-1}}\circ F_{w_{m}})\cdots(Q_{w_{1}}\circ F_{w_{2}}\circ\cdots\circ F_{w_{m}})

Then

Δν(uFw)=Qw(Δνu)Fw.\Delta_{\nu}(u\circ F_{w})=Q_{w}(\Delta_{\nu}u)\circ F_{w}.
Proof.

Use the theorem and iterate. ∎

To compute Δνu\Delta_{\nu}u on the cell FwSGF_{w}SG, zoom in by looking at uFwu\circ F_{w} as a function on SGSG, compute its Laplacian Δν(uFw)\Delta_{\nu}(u\circ F_{w}) and multiply by 1Qw\frac{1}{Q_{w}},

(Δνu)Fw=1QwΔν(uFw),(\Delta_{\nu}u)\circ F_{w}=\frac{1}{Q_{w}}\Delta_{\nu}(u\circ F_{w}),

and then zoom out via Fw1F_{w}^{-1}

Δνu|FwSG=(1QwΔν(uFw))Fw1=1QwFw1(Δν(uFw))Fw1.\Delta_{\nu}u|_{F_{w}SG}=\left(\frac{1}{Q_{w}}\Delta_{\nu}(u\circ F_{w})\right)\circ F_{w}^{-1}=\frac{1}{Q_{w}\circ F_{w}^{-1}}\left(\Delta_{\nu}(u\circ F_{w})\right)\circ F_{w}^{-1}.

3. Bounds of derivatives

By the definition of energy measures, we see that every energy measure is continuous; by the self similarity of SG, we see that it makes sense to consider the measure from cells to subcells. It is natural to study how the measure varies for a sequence of cells converging to a point.

Lemma 3.1.

For a harmonic function hh, if we let x=νh(FwF0SG)x=\nu_{h}(F_{w}F_{0}SG), y=νh(FwF1SG)y=\nu_{h}(F_{w}F_{1}SG), z=νh(FwF2SG)z=\nu_{h}(F_{w}F_{2}SG), then 14xyz014x-y-z\geq 0 with equality if and only if hh is skew symmetric (with respect to Fwq0F_{w}q_{0}) on the cell FwSGF_{w}SG.

Proof.

WLOG, let h(Fwq0)=1h(F_{w}q_{0})=1, h(Fwq1)=ah(F_{w}q_{1})=a, h(Fwq2)=0h(F_{w}q_{2})=0. Then, writing m=|w|m=|w| and a direct computation on the energies on each cell,

x=(53)m+1(625)(a23a+3);y=(53)m+1(625)(3a23a+1);z=(53)m+1(625)(a2+a+1)\begin{split}x&=\left(\frac{5}{3}\right)^{m+1}\left(\frac{6}{25}\right)\left(a^{2}-3a+3\right);\\ y&=\left(\frac{5}{3}\right)^{m+1}\left(\frac{6}{25}\right)\left(3a^{2}-3a+1\right);\\ z&=\left(\frac{5}{3}\right)^{m+1}\left(\frac{6}{25}\right)\left(a^{2}+a+1\right)\end{split}

so 14xyz=10(53)m(625)(a24a+4)14x-y-z=10\left(\frac{5}{3}\right)^{m}\left(\frac{6}{25}\right)\left(a^{2}-4a+4\right) and a24a+4=(a2)20a^{2}-4a+4=(a-2)^{2}\geq 0. Equality holds iff a=2a=2, in which case hh is skew symmetric. ∎

Lemma 3.2.

Let ww be a word. For

(xyz)=Ew(222),\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)=E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right),

we have

14xyz>0.14x-y-z>0.
Proof.

Since ν(SG)=6\nu(SG)=6, we have by symmetry ν(FiSG)=2\nu(F_{i}SG)=2 for i=0,1,2i=0,1,2. Thus

(xyz)=(ν(FwF0SG)ν(FwF1SG)ν(FwF2SG)).\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}[]{c}\nu(F_{w}F_{0}SG)\\ \nu(F_{w}F_{1}SG)\\ \nu(F_{w}F_{2}SG)\\ \end{array}\right).

Now Lemma 3.1 implies 14νi(FwF0SG)νi(FwF1SG)νi(FwF2SG)014\nu_{i}(F_{w}F_{0}SG)-\nu_{i}(F_{w}F_{1}SG)-\nu_{i}(F_{w}F_{2}SG)\geq 0 for all ii, so adding these up yields 14xyz014x-y-z\geq 0. To complete the proof, we show that 14νi(FwF0SG)νi(FwF1SG)νi(FwF2SG)>014\nu_{i}(F_{w}F_{0}SG)-\nu_{i}(F_{w}F_{1}SG)-\nu_{i}(F_{w}F_{2}SG)>0 for some ii. Suppose, on the contrary, that each hih_{i} is skew symmetric with respect to Fwq0F_{w}q_{0} on the cell FwSGF_{w}SG. Then there exist ci,aic_{i},a_{i} such that (hiFw(q0),hiFw(q1),hiFw(q2))=(ci,ci+ai,ciai)=ci(1,1,1)+ai(0,1,1)\left(h_{i}\circ F_{w}(q_{0}),h_{i}\circ F_{w}(q_{1}),h_{i}\circ F_{w}(q_{2})\right)=\left(c_{i},c_{i}+a_{i},c_{i}-a_{i}\right)=c_{i}(1,1,1)+a_{i}(0,1,-1), but this shows hih_{i} lie in a two-dimensional subspace, contradicting hi{h_{i}} being linearly independent. ∎

Lemma 3.3.

For the sequence {FwFimSG}m\{F_{w}F_{i}^{m}SG\}_{m}, which converges to the point Fw(qi)F_{w}(q_{i}), we have that ν(FwFimSG)=Θ((35)m)\nu(F_{w}F_{i}^{m}SG)=\Theta\left(\left(\frac{3}{5}\right)^{m}\right), (here Am=Θ(Bm)A_{m}=\Theta(B_{m}) means Am=O(Bm)A_{m}=O(B_{m}) and Bm=O(Am)B_{m}=O(A_{m})).

Proof.

We prove this statement for i=0i=0; the other cases are similar.

We know that

ν(FwF0mSG)=ν0(FwF0mSG)+ν1(FwF0mSG)+ν2(FwF0mSG)=(111)E0mEw((6/52/52/5)+(2/56/52/5)+(2/52/56/5))=(111)E0mEw(222).\begin{split}\nu(F_{w}F_{0}^{m}SG)&=\nu_{0}(F_{w}F_{0}^{m}SG)+\nu_{1}(F_{w}F_{0}^{m}SG)+\nu_{2}(F_{w}F_{0}^{m}SG)\\ &=\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}^{m}E_{w}\left(\left(\begin{array}[]{c}6/5\\ 2/5\\ 2/5\end{array}\right)+\left(\begin{array}[]{c}2/5\\ 6/5\\ 2/5\end{array}\right)+\left(\begin{array}[]{c}2/5\\ 2/5\\ 6/5\end{array}\right)\right)\\ &=\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}^{m}E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right).\\ \end{split}

Now we let

Ew(222)=(xyz)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)=\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)

and see from the diagonalization of E0mE_{0}^{m} that

ν(FwF0mSG)=(34x+98y+98z)(115)m+18(14xyz)(35)m=Θ(35)m\nu(F_{w}F_{0}^{m}SG)=\left(-\frac{3}{4}x+\frac{9}{8}y+\frac{9}{8}z\right)\left(\frac{1}{15}\right)^{m}+\frac{1}{8}\left(14x-y-z\right)\left(\frac{3}{5}\right)^{m}=\Theta\left(\frac{3}{5}\right)^{m}

since 14xyz>014x-y-z>0 by Lemma 3.2. ∎

We have proven the decay rates of the measures of a sequence of cells converging to a point which is not equivalent to a skew-symmetric cell. Surprisingly, in the case which the cell is equivalent to skew-symmetric at that point, we have the first condition for the Radon-Nikodym derivatives to be 0.

Lemma 3.4.

If hh is symmetric on the cell FwSGF_{w}SG with respect to the point Fwq0F_{w}q_{0}, then νh(F2F1mSG)=Θ((115)m)\nu_{h}(F_{2}F_{1}^{m}SG)=\Theta\left(\left(\frac{1}{15}\right)^{m}\right) and hence dνhdν(FwF1(q2))=0\frac{d\nu_{h}}{d\nu}(F_{w}F_{1}(q_{2}))=0. Conversely, if dνhdν(FwF1(q2))=0\frac{d\nu_{h}}{d\nu}(F_{w}F_{1}(q_{2}))=0, then hh is symmetric on the cell FwSGF_{w}SG with respect to the point Fwq0F_{w}q_{0}.

Proof.

We know that νh(FwSG)=(53)|w|νhFw\nu_{h}(F_{w}SG)=\left(\frac{5}{3}\right)^{|w|}\nu_{h\circ F_{w}}, so we may consider the case when ww is the empty word. We see from Lemmas 3.1 and 3.3 that the term (53)m\left(\frac{5}{3}\right)^{m} in νh(F1F2m)\nu_{h}(F_{1}F_{2}^{m}) vanishes if and only if hh is skew-symmetric on F1SGF_{1}SG with respect to the point F1F2m(q2)=F1(q2)F_{1}F_{2}^{m}(q_{2})=F_{1}(q_{2}), which is true if and only if hh is symmetric on SGSG with respect to the point q0q_{0}, in which case νh(F1F2mSG)=Θ((115)m)\nu_{h}(F_{1}F_{2}^{m}SG)=\Theta\left(\left(\frac{1}{15}\right)^{m}\right) and the denominator of dνhdν(x)=limνh(F2F1mSG)ν(F2F1mSG)\frac{d\nu_{h}}{d\nu}(x)=\lim\frac{\nu_{h}(F_{2}F_{1}^{m}SG)}{\nu(F_{2}F_{1}^{m}SG)} dominates. It follows that symmetry of hh around q0q_{0} is sufficient and necessary to have dνhdν(F1(q2))=0\frac{d\nu_{h}}{d\nu}(F_{1}(q_{2}))=0. ∎

Theorem 3.5.

Let hh be a harmonic function on SG with νh=aν0+bν1+cν2\nu_{h}=a\nu_{0}+b\nu_{1}+c\nu_{2} and hh^{\perp} be a harmonic function orthonormal to hh under the energy inner product. Also let CC be a cell in SGSG. Then
a)

infxCdνhdν(x)=0;\inf_{x\in C}\frac{d\nu_{h}}{d\nu}(x)=0;

b)

supxCdνhdν=23(a+b+c).\sup_{x\in C}\frac{d\nu_{h}}{d\nu}=\frac{2}{3}(a+b+c).

and if the maximum of dνhdν\frac{d\nu_{h}}{d\nu} is attained, then the minimum of dνhdν\frac{d\nu_{h^{\perp}}}{d\nu} is attained at the same point.

Proof.

We first prove a). If two of the values of hh on the boundary of CC are the same then hh is symmetric and the result follows from Lemma 3.4, so we assume all three values are distinct. Since we consider hh modulo constants and the conclusion is unchanged by replacing hh with a constant multiple chch, we may assume without loss of generality that hh is harmonic with h(q0)=0h(q_{0})=0, h(q1)=1h(q_{1})=1, h(q2)=a2h(q_{2})=a\geq 2. Let xx be the point on the edge from q0q_{0} to q1q_{1} where hh achieves its maximum. It was shown in [3] that if we identify this edge in the obvious fashion with the unit interval [0,1][0,1] oriented so q0q_{0} corresponds to 0 and q1q_{1} to 11, and define a map M:[2,)[0,1]M:[2,\infty)\to[0,1] by M(a)=xM(a)=x, then MM is continuous and strictly decreasing, and that limaM(a)\lim_{a\to\infty}M(a) gives the midpoint F1q0F_{1}q_{0}. It follows that if we have a harmonic function on a cell such that the maximum along a side occurs at the midpoint of that side then the function is symmetric under the reflection fixing the vertex opposite the side. This is applicable in the case when the location of our maximum is xVx\in V_{*} (equivalently a dyadic point of [0,1][0,1] under our identification), because then there is a word ww such that x=FwF1(q0)x=F_{w}F_{1}(q_{0}), and the function is harmonic on Fw(SG)F_{w}(SG) with its maximum along Fw[q0,q1]F_{w}[q_{0},q_{1}] at the midpoint of this side. The conclusion that hh is symmetric on Fw(SG)F_{w}(SG) allows us to apply Lemma 3.4 to deduce that dνhdν(x)=0\frac{d\nu_{h}}{d\nu}(x)=0.

Now suppose that xx is not a dyadic point. Since dyadic points are dense in the interval [0,1], we pick a increasing sequence xnx_{n} of dyadic points converging to xx. Because MM is decreasing and continuous, there exists a decreasing sequence of ana_{n} converging to aa satisfying M(an)=xnM(a_{n})=x_{n}.

Let ε>0\varepsilon>0 be given. A simple computation shows that the energy measure νh=aν0+(1a)ν1+(a2a)ν2\nu_{h}=a\nu_{0}+(1-a)\nu_{1}+(a^{2}-a)\nu_{2}. From the previous discussion, we can pick an ana_{n}(xnx_{n}, resp.), which will be denoted as aεa_{\varepsilon}(xεx_{\varepsilon}, resp.), satisfying |aaε|+|(1a)(1aε)|+|(a2a)(aε2aε)|<ε|a-a_{\varepsilon}|+|(1-a)-(1-a_{\varepsilon})|+|(a^{2}-a)-(a_{\varepsilon}^{2}-a_{\varepsilon})|<\varepsilon. We also denote the harmonic function with boundary values 0,1,aε0,1,a_{\varepsilon} as hεh_{\varepsilon}. The previous discussion shows that dνhεdν(xε)=0\frac{d\nu_{h_{\varepsilon}}}{d\nu}(x_{\varepsilon})=0. Now,

dνhdν(xε)=d(νhνhε)dν(xε)+dνhεdν(xε)=d(νhνhε)dν(xε)=(aaε)dν0dν(xε)+((1a)(1aε))dν1dν(xε)+((a2a)(aε2aε))dν2dν(xε)|aaε|+|(1a)(1aε)|+|(a2a)(aε2aε)|<ε\begin{split}&\frac{d\nu_{h}}{d\nu}(x_{\varepsilon})\\ =&\frac{d(\nu_{h}-\nu_{h_{\varepsilon}})}{d\nu}(x_{\varepsilon})+\frac{d\nu_{h_{\varepsilon}}}{d\nu}(x_{\varepsilon})\\ =&\frac{d(\nu_{h}-\nu_{h_{\varepsilon}})}{d\nu}(x_{\varepsilon})\\ =&\big{(}a-a_{\varepsilon}\big{)}\frac{d\nu_{0}}{d\nu}(x_{\varepsilon})+\big{(}(1-a)-(1-a_{\varepsilon})\big{)}\frac{d\nu_{1}}{d\nu}(x_{\varepsilon})+\big{(}(a^{2}-a)-(a_{\varepsilon}^{2}-a_{\varepsilon})\big{)}\frac{d\nu_{2}}{d\nu}(x_{\varepsilon})\\ \leq&|a-a_{\varepsilon}|+|(1-a)-(1-a_{\varepsilon})|+|(a^{2}-a)-(a_{\varepsilon}^{2}-a_{\varepsilon})|\\ <&\varepsilon\end{split}

where the penultimate inequality comes from the fact that dνidν1\frac{d\nu_{i}}{d\nu}\leq 1 since νi(C)ν(C)\nu_{i}(C)\leq\nu(C) for any cell CC.
And consequently,

infxSGdνhdν=0.\inf_{x\in SG}\frac{d\nu_{h}}{d\nu}=0.

On a general cell CC, h|Ch|_{C} is harmonic, so the same argument applies.

We then show b). An easy computation shows 13ν=νrh+νh\frac{1}{3}\nu=\nu_{rh}+\nu_{h^{\perp}}, where r2=1νh(SG)r^{2}=\frac{1}{\nu_{h}(SG)}. Since the total measure of νh\nu_{h} on SGSG is 2(a+b+c)2(a+b+c), we have

32(a+b+c)dνhdν+3dνhdν=1.\frac{3}{2(a+b+c)}\frac{d\nu_{h}}{d\nu}+3\frac{d\nu_{h^{\perp}}}{d\nu}=1.

It follows that

32(a+b+c)supxCdνhdν=13infxCdνhdν=1.\frac{3}{2(a+b+c)}\sup_{x\in C}\frac{d\nu_{h}}{d\nu}=1-3\inf_{x\in C}\frac{d\nu_{h^{\perp}}}{d\nu}=1.

Whence,

supxSGdνhdν=2(a+b+c)3.\sup_{x\in SG}\frac{d\nu_{h}}{d\nu}=\frac{2(a+b+c)}{3}.

Further, if the derivative attains its supremum, then the maximum occurs where 0 occurs in dνhdν\frac{d\nu_{h^{\perp}}}{d\nu}, that is, where the local extremum occurs in the smallest edge of the harmonic function hh^{\perp}. ∎

As we can see, on every cell of SG, the supremum and infimum are distance 2(a+b+c)3\frac{2(a+b+c)}{3} from each other, so the derivative is far from continuous. However, we will later recover limited continuity when we restrict to the edges of cells.

4. Characterzation of Positive Energy Measures

Within the 3-dimensional space of signed energy measures which measures are positive? The following result gives the answer.

Theorem 4.1.

Let PP be the set of (a,b,c)3(a,b,c)\in\mathbb{R}^{3} such that aν0+bν1+cν2a\nu_{0}+b\nu_{1}+c\nu_{2} is a positive energy measure. Then PP is the solid cone S={(a,b,c)3:ab+bc+ca0}S=\{(a,b,c)\in\mathbb{R}^{3}:ab+bc+ca\geq 0\}, and the boundary of PP is the set of (a,b,c)(a,b,c) such that aν0+bν1+cν2=νha\nu_{0}+b\nu_{1}+c\nu_{2}=\nu_{h} for some harmonic function hh.

Proof.

We first show that SPS\subset P. We know from [1] that the set of (x,y,z)(x,y,z) such that x=νh(F0SG),y=νh(F1SG),z=νh(F2SG)x=\nu_{h}(F_{0}SG),y=\nu_{h}(F_{1}SG),z=\nu_{h}(F_{2}SG) form the cone

1125(x+y+z)2=x2+y2+z2.\frac{11}{25}(x+y+z)^{2}=x^{2}+y^{2}+z^{2}.

A simple computation shows that

25(311131113)\frac{2}{5}\left(\begin{array}[]{ccc}3&1&1\\ 1&3&1\\ 1&1&3\end{array}\right)

transforms (a,b,c)(a,b,c) to the corresponding coefficients (x,y,z)(x,y,z). Hence, the set of (a,b,c)(a,b,c) that corresponds to some νh\nu_{h} is the cone

ab+bc+ca=0,ab+bc+ca=0,

which is the boundary of SS. Then a general point in SS has the form (a,b,c)+(δ0,δ1,δ2)(a,b,c)+(\delta_{0},\delta_{1},\delta_{2}) with all δj0\delta_{j}\geq 0, so it yields a positive measure νh+δ0ν0+δ1ν1+δ2ν2\nu_{h}+\delta_{0}\nu_{0}+\delta_{1}\nu_{1}+\delta_{2}\nu_{2} and so belongs to PP.

Now we show that the coefficients that come from harmonic functions (the boundary of SS) are on the boundary of PP. Consider the energy measure νhεν\nu_{h}-\varepsilon\nu for ε>0\varepsilon>0. Suppose, for contradiction, that this measure is positive. Then for any cell CC, (νhεν)(C)>0(\nu_{h}-\varepsilon\nu)(C)>0, so νh(C)ν(C)>ε>0\frac{\nu_{h}(C)}{\nu(C)}>\varepsilon>0 for all CC, but we know the infimum should be zero, so this is a contradiction. Therefore (aε,bε,cε)P(a-\varepsilon,b-\varepsilon,c-\varepsilon)\notin P and (a,b,c)(a,b,c) is on the boundary of PP. Since the two boundaries coincide, S=PS=P. ∎

Immediately, since the solid cone is convex, every positive energy measure is precisely a convex combination of two energy measures, each coming from a single harmonic function. But we can prove a little bit more.

Corollary 4.2.

Every positive energy measure of harmonic functions is a convex combination of positive energy measures of hh and hh^{\perp} for some harmonic function hh, and vice versa.

Proof.

For every harmonic function hh, νh+νh=cν\nu_{h}+\nu_{h^{\perp}}=c\nu for some c>0c>0. This shows νh\nu_{h}, νh\nu_{h^{\perp}}, ν\nu lie on the same two-dimensional subspace.

Suppose a two-dimensional subspace WW containing ν\nu is given. WW intersects the boundary of the cone SS stated in the theorem on two and only two lines which contain energy measures of a single harmonic function. The above paragraph proves that one of the two lines contains νh\nu_{h} for some hh while the other contains νh\nu_{h^{\perp}}.

Now, given any positive energy measure σ\sigma which is not a multiple of ν\nu (in which case, the conclusion is trivial), consider the two-dimensional subspace VV containing σ\sigma and ν\nu. By the previous reasoning and the fact that VSV\cap S is the closed convex hull of VSV\cap\partial S, there are harmonic functions hh and hh^{\perp} such that σ\sigma is the convex combination of νah\nu_{ah} and νbh\nu_{bh^{\perp}} for some positive constants aa and bb. It follows that for some 0t10\leq t\leq 1, for any r,s0r,s\geq 0,

σ=tr2νrah+(1t)s2νsbh.\sigma=tr^{-2}\nu_{rah}+(1-t)s^{-2}\nu_{sbh^{\perp}}.

Choosing r,sr,s such that tr2+(1t)s2=1tr^{-2}+(1-t)s^{-2}=1 and ra=sbra=sb concludes the proof. ∎

5. Limited Continuity

We have seen from the previous section that the derivative of an energy measure is not continuous; however, if we restrict the derivative to the set of vertices VV_{*}, it is continuous on the edges of every triangle. This can be seen in Figure 5.1.

Refer to caption
Figure 5.1. Radon-Nikodym Derivative of ν0\nu_{0} with red dots indicating minimum values and green dot indicating maximum value.

By (1.5), a simple computation shows that

(5.1) limm(53)mE0m=140(423314111411);limm(53)mE1m=140(114134231141);limm(53)mE2m=140(111411143342).\begin{split}\lim_{m\to\infty}\left(\frac{5}{3}\right)^{m}E_{0}^{m}&=\frac{1}{40}\left(\begin{array}[]{ccc}42&-3&-3\\ 14&-1&-1\\ 14&-1&-1\end{array}\right);\\ \lim_{m\to\infty}\left(\frac{5}{3}\right)^{m}E_{1}^{m}&=\frac{1}{40}\left(\begin{array}[]{ccc}-1&14&-1\\ -3&42&-3\\ -1&14&-1\end{array}\right);\\ \lim_{m\to\infty}\left(\frac{5}{3}\right)^{m}E_{2}^{m}&=\frac{1}{40}\left(\begin{array}[]{ccc}-1&-1&14\\ -1&-1&14\\ -3&-3&42\end{array}\right).\\ \end{split}

Notice that all of the above are rank-one matrices.

We denote A:=limm(53)mE2mA:=\lim_{m\to\infty}\left(\frac{5}{3}\right)^{m}E_{2}^{m} and Am:=(53)mE2m,Dm:=AAmA_{m}:=\left(\frac{5}{3}\right)^{m}E_{2}^{m},D_{m}:=A-A_{m}. We also know

EwA=Ew(113)(1114)E_{w}A=E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)

so

(1114)EwA=cw(1114).\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w}A=c_{w}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right).

In fact

cw=(1114)Ew(113).c_{w}=\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right).

We are going to prove, for any finite word ww consisting only 11 and 22, that (53)|w|cw\left(\frac{5}{3}\right)^{|w|}c_{w}  is bounded away from zero; to do this, we need the following lemmas.

Lemma 5.1.

Given any finite word ww consists of letters 11 and 22 only, and let

(xyz)=52Ew(ν2(F0SG)ν2(F1SG)ν2(F2SG))=Ew(113).\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)=\frac{5}{2}E_{w}\left(\begin{array}[]{c}\nu_{2}(F_{0}SG)\\ \nu_{2}(F_{1}SG)\\ \nu_{2}(F_{2}SG)\end{array}\right)=E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right).

Then

(1114)E2(xyz)(1114)E1(xyz).\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{2}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)\geq\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{1}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right).
Proof.

We prove this by induction. It is trivial for ww being the empty word. Suppose that the conclusion holds for |w|=m|w|=m, which is equivalent to saying x4y+11z0x-4y+11z\geq 0. Also assume for induction that zxz\geq x. Consider first the word w1w1. We have

(1114)(E2E1)E1(xyz)=35(1411)E1(xyz)=125(9x4y+21z)=125((x4y+11z)+10(zx))0.\begin{split}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)(E_{2}-E_{1})E_{1}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)&=\frac{3}{5}\left(\begin{array}[]{ccc}1&-4&11\end{array}\right)E_{1}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)\\ &=\frac{1}{25}(-9x-4y+21z)\\ &=\frac{1}{25}\big{(}(x-4y+11z)+10(z-x)\big{)}\\ &\geq 0.\end{split}

Furthermore, the third entry of E1(xyz)E_{1}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right) minus the first entry is

15(zx)0.15(z-x)\geq 0.

Now we turn to the second case concerning the word w2w2.

(1114)(E2E1)E2(xyz)=35(1411)E2(xyz)=15(3y+19z)=15((x4y+11z)+(zx)+y+7z)0.\begin{split}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)(E_{2}-E_{1})E_{2}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)&=\frac{3}{5}\left(\begin{array}[]{ccc}1&-4&11\end{array}\right)E_{2}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)\\ &=\frac{1}{5}(-3y+19z)\\ &=\frac{1}{5}\big{(}(x-4y+11z)+(z-x)+y+7z\big{)}\\ &\geq 0.\end{split}

Also, the third entry of E2(xyz)E_{2}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right) minus the first entry is

125(4x+y+11z)=125(y+4(zx)+7z)0.\frac{1}{25}\big{(}-4x+y+11z\big{)}=\frac{1}{25}\big{(}y+4(z-x)+7z\big{)}\geq 0.

The positivity of yy and zz comes from the fact that

Ew(113)=25(ν2(Fw0SG)ν2(Fw1SG)ν2(Fw2SG)).E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)=\frac{2}{5}\left(\begin{array}[]{c}\nu_{2}(F_{w0}SG)\\ \nu_{2}(F_{w1}SG)\\ \nu_{2}(F_{w2}SG)\end{array}\right).

This completes the induction. ∎

Lemma 5.2.

Under the same hypothesis and notations as in lemma 5.1, we have

(1114)E1E2Ew(113)(1114)E2E1Ew(113).\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{1}E_{2}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\geq\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{2}E_{1}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right).
Proof.

Fix an (m2)(m-2)-length word ww consisting only letters 11 and 22. The column vector (1, 1, 3) corresponds to the energy measure ν2\nu_{2} obtained from only h2h_{2}. By modulo a constant, we have, as shown in the figures of FwSGF_{w}SG, the values of h2h_{2} and energies on the cells of ν2\nu_{2}. Note that 0ab0\leq a\leq b.

[Uncaptioned image][Uncaptioned image]

A simple manipulation on matrices shows that

(1114)E2E1Ew(113)=15(111)E2E2E1Ew(113)(111)E2E1Ew(113)=52(15ν2(FwF1F2F2SG)ν2(FwF1F2SG)).\begin{split}&\hskip 2.0pt\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{2}E_{1}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\\ =&\hskip 2.0pt15\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{2}E_{2}E_{1}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)-\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{2}E_{1}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\\ =&\hskip 2.0pt\frac{5}{2}\left(15\nu_{2}(F_{w}F_{1}F_{2}F_{2}SG)-\nu_{2}(F_{w}F_{1}F_{2}SG)\right).\end{split}

Similarly,

(1114)E1E2Ew(113)=15(111)E2E1E2Ew(113)(111)E1E2Ew(113)=52(15ν2(FwF2F1F2SG)ν2(FwF2F1SG)).\begin{split}&\hskip 2.0pt\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{1}E_{2}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\\ =&\hskip 2.0pt15\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{2}E_{1}E_{2}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)-\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{1}E_{2}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\\ =&\hskip 2.0pt\frac{5}{2}\left(15\nu_{2}(F_{w}F_{2}F_{1}F_{2}SG)-\nu_{2}(F_{w}F_{2}F_{1}SG)\right).\end{split}

Whence, from the calculation in the figures,

(1114)E1E2Ew(113)(1114)E2E1Ew(113)=4625(53)m(6053a2+852ab+144b2)0.\begin{split}&\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{1}E_{2}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)-\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{2}E_{1}E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right)\\ =&\frac{4}{625}\left(\frac{5}{3}\right)^{m}(6053a^{2}+852ab+144b^{2})\geq 0.\end{split}

What we have proven in the above lemmas is that, for two level-mm cells FwSGF_{w}SG and FwSGF_{w^{\prime}}SG whose words consist only of the letters 11 and 22, we have ν2(FwSG)ν2(FwSG)\nu_{2}(F_{w}SG)\leq\nu_{2}(F_{w^{\prime}}SG), provided that FwSGF_{w}SG is to the left of FwF_{w^{\prime}}, in the orientation in which q1q_{1} is the left endpoint of [q1,q2][q_{1},q_{2}]. It also follows immediately that for those word ww of length mm,

cwc1,,1=155m+99415m+52(35)mc_{w}\geq c_{1,...,1}=\frac{15}{5^{m}}+\frac{99}{4\cdot 15^{m}}+\frac{5}{2}\cdot\left(\frac{3}{5}\right)^{m}

which implies (53)|w|cw\left(\frac{5}{3}\right)^{|w|}c_{w} is bounded away from zero, for every such ww.
We now turn to an upper bound for (53)|w|Ew\left(\frac{5}{3}\right)^{|w|}\|E_{w}\|, where \|\cdot\| is the operator 11-norm.


Lemma 5.3.

For any harmonic function ff and word ww with |w|=m|w|=m,

Fw(f)2Osc(f,SG)2(35)m.\mathcal{E}_{F_{w}}(f)\leq 2\hskip 1.0pt\textrm{Osc}(f,SG)^{2}\left(\frac{3}{5}\right)^{m}.
Proof.

By [8],

Osc(f,FwSG)(35)mOsc(f,SG).\textrm{Osc}(f,F_{w}SG)\leq\left(\frac{3}{5}\right)^{m}\textrm{Osc}(f,SG).

Since for fixed oscillation, the greatest energy is attained by a function which is symmetric on FwSGF_{w}SG, which has energy (53)m2(Osc(f,FwSG)2\left(\frac{5}{3}\right)^{m}2\left(\mathrm{Osc}(f,F_{w}SG\right)^{2}, we have

Fw(f)2Osc(f,SG)2(35)m.\mathcal{E}_{F_{w}}(f)\leq 2\hskip 1.0pt\textrm{Osc}(f,SG)^{2}\left(\frac{3}{5}\right)^{m}.

Lemma 5.4.

There exists a constant CC, independent of mm, such that

EwC(35)m\|E_{w}\|\leq C\left(\frac{3}{5}\right)^{m}

for all |w|=m|w|=m

Proof.

Consider (x,y,z)=1\|(x,y,z)\|=1. If we view x,y,zx,y,z as the energies of the level 1 cells, we know that (x,y,z)(x,y,z) corresponds to an energy measure a0ν0+a1ν1+a2ν2a_{0}\nu_{0}+a_{1}\nu_{1}+a_{2}\nu_{2}. Since (x,y,z)(x,y,z) and (a0,a1,a2)(a_{0},a_{1},a_{2}) differ only by a linear transformation, (a0,a1,a2)B\|(a_{0},a_{1},a_{2})\|\leq B for some constant BB for all such triples (a0,a1,a2)(a_{0},a_{1},a_{2}). Let a word ww be fixed, with |w|=m|w|=m. Since from Lemma 5.3 we know that νj(FwiSG)2(35)m+1\nu_{j}(F_{wi}SG)\leq 2\left(\frac{3}{5}\right)^{m+1}, we have

Ew(xyz)=i=02|j=02ajνj(FwiSG)|i=02j=02|aj||νj(FwiSG)|i=02j=02|aj|˙2(35)m+16B(35)m+1.\begin{split}\bigg{\|}E_{w}\left(\begin{array}[]{c}x\\ y\\ z\end{array}\right)\bigg{\|}&=\sum_{i=0}^{2}\bigg{|}\sum_{j=0}^{2}a_{j}\nu_{j}(F_{wi}SG)\bigg{|}\\ &\leq\sum_{i=0}^{2}\sum_{j=0}^{2}\big{|}a_{j}\big{|}\big{|}\nu_{j}(F_{wi}SG)\big{|}\\ &\leq\sum_{i=0}^{2}\sum_{j=0}^{2}\big{|}a_{j}\big{|}\dot{}2\left(\frac{3}{5}\right)^{m+1}\\ &\leq 6B\left(\frac{3}{5}\right)^{m+1}.\\ \end{split}

Taking supremum over all (x,y,z)=1\|(x,y,z)\|=1 and letting C=6B(35)C=6B\left(\frac{3}{5}\right) give

EwC(35)m.\|E_{w}\|\leq C\left(\frac{3}{5}\right)^{m}.

We now give the main result of this section, the continuity on the three edges of a cell.

Theorem 5.5.

Given an energy measure νf,g\nu_{f,g} and a cell FwSGF_{w}SG, the restriction of the derivative dνf,gdν\frac{d\nu_{f,g}}{d\nu} on the three edges restricted to the vertices VV_{*} is continuous.

Proof.

Since every point xx of VV_{*} that is on the edge of a cell is the junction point of two cells (boundary points excepted) it suffices to show that the restriction of the derivative dνf,gdν\frac{d\nu_{f,g}}{d\nu} on the edge Fw[q1,q2]F_{w}[q_{1},q_{2}] is continuous at Fwq2F_{w}q_{2}, where [q1,q2][q_{1},q_{2}] is the edge connecting q1q_{1} and q2q_{2}. Continuity on the other side of Fwq2F_{w}q_{2} follows from symmetry and arbitrary choice of cell. Let

X=(νf,g(Fw0SG)νf,g(Fw1SG)νf,g(Fw2SG)),Ξ=(αβγ)=(ν(Fw0SG)ν(Fw1SG)ν(Fw2SG)).X=\left(\begin{array}[]{c}\nu_{f,g}(F_{w0}SG)\\ \nu_{f,g}(F_{w1}SG)\\ \nu_{f,g}(F_{w2}SG)\\ \end{array}\right),\qquad\Xi=\left(\begin{array}[]{c}\alpha\\ \beta\\ \gamma\end{array}\right)=\left(\begin{array}[]{c}\nu(F_{w0}SG)\\ \nu(F_{w1}SG)\\ \nu(F_{w2}SG)\\ \end{array}\right).

The points of VV_{*} in the interval FwF2n[q1,q2]F_{w}F_{2}^{n}[q_{1},q_{2}] with the exception of FwF2n(q1)F_{w}F_{2}^{n}(q_{1}) are of the form FwF2nFwq2F_{w}F_{2}^{n}F_{w^{\prime}}q_{2}. We know that the Radon-Nikodym derivative at these points converges to dνf,gdν(Fwq2)\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{2}) uniformly in ww^{\prime}, establishing continuity of the derivative at Fwq2F_{w}q_{2}. To do so, fix ww^{\prime} with |w|=m|w^{\prime}|=m and observe that

dνf,gdν(FwF2nFwq2)=(1114)EwAnX(1114)EwAnΞ=(1114)Ew(ADn)X(1114)Ew(ADn)Ξ.\frac{d\nu_{f,g}}{d\nu}(F_{w}F_{2}^{n}F_{w^{\prime}}q_{2})=\frac{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}A_{n}X}{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}A_{n}\Xi}=\frac{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}(A-D_{n})X}{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}(A-D_{n})\Xi}.

Since AA is rank 11 we have

(1114)EwAX(1114)EwAΞ=(1114)X(1114)Ξ=dνf,gdν(Fwq2).\frac{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}AX}{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}A\Xi}=\frac{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)X}{\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)\Xi}=\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{2}).

Also, from Lemmas 5.1 and 5.2 we know (53)m(1114)EwAΞ52(14γαβ)>0\left(\frac{5}{3}\right)^{m}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w^{\prime}}A\Xi\geq\frac{5}{2}(14\gamma-\alpha-\beta)>0, while for sufficiently large nn independent of mm we have EwDnδ(35)m\|E_{w^{\prime}}D_{n}\|\leq\delta\left(\frac{3}{5}\right)^{m} because of Lemma 5.4 and the fact that AnAA_{n}\to A in operator norm. The result follows.

The above idea uses only the lower bound for (53)m(1114)Ew(113)\left(\frac{5}{3}\right)^{m}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{w}\left(\begin{array}[]{c}1\\ 1\\ 3\end{array}\right) and upper bound for (53)mEw\left(\frac{5}{3}\right)^{m}\|E_{w}\|.

6. Average Value of the Radon-Nikodym Derivative

We have shown how to calculate the values of the derivative on the vertices in VV_{*}; moreover, we have continuity on the edges of every triangle. We would like to relate the average of derivatives on the whole cell to a weighted average on three vertices of the cell; that is, to find, on a cell CC, positive real numbers {bj}j=02\{b_{j}\}_{j=0}^{2} whose sum is 11 satisfying

νf,g(C)ν(C)=1ν(C)Cdνf,gdν𝑑ν=rjCbjdνf,gdν(rj)\frac{\nu_{f,g}(C)}{\nu(C)}=\frac{1}{\nu(C)}\int_{C}{\frac{d\nu_{f,g}}{d\nu}d\nu}=\sum_{r_{j}\in\partial C}{b_{j}\frac{d\nu_{f,g}}{d\nu}(r_{j})}

for every energy measure of harmonic functions νf,g\nu_{f,g}. In this section, we will show that the coefficients bjb_{j} exist and discover some characterizations of them.

6.1. Existence

Recall from Theorem 2.1, that we have

(ν0ν1ν2)=M0(ν0ν1ν2)F01+M1(ν0ν1ν2)F11+M2(ν0ν1ν2)F21\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)=M_{0}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{0}^{-1}+M_{1}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{1}^{-1}+M_{2}\left(\begin{array}[]{c}\nu_{0}\\ \nu_{1}\\ \nu_{2}\end{array}\right)\circ F_{2}^{-1}

and

(ν0(FjA)ν1(FjA)ν2(FjA))=Mj(ν0(A)ν1(A)ν2(A)).\left(\begin{array}[]{c}\nu_{0}(F_{j}A)\\ \nu_{1}(F_{j}A)\\ \nu_{2}(F_{j}A)\end{array}\right)=M_{j}\left(\begin{array}[]{c}\nu_{0}(A)\\ \nu_{1}(A)\\ \nu_{2}(A)\end{array}\right).

Letting Mw=Mw1Mw2MwmM_{w}=M_{w_{1}}M_{w_{2}}\cdots M_{w_{m}} we see that

(ν0(FwA)ν1(FwA)ν2(FwA))=Mw(ν0(A)ν1(A)ν2(A)).\left(\begin{array}[]{c}\nu_{0}(F_{w}A)\\ \nu_{1}(F_{w}A)\\ \nu_{2}(F_{w}A)\end{array}\right)=M_{w}\left(\begin{array}[]{c}\nu_{0}(A)\\ \nu_{1}(A)\\ \nu_{2}(A)\end{array}\right).
Theorem 6.1.

On each cell C=FwSGC=F_{w}SG, there exist unique (b0,b1,b2)3(b_{0},b_{1},b_{2})\in\mathbb{R}^{3} with b0+b1+b2=1b_{0}+b_{1}+b_{2}=1 such that

AvgFwSGdνf,gdν=1ν(C)Cdνf,gdν𝑑ν=j=02bjdνf,gdν(Fw(qj))\textrm{Avg}_{F_{w}SG}\frac{d\nu_{f,g}}{d\nu}=\frac{1}{\nu(C)}\int_{C}\frac{d\nu_{f,g}}{d\nu}d\nu=\sum_{j=0}^{2}{b_{j}\frac{d\nu_{f,g}}{d\nu}\big{(}F_{w}(q_{j})\big{)}}

for all energy measure (not necessarily positive) νf,g\nu_{f,g} on SGSG. Explicitly,

(6.1) bj=16+12sum of column j of Mwsum of all entries of Mw.b_{j}=\frac{1}{6}+\frac{1}{2}\frac{\text{sum of column }j\text{ of }M_{w}}{\text{sum of all entries of }M_{w}}.
Proof.

We first introduce some notation. We let

z0=Mw(100),z1=Mw(010),z2=Mw(001),z=z0+z1+z2.\begin{split}z_{0}=M_{w}\left(\begin{array}[]{c}1\\ 0\\ 0\end{array}\right),z_{1}&=M_{w}\left(\begin{array}[]{c}0\\ 1\\ 0\end{array}\right),z_{2}=M_{w}\left(\begin{array}[]{c}0\\ 0\\ 1\end{array}\right),z=z_{0}+z_{1}+z_{2}.\end{split}

Now suppose νf,g=α0ν0+α1ν1+α2ν2\nu_{f,g}=\alpha_{0}\nu_{0}+\alpha_{1}\nu_{1}+\alpha_{2}\nu_{2}. Then we have

νf,g(FwSG)=2(α0α1α2)z,ν(FwSG)=2(111)z\nu_{f,g}(F_{w}SG)=2\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)z,\hskip 10.0pt\nu(F_{w}SG)=2\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z

so the average value of the Radon-Nikodym derivative of νf,g\nu_{f,g} on the cell FwSGF_{w}SG is

FwSGdνf,gdν𝑑νFwSG𝑑ν=νf,g(FwSG)ν(FwSG)=(α0α1α2)z(111)z.\frac{\int_{F_{w}SG}\frac{d\nu_{f,g}}{d\nu}d\nu}{\int_{F_{w}SG}d\nu}=\frac{\nu_{f,g}(F_{w}SG)}{\nu(F_{w}SG)}=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)z}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z}.

We also note that

dνf,gdν(Fwqj)=limmνf,g(FwFjmSG)ν(FwFjmSG)=limm(α0α1α2)Mw(dν0dν(qj)dν1dν(qj)dν2dν(qj))(111)Mw(dν0dν(qj)dν1dν(qj)dν2dν(qj))\begin{split}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{j})&=\lim_{m\to\infty}\frac{\nu_{f,g}(F_{w}F_{j}^{m}SG)}{\nu(F_{w}F_{j}^{m}SG)}\\ &=\lim_{m\to\infty}\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)M_{w}\left(\begin{array}[]{c}\frac{d\nu_{0}}{d\nu}(q_{j})\\ \frac{d\nu_{1}}{d\nu}(q_{j})\\ \frac{d\nu_{2}}{d\nu}(q_{j})\end{array}\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}\left(\begin{array}[]{c}\frac{d\nu_{0}}{d\nu}(q_{j})\\ \frac{d\nu_{1}}{d\nu}(q_{j})\\ \frac{d\nu_{2}}{d\nu}(q_{j})\end{array}\right)}\end{split}

and we know that dνjdν(qk)\frac{d\nu_{j}}{d\nu}(q_{k}) is equal to 23\frac{2}{3} if j=kj=k and 16\frac{1}{6} otherwise.

So

dνf,gdν(Fwq0)=(α0α1α2)Mw(231616)(111)Mw(231616)=(α0α1α2)(z0+13z)(111)(z0+13z);\begin{split}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{0})&=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)M_{w}\left(\begin{array}[]{c}\frac{2}{3}\\ \frac{1}{6}\\ \frac{1}{6}\end{array}\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}\left(\begin{array}[]{c}\frac{2}{3}\\ \frac{1}{6}\\ \frac{1}{6}\end{array}\right)}\\ &=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)\left(z_{0}+\frac{1}{3}z\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(z_{0}+\frac{1}{3}z\right)};\end{split}

and similarly

dνf,gdν(Fwq1)=(α0α1α2)(z1+13z)(111)(z1+13z);dνf,gdν(Fwq2)=(α0α1α2)(z2+13z)(111)(z2+13z).\begin{split}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{1})&=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)\left(z_{1}+\frac{1}{3}z\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(z_{1}+\frac{1}{3}z\right)};\\ \frac{d\nu_{f,g}}{d\nu}(F_{w}q_{2})&=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)\left(z_{2}+\frac{1}{3}z\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(z_{2}+\frac{1}{3}z\right)}.\end{split}

Now we suppose that we can write the average of the derivative on the whole cell as a weighted average of the value of the derivative on the boundary points; that is, we write

AvgFwSGdνf,gdν=(α0α1α2)z(111)z=b0dνf,gdν(Fwq0)+b1dνf,gdν(Fwq1)+b2dνf,gdν(Fwq2)=(α0α1α2)(j=02bj(zj+13z)(111)(zj+13z)).\begin{split}\textrm{Avg}_{F_{w}SG}\frac{d\nu_{f,g}}{d\nu}&=\frac{\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)z}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z}\\ &=b_{0}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{0})+b_{1}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{1})+b_{2}\frac{d\nu_{f,g}}{d\nu}(F_{w}q_{2})\\ &=\left(\begin{array}[]{ccc}\alpha_{0}&\alpha_{1}&\alpha_{2}\end{array}\right)\left(\sum_{j=0}^{2}{\frac{b_{j}\left(z_{j}+\frac{1}{3}z\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(z_{j}+\frac{1}{3}z\right)}}\right).\end{split}

Then it is implied that

(6.2) j=02bj(zj+13z)(111)(zj+13z)=z(111)z.\sum_{j=0}^{2}{\frac{b_{j}\left(z_{j}+\frac{1}{3}z\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(z_{j}+\frac{1}{3}z\right)}}=\frac{z}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z}.

We also see that multiplying both sides of this equation by (111)\left(\begin{array}[]{ccc}1&1&1\end{array}\right) gives b0+b1+b2=1b_{0}+b_{1}+b_{2}=1.

The solution to (6.2) is

bj=(111)(16z+12zj)(111)z,\begin{split}b_{j}&=\frac{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)\left(\frac{1}{6}z+\frac{1}{2}z_{j}\right)}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z},\end{split}

as can be seen by substitution. The solution is unique because the construction of the zj+13zz_{j}+\frac{1}{3}z from the invertible matrix MwM_{w} ensures that they are linearly independent.

Noting that (111)zj\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z_{j} is the sum of the entries in column jj of MwM_{w} and (111)z\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z is the sum of all entries of MwM_{w}. This yields (6.1). ∎

Later we will show that the bjb_{j} values lie between 0 and 23\frac{2}{3}; thus, the above theorem shows that we can relate the average value of the derivative on the whole cell to the weighted average value of the derivative at the vertices of the cell(one set of weights works for all energy measures). Since the bjb_{j} values depend on the word ww, from now on we will denote bjb_{j} on FwSGF_{w}SG as bj(w)b_{j}^{(w)}.

6.2. Characterization

We now have coefficients bj(w)b_{j}^{(w)} for the cell FwSGF_{w}SG to calculate the average of derivatives on the whole cell with the values on vertices of the cell. Next, we show they can be calculated recursively from cells to subcells, or alternatively using the Kusuoka measure.

The following theorem gives the first way to calculate all the bb’s by calculating them recursively.

Theorem 6.2.

Using the previous notations, we have the following relationship from cell to subcells:

b0(w0)=9b0(w)13b0(w)+b1(w)+b2(w)b1(w0)=2b0(w)+2b1(w)b2(w)13b0(w)+b1(w)+b2(w)b2(w0)=2b0(w)b1(w)+2b2(w)13b0(w)+b1(w)+b2(w)b0(w1)=2b0(w)+2b1(w)b2(w)b0(w)+13b1(w)+b2(w)b1(w1)=9b1(w)b0(w)+13b1(w)+b2(w)b2(w1)=b0(w)+2b1(w)+2b2(w)b0(w)+13b1(w)+b2(w)b0(w2)=2b0(w)b1(w)+2b2(w)b0(w)+b1(w)+13b2(w)b1(w2)=b0(w)+2b2(w)+2b1(w)b0(w)+b1(w)+13b2(w)b2(w2)=9b2(w)b0(w)+b1(w)+13b2(w)\begin{array}[]{ccc}b_{0}^{(w0)}=\frac{9b_{0}^{(w)}}{13b_{0}^{(w)}+b_{1}^{(w)}+b_{2}^{(w)}}&b_{1}^{(w0)}=\frac{2b_{0}^{(w)}+2b_{1}^{(w)}-b_{2}^{(w)}}{13b_{0}^{(w)}+b_{1}^{(w)}+b_{2}^{(w)}}&b_{2}^{(w0)}=\frac{2b_{0}^{(w)}-b_{1}^{(w)}+2b_{2}^{(w)}}{13b_{0}^{(w)}+b_{1}^{(w)}+b_{2}^{(w)}}\\ b_{0}^{(w1)}=\frac{2b_{0}^{(w)}+2b_{1}^{(w)}-b_{2}^{(w)}}{b_{0}^{(w)}+13b_{1}^{(w)}+b_{2}^{(w)}}&b_{1}^{(w1)}=\frac{9b_{1}^{(w)}}{b_{0}^{(w)}+13b_{1}^{(w)}+b_{2}^{(w)}}&b_{2}^{(w1)}=\frac{-b_{0}^{(w)}+2b_{1}^{(w)}+2b_{2}^{(w)}}{b_{0}^{(w)}+13b_{1}^{(w)}+b_{2}^{(w)}}\\ b_{0}^{(w2)}=\frac{2b_{0}^{(w)}-b_{1}^{(w)}+2b_{2}^{(w)}}{b_{0}^{(w)}+b_{1}^{(w)}+13b_{2}^{(w)}}&b_{1}^{(w2)}=\frac{-b_{0}^{(w)}+2b_{2}^{(w)}+2b_{1}^{(w)}}{b_{0}^{(w)}+b_{1}^{(w)}+13b_{2}^{(w)}}&b_{2}^{(w2)}=\frac{9b_{2}^{(w)}}{b_{0}^{(w)}+b_{1}^{(w)}+13b_{2}^{(w)}}\end{array}
Proof.

We know that, on FwSGF_{w}SG,

bj(w)=16+12cj(w)c0(w)+c1(w)+c2(w)b_{j}^{(w)}=\frac{1}{6}+\frac{1}{2}\frac{c_{j}^{(w)}}{c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)}}

where (c0(w)c1(w)c2(w))=(111)Mw\left(\begin{array}[]{ccc}c_{0}^{(w)}&c_{1}^{(w)}&c_{2}^{(w)}\end{array}\right)=\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}. Put aj=2(bj(w)16)a_{j}=2(b_{j}^{(w)}-\frac{1}{6}). We see that aj(c0(w)+c1(w)+c2(w))=cj(w)a_{j}(c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)})=c_{j}^{(w)}. Since

(c0(w)c1(w)c2(w))M0=115(9c0(w)+2c1(w)+2c2(w)2c1(w)c2(w)2c2(w)c1(w)),\left(\begin{array}[]{ccc}c_{0}^{(w)}&c_{1}^{(w)}&c_{2}^{(w)}\end{array}\right)M_{0}=\frac{1}{15}\left(\begin{array}[]{ccc}9c_{0}^{(w)}+2c_{1}^{(w)}+2c_{2}^{(w)}&2c_{1}^{(w)}-c_{2}^{(w)}&2c_{2}^{(w)}-c_{1}^{(w)}\end{array}\right),

we have

b0(w0)\displaystyle b_{0}^{(w0)} =16+129c0(w)+2c1(w)+2c2(w)9c0(w)+3c1(w)+3c2(w)\displaystyle=\frac{1}{6}+\frac{1}{2}\frac{9c_{0}^{(w)}+2c_{1}^{(w)}+2c_{2}^{(w)}}{9c_{0}^{(w)}+3c_{1}^{(w)}+3c_{2}^{(w)}}
=16+129a0+2a1+2a29a0+3a1+3a2.\displaystyle=\frac{1}{6}+\frac{1}{2}\frac{9a_{0}+2a_{1}+2a_{2}}{9a_{0}+3a_{1}+3a_{2}}.

Similarly, we have

b1(w0)=16+122a1a29a0+3a1+3a2b_{1}^{(w0)}=\frac{1}{6}+\frac{1}{2}\frac{2a_{1}-a_{2}}{9a_{0}+3a_{1}+3a_{2}}

and

b2(w0)=16+122a2a19a0+3a1+3a2.b_{2}^{(w0)}=\frac{1}{6}+\frac{1}{2}\frac{2a_{2}-a_{1}}{9a_{0}+3a_{1}+3a_{2}}.

With some computation from the definition of the aa values and using the fact that jbj(w)=1\sum_{j}b_{j}^{(w)}=1, we can relate bj(w)b_{j}^{(w)} and bj(w0)b_{j}^{(w0)} explicitly as in the statement of the theorem.

Observe that multiplying different MjM_{j}’s only differs by some permutation of subscript and entries, we have the rest of the results. ∎

The next theorem presents the second way to calculate bj(w)b_{j}^{(w)}.

Theorem 6.3.

The distance of bj(w)b_{j}^{(w)} from 1/31/3 is proportional to how “skewed” the Kusuoka measure is on the cell FwFjSGF_{w}F_{j}SG relative to FwSGF_{w}SG. Specifically,

15(bj(w)13)=14(ν(FwFjSG)ν(FwSG)13).\frac{1}{5}\left(b_{j}^{(w)}-\frac{1}{3}\right)=\frac{1}{4}\left(\frac{\nu(F_{w}F_{j}SG)}{\nu(F_{w}SG)}-\frac{1}{3}\right).
Proof.

By (6.1) we have

bj(w)=16+(111)zj(111)z.b_{j}^{(w)}=\frac{1}{6}+\frac{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z_{j}}{\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z}.

Note that (111)z=ν(FwSG)\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z=\nu(F_{w}SG). For simplicity take j=0j=0. Then

(111)z0=54(111)Mw(4/500)=54((111)Mw(6/52/52/5)(111)Mw(2/52/52/5))=54(ν(FwF0SG)15ν(FwSG))\begin{split}\left(\begin{array}[]{ccc}1&1&1\end{array}\right)z_{0}=&\frac{5}{4}\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}\left(\begin{array}[]{c}4/5\\ 0\\ 0\end{array}\right)\\ =&\frac{5}{4}\left(\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}\left(\begin{array}[]{c}6/5\\ 2/5\\ 2/5\end{array}\right)-\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}\left(\begin{array}[]{c}2/5\\ 2/5\\ 2/5\end{array}\right)\right)\\ =&\frac{5}{4}\left(\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG)\right)\end{split}

where the last equality comes from noting that 6/5=ν0(F0SG)6/5=\nu_{0}(F_{0}SG) and 2/5=ν1(F0SG)=ν2(F0SG)2/5=\nu_{1}(F_{0}SG)=\nu_{2}(F_{0}SG). More generally we have

bj(w)=54(215+ν(FwFjSG)15ν(FwSG)ν(FwSG))=54(115+ν(FwFjSG)ν(FwSG))b_{j}^{(w)}=\frac{5}{4}\left(\frac{2}{15}+\frac{\nu(F_{w}F_{j}SG)-\frac{1}{5}\nu(F_{w}SG)}{\nu(F_{w}SG)}\right)=\frac{5}{4}\left(\frac{-1}{15}+\frac{\nu(F_{w}F_{j}SG)}{\nu(F_{w}SG)}\right)

which gives

15(bj(w)13)=14(ν(FwFjSG)ν(FwSG)13).\frac{1}{5}\left(b_{j}^{(w)}-\frac{1}{3}\right)=\frac{1}{4}\left(\frac{\nu(F_{w}F_{j}SG)}{\nu(F_{w}SG)}-\frac{1}{3}\right).

As a consequence of the theorem, the bounds for (bj(w)13)(b_{j}^{(w)}-\frac{1}{3}) also give bounds for how skewed the Kusuoka measure is in a cell. We now provide some bounds for the coefficients bj(w)b_{j}^{(w)}.

Theorem 6.4.

a) The infimum of {bj(w)}\{b_{j}^{(w)}\} over all words ww is 0.
b) The supremum of {bj(w)}\{b_{j}^{(w)}\} over all words ww is 23\frac{2}{3}.

Proof.

We first prove a). We know that

115(1411)Ew(222)=115limm(53)mν(FwF0m)0\frac{1}{15}\left(\begin{array}[]{ccc}14&-1&-1\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)=\frac{1}{15}\lim_{m}\left(\frac{5}{3}\right)^{m}\nu(F_{w}F_{0}^{m})\\ \geq 0

and the inequality is sharp over all ww because every energy measure is non-atomic. So

115(1411)Ew(222)=(111)(E0115I)Ew(222)=(111)E0Ew(222)115(111)Ew(222)0\begin{split}&\frac{1}{15}\left(\begin{array}[]{ccc}14&-1&-1\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ =&\left(\begin{array}[]{ccc}1&1&1\end{array}\right)(E_{0}-\frac{1}{15}I)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ =&\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)-\frac{1}{15}\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ \geq&0\end{split}

and hence

(111)E0Ew(222)15(111)Ew(222)215(111)Ew(222)\begin{split}&\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{0}E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)-\frac{1}{5}\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ \geq&-\frac{2}{15}\left(\begin{array}[]{ccc}1&1&1\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\end{split}

which implies

ν(FwF0SG)15ν(FwSG)ν(FwSG)215\frac{\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG)}{\nu(F_{w}SG)}\geq-\frac{2}{15}

and so finally we get that

b0(w)=54(215+ν(FwF0SG)15ν(FwSG)ν(FwSG))0.b_{0}^{(w)}=\frac{5}{4}\left(\frac{2}{15}+\frac{\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG)}{\nu(F_{w}SG)}\right)\geq 0.

Since the inequality is sharp, zero is the infimum, completing a).
To prove b), we see that

(233)Ew(222)=15(1114)E1Ew(222)=15limm(53)mν(Fw,1F2m)0\begin{split}\left(\begin{array}[]{ccc}-2&3&3\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)&=\frac{1}{5}\left(\begin{array}[]{ccc}-1&-1&14\end{array}\right)E_{1}E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ &=\frac{1}{5}\lim_{m}\left(\frac{5}{3}\right)^{m}\nu(F_{w,1}F_{2}^{m})\geq 0\end{split}

and this inequality is sharp. So

5ν(FwF0SG)3ν(FwSG)=(111)(5E03I)Ew(222)=(233)Ew(222)0.\begin{split}5\nu(F_{w}F_{0}SG)-3\nu(F_{w}SG)&=\left(\begin{array}[]{ccc}1&1&1\end{array}\right)(5E_{0}-3I)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\\ &=\left(\begin{array}[]{ccc}2-3-3\end{array}\right)E_{w}\left(\begin{array}[]{c}2\\ 2\\ 2\end{array}\right)\leq 0.\end{split}

Whence

52(ν(FwF0SG)15ν(FwSG))ν(FwSG)\frac{5}{2}(\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG))\leq\nu(F_{w}SG)

which implies

52ν(FwF0SG)15ν(FwSG)ν(FwSG)1\frac{5}{2}\frac{\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG)}{\nu(F_{w}SG)}\leq 1

and so finally we have

b0(w)=16+54ν(FwF0SG)15ν(FwSG)ν(FwSG)23.b_{0}^{(w)}=\frac{1}{6}+\frac{5}{4}\frac{\nu(F_{w}F_{0}SG)-\frac{1}{5}\nu(F_{w}SG)}{\nu(F_{w}SG)}\leq\frac{2}{3}.

Since the inequality is sharp, this is the supremum. We have b). ∎

The previous results concern the bounds for a single value bj(w)b_{j}^{(w)}.It is natural to consider (b0(w),b1(w),b2(w))(b_{0}^{(w)},b_{1}^{(w)},b_{2}^{(w)}) as a vector in 3\mathbb{R}^{3}, lying in the plane {(x,y,z):x+y+z=1}\{(x,y,z):x+y+z=1\}. Surprisingly, these vectors all lie in the disk centered at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) of radius 16\frac{1}{\sqrt{6}}.

Theorem 6.5.

For all words ww,

(bj(w)13)2<16\sum\left(b_{j}^{(w)}-\frac{1}{3}\right)^{2}<\frac{1}{6}

and this inequality is sharp.

Proof.

Let (111)Mw=(c0(w)c1(w)c2(w)).\left(\begin{array}[]{ccc}1&1&1\end{array}\right)M_{w}=\left(\begin{array}[]{ccc}c_{0}^{(w)}&c_{1}^{(w)}&c_{2}^{(w)}\end{array}\right). Then we have from (6.1) that

bj(w)13=12cj(w)c0(w)+c1(w)+c2(w)16b_{j}^{(w)}-\frac{1}{3}=\frac{1}{2}\frac{c_{j}^{(w)}}{c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)}}-\frac{1}{6}

and so

(bj(w)13)2=(12cj(w)c0(w)+c1(w)+c2(w)16)2=112+14(cj(w))2(c0(w)+c1(w)+c2(w))216cj(w)c0(w)+c1(w)+c2(w)=14(cj(w))2(c0(w)+c1(w)+c2(w))2112.\begin{split}\sum\left(b_{j}^{(w)}-\frac{1}{3}\right)^{2}&=\sum\left(\frac{1}{2}\frac{c_{j}^{(w)}}{c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)}}-\frac{1}{6}\right)^{2}\\ &=\frac{1}{12}+\frac{1}{4}\sum\frac{(c_{j}^{(w)})^{2}}{(c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)})^{2}}-\frac{1}{6}\sum\frac{c_{j}^{(w)}}{c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)}}\\ &=\frac{1}{4}\sum\frac{(c_{j}^{(w)})^{2}}{(c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)})^{2}}-\frac{1}{12}.\end{split}

So we need to show that

14(cj(w))2(c0(w)+c1(w)+c2(w))214\frac{1}{4}\sum\frac{(c_{j}^{(w)})^{2}}{(c_{0}^{(w)}+c_{1}^{(w)}+c_{2}^{(w)})^{2}}\leq\frac{1}{4}

for which it suffices to show that

c0(w)c1(w)+c1(w)c2(w)+c0(w)c2(w)>0c_{0}^{(w)}c_{1}^{(w)}+c_{1}^{(w)}c_{2}^{(w)}+c_{0}^{(w)}c_{2}^{(w)}>0

which we will prove by induction on the length of ww. When |w|=1|w|=1, c0(w)c1(w)+c1(w)c2(w)+c0(w)c2(w)=1315115+1315115+115115>0c_{0}^{(w)}c_{1}^{(w)}+c_{1}^{(w)}c_{2}^{(w)}+c_{0}^{(w)}c_{2}^{(w)}=\frac{13}{15}\frac{1}{15}+\frac{13}{15}\frac{1}{15}+\frac{1}{15}\frac{1}{15}>0.

Now suppose it is true for ww with |w|=m|w|=m. We see that

(c0(w0)c1(w0)c2(w0))=(c0(w)c1(w)c2(w))M0=115(9c0(w)+2c1(w)+2c2(w)2c1(w)c2(w)2c2(w)c1(w))\begin{split}\left(\begin{array}[]{ccc}c_{0}^{(w0)}&c_{1}^{(w0)}&c_{2}^{(w0)}\end{array}\right)&=\left(\begin{array}[]{ccc}c_{0}^{(w)}&c_{1}^{(w)}&c_{2}^{(w)}\end{array}\right)M_{0}\\ &=\frac{1}{15}\left(\begin{array}[]{ccc}9c_{0}^{(w)}+2c_{1}^{(w)}+2c_{2}^{(w)}&2c_{1}^{(w)}-c_{2}^{(w)}&2c_{2}^{(w)}-c_{1}^{(w)}\end{array}\right)\end{split}

and so

152(c0(w0)c1(w0)+c1(w0)c2(w0)+c0(w0)c2(w0))=9c0(w)(c1(w)+c2(w))+2(c1(w))2+2(c2(w))2+4c1(w)c2(w)+5c1(w)c2(w)2(c1(w))22(c2(w))2=9(c0(w)c1(w)+c1(w)c2(w)+c0(w)c2(w))>0.\begin{split}15^{2}(c_{0}^{(w0)}c_{1}^{(w0)}+c_{1}^{(w0)}c_{2}^{(w0)}+c_{0}^{(w0)}c_{2}^{(w0)})&=9c_{0}^{(w)}(c_{1}^{(w)}+c_{2}^{(w)})+2(c_{1}^{(w)})^{2}+2(c_{2}^{(w)})^{2}\\ &+4c_{1}^{(w)}c_{2}^{(w)}+5c_{1}^{(w)}c_{2}^{(w)}-2(c_{1}^{(w)})^{2}-2(c_{2}^{(w)})^{2}\\ &=9(c_{0}^{(w)}c_{1}^{(w)}+c_{1}^{(w)}c_{2}^{(w)}+c_{0}^{(w)}c_{2}^{(w)})\\ &>0.\end{split}

As for the sharpness of the inequality, consider a word ww of length mm consisting of only the letter 0. We have that

b0(w)=2323(32m+1),b1(w)=b2(w)=13(32m+1)+16b_{0}^{(w)}=\frac{2}{3}-\frac{2}{3(3^{2m}+1)},b_{1}^{(w)}=b_{2}^{(w)}=\frac{1}{3(3^{2m}+1)}+\frac{1}{6}

and so as mm\to\infty we have limm(bj(w)13)2=16\lim_{m}\sum(b_{j}^{(w)}-\frac{1}{3})^{2}=\frac{1}{6}. In fact this is true for any sequence of words approaching an infinite word. ∎

Remark 6.6.

In Theorem 6.2, we saw that from FwSGF_{w}SG to Fw0SGF_{w0}SG, the properties of bj(w)b_{j}^{(w)} were translated to the properties of the rational transformation A0A_{0}

(a0a1a2)(9a0+2a1+2a29a0+3a1+3a22a1a29a0+3a1+3a22a2a19a0+3a1+3a2)\left(\begin{array}[]{c}a_{0}\\ a_{1}\\ a_{2}\end{array}\right)\mapsto\left(\begin{array}[]{c}\frac{9a_{0}+2a_{1}+2a_{2}}{9a_{0}+3a_{1}+3a_{2}}\\ \frac{2a_{1}-a_{2}}{9a_{0}+3a_{1}+3a_{2}}\\ \frac{2a_{2}-a_{1}}{9a_{0}+3a_{1}+3a_{2}}\end{array}\right)

and we have A1A_{1} and A2A_{2} for the maps from FwSGF_{w}SG to Fw1SGF_{w1}SG and Fw2SGF_{w2}SG. Thus it is worthwhile to look at the properties of aja_{j}. First, it is easy to see that aj=1\sum{a_{j}}=1. Also, the sum of squares of aja_{j} is strictly less than 11. To see this, using aj=2(bj(w)16)a_{j}=2\left(b_{j}^{(w)}-\frac{1}{6}\right), we have

(12aj16)2<16\sum{\left(\frac{1}{2}a_{j}-\frac{1}{6}\right)^{2}}<\frac{1}{6}

so

(aj13)2<23.\sum{\left(a_{j}-\frac{1}{3}\right)^{2}}<\frac{2}{3}.

Since aj=1\sum{a_{j}}=1, expanding the above gives

aj2<1.\sum{a_{j}^{2}}<1.

Finally, in terms of the ratios of Kusuoka measure, we have

ν(FwjSG)ν(FwSG)=25(aj+12).\frac{\nu(F_{wj}SG)}{\nu(F_{w}SG)}=\frac{2}{5}\left(a_{j}+\frac{1}{2}\right).

7. Energy Distribution

Let b(w)=(b0(w),b1(w),b2(w))b^{(w)}=(b_{0}^{(w)},b_{1}^{(w)},b_{2}^{(w)}). We have seen in Theorem 6.3 how the vector b(w)b^{(w)} relates to the splitting of Kusuoka measure in the cell FwSGF_{w}SG when it subdivides into three subcells at the next level. Theorem 6.2 gives a recursive algorithm to compute the b(w)b^{(w)} vectors. In this section we investigate the distribution of the collection of all b(w)b^{(w)} vectors as ww varies over all words of fixed length mm. We may write bw0=B0(bw)b^{w0}=B_{0}(b^{w}), b(w1)=B1(b(w))b^{(w1)}=B_{1}(b^{(w)}) and b(w2)=B2(b(w))b^{(w2)}=B_{2}(b^{(w)}) where the BjB_{j} are the rational maps given in Theorem 6.2. We may regard {B0,B1,B2}\{B_{0},B_{1},B_{2}\} as an iterated function system acting on the open disk described in Theorem 6.5.

Refer to caption
Figure 7.2. The map B0B_{0}
Refer to caption
Figure 7.3. Points of {b(w):|w|=14}\{b^{(w)}:|w|=14\}

In Figure 7.2 we illustrate the geometric structure of B0B_{0} by showing the image under B0B_{0} of a uniform polar coordinate grid on the disk. B1B_{1} and B2B_{2} are similar, but are rotated through angles 2π3\frac{2\pi}{3} and 2π3\frac{-2\pi}{3}.

To obtain {b(w):|w|=m}\{b^{(w)}:|w|=m\} we apply all 3m3^{m} iterates of {B0,B1,B2}\{B_{0},B_{1},B_{2}\} to the initial vector (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}). In Figure 7.3 we show the result for m=12m=12. To better understand this distribution of points in the disk we examine its angular and radial projections.

Introduce polar coordinates (r,θ)(r,\theta) for the disk with r<16r<\frac{1}{\sqrt{6}}(the origin corresponds to the vector b=(13,13,13)b=(\frac{1}{3},\frac{1}{3},\frac{1}{3})). The angular distribution at level mm is

Pm(A)=3m#{points with θA}.P_{m}(A)=3^{-m}\#\{\text{points with }\theta\in A\}.

By symmetry it suffices to understand the distribution for A[0,2π3]A\subseteq[0,\frac{2\pi}{3}]. In Figure 7.4 we show histograms (100 slices) of PmP_{m} on [0,2π3][0,\frac{2\pi}{3}] for m=11m=11 and m=13m=13; the values are normalized so that the average over all the values is 11.

Conjecture 7.1.

Let PmP_{m} be defined as above. The limit of PmP_{m} as mm\to\infty is an absolutely continuous (B0,B1,B2)(B_{0},B_{1},B_{2})-invariant measure.

Similarly, the radial distribution at level mm is

Qm(A)=3m#{points with rA}.Q_{m}(A)=3^{-m}\#\{\text{points with }r\in A\}.

In Figure 7.5 we show histograms (300 bins) of QmQ_{m} on [0,16][0,\frac{1}{\sqrt{6}}], for m=10m=10 and m=14m=14; the values are the ratios over the total number of bb vectors.

Conjecture 7.2.

Let QmQ_{m} be defined as above. The limit of QmQ_{m} as mm\to\infty is the delta measure on the boundary r=16r=\frac{1}{\sqrt{6}}.

We can say more about the mappings. Recall that we have defined

B0(x,y,z)=(9x13x+y+z,2x+2yz13x+y+z,2xy+2z13x+y+z);B_{0}(x,y,z)=\left(\frac{9x}{13x+y+z},\frac{2x+2y-z}{13x+y+z},\frac{2x-y+2z}{13x+y+z}\right);

in Theorem 6.2. For each word ww, we have B0(b(w))=b(w0)B_{0}(b^{(w)})=b^{(w0)}.

Refer to caption
(a) Level 11
Refer to caption
(b) Level 13
Figure 7.4. Angular Distribution
Refer to caption
(a) Level 10
Refer to caption
(b) Level 14
Figure 7.5. Radial Distribution

We first introduce the “polar coordinates” on the disk centered at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) on the plane {x+y+z=1}\{x+y+z=1\}. We know that (cos(θ+π6),sin(θ+π6),0)(\cos(\theta+\frac{\pi}{6}),\sin(\theta+\frac{\pi}{6}),0) is a parametrization of the unit circle lying on the xyx-y plane. Put

A=(12161312161302613)A=\left(\begin{array}[]{ccc}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[1.0pt] \frac{-1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[1.0pt] 0&\frac{-2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{array}\right)

which maps the unit circle lying on the xyx-y plane to the unit circle lying on the plane {x+y+z=0}\{x+y+z=0\}. It follows that a parametrization of the circle of radius rr centered at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) on the plane {x+y+z=1}\{x+y+z=1\} is

(7.1) (131313)+A(rcos(θ+π6)rsin(θ+π6)0)=(131313)+r(26cosθ12sinθ16cosθ12sinθ16cosθ).\left(\begin{array}[]{c}\frac{1}{3}\\[1.0pt] \frac{1}{3}\\[1.0pt] \frac{1}{3}\end{array}\right)+A\left(\begin{array}[]{c}r\cos(\theta+\frac{\pi}{6})\\ r\sin(\theta+\frac{\pi}{6})\\ 0\end{array}\right)=\left(\begin{array}[]{c}\frac{1}{3}\\[1.0pt] \frac{1}{3}\\[1.0pt] \frac{1}{3}\end{array}\right)+r\left(\begin{array}[]{c}\frac{2}{\sqrt{6}}\cos\theta\\[3.0pt] \frac{1}{\sqrt{2}}\sin\theta-\frac{1}{\sqrt{6}}\cos\theta\\[3.0pt] \frac{-1}{\sqrt{2}}\sin\theta-\frac{1}{\sqrt{6}}\cos\theta\end{array}\right).

With this parametrization, the image of radius rr under B0B_{0} is

(7.2) (131313)+16(106rcosθ+846rcosθ+592rsinθ56rcosθ446rcosθ+592rsinθ56rcosθ446rcosθ+5).\left(\begin{array}[]{c}\frac{1}{3}\\[1.0pt] \frac{1}{3}\\[1.0pt] \frac{1}{3}\end{array}\right)+\frac{1}{6}\left(\begin{array}[]{c}\frac{10\sqrt{6}r\cos\theta+8}{4\sqrt{6}r\cos\theta+5}\\[3.0pt] \frac{9\sqrt{2}r\sin\theta-5\sqrt{6}r\cos\theta-4}{4\sqrt{6}r\cos\theta+5}\\[3.0pt] \frac{-9\sqrt{2}r\sin\theta-5\sqrt{6}r\cos\theta-4}{4\sqrt{6}r\cos\theta+5}\end{array}\right).

Thus if we take r=16r=\frac{1}{\sqrt{6}}, we see that the image of the boundary circle is

(7.3) (131313)+16(10cosθ+84cosθ+533sinθ5cosθ44cosθ+533sinθ5cosθ44cosθ+5).\left(\begin{array}[]{c}\frac{1}{3}\\[1.0pt] \frac{1}{3}\\[1.0pt] \frac{1}{3}\end{array}\right)+\frac{1}{6}\left(\begin{array}[]{c}\frac{10\cos\theta+8}{4\cos\theta+5}\\[3.0pt] \frac{3\sqrt{3}\sin\theta-5\cos\theta-4}{4\cos\theta+5}\\[3.0pt] \frac{-3\sqrt{3}\sin\theta-5\cos\theta-4}{4\cos\theta+5}\end{array}\right).

Each BjB_{j} maps the disk onto itself injectively. We first show that the mappings map the disk into itself. By (7.1) and (7.2), we need to see, by considering B0B_{0}, whether to each pair (r,θ)(r,\theta), there exists a unique solution (γ,α)(\gamma,\alpha) satisfying

(106rcosθ+846rcosθ+592rsinθ56rcosθ446rcosθ+592rsinθ56rcosθ446rcosθ+5)=(2γ6cosα3γ2sinαγ6cosθ3γ2sinαγ6cosθ).\left(\begin{array}[]{c}\frac{10\sqrt{6}r\cos\theta+8}{4\sqrt{6}r\cos\theta+5}\\[3.0pt] \frac{9\sqrt{2}r\sin\theta-5\sqrt{6}r\cos\theta-4}{4\sqrt{6}r\cos\theta+5}\\[3.0pt] \frac{-9\sqrt{2}r\sin\theta-5\sqrt{6}r\cos\theta-4}{4\sqrt{6}r\cos\theta+5}\end{array}\right)=\left(\begin{array}[]{c}2\gamma\sqrt{6}\cos\alpha\\[3.0pt] 3\gamma\sqrt{2}\sin\alpha-\gamma\sqrt{6}\cos\theta\\[3.0pt] -3\gamma\sqrt{2}\sin\alpha-\gamma\sqrt{6}\cos\theta\end{array}\right).

The above reduces to

(7.4) γcosα=1656rcosθ+446rcosθ+5γsinα=3rsinθ46rcosθ+5.\begin{split}&\gamma\cos\alpha=\frac{1}{\sqrt{6}}\frac{5\sqrt{6}r\cos\theta+4}{4\sqrt{6}r\cos\theta+5}\\ &\gamma\sin\alpha=\frac{3r\sin\theta}{4\sqrt{6}r\cos\theta+5}.\end{split}

In particular,

(7.5) γ216=9(r216)(46rcosθ+5)2\gamma^{2}-\frac{1}{6}=\frac{9\left(r^{2}-\frac{1}{6}\right)}{(4\sqrt{6}r\cos\theta+5)^{2}}

so if r16r\leq\frac{1}{\sqrt{6}}, then γ16\gamma\leq\frac{1}{\sqrt{6}} and there is a solution (γ,α)(\gamma,\alpha) satisfying (7.4). But by (7.4)\eqref{eq:PtToPtOnTheDisk}, we see that actually (r,θ)(r,\theta) is also completely determined by (γ,α)(\gamma,\alpha); it follows that the BjB_{j} maps the disk onto itself injectively.

When we restrict the map on the boundary circle, we see from (7.5) that the circle is mapped onto itself and using the notations as before,

(7.6) cosα=5cosθ+44cosθ+5sinα=3sinθ4cosθ+5.\begin{split}&\cos\alpha=\frac{5\cos\theta+4}{4\cos\theta+5}\\ &\sin\alpha=\frac{3\sin\theta}{4\cos\theta+5}.\end{split}

These show that α\alpha corresponds to exactly one θ\theta; the map gjg_{j}, referring to the restriction of BjB_{j} to the boundary circle, given by gj(θ)=αg_{j}(\theta)=\alpha is well-defined. Differentiating both equations in (7.6), we have dg0(θ)dθ=34cosθ+5\frac{dg_{0}(\theta)}{d\theta}=\frac{3}{4\cos\theta+5}. It follows that

g0(θ)=0θ3dt4cost+5=2tan1(13tanθ2)g_{0}(\theta)=\int_{0}^{\theta}\frac{3dt}{4\cos t+5}=2\tan^{-1}\left(\frac{1}{3}\tan\frac{\theta}{2}\right)

and

g01(α)=2tan1(3tanα2).g_{0}^{-1}(\alpha)=2\tan^{-1}\left(3\tan\frac{\alpha}{2}\right).

Since g1g_{1} and g2g_{2} differ from g0g_{0} a rotation, we have

g1(θ)=2tan1(13tan(θ2π3))+2π3g11(α)=2tan1(3tan(α2π3))+2π3g2(θ)=2tan1(13tan(θ2+π3))2π3g21(α)=2tan1(3tan(α2+π3))2π3.\begin{split}g_{1}(\theta)=2\tan^{-1}\left(\frac{1}{3}\tan\left(\frac{\theta}{2}-\frac{\pi}{3}\right)\right)+\frac{2\pi}{3}\\ g_{1}^{-1}(\alpha)=2\tan^{-1}\left(3\tan\left(\frac{\alpha}{2}-\frac{\pi}{3}\right)\right)+\frac{2\pi}{3}\\ g_{2}(\theta)=2\tan^{-1}\left(\frac{1}{3}\tan\left(\frac{\theta}{2}+\frac{\pi}{3}\right)\right)-\frac{2\pi}{3}\\ g_{2}^{-1}(\alpha)=2\tan^{-1}\left(3\tan\left(\frac{\alpha}{2}+\frac{\pi}{3}\right)\right)-\frac{2\pi}{3}.\end{split}

We have the following theorem.

Theorem 7.3.

The mappings BjB_{j} (j=0,1,2)(j=0,1,2) map the disk U¯\overline{U} centered at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) of radius 16\frac{1}{\sqrt{6}} onto itself injectively. Further,

  • (i)

    the mappings map the boundary circle CC centered at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) of radius 16\frac{1}{\sqrt{6}} onto itself; and

  • (ii)

    If Ψ\Psi is the function in (7.1) which associates each (16,θ)(\frac{1}{\sqrt{6}},\theta)\in\mathbb{R} to CC, there are differentiable maps gjg_{j} satisfying Bj(Ψ(16,θ))=Ψ(16,gj(θ))B_{j}(\Psi(\frac{1}{\sqrt{6}},\theta))=\Psi(\frac{1}{\sqrt{6}},g_{j}(\theta)) given by

    g0(θ)=2tan1(13tanθ2)g1(θ)=2tan1(13tan(θ2π3))+2π3g2(θ)=2tan1(13tan(θ2+π3))2π3\begin{split}g_{0}(\theta)&=2\tan^{-1}\left(\frac{1}{3}\tan\frac{\theta}{2}\right)\\ g_{1}(\theta)&=2\tan^{-1}\left(\frac{1}{3}\tan\left(\frac{\theta}{2}-\frac{\pi}{3}\right)\right)+\frac{2\pi}{3}\\ g_{2}(\theta)&=2\tan^{-1}\left(\frac{1}{3}\tan\left(\frac{\theta}{2}+\frac{\pi}{3}\right)\right)-\frac{2\pi}{3}\end{split}
  • (iii)

    Each BjB_{j} has exactly two fixed points on the circle. The fixed points are, in “polar coordinates”, Ψ(16,0)\Psi(\frac{1}{\sqrt{6}},0), Ψ(16,π)\Psi(\frac{1}{\sqrt{6}},\pi) for B0B_{0}, Ψ(16,2π3)\Psi(\frac{1}{\sqrt{6}},\frac{2\pi}{3}), Ψ(16,π3)\Psi(\frac{1}{\sqrt{6}},\frac{-\pi}{3}) for B1B_{1} and Ψ(16,2π3)\Psi(\frac{1}{\sqrt{6}},\frac{-2\pi}{3}), Ψ(16,π3)\Psi(\frac{1}{\sqrt{6}},\frac{\pi}{3}) for B2B_{2}.

Proof.

We have already proved (i) and (ii). For (iii), we only need to prove for B0B_{0} and the other two follow. By (ii), Ψ(16,0)\Psi(\frac{1}{\sqrt{6}},0), Ψ(16,π)\Psi(\frac{1}{\sqrt{6}},\pi) are the only fixed points on the boundary circle. So it suffices to show that there is no fixed point in the open disk. Suppose, on the contrary, we have such a fixed point, say Ψ(r,θ)\Psi(r,\theta). Then we must have

16106rcosθ+846rcosθ+5=2r6cosθ\frac{1}{6}\frac{10\sqrt{6}r\cos\theta+8}{4\sqrt{6}r\cos\theta+5}=\frac{2r}{\sqrt{6}}\cos\theta

which shows, since r<16r<\frac{1}{\sqrt{6}},

cos2θ=16r2>1\cos^{2}\theta=\frac{1}{6r^{2}}>1

but this contradicts the fact that |cosθ|1|\cos\theta|\leq 1. ∎

By (ii), We can regard BjB_{j} and gjg_{j} on the boundary circle as the same map under the parametrization (7.1).

If we directly view the maps BjB_{j} as maps on the unit disk, we have an easy description on how they act on the unit disk.

Theorem 7.4.

Using the notation (r,θ)(r,\theta) and (γ,α)(\gamma,\alpha) as before. Let (x,y)=(6rcosθ,6rsinθ)(x,y)=\\ (\sqrt{6}r\cos\theta,\sqrt{6}r\sin\theta) and (x,y)=(6γcosθ,6γsinθ)(x^{\prime},y^{\prime})=(\sqrt{6}\gamma\cos\theta,\sqrt{6}\gamma\sin\theta) be on the unit disk. Then B0B_{0} maps every vertical line to a vertical line and every point (x,y)(x,y) with x>12x>-\frac{1}{2} closer the boundary circle. In other words, if we denote TT to be the triangle with vertices (1,0)(1,0), (12,32)(-\frac{1}{2},\frac{\sqrt{3}}{2}), (12,32)(-\frac{1}{2},\frac{-\sqrt{3}}{2}), then every point in TT is mapped closer to the boundary by all of the three mappings BjB_{j} and every point outside TT is mapped closer to the boundary by two of the mappings BjB_{j}.

Proof.

The map B0B_{0} on the unit disk is of the form

B0(x,y)=(x,y)=(5x+44x+5,3y4x+5)B_{0}(x,y)=(x^{\prime},y^{\prime})=\left(\frac{5x+4}{4x+5},\frac{3y}{4x+5}\right)

Fix a constant c[1,1]c\in[-1,1]. If x=cx=c, then x=5c+44c+5x^{\prime}=\frac{5c+4}{4c+5}; so B0B_{0}^{\prime} maps every vertical line to a vertical line. Also, by (7.5), we have

γ216=9(4x+5)2(r216);\gamma^{2}-\frac{1}{6}=\frac{9}{(4x+5)^{2}}\left(r^{2}-\frac{1}{6}\right);

thus if x>12x>-\frac{1}{2}, 9(4x+5)2<1\frac{9}{(4x+5)^{2}}<1 and hence

γ216<9(4x+5)2(r216).\gamma^{2}-\frac{1}{6}<\frac{9}{(4x+5)^{2}}\left(r^{2}-\frac{1}{6}\right).

The last assertion comes from the symmetry of BjB_{j}, since all of these mappings differ by a rotation of 2π3\frac{2\pi}{3} and 2π3\frac{-2\pi}{3} only. ∎

By Theorem 7.3, we can say more about the measure mentioned in Conjecture 7.1. We call the measure dλ(t)=f(t)dtd\lambda(t)=f(t)dt for some ff, λ[a,b]=λ(Bj[a,b])\lambda[a,b]=\lambda(B_{j}[a,b]) for all j=0,1,2j=0,1,2. Recall that it is (B0,B1,B2)(B_{0},B_{1},B_{2})-invariant, hence

abf=gi1[a,b]f(t)𝑑t=abf(gi1(t))(gi1)(t)𝑑t=abf(gi1(t))gi(gi1(t))𝑑t=13j=02abf(gj1(t))gj(gj1(t))𝑑t.\begin{split}\int_{a}^{b}f&=\int_{g_{i}^{-1}[a,b]}f(t)dt\\ &=\int_{a}^{b}f(g_{i}^{-1}(t))(g_{i}^{-1})^{\prime}(t)dt\\ &=\int_{a}^{b}\frac{f(g_{i}^{-1}(t))}{g_{i}^{\prime}(g_{i}^{-1}(t))}dt\\ &=\frac{1}{3}\sum_{j=0}^{2}\int_{a}^{b}\frac{f(g_{j}^{-1}(t))}{g_{j}^{\prime}(g_{j}^{-1}(t))}dt.\end{split}

It follows that the function ff can be characterized by one satisfying the relation

f(t)=13j=02f(gj1(t))gj(gj1(t)).f(t)=\frac{1}{3}\sum_{j=0}^{2}\frac{f(g_{j}^{-1}(t))}{g_{j}^{\prime}(g_{j}^{-1}(t))}.

If we consider gjg_{j} separately, we have

1g0(g01(t))=43cos(2tan1(3tant2))+53=354cost;1g1(g11(t))=354cos(t2π3);1g2(g21(t))=354cos(t+2π3).\begin{split}\frac{1}{g_{0}^{\prime}(g_{0}^{-1}(t))}&=\frac{4}{3}\cos\left(2\tan^{-1}\left(3\tan\frac{t}{2}\right)\right)+\frac{5}{3}=\frac{3}{5-4\cos t};\\ \frac{1}{g_{1}^{\prime}(g_{1}^{-1}(t))}&=\frac{3}{5-4\cos\left(t-\frac{2\pi}{3}\right)};\\ \frac{1}{g_{2}^{\prime}(g_{2}^{-1}(t))}&=\frac{3}{5-4\cos\left(t+\frac{2\pi}{3}\right)}.\end{split}

Finally, we give further experimental evidence for the conjectures. Conjecture 7.2 states that the radial distribution is the delta measure at r=16r=\frac{1}{\sqrt{6}}; the angular distribution on the whole disk should be the angular distribution on the boundary circle only. Pick three points (0,1)(0,1), (1,0)(-1,0), (12,12)(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) from the three arcs mentioned in Theorem 7.4 respectively and iterate them 1414 times by all of the three maps. We cut the whole circle into 800 bins and plot the histogram for the number of points (normalized so that the average over all the values is 11) in each bin for bins covering one sixth of the circle in Figure 7.6. This should be compared to Figure 7.4 which gives the level 13 histogram.

Refer to caption
Figure 7.6. Angular Distribution on Boundary

8. Acknowledgement

We are grateful to the referee for many useful suggestions.

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