Finitely Supported Measures on $S{L}_{2}(\mathbb{R})$ Which are Absolutely Continuous at Infinity

Abstract

We construct finitely supported symmetric probability measures on $S{L}_{2}(\mathbb{R})$ for which the Furstenberg measure on ${\mathbb{P}}_{1}(\mathbb{R})$ has a smooth density.

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1 Introduction

In this note, we give explicit examples of finitely supported symmetric probability measures ν on $S{L}_{2}(\mathbb{R})$ for which the corresponding Furstenberg measure μ on ${\mathbb{P}}_{1}(\mathbb{R})$ is absolutely continuous wrt to Haar measure dθ, and moreover $\frac{d\mu } {d\theta }$ is of class C ^r, with r any given positive integer. Probabilistic constructions of finitely supported (non-symmetric measures ν on $S{L}_{2}(\mathbb{R})$ with absolutely continuous Furstenberg measure appear in the paper [1], setting (in the negative) a conjecture from [4]. The construction in [1] may be viewed as a non-commutative analogue of the theory of random Bernoulli convolutions and uses methods from [5, 6].

It is not clear if this technique may produce Furstenberg measures with say C ¹-density. Our method also addresses the issue of obtaining a symmetric ν (raised in [4]), which seems problematic with the [1] technique.

Our starting point is a construction from [2] of certain Hecke operators on $S{L}_{2}(\mathbb{R})$ whose projective action exhibits a spectral gap. The mathematics underlying [2] is closely related to the paper [3] and makes essential use of results and techniques from arithmetic combinatorics. In particular, it should be pointed out that the spectral gap is not achieved by exploiting hyperbolicity, at least not in the usual way. Our measure ν has in fact a Lyapounov exponent that can be made arbitrary small, while the spectral gap (in an appropriate restricted sense) remains uniformly controlled (the size of supp ν becomes larger of course).

We believe that similar constructions are possible also in the $S{L}_{d}(\mathbb{R})$-setting, for d > 2 (cf. [4]). In fact, such Hecke operators can be produced using the construction from Lemmas 1 and 2 below in $S{L}_{2}(\mathbb{R})$ and considering a suitable family of $S{L}_{2}(\mathbb{R})$-embeddings in SL _d . We do not present the details here.

2 Preliminaries

We recall Lemmas 2.1 and 2.2 from [2].

Lemma 1.

Given ε > 0, there is $Q \in {\mathbb{Z}}_{+}$ and $\mathcal{G}\subset S{L}_{2}(\mathbb{R}) \cap \left ( \frac{1} {Q}Ma{t}_{2}(\mathbb{Z})\right )$ with the following properties

$$\frac{1} {\epsilon } < Q <{ \left (\frac{1} {\epsilon }\right )}^{{c}_{1} }$$

(1)

$$\vert \mathcal{G}\vert > {Q}^{{c}_{2} }$$

(2)

$$\text{ The elements of }\mathcal{G}\text{ are free generators of a free group}$$

(3)

$$\Vert g - 1\Vert < \epsilon \text{ for }g \in \mathcal{G}$$

(4)

Here c ₁, c ₂ are constants independent of ε.

Define the probability measure ν on $S{L}_{2}(\mathbb{R})$ as

$$\nu = \frac{1} {2\vert \mathcal{G}\vert }{\sum \nolimits }_{g\in \mathcal{G}}({\delta }_{g} + {\delta }_{{g}^{-1}}).$$

(5)

Denote also P _δ, δ > 0, an approximate identity on $S{L}_{2}(\mathbb{R})$. For instance, one may take ${P}_{\delta } = \frac{{1}_{{B}_{\delta }(1)}} {\vert {B}_{\delta }(1)\vert }$ where B _δ(1) is the ball of radius δ around 1 in $S{L}_{2}(\mathbb{R})$.

Lemma 2.

Fix τ > 0. Then we have

$$\Vert {\nu }^{(\mathcal{l})} {_\ast} {P{}_{ \delta }\Vert }_{\infty } < {\delta }^{-\tau }$$

(6)

provided

$$\mathcal{l} > {c}_{3}(\tau )\frac{\log 1/\delta } {\log 1/\epsilon }$$

(7)

and assuming δ small enough (depending on Q and τ).

3 Furstenberg Measure

Denote for $g \in S{L}_{2}(\mathbb{R})$ by τ _g the action on ${P}_{1}(\mathbb{R})$ that we identify with the circle $\mathbb{R}/\mathbb{Z} = \mathbb{T}$. Thus if $g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right ),ad-bc = 1$, then

$${e}^{i{\tau }_{g}(\theta )} = \frac{(a\cos \theta + b\sin \theta ) + i(c\cos \theta + d\sin \theta )} {{[{(a\cos \theta + b\sin \theta )}^{2} + {(c\cos \theta + d\sin \theta )}^{2}]}^{\frac{1} {2} }}.$$

(8)

Assume μ on ${P}_{1}(\mathbb{R})$ is ν-stationary, i.e.

$$\mu = \sum \nolimits \nu (g){g}_{{_\ast}}[\mu ].$$

(9)

4 A Restricted Spectral Gap

Take $\mathcal{G}$ as in Lemma 1 and $\nu = \frac{1} {2r}{ \sum \nolimits }_{g\in \mathcal{G}}({\delta }_{g} + {\delta }_{g-1})$ with $r = \vert \mathcal{G}\vert$.

Lemma 3.

There is some constant K > 0 (depending on ν), such that if $f \in {L}^{2}(\mathbb{T})$ satisfies

$$\Vert {f\Vert }_{2} \leq 1\text{ and }\hat{f}(n) = 0\text{ for }\ \vert n\vert < K$$

(10)

then

$${ \left \Vert \int \nolimits \nolimits (f \circ {\tau }_{g})d\nu \right \Vert }_{2} < \frac{1} {2}.$$

(11)

Proof.

Define ${\rho }_{g}f = {({\tau }_{g}^{\prime})}^{1/2}(f \circ {\tau }_{g})$, hence ρ is the projective representation. Since $\Vert 1 - g\Vert < \epsilon$, | τ _g ′ − 1 | ≲ ε and (11) will follow from

$${ \left \Vert \int \nolimits \nolimits ({\rho }_{g}f)\nu (dg)\right \Vert }_{2} < \frac{1} {3}.$$

(12)

Assume (12) fails. By almost orthogonality, there is $f \in {L}^{2}(\mathbb{T})$ such that

$$\text{ supp\,}\hat{f} \subset [{2}^{k},{2}^{k+1}]$$

(13)

$$\Vert {f\Vert }_{2} = 1$$

(14)

$${ \left \Vert \int \nolimits \nolimits ({\rho }_{g}f)\nu (dg)\right \Vert }_{2} > c\text{ (for some $c > 0$)}.$$

(15)

Let ℓ < k to be specified. From (15), since ν is symmetric,

$${ \left \Vert \int \nolimits \nolimits ({\rho }_{g}f){\nu }^{(\mathcal{l})}(dg)\right \Vert }_{ 2} > {c}^{\mathcal{l}}$$

(16)

and hence

$$\int \nolimits \nolimits \vert \langle {\rho }_{g}f,f\rangle \vert {\nu }^{(2\mathcal{l})}(dg) =\iint \vert \langle {\rho }_{ g}f,{\rho }_{h}f\rangle \vert {\nu }^{(\mathcal{l})}(dg){\nu }^{(\mathcal{l})}(dh) > {c}^{2\mathcal{l}}.$$

(17)

Take $\delta = 1{0}^{-k}$. Recalling (13), straightforward approximation permits us to replace in (16) the discrete measure ν^(ℓ) by ν^(ℓ) ∗ P _δ, where P _δ(δ > 0) denotes the approximate identity on $S{L}_{2}(\mathbb{R})$. Hence (17) becomes

$${\int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}\vert \langle {\rho }_{g}f,f\rangle \vert ({\nu }^{2\mathcal{l}} {_\ast} {P}_{ \delta })(g)dg + {2}^{-k} > {c}^{2\mathcal{l}}.$$

(18)

Fix a small constant τ > 0 and apply Lemma 2. This gives

$$\mathcal{l} \sim C(\tau )\frac{\log \frac{1} {\delta }} {\log \frac{1} {\epsilon }}$$

(19)

such that

$$\Vert {\nu }^{(\mathcal{l})} {_\ast} {P{}_{ \delta }\Vert }_{\infty } < {\delta }^{-\tau }.$$

(20)

Note that supp ν^(ℓ) is contained in a ball of radius at most (1 + ε)^ℓ, by (4).

Introduce a smooth function 0 ≤ ω ≤ 1 on $\mathbb{R},\omega = 1$ on $[-{(1 + \epsilon )}^{4\mathcal{l}},{(1 + \epsilon )}^{4\mathcal{l}}]$ and ω = 0 outside $[-2{(1 + \epsilon )}^{4\mathcal{l}},2{(1 + \epsilon )}^{4\mathcal{l}}]$.

Let ${\omega }_{1}(g) = \omega ({a}^{2} + {b}^{2} + {c}^{2} + {d}^{2})$ for $g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right )$.

From (20), the first term of (18) is bounded by

$${\delta }^{-\tau }{ \int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}\vert \langle {\rho }_{g}f,f\rangle \vert {\omega }_{1}(g)dg.$$

(21)

Note also that by assuming ε a sufficiently small constant, we can ensure that ℓ ≪ k and 2^− k < c ^2ℓ. Thus

$${\int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}\vert \langle {\rho }_{g}f,f\rangle \vert {\omega }_{1}(g)dg > \frac{1} {2}{\delta }^{\tau }{c}^{2\mathcal{l}}$$

(22)

and applying Cauchy-Schwarz

$$\begin{array}{rcl} & & {c}^{4\mathcal{l}}{\delta }^{2\tau }{(1 + \epsilon )}^{-6\mathcal{l}} \leq {\int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}\vert \langle {\rho }_{g}f,f\rangle {\vert }^{2}{\omega }_{ 1}(g)dg \\ & & = \left \vert {\int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}{ \int \nolimits \nolimits }_{\mathbb{T}}{ \int \nolimits \nolimits }_{\mathbb{T}}f(x)\overline{f(y)}\ \overline{f({\tau }_{g}x)}f({\tau }_{g}y){({\tau }_{g}^{\prime}(x))}^{1/2}{({\tau }_{ g}^{\prime}(y))}^{1/2}{\omega }_{ 1}(g)dgdxdy\right \vert \\ & &\leq {\int \nolimits \nolimits }_{\mathbb{T}}{ \int \nolimits \nolimits }_{\mathbb{T}}\vert f(x)\vert \,\vert f(y)\vert \left \vert {\int \nolimits \nolimits }_{S{L}_{2}(\mathbb{R})}f({\tau }_{g}x)\overline{f({\tau }_{g}y)}{({\tau }_{g}^{\prime}(x))}^{1/2}{\left ({\tau }_{ g}^{\prime}(y)\right )}^{\frac{1} {2} }{\omega }_{1}(g)dg\right \vert dxdy.\end{array}$$

(23)

Fix x ≠ y and consider the inner integral. If we restrict $g \in S{L}_{2}(\mathbb{R})$ s.t. τ _g x = θ (fixed), there is still an averaging in ψ = τ _g y that can be exploited together with (13). By rotations, we may assume $x = \theta = 0$. Write $g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right ) \in S{L}_{2}(\mathbb{R})$, $dg = \frac{dadbdc} {a}$ on the chart a ≠ 0. Since

$${e}^{i{\tau }_{g}x} = \frac{(a\cos x + b\sin x) + i(c\cos x + d\sin x)} {{[{(a\cos x + b\sin x)}^{2} + {(c\cos x + d\sin x)}^{2}]}^{1/2}}$$

the condition τ _g 0 = 0 means c = 0 and thus

$${e}^{i\psi } = {e}^{i{\tau }_{g}y} = \frac{(a\cos y + b\sin y) + \frac{i} {a}\sin y} {{[{(a\cos y + b\sin y)}^{2} +{ \frac{1} {{a}^{2}} \sin }^{2}y]}^{\frac{1} {2} }}.$$

Hence, fixing a

$$\frac{\partial \psi } {\partial b} = -{a\sin }^{2}\psi.$$

(24)

Also

$${\tau_g}^{\prime}(z) = \frac{\cos^2 \tau_g(z)}{{(a\cos z + b\sin z)^2}} = {a^2}\ \frac{\sin^2\tau_g(z)} {\sin^2}z$$

(25)

implying

$${\tau }_{g}^{\prime}(0) = \frac{1} {{a}^{2}}\text{ and }{\tau }_{g}^{\prime}(y) ={ \frac{{a{}^{2}\sin }^{2}\psi } {\sin }^{2}y}.$$

(26)

Substituting (24), (26) in (23) gives for the inner integral the bound

$$\begin{array}{rcl} & & \quad \frac{1} {\vert \sin (x - y)\vert }\iint d\theta \frac{da} {{a}^{2}} \vert f(\theta )\vert \\ & &\cdot \left \vert \int \nolimits \nolimits f(\psi ) \frac{1} {\vert \sin (\theta - \psi )\vert }\omega \left ({a}^{2} + \frac{1} {{a}^{2}} +{ \left (\frac{1} {a}\text{ cotg}(\psi - \theta ) - a\text{ cotg}(y - x)\right )}^{2}\right )d\psi \right \vert.\end{array}$$

(27)

The weight function restricts a to ${(1 + \epsilon )}^{-2\mathcal{l}} \lesssim \vert a\vert \lesssim {(1 + \epsilon )}^{2\mathcal{l}}$ and clearly

$$\vert\sin (\theta - \psi )\vert \gtrsim {(1 + \epsilon )}^{-4\mathcal{l}}\vert \sin (x - y)\vert.$$

(28)

If we restrict $\vert \sin (x - y)\vert > {2}^{-\frac{k} {10} }$, Assumption (13) gives a bound at most ${2}^{-k}\Vert {f\Vert }_{1}$ for the ψ-integral in (27). Indeed, if β is a smooth function vanishing on a neighborhood of 0 and | n | ∼ 2^k, partial integration implies that for any given A > 0

$$\int {e}^{-in\psi } \frac{1} {\sin (\theta - \psi )}\,\beta \left ({2}^{ \frac{k} {10} }(\theta - \psi )\right )d\psi \lesssim {2}^{-Ak}.$$

Thus

$$(<InternalRef RefID="Equ27">27</InternalRef>) < {2}^{k/10}{(1 + \epsilon )}^{2\mathcal{l}}\ {2}^{-k}\Vert {f\Vert }_{ 1}^{2}.$$

(29)

The contribution to (23) is at most

$${2}^{-k/2}{(1 + \epsilon )}^{2\mathcal{l}}\Vert {f\Vert }_{ 1}^{4}.$$

(30)

Next we consider, the contribution of $\vert \sin (x - y)\vert \leq {2}^{-\frac{k} {10} }$ to (23).

First, from (25), we have that

$$\vert {\tau }_{g}^{\prime}\vert \gtrsim {a}^{2} + \frac{1} {{a}^{2}} + {b}^{2} \lesssim \Vert {g\Vert }^{2} < {(1 + \epsilon )}^{4\mathcal{l}}.$$

By Cauchy-Schwarz, the inner integral in (23) is at most

$${(1 + \epsilon )}^{4\mathcal{l}}\left (\int \nolimits \nolimits \vert f({\tau }_{g}x){\vert }^{2}{\omega }_{ 1}(g)dg\right ) < {(1 + \epsilon )}^{10\mathcal{l}}\Vert {f\Vert }_{ 2}^{2}.$$

Hence, we obtain

$$\begin{array}{rcl} & & \left [{\int \nolimits \nolimits }_{\vert x-y\vert <{2}^{-k/10}}\vert f(x)\vert \,\vert f(y)\vert dxdy\right ]{(1 + \epsilon )}^{10\mathcal{l}}\Vert {f\Vert }_{ 2}^{2} \\ & & \qquad < {2}^{-k/20}{(1 + \epsilon )}^{10\mathcal{l}}\Vert {f\Vert }_{ 2}^{4}. \end{array}$$

(31)

From (30), (31),

$$(<InternalRef RefID="Equ23">23</InternalRef>) \leq {2}^{-k/20}{(1 + \epsilon )}^{10\mathcal{l}}$$

and hence, by (19)

$${2}^{k/10} < 10{0}^{k\tau }.{C}^{C(\tau ){(\log \frac{1} {\epsilon })}^{-1}k }.$$

(32)

Taking (in order) τ and ε small enough, a contradiction follows.

This proves Lemma 3.

5 Absolute Continuity of the Furstenberg Measure and Smoothness of the Density

Our aim is to establish the following.

Theorem.

Let μ be the stationary measure introduced in (9). Given $r \in {\mathbb{Z}}_{+}$ and taking ε in Lemma  1 small enough will ensure that $\frac{d\mu } {d\theta } \in {C}^{r}$.

This will be an immediate consequence of

Lemma 4.

Let k > k(ε) be sufficiently large and $f \in {L}^{\infty }(\mathbb{T}),\vert f\vert \leq 1$ such that $\mathrm{supp}\hat{f} \subset [{2}^{k-1},{2}^{k}]$ . Then

$$\vert \langle f,\mu \rangle \vert < {C}_{\epsilon }^{-k}$$

(33)

where ${C}_{\epsilon } \rightarrow ^{\epsilon \rightarrow 0}\infty$.

Proof.

Clearly, for any $\mathcal{l} \in {\mathbb{Z}}_{+}$

$$\vert \langle f,\mu \rangle \vert \leq {\left \Vert {\sum \nolimits }_{g}{\nu }^{(\mathcal{l})}(g)(f \circ {\tau }_{ g})\right \Vert }_{\infty }.$$

(34)

We will iterate Lemma 3 and let K = K(ε) satisfy (10), (11).

We assume 2^k > 10K ¹⁰. For m < ℓ and | n |  < K, we evaluate $\vert \widehat{{F}_{m}}(n)\vert$, denoting

$${F}_{m} ={ \sum \nolimits }_{g}{\nu }^{(m)}(g)(f \circ {\tau }_{ g}).$$

(35)

Clearly $\vert \widehat{{F}_{m}}(n)\vert {\leq \max }_{g\in \mathrm{supp}{\nu }^{(m)}}\vert {(f \circ {\tau }_{g})}^{\wedge }(n)\vert$ and by assumption on $\mathrm{supp}\hat{f}$

$$\begin{array}{rcl} \vert {(f \circ {\tau }_{g})}^{\wedge }(n)\vert & =& \left \vert \int \nolimits \nolimits f\left ({\tau }_{g}(x)\right ){e}^{-2\pi inx}dx\right \vert \\ &\leq & {2}^{k/2}\Vert {f\Vert {}_{ 2}\max }_{n^{\prime}\in [{2}^{k-1},{2}^{k}]}\left \vert \int \nolimits \nolimits {e}^{2\pi i(n^{\prime}{\tau }_{g}(x)-nx)}dx\right \vert \end{array}$$

. Performing a change of variables gives

$$\begin{array}{rcl} \left \vert \int \nolimits \nolimits {e}^{2\pi i(n^{\prime}{\tau }_{g}(x)-nx)}dx\right \vert \!\!& & = \left \vert \int \nolimits \nolimits {e}^{2\pi i(n^{\prime}y-n{\tau }_{{g}^{-1}}(y))}{\tau }_{{ g}^{{}^{-1}}}^{\prime}(y)dy\right \vert \\ & &{\ll }_{r}\Vert {e}^{-2\pi in{\tau }_{g-1} }{\tau {}_{g-1}^{\prime}\Vert }_{{C}^{r}}\vert n^{\prime}{\vert }^{-r} \\ & & {\ll }_{r} \frac{{K}^{r}} {\vert n^{\prime}{\vert }^{r}}{(1 + \epsilon )}^{2m(r+1)} {\ll }_{ r}{2}^{-\frac{3} {4} kr}{(1 + \epsilon )}^{2\mathcal{l}(r+1)}\end{array}$$

(36)

by partial integration and our assumptions. It follows from (36) that if ℓ satisfies

$$\mathcal{l} < \frac{k} {100\epsilon }$$

(37)

then for m < ℓ and k > k(r)

$${ \max }_{\vert n\vert <K}\vert \widehat{{F}_{m}}(n)\vert < {2}^{-\frac{kr} {2} }$$

(38)

(with r a fixed large integer).

Next, decompose

$${F}_{m} = {F}_{m}^{(1)} + {F}_{ m}^{(2)}\text{ where }{F}_{ m}^{(1)}(x) ={ \sum \nolimits }_{\vert n\vert <K}\widehat{{F}_{m}}(n){e}^{2\pi inx}.$$

Hence, by (38)

$$\Vert {F_m}^{(1)}{\Vert_ \infty } < 2K{2}^{-\frac{kr} {2} }.$$

(39)

Estimate using (39) and Lemma 3

$$\begin{array}{rcl} \Vert {F{}_{m+1}\Vert }_{2}& \leq &{ \left\Vert \int ({F}_{m}^{(1)} \circ {\tau }_{ g})d\nu \right\Vert }_{\infty } +{ \left\Vert \int ({F}_{m}^{(2)} \circ {\tau }_{ g})d\nu \right\Vert }_{2} \\ & \leq & \Vert {F_m}^{(1)}{\Vert_\infty } + \frac{1} {2}\Vert {F}_{m}^{(2)}{\Vert}_{ 2} \\ & \leq & 3K{2}^{-\frac{kr} {2} } + \frac{1} {2}\Vert {F{}_{m}\Vert }_{2}. \end{array}$$

(40)

Iteration of (40) implies by (37)

$$\Vert {F{}_{\mathcal{l}}\Vert }_{2} \leq 4K{2}^{-\frac{kr} {2} } + {2}^{-\mathcal{l}} \lesssim {2}^{-\frac{kr} {2} } + {2}^{- \frac{k} {100\epsilon } }.$$

(41)

Also

$$\vert {F}_{\mathcal{l}}^{\prime}\vert \leq \max\limits_{g\in \mathrm{supp}{\nu }^{(\mathcal{l})}}\Vert {(f \circ {\tau }_{g})^{\prime}\Vert }_{\infty }\leq \Vert {f^{\prime}\Vert }_{\infty }{(1 + \epsilon )}^{2\mathcal{l}} \lesssim {5}^{k}$$

(42)

and interpolation between (41), (42) implies for r (resp. ε) large (resp. small) enough

$$\Vert {F{}_{\mathcal{l}}\Vert }_{\infty }\lesssim {(41)}^{1/2}.{(42)}^{1/2} < {2}^{-\frac{kr} {5} } + {2}^{- \frac{k} {300\epsilon } }$$

(43)

provided k > k(ε, r).

In view of (34), this proves (33).

Remark.

For ν finitely supported (with positive Lyapounov exponent), one cannot obtain a Furstenberg measure μ that equals Haar measure on ${\mathbb{P}}_{1}(\mathbb{R}) \simeq \mathbb{T}$. Indeed, otherwise for any f on $\mathbb{T}$, we would have

$$\begin{array}{rcl} \hat{f}(0)& =& {\int \nolimits \nolimits }_{\mathbb{T}}fd\mu = \int \nolimits \nolimits \nu (dg)\left [\int \nolimits \nolimits (f \circ {\tau }_{g})d\mu \right ] \\ & =& \int \nolimits \nolimits \nu (dg)\left [\int \nolimits \nolimits f(x)({\tau }_{{g}^{{}^{-1}}})^{\prime}(x)dx\right ]. \end{array}$$

(44)

For $g \in S{L}_{2}(\mathbb{R})$,

$$\begin{array}{rcl} \int \nolimits \nolimits f(x)({\tau }_{{g}^{{}^{-1}}})^{\prime}(x)dx& =& \int \nolimits \nolimits f(\theta ){P}_{z}(2\theta )d\theta \\ & =& \hat{f}(0) +{ \sum \nolimits }_{n\not =0}\vert z{\vert }^{\vert n\vert }{e}^{2\pi in(Argz)}\hat{f}(-2n)\end{array}$$

(45)

for some $z \in D =\{ z \in \mathbb{C};\vert z\vert < 1\}$, with ${P}_{z}(\theta ) = \frac{1-\vert z{\vert }^{2}} {\vert 1-\bar{z}{e}^{i\theta }{\vert }^{2}}$ the Poisson kernel.

From (44), (45), taking $\nu ={ \sum \nolimits }_{j=1}^{r}{c}_{j}{\delta }_{{g}_{j}},{c}_{j} > 0$ and ∑c _j  = 1 and {z _j } the corresponding points in D, we get

$${\sum \nolimits }_{1}^{r}{c}_{ j}\vert {z}_{j}{\vert }^{n}{e}^{2\pi in(Arg{z}_{j})} = 0\text{ for all }n\not =0.$$

(46)

This easily implies that ${z}_{1} = \cdots = {z}_{r} = 0$. But then each g _j has unimodular spectrum and ν vanishing Lyapounov exponent.