---
abstract: |
    In many problems of classical analysis extremal configurations appear to exhibit complicated fractal structure. This makes it much harder to describe extremals and to attack such problems. Many of these problems are related to the [*multifractal analysis*]{} of [*harmonic measure*]{}.

    We argue that, searching for extremals in such problems, one should work with random fractals rather than deterministic ones. We introduce a new class of fractals [*random conformal snowflakes*]{} and investigate its properties developing tools to estimate spectra and showing that extremals can be found in this class. As an application we significantly improve known estimates from below on the extremal behaviour of harmonic measure, showing how to constuct a rather simple snowflake, which has a spectrum quite close to the conjectured extremal value.
author:
- 'D. Beliaev'
- 'S. Smirnov'
bibliography:
- 'snow.bib'
title: Random conformal snowflakes
---

Introduction
============

It became apparent during the last decade that extremal configurations in many important problems in classical complex analysis exhibit complicated fractal structure. This makes such problems more difficult to approach than similar ones, where extremal objects are smooth.

As an example one can consider the coefficient problem for univalent functions. Bieberbach formulated his famous conjecture arguing that the Köebe function, which maps a unit disc to a plane with a stright slit, is extremal. The Bieberbach conjecture was ultimately proved by de Branges in 1985 [@deBranges], while the sharp growth asymptotics was obtained by Littlewood [@Littlewood25] in 1925 by a much easier argument.

However, coefficient growth problem for bounded functions remains widely open, largely due to the fact that the extremals must be of fractal nature (cf [@CaJo]). This relates (see [@BeSmECM]) to a more general question of finding the [*universal multifractal spectrum*]{} of [*harmonic measure*]{} defined below, which includes many other problems, in particular conjectures of Brennan, Carleson and Jones, Kraetzer, Szegö, and Littlewood.

In this paper we report on our search for extremal fractals. We argue that one should study random fractals instead of deterministic ones. We introduce a new class of random fractals, [*random conformal snowflakes*]{}, investigate its properties, and as a consequence significantly improve known estimates from below for the multifractal spectra of harmonic measure.

Multifractal analysis of harmonic measure
-----------------------------------------

It became clear recently that appropriate language for many problems in geometric function theory is given by the [*multifractal analysis*]{} of [*harmonic measure*]{}. The concept of multifractal spectrum of a measure was introduced by Mandelbrot in 1971 in [@Mandelbrot72; @Mandelbrot74] in two papers devoted to the distribution of energy in a turbulent flow. We use the definitions that appeared in 1986 in a seminal physics paper [@HJKPS] by Halsey, Jensen, Kadanoff, Procaccia, Shraiman who tried to understand and describe scaling laws of physical measures on different fractals of physical nature (strange attractors, stochastic fractals like DLA, etc.).

There are various notions of spectra and several ways to make a rigorous definition. Two standard spectra are [*packing*]{} and [*dimension*]{} spectra. The packing spectrum of harmonic measure $\omega$ in a domain $\Omega$ with a compact boundary is defined as $$\pi_{\Omega}(t)=
\sup\,\Big\{q:\ \forall\delta>0~\exists~\delta-\mathrm{ packing }~\{B\}
~{\mathrm with}~\sum \mathrm{diam}(B)^t\omega(B)^q\,\ge\,1\Big\}\ ,$$ where $\delta$-packing is a collection of disjoint open sets whose diameters do not exceed $\delta$.

The [*dimension spectrum*]{} which is defined in terms of harmonic measure $\omega$ on the boundary of $\Omega$ (in the case of simply connected domain $\Omega$ harmonic measure is the image under the Riemann map $\phi$ of the normalised length on the unit circle). Dimension spectrum gives the dimension of the set of points, where harmonic measure satisfies a certain power law: $$f(\alpha)~:=~\mathrm{dim}\,
\Big\{z:~\omega\br{B(z,\delta)}\,\approx\,\delta^\alpha\,,~\delta\to0\Big\},~\alpha\ge\frac12~.$$ Here $\mathrm{dim}$ stands for the Hausdorff or Minkowski dimension, leading to possibly different spectra. The restriction $\alpha \ge 1/ 2$ is due to Beurling’s inequality. Of course in general there will be many points where measure behaves differently at different scales, so one has to add $\limsup$’s and $\liminf$’s to the definition above – consult [@Makarov] for details.

In our context it is more suitable to work with a modification of the packing spectrum which is specific for the harmonic measure on a two dimensional simply connected domain $\Omega$. In this case we can define the [*integral means spectrum*]{} as $$\beta_\phi(t)~:=~\limsup_{r\to1+}\frac{\log \int_{0}^{2\pi}|\phi'(re^{i\theta})|^t 
d\theta}{|\log(r-1)|},~t\in{{\mathbb R}}~,$$ where $\phi$ is a Riemann map from the complement of the unit disc onto a simply connected domain $\Omega$.

Connections between all these spectra for particular domains are not that simple, but the [*universal spectra*]{} $$\Pi(t)=\sup_\Omega \pi(t), \quad
F(\alpha)=\sup_\Omega f(\alpha), \ \ \mathrm{and} \ \ B(t)=\sup_\Omega\beta(t)$$ are related by Legendre-type transforms: $$\begin{aligned}
F(\alpha)&=&\inf_{0\le t \le 2} (\alpha \Pi(t)+t), \quad \alpha \ge 1~, \\
\Pi(t)&=&\sup_{\alpha\ge 1} \left(\frac{F(\alpha)-t)}{\alpha}\right), \quad 0\le 2\le  2~, \\
\Pi(t)&=&B(t)-t+1~.\end{aligned}$$ See Makarov’s survey [@Makarov] for details.

Random fractals
---------------

One of the main problems in the computation of the integral means spectrum (or other multifractal spectra) is the fact that the derivative of a Riemann map for a fractal domain depends on the argument in a very non regular way: $\phi'$ is a “fractal” object in itself. We propose to study random fractals to overcome this problem. For a random function $\phi$ it is natural to consider the [*average integral means spectrum:*]{} $$\begin{aligned}
\bar\beta(t)&=&\sup\brs{\beta: \int_1(r-1)^{\beta-1}\int_0^{2\pi}
{{\mathbb E}}\brb{|f'(r e^{i\theta})|^t}d \theta d r=\infty}
\\
&=&\inf\brs{\beta: \int_1(r-1)^{\beta-1}\int_0^{2\pi}
{{\mathbb E}}\brb{|f'(r e^{i\theta})|^t}d \theta d r<\infty}.\end{aligned}$$ The average spectrum does not have to be related to the spectra of a particular realization. We want to point out that even if $\phi$ has the same spectrum a.s. it does not guarantee that $\bar\beta(t)$ equal to the a.s. value of $\beta(t)$. Moreover, it can happen that $\bar\beta$ is not a spectrum of [*any*]{} particular domain.

But one can see that $\bar\beta(t)$ is bounded by the universal spectrum $B(t)$. Indeed, suppose that there is a random $f$ with $\bar\beta(t)>B+\epsilon$, hence for any $r$ there are particular realizations of $f$ with $\int |f'(z)|d\theta>(r-1)^{-B-\epsilon/2}$. Then by Makarov’s fractal approximation [@Makarov] there is a (deterministic) function $F$ such that $\beta_F(t)>B(t)$ which is impossible by the definition of $B(t)$.

For many classes of random fractals ${{\mathbb E}}|\phi'|^t$ (or its growth rate) does not depend on the argument. This allows us to drop the integration with respect to the argument and study the growth rate along any particular radius. Perhaps more importantly ${{\mathbb E}}|\phi'|$ is no longer a “fractal” function.

One can think that this is not a big advantage compared to the usual integral means spectrum: instead of averaging over different arguments we average over different realizations of a fractal. But most fractals are results of some kind of an iterative construction, which means that they are invariant under some (random) transformation. Thus ${{\mathbb E}}|\phi'|^t$ is a solution of some kind of equation. Solving this equation (or estimating its solutions) we can find $\bar\beta(t)$.

In this paper we want to show how one can employ these ideas. In the Section \[sec:def\] we introduce a new class of random fractals that we call random conformal snowflakes. In the Section \[sec:spectrum\] we show that $\bar\beta(t)$ for this class is related to the main eigenvalue of a particular integral operator. We also prove the fractal approximation for this class in the Section \[sec:approximation\]. In the Appendix \[sec:application\] we give an example of a snowflake and prove that for this snowflake $\bar\beta(1)>0.23$. This significantly improves previously known estimate $B(1)>0.17$ due to Pommerenke [@Pommerenke75].

Conformal snowflake {#sec:def}
===================

The construction of our conformal snowflake is similar to the construction in Pommerenke’s paper [@Pommerenke67lms]. The main difference is the introduction of the randomness.

By $\Sigma'$ we denote a class of all univalent functions $\phi:{{\mathbb D}}_-\to {{\mathbb D}}_-$ such that $\phi(\infty)=\infty$ and $\phi'(\infty)\in {{\mathbb R}}$. Let $\phi\in \Sigma'$ be a function with expansion at infinity $\phi(z)=b_1 z+\dots$, by ${{\mathrm cap\:}}\phi={{\mathrm cap\:}}\Omega$ we denote the logarithmic capacity of $\phi$ which is equal to $\log|b_1|$. We will also use the so called [*Koebe $n$-root transform*]{} which is defined as $$({K}\phi)(z)=({K}_n\phi)(z)=\sqrt[n]{\phi(z^n)}.$$ It is a well known fact that the Koebe transform is well defined and ${K}\phi\in\Sigma'$. It is easy to check that Koebe transform divides capacity by $n$ and that the capacity of a composition is the sum of capacities.

First we define the deterministic snowflake. To construct a snowflake we need a building block $\phi \in \Sigma'$ and an integer $k\ge 2$. Our snowflake will be the result of the following iterative procedure: we start with the building block and at $n$-th step we take a composition of our function and the $k^n$-root transform of the rotated building block.

Let $\phi \in \Sigma'$ and $\theta\in [0,2\pi]$. By $\phi_\theta(z)$ we denote the map whose range is the rotation of that for $\phi$, namely $e^{i\theta}\phi(z e^{-i\theta})$.

Let $\phi\in \Sigma'$, $k\ge 2$ be an integer number, and $\{\theta_n\}$ be a sequence of numbers from ${{\mathbb T}}$. Let $f_0(z)=\phi_{\theta_0}(z)$ and $$\begin{aligned}
f_{n}(z)=f_{n-1}(K_{k^n}\phi_{\theta_n}(z))=
\phi_{\theta_0}(\phi_{\theta_1}^{1/k}(\dots\phi_{\theta_n}^{1/k}(z^{k^n})\dots).\end{aligned}$$ The conformal snowflake $f$ is the limit of $f_n$. For simplicity $S={{\mathbb C}}\setminus f({{\mathbb D}}_-)$ and $g=f^{-1}$ are also called a snowflake.

Sometimes it is easier to work with a slightly different symmetric snowflake $$\bar f_{n}(z)=\phi_{\theta_1}^{1/k}(\dots\phi_{\theta_n}^{1/k}(z^{k^n})\dots)=
\Phi_1\circ \cdots \circ\Phi_n(z),$$ where $\Phi_j={K}_{k^j}\phi_{\theta_j}$. There are two equivalent ways to construct the symmetric snowflake from the usual one. One is to take the Koebe transform ${K}f_n$, another is to start with $f_0(z)=z$. It is easy to see that $f_n=\Phi_0\circ\cdots\circ\Phi_n.$

How this snowflake grows? This is easy to analyse looking at the evolution of $\bar f_n$. At every step we add $k^n$ equidistributed (according to the harmonic measure) small copies of the building block. But they are not exact copies, they are distorted a little bit by a conformal mapping.

Figures \[fbar\] and \[f\] show images of the first four functions $\bar f$ and $f$ with $k=2$ and the building block is a slit map (which adds a straight slit of length $4$).

![The third generation of a snowflake: $f_3$.[]{data-label="pic2"}](snow7313long.eps)

![The image of a small boundary arc under $\bar f_3$ with three Green’s lines.[]{data-label="pic3"}](snow7313locgreen.eps)

Let $f_n=\phi_{\theta_0}(\bar f_n(z))$ be the $n$-th approximation to the snowflake with a building block $\phi$ and $k\ge 2$. Then ${{\mathrm cap\:}}( f_n)$ and ${{\mathrm cap\:}}(\bar f_n)$ are bounded by (and converge to) ${{\mathrm cap\:}}(\phi)k/(k-1)$ and ${{\mathrm cap\:}}(\phi)/(k-1)$. \[l:boundedcapacity\]

This lemma follows immediately from the standard facts that $$\begin{aligned}
{{\mathrm cap\:}}(f\circ g)&=&{{\mathrm cap\:}}(f)+{{\mathrm cap\:}}(g),\\
{{\mathrm cap\:}}(K_n f)&=&{{\mathrm cap\:}}(f)/n.\end{aligned}$$

The conformal snowflake is well defined, namely let $f_n$ be an $n$-th approximation to a snowflake with a building block $\phi$ and $k\ge 2$. Then there is $f\in\Sigma'$ such that $f_n$ converge to $f$ uniformly on every compact subset of ${{\mathbb D}}_-$. \[thm:conv\]

Fix $\epsilon>0$. It is enough to prove that $\bar f_n$ converge uniformly on ${{\mathbb D}}_\epsilon=\{z: |z|\ge 1+\epsilon\}$. Suppose that $m>n$ so we can write $\bar f_m=\bar f_n \circ \Phi_{n,m}$ where $\Phi_{n,m}=\Phi_{n+1}\circ\cdots\circ \Phi_m$ and $$|\bar f_n(z)-\bar f_m(z)|=
|\bar f_n(z)-\bar f_n(\Phi_{n,m}(z))|\le 
\max_{\zeta \in {{\mathbb D}}_\epsilon}|\bar f_n'(\zeta)||z-\Phi_{n,m}(z)|.$$ By the Lemma \[l:boundedcapacity\] ${{\mathrm cap\:}}(\bar f_n)$ is uniformly bounded, hence by the growth theorem the derivative of $\bar f_n$ is uniformly bounded in ${{\mathbb D}}_\epsilon$. Thus it is enough to prove that $\Phi_{n,m}(z)$ converge uniformly to $z$.

Let $\phi(z)=b_1 z+\dots$ at infinity, then $\Phi_n(z)=b_1^{1/k^{n}}z+\dots$ Functions $\Phi_{n,m}$ have expansion $$b_1^{k^{-n}+\dots +k^{-m}} z+\dots=b_1^{(n,m)}z+\dots$$ Obviously, $b_1^{(n,m)}\to 1$ as $n\to\infty$. This proves that $\Phi_{n,m}(z)\to z$ uniformly on ${{\mathbb D}}_\epsilon$ hence $f_n$ converge uniformly. Uniform limit of a functions from $\Sigma'$ can be either a constant or a function from $\Sigma'$. Since ${{\mathrm cap\:}}(f_n)$ is uniformly bounded the limit can not be a constant.

Let $\phi\in\Sigma'$ and $k\ge 2$ be an integer number. The random conformal snowflake is a conformal snowflake defined by $\phi$, $k$, and $\{\theta_n\}$, where $\theta_n$ are independent random variables uniformly distributed on ${{\mathbb T}}$.

\[stationary\] Let $\phi\in\Sigma'$, $k\ge 2$ be an integer number, and $\psi=\phi^{-1}$. Let $f$ be a corresponding random snowflake and $g=f^{-1}$. Then the distribution of $f$ is invariant under the transformation $\Sigma'\times {{\mathbb T}}\mapsto \Sigma'$ which is defined by $$(f,\theta)\mapsto \phi_\theta({K}_k f).$$ In other words $$\begin{aligned}
f(z)&=&\phi_\theta(({K}_k f)(z))= \phi_\theta(f^{1/k}(z^k)),\\
g(z)&=&({K}_k g)(\psi_\theta(z))= g^{1/k}(\psi^k_\theta(z)),\end{aligned}$$ where $\theta$ is uniformly distributed on ${{\mathbb T}}$. Both equalities should be understood in the sense of distributions, i.e. distributions of both parts are the same.

Let $f$ be a snowflake generated by $\{\theta_n\}$. The probability distribution of the family of snowflakes is the infinite product of (normalised) Lebesgue measures on ${{\mathbb T}}$. By the definition $$f(z)=
\lim_{n\to\infty} \phi_{\theta_0}(\phi_{\theta_1}^{1/k}(\dots\phi_{\theta_n}^{1/k}(z^{k^n})\dots)$$ and $$\phi_\theta(({K}_k f)(z))=\lim_{n\to\infty} 
\phi_{\theta}(\phi_{\theta_0}^{1/k}(\dots\phi_{\theta_n}^{1/k}(z^{k^{n+1}})\dots),$$ hence $\phi_\theta({K}_k f)$ is just the snowflake defined by the sequence $\theta, \theta_0, \theta_1,\dots$. So the transformation $f(z)\mapsto\phi_\theta(({K}_k f)(z))$ is just a shift on the $[-\pi,\pi]^{\mathbb N}$. Obviously the product measure is invariant under the shift. This proves stationarity of $f$. Stationarity of $g$ follows immediately from stationarity of $f$.

There is another way to think about random snowflakes. Let $\mathcal M$ be a space of probability measures on $\Sigma'$. And let $T$ be a random transformation $f \mapsto \phi_\theta ({K}_k f)$, where $\theta$ is uniformly distributed on $[\pi,\pi]$. Obviously $T$ acts on $\mathcal M$. The distribution of a random snowflake is the only measure which is invariant under $T$. In some sense the random snowflake is an analog of a Julia set: it semi-conjugates $z^k$ and $\psi_\theta^k$.

  ------------------------------------ -------------------------------- -------------------------------------
                                        $\stackrel{f}{\longleftarrow}$   ![image](circle.eps){width="0.7cm"}
   ![image](snow24L.eps){width="3cm"}                                   
                                                                        
      $\uparrow \psi_\theta^k(z)$                                                  $\uparrow z^k$
                                                                        
                                        $\stackrel{f}{\longleftarrow}$   ![image](circle.eps){width="0.7cm"}
   ![image](snow24L.eps){width="3cm"}                                   
  ------------------------------------ -------------------------------- -------------------------------------

The random conformal snowflakes are also rotationally invariant, the exact meaning is given by the following theorem.

Let $\phi\in\Sigma'$, $k\ge 2$ and $g$ be the corresponding snowflake. Then $g$ is rotationally invariant, namely $g(z)$ and $e^{i\omega}g(e^{-i\omega}z)$ have the same distribution for any $\omega$.

Let $g_n(z)$ be the $n$th approximation to the snowflake defined by the sequence of rotations $\theta_0,\dots,\theta_n$. We claim that $\tilde g(z)=e^{i\omega}g(e^{-i\omega}z)$ is the approximation to the snowflake defined by $\tilde\theta_0,\dots,\tilde\theta_n$ where $\tilde\theta_j=\theta_j+\omega k^j$ (we add arguments $\mod 2 \pi$).

We prove this by induction. Obviously this is true for $\tilde g_0$. Suppose that it is true for $\tilde g_{n-1}$. By the definition of $g_n$ and assumption that $g_{n-1}(e^{-i\omega}z)=e^{-i\omega}\tilde g_{n-1}(z)$ we have that $$\begin{aligned}
e^{i\omega}g_n(e^{-i\omega}z)&=&e^{i\omega}e^{i\theta_n/k^n}
\psi^{1/k^n}(e^{-i\theta_n} g_{n-1}^{k^n}(e^{-i\omega}z))
\\
&=&
e^{i\tilde\theta_n/k^n}\psi^{1/k^n}(e^{i\tilde\theta_n}\tilde g_{n-1}^{k^n}(z))
=
\tilde g_n(z).\end{aligned}$$

Obviously $\tilde \theta_n$ are also independent and uniformly distributed on ${{\mathbb T}}]$, hence $\tilde g_n$ has the same distribution as $g_n$.

The distributions of $|g(z)|$ and $|g'(z)|$ depend on $|z|$ only. The same is true for $f$.

Spectrum of a conformal snowflake {#sec:spectrum}
=================================

As we discussed above, for random fractals it is more natural to consider the average spectrum $\bar \beta(t)$ instead of the usual spectrum $\beta(t)$. We will work with $\bar \beta(t)$ only and “spectrum” will always mean $\bar \beta(t)$.

We will write ${{\mathcal L}}$ for the class of functions on $(1,\infty)$ that are bounded on compact sets and integrable in the neighbourhood of $1$. In particular, these functions belong to $L^1[1,R]$ for any $1<R<\infty$.

Let $F(z)=F(|z|)=F(r)={{\mathbb E}}\brb{\,|g'(r)/g(r)|^\tau\log^{\sigma}|g(r)|}$ where $\tau=2-t$ and $\sigma=\beta-1$.

The $\bar\beta(t)$ spectrum of the snowflake is equal to $$\inf\brs{\beta: F(r)\in {{\mathcal L}}}.$$

By the definition $\bar\beta$ is the minimal value of $\beta$ such that $$\int_1\int_0^{2\pi}(r-1)^{\beta-1}{{\mathbb E}}|f(r e^{i\theta})|^t d\theta d r$$ is finite. We change variable to $w=f(z)=f(re^{i\theta})$ $$\begin{aligned}
\int\int {{\mathbb E}}\brb{|f'(re^{i\theta})|^t}(r-1)^{\beta-1} dr d\theta
&=&
\int \frac{{{\mathbb E}}\brb{|f'(z)|^t}(|z|-1)^{\beta-1}dm(z)}{r} 
\\&=&
\int {{\mathbb E}}\brb{\frac{|g'(w)|^{2-t}(|g(w)|-1)^{\beta-1}}{|g(w)|}}dm \, ,\end{aligned}$$ where $m$ is the Lebesgue measure. Note that $|g|$ is uniformly bounded and $g$ is rotationally invariant, hence the last integral is finite if and only if $$\int_1 |g'(r)|^\tau(|g(r)|-1)^\sigma d r<\infty.$$ Since $1<|g|$ is uniformly bounded we have that $|g'(r)|^\tau (|g(r)|-1)^\sigma$ is comparable up to an absolute constant to $$\br{\frac{|g'(r)|}{|g(r)|}}^\tau \log^\sigma|g(r)|.$$

If $F \in {{\mathcal L}}$ then it is a solution of the following equation: $$\label{eq}
F(r)=\frac{1}{k^\sigma}\int_{-\pi}^{\pi} F(|\psi^k(re^{i\theta})|)|\psi^{k-1}(re^{i\theta})
\psi'(re^{i\theta})|^\tau\frac{d\theta}{2\pi}.$$

By the Theorem \[stationary\] $g(z)$ and $g^{1/k}(\psi_\theta^{k}(z))$ have the same distribution, hence $$\begin{aligned}
F(r)&=&{{\mathbb E}}\brb{\left|g'(r)/g(r)\right|^\tau\log^\sigma |g(r)|} \\
&=& {{\mathbb E}}\brb{\left|\frac{(g^{1/k}(\psi_\theta^k(r)))'}{g^{1/k}(\psi_\theta^k(r))}\right|^\tau 
\log^\sigma|g^{1/k}(\psi_\theta^k(r))|}\\
&=&
{{\mathbb E}}\brb{\left|\frac{g'(\psi^k_\theta(r))}{g(\psi_\theta^k(r))}\right|^\tau\log^\sigma(g(\psi^k_\theta(r)))
\frac{|\psi'_\theta(r)\psi^{k-1}_\theta(r)|^\tau}{k^\sigma}},\end{aligned}$$ where $\theta$ has a uniform distribution. The expectation is the integral with respect to the joint distribution of $g$ and $\theta$, since they are independent this joint distribution is just a product measure. So we can write it as the double integral: first we take the expectation with respect to the distribution of $g$ and than with respect to the (uniform) distribution of $\theta$ $$\begin{aligned}
F(r)&=&\int_{-\pi}^\pi \br{\int  
\left|\frac{g'(\psi^k_\theta(r))}{g(\psi_\theta^k(r))}\right|^\tau\log^\sigma(g(\psi^k_\theta(r))) 
\frac{|\psi'_\theta(r)\psi^{k-1}_\theta(r)|^\tau}{k^\sigma}d g} \frac{d\theta}{2\pi}
\\
&=& \int_{-\pi}^\pi \br{\int  \left|\frac{g'(\psi^k_\theta(r))}
{g(\psi_\theta^k(r))}\right|^\tau\log^\sigma(g(\psi^k_\theta(r))) 
d g}\frac{|\psi'_\theta(r)\psi^{k-1}_\theta(r)|^\tau}{k^\sigma} \frac{d\theta}{2\pi}.\end{aligned}$$ The inner integral is equal to $F(\psi_\theta^k(r))=F(\psi^k(e^{-i\theta}r))$ by the definition of $F$, hence $$\begin{aligned}
F(r)&=&\int_{-\pi}^\pi F(\psi^k(e^{-i\theta}r))
\frac{|\psi'(e^{-i\theta}r)\psi^{k-1}(e^{-i\theta}r)|^\tau}{k^\sigma} \frac{d\theta}{2\pi}
\\
&=& \frac{1}{k^\sigma}\int_{-\pi}^\pi F(\psi^k(e^{i\theta}r))
|\psi'(e^{i\theta}r)\psi^{k-1}(e^{i\theta}r)|^\tau \frac{d\theta}{2\pi}\end{aligned}$$ which completes the proof.

This equation is the key ingredient in our calculations. One thinks about $F$ as the main eigenfunction of an integral operator. Hence the problem of finding the spectrum of the snowflake boils down to the question about the main eigenvalue of a particular integral operator. Usually it is not very difficult to estimate the latter.

This justifies the definition: $$Qf(r):=k \int_{-\pi}^{\pi} f (|\psi^k(re^{i\theta})|)\, |\psi^{k-1}(re^{i\theta})
\psi'(re^{i\theta})|^\tau\frac{d\theta}{2\pi}\, .$$ Using this notation we can rewrite (\[eq\]) as $$k^{\beta}F=QF.$$ Note that this is in fact an ordinary kernel operator, $|\psi|$ is a smooth function of $\theta$, hence we can change the variable and write it as an integral operator. As mentioned above, the study of a $F$ is closely related to the study of operator $Q$ and its eigenvalues. And our estimate of the spectrum is in fact estimate of the main eigenvalue.

Adjoint operator
----------------

First of all we want to find a formally adjoint operator. Let $\nu$ be a bounded function and $R>1$ such that $D_R \subset \psi^k(D_R)$ where $D_R=\brs{z:1<|z|<R}$.

$$\begin{aligned}
\int_1^R Qf(r)\nu(r)d r &=&
\int_1^R \nu(r)k \int_0^{2\pi} f(\psi^k(re^{i\theta}))\,
|\psi'(re^{i\theta})\psi^{k-1}(re^{i\theta})|^\tau\frac{d\theta}{2\pi}d r
\\
&=&
\int_{D_R}\frac{\nu(|z|)}{|z|}\frac{k}{ 2 \pi}f(\psi^k(z))
|\psi'(z)\psi^{k-1}(z)|^\tau d m(z),\end{aligned}$$

where $d m $ is the Lebesgue measure. Changing a variable to $w=\psi^k(z)$ we get $$\begin{aligned}
&&\int_{\psi^k(D_R)}\frac{\nu(\phi(w^{1/k}))}{|\phi(w^{1/k})|}
\frac{1}{k 2\pi}f(w)|\phi'(w^{1/k})w^{1/k-1}|^{2-\tau} d m(w)
\\
&\ge&
\int_1^R\int_0^{2\pi k}
\frac{r\nu(\phi(r^{1/k}e^{i\theta/k}))}{|\phi(r^{1/k}e^{i\theta/k})|}
f(r)r^{\frac{(1-k)(\tau-2)}{k}}|\phi'(r^{1/k}e^{i\theta/k})|^{2-\tau}
\frac{d\theta}{2\pi k}d r
\\
&=&
\int_1^R f(r)\int_0^{2\pi }
\frac{r\nu(\phi(r^{1/k}e^{i\theta}))}{|\phi(r^{1/k}e^{i\theta})|}
r^{\frac{(1-k)(\tau-2)}{k}}|\phi'(r^{1/k}e^{i\theta})|^{2-\tau}
\frac{d\theta}{2\pi}d r.\end{aligned}$$

So we define another operator $$P\nu(r):={r^{1-\frac{(k-1)(2-\tau)}{k}}}
\int_0^{2\pi} \frac{\nu(\phi(r^{1/k}e^{i\theta}))}{|\phi(r^{1/k}e^{i\theta})|}
|\phi'(r^{1/k}e^{i\theta})|^{2-\tau}\frac{d\theta}{2\pi}.
\label{defP}$$

Changing $2-\tau$ to $t$ we can rewrite (\[defP\]) as

$$P\nu(r):={r^{1-\frac{(k-1)t}{k}}}
\int_0^{2\pi} \frac{\nu(\phi(r^{1/k}e^{i\theta}))}{|\phi(r^{1/k}e^{i\theta})|}
|\phi'(r^{1/k}e^{i\theta})|^{t}\frac{d\theta}{2\pi}.$$

The inequality above can be written as $$\label{cover}
\int_1^R Q f(r)\nu(r)d r \ge
\int_1^R f(r)P\nu(r) d r.$$ We would like to note that for $R=\infty$ there is an equality since $\psi^K({{\mathbb D}}_-)$ covers ${{\mathbb D}}_-$ exactly $k$ times. In this case $$\label{conj}
\int_1^\infty Q f(r)\nu(r)d r =
\int_1^\infty f(r)P\nu(r) d r.$$ so operators $P$ and $Q$ are formally adjoint on $[1,\infty)$.

Operator $Q=Q(t)$ acts on ${{\mathcal L}}$ if $\int|\phi'(r e^{i\theta})|d \theta$ is bounded. If $t\ge 1$ then it also acts on $L^1(1,\infty)$.

Let $\nu=1$ in (\[conj\]). Then $$\int_1^\infty Q f(r)d r =
\int_1^\infty f(r)
\int_{-\pi}^{\pi} \frac{r^{1-\frac{(k-1)t}{k}}}{|\phi(r^{1/k}e^{i\theta})|}
|\phi'(r^{1/k}e^{i\theta})|^{t}\frac{d\theta}{2\pi}.$$ Let $r<R$, then $$\frac{r^{1-\frac{(k-1)t}{k}}}{|\phi(r^{1/k}e^{i\theta})|}<R^{1-\frac{(k-1)t}{k}},$$ so the second integral is bounded since $\int |\phi'|^t d\theta$ is bounded. This proves that $Q f$ is in ${{\mathcal L}}$. To prove that it acts on $L^1(1,\infty)$ we should consider the large values of $r$. At infinity $\phi(z)=c z+ \dots $ and $\phi'(z)=c+\dots$, hence $$\begin{aligned}
\frac{r^{1-\frac{(k-1)t}{k}}}{|\phi(r^{1/k}e^{i\theta})|}
|\phi'(r^{1/k}e^{i\theta})|^{t}\approx \frac{r^{1-\frac{(k-1)t}{k}} |c|^t}{|c|r^{1/k}}=
\mathrm{const}\, r^{1-\frac{(k-1)t)}{k}-\frac{1}{k}}
\\
=\mathrm{const}\, r^{\frac{(k-1)(1-t)}{k}}.
\end{aligned}$$ Thus the second integral is comparable (up to a universal constant) to $r^{\frac{(k-1)(1-t)}{k}}$, so it is bounded if and only if $t\ge 1$.

Note that the assumption on the integral of $|\phi'|$ is just a bit stronger than $\beta_\phi(t)=0$. We restrict ourselves to the building blocks that are smooth up to the boundary, for such building blocks this assumption is always true. Condition $t\ge 1$ is technical and due to the behavior at infinity which should be irrelevant. Introducing the weight at infinity we can get rid of this assumption.

Next we want to discuss how eigenvalues of $P$ and $Q$ are related to the spectrum of the snowflake. If $F$ is integrable then it is a solution of (\[eq\]) and using (\[cover\]) we can write $$\int_1^R F(r)\nu(r) =
\int_1^R \frac{Q F (r)}{k^{1+\sigma}}\nu(r)\ge
\int_1^R F(r) \nu(r) \frac{P\nu(r)}{\nu(r)k^{\sigma+1}}.$$ Suppose that $t$ is fixed. Let us fix a positive test function $\nu$. If $P\nu(r)\ge \nu(r)k^{\sigma+1}$ then we arrive at contradiction, this means that $F(r)$ for this particular pair of $\tau$ and $\sigma$ can not be integrable. Using this fact we can estimate $\bar\beta(t)$ from below. Hence any positive $\nu$ gives the lower bound of the spectrum. $$\label{betalog}
\bar\beta(t)\ge \min_{1\le r \le R}\log\left(\frac{P\nu(r)}{\nu(r)}\right)/\log k.$$ Obviously, the best choice of $\nu$ is an eigenfunction of $P$ corresponding to the maximal eigenvalue. This proves the following lemma:

Let $\lambda$ be the maximal eigenvalue of $P$ (on any interval $[1,R]$ such that $D_R \subset \psi^k(D_R)$) then $\bar\beta(t)\ge \log \lambda/\log k$.

Fractal approximation {#sec:approximation}
=====================

In this section we prove the fractal approximation by conformal snowflakes. Namely we show that for any $t$ one can construct a snowflake with building block which is smooth up to the boundary and $\bar\beta(t)$ arbitrary close to $B(t)$. The proof of this theorem is similar to the proof of the fractal approximation for standard snowflakes but it is less technical.

\[thm:approximation\] For any $\epsilon$ and $t$ there are a building block $\phi\in\Sigma' {{\mathrm cap\:}}C^\infty(\{|z|\ge 1\})$ and a positive integer $k$ that define the snowflake with $\bar\beta(t)>B(t)-\epsilon$.

We will use the following lemma.

\[l:polygon\] For any $\epsilon>0$, $t\in {{\mathbb R}}$ there is $A>0$ such that for any $\delta>0$ there is a function $\phi\in\Sigma' {{\mathrm cap\:}}C^\infty $ such that $$\int \left|\phi'(re^{i\theta})\right|^t d \theta >
A \br{ \frac{1}{\delta}}^{B(t)-\epsilon}$$ for $\delta>r-1$. Moreover, capacity of $\phi$ is bounded by a universal constant that does not depend on $\delta$.

There is a function $f$ with $\beta_f(t)>B(t)-\epsilon$. Hence there is a constant $A$ such that $$\int \left| f'(r e^{i\theta}) \right|^t d\theta > A (r-1)^{-B(t)+2\epsilon}.$$ The only problem is that this function is not smooth up to the boundary. Set $\phi(z)=f(s z)$. Obviously, $\phi(z) \rightrightarrows f(z)$ as $s \to 1$. If we fix a scale $\delta$ then there is $s$ sufficiently close to $1$ such that $\int|\phi'/\phi|^t d \theta > A \delta^{-B(t)+2\epsilon}/2$. But for $r<1+\delta$ the integral can not be smaller by the subharmonicity.

It is easy to see that $${{\mathrm cap\:}}(f_n)<{{\mathrm cap\:}}(f)={{\mathrm cap\:}}(\phi)/(1-1/k)<2{{\mathrm cap\:}}(\phi),$$ hence ${{\mathrm cap\:}}(f_n)$ and $|f_n(z)|$ for $|z|<2$ are bounded by the universal constants that depend on capacity of $\phi$ only and do not depend on $k$. It also follows that $|K_k f_n(z)|<1+c/k$ for $|z|<2$ and $c$ depending on ${{\mathrm cap\:}}(\phi)$ only.

Let us fix $t$ and let $\phi$ be a function from the Lemma \[l:polygon\] for $\delta=c/k$. By $I(f,\delta)$ we denote $$\int_{-\pi}^\pi \left| f'(re^{i\theta}) \right|^t d \theta,$$ where $r=\exp(\delta)$.

The $k$-root transform changes integral means in a simple way: $$I(K_k f,\delta/k)=\int \left|\frac{f'(r^k e^{i k\theta})}{f^{(k-1)/k}(r^k e^{i k\theta})}\right|^t r^{t(k-1)}d\theta.$$ As we mentioned before, the capacity of the snowflake is bounded by the universal constant, hence $|f|$ can be bounded by a universal constant. Thus $$I(K_k f,\delta/k)>\mathrm{const}\, I(f,\delta).$$

The function $f_{n+1}$ is a composition of a (random) function $\phi_\theta$ with $Kf_n$. The expectation of $I(f_{n+1},1/k^{n+1})$ conditioned on $f_n$ is $$\begin{aligned}
{{\mathbb E}}\brb{I(f_{n+1},1/k^{n+1})\mid f_n}&=&
\int \int |\phi_\theta'(K_k f_n(re^{i\xi}))|^t|(K_k f_n)'(re^{i\xi})|^t d\xi d \theta 
\\ 
&=&\int |(K_k f_n)'(re^{i\xi})|^t \int |\phi'(e^{-i\theta} K_k f_n(re^{i\xi}))|^t d \theta d \xi,\end{aligned}$$ where $r=\exp(1/k^{n+1})$. We know that $|K_k f_n(re^{i\xi})|-1<c/k$. By our choice of $\phi$ $$\int |\phi'(|K_k f_n(re^{i\xi})|e^{-i\theta})|^t d \theta> A \br{\frac{k}{c}}^{B(t)-\epsilon}.$$ So $$\begin{aligned}
{{\mathbb E}}\brb{I(f_{n+1},1/k^{n+1})} &>& A \br{\frac{k}{c}}^{B(t)-\epsilon}
{{\mathbb E}}\brb{I(K f_n,1/k^{n+1})}
\\
&>&A \br{\frac{k}{c}}^{B(t)-\epsilon} \mathrm{const}\, {{\mathbb E}}\brb{ I(f_n,1/k^n)}.\end{aligned}$$ Applying this inequality $n$ times we obtain $${{\mathbb E}}\brb{I(f_n,1/k^n)}>\mathrm{ const}^n\,  \br{\frac{k}{c}}^{n(B(t)-\epsilon)}.$$ So $$\frac{\log{{\mathbb E}}\brb{ I(f_n,1/k^n)}}{n\log k}>B(t)-2\epsilon$$ for sufficiently large $k$. This completes the proof.

Appendix: example of an estimate {#sec:application}
================================

The main purpose of this section is to show that using conformal snowflakes it is not very difficult to find good estimates. Particularly it means that if one of the famous conjectures mentioned in the introduction is wrong, then it should be possible to find a counterexample.

In this section we will give an example of a simple snowflake and estimate its spectrum at $t=1$. We could do essentially the same computations for other values of $t$, but $B(1)$ is of special interest because it is related to the coefficient problem and Littlewood conjecture (see [@BeSmECM] for details).

As a building block we use a very simple function: a straight slit map. We use the following scheme: first we define a building block and this gives us the operator $P$. By (\[betalog\]) any positive function $\nu$ gives us an estimate on the spectrum. To choose $\nu$, we find the first eigenvector of discretized operator $P$ and approximate it by a rational function. We compute $P\nu$ using Euler’s quadrature formula and estimate the error term. The minimum of $P\nu/\nu$ gives us the desired estimate of $\beta(1)$. For $t=1$ we give the rigorous estimate of the error term in the computation of $P\nu/nu$, for other values of $t$ we give approximate values (computed with less precision) without any estimates of the error terms.

Single slit domain
------------------

We use a straight slit functions. First we define the basic slit function $$\label{slit}
\phi(z,l)=\phi_l(z)=\mu_2\br{\frac{\sqrt{\mu_1^2(z s)+l^2/(4k+4)}}{\sqrt{1+l^2/(4l+4)}}},$$ where $s$ is a constant close to $1$, $\mu_1$ and $\mu_2$ are the Möbius transformation that maps ${{\mathbb D}}_-$ onto the right half plane and its inverse: $$\begin{aligned}
\mu_1(z)=\frac{z-1}{z+1},\\
\mu_2(z)=\frac{z+1}{z-1}.\end{aligned}$$ We also need the inverse function $$\psi(z,l)=\psi_l(z)=\phi(z,l)^{-1}.$$ The function $\phi$ first maps ${{\mathbb D}}_-$ onto the right half-plane, than we cut off a straight horizontal slit starting at the origin and map it back. The image $\phi_l({{\mathbb D}}_-)$ is ${{\mathbb D}}_-$ with a horizontal slit starting from $1$. The length of the slit is $l$. The derivative of a slit map has a singularities at points that are mapped to $1$. But if we take $s>1$ then these singularities are not in ${{\mathbb D}}_-$. We set $s=1.002$.

We study the snowflake generated by $\phi(z)=\phi_{73}(z)$ with $k=13$ (numbers $13$ and $73$ are found experimentally). Figure \[pic2\] show the image of the unit circle under $f_3$. The Figure \[pic3\] shows the image of a small arc under $\bar f_3$ and three Green’s lines.

First we have to find the critical radius $R$ such that $D_R\subset \psi^k(D_R)$. By symmetry of $\phi$, the critical radius is the only positive solution of $$\psi^k(x)=x.$$ This equation can not be solved explicitly, but we can solve it numerically (we don’t care about error term since we can take any greater value of $R$). The approximate value of $R$ is $ 76.1568$. To be on the safe side we fix $R=76.2$. The disc takes just a small portion of $\psi(D_R)$ which means that there is a huge overkill in the inequality (\[cover\]).

By (\[betalog\]) any positive function $\nu$ gives a lower bound of spectrum. And this estimate is sharp when $\nu$ is the main eigenfunction of $P$. So we have to find an “almost” eigenfunction of $P$.

Almost eigenfunction of operator P
----------------------------------

Even for such a simple building block we can not find the eigenfunction explicitly. Instead we look for some sort of approximation. The first idea is to substitute integral operator $P$ by its discretized version. Here we use the most simple and quite crude approximation.

Choose sufficiently large $N$ and $M$. Let $r_n=1+(R-1)n/N$ and $\theta_m=
2\pi m /M$. Instead of $P$ we have an $N\times N$ matrix with elements $$P_{n,n'}=\sum_{m} r_n^{1-t(k-1)/k}\frac{|\phi'(r_n^{1/k}e^{i\theta_{m}})|^t}
{|\phi(r_n^{1/k}e^{i\theta_{m}})|M},$$ where summation is over all indexes $m$ such that $r_{n'}$ is the nearest point to $|\phi(r_n^{1/k}e^{i\theta_m})|$. This defines the discretized operator $P_N$. Let $\lambda_N$ and $V_N$ be the main eigenvalue and the corresponding eigenvector. A priori, $\lambda_N$ should converge to $k^{\beta(t)}$, but it is not easy to prove and not clear how to find the rate of convergence. But this crude estimate gives us the fast test whether the pair $\phi$ and $k$ defines a snowflake with large spectrum or not (this is the way how we found $k=13$ and $l=73$).

Instead of proving convergence of $\lambda_N$ and estimating the error term we will study $V_N$ which is the discrete version of the eigenfunction. We approximate $V_N$ by a rational function of a relatively small degree (in our case $5$), or by any other simple function. In our case we find the rational function by the linear least square fitting. In any way we get a nice and simple function $\nu$ which is supposed to be close to the eigenfunction of $P$.

We would like to note that procedure, by which we obtained $\nu$, is highly non rigorous, but that does not matter since as soon as we have some explicit function $\nu$ we can plug it into $P$ and get the rigorous estimate of $\beta$.

In our case we take $N=1000$ and $M=500$. The logarithm of the first eigenvalue is $0.2321$ (it is $0.23492$ if we take $s=1$). Figure \[eigenvectors\] shows a plot with coordinates of the first eigenvector.

![image](eigenvector.eps){width="8cm"}

We scale this data from $[1,1000]$ to the interval $[1,R]$ and approximate by a rational function $\nu$: $$\begin{aligned}
\nu(x)=(7.1479+8.9280 x - 0.07765 x^2+ 1.733 \times 10^{-3} x^3 - 
\\ 2.0598 \times 10^{-5} x^4 + 
9.5353 \times 10^{-8}x^5)/( 2.7154+ 13.2845  x).
\end{aligned}$$ Figure \[nu\] shows the plot of $\nu$.

![An “almost” eigenfunction[]{data-label="nu"}](nu.eps){width="8cm"}

Estimates of derivatives
------------------------

To estimate $\beta(1)$ we have to integrate $\nu(|\phi|)|\phi'|/|\phi|$. It is easy to see that the main contribution to the derivative is given by a factor $|\phi'|/|\phi|$. Assume for a while that $s=1$. The fraction $|\phi'(z)/\phi(z)|$ can be written as $$\label{fraction}
\frac{|z-1|}{|z|\sqrt{|(z-z_1)(z-z_2)}},$$

where $$\begin{aligned}
z_1=\frac{-5033-292 i \sqrt{74}}{5625}\approx -0.894756 - 0.446556 i,
\\
z_2=\frac{-5033+292 i \sqrt{74}}{5625}\approx -0.894756 + 0.446556 i.
\end{aligned}$$

Singular points $z_1$ and $z_2$ are mapped to $1$ and $\phi'$ has a square root type singularity at these points. They will play essential role in all further calculations. We introduce notation $z_1=x+i y$ and $z_2=x-i y$, for $z$ we will use polar coordinates $z=r e^{i\theta}$.

We compute the integral of $f=|\nu(\phi)\phi'/\phi|$ using the Euler quadrature formula based on the trapezoid quadrature formula $$\int_0^{2\pi} f(x)dx \approx S_\epsilon^n(f)=
S_\epsilon(f) -\sum_{k=1}^{n-1} \gamma_{2k}\epsilon^{2k}\br{f^{(2k-1)}(2\pi)-f^{(2k-1)}(0)},$$ where $S_\epsilon(f)$ is a trapezoid quadrature formula with step $\epsilon$ and $\gamma_k=B_k/k!$ were $B_k$ is the Bernoulli number. The error term in the Euler formula is $$-\gamma_{2n}\max f^{(2n)}\epsilon^{2n}2\pi.
\label{error1}$$ In our case function $f$ is periodic and terms with higher derivatives vanish. This means that we can use (\[error1\]) for any $n$ as an estimate of the error in the trapezoid quadrature formula.

Function $\phi$ has two singular points: $z_1$ and $z_2$. Derivative of $\phi$ blows up near these points. This is why we introduce scaling factor $s$. We can write a power series of $\phi$ near $z_1$ (near $z_2$ situation is the same by the symmetry) $$\phi^{(k)}=c_{-k}(z-z_1)^{-k+1/2}+c_{-k+1}(z-z_1)^{-k+3/2}+\dots+c_0+\dots\ .$$ This means that for $s>1$ derivative can be estimated by $$|c_{-k}|(s-1)^{-k+1/2}+|c_{k+1}|(s-1)^{-k+3/2}+\dots\ .$$ Tail of this series can be estimated because series converges in a disc of a fixed radius (radius is $|z_1+1|$), and sum of tail can be estimated by a sum of a geometric progression. Writing these power series explicitly we find (for $s=1.002$) $$\begin{aligned}
{3}
|\phi'| & <  55, & \quad  
|\phi''| & <  11800, & \quad
|\phi^{(3)}| &<  8.69\times 10^6,\\
|\phi^{(4)}| & <  1.08 \times 10^{10},  &\quad
|\phi^{(5)}| & <  1.90 \times 10^{13}, &\quad 
|\phi^{(6)}| & <  4.25 \times 10^{16}, \\
|\phi^{(7)}| & <  1.17 \times 10^{20}.  &\quad &  & &\end{aligned}$$

The maximal values for first six derivatives of $\nu$ are $$\begin{aligned}
{3}
|\nu'|&<0.28, &\quad
|\nu''|&<0.45, &\quad
|\nu^{(3)}|&<1.12, \\
|\nu^{(4)}|&<3.69, &\quad
|\nu^{(5)}|&<15.3,  &\quad
|\nu^{(6)}|&<76.2.\end{aligned}$$

The derivative $\partial_\theta|\phi|$ can be estimated by $r |\phi'|$. We can write sixth derivative of $\nu(|\phi|)|\phi'|/|\phi|$ as a rational function of partial derivatives of $|\phi|$, $|\phi'|$, and $\nu$. Than we apply triangle inequality and plug in the above estimates. Finally we have $$\left|\frac{\partial\br{\frac{\nu(|\phi|)|\phi'|}{|\phi|}}}{\partial \theta^6}\right|<1.65\times 10^{21}.$$ Plugging $\epsilon=\pi/5000$ and estimate on sixth derivative into (\[error1\]) we find that error term in this case is less than $0.0034$.

Next we have to estimate modulus of continuity with respect to $r$. First we calculate $$\partial_r|z-(a+b i)|^2=2 r - 2 (a\cos\theta+b\sin\theta).$$ Applying this formula several times we find $$\begin{aligned}
\partial_r\br{\frac{|\phi'|}{|\phi|}}=
\partial_r \br{\frac{|z-1|}{r\sqrt{|z-z_1||z-z_2|}}} \le
\partial_r \br{\frac{|z-1|}{\sqrt{|z-z_1||z-z_2|}}} 
\\
=
\frac{r-\cos\theta}{|z-1|S}-\frac{r-x\cos\theta-y\sin\theta}{2|z-z_1|^2S}|z-1|-
\frac{r-x\cos\theta+y\sin\theta}{2|z-z_2|^2S}|z-1|,
\end{aligned}$$ where $S=\sqrt{|z-z_1||z-z_2|}$. Factoring out $$\frac{1}{2|z-1|\cdot|z-z_1|^{5/2}|z-z_2|^{5/2}}$$ we get

$$\begin{aligned}
2(r-\cos\theta)|z-z_1|^2|z-z_2|^2
-(r-x\cos\theta-y\sin\theta)|z-1|^2|z-z_2|^2
\\
-(r-x\cos\theta+y\sin\theta)|z-1|^2|z-z_1|^2
\\
=-2(r^2-1)(2\cos^2\theta r (x-1)+\cos\theta (r^2+1)(x-1)+2 r y^2).
\end{aligned}$$ This is a quadratic function with respect to $\cos\theta$. Taking values of $x$ and $y$ into account we can write it as $$\cos^2\theta+\cos\theta\br{r+\frac{1}{r}}\frac{1}{2}-\frac{592}{5625}~.$$ This quadratic function has two real roots. Their average is $-(r+1/r)/2<-1$, hence one root is definitely less than $-1$. The product of roots is a small negative number, which meant that the second root is positive and less than $1$. Simple calculation shows that this root decreases as $r$ grows. This means that the corresponding value of $\theta$ increases. Hence it attains its maximal value at $r=1.4$ and the maximal value is at most $1.48$. This gives us that the radial derivative of $|\phi'|/|\phi|$ can be positive only on the arc $\theta\in[-1.48,1.48]$. By subharmonicity it attains the maximal on the boundary of $\{z\mid 1<r<1.4,\ -1.48<\theta<1.48\}$. It is not very difficult to check that maximum is at $z=1.4$ and it is equal to $0.36$.

Let $$I(r)=\int_{-\pi}^\pi \nu(|\phi(r^{1/k}e^{i\theta})|
\left|\frac{\phi'(r^{1/k}e^{i\theta})}{\phi(r^{1/k}e^{i\theta})}\right| \frac{d \theta}{2\pi}.$$ The derivative is $$I'(r)=\frac{1}{k r^{1-1/k}}\br{\int_{-\pi}^\pi
 \nu'(|\phi|)\partial_r |\phi| \frac{|\phi'|}{|\phi|}\frac{d \theta}{2\pi}+
\int_{-\pi}^\pi \nu(|\phi|)\partial_r\br{\frac{|\phi'|}{|\phi|}}\frac{d \theta}{2\pi}}.$$ By the symmetry the first integral is zero. In the second integral $$\nu(|\phi|)\partial_r\br{\frac{|\phi'|}{|\phi|}}$$ can be positive only when $\theta\in[-1.48,1.48]$ and even in this case it is bounded by $0.36/(r^{1-1/k} k 2\pi)$. Hence $$I'(r)<2 \cdot 1.48 \cdot 0.36/(r^{1-1/k}k 2\pi)<0.0131 r^{1/k-1}.$$ If we compute values $I(r_1)$ and $I(r_2)$ (with precision $0.0034$) then the minimum of $P(\nu)/\nu$ on $[r_1,r_2]$ is at least $$r_1^{1/k}(\min\{I(r_1),I(r_2)\}-0.0034-0.0131(r_2-r_1) r^{1/k-1})/\nu(r_1).
\label{int}$$

![Values of $I(r)$[]{data-label="integral"}](int.eps)

![Plot of $\log(P\nu/\nu)/\log k$[]{data-label="log"}](log.eps)

We take $3000$ equidistributed points on $[1,R]$ and compute $I(r)$ at these points. The data for $I(r)$ is shown on the Figure \[integral\]. Applying the error estimate (\[int\]) we find a rigorous estimate from below of $P\nu/\nu$. The minimum of $P\nu/\nu$ is at least $1.8079$ which means that $$\beta(1)>0.2308.$$ The Figure \[log\] shows the plot of $\log(P\nu/\nu)/\log k$.

Estimates of spectrum for other values of $t$
---------------------------------------------

We also computed lower bound on spectrum of the same snowflake for other values of $t$. Below are given base $13$ logarithms of eigenvalues of discretized operator $P$ ($N=1000$, $M=400$), lower bounds of $\log(P\nu/\nu)/\log k$, $t^2/4$ and upper bound of the universal spectrum from [@HeSh; @MaPo]. For values of $t$ close to zero we can not find a function that gives us a positive lower bound.

  $t$    $\log_{13} \lambda$   $\beta(t)>$   $t^2/4$   $\beta(t)<$
  ------ --------------------- ------------- --------- -------------
  -2.0   0.6350                0.56          1         1.218
  -1.8   0.5348                0.48          0.81      1.042
  -1.6   0.4395                0.39          0.64      0.871
  -1.4   0.3502                0.31          0.49      0.706
  -1.2   0.2678                0.220         0.36      0.549
  -1.0   0.1936                0.152         0.25      0.403
  -0.8   0.1290                0.0925        0.16      0.272
  -0.6   0.0756                0.0430        0.09      0.159
  -0.4   0.0353                0.0050        0.04      0.072
  -0.2   0.0100                0             0.01      0.0179
  0.2    0.0105                0             0.01      0.031
  0.4    0.0387                0.0280        0.04      0.184
  0.6    0.0858                0.0795        0.09      0.276
  0.8    0.1515                0.1505        0.16      0.368
  1.0    0.234                 0.234         0.25      0.460
  1.2    0.334                 0.332         0.36      0.613
  1.4    0.448                 0.442         0.49      0.765
  1.6    0.576                 0.570         0.64      0.843
  1.8    0.713                 0.698         0.81      0.921
  2.0    0.859                 0.821         1         1
