On Rate of Convergence for Universality Limits

Abstract

Given a probability measure $\mu$ on the unit circle ${\mathbb {T}}$, consider the reproducing kernel $k_{\mu ,n}(z_1, z_2)$ in the space of polynomials of degree at most $n-1$ with the $L^2(\mu )$–inner product. Let $u, v \in {\mathbb {C}}$. It is known that under mild assumptions on $\mu$ near $\zeta \in \mathbb {T}$, the ratio $k_{\mu ,n}(\zeta e^{u/n}, \zeta e^{v/n})/k_{\mu ,n}(\zeta , \zeta )$ converges to a universal limit S(u, v) as $n \rightarrow \infty$. We give an estimate for the rate of this convergence for measures $\mu$ with finite logarithmic integral.

1 Introduction

Consider a probability measure $\mu$ on the unit circle ${\mathbb {T}}= \{z \in {\mathbb {C}}: \; |z| = 1\}$ of the complex plane, ${\mathbb {C}}$. Assume that the support of $\mu$ is an infinite subset of ${\mathbb {T}}$, so that monomials $z^k$, $k \geqslant 0$, are linearly independent in $L^2(\mu )$. For each integer $n \geqslant 1$, the set of polynomials of degree at most $n-1$,

$$\begin{aligned} {\mathcal {P}}_{n} = \mathop {\textrm{span}}\nolimits \{z^k,\; k=0,\ldots , n-1\}, \end{aligned}$$

can be viewed as the n-dimensional Hilbert space of analytic functions with respect to $L^2(\mu )$-inner product. Denote by $k_{\mu ,n}(z_1, z_2)$ the reproducing kernel at a point $z_2 \in {\mathbb {C}}$ in this space, i.e., $k_{\mu ,n}(\cdot , z_2) \in {\mathcal {P}}_{n}$ and

$$\begin{aligned} (p, k_{\mu ,n}(\cdot , z_2))_{L^2(\mu )} = p(z_2), \quad p \in {\mathcal {P}}_{n}. \end{aligned}$$

If $\mu = m$ is the Lebesgue measure on ${\mathbb {T}}$ normalized by $m({\mathbb {T}}) = 1$, the reproducing kernel has the following form:

$$\begin{aligned} k_{m,n}(z_1, z_2) = \frac{1 - \overline{z_2}^{n} z_1^{n}}{1 - \overline{z_2} z_1}. \end{aligned}$$

One might check that if $z_1 = \zeta e^{u/n}$, $z_2 = \zeta e^{v/n}$ for some $\zeta \in {\mathbb {T}}$ and $u, v \in {\mathbb {C}}$ (equivalently, $z_1$, $z_2$ are at a distance comparable to 1/n from $\zeta$), then

$$\begin{aligned} \frac{k_{m,n}(z_1, z_2)}{k_{m,n}(\zeta , \zeta )} = \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)} = \frac{e^{u+{\bar{v}}} - 1}{u+{\bar{v}}} + O\left( \frac{1}{n}\right) , \end{aligned}$$

where the remainder is uniform in (u, v) on compact subsets of ${\mathbb {C}}\times {\mathbb {C}}$. Such kind of behaviour of reproducing kernels is universal: under mild assumptions on a measure $\mu$ near $\zeta \in {\mathbb {T}}$, we have

$$\begin{aligned} \frac{k_{\mu ,n}(\zeta e^{u/n},\, \zeta e^{v/n})}{k_{\mu ,n}(\zeta , \zeta )} \rightarrow \frac{e^{u+{\bar{v}}} - 1}{u+{\bar{v}}}, \quad n \rightarrow +\infty , \end{aligned}$$

uniformly in (u, v) on compact subsets of ${\mathbb {C}}\times {\mathbb {C}}$. Universality of the limiting behaviour of ratios of reproducing kernels attracted major attention in recent years. Several essentially different approaches were developed. Let us mention some of them. First papers dealt with real analytic weights and used the Riemann-Hilbert method, see, e.g., Deift [6] or Kuijlaars and Vanlessen [9]. Lubinsky [10] found a way of reducing a wide class of universality problems to the study of asymptotic behaviour of $k_{\mu ,n}(z,z)$, $z \in {\mathbb {T}}$. The latter asymptotic behaviour has been previously identified for general measures of Szegő class by Máté et al. [12]. Global Szegő condition has been weakened to the local one by Findley [8]. Another approach, also pioneered by Lubinsky [11], is based on compactness of normal families of entire functions and properties of $\frac{\sin x}{x}$ kernel. An overview of this approach and further results can be found in Simon [15] and Totik [16]. Recently, Eichinger et al. [7] found yet another approach to universality based on spectral theory of canonical Hamiltonian systems. While this approach gives extremely general results (even the local Szegő condition can be omitted), it also involves a compactness argument as an essential element of the proof. Most of mentioned papers deal with measures on subsets of the real line due to motivation in the theory of random matrices. However, even in the simplified setting of measures on the unit circle, estimates for the rate of convergence

$$\begin{aligned} \frac{k_{\mu ,n}(z_1, z_2)}{k_{\mu ,n}(\zeta , \zeta )} \rightarrow \frac{e^{u+{\bar{v}}} - 1}{u+{\bar{v}}} \end{aligned}$$

(1.1)

are missing in the literature. In fact, the rate of convergence in (1.1) is not known even in the case where $\mu$ is an absolutely continuous measure on ${\mathbb {T}}$ with a smooth non-vanishing weight w. Indeed, compactness arguments, widely used for proving universality, cannot give bounds for the rate of convergence.

As an additional motivation of this work, we mention that Poltoratski [13] recently used universality in the proof of convergence of certain nonlinear Fourier transform (NLFT), and a subsequent development of this area, e.g., bounds for NLFT maximal operators, will require estimates for the convergence of universality limits.

In this paper, we estimate the rate of convergence in (1.1) for probability measures on the unit circle with finite logarithmic integral. For this we use an entropy function of a measure—a powerful instrument that recently found several applications in inverse problems [2, 3], scattering theory [5], and orthogonal polynomials [1, 4]. To be precise, let $\mu$ be a probability measure on ${\mathbb {T}}$, and let $\mu = w\,dm + \mu _{{\textbf{s}}}$ be its Radon-Nikodym decomposition into the absolutely continuous and singular parts. The measure $\mu$ is said to belong to the Szegő class ${\textrm{Sz}}({\mathbb {T}})$ if its logarithmic integral is finite:

$$\begin{aligned} \int _{{\mathbb {T}}}\log w\, dm > -\infty . \end{aligned}$$

Since $\log x \leqslant x$ for all $x>0$, the latter condition is equivalent to $\log w \in L^1(m)$. Let ${\mathbb {D}}= \{z \in {\mathbb {C}}:\; |z| < 1\}$ denote the open unit disk. For a measure $\mu \in {\textrm{Sz}}({\mathbb {T}})$, its entropy function is given by

$$\begin{aligned} {\mathcal {K}}_{\mu }(z) = \log \left( \int _{{\mathbb {T}}}\frac{1-|z|^2}{|1-\bar{\xi }z|^2}\,d\mu (\xi )\right) - \int _{{\mathbb {T}}}\log w(\xi )\frac{1-|z|^2}{|1-\bar{\xi }z|^2}\,dm(\xi ), \quad z \in {\mathbb {D}}. \end{aligned}$$

The function ${\mathcal {K}}_{\mu }$ is nonnegative in ${\mathbb {D}}$ by Jensen’s inequality. Moreover, for m-almost all $\zeta \in {\mathbb {T}}$, we have ${\mathcal {K}}_{\mu }(z) \rightarrow 0$ as z non-tangentially approaches $\zeta$. This follows from well-known properties of the Poisson kernel: we have

$$\begin{aligned} \lim _{r \rightarrow 1}\int _{{\mathbb {T}}}\frac{1-r^2}{|1-\bar{\xi }r\zeta |^2}\,d\mu _{{\textbf{s}}}(\xi ) = 0 \end{aligned}$$

(1.2)

for m-almost all $\zeta \in {\mathbb {T}}$, and

$$\begin{aligned}{} & {} \lim _{r \rightarrow 1}\int _{{\mathbb {T}}}w(\xi )\frac{1-r^2}{|1-\bar{\xi }r\zeta |^2}\,dm(\xi ) \rightarrow w(\zeta ), \end{aligned}$$

(1.3)

$$\begin{aligned}{} & {} \lim _{r \rightarrow 1}\int _{{\mathbb {T}}}\log w(\xi )\frac{1-r^2}{|1-\bar{\xi }r\zeta |^2}\,dm(\xi ) \rightarrow \log w(\zeta ), \end{aligned}$$

(1.4)

at each Lebesgue point $\zeta$ of functions $w, \log w \in L^1(m)$. In case (1.2)–(1.4) are satisfied, we have ${\mathcal {K}}_\mu (r\zeta ) \rightarrow 0$ as $r \rightarrow 1$. Let $B(\zeta , r) = \{z \in {\mathbb {C}}:\; |z-\zeta | \leqslant r\}$. The following theorem is the main result of the paper.

Theorem 1.1

Let $\mu \in {\textrm{Sz}}({\mathbb {T}})$, $A \geqslant 1$, $n \geqslant 10A$, $\zeta \in {\mathbb {T}}$. There exists $\varepsilon _0 > 0$ depending only on A, such that if $z_{1,2} \in B(\zeta , A/n)$ and ${\mathcal {K}}_{\mu }(\rho \zeta ) \leqslant \varepsilon _0$ for all $\rho \in [1-A/n, 1)$, then

$$\begin{aligned} \left| \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \leqslant c e^{4A}\sup _{\rho \in [1-\delta , 1) }\sqrt{{\mathcal {K}}_{\mu }(\rho \zeta )}, \end{aligned}$$

(1.5)

where $\delta = \max \limits _{k=1,2}|z_{k} - \zeta |$, and the constant $c>0$ is absolute.

Note that $\rho \in [1- A/n, 1)$ in Theorem 1.1 tends to 1 as $n \rightarrow \infty$, therefore, the right hand side of (1.5) tends to zero for all $\zeta$ satisfying (1.2)–(1.4). This gives a nontrivial bound for the rate of convergence in (1.1) for Lebesgue almost all $\zeta \in {\mathbb {T}}$. If $\mu$ has some regularity in a neighbourhood of $\zeta \in {\mathbb {T}}$, its entropy function can be explicitly estimated. For functions $f, g > 0$, we use notation $f \lesssim g$ (resp., $f \gtrsim g$) if $f \leqslant cg$ (resp., $f \leqslant cg$) for some constant c, and $f \sim g$ if both relations $f \lesssim g$ and $f \gtrsim g$ are satisfied.

Theorem 1.2

Let $\mu = w\,dm$ be an absolutely continuous probability measure in ${\textrm{Sz}}({\mathbb {T}})$ such that w is positive and continuous in a neighbourhood $I \subset {\mathbb {T}}$ of $\zeta \in {\mathbb {T}}$. Assume, moreover, that $|w(\xi ) - w(\zeta )| \sim |\xi - \zeta |^{s}$ for all $\xi \in I$ and some $s> 0$. Then we have

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta ) \sim {\left\{ \begin{array}{ll} 1-\rho , &{} s \in (1/2, +\infty ),\\ (1-\rho )|\log (1-\rho )|, &{} s =1/2,\\ (1-\rho )^{2s}, &{} s \in (0, 1/2), \end{array}\right. } \end{aligned}$$

for $\rho \in (0, 1)$ close enough to 1. The constants involved depend on s, the diameter of I, the value $w(\zeta )$, and the constants in the relation $|w(\xi ) - w(\zeta )| \sim |\xi - \zeta |^{s}$.

Let $\lambda \in {\mathbb {D}}$. For the absolutely continuous probability measure $\mu = \frac{1 - |\lambda |^2}{|1-\lambda \xi |^2}\,dm$, we have

$$\begin{aligned} {\mathcal {K}}(\mu , z)&= \log \mathop \textrm{Re}\left( \frac{1+\lambda z}{1-\lambda z}\right) - \log \left( \frac{1-|\lambda |^2}{|1-\lambda z|^2}\right) \\&= \log \frac{1-|\lambda z|^2}{|1-\lambda z|^2} - \log \left( \frac{1-|\lambda |^2}{|1-\lambda z|^2}\right) = \log \frac{1-|\lambda z|^2}{1-|\lambda |^2}, \end{aligned}$$

due to the fact that integration against the Poisson kernel corresponds to harmonic continuation into the unit disk. We see that ${\mathcal {K}}(\mu , (1-1/n)\zeta ) \sim 1/n$ as $n \rightarrow \infty$. Note that this agrees with bounds in Theorem 1.2 (we have $s=1$ for this measure). By Theorem 1.1, we then have

$$\begin{aligned} \left| \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \lesssim \frac{1}{\sqrt{n}} \end{aligned}$$

for all $z_1$, $z_2$ in $B(\zeta , 1/n)$ and large enough $n \geqslant 0$, uniformly in $\zeta \in {\mathbb {T}}$. As we will see in Sect. 4, in fact

$$\begin{aligned} \sup _{|z_{1,2}-\zeta | \leqslant 1/n}\left| \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \sim \frac{1}{n}. \end{aligned}$$

This shows that the bound in Theorem 1.1 is not sharp for smooth measures. It seems, however, that this bound cannot be improved in the setting of the whole class ${\textrm{Sz}}({\mathbb {T}})$ of measures with finite logarithmic integral, i.e., there is a measure $\mu \in {\textrm{Sz}}({\mathbb {T}})$ such that

$$\begin{aligned} \sup _{|z_{1,2} - 1| \leqslant 1/n}\left| \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(1, 1)} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \gtrsim \sqrt{{\mathcal {K}}_{\mu }(1-1/n)}, \end{aligned}$$

for all $n \geqslant 1$. In Sect. 5, we consider the absolutely continuous measures $w_s\,dm$ such that $w_s(e^{i\theta }) = c_{s}e^{|\theta |^{s}}$, $\theta \in [-\pi , \pi ]$, where the constant $c_s$ is chosen so that $\int _{{\mathbb {T}}} w_s \,dm = 1$. By Theorem 1.2, we have

$$\begin{aligned} {\mathcal {K}}_{w_s}(1 - 1/n) \sim n^{-2s}, \quad s \in (0, 1/2). \end{aligned}$$

We demonstrate numerically that for $x_n = 1- n^{-1}$, and fixed $s \in (0, 1/2)$, we have

$$\begin{aligned} \left| \frac{k_{w_s, n}(x_n, x_n)}{k_{w_s, n}(1, 1)} - \frac{1 - |x_n|^{2n}}{n(1 - |x_n|^2)}\right| \gtrsim n^{-s}. \end{aligned}$$

In other words, for each $s \in (0,1/2)$, we have

$$\begin{aligned} \sup _{|z_{1,2} - 1| \leqslant 1/n}\left| \frac{k_{w_s, n}(z_1, z_2)}{k_{w_s, n}(1, 1)} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \gtrsim \sqrt{{\mathcal {K}}_{w_s}(1-1/n)}, \end{aligned}$$

and estimate (1.5) in Theorem 1.1 is sharp on this class of examples. It remains an open problem to give a mathematical proof of this fact.

2 Proof of Theorem 1.1

Let $\mu$ be a probability measure supported on an infinite subset of ${\mathbb {T}}$, and let $\{\varphi _n\}_{n \geqslant 0}$ be the family of its orthonormal polynomials obtained by Gram–Schmidt orthogonalization of monomials $z^n$, $n \geqslant 0$, in $L^2(\mu )$. For a polynomial p of degree n, we set $p^*(z) = z^n\overline{p(1/ {\bar{z}})}$. Note that p is also a polynomial of degree at most n. The polynomials $\{\varphi _n^*\}_{n \geqslant 0}$ are called reflected orthonormal polynomials. We have the following recurrence relation (see formula (1.5.25), page 58, in [14]):

$$\begin{aligned} \varphi _{n} = \frac{z\varphi _{n-1} - \overline{a_{n-1}}\varphi ^{*}_{n-1}}{\sqrt{1-|a_{n-1}|^2}}, \quad n \geqslant 1. \end{aligned}$$

Here the recurrence coefficients, $a_n$, $n\geqslant 0$, belong to ${\mathbb {D}}$. It is also known (see Theorem 2.2.7, p. 124, in [14]) that the reproducing kernel in the n-dimensional space Hilbert space $({\mathcal {P}}_n, (\cdot , \cdot )_{L^2(\mu )})$ at $z_2 \in {\mathbb {C}}$ is given by

$$\begin{aligned} k_{\mu , n}(z_1, z_2) = \sum _{k=0}^{n-1}\overline{\varphi _{k}(z_2)}\varphi _{k}(z_1) = \frac{\overline{\varphi _{n}^*(z_2)}\varphi _{n}^*(z_1) - \overline{\varphi _{n}(z_2)}\varphi _{n}(z_1)}{1 - {\bar{z}}_2 z_1}. \end{aligned}$$

(2.1)

Note that $k_{\mu , n}(z_1, z_2)$ is indeed an element of ${\mathcal {P}}_n$, i.e., a polynomial with respect to $z_1$ of degree at most $n-1$. It will be convenient to use a different representation of the reproducing kernel. Take $n \geqslant 1$, $a \in {\mathbb {D}}$, and define

$$\begin{aligned} \tilde{\varphi }_{n} = \frac{z\varphi _{n-1} - {\bar{a}} \varphi ^*_{n-1}}{\sqrt{1-|a|^2}}, \quad \tilde{\varphi }_{n}^* = \frac{\varphi ^{*}_{n-1} - z a \varphi _{n-1}}{\sqrt{1-|a|^2}}. \end{aligned}$$

(2.2)

Lemma 2.1

For all $z_1, z_2 \in {\mathbb {C}}$, we have

$$\begin{aligned} \overline{\tilde{\varphi }_{n}^*(z_2)}\tilde{\varphi }_{n}^*(z_1) - \overline{\tilde{\varphi }_{n}(z_2)}\tilde{\varphi }_{n}(z_1) = \overline{\varphi _{n}^*(z_2)}\varphi _{n}^*(z_1) - \overline{\varphi _{n}(z_2)}\varphi _{n}(z_1) \end{aligned}$$

(2.3)

Proof

The proof is a direct computation. At first, let $z_1 = z_2 = z$. Then the left hand side in (2.3) is equal to

$$\begin{aligned} |\tilde{\varphi }_{n}^*(z)|^2 - |\tilde{\varphi }_{n}(z)|^2&= \frac{|\varphi _{n-1}^*(z) - za\varphi _{n-1}(z)|^2 - |z\varphi _{n-1}(z) - {\bar{a}}\varphi _{n-1}^*(z)|^2}{1-|a|^2}\\&= |\varphi _{n-1}^*(z)|^2 - |z\varphi _{n-1}(z)|^2, \end{aligned}$$

which does not depend on a. Taking a to be the recurrence coefficient $a_{n-1}$, we see that

$$\begin{aligned} |\tilde{\varphi }_{n}^*(z)|^2 - |\tilde{\varphi }_{n}(z)|^2 = |\varphi _{n}^*(z)|^2 - |\varphi _{n}(z)|^2. \end{aligned}$$

(2.4)

This relation holds for all $z \in {\mathbb {C}}$. Since functions in (2.3) are analytic in $z_1$ and anti-analytic in $z_2$, the lemma follows. $\square$

The following lemma is Corollary 4 in [4].

Lemma 2.2

For every $\lambda \in {\mathbb {D}}$ there is $a \in {\mathbb {D}}$ such that the corresponding polynomial $\tilde{\varphi }_{n}^*$ in (2.2) defines a probability measure $\nu _{n,\lambda } = |\tilde{\varphi }_{n}^*|^{-2}\,dm$ on ${\mathbb {T}}$ such that

$$\begin{aligned} {\mathcal {K}}_{\nu _{n,\lambda }}(\lambda ) \leqslant {\mathcal {K}}_{\mu }(\lambda ). \end{aligned}$$

(2.5)

In the rest of the paper, we use notation $\tilde{\varphi }_{n}^*$ for the polynomial from Lemma 2.2, where the value of the parameter $\lambda \in {\mathbb {D}}$ will be clear from the context.

Lemma 2.3

Let $\lambda \in {\mathbb {D}}$, and let $\tilde{\varphi }_{n}^*$ be the corresponding polynomial from Lemma 2.2. We have

$$\begin{aligned} \int _{{\mathbb {T}}}\left| \frac{\tilde{\varphi }_{n}^*(\lambda )}{\tilde{\varphi }_{n}^*(\xi )} - 1\right| ^2 \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm(\xi ) = e^{{\mathcal {K}}_{\nu _{n,\lambda }}(\lambda )} - 1. \end{aligned}$$

(2.6)

Proof

By (2.1) and Lemma 2.1, we have

$$\begin{aligned} \frac{|\tilde{\varphi }_{n}^*(z)|^2 - |\tilde{\varphi }_{n}(z)|^2}{1-|z|^2} = \sum _{k=0}^{n-1}|\varphi _{k}(z)|^2, \quad z \in {\mathbb {C}}. \end{aligned}$$

(2.7)

It follows that the function $|\tilde{\varphi }_{n}^*|^2 - |\tilde{\varphi }_{n}|^2$ is positive in ${\mathbb {D}}$ and is comparable to $1-|z|$ when z approaches ${\mathbb {T}}$. Therefore, $\tilde{\varphi }_{n}^*$ has no zeroes in ${\mathbb {D}}$. In fact, it has no zeroes in $\overline{{\mathbb {D}}} = {\mathbb {D}}\cup {\mathbb {T}}$ (if $\tilde{\varphi }_{n}^*(z_0) = 0$ at some $z_0 \in {\mathbb {T}}$, then $1-|z| \lesssim |\tilde{\varphi }_{n}^*|^2 - |\tilde{\varphi }_{n}|^2 \lesssim |\tilde{\varphi }_{n}^*|^2 \lesssim |z - z_0|^2$ near $z_0$, leading to a contradiction). It follows that the function $z \mapsto \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)}$ is analytic in a neighbourhood of $\overline{{\mathbb {D}}}$. Then the Poisson formula

$$\begin{aligned} u(\lambda ) = \int _{{\mathbb {T}}}u(\xi ) \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm(\xi ) \end{aligned}$$

(2.8)

for harmonic functions implies

$$\begin{aligned} \int _{{\mathbb {T}}}\left| \frac{\tilde{\varphi }_{n}^*(\lambda )}{\tilde{\varphi }_{n}^*(\xi )} - 1\right| ^2 \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm = \int _{{\mathbb {T}}}\left| \frac{\tilde{\varphi }_{n}^*(\lambda )}{\tilde{\varphi }_{n}^*(\xi )}\right| ^2 \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm - 1, \end{aligned}$$

(2.9)

after noting that the function $u = -2\mathop \textrm{Re}\Bigl (\frac{\tilde{\varphi }_{n}^*(\lambda )}{\tilde{\varphi }_{n}^*}\Bigr ) + 1$ is harmonic in a neighbourhood of $\overline{{\mathbb {D}}}$, $u(\lambda ) = -1$. Observe that

$$\begin{aligned} \int _{{\mathbb {T}}}\left| \frac{\tilde{\varphi }_{n}^*(\lambda )}{\tilde{\varphi }_{n}^*(\xi )}\right| ^2 \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm = |\tilde{\varphi }_{n}^*(\lambda )|^2 \int _{{\mathbb {T}}}\frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,d\nu _{n,\lambda } = e^{{\mathcal {K}}_{\nu _{n,\lambda }}(\lambda )},\nonumber \\ \end{aligned}$$

(2.10)

because $|\tilde{\varphi }_{n}^*(\lambda )|^2 = \exp \bigl (\int _{{\mathbb {T}}}\log |\tilde{\varphi }_{n}^*(\xi )|^2 \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm\bigr )$ (we use again formula (2.8), this time – for the harmonic function $u = \log |\tilde{\varphi }_{n}^*|^2$). The lemma now follows from (2.9) and (2.10). $\square$

Given two points $\xi _{\pm } \in {\mathbb {T}}$, $|\xi _+ - \xi _-| < 2$, and a number $r<1$, we denote by $\Gamma (\xi _{\pm }, r)$ the path in ${\mathbb {D}}$ formed by the union of two line segments $\{\rho \xi _{\pm }, \; \rho \in [r, 1)\}$ and the smaller arc of the circle $|z| = r$ with endpoints $r\xi _-$, $r\xi _{+}$. We also let $z^* = 1/{\bar{z}}$ for $z \in {\mathbb {C}}\setminus \{0\}$, and

$$\begin{aligned} \Gamma ^*(\xi _{\pm }, r) = \{z \in {\mathbb {C}}:\; z^* \in \Gamma (\xi _{\pm }, r)\}. \end{aligned}$$

The union $\Gamma (\xi _{\pm }, r) \cup \{\xi _\pm \} \cup \Gamma ^*(\xi _{\pm }, r)$ is then the boundary of a domain to be denoted by $\Omega (\xi _{\pm }, r)$. See Fig. 1.

Fig. 1

Objects that appear in the proof of Theorem 1.1. Here $\xi _- = e^{i\pi /5}$, $\xi _+ = e^{2i\pi /5}$, $r =3/4$, $\zeta = e^{1.43\pi /5}$, $\lambda = (1+r)\zeta /2.$

Lemma 2.4

Suppose that $h \in L^1(m)$, $\eta >0$, and $\lambda \in {\mathbb {D}}{\setminus } B(0, 3/4)$ are such that

$$\begin{aligned} \int _{{\mathbb {T}}}h(\xi ) \frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm(\xi ) \leqslant \eta . \end{aligned}$$

Then there are $\xi _{\pm } \in {\mathbb {T}}$ such that for every $f \in H^1$ satisfying $|f| \leqslant h$ on ${\mathbb {T}}$ we have $|f(z)| \lesssim \eta$ on $\Gamma (\xi _{\pm }, r)$, where r: $1-r = 2(1-|\lambda |)$. Moreover, $\xi _{\pm }$ are such that $\zeta = \lambda /|\lambda |$ belongs to the arc of ${\mathbb {T}}$ with endpoints $\xi _{\pm }$, and $1-r \leqslant |\xi _{\pm }-\zeta | \leqslant 2(1-r)$.

Proof

Let $\lambda \in {\mathbb {D}}{\setminus } B(0, 3/4)$, $r=2|\lambda |-1$, $\zeta = \lambda /|\lambda |$. Consider the arc of the unit circle $G = {\mathbb {T}}\cap B(\zeta ,2(1-r))$, and define $m_{G} = \frac{\chi _{G}}{m(G)}dm$. We have

$$\begin{aligned} \int _{{\mathbb {T}}} h \,dm_G \lesssim \eta . \end{aligned}$$

The set G is the metric space with respect to the usual distance in ${\mathbb {C}}$. The measure $m_G$ has doubling property on this space:

$$\begin{aligned} m_G(B(\xi , 2\rho )) \leqslant 4 m_G(B(\xi , \rho )) \end{aligned}$$

for every $\xi \in G$, $\rho > 0$. It follows that the weak norm of Hardy-Littlewood maximal operator on $L^1(G, m_G)$ is bounded by a constant that does not depend on G. In other words, for every $t>0$, $g \in L^1(G, m_G)$ we have

$$\begin{aligned} m_G\left( \left\{ \xi : \sup _{\rho>0}\frac{1}{m_G(B(\xi , \rho ))}\int _{B(\xi , \rho )} g \,dm_G > t\right\} \right) \lesssim \frac{\Vert g\Vert _{L^1(G, m_G)}}{t}, \end{aligned}$$

where the constant involved does not depend on G, t, g. Taking $g = h$, $t = \varepsilon ^{-1}\eta$ for some $\varepsilon > 0$, we obtain

$$\begin{aligned} m_G\left( \left\{ \xi : \sup _{\rho>0}\frac{1}{m_G(B(\xi , \rho ))}\int _{B(\xi , \rho )} h \,dm_G > \varepsilon ^{-1}\eta \right\} \right) \lesssim \varepsilon . \end{aligned}$$

It follows that

$$\begin{aligned} m_G\left( \left\{ \xi : \sup _{\rho \in [0, 1)}\int _{G} h(u) \frac{1-\rho ^2}{|1 - \rho {\bar{u}} \xi |^2}\,dm(u) > \varepsilon ^{-1}\eta \right\} \right) \lesssim \varepsilon . \end{aligned}$$

(2.11)

Indeed, this follows from the fact that for each $\xi \in {\mathbb {T}}$, $\rho \geqslant 0$, the Poisson kernel $u \mapsto \frac{1-\rho ^2}{|1 - \rho {\bar{u}} \xi |^2}$ can be uniformly approximated on ${\mathbb {T}}$ by positive convex combinations of functions of the form $\frac{\chi _{B(\xi , \delta )}}{m(B(\xi , \delta ))}$, $\delta > 0$. Therefore, if $\xi \in G$ is such that for some $\rho \in [0, 1)$ we have

$$\begin{aligned} \int _{G}h(u) \frac{1-\rho ^2}{|1 - \rho {\bar{u}} \xi |^2}\,dm(u) > \varepsilon ^{-1}\eta \end{aligned}$$

then

$$\begin{aligned} \frac{1}{m(B(\xi , \delta ))}\int _{B(\xi , \delta )}\chi _{G}(u) h(u)\,dm(u) > \varepsilon ^{-1}\eta \end{aligned}$$

for some $\delta > 0$, and so

$$\begin{aligned} \frac{1}{m_G(B(\xi , \delta ))}\int _{B(\xi , \delta )} h(u)\,dm_G(u) > \varepsilon ^{-1}\eta , \end{aligned}$$

proving (2.11). Let us now take $\varepsilon \in (0, 1)$ so small that the left hand side of (2.11) does not exceed 1/10. Then there are $\xi _{\pm } \in G$ such that $\zeta$ belongs to the arc of ${\mathbb {T}}$ with endpoints $\xi _{\pm }$, we have $|\xi _{\pm } - \zeta | \geqslant 1-r$, and, moreover,

$$\begin{aligned} \sup _{\rho \in [r, 1)}\int _{G} h \frac{1-\rho ^2}{|1 - \rho {\bar{u}} \xi _{\pm }|^2}\,dm \leqslant \varepsilon ^{-1}\eta . \end{aligned}$$

For $u \in {\mathbb {T}}\setminus G$, we have

$$\begin{aligned} \sup _{\rho \in [r, 1)}\frac{1-\rho ^2}{|1 - \rho {\bar{u}} \xi _{\pm }|^2}\,dm \lesssim \frac{1- |\lambda |^2}{|1 - {\bar{u}} \lambda |^2}. \end{aligned}$$

It follows that

$$\begin{aligned} |f(\rho \xi _{\pm })| \lesssim \int _{{\mathbb {T}}\setminus G} h \frac{1- |\lambda |^2}{|1 - {\bar{u}} \lambda |^2}\,dm + \int _{G} h \frac{1-\rho ^2}{|1 - \rho \bar{\xi }\xi _{\pm }|^2}\,dm \lesssim (1+\varepsilon ^{-1})\eta ,\nonumber \\ \end{aligned}$$

(2.12)

for every $\rho \in [r, 1)$ with absolute constants. We also note that

$$\begin{aligned} \sup _{\xi \in G}\frac{1-r^2}{|1 - r{\bar{u}} \xi |^2} \lesssim \frac{1- |\lambda |^2}{|1 - {\bar{u}} \lambda |^2}, \quad u \in {\mathbb {T}}, \end{aligned}$$

therefore,

$$\begin{aligned} |f(r\xi )| \lesssim \int _{{\mathbb {T}}} h(u) \frac{1- |\lambda |^2}{|1 - {\bar{u}} \lambda |^2}\,dm(u) \lesssim \eta , \end{aligned}$$

(2.13)

for $\xi \in G$. Collecting (2.12) and (2.13), we see that $|f(z)|\!\lesssim \!\eta$ for $z \in \Gamma (\xi _{\pm },\! r)$.$\square$

Define $\tilde{b}_n = \frac{\tilde{\varphi }_n}{\tilde{\varphi }_n^*}.$ Formula (2.7) shows that $|\tilde{b}_n|\leqslant 1$ on ${\mathbb {D}}$. In fact, $\tilde{b}_n$ is a Blaschke product of order n.

Lemma 2.5

Let $\lambda \in {\mathbb {D}}$, denote by $\eta = e^{{\mathcal {K}}_{\nu _{n,\lambda }}(\lambda )} - 1$ the number in the right hand side of (2.6). Set $\alpha = \overline{\tilde{\varphi }_n^*(\lambda )}/\tilde{\varphi }_n^*(\lambda )$. We have

$$\begin{aligned} \int _{{\mathbb {T}}}|\tilde{b}_n(\xi ) - \alpha \xi ^n|^2\frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm \lesssim \eta . \end{aligned}$$

Proof

Consider the sets

$$\begin{aligned} E_{l} = \left\{ \xi \in {\mathbb {T}}: \; \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(\xi )} - 1\right| ^2 > 1/4\right\} , \quad E_s = {\mathbb {T}}\setminus E_l. \end{aligned}$$

On $E_l$, the difference $|\tilde{b}_n(\xi ) - \alpha \xi ^n| \leqslant 2$ could be large, but the measure of this set is small. Let us use Chebyshev’s inequality and Lemma 2.3 to estimate the corresponding integral:

$$\begin{aligned} \int _{E_l}|\tilde{b}_n(\xi ) - \alpha \xi ^n|^2\frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2} \leqslant 4\int _{E_l}\frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}\,dm \lesssim \eta . \end{aligned}$$

On $E_s$, the difference $|\tilde{b}_n(\xi ) - \alpha \xi ^n|$ is small. Indeed, for $\xi \in E_s$, we write

$$\begin{aligned} |\tilde{b}_n - \alpha \xi ^n| = \left| \frac{\bar{\xi }^n\tilde{\varphi }_n(\xi )}{\tilde{\varphi }_n^*(\xi )} - \alpha \right| = \left| \overline{\frac{\tilde{\varphi }_n^*(\xi )}{\overline{\tilde{\varphi }_n^*(\lambda )}}} \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(\xi )} - 1 \right| . \end{aligned}$$

For $z \in {\mathbb {C}}$ such that $|1-z| < 1/2$, we have $|z^{-1}| \leqslant 2$, so $|\tilde{\varphi }_n^*(\xi )/\tilde{\varphi }_n^*(\lambda )| \leqslant 2$ on $E_s$. Then

$$\begin{aligned} \left| \frac{\tilde{\varphi }_n^*(\xi )}{\tilde{\varphi }_n^*(\lambda )} - 1\right| = \left| \frac{\tilde{\varphi }_n^*(\xi )}{\tilde{\varphi }_n^*(\lambda )}\right| \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(\xi )} - 1\right| \leqslant 2 \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(\xi )} - 1\right| . \end{aligned}$$

Using $ab -1 = a(b-1) + a-1$ for $a = \overline{\varphi _n^*(\xi )/\varphi _n^*(\lambda )}$, $b = \varphi _n^*(\lambda )/\varphi _n^*(\xi )$, we see that

$$\begin{aligned} |\tilde{b}_n - \alpha \xi ^n| \lesssim \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(\xi )} - 1\right| \end{aligned}$$

on $E_s$. The claim now follow from Lemma 2.3. $\square$

Lemma 2.6

Let $\lambda \in {\mathbb {D}}$, $A \geqslant 1$, $\eta = e^{{\mathcal {K}}_{\nu _{n,\lambda }}(\lambda )} - 1$. Define r so that $1-r = 2(1-|\lambda |)$, and assume that $1/2 \leqslant 1- A/n \leqslant r < 1$. There exists a number $\eta _0 \in (0,1)$ depending only on A, such that if $\eta \leqslant \eta _0$, then there are $\xi _{\pm } \in {\mathbb {T}}$ such that

$$\begin{aligned} |\tilde{b}_n(z) - \alpha z^n|{} & {} \lesssim e^{4A}\sqrt{\eta }, \end{aligned}$$

(2.14)

$$\begin{aligned} \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)} -1 \right|{} & {} \lesssim e^{2A}\sqrt{\eta }, \end{aligned}$$

(2.15)

for all $z \in \Gamma (\xi _{\pm }, r) \cup \Gamma ^*(\xi _{\pm }, r)$. Moreover, $\xi _{\pm }$ are such that $\zeta = \lambda /|\lambda |$ belongs to the arc of ${\mathbb {T}}$ with endpoints $\xi _{\pm }$, and $1-r \leqslant |\xi _{\pm }-\zeta | \leqslant 2(1-r)$.

Proof

Consider the function $h = |\tilde{b}_n - \alpha \xi ^n|^2 + \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*} -1 \right| ^2$ on ${\mathbb {T}}$. By Lemma 2.5 and Lemma 2.3, we have

$$\begin{aligned} \int _{{\mathbb {T}}}h(\xi )\frac{1-|\lambda |^2}{|1 - \bar{\xi }\lambda |^2}dm(\xi ) \lesssim \eta . \end{aligned}$$

Then, by Lemma 2.4, applied to the functions $(\tilde{b}_n(z) - \alpha z^n)^2$, $(\tfrac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)} -1)^2$, there exists a contour $\Gamma (\xi _{\pm }, r)$ such that

$$\begin{aligned} |\tilde{b}_n(z) - \alpha z^n|^2 \lesssim \eta , \quad \left| \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)} -1 \right| ^2 \lesssim \eta , \end{aligned}$$

for all $z \in \Gamma (\xi _{\pm }, r)$. Moreover, $\xi _{\pm }$ are such that $\zeta$ belongs to the arc of ${\mathbb {T}}$ with endpoints $\xi _{\pm }$, and $1-r \leqslant |\xi _{\pm }-\zeta | \leqslant 2(1-r)$. Choosing $\eta _0 \in (0,1)$ sufficiently small, one can guarantee that the left hand sides in these inequalities are smaller than $e^{-4A}/4$ for all $z \in \Gamma (\xi _{\pm }, r)$. In particular, for all $z \in \Gamma (\xi _{\pm }, r)$ we have

$$\begin{aligned} |z^n|&\geqslant r^n \geqslant (1-A/n)^n \geqslant e^{-2A},\\ |\tilde{b}_n(z)|&\geqslant |z^n| - |\alpha z^n - {\tilde{b}}_n| \geqslant e^{-2A} - e^{-2A}/2 \geqslant e^{-2A}/2, \end{aligned}$$

where we have used the elementary inequality $\log (1-x) \geqslant -2x$, $x \in [0, 1/2]$. Then the identity

$$\begin{aligned} \theta (1/{\bar{z}}) = \overline{1/\theta (z)}, \quad z \in {\mathbb {C}}, \end{aligned}$$

for the inner functions $\theta = \tilde{b}_n$, $\theta = \alpha z^n$, gives (2.14) on $\Gamma ^*(\xi _{\pm }, r)$:

$$\begin{aligned} |\tilde{b}_n(1/{\bar{z}}) - \alpha (1/{\bar{z}})^{n}| = \frac{|\tilde{b}_n(z) - \alpha z^{n}|}{|\tilde{b}_n(z)z^{n}|} \lesssim e^{4A}\sqrt{\eta }, \quad z \in \Gamma (\xi _{\pm }, r). \end{aligned}$$

To estimate $\bigl |\frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*} - 1\bigr |$ on $\Gamma ^*(\xi _{\pm }, r)$, we use relation

$$\begin{aligned} \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(1/\overline{z})} = \frac{\tilde{\varphi }_n^*(\lambda )}{\overline{\tilde{\varphi }_n(z)/z^n}} = \frac{{\bar{z}}^n}{\overline{\tilde{b}_n(z)}}\frac{\overline{\tilde{\varphi }_n^*(\lambda )}}{\overline{\tilde{\varphi }_n^*(z)}}\frac{\tilde{\varphi }_n^*(\lambda )}{\overline{\tilde{\varphi }_n^*(\lambda )}} = \overline{\left( \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)} \frac{\alpha z^n}{\tilde{b}_n(z)}\right) }. \end{aligned}$$

Then, formula $\overline{ab} - 1 = \overline{a(b-1) + a-1}$ for $a = \frac{\tilde{\varphi }_n^*(\lambda )}{\tilde{\varphi }_n^*(z)}$, $b=\frac{\alpha z^n}{\tilde{b}_n(z)}$ implies (2.15) on $\Gamma ^*(\xi _{\pm }, r)$. $\square$

Lemma 2.7

Let $\lambda , A, n, r, \zeta , \eta$ be as in Lemma 2.6 and let $z_{1,2} \in B(\zeta , 1-|\lambda |)$. Then

$$\begin{aligned} \left| \overline{\tilde{b}_n(z_1)}\tilde{b}_n(z_2) - \overline{z_1}^n z_2^n\right| \lesssim e^{4A}\sqrt{\eta } \cdot |1 - {\bar{z}}_1 z_2| \cdot n. \end{aligned}$$

Proof

By Lemma 2.6, relations (2.14), (2.15) hold on some contour $\Gamma (\xi _{\pm }, r) \cup \Gamma ^*(\xi _{\pm }, r)$, where $\xi _{\pm }$ are such that $\zeta = \lambda /|\lambda |$ belongs to the arc of ${\mathbb {T}}$ with endpoints $\xi _{\pm }$, and $1-r \leqslant |\xi _{\pm }-\zeta | \leqslant 2(1-r)$. By the maximum modulus principle, (2.14) holds in the bounded domain $\Omega (\xi _{\pm }, r)$ such that $\partial \Omega (\xi _{\pm }, r) = \Gamma (\xi _{\pm }, r) \cup \Gamma ^*(\xi _{\pm }, r) \cup \{\xi _{\pm }\}$. In particular, (2.14) holds for all $z \in B(\zeta , 1-r)$. Now pick two points $z_1, z$ such that

$$\begin{aligned} |z_1-\zeta | \leqslant 1-|\lambda |, \quad |z-\zeta | = 2(1-|\lambda |) = 1-r. \end{aligned}$$

We have

$$\begin{aligned} \left| \frac{\overline{\tilde{b}_n(z_1)}\tilde{b}_n(z) - \overline{z_1}^n z^n}{1-{\bar{z}}_1 z}\right|&\lesssim \frac{\bigl | (\tilde{b}_n(z_1) - \alpha z^n_1)(\tilde{b}_n(z)- \alpha z^n)\bigr |}{1-r} + \\&\quad +\, \frac{\bigl |\overline{z_1}^n(\tilde{b}_n(z)- \alpha z^n)\bigr |}{1-r} + \frac{\bigl |(\alpha z_1^n - \tilde{b}_n(z_1)) z^n \bigr |}{1-r}. \end{aligned}$$

Note that $|1-{\bar{z}}_1 z| \gtrsim 1-|\lambda | \geqslant \frac{1-r}{2} \gtrsim n^{-1},$ because $A \geqslant 1$. Recall that the maximum principle and (2.14) imply

$$\begin{aligned} |\tilde{b}_n(z_1) - \alpha z^n_1| \lesssim e^{2A}\sqrt{\eta }, \quad |\tilde{b}_n(z) - \alpha z^n| \lesssim e^{2A}\sqrt{\eta }. \end{aligned}$$

Next, we have

$$\begin{aligned} \max (|z_1|^n,|z|^n) \leqslant (1 + 1-r)^n \leqslant (1 + A/n)^{n} \leqslant e^{A}. \end{aligned}$$

Since $\eta \leqslant \eta _0 \leqslant 1$, we can conclude that

$$\begin{aligned} \left| \frac{\overline{\tilde{b}_n(z_1)}\tilde{b}_n(z) - \overline{z_1}^n z^n}{1-{\bar{z}}_1 z}\right| \lesssim e^{4A}\sqrt{\eta } \cdot n. \end{aligned}$$

The function $z \mapsto \frac{\overline{\tilde{b}_n(z_1)}\tilde{b}_n - \overline{z_1}^n z^n}{1-{\bar{z}}_1 z}$ is analytic, hence the same estimate holds for all $z \in B(\zeta , 1-r)$ by the maximum modulus principle. In particular, it holds for all points $z_{1}$, $z_2$ in $B(\zeta , 1-|\lambda |)$. The lemma follows. $\square$

Proof of Theorem 1.1

Let $\mu \in {\textrm{Sz}}({\mathbb {T}})$, $\zeta \in {\mathbb {T}}$, $A \geqslant 1$. Let us choose ${\tilde{N}}_{0}$ such that $A/{\tilde{N}}_{0} < 1/4$ and

$$\begin{aligned} \sup _{\rho \in [1 - A/{\tilde{N}}_{0}, 1)}e^{{\mathcal {K}}_{\mu }(\rho \zeta )} - 1 < \eta _0, \end{aligned}$$

(2.16)

where $\eta _0 \in (0, 1)$ is the number in Lemma 2.6. For $n \geqslant {\tilde{N}}_{0}$, consider $z_{1,2} \in B(\zeta , A/n)$ and let $\lambda \in {\mathbb {D}}$ be defined by

$$\begin{aligned} \lambda = (1-\max \limits _{k=1,2}|z_{k} - \zeta |)\zeta . \end{aligned}$$

Define $\eta = e^{{\mathcal {K}}_{\nu _{n,\lambda }}(\lambda )} - 1$ and observe that $\eta < \eta _0$ by (2.16) and Lemma 2.2. We are in assumptions of Lemma 2.6. Estimate (2.15) and the maximum modulus principle give

$$\begin{aligned} \frac{\tilde{\varphi }_n^*(z)}{\tilde{\varphi }_n^*(\lambda )} = \frac{1}{1 + R_1(z)}, \quad |R_1(z)| \lesssim e^{2A}\sqrt{\eta }, \end{aligned}$$

for all $z \in B(\zeta , 1-|\lambda |)$. By Lemma 2.7, for all $z_{1}, z_2 \in B(\zeta , 1-|\lambda |)$ we have

$$\begin{aligned} \frac{1 - \overline{\tilde{b}_n(z_1)}\tilde{b}_n(z_2)}{1 - \overline{z_1} z_2}&=\frac{1 - \overline{z_1}^n z_2^n}{1 - \overline{z_1} z_2} + R_2(z_1, z_2) \cdot n, \end{aligned}$$

with $|R_2(z_1, z_2)| \lesssim e^{4A}\sqrt{\eta }$, so

$$\begin{aligned} k_{\mu , n}(z_1, z_2)&= \left( \frac{1 - \overline{z_1^n} z_2^n}{n(1 - \overline{z_1} z_2)} + R_2(z_1, z_2)\right) \cdot \frac{|\tilde{\varphi }_{n}^*(\lambda )|^2 \cdot n}{(1+R_1(z_1))\overline{(1+R_1(z_2))}}. \end{aligned}$$

Taking $z_1 = z_2 = \zeta$, we get

$$\begin{aligned} k_{\mu , n}(\zeta , \zeta ) = \frac{1 + R_2(\zeta ,\zeta )}{|1+R_1(\zeta )|^2} \cdot |\tilde{\varphi }_{n}^*(\lambda )|^2 \cdot n. \end{aligned}$$

It remains to write

$$\begin{aligned} \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )} = \frac{\frac{1 - \overline{z_1}^n z_2^n}{n(1 - \overline{z_1} z_2)} + R_2(z_1, z_2)}{1+R_2(\zeta ,\zeta )}\frac{|1+R_1(\zeta )|^2}{(1+R_1(z_1))\overline{(1+R_1(z_2))}}. \end{aligned}$$

Note that

$$\begin{aligned} \max _{z \in B(\zeta , 1-|\lambda |)}|R_{1}(z)| + \max _{z_{1,2} \in B(\zeta , 1-|\lambda |)}|R_2(z_1, z_2)| \leqslant ce^{4A}\sqrt{\eta }, \end{aligned}$$

for an absolute constant c. We see that one can take $\varepsilon _0\in (0, \eta _0)$ so that $ce^{4A}\sqrt{\varepsilon _0} \leqslant 1/10$, then the required estimate will hold. $\square$

3 Proof of Theorem 1.2

Proof of Theorem 1.2

Let I be an arc of ${\mathbb {T}}$ with center $\zeta$ such that $|w(\xi ) - w(\zeta )| \sim |\xi - \zeta |^{s}$ for all $\xi \in I$ and some $s > 0$. By the assumption of the theorem, we have $w(\zeta ) > 0$. Replacing I by a smaller interval, we can assume that

$$\begin{aligned} \delta< w(\xi ) < 2\delta , \quad |\log w(\xi ) - \log w(\zeta )| \sim |\xi - \zeta |^{s}, \end{aligned}$$

for all $\xi \in I$ and some $\delta > 0$. Denote $u = \log w - \log w(\zeta )$. We have

$$\begin{aligned} \int _{I}e^{u(\xi )}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm = \int _{I}\bigl (1 + u(\xi ) + O(u^2)\bigr )\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm. \end{aligned}$$

We also have

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta )&= \log \left( \int _{{\mathbb {T}}}e^{u(\xi )}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm\right) - \int _{{\mathbb {T}}}u(\xi )\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm\\&\leqslant \int _{{\mathbb {T}}}(e^{u(\xi )}-1)\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm - \int _{{\mathbb {T}}}u(\xi )\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm\\&\lesssim (1 - \rho ) \int _{{\mathbb {T}}\setminus I}(|e^{u(\xi )}-1| + |u(\xi )|)\,dm + \int _{I}u^2(\xi )\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm. \end{aligned}$$

By construction, we have

$$\begin{aligned} \int _{I}u^2(\xi )\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm \lesssim \int _{I}|\xi -\zeta |^{2s}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm. \end{aligned}$$

(3.1)

Let us set $\varepsilon = 1-\rho$, assume that $\varepsilon < m(I)/2$, and estimate

$$\begin{aligned} \int _{I \cap B(\zeta ,\varepsilon )}|\xi -\zeta |^{2s}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm \lesssim \frac{1}{\varepsilon }\int _{I \cap B(\zeta ,\varepsilon )}|\xi -\zeta |^{2s}\,dm \lesssim \varepsilon ^{2s}, \end{aligned}$$

and

$$\begin{aligned} \int _{I \setminus B(\zeta ,\varepsilon )}|\xi -\zeta |^{2s}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm \sim (1-\rho )\int _{I \setminus B(\zeta ,\varepsilon )}|\xi -\zeta |^{2s-2}\,dm. \end{aligned}$$

We have

$$\begin{aligned} \int _{I \setminus B(\zeta ,\varepsilon )}|\xi -\zeta |^{2s-2}\,dm \sim {\left\{ \begin{array}{ll} 1, &{} s \in (1/2, +\infty ),\\ \log \frac{1}{\varepsilon }, &{} s =1/2,\\ \varepsilon ^{2s-1}, &{} s \in (0, 1/2), \end{array}\right. } \end{aligned}$$

(3.2)

as $\varepsilon \rightarrow 0$. The constants in these relations depend on s. We see that

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta ) \lesssim {\left\{ \begin{array}{ll} 1-\rho , &{} s \in (1/2, +\infty ),\\ (1-\rho )|\log (1-\rho )|, &{} s =1/2,\\ (1-\rho )^{2s}, &{} s \in (0, 1/2), \end{array}\right. } \end{aligned}$$

as $\rho \rightarrow 1$. On the other hand, for $\varepsilon = 1-\rho$ we have

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta ) \gtrsim \int _{{\mathbb {T}}}|u(\xi ) - c|^2\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm \end{aligned}$$

(3.3)

Let $c_1$, $c_2$ be such that $c_1|\xi - \zeta |^{s} \leqslant u(\xi ) \leqslant c_2|\xi - \zeta |^{s}$ for all $\xi \in I$. Choose $\delta \in (0, 1)$ so small that

$$\begin{aligned} c_1(1-\delta )^s - c_2\delta ^s > 0. \end{aligned}$$

From (3.3) we see that there are $\xi _1 \in B(\zeta , \delta \varepsilon )$, $\xi _2 \in B(\zeta , \varepsilon ) {\setminus } B(\zeta , (1-\delta )\varepsilon )$ such that

$$\begin{aligned} |u(\xi ) - c|^2 \lesssim {\mathcal {K}}_{\mu }(\rho \zeta ), \quad \xi = \xi _{1, 2}. \end{aligned}$$

Then

$$\begin{aligned} |u(\xi _1) - u(\xi _2)| \lesssim \sqrt{{\mathcal {K}}_{\mu }(\rho \zeta )}, \end{aligned}$$

and, simultaneously,

$$\begin{aligned} |u(\xi _1) - u(\xi _2)| \geqslant c_1((1-\delta )\varepsilon )^{s} - c_2(\delta \varepsilon )^{s} = \varepsilon ^{s}(c_1(1-\delta )^s - c_2\delta ^s) \gtrsim \varepsilon ^{s}. \end{aligned}$$

We see that ${\mathcal {K}}_{\mu }(\rho \zeta ) \gtrsim (1-\rho )^{2s}$ for all $\rho$ close enough to 1. A more accurate estimate is needed for $s \geqslant 1/2$. Let $\delta > 0$ be such that

$$\begin{aligned} c_1(1+\delta )^{s}/2 - c_2 > 0. \end{aligned}$$

Consider $j \in {\mathbb {Z}}$ satisfying $(1+\delta )^{j+3} < |I|/2$, and let $\xi _1$, $\xi _{2}$ be such that

$$\begin{aligned} (1+\delta )^{j}< |\xi _1-\zeta |< (1+\delta )^{j+1}, \quad (1+\delta )^{j+2}< |\xi _2-\zeta | < (1+\delta )^{j+3}. \end{aligned}$$

We claim that either $|u(\xi _2) - c| \gtrsim |\xi _2 - \zeta |^s$ or $|u(\xi _1) - c| \gtrsim |\xi _1 - \zeta |^s$ for all such $\xi _1$, $\xi _2$, j. Indeed, if $|u(\xi _2) - c| \leqslant c_1|\xi _2 - \zeta |^s/2$, then

$$\begin{aligned} |u(\xi _1) - c|&\geqslant |u(\xi _2)| - |u(\xi _1)| - |u(\xi _2) - c|\\&\geqslant c_1|\xi _2 - \zeta |^s - c_2|\xi _1 - \zeta |^s - c_1|\xi _2 - \zeta |^s/2\\&\gtrsim c_1(1+\delta )^{js+2s}/2 - c_2(1+\delta )^{js+s} \\&\gtrsim (c_1(1+\delta )^{s}/2 - c_2)(1+\delta )^{js+s}\\&\gtrsim (1+\delta )^{js+s} \gtrsim |\xi _1 - \zeta |^{s}, \end{aligned}$$

proving the claim. We see that for each $j \in {\mathbb {Z}}$ such that $(1+\delta )^{j+3} < |I|/2$, we have $|u(\xi ) - c| \gtrsim |\xi - \zeta |^{s}$ on a half of the set ${\mathbb {T}}\cap B(\zeta , (1+\delta )^{j+3}) {\setminus } B(\zeta , (1+\delta )^{j})$. It follows that

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta ) \gtrsim \int _{I}|\xi - \zeta |^{2s}\frac{1 - \rho ^2}{|1 - \rho \bar{\xi }\zeta |^2}\,dm, \end{aligned}$$

and estimate (3.2) shows that

$$\begin{aligned} {\mathcal {K}}_{\mu }(\rho \zeta ) \gtrsim {\left\{ \begin{array}{ll} 1-\rho , &{} s \in (1/2, +\infty ),\\ (1-\rho )|\log (1-\rho )|, &{} s =1/2,\\ (1-\rho )^{2s}, &{} s \in (0, 1/2). \end{array}\right. } \end{aligned}$$

$\square$

4 Example: The Poisson Kernel

In the following example, the asymptotic behaviour of ratios of reproducing kernels could be explicitly analysed.

Example

Let $\lambda \in {\mathbb {D}}\setminus \{0\}$. Consider the probability measure $\mu = \frac{1 - |\lambda |^2}{|1 - \lambda \xi |^2}\,dm$ on the unit circle ${\mathbb {T}}$. For $\zeta \in {\mathbb {T}}$, $u, v \in {\mathbb {R}}$, $z_1 = \zeta e^{u/n}$, $z_2 = \zeta e^{v/n}$, we have

$$\begin{aligned} \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )} = \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)} + \delta _n(u, v), \quad n \rightarrow \infty , \end{aligned}$$

where $\sup \nolimits _{|u|,\,|v| \leqslant 1}|\delta _n(u, v)|$ is comparable to 1/n.

Proof

We have (see Sect. 1.6 in [14])

$$\begin{aligned} \varphi _{n}(z) = \frac{z^n - \bar{\lambda }z^{n-1}}{\sqrt{1-|\lambda |^2}}, \quad \varphi ^*_{n}(z) = \frac{1 - \lambda z}{\sqrt{1-|\lambda |^2}}. \end{aligned}$$

It follows that

$$\begin{aligned} k_{\mu , n}(z, z) = \sum _{0}^{n-1}|\varphi _k(z)|^2 = \sum _{0}^{n-1}\frac{|z^{k}- \bar{\lambda }z^{k-1}|^2}{1 - |\lambda |^2} = \frac{|z - \bar{\lambda }|^2}{|z|^2(1 - |\lambda |^2)}\frac{1 - |z|^{2n}}{1-|z|^2}. \end{aligned}$$

Then

$$\begin{aligned} \frac{k_{\mu , n}(z, z)}{k_{\mu , n}(\zeta , \zeta )} = \frac{|z - \bar{\lambda }|^2}{|z|^2|\zeta - \bar{\lambda }|^2}\frac{1 - |z|^{2n}}{n(1-|z|^2)}, \end{aligned}$$

and we see that

$$\begin{aligned} \frac{k_{\mu , n}(z, z)}{k_{\mu , n}(\zeta , \zeta )} - \frac{k_{m, n}(z, z)}{k_{m, n}(\zeta , \zeta )} = \frac{1 - |z|^{2n}}{n(1-|z|^2)}\frac{|z - \bar{\lambda }|^2 - |z|^2|\zeta - \bar{\lambda }|^2}{|z|^2|\zeta - \bar{\lambda }|^2} \end{aligned}$$

is comparable to

$$\begin{aligned} |z - \bar{\lambda }|^2 - |z|^2|\zeta - \bar{\lambda }|^2&= (1 - |z|^2)(|\lambda |^2 + 2\mathop \textrm{Re}(\lambda (\zeta |z|^2-z)))\\&\gtrsim (1 - |z|^2)|\lambda |^2 \gtrsim \frac{1}{n}, \end{aligned}$$

if $\lambda \ne 0$, $|\zeta -z| \sim 1-|z| \sim 1/n$, and n is large. It follows that

$$\begin{aligned} \sup \limits _{|u|,\,|v| \leqslant 1}|\delta _n(u, v)| \gtrsim 1/n. \end{aligned}$$

Now take $z_1 = \zeta e^{u/n}$, $z_2 = \zeta e^{v/n}$, with $|u|,\,|v| \leqslant 1$. We have

$$\begin{aligned} k_{\mu , n}(z_1, z_2)&= \frac{1}{1-|\lambda |^2}\frac{(1 - \lambda z_1)\overline{(1 - \lambda z_2)} - z_1^n {\bar{z}}_2^n(1 - \bar{\lambda }/z_1)(1 - \lambda /{\bar{z}}_2)}{1 - z_1 {\bar{z}}_2 }\\&= \frac{(1 - \lambda z_1)\overline{(1 - \lambda z_2)}}{1-|\lambda |^2}\frac{1 - z_1^n {\bar{z}}_2^n\theta _\lambda (z_1, z_2)}{1 - z_1 {\bar{z}}_2 }, \end{aligned}$$

where

$$\begin{aligned} \theta _{\lambda }(z_1, z_2) = \frac{(1 - \bar{\lambda }/z_1)(1 - \lambda /{\bar{z}}_2)}{\overline{(1 - \lambda z_2)}(1 - \lambda z_1)} = 1 + O(1/n). \end{aligned}$$

In particular, we have

$$\begin{aligned} \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(\zeta , \zeta )}&= \frac{(1 - \lambda z_1)\overline{(1 - \lambda z_2)}}{|1 - \lambda \zeta |^2}\frac{1 - z_1^n {\bar{z}}_2^n\theta _\lambda (z_1, z_2)}{n(1 - z_1 {\bar{z}}_2)}. \end{aligned}$$

Here

$$\begin{aligned} \frac{(1 - \lambda z_1)\overline{(1 - \lambda z_2)}}{|1 - \lambda \zeta |^2} = 1 + O(1/n), \end{aligned}$$

and

$$\begin{aligned} \frac{1 - z_1^n {\bar{z}}_2^n\theta _\lambda (z_1, z_2)}{n(1 - z_1 {\bar{z}}_2)} = \frac{k_{m, n}(z_1, z_2)}{k_{m, n}(\zeta , \zeta )} + z_1^n {\bar{z}}_2^n\frac{\theta _\lambda (z_1, z_2)-1}{n(1 - z_1 {\bar{z}}_2)}. \end{aligned}$$

Note that $\theta _\lambda (z_1, z_2) =1$ in the case $z_1 {\bar{z}}_2 = 1$. Considering $z_1$, $z_2$ such that $|1 - z_1 {\bar{z}}_2| \sim 1/n$ and using maximum modulus principle for analytic functions, we see that

$$\begin{aligned} \left| z_1^n {\bar{z}}_2^n\frac{\theta _\lambda (z_1, z_2)-1}{n(1 - z_1 {\bar{z}}_2)}\right| \lesssim \frac{1}{n} \end{aligned}$$

for all $z_1$, $z_2$ such that $|z_{1,2} -\zeta | \lesssim 1/n$. It follows that $\sup \nolimits _{|u|,\,|v| \leqslant 1}|\delta _n(u, v)| \lesssim 1/n$.

5 Sharpness of Theorem 1.1

Sharpness of Theorem 1.1 is an open problem. As we have seen in the previous section, one can have

$$\begin{aligned} \sup _{|z_{1,2} - 1| \leqslant 1/n}\left| \frac{k_{\mu , n}(z_1, z_2)}{k_{\mu , n}(1, 1)} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| = o\left( \sqrt{{\mathcal {K}}_{\mu }(1-1/n)}\right) , \end{aligned}$$

if the measure $\mu$ is very regular. Our aim in this section is to present some numerical results for the measures of the form $w_s\,dm$, where $w_s(e^{i\theta }) = c_{s}e^{|\theta |^{s}}$, $\theta \in [-\pi , \pi ]$, and the constant $c_s$ is chosen so that $\int _{{\mathbb {T}}} w_s \,dm = 1$. We will see that

$$\begin{aligned} \sup _{|z_{1,2} - 1| \leqslant 1/n}\left| \frac{k_{w_s, n}(z_1, z_2)}{k_{w_s, n}(1, 1)} - \frac{1 - \overline{z_2}^n z_1^n}{n(1 - \overline{z_2} z_1)}\right| \gtrsim \sqrt{{\mathcal {K}}_{w_s}(1-1/n)}, \end{aligned}$$

in the case $s=0.1$, $s = 0.2$, and $s = 0.4$ (similar results can be obtained for other values of $s \in (0, 1/2)$). In particular, the upper bound in Theorem 1.1 in these cases coincides with the lower bound up to a multiplicative factor comparable to 1. We use the following MATLAB code to produce our examples (note that the weight $w_s$ and the orthogonal polynomials in the script are not normalized, because the normalization plays no role when we consider ratios of reproducing kernels).

$\square$

This script produces the array D consisting of differences

$$\begin{aligned} \texttt {D(n/20)} = \left| \frac{k_{w_s, n}(x_n, x_n)}{k_{w_s, n}(1, 1)} - \frac{1 - |x_n|^{2n}}{n(1 - |x_n|^2)}\right| , \quad n = 20, 40, 60, \ldots , 8000, \end{aligned}$$

for the value $s = 0.4$ of Hölder continuity index. One can plot the values of this array (see below) and observe that it behaves like $C_{\alpha }n^{-\alpha }$ for certain values $\alpha >0$, $C_{\alpha }>0$. To find these values, we observe that

$$\begin{aligned} \frac{C_{\alpha }(n-20)^{-\alpha }}{C_{\alpha }n^{-\alpha }} = \left( 1 - \frac{20}{n}\right) ^{-\alpha } = 1 + \frac{20\alpha }{n} + O(n^{-2}), \end{aligned}$$

which gives a reasonable recipe to compute the array alphaCand of candidates for the power index $\alpha$:

$$\begin{aligned} \texttt {alphaCand(n/20)} = \frac{n}{20}\left( \frac{D(n/20-1)}{D(n/20)} - 1\right) , \quad n=20, 40, \ldots , 8000. \end{aligned}$$

When printed, the last 7 elements of this array (corresponding to values $n=7880,7900$, ..., 8000) look as follows:

$$\begin{aligned} \ldots \quad 0.39361 \quad 0.39362 \quad 0.39363 \quad 0.39363 \quad 0.39364 \quad 0.39365 \quad 0.39365 \end{aligned}$$

which is pretty close to $s = 0.4$. So it seems plausible that

$$\begin{aligned} \left| \frac{k_{w_s, n}(x_n, x_n)}{k_{w_s, n}(1, 1)} - \frac{1 - |x_n|^{2n}}{n(1 - |x_n|^2)}\right| \geqslant \frac{C_{0.4}}{n^{0.4}} \end{aligned}$$

for all n large enough. We can even suggest a candidate for the constant $C_{0.4}$ from our numerical data. For this we use approximation

$$\begin{aligned} C_{0.4} \sim C_{\alpha } = N^{0.39365}D(N/20) = 0.03791..., \quad N = 8000. \end{aligned}$$

The plots of the resulting functions

$$\begin{aligned} f_1(n) = \left| \frac{k_{w_s, n}(x_n, x_n)}{k_{w_s, n}(1, 1)} - \frac{1 - |x_n|^{2n}}{n(1 - |x_n|^2)}\right| , \quad f_2(n) = C_{0.4}n^{-0.4}, \end{aligned}$$

$\square$ are given on Fig. 2. We also consider the cases $s = 0.1$ and $s = 0.2$.

Fig. 2

Graphs of functions $f_1$, $f_2$ for $s = 0.1$, $s = 0.2$, and $s = 0.4$

The graphs demonstrate the fact that $f_1 \geqslant f_2$ for $s \in \{0.1, 0.2, 0.4\}$ and large integers n. Similar numerical results can be obtained for other values $s \in (0,1/2)$.