# On rate of convergence for universality limits

Roman Bessonov<EMAIL_ADDRESS>St. Petersburg State University
Universitetskaya nab. 7-9, 199034 St. Petersburg, RUSSIA St. Petersburg
Department of Steklov Mathematical Institute Russian Academy of Sciences
Fontanka 27, 191023 St.Petersburg, RUSSIA

###### 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
$n\to\infty$. We give an estimate for the rate of this convergence for
measures $\mu$ with finite logarithmic integral.

###### Key words and phrases:

Szegő class, entropy, universality, reproducing kernels

###### 2010 Mathematics Subject Classification:

42C05, 46E22

The work is supported by grant RScF 19-71-30002-$\Pi$ of the Russian Science
Foundation.

##  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 $n-1$,

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

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

$(p,k_{\mu,n}(\cdot,z_{2}))_{L^{2}(\mu)}=p(z_{2}),\qquad p\in\mathcal{P}_{n}.$

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

$k_{m,n}(z_{1},z_{2})=\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{1-\overline{z_{2}}z_{1}}.$

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

$\frac{k_{m,n}(z_{1},z_{2})}{k_{m,n}(\zeta,\zeta)}=\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}=\frac{e^{u+\bar{v}}-1}{u+\bar{v}}+O\left(\frac{1}{n}\right),$

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

$\frac{k_{\mu,n}(\zeta e^{u/n},\,\zeta
e^{v/n})}{k_{\mu,n}(\zeta,\zeta)}\to\frac{e^{u+\bar{v}}-1}{u+\bar{v}},\qquad
n\to+\infty,$

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., P.
Deift [6] or A. Kuijlaars, M. Vanlessen [9]. D. 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 A. Máté, P. Nevai and V. Totik in [12]. Global Szegő condition has been
weakened to the local one by E. Findley in [8]. Another approach, also
pioneered by D. 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 B. Simon [15] and V. Totik
[16]. Recently, B. Eichinger, M. Lukic, and B. Simanek [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

$\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}\to\frac{e^{u+\bar{v}}-1}{u+\bar{v}}$
(1)

are missing in the literature. In fact, the rate of convergence in (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) 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 [4], [1]. To be precise, let $\mu$ be a probability
measure on $\mathbb{T}$, and let $\mu=w\,dm+\mu_{\mathbf{s}}$ be its Radon-
Nikodym decomposition into the absolutely continuous and singular parts. The
measure $\mu$ is said to belong to the Szegő class $\mathrm{Sz}(\mathbb{T})$
if its logarithmic integral is finite:

$\int_{\mathbb{T}}\log w\,dm>-\infty.$

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 $\mu\in\mathrm{Sz}(\mathbb{T})$, its entropy
function is given by

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

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
$\mathcal{K}_{\mu}(z)\to 0$ as $z$ non-tangentially approaches $\zeta$. This
follows from well-known properties of the Poisson kernel: we have

$\lim_{r\to
1}\int_{\mathbb{T}}\frac{1-r^{2}}{|1-\bar{\xi}r\zeta|^{2}}\,d\mu_{\mathbf{s}}(\xi)=0$
(2)

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

$\displaystyle\lim_{r\to
1}\int_{\mathbb{T}}w(\xi)\frac{1-r^{2}}{|1-\bar{\xi}r\zeta|^{2}}\,dm(\xi)\to
w(\zeta),$ (3) $\displaystyle\lim_{r\to 1}\int_{\mathbb{T}}\log
w(\xi)\frac{1-r^{2}}{|1-\bar{\xi}r\zeta|^{2}}\,dm(\xi)\to\log w(\zeta),$ (4)

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

######  Theorem 1.

Let $\mu\in\mathrm{Sz}(\mathbb{T})$, $A\geqslant 1$, $n\geqslant 10A$,
$\zeta\in\mathbb{T}$. 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

$\left|\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\leqslant
ce^{4A}\sup_{\rho\in[1-\delta,1)}\sqrt{\mathcal{K}_{\mu}(\rho\zeta)},$ (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 tends to $1$ as $n\to\infty$,
therefore, the right hand side of (5) tends to zero for all $\zeta$ satisfying
(2)-(4). This gives a nontrivial bound for the rate of convergence in (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 2.

Let $\mu=w\,dm$ be an absolutely continuous probability measure in
$\mathrm{Sz}(\mathbb{T})$ 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

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

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

$\displaystyle\mathcal{K}(\mu,z)$ $\displaystyle=\log\mathop{\rm
Re}\left(\frac{1+\lambda z}{1-\lambda
z}\right)-\log\left(\frac{1-|\lambda|^{2}}{|1-\lambda z|^{2}}\right)$
$\displaystyle=\log\frac{1-|\lambda z|^{2}}{|1-\lambda
z|^{2}}-\log\left(\frac{1-|\lambda|^{2}}{|1-\lambda
z|^{2}}\right)=\log\frac{1-|\lambda z|^{2}}{1-|\lambda|^{2}},$

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 $n\to\infty$. Note that this agrees
with bounds in Theorem 2 (we have $s=1$ for this measure). By Theorem 1, we
then have

$\left|\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\lesssim\frac{1}{\sqrt{n}}$

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 Section 4, in fact

$\sup_{|z_{1,2}-\zeta|\leqslant
1/n}\left|\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\sim\frac{1}{n}.$

This shows that the bound in Theorem 1 is not sharp for smooth measures. It
seems, however, that this bound cannot be improved in the setting of the whole
class $\mathrm{Sz}(\mathbb{T})$ of measures with finite logarithmic integral,
i.e., there is a measure $\mu\in\mathrm{Sz}(\mathbb{T})$ such that

$\sup_{|z_{1,2}-1|\leqslant
1/n}\left|\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(1,1)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\gtrsim\sqrt{\mathcal{K}_{\mu}(1-1/n)},$

for all $n\geqslant 0$. In Section 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 2, we have

$\mathcal{K}_{w_{s}}(1-1/n)\sim n^{-2s},\qquad s\in(0,1/2).$

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

$\left|\frac{k_{w_{s},n}(x_{n},x_{n})}{k_{w_{s},n}(1,1)}-\frac{1-|x_{n}|^{2n}}{n(1-|x_{n}|^{2})}\right|\gtrsim
n^{-s}.$

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

$\sup_{|z_{1,2}-1|\leqslant
1/n}\left|\frac{k_{w_{s},n}(z_{1},z_{2})}{k_{w_{s},n}(1,1)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\gtrsim\sqrt{\mathcal{K}_{w_{s}}(1-1/n)},$

and estimate (5) in Theorem 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

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]):

$\varphi_{n}=\frac{z\varphi_{n-1}-\overline{a_{n-1}}\varphi^{*}_{n-1}}{\sqrt{1-|a_{n-1}|^{2}}},\qquad
n\geqslant 1.$

Here the recurrence coefficients, $a_{n}$, $n\geqslant 0$, belong to
$\mathbb{D}$. It is also known (see Theorem 2.2.7, page 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

$k_{\mu,n}(z_{1},z_{2})=\sum_{k=0}^{n-1}\overline{\varphi_{k}(z_{2})}\varphi_{k}(z_{1})=\frac{\overline{\varphi_{n}^{*}(z_{2})}\varphi_{n}^{*}(z_{1})-\overline{\varphi_{n}(z_{2})}\varphi_{n}(z_{1})}{1-\bar{z}_{2}z_{1}}.$
(6)

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

$\tilde{\varphi}_{n}=\frac{z\varphi_{n-1}-\bar{a}\varphi^{*}_{n-1}}{\sqrt{1-|a|^{2}}},\qquad\tilde{\varphi}_{n}^{*}=\frac{\varphi^{*}_{n-1}-za\varphi_{n-1}}{\sqrt{1-|a|^{2}}}.$
(7)

######  Lemma 1.

For all $z_{1},z_{2}\in\mathbb{C}$, we have

$\overline{\tilde{\varphi}_{n}^{*}(z_{2})}\tilde{\varphi}_{n}^{*}(z_{1})-\overline{\tilde{\varphi}_{n}(z_{2})}\tilde{\varphi}_{n}(z_{1})=\overline{\varphi_{n}^{*}(z_{2})}\varphi_{n}^{*}(z_{1})-\overline{\varphi_{n}(z_{2})}\varphi_{n}(z_{1})$
(8)

Proof. The proof is a direct computation. At first, let $z_{1}=z_{2}=z$. Then
the left hand side in (8) is equal to

$\displaystyle|\tilde{\varphi}_{n}^{*}(z)|^{2}-|\tilde{\varphi}_{n}(z)|^{2}$
$\displaystyle=\frac{1}{1-|a|^{2}}\left(|\varphi_{n-1}^{*}(z)-za\varphi_{n-1}(z)|^{2}-|z\varphi_{n-1}(z)-\bar{a}\varphi_{n-1}^{*}(z)|^{2}\right)$
$\displaystyle=|\varphi_{n-1}^{*}(z)|^{2}-|z\varphi_{n-1}(z)|^{2},$

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

$|\tilde{\varphi}_{n}^{*}(z)|^{2}-|\tilde{\varphi}_{n}(z)|^{2}=|\varphi_{n}^{*}(z)|^{2}-|\varphi_{n}(z)|^{2}.$
(9)

This relation holds for all $z\in\mathbb{C}$. Since functions in (8) are
analytic in $z_{1}$ and anti-analytic in $z_{2}$, the lemma follows. ∎

The following lemma is Corollary 4 in [4].

######  Lemma 2.

For every $\lambda\in\mathbb{D}$ there is $a\in\mathbb{D}$ such that the
corresponding polynomial $\tilde{\varphi}_{n}^{*}$ in (7) defines a
probability measure $\nu_{n,\lambda}=|\tilde{\varphi}_{n}^{*}|^{-2}\,dm$ on
$\mathbb{T}$ such that

$\mathcal{K}_{\nu_{n,\lambda}}(\lambda)\leqslant\mathcal{K}_{\mu}(\lambda).$
(10)

In the rest of the paper, we use notation $\tilde{\varphi}_{n}^{*}$ for the
polynomial from Lemma 2, where the value of the parameter
$\lambda\in\mathbb{D}$ will be clear from the context.

######  Lemma 3.

Let $\lambda\in\mathbb{D}$, and let $\tilde{\varphi}_{n}^{*}$ be the
corresponding polynomial from Lemma 2. We have

$\int_{\mathbb{T}}\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}-1\right|^{2}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm(\xi)=e^{\mathcal{K}_{\nu_{n,\lambda}}(\lambda)}-1.$
(11)

Proof. By (6) and Lemma 1, we have

$\frac{|\tilde{\varphi}_{n}^{*}(z)|^{2}-|\tilde{\varphi}_{n}(z)|^{2}}{1-|z|^{2}}=\sum_{k=0}^{n-1}|\varphi_{k}(z)|^{2},\qquad
z\in\mathbb{C}.$ (12)

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

$u(\lambda)=\int_{\mathbb{T}}u(\xi)\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm(\xi)$
(13)

for harmonic functions implies

$\int_{\mathbb{T}}\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}-1\right|^{2}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm=\int_{\mathbb{T}}\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}\right|^{2}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm-1,$
(14)

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

$\int_{\mathbb{T}}\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}\right|^{2}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm=|\tilde{\varphi}_{n}^{*}(\lambda)|^{2}\int_{\mathbb{T}}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,d\nu_{n,\lambda}=e^{\mathcal{K}_{\nu_{n,\lambda}}(\lambda)},$
(15)

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 (13), this time – for the harmonic function
$u=\log|\tilde{\varphi}_{n}^{*}|^{2}$). The lemma now follows from (14) and
(15). ∎

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

$\Gamma^{*}(\xi_{\pm},r)=\\{z\in\mathbb{C}:\;z^{*}\in\Gamma(\xi_{\pm},r)\\}.$

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
Figure 1.

Figure 1. Objects that appear in the proof of Theorem 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 4.

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

$\int_{\mathbb{T}}h(\xi)\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm(\xi)\leqslant\eta.$

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

$\int_{\mathbb{T}}h\,dm_{G}\lesssim\eta.$

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:
$m_{G}(B(\xi,2\rho))\leqslant 4m_{G}(B(\xi,\rho))$ 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

$m_{G}\left(\left\\{\xi:\sup_{\rho>0}\frac{1}{m_{G}(B(\xi,\rho))}\int_{B(\xi,\rho)}g\,dm_{G}>t\right\\}\right)\lesssim\frac{\|g\|_{L^{1}(G,m_{G})}}{t},$

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

$m_{G}\left(\left\\{\xi:\sup_{\rho>0}\frac{1}{m_{G}(B(\xi,\rho))}\int_{B(\xi,\rho)}h\,dm_{G}>\varepsilon^{-1}\eta\right\\}\right)\lesssim\varepsilon.$

It follows that

$m_{G}\left(\left\\{\xi:\sup_{\rho\in[0,1)}\int_{G}h(u)\frac{1-\rho^{2}}{|1-\rho\bar{u}\xi|^{2}}\,dm(u)>\varepsilon^{-1}\eta\right\\}\right)\lesssim\varepsilon.$
(16)

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

$\int_{G}h(u)\frac{1-\rho^{2}}{|1-\rho\bar{u}\xi|^{2}}\,dm(u)>\varepsilon^{-1}\eta$

then

$\frac{1}{m(B(\xi,\delta))}\int_{B(\xi,\delta)}\chi_{G}(u)h(u)\,dm(u)>\varepsilon^{-1}\eta$

for some $\delta>0$, and so

$\frac{1}{m_{G}(B(\xi,\delta))}\int_{B(\xi,\delta)}h(u)\,dm_{G}(u)>\varepsilon^{-1}\eta,$

proving (16). Let us now take $\varepsilon\in(0,1)$ so small that the left
hand side of (16) 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,

$\sup_{\rho\in[r,1)}\int_{G}h\frac{1-\rho^{2}}{|1-\rho\bar{u}\xi_{\pm}|^{2}}\,dm\leqslant\varepsilon^{-1}\eta.$

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

$\sup_{\rho\in[r,1)}\frac{1-\rho^{2}}{|1-\rho\bar{u}\xi_{\pm}|^{2}}\,dm\lesssim\frac{1-|\lambda|^{2}}{|1-\bar{u}\lambda|^{2}}.$

It follows that

$|f(\rho\xi_{\pm})|\lesssim\int_{\mathbb{T}\setminus
G}h\frac{1-|\lambda|^{2}}{|1-\bar{u}\lambda|^{2}}\,dm+\int_{G}h\frac{1-\rho^{2}}{|1-\rho\bar{\xi}\xi_{\pm}|^{2}}\,dm\lesssim(1+\varepsilon^{-1})\eta,$
(17)

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

$\sup_{\xi\in
G}\frac{1-r^{2}}{|1-r\bar{u}\xi|^{2}}\lesssim\frac{1-|\lambda|^{2}}{|1-\bar{u}\lambda|^{2}},\qquad
u\in\mathbb{T},$

therefore,

$|f(r\xi)|\lesssim\int_{\mathbb{T}}h(u)\frac{1-|\lambda|^{2}}{|1-\bar{u}\lambda|^{2}}\,dm(u)\lesssim\eta,$
(18)

for $\xi\in G$. Collecting (17) and (18), we see that $|f(z)|\lesssim\eta$ for
$z\in\Gamma(\xi_{\pm},r)$. ∎

Define $\tilde{b}_{n}=\frac{\tilde{\varphi}_{n}}{\tilde{\varphi}_{n}^{*}}.$
Formula (12) shows that $|\tilde{b}_{n}|\leqslant 1$ on $\mathbb{D}$. In fact,
$\tilde{b}_{n}$ is a Blaschke product of order $n$.

######  Lemma 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 (11). Set
$\alpha=\overline{\tilde{\varphi}_{n}^{*}(\lambda)}/\tilde{\varphi}_{n}^{*}(\lambda)$.
We have

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

Proof. Consider the sets

$E_{l}=\left\\{\xi\in\mathbb{T}:\;\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}-1\right|^{2}>1/4\right\\},\qquad
E_{s}=\mathbb{T}\setminus E_{l}.$

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 3 to estimate the corresponding integral:

$\int_{E_{l}}|\tilde{b}_{n}(\xi)-\alpha\xi^{n}|^{2}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\leqslant
4\int_{E_{l}}\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}\,dm\lesssim\eta.$

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

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

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

$\left|\frac{\tilde{\varphi}_{n}^{*}(\xi)}{\tilde{\varphi}_{n}^{*}(\lambda)}-1\right|=\left|\frac{\tilde{\varphi}_{n}^{*}(\xi)}{\tilde{\varphi}_{n}^{*}(\lambda)}\right|\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}-1\right|\leqslant
2\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(\xi)}-1\right|.$

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

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

on $E_{s}$. The claim now follow from Lemma 3. ∎

######  Lemma 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

$\displaystyle|\tilde{b}_{n}(z)-\alpha z^{n}|$ $\displaystyle\lesssim
e^{4A}\sqrt{\eta},$ (19)
$\displaystyle\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(z)}-1\right|$
$\displaystyle\lesssim e^{2A}\sqrt{\eta},$ (20)

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 5 and Lemma 3, we have

$\int_{\mathbb{T}}h(\xi)\frac{1-|\lambda|^{2}}{|1-\bar{\xi}\lambda|^{2}}dm(\xi)\lesssim\eta.$

Then, by Lemma 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

$|\tilde{b}_{n}(z)-\alpha
z^{n}|^{2}\lesssim\eta,\qquad\left|\frac{\tilde{\varphi}_{n}^{*}(\lambda)}{\tilde{\varphi}_{n}^{*}(z)}-1\right|^{2}\lesssim\eta,$

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

$\displaystyle|z^{n}|$ $\displaystyle\geqslant
r^{n}\geqslant(1-A/n)^{n}\geqslant e^{-2A},$ $\displaystyle|\tilde{b}_{n}(z)|$
$\displaystyle\geqslant|z^{n}|-|\alpha z^{n}-\tilde{b}_{n}|\geqslant
e^{-2A}-e^{-2A}/2\geqslant e^{-2A}/2,$

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

$\theta(1/\bar{z})=\overline{1/\theta(z)},\qquad z\in\mathbb{C},$

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

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

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

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

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 (20) on
$\Gamma^{*}(\xi_{\pm},r)$. ∎

######  Lemma 7.

Let $\lambda,A,n,r,\zeta,\eta$ be as in Lemma 6 and let $z_{1,2}\in
B(\zeta,1-|\lambda|)$. Then

$\left|\overline{\tilde{b}_{n}(z_{1})}\tilde{b}_{n}(z_{2})-\overline{z_{1}}^{n}z_{2}^{n}\right|\lesssim
e^{4A}\sqrt{\eta}\cdot|1-\bar{z}_{1}z_{2}|\cdot n.$

Proof. By Lemma 6, relations (19), (20) 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, (19) 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, (19) holds for all $z\in B(\zeta,1-r)$. Now pick two points
$z_{1},z$ such that

$|z_{1}-\zeta|\leqslant 1-|\lambda|,\qquad|z-\zeta|=2(1-|\lambda|)=1-r.$

We have

$\displaystyle\left|\frac{\overline{\tilde{b}_{n}(z_{1})}\tilde{b}_{n}(z)-\overline{z_{1}}^{n}z^{n}}{1-\bar{z}_{1}z}\right|$
$\displaystyle\lesssim\frac{\bigl{|}(\tilde{b}_{n}(z_{1})-\alpha
z^{n}_{1})(\tilde{b}_{n}(z)-\alpha z^{n})\bigr{|}}{1-r}+$
$\displaystyle+\frac{\bigl{|}\overline{z_{1}}^{n}(\tilde{b}_{n}(z)-\alpha
z^{n})\bigr{|}}{1-r}+\frac{\bigl{|}(\alpha
z_{1}^{n}-\tilde{b}_{n}(z_{1}))z^{n}\bigr{|}}{1-r}.$

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 (19)
imply

$|\tilde{b}_{n}(z_{1})-\alpha z^{n}_{1}|\lesssim
e^{2A}\sqrt{\eta},\qquad|\tilde{b}_{n}(z)-\alpha z^{n}|\lesssim
e^{2A}\sqrt{\eta}.$

Next, we have

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

Since $\eta\leqslant\eta_{0}\leqslant 1$, we can conclude that

$\left|\frac{\overline{\tilde{b}_{n}(z_{1})}\tilde{b}_{n}(z)-\overline{z_{1}}^{n}z^{n}}{1-\bar{z}_{1}z}\right|\lesssim
e^{4A}\sqrt{\eta}\cdot n.$

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. ∎

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

$\sup_{\rho\in[1-A/\tilde{N}_{0},1)}e^{\mathcal{K}_{\mu}(\rho\zeta)}-1<\eta_{0},$
(21)

where $\eta_{0}\in(0,1)$ is the number in Lemma 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

$\lambda=(1-\max\limits_{k=1,2}|z_{k}-\zeta|)\zeta.$

Define $\eta=e^{\mathcal{K}_{\nu_{n,\lambda}}(\lambda)}-1$ and observe that
$\eta<\eta_{0}$ by (21) and Lemma 2. We are in assumptions of Lemma 6.
Estimate (20) and the maximum modulus principle give

$\frac{\tilde{\varphi}_{n}^{*}(z)}{\tilde{\varphi}_{n}^{*}(\lambda)}=\frac{1}{1+R_{1}(z)},\qquad|R_{1}(z)|\lesssim
e^{2A}\sqrt{\eta},$

for all $z\in B(\zeta,1-|\lambda|)$. By Lemma 7, for all $z_{1},z_{2}\in
B(\zeta,1-|\lambda|)$ we have

$\displaystyle\frac{1-\overline{\tilde{b}_{n}(z_{1})}\tilde{b}_{n}(z_{2})}{1-\overline{z_{1}}z_{2}}$
$\displaystyle=\frac{1-\overline{z_{1}}^{n}z_{2}^{n}}{1-\overline{z_{1}}z_{2}}+R_{2}(z_{1},z_{2})\cdot
n,$

with $|R_{2}(z_{1},z_{2})|\lesssim e^{4A}\sqrt{\eta}$, so

$\displaystyle k_{\mu,n}(z_{1},z_{2})$
$\displaystyle=\left(\frac{1-\overline{z_{1}^{n}}z_{2}^{n}}{n(1-\overline{z_{1}}z_{2})}+R_{2}(z_{1},z_{2})\right)\cdot\frac{|\tilde{\varphi}_{n}^{*}(\lambda)|^{2}\cdot
n}{(1+R_{1}(z_{1}))\overline{(1+R_{1}(z_{2}))}}.$

Taking $z_{1}=z_{2}=\zeta$, we get

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

It remains to write

$\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}=\frac{\frac{1-\overline{z_{1}}^{n}z_{2}^{n}}{n(1-\overline{z_{1}}z_{2})}+R_{2}(z_{1},z_{2})}{1+R_{2}(\zeta,\zeta)}\frac{|1+R_{1}(\zeta)|^{2}}{(1+R_{1}(z_{1}))\overline{(1+R_{1}(z_{2}))}}.$

Note that

$\max_{z\in B(\zeta,1-|\lambda|)}|R_{1}(z)|+\max_{z_{1,2}\in
B(\zeta,1-|\lambda|)}|R_{2}(z_{1},z_{2})|\leqslant ce^{4A}\sqrt{\eta},$

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.

##  3\. Proof of Theorem 2

Proof of Theorem 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

$\delta<w(\xi)<2\delta,\qquad|\log w(\xi)-\log w(\zeta)|\sim|\xi-\zeta|^{s},$

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

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

We also have

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

By construction, we have

$\int_{I}u^{2}(\xi)\frac{1-\rho^{2}}{|1-\rho\bar{\xi}\zeta|^{2}}\,dm\lesssim\int_{I}|\xi-\zeta|^{2s}\frac{1-\rho^{2}}{|1-\rho\bar{\xi}\zeta|^{2}}\,dm.$
(22)

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

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

and

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

We have

$\int_{I\setminus
B(\zeta,\varepsilon)}|\xi-\zeta|^{2s-2}\,dm\sim\begin{cases}1,&s\in(1/2,+\infty),\\\
\log\frac{1}{\varepsilon},&s=1/2,\\\
\varepsilon^{2s-1},&s\in(0,1/2),\end{cases}$ (23)

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

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

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

$\mathcal{K}_{\mu}(\rho\zeta)\gtrsim\int_{\mathbb{T}}|u(\xi)-c|^{2}\frac{1-\rho^{2}}{|1-\rho\bar{\xi}\zeta|^{2}}\,dm$
(24)

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

$c_{1}(1-\delta)^{s}-c_{2}\delta^{s}>0.$

From (24) 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

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

Then

$|u(\xi_{1})-u(\xi_{2})|\lesssim\sqrt{\mathcal{K}_{\mu}(\rho\zeta)},$

and, simultaneously,

$|u(\xi_{1})-u(\xi_{2})|\geqslant
c_{1}((1-\delta)\varepsilon)^{s}-c_{2}(\delta\varepsilon)^{s}=\varepsilon^{s}(c_{1}(1-\delta)^{s}-c_{2}\delta^{s})\gtrsim\varepsilon^{s}.$

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

$c_{1}(1+\delta)^{s}/2-c_{2}>0.$

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

$(1+\delta)^{j}<|\xi_{1}-\zeta|<(1+\delta)^{j+1},\qquad(1+\delta)^{j+2}<|\xi_{2}-\zeta|<(1+\delta)^{j+3}.$

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

$\displaystyle|u(\xi_{1})-c|$
$\displaystyle\geqslant|u(\xi_{2})|-|u(\xi_{1})|-|u(\xi_{2})-c|$
$\displaystyle\geqslant
c_{1}|\xi_{2}-\zeta|^{s}-c_{2}|\xi_{1}-\zeta|^{s}-c_{1}|\xi_{2}-\zeta|^{s}/2$
$\displaystyle\gtrsim c_{1}(1+\delta)^{js+2s}/2-c_{2}(1+\delta)^{js+s}$
$\displaystyle\gtrsim(c_{1}(1+\delta)^{s}/2-c_{2})(1+\delta)^{js+s}$
$\displaystyle\gtrsim(1+\delta)^{js+s}\gtrsim|\xi_{1}-\zeta|^{s},$

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

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

and estimate (23) shows that

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

This ends the proof of the theorem. ∎

##  4\. Example: the Poisson kernel

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

######  Example 1.

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

$\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}=\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}+\delta_{n}(u,v),\qquad
n\to\infty,$

where $\sup\limits_{|u|,\,|v|\leqslant 1}|\delta_{n}(u,v)|$ is comparable to
$1/n$.

Proof. We have (see Section 1.6 in [14])

$\varphi_{n}(z)=\frac{z^{n}-\bar{\lambda}z^{n-1}}{\sqrt{1-|\lambda|^{2}}},\qquad\varphi^{*}_{n}(z)=\frac{1-\lambda
z}{\sqrt{1-|\lambda|^{2}}}.$

It follows that

$k_{\mu,n}(z,z)=\sum_{0}^{n-1}|\varphi_{k}(z)|^{2}=\sum_{0}^{n-1}\frac{|z^{k}-\bar{\lambda}z^{k-1}|^{2}}{1-|\lambda|^{2}}=\frac{|z-\bar{\lambda}|^{2}}{|z|^{2}(1-|\lambda|^{2})}\frac{1-|z|^{2n}}{1-|z|^{2}}.$

Then

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

and we see that

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

is comparable to

$\displaystyle|z-\bar{\lambda}|^{2}-|z|^{2}|\zeta-\bar{\lambda}|^{2}$
$\displaystyle=(1-|z|^{2})(|\lambda|^{2}+2\mathop{\rm
Re}(\lambda(\zeta|z|^{2}-z)))$
$\displaystyle\gtrsim(1-|z|^{2})|\lambda|^{2}\gtrsim\frac{1}{n},$

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

$\sup\limits_{|u|,\,|v|\leqslant 1}|\delta_{n}(u,v)|\gtrsim 1/n.$

Now take $z_{1}=\zeta e^{u/n}$, $z_{2}=\zeta e^{v/n}$, with
$|u|,\,|v|\leqslant 1$. We have

$\displaystyle k_{\mu,n}(z_{1},z_{2})$
$\displaystyle=\frac{1}{1-|\lambda|^{2}}\frac{(1-\lambda
z_{1})\overline{(1-\lambda
z_{2})}-z_{1}^{n}\bar{z}_{2}^{n}(1-\bar{\lambda}/z_{1})(1-\lambda/\bar{z}_{2})}{1-z_{1}\bar{z}_{2}}$
$\displaystyle=\frac{(1-\lambda z_{1})\overline{(1-\lambda
z_{2})}}{1-|\lambda|^{2}}\frac{1-z_{1}^{n}\bar{z}_{2}^{n}\theta_{\lambda}(z_{1},z_{2})}{1-z_{1}\bar{z}_{2}},$

where

$\theta_{\lambda}(z_{1},z_{2})=\frac{(1-\bar{\lambda}/z_{1})(1-\lambda/\bar{z}_{2})}{\overline{(1-\lambda
z_{2})}(1-\lambda z_{1})}=1+O(1/n).$

In particular, we have

$\displaystyle\frac{k_{\mu,n}(z_{1},z_{2})}{k_{\mu,n}(\zeta,\zeta)}$
$\displaystyle=\frac{(1-\lambda z_{1})\overline{(1-\lambda
z_{2})}}{|1-\lambda\zeta|^{2}}\frac{1-z_{1}^{n}\bar{z}_{2}^{n}\theta_{\lambda}(z_{1},z_{2})}{n(1-z_{1}\bar{z}_{2})}.$

Here

$\frac{(1-\lambda z_{1})\overline{(1-\lambda
z_{2})}}{|1-\lambda\zeta|^{2}}=1+O(1/n),$

and

$\frac{1-z_{1}^{n}\bar{z}_{2}^{n}\theta_{\lambda}(z_{1},z_{2})}{n(1-z_{1}\bar{z}_{2})}=\frac{k_{m,n}(z_{1},z_{2})}{k_{m,n}(\zeta,\zeta)}+z_{1}^{n}\bar{z}_{2}^{n}\frac{\theta_{\lambda}(z_{1},z_{2})-1}{n(1-z_{1}\bar{z}_{2})}.$

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

$\left|z_{1}^{n}\bar{z}_{2}^{n}\frac{\theta_{\lambda}(z_{1},z_{2})-1}{n(1-z_{1}\bar{z}_{2})}\right|\lesssim\frac{1}{n}$

for all $z_{1}$, $z_{2}$ such that $|z_{1,2}-\zeta|\lesssim 1/n$. It follows
that $\sup\limits_{|u|,\,|v|\leqslant 1}|\delta_{n}(u,v)|\leqslant 1/n$. ∎

##  5\. Sharpness of Theorem 1.

Sharpness of Theorem 1 is an open problem. As we have seen in the previous
section, one can have

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

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

$\sup_{|z_{1,2}-1|\leqslant
1/n}\left|\frac{k_{w_{s},n}(z_{1},z_{2})}{k_{w_{s},n}(1,1)}-\frac{1-\overline{z_{2}}^{n}z_{1}^{n}}{n(1-\overline{z_{2}}z_{1})}\right|\gtrsim\sqrt{\mathcal{K}_{w_{s}}(1-1/n)},$

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 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).

    %% Clear window and variables
    clc
    clear vars

    %% Initialization

    % Maximal degree
    N=8000;

    % Step (N/s must be positive integer)
    step = 20;

    % Holder index
    s = 0.4;

    % Initial difference
    diffold = 0;

    % Auxiliary arrays
    MMNTS=zeros(N,1);              % array for moments
    D = zeros(1,N/step);           % array for differences
    alphaCand = zeros(1,N/step);   % array for Holder index candidates
    CalphaCand = zeros(1,N/step);  % array for C_alpha candidates

    %% Computing the moments of the weight e^{|x|^s}, x in [-pi, pi]
    for j=0:1:N-1
    fun = @(x) cos(j*x).*exp(abs(x).^s);     % the weight is even
    MMNTS(j+1) = integral(fun,-pi,pi);
    end

    %% Cycle for comparing the ratios of kernels for different n
    for n=step:step:N
    disp([’n = ’, num2str(n)]);
    % Define the reference point
    xn = 1-1/n;
    % Compute the free reproducing kernel of order n-1 at (xn, xn)
    RepKer0atxnxn =((xn).^(0:1:n-2))*((xn).^(0:1:n-2))’;
    % Compute the free reproducing kernel of order n-1 at (xn, xn)
    RepKer0at11 = n-1;
    % Define the Toeplitz matrix
    % and the basis vector e_n = (0,0..., 0,1)’
    j=0:1:n-1;
    moments = double(MMNTS(1:n));
    T = toeplitz(moments);
    eo = double(1:n == n)’;
    % Use MATLAB function invToeplitz() by John P. Cunningham for fast
    % invertion of the Toeplitz matrix T to compute coeffitients of OPs
    % see https://github.com/aecker/gpfa.git
    coefOP=((invToeplitz(T))*eo)’;
    coefRefOP = conj(flip(coefOP));
    % Compute value of OP and reflected OP at 1
    OPat1 = coefOP*ones(n,1);
    RefOPat1 = coefRefOP*ones(n,1);
    % Compute value of OP and reflected OP at xn
    OPatxn = coefOP*((xn).^(0:1:n-1))’;
    RefOPatxn = coefRefOP*((xn).^(0:1:n-1))’;
    % Compute coeffitients of reproducing kernel at 1
    % Numerator
    RepKerNumat1 = coefRefOP*conj(RefOPat1) - coefOP*conj(OPat1);
    % Denumerator
    RepKerDenat1 = [1, -1];
    % Kernel
    [RepKerat1, r1] = deconv(flip(RepKerNumat1), flip(RepKerDenat1));
    % Compute coeffitients of reproducing kernel at xn
    % Numerator
    RepKerNumatxn = coefRefOP*conj(RefOPatxn) - coefOP*conj(OPatxn);
    % Denumerator
    RepKerDenatxn = [1, -conj(xn)];
    % Kernel
    [RepKeratxn,rxn]=deconv(flip(RepKerNumatxn),flip(RepKerDenatxn));
    % Compute the value of reproducing kernel at (1,1)
    RepKerat11 = flip(RepKerat1)*ones(n-1,1);
    % Compute the value of reproducing kernel at (xn,xn)
    RepKeratxnxn = flip(RepKeratxn)*((xn).^(0:1:n-2))’;

    %% Compare the ratios
    ratio0 = double(RepKer0atxnxn/RepKer0at11);
    ratio1 = double(RepKeratxnxn/RepKerat11);
    diff = ratio0-ratio1;
    D(n/step) = abs(diff);
    % Assuming power-type law "C_alpha n^{-alpha}", find the candidate
    % for alpha for the current value of n
    alphaCand(n/step) = (n/step).*(diffold./diff - 1);
    % Assuming power-type law "C_alpha n^{-alpha}", find the candidate
    % for C_alpha for the current value of n
    CalphaCand(n/step) = diff.*(n.^alphaCand(n/step));
    diffold = diff;
    end

This script produces the array `D` consisting of differences

$\verb|D(n/20)|=\left|\frac{k_{w_{s},n}(x_{n},x_{n})}{k_{w_{s},n}(1,1)}-\frac{1-|x_{n}|^{2n}}{n(1-|x_{n}|^{2})}\right|,\qquad
n=20,40,60,\ldots,8000,$

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

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

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

$\verb|alphaCand(n/20)|=\frac{n}{20}\left(\frac{D(n/20-1)}{D(n/20)}-1\right),\qquad
n=20,40,\ldots,8000.$

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

$\ldots\quad 0.39361\quad 0.39362\quad 0.39363\quad 0.39363\quad 0.39364\quad
0.39365\quad 0.39365$

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

$\left|\frac{k_{w_{s},n}(x_{n},x_{n})}{k_{w_{s},n}(1,1)}-\frac{1-|x_{n}|^{2n}}{n(1-|x_{n}|^{2})}\right|\geqslant\frac{C_{0.4}}{n^{0.4}}$

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

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

The plots of the resulting functions

$f_{1}(n)=\left|\frac{k_{w_{s},n}(x_{n},x_{n})}{k_{w_{s},n}(1,1)}-\frac{1-|x_{n}|^{2n}}{n(1-|x_{n}|^{2})}\right|,\qquad
f_{2}(n)=C_{0.4}n^{-0.4},$

are given below. We also consider the cases $s=0.1$ and $s=0.2$.

Figure 2. Graphs of functions $f_{1}$, $f_{2}$ for $s=0.1$, $s=0.2$, $s=0.4$.

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 $s\in(0,1/2)$.

## References

  * [1] R. Bessonov.  Entropy function and orthogonal polynomials.  J. Approx. Theory, 272:Paper No. 105650, 16, 2021.
  * [2] R. Bessonov and S. Denisov.  A spectral Szegő theorem on the real line.  Advances in Mathematics, 359, 2020.
  * [3] R. Bessonov and S. Denisov.  De Branges canonical systems with finite logarithmic integral.  Anal. PDE, 14(5):1509–1556, 2021.
  * [4] R. Bessonov and S. Denisov.  Zero sets, entropy, and pointwise asymptotics of orthogonal polynomials.  J. Funct. Anal., 280(12):Paper No. 109002, 38, 2021.
  * [5] R. Bessonov and S. Denisov.  Szego condition, scattering, and vibration of Krein strings.  Inventiones mathematicae, doi:10.1007/s00222-023-01201-9, 2023.
  * [6] P. A. Deift.  Orthogonal polynomials and random matrices: a Riemann-Hilbert approach, volume 3 of Courant Lecture Notes in Mathematics.  New York University, Courant Institute of Mathematical Sciences, New York; American Mathematical Society, Providence, RI, 1999.
  * [7] B. Eichinger, M. Lukic, and B. Simanek.  An approach to universality using Weyl $m$-functions.  preprint arXiv:2108.01629, 21.
  * [8] E. Findley.  Universality for locally Szegő measures.  J. Approx. Theory, 155(2):136–154, 2008.
  * [9] A. B. J. Kuijlaars and M. Vanlessen.  Universality for eigenvalue correlations from the modified Jacobi unitary ensemble.  Int. Math. Res. Not., 30:575–1600, 2002.
  * [10] D. Lubinsky.  A new approach to universality limits involving orthogonal polynomials.  Ann. of Math. (2), 170(2):915–939, 2009.
  * [11] D. S. Lubinsky.  Universality limits in the bulk for arbitrary measures on compact sets.  J. Anal. Math., 106:373–394, 2008.
  * [12] A. Máté, P. Nevai, and V. Totik.  Szegő’s extremum problem on the unit circle.  Ann. of Math. (2), 134(2):433–453, 1991.
  * [13] A. Poltoratski.  Pointwise convergence of the non-linear Fourier transform.  Preprint arXiv:2103.13349, 2021.
  * [14] B. Simon.  Orthogonal Polynomials on the Unit Circle. Part 1: Classical Theory.  Colloquium Publications. American Mathematical Society, 2004.
  * [15] B. Simon.  The Christoffel-Darboux kernel.  In Perspectives in partial differential equations, harmonic analysis and applications, volume 79 of Proc. Sympos. Pure Math., pages 295–335. Amer. Math. Soc., Providence, RI, 2008.
  * [16] V. Totik.  Universality under Szegő’s condition.  Canad. Math. Bull., 59(1):211–224, 2016.