# Analytic capacities in Besov spaces

Anton Baranov  Department of Mathematics and Mechanics, St. Petersburg State
University, 28, Universitetskii prosp., 198504 Staryi Petergof, Russia
<EMAIL_ADDRESS>,  Michael Hartz  Fachrichtung Mathematik,
Universität des Saarlandes, 66123 Saarbrücken, Germany<EMAIL_ADDRESS>,
Ilgiz Kayumov  Department of Mathematics and Computer Science, St. Petersburg
State University, Russia, 14 line of the VO, 29B, 199178 St. Petersburg,
Russia<EMAIL_ADDRESS>and  Rachid Zarouf  Aix-Marseille University,
Laboratoire ADEF, Campus Universitaire de Saint-Jérôme, 52 Avenue Escadrille
Normandie Niemen, 13013 Marseille, France<EMAIL_ADDRESS>
###### Abstract.

We derive new estimates on analytic capacities of finite sequences in the unit
disc in Besov spaces with zero smoothness, which sharpen the estimates
obtained by N. K. Nikolski in 2005 and, for a range of parameters, are
optimal. The work is motivated both from the perspective of complex analysis
by the description of sets of zeros/uniqueness, and from the one of matrix
analysis/operator theory by estimates on norms of inverses.

M.H. was partially supported by the Emmy Noether Program of the German
Research Foundation (DFG Grant 466012782). The work of I.K. in Sections 5 and
6 was supported by Russian Science Foundation (grant 23-11-00153). The work of
R.Z. was supported by the pilot center Ampiric, funded by the France 2030
Investment Program operated by the Caisse des Dépôts.

##  1\. Introduction

Let $\mathbb{D}=\left\\{z\in\mathbb{C}:\>|z|<1\right\\}$ be the open unit
disk, let $\mathbb{T}=\left\\{z\in\mathbb{C}:\>|z|=1\right\\}$ be its boundary
and $\mathbb{D}_{*}=\mathbb{D}\setminus\\{0\\}$. We denote by
$\mathcal{H}ol(\mathbb{D})$ the space of analytic functions on $\mathbb{D}$,
equipped with the topology of local uniform convergence. Let $X$ be a Banach
space that is continuously contained in $\mathcal{H}ol(\mathbb{D})$ and that
contains the polynomials. Given a finite sequence
$\sigma=(\lambda_{1},\dots,\lambda_{N})\in\mathbb{D}_{*}^{N}$, Nikolski [9]
defined the $X$-zero capacity of $\sigma$ as

${\rm
cap}_{X}(\sigma)=\inf\\{\left|\\!\left|f\right|\\!\right|_{X}\>:\>f(0)=1,\>f|\sigma=0\\},$

where $f|\sigma=0$ means that $f(\lambda_{i})=0$ for all $i=1,\dots,N$ taking
into account possible multiplicities. Namely, if
$\sigma=(\lambda_{1},\dots,\lambda_{1},\lambda_{2},\dots,\lambda_{2},\dots,\lambda_{s},\dots,\lambda_{s})\in\mathbb{D}^{N}$,
where each $\lambda_{i}$ is repeated according to its multiplicity $m_{i}\geq
1,$ then $f|\sigma=0$ means that

$f(\lambda_{i})=f^{\prime}(\lambda_{i})=f^{\prime\prime}(\lambda_{i})=\dots=f^{(m_{i}-1)}(\lambda_{i})=0,\qquad
i=1,\dots,s.$

The latter quantity is closely related on one hand to the problem of
uniqueness sets for the function space $X$ and on the other hand to condition
numbers of large matrices and inverses, as observed by Nikolski [9, Section
1]. We briefly review these connections here.

###  1.1. Motivation from complex analysis: sets of zeros/uniqueness

From the point of view of complex analysis, the $X$-zero capacities are
closely related to the problem of characterizing uniqueness sets for the
function space $X$; here $\sigma$ is said to be a uniqueness set for $X$ if
$f\in X,\>f|\sigma=0\implies f=0$. Following [9], assume that the function
space $X$ satisfies the following Fatou property: if $f_{n}\in X$,
$\sup_{n}|\\!|f_{n}|\\!|_{X}<\infty$ and $\lim_{n\to\infty}f_{n}(z)=f(z)$ for
$z\in\mathbb{D}$, then $f\in X$. Then it is not hard to see that an infinite
sequence $\sigma=(\lambda_{i})_{i\geq 1}\in\mathbb{D}_{*}^{\infty}$ is a
uniqueness sequence for $X$ if and only if

(1.1) $\sup_{N}\\{{\rm cap}_{X}(\sigma_{N})\\}=\infty,$

where $\sigma_{N}=(\lambda_{i})_{i=1}^{N}$ is the truncation of $\sigma$ of
order $N$. For example, let $X$ be the algebra $H^{\infty}$ of bounded
holomorphic functions in $\mathbb{D}$ endowed with the norm
$|\\!|f|\\!|_{H^{\infty}}=\sup_{\zeta\in\mathbb{D}}|f(\zeta)|$. It is known
[9, Theorem 3.12] that given
$\sigma_{N}=(\lambda_{i})_{i=1}^{N}\in\mathbb{D}_{*}^{N}$,

(1.2) ${\rm
cap}_{H^{\infty}}(\sigma_{N})=\frac{1}{\prod_{i=1}^{N}|\lambda_{i}|}.$

Denoting by

$B=B_{\sigma_{N}}=\prod_{i=1}^{N}\frac{z-\lambda_{i}}{1-\overline{\lambda_{i}}z}$

the finite Blaschke product associated with $\sigma_{N},$ observe that the
right-hand side in (1.2) is achieved by the test function $f=B/B(0)$, which is
admissible for the conditions in the infimum defining the capacity of
$\sigma_{N}$. Thus, an application of the above criterion (1.1) leads to the
well-known Blaschke condition: an infinite sequence
$\sigma=(\lambda_{i})_{i\geq 1}\in\mathbb{D}_{*}^{\infty}$ is a uniqueness
sequence for $H^{\infty}$ if and only if

$\sum_{i\geq 1}(1-|\lambda_{i}|)=\infty.$

###  1.2. Motivation in operator theory/matrix analysis

Let $T$ be an invertible operator acting on a Banach space or an $N\times N$
invertible matrix with complex entries acting on $\mathbb{C}^{N}$ equipped
with some norm. We seek upper bounds on the norm of the inverse $T^{-1}$.
Assume that the minimal polynomial of $T$ is given by

$m(z)=m_{T}(z)=\prod_{i=1}^{N}(z-\lambda_{i}),$

where $\sigma=(\lambda_{i})_{i=1}^{N}\in\mathbb{D}_{*}^{N}$ and we assumed for
simplicity that $\deg m_{T}=N$. Following [9], assume that our Banach space
$X\subset\mathcal{H}ol(\mathbb{D})$ is an in fact an algebra, and write $A=X$.
Assume further that

  1. (1)

$T$ admits a $C$-functional calculus on $A$, i.e. there exists a bounded
homomorphism $f\mapsto f(T)$ extending the polynomial functional calculus and
a constant $C>0$ such that

$\left|\\!\left|f(T)\right|\\!\right|\leq
C\left|\\!\left|f\right|\\!\right|_{A},\qquad f\in A;$

  2. (2)

the shift operator $S:f\mapsto zf$, the backward shift operator
$S^{*}:f\mapsto\frac{f-f(0)}{z}$ and the generalized backward shift operators
$f\mapsto\frac{f-f(\lambda)}{z-\lambda}$ are bounded on $A$ for all
$\lambda\in\mathbb{D}$.

These assumptions are mild and satisfied by all the algebras $A$ considered
below. Noticing that the analytic polynomial $P=\frac{m(0)-m}{zm(0)}$
interpolates the function $\frac{1}{z}$ on $\sigma$ we observe that

$T^{-1}=P(T)=(P+mh)(T)$

for any $h\in A$. Applying assumption (1) to the above operator we obtain

$|\\!|T^{-1}|\\!|\leq C|\\!|P+mh|\\!|_{A}$

and taking the infimum over all $h\in A$ and using our assumptions on $A$, we
get

(1.3) $|\\!|T^{-1}|\\!|\leq
C\inf\Big{\\{}\left|\\!\left|g\right|\\!\right|_{A}:\
g|\sigma=P|\sigma=\frac{1}{z}\Big{|}\sigma\Big{\\}}.$

Now, if $f\in A$ satisfies $f(0)=1$ and $f|\sigma=0$, then $g:=S^{*}(1-f)$ is
admissible for the last infimum, and so

(1.4) $|\\!|T^{-1}|\\!|\leq C|\\!|S^{*}|\\!|_{A\to
A}\left(|\\!|1|\\!|_{A}+{\rm cap}_{A}(\sigma)\right).$

In particular (1.3) and (1.4) are applied (among other situations) in [9] to
the cases of:

  * •

Hilbert space contractions, $A$ the disc algebra and $C=1$;

  * •

Banach space contractions, $A$ the Wiener algebra of absolutely convergent
Taylor/Fourier series,

$A=W=\\{f=\sum_{k\geq
0}\hat{f}(k)z^{k}\in\mathcal{H}ol(\mathbb{D}):\>\left|\\!\left|f\right|\\!\right|_{W}=\sum_{k\geq
0}|\hat{f}(k)|<\infty\\},$

and once again $C=1$;

  * •

Tadmor–Ritt type matrices or power-bounded matrices on Hilbert spaces and $A$
the Besov algebra

$A=B_{\infty,1}^{0}=\left\\{f\in\mathcal{H}ol(\mathbb{D}):\>\left|\\!\left|f\right|\\!\right|_{B_{\infty,1}^{0}}=|f(0)|+\int_{0}^{1}\left|\\!\left|f_{\rho}^{\prime}\right|\\!\right|_{L^{\infty}(\mathbb{T})}{\rm
d}\rho<\infty\right\\},$

where $f_{\rho}(\zeta)=f(\rho\zeta),\>\zeta\in\mathbb{T}$.

### Outline of the paper

In Section 2 below we first review Nikolski’s upper estimates on ${\rm
cap}_{X}(\sigma)$ where $X$ is a general Besov space $\mathcal{B}_{p,q}^{s},$
$s\geq 0,\>(p,q)\in[1,\infty]^{2}$, see below for their definition. We also
relate the special case $(p,q)=(\infty,1)$ to applications in operator
theory/matrix analysis and especially to Schäffer’s question on norms of
inverses.

In Section 3 we formulate the main results of the paper. Theorem 2, which
corresponds to the special case $(p,q)=(\infty,1)$, exhibits an explicit
sequence $\sigma^{\star}$ for which we derive a quantitative lower bound on
${\rm cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma^{\star})$ and thereby $almost$
prove the sharpness of Nikolski’s upper bound in this case. Theorem 3 improves
Nikolski’s upper bounds on ${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)$ for a
range of parameters, while in Theorem 4 the sharpness of these new bounds is
discussed.

In Section 4 we prove Lemma 6, which is our main tool for bounding the
capacities from below. In Section 5 we prove Theorem 2. The proofs of the
lower bounds in Theorem 4 are provided in Section 6. Finally, in Section 7 we
prove the upper bounds stated in Theorem 3. The proof is based on estimates of
Besov norms of finite Blaschke products (Proposition 10) which may be of
independent interest.

##  2\. Known results and open questions

###  2.1. Capacities in Besov spaces

The case where $X$ is an analytic Besov space $X=\mathcal{B}_{p,q}^{s}$ is
considered in [9]. Let $s\geq 0,$ $1\leq p,\,q\leq\infty$ and let

$\mathcal{B}_{p,q}^{s}=\left\\{f\in\mathcal{H}ol(\mathbb{D}):\>\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,q}^{s}}^{*}=\left(\int_{0}^{1}((1-\rho)^{m-s-1/q}\left|\\!\left|f_{\rho}^{(m)}\right|\\!\right|_{L^{p}(\mathbb{T})})^{q}{\rm
d}\rho\right)^{1/q}<\infty\right\\},$

where $f_{\rho}^{(m)}(\zeta)=f^{(m)}(\rho\zeta),$ $m$ being a nonnegative
integer such that $m>s$ (the choice of $m$ is not essential and the norms for
different $m$-s are equivalent). We need to make the obvious modification for
$q=\infty$. The space $\mathcal{B}_{p,q}^{s}$ equipped with the norm

$\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,q}^{s}}=\sum_{k=0}^{m-1}|f^{(k)}(0)|+\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,q}^{s}}^{*}$

is a Banach space. We refer to [5, 11, 17] for general properties of Besov
spaces. Note that for $1\leq q<\infty$ we have $f_{\rho}\to f$ in the norm of
$\mathcal{B}_{p,q}^{s}$ as $\rho\to 1-$.

In the present paper we deal with Besov spaces with zero smoothness $s=0$. In
this case we take $m=1$ and

$\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,q}^{0}}^{*}=\left(\int_{0}^{1}(1-\rho)^{q-1}\left|\\!\left|f^{\prime}_{\rho}\right|\\!\right|_{L^{p}(\mathbb{T})}^{q}{\rm
d}\rho\right)^{1/q},\qquad 1\leq q<\infty,$
$\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,\infty}^{0}}^{*}=\sup_{0<\rho<1}(1-\rho)\|f^{\prime}_{\rho}\|_{L^{p}(\mathbb{T})}.$

Note that $\mathcal{B}_{\infty,\infty}^{0}$ coincides with the classical Bloch
space.

It is shown [9, Theorem 3.26] that given $1\leq p,\,q\leq\infty$, $s>0$ and
$\sigma\in\mathbb{D}_{*}^{N}$ the following upper estimate holds

${\rm cap}_{\mathcal{B}_{p,q}^{s}}(\sigma)\leq
c\frac{N^{s}}{\prod_{i=1}^{N}|\lambda_{i}|},$

where $c=c(s,q),$ and that if $s=0$ then

(2.1) ${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\leq c\frac{\left(\log
N\right)^{1/q}}{\prod_{i=1}^{N}|\lambda_{i}|},$

where $c>0$ is a numerical constant. It is also shown that for $s>0$ these
estimates are asymptotically sharp in the following sense [9, Theorem 3.31]:
there exist constants $c=c(s,p,q)>0$ and $K=K(s,p,q)>0$ such that for any
$\sigma=(\lambda_{1},\dots,\lambda_{N})\in\mathbb{D}_{*}^{N}$, $s>0$, $1\leq
p,q\leq\infty$,

${\rm cap}_{\mathcal{B}_{p,q}^{s}}(\sigma)\geq
c\frac{N^{s}}{\prod_{i=1}^{N}|\lambda_{i}|}\left(1+K-\prod_{i=1}^{N}(1+|\lambda_{i}|)\right).$

The sharpness of the upper bound in (2.1) is left as an open question in [9].

###  2.2. Norms of inverses and Schäffer’s question

Let $\left|\\!\left|\cdot\right|\\!\right|$ denote the operator norm induced
on $\mathcal{M}_{N}$ by a Banach space norm on $\mathbb{C}^{N}$. What is the
smallest constant $\mathcal{S}_{N}$ so that

$|\det{T}|\cdot\left|\\!\left|T^{-1}\right|\\!\right|\leq\mathcal{S}_{N}\left|\\!\left|T\right|\\!\right|{}^{N-1}$

holds for any invertible matrix $T\in\mathcal{M}_{N}$ and any operator norm
$\left|\\!\left|\cdot\right|\\!\right|$? Schäffer [14, Theorem 3.8] proved
that

$\mathcal{S}_{N}\leq\sqrt{eN},$

but he conjectured that $\mathcal{S}_{N}$ should in fact be bounded, as it is
the case for Hilbert space. This conjecture was disproved in the early 90’s by
E. Gluskin, M. Meyer, and A. Pajor [6]. Later, Queffélec [13] showed that the
$\sqrt{N}$ bound is essentially optimal for arbitrary Banach spaces, but both
arguments are non-constructive. An explicit construction giving a $\sqrt{N}$
lower bound was recently given in [16]. For a detailed account on the history
of Schäffer’s question, the reader is referred to [16]. A key tool in the
works cited above is the equality

(2.2)
$\displaystyle\mathcal{S}_{N}=\sup_{\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}^{N}}\prod_{i=1}^{N}|\lambda_{i}|\left({\rm
cap}_{W}\left(\lambda_{1},\dots,\lambda_{N}\right)-1\right),$

due to Gluskin, Meyer and Pajor. It connects Schäffer’s question to capacity
in the Wiener algebra and shows that (1.4) is essentially sharp in this case.

It is natural to consider Schäffer’s question for operator classes different
from Hilbert or Banach space contractions. In particular, following [9], we
may consider the following classes, which admit a Besov
$\mathcal{B}^{0}_{\infty,1}$-functional calculus.

  1. (1)

Power bounded operators on Hilbert space, i.e. operators $T$ on Hilbert space
satisfying

$\sup_{k\geq 0}\left|\\!\left|T^{k}\right|\\!\right|=C_{pb}<\infty.$

Peller [12] proved that $\left|\\!\left|f(T)\right|\\!\right|\leq
k_{G}C_{pb}^{2}\left|\\!\left|f\right|\\!\right|_{B_{\infty,1}^{0}}$ for every
analytic polynomial $f$, where $k_{G}$ is the Grothendieck constant. Combining
(1.4) with Nikolski’s upper estimate (2.1) for $q=1$, we obtain the upper
bounds

(2.3) $\left|\\!\left|T^{-1}\right|\\!\right|\leq c_{1}\cdot{\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}\left(\lambda_{1},\dots,\lambda_{N}\right)\leq
c_{3}\frac{k_{G}C_{pb}^{2}\log N}{\prod_{i=1}^{N}|\lambda_{i}|},$

where $c_{1}>0$ is an absolute constant and $(\lambda_{i})_{i=1}^{N}$ is the
sequence of eigenvalues of $T$.

  2. (2)

Tadmor–Ritt operators on Banach space, i.e. operators $T$ acting on a Banach
space and satisfying the resolvent estimate

$\sup_{|\zeta|>1}|\zeta-1|\left|\\!\left|(\zeta-T)^{-1}\right|\\!\right|=C_{TR}<\infty.$

According to P. Vitse’s functional calculus [18, Theorem 2.5] we have
$\left|\\!\left|f(T)\right|\\!\right|\leq
300C_{TR}^{5}\left|\\!\left|f\right|\\!\right|_{B_{\infty,1}^{0}}$ for every
analytic polynomial $f$, and following the same reasoning as above this yields

(2.4) $\left|\\!\left|T^{-1}\right|\\!\right|\leq c_{2}\cdot{\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}\left(\lambda_{1},\dots,\lambda_{N}\right)\leq
c_{2}\frac{300C_{TR}^{5}\log N}{\prod_{i=1}^{N}|\lambda_{i}|},$

where $c_{2}>0$ is an absolute constant. In fact, thanks to work of
Schwenninger [15], the dependence on $C_{TR}$ can be improved from
$C_{TR}^{5}$ to $C_{TR}(\log C_{TR}+1)$.

The sharpness of the right-hand side in (2.3) and (2.4) is an open question
both from the point of view of operators/matrices and from the one of
capacities. Note that we have the following (strict) inclusions:

(2.5) $W\subset B_{\infty,1}^{0}\subset H^{\infty}$

(see [5, 11] or [10, Section B.8.7]). Observe that $B_{\infty,1}^{0}$ is
actually contained in the disc algebra. From the perspective of capacities
(2.5) implies that for any sequence
$\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}_{*}^{N}$ we
have

(2.6) ${\rm cap}_{H^{\infty}}(\sigma)\leq c_{3}{\rm
cap}_{B_{\infty,1}^{0}}(\sigma)\leq c_{4}{\rm cap}_{W}(\sigma)$

where $c_{3},c_{4}>0$ are absolute constants. Observe that in view of (2.6)
and (2.2) any sequence $\sigma\in\mathbb{D}_{*}^{N}$ such that
$\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm cap}_{B_{\infty,1}^{0}}(\sigma)$ grows
unboundedly in $N$ will automatically give a counterexample to Schäffer’s
original question.

##  3\. Main results

Throughout this paper, we will use the following standard notation. For two
positive functions $f,g$ we say that $f$ is dominated by $g$, denoted by
$f\lesssim g$, if there is a constant $c>0$ such that $f\leq cg$ for all
admissible variables. We say that $f$ and $g$ are comparable, denoted by
$f\asymp g$, if both $f\lesssim g$ and $g\lesssim f$.

The main goals of this paper are to

  1. (1)

Provide an example of a sequence
$\sigma^{\star}=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}_{*}^{N}$
such that $\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma^{\star})$ $almost$ (up to a double
logarithmic factor) approaches of Nikolski’s upper bound $\log N$.

  2. (2)

Improve Nikolski’s upper bound (2.1) on $\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)$ identifying three regions of
$(p,q)\in[1,\infty]^{2}$ with a different behavior of this quantity (see
Theorem 3 below). For all $(p,q)$ with $p\neq\infty$ our estimates give a
smaller growth than the estimates in [9], and for a range of parameters,
namely for $1\leq q\leq p<\infty$ and $p\geq 2$, they are best possible.

###  3.1.  A lower estimate on ${\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma)$

Our approach to bounding ${\rm cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma)$ from
below uses duality. To estimate ${\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma)$ from below, we estimate the Besov
seminorm in $\mathcal{B}_{1,\infty}^{0}$ of finite Blaschke products from
above. The key inequality is

(3.1) ${\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma)\gtrsim\frac{1}{\prod_{i=1}^{N}|\lambda_{i}|}\frac{1-\prod_{i=1}^{N}|\lambda_{i}|^{2}}{\left|\\!\left|B\right|\\!\right|_{\mathcal{B}_{1,\infty}^{0}}^{*}},$

where $\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)$ is an arbitrary
sequence in $\mathbb{D}_{*}^{N}$, and $B=B_{\sigma}$ is the finite Blaschke
product associated to $\sigma$. To conclude we consider $n\geq 2$ and for
$k=1,\dots,n$ we put

$\sigma_{k}=(r_{k}^{(n)}e^{2i\pi
j/2^{k}})_{j=1}^{2^{k}}\in\mathbb{D}_{*}^{2^{k}},\qquad
r_{k}^{(n)}=\left(1-1/n\right)^{2^{-k}}.$

We put $N=\sum_{k=1}^{n}2^{k}\asymp 2^{n}$ and define the sequence
$\sigma^{\star}=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}_{*}^{N}$
by

(3.2) $\sigma^{\star}=(\sigma_{1},\,\sigma_{2},\,\dots,\sigma_{n}).$

Denoting by $B^{\star}$ the Blaschke product associated with $\sigma^{\star}$
we have

(3.3) $B^{\star}(z)=\prod_{k=1}^{n}\frac{z^{2^{k}}-a}{1-az^{2^{k}}},$

where $a=1-\frac{1}{n}.$ We will prove the following result.

######  Proposition 1.

The Blaschke product $B^{\star}$ satisfies

(3.4) $\|B^{\star}\|_{\mathcal{B}_{1,\infty}^{0}}^{*}\lesssim\frac{\log\log
N}{\log N}.$

Taking into account that $\prod_{j=1}^{N}|\lambda_{j}|\leq e^{-1}$ and
combining (3.1) with (3.4) we obtain the following theorem.

######  Theorem 2.

Let $\sigma^{\star}\in\mathbb{D}_{*}^{N}$ and $B^{\star}$ be defined by (3.2)
and (3.3). Then

$\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma^{\star})\gtrsim\frac{\log N}{\log\log
N}.$

As a consequence regarding Schäffer’s question, Theorem 2 implies (taking into
account (2.6)) that

$\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm cap}_{W}(\sigma^{*})\gtrsim\frac{\log
N}{\log\log N}.$

From this, following arguments in [16], one obtains another explicit
counterexample to Schäffer’s question, acting as multiplication by $z$ on the
quotient $W/B^{\star}W$ of the Wiener algebra. One can identify the dual space
of $W/B^{\star}W$ with the space of rational functions of degree at most $N$
with poles at $1/\bar{\lambda}_{j}$ for $j=1,\dots,N$, equipped with the
supremum norm of the Taylor coefficients. Then, as in [16, Theorem 8], one
obtains another explicit matrix that serves as a counterexample to Schäffer’s
question.

###  3.2.  Upper bounds on ${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)$ for
general values of $(p,q)\in[1,\infty]^{2}$

In the following statements the constants in $\lesssim$ relations may depend
on $p,q$, but not on $N$.

######  Theorem 3.

Given $(p,q)\in[1,\infty]^{2}$ and
$\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}_{*}^{N}$, the
following upper estimates on ${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)$ hold
depending on the region to which $(p,q)$ belongs.

1) If $(p,q)\in[1,2]^{2}$ (Region I), then

${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\lesssim\frac{(\log
N)^{1/q-1/2}}{\prod_{i=1}^{N}|\lambda_{i}|}.$

2) If $1\leq p\leq q\leq\infty$ and $q\geq 2$ (Region II), then

${\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\lesssim\frac{1}{\prod_{i=1}^{N}|\lambda_{i}|}.$

3) If $1\leq q\leq p\leq\infty$ and $p\geq 2$ (Region III), then

${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\lesssim\frac{(\log
N)^{1/q-1/p}}{\prod_{i=1}^{N}|\lambda_{i}|}.$

######  Remark.

The upper bound in part 2 of Theorem 3 is attained by any sequence
$\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}_{*}^{N}$ such
that $|\lambda_{i}|\geq 1-1/N$ for all $i=1,\dots,N$.

###  3.3.  Lower estimates on ${\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})$

In the following theorem we derive quantitive lower estimates on ${\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})$ for $1\leq q\leq p\leq\infty$.
This proves, in particular, the sharpness of Theorem 3 for $(p,q)$ in Region
III.

######  Theorem 4.

Let $\sigma^{\star}\in\mathbb{D}_{*}^{N}$ and $B^{\star}$ be defined by (3.2)
and (3.3), and let $(p,q)\in[1,\infty]^{2}$ be such that $1\leq q\leq
p\leq\infty$. Then

(3.5) $\displaystyle\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})$ $\displaystyle\gtrsim(\log
N)^{1/q-1/p},\qquad p<\infty,$ (3.6)
$\displaystyle\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{\infty,q}^{0}}(\sigma^{\star})$
$\displaystyle\gtrsim\frac{(\log N)^{1/q}}{\log\log N}.$

In particular, for $(p,q)$ in Region III and $p<\infty$,

(3.7) $\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})\asymp(\log N)^{1/q-1/p}.$

However, for $1\leq q\leq p<2$ there is still a certain gap between the upper
and lower estimates for the capacities:

$(\log N)^{1/q-1/p}\lesssim\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})\lesssim(\log N)^{1/q-1/2}.$

We will give a qualitative argument showing that, in case $1\leq q=p<2$, the
expression in the middle may be unbounded.

######  Theorem 5.

For each $N\in\mathbb{N}$ there exists a finite sequence
$\sigma_{N}\in\mathbb{D}_{*}^{N}$ such that for all $1\leq p<2$, we have

$\lim_{N\to\infty}\prod_{\lambda\in\sigma_{N}}|\lambda|\cdot{\rm
cap}_{\mathcal{B}^{0}_{p,p}}(\sigma_{N})=\infty.$

It will be convenient to extend the definition of ${\rm
cap}_{\mathcal{B}^{s}_{p,q}}(\sigma)$ to possibly infinite sequences $\sigma$
in the obvious way. The infimum over the empty set is understood to be
$+\infty$, so that ${\rm cap}_{\mathcal{B}^{s}_{p,q}}(\sigma)=+\infty$ in case
$\sigma$ is a uniqueness set for $\mathcal{B}_{p,q}^{s}$. Our approach to
bound ${\rm cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)$ from below is based on a
duality method. Namely, the key step of the proof is the following lemma:

######  Lemma 6.

Given $1\leq p,q\leq\infty$ and a finite sequence $\sigma$ in
$\mathbb{D}_{*}$, we have

$\prod_{\lambda\in\sigma}|\lambda|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\gtrsim\frac{1-\prod_{\lambda\in\sigma}|\lambda|^{2}}{\left|\\!\left|B_{\sigma}\right|\\!\right|_{\mathcal{B}_{p^{\prime},q^{\prime}}^{0}}^{*}},$

where $B_{\sigma}$ is the Blaschke product with the zero set $\sigma$ and
$p^{\prime},q^{\prime}$ are the exponents conjugate to $p,q$. The same
estimate is true for arbitrary Blaschke sequences $\sigma$ in $\mathbb{D}_{*}$
in case $1\leq p=q\leq 2$.

To prove the lower estimate (3.5) it remains to apply Lemma 6 to
$\sigma=\sigma^{\star}$ and estimate from above the Besov seminorm of
$B^{\star}$. Namely we prove the following.

######  Proposition 7.

If $1\leq p\leq q\leq\infty$, then

$\displaystyle\|B^{\star}\|_{\mathcal{B}_{p,q}^{0}}^{*}$
$\displaystyle\lesssim\frac{1}{(\log N)^{1/p-1/q}},\qquad p>1,$
$\displaystyle\|B^{\star}\|_{\mathcal{B}_{1,q}^{0}}^{*}$
$\displaystyle\lesssim\frac{\log\log N}{(\log N)^{1-1/q}}.$

The idea of the proof of Theorem 5 is also to use duality. In case $p=1$, the
dual norm turns out to be the Bloch semi-norm. An obstacle to this strategy is
a result of Baranov, Kayumov, and Nasyrov [4], according to which the Bloch
semi-norm of finite Blaschke products is bounded below by a universal
constant. Instead, we will work with infinite Blaschke products, and carry out
an approximation argument.

##  4\. Proof of Lemma 6

We first prove Lemma 6. Let $\left\langle\cdot,\,\cdot\right\rangle$ denote
the Cauchy sesquilinear form: given two functions $g\in H^{p}$ and $h\in
H^{p^{\prime}}$, let

$\langle h,g\rangle=\int_{\mathbb{T}}h(z)\overline{g(z)}\,dm(z),$

where $m$ denotes the normalized Lebesgue measure on $\mathbb{T}$. We require
the following basic duality result for Besov spaces.

######  Lemma 8.

Let $1\leq p,q\leq\infty$. There exists a constant $C\geq 0$ such that for all
functions $f$ and $g$ that are analytic in a neighborhood of
$\overline{\mathbb{D}}$, we have

$|\langle
f,g\rangle|\leq|f(0)||g(0)|+C\|f\|_{B^{0}_{p,q}}^{*}\|g\|_{B^{0}_{p^{\prime},q^{\prime}}}^{*},$

where $p^{\prime},q^{\prime}$ are the exponents conjugate to $p,q$.

###### Proof.

Denote by $(h,\,g)$ the scalar product on the Bergman space $A^{2}$ defined by

$(h,\,g)=\int_{\mathbb{D}}h(u)\overline{g(u)}{\rm d}\mathcal{A}(u),\qquad
h,\,g\in A^{2},$

where ${\rm d}\mathcal{A}(u)=\frac{{\rm d}x\,{\rm d}y}{\pi}$ is the normalized
planar Lebesgue measure on $\mathbb{D}$. We recall the simplest form of
Green’s formula,

(4.1)
$\left\langle\phi,\,\psi\right\rangle=(\phi^{\prime},\,S^{*}\psi)+\phi(0)\overline{\psi(0)},$

where $S^{*}$ is the backward shift operator $S^{*}f=(f-f(0))/z$ and
$\varphi,\psi$ are functions that are analytic in a neighborhood of
$\overline{\mathbb{D}}$. We will also need to use the following integral
formula. Recall that the fractional differentiation operator $D_{\alpha}$,
$-1<\alpha<\infty$, is defined by
$D_{\alpha}(z^{j})=\frac{\Gamma(j+2+\alpha)}{(j+1)!\Gamma(2+\alpha)}z^{j}$,
$j=0,\,1,\,2,\,\dots$, and extends linearly and continuously to the whole
space $\mathcal{H}ol(\mathbb{D})$. Then, for functions $f,g$ analytic in a
neighborhood of $\overline{\mathbb{D}}$ and $-1<\alpha<\infty$, we have

(4.2) $\int_{\mathbb{D}}f(u)\overline{g(u)}{\rm
d}\mathcal{A}(u)=(\alpha+1)\int_{\mathbb{D}}D_{\alpha}f(u)\overline{g(u)}\left(1-\left|u\right|^{2}\right)^{\alpha}{\rm
d}\mathcal{A}(u),$

see [7, Lemma 1.20].

Let $f,g$ be analytic in a neighborhood of $\overline{\mathbb{D}}$. Applying
(4.1) we get

$\left\langle f,\,g\right\rangle=(f^{\prime},\,S^{*}g)+f(0)\overline{g(0)}.$

Then we apply (4.2) to
$\overline{(f^{\prime},\,S^{*}g)}=(S^{*}g,\,f^{\prime})$ with $\alpha=1$:

$\displaystyle(S^{*}g,\,f^{\prime})$
$\displaystyle=2\int_{\mathbb{D}}D_{1}(S^{*}g)(u)\overline{f^{\prime}(u)}\left(1-\left|u\right|^{2}\right){\rm
d}\mathcal{A}(u)$
$\displaystyle=2\int_{0}^{1}\rho\left(1-\rho^{2}\right)\left(\int_{\mathbb{T}}D_{1}(S^{*}g)(\rho
z)\overline{f^{\prime}(\rho z)}{\rm d}m(z)\right){\rm d}\rho.$

By Hölder’s inequality

$\left|\int_{\mathbb{T}}D_{1}(S^{*}g)(\rho z)\overline{f^{\prime}(\rho z)}{\rm
d}m(z)\right|\leq\|f_{\rho}^{\prime}\|_{L^{p}}\|(D_{1}(S^{*}g))_{\rho}\|_{L^{p^{\prime}}}.$

Since $D_{1}(S^{*}g)=\frac{1}{2}(S^{*}g+g^{\prime})$ and
$(S^{*}g)(z)=\frac{1}{z}\int_{0}^{1}tg^{\prime}(tz)\,dt$, it follows that
$\|D_{1}(S^{*}g)\|_{L^{p^{\prime}}}\lesssim\|g^{\prime}\|_{L^{p^{\prime}}}$.
The preceding estimates therefore give

$|(S^{*}g,\,f^{\prime})|\lesssim\int_{0}^{1}\left(1-\rho\right)\|f_{\rho}^{\prime}\|_{L^{p}}\|g_{\rho}^{\prime}\|_{L^{p^{\prime}}}{\rm
d}\rho.$

Then (again by Hölder’s inequality) we get

$|(S^{*}g,\,f^{\prime})|\lesssim\left|\\!\left|f\right|\\!\right|_{\mathcal{B}_{p,q}^{0}}^{*}\left|\\!\left|g\right|\\!\right|_{\mathcal{B}_{p^{\prime},q^{\prime}}^{0}}^{*},$

as desired. ∎

###### Proof of Lemma 6.

Suppose first that $\sigma$ is a finite sequence in $\mathbb{D}_{*}$, say
$|\sigma|=N$. Let $f$ be a function that is analytic in a neighborhood of
$\overline{\mathbb{D}}$ such that $f(0)=1$ and $f\big{|}{\sigma}=0$. Then we
have (writing $B=B_{\sigma}$)

$\langle
f,B\rangle=\frac{f(0)}{B(0)}=\frac{1}{\prod_{\lambda\in\sigma}\lambda}.$

On the other hand, Lemma 8 shows that

$|\langle
f,B\rangle|\leq|f(0)||B(0)|+C\|f\|^{*}_{\mathcal{B}^{0}_{p,q}}\|B\|^{*}_{\mathcal{B}^{0}_{p^{\prime},q^{\prime}}}=\prod_{\lambda\in\sigma}|\lambda|+C\|f\|_{\mathcal{B}^{0}_{p,q}}^{*}\|B\|_{\mathcal{B}^{0}_{p^{\prime},q^{\prime}}}^{*}.$

Thus,

(4.3)
$\prod_{\lambda\in\sigma}|\lambda|\cdot\|f\|_{\mathcal{B}^{0}_{p,q}}^{*}\geq\frac{1-\prod_{\lambda\in\sigma}|\lambda|^{2}}{C\|B\|_{\mathcal{B}^{0}_{p^{\prime},q^{\prime}}}^{*}}.$

Now, let $f\in\mathcal{B}^{0}_{p,q}$ be an arbitrary function such that
$f(0)=1$ and $f\big{|}{\sigma}=0$. Let $0<r<1$ be such that
$\frac{1}{r}\sigma\subset\mathbb{D}$. Then $f_{r}$ vanishes on
$\frac{1}{r}\sigma$, hence by what has already been proved,

$r^{N}\prod_{\lambda\in\sigma}|\lambda|\cdot\|f_{r}\|_{\mathcal{B}^{0}_{p,q}}^{*}\geq\frac{1-r^{2N}\prod_{\lambda\in\sigma}|\lambda|^{2}}{C\|B_{\frac{1}{r}\sigma}\|_{\mathcal{B}^{0}_{p^{\prime},q^{\prime}}}^{*}}.$

Recall that
$\|f_{r}\|_{\mathcal{B}^{0}_{p,q}}^{*}\leq\|f\|_{\mathcal{B}^{0}_{p,q}}$.
Moreover, $B_{\frac{1}{r}\sigma}$ converges to $B_{\sigma}$ uniformly in a
neighborhood of $\overline{\mathbb{D}}$ as $r\to 1$. So taking the limit $r\to
1$, we conclude that (4.3) holds for arbitrary $f\in\mathcal{B}^{0}_{p,q}$
satisfying $f(0)=1$ and $f\big{|}\sigma=0$. Taking the infimum over all
admissible functions $f$, we obtain the lemma for finite sequences.

Let now $1\leq p=q\leq 2$ and let $\sigma$ be a possibly infinite Blaschke
sequence. Let $B=B_{\sigma}$ and let $f\in\mathcal{B}^{0}_{p,p}$ be a function
vanishing on $\sigma$ with $f(0)=1$. We apply Lemma 8 to the functions $f_{r}$
and $B_{r}$ to obtain the bound

$|\langle
f_{r},B_{r}\rangle|\leq|B(0)|+C\|f\|_{B^{0}_{p,p}}^{*}\|B\|^{*}_{B^{0}_{q,q}}$

for all $r<1$.

The classical Littlewood–Paley inequality shows that
$\mathcal{B}^{0}_{p,p}\subset H^{p}\subset H^{1}$ (see [8, Theorem 6]), so
$f_{r}\to f$ in the norm of $H^{1}$. Moreover, $B\in H^{\infty}$ and $B_{r}\to
B$ weak-$*$ in $H^{\infty}$. From this, it follows that

$\langle f,B\rangle-\langle f_{r},B_{r}\rangle=\langle
f,B-B_{r}\rangle+\langle f-f_{r},B_{r}\rangle\to 0$

as $r\to 1$. Thus,

$|\langle
f,B\rangle|\leq|B(0)|+C\|f\|_{\mathcal{B}^{0}_{p,p}}^{*}\|B\|_{\mathcal{B}^{0}_{p^{\prime},p^{\prime}}}^{*}.$

Using that $f\in H^{1}$ vanishes on $\sigma$, we may factor $f=Bg$ for some
$g\in H^{1}$. Then

$\langle f,B\rangle=\langle g,1\rangle=g(0)=\frac{1}{B(0)}.$

Combining the last two formulas and taking the infimum over all admissible
$f\in\mathcal{B}^{0}_{p,p}$ again yields the desired inequality. ∎

##  5\. Proof of Theorem 2

###  5.1. Proof of Proposition 1

For simplicity we write $B$ instead of $B^{\star}$ throughout the proof. Then
$N={\rm deg}\,B\asymp 2^{n}$. For the zeros $z_{1},\dots z_{N}$ of $B$ we have

$\prod_{j=1}^{N}|z_{j}|=a^{n}<e^{-1}.$

Since for $z\in\mathbb{D}$, $|z|=r$,

(5.1)
$|B^{\prime}(z)|\leq\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{1-a^{2}}{|1-az^{2^{k}}|^{2}},$

we have

$(1-r)\int_{0}^{2\pi}|B^{\prime}(re^{it})|dt\leq
2\pi\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{(1-r)(1-a^{2})}{1-a^{2}r^{2^{k+1}}}\lesssim\frac{1}{n}\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{1-r}{1-ar^{2^{k}}}.$

Let us first estimate this quantity for $0\leq r\leq\frac{1}{2}$. In this case

$\frac{1}{n}\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{1-r}{1-ar^{2^{k}}}\lesssim\frac{1}{n}\sum_{k=1}^{n}2^{k-2^{k}}\lesssim\frac{1}{n}.$

From now one we assume that $r=1-\frac{1}{2^{s}}$, where $s\geq 1$, and we
write

$\frac{1}{n}\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{1-r}{1-ar^{2^{k}}}=\frac{1}{n}\sum_{k=1}^{[s]}2^{k}r^{2^{k}-1}\frac{1-r}{1-ar^{2^{k}}}+\frac{1}{n}\sum_{k=[s]+1}^{n}2^{k}r^{2^{k}-1}\frac{1-r}{1-ar^{2^{k}}}=S_{1}+S_{2}.$

Since $(1-x)^{t}<e^{-tx}$, $0<x<1$, $t>0$, we have

$r^{2^{k}}=\bigg{(}1-\frac{1}{2^{s}}\bigg{)}^{2^{k}}<e^{-2^{k-s}}.$

Therefore, for $k\geq[s]+1$ we have $r^{2^{k}}<e^{-1}$ and so

$S_{2}\lesssim\frac{1}{n}\sum_{k=[s]+1}^{n}2^{k-s}e^{-2^{k-s}}\lesssim\frac{1}{n}.$

For $k\leq[s]$ we use the inequality

$r^{-2^{k}}-a>e^{2^{k-s}}-1+\frac{1}{n}>2^{k-s}+\frac{1}{n}.$

Thus,

$S_{1}\lesssim\frac{1}{n}\sum_{k=1}^{[s]}2^{k}\frac{1-r}{r^{-2^{k}}-a}<\frac{1}{n}\sum_{k=1}^{[s]}\frac{2^{k-s}}{2^{k-s}+\frac{1}{n}}.$

We split this sum into two more sums, over $k$ such that $2^{k-s}<\frac{1}{n}$
and $2^{k-s}\geq\frac{1}{n}$. Then we have

$S_{1}\lesssim\frac{1}{n}\sum_{1\leq k<s-\frac{\log n}{\log 2}}2^{k-s}\cdot
n+\frac{1}{n}\sum_{s-\frac{\log n}{\log 2}\leq k\leq[s]}2^{k-s}\cdot
2^{s-k}\lesssim 2^{-\frac{\log n}{\log 2}}+\frac{\log n}{n\log
2}\lesssim\frac{1}{n}+\frac{\log n}{n\log 2}.$

Thus, we have shown that

$\|B\|_{B_{1,\infty}^{0}}^{*}\lesssim\frac{1}{n}+\frac{\log n}{n\log
2}\lesssim\frac{\log\log N}{\log N}.$

∎

###  5.2. Proof of Theorem 2

Applying Lemma 6 to $\sigma=\sigma^{\star}$ with $(p,q)=(\infty,1)$ we obtain

$\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{\infty,1}^{0}}(\sigma^{\star})\gtrsim\frac{1}{\left|\\!\left|B\right|\\!\right|_{\mathcal{B}_{1,\infty}^{0}}^{*}}$

because $\prod_{i=1}^{N}|\lambda_{i}|=a^{n}<e^{-1}.$ It remains to apply
Proposition 1. ∎

##  6\. Proofs of Theorem 4 and Theorem 5

###  6.1. Proof of Proposition 7

As in the proof of Proposition 1, for simplicity we write $B$ instead of
$B^{\star}$ throughout the proof.
Step 1: the case $q=\infty$. Note that the case $p=1$ is already covered by
Proposition 1. We start with the case $1<p\leq 2$. We have to prove that

(6.1) $\sup_{0\leq
r<1}(1-r)\left(\int_{0}^{2\pi}|B^{\prime}{}(re^{it})|^{p}dt\right)^{1/p}\lesssim\frac{1}{(\log
N)^{1/p}}.$

It follows from (5.1) that

$I:=(1-r)^{p}\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\lesssim\int_{0}^{2\pi}\left(\frac{1}{n}\sum_{k=1}^{n}2^{k}r^{2^{k}-1}\frac{(1-r)}{|1-ar^{2^{k}e^{i2^{k}t}}|^{2}}\right)^{p}dt.$

Since for $0<p/2\leq 1$ and any $a_{k}\geq 0$ one has

$\Big{(}\sum_{k}a_{k}\Big{)}^{p}\leq\Big{(}\sum_{k}a_{k}^{p/2}\Big{)}^{2},$

we conclude that

$\displaystyle I$
$\displaystyle\leq\frac{1}{n^{p}}\int_{0}^{2\pi}\left(\sum_{k=1}^{n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{|1-ar^{2^{k}}e^{i2^{k}t}|^{p}}\right)^{2}dt$
$\displaystyle\lesssim\frac{1}{n^{p}}\int_{0}^{2\pi}\sum_{k\leq j\leq
n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}(2^{j}r^{2^{j}-1}(1-r))^{p/2}}{|1-ar^{2^{k}}e^{i2^{k}t}|^{p}|1-ar^{2^{j}}e^{i2^{j}t}|^{p}}dt$
$\displaystyle\lesssim\frac{1}{n^{p}}\int_{0}^{2\pi}\sum_{k\leq j\leq
n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}(2^{j}r^{2^{j}-1}(1-r))^{p/2}}{|1-ar^{2^{k}}e^{i2^{k}t}|^{p}(1-ar^{2^{j}})^{p}}dt.$

After integration with respect to $t$ and using a well-known estimate of
Forelli and Rudin (see [7, Theorem 1.7]) we get

$\displaystyle I$ $\displaystyle\lesssim\frac{1}{n^{p}}\sum_{k\leq j\leq
n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}(2^{j}r^{2^{j}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1}(1-ar^{2^{j}})^{p}}$
$\displaystyle\lesssim\frac{1}{n^{p}}\sum_{k\leq j\leq
n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}(2^{j}r^{2^{j}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}(1-ar^{2^{j}})^{p-1/2}}$
$\displaystyle\lesssim\frac{1}{n^{p}}\left(\sum_{k=1}^{n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}}\right)^{2}.$

Thus, we need to show that

$S=\frac{1}{n^{p/2}}\sum_{k=1}^{n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}}\leq\frac{1}{\sqrt{n}}.$

If $r\leq 1/2$, then, clearly, $S\lesssim n^{-p/2}\leq n^{-1/2}$. Now, let
$r=1-1/2^{s}$ where $s\geq 1$. If $k\geq s+1$, then
$r^{2^{k}}<e^{-2^{k-s}}\leq 1/e$ and

$\frac{1}{n^{p/2}}\sum_{k=[s]+1}^{n}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}}\lesssim\frac{1}{n^{p/2}}\sum_{k=[s]+1}^{n}(2^{k-s}e^{-2^{k-s}})^{p/2}\lesssim\frac{1}{n^{p/2}}.$

Note that $r^{2^{k}}=(1-1/2^{s})^{2^{k}}\geq(1-1/2)^{2}$ for $k\in[1,s]$ and,
therefore, as in the proof of Proposition 1,

$|1-ar^{2^{k}}|\gtrsim r^{-2^{k}}-a=r^{-2^{k}}-1+1/n\geq 2^{k-s}+1/n.$

As in the proof of Proposition 1 we split the sum into two parts. For $1\leq
k<s-\frac{\log n}{\log 2}$ we have $2^{k-s}<1/n$ and, therefore,

$\displaystyle\frac{1}{n^{p/2}}\sum_{k<s-\log n/\log
2}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}}$
$\displaystyle\lesssim\frac{1}{n^{p/2}}n^{p-1/2}\sum_{k<s-\log n/\log
2}2^{(k-s)p/2}$ $\displaystyle\lesssim n^{p/2-1/2}2^{(-\log n/\log
2)p/2}=n^{-1/2}.$

Finally, for $s-\frac{\log n}{\log 2}\leq k\leq[s]$ we have $2^{k-s}\geq 1/n$
and so

$\displaystyle\frac{1}{n^{p/2}}\sum_{s-\log n/\log 2\leq
k\leq[s]}\frac{(2^{k}r^{2^{k}-1}(1-r))^{p/2}}{(1-ar^{2^{k}})^{p-1/2}}$
$\displaystyle\lesssim\frac{1}{n^{p/2}}\sum_{s-\log n/\log 2\leq
k\leq[s]}2^{(k-s)p/2}2^{(s-k)(p-1/2)}$ $\displaystyle\lesssim n^{-p/2}2^{(\log
n/\log 2)(p-1)/2}=n^{-1/2}.$

Thus, we have shown that $S\leq n^{-1/2}$ for $1<p\leq 2$, and so

$I=(1-r)^{p}\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\lesssim
S^{2}\lesssim\frac{1}{\log N}.$

The estimate remains true for $p>2$ since by the Schwarz–Pick inequality, we
have $|B^{\prime}(re^{it})|^{p}\leq(1-r^{2})^{2-p}|B^{\prime}(re^{it})|^{2}$.
Step 2: the case $1\leq p\leq q<\infty$. We have to show that

$\int_{0}^{1}(1-r)^{q-1}\left(\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\right)^{q/p}dr\lesssim\frac{1}{(\log
N)^{q/p-1}}$

(respectively $\lesssim\frac{(\log\log N)^{q}}{(\log N)^{q-1}}$ in case
$p=1$). It follows from (6.1) that for $1<p<\infty$

$\int_{0}^{1-1/N^{2}}(1-r)^{q-1}\left(\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\right)^{q/p}dr\lesssim\frac{1}{(\log
N)^{q/p}}\int_{0}^{1-1/N^{2}}\frac{dr}{1-r}\lesssim\frac{1}{(\log
N)^{q/p-1}},$

while for $p=1$ we have by Proposition 1 that

$\int_{0}^{1-1/N^{2}}(1-r)^{q-1}\left(\int_{0}^{2\pi}|B^{\prime}(re^{it})|dt\right)^{q}dr\lesssim\frac{(\log\log
N)^{q}}{(\log N)^{q-1}}.$

On the other hand, note that, since
$\int_{0}^{2\pi}|B^{\prime}(re^{it})|dt\leq\int_{0}^{2\pi}|B^{\prime}(e^{it})|dt=2\pi
N$, we have by the Schwarz–Pick inequality

$\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\leq\frac{1}{(1-r^{2})^{p-1}}\int_{0}^{2\pi}|B^{\prime}(re^{it})|dt\leq\frac{2\pi
N}{(1-r^{2})^{p-1}}.$

Therefore,

$\displaystyle\int_{1-1/N^{2}}^{1}(1-r)^{q-1}$
$\displaystyle\left(\int_{0}^{2\pi}|B^{\prime}(re^{it})|^{p}dt\right)^{q/p}dr\lesssim
N^{q/p}\int_{1-1/N^{2}}^{1}(1-r)^{q-1-q(p-1)/p}dr$
$\displaystyle=N^{q/p}\int_{1-1/N^{2}}^{1}(1-r)^{q/p-1}\lesssim
N^{-q/p}=o\Big{(}\frac{1}{(\log N)^{q/p-1}}\Big{)}.$

Combining the above estimates we come to the conclusion of the proposition. ∎

###  6.2. Proof of Theorem 4

As in the proof of Theorem 2, we apply Lemma 6 to $\sigma=\sigma^{\star}$.
This gives

$\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})\gtrsim\frac{1}{\left|\\!\left|B\right|\\!\right|_{\mathcal{B}_{p^{\prime},q^{\prime}}^{0}}^{*}}$

again because $\prod_{i=1}^{N}|\lambda_{i}|=a^{n}<e^{-1}.$ It remains to apply
Proposition 7 (with $p^{\prime},q^{\prime}$ in place of $p,q$) to prove the
lower bounds (3.5) and (3.6). The upper estimate in (3.7) follows from Theorem
3. ∎

###  6.3. Proof of Theorem 5

To pass from infinite to finite Blaschke sequences, we require the following
continuity property of capacity.

######  Lemma 9.

Let $1\leq p,q\leq\infty$. Let $\sigma\subset\mathbb{D}\setminus\\{0\\}$ be an
infinite sequence. For $N\in\mathbb{N}$, let $\sigma_{N}$ consist of the first
$N$ points of $\sigma$. Then

${\rm cap}_{\mathcal{B}^{0}_{p,q}}(\sigma)=\lim_{N\to\infty}{\rm
cap}_{\mathcal{B}^{0}_{p,q}}(\sigma_{N}).$

###### Proof.

For simplicity, we abbreviate ${\rm cap}={\rm cap}_{\mathcal{B}^{0}_{p,q}}$.
The inequality ${\rm cap}(\sigma_{N})\leq{\rm cap}(\sigma)$ is trivial, so it
suffices to show that ${\rm cap}(\sigma)\leq\liminf_{N\to\infty}{\rm
cap}(\sigma_{N})$. Clearly, we may assume that the limit inferior is finite.

If $c>\liminf_{N\to\infty}{\rm cap}(\sigma_{N})$, then by definition of
capacity, there exist a sequence $(N_{k})$ tending to infinity and functions
$f_{N_{k}}\in\mathcal{B}^{0}_{p,q}$ such that $f_{N_{k}}$ vanishes on
$\sigma_{N_{k}}$, $f_{N_{k}}(0)=1$ and $\|f\|_{\mathcal{B}^{0}_{p,q}}\leq c$
for all $k$. Then $(f_{N_{k}})_{k}$ is a normal family, so a subsequence
converges locally uniformly on $\mathbb{D}$ to a holomorphic function $f$. By
Fatou’s lemma, $f\in\mathcal{B}^{0}_{p,q}$ with
$\|f\|_{\mathcal{B}^{0}_{p,q}}\leq c$, and $f$ vanishes on $\sigma$ and
$f(0)=1$. Thus, ${\rm cap}(\sigma)\leq c$. ∎

###### Proof of Theorem 5.

Let $2<q<\infty$. Then, for
$f\in\mathcal{B}^{0}_{\infty,\infty}\cap\mathcal{B}^{0}_{2,2}$, we have

$\int_{0}^{1}(1-\rho)^{q-1}\|f^{\prime}_{\rho}\|_{L^{q}}^{q}\,d\rho\leq\int_{0}^{1}(1-\rho)^{q-2}\|f^{\prime}_{\rho}\|_{L^{\infty}}^{q-2}(1-\rho)\|f_{\rho}^{\prime}\|_{L^{2}}^{2}\,d\rho\leq{\|f\|^{*}}^{q-2}_{\mathcal{B}^{0}_{\infty,\infty}}{\|f\|^{*}}^{2}_{\mathcal{B}^{0}_{2,2}}.$

Recalling that $\mathcal{B}^{0}_{2,2}=H^{2}$ with equivalence of norms, we
conclude that there exists a constant $C>0$ so that for any Blaschke product
$B$, we have

$\|f\|^{*}_{\mathcal{B}^{0}_{q,q}}\leq
C^{1/q}{\|f\|^{*}}^{1-\frac{2}{q}}_{\mathcal{B}^{0}_{\infty,\infty}}.$

(This also follows from the Littlewood–Paley inequality.) Note that the
inequality trivially holds for $q=\infty$ as well.

Let $\varepsilon>0$. It follows from a theorem of Aleksandrov, Anderson, and
Nicolau [1, Theorem 2] that there exists an infinite Blaschke product $B$ with
$\|B\|^{*}_{\mathcal{B}^{0}_{\infty,\infty}}\leq\varepsilon$. By precomposing
$B$ with a conformal automorphism, we may assume that $|B(0)|=\frac{1}{2}$. By
the preceding estimate with $q=p^{\prime}$, we have
$\|B\|^{*}_{\mathcal{B}^{0}_{p^{\prime},p^{\prime}}}\leq
C^{1/p^{\prime}}\varepsilon^{1-\frac{2}{p^{\prime}}}$, so Lemma 6 implies that
there exists a constant $c>0$ such that

$\frac{1}{2}{\rm cap}_{\mathcal{B}^{0}_{p,p}}(\sigma)\geq
c\varepsilon^{\frac{2}{p^{\prime}}-1}.$

We distinguish two cases. If ${\rm
cap}_{\mathcal{B}^{0}_{p,p}}(\sigma)<\infty$, then Lemma 9 yields a finite
subsequence $\sigma^{\prime}\subset\sigma$ such that ${\rm
cap}_{\mathcal{B}^{0}_{p,p}}(\sigma^{\prime})\geq\frac{1}{2}{\rm
cap}_{\mathcal{B}^{0}_{p,p}}(\sigma)$. Note that
$\prod_{\lambda\in\sigma^{\prime}}|\lambda|\geq\prod_{\lambda\in\sigma}|\lambda|=1/2$,
so

$\prod_{\lambda\in\sigma^{\prime}}|\lambda|\cdot{\rm
cap}_{\mathcal{B}^{0}_{p,p}}(\sigma^{\prime})\geq
c\varepsilon^{\frac{2}{p^{\prime}}-1}.$

If ${\rm cap}_{\mathcal{B}^{0}_{p,p}}(\sigma)=\infty$, then Lemma 9 directly
yields a finite sequence $\sigma^{\prime}\subset\sigma$ satisfying the last
inequality. Since $\varepsilon>0$ was arbitrary, the result follows. ∎

##  7\. Proofs of the upper bounds

An obvious choice of a test function for an estimate of the capacity from
above is
$f=B_{\sigma}/B_{\sigma}(0)=(-1)^{N}\big{(}\prod_{i=1}^{N}\lambda_{i}\big{)}^{-1}B_{\sigma}$,
$\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)$. Surprisingly, this
function gives sharp estimates in many situations. To use $f$ as a test
function in the proof of Theorem 3 we need first to obtain some estimates for
the norms of finite Blaschke products in Besov spaces which may be of
independent interest.

###  7.1. Estimates for the $\mathcal{B}_{p,q}^{0}$ norms of finite Blaschke
products

######  Proposition 10.

Let $(p,q)\in[1,\infty]^{2}$,
$\sigma=\left(\lambda_{1},\dots,\lambda_{N}\right)\in\mathbb{D}^{N}$ and
$B=B_{\sigma}$.

1) If $(p,q)\in[1,2]^{2}$, then

$\|B\|_{\mathcal{B}_{p,q}^{0}}\lesssim(\log N)^{1/q-1/2}.$

2) If $1\leq p\leq q\leq\infty$ and $q\geq 2$, then

$\|B\|_{\mathcal{B}_{p,q}^{0}}\lesssim 1.$

3) If $1\leq q\leq p<\infty$ and $p\geq 2$, then

$\|B\|_{\mathcal{B}_{p,q}^{0}}\lesssim(\log N)^{1/q-1/p}.$

4) If $1\leq q<\infty$ and $p=\infty$, then

$\|B\|_{\mathcal{B}_{p,q}^{0}}\lesssim N^{1/q}.$

The constants in the relations $\lesssim$ depend only on $p,q$, but not on $N$
and $\sigma$. Moreover, in all inequalities the dependence on the growth on
$N$ is best possible.

Note that there is an essential difference between the case $p<\infty$ and
$p=\infty$, where the growth is much faster.

In the proof of Proposition 10 we will need several simple estimates, the
first of which can be found in [3]. We will give their proofs for the sake of
completeness.

######  Lemma 11.

Let $B$ be a Blaschke product of degree $N$. Then

(7.1) $\int_{0}^{1}(1-\rho)^{p-1}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dtd\rho\lesssim\begin{cases}(\log N)^{1-p/2},&1\leq p\leq 2,\\\
1,&2\leq p<\infty,\end{cases}$

and, for $\rho\in[0,1)$ and $1\leq p<\infty$,

(7.2) $\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dt\lesssim\frac{N}{(1-\rho)^{p-1}}.$

###### Proof.

Since $\int_{0}^{2\pi}|B^{\prime}(\rho e^{it})|dt\leq 2\pi N$, $\rho\in[0,1]$,
we conclude that

$\int_{1-1/N}^{1}\int_{0}^{2\pi}|B^{\prime}(\rho e^{it})|dtd\rho\lesssim 1.$

Note that

(7.3) $\int_{0}^{1}\rho(1-\rho^{2})\int_{0}^{2\pi}|f^{\prime}(\rho
e^{it})|^{2}dtd\rho=\pi\sum_{n=1}^{\infty}\frac{n}{n+1}|a_{n}|^{2}\leq\pi\|f\|_{H^{2}}^{2}\leq\pi\|f\|_{H^{\infty}}^{2}$

for any function $f(z)=\sum_{n\geq 0}a_{n}z^{n}$ in the Hardy space $H^{2}$.
Therefore,

$\displaystyle\int_{0}^{1-1/N}\int_{0}^{2\pi}|B^{\prime}(\rho e^{it})|dtd\rho$
$\displaystyle\lesssim$
$\displaystyle\bigg{(}\int_{0}^{1-1/N}\int_{0}^{2\pi}(1-\rho)|B^{\prime}(\rho
e^{it})|^{2}dtd\rho\bigg{)}^{1/2}\bigg{(}\int_{0}^{1-1/N}\int_{0}^{2\pi}\frac{dtd\rho}{1-\rho}\bigg{)}^{1/2}$
$\displaystyle\lesssim$ $\displaystyle(\log N)^{1/2}.$

Thus, the inequality is already proved for $p=2$ (simply apply (7.3) to $B$)
and $p=1$. For $1<p<2$ inequality (7.1) follows from the Hölder inequality
with exponents $(p-1)^{-1}$ and $(2-p)^{-1}$ (note that $p=2(p-1)+2-p$) and
the estimates for exponents 1 and 2. Finally, for $p>2$ it follows from the
estimate $(1-|z|^{2})|B^{\prime}(z)|\leq 1$, $z\in\mathbb{D}$, that
$(1-\rho)^{p-1}\|B^{\prime}_{\rho}\|_{L^{p}(\mathbb{T})}^{p}\leq(1-\rho)\|B^{\prime}_{\rho}\|_{L^{2}(\mathbb{T})}^{2}$
and we can again apply (7.3).

The estimate (7.2) is obvious:

$\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dt\lesssim\frac{1}{(1-\rho)^{p-1}}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|dt\lesssim\frac{N}{(1-\rho)^{p-1}}.$

∎

###### Proof of Proposition 10.

Let $1\leq p<\infty$. Then it follows from (7.1) that

$(\|B\|_{\mathcal{B}_{p,p}^{0}}^{*})^{p}=2\pi\int_{0}^{1}(1-\rho)^{p-1}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dtd\rho\lesssim\begin{cases}(\log N)^{1-p/2},&1\leq p\leq 2\\\
1,&p\geq 2.\end{cases}$

Also trivially $\|B\|_{\mathcal{B}_{\infty,\infty}^{0}}\lesssim 1$.

Now let $1\leq q\leq p<\infty$. We write

$\displaystyle\frac{1}{2\pi}(\|B\|_{\mathcal{B}_{p,q}^{0}}^{*})^{q}$
$\displaystyle=\int_{0}^{1-1/N}(1-\rho)^{q-1}\bigg{(}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dt\bigg{)}^{q/p}d\rho$
$\displaystyle+\int_{1-1/N}^{1}(1-\rho)^{q-1}\bigg{(}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dt\bigg{)}^{q/p}d\rho=I_{1}+I_{2}.$

We show that $I_{2}\lesssim 1$. Indeed, applying (7.2) we get

$I_{2}\lesssim
N^{q/p}\int_{1-1/N}^{1}(1-\rho)^{q-1}(1-\rho)^{-q(1-1/p)}d\rho=N^{q/p}\int_{1-1/N}^{1}(1-\rho)^{-1+q/p}d\rho\lesssim
1.$

To estimate $I_{1}$, we apply the Hölder inequality with exponents $p/q$ and
$p/(p-q)$ to get (with an obvious modification for $p=q$)

$I_{1}\leq\bigg{(}\int_{0}^{1-1/N}(1-\rho)^{p-1}\int_{0}^{2\pi}|B^{\prime}(\rho
e^{it})|^{p}dtd\rho\bigg{)}^{\frac{q}{p}}\bigg{(}\int_{0}^{1-1/N}\frac{d\rho}{1-\rho}\bigg{)}^{\frac{p-q}{p}}.$

Hence, for $1<p\leq 2$,

$I_{1}\lesssim(\log N)^{\frac{q}{p}(1-\frac{p}{2})+\frac{p-q}{p}}=(\log
N)^{1-\frac{q}{2}},$

while for $p>2$

$I_{1}\lesssim(\log N)^{\frac{p-q}{p}}.$

Thus, we have proved 3) and 1) for the case $p\geq q$.

If $1\leq p<q<\infty$ we simply have

$\displaystyle\|B\|_{\mathcal{B}_{p,q}^{0}}^{*}$
$\displaystyle=\bigg{(}\int_{0}^{1}(1-\rho)^{q-1}\|B^{\prime}_{\rho}\|_{L^{p}(\mathbb{T})}^{q}d\rho\bigg{)}^{1/q}$
$\displaystyle\lesssim\bigg{(}\int_{0}^{1}(1-\rho)^{q-1}\|B^{\prime}_{\rho}\|_{L^{q}(\mathbb{T})}^{q}d\rho\bigg{)}^{1/q}\lesssim\begin{cases}(\log
N)^{1/q-1/2},&1\leq q\leq 2,\\\ 1,&q\geq 2.\end{cases}$

The case $q=\infty$ is trivial by the Schwarz–Pick lemma. The proof of the
statements 1)–3) is completed.

4) Consider the case $p=\infty$. If
$B(z)=\prod_{j=1}^{N}\frac{z-\lambda_{j}}{1-\bar{\lambda}_{j}z}$, then

$B^{\prime}(z)=\sum_{j=1}^{N}\hat{B}_{j}(z)\frac{1-|\lambda_{j}|^{2}}{(1-\bar{\lambda}_{j}z)^{2}},$

where $\hat{B}_{j}(z)=\prod_{k\neq
j}\frac{z-\lambda_{k}}{1-\bar{\lambda}_{k}z}$. Hence,

$\|B^{\prime}_{\rho}\|_{\infty}\lesssim\sum_{j=1}^{N}\frac{1-|\lambda_{j}|^{2}}{(1-|\lambda_{j}|\rho)^{2}}$

and, using again the fact that
$\|B^{\prime}_{\rho}\|_{\infty}\leq(1-\rho)^{-1}$, we get

$(\|B\|_{\mathcal{B}_{\infty,q}^{0}}^{*})^{q}=\int_{0}^{1}(1-\rho)^{q-1}\|B^{\prime}_{\rho}\|_{\infty}^{q}d\rho\leq\int_{0}^{1}\|B^{\prime}_{\rho}\|_{\infty}d\rho\lesssim
N.$

Let us show that all estimates are sharp. The growth $(\log N)^{1/q-1/p}$ in
the case 3) is achieved by the product $B^{\star}$ defined by (3.3). Indeed,
for $1\leq q\leq p<\infty$,

$\|B^{\star}\|_{\mathcal{B}_{p,q}^{0}}^{*}\geq\prod_{i=1}^{N}|\lambda_{i}|\cdot{\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma^{\star})\gtrsim(\log N)^{1/q-1/p}$

by Theorem 4. In the case 2) the optimality of the estimate can be already
seen on $B(z)=z^{N}$.

For the case $1\leq p,q\leq 2$ one can use an example of a Blaschke product
constructed in [3]: there exists a Blaschke product of order $N$ such that

$\int_{0}^{1-1/N}\int_{0}^{2\pi}|B^{\prime}(\rho e^{it})|dtd\rho\geq
c\sqrt{\log N},$

where $c>0$ is an absolute constant; see the end of Section 2 in [3]. This
construction is based on deep results of R. Bañuelos and C. N. Moore [2]
related to Makarov’s law of the iterated logarithm. An easy application of the
Hölder inequality shows that for $1\leq p,q\leq 2$

$\int_{0}^{1-1/N}(1-\rho)^{q-1}\|B^{\prime}_{\rho}\|^{q}_{L^{p}(\mathbb{T})}d\rho\gtrsim(\log
N)^{1-q/2}.$

Finally, let us show that the estimate in the case 4) also is best possible.
Take $\lambda_{j}=1-2^{-j}$. Since the sequence $(\lambda_{j})$ is an
interpolating sequence for $H^{\infty}$, there exists $\delta>0$ such that

$\prod_{k\neq
j}\Big{|}\frac{\lambda_{k}-\lambda_{j}}{1-\overline{\lambda_{k}}\lambda_{j}}\Big{|}\geq\delta$

for all $j$. Thus, if $B$ denotes the Blaschke product with zeros
$\lambda_{1},\ldots,\lambda_{N}$, then

$|B^{\prime}(\lambda_{j})|\geq\frac{\delta}{1-|\lambda_{j}|^{2}}\gtrsim 2^{j}$

for $j=1,\ldots,N$. It follows that $\|B^{\prime}_{\rho}\|_{\infty}\geq 2^{j}$
for $\rho\geq\lambda_{j}$ and thus

$\|B\|_{\mathcal{B}_{\infty,q}^{0}}^{q}\gtrsim\sum_{j=1}^{N-1}\int_{\lambda_{j}}^{\lambda_{j+1}}(1-\rho)^{q-1}\|B^{\prime}_{\rho}\|_{\infty}^{q}d\rho\gtrsim
N.$

∎

###  7.2. Proof of Theorem 3

Consider the simplest test function

$f=\frac{B_{\sigma}}{B_{\sigma}(0)}=(-1)^{N}\frac{B_{\sigma}}{\prod_{i=1}^{N}\lambda_{i}}.$

Then

${\rm
cap}_{\mathcal{B}_{p,q}^{0}}(\sigma)\leq\frac{\|B_{\sigma}\|_{\mathcal{B}_{p,q}^{0}}}{\prod_{i=1}^{N}|\lambda_{i}|},$

and the statement follows from Proposition 10 in all cases except $p=\infty$
(in Region III), where an application of Proposition 10 will lead to a
substantially worse growth order.

To treat the case $p=\infty$ we use the test function from [9]. Put $r=1-1/N$
and consider the finite Blaschke product $\tilde{B}$ with zeros
$r\lambda_{1},\dots,r\lambda_{N}$,

$\tilde{B}(z)=\prod_{i=1}^{n}\frac{z-r\lambda_{i}}{1-r\overline{\lambda_{i}}z}.$

Let

$f(z)=(-1)^{N}\frac{\tilde{B}(rz)}{r^{N}\prod_{i=1}^{N}\lambda_{i}}.$

Clearly $f$ satisfies $f(0)=1$ and $f(\lambda_{i})=0$ for $i=1,\dots,n.$ Since
$r^{N}\asymp 1$, we have

$\prod_{i=1}^{N}|\lambda_{i}|\cdot|f^{\prime}(\rho
e^{it})|\lesssim|\tilde{B}^{\prime}(r\rho e^{it})|\lesssim\frac{1}{1-r\rho}.$

Then, for $1\leq q<\infty$,

$\bigg{(}\prod_{i=1}^{N}|\lambda_{i}|\bigg{)}^{q}\cdot\|f\|_{\mathcal{B}_{\infty,q}^{0}}^{q}\lesssim\int_{0}^{1-1/N}\frac{d\rho}{1-\rho}+\int_{1-1/N}^{1}\frac{d\rho}{1-r}\lesssim\log
N.$

Theorem 3 is proved. ∎

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