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# On the Fourier coefficients of powers
of a finite Blaschke product

Alexander Borichev  Aix-Marseille Université, CNRS, Centrale Marseille, I2M,
Marseille, France<EMAIL_ADDRESS>,  Karine Fouchet  Aix-
Marseille Université, Laboratoire ADEF, Campus Universitaire de Saint-Jérôme,
52 Avenue Escadrille Normandie Niemen, 13013 Marseille, France  Aix-Marseille
Université, CNRS, Centrale Marseille, I2M, Marseille, France
<EMAIL_ADDRESS>and  Rachid Zarouf  Aix-Marseille Université,
Laboratoire ADEF, Campus Universitaire de Saint-Jérôme, 52 Avenue Escadrille
Normandie Niemen, 13013 Marseille, France<EMAIL_ADDRESS>
###### Abstract.

Given a finite Blaschke product $B$ we prove asymptotically sharp estimates on
the $\ell^{\infty}$-norm of the sequence of the Fourier coefficients of
$B^{n}$ as $n$ tends to $\infty$. We provide constructive examples which show
that our estimates are sharp. As an application we construct a sequence of
$n\times n$ invertible matrices $T$ with arbitrary spectrum in the unit disk
and such that the quantity $|\det{T}|\cdot\|T^{-1}\|\cdot\|T\|^{1-n}$ grows as
a power of $n$. This is motivated by Schäffer’s question on norms of inverses.

###### Key words and phrases:

Fourier coefficients, finite Blaschke products, van der Corput lemma,
condition number of matrices.

###### 2020 Mathematics Subject Classification:

30J10, 42A16, 41A60, 15A60.

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 $\mathbb{D}$ be the unit disk of the complex plane, $m\geq 1$, let
$\sigma=\left(\lambda_{1},\dots,\lambda_{m}\right)\in\mathbb{D}^{m}$ and let
$B$ be the finite Blaschke product of degree $m\geq 1$ associated to $\sigma$:

$B(z)=\prod_{j=1}^{m}b_{\lambda_{j}}(z),$

where $b_{\lambda}(z)=\frac{z-\lambda}{1-\bar{\lambda}z}$ is the Blaschke
factor corresponding to $\lambda\in\mathbb{D}$. Here and later on we omit the
standard unimodular factor $\bar{\lambda}/|\lambda|$ which is of no importance
for the questions we study here. Furthermore, we assume that $0\not\in\sigma$.
For $k\geq 0$ we consider the $k^{{\rm th}}$-Fourier coefficient of $B^{n}$
defined by:

$\widehat{B^{n}}(k)=\frac{1}{2\pi}\int_{0}^{2\pi}B^{n}(e^{{\rm
i}\theta})e^{-\rm ik\theta}\,d\theta.$

We study the asymptotic behavior of the $\ell^{\infty}$-norm of the sequence
$\bigl{(}\widehat{B^{n}}(k)\bigr{)}_{k\geq 0}$:

$\|\widehat{B^{n}}\|_{\ell^{\infty}}:=\sup_{k\geq 0}|\widehat{B^{n}}(k)|,$

as $n$ grows large. We use some standard notation from asymptotic analysis.
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 $f\leq cg$. We say
that $f$ and $g$ are comparable, denoted by $f\asymp g$, if both $f\lesssim g$
and $g\lesssim f$.
The case $B=b_{\lambda}$, $\lambda$ being arbitrary in
$\mathbb{D}\setminus\\{0\\}$, is well-studied [1, 16, 2, 10] and it is known
that

(1.1) $\|\widehat{b_{\lambda}^{n}}\|_{\ell^{\infty}}\asymp n^{-1/3}.$

We mention in passing that Y. Meyer recently rediscovered the upper bound in
(1.1) in view of applying it to some sparse crystalline measure constructions
[10]. Precise asymptotic formulas for the $k^{{\rm th}}$ Fourier coefficients
of $b_{\lambda}^{n}$, $k\in[0,\infty)$ as $n\rightarrow\infty$ have been
recently obtained in [2] where the authors distinguish several regions of
different asymptotic behavior of $\widehat{b_{\lambda}^{n}}(k)$ in terms of
$k$ and $n.$ These asymptotic formulas are applied to the construction of
strongly annular functions with Taylor coefficients satisfying sharp summation
properties, which improves and generalizes the results in [1].
Information about those coefficients has also recently been exploited in
operator theory [8] to identify decreasing weights for which composition
operators are bounded on weighted Hardy spaces of Hilbert type. Observe also
that a connection was earlier established in [3] between the asymptotics of
the Fourier coefficients of $B^{n}$ and the boundedness of the composition
operator $\mathcal{C}_{B}$, $\mathcal{C}_{B}(f)=f\circ B$ on the analytic
Beurling–Sobolev space of analytic functions in $\mathbb{D}$ whose sequence of
Taylor coefficients belongs to $l^{p},$ $p\in[1,\infty]$. Generally speaking,
to verify whether $\mathcal{C}_{B}$ is a bounded linear operator from one
Banach space $X$ of analytic functions into another, say $Y$, it is often
enough to know the asymptotic behavior of $\|B^{n}\|_{Y}$. It is shown in
particular in [3] that for any finite Blaschke product $B$

(1.2) $\|\widehat{B^{n}}\|_{\ell^{p}}\asymp n^{\frac{2-p}{2p}}\
\textnormal{for}\ p\in[1,2]$

and that

(1.3) $\|\widehat{B^{n}}\|_{\ell^{p}}\gtrsim n^{\frac{2-p}{2p}}\
\textnormal{for}\ p\in[2,\infty].$

The study of the case $p=1$ in (1.2) with a single Blaschke factor
$B=b_{\lambda}$ was probably initiated by J.-P. Kahane [6]. Applying van der
Corput type estimates on $\widehat{b_{\lambda}^{n}}(k)$ [6, p. 253], he proved
that $\|\widehat{b_{\lambda}^{n}}\|_{\ell^{1}}\asymp n^{-1/2}$ which is in
line with (1.2). Notice that his motivation [6, Theorem 1] was different from
the above: he generalized a theorem by Z. K. Leibenson [7], which is a special
case of a theorem [13, Theorem 4.1.3] about homomorphisms of group algebras
due to P. T. Cohen. The exact asymptotic $n$-dependency of
$\|\widehat{b_{\lambda}^{n}}\|_{\ell^{p}}$ for $p\in[1,\infty]$ was described
in [16] and a change of asymptotic behavior at $p=4$ was in particular
discovered [16, 2]. In this paper we focus on the case $p=\infty$: in this
case the estimate (1.3) reads

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\gtrsim n^{-1/2}$

for any finite Blaschke product $B$. We will prove that
$\|\widehat{B^{n}}\|_{\ell^{\infty}}$ is never of order $n^{-1/2}$. More
precisely, for any finite Blaschke product $B$, there exists an integer $N\geq
3$ such that

(1.4) $\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/N}.$

For every integer $N\geq 3$ we will exhibit an example of a Blaschke product
$B$ of degree $N$ satisfying estimate (1.4). It is known [16] that for $N=3$
every Blaschke product $B(z)\not=z$ of degree 1 satisfies (1.4). For $N=5$ –
respectively $N=7$ – we produce explicit examples of Blaschke products of
degree 2 – respectively degree 4 – such that (1.4) holds.
Finally we apply (1.4) to construct a new class of $n\times n$ invertible
matrices $T$ with arbitrary spectrum in $\mathbb{D}$ such that

(1.5) $\frac{|\det{T}|\cdot\|T^{-1}\|}{\|T\|^{n-1}}\gtrsim n^{1/N}$

for some integer $N\geq 3.$ This is motivated by Schäffer’s question on the
norms of $n\times n$ invertible matrices, see Section 3.

### Outline of the paper

The paper is organized as follows. In Section 2 we establish sharp estimates
on the $\ell^{\infty}$-norm of the sequence $(\widehat{B^{n}}(k))_{k\geq 0}$
for large $n$, where $B$ is an arbitrary Blaschke product, see Theorem 1.
Concrete examples of finite Blaschke products achieving estimate (1.4) are
provided in Theorem 2. In Section 3 we state Theorem 3, which extends the work
[16, 17, 18] already undertaken to answer a question raised by Schäffer on
norms of inverse matrices. In Theorem 3 we exhibit a sequence of $n\times n$
matrices $T$ satisfying (1.5), which we obtain as an application of the
techniques used to prove Theorem 1. Section 4 is devoted to the proof of
Theorem 1 which combines the use of two van der Corput lemmata with a
stationary phase type argument. In Section 5 we prove Theorem 2 and in
particular we construct finite Blaschke products $B$ of degree 2 –
respectively 4 – such that $\|\widehat{B^{n}}\|_{\ell^{\infty}}$ grows as
$n^{-1/5}$ – respectively $n^{-1/7}$. Finally in Section 6 we prove Theorem 3
combining a duality method with the techniques that lead to the proof of
Theorem 1. An alternative proof of Theorem 3 using Theorem 1 is provided in
Remark 9.

##  2\. Notation and statement of our results

In this section we give sharp asymptotics for the $\ell^{\infty}$-norm of the
sequence $\bigl{(}\widehat{B^{n}}(k)\bigr{)}_{k\geq 0}$, see Theorem 1 below,
and constructive examples that achieve these asymptotics, see Theorem 2 below.
We denote by $\psi_{B}(\theta)$ the continuous argument (determined modulo
$2\pi$) of $B(e^{{\rm i}\theta})$:

$B(e^{{\rm i}\theta})=\exp\left({\rm i}\psi_{B}(\theta)\right)$

and by $(\xi_{\ell})_{\ell=1}^{s}$ the sequence of (consecutive) zeros of
$\psi_{B}^{\prime\prime}$ on $[0,2\pi)$ with respective multiplicities
$(N_{\ell}-2)_{\ell=1}^{s}$, $N_{\ell}\geq 3$. This means that

$0\leq\xi_{1}<\xi_{2}<\ldots<\xi_{s}<2\pi$

and that for any $\ell=1,\ldots,s$

$\psi^{\prime\prime}_{B}(\xi_{\ell})=\ldots=\psi_{B}^{(N_{\ell}-1)}(\xi_{\ell})=0,\qquad\psi_{B}^{(N_{\ell})}(\xi_{\ell})\not=0.$

Without loss of generality (using a simple rotation if necessary), we can
assume that $\xi_{1}>0$.

######  Theorem 1.

Let $B$ be a finite Blaschke product, and let $\psi_{B}$,
$(\xi_{\ell})_{\ell=1}^{s}$, $(N_{\ell})_{\ell=1}^{s}$ be defined as above.
Then we have

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/N},$

where $N=\max_{1\leq\ell\leq s}N_{\ell}$.

It is known [2] that if $B=b_{\lambda}$ is the Blaschke factor associated to
any fixed $\lambda\in\mathbb{D}\setminus\\{0\\},$ then $N=3$ and

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/3},$

which proves the sharpness of Theorem 1 for $N=3$. Here we establish the
following result.

######  Theorem 2.

For every $N\geq 3$, there exists a Blaschke product $B$ of degree $N$ such
that

(2.1) $\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/N}.$

Moreover:
(1) For $N=3$ every Blaschke product of degree 1, different from identity,
satisfies (2.1).
(2) For $N=5$ there exists a Blaschke product of degree 2 such that (2.1)
holds.
(3) For $N=7$ there exists a Blaschke product of degree 4 such that (2.1)
holds.

######  Remark.

The following two questions remain open:
(1) Every nontrivial Blaschke product $B$ of degree $1$ satisfies (2.1) with
$N=3$. In the case $N=5$, a Blaschke product of degree $2$ satisfies (2.1),
and in the case $N=7$, a Blaschke product of degree $4$ satisfies (2.1). What
is the minimal degree of a Blaschke product $B_{N}$ such that (2.1) holds for
a fixed $N\geq 6$?
(2) Is there an infinite Blaschke product $B$ such that for $n\geq 1$ we have

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp\frac{1}{\log(n+1)}\,?$

##  3\. Application to operator theory

A well-established circle of questions in Operator Theory concerns estimating
the norm of the inverse of an invertible $n\times n$ matrix $T$ on an
$n$-dimensional Banach space $X$. In the early 1970’s, B. L. Van der Waerden,
W. A. Coppel and J. J. Schäffer [14] studied the smallest quantity $C$,
denoted $C_{\mathcal{B}}(n)$, such that

$|\det{T}|\cdot\|T^{-1}\|\leq C\|T\|^{n-1}$

for any invertible $T$ and for any $X$. In 1970, Schäffer [14] proved the
estimate $C_{\mathcal{B}}(n)\leq\sqrt{en}$, and showed that the inequality

(3.1) $|\det T|\cdot\|T^{-1}\|\leq 2\|T\|^{n-1}$

holds for any invertible $T$ acting on $\mathbb{C}^{n}$ endowed with the
$\ell^{1}$-norm or with the $\ell^{\infty}$-norm (and that (3.1) is optimal in
both cases). This led him to make the conjecture – nowadays known as
Schäffer’s conjecture – that $C_{\mathcal{B}}(n)=2$, $n\in\mathbb{N}$. The
latter was disproved first by E. Gluskin, M. Meyer and A. Pajor [4], J.
Bourgain [4] 111The same article [4] contains an appendix with a stronger
estimate due to Bourgain. and later by H. Queffélec [12] who also proved that
$C_{\mathcal{B}}(n)\gtrsim\sqrt{n}$. The above mentioned results make use of
the following analytic expression for $C_{\mathcal{B}}(n)$, given in [4], in
terms of a “max-min-type” optimization problem :

$C_{\mathcal{B}}(n)=\sup_{(\lambda_{1},\ldots,\lambda_{n})\in\mathbb{D}^{n}}\Phi(\lambda_{1},\ldots,\lambda_{n}),$

where

$\Phi(\lambda_{1},\ldots,\lambda_{n})\\\
:=\inf\biggl{\\{}\sum_{k=1}^{\infty}|a_{k}|:f(z)=\prod_{j=1}^{n}\lambda_{j}+\sum_{k=1}^{\infty}a_{k}z^{k},\,f(\lambda_{j})=0,\,j=1,\ldots,n\biggr{\\}}.$

The proof by Queffélec makes use of Bourgain’s lower bound on $\Phi$ [18,
Inequality (2.2)], which he combines with a number theoretic argument to prove
the existence of a family $(\lambda_{1},\ldots,\lambda_{n})$ on the circle of
radius $1-1/n$ such that
$\Phi(\lambda_{1},\ldots,\lambda_{n})\gtrsim\sqrt{n}$.
The question of finding a concrete exemple of a sequence
$(\lambda_{1},\ldots,\lambda_{n})$ in $\mathbb{D}^{n}$ such that
$\Phi(\lambda_{1},\ldots,\lambda_{n})\gtrsim\sqrt{n}$ has been recently
addressed by O. Szehr–R. Zarouf [18] who proved that for every
$\lambda\in\mathbb{D}\setminus\\{0\\}$ we have
$\Phi(\lambda,\ldots,\lambda)\gtrsim\sqrt{n}$, where $\lambda$ is repeated $n$
times. As a consequence, they construct in [18] an explicit class of
counterexamples to Schäffer’s conjecture: a sequence of invertible lower
triangular Toeplitz matrices $T_{\lambda}\in\mathcal{M}_{n}$ with singleton
spectrum $\\{\lambda\\}\subset\mathbb{D}\backslash\\{0\\}$ such that

$|\lambda|^{n}\|T_{\lambda}^{-1}\|\geq
c(\lambda)\sqrt{n}\|T_{\lambda}\|^{n-1},$

where $c(\lambda)>0$ depends only on $\lambda$.

Furthermore, Gluskin–Meyer–Pajor [4, p. 2, lines 25–26] mention that it is of
interest to find concrete examples $(\lambda_{1},\ldots,\lambda_{n})$ for
which $\Phi(\lambda_{1},\ldots,\lambda_{n})$ grows. We will combine the
approach in [18] with Theorem 1 to prove that given a fixed $m\geq 1$ and an
arbitrary sequence
$(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in\mathbb{D}^{m}$, the sequence

$(\lambda_{1},\ldots,\lambda_{1},\lambda_{2},\ldots,\lambda_{2},\ldots,\lambda_{m},\ldots,\lambda_{m})\in\mathbb{D}^{n\times
m}$

(i.e. each $\lambda_{i}$ is repeated according to its multiplicity $n$)
satisfies the estimate

$\Phi(\lambda_{1},\ldots,\lambda_{1},\lambda_{2},\ldots,\lambda_{2},\ldots,\lambda_{m},\ldots,\lambda_{m})\gtrsim
n^{1/N}$

for some integer $N\geq 3$, as $n$ tends to infinity. As a consequence, we
obtain a family of invertible lower triangular matrices $T\in\mathcal{M}_{nm}$
with arbitrary spectrum $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})$ such
that

$|\det T|\cdot\|T^{-1}\|\gtrsim n^{1/N}\|T\|^{n-1}$

for some integer $N\geq 3$, as $n$ tends to infinity. Our construction of $T$
is given in Section 3.2.

###  3.1. Background

We denote by $H(\mathbb{D})$ the space of the functions analytic in the open
unit disk $\mathbb{D}$, and by $H^{\infty}$ the Banach algebra of bounded
analytic functions in $\mathbb{D}$, endowed with the supremum norm. We recall
that the Hardy space $H^{2}$ is defined as the subspace of $H(\mathbb{D})$
consisting of the functions $f$ such that

$\|f\|_{H^{2}}^{2}:=\sup_{0\leq
r<1}\int_{\mathbb{T}}\left|f(rz)\right|^{2}\,d\nu(z)<\infty,$

where $\nu$ is the normalized Lebesgue measure on the unit circle
$\mathbb{T}:=\left\\{z\in\mathbb{C}:\,|z|=1\right\\}$. Let $\sigma$ be a
finite sequence of points in $\mathbb{D}.$ The finite Blaschke product
$B=B_{\sigma}$ corresponding to $\sigma$ is defined by

$B=B_{\sigma}=\prod_{\lambda\in\sigma}b_{\lambda},$

where $b_{\lambda}=\frac{z-\lambda}{1-\overline{\lambda}z}$ is the Blaschke
factor corresponding to $\lambda\in\mathbb{D}$. Then one defines the model
space $K_{B}$ as the finite dimensional subspace of $H^{2}$ given by

$K_{B}:=\left(BH^{2}\right)^{\perp}=H^{2}\ominus BH^{2}.$

Let $\sigma=(\lambda_{1},\dots,\lambda_{q})\in\mathbb{D}^{q}$. We set
$f_{k}=\dfrac{1}{1-\overline{\lambda_{k}}z}$, $k=1,\ldots,q$. Observe that
$\|f_{k}\|_{H^{2}}=\left(1-|\lambda_{k}|^{2}\right)^{-1/2}$. The family
$(f_{k})_{1\leq k\leq n}$ is a basis of $K_{B}$. Furthermore, the family
$(e_{k})_{1\leq k\leq n}$ given by

$e_{1}=\frac{f_{1}}{\|f_{1}\|_{H^{2}}}\,\quad\mbox{and}\quad
e_{k}=\frac{f_{k}}{\|f_{k}\|_{H^{2}}}{\displaystyle\prod_{j=1}^{k-1}}b_{\lambda_{j}},\quad
k=2,\ldots,q,$

is an orthonormal basis of $K_{B}$ (known as the Malmquist–Walsh basis, see
[11]).

The backward shift operator $S:f\mapsto(f-f(0))/z$ acts on $K_{B}$.

###  3.2. Constructive counterexamples to Schäffer’s conjecture

A We put

$\sigma=(\lambda_{1},\dots,\lambda_{1},\lambda_{2},\dots,\lambda_{2},\dots,\lambda_{m},\dots,\lambda_{m})\in\mathbb{D}^{nm},$

where distinct $\lambda_{i}$ are arbitrary in $\mathbb{D}\backslash\\{0\\}$
and are repeated $n$ times. We consider the Malmquist–Walsh basis of $K_{B}$
given by

$e_{sn+t}=\biggl{(}\prod_{j=1}^{s}b^{n}_{\lambda_{j}}\biggr{)}\frac{f_{s+1}}{\|f_{s+1}\|_{H^{2}}}b_{\lambda_{s+1}}^{t-1},\qquad
0\leq s<m,\,1\leq t\leq n.$

The backward shift operator $S$ has a lower triangular matrix with respect to
this basis, with diagonal
$(\lambda_{1},\dots,\lambda_{1},\dots,\lambda_{m},\dots,\lambda_{m})$, see
[15] for the entry-wise description of this matrix. Hence,

(3.2) $\det S=\prod_{j=1}^{m}\lambda^{n}_{j}.$

We will use the duality method from [18] and combine it with our upper
estimates on the Fourier coefficients of the $n^{\text{th}}$ power of a finite
Blaschke product $B$ (see Theorem 1) to show that there exists a Banach space
norm on $K_{B}$ such that the operator $S$ with spectrum
$(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})$ acting there, satisfies the
asymptotic relation $|\det S|\cdot\|S^{-1}\|\cdot\|S\|^{1-n}\gtrsim n^{1/N}$.

Following [4], we choose the norm $\|\cdot\|_{\ell^{\infty}_{A}}$ on $K_{B}$:

$\|f\|_{\ell^{\infty}_{A}}=\|\widehat{f}\|_{\ell^{\infty}}.$

Then $\|S\|_{\ell^{\infty}_{A}\to\ell^{\infty}_{A}}\leq 1$. Furthermore, if
$\dim K_{B}>1$, then $\|S\|_{\ell^{\infty}_{A}\to\ell^{\infty}_{A}}=1$. We are
now ready to state the corresponding result.

######  Theorem 3.

Let

$\sigma=(\lambda_{1},\ldots,\lambda_{1},\lambda_{2},\ldots,\lambda_{2},\ldots,\lambda_{m},\ldots,\lambda_{m})\in\mathbb{D}^{nm},$

where distinct $\lambda_{i}$ are arbitrary in $\mathbb{D}\backslash\\{0\\}$
and are repeated $n$ times. There exists an integer $N\geq 3$ such that

  1. (i)

$|\det
S|\cdot\|S^{-1}\|_{\ell^{\infty}_{A}\to\ell^{\infty}_{A}}\gtrsim{n^{1/N}}$
and

  2. (ii)

$\Phi(\lambda_{1},\ldots,\lambda_{1},\ldots,\lambda_{m},\ldots,\lambda_{m})\gtrsim
n^{1/N}$.

Here $N$ is the integer associated with the Blaschke product
$B=\prod_{j=1}^{m}b_{\lambda_{j}}$ in Theorem 1.

##  4\. Proof of Theorem 1

The proof of Theorem 1 makes use of the following van der Corput lemmata, see,
for example, [5, Lemma 2.1 and Lemma 2.2, p. 56].

######  Lemma 4.

Let $g$ be a real function continuously differentiable on the interval
$[a,b]\subset\mathbb{R}$ such that $g$ and $g^{\prime}$ are monotone and
$g^{\prime}$ does not vanish on $[a,b].$ Then

$\left|\int_{a}^{b}e^{{\rm
i}g(t)}\,dt\right|\leq\frac{2}{|g^{\prime}(a)|}+\frac{2}{|g^{\prime}(b)|}.$

######  Lemma 5.

Let $F(x)$ be a real, twice differentiable function in $[a,b]$ such that
$F^{\prime\prime}(x)\geq\mu>0$ or $F^{\prime\prime}(x)\leq-\mu<0$. Let $G(x)$
be a positive monotonic function in $[a,b]$ such that $G(x)\leq M$. Then

$\left|\int_{a}^{b}G(x)e^{{\rm i}F(x)}\,dx\right|\leq\frac{8M}{\sqrt{\mu}}.$

###### Proof of Theorem 1..

For $j=1,\ldots,m$ we write $\lambda_{j}=\rho_{j}\exp({\rm i}\theta_{j})$,
where $\rho_{j}\in(0,1)$ and $\theta_{j}\in(-\pi,\pi]$. A direct computation
shows that

$\psi^{\prime}_{B}(\theta)=|B^{\prime}(e^{{\rm
i}\theta})|=\Bigl{|}\frac{B^{\prime}(e^{{\rm i}\theta})}{B(e^{{\rm
i}\theta})}\Bigr{|}=\biggl{|}\sum_{1\leq j\leq
m}\frac{b^{\prime}_{\lambda_{j}}(e^{{\rm i}\theta})}{b_{\lambda_{j}}(e^{{\rm
i}\theta})}\biggr{|}.$

Since

$\frac{zb^{\prime}_{\lambda}(z)}{b_{\lambda}(z)}=\Re\frac{1+\overline{\lambda}z}{1-\overline{\lambda}z}\geq
0,\qquad\lambda\in\mathbb{D},\,z\in\mathbb{T},$

we have

(4.1) $\psi^{\prime}_{B}(\theta)=\Re\sum_{1\leq j\leq
m}\frac{1+\overline{\lambda_{j}}e^{{\rm
i}\theta}}{1-\overline{\lambda_{j}}e^{{\rm i}\theta}}.$

Therefore,

$\psi_{B}^{\prime}(\theta)=\sum_{j=1}^{m}\frac{1-\rho_{j}^{2}}{1+\rho_{j}^{2}-2\rho_{j}\cos\left(\theta-\theta_{j}\right)}.$

The $k^{th}$-Fourier coefficient of $B^{n}$ is given by

$\widehat{B^{n}}(k)=\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left({\rm
i}\left(n\psi_{B}(\theta)-k\theta\right)\right)\,d\theta.$

We put

$f(\theta)=\psi_{B}(\theta)-\frac{k}{n}\theta,$

and given $\varepsilon>0$ we write (recall that $\xi_{1}>0$)

$\int_{0}^{2\pi}\exp\left({\rm i}n\,f(\theta)\right){\rm
d}\theta=\sum_{\ell=1}^{s}\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta\\\
+\sum_{\ell=1}^{s-1}\int_{\xi_{\ell}+\varepsilon}^{\xi_{\ell+1}-\varepsilon}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta+\int_{0}^{\xi_{1}-\varepsilon}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta\\\
+\int_{\xi_{s}+\varepsilon}^{2\pi}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta.$

We put

$\varepsilon=n^{-1/N}\rightarrow 0,\qquad n\rightarrow\infty.$

Clearly

$\biggl{|}\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta\biggr{|}\leq 2\varepsilon,$

and therefore the first sum is bounded from above by

$\biggl{|}\sum_{\ell=1}^{s}\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta\biggr{|}\lesssim n^{-1/N}.$

Writing the Taylor expansion (of order $N_{\ell}-2$) of
$\psi_{B}^{\prime\prime}$ near $\xi_{\ell}$ we obtain that

$\psi_{B}^{\prime\prime}(\theta)=\frac{\psi_{B}^{(N_{\ell})}(\xi_{\ell})}{(N_{\ell}-2)!}\left(\theta-\xi_{\ell}\right)^{N_{\ell}-2}\left(1+\mathcal{O}\left(\theta-\xi_{\ell}\right)\right),\qquad\theta\to\xi_{\ell}.$

Therefore, if $\theta$ is close to $\xi_{\ell}$ but satisfies
$|\theta-\xi_{\ell}|\geq\varepsilon$, then

(4.2) $|\psi_{B}^{\prime\prime}(\theta)|\gtrsim n^{-(N_{l}-2)/N}.$

Since $\psi_{B}^{\prime\prime}$ does not vanish on the intervals
$(0,\xi_{1}-\varepsilon)$, $(\xi_{s}+\varepsilon,2\pi)$ and
$(\xi_{\ell}+\varepsilon,\xi_{\ell+1}-\varepsilon)$ for $\ell=1,\ldots,s-1$,
we obtain that (4.2) holds everywhere outside of the intervals
$[\xi_{\ell}-\varepsilon,\xi_{\ell}+\varepsilon]$. Now we can apply Lemma 5 to
the remaining integrals

$\int_{\xi_{\ell}+\varepsilon}^{\xi_{\ell+1}-\varepsilon}e^{{\rm
i}n\,f(\theta)}\,d\theta,\qquad\int_{0}^{\xi_{1}-\varepsilon}e^{{\rm
i}n\,f(\theta)}\,d\theta,\qquad\int_{\xi_{s}+\varepsilon}^{2\pi}e^{{\rm
i}n\,f(\theta)}\,d\theta.$

Since

$|nf^{\prime\prime}|\gtrsim n^{2/N}$

on these intervals, these integrals are also
$\mathcal{O}\left(n^{-1/N}\right)$. Thus,

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\lesssim n^{-1/N}.$

Let $r\in\\{1,\ldots,s\\}$ be such that $N_{r}=N$ and define

$\mathcal{L}=\bigl{\\{}\ell:1\leq\ell\leq
s,\,N_{\ell}=N,\,\psi_{B}^{\prime}(\xi_{\ell})=\psi_{B}^{\prime}(\xi_{r})\bigr{\\}},\quad
D=\mathop{\rm Card}\mathcal{L}.$

We will show that the maximal value of $|\widehat{B^{n}}(k)|$ for $k\geq 0$ is
(asymptotically as $n$ grows large) attained at
$k=k_{d}=k_{d}(n)=[n\psi_{B}^{\prime}(\xi_{r})]+d$, $0\leq d<D$ (where
$\left[A\right]$ means the integer part of $A$), that is,

$\max_{0\leq d<D}|\widehat{B^{n}}(k_{d}(n))|\asymp n^{-1/N}.$

Recall that the $k_{d}^{th}$-Fourier coefficient of $B^{n}$ is given by

$\widehat{B^{n}}(k_{d})=\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta,$

where

$f_{d}(\theta)=\psi_{B}(\theta)-\frac{k_{d}}{n}\theta.$

Let $(\varepsilon_{\ell})_{\ell=1}^{s}$ be a sequence of nonnegative numbers.
Each $\varepsilon_{\ell}$ for $\ell=1,\ldots,s$ will be chosen below over the
proof, depending on the nature of $\xi_{\ell}$. Let $\tau_{0}=0$,
$\tau_{\ell}\in(\xi_{\ell}+\varepsilon_{\ell},\xi_{\ell+1}-\varepsilon_{\ell+1})$,
$1\leq\ell\leq s-1$, $\tau_{s}=2\pi$,
$J_{\ell}=[\xi_{\ell}-\varepsilon_{\ell},\xi_{\ell}+\varepsilon_{\ell}]$,
$1\leq\ell\leq s$. Then

$[0,2\pi)\setminus\bigsqcup_{1\leq\ell\leq s}J_{\ell}=\bigsqcup_{1\leq\ell\leq
s}[\tau_{\ell-1},\xi_{\ell}-\varepsilon_{\ell})\sqcup(\xi_{\ell}+\varepsilon_{\ell},\tau_{\ell})=:\bigsqcup_{1\leq\ell\leq
s}(J^{\prime}_{\ell}\sqcup J^{\prime\prime}_{\ell}).$

We have

$\int_{0}^{2\pi}e^{{\rm i}n\,f(\theta)}\,d\theta=\sum_{1\leq\ell\leq
s}\int_{J_{\ell}}e^{{\rm i}n\,f(\theta)}\,d\theta+\sum_{1\leq\ell\leq
s}\int_{J^{\prime}_{\ell}\sqcup J^{\prime\prime}_{\ell}}e^{{\rm
i}n\,f(\theta)}\,d\theta.$

We divide our argument into three steps in order to show that the main
contribution of $\int_{0}^{2\pi}\exp\left({\rm i}n\,f(\theta)\right)\,d\theta$
is due to $J_{\ell}$, $\ell\in\mathcal{L}$.

The content of Step 1 is computing an asymptotic formula for
$\Sigma=\Sigma_{d}$, the sum of the integrals

$I_{\ell,d}=\int_{J_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta,\qquad\ell\in\mathcal{L},$

with $0\leq d<D$, for suitable choices of $\varepsilon_{\ell}$. In Step 2 we
estimate from above the integrals

$\int_{J_{\ell}}\exp\left({\rm i}n\,f(\theta)\right)\,d\theta,$

for $1\leq\ell\leq s$, $\ell\not\in\mathcal{L}$, and show that they are
asymptotically much smaller than $\Omega=\max_{0\leq d<D}|\Sigma_{d}|$, again
for suitable choices of $\varepsilon_{\ell}$. Finally the goal of Step 3 is to
show – in the same spirit as in Step 2 – that the integrals

$\int_{J^{\prime}_{\ell}}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta,\qquad\int_{J^{\prime\prime}_{\ell}}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta,$

are also asymptotically much smaller than $\Omega$.

Step 1. To make the notation less cluttered we set $\xi=\xi_{\ell}$,
$F=F_{d}=nf_{d}$, $\varepsilon=\varepsilon_{\ell}$, so that

$I=I_{\ell}=I_{\ell,d}=\int_{\xi-\varepsilon}^{\xi+\varepsilon}\exp\left({\rm
i}F_{d}(\theta)\right)\,d\theta.$

Without loss of generality we may assume that

$\psi_{B}^{(N)}(\xi)>0.$

In the opposite case the argument is analogous.

Furthermore, we fix $\delta\in(0,1/(2N^{2}))$ and set

(4.3) $\varepsilon=n^{\delta-\frac{1}{N}}.$

This choice of $\varepsilon$ is motivated below in the proofs of formulas
(4.4) and (4.5).

We will first establish that:

(a) If $N$ is even, then

(4.4)
$I_{\ell,d}=\frac{2}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\times\\\
\times\exp\left({\rm i}n\psi_{B}(\xi)-{\rm i}k_{d}\xi+\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right).$

(b) If $N$ is odd, then

(4.5)
$I_{\ell,d}=\frac{2}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\times\\\
\times\exp\left({\rm i}n\psi_{B}(\xi)-{\rm
i}k_{d}\xi\right)\cos\left(\frac{\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right).$

Here and later on, $\Gamma$ is the Gamma function.

Writing the Taylor expansion of $\psi_{B}$ near $\xi$ we obtain that

$\psi_{B}(\theta)=\psi_{B}(\xi)+(\theta-\xi)\psi_{B}^{\prime}(\xi)\\\
+\frac{(\theta-\xi)^{N}}{N!}\psi_{B}^{(N)}(\xi)+\frac{(\theta-\xi)^{N+1}}{(N+1)!}\psi_{B}^{(N+1)}(\xi+t_{\theta}(\theta-\xi)),$

for some $t_{\theta}\in(0,1)$. Therefore,

$\psi_{B}(\theta)-\frac{k}{n}\theta=\psi_{B}(\xi)-\frac{k}{n}\xi+(\theta-\xi)\left(\psi_{B}^{\prime}(\xi)-\frac{k}{n}\right)+\frac{(\theta-\xi)^{N}}{N!}\psi_{B}^{(N)}(\xi)\\\
+\frac{(\theta-\xi)^{N+1}}{(N+1)!}\psi_{B}^{(N+1)}(\xi+t_{\theta}(\theta-\xi)),$

and

$F(\theta)=F(\xi)+(\theta-\xi)F^{\prime}(\xi)+\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\\\
+\frac{(\theta-\xi)^{N+1}}{(N+1)!}F^{(N+1)}(\xi+t_{\theta}(\theta-\xi)).$

Thus, going back to the integral $I$ we obtain that if $n$ is sufficiently
large, then

$I=\exp\left({\rm
i}F(\xi)\right)\cdot\int_{\xi-\varepsilon}^{\xi+\varepsilon}\exp\left({\rm
i}\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\right)\times\\\ \times\exp\left({\rm
i}(\theta-\xi)F^{\prime}(\xi)+{\rm
i}\frac{(\theta-\xi)^{N+1}}{(N+1)!}F^{(N+1)}(\xi+t_{\theta}(\theta-\xi))\right)\,d\theta.$

Furthermore,

$\exp\left({\rm i}(\theta-\xi)F^{\prime}(\xi)+{\rm
i}\frac{(\theta-\xi)^{N+1}}{(N+1)!}F^{(N+1)}(\xi+t_{\theta}(\theta-\xi))\right)\\\
=1+\mathcal{O}\left(|(\theta-\xi)F^{\prime}(\xi)|+\frac{|\theta-\xi|^{N+1}}{(N+1)!}|F^{(N+1)}(\xi+t_{\theta}(\theta-\xi))|\right)\\\
=1+\mathcal{O}\left(n|\theta-\xi|\left(|f^{\prime}(\xi)|+|\theta-\xi|^{N}\right)\right),$

for $\theta$ in a small neighborhood of $\xi$. This gives

(4.6) $\exp\left(-{\rm i}F(\xi)\right)I\\\
=\int_{\xi-\varepsilon}^{\xi+\varepsilon}\exp\left({\rm
i}\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\right)\,d\theta+\mathcal{O}\left(n\varepsilon^{2}|f^{\prime}(\xi)|+n\varepsilon^{N+2}\right).$

Let us verify that with our choice of $\varepsilon$ we have

  * (i)

$n\varepsilon^{2}|f^{\prime}(\xi)|=o\left(n^{-1/N}\right)$,

  * (ii)

$n\varepsilon^{N+2}=o\left(n^{-1/N}\right)$,

  * (iii)

$n\varepsilon^{N}\rightarrow\infty$ as $n\rightarrow\infty$.

Since $k\leq n\psi_{B}^{\prime}(\xi)<k+D$, we have $n|f^{\prime}(\xi)|\lesssim
1$. Now, (i)–(iii) follow from (4.3).

Thus,

(4.7) $n\varepsilon^{2}|f^{\prime}(\xi)|+n\varepsilon^{N+2}=o(n^{-1/N}).$

Next we consider

$J=\int_{\xi-\varepsilon}^{\xi+\varepsilon}\exp\left({\rm
i}\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\right)\,d\theta\\\ \\\
=\int_{\xi}^{\xi+\varepsilon}\exp\left({\rm
i}\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\right)\,d\theta+\int_{\xi-\varepsilon}^{\xi}\exp\left({\rm
i}\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi)\right)\,d\theta\\\ \\\
=J_{1}+J_{2}.$

First we estimate $J_{1}$. Changing the variable

$u=\frac{(\theta-\xi)^{N}}{N!}F^{(N)}(\xi),$

we obtain that

$J_{1}=\frac{1}{N}\left(\frac{N!}{F^{(N)}(\xi)}\right)^{1/N}\int_{0}^{\frac{\varepsilon^{N}}{N!}F^{(N)}(\xi)}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\,du\\\ \\\
=\frac{1}{N}\left(\frac{N!}{F^{(N)}(\xi)}\right)^{1/N}\left(\int_{0}^{\infty}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\,du-\int_{\frac{\varepsilon^{N}}{N!}F^{(N)}(\xi)}^{\infty}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\,du\right)\\\ \\\
\\!=\frac{1}{N}\left(\frac{N!}{F^{(N)}(\xi)}\right)^{1/N}\\!\\!\left(\exp\left(\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+\mathcal{O}\left(\frac{1}{(\varepsilon^{N}F^{(N)}(\xi))^{1-1/N}}\right)\right).$

Here we use that

$\int_{0}^{\infty}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\,du=\exp\left(\frac{{\rm i}\pi}{2N}\right)\Gamma(1/N),$

and (via integration by parts, with $\gamma>0$) that

$-\int_{\gamma}^{\infty}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\,du=\left[{\rm i}\frac{\exp\left({\rm
i}u\right)}{u^{1-1/N}}\right]_{\gamma}^{\infty}+\left(1-\frac{1}{N}\right){\rm
i}\int_{\gamma}^{\infty}\frac{\exp\left({\rm i}u\right)}{u^{2-1/N}}\,du\\\
=\mathcal{O}\left(\frac{1}{\gamma^{1-1/N}}\right).$

Here $\gamma=\frac{\varepsilon^{N}}{N!}F^{(N)}(\xi)$ and thus
$\gamma^{(N-1)/N}\asymp n^{\delta(N-1)}$.

This gives

$J_{1}=\frac{1}{N}\left(\frac{N!}{F^{(N)}(\xi)}\right)^{1/N}\left(\exp\left(\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+\mathcal{O}\left(\frac{1}{n^{\delta(N-1)}}\right)\right)\\\
\\\
=\frac{1}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\exp\left(\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right).$

If $N$ is even, then $J_{2}=J_{1}$,

$J=\frac{2}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\exp\left(\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right),$

and asymptotic formula (4.4) follows from the above equality combined with
(4.6) and (4.7).

For odd $N$ we obtain in an analogous way that

$J_{2}=\frac{1}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\exp\left(-\frac{{\rm
i}\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right).$

Thus, if $N$ is odd, then

$J=\frac{2}{N}\left(\frac{N!}{n\psi_{B}^{(N)}(\xi)}\right)^{1/N}\cos\left(\frac{\pi}{2N}\right)\Gamma(1/N)+o\left(\frac{1}{n^{1/N}}\right),$

and asymptotic formula (4.5) follows from the above equality combined with
(4.6) and (4.7).

By (4.4) and (4.5) we obtain that

$\Sigma=\sum_{\ell\in\mathcal{L}}I_{\ell,d}=\sum_{\ell\in\mathcal{L}}c_{\ell}\exp(-{\rm
i}d\xi_{\ell}),\qquad 0\leq d<D.$

The points $\xi_{\ell}$ are pairwise disjoint, and hence the square matrix
$\bigl{(}e^{-{\rm i}d\xi_{\ell}}\bigr{)}_{\ell\in\mathcal{L},\,0\leq d<D}$ is
invertible. Since it does not depend on $n$, we conclude that

$\Omega=\max_{0\leq
d<D}\Bigl{|}\sum_{\ell\in\mathcal{L}}I_{\ell,d}\Bigr{|}\gtrsim\sum_{\ell\in\mathcal{L}}|c_{\ell}|\asymp
n^{-1/N}.$

Step 2. Now we deal with the $s-D$ integrals

$\int_{\xi_{\ell}-\varepsilon_{\ell}}^{\xi_{\ell}+\varepsilon_{\ell}}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta,$

$1\leq\ell\leq s$, $N_{\ell}\neq N$ or
$\psi^{\prime}_{B}(\xi_{\ell})\neq\psi^{\prime}_{B}(\xi_{r})$, and choose
$\varepsilon_{\ell}$ in such a way that the corresponding integrals are
$o\left(\frac{1}{n^{1/N}}\right)$. We distinguish the following two cases.

(1) If $N_{\ell}=N$, then the integral

$\int_{\xi_{\ell}-\varepsilon_{\ell}}^{\xi_{\ell}+\varepsilon_{\ell}}\exp\left({\rm
i}n\,f(\theta)\right)\,d\theta$

is treated as follows. By continuity of $\psi_{B}^{\prime}$ at the point
$\xi_{\ell}$ there exists $\eta_{\ell}>0$ (independent of $n$) such that

$|\theta-\xi_{\ell}|\leq\eta_{\ell}\implies|\psi_{B}^{\prime}(\xi_{\ell})-\psi_{B}^{\prime}(\theta)|\leq|\psi_{B}^{\prime}(\xi_{\ell})-\psi_{B}^{\prime}(\xi_{r})|/2.$

We choose $\varepsilon_{\ell}=\eta_{\ell}$ and observe that

$n|f^{\prime}_{d}(\theta)|\geq
n|\psi_{B}^{\prime}(\xi_{\ell})-\psi_{B}^{\prime}(\xi_{r})|/2-D\gtrsim
n,\qquad|\theta-\xi_{\ell}|\leq\varepsilon_{\ell},\,0\leq d<D.$

Moreover, $f=f_{d}$ is monotonic over the interval
$[\xi_{\ell}-\varepsilon_{\ell},\xi_{\ell}+\varepsilon_{\ell}]$ and
$f^{\prime}$ is monotonic over $[\xi_{\ell}-\varepsilon_{\ell},\xi_{\ell}]$
and $[\xi_{\ell},\xi_{\ell}+\varepsilon_{\ell}]$ (because
$f^{\prime\prime}=\psi_{B}^{\prime\prime}$ vanishes at $\xi_{\ell}$ and
nowhere else on the interval
$[\xi_{\ell}-\varepsilon_{\ell},\xi_{\ell}+\varepsilon_{\ell}]$). The
assumptions of Lemma 4 are therefore satisfied and an application of this
lemma gives:

$\int_{\xi_{\ell}-\varepsilon_{\ell}}^{\xi_{\ell}+\varepsilon_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta=\mathcal{O}\left(\frac{1}{n}\right).$

(2) If $N_{\ell}<N$, then the situation is even simpler: we set
$\varepsilon_{\ell}=n^{-1/N_{\ell}}$, and estimate directly

$\biggl{|}\int_{\xi_{\ell}-\varepsilon_{\ell}}^{\xi_{\ell}+\varepsilon_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta\biggr{|}\lesssim n^{-1/N_{\ell}}.$

Step 3. We need to verify that

$\biggl{|}\int_{J^{\prime}_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta\biggr{|}+\biggl{|}\int_{J^{\prime\prime}_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta\biggr{|}\\\ =o(n^{-1/N}),\quad
1\leq\ell\leq s,\,0\leq d<D.$

Given $J^{\prime}_{\ell}$ (or $J^{\prime\prime}_{\ell}$ with an analogous
argument), we consider the following cases.

(i) $\ell\in\mathcal{L}$. Then $\varepsilon_{\ell}=n^{\delta-(1/N)}$, and
considering the Taylor expansion (of order $N-2$) of $\psi_{B}^{\prime\prime}$
near $\xi_{\ell}$, we obtain for large $n$ and for $\theta$ close to
$\xi_{\ell}$ satisfying $|\theta-\xi_{\ell}|\geq\varepsilon_{\ell}$ that

$|\psi_{B}^{\prime\prime}(\theta)|\gtrsim\varepsilon_{\ell}^{N-2}=n^{\delta(N-2)-(N-2)/N}.$

Since $\psi_{B}^{\prime\prime}$ does not vanish on the interval
$J^{\prime}_{\ell}$, we conclude that

$n|f_{d}^{\prime\prime}(\theta)|\gtrsim n^{(2/N)+\delta(N-2)},\qquad\theta\in
J^{\prime}_{\ell},\,0\leq d<D.$

An application of Lemma 5 gives

$\int_{J^{\prime}_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta=o(n^{-1/N}).$

(ii) $\ell\not\in\mathcal{L}$ and $N_{\ell}=N$. Since $\varepsilon_{\ell}$
does not depend on $n$, we have

$|\psi_{B}^{\prime\prime}(\theta)|\gtrsim 1,\qquad\theta\in
J^{\prime}_{\ell},$

and

$n|f_{d}^{\prime\prime}(\theta)|\gtrsim n,\qquad\theta\in
J^{\prime}_{\ell},\,0\leq d<D.$

Therefore, by Lemma 5, we have

$\int_{J^{\prime}_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta=\mathcal{O}\left(n^{-1/2}\right).$

(iii) $N_{\ell}<N$. Here $\varepsilon_{\ell}=n^{-1/N_{\ell}}$, and, arguing as
above, we conclude that

$n|f_{d}^{\prime\prime}(\theta)|\gtrsim n^{2/N_{\ell}},\qquad\theta\in
J^{\prime}_{\ell},\,0\leq d<D,$

and

$\int_{J^{\prime}_{\ell}}\exp\left({\rm
i}n\,f_{d}(\theta)\right)\,d\theta=\mathcal{O}\left(n^{-1/N_{\ell}}\right).$

Summing up, Steps 1–3 give us that

$\max_{0\leq d<D}|\widehat{B^{n}}(k_{d})|\gtrsim n^{-1/N},$

which completes the proof. ∎

##  5\. Proof of Theorem 2

###  5.1. The general case $N\geq 3$

###### Proof of Theorem 2, formula (2.1)..

Let $N\geq 3$ and let $B=\prod_{1\leq j\leq N}b_{\lambda_{j}}$ be a finite
Blaschke product. The argument $\psi_{B}(\theta)$ of $B(e^{{\rm i}\theta})$ is
determined modulo $2\pi$. Furthermore, $\psi^{\prime}_{B}$ is a real analytic
$2\pi$-periodic function, and by (4.1), we have

$\psi^{\prime}_{B}(\theta)=\Re\sum_{1\leq j\leq
N}\frac{1+\overline{\lambda_{j}}e^{{\rm
i}\theta}}{1-\overline{\lambda_{j}}e^{{\rm i}\theta}}.$

Since $\psi^{\prime}_{B}$ is real analytic, the function
$\psi^{\prime\prime}_{B}$ has only finite number of zeros on $[0,2\pi]$. By
Theorem 1, to obtain that

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/N},$

it suffices to verify that
$\psi^{\prime\prime}_{B}(0)=\ldots=\psi_{B}^{(N-1)}(0)=0$,
$\psi_{B}^{(N)}(0)\not=0$, and $\psi^{\prime\prime}_{B}$ has no zeros of order
$N-1$.

Set $u=\exp(\frac{2\pi{\rm i}}{N})$. For $t\in(0,1/(2\pi))$ to be chosen later
on, we set $\zeta=t\exp(\frac{\pi{\rm i}(N-1)}{2N})$ and

$\lambda_{j}=\frac{\overline{\zeta u^{j}}}{1+\overline{\zeta
u^{j}}}\in\mathbb{D},\qquad 1\leq j\leq N.$

Then

$\displaystyle\psi^{\prime}_{B}(\theta)$ $\displaystyle=-N+2\Re\sum_{1\leq
j\leq N}\frac{1}{1-\overline{\lambda_{j}}e^{{\rm i}\theta}}$
$\displaystyle=-N+2\Re\sum_{1\leq j\leq N}\frac{1+\zeta u^{j}}{1-\zeta
u^{j}(e^{{\rm i}\theta}-1)}$ $\displaystyle=-N+2\Re\sum_{s\geq 0}\sum_{1\leq
j\leq N}(1+\zeta u^{j})\zeta^{s}u^{js}(e^{{\rm i}\theta}-1)^{s}$
$\displaystyle=N+2N\Re\sum_{k\geq 1}\zeta^{kN}e^{{\rm i}\theta}(e^{{\rm
i}\theta}-1)^{kN-1}$ $\displaystyle=N+2N\sum_{k\geq 1}t^{kN}\Re\Bigl{(}{\rm
i}^{k(N-1)}e^{{\rm i}\theta}(e^{{\rm i}\theta}-1)^{kN-1}\Bigr{)}.$

Furthermore,

$\displaystyle\psi^{\prime\prime}_{B}(\theta)$ $\displaystyle=2N\Re\sum_{k\geq
1}\zeta^{kN}{\rm i}e^{{\rm i}\theta}(kNe^{{\rm i}\theta}-1)(e^{{\rm
i}\theta}-1)^{kN-2}$ $\displaystyle=2N\sum_{k\geq 1}t^{kN}\Re\Bigl{(}{\rm
i}^{k(N-1)+1}e^{{\rm i}\theta}(kNe^{{\rm i}\theta}-1)(e^{{\rm
i}\theta}-1)^{kN-2}\Bigr{)}.$

It is clear that (independently of $t$) we have
$\psi^{\prime\prime}_{B}(0)=\ldots=\psi_{B}^{(N-1)}(0)=0$. Since ${\rm
i}^{2N-2}\in\mathbb{R}$, for some $c=c(t)\not=0$ we have
$\psi^{\prime\prime}_{B}(\theta)\sim c\theta^{N-2}$ at $0$ and, hence,
$\psi_{B}^{(N)}(0)\not=0$.

Set

$\displaystyle h(\theta)$ $\displaystyle={\rm i}^{N}e^{{\rm i}\theta}(Ne^{{\rm
i}\theta}-1)(e^{{\rm i}\theta}-1)^{N-2}$ $\displaystyle=(-1)^{N-1}e^{{\rm
i}\theta(N+2)/2}(N-e^{-{\rm i}\theta})\cdot 2^{N-2}(\sin(\theta/2))^{N-2}.$

Then $h(0)=0$ if and only if $\theta\in 2\pi\mathbb{Z}$, and

$\frac{h^{\prime}(\theta)}{h(\theta)}=\frac{N+2}{2}{\rm i}+\frac{{\rm
i}e^{-{\rm i}\theta}}{N-e^{-{\rm
i}\theta}}+\frac{N-2}{2}\cdot\frac{\cos(\theta/2)}{\sin(\theta/2)}.$

Hence,

$\Im\frac{h^{\prime}(\theta)}{h(\theta)}\geq\frac{N+2}{2}-\frac{1}{N-1}>0,\qquad
0<|\theta|\leq\pi.$

Thus, at every point $\theta\in]0,2\pi[$ we have

$|\Re h(\theta)|+|\Re h^{\prime}(\theta)|>0.$

Since $\Re h^{(N-2)}(0)=(-1)^{N-1}(N-1)!\not=0$, we have

$\sum_{s=0}^{N-2}|\Re h^{(s)}(\theta)|>0,\qquad|\theta|\leq\pi.$

By continuity, we can find $\delta>0$ such that

$\sum_{s=0}^{N-2}|\Re h^{(s)}(\theta)|\geq\delta,\qquad|\theta|\leq\pi.$

Now we use that

$\psi^{\prime\prime}_{B}(\theta)=2Nt^{N}\Re h(\theta)\\\ +2Nt^{2N}\sum_{k\geq
2}t^{(k-2)N}\Re\Bigl{(}{\rm i}^{N}e^{{\rm i}\theta}(kNe^{{\rm
i}\theta}-1)(e^{{\rm i}\theta}-1)^{kN-2}\Bigr{)}.$

Therefore, we can fix a small positive $t$, $t<1/(2\pi)$, such that

$\sum_{s=2}^{N}|\psi_{B}^{(s)}(\theta)|\geq\delta
Nt^{N},\qquad|\theta|\leq\pi.$

Now, $\psi^{\prime\prime}_{B}$ has no zeros of order $N-1$. Thus, by Theorem
1, for every $N\geq 3$, we have constructed a Blaschke product $B_{N}$ of
order $N$ such that $\|\widehat{B_{N}^{n}}\|_{\ell^{\infty}}\asymp n^{-1/N}$.
∎

###  5.2. The case $N=5$

In this section we give two explicit examples of finite Blaschke product $B$
of degree 2 satisfying the estimate

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/5}.$

In the first example, the zeros of $B$ are of the same modulus, while in the
second example they are on the same diameter of the unit disk.

####  5.2.1. Example 1

Let $w\in\mathbb{C}\setminus\\{1\\}$ be such that $\Re w>0$,

(5.1) $\Re(w)=\Re(w^{3})\not=\Re(w^{5})$

and

(5.2) $\Re(w^{-1})\not=\Re(w^{-3}).$

Set $w_{1}=w$, $w_{2}=\overline{w}$,

$\lambda_{j}=\frac{w_{j}-1}{w_{j}+1},\qquad j=1,2,$

and consider

$B=\prod_{1\leq j\leq 2}b_{\lambda_{j}}.$

For example, we can choose

$w=2+{\rm i}.$

Then

$\displaystyle\Re(w)=\Re(w^{3})=2,$
$\displaystyle\Re(w^{5})=-38,\quad\Re(w^{-1})=\frac{2}{5},\quad\Re(w^{-3})=\frac{2}{125}.$

######  Proposition 6.

We have

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/5}.$

###### Proof.

As above, the argument $\psi_{B}(\theta)$ of $B(e^{{\rm i}\theta})$
(determined modulo $2\pi$) satisfies the relation

$\psi^{\prime}_{B}(\theta)=\Re\sum_{1\leq j\leq 2}\frac{1+\lambda_{j}e^{{\rm
i}\theta}}{1-\lambda_{j}e^{{\rm i}\theta}}.$

Furthermore,

$\displaystyle\psi^{\prime\prime}_{B}(\theta)$ $\displaystyle=2\Re\sum_{1\leq
j\leq 2}\frac{{\rm i}\lambda_{j}e^{{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{2}},$ $\displaystyle\psi^{\prime\prime\prime}_{B}(\theta)$
$\displaystyle=-2\Re\sum_{1\leq j\leq 2}\frac{\lambda_{j}e^{{\rm
i}\theta}+\lambda^{2}_{j}e^{2{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{3}}.$

Set

$e^{{\rm i}\theta}=\frac{1-{\rm i}x}{1+{\rm i}x},\qquad x\in\mathbb{R},$

and define $h(x)=\psi^{\prime\prime}_{B}(\theta)$. Then $h$ has a zero of
order $N$ at $x_{0}\in\mathbb{R}$ if and only if $\psi^{\prime\prime}_{B}$ has
a zero of order $N$ at $\theta_{0}$, $e^{{\rm i}\theta_{0}}=(1-{\rm
i}x_{0})/(1+{\rm i}x_{0})$.

By Theorem 1, we need to verify that

$h(0)=h^{\prime}(0)=h^{\prime\prime}(0)=0,\quad
h^{\prime\prime\prime}(0)\not=0,$

$h$ has no zeros of order $3$ except at the origin, and
$\psi^{\prime\prime\prime}_{B}(\pi)\not=0$.

First,

$h(x)=2\Re\sum_{1\leq j\leq 2}\frac{{\rm i}\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm
i}x}{1+{\rm i}x}}{(1-\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm i}x}{1+{\rm
i}x})^{2}}=\frac{1+x^{2}}{2}\Re\sum_{1\leq j\leq 2}\frac{{\rm
i}(w^{2}_{j}-1)}{(1+{\rm i}w_{j}x)^{2}}\\\
=\frac{1+x^{2}}{4}\Bigl{(}\frac{{\rm i}(w^{2}-1)}{(1+{\rm
i}wx)^{2}}+\frac{{\rm i}(\overline{w^{2}}-1)}{(1+{\rm
i}\overline{w}x)^{2}}-\frac{{\rm i}(\overline{w^{2}}-1)}{(1-{\rm
i}\overline{w}x)^{2}}-\frac{{\rm i}(w^{2}-1)}{(1-{\rm i}wx)^{2}}\Bigr{)}\\\
=x(1+x^{2})\sum_{1\leq j\leq
2}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x^{2})^{2}}.$

Therefore, $h(0)=0$. Denote

$h_{1}(x)=\sum_{1\leq j\leq 2}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x)^{2}}.$

Then $h(x)=x(1+x^{2})h_{1}(x^{2})$. We need to verify that

$h_{1}(0)=0,\quad h_{1}^{\prime}(0)\not=0.$

By (5.1),

$h_{1}(0)=\sum_{1\leq j\leq 2}(w^{3}_{j}-w_{j})=0,$

and

$h^{\prime}_{1}(0)=-2\sum_{1\leq j\leq 2}(w^{5}_{j}-w^{3}_{j})\not=0.$

If the function $h$ has a zero of order $3$ at $x_{0}\not=0$, then $h_{1}$ has
a zero of order $3$ at $y_{0}=\sqrt{|x_{0}|}\not=0$. We have

$h_{1}(x)=\frac{Q(x)}{\prod_{1\leq j\leq 2}(1+w_{j}^{2}x)^{2}},$

where $Q$ is a polynomial of degree at most $2$, and, hence, $Q$ cannot have a
zero of order $3$ at $y_{0}$.

Finally, by (5.2), we have

$\psi^{\prime\prime\prime}_{B}(\pi)=2\Re\sum_{1\leq j\leq
2}\frac{\lambda_{j}-\lambda^{2}_{j}}{(1+\lambda_{j})^{3}}=2\Re\sum_{1\leq
j\leq
2}\frac{\frac{w_{j}-1}{w_{j}+1}(1-\frac{w_{j}-1}{w_{j}+1})}{(1+\frac{w_{j}-1}{w_{j}+1})^{3}}\\\
=\frac{1}{2}\Re\sum_{1\leq j\leq
2}\frac{w^{2}_{j}-1}{w^{3}_{j}}=\Re\bigl{(}w^{-1}-w^{-3}\bigr{)}\not=0.$

∎

####  5.2.2. Example 2

Here we give another example of a finite Blaschke product $B$ of degree 2 such
that $\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/5}$.

Let $w_{1},w_{2}\in(0,\infty)\setminus\\{1\\}$ be such that

(5.3) $w_{1}+w_{2}=w^{3}_{1}+w^{3}_{2}\not=w^{5}_{1}+w^{5}_{2}$

and

(5.4) $w^{-1}_{1}+w^{-1}_{2}\not=w^{-3}_{1}+w^{-3}_{2}.$

Set

$\lambda_{j}=\frac{w_{j}-1}{w_{j}+1},\qquad j=1,2,$

and consider

$B=\prod_{1\leq j\leq 2}b_{\lambda_{j}}.$

For example, we can choose

$w_{1}=\frac{1}{2},\quad w_{2}=\frac{1+\sqrt{13}}{4}.$

Then

$\displaystyle w_{1}+w_{2}=w^{3}_{1}+w^{3}_{2}=\frac{3+\sqrt{13}}{4},\quad
w^{5}_{1}+w^{5}_{2}=\frac{63+19\sqrt{13}}{64},$ $\displaystyle
w^{-1}_{1}+w^{-1}_{2}=\frac{5+\sqrt{13}}{3},\quad
w^{-3}_{1}+w^{-3}_{2}=\frac{176+16\sqrt{13}}{27}.$

######  Proposition 7.

We have

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/5}.$

###### Proof.

As above,

$\displaystyle\psi^{\prime}_{B}(\theta)$ $\displaystyle=\Re\sum_{1\leq j\leq
2}\frac{1+\lambda_{j}e^{{\rm i}\theta}}{1-\lambda_{j}e^{{\rm i}\theta}},$
$\displaystyle\psi^{\prime\prime}_{B}(\theta)$ $\displaystyle=2\Re\sum_{1\leq
j\leq 2}\frac{{\rm i}\lambda_{j}e^{{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{2}},$ $\displaystyle\psi^{\prime\prime\prime}_{B}(\theta)$
$\displaystyle=-2\Re\sum_{1\leq j\leq 2}\frac{\lambda_{j}e^{{\rm
i}\theta}+\lambda^{2}_{j}e^{2{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{3}}.$

Set

$e^{{\rm i}\theta}=\frac{1-{\rm i}x}{1+{\rm i}x},\qquad x\in\mathbb{R},$

and define $h(x)=\psi^{\prime\prime}_{B}(\theta)$.

As above, by Theorem 1, we need to verify that

$h(0)=h^{\prime}(0)=h^{\prime\prime}(0)=0,\quad
h^{\prime\prime\prime}(0)\not=0,$

$h$ has no zeros of order $3$ except at the origin, and
$\psi^{\prime\prime\prime}_{B}(\pi)\not=0$.

First,

$h(x)=2\Re\sum_{1\leq j\leq 2}\frac{{\rm i}\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm
i}x}{1+{\rm i}x}}{(1-\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm i}x}{1+{\rm
i}x})^{2}}=\frac{1+x^{2}}{2}\Re\sum_{1\leq j\leq 2}\frac{{\rm
i}(w^{2}_{j}-1)}{(1+{\rm i}w_{j}x)^{2}}\\\
=\frac{1+x^{2}}{4}\Bigl{(}\frac{{\rm i}(w_{1}^{2}-1)}{(1+{\rm
i}w_{1}x)^{2}}+\frac{{\rm i}(w_{2}^{2}-1)}{(1+{\rm i}w_{2}x)^{2}}-\frac{{\rm
i}(w_{1}^{2}-1)}{(1-{\rm i}w_{1}x)^{2}}-\frac{{\rm i}(w_{2}^{2}-1)}{(1-{\rm
i}w_{2}x)^{2}}\Bigr{)}\\\ =x(1+x^{2})\sum_{1\leq j\leq
2}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x^{2})^{2}}.$

Therefore, $h(0)=0$. Denote

$h_{1}(x)=\sum_{1\leq j\leq 2}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x)^{2}}.$

As above, $h(x)=x(1+x^{2})h_{1}(x^{2})$, and we need to verify that

$h_{1}(0)=0,\quad h_{1}^{\prime}(0)\not=0.$

By (5.3),

$h_{1}(0)=\sum_{1\leq j\leq 2}(w^{3}_{j}-w_{j})=0,$

and

$h^{\prime}_{1}(0)=-2\sum_{1\leq j\leq 2}(w^{5}_{j}-w^{3}_{j})\not=0.$

If the function $h$ has a zero of order $3$ at $x_{0}\not=0$, then $h_{1}$ has
a zero of order $3$ at $y_{0}=\sqrt{|x_{0}|}\not=0$. We have

$h_{1}(x)=\frac{Q(x)}{\prod_{1\leq j\leq 2}(1+w_{j}^{2}x)^{2}},$

where $Q$ is a polynomial of degree at most $2$, and, hence, $Q$ cannot have a
zero of order $3$ at $y_{0}$.

Finally, by (5.4), we have

$\psi^{\prime\prime\prime}_{B}(\pi)=2\sum_{1\leq j\leq
2}\frac{\lambda_{j}-\lambda^{2}_{j}}{(1+\lambda_{j})^{3}}=2\sum_{1\leq j\leq
2}\frac{\frac{w_{j}-1}{w_{j}+1}(1-\frac{w_{j}-1}{w_{j}+1})}{(1+\frac{w_{j}-1}{w_{j}+1})^{3}}\\\
=\frac{1}{2}\sum_{1\leq j\leq
2}\frac{w^{2}_{j}-1}{w^{3}_{j}}=\frac{1}{2}\sum_{1\leq j\leq
2}\bigl{(}w_{j}^{-1}-w_{j}^{-3}\bigr{)}\not=0.$

∎

###  5.3. The case $N=7$

Here we give an explicit example of a finite Blaschke product $B$ of degree 4
such that $\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/7}$.

Let $w_{1},w_{2}\in\mathbb{C}\setminus\\{1\\}$ be such that $\Re w_{1},\Re
w_{2}>0$,

(5.5)
$\Re(w_{1}+w_{2})=\Re(w^{3}_{1}+w^{3}_{2})=\Re(w_{1}^{5}+w_{2}^{5})\not=\Re(w^{7}_{1}+w^{7}_{2})$

and

(5.6) $\Re(w^{-1}_{1}+w^{-1}_{2})\not=\Re(w^{-3}_{1}+w^{-3}_{2}).$

Set

$\displaystyle w_{j+2}=\overline{w_{j}},\qquad j=1,2,$
$\displaystyle\lambda_{j}=\frac{w_{j}-1}{w_{j}+1},\qquad 1\leq j\leq 4,$

and consider

$B=\prod_{1\leq j\leq 4}b_{\lambda_{j}}.$

For example, we can choose

$w_{1}=1+\frac{2{\rm i}}{\sqrt{3}},\quad w_{2}=2+\frac{{\rm i}}{\sqrt{3}}.$

Then

$\displaystyle\Re(w_{1}+w_{2})=\Re(w^{3}_{1}+w^{3}_{2})=\Re(w_{1}^{5}+w_{2}^{5})=3,$
$\displaystyle\Re(w^{7}_{1}+w^{7}_{2})=-\frac{421}{9},\quad\Re(w^{-1}_{1}+w^{-1}_{2})=\frac{81}{91},$
$\displaystyle\Re(w^{-3}_{1}+w^{-3}_{2})=-\frac{122391}{753571}.$

######  Proposition 8.

We have

$\|\widehat{B^{n}}\|_{\ell^{\infty}}\asymp n^{-1/7}.$

###### Proof.

As above,

$\displaystyle\psi^{\prime}_{B}(\theta)$ $\displaystyle=\Re\sum_{1\leq j\leq
4}\frac{1+\lambda_{j}e^{{\rm i}\theta}}{1-\lambda_{j}e^{{\rm i}\theta}},$
$\displaystyle\psi^{\prime\prime}_{B}(\theta)$ $\displaystyle=2\Re\sum_{1\leq
j\leq 4}\frac{{\rm i}\lambda_{j}e^{{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{2}},$ $\displaystyle\psi^{\prime\prime\prime}_{B}(\theta)$
$\displaystyle=-2\Re\sum_{1\leq j\leq 4}\frac{\lambda_{j}e^{{\rm
i}\theta}+\lambda^{2}_{j}e^{2{\rm i}\theta}}{(1-\lambda_{j}e^{{\rm
i}\theta})^{3}}.$

Set

$e^{{\rm i}\theta}=\frac{1-{\rm i}x}{1+{\rm i}x},\qquad x\in\mathbb{R},$

and define $h(x)=\psi^{\prime\prime}_{B}(\theta)$. Then $h$ has a zero of
order $N$ at $x_{0}\in\mathbb{R}$ if and only if $\psi^{\prime\prime}_{B}$ has
a zero of order $N$ at $\theta_{0}$, $e^{{\rm i}\theta_{0}}=(1-{\rm
i}x_{0})/(1+{\rm i}x_{0})$.

We need to verify that

$h(0)=h^{\prime}(0)=h^{\prime\prime}(0)=h^{\prime\prime\prime}(0)=h^{(4)}(0)=0,\quad
h^{(5)}(0)\not=0,$

$h$ has no zeros of order $5$ except at the origin, and
$\psi^{\prime\prime\prime}_{B}(\pi)\not=0$.

As above,

$h(x)=2\Re\sum_{1\leq j\leq 4}\frac{{\rm i}\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm
i}x}{1+{\rm i}x}}{(1-\frac{w_{j}-1}{w_{j}+1}\frac{1-{\rm i}x}{1+{\rm
i}x})^{2}}=\frac{1+x^{2}}{2}\Re\sum_{1\leq j\leq 4}\frac{{\rm
i}(w^{2}_{j}-1)}{(1+{\rm i}w_{j}x)^{2}}\\\ =\frac{1+x^{2}}{4}\sum_{1\leq j\leq
2}\Bigl{(}\frac{{\rm i}(w^{2}_{j}-1)}{(1+{\rm i}w_{j}x)^{2}}+\frac{{\rm
i}(\overline{w^{2}_{j}}-1)}{(1+{\rm i}\overline{w_{j}}x)^{2}}-\frac{{\rm
i}(\overline{w^{2}_{j}}-1)}{(1-{\rm i}\overline{w_{j}}x)^{2}}-\frac{{\rm
i}(w^{2}_{j}-1)}{(1-{\rm i}w_{j}x)^{2}}\Bigr{)}\\\ =x(1+x^{2})\sum_{1\leq
j\leq 4}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x^{2})^{2}}.$

Therefore, $h(0)=0$. Denote

$h_{1}(x)=\sum_{1\leq j\leq 4}\frac{w_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x)^{2}}.$

Then $h(x)=x(1+x^{2})h_{1}(x^{2})$. We need to verify that

$h_{1}(0)=h_{1}^{\prime}(0)=0,\quad h_{1}^{\prime\prime}(0)\not=0.$

By (5.5),

$h_{1}(0)=\sum_{1\leq j\leq 4}(w^{3}_{j}-w_{j})=0.$

Furthermore,

$h^{\prime}_{1}(x)=-2\sum_{1\leq j\leq
4}\frac{w^{3}_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x)^{3}}.$

Again by (5.5),

$h^{\prime}_{1}(0)=-2\sum_{1\leq j\leq 4}(w^{5}_{j}-w^{3}_{j})=0.$

Next,

$h^{\prime\prime}_{1}(x)=6\sum_{1\leq j\leq
4}\frac{w^{5}_{j}(w^{2}_{j}-1)}{(1+w_{j}^{2}x)^{4}},$

and by (5.6),

$h^{\prime\prime}_{1}(0)=6\sum_{1\leq j\leq 4}w^{5}_{j}(w^{2}_{j}-1)\not=0.$

If the function $h$ has a zero of order $5$ at $x_{0}\not=0$, then $h_{1}$ has
a zero of order $5$ at $y_{0}=\sqrt{|x_{0}|}\not=0$. We have

$h_{1}(x)=\frac{Q(x)}{\prod_{1\leq j\leq 4}(1+w_{j}^{2}x)^{2}},$

where $Q$ is a polynomial of degree at most $6$. Since
$h_{1}(0)=h_{1}^{\prime}(0)=0$, $Q$ cannot have a zero of order $5$ at
$y_{0}$.

Finally, by (5.6), we have

$\psi^{\prime\prime\prime}_{B}(\pi)=2\Re\sum_{1\leq j\leq
4}\frac{\lambda_{j}-\lambda^{2}_{j}}{(1+\lambda_{j})^{3}}=2\Re\sum_{1\leq
j\leq
4}\frac{\frac{w_{j}-1}{w_{j}+1}(1-\frac{w_{j}-1}{w_{j}+1})}{(1+\frac{w_{j}-1}{w_{j}+1})^{3}}\\\
=\frac{1}{2}\Re\sum_{1\leq j\leq
4}\frac{w^{2}_{j}-1}{w^{3}_{j}}=\Re\sum_{1\leq j\leq
2}\bigl{(}w^{-1}_{j}-w^{-3}_{j}\bigr{)}\not=0.$

∎

##  6\.  Proof of Theorem 3

We recall that

$e_{nm}=\frac{(1-|\lambda_{m}|^{2})^{1/2}}{z-\lambda_{m}}B^{n}.$

We set $g=S^{-1}e_{nm}$ so that $g=g(0)+ze_{nm}$. Then

(6.1)
$\|S^{-1}\|_{\ell^{\infty}_{A}\to\ell^{\infty}_{A}}\geq\frac{\|\widehat{S^{-1}e_{nm}}\|_{\ell^{\infty}}}{\|\widehat{e_{nm}}\|_{\ell^{\infty}}}=\frac{\|\widehat{g}\|_{\ell^{\infty}}}{\|\widehat{e_{nm}}\|_{\ell^{\infty}}}\geq\frac{|g(0)|}{\|\widehat{e_{nm}}\|_{\ell^{\infty}}}.$

Let us first concentrate on estimating $\|\widehat{e_{nm}}\|_{\ell^{\infty}}$.
We follow the same steps as in the proof of Theorem 1. We set
$d_{nm}=\frac{e_{nm}}{(1-|\lambda_{m}|^{2})^{1/2}}$. We have

$\widehat{d_{nm}}(k)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{B^{n}(e^{{\rm
i}\theta})}{e^{{\rm i}\theta}-\lambda_{m}}e^{-{\rm
i}k\theta}\,d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta,$

where $f(\theta)=\psi(\theta)-\frac{k}{n}\theta$ and
$\psi(\theta)=\arg(B(e^{{\rm i}\theta}))$. Using the same notation as in
Theorem 1, we consider the sequence $(\xi_{\ell})_{\ell=1}^{s}$ of consecutive
zeros of $\psi^{\prime\prime}$ on $[0,2\pi)$ with respective multiplicities
$(N_{\ell}-2)_{\ell=1}^{s}$, $N_{\ell}\geq 3$. We have:

$\widehat{d_{nm}}(k)=\sum_{\ell=1}^{s}\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\,d\theta+\sum_{\ell=1}^{s-1}\int_{\xi_{\ell}+\varepsilon}^{\xi_{\ell+1}-\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta\\\
+\int_{0}^{\xi_{1}-\varepsilon}\frac{e^{{\rm i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\,d\theta+\int_{\xi_{s}+\varepsilon}^{2\pi}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta.$

From now on we put

$\varepsilon=n^{-1/N}\rightarrow 0,\qquad n\rightarrow\infty,$

where $N=\max_{1\leq\ell\leq s}N_{\ell}$. Clearly

$\left|\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\,d\theta\right|\leq\frac{2\varepsilon}{\bigl{|}1-|\lambda_{m}|\bigr{|}},$

and therefore the first sum is bounded from above as

$\left|\sum_{\ell=1}^{s}\int_{\xi_{\ell}-\varepsilon}^{\xi_{\ell}+\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta\right|\lesssim
n^{-1/N}.$

Furthermore, $f^{\prime\prime}=\psi^{\prime\prime}$ does not vanish on the
intervals $(0,\xi_{1}-\varepsilon)$, $(\xi_{s}+\varepsilon,2\pi)$ and
$(\xi_{\ell}+\varepsilon,\xi_{\ell+1}-\varepsilon)$ for $\ell=1,\ldots,s-1$.
Writing the Taylor expansion (of order $N_{\ell}-2$) of $\psi^{\prime\prime}$
near $\xi_{\ell}$ we find

$\psi^{\prime\prime}(\theta)=\frac{\psi^{(N_{\ell})}(\xi_{\ell})}{(N_{\ell}-2)!}\left(\theta-\xi_{\ell}\right)^{N_{\ell}-2}\left(1+\mathcal{O}\left(\theta-\xi_{\ell}\right)\right),$

and it follows that if $\theta$ is close to $\xi_{\ell}$ but satisfies
$|\theta-\xi_{\ell}|\geq\varepsilon$, then

$|\psi^{\prime\prime}_{B}(\theta)|\gtrsim n^{-(N_{\ell}-2)/N},$

for $n$ large enough. Below we apply Lemma 5 to the remaining integrals:

(6.2) $\int_{\xi_{l}+\varepsilon}^{\xi_{l+1}-\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\,d\theta,\quad\int_{0}^{\xi_{1}-\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\,d\theta,\quad\int_{\xi_{s}+\varepsilon}^{2\pi}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta.$

We describe how to get an upper estimate on the first integral in (6.2), the
argument for two others being similar. We have

$\biggl{|}\int_{\xi_{l}+\varepsilon}^{\xi_{l+1}-\varepsilon}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\,d\theta\biggr{|}\\\
\leq\biggl{|}\int_{\xi_{l}+\varepsilon}^{\xi_{l+1}-\varepsilon}\Re\biggl{(}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm
i}\theta}-\lambda_{m}}\biggr{)}\,d\theta\biggr{|}+\biggl{|}\int_{\xi_{l}+\varepsilon}^{\xi_{l+1}-\varepsilon}\Im\biggl{(}\frac{e^{{\rm
i}n\,f(\theta)}}{e^{{\rm i}\theta}-\lambda_{m}}\biggr{)}\,d\theta\biggr{|}.$

We set $\lambda_{m}=\rho_{m}\exp({\rm i}\theta_{m})$ and

$G(\theta)=\Re\left(\frac{1}{e^{{\rm
i}\theta}-\lambda_{m}}\right)=\frac{\cos{\theta}-\rho_{m}\cos\theta_{m}}{(\cos{\theta}-\rho_{m}\cos\theta_{m})^{2}+(\sin{\theta}-\rho_{m}\sin\theta_{m})^{2}}.$

The function $G$ is bounded on $[0,2\pi]$ and it vanishes twice on $[0,2\pi)$.
Furthermore, $G^{\prime}$ has a finite number of zeros on $[0,2\pi)$.
Therefore, we can split every interval of integration into a (uniformly
bounded in $n$) number of intervals so that $G$ is of constant sign and
monotonic on each of them. Then

$|F^{\prime\prime}|\gtrsim n^{2/N}$

on these intervals and therefore, applying Lemma 5, we obtain that the
corresponding integrals are $\mathcal{O}\left(n^{-1/N}\right)$. Applying the
same steps with $G(\theta)=\Bigl{(}\dfrac{1}{e^{{\rm
i}\theta}-\lambda_{m}}\Bigr{)}$, and estimating in the same way the remaining
integrals in (6.2), we conclude that
$\|\widehat{e_{nm}}\|_{\ell^{\infty}}\lesssim n^{-1/N}$.
Now we return to (6.1). It remains to estimate $|g(0)|$. We use that
$g=S^{-1}e_{nm}\in K_{B}$, and that

$K_{B}=\Biggl{\\{}\frac{P(z)}{\prod\limits_{i=1}^{m}(1-\bar{\lambda}_{i}z)^{n}}:\deg
P\leq nm-1\Biggr{\\}}.$

Thus we can determine $g(0)$ by the relation

(6.3)
$g(0)+(1-|\lambda_{m}|^{2})^{1/2}\frac{z(z-\lambda_{m})^{n-1}}{(1-\bar{\lambda}_{m}z)^{n}}\prod_{j=1}^{m-1}\left(\frac{z-\lambda_{j}}{1-\bar{\lambda}_{j}z}\right)^{n}\in
K_{B}.$

Since

$\frac{z-\lambda}{1-\bar{\lambda}z}=-\frac{1}{\bar{\lambda}}+\frac{(1/\bar{\lambda})-\lambda}{1-\bar{\lambda}z},\qquad\frac{z}{1-\bar{\lambda}z}=-\frac{1}{\bar{\lambda}}+\frac{1/\bar{\lambda}}{1-\bar{\lambda}z},$

relation (6.3) implies that

$g(0)+(1-|\lambda_{m}|^{2})^{1/2}\prod_{j=1}^{m}\frac{1}{(-\bar{\lambda}_{j})^{n}}+\frac{P(z)}{\prod_{j=1}^{m}(1-\bar{\lambda}_{j}z)^{n}}\in
K_{B},$

where $\deg P\leq nm-1$. Therefore, the condition $g\in K_{B}$ implies that

$g(0)=-(1-|\lambda_{m}|^{2})^{1/2}\prod_{j=1}^{m}\frac{1}{(-\bar{\lambda}_{j})^{n}}.$

Thus, (6.1) yields

(6.4)
$\|S^{-1}\|_{\ell^{\infty}_{A}\to\ell^{\infty}_{A}}\gtrsim\frac{n^{1/N}}{\prod_{j=1}^{m}|\lambda_{j}|^{n}},$

which completes the proof of part (i) because of (3.2).

To prove part (ii), we use the analytic expression of $\Phi$ established by
Gluskin–Meyer–Pajor [4]:

$\Phi(\lambda_{1},\dots,\lambda_{1},\dots,\lambda_{m},\dots,\lambda_{m})=\sup\\{|\det
T|\cdot\|T^{-1}\|\\},$

where we take the supremum by all norms in $\mathbb{C}^{nm}$ and all $T$ such
that $\|T\|\leq 1$ in the induced norm and
$\sigma_{T}=\\{\lambda_{1},\ldots,\lambda_{m}\\}$ with multiplicity $n$ of
every eigenvalue. We conclude by setting $T=S$ and using (3.2) and (6.4). ∎

######  Remark 9.

Here we establish part (ii) of Theorem 3 as a direct application of Theorem 1.
To this aim we reproduce and adapt the duality method used to prove the main
result of [18]. We recall the definition of the Wiener algebra, which is the
subset of $H(\mathbb{D})$ of absolutely convergent Fourier series,

$\displaystyle W:=\Bigl{\\{}f=\sum_{k\geq
0}\hat{f}(k)z^{k}:\|f\|_{W}:=\sum_{k\geq 0}|\hat{f}(k)|<\infty\Bigr{\\}}.$

Let $B$ be the finite Blaschke product with simple zeros at
$\lambda_{1},\dots,\lambda_{m}$. It is easily verified that if $f$ is in $W$
and $f(\lambda)=0$ with $\lambda\in\mathbb{D}$, then

$\Bigl{\|}\frac{f}{z-\lambda}\Bigr{\|}_{W}\leq\frac{\|f\|_{W}}{1-|\lambda|}.$

In particular, $\frac{f}{z-\lambda}$ is also in $W$. Therefore, for any $h\in
W$ with

$h(\lambda_{j})=h^{\prime}(\lambda_{j})=\ldots=h^{(n-1)}(\lambda_{j})=0,\qquad
j=1,\dots,m,$

we have $h/B^{n}\in W$. With this observation we rewrite the
Gluskin–Meyer–Pajor expression for $\Phi$ as follows:

(6.5)
$\Phi(\lambda_{1},\dots,\lambda_{1},\dots,\lambda_{m},\dots,\lambda_{m})\\\
=\inf\biggl{\\{}\|h\|_{W}-|h(0)|:h\in
B^{n}W,\,h(0)=\prod_{j=1}^{m}\lambda_{j}^{n}\biggr{\\}}.$

Let $L^{2}(\partial\mathbb{D})$ be the usual $L^{2}$ space on the unit circle
$\partial\mathbb{D}$, equipped with the standard scalar product

$\left\langle f,g\right\rangle:=\int_{-\pi}^{\pi}f(e^{{\rm
i}\varphi})\overline{g(e^{{\rm i}\varphi})}\,\frac{d\varphi}{2\pi}.$

For $f=\sum_{k}\hat{f}(k)z^{k}$, $g=\sum_{k}\hat{g}(k)z^{k}\in H^{2}$, it is
well known [9] that the $L^{2}(\partial\mathbb{D})$ scalar product can be
written as

$\left\langle f,\,g\right\rangle=\sum_{j\geq
0}\hat{f}(j)\overline{\hat{g}(j)}.$

Let $h=B^{n}g$ with $g\in W$ and $h(0)=\prod_{j=1}^{m}\lambda_{j}^{n}$. Then

$\|h\|_{W}\cdot\|B^{n}\|_{\ell^{\infty}_{A}}\geq\bigl{|}\bigl{\langle}h,B^{n}\bigr{\rangle}\bigr{|}=\bigl{|}\bigl{\langle}g,1\bigr{\rangle}\bigr{|}=|g(0)|=\Bigl{|}\frac{h(0)}{B^{n}(0)}\Bigr{|}=1.$

It follows that any candidate function $h$ in (6.5) satisfies

$\|h\|_{W}\geq\frac{1}{\|B^{n}\|_{\ell^{\infty}_{A}}},$

and consequently

$\Phi\left(\lambda_{1},\dots,\lambda_{1},\dots,\lambda_{m},\dots,\lambda_{m}\right)\geq\frac{1}{\|B^{n}\|_{\ell^{\infty}_{A}}}-\prod_{j=1}^{m}|\lambda_{j}|^{n}.$

It remains to apply Theorem 1 to obtain that

$\Phi(\lambda_{1},\dots,\lambda_{1},\dots,\lambda_{m},\dots,\lambda_{m})\gtrsim
n^{1/N},\qquad n\to\infty.$

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