# Characterization of weighted Hardy spaces on which all composition operators
are bounded

Pascal Lefèvre, Daniel Li,
Hervé Queffélec, Luis Rodríguez-Piazza

Abstract. We give a complete characterization of the sequences
$\beta=(\beta_{n})$ of positive numbers for which all composition operators on
$H^{2}(\beta)$ are bounded, where $H^{2}(\beta)$ is the space of analytic
functions $f$ on the unit disk $\mathbb{D}$ such that
$\sum_{n=0}^{\infty}|a_{n}|^{2}\beta_{n}<+\infty$ if
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. We prove that all composition operators
are bounded on $H^{2}(\beta)$ if and only if $\beta$ is essentially decreasing
and slowly oscillating. We also prove that every automorphism of the unit disk
induces a bounded composition operator on $H^{2}(\beta)$ if and only if
$\beta$ is slowly oscillating. We give applications of our results.

MSC 2010. primary: 47B33 ; secondary: 30H10

Key-words. automorphism of the unit disk; composition operator; slowly
oscillating sequence; weighted Hardy space

##  1 Introduction

Let $\beta=(\beta_{n})_{n\geq 0}$ be a sequence of positive numbers such that

(1.1) $\liminf_{n\to\infty}\beta_{n}^{1/n}\geq 1\,.$

The associated weighted Hardy space $H^{2}(\beta)$ is the Hilbertian space of
analytic functions $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ such that

(1.2) $\|f\|^{2}:=\sum_{n=0}^{\infty}|a_{n}|^{2}\beta_{n}<\infty\,.$

Condition (1.1) is equivalent to the inclusion $H^{2}(\beta)\subseteq{\cal
H}ol(\mathbb{D})$. Indeed, if (1.1) holds, we have $H^{2}(\beta)\subseteq{\cal
H}ol(\mathbb{D})$ since $|a_{n}|^{2}\beta_{n}$ is bounded and thanks to the
Hadamard formula. Conversely, testing the inclusion
$H^{2}(\beta)\subseteq{\cal H}ol(\mathbb{D})$ on the function
$f(z)=\sum_{n=1}^{\infty}\frac{1}{n\sqrt{\beta(n)}}z^{n}\in H^{2}(\beta)$, we
get (1.1) from the Hadamard formula.

Condition (1.1) will therefore be assumed throughout this paper, without
repeating it.

When $\beta_{n}\equiv 1$, we recover the usual Hardy space $H^{2}$; the
Bergman space corresponds to $\beta_{n}=1/(n+1)$, and the Dirichlet space to
$\beta_{n}=n+1$.

Recall that a symbol is a (non constant) analytic self-map
$\varphi\colon\mathbb{D}\to\mathbb{D}$, and the associated composition
operator $C_{\varphi}\colon H^{2}(\beta)\to{\cal H}ol(\mathbb{D})$ is defined
as:

(1.3) $C_{\varphi}(f)=f\circ\varphi\,.$

An important question in the theory is to decide when $C_{\varphi}$ is bounded
on $H^{2}(\beta)$, i.e. when $C_{\varphi}\colon H^{2}(\beta)\to H^{2}(\beta)$.

This question appears in the literature in several places. For instance, it is
Problem 1 in the thesis of Nina Zorboska [23, Pb 1, p. 49]. This thesis
contains many interesting results, in particular Proposition 3.1 and
Proposition 4.2 of the present paper (actually we discovered the content of
Zorboska’s thesis once the present paper was almost finished). See also
Question 36 raised by Deddens in [19, p. 122.c].

When $H^{2}(\beta)$ is the usual Hardy space $H^{2}$ (i.e. when
$\beta_{n}\equiv 1$), it is well-known, as a consequence of the Littlewood
subordination principle (see [14]), that all symbols generate bounded
composition operators (see [18, pp. 13–17]). On the other hand, for the
Dirichlet space, corresponding to $\beta_{n}=n+1$, not all composition
operators are bounded since there exist symbols $\varphi$ not belonging to the
Dirichlet space (e.g. any infinite Blaschke product).

Note that, by definition of the norm of $H^{2}(\beta)$, all rotations
$R_{\theta}$, defined by $R_{\theta}(z)={\rm e}^{i\theta}z$, with
$\theta\in\mathbb{R}$, induce bounded and surjective composition operators on
$H^{2}(\beta)$ and send isometrically $H^{2}(\beta)$ into itself.

Our goal in this paper is characterizing the sequences $\beta$ for which all
composition operators act boundedly on the space $H^{2}(\beta)$, i.e. send
$H^{2}(\beta)$ into itself.

In Shapiro’s presentation for the Hardy space $H^{2}$, the main point is the
case $\varphi(0)=0$ and a subordination principle for subharmonic functions
(Littlewood’s subordination principle). The case of automorphisms is claimed
simple, using an integral representation for the norm and some change of
variable. For general weights $\beta$, the situation is different, as we will
see in this paper, and it turns out that the conditions on $\beta$ for the
boundedness of the composition operators $C_{\varphi}$ on $H^{2}(\beta)$ are
not the same depending on whether we consider the class of all symbols such
that $\varphi(0)=0$, or the class of symbols $\varphi=T_{a}$, where

(1.4) $T_{a}(z)=\frac{a+z}{1+\bar{a}\,z}$

for $a\in\mathbb{D}$.

It is clear that when these two classes of composition operators are bounded,
then all composition operators are bounded. Recall that every symbol $\varphi$
can be written as the composition $\varphi=T_{a}\circ\psi$ where $\psi(0)=0$
and $a=\varphi(0)$; and then $C_{\varphi}=C_{\psi}\circ C_{T_{a}}$.

In many occurrences, the weight $\beta$ is defined as

(1.5) $\beta_{n}=\int_{0}^{1}t^{n}\,d\sigma(t)$

where $\sigma$ is a positive measure on $(0,1)$; more specifically the
following definition is often used: let $G\colon(0,1)\to\mathbb{R}_{+}$ be an
integrable function and let $H^{2}_{G}$ be the space of analytic functions
$f\colon\mathbb{D}\to\mathbb{C}$ such that:

(1.6)
$\|f\|_{H^{2}_{G}}^{2}:=\int_{\mathbb{D}}|f(z)|^{2}\,G(1-|z|^{2})\,dA(z)<\infty\,.$

Such weighted Bergman type spaces are used, for instance, in [10], [11] and in
[13]. We have $H^{2}_{G}=H^{2}(\beta)$ with:

(1.7)
$\beta_{n}=2\int_{0}^{1}r^{2n+1}G(1-r^{2})\,dr=\int_{0}^{1}t^{n}\,G(1-t)\,dt\,,$

and the sequence $\beta=(\beta_{n})_{n}$ is _non-increasing_ (actually, the
above representation (1.5) is equivalent, by the Hausdorff moment theorem, to
a high regularity of the sequence $\beta$, namely its _complete monotony_).

When the weight $\beta$ is non-increasing (or more generally, essentially
decreasing), all the symbols vanishing at the origin induce a bounded
composition operator. This was proved by C. Cowen [4, Corollary, page 31],
using Hadamard multiplication. We can also use Kacnel’son’s theorem (see [2]
or [12, Theorem 3.12]). Actually that follows from an older theorem of Goluzin
[8] (see [5, Theorem 6.3]), which itself uses a self-refinement observed by
Rogosinski of Littlewood’s principle ([5, Theorem 6.2]).

For weights defined as in (1.5), we dispose of integral representations for
the norm in $H^{2}(\beta)$, and, as in the Hardy space case, this integral
representation rather easily allows us to decide when the boundedness of
$C_{T_{a}}$ on $H^{2}(\beta)$ occurs. This is not always the case, as shown by
T. Kriete and B. MacCluer in [11]. They consider spaces of Bergman type
$A^{2}_{\widetilde{G}}:=H^{2}_{G}$, where $\widetilde{G}(r)=G(1-r^{2})$,
defined as the spaces of analytic functions in $\mathbb{D}$ such that
$\int_{\mathbb{D}}|f(z)|^{2}\,{\widetilde{G}}(|z|)\,dA<\infty$, for a positive
non-increasing continuous function ${\widetilde{G}}$ on $[0,1)$. They prove
[11, Theorem 3] that for

$\qquad\qquad\qquad{\widetilde{G}}(r)=\exp\,\bigg{(}-B\,\frac{1}{(1-r)^{\alpha}}\bigg{)}\,,\qquad
B>0\,,\ 0<\alpha\leq 2\,,$

and

$\qquad\qquad\qquad\qquad\varphi(z)=z+t(1-z)^{\gamma}\,,\qquad\qquad
1<\gamma\leq 3\,,\ 0<t<2^{1-\gamma}\,,$

then $\varphi$ is a symbol and $C_{\varphi}$ is bounded on
$A^{2}_{\widetilde{G}}$ if and only if $\gamma\geq\alpha+1$.

Here

$\beta_{n}=\int_{0}^{1}t^{n}{\rm
e}^{-B/(1-\sqrt{t})^{\alpha}}\,dt\lesssim\exp(-c\,n^{\alpha/(\alpha+1)})\,.$

We point out that $\beta$ is non-increasing, so for every symbol $\varphi$
fixing the origin, the composition operator $C_{\varphi}$ is bounded.
Nevertheless, choosing $\gamma<\alpha+1$, there exist symbols inducing an
unbounded composition operator, hence not all the $C_{T_{a}}$ are bounded.
Actually, for every $\alpha\in(0,2]$, no $C_{T_{a}}$ is bounded because
$\beta$ has no polynomial lower estimate (see Proposition 4.5 below).

Let us describe the contents of this paper. In Section 2, we introduce several
notion of growth or regularity for a sequence $\beta$: essentially decreasing,
polynomial decay and polynomial growth, slow oscillation, and give some
connections between them. In Section 3, we consider the composition operators
whose symbol vanishes at the origin. We show that in order for all these
operators to be bounded, it is necessary that $\beta$ be bounded above. We
show that $\beta$ is essentially decreasing if and only if all these operators
are bounded and $\sup_{\varphi(0)=0}\|C_{\varphi}\|<+\infty$. In Theorem 3.3,
we give a sufficient condition for having all the composition operators
$C_{\varphi}$ with $\varphi(0)=0$ bounded, allowing us to give an example of a
sequence $\beta$ for which this happens though
$\sup_{\varphi(0)=0}\|C_{\varphi}\|=+\infty$ (Theorem 3.7). In Section 4, we
prove that all $C_{T_{a}}$ are bounded on $H^{2}(\beta)$ if and only if
$\beta$ is slowly oscillating (Theorem 4.6 and Theorem 4.9). We hence have:

######  Theorem 1.1.

Let $\beta$ be a sequence of positive numbers, and let

$T_{a}(z)=\frac{a+z}{1+\bar{a}\,z}$

for $a\in\mathbb{D}$. The following assertions are equivalent:

  * $1)$

for some $a\in\mathbb{D}\setminus\\{0\\}$, the map $T_{a}$ induces a bounded
composition operator $C_{T_{a}}$ on $H^{2}(\beta)$;

  * $2)$

for all $a\in\mathbb{D}$, the maps $T_{a}$ induce bounded composition
operators $C_{T_{a}}$ on $H^{2}(\beta)$;

  * $3)$

$\beta$ is slowly oscillating.

The deep implication is $2)\Rightarrow 3)$. Its proof requires some sharp
estimates on the mean of Taylor coefficients of $T_{a}$ for $a$ belonging to a
subinterval of $(0,1)$. Once we found the equivalence of $1)$ and $2)$, we
realized that it already appeared in Zorboska’s thesis [23].

In Section 5, we show (Theorem 5.1) that if $\beta$ is slowly oscillating, and
moreover all composition operators are bounded on $H^{2}(\beta)$, then $\beta$
is essentially decreasing. We thus obtain the following theorem.

######  Theorem 1.2.

Let $\beta$ be a sequence of positive numbers. Then all composition operators
on $H^{2}(\beta)$ are bounded if and only if $\beta$ is essentially decreasing
and slowly oscillating.

For the notion of essentially decreasing and slowly oscillating sequences, see
Definition 2.1 and Definition 2.2

We end the paper with some results about multipliers.

A first version of this paper, not including the complete characterization
given here, was put on arXiv on 30 November 2020 (and a second version on 21
March 2022) under the title _Boundedness of composition operators on general
weighted Hardy spaces of analytic functions_.

##  2 Definitions, notation, and preliminary results

The open unit disk of $\mathbb{C}$ is denoted $\mathbb{D}$ and we write
$\mathbb{T}$ its boundary $\partial\mathbb{D}$. We set $e_{n}(z)=z^{n}$,
$n\geq 0$.

The weighted Hardy space $H^{2}(\beta)$ defined in the introduction is a
Hilbert space with the canonical orthonormal basis

(2.1) $\qquad\qquad e^{\beta}_{n}(z)=\frac{1}{\sqrt{\beta_{n}}}\,z^{n}\,,\quad
n\geq 0\,,$

and the reproducing kernel $K_{w}$ given for all $w\in\mathbb{D}$ by

(2.2) $K_{w}(z)=\sum_{n=0}^{\infty}e^{\beta}_{n}(z)\,\mkern
1.5mu\overline{\mkern-1.5mue^{\beta}_{n}(w)\mkern-1.5mu}\mkern
1.5mu=\sum_{n=0}^{\infty}\frac{1}{\beta_{n}}\,\mkern
1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{n}\,z^{n}\,.$

Note that $H^{2}$ is continuously embedded in $H^{2}(\beta)$ if and only if
$\beta$ is bounded above. In particular, this is the case when $\beta$ is non-
increasing. In this paper, we need a slightly more general notion.

######  Definition 2.1.

A sequence of positive numbers $\beta=(\beta_{n})_{n\geq 0}$ is said to be
_essentially decreasing_ if, for some constant $C\geq 1$, we have, for all
$m\geq n\geq 0$:

(2.3) $\beta_{m}\leq C\,\beta_{n}\,.$

Note that saying that $\beta$ is essentially decreasing means that the shift
operator on $H^{2}(\beta)$ is power bounded.

If $\beta$ is essentially decreasing, and if we set:

$\widetilde{\beta}_{n}=\sup_{m\geq n}\beta_{m}\,,$

the sequence $\widetilde{\beta}=(\widetilde{\beta}_{n})$ is non-increasing and
we have $\beta_{n}\leq\widetilde{\beta}_{n}\leq C\,\beta_{n}$. In particular,
$H^{2}(\beta)=H^{2}(\widetilde{\beta})$ (with equivalent norms) and $H^{2}$ is
continuously embedded in $H^{2}(\beta)$.

######  Definition 2.2.

We say that a sequence $\beta$ is _slowly oscillating_ if there are positive
constants $c<1<C$ such that

(2.4) $\qquad c\leq\frac{\beta_{m}}{\beta_{n}}\leq C\quad\text{when }n/2\leq
m\leq 2n\,.$

We may remark that this is equivalent to the existence of some function
$\rho\colon(0,\infty)\to(0,\infty)$ which is bounded above on each compact
subset of $(0,\infty)$ and for which $\beta_{m}/\beta_{n}\leq\rho(m/n)$,
equivalently

$\frac{1}{\rho(n/m)}\leq\frac{\beta_{m}}{\beta_{n}}\leq\rho(m/n).$

######  Definition 2.3.

The sequence of positive numbers $\beta=(\beta_{n})$ is said to have
polynomial decay if there are positive constants $c$ and $\alpha$ such that,
for all integers $n\geq 1$:

(2.5) $\beta_{n}\geq c\,n^{-\alpha}\,.$

That means that $H^{2}(\beta)$ is continuously embedded in the weighted
Bergman space ${\mathfrak{B}}^{2}_{\alpha-1}$ of the analytic functions
$f\colon\mathbb{D}\to\mathbb{C}$ such that

$\|f\|_{{\mathfrak{B}}^{2}_{\alpha-1}}^{2}:=\alpha\int_{\mathbb{D}}|f(z)|^{2}(1-|z|^{2})^{\alpha-1}\,dA(z)<\infty$

since ${\mathfrak{B}}^{2}_{\alpha-1}=H^{2}(\gamma)$ with $\gamma_{n}\approx
n^{-\alpha}$.

######  Definition 2.4.

The sequence of positive numbers $\beta=(\beta_{n})$ is said to have
_polynomial growth_ if there are positive constants $C$ and $\gamma$ such
that, for all integers $n\geq 1$:

(2.6) $\beta_{n}\leq C\,n^{\gamma}\,.$

The following simple proposition links those notions.

######  Proposition 2.5.

$1)$ Every slowly oscillating sequence $\beta$ has polynomial decay and
growth.

$2)$ There are sequences that are essentially decreasing, and with polynomial
decay, but not slowly oscillating.

$3)$ There are bounded sequences that are slowly oscillating, but not
essentially decreasing.

###### Proof.

$1)$ This is clear, because, for some $c\in(0,1)$, if $2^{j}\leq n<2^{j+1}$,
then

$\beta_{n}\geq c\,\beta_{2^{j}}\geq c^{j+1}\beta_{1}\geq
c\,\beta_{1}\,n^{-\alpha}\,,$

with $\alpha=\log(1/c)/\log 2$; and, for some $C>1$,

$\beta_{n}\leq C\,\beta_{2^{j}}\leq C^{j+1}\beta_{1}\leq
C\beta_{1}\,n^{\gamma}\,,$

with $\gamma=\log C/\log 2$.

$2)$ Let $\delta>0$. We set $\beta_{0}=\beta_{1}=1$ and for $n\geq 2$:

$\beta_{n}=\frac{1}{(k!)^{\delta}}\quad\text{when }k!<n\leq(k+1)!\,.$

The sequence $\beta$ is non-increasing.

For $n$ and $k$ as above, we have:

$\beta_{n}=\frac{1}{(k!)^{\delta}}\geq\frac{1}{n^{\delta}}\,;$

hence $\beta$ has arbitrarily slow polynomial decay. However we have, for
$k\geq 2$:

$\frac{\beta_{2(k!)}}{\beta_{k!}}=\frac{(k!)^{-\delta}}{[(k-1)!]^{-\delta}}=\frac{1}{k^{\delta}}\mathop{\longrightarrow}\limits_{k\to\infty}0\,,$

so $\beta$ is not slowly oscillating.

$3)$ We define $\beta_{n}$ as follows. Let $(a_{k})$ be an increasing sequence
of positive square integers such that $\lim_{k\to\infty}a_{k+1}/a_{k}=\infty$,
for example $a_{k}=4^{k^{2}}$, and let $b_{k}=\sqrt{a_{k}a_{k+1}}\,$; with our
choice, this is an integer and we clearly have $a_{k}<b_{k}<a_{k+1}$. We set:

$\displaystyle\beta_{n}=\left\\{\begin{array}[]{ll}\,a_{k}/n&\text{for
}a_{k}\leq n<b_{k}\\\ &\\\ (a_{k}/b_{k}^{2})\,n=(1/a_{k+1})\,n&\text{for
}b_{k}\leq n<a_{k+1}\,.\end{array}\right.$

This sequence $(\beta_{n})$ is slowly oscillating by construction. Indeed,
since the other cases are obvious, it suffices to check that for $a_{k}\leq
n/2<b_{k}\leq n<a_{k+1}$, the quotient $\beta_{m}/\beta_{n}$ remains lower and
upper bounded when $n/2\leq m\leq n$ (it will then be automatically also
satisfied when $n\leq m\leq 2n$). But for $n/2\leq m<b_{k}$, we have

$\frac{\beta_{m}}{\beta_{n}}=\frac{a_{k}/m}{n/a_{k+1}}=\frac{a_{k}a_{k+1}}{mn}=\frac{b_{k}^{2}}{mn}\,\raise
1.0pt\hbox{,}$

which is $\leq 2\,b_{k}^{2}/n^{2}\leq 2$ and $\geq
b_{k}^{2}/n^{2}\geq(n/2)^{2}/n^{2}=1/4$; and for $b_{k}\leq m$, we have

$\frac{\beta_{m}}{\beta_{n}}=\frac{m/a_{k+1}}{n/a_{k+1}}=\frac{m}{n}\in\Big{[}\frac{1}{2}\,\raise
1.5pt\hbox{,}\,\raise 0.5pt\hbox{$1$}\Big{]}\,.$

However, though $(\beta_{n})$ is bounded, since $\beta_{n}\leq 1$ for
$a_{k}\leq n<b_{k}$ and, for $b_{k}\leq n<a_{k+1}$,

$\beta_{n}\leq\beta_{a_{k+1}-1}=\frac{1}{a_{k+1}}\,(a_{k+1}-1)\leq 1\,,$

it is not essentially decreasing, since

$\frac{\beta_{a_{k+1}-1}}{\beta_{b_{k}}}=\frac{1}{\sqrt{a_{k}a_{k+1}}}\,(a_{k+1}-1)\sim\sqrt{\frac{a_{k+1}}{a_{k}}}\mathop{\longrightarrow}\limits_{k\to\infty}\infty\,.\qed$

Now we are going to recall some well known facts about matrix representation
of an operator $T$ defined on a Hilbert space with an orthonormal basis $({\rm
e}_{n})_{n\geq 0}$ and how it is translated in our framework.

The entry $a_{m,n}$ (where $m,n\geq 0$) is defined by the $m^{th}$ coordinate
of $T({\rm e}_{n})$:

$a_{m,n}={\rm e}_{m}^{\ast}\big{(}T({\rm e}_{n})\big{)}$

where ${\rm e}_{k}^{\ast}(x)$ stands for the $k^{th}$ coordinate of the vector
$x$.

We shall use the notation $\widehat{f}(k)$ for the $k^{th}$ Fourier
coefficient of a function $f\in L^{1}(-\pi,\pi)$:

$\widehat{f}(k)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\,{\rm e}^{-ikt}\,dt\;.$

Let us point out that when the operator is the composition operator
$C_{\varphi}$ associated to the symbol $\varphi$, viewed on $H^{2}(\beta)$,
its matrix representation in the basis $\big{(}e^{\beta}_{n}\big{)}_{n\geq 0}$
has an entry $(m,n)$ which can be written as

$(e^{\beta}_{m})^{\ast}\big{(}C_{\varphi}(e^{\beta}_{n})\big{)}=\frac{\sqrt{\beta_{m}}}{\sqrt{\beta_{n}}}e_{m}^{\ast}\big{(}\varphi^{n}\big{)}=\frac{\sqrt{\beta_{m}}}{\sqrt{\beta_{n}}}\,\widehat{\varphi^{n}}(m)$

since the $m^{th}$ Taylor coefficient of $\varphi^{n}$ coincides with its
$m^{th}$ Fourier coefficient.

We say that the reproducing kernels $K_{w}$ have a _slow growth_ if

(2.7) $\|K_{w}\|\leq\frac{C}{(1-|w|)^{s}}$

for positive constants $C$ and $s$. We have the following equivalence.

######  Proposition 2.6.

The sequence $\beta$ has polynomial decay if and only if the reproducing
kernels $K_{w}$ of $H^{2}(\beta)$ have a slow growth.

###### Proof.

Assume that the reproducing kernels have a slow growth. Since

$\|K_{w}\|^{2}=\sum_{k=0}^{\infty}\frac{\ |w|^{2k}}{\beta_{k}}\,\raise
1.0pt\hbox{,}$

we get, for any $k\geq 2$:

$\frac{\ |w|^{2k}}{\beta_{k}}\leq\frac{C^{2}}{(1-|w|)^{2s}}\,\cdot$

Taking $w=1-\frac{1}{k}\,$, we obtain $\beta_{k}\geq C^{\prime}\,k^{-2s}$.

For the necessity, we only have to see that:

$\|K_{w}\|^{2}=\frac{1}{\beta_{0}}+\sum_{n=1}^{\infty}\frac{\,\,\,|w|^{2n}}{\beta_{n}}\leq\frac{1}{\beta_{0}}+\delta^{-1}\sum_{n=1}^{\infty}n^{\alpha}|w|^{2n}\leq\frac{C}{(1-|w|^{2})^{\alpha+1}}\,\cdot\qed$

##  3 Boundedness of composition operators whose symbol vanishes at the origin

###  3.1 Necessary conditions

We begin with this simple observation (see [23, Proposition 3.1]).

######  Proposition 3.1.

If all composition operators with symbol vanishing at $0$ are bounded on
$H^{2}(\beta)$, then the sequence $\beta$ is bounded above.

###### Proof.

Let $f\in H^{\infty}$. Write $f=A\,\varphi+f(0)$ where $A$ is a constant and
$\varphi$ a symbol vanishing at $0$. We have $\varphi=C_{\varphi}(z)\in
H^{2}(\beta)$, by hypothesis. So that $f\in H^{2}(\beta)$ and
$H^{\infty}\subseteq H^{2}(\beta)$. It follows (by the closed graph theorem,
since the convergence in norm implies pointwise convergence) that there exists
a constant $M$ such that $\|f\|_{H^{2}(\beta)}\leq M\,\|f\|_{\infty}$ for all
$f\in H^{\infty}$. Testing that with $f(z)=z^{n}$, we get $\beta_{n}\leq
M^{2}$. ∎

Let us point out that boundedness of $\beta_{n}$ does not suffice. For
example, let $(\beta_{n})$ be a sequence such that
$\beta_{4k+2}/\beta_{2k+1}\mathop{\longrightarrow}\limits_{k\to\infty}\infty$
(for instance $\beta_{2k}=1$ and $\beta_{2k+1}=1/(k+1)$); if
$\varphi(z)=z^{2}$, then
$\|C_{\varphi}(z^{2n+1})\|^{2}=\|z^{2(2n+1)}\|^{2}=\beta_{2(2n+1)}$; since
$\|z^{2n+1}\|^{2}=\beta_{2n+1}$, the operator $C_{\varphi}$ is not bounded on
$H^{2}(\beta)$.

A partial characterization is given in the next proposition.

######  Proposition 3.2.

The following assertions are equivalent:

$1)$ all symbols $\varphi$ such that $\varphi(0)=0$ induce bounded composition
operators $C_{\varphi}$ on $H^{2}(\beta)$ and

(3.1) $\sup_{\varphi(0)=0}\|C_{\varphi}\|<\infty\,;$

$2)$ $\beta$ is an essentially decreasing sequence.

Of course, by the uniform boundedness principle, (3.1) is equivalent to:

$\qquad\qquad\sup_{\varphi(0)=0}\|f\circ\varphi\|<\infty\quad\text{for all
}f\in H^{2}(\beta)\,.$

Let us point out an important fact: we shall see in Theorem 3.7 that there are
weights $\beta$ for which all composition operators $C_{\varphi}$ with
$\varphi(0)=0$ are bounded, but $\sup_{\varphi(0)=0}\|C_{\varphi}\|=+\infty$.

###### Proof.

$2)\Rightarrow 1)$ We may assume that $\beta$ is non-increasing. Then the
Goluzin-Rogosinski theorem ([5, Theorem 6.3]) gives the result; in fact,
writing $f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$ and
$(C_{\varphi}f)(z)=\sum_{n=0}^{\infty}d_{n}z^{n}$, it says that

$\sum_{0\leq k\leq n}|d_{k}|^{2}\leq\sum_{0\leq k\leq
n}|c_{k}|^{2}\quad\forall n\geq 0,$

and hence, by Abel summation:

$\|C_{\varphi}f\|^{2}=\sum_{n=0}^{\infty}|d_{n}|^{2}\beta_{n}\leq\sum_{n=0}^{\infty}|c_{n}|^{2}\beta_{n}=\|f\|^{2}\,,$

leading to $C_{\varphi}$ bounded and $\|C_{\varphi}\|\leq 1$. This result was
also proved by C. Cowen [4, Corollary of Theorem 7]. Alternatively, we can use
a result of Kacnel’son ([9]; see also [2], [3, Corollary 2.2], or [12, Theorem
3.12]).

$1)\Rightarrow 2)$ Set $M=\sup_{\varphi(0)=0}\|C_{\varphi}\|$. Let $m>n$, and
take

$\varphi(z)=\varphi_{m,n}(z)=z\,\bigg{(}\frac{1+z^{m-n}}{2}\bigg{)}^{1/n}\,.$

Then $\varphi(0)=0$ and $[\varphi(z)]^{n}=\frac{z^{n}+z^{m}}{2}\,$; hence

$\frac{1}{4}\,(\beta_{n}+\beta_{m})=\|\varphi^{n}\|^{2}=\|C_{\varphi}(e_{n})\|^{2}\leq\|C_{\varphi}\|^{2}\|e_{n}\|^{2}\leq
M^{2}\,\beta_{n}\,,$

so $\beta$ is essentially decreasing. ∎

Remark. Let us mention the following example. For $0<r<1$, let
$\beta_{n}=\pi\,n\,r^{2n}$, for $n\geq 1$ and $\beta_{0}=1$. This sequence is
eventually decreasing, so it is essentially decreasing. The quantity
$\|f\|^{2}_{H^{2}(\beta)}-|f(0)|^{2}$ is the area of the part of the Riemann
surface on which $r\mathbb{D}$ is mapped by $f$. E. Reich [16], generalizing
Goluzin’s result [8] (see [5, Theorem 6.3]), proved that for all symbols
$\varphi$ such that $\varphi(0)=0$, the composition operator $C_{\varphi}$ is
bounded on $H^{2}(\beta)$ and

$\|C_{\varphi}\|\leq\sup_{n\geq 1}\sqrt{n}\,r^{n-1}\leq\frac{1}{\sqrt{2\,{\rm
e}}}\,\frac{1}{r\sqrt{\log(1/r)}}\,\cdot$

For $0<r<1/\sqrt{2}$, Goluzin’s theorem asserts that $\|C_{\varphi}\|\leq 1$.

Note that this sequence $\beta$ is not slowly oscillating since
$\beta_{2n}/\beta_{n}=2\,r^{2n}$. Hence, from Theorem 4.9 below, we get that
no composition operator $C_{T_{a}}$ is bounded on $H^{2}(\beta)$.

However, that the weight $\beta$ is essentially decreasing is not necessary
for the boundedness of all composition operators $C_{\varphi}$, with symbol
$\varphi$ vanishing at $0$, as we will see later (Theorem 3.7).

###  3.2 Sufficient condition

######  Theorem 3.3.

Let $\beta=(\beta_{n})_{n=0}^{\infty}$ be a sequence of positive numbers with
polynomial decay and growth. Assume that $\beta$ is _weakly decreasing_ ,
i.e.:

(3.2) $\begin{split}\text{For every }\delta>0,&\text{ there exists }\text{a
positive constant }C=C(\delta)\text{ such that }\\\
&\hskip-8.00003pt\beta_{m}\leq C\,\beta_{n}\quad\text{whenever
}m>(1+\delta)\,n\,.\end{split}$

Then, for all symbols $\varphi\colon\mathbb{D}\to\mathbb{D}$ vanishing at $0$,
the composition operator $C_{\varphi}$ is bounded on $H^{2}(\beta)$.

Let us point out that (3.2) implies that $\beta$ is bounded. For a
counterexample with an exponential weight, see [23, Ex. 1, p. 14-15].

To prove Theorem 3.3, we need several lemmas.

######  Lemma 3.4.

Let $\varphi\colon\mathbb{D}\to\mathbb{D}$ be an analytic self-map such that
$\varphi(0)=0$ and $|\varphi^{\prime}(0)|<1$. Then there exists $\rho>0$ such
that, for every integers $n$ and $m$,

$|\widehat{\varphi^{n}}(m)|\leq\exp\Big{(}-\frac{1}{2}\,[(1+\rho)\,n-m]\Big{)}\,.$

###### Proof.

Since $\varphi(0)=0$, we can write $\varphi(z)=z\,\varphi_{1}(z)$. Since
$|\varphi^{\prime}(0)|<1$, we have $\varphi_{1}\colon\mathbb{D}\to\mathbb{D}$.
Let $M(r)=\sup_{|z|=r}|\varphi_{1}(z)|$. Cauchy’s inequalities say that
$|\widehat{\varphi_{1}^{n}}(m)|\leq[M(r)]^{n}/r^{m}$. We have $M(r)<1$, so
there exists a positive number $\rho=\rho(r)$ such that $M(r)=r^{\rho}$. We
get:

$|\widehat{\varphi^{n}}(m)|=|\widehat{\varphi_{1}^{n}}(m-n)|\leq\frac{r^{\rho
n}}{r^{m-n}}=r^{(1+\rho)\,n-m}\,,$

and the result follows, by taking $r={\rm e}^{-1/2}$. ∎

The next lemma is a variant of the following result of V. È Kacnel’son ([9];
see also [2], [3, Corollary 2.2], or [12, Theorem 3.12]).

######  Theorem 3.5 (V. È. Kacnel’son).

Let $H$ be a separable complex Hilbert space and $(e_{i})_{i\geq 0}$ a fixed
orthonormal basis of $H$. Let $M\colon H\to H$ be a bounded linear operator.
We assume that the matrix of $M$ with respect to this basis is lower-
triangular: $\langle Me_{j}\mid e_{i}\rangle=0$ for $i<j$.

Let $(\gamma_{j})_{j\geq 0}$ be a non-decreasing sequence of positive real
numbers and $\Gamma$ the (possibly unbounded) diagonal operator such that
$\Gamma(e_{j})=\gamma_{j}e_{j}$, $j\geq 0$. Then the operator
$\Gamma^{-1}M\,\Gamma\colon H\to H$ is bounded and moreover:

$\|\Gamma^{-1}M\,\Gamma\|\leq\|M\|\,.$

This variant is used implicitly in [12, page 13].

######  Lemma 3.6.

Let $A\colon\ell_{2}\to\ell_{2}$ be a bounded operator represented by the
matrix $\big{(}a_{m,n}\big{)}_{m,n}$, i.e. $a_{m,n}=\langle
A\,e_{n},e_{m}\rangle$, where $(e_{n})_{n\geq 0}$ is the canonical basis of
$\ell_{2}$.

Let $(d_{n})$ be a sequence of positive numbers such that, for every $m$ and
$n$:

(3.3) $d_{m}<d_{n}\quad\Longrightarrow\quad a_{m,n}=0\,.$

Then, $D$ being the (possibly unbounded) diagonal operator with entries
$d_{n}$, we have:

$\|D^{-1}AD\|\leq\|A\|\,.$

The proof is the same as that of Kacnel’son’s theorem, but we reproduce it for
the convenience of the reader. Actually we propose two different proofs.

###### Proof 1..

Let $\mathbb{C}_{0}$ be the right-half plane
$\mathbb{C}_{0}=\\{z\in\mathbb{C}\,;\ {\mathfrak{R}}{\rm e}\,z>0\\}$. We set
$H_{N}={\rm span}\,\\{e_{n}\,;\ n\leq N\\}$ and

$A_{N}=P_{N}AJ_{N}\,,$

where $P_{N}$ is the orthogonal projection from $\ell_{2}$ onto $H_{N}$ and
$J_{N}$ the canonical injection from $H_{N}$ into $\ell_{2}$. We consider, for
$z\in\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern
1.5mu$:

$A_{N}(z)=D^{-z}A_{N}D^{z}\colon H_{N}\to H_{N}\,,$

where $D^{z}(e_{n})=d_{n}^{\,z}e_{n}$.

If $\big{(}a_{m,n}(z)\big{)}_{m,n}$ is the matrix of $A_{N}(z)$ on the basis
$\\{e_{n}\,;\ n\leq N\\}$ of $H_{N}$, we clearly have:

$a_{m,n}(z)=a_{m,n}(d_{n}/d_{m})^{z}\,.$

In particular, we have, thanks to (3.3):

$a_{m,n}(z)=0\quad\text{if }d_{m}<d_{n}\,,$

and

$\qquad\qquad\quad|a_{m,n}(z)|\leq\sup_{k,l}|a_{k,l}|:=M\,,\qquad\text{for all
}z\in\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern
1.5mu\,.$

Since $\|A_{N}(z)\|^{2}\leq\|A_{N}(z)\|_{HS}^{2}=\sum_{m,n\leq
N}|a_{m,n}(z)|^{2}\leq(N+1)^{2}M^{2}$, we get:

$\qquad\qquad\|A_{N}(z)\|\leq(N+1)\,M\qquad\text{for all }z\in\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern 1.5mu\,.$

Let us consider the function $u\colon\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern 1.5mu\to\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern 1.5mu$ defined
by:

(3.4) $u_{N}(z)=\|A_{N}(z)\|\,.$

This function $u_{N}$ is continuous on $\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern 1.5mu$, bounded
above by $(N+1)M$, and subharmonic in $\mathbb{C}_{0}$. Moreover, thanks to
(3.3), the maximum principle gives:

$\sup_{\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern
1.5mu}u_{N}(z)=\sup_{\partial\mathbb{C}_{0}}u_{N}(z)\,.$

Since $\|D^{z}\|=\|D^{-z}\|=1$ for $z\in\partial\mathbb{C}_{0}$, we have
$\|A_{N}(z)\|\leq\|A_{N}\|$ for $z\in\partial\mathbb{C}_{0}$, and we get:

$\sup_{\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{C}_{0}\mkern-1.5mu}\mkern
1.5mu}u_{N}(z)\leq\|A_{N}\|\leq\|A\|\,.$

In particular $u_{N}(1)\leq\|A\|$, and, letting $N$ going to infinity, we
obtain that $\|D^{-1}AD\|\leq\|A\|$.

_Proof 2._ Since $d_{n}$ is positive, we can write $d_{n}={\rm e}^{-\rho_{n}}$
where $\rho_{n}\in\mathbb{R}$. Let $x=(x_{n})_{n\geq 0}$ and $y=(y_{n})_{n\geq
0}\in\ell^{2}$ with finite support, we are interested in controlling the sum

$S=\sum_{m,n}a_{m,n}\frac{d_{n}}{d_{m}}\,x_{n}\overline{y_{m}}$

which can also be written

$S=\sum_{m,n}a_{m,n}{\rm e}^{-|\rho_{n}-\rho_{m}|}\,x_{n}\overline{y_{m}}.$

since the non trivial part of the sum runs over the pairs $(m,n)$ such that
$d_{m}\geq d_{n}$ i.e. $\rho_{n}\geq\rho_{m}$.

Now we introduce the function $\displaystyle f(t)=\frac{1}{\pi(1+t^{2})}$ for
$t\in\mathbb{R}$, which is positive and belongs to the unit ball of
$L^{1}(\mathbb{R})$. Moreover, its Fourier transform satisfies, for every
$x\in\mathbb{R}$,

$\mathcal{F}(f)(-x)=\int_{\mathbb{R}}f(t){\rm e}^{ixt}dt={\rm e}^{-|x|}\;.$

We get

$S=\int_{\mathbb{R}}f(t)\bigg{(}\sum_{m,n}a_{m,n}x_{n}{\rm
e}^{i\rho_{n}t}\overline{y_{m}{\rm
e}^{i\rho_{m}t}}\bigg{)}\,dt=\int_{\mathbb{R}}f(t)\langle
A(x(t)),y(t)\rangle_{\ell^{2}}\,dt$

where

$x(t)=\big{(}x_{n}{\rm e}^{i\rho_{n}t}\big{)}_{n\geq 0}\quad\text{and}\quad\
y(t)=\big{(}y_{n}{\rm e}^{i\rho_{n}t}\big{)}_{n\geq 0}.$

We obtain

$|S|\leq\int_{\mathbb{R}}f(t)\|A\|\,\|x(t)\|\,\|y(t)\|\,dt=\int_{\mathbb{R}}f(t)\|A\|\,\|x\|\,\|y\|\,dt=\|A\|\|x\|\|y\|$

since $\|f\|_{L^{1}(\mathbb{R})}=1$.

Since $x$ and $y$ are arbitrary, this proves
$\displaystyle\|D^{-1}AD\|\leq\|A\|$. ∎

###### Proof of Theorem 3.3.

First, if $|\varphi^{\prime}(0)|=1$, we have $\varphi(z)=\alpha\,z$ for some
$\alpha$ with $|\alpha|=1$, and the result is trivial.

So, we assume that $|\varphi^{\prime}(0)|<1$. Then, by Lemma 3.4, there exists
$\rho>0$ such that, for all $m$, $n$:

$|\widehat{\varphi^{n}}(m)|\leq\exp\Big{(}-\frac{1}{2}\,[(1+\rho)\,n-m]\Big{)}\,.$

Since $\varphi(0)=0$, we also know that $\widehat{\varphi^{n}}(m)=0$ if $m<n$.

Take $\delta=\rho/2$ and use property (3.2): there exists $M\geq 1$ such that:

$\qquad\frac{\beta_{m}}{\beta_{n}}\leq M\quad\text{when
}m\geq(1+\delta)\,n\,.$

Define now a new sequence $\gamma=(\gamma_{n})$ as:

$\gamma_{n}=\max\bigg{\\{}\beta_{n},\sup_{m>(1+\delta)\,n}\beta_{m}\bigg{\\}}\,.$

We have:

1) $\beta_{n}\leq\gamma_{n}\leq M\,\beta_{n}$;

2) $\gamma_{m}\leq\gamma_{n}$ if $m\geq(1+\delta)\,n$.

Item 1) implies that $H^{2}(\gamma)=H^{2}(\beta)$, and we are reduced to prove
that $C_{\varphi}\colon H^{2}(\gamma)\to H^{2}(\gamma)$ is bounded.

Let
$A=\big{(}a_{m,n}\big{)}_{m,n}=\big{(}\widehat{\varphi^{n}}(m)\big{)}_{m,n}$.
We have to prove that

$B=\big{(}\gamma_{m}^{1/2}\gamma_{n}^{-1/2}a_{m,n}\big{)}_{m,n}$

represents a bounded operator on $\ell_{2}$.

Define the matrix

$A_{1}=\big{(}a_{m,n}\mathds{1}_{\\{(m,n)\,;\
m\leq(1+\delta)\,n\\}}\big{)}_{m,n}$

and set $A_{2}=A-A_{1}$. Define analogously $B_{1}$ and $B_{2}=B-B_{1}$.

Then $A_{1}$ is a Hilbert-Schmidt operator, because (recall that $a_{m,n}=0$
if $m<n$), we have

$\displaystyle\sum_{n=1}^{\infty}\sum_{m=1}^{(1+\delta)\,n}|a_{m,n}|^{2}$
$\displaystyle\leq\sum_{n=1}^{\infty}\sum_{m=n}^{(1+\delta)\,n}\exp\big{(}-[(1+\rho)\,n-m]\big{)}$
$\displaystyle\leq\sum_{n=1}^{\infty}(\delta\,n+1)\,\exp(-\delta\,n)<\infty\,.$

Now, $\beta$ is bounded and has polynomial decay, so, for some positive
constants $C_{1}$, $C_{2}$, and $\alpha$, we have:

$\displaystyle\sum_{n=1}^{\infty}\sum_{m=n}^{(1+\delta)\,n}\frac{\gamma_{m}}{\gamma_{n}}\,|a_{m,n}|^{2}$
$\displaystyle\leq\sum_{n=1}^{\infty}\sum_{m=n}^{(1+\delta)\,n}\frac{C_{1}}{n^{-\alpha}}\,\exp(-\delta\,n)$
$\displaystyle\leq\sum_{n=1}^{\infty}C_{2}(1+\delta)n^{\alpha+1}\exp(-\delta\,n)<\infty\,,$

meaning that $B_{1}$ is a Hilbert-Schmidt operator.

Since $A$ is bounded, it follows that $A_{2}=A-A_{1}$ is bounded. Remark that,
writing $A_{2}=\big{(}\alpha_{m,n}\big{)}_{m,n}$, we have, with
$d_{n}=1/\sqrt{\gamma_{n}}$:

$d_{m}<d_{n}\quad\Longrightarrow\quad\gamma_{m}>\gamma_{n}\quad\Longrightarrow\quad
m<(1+\delta)\,n\quad\Longrightarrow\quad\alpha_{m,n}=0\,.$

Hence we can apply Lemma 3.6 to the matrix $A_{2}$, and it ensues that $B_{2}$
is bounded, and therefore that $B=B_{1}+B_{2}$ is bounded as well, as wanted.
∎

As a corollary of Theorem 3.3, we can provide the following example.

######  Theorem 3.7.

There exists a bounded sequence $\beta$, with polynomial decay, but which is
_not essentially decreasing_ , and for which every composition operator with
symbol vanishing at $0$ is bounded on $H^{2}(\beta)$.

We hence have $\sup_{\varphi(0)=0}\|C_{\varphi}\|=+\infty$.

It should be noted that for this weight, the composition operators are not all
bounded, as we will see in Proposition 4.10.

###### Proof.

Define $\beta_{n}=1$ for $n\leq 3!$, and, for $k\geq 3$:

$\left\\{\begin{array}[]{lcl}\beta_{n}=&\displaystyle\frac{1}{k!}&\text{for
}k!<n\leq(k+1)!-2\text{ and for }n=(k+1)!\\\ \\\
\beta_{n}=&\displaystyle\frac{1}{(k+1)!}&\text{for
}n=(k+1)!-1\,.\end{array}\right.$

Note that, for $m>n$, we have $\beta_{m}>\beta_{n}$ only if $n=(k+1)!-1$ and
$m=(k+1)!=n+1$, for some $k\geq 3$.

However $\beta$ is not essentially decreasing since, for every $k\geq 3$, we
have $\beta_{n+1}/\beta_{n}=k+1$ if $n=(k+1)!-1$.

The sequence $\beta$ has a polynomial decay because $\beta_{n}\geq 1/(2\,n)$
for all $n\geq 1$. In fact, for $k\geq 3$, we have $\beta_{n}\geq(k+1)/n\geq
1/n$ if $k!<n\leq(k+1)!-2$ or if $n=(k+1)!$; and for $n=(k+1)!-1$, we have
$n\,\beta_{n}=[(k+1)!-1]/(k+1)!\geq 1/2$. It has a polynomial growth since it
is bounded above, by $1$.

Now, it remains to check (3.2) in order to apply Theorem 3.3 and finish the
proof of Theorem 3.7. Note first that we have $\beta_{m}/\beta_{n}\leq 1$ if
$m\geq n+2$. Next, for given $\delta>0$, there exists an integer $N$ such that
$(1+\delta)\,n\geq n+2$ for every $n\geq N$, so $\beta_{m}/\beta_{n}\leq 1$ if
$m\geq(1+\delta)\,n$ and $n\geq N$. It suffices to take $C=\max_{1\leq n\leq
N}\beta_{n+1}/\beta_{n}$ to obtain (3.2). The last assertion follows from
Proposition 3.2. ∎

##  4 Boundedness of composition operators of symbol $T_{a}$

Recall that for $a\in\mathbb{D}$, we defined

(4.1) $\qquad\quad T_{a}(z)=\frac{a+z}{1+\bar{a}\,z}\,\raise
1.0pt\hbox{,}\quad z\in\mathbb{D}\,.$

It is well-known that $T_{a}$ is an automorphism of $\mathbb{D}$ and that
$T_{a}(0)=a$ and $T_{a}(-a)=0$.

Though we do not really need this, we may remark that $(T_{a})_{a\in(-1,1)}$
is a group and $(T_{a})_{a\in(0,1)}$ is a semigroup. It suffices to see that
$T_{a}\circ T_{b}=T_{a\ast b}$, with:

(4.2) $a\ast b=\frac{a+b}{1+ab}\,\cdot$

In this section, we are going to prove a necessary and sufficient condition in
order that all composition operators $C_{T_{a}}$ for $a\in\mathbb{D}$ are
bounded on $H^{2}(\beta)$. Namely, we have the following theorem, the proof of
which will occupy Section 4.2 and Section 4.3.

######  Theorem 4.1.

All composition operators $C_{T_{a}}$, with $a\in\mathbb{D}$ are bounded on
$H^{2}(\beta)$ if and only if $\beta$ is slowly oscillating.

Before that, let us note the following fact (see also [23, Proposition 3.6]).
Recall that if $\varphi$ and $\psi$ are two symbols, then $C_{\varphi}\circ
C_{\psi}=C_{\psi\circ\varphi}$.

######  Proposition 4.2.

If $C_{T_{a}}$ is bounded on $H^{2}(\beta)$ for some
$a\in\mathbb{D}\setminus\\{0\\}$, then $C_{T_{b}}$ is bounded on
$H^{2}(\beta)$ for all $b\in\mathbb{D}$.

Moreover the maps $C_{T_{b}}$ are uniformly bounded on the compact subsets of
$\mathbb{D}$.

We decompose the proof into lemmas. The first one was first proved in [23]
(see also [6, Proposition 2.1]), and follows from the fact that if
$b=\rho\,{\rm e}^{i\theta}$ and $R_{\theta}$ is the rotation
$R_{\theta}(z)={\rm e}^{i\theta}z$, which induces a unitary operator
$C_{R_{\theta}}$ on $H^{2}(\beta)$, then $T_{b}=R_{\theta}\circ T_{\rho}\circ
R_{-\theta}$ and $C_{T_{b}}=C_{R_{-\theta}}\circ C_{T_{\rho}}\circ
C_{R_{\theta}}$.

######  Lemma 4.3.

The composition operator $C_{T_{b}}$ is bounded if and only if $C_{T_{|b|}}$
is bounded, with equal norms.

######  Lemma 4.4.

Let $r\in(0,1)$ such that $C_{T_{r}}$ is bounded. For any $b\in\mathbb{D}$
satisfying $|b|\leq\frac{2r}{1+r^{2}},$ $C_{T_{b}}$ is bounded and we have
$\|C_{T_{b}}\|\leq\|C_{T_{r}}\|^{2}$.

###### Proof.

Let $S$ be the circle $C(0,r)$ and $u\colon S\to\mathbb{R}_{+}$ be the
continuous function defined by

(4.3) $u(s)=\bigg{|}\frac{s+r}{1+\overline{s}r}\bigg{|}\,\cdot$

By connectedness, $u(S)$ contains the segment
$\big{[}0,\frac{2r}{1+r^{2}}\big{]}=[u(-r),u(r)]$. Let now $b\in
D\big{(}0,\frac{2r}{1+r^{2}}\big{)}$. By the above, there exists $s\in S$ such
that $|b|=u(s)$. That means that
$|T_{b}(0)|=|b|=|u(s)|=|T_{s}(r)|=|(T_{s}\circ T_{r})(0)|$. Therefore,
$T_{b}(0)={\rm e}^{i\alpha}(T_{s}\circ T_{r})(0)$ for some
$\alpha\in\mathbb{R}$, and hence, by Schwarz’s lemma, there is some
$\theta\in\mathbb{R}$ such that $T_{b}=R_{\alpha}\circ T_{s}\circ T_{r}\circ
R_{\theta}$. We then have $C_{T_{b}}=C_{R_{\theta}}\circ C_{T_{r}}\circ
C_{T_{s}}\circ C_{R_{\alpha}}$. Since $C_{R_{\theta}}$ and $C_{R_{\alpha}}$
are unitary, we get using Lemma 4.3 for $C_{T_{s}}$:

$\|C_{T_{b}}\|=\|C_{T_{r}}\circ
C_{T_{s}}\|\leq\|C_{T_{r}}\|\,\|C_{T_{s}}\|=\|C_{T_{r}}\|^{2}\,.\qed$

###### Proof of Proposition 4.2.

It suffices to use Lemma 4.3 and Lemma 4.4 and do an iteration, in noting that
if $r_{0}=|a|>0$ and $r_{n+1}=\frac{2r_{n}}{1+r_{n}^{2}}=r_{n}\ast r_{n}$,
then $(r_{n})_{n\geq 0}$ increases to $1$. ∎

###  4.1 An elementary necessary condition

We begin by an elementary necessary condition. It is implied by Theorem 4.9,
but its statement deserves to be pointed out. Moreover, its proof is simple
and highlights the role of the reproducing kernel.

######  Proposition 4.5.

Let $a\in(0,1)$ and assume that $T_{a}$ induces a bounded composition operator
on $H^{2}(\beta)$. Then $\beta$ has polynomial decay.

###### Proof.

Since

$\|K_{x}\|^{2}=\sum_{n=0}^{\infty}\frac{\ x^{2n}}{\beta_{n}}\,\raise
1.0pt\hbox{,}$

we have $\|K_{x}\|\leq\|K_{y}\|$ for $0\leq x\leq y<1$.

We define by induction a sequence $(u_{n})_{n\geq 0}$ with:

$u_{0}=0\qquad\text{and}\qquad u_{n+1}=T_{a}(u_{n})\,.$

Since $T_{a}(1)=1$ (recall that $a\in(0,1)$), we have:

$1-u_{n+1}=\int_{u_{n}}^{1}T_{a}^{\prime}(t)\,dt=\int_{u_{n}}^{1}\frac{1-a^{2}}{(1+at)^{2}}\,dt\,;$

hence

$\frac{1-a}{1+a}\,(1-u_{n})\leq 1-u_{n+1}\leq(1-a^{2})(1-u_{n})\,.$

Let $0<x<1$. We can find $N\geq 0$ such that $u_{N}\leq x<u_{N+1}$. Then:

$1-x\leq 1-u_{N}\leq(1-a^{2})^{N}\,.$

On the other hand, since $C_{T_{a}}^{{}^{\,\ast}}K_{z}=K_{T_{a}(z)}$ for all
$z\in\mathbb{D}$, we have:

$\|K_{x}\|\leq\|K_{u_{N+1}}\|\leq\|C_{T_{a}}\|\,\|K_{u_{N}}\|\leq\|C_{T_{a}}\|^{N+1}\|K_{u_{0}}\|=\frac{1}{\sqrt{\beta_{0}}}\|C_{T_{a}}\|^{N+1}\,.$

Let $s\geq 0$ such that $(1-a^{2})^{-s}=\|C_{T_{a}}\|$. We obtain:

(4.4) $\|K_{x}\|\leq\frac{1}{\sqrt{\beta_{0}}(1-x)^{s}}\|C_{T_{a}}\|\,.$

We get the result by using Proposition 2.6. ∎

Remarks. 1) For example, when
$\beta_{n}=\exp\big{[}-c\,\big{(}\log(n+1)\big{)}^{2}\big{]}$, with $c>0$, no
$T_{a}$ induces a bounded composition operator on $H^{2}(\beta)$, though
$C_{\varphi}$ is bounded for all symbols $\varphi$ with $\varphi(0)=0$, since
$\beta$ is decreasing, as we saw in Proposition 3.2.

2) For the Dirichlet space ${\cal D}^{2}$, we have $\beta_{n}=n+1$, but all
the maps $T_{a}$ induce bounded composition operators on ${\cal D}^{2}$ (see
[12, Remark before Theorem 3.12]). In this case $\beta$ has polynomial growth
though it is not bounded above.

3) However, even for decreasing sequences, a polynomial decay for $\beta$ is
not enough for some $T_{a}$ to induce a bounded composition operator. Indeed,
we saw in Proposition 2.5 an example of a decreasing sequence $\beta$ with
polynomial decay but not slowly oscillating, and we will see in Theorem 4.9
that this condition is needed for having some $T_{a}$ inducing a bounded
composition operator.

4) In [6], Eva Gallardo-Gutiérrez and Jonathan Partington give estimates for
the norm of $C_{T_{a}}$, with $a\in(0,1)$, when $C_{T_{a}}$ is bounded on
$H^{2}(\beta)$. More precisely, they proved that if $\beta$ is bounded above
and $C_{T_{a}}$ is bounded, then

$\|C_{T_{a}}\|\geq\bigg{(}\frac{1+a}{1-a}\bigg{)}^{\sigma}\,\raise
1.0pt\hbox{,}$

where $\sigma=\inf\\{s\geq 0\,;\ (1-z)^{-s}\notin H^{2}(\beta)\\}$, and

$\|C_{T_{a}}\|\leq\bigg{(}\frac{1+a}{1-a}\bigg{)}^{\tau}\,\raise
1.0pt\hbox{,}$

where $\tau=\frac{1}{2}\,\sup{\mathfrak{R}}{\rm e}\,W(A)$, with $A$ the
infinitesimal generator of the continuous semigroup $(S_{t})$ defined as
$S_{t}=C_{T_{\tanh t}}$, namely $(Af)(z)=f^{\prime}(z)(1-z^{2})$, and $W(A)$
its numerical range.

For $\beta_{n}=1/(n+1)^{\nu}$ with $0\leq\nu\leq 1$, the two bounds coincide,
so they get $\|C_{T_{a}}\|=\big{(}\frac{1+a}{1-a}\big{)}^{(\nu+1)/2}$.

###  4.2 Sufficient condition

The following sufficient condition explains in particular why all composition
operators $C_{T_{a}}$ are bounded on the Dirichlet space.

######  Theorem 4.6.

If $\beta$ is slowly oscillating, then all symbols that extend analytically in
a neighborhood of $\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{D}\mkern-1.5mu}\mkern 1.5mu$ induce a
bounded composition operator on $H^{2}(\beta)$.

In particular, all $C_{T_{a}}$ for $a\in\mathbb{D}$ are bounded on
$H^{2}(\beta)$.

To prove Theorem 4.6, we begin by a very elementary fact.

######  Lemma 4.7.

Let $\varphi:\mathbb{D}\to\mathbb{D}$ have an analytic extension to an open
neighborhood $\Omega$ of $\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{D}\mkern-1.5mu}\mkern 1.5mu$. Then, there
are a constant $b>0$ and an integer $\lambda>1$ such that

$|\widehat{\varphi^{n}}(m)|\leq\left\\{\begin{array}[]{lcl}{\rm
e}^{-bn}&\text{if}&n\geq\lambda m\,,\\\ {\rm e}^{-bm}&\text{if}&m\geq\lambda
n\,.\end{array}\right.$

###### Proof.

Let $R>1$ such that $\overline{D(0,R)}\subseteq\Omega$. For $0<r\leq R$, we
set

$M(r)=\sup_{|z|=r}|\varphi(z)|\,.$

Take any $r\in(0,1)$, for instance $r={\rm e}^{-1}$. We have $M(r)<1$, so we
can write $M(r)={\rm e}^{-\rho}$, for some positive $\rho$.

Cauchy’s inequalities give:

$|\widehat{\varphi^{n}}(m)|\leq\frac{[M(r)]^{n}}{r^{m}}={\rm e}^{m-\rho n}\,.$

Choose $\lambda_{1}=\max(2,2/\rho)$ and $b_{1}=\rho-\lambda_{1}^{-1}$ then
$|\widehat{\varphi^{n}}(m)|\leq{\rm e}^{-b_{1}n}$ if $n\geq\lambda_{1}m$.

For the second inequality, write $R=:{\rm e}^{\beta}$, with $\beta>0$. Let
$\alpha>0$ with $M(R)\leq{\rm e}^{\alpha}.$ Cauchy’s inequalities again give:

$|\widehat{\varphi^{n}}(m)|\leq\frac{[M(R)]^{n}}{R^{m}}\leq{\rm e}^{\alpha
n-\beta m}\,.$

Choose $\lambda_{2}=\max(2,2\alpha/\beta)$ and
$b_{2}=\beta-\alpha\lambda_{2}^{-1}$ then $|\widehat{\varphi^{n}}(m)|\leq{\rm
e}^{-b_{2}m}$ if $m\geq\lambda_{2}n$. We get the conclusion taking
$b=\min(b_{1},b_{2})$ and choosing an integer
$\lambda\geq\max(\lambda_{1},\lambda_{2})$.∎

######  Lemma 4.8.

Let $(\beta_{n})$ be a slowly oscillating sequence of positive numbers. Let
$A=(a_{m,n})_{m,n}$ be the matrix of a bounded operator on $\ell_{2}$. Assume
that, for some integer $\lambda>1$, and some constants $c$, $b$, we have:

$1)$ $|a_{m,n}|\leq c\,{\rm e}^{-bn}$ when $n\geq\lambda m$;

$2)$ $|a_{m,n}|\leq c\,{\rm e}^{-bm}$ when $m\geq\lambda n$.

Then the matrix
$\displaystyle\widetilde{A}=\bigg{(}a_{m,n}\,\sqrt{\frac{\beta_{m}}{\beta_{n}}}\bigg{)}_{m,n}$
also defines a bounded operator on $\ell_{2}$.

###### Proof.

In the sequel $\|\cdot\|$ stands for the $\ell^{2}$-norm.

Since $\beta$ is slowly oscillating, it has polynomial decay and growth: for
some $\alpha,\gamma>0$ and $\delta\in(0,1)$, we have
$\delta\,(n+1)^{-\alpha}\leq\beta_{n}\leq\delta^{-1}(n+1)^{\gamma}$.

The matrix $\widetilde{A}$ is Hilbert-Schmidt far from the diagonal since

$\displaystyle\sum_{n=1}^{\infty}\sum_{\lambda
m<n}|a_{m,n}|^{2}\,\frac{\beta_{m}}{\beta_{n}}$
$\displaystyle\lesssim\sum_{n=1}^{\infty}\sum_{\lambda
m<n}\,(n+1)^{\alpha+\gamma}|a_{m,n}|^{2}$
$\displaystyle\lesssim\sum_{n=1}^{\infty}(n+1)^{\alpha+\gamma+1}{\rm
e}^{-2bn}<+\infty\,,$

and

$\displaystyle\sum_{n=0}^{\infty}\sum_{m>\lambda
n}|a_{m,n}|^{2}\,\frac{\beta_{m}}{\beta_{n}}$
$\displaystyle\lesssim\sum_{n=0}^{\infty}\sum_{m>\lambda
n}(n+1)^{\alpha+\gamma}|a_{m,n}|^{2}$
$\displaystyle\lesssim\sum_{n=0}^{\infty}(n+1)^{\alpha+\gamma}\bigg{(}\sum_{m>\lambda
n}{\rm e}^{-2bm}\bigg{)}<+\infty\,.$

Since $\beta_{m}/\beta_{n}$ remains bounded from above and below around the
diagonal, the matrix $\widetilde{A}$ behaves like $A$ near the diagonal. More
precisely, if $I$, $J$ are blocks of integers such that $(m,n)\in I\times J$
implies that $n/\lambda^{2}\leq m\leq\lambda^{2}n$, then, with obvious
notations, the slow oscillation of $\beta$ gives, for some $C>0$:

$\displaystyle\Bigg{|}\sum_{(m,n)\in I\times
J}a_{m,n}\,x_{n}\overline{y_{m}}\,\sqrt{\frac{\beta_{m}}{\beta_{n}}}\,\Bigg{|}$
$\displaystyle\leq\|A\|\,\bigg{(}\sum_{(m,n)\in I\times
J}|x_{n}|^{2}\,|y_{m}|^{2}\,\frac{\beta_{m}}{\beta_{n}}\bigg{)}^{1/2}$
$\displaystyle\leq C^{1/2}\|A\|\,\|P_{J}x\|\,\|P_{I}y\|\,.$

For $k=0,1,2,\ldots\,,$ let $J_{k}=[\lambda^{k},\lambda^{k+1}[\,$ and, for
$k=1,2,\ldots\,$, we define $I_{k}=[\lambda^{k-1},\lambda^{k+2}[$. We also
define $I_{0}=[0,\lambda^{2}[$.

We define the matrix $R$, whose entries are

$\displaystyle\phantom{+1}r_{m,n}$
$\displaystyle=\left\\{\begin{array}[]{lcl}\displaystyle\sqrt{\frac{\beta_{m}}{\beta_{n}}}\,a_{m,n}&\text{if}&\displaystyle(m,n)\in\bigcup_{k=0}^{\infty}(I_{k}\times
J_{k})\\\ \quad 0&\text{elsewhere.}\end{array}\right.$

Let $H_{k}$ be the subspace of the sequences $(x_{n})_{n\geq 0}$ in $\ell_{2}$
such that $x_{n}=0$ for $n\notin I_{k}$, i.e. $H_{k}={\rm span}\,\\{e_{n}\,;\
n\in I_{k}\\}$, and ${\widetilde{H}}_{k}={\rm span}\,\\{e_{n}\,;\ n\in
J_{k}\\}$. Let $P_{k}$ be (the matrix of) the orthogonal projection of
$\ell_{2}$ with range $H_{k}$ and $Q_{k}$ that with range
${\widetilde{H}}_{k}$. Then $R_{k}=P_{k}AQ_{k}$ is the matrix with entries
$a_{m,n}$ when $(m,n)\in I_{k}\times J_{k}$ and $0$ elsewhere. By the above
discussion, we have

$|(R_{k}x\mid y)|\leq C^{1/2}\|A\|\,\|Q_{k}x\|\,\|P_{k}y\|\,.$

Point out that, for every $y\in\ell^{2}$, we have
$\displaystyle\sum\|P_{k}y\|^{2}\leq 3\|y\|^{2}$ since each integer belongs to
at most $3$ intervals $I_{k}$.

In the same way, for every $x\in\ell^{2}$, we have
$\displaystyle\sum\|Q_{k}x\|^{2}\leq\|x\|^{2}$ since the subspaces
$\widetilde{H}_{k}$ are orthogonal.

Summing up over $k$, we get the boundedness of
$R=\\!\sum_{k=0}^{\infty}R_{k}$.

Now let us check when the entries of $R$ do not coincide with the entries of
$\widetilde{A}$. Actually it happens when $(m,n)$ does not belong to the union
of the $I_{k}\times J_{k}$. When $n\geq 1$, it means that $n$ belongs to some
$J_{p}$ but $m\notin I_{p}$: either $m<\lambda^{p-1}$ or $m\geq\lambda^{p+2}$,
hence either $m/n<\lambda^{-1}$ or $m/n>\lambda$. Therefore the non zero
entries $(m,n)$ of $\widetilde{A}-R$ satisfy either $n>\lambda m$ or
$m>\lambda n$.

That ends the proof since we have seen at the beginning that $\widetilde{A}-R$
is Hilbert-Schmidt. ∎

Remark. The proof shows that, instead of $1)$ and $2)$, it is enough to have:

$\sum_{m<C_{1}n}n^{\alpha+1}|a_{m,n}|^{2}<\infty\quad\text{and}\quad\sum_{m>C_{2}n}m^{\alpha}|a_{m,n}|^{2}<\infty\,.$

Moreover the proof also shows that when $\beta$ is slowly oscillating, if we
set $E=\\{(m,n)\,;\ C_{1}n\leq m\leq C_{2}n\\}$, for some $C_{1}$,$C_{2}>0$,
then the matrix
$\big{(}\sqrt{\beta_{m}/\beta_{n}}\,\mathds{1}_{E}(m,n)\big{)}$ is a Schur
multiplier over _all_ the bounded matrices, while Kacnel’son’s theorem
(Theorem 3.5) says that, if $\gamma=(\gamma_{n})$ is non-increasing, the
matrix $(\gamma_{m}/\gamma_{n})$ is a Schur multiplier of all bounded _lower-
triangular_ matrices.

###### Proof of Theorem 4.6.

Thanks to Lemma 4.7, the hypotheses of Lemma 4.8 are fulfilled by the matrix
whose entries are $a_{m,n}=\widehat{\varphi^{n}}(m)$. It follows (with the
notations of Lemma 4.8) that $\widetilde{A}$ is bounded on $\ell^{2}$, which
means exactly that $T_{a}$ is bounded on $H^{2}(\beta)$. ∎

###  4.3 Necessary condition

The main theorem of this section is the following.

######  Theorem 4.9.

If the composition operator $C_{T_{a}}$ is bounded on $H^{2}(\beta)$ for some
$a\in\mathbb{D}\setminus\\{0\\}$, then $\beta$ is slowly oscillating.

Let us give right now a corollary of this result.

######  Proposition 4.10.

For the weight $\beta$ constructed in the proof of Theorem 3.7, no
automorphism $T_{a}$ with $0<a<1$ can be bounded.

###### Proof.

Indeed, it is clear that $\beta$ is not slowly oscillating, since

$\beta_{(k+1)!-1}/\beta_{(k+1)!}=1/(k+1)\mathop{\longrightarrow}\limits_{k\to\infty}0\,.\qed$

To prove Theorem 4.9, we need estimates on the Taylor coefficients of
$T_{a}^{n}$. Actually the Taylor coefficients of $T_{a}^{n}$ are the Fourier
coefficients of $x\in\mathbb{R}\mapsto T_{a}^{n}({\rm e}^{ix})$ and we shall
denote them with the same notation $\widehat{T_{a}^{n}}$. Sharp such estimates
are given in the papers [20] and [21], and we thank R. Zarouf for interesting
informations in this respect. Our method, using stationary phase and the van
der Corput lemma, is a variant of that used in [20], [21], and goes back at
least to [7]. However, we need minorations of $|\widehat{T_{a}^{n}}(m)|$ when
$m$ is close to $n$, and Szehr and Zarouf’s estimates show that this quantity
oscillates and, for individual $a$, can be too small for our purpose, so we
cannot use them, and have to prove an estimate _in mean_ for $a$ in some
subinterval of $(0,1)$.

We begin with a standard fact, that we give with its proof for the convenience
of the reader.

######  Lemma 4.11.

Let $a\in(0,1)$ and let

$P_{-a}(x)=\frac{1-a^{2}}{1+2\,a\cos x+a^{2}}$

be the Poisson kernel at the point $-a$. Then, for all $x\in[-\pi,\pi]$:

(4.5) $T_{a}({\rm e}^{ix})=\exp\big{[}i\,V_{a}(x)\big{]}\,,$

where

(4.6) $V_{a}(x)=\int_{0}^{x}P_{-a}(t)\,dt\,.$

###### Proof.

For $t\in[-\pi,\pi]$, write:

$\psi(t):=\frac{{\rm e}^{it}+a}{1+a\,{\rm e}^{it}}=\exp\,(i\,v(t))\,,$

with $v$ a real-valued, ${\cal C}^{1}$-function on $[-\pi,\pi]$ such that
$v(0)=0$. This is possible since $|\psi({\rm e}^{it})|=1$ and $\psi(0)=1$.
Differentiating both sides with respect to $t$, we get:

$i\,{\rm e}^{it}\frac{1-a^{2}}{(1+a\,{\rm
e}^{it})^{2}}=i\,v^{\prime}(t)\,\frac{{\rm e}^{it}+a\,}{1+a\,{\rm
e}^{it}}\,\cdot$

This implies

$v^{\prime}(t)=\frac{1-a^{2}}{|1+a\,{\rm e}^{it}|^{2}}=P_{-a}(t)\,,$

and the result follows since $v(0)=0=V_{a}(0)$. ∎

Let us note that, with $V_{a}$ the function of Lemma 4.11, the Fourier
formulas give, since $\widehat{T_{a}^{n}}(m)$ is real, or since $nV_{a}(x)-mx$
is odd:

(4.7)
$2\pi\,\widehat{T_{a}^{n}}(m)=\int_{-\pi}^{\pi}\exp(i[nV_{a}(x)-mx])\,dx=2\,{\mathfrak{R}}{\rm
e}\,I_{m,n}\,,$

where

(4.8) $I_{m,n}=\int_{0}^{\pi}\exp i[nV_{a}(x)-mx]\,dx\,.$

Now the main ingredient for proving Theorem 4.9 is the following.

######  Proposition 4.12.

Let $I:=\big{[}\frac{1}{2}\raise 1.0pt\hbox{,}\frac{2}{3}\big{]}$. There exist
constants $\alpha>1$, e.g. $\alpha=5/4$, and $\delta\in(0,1/2)$ such that, for
$n$ large enough ($n\geq n_{0}$), it holds

(4.9)
$\qquad\qquad\int_{I}|\widehat{T_{a}^{n}}(m)|^{2}\,da\geq\frac{\delta}{n}\qquad\text{for
all }m\in[\alpha^{-1}n,\alpha\,n]\,.\qquad{}$

###### Proof.

We will set once and for all

(4.10) $q=\frac{m}{n}\,\raise 1.0pt\hbox{,}$

so that $\alpha^{-1}\leq q\leq\alpha$ where $\alpha=5/4$ (say). We will only
consider pairs $(a,q)$ satisfying

(4.11) $a\in I=\bigg{[}\frac{1}{2}\,\raise
1.5pt\hbox{,}\,\frac{2}{3}\bigg{]}\,,\qquad q\in
J:=\bigg{[}\frac{4}{5}\,\raise 1.5pt\hbox{,}\,\frac{5}{4}\bigg{]}\,\cdot$

Such pairs will be called _admissible_.

With these notations, we set, for $0\leq x\leq\pi$:

(4.12) $F_{q}(x)=V_{a}(x)-\frac{m}{n}x=\int_{0}^{x}P_{-a}(t)\,dt-qx\,,$

where $P_{-a}$ is the Poisson kernel at $-a$. We have

$F^{\prime}_{q}(x)=\frac{(1-a^{2})}{1+2a\cos x+a^{2}}-q\,.$

and the unique (if some) critical point $x_{q}=x_{q}(a)$ of $F_{q}$ in
$[0,\pi]$ is given by $P_{-a}(x_{q})=q$, that is

(4.13) $\cos
x_{q}=\frac{1}{q}\,\frac{1-a^{2}}{2a}-\frac{1+a^{2}}{2a}=:h_{q}(a)\,.$

We now proceed through a series of simple lemmas, and begin by estimates on
$h_{q}$ and $x_{q}$.

######  Lemma 4.13.

There are positive constants $C>1$ and $\delta\in(0,1/2)$ such that, for every
admissible couple $(a,q)$, we have:

(4.14) $\qquad|h_{q}(a)|\leq
1-\delta\quad\text{and}\quad|h^{\prime}_{q}(a)|\leq C\,,$

so, there is one critical point $x_{q}(a)$ satisfying

(4.15) $\qquad\delta\leq x_{q}(a)\leq\pi-\delta\quad\text{and}\quad\sin
x_{q}(a)\geq\delta\,;$

moreover

(4.16) $\qquad|x^{\prime}_{q}(a)|\leq
C\quad\text{and}\quad\delta\leq|P^{\prime}_{-a}(x_{q})|\leq C\,.$

###### Proof.

We have

$h_{q}(a)=\bigg{(}\frac{1}{q}\,\frac{1-a^{2}}{2a}\bigg{)}+\bigg{(}-\frac{1+a^{2}}{2a}\bigg{)}=:u(a)+v(a)\,,$

with $u$ and $v$ respectively decreasing and increasing on $[0,1]$, and with
$v\leq 0$, so that we have, for $q\in J$:

$h_{q}(a)\leq u\Big{(}\frac{1}{2}\Big{)}=\frac{3}{4q}\leq\frac{15}{16}\,\cdot$

Similarly:

$h_{q}(a)\geq
u\Big{(}\frac{2}{3}\Big{)}+v\Big{(}\frac{1}{2}\Big{)}=\frac{5}{12q}-\frac{5}{4}\geq\frac{1}{3}-\frac{5}{4}=-\frac{11}{12}\,\cdot$

Next,
$2h^{\prime}_{q}(a)=\big{(}1-\frac{1}{q}\big{)}\,\frac{1\,}{a^{2}}-\big{(}1+\frac{1}{q}\big{)}\,$;
hence $|h^{\prime}_{q}(a)|\leq C$. So that, writing $x_{q}=x_{q}(a)=\arccos
h_{q}(a)$, we get, with another constant $C>0$,

$|x^{\prime}_{q}(a)|=\frac{|h^{\prime}_{q}(a)|}{\sqrt{1-h_{q}(a)^{2}}}\leq
C\,,$

since $h_{q}(a)^{2}\leq 1-\delta$. Finally, $1/9\leq(1-a)^{2}\leq 1+2a\cos
x_{q}+a^{2}\leq 4$ and since

$P^{\prime}_{-a}(x_{q})=\frac{2a(1-a^{2})\sin x_{q}}{(1+2a\cos
x_{q}+a^{2})^{2}}\,\raise 1.0pt\hbox{,}$

we get the final estimates, ending the proof. ∎

Back to Proposition 4.12.

We saw in (4.7) that the value of $a_{m,n}:=\widehat{T_{a}^{n}}(m)$ is given
by the formula

(4.17) $a_{m,n}=\frac{1}{\pi}\,{\mathfrak{R}}{\rm e}\,I_{m,n}\,.$

We have the following estimate, whose proof is postponed (recall that $q=m/n$,
and $x_{q}=x_{q}(a)$).

######  Proposition 4.14.

We have

(4.18) $I_{m,n}=\sqrt{2\pi}\,n^{-1/2}\,\,\frac{{\rm
e}^{i\,[nF_{q}(x_{q})+\pi/4]}}{\sqrt{|P^{\prime}_{-a}(x_{q})|}}+O\,(n^{-3/5})\,,$

where the $O$ only depends on $a$, and so is absolute as long as $(a,q)$ is
admissible.

Note that $3/5>1/2$. We hence have:

(4.19)
$a_{m,n}=\sqrt{\frac{2}{\pi}}\,n^{-1/2}\,\,\frac{\cos\,[\pi/4+nF_{q}(x_{q})]}{\sqrt{|P_{-a}^{\prime}(x_{q})|}}+O\,(n^{-3/5})\,,$

It will be convenient to introduce $\varphi_{q}(a)$, by settting

(4.20) $F_{q}\big{(}x_{q}(a)\big{)}=\varphi_{q}(a)\,.$

Then, since $1/2+3/5=11/10$, $\cos^{2}(\pi/4+x)=\frac{1-\sin 2x}{2}$ and
$|P^{\prime}_{-a}(x_{q})|\geq\delta$ by Lemma 4.13:

$a_{m,n}^{\,2}=\frac{1}{\pi}\,n^{-1}\,\frac{1-\sin\,[2n\,\varphi_{q}(a)]}{|P_{-a}^{\prime}(x_{q})|}+O\,(n^{-11/10})\,,$

implying, since $|P^{\prime}_{-a}(x_{q})|\leq C$ by Lemma 4.13 (again for
$(a,q)$ admissible), and changing $\delta$:

(4.21)
$a_{m,n}^{\,2}\geq\delta\,n^{-1}(1-\sin\,[2n\,\varphi_{q}(a)])+O\,(n^{-11/10})\,.$

We will also need estimates on the derivatives of $\varphi_{q}(a)$.

######  Lemma 4.15.

If $(a,q)$ is admissible, then $\varphi_{q}$ decreases on $I$ and moreover:

$1)$ $\quad|\varphi^{\prime}_{q}(a)|\geq\delta$;

$2)$ $|\varphi^{\prime\prime}_{q}(a)|\leq C$.

###### Proof.

Note, in passing, that, with $x=x_{q}(a)\in[0,\pi]$ (thanks to (4.12)):

$\varphi_{q}(a)=\int_{0}^{x}[P_{-a}(t)-P_{-a}(x)]\,dt\leq 0\,,$

since the integrand is negative. Next, if $f$ and $g$ are real ${\cal
C}^{1}$-functions and

$\Phi(a)=\int_{0}^{f(a)}g(a,t)\,dt\,,$

the chain rule gives

$\Phi^{\prime}(a)=f^{\prime}(a)\,g\big{(}a,f(a)\big{)}+\int_{0}^{f(a)}\frac{\partial
g}{\partial a}\,(a,t)\,dt\,.$

With $g(a,t)=P_{-a}(t)$ and $f(a)=x_{q}(a)$, we get, remembering that
$x_{q}(a)$ is critical for $F_{q}$:

$\displaystyle\varphi^{\prime}_{q}(a)$
$\displaystyle=\big{[}P_{-a}\big{(}x_{q}(a)\big{)}-q\big{]}\,x_{q}^{\prime}(a)+\int_{0}^{x_{q}(a)}\frac{\partial
P_{-a}}{\partial a}\,(a,t)\,dt$
$\displaystyle=\int_{0}^{x_{q}(a)}\frac{\partial P_{-a}}{\partial
a}\,(a,t)\,dt\,.$

But $P_{-a}(t)=1+2\sum_{k=1}^{\infty}(-a)^{k}\cos kt$, so that

$\varphi^{\prime}_{q}(a)=\int_{0}^{x_{q}(a)}\bigg{(}-2\sum_{k=1}^{\infty}k\,(-a)^{k-1}\cos
kt\bigg{)}\,dt=\frac{2}{a}\,\sum_{k=0}^{\infty}(-a)^{k}\sin\,[kx_{q}(a)]\,,$

that is

(4.22) $\varphi^{\prime}_{q}(a)=\frac{2}{a}\,{\mathfrak{I}}{\rm
m}\,\frac{1}{1+a\,{\rm e}^{ix_{q}(a)}}=\frac{-2\sin x_{q}(a)}{1+2a\cos
x_{q}(a)+a^{2}}<0\,.$

Now, (4.15) gives $1)$.

Since $|x^{\prime}_{q}(a)|\leq C$ by Lemma 4.13, the chain rule and (4.22)
clearly give the uniform boundedness of $|\varphi^{\prime\prime}_{q}(a)|$ when
$(a,q)$ is admissible, and this ends the proof. ∎

Lemma 4.13 and Lemma 4.15 will now be exploited through a simple variant of
the van der Corput inequalities.

######  Lemma 4.16.

Let $f\colon[A,B]\to\mathbb{R}$, with $A<B$, be a ${\cal C}^{2}$-function
satisfying $|f^{\prime}|\geq\delta$ and $|f^{\prime\prime}|\leq C$, and let us
put $M=\int_{A}^{B}{\rm e}^{inf(x)}\,dx$. Then

$|M|\leq\frac{2}{n\delta}+\frac{C(B-A)}{n\,\delta^{2}}\,\cdot$

###### Proof.

Write

${\rm e}^{inf}=\frac{({\rm e}^{inf})^{\prime}}{inf^{\prime}}$

and integrate by parts to get

$M=\bigg{[}\frac{{\rm
e}^{inf}}{inf^{\prime}}\bigg{]}_{A}^{B}-\frac{i}{n}\int_{A}^{B}{\rm
e}^{inf(x)}\frac{f^{\prime\prime}(x)}{[f^{\prime}(x)]^{2}}\,dx=:M_{1}+M_{2}\,,$

with $|M_{1}|\leq\frac{2}{n\delta}$ and
$|M_{2}|\leq\frac{B-A}{n}\,\frac{C}{\delta^{2}}$. This gives the result. ∎

_End of proof of Proposition 4.12_. The preceding lemma can be applied with
$A=1/2$, $B=2/3$, $f=\varphi_{q}$ and $n$ changed into $2n$, since Lemma 4.15
shows that this $f$ meets the assumptions of Lemma 4.16. This gives us,
uniformly with respect to $(a,q)$ admissible:

(4.23)
$\bigg{|}\int_{I}\sin\,[2n\varphi_{q}(a)]\,da\bigg{|}\leq\bigg{|}\int_{I}{\rm
e}^{2in\varphi_{q}(a)}\,da\bigg{|}\leq\frac{C}{n}\,\cdot$

Now, integrating (4.21) on $I$ and using (4.23) give for some numerical
$\delta\in(0,1/2)$ (recall that $a_{m,n}=\widehat{T_{a}^{n}}(m)$):

$\int_{I}|\widehat{T_{a}^{n}}(m)|^{2}\,da\geq\delta
n^{-1}+O\,(n^{-2})+O\,(n^{-11/10})\geq(\delta/2)\,n^{-1}$

for $n\geq n_{0}$ and $\alpha^{-1}\leq m/n\leq\alpha$. This ends the proof of
Proposition 4.12. ∎

###### Proof of Theorem 4.9.

We know, by Proposition 4.2, that $C_{T_{a}}$ is bounded for all
$a\in\mathbb{D}$, and that, thanks to Lemma 4.4,

$K:=\sup_{1/2\leq a\leq 2/3}\|C_{T_{a}}\|<+\infty\,.$

Matricially, this writes, for all $a\in(1/2,2/3)$:

$\bigg{\|}\bigg{(}\widehat{T_{a}^{n}}(m)\,\sqrt{\frac{\beta_{m}}{\beta_{n}}}\bigg{)}_{m,n}\bigg{\|}\leq
K\,.$

In particular, for every $n\geq 1$, we have, considering the columns and rows
of the previous matrix:

$\sum_{m=1}^{\infty}|\widehat{T_{a}^{n}}(m)|^{2}\,\frac{\beta_{m}}{\beta_{n}}\leq
K^{2}\,,\quad\text{i.
e.}\quad\sum_{m=1}^{\infty}|\widehat{T_{a}^{n}}(m)|^{2}\,\beta_{m}\leq
K^{2}\,\beta_{n}\,,$

and, for every $m\geq 1$:

$\sum_{n=1}^{\infty}|\widehat{T_{a}^{n}}(m)|^{2}\,\frac{\beta_{m}}{\beta_{n}}\leq
K^{2}\,,\quad\text{i.
e.}\quad\sum_{n=1}^{\infty}|\widehat{T_{a}^{n}}(m)|^{2}\,\frac{1}{\beta_{n}}\leq\frac{K^{2}}{\beta_{m}}\,\cdot$

In particular, for every $n\geq 1$:

(4.24) $\sum_{(4/5)n\leq
j\leq(5/4)n}|\widehat{T_{a}^{n}}(j)|^{2}\,\beta_{j}\leq
K^{2}\,\beta_{n}\,\cdot$

and, for every $m\geq 1$:

(4.25) $\sum_{(4/5)m\leq
k\leq(5/4)m}|\widehat{T_{a}^{k}}(m)|^{2}\,\frac{1}{\beta_{k}}\leq\frac{K^{2}}{\beta_{m}}\,\cdot$

Integrating on $a\in(1/2,2/3)$, and using Proposition 4.12, we get, from
(4.24), for $n$ large enough,

(4.26) $\frac{\delta}{n}\sum_{(4/5)n\leq
j\leq(5/4)n}\beta_{j}\leq\frac{K^{2}}{6\,}\,\beta_{n}$

and, from (4.25), for $m$ large enough, we have both

(4.27) $\frac{\delta}{m}\sum_{(4/5)m\leq k\leq
m}\frac{1}{\beta_{k}}\leq\frac{5K^{2}}{24}\,\frac{1}{\beta_{m}}\,\cdot$

and

(4.28) $\frac{\delta}{m}\sum_{m\leq
k\leq(5/4)m}\frac{1}{\beta_{k}}\leq\frac{5K^{2}}{24}\,\frac{1}{\beta_{m}}\,\cdot$

Since the harmonic mean (over the sets of integers $[4m/5,m]$ and $[m,5m/4]$,
which have cardinality $\approx n\approx m$) is less than the arithmetical
mean, we obtain from (4.27) and (4.28) both

(4.29) $\beta_{m}\leq\frac{125}{24\delta}\,\frac{K^{2}}{m}\sum_{(4/5)m\leq
k\leq m}\beta_{k}\,.$

and

(4.30) $\beta_{m}\leq\frac{10}{3\delta}\,\frac{K^{2}}{m}\sum_{m\leq
k\leq(5/4)m}\beta_{k}\,.$

Now assume that $n\leq m\leq(5/4)n$. From (4.29), we have

$\beta_{m}\lesssim\,\frac{1}{m}\sum_{(4/5)m\leq k\leq
m}\beta_{k}\lesssim\frac{1}{n}\sum_{(4/5)n\leq
k\leq(5/4)n}\beta_{k}\lesssim\beta_{n}$

thanks to (4.26).
From (4.30) and (4.26), we treat the case $(4/5)n\leq m\leq n$ in the same
way.
We conclude that for some constant $c>0$, we have for $n$ and $m$ large enough
satisfying $(4/5)n\leq m\leq(5/4)n$:

(4.31) $\beta_{m}\leq c\,\beta_{n}\,,$

which means that $\beta$ is slowly oscillating. ∎

###### Proof of Proposition 4.14.

We will use a variant of [22, Lemma 4.6, p. 72] on the van der Corput’s
version of the stationary phase method. A careful reading of the proof in [22,
page 72] gives the version below, which only needs local estimates on the
second derivative $F^{\prime\prime}$, as occurs in our situation. For sake of
completeness, we will give a proof, however postponed to the Appendix.

######  Proposition 4.17 (Stationary phase).

Let $F$ be a real function with continuous derivatives up to the third order
on the interval $[A,B]$ and $F^{\prime\prime}>0$ throughout $]A,B[\,$. Assume
that there is a (unique) point $c$ in $]A,B[$ such that $F^{\prime}(c)=0$, and
that, for some positive numbers $\lambda_{2}$, $\lambda_{3}$, and $\eta$, the
following assertions hold:

  1. $1)$

$[c-\eta,c+\eta]\subseteq[A,B]$;

  2. $2)$

$F^{\prime\prime}(x)\geq\lambda_{2}$ for all $x\in[c-\eta,c+\eta]$;

  3. $3)$

$|F^{\prime\prime\prime}(x)|\leq\lambda_{3}$ for all $x\in[A,B]$.

Then:

(4.32) $\int_{A}^{B}{\rm e}^{iF(x)}\,dx=\sqrt{2\pi}\,\,\frac{\,{\rm
e}^{i[F(c)+\pi/4]}\,}{F^{\prime\prime}(c)^{1/2}}+O\,\bigg{(}\frac{1}{\eta\lambda_{2}}+\eta^{4}\lambda_{3}\bigg{)}\,,$

where the $O$ involves an absolute constant.

We will show that Proposition 4.17 is applicable with $F=nF_{q}$ and

$[A,B]=[0,\pi]\,,\quad
c=x_{q}\,,\quad\lambda_{2}=\kappa_{0}\,n\,,\quad\lambda_{3}=C_{0}n\,,\quad\eta=(\lambda_{2}\lambda_{3})^{-1/5}\,.$

The parameter $\eta$ is chosen in order to make both error terms in
Proposition 4.17 equal: $\frac{1}{\eta\lambda_{2}}=\eta^{4}\lambda_{3}$; so:

$\eta=\kappa\,n^{-2/5}$

and

(4.33)
$\frac{1}{\eta\lambda_{2}}+\eta^{4}\lambda_{3}=\widetilde{\kappa}\,n^{-3/5}=O\,(n^{-3/5})$

(with $\kappa=(\kappa_{0}C_{0})^{-1/5}$ and
$\widetilde{\kappa}=2/\kappa_{0}\kappa$).

The slight technical difficulty encountered here is that
$F_{q}^{\prime\prime}(x)$ vanishes at $0$ and $\pi$. But Proposition 4.17
covers this case. We have

$F^{\prime\prime}(x)=nF_{q}^{\prime\prime}(x)=nP^{\prime}_{-a}(x)=2\,a(1-a^{2})\,\frac{\sin
x}{(1+2a\cos x+a^{2})^{2}}\,n\,,$

and there are some positive (and absolute) constants $\kappa_{0}$ and $\sigma$
such that

(4.34) $F^{\prime\prime}(x)\geq\kappa_{0}\,n=\lambda_{2}\quad\text{for
}x\in[\sigma,\pi-\sigma]\,.$

Now (for $n$ large enough),
$[x_{q}-\eta,x_{q}+\eta]\subseteq[\sigma,\pi-\sigma]$. Hence the assumptions
$1)$ and $2)$ of Proposition 4.17 are satisfied.

Finally, since $F(x)=nF_{q}(x)=n[V_{a}(x)-qx]$,
$F^{\prime\prime\prime}=nF_{q}^{\prime\prime\prime}=nV_{a}^{\prime\prime\prime}=nP_{-a}^{\prime\prime}$,
we have, for all $x\in[0,\pi]$ and $(a,q)$ admissible:

$|F^{\prime\prime\prime}(x)|\leq C_{0}\,n=\lambda_{3}\,,$

where $C_{0}$ is absolute and assertion $3)$ of Proposition 4.17 holds.

With (4.33) this ends the proof of (4.18), once we remarked that
$nV_{a}^{\prime\prime}(x_{q})=F^{\prime\prime}(x_{q})$. ∎

##  5 Boundedness of all composition operators

In this section, we characterize all the sequences $\beta$ for which all
composition operators are bounded on $H^{2}(\beta)$. The main remaining step
is the following theorem.

######  Theorem 5.1.

Assume that all composition operators $C_{\varphi}$ are bounded on
$H^{2}(\beta)$. Then $\beta$ is essentially decreasing.

As an immediate consequence, we obtain Theorem 1.2.

###### Proof of Theorem 1.2.

Assume that $\beta$ is essentially decreasing and slowly oscillating. All
composition operators $C_{\psi}$ with $\psi(0)=0$ are bounded on
$H^{2}(\beta)$ (see the Introduction or Proposition 3.2). Since $\beta$ is
slowly oscillating, all the composition operators $C_{T_{a}}$, with
$a\in\mathbb{D}$, are bounded thanks to Theorem 4.6. Now it is very classical
that we can get the boundedness of every composition operators. Indeed given a
symbol $\varphi$, the symbol $\psi=T_{a}\circ\varphi$ fixes the origin for
$a=-\varphi(0)$. Since $C_{\varphi}=C_{\psi}\circ C_{T_{-a}}$, the conclusion
follows.

Assume that all composition operators are bounded on $H^{2}(\beta)$, in
particular, the $C_{T_{a}}$ ones are so, and $\beta$ is slowly oscillating,
thanks to Theorem 4.9. It also follows from Theorem 5.1 that $\beta$ is
essentially decreasing. ∎

We will use the following elementary, but crucial, lemma.

######  Lemma 5.2.

Let $u$ be a function analytic in an open neighborhood $\Omega$ of
$\overline{\mathbb{D}}$. Then, for every $\varepsilon>0$, there exists an
integer $N\geq 1$ such that

(5.1)
$\sum_{j=Np}^{\infty}|\widehat{u^{p}}(j)|^{2}\leq\varepsilon\,,\quad\forall
p\geq 1\,.$

###### Proof.

From Lemma 4.7, we know that there exist some integer $\lambda>1$ and a
constant $b>0$, such that $|\widehat{u^{p}}(j)|\leq{\rm e}^{-bj}$ when
$j\geq\lambda p$. Therefore, for any $N\geq\lambda$, we have

$\sum_{j=Np}^{\infty}|\widehat{u^{p}}(j)|^{2}\leq\big{(}1-{\rm
e}^{-2b}\big{)}^{-1}{\rm e}^{-2bNp}\leq\big{(}1-{\rm e}^{-2b}\big{)}^{-1}{\rm
e}^{-2bN}\leq\varepsilon\,,$

as soon as $N$ is chosen large enough. ∎

###### Proof of Theorem 5.1.

Thanks to Theorem 4.9, we know that $\beta$ is slowly oscillating. Therefore
we shall assume that the sequence $\beta$ is slowly oscillating but is not
essentially decreasing. We are going to construct an analytic function
$\varphi\colon\mathbb{D}\to\mathbb{D}$ such that the composition operator
$C_{\varphi}$ is not bounded on $H^{2}(\beta)$. This function $\varphi$ will
be a Blaschke product, of the form

$\varphi(z)=\prod_{k=1}^{\infty}T_{a_{k}}(z^{n_{k}})=\prod_{k=1}^{\infty}\frac{\\!z^{n_{k}}+a_{k}}{\,1+a_{k}z^{n_{k}}}\,\raise
1.0pt\hbox{,}$

for a sequence of numbers $a_{k}\in(0,1)$ such that $\sum_{k\geq
1}(1-a_{k})<\infty$ and a sequence of positive integers $n_{k}$ increasing to
infinity. We recall that $T_{a}(z)=\frac{z+a}{1+az}$ for $0<a<1$. Then,

$|T_{a_{k}}(z^{n_{k}})-1|\leq\frac{2(1-a_{k})}{1-|z|}$

and $\varphi$ is a locally uniformly convergent infinite product. Observe that
$\varphi$ is indeed a convergent Blaschke product, with $n_{k}$ zeroes of
modulus $a_{k}^{1/n_{k}},\ k=1,2,\ldots,$ and, setting $a_{k}={\rm
e}^{-\varepsilon_{k}}$,
$\sum_{k}n_{k}(1-a_{k}^{1/n_{k}})\leq\sum_{k}n_{k}(\varepsilon_{k}/n_{k})=\sum_{k}\varepsilon_{k}<\infty$.

These sequences will be constructed by induction, together with another
sequence of integers $(m_{k})_{k\geq 1}$.

Since $\beta$ is not essentially decreasing, there exist integers
$n_{1}>m_{1}\geq 4$ such that $\beta_{n_{1}}\geq 2\,\beta_{m_{1}}$. We start
with

$a_{1}=1-\frac{1}{m_{1}}\geq\frac{3}{4}\,\cdot$

Using Lemma 5.2 with $u=T_{a_{1}}$, we get $N_{0}\geq 1$ such that

$\qquad\sum_{j=N_{0}m}^{\infty}|\widehat{T_{a_{1}}^{m}}(j)|^{2}\leq
2^{-15}\,,\quad\forall m\geq 1\,.$

Assume now that we have constructed increasing sequences of integers

$m_{1},m_{2},\ldots,m_{k},\quad n_{1},n_{2},\ldots,n_{k},\quad
N_{0},N_{1},\ldots,N_{k-1}$

such that, for $1\leq l\leq k-1$, we have

$m_{l+1}\geq 4m_{l}\quad\text{and}\quad n_{l+1}\geq 4n_{l}$

and, for $1\leq l\leq k$:

$n_{l}\geq N_{l-1}m_{l}\quad\text{and}\quad\beta_{n_{l}}\geq
2^{l}\beta_{m_{l}}$

and

$\sum_{j=N_{l-1}m_{l}}^{\infty}|\widehat{\varphi_{l}^{m}}(j)|^{2}\leq
2^{-15}\,,$

where

$a_{l}=1-\frac{1}{m_{l}}\quad\text{and}\quad\varphi_{l}(z)=T_{a_{l}}(z^{n_{l}})\,.$

We then apply Lemma 5.2 again, to the function
$u=u_{k}=\varphi_{1}\cdots\varphi_{k}$. We get $N_{k}>N_{k-1}$ such that

(5.2) $\quad\sum_{j=N_{k}m}^{\infty}|\widehat{u_{k}^{m}}(j)|^{2}\leq
2^{-15}\,,\quad\forall m\geq 1\,.$

Since $\beta$ is not essentially decreasing but is slowly oscillating, there
exist $m_{k+1}\geq 4m_{k}$ and $n_{k+1}\geq 4n_{k}$ such that

$n_{k+1}\geq N_{k}m_{k+1}\quad\text{and}\quad\beta_{n_{k+1}}\geq
2^{k+1}\beta_{m_{k+1}}\,.$

We set

$a_{k+1}=1-\frac{1}{m_{k+1}}\quad\text{and}\quad\varphi_{k+1}(z)=T_{a_{k+1}}(z^{n_{k+1}})\,.$

That ends the induction.

Now, since
$\sum_{k=1}^{\infty}(1-a_{k})=\sum_{k=1}^{\infty}\frac{1}{m_{k}}\leq\sum_{k=1}^{\infty}4^{-k}=1/3<+\infty$,
the infinite product $\varphi=\prod_{k\geq 1}\varphi_{k}$ converges uniformly
on compact subsets of $\mathbb{D}$.

To show that the composition operator $C_{\varphi}$ is not bounded on
$H^{2}(\beta)$, it suffices to show that, for some constant $c_{1}>0$, we
have, for all $k\geq 2$:

(5.3) $\sum_{j=n_{k}}^{2n_{k}}|\widehat{\varphi^{m_{k}}}(j)|^{2}\geq c_{1}\,.$

Indeed, since $\beta$ is slowly oscillating, there is a positive constant
$\delta<1$ such that

$\qquad\beta_{j}\geq\delta\,\beta_{n_{k}}\quad\text{for}\quad
j=n_{k},n_{k}+1,\ldots,2n_{k}\,.$

Then, if we set $e_{k}(z)=z^{m_{k}}$, we have, since
$C_{\varphi}(e_{k})=\varphi^{m_{k}}$:

$\frac{\|C_{\varphi}(e_{k})\|_{H^{2}(\beta)}^{2}}{\|e_{k}\|_{H^{2}(\beta)}^{2}}\geq\frac{\sum_{j=n_{k}}^{2n_{k}}|\widehat{\varphi^{m_{k}}}(j)|^{2}\beta_{j}}{\beta_{m_{k}}}\geq\frac{c_{1}\delta\,\beta_{n_{k}}}{\beta_{m_{k}}}\geq
2^{k}c_{1}\delta\mathop{\longrightarrow}\limits_{k\to\infty}+\infty\,,$

so that $C_{\varphi}$ is not bounded on $H^{2}(\beta)$.

We now have to show (5.3). Let us agree to write formally, for an analytic
function $f(z)=\sum_{k=0}^{\infty}f_{k}z^{k}$ and an arbitrary positive
integer $p$:

$f(z)=\sum_{k=0}^{p}f_{k}z^{k}+O(z^{p+1}).$

For that, we set

$G_{k}(z)=\prod_{l=k+1}^{\infty}\varphi_{l}(z)=\prod_{l=k+1}^{\infty}a_{l}+O\,(z^{n_{k+1}})\,.$

We have, for $k\geq 2$:

$\varphi(z)=v_{k}(z)\,\varphi_{k}(z)\,G_{k}(z)\,,$

where $v_{k}=\varphi_{1}\cdots\varphi_{k-1}$.

Remark now that, for $0<a<1$, we have

$T_{a}(z)=a+(1-a^{2})z+O\,(z^{2})\,,$

so

$\varphi_{k}(z)=T_{a_{k}}(z^{n_{k}})=a_{k}+(1-a_{k}^{2})z^{n_{k}}+O\,(z^{2n_{k}})\,.$

Then

(5.4)
$[G_{k}(z)]^{m_{k}}=\bigg{(}\prod_{l=k+1}^{\infty}a_{l}\bigg{)}^{m_{k}}+O\,(z^{n_{k+1}})$

and

(5.5)
$[\varphi_{k}(z)]^{m_{k}}=a_{k}^{m_{k}}+(1-a_{k}^{2})m_{k}a_{k}^{m_{k}-1}z^{n_{k}}+O\,(z^{2n_{k}})\,.$

But

(5.6) $a_{k}^{m_{k}-1}=\Big{(}1-\frac{1}{m_{k}}\Big{)}^{m_{k}-1}\geq{\rm
e}^{-1}:=c_{2}$

and

(5.7) $(1-a_{k}^{2})m_{k}a_{k}^{m_{k}-1}\geq(1-a_{k})m_{k}a_{k}^{m_{k}-1}\geq
c_{2}\,.$

Moreover, since $1-x\geq{\rm e}^{-2x}$ for $0\leq x\leq 1/2$, we have:

(5.8) $\displaystyle\bigg{(}\prod_{l=k+1}^{\infty}a_{l}\bigg{)}^{m_{k}}$
$\displaystyle\geq\exp\,\bigg{(}-2\Big{(}\sum_{l=k+1}^{\infty}\frac{1}{m_{l}}\Big{)}m_{k}\bigg{)}$
$\displaystyle\geq\exp\,\bigg{(}-2\sum_{l=1}^{\infty}4^{-l}\bigg{)}=\exp(-2/3):=c_{3}\,.$

Afterwards, by (5.2), we have

(5.9) $\sum_{j=N_{k-1}m_{k}}^{\infty}|\widehat{v_{k}^{m_{k}}}(j)|^{2}\leq
2^{-15}\,.$

Set $v_{k}^{m_{k}}=g_{1}+g_{2}$, with

$\left\\{\begin{array}[]{rl}g_{1}(z)&=\displaystyle\
\sum_{j=0}^{N_{k-1}m_{k}}\widehat{v_{k}^{m_{k}}}(j)\,z^{j}\vskip 3.0pt plus
1.0pt minus 1.0pt\\\
g_{2}(z)&=\displaystyle\sum_{j>N_{k-1}m_{k}}\widehat{v_{k}^{m_{k}}}(j)\,z^{j}\,.\end{array}\right.$

By (5.9), we have, with $\|\cdot\|_{2}=\|\cdot\|_{L^{2}(\mathbb{T})}$:

$\|g_{2}\|_{2}^{2}=\sum_{j>N_{k-1}m_{k}}|\widehat{v_{k}^{m_{k}}}(j)|^{2}\leq
2^{-15}\,.$

Besides, since $\varphi_{k}$ is inner as a product of inner functions, we have
$|v_{k}(z)|=1$ for all $z\in\mathbb{T}$, so

$\|g_{1}\|_{2}^{2}=\|v_{k}\|_{2}^{2}-\|g_{2}\|_{2}^{2}\geq 1-2^{-15}\,.$

Now,
$\varphi^{m_{k}}=v_{k}^{m_{k}}\varphi_{k}^{m_{k}}G_{k}^{m_{k}}=F_{1}+F_{2}$,
with

$F_{1}=g_{1}\varphi_{k}^{m_{k}}G_{k}^{m_{k}}\quad\text{and}\quad
F_{2}=g_{2}\varphi_{k}^{m_{k}}G_{k}^{m_{k}}\,.$

Using (5.4), (5.5), (5.7) and (5.8), we get

$\displaystyle\sum_{j=n_{k}}^{n_{k}+N_{k-1}m_{k}}|\widehat{F_{1}}(j)|^{2}$
$\displaystyle=\bigg{(}\prod_{l=k+1}^{\infty}a_{l}\bigg{)}^{2m_{k}}\big{[}(1-a_{k}^{2})m_{k}a_{k}^{m_{k}-1}\big{]}^{2}\sum_{j=0}^{N_{k-1}m_{k}}|\widehat{g_{1}}(j)|^{2}$
$\displaystyle\geq(1-2^{-15})\,c_{2}^{2}c_{3}^{2}\,.$

As

$\|F_{2}\|_{2}^{2}\leq\|g_{2}\|_{2}^{2}\,\|\varphi_{k}^{m_{k}}\|_{\infty}^{2}\,\|G_{k}^{m_{k}}\|_{\infty}^{2}\leq
2^{-15}\,,$

we get, using the inequality $|a+b|^{2}\geq\frac{1}{2}\,|a|^{2}-|b|^{2}$:

$\displaystyle\sum_{j=n_{k}}^{2n_{k}}|\widehat{\varphi^{m_{k}}}(j)|^{2}$
$\displaystyle\geq\sum_{j=n_{k}}^{n_{k}+N_{k-1}m_{k}}|\widehat{F_{1}}(j)+\widehat{F_{2}}(j)|^{2}$
$\displaystyle\geq\frac{1}{2}\sum_{j=n_{k}}^{n_{k}+N_{k-1}m_{k}}|\widehat{F_{1}}(j)|^{2}\
-\sum_{j=n_{k}}^{n_{k}+N_{k-1}m_{k}}|\widehat{F_{2}}(j)|^{2}$
$\displaystyle\geq\frac{1}{2}\,(1-2^{-15})\,c_{2}^{2}\,c_{3}^{2}-2^{-15}=\frac{1}{2}\,(1-2^{-15})\,{\rm
e}^{-10/3}-2^{-15}$ $\displaystyle\geq 2^{-9}-2^{-15}>0\,.$

The proof is now complete. ∎

##  6 Some results on multipliers

In this section, we give some results on the multipliers on $H^{2}(\beta)$,
which show how the different notions of regularity for $\beta$ intervene.

The set $\mathcal{M}\big{(}H^{2}(\beta)\big{)}$ of multipliers of
$H^{2}(\beta)$ is by definition the vector space of functions $h$ analytic on
$\mathbb{D}$ and such that $hf\in H^{2}(\beta)$ for all $f\in H^{2}(\beta)$.
When $h\in\mathcal{M}\big{(}H^{2}(\beta)\big{)}$, the operator $M_{h}$ of
multiplication by $h$ is bounded on $H^{2}(\beta)$ by the closed graph
theorem. The space $\mathcal{M}\big{(}H^{2}(\beta)\big{)}$ equipped with the
operator norm is a Banach space. We note the obvious property:

(6.1) $\mathcal{M}\big{(}H^{2}(\beta)\big{)}\hookrightarrow
H^{\infty}\quad\text{contractively.}$

Indeed, if $h\in\mathcal{M}\big{(}H^{2}(\beta)\big{)}$, we easily get for all
$w\in\mathbb{D}$:

$M_{h}^{\ast}(K_{w})=\mkern 1.5mu\overline{\mkern-1.5muh(w)\mkern-1.5mu}\mkern
1.5muK_{w}\,;$

so that taking norms and simplifying, we are left with $|h(w)|\leq\|M_{h}\|$,
showing that $h\in H^{\infty}$ with $\|h\|_{\infty}\leq\|M_{h}\|$.

######  Proposition 6.1.

We have ${\cal M}\big{(}H^{2}(\beta)\big{)}=H^{\infty}$ isomorphically if and
only if $\beta$ is essentially decreasing.

###### Proof.

The sufficient condition is proved in [12, beginning of the proof of
Proposition 3.16]. For the necessity, we then have
$\|M_{h}\|\approx\|h\|_{\infty}$ for every $h\in H^{\infty}$ by the Banach
isomorphism theorem. Now, for $m>n$ (recall that $e_{n}(z)=z^{n}$):

$e_{m}(z)=z^{m-n}z^{n}=(M_{e_{m-n}}e_{n})(z)\,;$

so, since $\|M_{e_{m-n}}\|\leq C\,\|e_{m-n}\|_{\infty}=C$ for some positive
constant $C$:

$\beta_{m}=\|e_{m}\|^{2}\leq C^{2}\,\|e_{n}\|^{2}=C^{2}\,\beta_{n}\,.\qed$

In [12, Section 3.6], we gave the following notion of _admissible_ Hilbert
space of analytic functions.

######  Definition 6.2.

A Hilbert space $H$ of analytic functions on $\mathbb{D}$, containing the
constants, and with reproducing kernels $K_{a}$, $a\in\mathbb{D}$, is said to
be _admissible_ if:

  1. $(i)$

$H^{2}$ is continuously embedded in $H$;

  2. $(ii)$

$\mathcal{M}(H)=H^{\infty}$;

  3. $(iii)$

the automorphisms of $\mathbb{D}$ induce bounded composition operators on $H$;

  4. $(iv)$

$\displaystyle\frac{\|K_{a}\|_{H}}{\|K_{b}\|_{H}}\leq
h\bigg{(}\frac{1-|b|}{1-|a|}\bigg{)}$ for $a,b\in\mathbb{D}$, where
$h\colon\mathbb{R}^{+}\to\mathbb{R}^{+}$ is a non-decreasing function.

We proved in that paper that every weighted Hilbert space $H^{2}(\beta)$ with
$\beta$ non-increasing is admissible, under the additional hypothesis that the
automorphisms of $\mathbb{D}$ induce bounded composition operators. In view of
Theorem 4.6, we get the following result.

######  Proposition 6.3.

Let $\beta$ be a weight.

  1. $1)$

If $\beta$ is essentially decreasing, then we have $(i),(ii),(iii)$ in
Definition 6.2.

  2. $2)$

If $\beta$ is slowly oscillating, then we have $(iv)$ in Definition 6.2.

Let us give a different proof from the one in [12].

###### Proof.

$1)$ Let us assume that $\beta$ is essentially decreasing. Then item $(i)$
holds, as well as item $(ii)$, by Proposition 6.1. Item $(iii)$ is Theorem
4.6.

$2)$ Now we assume that $\beta$ is slowly oscillating.

Let $0<s<r<1$.

Without loss of generality, we may assume that $r,s\geq 1/2$. It is enough to
prove:

(6.2) $\|K_{r}\|^{2}\leq C\,\|K_{r^{2}}\|^{2}$

for some constant $C>1$. Indeed, iteration of (6.2) gives:

$\|K_{r}\|^{2}\leq C^{k}\,\|K_{r^{2^{k}}}\|^{2}$

and if $k$ is the smallest integer such that $r^{2^{k}}\leq s$, we have
$2^{k-1}\log r>\log s$ and $2^{k}\leq D\,\frac{1-s}{1-r}$ where $D$ is a
numerical constant. Writing $C=2^{\alpha}$ with $\alpha>1$, we obtain:

$\bigg{(}\frac{\|K_{r}\|}{\|K_{s}\|}\bigg{)}^{2}\leq
C^{k}=(2^{k})^{\alpha}\leq
D^{\alpha}\bigg{(}\frac{1-s}{1-r}\bigg{)}^{\alpha}\,.$

To prove (6.2), we pick some $M>1$ such that $\beta_{2n}\geq M^{-1}\beta_{n}$
and $\beta_{2n-1}\geq M^{-1}\beta_{n}$, for all $n\geq 1$, since $\beta$ is
slowly oscillating. Write $t=r^{2}$. We have:

$\|K_{r}\|^{2}=\frac{1}{\beta_{0}}+\sum_{n=1}^{\infty}\frac{t^{2n}}{\beta_{2n}}+\sum_{n=1}^{\infty}\frac{t^{2n-1}}{\beta_{2n-1}}\,\raise
1.0pt\hbox{,}$

implying, since $t^{2n-1}\leq 4\,t^{2n}$:

$\|K_{r}\|^{2}\leq\frac{1}{\beta_{0}}+M\sum_{n=1}^{\infty}\frac{t^{2n}}{\beta_{n}}+4M\sum_{n=1}^{\infty}\frac{t^{2n}}{\beta_{n}}\leq
5M\|K_{t}\|^{2}\,.\qed$

The notion of admissible Hilbert space $H$ is useful for the set of
conditional multipliers:

$\mathcal{M}(H,\varphi)=\\{w\in H\,;\ w\,(f\circ\varphi)\in H\text{ for all
}f\in H\\}\,.$

As a corollary of [12, Theorem 3.18] we get:

######  Corollary 6.4.

Let $\beta$ be essentially decreasing and slowly oscillating. Then:

  1. $1)$

$\mathcal{M}(H^{2},\varphi)\subseteq\mathcal{M}\big{(}H^{2}(\beta),\varphi\big{)}$;

  2. $2)$

$\mathcal{M}\big{(}H^{2}(\beta),\varphi\big{)}=H^{2}(\beta)$ if and only if
$\|\varphi\|_{\infty}<1$;

  3. $3)$

$\mathcal{M}\big{(}H^{2}(\beta),\varphi\big{)}=H^{\infty}$ if and only if
$\varphi$ is a finite Blaschke product.

We add here as another application of our results an answer to a question
appearing in Problem 5 in Zorboska’s thesis [23].

######  Theorem 6.5.

Let $\beta$ be a weight such that $H^{2}(\beta)$ is disc-automorphism
invariant and a symbol $\varphi$ inducing a compact composition operator on
$H^{2}(\beta)$. Then the Denjoy-Wolff point of $\varphi$ must be in
$\mathbb{D}$.

In other words, $\varphi$ has a fixed point in $\mathbb{D}$.

In the statement, “$H^{2}(\beta)$ is disc-automorphism invariant” means that
for all the automorphisms $T_{a}$, where $a\in\mathbb{D}$, $C_{T_{a}}$ is
bounded on $H^{2}(\beta)$ (equivalently it is bounded for at least one
$a\in\mathbb{D}\setminus\\{0\\}$).

For the definition of the Denjoy-Wolff point, we refer to [18].

###### Proof.

From Theorem 4.9, we know that $\beta$ is slowly oscillating, and from
Proposition 6.3, we know that

(6.3)
$\qquad\quad\frac{\|K_{a}\|_{H^{2}(\beta)}}{\|K_{b}\|_{H^{2}(\beta)}}\leq
h\bigg{(}\frac{1-|b|}{1-|a|}\bigg{)}\qquad\hbox{for every }a,b\in\mathbb{D},$

where $h\colon\mathbb{R}^{+}\to\mathbb{R}^{+}$ is a non-decreasing function.

Now we split the proof into two cases.

If $\sum\dfrac{1}{\beta_{n}}<\infty$, then $H^{2}(\beta)\subset A(\mathbb{D})$
(continuously) thanks to the Cauchy-Schwarz inequality. It follows from a
result due to Shapiro [17, Theorem 2.1] that $\|\varphi\|_{\infty}<1$ and the
conclusion follows obviously in this case.

If $\sum\dfrac{1}{\beta_{n}}=\infty$, then the normalized reproducing kernel
$\dfrac{K_{z}}{\|K_{z}\|}$ is weakly converging to $0$ when $|z|\to 1^{-}$,
since $\|K_{z}\|\rightarrow+\infty$.

Since $C_{\varphi}$ is compact, $C_{\varphi}^{\ast}$ is compact as well and we
get

$\dfrac{K_{\varphi(z)}}{\|K_{z}\|}\longrightarrow 0\qquad\hbox{when }|z|\to
1^{-}$

and equivalently

$\dfrac{\|K_{z}\|}{\|K_{\varphi(z)}\|}\longrightarrow+\infty\qquad\hbox{when
}|z|\to 1^{-}\;.$

But, from (6.3), we get

$h\bigg{(}\frac{1-|\varphi(z)|}{1-|z|}\bigg{)}\longrightarrow+\infty\qquad\hbox{when
}|z|\to 1^{-}$

hence, since $h$ is non-decreasing,

$\frac{1-|\varphi(z)|}{1-|z|}\longrightarrow+\infty\qquad\hbox{when }|z|\to
1^{-}\,.$

By the Denjoy-Wolff theorem [18], the conclusion follows in this case too. ∎

##  7 Appendix

In this appendix, we give the proof of Proposition 4.17.

The following lemma can be found in [15, Lemma 1, page 47].

######  Lemma 7.1.

Let $F\colon[u,v]\to\mathbb{R}$, with $u<v$, be a ${\cal C}^{2}$\- function
with $F^{\prime\prime}>0$, and $F^{\prime}$ not vanishing on $[u,v]$. Let

$J=\int_{u}^{v}e^{iF(x)}\,dx\,.$

Then:

  * a)

if $F^{\prime}>0$ on $[u,v]$, then $|J|\leq\frac{2}{F^{\prime}(u)}\,$;

  * b)

if $F^{\prime}<0$ on $[u,v]$, then $|J|\leq\frac{2}{|F^{\prime}(v)|}\,\cdot$

###### Proof of Proposition 4.17.

Write now the integral $I$ of Proposition 4.17 on $[A,B]$ as
$I=I_{1}+I_{2}+I_{3}$ with:

$I_{1}=\int_{A}^{c-\eta}{\rm e}^{iF(x)}\,dx\,,\quad
I_{2}=\int_{c-\eta}^{c+\eta}{\rm e}^{iF(x)}\,dx\,,\quad
I_{3}=\int_{c+\eta}^{B}{\rm e}^{iF(x)}\,dx\,.$

Lemma 7.1 with $u=A$ and $v=c-\eta$ implies, since $F^{\prime}<0$ on
$[A,c-\eta]$:

(7.1)
$|I_{1}|\leq\frac{2}{|F^{\prime}(c-\eta)|}\leq\frac{2}{\eta\lambda_{2}}\,\raise
1.0pt\hbox{,}$

where, for the last inequality, we just have to write

$|F^{\prime}(c-\eta)|=F^{\prime}(c)-F^{\prime}(c-\eta)=\eta\,F^{\prime\prime}(\xi)$

for some $\xi\in[c-\eta,c]$ so that $F^{\prime\prime}(\xi)\geq\lambda_{2}$.

Similarly, Lemma 7.1 with $u=c+\eta$ and $v=B$ implies

(7.2)
$|I_{3}|\leq\frac{2}{F^{\prime}(c+\eta)}\leq\frac{2}{\eta\lambda_{2}}\,\cdot$

We can now estimate $I_{2}$. The Taylor formula shows that

$F(x)=F(c)+\frac{(x-c)^{2}}{2}F^{\prime\prime}(c)+R\,,$

with

$|R|\leq\frac{|x-c|^{3}}{6}\,\lambda_{3}\,.$

Hence

$I_{2}={\rm
e}^{iF(c)}\int_{0}^{\eta}2\exp\bigg{(}\frac{i}{2}\,x^{2}F^{\prime\prime}(c)\bigg{)}\,dx+S$

with

$|S|\leq\lambda_{3}\int_{0}^{\eta}\frac{x^{3}}{3}\,dx=\frac{\,\eta^{4}}{12}\,\lambda_{3}\,.$

Finally, set

$K=\int_{0}^{\eta}2\exp\bigg{(}\frac{i}{2}\,x^{2}F^{\prime\prime}(c)\bigg{)}\,dx\,.$

We make the change of variable
$x=\sqrt{\frac{2}{F^{\prime\prime}(c)}}\,\sqrt{t}$. Recall that
$\int_{0}^{\infty}\frac{{\rm e}^{it}}{\sqrt{t}}dt=\sqrt{\pi}\,{\rm
e}^{i\pi/4}$ is the classical Fresnel integral, and that an integration by
parts gives, for $m>0$:

$\bigg{|}\int_{m}^{\infty}\frac{{\rm
e}^{it}}{\sqrt{t}}\,dt\bigg{|}\leq\frac{2}{\sqrt{m}}\,\cdot$

Therefore, with $m=\frac{\,\eta^{2}}{2}\,F^{\prime\prime}(c)$:

$K=\sqrt{\frac{2}{F^{\prime\prime}(c)}}\int_{0}^{m}\frac{{\rm
e}^{it}}{\sqrt{t}}\,dt=\sqrt{\frac{2\pi}{F^{\prime\prime}(c)}}\,{\rm
e}^{i\pi/4}+R_{m}\,,$

with

$|R_{m}|\leq
C\,\sqrt{\frac{1}{F^{\prime\prime}(c)}}\,\frac{1}{\sqrt{m}}\leq\frac{C}{\eta\lambda_{2}}\,\cdot$

All in all, we proved that

(7.3)
$I_{2}=\sqrt{\frac{2\pi}{F^{\prime\prime}(c)}}\,\exp\big{[}i(F(c)+\pi/4)\big{]}+O\,\bigg{(}\frac{1}{\eta\lambda_{2}}+\eta^{4}\lambda_{3}\bigg{)}\,\cdot$

and the same estimate holds for $I$, thanks to (7.1) and (7.2).

We have hence proved Proposition 4.17. ∎

Acknowledgements. L. Rodríguez-Piazza is partially supported by the projects
PGC2018-094215-B-I00 and PID2022-136320NB-I00 (Spanish Ministerio de Ciencia,
Innovación y Universidades, and FEDER funds). Parts of this paper was made
when he visited the Université d’Artois in Lens and the Université de Lille in
January 2020 and September 2022. It is his pleasure to thank all his
colleagues in these universities for their warm welcome.

Parts of this paper were made during an invitation of the second-named and
third-named authors by the IMUS at the Universidad de Sevilla; it is their
pleasure to thank all people in that University who made this stay possible
and very pleasant, and especially Manuel Contreras.

The third-named author was partly supported by the Labex CEMPI (ANR-
LABX-0007-01).

This work is also partially supported by the grant ANR-17-CE40-0021 of the
French National Research Agency ANR (project Front).

We warmly thank R. Zarouf for useful discussions and informations.

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Pascal Lefèvre and Daniel Li
Univ. Artois, UR 2462, Laboratoire de Mathématiques de Lens (LML), F-62300
LENS, FRANCE
<EMAIL_ADDRESS>–<EMAIL_ADDRESS>
Hervé Queffélec
Univ. Lille Nord de France, USTL, Laboratoire Paul Painlevé U.M.R. CNRS 8524 &
Fédération Mathématique des Hauts-de-France FR 2037 CNRS, F-59655 VILLENEUVE
D’ASCQ Cedex, FRANCE
<EMAIL_ADDRESS>
Luis Rodríguez-Piazza
Universidad de Sevilla, Facultad de Matemáticas, Departamento de Análisis
Matemático & IMUS, Calle Tarfia s/n 41012 SEVILLA, SPAIN
<EMAIL_ADDRESS>