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**Boundedness of composition operators on general** **weighted Hardy spaces of analytic functions**

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

**To cite this version:**

Pascal Lefèvre, Daniel Li, Hervé Queffélec, Luis Rodríguez-Piazza. Boundedness of composition op- erators on general weighted Hardy spaces of analytic functions. 2020. �hal-03029931�

### Boundedness of composition operators on general weighted Hardy spaces of analytic

### functions

Pascal Lefèvre, Daniel Li,

Hervé Queffélec, Luis Rodríguez-Piazza

November 29, 2020

Abstract. We characterize the (essentially) decreasing sequences of positive
numbersβ = (βn)for which all composition operators onH^{2}(β)are bounded,
where H^{2}(β) is the space of analytic functions f in the unit disk such that
P∞

n=0|cn|^{2}βn<∞iff(z) =P∞

n=0cnz^{n}. We also give conditions for the bound-
edness whenβ is not assumed essentially decreasing.

MSC 2010primary: 47B33 ; secondary: 30H10

Key-wordsautomorphism of the unit disk; composition operator;∆2-condition;

multipliers; weighted Hardy space

### 1 Introduction

Letβ= (βn)n≥0 be a sequence of positive numbers such that

(1.1) lim inf

n→∞ β_{n}^{1/n}≥1.

The associated weighted Hardy space H^{2}(β) is the set of analytic functions
f(z) =P_{∞}

n=0anz^{n} such that

(1.2) kfk^{2}:=

X∞ n=0

|an|^{2}βn<∞.

In view of (1.1),H^{2}(β)is a Hilbert space of analytic functions onD with the
canonical orthonormal basis

(1.3) e^{β}_{n}(z) = 1

√βn

z^{n}, n≥0,

and the reproducing kernelKw given for allw∈Dby

(1.4) Kw(z) =

X∞ n=0

e^{β}_{n}(z)e^{β}n(w) =
X∞
n=0

1 βn

w^{n}z^{n}.

Indeed, we have:

X∞ k=0

anw^{n}
≤

X∞ k=0

βk|ak|^{2}

^{1/2}X∞
k=0

1
βk |w|^{2k}

^{1/2}

<∞, thanks to (1.1).

Note that (1.1) is necessary for H^{2}(β) to consist of analytic functions in
D. Indeed the fact that P

n≥1 1 n√

βnz^{n} belongs toH^{2}(β)and is analytic inD
implies (1.1).

Whenβn≡1, we recover the usual Hardy spaceH^{2}.

Note thatH^{2}is continuously embedded inH^{2}(β)if and only ifβis bounded
above. In particular, this is the case whenβ is non-increasing.

Most of works with weighted Hardy spaces concern the case

(1.5) βn=

Z 1 0

t^{n}dσ(t)

whereσis a positive measure on(0,1). More specifically the following definition
is often used. LetG: (0,1)→R_{+} be an integrable function and letH_{G}^{2} be the
space of analytic functionsf:D→Csuch that:

(1.6) kfk^{2}H_{G}^{2} :=

Z

D|f(z)|^{2}G(1− |z|^{2})dA(z)<∞.

Such weighted Bergman type spaces are used, for instance, in [11], [10] and
in [14]. We haveH_{G}^{2} =H^{2}(β)with:

(1.7) βn= 2 Z 1

0

r^{2n+1}G(1−r^{2})dr=
Z 1

0

t^{n}G(1−t)dt ,
and the sequenceβ = (βn)n is non-increasing.

Recall that a *symbol* is a (non constant) analytic self-map ϕ:D→D, and
the associated composition operatorCϕ:H^{2}(β)→ Hol(D)is defined (formally)
as:

(1.8) Cϕ(f) =f◦ϕ .

An important question in the theory is to decide when Cϕ is bounded on
H^{2}(β), i.e. whenCϕ:H^{2}(β)→H^{2}(β).

Whenβn≡1, i.e. whenH^{2}(β)is the usual Hardy spaceH^{2}, it is known ([18,
pp. 13–17]) that all symbols generate bounded composition operators. But in
Shapiro’s presentation, the main point is the caseϕ(0) = 0and a subordination
principle for subharmonic functions. The case of automorphisms is claimed sim-
ple, using an integral representation for the norm and some change of variable.

Whenβ is defined as in (1.5), one disposes of integral representations for the

norm inH^{2}(β), and, as in in the Hardy space case, this integral representation
rather easily gives the boundedness ofCTa onH^{2}, where

(1.9) Ta(z) = a+z

1 + ¯a z

fora ∈D. But the above representation (1.5) is equivalent, by the Hausdorff
moment theorem, to a high regularity of the sequence β, namely its *complete*
*monotony. When integral representations fail, we have to work with bare hands.*

If the symbol vanishes at the origin, Kacnel’son’s theorem (see Theorem 3.6
below) gives a positive answer whenβ is essentially decreasing (see [2] or [13,
Theorem 3.12]). Actually that follows from an older theorem of Goluzin [8] (see
[5, Theorem 6.3]), which itself uses a refinement by Rogosinski of Littlewood’s
principle ([5, Theorem 6.2]). So that the main issue remains the boundedness
ofCTa. A polynomial minoration (see Definition 2.3 below) for β is necessary
for anyCTa to be bounded on H^{2}(β)(Proposition 2.5) and we showed in [13,
end of Section 3] that for βn = exp(−√n), CTa is never bounded on H^{2}(β).

But this polynomial minoration is not sufficient, as we will see in Theorem 1.1 below, which also evidences the basic role of the mapsTa in the question.

So, our goal in this paper is characterizing the sequences β for which all
composition operators act boundedly on the spaceH^{2}(β), i.e. sendH^{2}(β)into
itself. Eventually, we will obtain in Theorem 3.4 and Theorem 4.1 the following
result, where the∆2-condition is defined in (2.2).

Theorem 1.1. *Let*β *be an essentially decreasing sequence of positive numbers.*

*The following assertions are equivalent:*

1) *all composition operators are bounded on*H^{2}(β);

2) *all maps*Ta*, for*0< a <1, induce bounded composition operatorsCTa *on*
H^{2}(β);

3) *for some* a ∈(0,1), the map Ta *induces a bounded composition operator*
CTa *on* H^{2}(β);

4) β *satisfies the*∆2*-condition.*

Note that, by definition of the norm of H^{2}(β), all rotations Rθ, θ ∈ R,
induce bounded composition operators onH^{2}(β)and send isometricallyH^{2}(β)
into itself.

However, we construct a weightβwhich is not essentially decreasing and for which all composition operators with symbol vanishing at0are bounded (The- orem 3.3), though no mapTa with0 < a <1 induces a bounded composition operator (Proposition 4.4).

For spaces of Bergman type A^{2}_{e}

G := H_{G}^{2}, whereG(r) =e G(1−r^{2}), defined
as the spaces of analytic functions in D such that R

D|f(z)|^{2}G(e |z|)dA < ∞,
for a positive non-increasing continuous functionGe on [0,1), Kriete and Mac-
Cluer studied in [11] some analogous problems. They proved, in particular [11,

Theorem 3] that, for:

G(r) = expe

−B 1
(1−r)^{α}

, B >0, 0< α≤2, and

ϕ(z) =z+t(1−z)^{β}, 1< β≤3, 0< t <2^{1}^{−}^{β},
thenCϕ is bounded onA^{2}_{e}

G if and only ifβ ≥α+ 1.

Here

βn = Z 1

0

t^{n}e^{−}^{B/(1}^{−}^{√}^{t)}^{α}dt;

and, sinceβn ≈ exp(−c n^{α/(α+1)}), the sequence (βn) does not satisfy the ∆2-
condition, accordingly to our Theorem 3.4 below.

### 2 Definitions, notation, and preliminary results

The open unit disk ofCis denotedDand we writeTits boundary∂D.
We seten(z) =z^{n},n≥0.

As said in the introduction,H^{2}is continuously embedded inH^{2}(β)whenβ
is non-increasing. In this paper, we need a slightly more general notion.

Definition 2.1. *A sequence of positive numbers* β = (βn) *is said* essentially
decreasing*if, for some constant* C≥1, we have, for all m≥n≥0:

(2.1) βm≤C βn.

Note that saying thatβis essentially decreasing means that the shift operator
onH^{2}(β)is power bounded.

Ifβ is essentially decreasing, and if we set:

βen = sup

m≥n

βm,

the sequence βe = (βen) is non-increasing and we have βn ≤ βen ≤ C βn. In
particular, the spaceH^{2}(β)is isomorphic toH^{2}(β)e and H^{2}(β)is continuously
embedded inH^{2}.

Definition 2.2. *The sequence of positive numbers* β = (βn) *is said to satisfy*
*the* ∆2-condition*if there is a positive constant*δ <1 *such that, for all integers*
n≥0:

(2.2) β2n≥δ βn.

Definition 2.3. *The sequence of positive numbers* β = (βn) *is said to have a*
polynomial minoration*if there are positive constants*δ*and*α*such that, for all*
*integers*n≥1:

(2.3) βn ≥δ n^{−}^{α}.

That means thatH^{2}(β)is continuously embedded in the weighted Bergman
spaceB^{2}_{α}of the analytic functionsf:D→Csuch that

kfk^{2}B^{2}_{α}:= (α+ 1)
Z

D|f(z)|^{2}(1− |z|^{2})^{α}dA(z)<∞
sinceB^{2}_{α}=H^{2}(γ)withγn ≈n^{−}^{α}.

The following simple proposition links those notions.

Proposition 2.4. *Let* β *be an essentially decreasing sequence of positive num-*
*bers. Then if*β *satisfies the*∆2*-condition, it has a polynomial minoration.*

*The converse does not hold.*

*Proof.* 1) Assume βm ≤C βn for m ≥n and β2p ≥e^{−}^{A}βp. Let now nbe an
integer≥2, andk≥1the smallest integer such that2^{k}≥n, so thatk≤alogn
withaa positive constant. We get:

βn ≥C^{−}^{1}β2^{k}≥C^{−}^{1}e^{−}^{kA}β1≥C^{−}^{1}β1e^{−}^{aA}^{log}^{n}=:ρ n^{−}^{α},
withρ=C^{−}^{1}β1andα=aA.

2) Letδ >0. We setβ0=β1= 1and forn≥2:

βn= 1

(k!)^{δ} whenk!< n≤(k+ 1)!.
The sequenceβ is non-increasing.

Fornandkas above, we have:

βn= 1
(k!)^{δ} ≥ 1

n^{δ};

henceβ has arbitrary polynomial minoration. However we have, fork≥2:

β2k!

βk!

= (k!)^{−}^{δ}

[(k−1)!]^{−}^{δ} = 1
k^{δ} −→

k→∞0, soβ fails to satisfy the∆2-condition.

Fora∈D, we define:

(2.4) Ta(z) = a+z

1 + ¯a z, z∈D.

Recall thatTais an automorphism ofDand thatTa(0) =aandTa(−a) = 0.

Though we do not need this, we may remark that (Ta)a∈(−1,1) is a group and(Ta)a∈(0,1)is a semigroup. It suffices to see thatTa◦Tb=Ta∗b, with:

(2.5) a∗b= a+b

1 +ab·

Proposition 2.5. *Let*a∈(0,1)*and assume that*Ta *induces a bounded compo-*
*sition operator on*H^{2}(β). Then β *has a polynomial minoration.*

Remarks. 1) For example, when βn = exp

−c log(n+ 1)^{2}, with c > 0,
noTa induces a bounded composition operator on H^{2}(β), though all symbols
ϕwithϕ(0) = 0are bounded, sinceβ is decreasing, as said by the forthcoming
Proposition 3.2.

2) For the Dirichlet spaceD^{2}, we haveβn =n+1, but all the mapsTainduce
bounded composition operators onD^{2}(see [13, Remark before Theorem 3.12]).

In this caseβ has a polynomial minoration though it is not bounded above.

3) However, even for decreasing sequences, a polynomial minoration forβ is not enough for someTa to induce a bounded composition operator. Indeed, we saw in Proposition 2.4 an example of a decreasing sequenceβ with polynomial minoration, but not sharing the∆2-condition, and we will see in Theorem 4.1 that the∆2-condition is needed for having some Ta inducing a bounded com- position operator.

4) In [7], Eva Gallardo-Gutiérrez and Jonathan Partington give estimates
for the norm of CTa, with a ∈ (0,1), when CTa is bounded on H^{2}(β). More
precisely, they proved that ifβ is bounded above andCTa is bounded, then

kCTak ≥

1 +a 1−a

σ

,
whereσ= inf{s≥0 ; (1−z)^{−}^{s}∈/ H^{2}(β)}, and

kCTak ≤

1 +a 1−a

τ

,

whereτ =^{1}_{2} supReW(A), withAthe infinitesimal generator of the continuous
semigroup(St) defined as St =CTtanht, namely (Af)(z) = f^{′}(z)(1−z^{2}), and
W(A)its numerical range.

Forβn = 1/(n+ 1)^{ν} with 0≤ν ≤1, the two bounds coincide, so they get
kCTak= ^{1+a}_{1}_{−}_{a}^{(ν+1)/2}.

*Proof of Proposition 2.5.* Since

kKxk^{2}=
X∞
n=0

x^{2n}
βn

, we havekKxk ≤ kKyk for0≤x≤y <1.

We define by induction a sequence(un)n≥0 with:

u0= 0 and un+1=Ta(un). SinceTa(1) = 1(recall thata∈(0,1)), we have:

1−un+1= Z 1

un

T_{a}^{′}(t)dt=
Z 1

un

1−a^{2}
(1 +at)^{2}dt;

hence

1−a

1 +a(1−un)≤1−un+1≤(1−a^{2})(1−un).
Let0< x <1. We can findN ≥0 such thatuN ≤x < uN+1. Then:

1−x≤1−uN ≤(1−a^{2})^{N}.

On the other hand, sinceC_{T}^{∗}_{a}Kz=KTa(z) for allz∈D, we have:

kKxk ≤ kKuN+1k ≤ kCTak kKuNk ≤ kCTak^{N}^{+1}kKu0k=kCTak^{N+1}.
Lets≥0 such that(1−a^{2})^{−}^{s}=kCTak. We obtain:

(2.6) kKxk ≤ kCTak 1

(1−x)^{s}·
But

kKxk^{2}=
X∞
k=0

x^{2k}
βk

;

so we get, for anyk≥2:

x^{2k}

βk ≤ kCTak^{2} 1
(1−x)^{2s}·
Takingx= 1−k^{1}, we obtainβk ≥C k^{−}^{2s}.

Remark. We saw in the proof of Proposition 2.5 that if CTa is bounded on
H^{2}(β) for some a∈ (0,1), then the reproducing kernels Kw have, by (2.6), a
*slow growth:*

(2.7) kKwk ≤ C

(1− |w|)^{s}

for positive constantsC ands. Actually, we have the following equivalence.

Proposition 2.6. *The sequence* β *has a polynomial minoration if and only if*
*the reproducing kernels*Kw *of* H^{2}(β)*have a slow growth.*

*Proof.* The sufficiency is easy and seen at the end of the proof of Proposition 2.5.

For the necessity, we only have to see that:

kKwk^{2}= 1
β0

+ X∞ n=1

|w|^{2n}
βn ≤ 1

β0

+δ^{−}^{1}
X∞
n=1

n^{α}|w|^{2n} ≤ C
(1− |w|^{2})^{α+1} ·

### 3 Boundedness of composition operators

We study in this section conditions ensuring that all composition operators
onH^{2}(β)are bounded.

3.1 Conditions on the weight

We begin with this simple observation.

Proposition 3.1. *If all composition operators, and even if all composition op-*
*erators with symbol vanishing at*0, are bounded onH^{2}(β), then the sequenceβ
*is bounded above.*

*Proof.* Ifβis not bounded above, there is a subsequence(βnk), withnk≥1, such
that βnk ≥4^{k}. Thenϕ(z) = P∞

k=12^{−}^{k}z^{n}^{k} defines a symbol, since |ϕ(z)| <1,
andϕ(0) = 0; but:

kCϕ(e1)k^{2}=kϕk^{2}=
X∞
k=1

4^{−}^{k}βnk=∞.

For symbols vanishing at0, we have the following characterization.

Proposition 3.2. *The following assertions are equivalent:*

1) *all symbols* ϕ *such that* ϕ(0) = 0 *induce bounded composition operators*
Cϕ *on* H^{2}(β)*and*

(3.1) sup

ϕ(0)=0kCϕk<∞;
2) β *is an essentially decreasing sequence.*

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

sup

ϕ(0)=0kf◦ϕk<∞ for allf ∈H^{2}(β).

*Proof.* 2) ⇒ 1) We may assume that β is non-increasing. Then the Goluzin-
Rogosinski theorem ([5, Theorem 6.3]) gives the result; in fact, writingf(z) =
P∞

n=0cnz^{n} and(Cϕf)(z) =P∞

n=0dnz^{n}, it says that:

kCϕfk^{2}=|d0|^{2}β0+
X∞
n=1

|dn|^{2}βn≤ |c0|^{2}β0+
X∞
n=1

|cn|^{2}βn=kfk^{2},

leading to Cϕ bounded and kCϕk ≤ 1. Alternatively, we can use a result of Kacnel’son ([9]; see also [2], [3, Corollary 2.2], or [13, Theorem 3.12]). This result was also proved by C. Cowen [4, Corollary of Theorem 7].

1)⇒2)SetM = sup_{ϕ(0)=0}kCϕk. Letm > n, and take:

ϕ(z) =z

1 +z^{m}^{−}^{n}
2

1/n

.

Thenϕ(0) = 0and[ϕ(z)]^{n} =^{z}^{n}^{+z}_{2} ^{m}; hence
1

4(βn+βm) =kϕ^{n}k^{2}=kCϕ(en)k^{2}≤ kCϕk^{2}kenk^{2}≤M^{2}βn,
soβ is essentially decreasing.

For example, let (βn) such thatβ2k+2/β2k+1 −→

k→∞∞ (for instanceβ2k = 1
and β2k+1 = 1/(k+ 1)); if ϕ(z) = z^{2}, then kCϕ(z^{2n+1})k^{2} = kz^{2(2n+1)}k^{2} =
β2(2n+1); sincekz^{2n+1}k^{2}=β2n+1, Cϕ is not bounded onH^{2}(β).

A more interesting example is the following. For0< r <1, letβn=π n r^{2n}.
This sequence is eventually decreasing, so it is essentially decreasing. The square
of the normkfk^{2}H^{2}(β) is the area of the part of the Riemann surface on which
rD is mapped by f. E. Reich [17], generalizing Goluzin’s result [8] (see [5,
Theorem 6.3]), proved that for all symbolsϕsuch thatϕ(0) = 0, the composition
operatorCϕis bounded onH^{2}(β)and

kCϕk ≤sup

n≥1

√n r^{n}^{−}^{1}≤ 1

√2 e 1 rp

log(1/r)· For0< r <1/√

2, Goluzin’s theorem asserts thatkCϕk ≤1.

Note that this sequenceβ does not satisfy the∆2-condition sinceβ2n/βn =
2r^{2n}, Theorem 4.1 below states that no composition operatorCTa is bounded.

However that the weightβ is essentially decreasing is not necessary for the boundedness of all composition operatorsCϕ, with symbolϕvanishing at0. as stated by the following theorem, whose proof will be given in Section 3.3.

Theorem 3.3. *There exists a bounded sequence* β, with a polynomial minora-
*tion, but which is* not essentially decreasing, for which every composition oper-
*ator with symbol vanishing at*0 *is bounded on*H^{2}(β).

It should be noted that for this weight, the composition operators are not all bounded, as we will see in Proposition 4.4.

3.2 Boundedness of composition operators I

We now have one of the the main results of this section.

Theorem 3.4. *Let*H^{2}(β)*be a weighted Hardy space with*β = (βn)*essentially*
*decreasing and satisfying the* ∆2*-condition. Then all composition operators on*
H^{2}(β)*are bounded.*

For the proof, we need a lemma.

Lemma 3.5. *For*0< a <1, we write:

(3.2) (Taz)^{n}=

X∞ m=0

am,nz^{m}.

*Then, for every*α >0, there exists a positive constantρ, depending on a, such
*that:*

(3.3) |am,n| ≤e^{−}^{α(ρn}^{−}^{m)}.

Note that this estimate is interesting only whenm < ρ nsince we know that

|am,n| ≤ kTan

k∞= 1for allmandn.

*Proof.* For0< r <1, let:

M(r) = sup

|z|=r|Ta(z)|= sup

|z|=r

z+a

1 +az ·

We haveM(r)<1, so we can writeM(r) =r^{ρ}, for some positiveρ=ρ(a).

The Cauchy inequalities give:

|am,n| ≤ [M(r)]^{n}

r^{m} =r^{ρn}^{−}^{m},
and we obtain the result by takingr= e^{−}^{α}.

*Proof of Theorem 3.4.* We may, and do, assume thatβ is non-increasing.

Proposition 3.2 gives the result when ϕ(0) = 0.

It remains to show that allCTa, a∈D, are bounded. Indeed, if a=ϕ(0)
andψ=T_{−}a◦ϕ, then ψ(0) = 0andϕ=Ta◦ψ, soCϕ=Cψ◦CTa. Moreover,
we have only to show that whena∈[0,1). Indeed, ifa∈Danda=|a|e^{iθ}, we
haveTa=Rθ◦T_{|}a|◦R_{−}θ, soCTa=CR−θ◦CT|a|◦CRθ.

We consider the matrices

A= (am,n)m,n≥0 and Aβ=

sβm

βn

am,n

m,n≥0

.

SinceCTaen =Tan, the formula (3.2) shows that Ais the matrix of CTa in
H^{2} with respect to the basis (en)n≥0. On the other hand, Aβ is the matrix of
CTa in H^{2}(β) with respect to the basis(e^{β}_{n})n≥0. We note that Aβ =BAB^{−}^{1},
whereB is the diagonal matrix with values√

β0,√

β1, . . .on the diagonal.

SinceCTa is a bounded composition operators on H^{2}, the matrixAdefines
a bounded operator onℓ2. We have to show thatAβ also, i.e. kAβk<∞.

For that purpose, we splitAandAβ into several sub-matrices.

Let N be an integer such thatN ≥ 2/ρ, where ρ is defined in Lemma 3.5 (actually, the proof of that lemma shows that we can take ρ such that 1/ρ is

an integer, so we could takeN = 2/ρ). Let I0= [0, N[ J0= [N,+∞[ and for k= 1,2, . . .:

Ik= [N^{k}, N^{k+1}[ and Jk = [N^{k+1},+∞[.
We define the matricesDβ andRβ, whose entries are respectively:

dm,n=

sβm

βn

am,n if (m, n)∈ [∞ k=0

(Ik×Ik)

0 elsewhere;

and

rm,n=

sβm

βn

am,n if (m, n)∈ [∞ k=0

(Ik×Ik+1)

0 elsewhere.

We also define the matrixSβ with entries:

sm,n=

sβm

βn am,n if (m, n)∈ [∞ k=0

(Jk×Ik)

0 elsewhere.

Matrices D,R, andS are constructed in the same way fromA and we set U =A−(D+R+S).

Now, letHkbe the subspace of the sequences(xn)n≥0inℓ2such thatxn= 0 for n /∈ Ik, i.e. Hk = span{ek; k ∈ Ik}, and let Pk be (the matrix of) the orthogonal projection ofℓ2with rangeHk. We have:

D= X∞ k=0

PkAPk and R= X∞ k=0

PkAPk+1,

whereDk =PkAPk is the matrix with entriesam,nwhen(m, n)∈Ik×Ik and 0 elsewhere, and Rk = PkAPk+1 the matrix with entries am,n when (m, n)∈ Ik×Ik+1 and0elsewhere.

D0 R0

D1 R1

## U

S0 S1 D2 R2

S2