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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 onH2(β)are bounded, where H2(β) is the space of analytic functions f in the unit disk such that P∞
n=0|cn|2βn<∞iff(z) =P∞
n=0cnzn. 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→∞ βn1/n≥1.
The associated weighted Hardy space H2(β) is the set of analytic functions f(z) =P∞
n=0anzn such that
(1.2) kfk2:=
X∞ n=0
|an|2βn<∞.
In view of (1.1),H2(β)is a Hilbert space of analytic functions onD with the canonical orthonormal basis
(1.3) eβn(z) = 1
√βn
zn, n≥0,
and the reproducing kernelKw given for allw∈Dby
(1.4) Kw(z) =
X∞ n=0
eβn(z)eβn(w) = X∞ n=0
1 βn
wnzn.
Indeed, we have:
X∞ k=0
anwn ≤
X∞ k=0
βk|ak|2
1/2X∞ k=0
1 βk |w|2k
1/2
<∞, thanks to (1.1).
Note that (1.1) is necessary for H2(β) to consist of analytic functions in D. Indeed the fact that P
n≥1 1 n√
βnzn belongs toH2(β)and is analytic inD implies (1.1).
Whenβn≡1, we recover the usual Hardy spaceH2.
Note thatH2is continuously embedded inH2(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
tndσ(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 letHG2 be the space of analytic functionsf:D→Csuch that:
(1.6) kfk2HG2 :=
Z
D|f(z)|2G(1− |z|2)dA(z)<∞.
Such weighted Bergman type spaces are used, for instance, in [11], [10] and in [14]. We haveHG2 =H2(β)with:
(1.7) βn= 2 Z 1
0
r2n+1G(1−r2)dr= Z 1
0
tnG(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:H2(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 H2(β), i.e. whenCϕ:H2(β)→H2(β).
Whenβn≡1, i.e. whenH2(β)is the usual Hardy spaceH2, 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 inH2(β), and, as in in the Hardy space case, this integral representation rather easily gives the boundedness ofCTa onH2, 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 H2(β)(Proposition 2.5) and we showed in [13, end of Section 3] that for βn = exp(−√n), CTa is never bounded on H2(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 spaceH2(β), i.e. sendH2(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. Letβ be an essentially decreasing sequence of positive numbers.
The following assertions are equivalent:
1) all composition operators are bounded onH2(β);
2) all mapsTa, for0< a <1, induce bounded composition operatorsCTa on H2(β);
3) for some a ∈(0,1), the map Ta induces a bounded composition operator CTa on H2(β);
4) β satisfies the∆2-condition.
Note that, by definition of the norm of H2(β), all rotations Rθ, θ ∈ R, induce bounded composition operators onH2(β)and send isometricallyH2(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 A2e
G := HG2, whereG(r) =e G(1−r2), defined as the spaces of analytic functions in D such that R
D|f(z)|2G(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 <21−β, thenCϕ is bounded onA2e
G if and only ifβ ≥α+ 1.
Here
βn = Z 1
0
tne−B/(1−√t)αdt;
and, sinceβn ≈ exp(−c nα/(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) =zn,n≥0.
As said in the introduction,H2is continuously embedded inH2(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 decreasingif, 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 onH2(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 spaceH2(β)is isomorphic toH2(β)e and H2(β)is continuously embedded inH2.
Definition 2.2. The sequence of positive numbers β = (βn) is said to satisfy the ∆2-conditionif there is a positive constantδ <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 minorationif there are positive constantsδandαsuch that, for all integersn≥1:
(2.3) βn ≥δ n−α.
That means thatH2(β)is continuously embedded in the weighted Bergman spaceB2αof the analytic functionsf:D→Csuch that
kfk2B2α:= (α+ 1) Z
D|f(z)|2(1− |z|2)αdA(z)<∞ sinceB2α=H2(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 ifβ 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−Aβp. Let now nbe an integer≥2, andk≥1the smallest integer such that2k≥n, so thatk≤alogn withaa positive constant. We get:
βn ≥C−1β2k≥C−1e−kAβ1≥C−1β1e−aAlogn=:ρ n−α, withρ=C−1β1andα=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
nδ;
henceβ has arbitrary polynomial minoration. However we have, fork≥2:
β2k!
βk!
= (k!)−δ
[(k−1)!]−δ = 1 kδ −→
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. Leta∈(0,1)and assume thatTa induces a bounded compo- sition operator onH2(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 H2(β), though all symbols ϕwithϕ(0) = 0are bounded, sinceβ is decreasing, as said by the forthcoming Proposition 3.2.
2) For the Dirichlet spaceD2, we haveβn =n+1, but all the mapsTainduce bounded composition operators onD2(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 H2(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∈/ H2(β)}, and
kCTak ≤
1 +a 1−a
τ
,
whereτ =12 supReW(A), withAthe infinitesimal generator of the continuous semigroup(St) defined as St =CTtanht, namely (Af)(z) = f′(z)(1−z2), and W(A)its numerical range.
Forβn = 1/(n+ 1)ν with 0≤ν ≤1, the two bounds coincide, so they get kCTak= 1+a1−a(ν+1)/2.
Proof of Proposition 2.5. Since
kKxk2= X∞ n=0
x2n 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
Ta′(t)dt= Z 1
un
1−a2 (1 +at)2dt;
hence
1−a
1 +a(1−un)≤1−un+1≤(1−a2)(1−un). Let0< x <1. We can findN ≥0 such thatuN ≤x < uN+1. Then:
1−x≤1−uN ≤(1−a2)N.
On the other hand, sinceCT∗aKz=KTa(z) for allz∈D, we have:
kKxk ≤ kKuN+1k ≤ kCTak kKuNk ≤ kCTakN+1kKu0k=kCTakN+1. Lets≥0 such that(1−a2)−s=kCTak. We obtain:
(2.6) kKxk ≤ kCTak 1
(1−x)s· But
kKxk2= X∞ k=0
x2k βk
;
so we get, for anyk≥2:
x2k
βk ≤ kCTak2 1 (1−x)2s· Takingx= 1−k1, we obtainβk ≥C k−2s.
Remark. We saw in the proof of Proposition 2.5 that if CTa is bounded on H2(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 kernelsKw of H2(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:
kKwk2= 1 β0
+ X∞ n=1
|w|2n βn ≤ 1
β0
+δ−1 X∞ n=1
nα|w|2n ≤ C (1− |w|2)α+1 ·
3 Boundedness of composition operators
We study in this section conditions ensuring that all composition operators onH2(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 at0, are bounded onH2(β), then the sequenceβ is bounded above.
Proof. Ifβis not bounded above, there is a subsequence(βnk), withnk≥1, such that βnk ≥4k. Thenϕ(z) = P∞
k=12−kznk defines a symbol, since |ϕ(z)| <1, andϕ(0) = 0; but:
kCϕ(e1)k2=kϕk2= X∞ k=1
4−kβnk=∞.
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 H2(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 ∈H2(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=0cnzn and(Cϕf)(z) =P∞
n=0dnzn, it says that:
kCϕfk2=|d0|2β0+ X∞ n=1
|dn|2βn≤ |c0|2β0+ X∞ n=1
|cn|2βn=kfk2,
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 = supϕ(0)=0kCϕk. Letm > n, and take:
ϕ(z) =z
1 +zm−n 2
1/n
.
Thenϕ(0) = 0and[ϕ(z)]n =zn+z2 m; hence 1
4(βn+βm) =kϕnk2=kCϕ(en)k2≤ kCϕk2kenk2≤M2βn, 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) = z2, then kCϕ(z2n+1)k2 = kz2(2n+1)k2 = β2(2n+1); sincekz2n+1k2=β2n+1, Cϕ is not bounded onH2(β).
A more interesting example is the following. For0< r <1, letβn=π n r2n. This sequence is eventually decreasing, so it is essentially decreasing. The square of the normkfk2H2(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 onH2(β)and
kCϕk ≤sup
n≥1
√n rn−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 = 2r2n, 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 at0 is bounded onH2(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. LetH2(β)be a weighted Hardy space withβ = (βn)essentially decreasing and satisfying the ∆2-condition. Then all composition operators on H2(β)are bounded.
For the proof, we need a lemma.
Lemma 3.5. For0< a <1, we write:
(3.2) (Taz)n=
X∞ m=0
am,nzm.
Then, for everyα >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) =rρ, for some positiveρ=ρ(a).
The Cauchy inequalities give:
|am,n| ≤ [M(r)]n
rm =rρn−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|eiθ, 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 H2 with respect to the basis (en)n≥0. On the other hand, Aβ is the matrix of CTa in H2(β) with respect to the basis(eβn)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 H2, 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= [Nk, Nk+1[ and Jk = [Nk+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
