A bounded linear operator T on a Hilbert space ℋ, satisfying ∥T2h∥2+∥h∥2≥2∥Th∥2 for every h∈ℋ, is called a convex operator. In this paper, we give necessary and sufficient conditions under which a convex
composition operator on a large class of weighted Hardy spaces is an isometry. Also, we discuss convexity of multiplication operators.

1. Introduction and Preliminaries

We denote by B(ℋ) the space of all bounded linear operators on a Hilbert space ℋ. An operator T∈B(ℋ) is said to be convex, if

∥T2h∥2+∥h∥2≥2∥Th∥2
for each h∈ℋ. Note that if T is a convex operator then the sequence (∥Tnh∥2)n∈N forms a convex sequence for every h∈ℋ. Taking ΔT=T*T-I, it is easily seen that T is a convex operator if and only if T*ΔTT≥ΔT.

A weighted Hardy space is a Hilbert space of analytic functions on the open unit disc D for which the sequence (zj)j=0∞ forms a complete orthogonal set of nonzero vectors. It is usually assumed that ∥1∥=1. Writing β(j)=∥zj∥, this space is denoted by H2(β) and its norm is given by

∥∑j=0∞ajzj∥2=∑j=0∞|aj|2β(j)2.

Let φ be an analytic map of the open unit disc D into itself, and define Cφ(f)=f∘φ whenever f is analytic on D. The function φ is called the symbol of the composition operator. For a positive integer n, the nth iterate of φ, denoted by φn, is the function obtained by composing φ with itself n times; also φ0 is defined to be the identity function. Denote the reproducing kernel at z∈D, for the space H2(β), by Kz. Then 〈f,Kz〉=f(z) for every f∈H2(β). It is known that Cφ*(Kz)=Kφ(z) for all z in D. The generating function for H2(β) is the function given by

k(z)=∑j=0∞zjβ(j)2.
This function is analytic on D. Moreover, if w∈D then Kw(z)=k(w̅z) and ∥Kw∥2=k(|w|2) (see [1]).

Recently, there has been a great interest in studying operator theoretic properties of composition and weighted composition operators, see, for example, monographs [1, 2], papers [3–16], as well as the reference therein.

Isometric operators on weighted Hardy spaces, especially those that are composition operators are discussed by many authors. Isometries of the Hardy space H2 among composition operators are characterized in [17, page 444], [18] and [12, page 66]. Indeed, it is shown that the only composition operators on H2 that are isometries are the ones induced by inner functions vanishing at the origin. Bayart [5] generalized this result and showed that every composition operator on H2 which is similar to an isometry is induced by an inner function with a fixed point in the unit disc. The surjective isometries of Hp, 1≤p<∞ that are weighted composition operators have been described by Forelli [19]. Carswell and Hammond [6] proved that the isometric composition operators of the weighted Bergman space Aα2 are the rotations. Cima and Wogen [20] have characterized all surjective isometries of the Bloch space. Furthermore, the identification of all isometric composition operators on the Bloch space is due to Colonna [8]. Some related results can be found also in [3, 4, 6, 21–25].

Herein, we are interested in studying the convexity of composition and multiplication operators acting on a weighted Hardy space H2(β). First, we give some preliminary facts on convex operators. Next, we will offer necessary and sufficient conditions under which a convex composition operator may be isometry on a large class of weighted Hardy spaces containing Hardy, Bergman, and Dirichlet spaces. We also discuss on convexity of the adjoint of a composition operator. Finally, we will obtain similar results for multiplication operators and their adjoints. For a good reference on isometric multiplication operators the reader can see [3].

Throughout this paper, T is assumed to be a bounded linear operator on a Hilbert space ℋ. It is easy to see that for every convex operator T, the sequence (T*nΔTTn)n forms an increasing sequence. We use this fact to prove the following theorem.

Theorem 1.1.

If T is a convex operator then so is every nonnegative integer power of T.

Proof.

We argue by using mathematical induction. The convexity of T implies that the result holds for k=1. Suppose that T*nΔTnTn≥ΔTn, then
T*n+1ΔTn+1Tn+1-ΔTn+1=T*n+1(T*ΔTnT+ΔT)Tn+1-ΔTn+1=T*2(T*nΔTnTn)T2+T*n+1ΔTTn+1-ΔTn+1≥T*2ΔTnT2+T*nΔTTn-ΔTn+1=T*2(T*nTn-I)T2+T*nΔTTn-T*(T*nTn-I)T-ΔT=T*n(T*2T2)Tn-T*2T2+T*nΔTTn-T*n(T*T)Tn+T*T-ΔT=T*n(T*2T2-I)Tn-T*2T2+I≥2T*nΔTTn-T*2T2+I≥2T*ΔTT-T*2T2+I=T*ΔTT-ΔT≥0.
So the result holds for k=n+1.

Proposition 1.2.

If T is a convex operator, then for every nonnegative integer n,
T*nTn≥nΔT+I.

Proof.

We give the assertion by using mathematical induction on n. The result is clearly true for n=1. Suppose that T*nTn≥nΔT+I. Thus,
T*n+1Tn+1=T*(T*nTn)T≥T*(nΔT+I)T=nT*ΔTT+T*T=n(T*2T2-2T*T+I)+nT*T+T*T-nI≥(n+1)T*T-nI=(n+1)ΔT+I.
So the result holds for k=n+1.

Proposition 1.3.

Let T∈ℬ(ℋ) be a convex operator and let h∈ℋ be such that supk≥0∥Tkh∥<∞. If ΔT≥0, then ∥Th∥=∥h∥.

Proof.

By applying Proposition 1.2, we observe that for every nonnegative integer n,
n〈ΔTh,h〉+∥h∥2≤∥Tnh∥2≤supk≥0∥Tkh∥2<∞.
Letting n→∞, the positivity of ΔT implies that ΔTh=0; hence, ∥Th∥=∥h∥.

Proposition 1.4.

Let {en}n=0∞ be an orthonormal basis for ℋ and let T∈ℬ(ℋ) be a convex operator satisfying ΔT≥0. Suppose that there is a nonnegative integer i and a scalar αi with 0<|αi|≤1 so that Tei=αiei, then ℳ=∨n≠i{en} is an invariant subspace for T.

Proof.

Using Proposition 1.2, we see that
∥ei∥2≥∥αinei∥2=∥Tnei∥2=〈T*nTnei,ei〉≥n〈ΔTei,ei〉+∥ei∥2
for every n≥0. Let n→∞. Since ΔT is a positive operator, we conclude that ΔTei=0. Consequently, T*ei=(1/αi)T*Tei=(1/αi)ei. Now, if f∈ℳ then 〈Tf,ei〉=0; hence, Tf∈ℳ.

2. Composition Operators

Our purpose in this section is to discuss on convex composition operators on a weighted Hardy space. Recall that an operator T in B(ℋ) is an isometry, if ΔT=0. At first, we give an example of a nonisometric composition operator T on a weighted Hardy space such that T*ΔTT≥ΔT≥0. For simplicity of notation, ΔCφ is denoted by Δφ.

Example 2.1.

Consider the weighted Hardy space H2(β) with weight sequence (β(n))n given by β(n)=n+1. Define φ:D→D by φ(z)=z2. It is easily seen that Cφ(H2(β))⊆H2(β), and an application of the closed graph theorem shows that Cφ is bounded. Now, a simple calculation shows that
〈(Cφ*ΔφCφ-Δφ)(zk),zk〉=∥Cφ2zk∥2-2∥Cφzk∥2+∥zk∥2>0
for all k≥0; besides
〈Δφzk,zk〉=∥Cφzk∥2-∥zk∥2
which is positive for all k≥1, and zero whenever k=0. It follows that Cφ*ΔφCφ≥Δφ≥0, but Cφ is not an isometry.

Proposition 2.2.

Suppose that T:H2(β)→H2(β) is a convex operator satisfying T1=1 and ΔT≥0, then
M={f∈H2(β):f(0)=0}
is a nontrivial invariant subspace of T.

Proof.

Clearly M is a nontrivial closed subspace of T. To show that M is invariant for T, apply Proposition 1.4 for the Hilbert space ℋ=H2(β), the orthonormal basis {en}n given by en=zn/β(n),i=0 and α0=1.

Example 2.3.

Consider the Bergman space A2(D) consisting of all analytic functions f on the open unit disc D, for which
∥f∥2=∫D|f(z)|2dA(z)<∞,
where dA(z) is the normalized Lebesgue area measure on D. If f∈A2(D) is represented by f(z)=∑n=0∞anzn, then
∥f∥2=∑n=0∞|an|2n+1.
Also, {zk}k forms an orthogonal basis for A2(D). Fix nonnegative integers k and n, and observe that
∥Cφnzk∥2=∥φkn∥2=∫D|φkn(z)|2dA(z)≤∫DdA(z)=1.
Thus, Proposition 1.3 implies that Cφ*ΔφCφ≥Δφ≥0 if and only if Cφ is an isometry. In this case, taking T=Cφ and f(z)=z in Proposition 2.2, we conclude that φ(0)=0; thus, the Schwarz lemma implies that |φ(z)|≤|z| for all z∈D. On the other hand, if f(z)=z then
∫D|φ(z)|2dA(z)=∥Cφf∥2=∥f∥2=∫D|z|2dA(z),
and so |φ(z)|=|z| almost everywhere with respect to the area measure. Hence, φ(z)=eiθz for some θ∈[0,2π).

Example 2.4.

Consider the Hardy space H2(D). If φ is an analytic self-map of the unit disc, then φ induces a bounded composition operator, and ∥Cφnzk∥≤1 for all nonnegative integers n and k. Thus, by Proposition 1.3, Cφ*ΔφCφ≥Δφ≥0 if and only if Cφ is an isometry.

Recall that the Dirichlet space 𝒟 is the set of all functions analytic on D whose derivatives lie in the Bergman space A2(D). The Dirichlet norm is defined by

∥f∥𝒟2=|f(0)|2+∫D|f′(z)|2dA(z).
If φ is a univalent self-map of D, then Cφ is bounded on 𝒟 [2, page 18]. Also, the area formula [1, page 36], shows that

∥Cφf∥𝒟2=|(foφ)(0)|2+∫D|f′(z)|2nφ(z)dA(z),
where nφ(z) is, as usual, the counting function defined as the cardinality of the set {w∈D:φ(w)=z}.

In the next theorem, we characterize all convex composition operators Cφ on 𝒟 satisfying Δφ≥0. Note that we cannot use Proposition 1.3 for the Dirichlet space, thanks to the fact that in general the positive powers of Cφ are not uniformly bounded on the zi's.

Theorem 2.5.

If Cφ is convex on the Dirichlet space 𝒟, then Δφ≥0 if and only if Cφ is an isometry.

Proof.

One implication is clear. Suppose that Δφ is a positive operator, and take T=Cφ in Proposition 2.2. Since the identity function is in the subspace M={f∈𝒟:f(0)=0}, we conclude that φ(0)=0. Thus, in light of (2.9), to show that Cφ is an isometry it is sufficient to prove that
∫D|f′(z)|(1-nφ)(z)dA(z)=0,∀f∈𝒟.
Let f be any function in the Dirichlet space 𝒟. Then
0≤〈(Cφ*ΔφCφ-Δφ)(f),f〉=∫D|f′(z)|2(nφ2-2nφ+1)(z)dA(z).
Furthermore,
0≤〈Δφf,f〉=∫D|f′(z)|2(nφ-1)(z)dA(z).
By summing up these two relations we get
∫D|f′(z)|2(nφ2-nφ)(z)dA(z)≥0.
But nφ2(z)≤nφ(z), and so
∫D|f′(z)|2(nφ2-nφ)(z)dA(z)=0,∀f∈𝒟.
This, in turn, implies that nφ2(z)=nφ(z) almost everywhere. Substituting this in (2.11), and then considering (2.12) the assertion will be completed.

Observe that if φ(0)=0, nφ2-2nφ+1≥0 almost everywhere, and Cφ is bounded on 𝒟 then it is convex. Indeed,

In the next theorem, we turn to the adjoint of a composition operator and give necessary and sufficient conditions under which a convex operator Cφ* is an isometry.

Theorem 2.6.

Let φ be an analytic self-map of D with φ(0)=0. If Cφ* is a convex operator on H2(β), then it is an isometry if and only if ΔCφ*≥0.

Proof.

Suppose that ΔCφ*≥0, and assume that φ is not the identity or an elliptic automorphism. By the Denjoy-Wolff theorem φn converges uniformly to zero on compact subsets of D [1], and so for every z∈D,
limn→∞∥Kφn(z)∥=∥K0∥.
Proposition 1.2 coupled with the fact that Cφ*nKz=Kφn(z) implies that for all z∈D and all nonnegative integers n,
∥Kφn(z)∥2≥n(∥Kφ(z)∥2-∥Kz∥2)+∥Kz∥2.
Furthermore, the positivity of ΔT shows that ∥Kφ(z)∥≥∥Kz∥. Thus, in light of (2.16) and (2.17) we conclude that ∥Kz∥=∥Kφ(z)∥ for all z∈D, and so ∥Kz∥=∥Kφn(z)∥ for every positive integer n. Consequently, ∥Kz∥=∥K0∥ for all z∈D. It follows that
1=∥K0∥2=∥Kz∥2=k(|z|2)=1+∑j=1∞(|z|2)jβ(j)2,forz∈D.
This contradiction shows that φ is the identity or an elliptic automorphism. Thus, there is a θ∈[0,2π) so that φ(z)=eiθz for all z∈D. Now, if ω∈D then
Cφ*Kω(z)=Kφ(ω)(z)=k(φ(ω)¯z)=Kω(e-iθz)=Kω(φ-1(z))=Cφ-1Kω(z).

It follows that Cφ*=Cφ-1. But it is easily seen that ∥Cφ-1f∥=∥f∥ for every f∈H2(β). Hence, Cφ* is an isometry. The converse is obvious.

3. Multiplication Operators

This section deals with convex multiplication operators on a weighted Hardy space. Recall that a multiplier of H2(β) is an analytic function φ on D such that φH2(β)⊆H2(β). The set of all multipliers of H2(β) is denoted by M(H2(β)). It is known that M(H2(β))⊆H∞. In fact, if φ∈M(H2(β)) and f is the constant function 1 then for every positive integer n and for every z∈D we have

|φ(z)|=|〈Mφnf,Kz〉|1/n≤∥Mφnf∥1/n∥Kz∥1/n≤∥Mφ∥∥Kz∥1/n.
Now, letting n→∞, we conclude that φ is bounded. This coupled with the fact that φ∈H2(β) implies that φ∈H∞. If φ is a multiplier, then the multiplication operator Mφ, defined by Mφf=φf, is bounded on H2(β). Also note that for each λ∈D, Mφ*Kλ=φ(λ)¯Kλ.

In what follows, the operator Mφ is assumed to be convex. First, we present an example of a nonisometric convex multiplication operator T with ΔT≥0.

Example 3.1.

Consider the weighted Hardy space H2(β) with weight sequence (β(n))n given by β(n)=n+1. Define the mapping φ on D by φ(z)=z2. Obviously, Mφ is bounded. Furthermore, it is easy to see that for every nonnegative integer k,
∥Mφ2zk∥2-2∥Mφzk∥2+∥zk∥2>0,∥Mφzk∥>∥zk∥.
Consequently, Mφ is convex but not an isometry. Besides, ΔMφ is a positive operator.

Theorem 3.2.

Let H∞ consist of all multipliers of H2(β), and let φ∈H∞ be such that ∥φ∥∞≤1. If T=Mφ or T=Mφ* then T*ΔTT≥ΔT≥0 if and only if T is an isometry.

Proof.

Suppose that T is Mφ or Mφ* and T*ΔTT≥ΔT≥0. Define the linear mapping S:H∞→ℬ(H2(β)) by S(ψ)=Mψ. An application of the closed graph theorem implies that S is bounded. Therefore, there is c>0 such that for all ψ∈H∞,
∥Mψ∥≤c∥ψ∥∞.
It follows that for every f∈H2(β) and every nonnegative integer n,
∥Mφnf∥≤c∥φn∥∞∥f∥≤c∥f∥.
Thus, supn≥0∥Mφnf∥<∞ for every f∈H2(β). Since ∥Mψ*∥=∥Mψ∥ for all ψ∈H∞, by a similar method one can show that supn≥0∥Mφ*nf∥<∞ for all f∈H2(β). Therefore, the result follows from Proposition 1.3.

Example 3.3.

Let ℋ be the Bergman space or the Hardy space and let T be Mφ or its adjoint on ℋ. It is well known that M(ℋ)=H∞. So if φ is a multiplier with ∥φ∥∞≤1, then by applying the preceding theorem, we observe that T*ΔTT≥ΔT≥0 if and only if T is an isometry.

We remark herein that if φ(z)=z and T=Mφ on the Dirichlet space 𝒟, then it is easily seen that T*ΔTT≥ΔT≥0 but T is not an isometry.

Acknowledgments

The authors would like to thank Dr. Faghih Ahmadi for her assistance and the referee for a number of helpful comments and suggestions. This research was in part supported by a grant no. (88-GR-SC-27) from Shiraz University Research Council.

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