An Integral-Type Operator from H

MALAYSIANMATHEMATICAL

SCIENCESSOCIETY http://math.usm.my/bulletin

An Integral-Type Operator from H

to Zygmund-Type Spaces

XIANGLINGZHU

Department of Mathematics, Jiaying University, Meizhou, Guangdong, 514015, China [email protected]; [email protected]

Abstract. Letg∈H(D),nbe a nonnegative integer andϕbe an analytic self-map ofD.

We study the boundedness and compactness of the integral operatorCⁿ_ϕ,gdefined by (C_ϕ,gⁿ f)(z) =

Zz 0

f⁽ⁿ⁾(ϕ(ξ))g(ξ)dξ, z∈D, f∈H(D), fromH^∞to Zygmund-type spaces on the unit disk.

2010 Mathematics Subject Classification: Primary: 47B38; Secondary: 30H05

Keywords and phrases: Integral-type operator, composition operator, Zygmund-type space.

1. Introduction

A positive continuous functionφon[0,1)is called normal if there exist positive numberss andt,0<s<t,andδ∈[0,1)such that

φ(r)

(1−r)^s is decreasing on [δ,1)and lim

r→1

φ(r) (1−r)^s =0;

φ(r)

(1−r)^t is increasing on [δ,1) and lim

r→1

φ(r) (1−r)^t =∞

(see, e.g. [11]). From now on we always assume thatω andµ are normal functions and non-negative functions on[0,1)such thatω(t_n),µ(t_n)>0 for some sequence{t_n}^∞₀ ⊂[0,1) with limn→∞t_n=1.

LetDbe the open unit disk in the complex planeC, andH(D)be the class of all analytic functions onD. We denote byH^∞=H^∞(D)the bounded analytic function space onD. An

f ∈H(D)is said to belong to the Zygmund-type space, denoted byZµ, if sup

z∈D

µ(|z|)|f⁰⁰(z)|<∞.

Under the norm

kfk_Zµ =|f(0)|+|f⁰(0)|+sup

z∈D

µ(|z|)|f⁰⁰(z)|,

Communicated byMohammad Sal Moslehian.

Received:August 23, 2010;Revised:October 12, 2010.

it is easy to see thatZµ is a Banach space. The little Zygmund-type spaceZµ,0is defined to be the subspace ofZµconsisting of those f ∈Zµ such that

lim

|z|→1µ(|z|)|f⁰⁰(z)|=0.

Whenµ(r) = (1−r²), the induced spacesZµ andZµ,0 become the classical Zygmund space and the little Zygmund space respectively (see [2, 4, 6]).

Letϕbe an analytic self-map ofD. The composition operatorCϕis defined by (C_ϕf)(z) =f(ϕ(z)), f ∈H(D).

It will be of interest to provide a function theoretic characterization of when ϕ induces a bounded or compact composition operator between spaces of analytic functions. The composition operator has been studied by many researchers on various spaces (see, e.g., [1, 18] and the references therein).

Letg∈H(D)andϕbe an analytic self-map ofD. In [6], the authors defined and studied the generalized composition operator as follows

(C_ϕ^gf)(z) = Z z

0

f⁰(ϕ(ξ))g(ξ)dξ, f∈H(D),z∈D.

The boundedness and compactness of the generalized composition operator on Zygmund spaces and Bloch spaces were investigated in [6]. In [3], Li studied another type Volterra composition operator between weighted Bergman spaces and Bloch spaces. In [22], the author of this paper generalized the operatorC_ϕ^gto the unit ball and studied the boundedness and compactness of the corresponding operator on some function spaces. Some related results can be found, e.g., in [5, 7–9, 12–17, 19–22].

Here we generalize the generalized composition operatorC_ϕ^gfrom another point of view.

Letg∈H(D),nbe a nonnegative integer andϕbe an analytic self-map ofD. We define (C_ϕ,gⁿ f)(z) =

Z z 0

f⁽ⁿ⁾(ϕ(ξ))g(ξ)dξ, z∈D, f ∈H(D).

Whenn=1, thenC_ϕ,g¹ is the generalized composition operator defined by Li and Stevi´c in [6]. Whenn=0, thenC⁰_ϕ,gis the Volterra composition operator defined by Li in [3]. To the best of our knowledge, the operatorCⁿ_ϕ,gis studied in the present paper for the first time.

The purpose of this paper is to study the operatorC_ϕ,gⁿ . The boundedness and compact- ness of the operatorC_ϕ,gⁿ fromH^∞to Zygmund-type spaces are completely characterized.

Throughout the paper,Cdenotes a positive constant which may differ from one occurrence to the other.

2. Main results and proofs

In this section, we give the main results and proofs. Before proving the main results, it is necessary to give some lemmas. By standard arguments (see for e.g. [1, Proposition 3.11]), the following lemma follows.

Lemma 2.1. Let g∈H(D), n be a nonnegative integer andϕ be an analytic self-map of D. Then Cⁿ_ϕ,g:H^∞→Zµ is compact if and only if C_ϕ,gⁿ :H^∞→Zµis bounded and for any bounded sequence(f_k)_k∈Nin H^∞which converges to zero uniformly on compact subsets of Das k→∞, we havekCⁿ_ϕ,gf_kk_Zµ →0as k→∞.

Lemma 2.2. A closed set K inZµ,0is compact if and only if it is bounded and satisfies

|z|→1lim sup

f∈K

µ(|z|)|f⁰⁰(z)|=0.

(2.1)

Proof. The proof is similar to that of [10, Lemma 1], and the details are omitted here.

By the Cauch integral formula, we have

Lemma 2.3. Let f ∈H^∞. Then for each m∈N, there is a positive constant C independent of f such that

sup

z∈D

(1− |z|²)^m|f^(m)(z)| ≤Ckfk_∞. (2.2)

Now we are in a position to state and prove the main results of this paper.

Theorem 2.1. Let g∈H(D), n be a nonnegative integer andϕbe an analytic self-map of D. Then Cⁿ_ϕ,g:H^∞→Zµ is bounded if and only if

sup

z∈D

µ(|z|)|ϕ⁰(z)||g(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ <∞ and sup

z∈D

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ<∞.

(2.3)

Proof. Suppose thatC_ϕ,gⁿ :H^∞→Zµ is bounded, i.e., there exists a constantC such that kC_ϕ,gⁿ fk_Zµ ≤Ckfk_∞ for all f ∈H^∞. Taking f(z) =zⁿ and f(z) =zⁿ⁺¹, and using the boundedness of the functionϕ(z), we get

sup

z∈D

µ(|z|)|g⁰(z)|<∞, (2.4)

and

sup

z∈D

µ(|z|)|g(z)||ϕ⁰(z)|<∞.

(2.5)

Forw∈D, set

h_w(z) =1− |w|² 1−wz − 1

n+1

(1− |w|²)² (1−wz)² .

It is easy to check thath_w∈H^∞,kh_wk_∞<(2n+6)/(n+1)for everyw∈D, h⁽ⁿ⁾_ϕ(λ)(ϕ(λ)) =0 and |h⁽ⁿ⁺¹⁾_ϕ(λ) (ϕ(λ))|= n!|ϕ(λ)|ⁿ⁺¹

(1− |ϕ(λ)|²)ⁿ⁺¹. It follows that

∞>kC_ϕ,gⁿ h_ϕ(λ)k_Zµ≥n!µ(|λ|)|g(λ)||ϕ⁰(λ)||ϕ(λ)|ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹ (2.6)

for everyλ ∈D.

For any fixedr∈(0,1), from (2.6) we have sup

|ϕ(λ)|>r

µ(|λ|)|g(λ)||ϕ⁰(λ)|

(1− |ϕ(λ)|²)ⁿ⁺¹ ≤ sup

|ϕ(λ)|>r

1 rⁿ⁺¹

µ(|λ|)|g(λ)||ϕ⁰(λ)||ϕ(λ)|ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹

≤CkCⁿ_ϕ,gk_H∞→Zµ <∞.

(2.7) By (2.5),

sup

|ϕ(λ)|≤r

µ(|λ|)|g(λ)||ϕ⁰(λ)|

(1− |ϕ(λ)|²)ⁿ⁺¹ ≤ 1

(1−r²)ⁿ⁺¹ sup

|ϕ(λ)|≤r

µ(|λ|)|g(λ)||ϕ⁰(λ)|<∞.

(2.8)

Therefore, (2.7) and (2.8) yield the first inequality of (2.3).

Next, set f_w(z) = (1− |w|²)/(1−wz).Then f_w∈H^∞and sup_w∈Dkf_wk∞≤2.Hence,

∞>2kC_ϕ,gⁿ k_H∞→Zµ ≥ kC_ϕ,gⁿ f_ϕ(λ)k_Zµ

≥sup

z∈D

µ(|z|)|(Cⁿ_ϕ,gf_ϕ(λ))⁰⁰(z)|

=sup

z∈D

µ(|z|)

f_ϕ(λ⁽ⁿ⁺¹⁾₎ (ϕ(z))g(z)ϕ⁰(z) +f_ϕ(λ⁽ⁿ⁾₎(ϕ(z))g⁰(z)

≥µ(|λ|)

n!g⁰(λ)(ϕ(λ))ⁿ

(1− |ϕ(λ)|²)ⁿ +(n+1)!g(λ)ϕ⁰(λ)(ϕ(λ))ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹

≥µ(|λ|)

n!g⁰(λ)(ϕ(λ))ⁿ (1− |ϕ(λ)|²)ⁿ

−µ(|λ|)

(n+1)!g(λ)ϕ⁰(λ)(ϕ(λ))ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹

=n!µ(|λ|)|g⁰(λ)||ϕ(λ)|ⁿ

(1− |ϕ(λ)|²)ⁿ −(n+1)!µ(|λ|)|g(λ)||ϕ⁰(λ)||ϕ(λ)|ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹

for everyλ ∈D. Therefore µ(|λ|)|g⁰(λ)||ϕ(λ)|ⁿ

(1− |ϕ(λ)|²)ⁿ ≤ 2

n!kC_ϕ,gⁿ k_H∞→Zµ+(n+1)µ(|λ|)|g(λ)||ϕ⁰(λ)||ϕ(λ)|ⁿ⁺¹ (1− |ϕ(λ)|²)ⁿ⁺¹ . (2.9)

From (6) and (9), we get

sup

λ∈D

µ(|λ|)|g⁰(λ)||ϕ(λ)|ⁿ (1− |ϕ(λ)|²)ⁿ <∞.

(2.10)

Combining (2.10) with (2.4), similar to the former proof, we get the second inequality of (2.3).

For the converse, suppose that (2.3) holds. For any f∈H^∞, by Lemma 2.3, we have µ(|z|)|(C_ϕ,gⁿ f)⁰⁰(z)|=µ(|z|)|(f⁽ⁿ⁾(ϕ)g)⁰(z)|

≤µ(|z|)|g(z)||ϕ⁰(z)||f⁽ⁿ⁺¹⁾(ϕ(z))|+µ(|z|)|g⁰(z)||f⁽ⁿ⁾(ϕ(z))|

≤Cµ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ kfk_∞+C µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿkfk_∞. Moreover,|(C_ϕ,gⁿ f)(0)|=0 and

|(C_ϕ,gⁿ f)⁰(0)|=|f⁽ⁿ⁾(ϕ(0))g(0)| ≤ |g(0)|

(1− |ϕ(0)|²)ⁿkfk_∞. From (2.3), we see that

kC_ϕ,gⁿ fk_Zµ =|(Cⁿ_ϕ,gf)(0)|+|(Cⁿ_ϕ,gf)⁰(0)|+sup

z∈D

µ(|z|)|(C_ϕ,gⁿ f)⁰(z)|<∞.

ThereforeC_ϕ,gⁿ :H^∞→Zµ is bounded. The proof of the theorem is complete.

Theorem 2.2. Let g∈H(D), n be a nonnegative integer andϕbe an analytic self-map of D. Then Cⁿ_ϕ,g:H^∞→Zµ is compact if and only if C_ϕ,gⁿ :H^∞→Zµis bounded,

|ϕ(z)|→1lim

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ =0 and lim

|ϕ(z)|→1

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ =0.

(2.11)

Proof. Suppose thatC_ϕ,gⁿ :H^∞→Zµ is compact. ThenC_ϕ,gⁿ :H^∞→Zµ is bounded. Let (z_k)_k∈Nbe a sequence inDsuch that|ϕ(z_k)| →1 ask→∞. Set

h_k(z) =1− |ϕ(z_k)|² 1−ϕ(z_k)z − 1

n+1

(1− |ϕ(z_k)|²)²

(1−ϕ(z_k)z)² , k∈N.

Notice thath_k is a sequence inH^∞and converges to 0 uniformly on compact subsets ofD ask→∞,

h⁽ⁿ⁾_k (ϕ(z_k)) =0 and |h⁽ⁿ⁺¹⁾_k (ϕ(z_k))|= n!|ϕ(z_k)|ⁿ⁺¹ (1− |ϕ(z_k)|²)ⁿ⁺¹.

The compactness ofC_ϕ,gⁿ :H^∞→Zµ implies lim_k→∞kCⁿ_ϕ,gh_kk_Zµ =0.On the other hand, similar to the proof of Theorem 2.1, we have

n!µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹

(1− |ϕ(z_k)|²)ⁿ⁺¹ ≤ kC_ϕ,gⁿ h_kk_Zµ, i.e. we get

k→∞lim

µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹ (1− |ϕ(z_k)|²)ⁿ⁺¹ =lim

k→∞kC_ϕ,gⁿ h_kk_Zµ =0.

Therefore lim

|ϕ(z_k)|→1

µ(|z_k|)|g(z_k)||ϕ⁰(z_k)|

(1− |ϕ(z_k)|²)ⁿ⁺¹ = lim

|ϕ(z_k)|→1

µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹ (1− |ϕ(z_k)|²)¹⁺ⁿ

=lim

k→∞

µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹ (1− |ϕ(z_k)|²)ⁿ⁺¹ =0.

(2.12)

Next, set

f_k(z) =1− |ϕ(z_k)|²

1−ϕ(z_k)z , k∈N.

Thenf_k∈H^∞andf_k→0 uniformly on compact subsets ofDask→∞. SinceC_ϕ,gⁿ :H^∞→ Zµis compact, we have lim_k→∞kC_ϕ,gⁿ f_kk_Zµ=0.On the other hand, we have

kC_ϕ,gⁿ f_kk_Zµ≥n!µ(|z_k|)|g⁰(z_k)||ϕ(z_k)|ⁿ

(1− |ϕ(z_k)|²)ⁿ −(n+1)!µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹ (1− |ϕ(z_k)|²)ⁿ⁺¹ , which implies that

|ϕ(zlim_k)|→1

(n+1)µ(|z_k|)|g(z_k)||ϕ⁰(z_k)||ϕ(z_k)|ⁿ⁺¹

(1− |ϕ(z_k)|²)¹⁺ⁿ = lim

|ϕ(z_k)|→1

µ(|z_k|)|g⁰(z_k)||ϕ(z_k)|ⁿ (1− |ϕ(z_k)|²)ⁿ , if one of these two limits exists. From the last equality and (2.12), we have

lim

|ϕ(z_k)|→1

µ(|z_k|)|g⁰(z_k)|

(1− |ϕ(z_k)|²)ⁿ = lim

|ϕ(z_k)|→1

µ(|z_k|)|g⁰(z_k)||ϕ(z_k)|ⁿ (1− |ϕ(z_k)|²)ⁿ =0.

(2.13)

From (2.12) and (2.13) we obtain the desired results.

Conversely, suppose thatC_ϕ,gⁿ :H^∞→Zµis bounded and (2.11) holds. Assume(f_k)k∈N

is a sequence inH^∞such thatf_kconverges to 0 uniformly on compact subsets ofDask→∞.

By the assumption, for anyε>0, there exists aδ∈(0,1), µ(|z|)|ϕ⁰(z)||g(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ <ε and µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ<ε, (2.14)

whenδ<|ϕ(z)|<1. By the boundedness ofC_ϕ,gⁿ :H^∞→Zµand the proof of Theorem 2.1, C1=sup

z∈D

µ(|z|)|g⁰(z)|<∞ and C2=sup

z∈D

µ(|z|)|g(z)||ϕ⁰(z)|<∞.

(2.15)

LetK={z∈D:|ϕ(z)| ≤δ}. Then by (2.14) and (2.15), we have that sup

z∈D

µ(|z|)|(Cⁿ_ϕ,gf_k)⁰⁰(z)|

≤sup

z∈K

µ(|z|)|g(z)||ϕ⁰(z)||f_k⁽ⁿ⁺¹⁾(ϕ(z))|+sup

z∈K

µ(|z|)|g⁰(z)||f_k⁽ⁿ⁾(ϕ(z))|

+C sup

z∈D\K

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ kf_kk_∞+C sup

z∈D\K

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿkf_kk_∞

≤C2sup

z∈K

|f_k⁽ⁿ⁺¹⁾(ϕ(z))|+C₁sup

z∈K

|f_k⁽ⁿ⁾(ϕ(z))|+Cεkf_kk_∞, i.e. we get

kCⁿ_ϕ,gf_kk_Zµ ≤C₂ sup

|w|≤δ

|f_k⁽ⁿ⁺¹⁾(w)|+C₁ sup

|w|≤δ

|f_k⁽ⁿ⁾(w)|

+Cεkf_kk_∞+|g(0)||f_k⁽ⁿ⁾(ϕ(0))|.

(2.16)

Since f_kconverges to 0 uniformly on compact subsets ofDask→∞, Cauchy’s estimate gives that f_k⁽ⁿ⁾→0 ask→∞on compact subsets ofD. Hence, lettingk→∞in (2.16), and using the fact thatεis an arbitrary positive number, we obtain limk→∞kC_ϕ,gⁿ f_kk_Zµ =0.

Applying Lemma 2.1 the result follows.

Theorem 2.3. Let g∈H(D), n be a nonnegative integer andϕbe an analytic self-map of D. Then Cⁿ_ϕ,g:H^∞→Zµ,0is compact if and only if

lim

|z|→1

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ =0 and lim

|z|→1

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ=0.

(2.17)

Proof. Assume thatCⁿ_ϕ,g:H^∞→Zµ,0is compact. ThenC_ϕ,gⁿ :H^∞→Zµ is compact and C_ϕ,gⁿ :H^∞→Zµ,0is bounded. Takingf(z) =zⁿandf(z) =zⁿ⁺¹, and using the boundedness ofC_ϕ,gⁿ :H^∞→Zµ,0and the functionϕ(z), we get

|z|→1lim µ(|z|)|g⁰(z)|=0 (2.18)

and

lim

|z|→1µ(|z|)|g(z)||ϕ⁰(z)|=0.

(2.19)

Ifkϕk_∞<1,from (2.18) and (2.19) we get lim

|z|→1

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ ≤ 1

(1− kϕk²_∞)ⁿ⁺¹ lim

|z|→1µ(|z|)|g(z)||ϕ⁰(z)|=0 and

lim

|z|→1

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ≤ 1

(1− kϕk²_∞)ⁿ lim

|z|→1µ(|z|)|g⁰(z)|=0.

The result follows.

Now we assume thatkϕk_∞=1.By the compactness ofC_ϕ,gⁿ :H^∞→Zµand Theorem 2.2 we have

lim

|ϕ(z)|→1

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ =0 (2.20)

and

|ϕ(z)|→1lim

µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ=0.

(2.21)

From (2.18) and (2.21), for anyε>0, there exists anr∈(0,1)such that µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ <ε whenr<|ϕ(z)|<1 and there exists aσ∈(0,1)such that

µ(|z|)|g⁰(z)| ≤ε(1−r²)ⁿ

whenσ<|z|<1.Therefore, whenσ<|z|<1 andr<|ϕ(z)|<1, we have µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ <ε.

(2.22)

On the other hand, whenσ<|z|<1 and|ϕ(z)| ≤r, we have µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ < 1

(1−r²)ⁿµ(|z|)|g⁰(z)|<ε.

(2.23)

Combining (2.22) and (2.23), we obtain the second equality of (2.17). Similar to the above proof we get the first equality of (2.17).

Conversely suppose that (2.17) holds. Letf ∈H^∞. We have µ(|z|)|(C_ϕ,gⁿ f)⁰⁰(z)| ≤C

µ(|z|)|g(z)||ϕ⁰(z)|

(1− |ϕ(z)|²)ⁿ⁺¹ + µ(|z|)|g⁰(z)|

(1− |ϕ(z)|²)ⁿ

kfk_∞.

Taking the supremum in this inequality over all f ∈H^∞such thatkfk_∞≤1,apply (2.17) we obtain

|z|→1lim sup

kfk_∞≤1

µ(|z|)|(C_ϕ,gⁿ f)⁰⁰(z)|=0.

The result follows from Lemma 2.2.

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