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Weighted composition followed and proceeded by differentiation operators fromZygmund spaces to Bloch-type spaces

Abstract

The boundedness and compactness of the weighted composition followed andproceeded by differentiation operators from Zygmund spaces to Bloch-type spacesand little Bloch-type spaces are characterized.

MSC: 47B38, 30H30, 32C15.

1 Introduction

Let Δ={z:|z|<1} be the open unit disc in the complex plane ,and let H(Δ) be the class of all analytic functions on Δ. Theα-Bloch space B α (0<α<) is, by definition, the set of all functionsf in H(Δ) such that

f B α = | f ( 0 ) | + sup z Δ ( 1 | z | 2 ) α | f ( z ) | <.
(1.1)

Under the above norm, B α is a Banach space. When α=1, B 1 =B is the well-known Bloch space. Let B 0 α denote the subspace of B α , i.e.,

B 0 α = { f : ( 1 | z | 2 ) α | f ( z ) | 0  as  | z | 1 , f B α } .

This space is called the little α-Bloch space (see [1]).

Assume that μ is a positive continuous function on[0,1), having the property that there exist positivenumbers s and t, 0<s<t, and δ[0,1), such that

μ ( r ) ( 1 r ) s  is decreasing on  [ δ , 1 ) , lim r 1 μ ( r ) ( 1 r ) s = 0 , μ ( r ) ( 1 r ) t  is increasing on  [ δ , 1 ) , lim r 1 μ ( r ) ( 1 r ) t = .

Then μ is called a normal function (see [2]).

Denote (see, e.g., [35])

B μ = { f : f B μ = | f ( 0 ) | + sup z Δ μ ( | z | ) | f ( z ) | < , f H ( Δ ) } .
(1.2)

It is well known that B μ is a Banach space with the norm B μ (see [4]).

Let B μ , 0 denote the subspace of B μ , i.e.,

B μ , 0 = { f : μ ( | z | ) | f ( z ) | 0  as  | z | 1 , f B μ } .

This space is called the little Bloch-type space. When μ(r)= ( 1 r 2 ) α , the induced space B μ becomes the α-Bloch space B α .

An f in H(Δ) is said to belong to the Zygmund space, denoted by, if

sup | f ( e i ( θ + h ) ) + f ( e i ( θ h ) ) 2 f ( e i θ ) | h <,

where the supremum is taken over all e i θ Δ and h>0. By Theorem 5.3 in [6], we see that fZ if and only if

f Z = | f ( 0 ) | + | f ( 0 ) | + sup z Δ ( 1 | z | 2 ) | f ( z ) | <.
(1.3)

It is easy to check that is a Banach space under the above norm. For everyfZ, by using a result in [7], we have

| f ( z ) | C f Z ln e 1 | z | 2 .
(1.4)

Let Z 0 denote the subspace of consisting of thosefZ for which

lim | z | 1 ( 1 | z | 2 ) | f ( z ) | =0.

The space Z 0 is called the little Zygmund space. For thecorresponding n-dimensional Zygmund space see, e.g., [8] and [9].

Let φ be a nonconstant analytic self-map of Δ, and letϕ be an analytic function in Δ. We define the linearoperators

ϕ C φ Df=ϕ ( f φ ) =ϕ f (φ)andϕD C φ f=ϕ ( f φ ) =ϕ f (φ) φ ,for fH(Δ).

They are called weighted composition followed and proceeded by differentiationoperators, respectively, where C φ and D are composition and differentiationoperators respectively. Associated with φ is the composition operator C φ f=fφ and weighted composition operatorϕ C φ f=ϕfφ for ϕH(Δ) and fH(Δ). It is interesting to provide a function theoreticcharacterization for φ inducing a bounded or compact compositionoperator, weighted composition operator and related ones on various spaces (see,e.g., [1019]). For example, it is well known that C φ is bounded on the classical Hardy, Bloch and Bergmanspaces. Operators D C φ and C φ D as well as some other products of linear operatorswere studied, for example, in [2029] (see also the references therein). There has been some considerablerecent interest in investigation various type of operators from or to Zygmund typespaces (see, [7, 11, 23, 3037]).

In this paper, we investigate the operators ϕD C φ and ϕ C φ D from Zygmund spaces to Bloch-type spaces and littleBloch-type spaces. Some sufficient and necessary conditions for the boundedness andcompactness of these operators are given.

Throughout this paper, constants are denoted by C, they are positive and maydiffer from one occurrence to the other. The notation AB means that there is a positive constant Csuch that B C ACB.

2 Main results and proofs

In this section, we state and prove our main results. In order to formulate our mainresults, we quote several lemmas which will be used in the proofs of the mainresults in this paper. The following lemma can be proved in a standard way (see,e.g., Proposition 3.11 in [10]). Hence we omit the details.

Lemma 2.1 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. ThenϕD C φ (or ϕ C φ D):Z(or  Z 0 ) B μ is compact if and only ifϕD C φ (or ϕ C φ D):Z(or  Z 0 ) B μ is bounded and for any bounded sequence { f n } n N in(or Z 0 ) which converges to zero uniformly on compactsubsets of Δ asn, and ϕ D C φ f n B μ 0 (or ϕ C φ D f n B μ 0) asn.

Lemma 2.2[25]

A closed set of B μ , 0 is compact if and only if it is bounded and satisfied

lim | z | 1 sup f K μ ( | z | ) | f ( z ) | =0.

Theorem 2.3 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕD C φ :Z B μ is bounded;

  2. (ii)

    ϕD C φ : Z 0 B μ is bounded;

  3. (iii)
    sup z Δ μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 <
    (2.1)

and

sup z Δ μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 <.
(2.2)

Proof of Theorem 2.3 (i) (ii). This implication is obvious.

  1. (ii)

    (iii). Assume that ϕD C φ : Z 0 B μ is bounded, i.e., there exists a constant C such that

    ϕ D C φ f B μ C f Z

for all f Z 0 . Taking the functions f(z)=z Z 0 and f(z)= z 2 Z 0 respectively, we get

sup z Δ μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | < , sup z Δ μ ( | z | ) | ( ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ) φ ( z ) + ϕ ( z ) ( φ ( z ) ) 2 | < .
(2.3)

Using these facts and the boundedness of function φ, we have

sup z Δ μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | <.
(2.4)

Set

h(z)=(z1) [ ( 1 + ln 1 1 z ) 2 + 1 ]

and

h a (z)= h ( a ¯ z ) a ¯ ( ln 1 1 | a | 2 ) 1
(2.5)

for aΔ{0}. It is known that h a Z 0 (see [7]). Since

h a (z)= ( ln 1 1 a ¯ z ) 2 ( ln 1 1 | a | 2 ) 1
(2.6)

and

h a (z)= 2 a ¯ 1 a ¯ z ( ln 1 1 a ¯ z ) ( ln 1 1 | a | 2 ) 1 ,
(2.7)

for |φ(λ)|> 1 2 , we have

C ϕ D C φ Z 0 B μ ϕ D C φ h φ ( λ ) B μ μ ( | λ | ) | ϕ ( λ ) φ ( λ ) + ϕ ( λ ) φ ( λ ) | ln 1 1 | φ ( λ ) | 2 2 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 φ ( λ ) | 1 | φ ( λ ) | 2 .

Hence

μ ( | λ | ) | ϕ ( λ ) φ ( λ ) + ϕ ( λ ) φ ( λ ) | ln 1 1 | φ ( λ ) | 2 C ϕ D C φ Z 0 B μ + 2 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 φ ( λ ) | 1 | φ ( λ ) | 2 .
(2.8)

For aΔ{0}, set

f a (z)= h ( a ¯ z ) a ¯ ( ln 1 1 | a | 2 ) 1 0 z ln 1 1 a ¯ ω dω.
(2.9)

Then f a Z 0 . It is easy to see that

f a (z)= ( ln 1 1 a ¯ z ) 2 ( ln 1 1 | a | 2 ) 1 ln 1 1 a ¯ z , f a (a)=0,

and

f a (z)= 2 a ¯ 1 a ¯ z ( ln 1 1 a ¯ z ) ( ln 1 1 | a | 2 ) 1 a ¯ 1 a ¯ z , f a (a)= a ¯ 1 | a | 2 .

Therefore

C ϕ D C φ Z 0 B μ ϕ D C φ f φ ( λ ) B μ μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 φ ( λ ) | 1 | φ ( λ ) | 2 .
(2.10)

From (2.8) and (2.10), we have

sup | φ ( λ ) | > 1 2 μ ( | λ | ) | ϕ ( λ ) φ ( λ ) + ϕ ( λ ) φ ( λ ) | ln 1 1 | φ ( λ ) | 2 C ϕ D C φ Z 0 B μ <.
(2.11)

On the other hand, from the first inequality in (2.3), we have

sup | φ ( λ ) | 1 2 μ ( | λ | ) | ϕ ( λ ) φ ( λ ) + ϕ ( λ ) φ ( λ ) | ln 1 1 | φ ( λ ) | 2 sup λ Δ μ ( | λ | ) | ϕ ( λ ) φ ( λ ) + ϕ ( λ ) φ ( λ ) | ln 4 3 < .
(2.12)

Hence, from (2.3), (2.11), and (2.12), we obtain (2.2). Further, from (2.10), wehave

sup | φ ( λ ) | > 1 2 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 | 1 | φ ( λ ) | 2 sup | φ ( λ ) | > 1 2 2 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 φ ( λ ) | 1 | φ ( λ ) | 2 C ϕ D C φ Z 0 B μ < .
(2.13)

On the other hand, by (2.4), we have

sup | φ ( λ ) | 1 2 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 | 1 | φ ( λ ) | 2 sup | φ ( λ ) | 1 2 4 3 μ ( | λ | ) | ϕ ( λ ) ( φ ( λ ) ) 2 | <.
(2.14)

Combining (2.13) and (2.14), (2.1) follows.

  1. (iii)

    (i). Assume that (2.1) and (2.2) hold. Then, for every fZ, from (1.4), we have

    μ ( | z | ) | ( ϕ D C φ f ) ( z ) | = μ ( | z | ) | ϕ ( z ) φ ( z ) f ( φ ( z ) ) + ϕ ( z ) [ f ( φ ( z ) ) ( φ ( z ) ) 2 + f ( φ ( z ) ) φ ( z ) ] | μ ( | z | ) | ϕ ( z ) f ( φ ( z ) ) ( φ ( z ) ) 2 | + μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] f ( φ ( z ) | μ ( | z | ) C | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 f Z + μ ( | z | ) C | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | × ln e 1 | φ ( z ) | 2 f Z .
    (2.15)

Taking the supremum in (2.15) for zΔ, and employing (2.1) and (2.2), we deduce thatϕD C φ :Z B μ is bounded. The proof of Theorem 2.3 iscompleted. □

Theorem 2.4 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕD C φ :Z B μ is compact;

  2. (ii)

    ϕD C φ : Z 0 B μ is compact;

  3. (iii)

    ϕD C φ :Z B μ is bounded,

    lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 =0
    (2.16)

and

lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 =0.
(2.17)

Proof of Theorem 2.4 (i) (ii). This implication is clear.

  1. (ii)

    (iii). Assume that ϕD C φ : Z 0 B μ is compact. Then it is clear that ϕD C φ : Z 0 B μ is bounded. By Theorem 2.3 we know that ϕD C φ :Z B μ is bounded. Let ( z n ) n N be a sequence in Δ such that |φ( z n )|1 as n and φ( z n )0, nN (if such a sequence does not exist then (2.16) and (2.17) are vacuously satisfied). Set

    h n (z)= h ( φ ( z n ) ¯ z ) φ ( z n ) ¯ ( ln 1 1 | φ ( z n ) | 2 ) 1 ,nN.
    (2.18)

Then from the proof of Theorem 2.3, we see that h n Z 0 for each nN. Moreover h n 0 uniformly on compact subsets of Δ asn and

h n ( φ ( z n ) ) =ln 1 1 | φ ( z n ) | 2 , h n ( φ ( z n ) ) =ln 2 φ ( z n ) ¯ 1 | φ ( z n ) | 2 .

Since ϕD C φ : Z 0 B μ is compact, by Lemma 2.1, we have

lim n ϕ D C φ h n B μ =0.

On the other hand, similar to the proof of Theorem 2.3, we have

ϕ D C φ h n B μ | 2 μ ( | z n | ) | ϕ ( z n ) ( φ ( z n ) ) 2 | | φ ( z n ) | 1 | φ ( z n ) | 2 μ ( | z n | ) | × ϕ ( z n ) φ ( z n ) + ϕ ( z n ) φ ( z n ) | ln 1 1 | φ ( z n ) | 2 | ,

which implies that

lim n 2 μ ( | z n | ) | ϕ ( z n ) ( φ ( z n ) ) 2 | | φ ( z n ) | 1 | φ ( z n ) | 2 = lim n μ ( | z n | ) | ϕ ( z n ) φ ( z n ) + ϕ ( z n ) φ ( z n ) | ln 1 1 | φ ( z n ) | 2 ,
(2.19)

if one of these two limits exists.

Next, set

f n ( z ) = h ( φ ( z n ) ¯ z ) φ ( z n ) ¯ ( ln 1 1 | φ ( z n ) | 2 ) 1 0 z ln 3 1 1 φ ( z n ) ¯ ω d ω ( ln 1 1 | φ ( z n ) | 2 ) 2 .
(2.20)

Then f n Z 0 and f n converges to 0 uniformly on compact subsets of Δas n (see [7]). Since

f n (z)= ( ln 1 1 φ ( z n ) ¯ z ) 2 ( ln 1 1 | φ ( z n ) | 2 ) 1 ( ln 1 1 φ ( z n ) ¯ z ) 3 ( ln 1 1 | φ ( z n ) | 2 ) 2 ,

we have f n (φ( z n ))=0, for every nN and

f n ( φ ( z n ) ) = φ ( z n ) ¯ 1 | φ ( z n ) | 2 .

By using these facts, since ϕD C φ : Z 0 B μ is compact, and from Lemma 2.1, we find that

0 lim n μ ( | z n | ) | ϕ ( z n ) ( φ ( z n ) ) 2 | | φ ( z n ) | 1 | φ ( z n ) | 2 lim n ϕ D C φ f n B μ =0.

Therefore

lim n μ ( | z n | ) | ϕ ( z n ) ( φ ( z n ) ) 2 | 1 | φ ( z n ) | 2 = lim n μ ( | z n | ) | ϕ ( z n ) ( φ ( z n ) ) 2 φ ( z n ) | 1 | φ ( z n ) | 2 =0,

which implies (2.16). From this and (2.19), we have

lim n μ ( | z n | ) | ϕ ( z n ) φ ( z n ) + ϕ ( z n ) φ ( z n ) | ln 1 1 | φ ( z n ) | 2 =0.
(2.21)

From (2.21), it follows that lim n μ(| z n |)|ϕ( z n ) φ ( z n )+ ϕ ( z n ) φ ( z n )|=0, which altogether imply (2.17).

  1. (iii)

    (i). Suppose that ϕD C φ :Z B μ is bounded and conditions (2.16) and (2.17) hold. From Theorem 2.3, it follows that

    C 1 = sup z Δ μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | < , C 2 = sup z Δ μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | < .
    (2.22)

By the assumption, for every ε>0, there is a δ(0,1), such that

μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 < ε and μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 < ε ,
(2.23)

whenever δ<|φ(z)|<1.

Assume that ( f k ) k N is a sequence in such that sup k N f k Z L and f k converges to 0 uniformly on compact subsets of Δas k. Let K={zΔ:|φ(z)|δ}. Then by (1.4), (2.22), and (2.23), we have

sup z Δ μ ( | z | ) | ( ϕ D C φ f k ) ( z ) | = sup z Δ μ ( | z | ) | ϕ ( z ) φ ( z ) f k ( φ ( z ) ) + ϕ ( z ) [ f k ( φ ( z ) ) ( φ ( z ) ) 2 + f k ( φ ( z ) ) φ ( z ) ] | sup z Δ μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 f ( φ ( z ) ) | + sup z Δ μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] f ( φ ( z ) ) | sup z K μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 f ( φ ( z ) ) | + sup z K μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] f ( φ ( z ) ) | + sup z Δ K μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 f ( φ ( z ) ) | + sup z Δ K μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] f ( φ ( z ) ) | sup z K μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 f ( φ ( z ) ) | + sup z K μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] f ( φ ( z ) ) | + sup z Δ K μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 f k Z + C sup z Δ K μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 f k Z C 2 sup | ω | δ | f k ( ω ) | + C 1 sup | ω | δ | f k ( ω ) | + ( C + 1 ) ε f k Z ,

i.e. we obtain

ϕ D C φ f k B μ C 2 sup | ω | δ | f k ( ω ) | + C 1 sup | ω | δ | f k ( ω ) | + ( C + 1 ) ε f k Z + | ϕ ( 0 ) | | f k ( φ ( 0 ) ) | | φ ( 0 ) | .
(2.24)

Since f k converges to 0 uniformly on compact subsets of Δas k, from Cauchy’s estimate, it follows that f k 0 and f k 0 as k on compact subsets of Δ. Hence, lettingk in (2.24), and using the fact that ε isan arbitrary positive number, we obtain

lim k ϕ D C φ f k B μ =0.

By combining this with Lemma 2.1 the result easily follows. The proof ofTheorem 2.4 is completed. □

Theorem 2.5 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. ThenϕD C φ : Z 0 B μ , 0 is bounded if and only ifϕD C φ : Z 0 B μ is bounded and

lim | z | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | =0and lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | =0.
(2.25)

Proof of Theorem 2.5 Assume that ϕD C φ : Z 0 B μ , 0 is bounded. Then, it is clear thatϕD C φ : Z 0 B μ is bounded. Taking the test functionsf(z)=z and f(z)= z 2 respectively, we obtain (2.25).

Conversely, assume that ϕD C φ : Z 0 B μ is bounded and (2.25) holds. Then for each polynomialp, we have

μ ( | z | ) | ( ϕ D C φ p ) ( z ) | μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 p ( φ ( z ) ) | + μ ( | z | ) | [ ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) ] p ( φ ( z ) ) | .
(2.26)

In view of the facts

sup ω Δ | p ( ω ) | <, sup ω Δ | p ( ω ) | <,

from (2.25) and (2.26), it follows that ϕD C φ p B μ , 0 . Since the set of all polynomials is dense in Z 0 (see [23]), it follows that for every f Z 0 , there is a sequence of polynomials ( p n ) n N such that f p n Z 0 as n. Hence

ϕ D C φ f ϕ D C φ p n B μ ϕ D C φ Z 0 B μ f p n Z 0

as n. Since the operator ϕD C φ : Z 0 B μ is bounded, we have ϕD C φ ( Z 0 ) B μ , 0 , which implies the boundedness ofϕD C φ : Z 0 B μ , 0 . □

Theorem 2.6 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕD C φ :Z B μ , 0 is compact;

  2. (ii)

    ϕD C φ : Z 0 B μ , 0 is compact;

  3. (iii)
    lim | z | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 =0
    (2.27)

and

lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 =0.
(2.28)

Proof of Theorem 2.6 (i) (ii). This implication is trivial.

  1. (ii)

    (iii). Assume that ϕD C φ : Z 0 B μ , 0 is compact. Then ϕD C φ : Z 0 B μ , 0 is bounded. From the proof of Theorem 2.5, we have

    lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | =0
    (2.29)

and

lim | z | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | =0.
(2.30)

Hence, if φ <1, from (2.29) and (2.30), we obtain

lim | z | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 1 1 φ 2 lim | z | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | =0

and

lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 ln e 1 φ 2 lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | = 0 ,

from which the result follows in this case.

Now assume that φ =1. Let ( z k ) k N be a sequence such that |φ( z k )|1 as k. Since ϕD C φ : Z 0 B μ is compact, by Theorem 2.4, we have

lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 =0
(2.31)

and

lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 =0.
(2.32)

From (2.30) and (2.31), it follows that for every ε>0, there exists an r(0,1) such that μ(|z|) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 <ε, when r<|φ(z)|<1, and there exists a σ(0,1) such that μ(|z|)|ϕ(z) ( φ ( z ) ) 2 |ε(1 r 2 ), when σ<|z|<1. Therefore, when σ<|z|<1 and r<|φ(z)|<1, we have

μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 <ε.
(2.33)

On the other hand, if σ<|z|<1 and |φ(z)|r, we obtain

μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 <μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 r 2 <ε.
(2.34)

Inequality (2.33) together with (2.34) gives the (2.27). Similarly, (2.29) and(2.32) imply (2.28).

  1. (iii)

    (i). Let fZ. Then we have

    μ ( | z | ) | ( ϕ D C φ f ) ( z ) | C [ μ ( | z | ) | ϕ ( z ) ( φ ( z ) ) 2 | 1 | φ ( z ) | 2 + μ ( | z | ) | ϕ ( z ) φ ( z ) + ϕ ( z ) φ ( z ) | ln e 1 | φ ( z ) | 2 ] f Z .

Taking the supremum in this inequality over all fZ such that f Z 1, then letting |z|1, and using (2.27) and (2.28), we obtain

lim | z | 1 sup f Z 1 μ ( | z | ) | ( ϕ D C φ f ) ( z ) | =0.

From Lemma 2.2 it follows that the operator ϕD C φ :Z B μ , 0 is compact. □

Similarly to the proofs of Theorems 2.3-2.6, we can get the following results; weomit the proof.

Theorem 2.7 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕ C φ D:Z B μ is bounded;

  2. (ii)

    ϕ C φ D: Z 0 B μ is bounded;

  3. (iii)
    sup z Δ μ ( | z | ) | ϕ ( z ) φ ( z ) | 1 | φ ( z ) | 2 <

and

sup z Δ μ ( | z | ) | ϕ ( z ) | ln e 1 | φ ( z ) | 2 <.

Theorem 2.8 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕ C φ D:Z B μ is compact;

  2. (ii)

    ϕ C φ D: Z 0 B μ is compact;

  3. (iii)

    ϕ C φ D:Z B μ is bounded,

    lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) | 1 | φ ( z ) | 2 =0

and

lim | φ ( z ) | 1 μ ( | z | ) | ϕ ( z ) | ln e 1 | φ ( z ) | 2 =0.

Theorem 2.9 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Thenϕ C φ D: Z 0 B μ , 0 is bounded if and only ifϕ C φ D: Z 0 B μ is bounded and

lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) | =0and lim | z | 1 μ ( | z | ) | ϕ ( z ) | =0.

Theorem 2.10 Let φ be an analytic self-map of Δ, and let ϕ be an analytic function in Δ. Suppose that μ is normal. Then the following statements are equivalent.

  1. (i)

    ϕ C φ D:Z B μ , 0 is compact;

  2. (ii)

    ϕ C φ D: Z 0 B μ , 0 is compact;

  3. (iii)
    lim | z | 1 μ ( | z | ) | ϕ ( z ) φ ( z ) | 1 | φ ( z ) | 2 =0

and

lim | z | 1 μ ( | z | ) | ϕ ( z ) | ln e 1 | φ ( z ) | 2 =0.

References

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Acknowledgements

The authors would like to thank the referees for their valuable suggestions whichgreatly improve the present article. The work was supported by the UnitedTechnology Foundation of Science and Technology Department of Guizhou Provinceand Guizhou Normal University (Grant No. LKS[2012]12), and National NaturalScience Foundation of China (Grant No. 11171080, Grant No. 11171277).

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Long, J., Qiu, C. & Wu, P. Weighted composition followed and proceeded by differentiation operators fromZygmund spaces to Bloch-type spaces. J Inequal Appl 2014, 152 (2014). https://doi.org/10.1186/1029-242X-2014-152

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