# Products of Composition and Differentiation Operators from Bloch intoQ

Authors: Shunlai Wang; Taizhong Zhang
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

The boundedness and compactness of the product of differentiation and composition operators from Bloch spaces intoQ
K spaces are discussed in this paper.


---

## Body

## 1. Introduction and Motivation

LetΔ be the open unit disk in the complex plane and let H
(
Δ
) be the class of all analytic functions on Δ. Let d
A
(
z
) be the Euclidean area element on Δ. The Bloch space B on Δ is the space of all analytic functions f on Δ such that (1)
f
B
=
f
0
+
s
u
p
z
∈
Δ
1
-
z
2
f
′
z
<
∞
.Under the above norm, B is a Banach space. Let B
0 denote the subspace of B consisting of those f
∈
B for which (
1
-
z
2
)
f
′
z
→
0 as z
→
1. This space is called the little Bloch space.Throughout this paper, we assume thatK
:
[
0
,
∞
)
→
[
0
,
∞
) is a nondecreasing and right-continuous function. A function f
∈
H
(
Δ
) is said to belong to Q
K space (see [1]) if (2)
f
K
2
=
s
u
p
a
∈
Δ
∬
Δ
f
′
z
K
g
z
,
a
d
A
z
<
∞
,where g
(
z
,
a
) is the Green function with logarithmic singularity at a; that is, g
(
z
,
a
)
=
log
⁡
(
1
/
|
φ
a
(
z
)
|
) (φ
a is a conformal automorphism defined by φ
a
(
z
)
=
(
a
-
z
)
/
(
1
-
a
z
¯
) for a
∈
Δ). Q
K is a Banach space under the norm (3)
f
Q
K
2
=
f
0
+
f
K
.From [1], we know that Q
K
⊆
B if (4)
∫
0
1
/
e
K
-
log
⁡
r
r
d
r
<
∞
.Letφ denote a nonconstant analytic self-map of Δ. Associated with φ is the composition operator C
φ defined by C
φ
(
f
)
=
f
∘
φ for f
∈
H
(
Δ
). The problem of characterizing the boundedness and compactness of composition operators on many Banach spaces of analytic functions has attracted lots of attention recently, for example, [2] and the reference therein.LetD be the differentiation operator on H
(
Δ
); then we have D
f
(
z
)
=
f
′
(
z
). For f
∈
H
(
Δ
), the products of differentiation and composition operators D
C
φ and C
φ
D are defined by (5)
D
C
φ
f
=
f
∘
φ
′
=
f
′
φ
φ
′
,
C
φ
D
f
=
f
′
φ
,
f
∈
H
Δ
.Operators C
φ
D as well as some other products of linear operators were studied, for example, in [3–9] (see also the references therein).Recall that a linear operatorT
:
X
→
Y is said to be bounded if there exists a constant M
>
0 such that T
(
f
)
Y
≤
M
f
X for all maps f
∈
X. And X
→
Y is compact if it takes bounded sets in X to sets in Y which have compact closure. For Banach spaces X and Y of H
(
Δ
), T is compact from X to Y if and only if for each sequence {
x
n
} in X; the sequence {
T
x
n
}
∈
Y contains a subsequence converging to some limit in Y.Considering the definition ofQ
K spaces and Q
K
⊆
B with some conditions, it is difficult to study the operator C
φ
D from Bloch spaces to Q
K spaces. In this paper, some sufficient and necessary conditions for the boundedness and compactness of this operator are given.


## 2. The Boundedness

Lemma 1 (see [10]).
If allf
∈
B, then (6)
s
u
p
z
∈
Δ
1
-
z
2
n
f
n
+
f
0
+
f
′
0
+
f
′
′
0
+
⋯
+
f
n
-
1
0
≈
f
B
,
n
=
1,2
,
…
.Theorem 2.
Letφ be an analytic self-map of Δ. Suppose K is a nondecreasing and right-continuous function on [
0
,
+
∞
) such that (7)
∫
0
1
/
e
K
-
log
⁡
r
r
d
r
<
+
∞
.Then the following statements are equivalent:(a)
C
φ
D
:
B
→
Q
K is bounded.
(b)
C
φ
D
:
B
0
→
Q
K is bounded.
(c)
(8)
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
2
1
-
φ
z
2
4
K
g
z
,
a
d
A
z
<
∞
.Proof.
(c)
⇒
(a). Suppose that (
c
) holds. For any z
∈
Δ and f
(
z
)
∈
B, we have (9)
C
φ
D
f
z
K
2
=
s
u
p
a
∈
Δ
∬
Δ
f
′
φ
′
2
K
g
z
,
a
d
A
z
=
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
f
′
′
φ
2
K
g
z
,
a
d
A
z
≤
f
B
2
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
2
1
-
φ
z
2
4
K
g
z
,
a
d
A
z
<
+
∞
.Thus (
a
) holds.
(a)
⇒
(b). It is obvious.
(b)
⇒
(c). Assume (
b
) holds; that is, there exists a constant C such that C
φ
D
f
(
z
)
K
≤
C
f
B for all f
∈
B
0. Conversely, suppose that C
φ
D
:
B
→
Q
K is bounded. Fix w
∈
Δ and assume that φ
(
w
)
≠
0. Consider the function f defined by (10)
f
w
z
=
1
-
φ
w
2
2
1
-
φ
w
¯
z
2
,for z
∈
Δ; then (11)
f
w
′
z
=
2
φ
w
¯
1
-
φ
w
2
2
1
-
φ
w
¯
z
3
,for z
∈
Δ. Since(12)
f
w
′
z
≤
2
1
-
φ
w
2
2
1
-
φ
w
z
3
≤
2
1
+
φ
w
2
2
1
-
z
≤
8
1
-
zfor all z
∈
Δ, f
w
∈
B. Furthermore, it is clear that f
w
∈
B
0, since (13)
f
w
0
=
1
-
φ
w
2
2
≤
4
.Thus (14)
M
1
=
sup
⁡
f
w
B
:
w
∈
Δ
≤
12
.Note that (15)
f
w
′
′
φ
w
=
2
φ
w
1
-
φ
w
2
2
.For this function f
w and this point w we have (16)
C
φ
D
f
w
′
w
=
φ
′
w
f
w
′
′
φ
w
=
2
φ
′
w
φ
w
2
1
-
φ
w
2
2
.So (17)
s
u
p
a
∈
Δ
∬
Δ
4
φ
′
w
2
φ
w
4
1
-
φ
w
2
4
K
g
z
,
a
d
A
z
≤
C
φ
D
f
w
K
2
≤
M
1
C
φ
D
B
→
Q
K
2
<
+
∞
,for all w
∈
Δ. Then we can imply(18)
s
u
p
a
∈
Δ
∬
φ
w
>
1
/
2
φ
′
w
2
1
-
φ
w
2
4
K
g
z
,
a
d
A
z
≤
16
s
u
p
a
∈
Δ
∬
φ
w
>
1
/
2
φ
′
w
2
φ
w
4
1
-
φ
w
2
4
K
g
z
,
a
d
A
z
≤
C
φ
D
f
w
K
2
≤
M
1
C
φ
D
B
→
Q
K
2
<
+
∞
.On the other hand, we note the functions f
(
z
)
≡
z
2, which belong to B
0, and we get (19)
2
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
2
K
g
z
,
a
d
A
z
<
∞
.
Then(20)
s
u
p
a
∈
Δ
∬
φ
w
<
1
/
2
φ
′
w
2
1
-
φ
w
2
4
K
g
z
,
a
d
A
z
≤
4
3
4
s
u
p
a
∈
Δ
∬
φ
w
<
1
/
2
φ
′
z
2
K
g
z
,
a
d
A
z
≤
∞
.So, we get (21)
s
u
p
a
∈
Δ
∬
Δ
φ
′
w
2
1
-
φ
w
2
4
K
g
z
,
a
d
A
<
∞
.By the arbitrary of w, we have (22)
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
1
-
φ
z
2
4
K
g
z
,
a
d
A
<
∞
,for all z
∈
Δ.
This completes the proof of this theorem.


## 3. The Compactness

The following lemma can be proved similarly to [11].Lemma 3.
Letφ be an analytic self-map of Δ. Then C
φ
D
:
B
→
Q
K (or B
0
→
Q
K) is compact if and only if C
φ
D
:
B
→
Q
K (or B
0
→
Q
K) is bounded and for any bounded sequence (
f
n
)
n
∈
N in B which converges to zero uniformly on compact subsets of Δ; one has C
φ
D
f
n
Q
K
→
0
a
s
n
→
∞.Lemma 4.
Letφ be an analytic self-map of Δ. Suppose K is a nondecreasing and right-continuous function on [
0
,
+
∞
) such that (23)
∫
0
1
/
e
K
-
log
⁡
r
r
d
r
<
+
∞
.If C
φ
D
:
B
(
B
0
)
→
Q
K is compact, then for any ϵ
>
0 there exists a δ
,
0
<
δ
<
1 such that, for all f in E, (24)
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
′
z
2
K
g
z
,
a
d
A
z
<
ϵholds whenever δ
<
r
<
1, where E is the unit ball of B
(
B
0
).Proof.
Forf
∈
B
0, let f
t
(
z
)
=
f
(
t
z
)
(
0
<
t
<
1
). Then f
t
∈
B
0, and f
t
→
f uniformly on compact subsets of Δ as t
→
1. Since C
φ
D is compact, (
C
φ
D
f
t
-
C
φ
D
f
)
(
z
)
→
0 as t
→
1. That is, for given ε
>
0, there exists t
∈
(
0,1
) such that (25)
s
u
p
a
∈
Δ
∬
Δ
C
φ
D
f
t
′
-
C
φ
D
f
z
2
K
g
z
,
a
d
A
z
<
ϵ
.For r
(
0
<
r
<
1
), the triangle equality gives (26)
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
′
z
2
K
g
z
,
a
d
A
z
≤
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
t
′
z
2
K
g
z
,
a
d
A
z
+
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
t
′
-
C
φ
D
f
z
2
K
g
z
,
a
d
A
z
≤
s
u
p
a
∈
Δ
∬
φ
z
>
r
f
t
′
′
φ
z
φ
′
z
2
K
g
z
,
a
d
A
z
+
ε
≤
f
t
′
′
∞
2
s
u
p
a
∈
Δ
∬
φ
z
>
r
φ
′
z
2
K
g
z
,
a
d
A
z
+
ε
.Then, we prove that for that given ε
>
0 and f
t
′
′
∞
2
>
0 there exists a δ
∈
(
0,1
) such that if δ
<
r
<
1, (27)
f
t
′
′
∞
2
s
u
p
a
∈
Δ
∬
φ
z
>
r
φ
′
z
2
K
g
z
,
a
d
A
z
<
ε
.Choose f
n
(
z
)
=
z
n
∈
B
0, and we have n
φ
n
-
1
∈
Q
K. Since C
φ
D is compact, l
i
m
n
→
∞
n
φ
n
-
1
=
0. Thus, for given ε
>
0 and f
t
′
′
∞
2
>
0, there exists an N
∈
N such that(28)
f
t
′
′
∞
2
s
u
p
a
∈
δ
∬
Δ
n
2
n
-
1
2
φ
n
-
2
z
2
φ
′
z
2
K
g
z
,
a
d
A
z
<
ε
,whenever n
≥
N. Hence, for 0
<
r
<
1,(29)
N
2
N
-
1
2
s
u
p
a
∈
δ
∬
Δ
φ
N
-
2
z
2
φ
′
z
2
K
g
z
,
a
d
A
z
≥
N
2
N
-
1
2
s
u
p
a
∈
Δ
∬
φ
z
>
r
φ
N
-
2
z
2
φ
′
z
2
K
g
z
,
a
d
A
z
≥
N
2
N
-
1
2
r
2
N
-
2
s
u
p
a
∈
Δ
∬
φ
z
>
r
φ
′
z
2
K
g
z
,
a
d
A
z
.Therefore, for r
>
(
N
2
-
N
)
-
1
/
N
-
2, (30)
f
t
′
′
∞
2
s
u
p
a
∈
Δ
∬
φ
z
>
r
φ
′
z
2
K
g
z
,
a
d
A
z
<
ε
.Thus we have already proved that, for any ε
>
0 and f
∈
B
0, there exists a δ
=
δ
(
ε
,
f
) such that (31)
s
u
p
a
∈
Δ
∬
φ
z
>
r
f
′
′
φ
z
φ
′
z
2
K
g
z
,
a
d
A
z
<
ϵholds whenever δ
<
r
<
1.
We finish our proof by showing that the aboveδ
=
δ
(
ε
,
f
), in fact, is independent of f
∈
B
0. Since C
φ
D is compact, C
φ
D
(
E
) is relatively compact in Q
K. It means that there is a finite collection of functions f
1
,
f
2
,
…
,
f
m in E such that, for any ε
>
0 and f
∈
E, we can find f
k
(
1
≤
k
≤
m
) satisfying (32)
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
k
′
-
C
φ
D
f
z
2
K
g
z
,
a
d
A
z
<
ε
.On the other hand, if m
a
x
1
≤
k
≤
m
δ
k
(
ε
,
f
k
)
=
δ
<
r
<
1, we have from the previous observation that, for all k
=
1,2
,
…
,
m, (33)
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
k
′
z
2
K
g
z
,
a
d
A
z
<
ε
.By the triangle inequality we obtain that (34)
s
u
p
a
∈
Δ
∬
φ
z
>
r
C
φ
D
f
′
z
2
K
g
z
,
a
d
A
z
<
εholds whenever δ
<
r
<
1. The proof is complete.Theorem 5.
Letφ be an analytic self-map of Δ. Suppose K is a nondecreasing and right-continuous function on [
0
,
+
∞
) such that (35)
∫
0
1
/
e
K
-
log
⁡
r
r
d
r
<
+
∞
.Then the following statements are equivalent:(a)
C
φ
D
:
B
→
Q
K is compact.
(b)
C
φ
D
:
B
0
→
Q
K is compact.
(c)
C
φ
D
:
B
→
Q
K is bounded: (36)
l
i
m
t
→
1
s
u
p
a
∈
Δ
∬
φ
z
>
t
φ
′
z
2
1
-
φ
z
2
4
K
g
z
,
a
d
A
z
=
0
.Proof.
(c)
⇒
(a). Assume (
c
) holds. Without loss of generality, let {
f
n
}
n
∈
N be a sequence in E which converges to 0 uniformly on compact subsets of Δ, as n
→
+
∞, where E is the unit ball of B. By Cauchy’s estimate, we know that {
f
n
′
′
}
n
∈
N also converges to 0 uniformly on compact subsets of Δ. For the sufficiency we will be verifying that {
C
φ
D
f
n
} converges to 0 in Q
K norm. By the assumption, for any ε
>
0, there exists a t
0
∈
(
0,1
) such that (37)
s
u
p
a
∈
Δ
∬
φ
z
>
t
0
φ
′
z
2
1
-
φ
z
2
4
K
g
z
,
a
d
A
z
<
ε
.Let T
=
z
∈
Δ
:
φ
z
≤
t
0; then we have (38)
C
φ
D
f
n
2
=
s
u
p
a
∈
Δ
∬
Δ
φ
′
z
f
n
′
′
φ
z
2
K
g
z
,
a
d
A
z
=
s
u
p
a
∈
Δ
∬
T
φ
′
z
f
n
′
′
φ
z
2
K
g
z
,
a
d
A
z
+
s
u
p
a
∈
Δ
∬
Δ
/
T
φ
′
z
f
n
′
′
φ
z
2
K
g
z
,
a
d
A
z
≤
s
u
p
f
n
′
′
φ
z
2
:
z
∈
T
φ
z
K
+
f
n
B
2
s
u
p
a
∈
Δ
∬
Δ
/
T
φ
′
z
2
1
-
φ
z
2
4
K
g
z
,
a
d
A
z
≤
sup
⁡
f
n
′
′
φ
z
2
:
z
∈
T
φ
z
K
+
f
n
B
2
·
ε
.By C
φ
D is bounded, we know that φ
∈
Q
K. It follows that C
φ
D
f
n
→
0 since that s
u
p
{
f
n
′
′
φ
z
2
:
z
∈
T
}
→
0 as n
→
∞. By Lemma 1, we can obtain that C
φ
D
:
B
→
Q
K is compact.
(a)
⇒
(b). It is obvious.
(b)
⇒
(c). Suppose that C
φ
D
:
B
0
→
Q
K is compact. Then it is clear that C
φ
D
:
B
0
→
Q
K is bounded. We know that f
θ
(
z
)
=
1
/
2
log
⁡
1
/
1
-
e
-
i
θ
z
∈
B for all θ
∈
[
0,2
π
). Choose a sequence {
λ
n
} in Δ which converges to 1 as n
→
∞, and let f
θ
,
n
=
f
θ
(
λ
n
z
) for n
∈
N. Thus f
θ
,
n in E for all n
∈
N and θ
∈
[
0,2
π
), where E is the unit ball of B
0. By Lemma 4, for any ε
>
0
(39)
s
u
p
a
∈
Δ
∬
φ
z
>
t
λ
n
2
φ
′
z
2
1
-
e
-
i
θ
λ
n
φ
z
4
K
g
z
,
a
d
A
z
<
εholds for any n
∈
N and θ
∈
[
0,2
π
). That is, (40)
l
i
m
t
→
1
s
u
p
a
∈
Δ
∬
φ
z
>
t
λ
n
2
φ
′
z
2
1
-
e
-
i
θ
λ
n
φ
z
4
K
g
z
,
a
d
A
z
=
0
.Thus, we obtain (
c
) by integrating, with respect to θ, the Fubini theorem, the Poisson formula, and the Fatou’s lemma.



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*Source: 5084794-2016-03-03.xml*