MATEMATIQKI VESNIK
originalni nauqni rad
research paper
65, 4 (2013), 481–487
December 2013
VOLTERRA TYPE OPERATORS FROM WEIGHTED
HARDY SPACES TO BLOCH SPACES
Xiangling Zhu
Abstract. Let H(D) denote the space of all analytic functions on the unit disk
this paper we consider the following Volterra type operator
Z
D of C. In
z
f (ξ)g 0 (ξ) dξ,
Jg (f )(z) =
f ∈ H(D), z ∈ D.
0
The boundedness and compactness of the operator Jg from the weighted Hardy space to a Bloch
space are studied.
1. Introduction
Let D be the unit disk of complex plane C, and H(D) the class of functions
analytic in D. Recall that an f ∈ H(D) is said to belong to the Bloch space B if
kf kb = sup(1 − |z|2 )|f 0 (z)| < ∞.
z∈D
With the norm kf kB = |f (0)| + kf kb , B is a Banach space. Let B0 be the space
which consists of all f ∈ B satisfying
lim (1 − |z|2 )|f 0 (z)| = 0.
|z|→1
This space is called the little Bloch space.
Throughout this paper, we assume that {β(n)}∞
n=0 is a sequence of positive
numbers such that
∞
P
β(0) = 1, lim inf β(n)1/n = 1 and
1/(β(n))2 = ∞.
n→∞
n=0
2010 AMS Subject Classification: 47B38, 32A18
Keywords and phrases: Volterra type operator; weighted Hardy space; Bloch space; boundedness; compactness.
The project supported by the Foundation for Distinguished Young Talents in Higher Education of Guangdong, China(No.LYM11117), the Guangdong Natural Science Foundation (No.
10451401501004305) and National Natural Science Foundation of China (No. 11001107).
481
482
Xiangling Zhu
The
Hardy space, denoted by H 2 (β), is defined to be the set of all f (z) =
P∞ weighted
n
n=0 an z ∈ H(D) such that
kf k2H 2 (β) =
∞
P
n=0
|an |2 (β(n))2 < ∞.
It is clear that H 2 (β) is a Hilbert space on D with the inner product given by
hf, gi =
∞
P
n=0
an bn (β(n))2 ,
P∞
P∞
where f (z) = n=0 an z n and g(z) = n=0 bn z n are in H 2 (β). Some well-known
special cases of this type of Hilbert space are, the Hardy space H 2 with weights
β(n) ≡ 1, the Bergman space A2 with weights β(n) = (n+1)−1/2 , and the Dirichlet
space D with weights β(n) = (n + 1)1/2 for all n. See [4] for more details of the
weighted Hardy space.
Suppose that g ∈ H(D). The integral operator
Z z
Jg f (z) =
f (ξ)g 0 (ξ) dξ, z ∈ D,
(1)
0
was introduced by Pommerenke in [12] and is called the Volterra type operator (see
[13]).
In [12], Pommerenke showed that Jg is bounded on the Hardy space H 2 if and
only if g ∈ BM OA. The boundedness and compactness of Jg between some spaces
of analytic functions, as well as their n-dimensional extensions on the unit ball in
Cn , were investigated in [1–3,5–9,11,14–20] (see also the related references therein).
In this paper, we study the operator Jg from the weighted Hardy space to
Bloch space. Some sufficient and necessary conditions for the operator Jg to be
bounded and compact are given.
Throughout the paper, constants are denoted by C, they are positive and may
not be the same in every occurrence. The notation a ³ b means that there is a
positive constant C such that C −1 b ≤ a ≤ Cb.
2. Main results and proofs
In this section, we give our main results and their proofs. Before stating these
results, we need some auxiliary results, which are incorporated in the lemmas which
follow.
Lemma 1. Assume that g ∈ H(D). Then Tg : H 2 (β) → B is compact if and
only if Tg : H 2 (β) → B is bounded and for any bounded sequence {fk }k∈N in H 2 (β)
which converges to zero uniformly on compact subsets of D as k → ∞, we have
kTg fk kB → 0 as k → ∞.
The proof of Lemma 1 follows by standard arguments (see, for example, Proposition 3.11 of [4]). Hence, we omit the details.
483
Volterra type operators
Lemma 2. [10] A closed set K in B0 is compact if and only if it is bounded
and satisfies
lim sup (1 − |z|2 )|f 0 (z)| = 0.
(2)
|z|→1 f ∈K
Lemma 3. Let f ∈ H 2 (β). Then
s
∞ |z|2n
P
.
2
n=0 β (n)
|f (z)| ≤ kf kH 2 (β)
P∞
n
n
2
Proof. For w ∈ D, define Kw (z) = n=0 βw
2 (n) z . Then Kw ∈ H (β). Let
P∞
n
f (z) = n=0 an z . From [4, p. 16], we see that
f (w) = hf, Kw i =
and
s
kKw kH 2 (β) =
∞
∞ a wn
P
P
n
β 2 (n) =
an wn
2
n=0 β (n)
n=0
∞ |w|2n
P
(β 2 (n)) =
4
n=0 β (n)
s
∞ |w|2n
P
< ∞.
2
n=0 β (n)
(3)
(4)
Then the desired result follows from (3) and (4).
Now we are in a position to state and prove our main results.
Theorem 1. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : H 2 (β) → B is bounded;
(ii) The operator Tg : H 2 (β) → B0 is bounded;
s
(iii)
2
0
M := sup(1 − |z| )|g (z)|
z∈D
∞ |z|2n
P
< ∞.
2
n=0 β (n)
Proof. (ii) ⇒ (i). It is obvious.
(i) ⇒ (iii). Assume that Tg : H 2 (β) → B is bounded. For a ∈ D, set
µ ∞
¶
∞ an z n
P |a|2n −1/2
P
fa (z) =
.
2
2
n=0 β (n) n=0 β (n)
(5)
(6)
It is easy to see that fa ∈ H 2 (β) and supa∈D kfa kH 2 (β) = 1. We have
∞ > kTg fa kB = sup(1 − |z|2 )|(Tg fa )0 (z)|
z∈D
2
0
= sup(1 − |z| )|g (z)||fa (z)|
z∈D
s
∞ |a|2n
P
2
0
,
≥ (1 − |a| )|g (a)|
2
n=0 β (n)
which implies (5).
(7)
484
Xiangling Zhu
(iii) ⇒ (ii). Assume that (5) holds. Then, for any f ∈ H 2 (β), by Lemma 3
we have
(1 − |z|2 )|(Tg f )0 (z)| = (1 − |z|2 )|g 0 (z)||f (z)|
s
∞ |z|2n
P
≤ (1 − |z|2 )|g 0 (z)|
kf kH 2 (β) .
2
n=0 β (n)
(8)
Taking the supremum in (8) over D and using the condition (5), we see that Tg :
H 2 (β) → B is bounded.
P∞
Since we assume n=0 1/(β(n))2 = ∞, (5) implies that g ∈ B0 . Then, for
each polynomial p(z), we have that
(1 − |z|2 )|(Tg p)0 (z)| = (1 − |z|2 )|g 0 (z)p(z)| ≤ kpk∞ (1 − |z|2 )|g 0 (z)|,
from which it follows that Tg p ∈ B0 . Since the set of all polynomials is dense in
H 2 (β) (see [4]), we have that for every f ∈ H 2 (β) there is a sequence of polynomials
{pk }k∈N such that kf − pk kH 2 (β) → 0, as k → ∞. Hence
kTg f − Tg pk kB ≤ kTg kH 2 (β)→B kf − pk kH 2 (β) → 0
as k → ∞, since Tg : H 2 (β) → B is bounded. Since B0 is a closed subset of B,
we obtain Tg (H 2 (β)) ⊆ B0 . Therefore Tg : H 2 (β) → B0 is bounded. The proof is
finished.
Theorem 2. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : H 2 (β) → B is compact;
(ii) The operator Tg : H 2 (β) → B0 is compact;
s
(iii)
2
0
lim (1 − |z| )|g (z)|
|z|→1
∞ |z|2n
P
= 0.
2
n=0 β (n)
(9)
Proof. (ii) ⇒ (i). It is obvious.
(i) ⇒ (iii). Assume that Tg : H 2 (β) → B is compact. Let {zk }k∈N be a
sequence in D such that limk*∞ |zk | = 1 (if such a sequence does not exist (9) is
automatically satisfied). Set
∞ z nzn
P
k
fk (z) =
2 (n)
β
n=0
µ
∞ |z |2n
P
k
2 (n)
β
n=0
¶−1/2
, k ∈ N.
(10)
It is easy to see that supk∈N kfk kH 2 (β) < ∞ and fk → 0 uniformly on compact
subsets of D as k → ∞. By Lemma 1,
lim kTg fk kB = 0.
k→∞
(11)
Volterra type operators
485
In addition, we have
kTg fk kB = sup(1 − |z|2 )|g 0 (z)| |fk (z)|
z∈D
s
∞ |z |2n
P
k
2
0
≥ (1 − |zk | )|g (zk )|
,
2
n=0 β (n)
(12)
which together with (11) implies that
s
lim (1 − |zk |2 )|g 0 (zk )|
k→∞
∞ |z |2n
P
k
= 0.
2
n=0 β (n)
This proves that (9) holds.
(iii) ⇒ (ii). Assume that (9) holds. From Theorem 1, we see that Tg :
H 2 (β) → B0 is bounded. Let f ∈ H 2 (β). From the proof of Theorem 1, we have
that
s
∞ |z|2n
P
(1 − |z|2 )|(Tg f )0 (z)| ≤ (1 − |z|2 )|g 0 (z)|
kf kH 2 (β) .
2
n=0 β (n)
Taking the supremum in the above inequality over all f ∈ H 2 (β) such that
kf kH 2 (β) ≤ 1, then letting |z| → 1, by (9) it follows that
lim
|z|→1 kf k
sup
(1 − |z|2 )|(Tg f )0 (z)| = 0.
H 2 (β) ≤1
From this and by employing Lemma 2, we see that Tg : H 2 (β) → B0 is compact.
The proof is completed.
Let β(n) = (n + 1)−1/2 . Then
∞ |z|2n
∞
P
P
1
=
.
(n + 1)|z|2n ³
2
(1 − |z|2 )2
n=0 β (n)
n=0
From Theorems 1–2 and the last formula, we have the following two corollaries.
Corollary 3. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : A2 → B is bounded;
(ii) The operator Tg : A2 → B0 is bounded;
(iii) supz∈D |g 0 (z)| < ∞.
Corollary 4. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : A2 → B is compact;
(ii) The operator Tg : A2 → B0 is compact;
(iii) g is a constant.
486
Xiangling Zhu
Let β(n) ≡ 1. Then
∞ |z|2n
∞
P
P
1
=
|z|2n =
.
2 (n)
β
1
−
|z|2
n=0
n=0
From Theorems 1–2 and the last formula, we have the following two corollaries.
Corollary 5. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : H 2 → B is bounded;
(ii) The operator Tg : H 2 → B0 is bounded;
p
(iii) supz∈D 1 − |z|2 |g 0 (z)| < ∞.
Corollary 6. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : H 2 → B is compact;
(ii) The operator Tg : H 2 → B0 is compact;
p
(iii) lim|z|→1 1 − |z|2 |g 0 (z)| = 0.
Let β(n) = (n + 1)1/2 . Then
∞ |z|2n
∞ |z|2n
P
P
e
=
³ ln 1−|z|
2.
2
n=0 n + 1
n=0 β (n)
From Theorems 1–2 and the last formula, we have the following two corollaries.
Corollary 7. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : D → B is bounded;
(ii) The operator Tg : D → B0 is bounded;
r
(iii)
2
0
sup(1 − |z| )|g (z)|
ln
z∈D
e
< ∞.
1 − |z|2
Corollary 8. Assume that g ∈ H(D). Then the following statements are
equivalent.
(i) The operator Tg : D → B is compact;
(ii) The operator Tg : D → B0 is compact;
r
(iii)
lim (1 − |z|2 )|g 0 (z)|
|z|→1
ln
e
= 0.
1 − |z|2
Volterra type operators
487
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(received 14.11.2011; in revised form 24.05.2012; available online 10.09.2012)
Department of Mathematics, JiaYing University, 514015, Meizhou, GuangDong, China
E-mail: xiangling-zhu@163.com