Generalized integration operators from Hardy spaces

Qu Huiying

(School of Mathematics & Statistics,Jiangsu Normal University,Xuzhou 221116,Jiangsu,China)

0 Introduction

Let D denote the open unit disk of the complex plane C and H(D) the space of all analytic functions in D.

A positive continuous functionνon the interval [0,1) is called normal[1], if there areδ∈[0,1) anda,b, 0

is decreasing on [δ,1)

If we say that a functionν: D→[0,1) is normal, we also assume that it is radial, i.e.ν(z)=ν(|z|),z∈D.

For 0

For 0

Letμbe a weight, that is,μis a positive continuous function on D. The Bloch-type Bμconsists of allf∈H(D) such that

With the norm ‖f‖Bμ=|f(0)|+bμ(f), it becomes a Banach space. The little Bloch-type space Bμ,0is a subspace of Bμconsisting of thosef∈Bμ, such that

1 Preliminary material

Here we quote some auxiliary results which will be used in the proofs of the main results in this paper.

Lemma1[2]Forp>1, there exists a constantC(p) such that

Lemma2[16]Suppose that 0

for everyz∈D and all nonnegativen=0,1,2,….

Lemma4A closed setKin Bμ,0is compact if and only ifKis bounded and satisfies

The proof is similar to that of Lemma 1 in [17], so we omit it.

2 Boundedness and compactness of from Hp (0

(1)

ProofAssume that (1) holds. Then for everyz∈D andf∈Hp, by Lemma 2, we have

(2)

For a fixedω∈D, set

we get that

From Lemma 1, we have

Hence,

(3)

Forω∈D, by (3) we have

≤C<∞,

(4)

and from (3), we obtain that

≤C<∞.

(5)

Thus combining with (4), (5), we get the condition (1).

(6)

and

(7)

We assume that ‖fk‖p≤1. From (7), we have for anyε>0, there existsρ∈(0,1), whenρ<|φ(z)|<1, we have

(8)

≤(M+C)ε.

(9)

We can use the test functions

Note that

(10)

From (10) and |φ(zk)|→1, it follows that

and consequently (7) holds.

(11)

(12)

Then for anyf∈Hp, from Lemma 2, we obtain that

then (12) holds.

From (7), it follows that for everyε>0, there existsδ∈(0,1), such that

(13)

whenδ<|φ(z)|<1. Using (12), we see that there existsτ∈(0,1) such that

(14)

whenτ<|z|<1. Therefore whenτ<|z|<1 andδ<|φ(z)|<1, by (13), we have

(15)

On the other hand, whenτ<|z|<1 and |φ(z)|≤δ, by(14), we obtain

(16)

From (15) and (16), (11) holds. The proof is completed.

AcknowledgmentsThe author thanks the referee(s) for carefully reading the manuscript and making several useful suggestions for improvement. The author is also indebted to Professor Liu Yongmin, who gave him kind encouragement and useful instructions all through his writing.

:

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