Journal of Function Spaces
Volume 2015 (2015), Article ID 250358, 5 pages
http://dx.doi.org/10.1155/2015/250358

Research Article

Invertible Weighted Composition Operators on the Fock Space of

Liankuo Zhao

School of Mathematics and Computer Science, Shanxi Normal University, Linfen 041004, China

Received 8 March 2015; Accepted 2 June 2015

Academic Editor: John R. Akeroyd

Copyright © 2015 Liankuo Zhao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We give a complete characterization of bounded invertible weighted composition operators on the Fock space of .

1. Introduction

It seems that there are simple forms for the weighted composition operators on the Fock space as implied in [1], where bounded and compact weighted composition operators on the Fock space of complex plane are characterized. Following the ideas in [1], in this paper a complete characterization of bounded invertible weighted composition operators on Fock space of is given.

Recall that the Fock space is the space of analytic functions on for which

where is the usual Lebesgue measure on and denotes the norm for . is a reproducing kernel Hilbert space with inner product

and reproducing kernel function

where denotes the inner product for and . Note that it is unnecessary to distinguish the symbols of inner product in and inner product in .

Let be the normalization of ; then

For analytic function on and analytic self-mapping on , the weighted composition operator on is defined as

For an operator on , denote by the norm of .

We have the following main result.

Theorem 1. Let be an analytic function on and let be an analytic self-mapping on . Then is a bounded invertible operator on if and only if

for some invertible operator on with , , and there exist positive constants such that

Weighted composition operators on various function spaces have been studied intensively and extensively, which reflects the perfect combination of operator theory and function theory. For related topic of (weighted) composition operators on the Fock space, see [2–9] and so forth.

2. Proof of the Main Results

In this section, we present the proof of the main results. First, we list some known results.

Lemma 2. Let be analytic functions on and let be analytic self-mapping on . If are bounded operators on , then

Lemma 3. Let be an analytic function on and let be an analytic self-mapping on . If is a bounded operator on , then, for ,

For , denote , .

Lemma 4 (see [8, Proposition 2.3]). is a unitary operator on and .

The following lemma is a modification of Proposition  2.1 in [1] since the reproducing kernel function in this paper is a little different from the reproducing kernel function in [1].

Lemma 5. Let , be entire functions on with . If there exists a positive constant such that

then

for some constants , with . If , then for some nonzero constant and .

Let be an analytic function on ; for any , denote

where is called the slice function of in and is an analytic function on .

Now, we extend Lemma 5 to the case of in some sense.

Lemma 6. Let be an analytic function on with and let be an analytic self-mapping on . If there exists a positive constant such that

then

where is an operator on with and .
Moreover, there exists a constant such that

whenever for .
In particular, when is a unitary operator on , then there exists a constant such that

with .

Proof. Since is an analytic mapping, assume

where , are analytic functions on .
By formula (13), for any ,

where .
For any , we have

since

Since , it follows from Lemma 5 that there exist constants such that

Let in the formula above; then

which implies that is a constant. Assume

then

Let , , be the homogeneous expansion of , where is homogeneous of degree ; then

Comparing formulas (24) and (25), it follows from the arbitrary of that

By the arbitrary of and homogeneity of , we have

Since is homogeneous of degree , assume

then

Let , ; then

The following reasoning is similar as Proposition  2.1 in [1].
Taking logarithms in both sides of formula (18), we have

Put and, integrating with respect to on , we obtain

Since is subharmonic,

So

Since

where , we have

By formula (34) again, we get

Since , it follows from the arbitrary of that

for all , which implies that

When for some , without loss of generality, assume that ; then and

It follows from (18) that

which implies that is a bounded analytic function on . By Liouville theorem, there exists a constant such that

Let ; then , which implies that is a constant. Assume ; then

If is a unitary operator on , then, for any , . Taking , such that , then

Proposition 7. Let be a nonzero analytic function on and let be an analytic self-mapping on . If is a bounded operator on , then there exists an operator on , , , such that

In particular, when is invertible, condition (45) is sufficient also.

Proof. If is a bounded operator on , then there exists a positive constant such that, for any ,

Since and , we have

Take such that . Since is a bounded operator on , so is . Denote , ; then

Since is a bounded operator on also, we obtain

Note that . It follows from Lemma 6 that

for some operator on with and .
Since , we have

with .
When is invertible, we have

If satisfy condition (45), then, for any ,

which implies that is a bounded operator on and

Now we restate the main result and present the proof.

Theorem 8. Let be an analytic function on and let be an analytic self-mapping on . Then is a bounded invertible operator on if and only if there exist an invertible operator on , with , , and positive constants , such that

Proof. Assume that is a bounded invertible operator on . By the boundedness of , it follows from Proposition 7 that

for some operator on with , and there exists a positive constant such that

Since is a bounded invertible operator on , so is . Hence there exists a positive constant such that, for any ,

It follows from and that

which implies that has no zeroes in .
Let with ; then

which implies that

Since , we have

and hence . So . It follows that is an injective mapping on .
Since , is an injective mapping on also and hence a bijection, which implies that is invertible on . So

Taking to place in formula (59), we have

It follows from Lemma 6 that and from Proposition 7 that is a bounded operator on .
Since , , and , the identity operator on , we have

On the other hand, if satisfy the condition stated in the theorem, by Proposition 7, and are bounded operators on . Direct computation shows that

the identity operator on .
So is a bounded invertible operator on .

For , combined with Lemma 5, [8, Corollary 1.2], we have the following corollary.

Corollary 9. Let be entire functions on . Then is a bounded invertible operator on of if and only if is a nonzero constant multiple of a unitary operator on .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This work is supported by NSFC (11471189, 11201274).

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