Abstract

Numerical ranges of normal weighted composition operators on the Fock space of are completely characterized. The main result shows that numerical ranges of such operators are closely related to their composition symbols.

1. Introduction

Inspired by results on numerical ranges of (weighted) composition operators on Hardy space [1], in this paper we give a complete characterization of numerical ranges of normal weighted composition operators on the Fock space of , the -dimensional complex Euclidean space.

Recall that for a bounded operator on a complex Hilbert space , the numerical range of is defined as

The Fock space over is the space of analytic functions on withwhere is the norm for and denotes usual Lebesgue measure on . It is well known that is a reproducing kernel Hilbert space with inner productand reproducing kernel functionswhere denotes the inner product for and . Usually, denotes the normalization of ,

A weighted composition operator on with an analytic function on and an analytic self-map of is defined asWeighted composition operators on have been studied extensively. It is noteworthy that weighted composition operators on of are closely related to some important transforms on [2].

In [3], normal weighted composition operators on are characterized completely. We have the following result.

Theorem 1. Let be a nonzero analytic function on and be an analytic self-map of . Then is a bounded normal operator on if and only ifwhere is a normal operator on with ,   with ,  , and is a nonzero constant. Furthermorewhenever for

For the presentation of our main result, we make some notations.

Let be an operator on . means the norm of . and . For ,  . is the identity operator on

For ,   ( denotes the set of nonnegative integers), let

For a set , let and denote the closure of and the convex set generated by , respectively. For ,  . is the open unit disk of and is the unit circle of . For ,  

Based on Theorem 1, our main result reads as follows.

Theorem 2. Let and such that is a bounded normal operator on . Let ,   be eigenvalues of
(1) If , thenwith
(2) If and with ,  , thenwith

In [4], the spectrum of normal weighted composition operators on is considered. Although the spectrum of a normal operator is closely related to its numerical ranges, it seems that a complete characterization of spectrum of normal weighted composition operators on is a little more difficult. For more results on numerical ranges of (weighted) composition operators on Hardy space, see [58].

2. Proof of Main Result

In this section, we give the proof of our main result, Theorem 2. First, we recall some known results on numerical ranges of operators and a class of very important unitary operators on the Fock space.

Let be a bounded operator on a complex Hilbert space . Denote with the spectrum of and the eigenvalues of . if and only if there exists such that , and  is called an eigenvector of corresponding to the eigenvalue . It is well known that If is a normal operator, then eigenvectors that correspond to distinct eigenvalues are orthogonal. Moreover, for a normal operator ,   [9, Theorem 1.4-4].

For , denote . Then is a unitary operator on and [10, Proposition 2.3].

For convenience, we make some notations.

Let be an analytic function on and be an analytic mapping of DenoteandThenand is unitarily equivalent to .

Let Thenand

Now we consider normal weighted composition operators on from a different view, which will simplify the characterization of such operators’ numerical ranges.

For an operator on , let denote the composition operator defined by

Theorem 3. Let be a bounded normal operator on with
(1) If , then is unitarily equivalent to with
(2) If , then for any ,   is unitarily equivalent to .
(3) If and with ,  , then is unitarily equivalent to , where ,   and

Proof. (1) If , then there exists such thatSince is bounded and normal, ,   is normal and by Theorem 1. So we haveI.e., . It follows thatFor any ,   and by Proposition 3.1 in [11, Ch I], . Since ,   By Theorem 1, ButSo Hencewhich implies thatIt follows from (18) and (21) thatBy formulas (14) and (15), and So is unitarily equivalent to
(2) If , then and there exist and such thatObviously,
For any , let ,   is the imaginary unit. Notice by Proposition 3.1 in [11, Ch I], so we haveBy Theorem 1, SoBy formulas (14) and (15),So is unitarily equivalent to .
(3) Since , there exists such thatBy Theorem 1, is normal and . We havesince by Proposition 3.1 in [11, Ch I], which implies that HenceOn the other hand, for any , by Ch 1, Proposition 3.1 in [11],So, by Theorem 1,ButSo we have , which implies thatIt follows from (29) and (33) thatBy formulas (14) and (15),So is unitarily equivalent to

For an analytic function on , it is well known that the composition operator on is a normal operator if and only if for some normal operator on with In fact, such operators are diagonalizable. In [8], the numerical ranges of such operators on Hardy space of are studied.

Lemma 4. Let be a normal operator on with and . Then on ,

Proof. Since is a normal operator on , there exists a unitary operator on such thatThen and
For any , let ; thenwhere Since ,  . It follows that . But . So .

Theorem 5. Let be a bounded normal operator on with
(1) If , then with
(2) If and with ,  , then , where

Proof. (1) The conclusion follows from Theorem 3 (1) and Lemma 4.
(2) By Theorem 3(3), is unitarily equivalent to withand .
For any ,Since ,   Therefore,It follows from Lemma 4 that Hence

The following result is well known.

Lemma 6. Let be a normal operator on ,   and . Then on ,where

Theorem 7. Let be a normal operator on ,  , and . Then on ,

Proof. Note that is a normal operator on ; we haveIf , then . In this case, . By formula (43),If , then . In this case, . By formula (43),In the following, we assume We take the proof into several parts.
Case I. There exists , such that ;   is not a root of .
It is well known that is dense in . So we havethe last equality follows from Lemma 4.
For any , if , then by [1, lemma 2.3], since . SoTherefore
Case II. .
In this case, is a positive operator on Since , there exists ,  . Obviously, . Since , we haveIf there exists such that , then and hence . Therefore So
Case III. For any , either or is a root of , and there exists , .
Note that, in this case, . Since or is a root of and ,So we obtainIt follows that

Lastly, we give the proof of Theorem 2. Some ideas are derived from [1, 8].

Proof. (1) Since unitarily equivalent operators have the same numerical ranges, the conclusion follows from Theorems 3(1), and 7 and Lemma 6.
(2) Since is normal, there exists such that By Theorem 3(2), for any ,  . It follows thatSo we haveIf there exists such that , then by [1, Lemma 2.3] and Theorem 3(2), for any ,  , which implies that there is an uncountable collection of orthogonal vectors in since is normal and eigenvalues of that correspond to distinct eigenvalues are orthogonal, a contradiction. Soand is dense in . Since is convex, we must have Henceby Theorem 5(2).

Remark 8. In Theorem 2(2), we can choose such that , and then since . Sowhere

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work is supported by NSFC (11771401) and CSC (201608140025). Partial work in this paper was done while the second author visited the Department of Mathematics and Statistics at the State University of New York at Albany from 2016.12 to 2017.12. He wishes to thank SUNY at Albany for hosting his visit.