# Invertible Weighted Composition Operators on the Fock Space ofC

Authors: Liankuo Zhao
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

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


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## Body

## 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 C are characterized. Following the ideas in [1], in this paper a complete characterization of bounded invertible weighted composition operators on Fock space of C
N is given.Recall that the Fock spaceF
2 is the space of analytic functions f on C
N for which(1)
f
2
=
1
2
π
N
∫
C
N
f
z
2
e
-
z
2
/
2
d
m
2
N
z
,where d
m
2
N is the usual Lebesgue measure on C
N and |
z
| denotes the norm for z
∈
C
N. F
2 is a reproducing kernel Hilbert space with inner product(2)
f
,
g
=
1
2
π
N
∫
C
N
f
z
g
z
¯
e
-
z
2
/
2
d
m
2
N
z
,
f
,
g
∈
F
2and reproducing kernel function(3)
K
w
z
=
e
x
p
z
,
w
2
,
w
,
z
∈
C
N
,where 〈
z
,
w
〉 denotes the inner product for z
,
w
∈
C
N and |
z
|
2
=
〈
z
,
z
〉. Note that it is unnecessary to distinguish the symbols of inner product in F
2 and inner product in C
N.Letk
w be the normalization of K
w; then(4)
k
w
z
=
e
x
p
z
,
w
2
-
w
2
4
.For analytic functionψ on C
N and analytic self-mapping φ on C
N, the weighted composition operator C
ψ
,
φ on F
2 is defined as(5)
C
ψ
,
φ
f
=
ψ
f
∘
φ
,
f
∈
F
2
.For an operatorA on C
N, denote by |
A
| the norm of A.We have the following main result.Theorem 1.
Letψ be an analytic function on C
N and let φ be an analytic self-mapping on C
N. Then C
ψ
,
φ is a bounded invertible operator on F
2 if and only if(6)
φ
z
=
A
z
+
bfor some invertible operator A on C
N with |
A
|
=
1, b
∈
C
N, and there exist positive constants M
,
L such that(7)
L
≤
ψ
z
2
e
x
p
φ
z
2
-
z
2
2
≤
M
,
z
∈
C
N
.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.
Letψ
1
,
…
,
ψ
n be analytic functions on C
N and let φ
1
,
…
,
φ
n be analytic self-mapping on C
N. If C
ψ
1
,
φ
1
,
…
,
C
ψ
n
,
φ
n are bounded operators on F
2, then (8)
C
ψ
1
,
φ
1
C
ψ
2
,
φ
2
⋯
C
ψ
n
,
φ
n
=
C
ψ
1
ψ
2
∘
φ
1
⋯
ψ
n
∘
φ
n
-
1
∘
⋯
∘
φ
1
,
φ
n
∘
φ
n
-
1
∘
⋯
∘
φ
1
.Lemma 3.
Letψ be an analytic function on C
N and let φ be an analytic self-mapping on C
N. If C
ψ
,
φ is a bounded operator on F
2, then, for z
∈
C
N, (9)
C
ψ
,
φ
∗
K
z
=
ψ
z
¯
K
φ
z
.Forp
∈
C
N, denote φ
p
(
z
)
=
z
-
p, U
p
=
C
k
p
,
φ
p.Lemma 4 (see [8, Proposition 2.3]).
U
p is a unitary operator on F
2 and U
p
-
1
=
U
-
p.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.
Letψ, φ be entire functions on C with ψ
≠
0. If there exists a positive constant M such that(10)
ψ
u
2
e
x
p
φ
u
2
-
u
2
2
≤
M
,
u
∈
C
,then(11)
φ
u
=
a
u
+
bfor some constants a, b with |
a
|
≤
1. If |
a
|
=
1, then ψ
(
u
)
=
s
k
c
(
u
) for some nonzero constant s and c
=
-
a
-
b.Letf be an analytic function on C
N; for any ξ
∈
S
=
{
z
∈
C
N
,
|
z
|
=
1
}, denote (12)
f
ξ
u
=
f
u
ξ
,
u
∈
C
,where f
ξ is called the slice function of f in ξ and f
ξ is an analytic function on C.Now, we extend Lemma5 to the case of C
N in some sense.Lemma 6.
Letψ be an analytic function on C
N with ψ
(
0
)
≠
0 and let φ be an analytic self-mapping on C
N. If there exists a positive constant M such that (13)
ψ
z
2
e
x
p
φ
z
2
-
z
2
2
≤
M
,
z
∈
C
N
,then (14)
φ
z
=
A
z
+
b
,where A is an operator on C
N with |
A
|
≤
1 and b
∈
C
N.
Moreover, there exists a constants
∈
C such that (15)
ψ
ξ
u
=
s
exp
⁡
-
u
A
ξ
,
b
2
-
b
2
4
,
u
∈
Cwhenever |
A
ξ
|
=
|
ξ
| for ξ
∈
C
N.
In particular, whenA is a unitary operator on C
N, then there exists a constant s
∈
C such that(16)
ψ
z
=
s
k
c
z
,with c
=
-
A
∗
b.Proof.
Sinceφ
:
C
N
→
C
N is an analytic mapping, assume(17)
φ
z
=
φ
1
z
,
…
,
φ
N
z
,
z
∈
C
N
,where φ
j
,
1
≤
j
≤
N, are analytic functions on C
N.
By formula (13), for any ξ
∈
S,(18)
ψ
ξ
u
2
e
x
p
φ
ξ
u
2
-
ξ
u
2
2
≤
M
,
u
∈
C
,where φ
ξ
(
u
)
=
(
φ
1
,
ξ
(
u
)
,
…
,
φ
N
,
ξ
(
u
)
).
For anyj
,
1
≤
j
≤
N, we have(19)
ψ
ξ
u
2
e
x
p
φ
j
,
ξ
u
2
-
u
2
2
≤
M
,
u
∈
C
,since(20)
φ
j
,
ξ
u
2
≤
φ
1
,
ξ
u
2
+
⋯
+
φ
N
,
ξ
u
2
=
φ
ξ
u
2
.
Sinceψ
ξ
(
0
)
=
ψ
(
0
)
≠
0, it follows from Lemma 5 that there exist constants A
j
(
ξ
)
,
b
j
(
ξ
)
∈
C such that (21)
φ
j
,
ξ
u
=
A
j
ξ
u
+
b
j
ξ
,
u
∈
C
.Let u
=
0 in the formula above; then(22)
b
j
ξ
=
φ
j
,
ξ
0
=
φ
j
0
,which implies that b
j
(
ξ
) is a constant. Assume(23)
b
j
=
b
j
ξ
=
φ
j
0
;then(24)
φ
j
,
ξ
u
=
A
j
ξ
u
+
b
j
,
u
∈
C
.
Letφ
j
(
z
)
=
∑
n
=
0
∞
‍
φ
j
,
n
(
z
), z
∈
C
N, be the homogeneous expansion of φ
j, where φ
j
,
n is homogeneous of degree n; then(25)
φ
j
,
ξ
u
=
φ
j
u
ξ
=
∑
n
=
0
∞
φ
j
,
n
u
ξ
=
∑
n
=
0
∞
u
n
φ
j
,
n
ξ
,
u
∈
C
.
Comparing formulas (24) and (25), it follows from the arbitrary of u that (26)
φ
j
,
0
ξ
=
b
j
,
A
j
ξ
=
φ
j
,
1
ξ
,
φ
j
,
n
ξ
=
0
,
n
≥
2
.By the arbitrary of ξ and homogeneity of φ
j
,
n
,
0
≤
n
<
∞, we have(27)
φ
j
,
0
=
b
j
,
φ
j
,
n
=
0
,
n
≥
2
.
Sinceφ
j
,
1 is homogeneous of degree 1, assume(28)
φ
j
,
1
z
=
a
j
,
1
z
1
+
⋯
+
a
j
,
N
z
N
,
z
=
z
1
,
…
,
z
N
∈
C
N
,
a
j
,
m
∈
C
,
1
≤
m
≤
N
;then(29)
φ
j
z
=
a
j
,
1
z
1
+
⋯
+
a
j
,
N
z
N
+
b
j
,
z
=
z
1
,
…
,
z
N
∈
C
N
.
LetA
=
(
a
j
,
m
)
1
≤
j
,
m
≤
N, b
=
(
b
1
,
…
,
b
N
); then (30)
φ
z
=
A
z
+
b
,
z
∈
C
N
.
The following reasoning is similar as Proposition  2.1 in [1].
Taking logarithms in both sides of formula (18), we have(31)
4
log
⁡
ψ
ξ
u
+
φ
ξ
u
2
-
u
2
≤
2
log
⁡
M
,
u
∈
C
.Put u
=
r
e
i
θ and, integrating with respect to θ on [
-
π
,
π
], we obtain(32)
∫
-
π
π
φ
ξ
r
e
i
θ
2
d
θ
2
π
-
r
2
+
4
∫
-
π
π
log
⁡
ψ
ξ
r
e
i
θ
d
θ
2
π
≤
2
log
⁡
M
.Since log
⁡
|
ψ
ξ
| is subharmonic,(33)
log
⁡
ψ
0
=
log
⁡
ψ
ξ
0
≤
∫
-
π
π
log
⁡
ψ
ξ
r
e
i
θ
d
θ
2
π
.So(34)
∫
-
π
π
φ
ξ
r
e
i
θ
2
d
θ
2
π
-
r
2
+
4
log
⁡
ψ
0
≤
2
log
⁡
M
.
Since(35)
φ
ξ
r
e
i
θ
2
=
φ
r
e
i
θ
ξ
2
=
A
r
e
i
θ
ξ
+
b
2
=
r
e
i
θ
A
ξ
+
b
2
=
r
e
i
θ
a
1,1
ξ
1
+
⋯
+
a
1
,
N
ξ
N
+
b
1
2
+
⋯
+
r
e
i
θ
a
N
,
1
ξ
1
+
⋯
+
a
N
,
N
ξ
N
+
b
N
2
,where ξ
=
(
ξ
1
,
…
,
ξ
N
)
∈
S, we have (36)
∫
-
π
π
φ
ξ
r
e
i
θ
2
d
θ
2
π
=
r
2
a
1,1
ξ
1
+
⋯
+
a
1
,
N
ξ
N
2
+
b
1
2
+
⋯
+
r
2
a
N
,
1
ξ
1
+
⋯
+
a
N
,
N
ξ
N
2
+
b
N
2
=
r
2
A
ξ
2
+
b
2
.By formula (34) again, we get(37)
A
ξ
2
-
1
r
2
+
b
2
+
4
log
⁡
ψ
0
≤
2
log
⁡
M
.Since ψ
(
0
)
≠
0, it follows from the arbitrary of r that(38)
A
ξ
2
-
1
≤
0for all ξ
∈
S, which implies that(39)
A
≤
1
.
WhenA
ξ
=
ξ for some ξ
∈
C
N, without loss of generality, assume that ξ
∈
S; then A
ξ
=
ξ
=
1 and(40)
exp
⁡
A
u
ξ
+
b
2
-
u
2
2
=
exp
⁡
u
A
ξ
,
b
+
b
,
u
A
ξ
2
+
b
2
2
=
exp
⁡
u
A
ξ
,
b
2
+
b
2
4
2
.It follows from (18) that(41)
ψ
ξ
u
exp
⁡
u
A
ξ
,
b
2
+
b
2
4
2
=
ψ
ξ
u
2
exp
⁡
A
u
ξ
+
b
2
-
u
2
2
≤
M
,which implies that ψ
ξ
(
u
)
e
x
p
(
u
〈
A
ξ
,
b
〉
/
2
+
|
b
|
2
/
4
) is a bounded analytic function on C. By Liouville theorem, there exists a constant s
(
ξ
)
∈
C such that(42)
ψ
ξ
u
exp
⁡
u
A
ξ
,
b
2
+
b
2
4
=
s
ξ
.Let u
=
0; then s
(
ξ
)
=
ψ
ξ
(
0
)
e
|
b
|
2
/
4
=
ψ
(
0
)
e
|
b
|
2
/
4, which implies that s
(
ξ
) is a constant. Assume s
=
s
(
ξ
); then(43)
ψ
ξ
u
=
s
exp
⁡
-
u
A
ξ
,
b
2
-
b
2
4
,
u
∈
C
.
IfA is a unitary operator on C
N, then, for any z
∈
C
N, |
A
z
|
=
|
z
|. Taking ξ
∈
S, u
∈
C such that z
=
u
ξ, then (44)
ψ
z
=
ψ
u
ξ
=
ψ
ξ
u
=
s
exp
⁡
-
A
u
ξ
,
b
2
-
b
2
4
=
s
exp
⁡
z
,
-
A
∗
b
2
-
A
∗
b
2
4
=
s
k
-
A
∗
b
z
.Proposition 7.
Letψ be a nonzero analytic function on C
N and let φ be an analytic self-mapping on C
N. If C
ψ
,
φ is a bounded operator on F
2, then there exists an operator A on C
N, |
A
|
≤
1, b
∈
C
N, such that (45)
φ
z
=
A
z
+
b
,
sup
z
∈
C
N
⁡
ψ
z
2
exp
⁡
φ
z
2
-
z
2
2
<
∞
.In particular, when A is invertible, condition (45) is sufficient also.Proof.
IfC
ψ
,
φ is a bounded operator on F
2, then there exists a positive constant M such that, for any z
∈
C
N,(46)
C
ψ
,
φ
∗
K
z
2
≤
M
K
z
2
.Since C
ψ
,
φ
∗
K
z
=
ψ
(
z
)
¯
K
φ
(
z
) and K
z
2
=
e
|
z
|
2
/
2, we have(47)
ψ
z
2
exp
⁡
φ
z
2
-
z
2
2
≤
M
,
z
∈
C
N
.
Takep
∈
C
N such that ψ
(
-
p
)
≠
0. Since C
ψ
,
φ is a bounded operator on F
2, so is U
p
C
ψ
,
φ. Denote ψ
1
=
k
p
·
(
ψ
∘
φ
p
), φ
1
=
φ
∘
φ
p; then (48)
U
p
C
ψ
,
φ
=
C
ψ
1
,
φ
1
.Since C
ψ
1
,
φ
1 is a bounded operator on F
2 also, we obtain(49)
sup
z
∈
C
N
⁡
ψ
1
z
2
exp
⁡
φ
1
z
2
-
z
2
2
<
∞
.Note that ψ
1
(
0
)
=
k
p
(
0
)
ψ
(
-
p
)
≠
0. It follows from Lemma 6 that(50)
φ
1
z
=
A
z
+
d
,
z
∈
C
Nfor some operator A on C
N with |
A
|
≤
1 and d
∈
C
N.
Sinceφ
(
z
)
=
(
φ
1
∘
φ
-
p
)
(
z
), we have (51)
φ
z
=
A
z
+
bwith b
=
A
p
+
d.
WhenA is invertible, we have(52)
φ
-
1
w
=
A
-
1
w
-
A
-
1
b
,
w
∈
C
N
.If ψ
,
φ satisfy condition (45), then, for any f
∈
F
2,(53)
C
ψ
,
φ
f
2
=
1
2
π
N
∫
C
N
ψ
z
2
f
φ
z
2
·
e
-
z
2
/
2
d
m
2
N
z
=
1
2
π
N
∫
C
N
f
φ
z
2
ψ
z
2
·
e
φ
z
2
-
z
2
/
2
e
-
φ
z
2
/
2
d
m
2
N
z
≤
1
2
π
N
·
sup
z
∈
C
N
⁡
ψ
z
2
e
φ
z
2
-
z
2
/
2
∫
C
N
f
φ
z
2
·
e
-
φ
z
2
/
2
d
m
2
N
z
=
det
⁡
A
-
1
2
2
π
N
·
sup
z
∈
C
N
⁡
ψ
z
2
e
φ
z
2
-
z
2
/
2
∫
C
N
f
w
2
·
e
-
w
2
/
2
d
m
2
N
w
=
det
⁡
A
-
1
2
·
sup
z
∈
C
N
⁡
ψ
z
2
e
φ
z
2
-
z
2
/
2
f
2
,which implies that C
ψ
,
φ is a bounded operator on F
2 and(54)
C
ψ
,
φ
2
≤
det
⁡
A
-
1
2
sup
z
∈
C
N
⁡
ψ
z
2
e
φ
z
2
-
z
2
/
2
.Now we restate the main result and present the proof.Theorem 8.
Letψ be an analytic function on C
N and let φ be an analytic self-mapping on C
N. Then C
ψ
,
φ is a bounded invertible operator on F
2 if and only if there exist an invertible operator A on C
N, with |
A
|
=
1, b
∈
C
N, and positive constants M, L such that(55)
φ
z
=
A
z
+
b
,
L
≤
ψ
z
2
exp
⁡
φ
z
2
-
z
2
2
≤
M
,
z
∈
C
N
.Proof.
Assume thatC
ψ
,
φ is a bounded invertible operator on F
2. By the boundedness of C
ψ
,
φ, it follows from Proposition 7 that(56)
φ
z
=
A
z
+
bfor some operator A on C
N with |
A
|
≤
1, and there exists a positive constant M such that(57)
ψ
z
2
exp
⁡
φ
z
2
-
z
2
2
≤
M
,
z
∈
C
N
.
SinceC
ψ
,
φ is a bounded invertible operator on F
2, so is C
ψ
,
φ
∗. Hence there exists a positive constant L such that, for any z
∈
C
N,(58)
C
ψ
,
φ
∗
K
z
2
≥
L
K
z
2
.It follows from C
ψ
,
φ
∗
K
z
=
ψ
(
z
)
¯
K
φ
(
z
) and K
z
2
=
e
|
z
|
2
/
2 that(59)
ψ
z
2
exp
⁡
φ
z
2
-
z
2
2
≥
L
,
z
∈
C
N
,which implies that ψ has no zeroes in C
N.
Letz
,
w
∈
C
N with φ
(
z
)
=
φ
(
w
); then (60)
C
ψ
,
φ
∗
K
z
ψ
z
¯
-
K
w
ψ
w
¯
=
K
φ
z
-
K
φ
w
=
0
,which implies that(61)
K
z
ψ
z
¯
-
K
w
ψ
w
¯
=
0
.Since K
z
(
0
)
=
K
w
(
0
)
=
1, we have(62)
ψ
z
=
ψ
w
,and hence K
z
=
K
w. So z
=
w. It follows that φ is an injective mapping on C
N.
Sinceφ
(
z
)
=
A
z
+
b, A is an injective mapping on C
N also and hence a bijection, which implies that A is invertible on C
N. So (63)
φ
-
1
w
=
A
-
1
w
-
A
-
1
b
,
w
∈
C
N
.Taking φ
-
1
(
w
) to place z in formula (59), we have(64)
1
ψ
φ
-
1
w
2
exp
⁡
φ
-
1
w
2
-
w
2
2
≤
1
L
,
w
∈
C
N
.It follows from Lemma 6 that |
A
-
1
|
≤
1 and from Proposition 7 that C
1
/
(
ψ
∘
φ
-
1
)
,
φ
-
1 is a bounded operator on F
2.
Since|
A
|
≤
1, |
A
-
1
|
≤
1, and A
A
-
1
=
I, the identity operator on C
N, we have (65)
A
=
A
-
1
=
1
.
On the other hand, ifψ
,
φ satisfy the condition stated in the theorem, by Proposition 7, C
ψ
,
φ and C
1
/
(
ψ
∘
φ
-
1
)
,
φ
-
1 are bounded operators on F
2. Direct computation shows that(66)
C
ψ
,
φ
C
1
/
ψ
∘
φ
-
1
,
φ
-
1
=
C
1
/
ψ
∘
φ
-
1
,
φ
-
1
C
ψ
,
φ
=
I
,the identity operator on F
2.
SoC
ψ
,
φ is a bounded invertible operator on F
2.ForN
=
1, combined with Lemma 5, [8, Corollary 1.2], we have the following corollary.Corollary 9.
Letψ
,
φ be entire functions on C. Then C
ψ
,
φ is a bounded invertible operator on F
2 of C if and only if C
ψ
,
φ is a nonzero constant multiple of a unitary operator on F
2.



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*Source: 250358-2015-06-17.xml*