# Boundedness of the Segal-Bargmann Transform on Fractional Hermite-Sobolev Spaces

Authors: Hong Rae Cho; Hyunil Choi; Han-Wool Lee
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

Lets
∈
R and 2
≤
p
≤
∞. We prove that the Segal-Bargmann transform B is a bounded operator from fractional Hermite-Sobolev spaces W
H
s
,
p
R
n to fractional Fock-Sobolev spaces F
R
s
,
p.


---

## Body

## 1. Introduction

In quantum mechanics, the Schrödinger equation is a partial differential equation that describes how the quantum state of some physical system changes with time. The most famous example is the nonrelativistic Schrödinger equation for a single particle moving in a potential:(1)
-
1
ħ
∂
∂
t
Ψ
x
,
t
=
-
ħ
2
2
m
Δ
+
V
x
,
t
Ψ
x
,
t
,where m is the particle’s mass, ħ is the Planck constant, V is its potential energy, and Ψ is the wave function.LetH be the most basic Schrödinger operator in R
n
,
n
≥
1, the Hermite operator (or the harmonic oscillator):(2)
H
=
-
Δ
+
x
2
.Then the Schrödinger equation can be written by (3)
-
1
∂
Ψ
∂
t
=
H
Ψ
.This is an important model in quantum mechanics (see, e.g., [1]).Fors
∈
R, we define the fractional Hermite operator H
s
=
-
Δ
+
x
s of order s. Let 0
<
p
≤
∞. The Hermite-Sobolev space W
H
s
,
p
R
n of fractional order s is the space of all tempered distributions for which the distribution H
s
/
2
f is given by an L
p function on R
n.LetC
n be the complex n-space and let d
V be the ordinary volume measure on C
n. If z
=
z
1
,
…
,
z
n and w
=
w
1
,
…
,
w
n are points in C
n, we write(4)
z
·
w
¯
=
∑
j
=
1
n
z
j
w
¯
j
,
z
=
z
·
z
¯
1
/
2
.For any 0
<
p
≤
∞ the Fock space F
p denotes the space of entire functions f on C
n such that the function f
z
e
-
1
/
4
z
2 is in L
p
C
n
,
d
V. We define (5)
f
F
p
=
p
4
π
n
∫
C
n
f
z
e
-
1
/
4
z
2
p
d
V
z
1
/
p
.For p
=
∞ the norm in F
∞ is defined by(6)
f
F
∞
=
sup
⁡
f
z
e
-
1
/
4
z
2
:
z
∈
C
n
.Let(7)
A
j
f
z
=
2
∂
∂
z
j
f
z
,
A
j
∗
f
z
=
z
j
f
z
,
1
≤
j
≤
n
,
f
∈
F
p
.Both A
j and A
j
∗, as defined above, are densely defined linear operators on F
p (unbounded though). We consider the radial derivative R defined by (8)
R
≔
1
2
∑
j
=
1
n
A
j
A
j
∗
+
A
j
∗
A
j
.Let s be a real number and 0
<
p
≤
∞. The fractional Fock-Sobolev space F
R
s
,
p of order s is the space of all entire functions for which R
s
/
2
f is given by an F
p function.The Segal-Bargmann transformB is defined by(9)
B
f
z
=
1
π
n
/
4
∫
R
n
f
x
e
x
·
z
-
1
/
2
x
2
-
1
/
4
z
·
z
d
V
x
,where d
V
x is the volume measure on R
n. It is well-known that the Segal-Bargmann transform is a unitary isomorphism between L
2
R
n and F
2 [2, 3].We prove that the radial derivativeR has a parallel behavior to the Hermite operator H. In particular, R is densely defined, positive, self-adjoint and has the discrete spectrum; it generates a diffusion semigroup. Moreover, we show that the Segal-Bargmann transform intertwines fractional Hermite-Sobolev spaces with fractional Fock-Sobolev spaces as follows.Theorem 1.
Lets
∈
R and 2
≤
p
≤
∞. Then the Segal-Bargmann transform B
:
W
H
s
,
p
R
n
→
F
R
s
,
p is bounded.


## 2. Fractional Hermite-Sobolev Spaces

In one dimension, the Hermite polynomialsH
k are defined by (10)
H
k
x
=
e
x
2
d
k
d
x
k
e
-
x
2
,
x
∈
R
,and by normalization we obtain the Hermite functions (11)
h
k
x
=
π
2
k
k
!
-
1
/
2
e
-
x
2
/
2
-
1
k
H
k
x
,
x
∈
R
.Note that(12)
-
d
2
d
x
2
+
x
2
e
-
1
/
2
x
2
H
k
x
=
2
k
+
1
e
-
1
/
2
x
2
H
k
x
,
x
∈
R
.In higher dimensions, for each multi-indexα
=
α
1
,
…
,
α
n
∈
N
0
n, the Hermite functions h
α are defined by (13)
h
α
x
=
∏
j
=
1
n
h
α
j
x
j
,
x
=
x
1
,
…
,
x
n
∈
R
n
.Here, N
0
=
N
∪
0 is the set of nonnegative integer. By (12), we know that these are the eigenfunctions of the Hermite operator defined in (2). In fact,(14)
H
h
α
=
2
α
+
n
h
α
.Moreover, h
α
:
α
∈
N
0
n is an orthonormal basis for L
2
R
n.LetH be the space of finite linear combinations of Hermite functions(15)
f
=
∑
α
≤
N
f
,
h
α
h
α
,where(16)
f
,
h
α
=
∫
R
n
f
x
h
α
x
d
V
x
.The space H is dense in L
2
R
n, and so, by the orthonormality of the Hermite functions,(17)
f
L
2
R
n
=
∑
α
∈
N
0
n
f
,
h
α
2
1
/
2
.Fors
∈
R, we define the fractional Hermite operator H
s
=
-
Δ
+
x
s of order s. For f
∈
S
R
n, the Hermite series expansion(18)
∑
α
∈
N
0
n
f
,
h
α
h
αconverges to f uniformly in R
n (and also in L
2
R
n), since h
α
L
∞
R
n
≤
C, for all α
∈
N
0
n, and each m
∈
N, and we have (see [4])(19)
f
,
h
α
≤
H
m
f
L
2
R
n
2
α
+
n
-
m
.Definition 2.
Lets
∈
R and f
∈
S
R
n. One defines the fractional Hermite operator H
s by(20)
H
s
f
=
∑
α
∈
N
0
n
2
α
+
n
s
f
,
h
α
h
α
.The fractional Hermite operatorsH
s were introduced in [5].Definition 3.
Lets
∈
R and 0
<
p
≤
∞. The fractional Hermite-Sobolev space W
H
s
,
p
R
n of order s is the space of all tempered distributions for which the distribution H
s
/
2
f is given by an L
p function on R
n. The fractional Hermite-Sobolev norm of order s is defined accordingly,(21)
f
W
H
s
,
p
R
n
=
H
s
/
2
f
L
p
R
n
.The fractional Hermite-Sobolev spacesW
s
,
p
R
n of order s were introduced in [6].


## 3. Radial Derivative

We consider the radial derivativeR defined on(22)
D
o
m
R
=
f
∈
F
2
:
R
f
∈
F
2by(23)
R
≔
1
2
∑
j
=
1
n
A
j
A
j
∗
+
A
j
∗
A
j
,where(24)
A
j
f
z
=
2
∂
∂
z
j
f
z
,
A
j
∗
f
z
=
z
j
f
z
,
1
≤
j
≤
n
,
f
∈
F
2
.We have(25)
R
=
2
∑
j
=
1
n
z
j
∂
∂
z
j
+
n
.The following example tells us that D
o
m
R
⊊
F
2. Thus R is an unbounded operator on F
2.Example 4.
Let(26)
f
z
=
∑
k
=
0
∞
z
1
k
2
k
k
+
1
k
!
.Then f
∈
F
2, but R
f
∉
F
2.Proof.
Note that(27)
f
F
2
2
=
1
2
π
n
∑
k
=
0
∞
∫
C
n
z
1
k
2
k
k
+
1
k
!
2
e
-
1
/
2
z
2
d
V
z
=
∑
k
=
0
∞
1
k
+
1
2
=
ζ
2
<
∞
,where ζ
· is the Riemann zeta function. However, we have(28)
R
f
F
2
2
=
∑
k
=
0
∞
2
k
+
n
z
1
k
2
k
k
+
1
k
!
F
2
2
=
∑
k
=
0
∞
2
k
+
n
2
k
+
1
2
=
∞
.Lemma 5.
R is a positive, self-adjoint operator on D
o
m
R.Proof.
LetP
C
n be the set of all holomorphic polynomials on C
n. We know that P
C
n is dense in F
2 and R is self-adjoint on P
C
n. Hence D
o
m
R is the domain of its unique self-adjoint extension.
Note that(29)
f
,
R
f
F
2
=
2
∑
j
=
1
n
∂
f
∂
z
j
F
2
2
+
n
f
F
2
2
≥
n
f
F
2
2
,
∀
f
∈
D
o
m
R
.Thus R is positive.Lemma 6.
R has the discrete spectrum σ
R
=
2
α
+
n
:
α
∈
N
0
n.Proof.
By (29), we have σ
R
⊆
n
,
∞.
We define(30)
e
α
z
=
z
α
z
α
F
2
=
z
α
2
α
α
!
.Then e
α
:
α
∈
N
0
n is an orthonormal basis for F
2. It is easy to see that 2
α
+
n
:
α
∈
N
0
n is the set of all eigenvalues.
Letλ
∈
n
,
∞
∖
2
α
+
n
:
α
∈
N
0
n. First, we show that λ
I
-
R
:
D
o
m
R
→
F
2 is injective and surjective.
Suppose thatλ
I
-
R
f
=
λ
I
-
R
f
~. Then(31)
0
=
λ
I
-
R
f
-
λ
I
-
R
f
~
=
∑
α
∈
N
0
n
λ
-
2
α
+
n
f
-
f
~
,
e
α
e
α
.This implies f
=
f
~. Thus λ
I
-
R
:
D
o
m
R
→
F
2 is injective.
Forf
∈
F
2 let(32)
f
z
=
∑
α
∈
N
0
n
c
α
e
α
zbe the orthonormal decomposition of f. We define(33)
g
=
1
λ
f
+
1
λ
∑
α
∈
N
0
n
2
α
+
n
λ
-
2
α
+
n
c
α
e
α
z
.Since(34)
φ
N
=
∑
α
=
0
N
2
α
+
n
λ
-
2
α
+
n
c
α
e
α
zis a Cauchy sequence in F
2, the series in (33) converges in F
2. Hence(35)
g
=
1
λ
f
+
1
λ
∑
α
=
0
∞
2
α
+
n
λ
-
2
α
+
n
c
α
e
α
zis a well-defined element of F
2 and it satisfies λ
I
-
R
g
=
f. This means that λ
I
-
R
:
D
o
m
R
→
F
2 is surjective.
Moreover,(36)
λ
I
-
R
-
1
f
F
2
≤
1
λ
f
F
2
+
1
λ
β
f
F
2
=
1
λ
1
+
β
f
F
2
,where β
=
s
u
p
α
∈
N
0
n
2
α
+
n
/
λ
-
2
α
+
n. Hence λ
I
-
R
-
1 is bounded and so σ
R
=
2
α
+
n
:
α
∈
N
0
n.Forf
∈
F
2 let (37)
f
z
=
∑
α
∈
N
0
n
c
α
e
α
zbe the orthonormal decomposition of f. Associated with the operator R is a semigroup B
t
t
≥
0 defined by the expansion(38)
B
t
f
z
=
∑
α
∈
N
0
n
e
-
2
α
+
n
t
c
α
e
α
z
.We can check that u
z
,
t
≔
B
t
f
z is the solution of the heat-type equation:(39)
∂
t
+
R
u
=
0
on
C
n
×
0
,
∞
,
u
·
,
0
=
f
on
C
n
.It is easy to see that(40)
B
t
f
F
2
2
≤
e
-
2
n
t
f
F
2
2
.Thus B
t is contractive.Proposition 7.
B
t
t
≥
0 is a strongly continuous semigroup.Proof.
We note that(41)
B
t
f
-
f
F
2
2
=
∑
α
∈
N
0
n
e
-
2
α
+
n
t
-
1
2
c
α
2
=
∑
k
=
0
∞
e
-
2
k
+
n
t
-
1
2
∑
α
=
k
c
α
2
.For k
∈
N
0 and X
⊂
N
0 we define δ
k
X by (42)
δ
k
X
=
1
,
if
k
∈
X
,
0
,
if
k
∉
X
.Then(43)
l
i
m
t
→
0
+
B
t
f
-
f
F
2
2
=
l
i
m
t
→
0
+
∑
k
=
0
∞
e
-
2
k
+
n
t
-
1
2
∑
α
=
k
c
α
2
=
l
i
m
t
→
0
+
∫
0
∞
e
-
2
λ
+
n
t
-
1
2
d
ν
λ
,where ν is a discrete measure defined by(44)
ν
=
∑
k
=
0
∞
∑
α
=
k
c
α
2
δ
k
.By Lebesgue dominate convergence theorem, we have (45)
l
i
m
t
→
0
+
B
t
f
-
f
F
2
2
=
∫
0
∞
l
i
m
t
→
0
+
e
-
2
λ
+
n
t
-
1
2
d
ν
λ
=
0
.Hence B
t
t
≥
0 is a strongly continuous semigroup.Proposition 8.
-
R is the infinitesimal generator of B
t
t
≥
0. That is,(46)
l
i
m
t
→
0
+
B
t
f
-
f
t
=
-
R
f
.Proof.
By using the previous discrete measureν, it follows that(47)
B
t
f
-
f
t
-
-
R
f
F
2
2
=
∫
0
∞
e
-
2
λ
+
n
t
-
1
t
+
2
λ
+
n
2
d
ν
λ
.Taking limit on both sides and by Lebesgue dominate convergence theorem, (48)
l
i
m
t
→
0
+
B
t
f
-
f
t
-
-
R
f
F
2
2
=
l
i
m
t
→
0
+
∫
0
∞
e
-
2
λ
+
n
t
-
1
t
+
2
λ
+
n
2
d
ν
λ
=
∫
0
∞
l
i
m
t
→
0
+
e
-
2
λ
+
n
t
-
1
t
+
2
λ
+
n
2
d
ν
λ
=
0
.Thus we get the result.By Proposition8, we have (49)
B
t
=
e
-
t
R
.


## 4. Fractional Fock-Sobolev Spaces

SinceR has discrete spectrum 2
α
+
n
:
α
∈
N
0
n, by using the spectral theorem, we define the fractional radial derivative R
s for s
∈
R as follows.Definition 9.
Lets
∈
R. For f
∈
F
2 let(50)
f
z
=
∑
α
∈
N
0
n
c
α
e
α
zbe the orthonormal decomposition of f. By the spectral theorem, R
s is given by(51)
R
s
f
z
=
∑
α
∈
N
0
n
2
α
+
n
s
c
α
e
α
z
,
f
∈
D
o
m
R
s
.Definition 10.
Lets be a real number and 0
<
p
≤
∞. The fractional Fock-Sobolev space F
R
s
,
p of order s is the space of all entire functions for which R
s
/
2
f is given by an F
p function. The fractional Fock-Sobolev norm of f of order s is defined accordingly, (52)
f
F
R
s
,
p
=
R
s
/
2
f
F
p
.We refer the reader to [7–10] for other Fock-Sobolev spaces.


## 5.L
p-Boundedness of the Segal-Bargmann Transform

The Hermite operatorH is self-adjoint on the set of infinitely differentiable functions with compact support C
c
∞
R
n, and it can be factorized as(53)
H
=
1
2
∑
j
=
1
n
a
j
a
j
†
+
a
j
†
a
j
,where(54)
a
j
=
∂
∂
x
j
+
x
j
,
a
j
†
=
-
∂
∂
x
j
+
x
j
,
1
≤
j
≤
n
.Lemma 11.
For eachj
=
1
,
…
,
n, one has (55)
B
a
j
f
=
A
j
B
f
,
B
a
j
†
f
=
A
j
∗
B
f
.Proof.
Letf
∈
C
c
∞
R
n. By the integration by parts, we have(56)
B
∂
∂
x
j
f
z
=
1
π
n
/
4
∫
R
n
∂
f
∂
x
j
x
e
x
·
z
-
1
/
2
x
2
-
1
/
4
z
·
z
d
V
x
=
-
z
j
B
f
+
B
x
j
f
.This gives(57)
B
a
j
†
f
=
A
j
∗
B
f
.
We differentiate(58)
B
f
z
=
1
π
n
/
4
∫
R
n
f
x
e
x
·
z
-
1
/
2
x
2
-
1
/
4
z
·
z
d
V
xunder the integral sign to obtain(59)
A
j
B
f
z
=
1
π
n
/
4
∫
R
n
2
x
j
-
z
j
f
x
e
x
·
z
-
1
/
2
x
2
-
1
/
4
z
·
z
d
V
x
.This gives(60)
A
j
B
f
=
2
B
x
j
f
-
A
j
∗
B
f
.By (57) and (60), it follows that(61)
A
j
B
f
=
B
a
j
f
.Corollary 12.
Consider(62)
B
H
=
R
B
.Proof.
By Lemma11, we have(63)
B
H
f
=
1
2
∑
j
=
1
n
A
j
A
j
∗
+
A
j
∗
A
j
B
f
=
R
B
.Proposition 13.
Lets
∈
R. Then(64)
B
H
s
=
R
s
B
.Proof.
We define(65)
e
α
z
=
z
α
z
α
F
2
.Then e
α
:
α
∈
N
0
n is an orthonormal basis for F
2 and B
h
α
=
e
α. For f
∈
S
R
n we have(66)
H
s
f
=
∑
α
∈
N
0
n
2
α
+
n
s
f
,
h
α
h
αand so(67)
B
H
s
f
=
∑
α
∈
N
0
n
2
α
+
n
s
f
,
h
α
e
α
.Since B is a unitary isomorphism, we have f
,
h
α
=
B
f
,
e
α. Hence (68)
B
H
s
f
=
∑
α
∈
N
0
n
2
α
+
n
s
B
f
,
e
α
e
α
=
R
s
B
f
.Thus we get the result.We consider the mapping property of the Segal-Bargmann transformB as a map from L
p
R
n to F
p for p
∈
[
2
,
∞
]. Note that one-dimensional case is in [11].Theorem 14.
Consider(69)
B
f
F
∞
≤
4
π
n
/
4
f
L
∞
R
n
.Proof.
We have(70)
B
f
z
≤
1
π
n
/
4
e
z
2
/
4
s
u
p
x
∈
R
n
f
x
∫
R
n
e
Re
⁡
z
·
x
-
1
/
2
x
2
-
1
/
4
Re
⁡
z
·
z
-
z
2
/
4
d
V
x
.Note that(71)
Re
⁡
z
2
=
1
2
z
2
+
Re
⁡
z
·
z
.Hence(72)
Re
⁡
z
·
x
-
1
2
x
2
-
1
4
Re
⁡
z
·
z
-
z
2
4
=
Re
⁡
z
·
x
-
1
2
x
2
-
1
2
Re
⁡
z
2
=
-
1
2
Re
⁡
z
-
x
2and so(73)
B
f
z
≤
1
π
n
/
4
e
z
2
/
4
s
u
p
x
∈
R
n
f
x
∫
R
n
e
-
1
/
2
Re
⁡
z
-
x
2
d
V
x
=
4
π
n
/
4
e
z
2
/
4
s
u
p
x
∈
R
n
f
x
.Thus we get the result.The following Stein-Weiss interpolation theorem is well-known. See, for example, [3, 12].Lemma 15.
Letw
,
w
0, and w
1 be positive weight functions on a measure space X
,
d
λ. If 1
≤
p
0
≤
p
1
≤
∞ and 0
≤
θ
≤
1, then (74)
L
p
0
X
,
w
0
d
λ
,
L
p
1
X
,
w
1
d
λ
θ
=
L
p
X
,
w
d
λwith equal norms, where(75)
1
p
=
1
-
θ
p
0
+
θ
p
1
,
w
1
/
p
=
w
0
1
-
θ
/
p
0
w
1
θ
/
p
1
.Theorem 16.
Let2
≤
p
≤
∞. There exists C
>
0 such that(76)
B
f
F
p
≤
C
f
L
p
R
n
.Proof.
TheL
2-boundedness is followed by the unitary isomorphism of the Segal-Bargmann transform. In Theorem 14, we proved the L
∞-boundedness of the Segal-Bargmann transform. By Lemma 15, we have the required result.By Proposition13 and Theorem 16, we have the following result.Theorem 17.
Lets
∈
R and 2
≤
p
≤
∞. Then the Segal-Bargmann transform B
:
W
H
s
,
p
R
n
→
F
R
s
,
p is bounded.



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*Source: 9176914-2017-01-19.xml*