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# Advances in Pure and Applied Mathematics

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Volume 10, Issue 1

# On the images of Sobolev spaces under the Schrödinger semigroup

Sivaramakrishnan C
/ Sukumar D
/ Venku Naidu Dogga
• Corresponding author
• Department of Mathematics, Indian Institute of Technology Hyderabad, Telangana – 502285, India
• Email
• Other articles by this author:
Published Online: 2018-01-20 | DOI: https://doi.org/10.1515/apam-2016-0116

## Abstract

In this article, we consider the Schrödinger semigroup for the Laplacian Δ on ${ℝ}^{n}$, and characterize the image of a Sobolev space in ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$ under this semigroup as weighted Bergman space (up to equivalence of norms). Also we have a similar characterization for Hermite Sobolev spaces under the Schrödinger semigroup associated to the Hermite operator H on ${ℝ}^{n}$.

MSC 2010: 46E35; 46F12; 47D06; 35B65; 35J10

## 1 Introduction

It is well known that the classical Bargmann transform is an isometric isomorphism of ${L}^{2}\left({ℝ}^{n}\right)$ onto the space of square integrable holomorphic functions $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{e}^{-{|z|}^{2}}dz\right)$; we refer to [1, 3] for further details. In [4], Hall generalized the concept of the Bargmann transform for the general compact Lie group K. He called this transform a heat-kernel transform. Also Hall characterized the image of ${L}^{2}\left(K\right)$ under the heat-kernel transform as a weighted Bergman space. Later, the topic became popular and attracted many mathematicians; see for instance [8, 7]. For the heat kernel transform associated to the Hermite and special Hermite semigroups, we refer to [2, 8]. The same kind of results are known for Sobolev spaces; we refer to [5, 10, 11].

In [6], Hayashi and Saitoh considered the Schrödinger semigroup for the Laplacian Δ on $ℝ$ and showed that ${e}^{it\mathrm{\Delta }}f$ can be extended as an entire function for $f\in {L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$. Also they characterized the range of ${e}^{it\mathrm{\Delta }}$ as a weighted Bergman space. In [9], Parui et al. generalized the result of Huyashi and Saitoh to ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$ and gave a completely different and simple proof for the image characterization with the help of the classical Bargmann transform. The Schrödinger equation associated to the Laplacian on ${ℝ}^{n}$ is given by

$i{u}_{t}=-\mathrm{\Delta }u,u\left(x,0\right)=f\left(x\right)\in {L}^{2}\left({ℝ}^{n}\right).$

The solution of the Schrödinger equation is given by the unitary group ${e}^{it\mathrm{\Delta }}$, which is unitary on ${L}^{2}\left({ℝ}^{n}\right)$. For a non-zero real number t, on a dense subspace the operator ${e}^{it\mathrm{\Delta }}$ is given by

${e}^{it\mathrm{\Delta }}f\left(x\right)={c}_{t}^{n}{\int }_{{ℝ}^{n}}f\left(u\right){e}^{\frac{i}{4t}{\left(x-u\right)}^{2}}𝑑u.$

Here the constant ${c}_{t}$ is given by ${c}_{t}={\left(4\pi it\right)}^{-1/2}$. Since ${e}^{it\mathrm{\Delta }}$ is unitary, ${e}^{it\mathrm{\Delta }}f$ cannot be extended as an entire function for all $f\in {L}^{2}\left({ℝ}^{n}\right)$. Further, if we assume enough decay on f, then we can extend ${e}^{it\mathrm{\Delta }}f$ as an entire function on ${ℂ}^{n}$. The following theorem is one in this direction and it also gives the image characterization.

#### Theorem 1.1 ([9]).

For a non-zero real number t and $f\mathrm{\in }{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}{e}^{{v}^{\mathrm{2}}}\mathit{}d\mathit{}v\mathrm{\right)}$, the function ${e}^{i\mathit{}t\mathit{}\mathrm{\Delta }}\mathit{}\mathrm{\left(}f\mathrm{\right)}$ can be extended as an analytic function. Moreover,

${\int }_{{ℂ}^{n}}{|{e}^{it\mathrm{\Delta }}\left(f\right)\left(x+iy\right)|}^{2}{e}^{\frac{xy}{t}}{e}^{\frac{-{y}^{2}}{4{t}^{2}}}𝑑x𝑑y={\left(2\sqrt{\pi }t\right)}^{n}{\int }_{{ℝ}^{n}}{|f\left(v\right)|}^{2}{e}^{{v}^{2}}𝑑v.$

Conversely, if

$F\in \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{e}^{\frac{xy}{t}-\frac{{y}^{2}}{4{t}^{2}}}dxdy\right),$

then there exists $f\mathrm{\in }{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}{e}^{{v}^{\mathrm{2}}}\mathit{}d\mathit{}v\mathrm{\right)}$ such that ${e}^{i\mathit{}t\mathit{}\mathrm{\Delta }}\mathit{}\mathrm{\left(}f\mathrm{\right)}\mathrm{=}F$.

#### Remark 1.2.

From Theorem 1.1 we have the following:

• (i)

${e}^{it\mathrm{\Delta }}:{L}^{2}\left({ℝ}^{n},{e}^{{v}^{2}}dv\right)\to \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(x+iy\right)dxdy\right)$ is unitary, where

${w}_{t}\left(x+iy\right)=\frac{1}{{\left(2\sqrt{\pi }t\right)}^{n}}{e}^{\frac{xy}{t}-\frac{{y}^{2}}{4{t}^{2}}}.$

• (ii)

Given any $F\in \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(x+iy\right)dxdy\right)$, there exists a non-negative constant c (depends only on $t,n,F$) such that

(1.1)

In this article, we wish to characterize the images of Sobolev spaces under the Schrödinger semigroup associated to the Laplacian Δ and the Hermite operator H on ${ℝ}^{n}$ as a weighted Bergman space (up to equivalence of norms).

We have organized the article as follows: Section 2 contains preliminaries. In Section 3, we will give some results on Sobolev spaces, which will help us to prove the main results. We present our main result associated to the Laplacian in Section 4. In Section 5, we will characterize the images of Hermite Sobolev spaces under the Schrödinger semigroup associated to the Hermite operator (up to equivalence of norms).

## 2 Preliminaries

We begin with the definition of Hermite polynomials ${H}_{k}\left(u\right)$, where k is a non-negative integer and $u\in ℝ$. These are defined by

${H}_{k}\left(u\right)={\left(-1\right)}^{k}{e}^{{u}^{2}}\frac{{d}^{k}}{d{u}^{k}}\left({e}^{-{u}^{2}}\right).$

By setting

${\varphi }_{k}\left(u\right)={\left({2}^{k}k!{\pi }^{\frac{1}{2}}\right)}^{-\frac{1}{2}}{H}_{k}\left(u\right){e}^{\frac{-{u}^{2}}{2}}$

for $k=0,1,2,\mathrm{\dots }$, the family $\left\{{\varphi }_{k}:k=0,1,2,\mathrm{\dots }\right\}$ forms an orthonormal basis for ${L}^{2}\left(ℝ\right)$. It is easy to see that the ${\varphi }_{k}$’s satisfy the following relations:

$\frac{d}{du}{\varphi }_{k}\left(u\right)={\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\varphi }_{k-1}\left(u\right)-{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\varphi }_{k+1}\left(u\right),$(2.1)$u{\varphi }_{k}\left(u\right)={\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\varphi }_{k+1}\left(u\right)+{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\varphi }_{k-1}\left(u\right).$(2.2)

#### Definition 2.1.

For $s>0$ the scaled Hermite polynomial is defined by

${H}_{k}^{s}\left(u\right)={\left(-1\right)}^{k}\frac{{d}^{k}}{d{u}^{k}}\left({e}^{-s{u}^{2}}\right){e}^{s{u}^{2}}$

for $u\in ℝ$ and $k\in ℕ$.

So by definition, ${H}_{k}^{s}\left(u\right)$ is a polynomial in u of degree k. Let us write ${H}_{k}^{s}\left(u\right)={\sum }_{r=0}^{k}{a}_{r}^{\left(k\right),s}{u}^{r}$. From the definition it is clear that in ${H}_{k}^{s}\left(u\right)$ the only even powers of u occur if k is even, and only odd powers of u occur if k is odd. The following proposition gives the coefficients of ${H}_{k}^{s}$.

#### Proposition 2.2.

The coefficient ${a}_{r}^{\mathrm{\left(}k\mathrm{\right)}\mathrm{,}s}$ of ${u}^{r}$ in the scaled Hermite polynomial ${H}_{k}^{s}$ is given by the relation

#### Proof.

This follows from the Rodrigues formula for Hermite polynomials. ∎

Let us introduce some multi-index notations:

• ${ℕ}^{n}=ℕ×\mathrm{\cdots }×ℕ$ (n times), where $ℕ:=\left\{0,1,2,\mathrm{\dots }\right\}$.

• For $k\in \left\{1,2,\mathrm{\dots },n\right\}$, define

${e}_{k}:=\left(0,\mathrm{\dots },0,\stackrel{k\text{-th place}}{\stackrel{⏞}{1}},0,\mathrm{\dots },0\right)\in {ℕ}^{n}.$

• For $\alpha =\left({\alpha }_{1},{\alpha }_{2},\mathrm{\dots },{\alpha }_{n}\right)\in {ℕ}^{n}$, define $|\alpha |:={\alpha }_{1}+\mathrm{\cdots }+{\alpha }_{n}$.

• $\alpha !:={\alpha }_{1}!\mathrm{\cdots }{\alpha }_{n}!$.

• For $\alpha ,\beta \in {ℕ}^{n}$, we say $\alpha \le \beta$ if ${\alpha }_{k}\le {\beta }_{k}$ for $k=1,2,\mathrm{\dots },n$.

• ${z}^{\alpha }:={z}_{1}^{{\alpha }_{1}}\mathrm{\cdots }{z}_{n}^{{\alpha }_{n}}$ and ${z}^{2}:={z}_{1}^{2}+\mathrm{\cdots }+{z}_{n}^{2}$ for $z=\left({z}_{1},\mathrm{\dots },{z}_{n}\right)\in {ℂ}^{n}$ and $\alpha \in {ℕ}^{n}$.

• $zw:={z}_{1}{w}_{1}+{z}_{2}{w}_{2}+\mathrm{\cdots }+{z}_{n}{w}_{n}$ for $z=\left({z}_{1},\mathrm{\dots },{z}_{n}\right),w=\left({w}_{1},\mathrm{\dots },{w}_{n}\right)\in {ℂ}^{n}$.

• ${D}^{\alpha }:={D}^{{\alpha }_{1}}{D}^{{\alpha }_{2}}\mathrm{\cdots }{D}^{{\alpha }_{n}}$. Here

${D}^{{\alpha }_{k}}=\frac{{\partial }^{{\alpha }_{k}}}{\partial {x}_{k}^{{\alpha }_{k}}}$

for $k=1,2,\mathrm{\dots },n$.

#### Definition 2.3.

Let $F:{ℂ}^{n}\to ℂ$ be a function and let $a=\left({a}_{1},\mathrm{\dots },{a}_{n}\right)\in {ℂ}^{n}$. Denote

We say that F is holomorphic if for every $a\in {ℂ}^{n}$ the map ${F}_{a,j}:ℂ\to ℂ$ is holomorphic for $j=1,\mathrm{\dots },n$.

Let us denote . Let $\rho \left(z\right)$ be a strictly positive continuous function on ${ℂ}^{n}$. We define the weighted Bergman space as follows:

$\mathcal{ℋ}{\mathcal{ℒ}}_{\rho }^{2}:=\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},\rho \left(z\right)dz\right)=\left\{f\in \mathcal{𝒪}\left({ℂ}^{n}\right):{\int }_{{ℂ}^{n}}{|f\left(z\right)|}^{2}\rho \left(z\right)𝑑x<\mathrm{\infty }\right\}.$

The inner product on $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},\rho \left(z\right)dz\right)$ is given by

${〈F,G〉}_{\mathcal{ℋ}{\mathcal{ℒ}}_{\rho }^{2}}:={\int }_{{ℂ}^{n}}F\left(z\right)\overline{G\left(z\right)}\rho \left(z\right)𝑑z.$

It is known that $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},\rho \left(z\right)dz\right)$ is the reproducing kernel Hilbert space. The well-known reproducing kernel Hilbert space is a Fock space and it is defined by

${\mathcal{ℱ}}_{r}^{2}:={\mathcal{ℱ}}^{2}\left({ℂ}^{n},\frac{r}{\pi }{e}^{-r{|z|}^{2}}dxdy\right)=\left\{F\in \mathcal{𝒪}\left({ℂ}^{n}\right):{\parallel F\parallel }_{r,2}^{2}<\mathrm{\infty }\right\},r>0,$

where

${\parallel F\parallel }_{r,2}^{2}={\left(\frac{r}{\pi }\right)}^{n}{\int }_{{ℂ}^{n}}{|F\left(z\right)|}^{2}{e}^{-r{|z|}^{2}}𝑑z.$

The set

$\left\{{𝐞}_{\alpha }^{r}\left(z\right)={\left(\frac{{r}^{|\alpha |}}{\alpha !}\right)}^{\frac{1}{2}}{z}^{\alpha }:\alpha \in {ℕ}^{n}\right\}$

forms an orthonormal basis for ${\mathcal{ℱ}}_{r}^{2}$. The reproducing kernel for ${\mathcal{ℱ}}_{r}^{2}$ is given by ${K}_{r}\left(z,w\right)={e}^{rz\overline{w}}$.

## 3 Sobolev space and its properties

We define Sobolev spaces in ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$ in the following way.

#### Definition 3.1.

Let $m>0$ be an integer. The Sobolev space ${W}^{m,2}\left({ℝ}^{n}\right)$ in ${L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$ is defined by

The inner product on ${W}^{m,2}\left({ℝ}^{n}\right)$ is defined for $f,g\in {W}^{m,2}\left({ℝ}^{n}\right)$ by

${〈f,g〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}=\sum _{|\alpha |\le m}{〈{D}^{\alpha }f,{D}^{\alpha }g〉}_{{L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)}.$

With respect to the above inner product, ${W}^{m,2}\left({ℝ}^{n}\right)$ becomes a Hilbert space. From now onwards, we consider for any $f,g\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$ the inner product $〈f,g〉$ and the norm $\parallel f\parallel$ given by

$〈f,g〉:={〈f,g〉}_{{L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)}\mathit{ }\text{and}\mathit{ }\parallel f\parallel :={\parallel f\parallel }_{{L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)},$

respectively. Also we assume the following notation: for $f\in {W}^{m,2}\left({ℝ}^{n}\right)$, the Sobolev norm of f is denoted by ${\parallel f\parallel }_{{W}_{m}}={\parallel f\parallel }_{{W}^{m,2}\left({ℝ}^{n}\right)}$.

#### Note 3.2.

For $m\ge 1$ and $f\in {W}^{m,2}\left(ℝ\right)$, we have

$\underset{|x|\to \mathrm{\infty }}{lim}|{D}^{k}f\left(x\right)p\left(x\right)|=0$(3.1)

for any polynomial p on $ℝ$ and $k\in ℕ$ with $k\le m-1$.

This can be seen as follows: First, we consider the case $m=1$; the general case follows similarly. Using the Sobolev embedding lemma, we have

${\left(f\left(t\right){e}^{\frac{{t}^{2}}{2}}\right)}^{\prime }=tf\left(t\right){e}^{\frac{{t}^{2}}{2}}+{f}^{\prime }\left(t\right){e}^{\frac{{t}^{2}}{2}}.$

Integrate both sides from a to b and use the Schwarz inequality to estimate the integral on the right:

${\left(|f\left(t\right){e}^{\frac{{t}^{2}}{2}}|\right)}_{a}^{b}\le {\left(\frac{{b}^{3}-{a}^{3}}{3}{\int }_{a}^{b}{|f\left(t\right)|}^{2}{e}^{{t}^{2}}𝑑t\right)}^{\frac{1}{2}}+{\left(\left(b-a\right){\int }_{a}^{b}{|{f}^{\prime }\left(t\right)|}^{2}{e}^{{t}^{2}}𝑑t\right)}^{\frac{1}{2}}.$

From the above equation we can find $\stackrel{~}{B}>0$ such that

#### Lemma 3.3.

The map $P\mathrm{:}{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}d\mathit{}u\mathrm{\right)}\mathrm{\to }{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}{e}^{{u}^{\mathrm{2}}}\mathit{}d\mathit{}u\mathrm{\right)}$ defined by $P\mathit{}\mathrm{\left(}f\mathrm{\right)}\mathit{}\mathrm{\left(}u\mathrm{\right)}\mathrm{=}f\mathit{}\mathrm{\left(}u\mathrm{\right)}\mathit{}{e}^{\mathrm{-}{u}^{\mathrm{2}}\mathrm{/}\mathrm{2}}$ is unitary.

#### Proof.

This is obvious. ∎

Let ${\psi }_{k}\left(u\right)={\varphi }_{k}\left(u\right){e}^{-{u}^{2}/2}$ for $k\in ℕ$. Since $\left\{{\varphi }_{k}:k\in ℕ\right\}$ forms an orthonormal basis for ${L}^{2}\left(ℝ\right)$, from Lemma 3.3 we can see that $\left\{{\psi }_{k}:k\in ℕ\right\}$ is an orthonormal basis for ${L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$. The relation between the ${\psi }_{k}$’s and their derivatives is as follows: Using equations (2.1) and (2.2), we have

$\frac{d}{du}{\psi }_{k}\left(u\right)=\frac{d}{du}{\varphi }_{k}\left(u\right){e}^{-\frac{{u}^{2}}{2}}-u{\varphi }_{k}\left(u\right){e}^{-\frac{{u}^{2}}{2}}$$=\left\{{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\varphi }_{k-1}\left(u\right)-{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\varphi }_{k+1}\left(u\right)\right\}{e}^{-\frac{{u}^{2}}{2}}-\left\{{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\varphi }_{k-1}\left(u\right)+{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\varphi }_{k+1}\left(u\right)\right\}{e}^{-\frac{{u}^{2}}{2}}$$=-2{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\varphi }_{k+1}\left(u\right){e}^{-\frac{{u}^{2}}{2}}$$=-2{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\psi }_{k+1}\left(u\right).$

In general, for $r\in ℕ$,

$\frac{{d}^{r}}{d{u}^{r}}{\psi }_{k}\left(u\right)={\left(-2\right)}^{r}{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}\mathrm{\cdots }{\left(\frac{k+r}{2}\right)}^{\frac{1}{2}}{\psi }_{k+r}\left(u\right).$(3.2)

Also for $k\in ℕ$,

$u{\psi }_{k}={\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\psi }_{k+1}+{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\psi }_{k-1}.$(3.3)

For $\alpha =\left({\alpha }_{1},{\alpha }_{2},\mathrm{\dots },{\alpha }_{n}\right)\in {ℕ}^{n}$ and $u=\left({u}_{1},{u}_{2},\mathrm{\dots },{u}_{n}\right)\in {ℝ}^{n}$, we define

${\mathrm{\Psi }}_{\alpha }\left(u\right)=\prod _{k=1}^{n}{\psi }_{{\alpha }_{k}}\left({u}_{k}\right).$

Then for any $\beta \in {ℕ}^{n}$,

${D}^{\beta }{\mathrm{\Psi }}_{\alpha }\left(u\right)={\left(-2\right)}^{|\beta |}\prod _{k=1}^{n}\prod _{p=1}^{{\beta }_{k}}{\left(\frac{{\alpha }_{k}+p}{2}\right)}^{\frac{1}{2}}{\mathrm{\Psi }}_{\alpha +\beta }\left(u\right).$

Thus we have the following theorem.

#### Theorem 3.4 ([12]).

The set $\mathrm{\left\{}{\mathrm{\Psi }}_{\alpha }\mathrm{:}\alpha \mathrm{\in }{\mathrm{N}}^{n}\mathrm{\right\}}$ forms an orthonormal basis for ${L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}{e}^{{u}^{\mathrm{2}}}\mathit{}d\mathit{}u\mathrm{\right)}$.

Hence given any $f\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$, we have

$f=\sum _{\alpha \in {ℕ}^{n}}〈f,{\mathrm{\Psi }}_{\alpha }〉{\mathrm{\Psi }}_{\alpha }=\sum _{k=0}^{\mathrm{\infty }}\sum _{|\alpha |=k}〈f,{\mathrm{\Psi }}_{\alpha }〉{\mathrm{\Psi }}_{\alpha }.$

#### Lemma 3.5.

If $f\mathrm{\in }{W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$, then ${D}^{\beta }\mathit{}f\mathrm{=}{\mathrm{\sum }}_{\alpha \mathrm{\in }{\mathrm{N}}^{n}}\mathrm{〈}f\mathrm{,}{\mathrm{\Psi }}_{\alpha }\mathrm{〉}\mathit{}{D}^{\beta }\mathit{}{\mathrm{\Psi }}_{\alpha }$ for all $\mathrm{|}\beta \mathrm{|}\mathrm{\le }m$.

#### Proof.

For $k\in ℕ,x\in ℝ$ and by using equations (3.2) and (3.3), we have

$D\left({\psi }_{k}\left(x\right){e}^{{x}^{2}}\right)=\left(D\left({\psi }_{k}\right)\left(x\right)+2x{\psi }_{k}\left(x\right)\right){e}^{{x}^{2}}$$=\left(\left(-2\right){\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\psi }_{k+1}\left(x\right)+2{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}{\psi }_{k+1}\left(x\right)+2{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\psi }_{k-1}\left(x\right)\right){e}^{{x}^{2}}$$=2{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\psi }_{k-1}\left(x\right){e}^{{x}^{2}}.$

For each $r\in ℕ$, inductively we can see that

${D}^{r}\left({\psi }_{k}\left(x\right){e}^{{x}^{2}}\right)={D}^{r-1}\left(D\left({\psi }_{k}\right)\left(x\right){e}^{{x}^{2}}\right)+{D}^{r-1}\left(2x{\psi }_{k}\left(x\right){e}^{{x}^{2}}\right)$$=2{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{D}^{r-1}\left({\psi }_{k-1}\left(x\right){e}^{{x}^{2}}\right)$$={2}^{r}{\left(\frac{k}{2}\right)}^{\frac{1}{2}}\mathrm{\cdots }{\left(\frac{k-\left(r-1\right)}{2}\right)}^{\frac{1}{2}}{\psi }_{k-r}\left(x\right){e}^{{x}^{2}}.$(3.4)

Let $\beta \in {ℕ}^{n}$ with $|\beta |\le m$, $f\in {W}^{m,2}\left({ℝ}^{n}\right)$ and $i\in \left\{1,2,\mathrm{\dots },n\right\}$. Then from equations (3.1) and (3.4) we have

$〈{D}^{{\beta }_{i}{e}_{i}}f,{\mathrm{\Psi }}_{\alpha }〉={\int }_{{ℝ}^{n}}{D}^{{\beta }_{i}{e}_{i}}f\left(u\right){\mathrm{\Psi }}_{\alpha }\left(u\right){e}^{{u}^{2}}𝑑u$$={\left(-1\right)}^{{\beta }_{i}}{\int }_{{ℝ}^{n}}f\left(u\right){D}^{{\beta }_{i}{e}_{i}}\left({\mathrm{\Psi }}_{\alpha }\left(u\right){e}^{{u}^{2}}\right)𝑑u$$={\left(-1\right)}^{{\beta }_{i}}{\int }_{{ℝ}^{n}}f\left(u\right)\left\{\prod _{j\ne i}{\psi }_{{\alpha }_{j}}\left({u}_{j}\right){e}^{{u}_{j}^{2}}\right\}{D}^{{\beta }_{i}{e}_{i}}\left({\psi }_{{\alpha }_{i}}\left({u}_{i}\right){e}^{{u}_{i}^{2}}\right)𝑑u$$={\left(-2\right)}^{{\beta }_{i}}\prod _{p=1}^{{\beta }_{i}}{\left(\frac{{\alpha }_{i}-\left(p-1\right)}{2}\right)}^{\frac{1}{2}}〈f,{\mathrm{\Psi }}_{\alpha -{\beta }_{i}{e}_{i}}〉.$

Hence,

${D}^{{\beta }_{i}{e}_{i}}f=\sum _{\alpha \in {ℕ}^{n}}〈{D}^{{\beta }_{i}{e}_{i}}f,{\mathrm{\Psi }}_{\alpha }〉{\mathrm{\Psi }}_{\alpha }$$=\sum _{\alpha \in {ℕ}^{n}}{\left(-2\right)}^{{\beta }_{i}}\prod _{p=1}^{{\beta }_{i}}{\left(\frac{{\alpha }_{i}-\left(p-1\right)}{2}\right)}^{\frac{1}{2}}〈f,{\mathrm{\Psi }}_{\alpha -{\beta }_{i}{e}_{i}}〉{\mathrm{\Psi }}_{\alpha }$$=\sum _{\alpha \in {ℕ}^{n}}{\left(-2\right)}^{{\beta }_{i}}\prod _{p=1}^{{\beta }_{i}}{\left(\frac{{\alpha }_{i}+p}{2}\right)}^{\frac{1}{2}}〈f,{\mathrm{\Psi }}_{\alpha }〉{\mathrm{\Psi }}_{\alpha +{\beta }_{i}}$$=\sum _{\alpha \in {ℕ}^{n}}〈f,{\mathrm{\Psi }}_{\alpha }〉{D}^{{\beta }_{i}{e}_{i}}{\mathrm{\Psi }}_{\alpha }.$

If we apply the above step repeatedly, we have

${D}^{\beta }f=\sum _{\alpha \in {ℕ}^{n}}〈f,{\mathrm{\Psi }}_{\alpha }〉{D}^{\beta }{\mathrm{\Psi }}_{\alpha }.\mathit{∎}$

With the help of Lemma 3.5 we can see an important result which does not hold for classical Sobolev spaces.

#### Lemma 3.6.

If $f\mathrm{\in }{W}^{\mathrm{1}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{\right)}$, then $u\mathit{}f\mathrm{\in }{L}^{\mathrm{2}}\mathit{}\mathrm{\left(}\mathrm{R}\mathrm{,}{e}^{{u}^{\mathrm{2}}}\mathit{}d\mathit{}u\mathrm{\right)}$.

#### Proof.

Let $f\in {W}^{1,2}\left(ℝ\right)$. From Lemma 3.5 we can see that

$\frac{d}{du}f=\sum _{k=0}^{\mathrm{\infty }}〈f,{\psi }_{k}〉\frac{d}{du}{\psi }_{k}=\sum _{k=0}^{\mathrm{\infty }}-2{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k+1}.$(3.5)

Now let us consider the N-th partial sum of ${\sum }_{k=0}^{\mathrm{\infty }}〈f,{\psi }_{k}〉{\psi }_{k}$, that is,

$\sum _{k=0}^{N}〈f,{\psi }_{k}〉u{\psi }_{k}=\sum _{k=1}^{N}{\left(\frac{k}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k-1}+\sum _{k=0}^{n}{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k+1}.$

For $N\in ℕ$, let us define

${A}_{N}f=\sum _{k=1}^{N}{\left(\frac{k}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k-1}\mathit{ }\text{and}\mathit{ }{B}_{N}f=\sum _{k=0}^{N}{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k+1}.$

Since $f\in {W}^{1,2}\left(ℝ\right)$ and by equation (3.5), ${A}_{N}f$ converges to

$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k-1},$

and ${B}_{N}f$ converges to

$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{2}\right)}^{\frac{1}{2}}〈f,{\psi }_{k}〉{\psi }_{k+1}$

as $N\to \mathrm{\infty }$ in ${L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$. So ${\sum }_{k=0}^{\mathrm{\infty }}〈f,{\psi }_{k}〉u{\psi }_{k}$ converges in ${L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$. Since $f={\sum }_{k=0}^{\mathrm{\infty }}〈f,{\psi }_{k}〉{\psi }_{k}$ converges in ${L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$, there is a subsequence, say ${\sum }_{k=0}^{{N}_{k}}〈f,{\psi }_{k}〉{\psi }_{k}$ converges to f pointwise a.e. on $ℝ$. This gives us

This shows that $uf={\sum }_{k=0}^{\mathrm{\infty }}〈f,{\psi }_{k}〉u{\psi }_{k}$ and $uf\in {L}^{2}\left(ℝ,{e}^{{u}^{2}}du\right)$ for $f\in {W}^{1,2}\left(ℝ\right)$. ∎

The above result can be extended to higher dimensions, and the result is as follows.

#### Theorem 3.7.

If $m\mathrm{\in }\mathrm{N}$, $\alpha \mathrm{\in }{\mathrm{N}}^{n}$ with $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }m$, and $f\mathrm{\in }{W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$, then ${u}^{\alpha }\mathit{}f\mathrm{\in }{W}^{m\mathrm{-}\mathrm{|}\alpha \mathrm{|}\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$.

The orthonormal basis $\left\{{\mathrm{\Psi }}_{\alpha }:\alpha \in {ℕ}^{n}\right\}$ for ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$ is also an orthonormal basis for ${W}^{m,2}\left({ℝ}^{n}\right)$ after multiplying each function with an appropriate normalizing constant.

#### Theorem 3.8.

The set $\mathrm{\left\{}{\mathrm{\Psi }}_{\alpha }\mathrm{/}{\mathrm{\parallel }{\mathrm{\Psi }}_{\alpha }\mathrm{\parallel }}_{{W}_{m}}\mathrm{:}\alpha \mathrm{\in }{\mathrm{N}}^{n}\mathrm{\right\}}$ forms a complete orthonormal system in ${W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$.

#### Proof.

Let $\alpha \ne \beta$. Then

${〈{\mathrm{\Psi }}_{\alpha },{\mathrm{\Psi }}_{\beta }〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}=\sum _{|\gamma |\le m}〈{D}^{\gamma }{\mathrm{\Psi }}_{\alpha },{D}^{\gamma }{\mathrm{\Psi }}_{\beta }〉$$=\sum _{|\gamma |\le m}{\left(2\right)}^{2|\gamma |}\prod _{k=1}^{n}\prod _{p=1}^{{\gamma }_{k}}{\left(\frac{{\alpha }_{k}+p}{2}\right)}^{\frac{1}{2}}{\left(\frac{{\beta }_{k}+p}{2}\right)}^{\frac{1}{2}}〈{\mathrm{\Psi }}_{\alpha +\gamma },{\mathrm{\Psi }}_{\beta +\gamma }〉$$=0.$

For completeness, let $f\in {W}^{m,2}\left({ℝ}^{n}\right)$ such that ${〈f,{\mathrm{\Psi }}_{\alpha }〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}=0$ for all $\alpha \in {ℕ}^{n}$. Then

$0={〈f,{\mathrm{\Psi }}_{\alpha }〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}$$=\sum _{|\gamma |\le m}〈{D}^{\gamma }f,{D}^{\gamma }{\mathrm{\Psi }}_{\alpha }〉$$=\sum _{|\gamma |\le m}\sum _{\beta \in {ℕ}^{n}}〈f,{\mathrm{\Psi }}_{\beta }〉〈{D}^{\gamma }{\mathrm{\Psi }}_{\beta },{D}^{\gamma }{\mathrm{\Psi }}_{\alpha }〉$$=〈f,{\mathrm{\Psi }}_{\alpha }〉\sum _{|\gamma |\le m}{\left(2\right)}^{2|\gamma |}\prod _{k=1}^{n}\prod _{p=1}^{{\gamma }_{k}}\left(\frac{{\alpha }_{k}+p}{2}\right).$

The above relation forces that $〈f,{\mathrm{\Psi }}_{\alpha }〉=0$ for all $\alpha \in {ℕ}^{n}$. This implies that $f=0$. Thus the theorem follows. ∎

#### Note 3.9.

Here for $\alpha \in {ℕ}^{n}$,

${\parallel {\mathrm{\Psi }}_{\alpha }\parallel }_{{W}_{m}}^{2}=\sum _{|\gamma |\le m}{2}^{2|\gamma |}\prod _{k=1}^{n}\prod _{q=1}^{{\gamma }_{k}}\left(\frac{{\alpha }_{k}+q}{2}\right).$

Now let us take

${\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\left(u\right)={\mathrm{\Psi }}_{\alpha }\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}$

for $\alpha \in {ℕ}^{n}$ and $u\in {ℝ}^{n}$. Set $\mathcal{ℬ}=\left\{{\stackrel{~}{\mathrm{\Psi }}}_{\alpha }:\alpha \in {ℕ}^{n}\right\}$ and $\mathcal{𝒜}=\left\{{\mathrm{\Psi }}_{\alpha }:\alpha \in {ℕ}^{n}\right\}$. Observe that $\mathcal{𝒜}$ and $\mathcal{ℬ}$ form an orthonormal basis for ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$. The reason to consider the collection $\mathcal{ℬ}$ is as follows: this collection behaves nicely with the Schrödinger semigroup, and in fact, the Schrödinger semigroup associated to Δ on ${ℝ}^{n}$ takes the collection $\mathcal{ℬ}$ into some nice functions in $\mathcal{𝒪}\left({ℂ}^{n}\right)$. Those nice functions will help us to characterize the image of a Sobolev space under the Schrödinger semigroup. We will discuss the above ideas in the next section. The set $\mathcal{𝒜}$ is an orthogonal system in ${W}^{m,2}\left({ℝ}^{n}\right)$, but $\mathcal{ℬ}$ is not an orthogonal system in ${W}^{m,2}\left(ℝ\right)$. For instance, if $m=n=1$, then

${\stackrel{~}{\psi }}_{k}^{\prime }\left(u\right)={\psi }_{k}^{\prime }\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}-\frac{i}{2t}u{\psi }_{k}\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}$$=\left(-1\right)\left(2+\frac{i}{2t}\right){\stackrel{~}{\psi }}_{k+1}\left(u\right)-\frac{i}{2t}{\left(\frac{k}{2}\right)}^{\frac{1}{2}}{\stackrel{~}{\psi }}_{k-1}\left(u\right).$

So for $k\ne p\in ℕ$ and $k,

${〈{\stackrel{~}{\psi }}_{k},{\stackrel{~}{\psi }}_{p}〉}_{{W}^{1,2}\left(ℝ\right)}=〈{\stackrel{~}{\psi }}_{k},{\stackrel{~}{\psi }}_{p}〉+〈{\stackrel{~}{\psi }}_{k}^{\prime },{\stackrel{~}{\psi }}_{p}^{\prime }〉$

#### Remark 3.10.

Let $m\ge 1$ and $f\in {W}^{m,2}\left({ℝ}^{n}\right)$. Then we have the following:

• (i)

For each $k\in \left\{1,2,\mathrm{\dots },n\right\}$,

${D}^{{e}_{k}}\left(f\left(u\right){e}^{\frac{-i}{4t}{u}^{2}}\right)=\left\{{D}^{{e}_{k}}f\left(u\right)-\frac{i}{2t}{u}_{k}f\left(u\right)\right\}{e}^{\frac{-i}{4t}{u}^{2}}.$

By Lemma 3.6 we have ${u}_{k}f\left(u\right)\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$. So $f\left(u\right){e}^{-i/\left(4t\right){u}^{2}}\in {W}^{1,2}\left({ℝ}^{n}\right)$.

• (ii)

Similarly, we can see that $f\left(u\right){e}^{-i/\left(4t\right){u}^{2}}\in {W}^{m,2}\left({ℝ}^{n}\right)$.

• (iii)

For $\alpha \le \beta \in {ℕ}^{n}$, we have ${\parallel {D}^{\alpha }f\parallel }^{2}\le {\parallel {D}^{\beta }f\parallel }^{2}$.

Using Remark 3.10, we can define the transformation $\mathcal{𝒮}:{W}^{m,2}\left({ℝ}^{n}\right)\to {W}^{m,2}\left({ℝ}^{n}\right)$ by $\mathcal{𝒮}\left(f\right)\left(u\right)=f\left(u\right){e}^{-i/\left(4t\right){u}^{2}}$. Let us take

Then, by Theorem 3.8, ${\mathcal{𝒟}}_{n}$ forms a dense subspace of ${W}^{m,2}\left({ℝ}^{n}\right)$.

#### Theorem 3.11.

The map $\mathcal{S}$ is a bounded invertible map.

#### Proof.

We use mathematical induction on m to prove this theorem. First let $m=1$ and $f\in {\mathcal{𝒟}}_{n}$ with

$f=\sum _{|\alpha |\le p}{a}_{\alpha }{\mathrm{\Psi }}_{\alpha },$

where ${a}_{\alpha }\in ℂ$ and $p\in ℕ$.

For $k\in \left\{1,2,\mathrm{\dots },n\right\}$,

${\parallel {u}_{k}f\parallel }^{2}={\parallel \sum _{|\alpha |\le p}{a}_{\alpha }{u}_{k}{\mathrm{\Psi }}_{\alpha }\parallel }^{2}$$={\parallel \sum _{|\alpha |\le p}{a}_{\alpha }{\left(\frac{{\alpha }_{k+1}}{2}\right)}^{\frac{1}{2}}{\mathrm{\Psi }}_{\alpha +{e}_{k}}+\sum _{1\le |\alpha |\le p}{\left(\frac{{\alpha }_{k}}{2}\right)}^{\frac{1}{2}}{a}_{\alpha }{\mathrm{\Psi }}_{\alpha -{e}_{k}}\parallel }^{2}$$\le {\left\{\parallel \sum _{|\alpha |\le p}{a}_{\alpha }{\left(\frac{{\alpha }_{k+1}}{2}\right)}^{\frac{1}{2}}{\mathrm{\Psi }}_{\alpha +{e}_{k}}\parallel +\parallel \sum _{1\le |\alpha |\le p}{\left(\frac{{\alpha }_{k}}{2}\right)}^{\frac{1}{2}}{a}_{\alpha }{\mathrm{\Psi }}_{\alpha -{e}_{k}}\parallel \right\}}^{2}$$={\left\{{\left(\sum _{|\alpha |\le p}\frac{{\alpha }_{k+1}}{2}{|{a}_{\alpha }|}^{2}\right)}^{\frac{1}{2}}+{\left(\sum _{1\le |\alpha |\le p}\frac{{\alpha }_{k}}{2}{|{a}_{\alpha }|}^{2}\right)}^{\frac{1}{2}}\right\}}^{2}$$\le {\left\{2{\left(\sum _{|\alpha |\le p}\frac{{\alpha }_{k+1}}{2}{|{a}_{\alpha }|}^{2}\right)}^{\frac{1}{2}}\right\}}^{2}$$={\parallel {D}^{{e}_{k}}f\parallel }^{2}.$

Using the above inequality, we have

${\parallel \mathcal{𝒮}f\parallel }_{{W}_{1}}^{2}=\sum _{k=0}^{n}{\parallel {D}^{{e}_{k}}\left(f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\parallel }^{2}$$={\parallel f\parallel }^{2}+\sum _{k=1}^{n}{\parallel {D}^{{e}_{k}}f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}-\frac{i}{2t}{u}_{k}f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel }^{2}$$\le {\parallel f\parallel }^{2}+\sum _{k=0}^{n}{\left(\parallel {D}^{{e}_{k}}f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel +|\frac{1}{2t}|\parallel {u}_{k}f\parallel \right)}^{2}$$\le {\parallel f\parallel }^{2}+{\left(1+|\frac{1}{2t}|\right)}^{2}\sum _{k=1}^{n}\parallel {D}^{{e}_{k}}f{\parallel }^{2}$$\le {B}_{1}{\parallel f\parallel }_{{W}_{1}}^{2},$

where ${B}_{1}={\left(1+|\frac{1}{2t}|\right)}^{2}$.

So, $\mathcal{𝒮}:{W}^{1,2}\left({ℝ}^{n}\right)\to {W}^{1,2}\left({ℝ}^{n}\right)$ is a bounded map. Now let $m\in ℕ$. For every α with $|\alpha |=m$ there exist $k\left(\alpha \right)\in \left\{1,2,\mathrm{\dots },n\right\}$ such that ${\alpha }_{k\left(\alpha \right)}\ne 0$. Then observe that for $f\in {\mathcal{𝒟}}_{n}$,

${\parallel {u}_{k\left(\alpha \right)}f\parallel }_{{W}_{m-1}}^{2}=\sum _{|\beta |\le m-1}{\parallel {D}^{\beta }\left({u}_{k\left(\alpha \right)}f\right)\parallel }^{2}$$=\sum _{|\beta |\le m-1}{\parallel {u}_{k\left(\alpha \right)}{D}^{\beta }\left(f\right)+{\beta }_{k\left(\alpha \right)}{D}^{\beta -{e}_{k\left(\alpha \right)}}f\parallel }^{2}$$\le \sum _{|\beta |\le m-1}{\left\{\parallel {u}_{k\left(\alpha \right)}{D}^{\beta }\left(f\right)\parallel +{\beta }_{k\left(\alpha \right)}\parallel {D}^{\beta -{e}_{k\left(\alpha \right)}}f\parallel \right\}}^{2}$$\le \sum _{|\beta |\le m-1}{\left\{\parallel {D}^{\beta +{e}_{k\left(\alpha \right)}}\left(f\right)\parallel +{\beta }_{k\left(\alpha \right)}\parallel {D}^{\beta +{e}_{k\left(\alpha \right)}}f\parallel \right\}}^{2}$$\le m\sum _{|\beta |\le m-1}{\parallel {D}^{\beta +{e}_{k\left(\alpha \right)}}\left(f\right)\parallel }^{2}\le m{\parallel f\parallel }_{{W}_{m}}^{2}.$(3.6)

Also we can see that

${D}^{\alpha }\left(f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)={D}^{\alpha -{e}_{k\left(\alpha \right)}}\left\{\left({D}^{{e}_{k\left(\alpha \right)}}f\right)\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}-\frac{i}{2t}\left({u}_{k\left(\alpha \right)}f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\right\}$$={D}^{\alpha -{e}_{k\left(\alpha \right)}}\left\{\mathcal{𝒮}\left({D}^{{e}_{k\left(\alpha \right)}}f\right)\left(u\right)-\frac{i}{2t}\mathcal{𝒮}\left({u}_{k\left(\alpha \right)}f\right)\left(u\right)\right\}.$(3.7)

Using induction on m and equations (3.6) and (3.7), we have

${\parallel \mathcal{𝒮}f\parallel }_{{W}_{m}}^{2}=\sum _{|\alpha |\le m-1}{\parallel {D}^{\alpha }\left(S\left(f\right)\right)\parallel }^{2}+\sum _{|\alpha |=m}{\parallel {D}^{\alpha }\left(f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\parallel }^{2}$$={\parallel S\left(f\right)\parallel }_{{W}_{m-1}}^{2}+\sum _{|\alpha |=m}{\parallel {D}^{\alpha -{e}_{k\left(\alpha \right)}}\left\{\mathcal{𝒮}\left({D}^{{e}_{k\left(\alpha \right)}}f\right)-\frac{i}{2t}\mathcal{𝒮}\left({u}_{k\left(\alpha \right)}f\right)\right\}\parallel }^{2}$$\le {\parallel S\left(f\right)\parallel }_{{W}_{m-1}}^{2}+\sum _{|\alpha |=m}{\left\{\parallel {D}^{\alpha -{e}_{k\left(\alpha \right)}}\mathcal{𝒮}\left({D}^{{e}_{k\left(\alpha \right)}}f\right)\parallel +\frac{1}{2t}\parallel {D}^{\alpha -{e}_{k\left(\alpha \right)}}\mathcal{𝒮}\left({u}_{k\left(\alpha \right)}f\right)\parallel \right\}}^{2}$$\le {\parallel S\left(f\right)\parallel }_{{W}_{m-1}}^{2}+{\left\{{\parallel \mathcal{𝒮}\left({D}^{{e}_{k\left(\alpha \right)}}f\right)\parallel }_{{W}_{m-1}}+\frac{1}{2t}{\parallel \mathcal{𝒮}\left({u}_{k\left(\alpha \right)}f\right)\parallel }_{{W}_{m-1}}\right\}}^{2}$$\le {B}_{m-1}{\parallel f\parallel }_{{W}_{m-1}}^{2}+{B}_{m-1}{\left\{{\parallel {D}^{{e}_{k\left(\alpha \right)}}f\parallel }_{{W}_{m-1}}+\frac{1}{2t}{\parallel {u}_{k\left(\alpha \right)}f\parallel }_{{W}_{m-1}}\right\}}^{2}$$\le {B}_{m-1}{\parallel f\parallel }_{{W}_{m-1}}^{2}+{B}_{m-1}{\left(1+\sqrt{m}\frac{1}{2t}\right)}^{2}{\parallel f\parallel }_{{W}_{m}}^{2}$$\le {B}_{m-1}\left\{1+{\left(1+\sqrt{m}\frac{1}{2t}\right)}^{2}\right\}{\parallel f\parallel }_{{W}_{m}}^{2}.$

This implies that $\mathcal{𝒮}:{W}^{m,2}\left({ℝ}^{n}\right)\to {W}^{m,2}\left({ℝ}^{n}\right)$ is a bounded map. It is clear that $\mathcal{𝒮}$ is one-to-one and onto from the definition. Hence the theorem is proved. ∎

#### Note 3.12.

We have just proved that if $f\in {W}^{m,2}\left({ℝ}^{n}\right)$, then $\parallel {u}_{k}f\parallel \le \parallel {D}^{{e}_{k}}f\parallel$ for $k=1,2,\mathrm{\dots },n$. In a similar way, it is easy to show that $\parallel {u}^{\alpha }f\parallel \le \parallel {D}^{\alpha }f\parallel$ for all $|\alpha |\le m$.

#### Remark 3.13.

Using Theorem 3.11, we can see that for each $m\in ℕ$ the set $\mathcal{ℬ}$ forms a Schauder basis for ${W}^{m,2}\left({ℝ}^{n}\right)$.

## 4 Image of Sobolev spaces under the Schrödinger semigroup

In this section, we will prove our main result. First we define the holomorphic Sobolev space ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ to be the image of ${W}^{m,2}\left({ℝ}^{n}\right)$ under ${e}^{it\mathrm{\Delta }}$. The space ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ is made into a Hilbert space by transferring the Hilbert space structure of ${W}^{m,2}\left({ℝ}^{n}\right)$ to ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ so that the Schrödinger semigroup ${e}^{it\mathrm{\Delta }}$ is an isometric isomorphism from ${W}^{m,2}\left({ℝ}^{n}\right)$ onto ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$. This means that

${〈F,G〉}_{{W}_{t}^{m,2}\left({ℂ}^{n}\right)}:={〈f,g〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}$

whenever $F={e}^{it\mathrm{\Delta }}f$ and $G={e}^{it\mathrm{\Delta }}g$. We wish to characterize ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ as a weighted Bergman space. The following theorem gives a relationship between $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(x+iy\right)dxdy\right)$ and the classical Fock space.

#### Theorem 4.1.

For $r\mathrm{=}\frac{\mathrm{1}}{\mathrm{8}\mathit{}{t}^{\mathrm{2}}}$, the map

$U:\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(x+iy\right)dxdy\right)\to {\mathcal{ℱ}}_{\frac{1}{8{t}^{2}}}^{2}$

defined for $F\mathrm{\in }\mathcal{H}\mathit{}{\mathcal{L}}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{C}}^{n}\mathrm{,}{w}_{t}\mathit{}\mathrm{\left(}z\mathrm{\right)}\mathit{}d\mathit{}z\mathrm{\right)}$ by

$U\left(F\right)\left(z\right)={\left(4t\sqrt{\pi }\right)}^{\frac{n}{2}}F\left(z\right){e}^{\left(\frac{-i}{4t}+\frac{1}{16{t}^{2}}\right){z}^{2}}$

is a unitary map.

#### Proof.

This is an obvious verification. ∎

#### Remark 4.2.

• (i)

From Theorem 4.1 we can conclude that the set

$\left\{{\mathrm{Υ}}_{\alpha }^{t}\left(z\right)={\left(4t\sqrt{\pi }\right)}^{-\frac{n}{2}}{𝐞}_{\alpha }^{\frac{1}{8{t}^{2}}}\left(z\right){e}^{\left(\frac{i}{4t}-\frac{1}{16{t}^{2}}\right){z}^{2}}:\alpha \in {ℕ}^{n}\right\}$

forms an orthonormal basis for $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left(ℂ,{w}_{t}\left(x+iy\right)dxdy\right)$.

• (ii)

Also for $\gamma ,\alpha \in {ℕ}^{n}$,

${z}^{\gamma }{\mathrm{Υ}}_{\alpha }^{t}\left(z\right)={\left(4t\right)}^{|\gamma |}\prod _{j=1}^{n}\prod _{q=1}^{{\gamma }_{j}}{\left(\frac{{\alpha }_{j}+q}{2}\right)}^{\frac{1}{2}}{\mathrm{Υ}}_{\alpha +\gamma }^{t}\left(z\right).$(4.1)

The following lemma gives the pointwise estimate of ${e}^{it\mathrm{\Delta }}\left(f\right)$ for $f\in {W}^{m,2}\left({ℝ}^{n}\right)$.

#### Lemma 4.3.

If $f\mathrm{\in }{W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$, then there exists $C\mathrm{>}\mathrm{0}$ (depends only on $m\mathrm{,}n\mathrm{,}t$ and f) such that for any $z\mathrm{\in }{\mathrm{C}}^{n}$,

${|{e}^{it\mathrm{\Delta }}\left(f\right)\left(z\right)|}^{2}\le C\frac{{e}^{\frac{-xy}{t}+\frac{{y}^{2}}{4{t}^{2}}}}{1+{|{z}^{\alpha }|}^{2}}$

for $\mathrm{|}\alpha \mathrm{|}\mathrm{\le }m$.

#### Proof.

For $f\in {W}^{m,2}\left(ℝ\right)$, $k\in \left\{1,2,\mathrm{\dots },n\right\}$ and $z\in {ℂ}^{n}$, consider

${e}^{it\mathrm{\Delta }}\left({D}^{{e}_{k}}f\right)\left(z\right)={c}_{t}^{n}{\int }_{{ℝ}^{n}}{D}^{{e}_{k}}f\left(u\right){e}^{\frac{i}{4t}{\left(z-u\right)}^{2}}𝑑u$$=-\frac{i{c}_{t}^{n}}{2t}{\int }_{{ℝ}^{n}}\left({u}_{k}-{z}_{k}\right)f\left(u\right){e}^{\frac{i}{4t}{\left(z-u\right)}^{2}}𝑑u$$={c}_{t}^{n}\left\{\frac{i}{2t}{z}_{k}{e}^{it\mathrm{\Delta }}\left(f\right)\left(z\right)-\frac{i}{2t}{e}^{it\mathrm{\Delta }}\left({u}_{k}f\right)\left(z\right)\right\}.$

From the above equation we obtain

${z}_{k}{e}^{it\mathrm{\Delta }}\left(f\right)\left(z\right)=\frac{2t}{i}{e}^{it\mathrm{\Delta }}\left(\frac{1}{{c}_{t}^{n}}{D}^{{e}_{k}}f+\frac{i}{2t}{u}_{k}f\right)\left(z\right).$

Now the assumption on f tells us that for $|\alpha |\le m$,

${\left(\frac{1}{{c}_{t}^{n}}{D}^{{e}_{1}}+\frac{i}{2t}{u}_{1}\right)}^{{\alpha }_{1}}\mathrm{\cdots }{\left(\frac{1}{{c}_{t}^{n}}{D}^{{e}_{1}}+\frac{i}{2t}{u}_{n}\right)}^{{\alpha }_{n}}f\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right).$

This implies that ${z}^{\alpha }{e}^{it\mathrm{\Delta }}\left(f\right)$ is in $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$. Moreover, using pointwise estimates of functions in $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$ given in (1.1), there exists a constant (depends only on $f,m,n$ and t) $C>0$ such that

${|{e}^{it\mathrm{\Delta }}\left(f\right)\left(z\right)|}^{2}\le C\frac{{e}^{\frac{-xy}{t}+\frac{{y}^{2}}{4{t}^{2}}}}{1+{|{z}^{\alpha }|}^{2}}$

for all $z\in {ℂ}^{n}$. ∎

Let

${u}_{t}^{m}\left(z\right)=\sum _{|\gamma |\le m}{\left(\frac{1}{2t}\right)}^{2|\gamma |}{|{z}^{\gamma }|}^{2}{w}_{t}\left(z\right)$

and consider the weighted Bergman space $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$. Using pointwise estimate given in Lemma 4.3, we can compare the spaces ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ and $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$.

#### Lemma 4.4.

The set

$\mathrm{\Omega }=\left\{\frac{{\mathrm{Υ}}_{\alpha }^{t}}{{\parallel {\mathrm{Υ}}_{\alpha }^{t}\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t}^{m}}^{2}}}:\alpha \in {ℕ}^{n}\right\}$

forms a complete orthonormal basis for $\mathcal{H}\mathit{}{\mathcal{L}}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{C}}^{n}\mathrm{,}{u}_{t}^{m}\mathit{}\mathrm{\left(}z\mathrm{\right)}\mathit{}d\mathit{}z\mathrm{\right)}$.

#### Proof.

We know that $\left\{{\mathrm{Υ}}_{\alpha }^{t}:\alpha \in {ℕ}^{n}\right\}$ forms an orthonormal basis for $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$. Using this, we can prove that Ω is a complete orthonormal set in $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$. Since ${w}_{t}\left(z\right)\le {u}_{t}^{m}\left(z\right)$, this implies that $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)\subset \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left(ℂ,{w}_{t}\left(z\right)dz\right)$. Now for $\alpha \ne \beta \in {ℕ}^{n}$ and using equation (4.1),

${〈{\mathrm{Υ}}_{\alpha }^{t},{\mathrm{Υ}}_{\beta }^{t}〉}_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t}^{m}}^{2}}={\int }_{{ℂ}^{n}}{\mathrm{Υ}}_{\alpha }^{t}\left(z\right)\overline{{\mathrm{Υ}}_{\beta }^{t}\left(z\right)}\sum _{|\gamma |\le m}{\left(\frac{1}{2t}\right)}^{2|\gamma |}{|{z}^{\gamma }|}^{2}{w}_{t}\left(z\right)dz$$=\sum _{|\gamma |\le m}{\left(\frac{1}{2t}\right)}^{2|\gamma |}{\int }_{{ℂ}^{n}}{z}^{\gamma }{\mathrm{Υ}}_{\alpha }^{t}\left(z\right)\overline{{z}^{\gamma }{\mathrm{Υ}}_{\beta }^{t}\left(z\right)}{w}_{t}\left(z\right)𝑑z$$=\sum _{|\gamma |\le m}{2}^{2|\gamma |}\prod _{j=1}^{n}\prod _{q=1}^{{\gamma }_{j}}{\left(\frac{{\alpha }_{j}+q}{2}\right)}^{\frac{1}{2}}{\left(\frac{{\beta }_{j}+q}{2}\right)}^{\frac{1}{2}}{〈{\mathrm{Υ}}_{\alpha +\gamma }^{t},{\mathrm{Υ}}_{\beta +\gamma }^{t}〉}_{\mathcal{ℋ}{\mathcal{ℒ}}_{{w}_{t}}^{2}}$$=0.$

This proves the orthogonality. The completeness of Ω in $\mathcal{ℋ}{\mathcal{ℒ}}^{m,2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$ easily follows from the completeness of Ω in $\mathcal{ℋ}{\mathcal{ℒ}}^{m,2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$. ∎

The following proposition and theorem show that the norms of the spaces ${W}_{t}^{m,2}\left({ℂ}^{n}\right)$ and $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$ are equivalent.

#### Proposition 4.5.

For $p\mathrm{\in }\mathrm{N}$ and $x\mathrm{\in }\mathrm{R}$, we have

${e}^{it\mathrm{\Delta }}\left({u}^{p}{e}^{-{u}^{2}}{e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={\left(\frac{2t}{i}\right)}^{p}{c}_{t}{\left(-1\right)}^{p}\frac{{d}^{p}}{d{x}^{p}}\left({e}^{-\frac{1}{16{t}^{2}}{x}^{2}}\right){e}^{\frac{i}{4t}{x}^{2}}.$

Furthermore,

${e}^{it\mathrm{\Delta }}\left({\psi }_{k}\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={c}_{t}\sqrt{\pi }{\left(\frac{1}{{2}^{k}k!\sqrt{\pi }}\right)}^{\frac{1}{2}}{\left(\frac{1}{2ti}\right)}^{k}{x}^{k}{e}^{\left(\frac{i}{4t}-\frac{1}{16{t}^{2}}\right){x}^{2}}.$

#### Proof.

For $x\in ℝ$ and $s=\frac{1}{16{t}^{2}}$,

${e}^{it\mathrm{\Delta }}\left({u}^{p}{e}^{-{u}^{2}-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={c}_{t}{\int }_{ℝ}{u}^{p}{e}^{-{u}^{2}-\frac{i}{4t}{u}^{2}}{e}^{\frac{i}{4t}{\left(x-u\right)}^{2}}𝑑u$$={c}_{t}{\left(-\frac{2t}{i}\right)}^{p}{e}^{\frac{i}{4t}{x}^{2}}{\int }_{ℝ}{e}^{-{u}^{2}}\frac{{d}^{p}}{d{x}^{p}}\left({e}^{-\frac{i}{2t}xu}\right)𝑑u$$={c}_{t}{\left(-\frac{2t}{i}\right)}^{p}{e}^{\frac{i}{4t}{x}^{2}}\frac{{d}^{p}}{d{x}^{p}}\left({\int }_{ℝ}{e}^{-{u}^{2}}{e}^{-\frac{i}{2t}xu}𝑑u\right)$$=\sqrt{\pi }{\left(\frac{2t}{i}\right)}^{p}{c}_{t}{\left(-1\right)}^{p}\frac{{d}^{p}}{d{x}^{p}}\left({e}^{-\frac{1}{16{t}^{2}}{x}^{2}}\right){e}^{\frac{i}{4t}{x}^{2}}$$=\sqrt{\pi }{\left(\frac{2t}{i}\right)}^{p}{c}_{t}{H}_{p}^{s}\left(x\right){e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}.$

Now for $k\in ℕ$,

${e}^{it\mathrm{\Delta }}\left({\psi }_{k}\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={\left(\frac{1}{{2}^{k}k!\sqrt{\pi }}\right)}^{\frac{1}{2}}{e}^{it\mathrm{\Delta }}\left({H}_{k}\left(u\right){e}^{-{u}^{2}}{e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right).$

First assume that k is even. So,

${e}^{it\mathrm{\Delta }}\left({H}_{k}\left(u\right){e}^{-{u}^{2}}{e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)=\sum _{p=0}^{\frac{k}{2}}{a}_{k-2p}^{\left(k\right)}{e}^{it\mathrm{\Delta }}\left({u}^{k-2p}{e}^{-{u}^{2}}{e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)$$={c}_{t}\sqrt{\pi }\sum _{p=0}^{\frac{k}{2}}{a}_{k-2p}^{\left(k\right)}{\left(\frac{2t}{i}\right)}^{k-2p}{H}_{k-2p}^{s}\left(x\right){e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}$$={c}_{t}\sqrt{\pi }\left\{\sum _{p=0}^{\frac{k}{2}}{b}_{k-2p}{x}^{k-2p}\right\}{e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}.$

Here ${b}_{k-2p}$ is a coefficient of ${x}^{k-2p}$ in the above polynomial, and it is given by

${b}_{k-2p}=\sum _{r=0}^{p}{a}_{k-2r}^{\left(k\right)}{\left(\frac{2t}{i}\right)}^{k-2r}{a}_{k-2p}^{\left(k-2r\right),s}.$(4.2)

Claim: ${b}_{k\mathrm{-}\mathrm{2}\mathit{}p}\mathrm{=}\mathrm{0}$ for all $\mathrm{1}\mathrm{\le }p$. From Proposition 2.2 we have

${a}_{k-2r}^{\left(k\right)}={2}^{k-2r}{\left(-1\right)}^{r}\frac{k!}{\left(k-2r\right)!r!}={a}_{k}^{\left(k\right)}{2}^{-2r}{\left(-1\right)}^{r}\frac{k!}{\left(k-2r\right)!r!}$

and

${a}_{k-2p}^{\left(k-2r\right),s}={\left(2s\right)}^{k-2p}{\left(s\right)}^{p-r}{\left(-1\right)}^{p-r}\frac{\left(k-2r\right)!}{\left(k-2p\right)!\left(p-r\right)!}$$={a}_{k-2p}^{\left(k\right),s}{s}^{-r}{\left(-1\right)}^{-r}\frac{\left(p-r+1\right)\mathrm{\cdots }p}{\left(k-2r+1\right)\mathrm{\cdots }k}.$

Using the above equations, we can see that equation (4.2) is equal to zero. In fact,

${b}_{k-2p}={a}_{k}^{\left(k\right)}{\left(\frac{2t}{i}\right)}^{k}\left\{\sum _{r=0}^{p}{\left(\frac{2t}{i}\right)}^{-2r}{\left(\frac{1}{16{t}^{2}}\right)}^{-r}{2}^{-2r}\frac{\left(p-r+1\right)\mathrm{\cdots }p}{r!}\right\}{a}_{k-2p}^{\left(k\right),s}$$={a}_{k}^{\left(k\right)}{\left(\frac{2t}{i}\right)}^{k}\left\{\sum _{r=0}^{p}{i}^{2r}\frac{\left(p-r+1\right)\mathrm{\cdots }p}{r!}\right\}{a}_{k-2p}^{\left(k\right),s}$$=0$

because

$\sum _{r=0}^{p}{i}^{2r}\frac{\left(p-r+1\right)\mathrm{\cdots }p}{r!}={\left(1-1\right)}^{p}=0$

for all $1\le p\le \frac{k}{2}$. This implies that

${e}^{it\mathrm{\Delta }}\left({H}_{k}\left(u\right){e}^{-{u}^{2}}{e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={c}_{t}\sqrt{\pi }{a}_{k}^{\left(k\right)}{\left(\frac{2t}{i}\right)}^{k}{a}_{k}^{\left(k\right),s}{x}^{k}{e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}$$={c}_{t}\sqrt{\pi }{2}^{k}{\left(\frac{2t}{i}\right)}^{k}{\left(2\frac{1}{16{t}^{2}}\right)}^{k}{x}^{k}{e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}$$={c}_{t}\sqrt{\pi }{\left(\frac{1}{i}\right)}^{k}{\left(\frac{1}{2t}\right)}^{k}{x}^{k}{e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}.$

If k is odd, the proof follows in similar steps. So, for each $k\in ℕ$, we have

${e}^{it\mathrm{\Delta }}\left({\psi }_{k}\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\left(x\right)={c}_{t}\sqrt{\pi }{\left(\frac{1}{{2}^{k}k!\sqrt{\pi }}\right)}^{\frac{1}{2}}{\left(\frac{1}{2ti}\right)}^{k}{x}^{k}{e}^{-\frac{1}{16{t}^{2}}{x}^{2}+\frac{i}{4t}{x}^{2}}.\mathit{∎}$

Now define another operator ${\mathcal{𝒯}}_{t}:{W}^{m,2}\left({ℝ}^{n}\right)\to \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$ by

${\mathcal{𝒯}}_{t}\left(f\right)\left(z\right)={e}^{it\mathrm{\Delta }}\left(f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\left(z\right).$

This will help us to prove norm equivalence between the spaces ${W}^{m,2}\left({ℂ}^{n}\right)$ and $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$.

#### Theorem 4.6.

The operator ${\mathcal{T}}_{t}$ is a unitary map. Furthermore, for any $m\mathrm{\in }\mathrm{N}$ and $f\mathrm{\in }{W}_{m}$ there exist ${M}_{\mathrm{1}}\mathrm{,}{M}_{\mathrm{2}}\mathrm{>}\mathrm{0}$ depending on $m\mathrm{,}t$ and n such that

${M}_{1}{\parallel f\parallel }_{{W}_{m}}^{2}\le {\parallel {e}^{it\mathrm{\Delta }}f\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t}^{m}}^{2}}\le {M}_{2}{\parallel f\parallel }_{{W}_{m}}^{2}.$

#### Proof.

We know that ${e}^{it\mathrm{\Delta }}:{L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)\to \mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$ is unitary. Therefore, $\left\{{e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right):\alpha \in {ℕ}^{n}\right\}$ is a complete orthonormal set in $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{w}_{t}\left(z\right)dz\right)$, as $\mathcal{ℬ}=\left\{{\stackrel{~}{\mathrm{\Psi }}}_{\alpha }:\alpha \in {ℕ}^{n}\right\}$ is a complete orthonormal set in ${L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right)$. Since ${\mathrm{\Psi }}_{\alpha }={\prod }_{k=1}^{n}{\psi }_{{\alpha }_{k}}$, by Proposition 4.5 we have

${e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)\left(x\right)={c}_{t}^{n}{\sqrt{\pi }}^{n}{\left(\frac{1}{{2}^{|\alpha |}\alpha !{\left(\pi \right)}^{\frac{n}{2}}}\right)}^{\frac{1}{2}}{\left(\frac{1}{2ti}\right)}^{|\alpha |}{x}^{\alpha }{e}^{\left(-\frac{1}{16{t}^{2}}+\frac{i}{4t}\right){x}^{2}}$

for every $x\in {ℝ}^{n}$ and $\alpha \in {ℕ}^{n}$. Extending this analytically to ${ℂ}^{n}$, for $\alpha \in {ℕ}^{n}$ and $z\in {ℂ}^{n}$ we get

${e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)\left(z\right)={\left(4i\pi t\right)}^{-\frac{n}{2}}{\left(\sqrt{\pi }\right)}^{\frac{n}{2}}{\left(\frac{1}{{2}^{|\alpha |}\alpha !}\right)}^{\frac{1}{2}}{\left(\frac{1}{2ti}\right)}^{|\alpha |}{z}^{\alpha }{e}^{\left(\frac{i}{4t}-\frac{1}{16{t}^{2}}\right){z}^{2}}.$

The function ${e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)$ is nothing but a constant multiple of ${\mathrm{Υ}}_{\alpha }^{t}$, that is,

${\mathcal{𝒯}}_{t}\left({\mathrm{\Psi }}_{\alpha }\right)={e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)={i}^{-\frac{n}{2}}{\left(\frac{1}{i}\right)}^{|\alpha |}{\mathrm{Υ}}_{\alpha }^{t}.$

But we have

${\parallel {\mathrm{\Psi }}_{\alpha }\parallel }_{{W}_{m}}^{2}=\sum _{|\gamma |\le m}{\parallel {D}^{\gamma }{\mathrm{\Psi }}_{\alpha }\parallel }^{2}=\sum _{|\gamma |\le m}{2}^{2|\gamma |}\prod _{k=1}^{n}\prod _{q=1}^{{\gamma }_{k}}\left(\frac{{\alpha }_{k}+q}{2}\right)$

and

${\parallel {e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t}^{m}}^{2}}^{2}={\int }_{{ℂ}^{n}}{|{e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)\left(z\right)|}^{2}\sum _{|\gamma |\le m}{\left(\frac{1}{2t}\right)}^{2|\gamma |}{|{z}^{\gamma }|}^{2}{w}_{t}\left(z\right)dz$$=\sum _{|\gamma |\le m}{\left(\frac{1}{2t}\right)}^{2|\gamma |}{\parallel {z}^{\gamma }{e}^{it\mathrm{\Delta }}\left({\stackrel{~}{\mathrm{\Psi }}}_{\alpha }\right)\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{w}_{t}}^{2}}^{2}$$=\sum _{|\gamma |\le m}{\left(\frac{1}{t}\right)}^{2|\gamma |}{\parallel {\left(2t\right)}^{|\gamma |}\prod _{j=1}^{n}\prod _{q=1}^{{\gamma }_{j}}{\left(\frac{{\alpha }_{i}+q}{2}\right)}^{\frac{1}{2}}{\mathrm{Υ}}_{\alpha +\gamma }^{t}\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{w}_{t}}^{2}}^{2}$$=\sum _{|\gamma |\le m}{\left(2\right)}^{2|\gamma |}\prod _{j=1}^{n}\prod _{q=1}^{{\gamma }_{j}}\left(\frac{{\alpha }_{i}+q}{2}\right)$$={\parallel {\mathrm{\Psi }}_{\alpha }\parallel }_{{W}_{m}}^{2}.$

So ${\mathcal{𝒯}}_{t}$ takes a complete orthonormal set in ${W}^{m,2}\left({ℝ}^{n}\right)$ into a complete orthonormal set in $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$. This implies that

(4.3)

By Theorem 3.11, for any $f\in {W}^{m,2}\left({ℝ}^{n}\right)$ there exist ${M}_{1},{M}_{2}>0$ such that

${M}_{1}{\parallel f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel }_{{W}_{m}}^{2}\le {\parallel f\parallel }_{{W}_{m}}^{2}\le {M}_{2}{\parallel f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel }_{{W}_{m}}^{2}.$(4.4)

From equations (4.3) and (4.4) we can see that

${M}_{1}{\parallel f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel }_{{W}_{m}}^{2}\le {\parallel {e}^{it\mathrm{\Delta }}\left(f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\right)\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t}^{m}}^{2}}^{2}\le {M}_{2}{\parallel f\left(u\right){e}^{-\frac{i}{4t}{u}^{2}}\parallel }_{{W}_{m}}^{2}.$

The above equation is true for any $f\in {W}^{m,2}\left({ℝ}^{n}\right)$. Hence the theorem follows. ∎

So the image of the Sobolev space ${W}^{m,2}\left({ℝ}^{n}\right)$ under ${e}^{it\mathrm{\Delta }}$ is identified with the weighted Bergman space $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t}^{m}\left(z\right)dz\right)$ up to equivalence of norms.

## 5 Image of Sobolev spaces under the Schrödinger semigroup associated to the Hermite operator

Consider the Schrödinger group ${e}^{itH}$ associated to the Hermite operator $H=-\mathrm{\Delta }+{|x|}^{2}$ on ${ℝ}^{n}$. The spectral decomposition of H is given by

$H=\sum _{k=0}^{\mathrm{\infty }}\left(2k+n\right){P}_{k},$

where ${P}_{k}f={\sum }_{|\alpha |=k}〈f,{\mathrm{\Phi }}_{\alpha }〉{\mathrm{\Phi }}_{\alpha }$ with ${\mathrm{\Phi }}_{\alpha }$ being the normalized Hermite functions on ${ℝ}^{n}$. Using Mehler’s formula, we have

${e}^{itH}f\left(x\right)={\int }_{{ℝ}^{n}}{K}_{it}\left(x,u\right)f\left(u\right)𝑑u,$

where

${K}_{it}\left(x,u\right)={\left(2\pi i\mathrm{sin}2t\right)}^{-\frac{n}{2}}{e}^{\frac{i}{4}\left(\mathrm{cot}2t\right)\left({|x|}^{2}+{|u|}^{2}\right)-i\left(\mathrm{csc}2t\right)xu}.$

This ${K}_{it}$ is well defined for $2t\ne k\pi$ for $k\in ℤ$. In [9], Parui et al. proved the following theorem.

#### Theorem 5.1.

For any $t\mathrm{\ne }\frac{k}{\mathrm{2}}$, $k\mathrm{\in }\mathrm{Z}$, ${e}^{i\mathit{}t\mathit{}H}$ defines an isometric isomorphism between ${L}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{,}{e}^{{u}^{\mathrm{2}}}\mathit{}d\mathit{}u\mathrm{\right)}$ and

$\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{\left(2\pi |\mathrm{sin}2t|\right)}^{-n}{e}^{-\left({\mathrm{csc}}^{2}2t\right){y}^{2}+\left(\mathrm{cot}2t\right)xy}dxdy\right).$

The Hermite operator H can be written as

$H=\frac{1}{2}\sum _{k=1}^{n}\left({A}_{k}{A}_{k}^{*}+{A}_{k}^{*}{A}_{k}\right).$

Here ${A}_{k}=\frac{\partial }{\partial {u}_{k}}+{u}_{k}$ and ${A}_{k}^{*}=-\frac{\partial }{\partial {u}_{k}}+{u}_{k}$ for $k=1,2,\mathrm{\dots },n$. For an integer $m\ge 1$, we define the Hermite Sobolev space ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ as follows:

${W}_{H}^{m,2}\left({ℝ}^{n}\right)=\left\{f\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right):{A}^{\alpha }{\overline{A}}^{\beta }f\in {L}^{2}\left({ℝ}^{n},{e}^{{u}^{2}}du\right),|\alpha |+|\beta |\le m\right\}.$

Here ${A}^{\alpha }={A}_{1}^{{\alpha }_{1}}\mathrm{\cdots }{A}_{n}^{{\alpha }_{n}}$ and ${\overline{A}}^{\beta }={\left({A}_{1}^{*}\right)}^{{\alpha }_{1}}\mathrm{\cdots }{\left({A}_{n}^{*}\right)}^{{\alpha }_{n}}$. For $f,g\in {W}_{H}^{m,2}\left({ℝ}^{n}\right)$, we define the inner product and norm on the space ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ as follows:

${〈f,g〉}_{{W}_{H}^{m,2}\left({ℝ}^{n}\right)}:=\sum _{|\alpha |+|\beta |\le m}〈{A}^{\alpha }{\overline{A}}^{\beta }f,{A}^{\alpha }{\overline{A}}^{\beta }g〉,$${\parallel f\parallel }_{{W}_{H}^{m,2}\left({ℝ}^{n}\right)}^{2}\mathrm{ }:=\sum _{|\alpha |+|\beta |\le m}{\parallel {A}^{\alpha }{\overline{A}}^{\beta }f\parallel }^{2}.$

Then ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ becomes a Hilbert space with respect to the above inner product. Note that ${A}_{k}+{A}_{k}^{*}=2{u}_{k}$ and ${A}_{k}-{A}_{k}^{*}=2\frac{\partial }{\partial {u}_{k}}$ for $k=1,2,\mathrm{\dots },n$. By using this relation, it is easy to see that ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ and ${W}^{m,2}\left({ℝ}^{n}\right)$ represent the same vector space. From Note 3.12 it is easy to see that there exists $B>0$ (depending only on m) such that

${\parallel f\parallel }_{{W}_{H}^{m,2}\left({ℝ}^{n}\right)}\le B{\parallel f\parallel }_{{W}^{m,2}\left({ℝ}^{n}\right)}$

for any $f\in {W}^{m,2}\left({ℝ}^{n}\right)$. That is, the identity map $i:{W}^{m,2}\left({ℝ}^{n}\right)\to {W}_{H}^{m,2}\left({ℝ}^{n}\right)$ is continuous and onto. From the bounded inverse theorem it follows that the norms ${\parallel \cdot \parallel }_{{W}_{H}^{m,2}\left({ℝ}^{n}\right)}$ and ${\parallel \cdot \parallel }_{{W}^{m,2}\left({ℝ}^{n}\right)}$ are equivalent. Hence characterizing the image of ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ under ${e}^{itH}$ is equivalent to characterizing the image of ${W}^{m,2}\left({ℝ}^{n}\right)$ under ${e}^{itH}$.

Since the linear map ${e}^{itH}:{W}^{m,2}\left({ℝ}^{n}\right)\to \mathcal{𝒪}\left({ℂ}^{n}\right)$ is one to one, we can give an inner product on

${W}_{t,H}^{m,2}\left({ℂ}^{n}\right):={e}^{itH}\left({W}^{m,2}\left({ℝ}^{n}\right)\right)$

in the following way: For $f,g\in {W}^{m,2}\left({ℝ}^{n}\right)$, we define $〈{e}^{itH}f,{e}^{itH}g〉:={〈f,g〉}_{{W}^{m,2}\left({ℝ}^{n}\right)}$. From this we can see that ${W}_{t,H}^{m,2}\left({ℂ}^{n}\right)$ is a Hilbert space and

${e}^{itH}:{W}^{m,2}\left({ℝ}^{n}\right)\to {W}_{t,H}^{m,2}\left({ℂ}^{n}\right)$

is unitary.

Let $\stackrel{~}{\mathcal{𝒮}}f\left(u\right)=f\left(u\right){e}^{-i/4\mathrm{cot}2t{u}^{2}}$ for $f\in {W}^{m,2}\left({ℝ}^{n}\right)$. Then $\stackrel{~}{\mathcal{𝒮}}$ becomes a bounded invertible operator on ${W}^{m,2}\left({ℝ}^{n}\right)$. For $z=x+iy\in {ℂ}^{n}$, let us denote

${w}_{t,H}\left(z\right)={\left(2\pi |\mathrm{sin}2t|\right)}^{-n}{e}^{-\left({\mathrm{csc}}^{2}2t\right){y}^{2}+\left(\mathrm{cot}2t\right)xy},$${u}_{t,H}^{m}\left(z\right)=\sum _{|\gamma |\le m}{\left(\mathrm{csc}2t\right)}^{2|\gamma |}{|{z}^{\alpha }|}^{2}{w}_{t,H}\left(z\right).$

Consider the weighted Bergman space

$\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t,H}^{m}}^{2}:=\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t,H}^{m}\left(z\right)dz\right).$

The proofs of the following proposition and theorem are similar to the ones of Proposition 4.5 and Theorem 4.6, respectively.

#### Proposition 5.2.

For $z\mathrm{\in }{\mathrm{C}}^{n}$ and $\alpha \mathrm{\in }{\mathrm{N}}^{n}$, we have

$\left({e}^{itH}\stackrel{~}{\mathcal{𝒮}}{\mathrm{\Psi }}_{\alpha }\right)\left(z\right)={\left(2\pi i\mathrm{sin}2t\right)}^{-\frac{n}{2}}{\pi }^{\frac{n}{4}}{\left(\frac{1}{{2}^{|\alpha |}\alpha !}\right)}^{\frac{1}{2}}{\left(-i\mathrm{csc}2t\right)}^{|\alpha |}{z}^{\alpha }{e}^{\left(\frac{i}{4}\mathrm{cot}2t-\frac{1}{4}{\mathrm{csc}}^{2}2t\right){z}^{2}}.$

Moreover, the transform ${e}^{i\mathit{}t\mathit{}H}\mathit{}\stackrel{\mathrm{~}}{\mathcal{S}}\mathrm{:}{W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}\mathrm{\to }\mathcal{H}\mathit{}{\mathcal{L}}^{\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{C}}^{n}\mathrm{,}{u}_{t\mathrm{,}H}^{m}\mathit{}\mathrm{\left(}z\mathrm{\right)}\mathit{}d\mathit{}z\mathrm{\right)}$ is unitary.

#### Theorem 5.3.

For any $f\mathrm{\in }{W}^{m\mathrm{,}\mathrm{2}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{n}\mathrm{\right)}$, there exists ${\stackrel{\mathrm{~}}{M}}_{\mathrm{1}}\mathrm{>}\mathrm{0}$ and ${\stackrel{\mathrm{~}}{M}}_{\mathrm{2}}\mathrm{>}\mathrm{0}$ such that

${\stackrel{~}{M}}_{1}{\parallel f\parallel }_{{W}_{m}}^{2}\le {\parallel {e}^{itH}f\parallel }_{\mathcal{ℋ}{\mathcal{ℒ}}_{{u}_{t,H}^{m}}^{2}}\le {\stackrel{~}{M}}_{2}{\parallel f\parallel }_{{W}_{m}}^{2}.$

So the image of the Sobolev space ${W}_{H}^{m,2}\left({ℝ}^{n}\right)$ under ${e}^{itH}$ is identified with the weighted Bergman space $\mathcal{ℋ}{\mathcal{ℒ}}^{2}\left({ℂ}^{n},{u}_{t,H}^{m}\left(z\right)dz\right)$ up to equivalence of norms.

## Acknowledgements

The authors wish to thank G. B. Folland for giving clarification to their questions related to weighted Sobolev spaces.

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Accepted: 2017-12-27

Published Online: 2018-01-20

Published in Print: 2019-01-01

The first author thanks University Grant Commission, India, for financial support.

Citation Information: Advances in Pure and Applied Mathematics, Volume 10, Issue 1, Pages 65–79, ISSN (Online) 1869-6090, ISSN (Print) 1867-1152,

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