Gaussian measures, Hermite polynomials, and the Ornstein-Uhlenbeck semigroup

1 Definitions

For a topological space $X$, we denote by ${ℬ}_X$ the Borel $\sigma$-algebra of $X$.

We write $\overline{ℝ}=ℝ\cup \{-\mathrm{\infty },\mathrm{\infty }\}$. With the order topology, $\overline{ℝ}$ is a compact metrizable space, and $ℝ$ has the subspace topology inherited from $\overline{ℝ}$, namely the inclusion map is an embedding $ℝ\to \overline{ℝ}$. It follows that

$${ℬ}_{ℝ}=\{E\cap ℝ:E\in {ℬ}_{\overline{ℝ}}\}.$$

If $ℱ$ is a collection of functions $X\to \overline{ℝ}$ on a set $X$, we define $\bigvee ℱ:X\to \overline{ℝ}$ and $\bigwedge ℱ:X\to \overline{ℝ}$ by

$$\left(\bigvee ℱ\right)(x)=sup\{f(x):f\in ℱ\},x\in X$$

and

$$\left(\bigwedge ℱ\right)(x)=inf\{f(x):f\in ℱ\},x\in X.$$

If $X$ is a measurable space and $ℱ$ is a countable collection of measurable functions $X\to \overline{ℝ}$, it is a fact that $\bigwedge ℱ$ and $\bigvee ℱ$ are measurable $X\to \overline{ℝ}$.

2 Kolmogorov’s inequality

Kolmogorov’s inequality is the following.

Theorem 1 (Kolmogorov’s inequality).

Suppose that $(\mathrm{\Omega },𝒮,P)$ is a probability space, that $X_1,\mathrm{\dots },X_n\in L^2(P)$, that $E(X_1)=0,\mathrm{\dots },E(X_n)=0$, and that $X_1,\mathrm{\dots },X_n$ are independent. Let

$$S_k(\omega )=\sum _{j=1}^k{X}_j(\omega ),\omega \in \mathrm{\Omega },$$

for $1\le k\le n$. Then for any $\lambda >0$,

$$P\left(\{\omega \in \mathrm{\Omega }:\underset{k=1}{\overset{n}{\bigvee }}|S_k(\omega )|\ge \lambda \}\right)\le \frac{1}{{\lambda }^2}\sum _{j=1}^{n}V(X_j)=\frac{1}{{\lambda }^2}V(S_n).$$

3 Gaussian measures on R

For real $a$ and $\sigma >0$, one computes that

$$\frac{1}{\sigma \sqrt{2\pi }}{\int }_{ℝ}\exp \left(-\frac{{(t-a)}^2}{2{\sigma }^2}\right)𝑑t=1.$$ (1)

Suppose that $\gamma$ is a Borel probability measure on $ℝ$. If

$$\gamma ={\delta }_a$$

for some $a\in ℝ$ or has density

$$p(t,a,{\sigma }^2)=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{{(t-a)}^2}{2{\sigma }^2}\right),t\in ℝ,$$

for some $a\in ℝ$ and some $\sigma >0$, with respect to Lebesgue measure on $ℝ$, we say that $\gamma$ is a Gaussian measure. We say that ${\delta }_a$ is a Gaussian measure with mean $a$ and variance $\mathrm{0}$, and that a Gaussian measure with density $p(\cdot ,a,{\sigma }^2)$ has mean $a$ and variance ${\sigma }^{\mathrm{2}}$. A Gaussian measure with mean $0$ and variance $1$ is said to be standard.

One calculates that the characteristic function of a Gaussian measure $\gamma$ with density $p(\cdot ,a,{\sigma }^2)$ is

$$\tilde{\gamma }(y)={\int }_{ℝ}\exp (iyx)𝑑\gamma (x)=\exp \left(iay-\frac{1}{2}{\sigma }^2{y}^2\right),y\in ℝ.$$ (2)

The cumulative distribution function of a standard Gaussian measure $\gamma$ is, for $t\in ℝ$,

$$\mathrm{\Phi }(t)=\gamma (-\mathrm{\infty },t]={\int }_{-\mathrm{\infty }}^t𝑑\gamma (s)={\int }_{-\mathrm{\infty }}^{t}p(s,0,1)𝑑s={\int }_{-\mathrm{\infty }}^t\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{s^2}{2}\right)𝑑s.$$

We define $\mathrm{\Phi }(-\mathrm{\infty })=0$ and also define

$$\mathrm{\Phi }(\mathrm{\infty })={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{s^2}{2}\right)𝑑s=1,$$

using (1).

$\mathrm{\Phi }:\overline{ℝ}\to [0,1]$ is strictly increasing, thus ${\mathrm{\Phi }}^{-1}:[0,1]\to \overline{ℝ}$ makes sense, and is itself strictly increasing. Then $1-\mathrm{\Phi }$ is strictly decreasing. By (1),

$1-\mathrm{\Phi }(t)$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s-{\displaystyle {\int }_{-\mathrm{\infty }}^t}{\displaystyle \frac{1}{\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$={\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s.$

The following lemma gives an estimate for $1-\mathrm{\Phi }(t)$ that tells us something substantial as $t\to +\mathrm{\infty }$, beyond the immediate fact that $(1-\mathrm{\Phi })(\mathrm{\infty })=1-\mathrm{\Phi }(\mathrm{\infty })=0$.

Lemma 2.

For $t>0$,

$$\frac{1}{\sqrt{2\pi }}\left(\frac{1}{t}-\frac{1}{t^3}\right)e^{-t^2/2}\le 1-\mathrm{\Phi }(t)\le \frac{1}{\sqrt{2\pi }}\frac{1}{t}{e}^{-t^2/2}.$$

Proof.

Integrating by parts,

$1-\mathrm{\Phi }(t)$ $={\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$={\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{1}{s\sqrt{2\pi }}}\cdot s\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$=-{{\displaystyle \frac{1}{s\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)|}_t^{\mathrm{\infty }}-{\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{1}{s^2\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$\le {\displaystyle \frac{1}{t\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right).$

On the other hand, using the above work and again integrating by parts,

$1-\mathrm{\Phi }(t)$ $={\displaystyle \frac{1}{t\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right)-{\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{1}{s^3\sqrt{2\pi }}}\cdot s\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$={\displaystyle \frac{1}{t\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right)+{{\displaystyle \frac{1}{s^3\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)|}_t^{\mathrm{\infty }}$

$+{\displaystyle {\int }_t^{\mathrm{\infty }}}{\displaystyle \frac{3}{s^4\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{s^2}{2}}\right)𝑑s$

$\ge {\displaystyle \frac{1}{t\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right)-{\displaystyle \frac{1}{t^3\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right).$

∎

The following theorem shows that if the variances of a sequence of independent centered random variables are summable then the sequence of random variables is summable almost surely.

Theorem 3.

Suppose that ${\xi }_j\in L^2(\mathrm{\Omega },𝒮,P)$, $j\ge 1$, are independent random variables each with mean $0$. If , then ${\sum }_{j=1}^{\mathrm{\infty }}{\xi }_j$ converges almost surely.

Proof.

Define $S_n:\mathrm{\Omega }\to ℝ$ by

$$S_n(\omega )=\sum _{j=1}^n{\xi }_j(\omega ),$$

define $Z_n:\mathrm{\Omega }\to [0,\mathrm{\infty }]$ by

$$Z_n=\underset{j=1}{\overset{\mathrm{\infty }}{\bigvee }}|S_{n+j}-S_n|,$$

and define $Z:\mathrm{\Omega }\to [0,\mathrm{\infty }]$ by

$$Z=\underset{n=1}{\overset{\mathrm{\infty }}{\bigwedge }}{Z}_n.$$

If $S_n(\omega )$ converges and $ϵ>0$, there is some $n$ such that for all $j\ge 1$, and so $Z_n(\omega )\le ϵ$ and $Z(\omega )\le ϵ$. Therefore, if $S_n(\omega )$ converges then $Z(\omega )=0$. On the other hand, if $Z(\omega )=0$ and $ϵ>0$, there is some $n$ such that , hence for all $j\ge 1$. That is, $S_n(\omega )$ is a Cauchy sequence in $ℝ$, and hence converges. Therefore

$$\{\omega \in \mathrm{\Omega }:S_n(\omega )\text{ converges}\}=\{\omega \in \mathrm{\Omega }:Z(\omega )=0\}.$$ (3)

Let $ϵ>0$. For any $n$ and $k$, using Kolmogorov’s inequality with $X_j={\xi }_{n+j}$ for $j=1,\mathrm{\dots },k$,

$$P(\underset{j=1}{\overset{k}{\bigvee }}|S_{n+j}-S_n|\ge ϵ)\le \frac{1}{{ϵ}^2}\sum _{j=1}^{k}V(X_j)\le \frac{1}{{ϵ}^2}\sum _{j=n+1}^{\mathrm{\infty }}V({\xi }_j).$$

Because this is true for each $k$, it follows that

$$P(Z_n\ge ϵ)\le \frac{1}{{ϵ}^2}\sum _{j=n+1}^{\mathrm{\infty }}V({\xi }_j),$$

hence, for each $n$,

$$P(Z\ge ϵ)\le P(Z_n\ge ϵ)\le \frac{1}{{ϵ}^2}\sum _{j=n+1}^{\mathrm{\infty }}V({\xi }_j).$$

Because , ${\sum }_{j=n+1}^{\mathrm{\infty }}V({\xi }_j)\to 0$ as $n\to \mathrm{\infty }$, so

$$P(Z\ge ϵ)=0.$$

Because this is true for all $ϵ>0$, we get $P(Z>0)=0$, i.e. $P(Z=0)=1$. By (3), this means that $S_n$ converges almost surely. ∎

The following theorem gives conditions under which the converse of the above theorem holds.

Theorem 4.

Suppose that ${\xi }_j\in L^2(\mathrm{\Omega },𝒮,P)$, $j\ge 1$, are independent random variables each with mean $0$, and let $S_n={\sum }_{j=1}^n{\xi }_j$. If

(4)

and there is some $\beta \in [0,\mathrm{\infty })$ such that ${\bigvee }_{j=1}^{\mathrm{\infty }}|{\xi }_j|\le \beta$ almost surely, then

Proof.

By (4), there is some $\alpha \in [0,\mathrm{\infty })$ such that $P(A)>0$, for

$$A=\{\omega \in \mathrm{\Omega }:\underset{n=1}{\overset{\mathrm{\infty }}{\bigvee }}|S_n(\omega )|\le \alpha \}.$$

For $p\ge 1$, let

$$A_p=\{\omega \in \mathrm{\Omega }:\underset{n=1}{\overset{p}{\bigvee }}|S_n(\omega )|\le \alpha \},$$

which satisfies $A_p\downarrow A$ as $p\to \mathrm{\infty }$. For each $p$, the random variables ${\chi }_{A_p}{S}_p$ and ${\xi }_{p+1}$ are independent and the random variables ${\chi }_{A_p}$ and ${\xi }_{p+1}^2$ are independent, whence

$E({\chi }_{A_p}{S}_{p+1}^2)$ $=E({\chi }_{A_p}(S_p+{\xi }_{p+1})(S_p+{\xi }_{p+1}))$

$=E({\chi }_{A_p}{S}_p^2+2{\chi }_{A_p}{S}_p{\xi }_{p+1}+{\chi }_{A_p}{\xi }_{p+1}^2)$

$=E({\chi }_{A_p}{S}_p^2)+2E({\chi }_{A_p}{S}_p)E({\xi }_{p+1})+E({\chi }_{A_p})E({\xi }_{p+1}^2)$

$=E({\chi }_{A_p}{S}_p^2)+P(A_p)V({\xi }_{p+1})$

$\ge E({\chi }_{A_p}{S}_p^2)+P(A)V({\xi }_{p+1}).$

Set $B_p=A_p\setminus A_{p+1}$. For $\omega \in A_p$, $|S_p(\omega )|\le \alpha$, and for almost all $\omega \in \mathrm{\Omega }$, $|{\xi }_{p+1}(\omega )|\le \beta$, so for almost all $\omega \in B_p$,

$$|S_{p+1}(\omega )|\le |S_p(\omega )|+|{\xi }_{p+1}(\omega )\le \alpha +\beta ,$$

hence

$P(A)V({\xi }_{p+1})$ $\le E(({\chi }_{B_p}+{\chi }_{A_{p+1}})S_{p+1}^2)-E({\chi }_{A_p}{S}_p^2)$

$=E({\chi }_{B_p}{S}_{p+1}^2)+E({\chi }_{A_{p+1}}{S}_{p+1}^2)-E({\chi }_{A_p}{S}_p^2)$

$\le P(B_p){(\alpha +\beta )}^2+E({\chi }_{A_{p+1}}{S}_{p+1}^2)-E({\chi }_{A_p}{S}_p^2).$

Adding the inequalities for $p=1,2,\mathrm{\dots },n-1$, because $B_p$ are pairwise disjoint,

$P(A){\displaystyle \sum _{p=1}^{n-1}}V({\xi }_{p+1})$ $={(\alpha +\beta )}^2{\displaystyle \sum _{p=1}^{n-1}}P(B_p)+E({\chi }_{A_n}{S}_n^2)-E({\chi }_{A_1}{S}_1^2)$

$\le {(\alpha +\beta )}^2+E({\chi }_{A_n}{S}_n^2)$

$\le {(\alpha +\beta )}^2+{\alpha }^2.$

Because this is true for all $n$ and $P(A)>0$,

and with this completes the proof. ∎

4 Rⁿ

If $\mu$ is a finite Borel measure on ${ℝ}^n$, we define the characteristic function of $\mu$ by

$$\tilde{\mu }(y)={\int }_{{ℝ}^n}{e}^{i⟨y,x⟩}𝑑\mu (x),y\in {ℝ}^n.$$

A Borel probability measure $\gamma$ on ${ℝ}^n$ is said to be Gaussian if for each $f\in {({ℝ}^n)}^{*}$, the pushforward measure $f_{*}\gamma$ on $ℝ$ is a Gaussian measure on $ℝ$, where

$$(f_{*}\gamma )(E)=\gamma (f^{-1}(E))$$

for $E$ a Borel set in $ℝ$.

We now give a characterization of Gaussian measures on ${ℝ}^n$ and their densities. In the following theorem, the vector $a\in {ℝ}^n$ is called the mean of $\gamma$ and the linear transformation $K\in ℒ({ℝ}^n)$ is called the covariance operator of $\gamma$. When $a=0\in {ℝ}^n$ and $K={\mathrm{id}}_{{ℝ}^n}$, we say that $\gamma$ is standard.

Theorem 5.

A Borel probability measure $\gamma$ on ${ℝ}^n$ is Gaussian if and only if there is some $a\in {ℝ}^n$ and some positive semidefinite $K\in ℒ({ℝ}^n)$ such that

$$\tilde{\gamma }(y)=\exp \left(i⟨y,a⟩-\frac{1}{2}⟨Ky,y⟩\right),y\in {ℝ}^n.$$ (5)

If $\gamma$ is a Gaussian measure whose covariance operator $K$ is positive definite, then the density of $\gamma$ with respect to Lebesgue measure on ${ℝ}^n$ is

$$x\mapsto \frac{1}{\sqrt{{(2\pi )}^{n}detK}}\exp \left(-\frac{1}{2}⟨K^{-1}(x-a),x-a⟩\right),x\in {ℝ}^n.$$

Proof.

Suppose that (5) is satisfied. Let $f\in {({ℝ}^n)}^{*}$, i.e. a linear map ${ℝ}^n\to ℝ$, and put $\nu =f_{*}\gamma$. Using the change of variables formula, the characteristic function of $\nu$ is

$\tilde{\nu }(t)={\displaystyle {\int }_{ℝ}}{e}^{its}𝑑\nu (s)={\displaystyle {\int }_{{ℝ}^n}}{e}^{itf(x)}𝑑\gamma (x),t\in ℝ.$

Let $v$ be the unique element of ${ℝ}^n$ such that $f(x)=⟨v,x⟩$ for all $x\in {ℝ}^n$. Then

$$\tilde{\nu }(t)={\int }_{{ℝ}^n}{e}^{i⟨tv,x⟩}𝑑\gamma (x)=\tilde{\gamma }(tv).$$

so by (5),

$$\tilde{\nu }(t)=\exp \left(i⟨tv,a⟩-\frac{1}{2}⟨Ktv,tv⟩\right)=\exp \left(if(a)t-\frac{1}{2}⟨Kv,v⟩t^2\right).$$

This implies that $\nu$ is a Gaussian measure on $ℝ$ with mean $f(a)$ and variance $⟨Kv,v⟩$: if $⟨Kv,v⟩=0$ then $\nu ={\delta }_{f(a)}$, and if $⟨Kv,v⟩>0$ then $\nu$ has density

$$\frac{1}{\sqrt{⟨Kv,v⟩}\sqrt{2\pi }}\exp \left(-\frac{{(s-f(a))}^2}{2⟨Kv,v⟩}\right),s\in ℝ,$$

with respect to Lebesgue measure on $ℝ$. That is, for any $f\in {({ℝ}^n)}^{*}$, the pushforward measure $f_{*}\gamma$ is a Gaussian measure on $ℝ$, which is what it means for $\gamma$ to be a Gaussian measure on ${ℝ}^n$.

Suppose that $\gamma$ is Gaussian and let $f\in {({ℝ}^n)}^{*}$. Then the pushforward measure $f_{*}\gamma$ is a Gaussian measure on $ℝ$. Let $a(f)$ be the mean of $f_{*}\gamma$ and let ${\sigma }^2(f)$ be the variance of $f_{*}\gamma$, and let $v_f$ be the unique element of ${ℝ}^n$ such that $f(x)=⟨x,v_f⟩$ for all $x\in {ℝ}^n$. Using the change of variables formula,

$$a(f)={\int }_{ℝ}td(f_{*}\gamma )(t)={\int }_{{ℝ}^n}f(x)𝑑\gamma (x)$$

and

${\sigma }^2(f)$ $={\displaystyle {\int }_{ℝ}}{(t-a(f))}^{2}d(f_{*}\gamma )(t)$

$={\displaystyle {\int }_{{ℝ}^n}}{(f(x)-a(f))}^2𝑑\gamma (x)$

$={\displaystyle {\int }_{{ℝ}^n}}(f{(x)}^2-2f(x)a(f)+a{(f)}^2)𝑑\gamma (x).$

Because $f\mapsto a(f)$ is linear ${({ℝ}^n)}^{*}\to ℝ$, there is a unique $a\in {ℝ}^n={({ℝ}^n)}^{**}$ such that

$$a(f)=⟨v_f,a⟩,f\in {({ℝ}^n)}^{*}.$$

For $f,g\in {({ℝ}^n)}^{*}$,

${\sigma }^2(f+g)$ $={\displaystyle {\int }_{{ℝ}^n}}(f{(x)}^2+2f(x)g(x)+g{(x)}^2$

$-2f(x)a(f)-2f(x)a(g)-2g(x)a(f)-2g(x)a(g)$

$+a{(f)}^2+2a(f)a(g)+a{(g)}^2)d\gamma (x),$

so

${\sigma }^2(f+g)-{\sigma }^2(f)-{\sigma }^2(g)$ $={\displaystyle {\int }_{{ℝ}^n}}(2f(x)g(x)-2f(x)a(g)-2g(x)a(f)$

$+2a(f)a(g))d\gamma (x).$

$B(f,g)=\frac{1}{2}({\sigma }^2(f+g)-{\sigma }^2(f)-{\sigma }^2(g))$ is a symmetric bilinear form on ${ℝ}^n$, and

$$B(f,f)=2{\int }_{{ℝ}^n}{(f(x)-a(f))}^2𝑑\gamma (x)\ge 0,$$

namely, $B$ is positive semidefinite. It follows that there is a unique positive semidefinite $K\in ℒ({ℝ}^n)$ such that $B(f,g)=⟨K{v}_f,v_g⟩$ for all $f,g\in {({ℝ}^n)}^{*}$. For $y\in {ℝ}^n$ and for $v_f=y$, using the change of variables formula, using the fact that $f_{*}\gamma$ is a Gaussian measure on $ℝ$ with mean

$$a(f)=⟨v_f,a⟩=⟨y,a⟩$$

and variance

$${\sigma }^2(f)=B(f,f)=⟨K{v}_f,v_f⟩=⟨Ky,y⟩$$

and using (2),

$\tilde{\gamma }(y)$ $={\displaystyle {\int }_{{ℝ}^n}}{e}^{if(x)}𝑑\gamma (x)$

$={\displaystyle {\int }_{ℝ}}{e}^{it}d(f_{*}\gamma )(t)$

$=\exp \left(i⟨y,a⟩\cdot 1-{\displaystyle \frac{1}{2}}⟨Ky,y⟩\cdot 1^2\right)$

$=\exp \left(i⟨y,a⟩-{\displaystyle \frac{1}{2}}⟨Ky,y⟩\right),$

which shows that (5) is satisfied.

Suppose that $\gamma$ is a Gaussian measure and further that the covariance operator $K$ is positive definite. By the spectral theorem, there is an orthonormal basis $\{e_1,\mathrm{\dots },e_n\}$ for ${ℝ}^n$ such that $⟨K{e}_j,e_j⟩>0$ for each $1\le j\le n$. Write $⟨K{e}_j,e_j⟩={\sigma }_j^2$, and for $y\in {ℝ}^n$ set $y_j=⟨y,e_j⟩$, with which $y=y_1{e}_1+\mathrm{\cdots }+y_n{e}_n$ and then

$⟨Ky,y⟩$ $=⟨y_{1}K{e}_1+\mathrm{\cdots }+y_{n}K{e}_n,y_1{e}_1+\mathrm{\cdots }+y_n{e}_n⟩$

$=⟨y_1{\sigma }_1^2{e}_1+\mathrm{\cdots }+y_n{\sigma }_n^2{e}_n,y_1{e}_1+\mathrm{\cdots }+y_n{e}_n⟩$

$={\sigma }_1^2{y}_1^2+\mathrm{\cdots }+{\sigma }_n^2{y}_n^2.$

And

$$⟨y,a⟩=⟨y_1{e}_1+\mathrm{\cdots }+y_n{e}_n,a_1{e}_1+\mathrm{\cdots }+a_n{e}_n⟩=a_1{y}_1+\mathrm{\cdots }+a_n{y}_n.$$

Let ${\gamma }_j$ be the Gaussian measure on $ℝ$ with mean $a_j$ and variance ${\sigma }_j^2$. Because ${\sigma }_j^2>0$, the measure ${\gamma }_j$ has density $p(\cdot ,a_j,{\sigma }_j^2)$ with respect to Lebesgue measure on $ℝ$, and thus

$\tilde{\gamma }(y)$ $=\exp \left(i⟨y,a⟩-{\displaystyle \frac{1}{2}}⟨Ky,y⟩\right)$

$=\exp \left(i{\displaystyle \sum _{j=1}^n}{a}_j{y}_j-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{j=1}^n}{\sigma }_j^2{y}_j^2\right)$

$={\displaystyle \prod _{j=1}^n}\exp \left(i{a}_j{y}_j-{\displaystyle \frac{1}{2}}{\sigma }_j^2{y}_j^2\right)$

$={\displaystyle \prod _{j=1}^n}\widetilde{{\gamma }_j}(y_j)$

$={\displaystyle \prod _{j=1}^n}{\displaystyle {\int }_{ℝ}}\exp (i{y}_{j}t)𝑑{\gamma }_j(t)$

$={\displaystyle \prod _{j=1}^n}{\displaystyle {\int }_{ℝ}}\exp (i{y}_{j}t)p(t,a_j,{\sigma }_j^2)𝑑t$

$={\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \prod _{j=1}^n}\exp (i{y}_j{x}_j)p(x_j,a_j,{\sigma }_j^2)dx$

$={\displaystyle {\int }_{{ℝ}^n}}{e}^{i⟨y,x⟩}{\displaystyle \prod _{j=1}^n}p(x_j,a_j,{\sigma }_j^2)dx.$

This implies that $\gamma$ has density

$$x\mapsto \prod _{j=1}^{n}p(x_j,a_j,{\sigma }_j^2),x\in {ℝ}^n,$$

with respect to Lebesgue measure on ${ℝ}^n$. Moreover,

$⟨K^{-1}(x-a),x-a⟩$ $=⟨{\displaystyle \sum _{j=1}^n}{\sigma }_j^{-2}(x_j-a_j)e_j,{\displaystyle \sum _{j=1}^n}(x_j-a_j)e_j⟩$

$={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{{(x_j-a_j)}^2}{{\sigma }_j^2}},$

so we have, as $detK={\prod }_{j=1}^n{\sigma }_j^2$,

${\displaystyle \prod _{j=1}^n}p(x_j,a_j,{\sigma }_j^2)$ $={\displaystyle \prod _{j=1}^n}{\displaystyle \frac{1}{{\sigma }_j\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{{(x_j-a_j)}^2}{2{\sigma }_j^2}}\right)$

$={\displaystyle \frac{1}{\sqrt{{(2\pi )}^{n}detK}}}\exp \left(-{\displaystyle \frac{1}{2}}⟨K^{-1}(x-a),x-a⟩\right).$

∎

Because $ℝ$ is a second-countable topological space, the Borel $\sigma$-algebra ${ℬ}_{{ℝ}^n}$ is equal to the product $\sigma$-algebra ${\otimes }_{j=1}^n{ℬ}_{ℝ}$. The density of the standard Gaussian measure ${\gamma }_n$ with respect to Lebesgue measure on ${ℝ}^n$ is, by Theorem 5,

$$x\mapsto \frac{1}{\sqrt{{(2\pi )}^n}}\exp \left(-\frac{1}{2}⟨x,x⟩\right),x\in {ℝ}^n.$$

It follows that ${\gamma }_n$ is equal to the product measure ${\prod }_{j=1}^n{\gamma }_1$, and thus that the probability space $({ℝ}^n,{ℬ}_{{ℝ}^n},{\gamma }_n)$ is equal to the product ${\prod }_{j=1}^n(ℝ,{ℬ}_{ℝ},{\gamma }_1)$.

For $f_1,\mathrm{\dots },f_n\in L^2({\gamma }_1)$, we define $f_1\otimes \mathrm{\cdots }\otimes f_n\in L^2({\gamma }_n)$, called the tensor product of $f_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{f}_n$, by

$$(f_1\otimes \mathrm{\cdots }\otimes f_n)(x)=\prod _{j=1}^n{f}_j(x_j),x=(x_1,\mathrm{\dots },x_n)\in {ℝ}^n.$$

It is straightforward to check that for $f_1,\mathrm{\dots },f_n,g_1,\mathrm{\dots },g_n\in L^2({\gamma }_1)$,

$${⟨f_1\otimes \mathrm{\cdots }\otimes f_n,g_1\otimes \mathrm{\cdots }\otimes g_n⟩}_{L^2({\gamma }_n)}=\prod _{j=1}^n{⟨f_j,g_j⟩}_{L^2({\gamma }_1)}.$$

One proves that the linear span of the collection of all tensor products is dense in $L^2({\gamma }_n)$, and that $\{v_k:k\ge 0\}$ is an orthonormal basis for $L^2({\gamma }_1)$, then

$$\{v_{k_1}\otimes \mathrm{\cdots }\otimes v_{k_n}:(k_1,\mathrm{\dots },k_n)\in {\mathbb{Z}}_{\ge 0}^n\}$$ (6)

is an orthonormal basis for $L^2({\gamma }_n)$.

We will later use the following statement about centered Gaussian measures.

Theorem 6.

Let $\gamma$ be a Gaussian measure on ${ℝ}^n$ with mean $0$ and let $\theta \in ℝ$. Then the pushforward of the product measure $\gamma \times \gamma$ on ${ℝ}^n\times {ℝ}^n$ under the mapping $(u,v)\mapsto u\sin \theta +v\cos \theta$, ${ℝ}^n\times {ℝ}^n\to {ℝ}^n$, is equal to $\gamma$.

Proof.

Let $\mu$ be the pushforward of $\gamma \times \gamma$ under the above mapping. and let $K\in ℒ({ℝ}^n)$ be the covariance operator of $\gamma$. For $y\in {ℝ}^n$, using the change of variables formula,

${\displaystyle {\int }_{{ℝ}^n}}\exp \left(i⟨y,x⟩\right)𝑑\mu (x)$ $={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}\exp \left(i⟨y,u\sin \theta +v\cos \theta ⟩\right)d(\gamma \times \gamma )(u,v)$

$=\left({\displaystyle {\int }_{{ℝ}^n}}\exp \left(i⟨y\sin \theta ,u⟩\right)𝑑\gamma (u)\right)$

$\cdot \left({\displaystyle {\int }_{{ℝ}^n}}\exp \left(i⟨y\cos \theta ,v⟩\right)𝑑\gamma (v)\right)$

$=\tilde{\gamma }(y\sin \theta )\tilde{\gamma }(y\cos \theta ).$

By Theorem 5,

$\tilde{\gamma }(y\sin \theta )\tilde{\gamma }(y\cos \theta )$ $=\exp \left(-{\displaystyle \frac{1}{2}}⟨Ky\sin \theta ,y\sin \theta ⟩\right)\exp \left(-{\displaystyle \frac{1}{2}}⟨Ky\cos \theta ,y\cos \theta ⟩\right)$

$=\exp \left(-{\displaystyle \frac{1}{2}}{\sin }^2(\theta )⟨Ky,y⟩-{\displaystyle \frac{1}{2}}{\cos }^2(\theta )⟨Ky,y⟩\right)$

$=\exp \left(-{\displaystyle \frac{1}{2}}⟨Ky,y⟩\right).$

Thus, the characteristic function of $\mu$ is

$$\tilde{\mu }(y)=\exp \left(-\frac{1}{2}⟨Ky,y⟩\right),y\in {ℝ}^n,$$

which implies that $\mu$ is equal to the Gaussian measure with mean $0$ and covariance operator $K$, i.e., $\mu =\gamma$. ∎

5 Hermite polynomials

For $k\ge 0$, we define the Hermite polynomial $H_k$ by

$$H_k(t)=\frac{{(-1)}^k}{\sqrt{k!}}\exp \left(\frac{t^2}{2}\right)\frac{d^k}{d{t}^k}\exp \left(-\frac{t^2}{2}\right),t\in ℝ.$$

It is apparent that $H_k(t)$ is a polynomial of degree $k$.

Theorem 7.

For real $\lambda$ and $t$,

$$\exp \left(\lambda t-\frac{1}{2}{\lambda }^2\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{\sqrt{k!}}{H}_k(t){\lambda }^k.$$

Proof.

For $u\in ℂ$, let $g(u)=\exp \left(-\frac{1}{2}{u}^2\right)$. For $t\in ℝ$,

$g(u)$ $={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{g^{(k)}(t)}{k!}}{(u-t)}^k$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{\sqrt{k!}}{{(-1)}^k}}\exp \left(-{\displaystyle \frac{t^2}{2}}\right)H_k(t){\displaystyle \frac{1}{k!}}{(u-t)}^k$

$=\exp \left(-{\displaystyle \frac{t^2}{2}}\right){\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(-1)}^k}{\sqrt{k!}}}{H}_k(t){(u-t)}^k.$

Therefore, for real $\lambda$ and $t$,

$\exp \left(\lambda t-{\displaystyle \frac{1}{2}}{\lambda }^2\right)$ $=\exp \left({\displaystyle \frac{1}{2}}{t}^2-{\displaystyle \frac{1}{2}}{(\lambda -t)}^2\right)$

$=\exp \left({\displaystyle \frac{1}{2}}{t}^2\right)g(\lambda -t)$

$=\exp \left({\displaystyle \frac{1}{2}}{t}^2\right)g(t-\lambda )$

$=\exp \left({\displaystyle \frac{1}{2}}{t}^2\right)\exp \left(-{\displaystyle \frac{t^2}{2}}\right){\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(-1)}^k}{\sqrt{k!}}}{H}_k(t){(-\lambda )}^k$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{k!}}}{H}_k(t){\lambda }^k.$

∎

Theorem 8.

Let ${\gamma }_1$ be the standard Gaussian measure on $ℝ$, with density $p(t,0,1)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{t^2}{2}\right)$. Then

$$\{H_k:k\ge 0\}$$

is an orthonormal basis for $L^2({\gamma }_1)$.

Proof.

For $\lambda ,\mu \in ℝ$, on the one hand, using (1) with $a=\lambda +\mu$ and $\sigma =1$,

On the other hand, using Theorem 7,

Therefore

$$\sum _{k,l\ge 0}\frac{1}{\sqrt{k!l!}}{\lambda }^k{\mu }^l{⟨H_k,H_l⟩}_{L^2({\gamma }_1)}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}{\lambda }^k{\mu }^k.$$

From this, we get that if $k\ne l$ then $\frac{1}{\sqrt{k!l!}}{⟨H_k,H_l⟩}_{L^2({\gamma }_1)}=0$, i.e.

$${⟨H_k,H_l⟩}_{L^2({\gamma }_1)}=0.$$

If $k=l$, then $\frac{1}{\sqrt{k!l!}}{⟨H_k,H_l⟩}_{L^2({\gamma }_1)}=\frac{1}{k!}$, i.e.

$${⟨H_k,H_k⟩}_{L^2({\gamma }_1)}=1.$$

Therefore, $\{H_k:k\ge 0\}$ is an orthonormal set in $L^2({\gamma }_1)$.

Suppose that $f\in L^2({\gamma }_1)$ satisfies ${⟨f,H_k⟩}_{L^2({\gamma }_1)}=0$ for each $k\ge 0$. Because $H_k(t)$ is a polynomial of degree $k$, for each $k\ge 0$ we have

$$\mathrm{span}\{H_0,H_1,H_2,\mathrm{\dots },H_k\}=\mathrm{span}\{1,t,t^2,\mathrm{\dots },t^k\}.$$

Hence for each $k\ge 0$, ${⟨f,t^k⟩}_{L^2({\gamma }_1)}=0$. One then proves that $\mathrm{span}\{1,t,t^2,\mathrm{\dots }\}$ is dense in $L^2({\gamma }_1)$, from which it follows that the linear span of the Hermite polynomials is dense in $L^2({\gamma }_1)$ and thus that they are an orthonormal basis. ∎

Lemma 9.

For $k\ge 1$,

$$H_k^{\prime }(t)=\sqrt{k}{H}_{k-1}(t),H_k^{\prime }(t)=t{H}_k(t)-\sqrt{k+1}{H}_{k+1}(t).$$

Proof.

Theorem 7 says

$$\exp \left(\lambda t-\frac{1}{2}{\lambda }^2\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{\sqrt{k!}}{H}_k(t){\lambda }^k.$$

On the one hand,

${\displaystyle \frac{d}{dt}}\exp \left(\lambda t-{\displaystyle \frac{1}{2}}{\lambda }^2\right)$ $=\lambda \exp \left(\lambda t-{\displaystyle \frac{1}{2}}{\lambda }^2\right)$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{k!}}}{H}_k(t){\lambda }^{k+1}$

$={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\sqrt{(k-1)!}}}{H}_{k-1}(t){\lambda }^k.$

On the other hand,

$$\frac{d}{dt}\exp \left(\lambda t-\frac{1}{2}{\lambda }^2\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{\sqrt{k!}}{H}_k^{\prime }(t){\lambda }^k.$$

Therefore, $H_0^{\prime }(t)=0$, and for $k\ge 1$,

$$\frac{1}{\sqrt{(k-1)!}}{H}_{k-1}(t)=\frac{1}{\sqrt{k!}}{H}_k^{\prime }(t),$$

i.e.,

$$H_k^{\prime }(t)=\sqrt{k}{H}_{k-1}(t).$$

∎

For $\alpha =(k_1,\mathrm{\dots },k_n)\in {\mathbb{Z}}_{\ge 0}^n$, we define the Hermite polynomial $H_{\alpha }$ by

$$H_{\alpha }(x)=H_{k_1}(x_1)\mathrm{\cdots }{H}_{k_n}(x_n),x=(x_1,\mathrm{\dots },x_n)\in {ℝ}^n.$$

Because the collection of all Hermite polynomials $H_k$ is an orthonormal basis for the Hilbert space $L^2({\gamma }_1)$, following (6) we have that the collection of all Hermite polynomials $H_{\alpha }$ is an orthonormal basis for the Hilbert space $L^2({\gamma }_n)$.

Theorem 10.

For ${\gamma }_n$ the standard Gaussian measure on ${ℝ}^n$, with mean $0\in {ℝ}^n$ and covariance operator ${\mathrm{id}}_{{ℝ}^n}$, the collection

$$\{H_{\alpha }:\alpha \in {\mathbb{Z}}_{\ge 0}^n\}$$

is an orthonormal basis for $L^2({\gamma }_n)$.

For $\alpha =(k_1,\mathrm{\dots },k_n)\in {\mathbb{Z}}_{\ge 0}^n$, write $|\alpha |=k_1+\mathrm{\cdots }+k_n$. For $k\ge 0$, we define

$${𝒳}_k=\mathrm{span}\{H_{\alpha }:|\alpha |=k\},$$

which is a subspace of $L^2({\gamma }_n)$ of dimension

$$\left(\genfrac{}{}{0pt}{}{k+n-1}{k}\right).$$

As ${𝒳}_k$ is a finite dimensional subspace of $L^2({\gamma }_n)$, it is closed. $L^2({\gamma }_n)$ is equal to the orthogonal direct sum of the ${𝒳}_k$:

$$L^2({\gamma }_n)=\underset{k=0}{\overset{\mathrm{\infty }}{\oplus }}{𝒳}_k.$$

Let

$$I_k:L^2({\gamma }_n)\to {𝒳}_k$$

be the orthogonal projection onto ${𝒳}_k$.

6 Ornstein-Uhlenbeck semigroup

Let $\gamma$ be a Gaussian measure on ${ℝ}^n$ with mean $0$ and covariance operator $K$. For $t\ge 0$, we define $M_t:{ℝ}^n\times {ℝ}^n\to {ℝ}^n$ by

$$M_t(u,v)=e^{-t}u+\sqrt{1-e^{-2t}}v,(x,y)\in {ℝ}^n\times {ℝ}^n.$$

By Theorem 6, $M_t{}_{*}(\gamma \times \gamma )=\gamma$. Therefore, for $p\ge 1$ and $f\in L^p(\gamma )$, using the change of variables formula,

$${\int }_{{ℝ}^n}{|f(x)|}^p𝑑\gamma (x)={\int }_{{ℝ}^n\times {ℝ}^n}{|f(M_t(u,v))|}^{p}d(\gamma \times \gamma )(u,v).$$

Applying Fubini’s theorem, the function

$$u\mapsto {\int }_{{ℝ}^n}{|f(M_t(u,v))|}^p𝑑\gamma (v)={\int }_{{ℝ}^n}{|f(e^{-t}u+\sqrt{1-e^{-2t}}v)|}^p𝑑\gamma (v)$$

belongs to $L^1(\gamma )$. We define the Ornstein-Uhlenbeck semigroup $\mathrm{\{}{T}_t\mathrm{:}t\mathrm{\ge }\mathrm{0}\mathrm{\}}$ on $L^p\mathbf{}\mathrm{(}\gamma \mathrm{)}$, $p\ge 1$, by

$$T_t(f)(u)={\int }_{{ℝ}^n}f(M_t(u,v))𝑑\gamma (v)={\int }_{{ℝ}^n}f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)𝑑\gamma (v),$$

for $u\in {ℝ}^n$.

Theorem 11.

Let $\gamma$ be a Gaussian measure on ${ℝ}^n$ with mean $0$. If $f\in L^1(\gamma )$, then

$${\int }_{{ℝ}^n}(T_{t}f)(x)𝑑\gamma (x)={\int }_{{ℝ}^n}f(x)𝑑\gamma (x).$$

Proof.

Using Fubini’s theorem, then the change of variables formula, then Theorem 6,

${\displaystyle {\int }_{{ℝ}^n}}(T_{t}f)(u)𝑑\gamma (u)$ $={\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}f(M_t(u,v))𝑑\gamma (v)\right)𝑑\gamma (u)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}f(M_t(u,v))d(\gamma \times \gamma )(u,v)$

$={\displaystyle {\int }_{{ℝ}^n}}f(x)d(M_t{}_{*}(\gamma \times \gamma ))(x)$

$={\displaystyle {\int }_{{ℝ}^n}}f(x)𝑑\gamma (x).$

∎

Theorem 12.

Let $\gamma$ be a Gaussian measure on ${ℝ}^n$ with mean $0$. For $p\ge 1$ and $t\ge 0$, $T_t$ is a bounded linear operator $L^p(\gamma )\to L^p(\gamma )$ with operator norm $1$.

Proof.

For $f\in L^p(\gamma )$, using Jensen’s inequality and then Theorem 11,

${\parallel T_{t}f\parallel }_{L^p(\gamma )}^p$ $={\displaystyle {\int }_{{ℝ}^n}}{\left|{\displaystyle {\int }_{{ℝ}^n}}f(M_t(u,v))𝑑\gamma (v)\right|}^p𝑑\gamma (u)$

$\le {\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}{|f(M_t(u,v))|}^p𝑑\gamma (v)\right)𝑑\gamma (u)$

$={\displaystyle {\int }_{{ℝ}^n}}{T}_t({|f|}^p)(u)𝑑\gamma (u)$

$={\displaystyle {\int }_{{ℝ}^n}}{|f|}^p(u)𝑑\gamma (u)$

$={\parallel f\parallel }_{L^p(\gamma )}^p,$

i.e. ${\parallel T_{t}f\parallel }_{L^p(\mu )}\le {\parallel f\parallel }_{L^p(\mu )}$. This shows that the operator norm of $T_t$ is $\le 1$. But, as $\gamma$ is a probability measure,

$$T_{t}1={\int }_{{ℝ}^n}1𝑑\gamma (v)=1,$$

so $T_t$ has operator norm $1$. ∎

For a Banach space $E$, we denote by $ℬ(E)$ the set of bounded linear operators $E\to E$. The strong operator topology on $E$ is the coarsest topology on $E$ such that for each $x\in E$, the map $A\mapsto Ax$ is continuous $ℬ(E)\to 𝔼$. To say that a map $Q:[0,\mathrm{\infty })\to ℬ(E)$ is strongly continuous means that for each $t\in [0,\mathrm{\infty })$, $Q(s)\to Q_t$ in the strong operator topology as $s\to t$, i.e., for each $x\in E$, $Q(s)x\to Q(t)x$ in $E$.

A one-parameter semigroup in $\mathrm{B}\mathbf{}\mathrm{(}E\mathrm{)}$ is a map $Q:[0,\mathrm{\infty })\to ℬ(E)$ such that (i) $Q(0)={\mathrm{id}}_E$ and (ii) for $s,t\ge 0$, $Q(s+t)=Q(s)\circ Q(t)$. For a one-parameter semigroup to be strongly continuous, one proves that it is equivalent that $Q(t)\to {\mathrm{id}}_E$ in the strong operator topology as $t\downarrow 0$, i.e. for each $x\in E$, $Q(t)x\to x$.

We now establish that the $\{T_t:t\ge 0\}$ is indeed a one-parameter semigroup and that it is strongly continuous.

Theorem 13.

Suppose $\mu$ is a Gaussian measure on ${ℝ}^n$ with mean $0$ and let $p\ge 1$. Then $\{T_t:t\ge 0\}$ is a strongly continuous one-parameter semigroup in $ℬ(L^p(\gamma ))$.

Proof.

For $f\in L^p(\gamma )$, because $\gamma$ is a probability measure,

$$T_0(f)(u)={\int }_{{ℝ}^n}f(u)𝑑\gamma (v)=f(u),$$

hence $T_0={\mathrm{id}}_{L^p(\mu )}$. For $s,t\ge 0$, define $P:{ℝ}^n\times {ℝ}^n\to {ℝ}^n$ by

$$P(u,v)=e^{-s}\frac{\sqrt{1-e^{-2t}}}{\sqrt{1-e^{-2t-2s}}}u+\frac{\sqrt{1-e^{-2s}}}{\sqrt{1-e^{-2t-2s}}}v.$$

By Theorem 6, $P_{*}(\gamma \times \gamma )=\gamma$, whence

hence $T_t\circ T_s=T_{s+t}$. This establishes that $\{T_t:t\ge 0\}$ is a semigroup.

For $f\in C_b({ℝ}^n)$, $u\in {ℝ}^n$, and $v\in {ℝ}^n$, as $t\downarrow 0$ we have

$$f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)-f(u)\to 0,$$

thus by the dominated convergence theorem, since

$$\left|f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)-f(u)\right|\le 2{\parallel f\parallel }_{\mathrm{\infty }}$$

and $\gamma$ is a probability measure, we have

$${\int }_{{ℝ}^n}\left(f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)-f(u)\right)𝑑\gamma (v)\to 0,$$

and hence

$(T_{t}f-T_{0}f)(u)$ $={\displaystyle {\int }_{{ℝ}^n}}f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)𝑑\gamma (v)-{\displaystyle {\int }_{{ℝ}^n}}f(u)𝑑\gamma (v)$

$={\displaystyle {\int }_{{ℝ}^n}}\left(f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)-f(u)\right)𝑑\gamma (v)$

$\to 0.$

Because this is true for each $u\in {ℝ}^n$ and

$$|(T_{t}f-T_{0}f)(u)|\le {\int }_{{ℝ}^n}2{\parallel f\parallel }_{\mathrm{\infty }}𝑑\gamma (v)=2{\parallel f\parallel }_{\mathrm{\infty }},$$

by the dominated convergence theorem we then have

$${\parallel T_{t}f-T_{0}f\parallel }_{L^p(\gamma )}\to 0.$$ (7)

Now let $f\in L^p(\gamma )$. There is a sequence $f_j\in C_b({ℝ}^n)$ satisfying ${\parallel f_j-f\parallel }_{L^p(\gamma )}\to 0$, with ${\parallel f_j\parallel }_{L^p(\gamma )}\le 2{\parallel f\parallel }_{L^p(\gamma )}$ for all $j$. For any $t\ge 0$,

${\parallel T_{t}f-T_{0}f\parallel }_{L^p(\gamma )}$ $\le {\parallel T_{t}f-T_t{f}_j\parallel }_{L^p(\gamma )}+{\parallel T_t{f}_j-T_0{f}_j\parallel }_{L^p(\gamma )}+{\parallel T_0{f}_j-T_{0}f\parallel }_{L^p(\gamma )}$

$={\parallel T_t(f-f_j)\parallel }_{L^p(\gamma )}+{\parallel T_{t}f-T_0{f}_j\parallel }_{L^p(\gamma )}+{\parallel f_j-f\parallel }_{L^p(\gamma )}$

$\le {\parallel f-f_j\parallel }_{L^p(\gamma )}+{\parallel T_{t}f-T_0{f}_j\parallel }_{L^p(\gamma )}+{\parallel f_j-f\parallel }_{L^p(\gamma )}.$

Let $ϵ>0$ and let $j$ be so large that . Because $f_j\in C_b({ℝ}^n)$, by (7) there is some $\delta >0$ such that when , . Then when ,

$${\parallel T_{t}f-T_{0}f\parallel }_{L^p(\gamma )}\le ϵ+ϵ+ϵ,$$

which shows that for each $f\in L^p(\gamma )$, ${\parallel T_{t}f-T_{0}f\parallel }_{L^p(\gamma )}$ as $t\downarrow 0$, which suffices to establish that $\{T_t:t\ge 0\}$ is strongly continuous $[0,\mathrm{\infty })\to ℬ(L^p(\gamma ))$. ∎

For $t>0$, we define $L_t\in ℬ(L^p(\gamma ))$ by

$$L_{t}f=\frac{1}{t}(T_{t}f-f),f\in L^p(\gamma ).$$

We define $𝒟(L)$ to be the set of those $f\in L^p(\gamma )$ such that $L_{t}f$ converges to some element of $L^p(\gamma )$ as $t\downarrow 0$, and we define $L:𝒟(L)\to L^p(\gamma )$. This is the infinitesimal generator of the semigroup $\{T_t:t\ge 0\}$, and the infinitesimal generator $L$ of the Ornstein-Uhlenbeck semigroup is called the Ornstein-Uhlenbeck operator. Because the Ornstein-Uhlenbeck semigroup is strongly continuous, we get the following.

Theorem 14.

Suppose $\mu$ is a Gaussian measure on ${ℝ}^n$ with mean $0$, let $p\ge 1$, and let $L$ be the infinitesimal generator of the Ornstein-Uhlenbeck semigroup $\{T_t:t\ge 0\}$. Then:

1. $𝒟(L)$ is a dense linear subspace of $L^p(\gamma )$ and $L:𝒟(L)\to L^p(\gamma )$ is a closed operator.

2. For each $f\in 𝒟(L)$ and for each $t\ge 0$,

   $$\frac{d}{dt}(T_{t}f)=(L\circ T_t)f=(T_t\circ L)f.$$

3. For $f\in L^p(\gamma )$ and $K$ a compact subset of $[0,\mathrm{\infty })$, $(\exp (t{L}_{ϵ})f\to T_{t}f$ as $ϵ\downarrow 0$ uniformly for $t\in K$.

4. For $\lambda \in ℂ$ with $\mathrm{Re}\lambda >0$, $R(\lambda ):L^p(\gamma )\to L^p(\gamma )$ defined by

   $$R(\lambda )f={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}{T}_{t}f𝑑t,f\in L^p(\gamma ),$$

   belongs to $ℬ(L^p(\gamma ))$, the range of $R(\lambda )$ is equal to $𝒟(L)$, and

   $$((\lambda I-L)\circ R(\lambda ))f=f,f\in L^p(\gamma ),(R(\lambda )\circ (\lambda I-L))f=𝒟(L),$$

   where $I$ is the identity operator on $L^p(\gamma )$.

We remind ourselves that if $H$ is a Hilbert space with inner product $⟨\cdot ,\cdot ⟩$, an element $A$ of $ℬ(H)$ is said to be a positive operator when $⟨Ax,x⟩\ge 0$ for all $x\in H$. We prove that each $T_t$ is a positive operator on the Hilbert space $L^2(\gamma )$.

Theorem 15.

Suppose $\mu$ is a Gaussian measure on ${ℝ}^n$ with mean $0$. For each $t\ge 0$, $T_t\in ℬ(L^2(\mu ))$ is a positive operator.

Proof.

For $t\ge 0$, define $N_t:{ℝ}^n\times {ℝ}^n\to {ℝ}^n\times {ℝ}^n$ by

$$O_t(x,y)=(e^{-t}x+\sqrt{1-e^{-2t}}y,-\sqrt{1-e^{-2t}}x+e^{-t}y),(x,y)\in {ℝ}^n\times {ℝ}^n,$$

whose transpose is the linear operator $N_t^{*}:{ℝ}^n\times {ℝ}^n\to {ℝ}^n\times {ℝ}^n$ defined by

$$O_t^{*}(u,v)=(e^{-t}u-\sqrt{1-e^{-2t}}v,\sqrt{1-e^{-2t}}u+e^{-t}v),(u,v)\in {ℝ}^n\times {ℝ}^n.$$

For $(x,y)\in {ℝ}^n\times {ℝ}^n$, we calculate

which shows that $O_t{}_{*}(\gamma \times \gamma )$ and $\gamma \times \gamma$ have equal characteristic functions and hence are themselves equal.

For $f,g\in L^2(\gamma )$ and $t\ge 0$,

${⟨T_{t}f,g⟩}_{L^2(\gamma )}$ $={\displaystyle {\int }_{{ℝ}^n}}(T_{t}f)(x)g(x)𝑑\mu (x)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}f\left(e^{-t}x+\sqrt{1-e^{-2t}}y\right)g(x)d(\gamma \times \gamma )(x,y)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}(f\circ {\pi }_1\circ O_t)(x,y)(g\circ {\pi }_1\circ O_t^{-1}\circ O_t)(x,y)d(\gamma \times \gamma )(x,y)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}(f\circ {\pi }_1)(u,v)(g\circ {\pi }_1\circ O_t^{-1})(u,v)d(O_t{}_{*}(\gamma \times \gamma ))(u,v)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}(f\circ {\pi }_1)(u,v)(g\circ {\pi }_1\circ O_t^{-1})(u,v)d(\gamma \times \gamma )(u,v)$

$={\displaystyle {\int }_{{ℝ}^n\times {ℝ}^n}}f(u)g\left(e^{-t}u-\sqrt{1-e^{-2t}}v\right)d(\gamma \times \gamma )(u,v)$

$={\displaystyle {\int }_{{ℝ}^n}}f(u)\left({\displaystyle {\int }_{{ℝ}^n}}g(M_t(u,-v))𝑑\gamma (v)\right)𝑑\gamma (u)$

$={\displaystyle {\int }_{{ℝ}^n}}f(u)\left({\displaystyle {\int }_{{ℝ}^n}}g(M_t(u,v))𝑑\gamma (v)\right)𝑑\gamma (u)$

$={\displaystyle {\int }_{{ℝ}^n}}f(u)(T_{t}g)(u)𝑑\gamma (u)$

$={⟨f,T_{t}g⟩}_{L^2(\gamma )},$

which establishes that $T_t$ is a self-adjoint operator on $L^2(\gamma )$.

Furthermore, using that $T_t=T_{t/2}\circ T_{t/2}$ and that $T_{t/2}$ is self-adjoint,

$${⟨T_{t}f,f⟩}_{L^2(\gamma )}={⟨T_{t/2}{T}_{t/2}f,f⟩}_{L^2(\gamma )}={⟨T_{t/2}f,T_{t/2}^{*}f⟩}_{L^2(\gamma )}={⟨T_{t/2}f,T_{t/2}f⟩}_{L^2(\gamma )},$$

which is $\ge 0$, which establishes that $T_t$ is a positive operator on $L^2(\gamma )$. ∎

We now write the Ornstein-Uhlenbeck semigroup using the orthogonal projections $I_k:L^2({\gamma }_n)\to {𝒳}_k$, where ${\gamma }_n$ is the standard Gaussian measure on ${ℝ}^n$.

Theorem 16.

For each $t\ge 0$ and $f\in L^2({\gamma }_n)$,

$$T_{t}f=\sum _{k=0}^{\mathrm{\infty }}{e}^{-kt}{I}_k(f).$$

Proof.

Define $S_t:L^2({\gamma }_n)\to L^2({\gamma }_n)$ by $S_{t}f={\sum }_{k=0}^{\mathrm{\infty }}{e}^{-kt}{I}_k(f)$, which satisfies, using that the subspaces ${𝒳}_k$ are pairwise orthogonal,

$${\parallel S_{t}f\parallel }_{L^2({\gamma }_n)}^2=\sum _{k=0}^{\mathrm{\infty }}{e}^{-kt}{\parallel I_k(f)\parallel }_{L^2({\gamma }_n)}^2\le \sum _{k=0}^{\mathrm{\infty }}{\parallel I_k(f)\parallel }_{L^2({\gamma }_n)}^2={\parallel f\parallel }_{L^2({\gamma }_n)}^2,$$

so $S_t\in ℬ(L^2({\gamma }_n))$. To prove that $T_t=S_t$, it suffices to prove that $T_t{H}_{\alpha }=S_t{H}_{\alpha }$ for each Hermite polynomial, which are an orthonormal basis for $L^2({\gamma }_n)$. For $\alpha =(k_1,\mathrm{\dots },k_n)$ with $k=|\alpha |=k_1+\mathrm{\cdots }+k_n$,

$$S_t{H}_{\alpha }=e^{-kt}{H}_{\alpha },$$

and

$(T_t{H}_{\alpha })(x)$ $={\displaystyle {\int }_{{ℝ}^n}}{H}_{\alpha }\left(e^{-t}x+\sqrt{1-e^{-2t}}y\right)𝑑{\gamma }_n(y)$

$={\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \prod _{j=1}^n}{H}_{k_j}\left(e^{-t}{x}_j+\sqrt{1-e^{-2t}}{y}_j\right)d{\gamma }_n(y)$

$={\displaystyle \prod _{j=1}^n}{\displaystyle {\int }_{ℝ}}{H}_{k_j}\left(e^{-t}{x}_j+\sqrt{1-e^{-2t}}{y}_j\right)𝑑{\gamma }_1(y_j).$

To prove that $T_t{H}_{\alpha }=e^{-kt}{H}_{\alpha }$, it thus suffices to prove that for any $t$, for any $k_j$, and for any $x_j$,

$${\int }_{ℝ}{H}_{k_j}\left(e^{-t}{x}_j+\sqrt{1-e^{-2t}}{y}_j\right)𝑑{\gamma }_1(y_j)=e^{-k_{j}t}{H}_{k_j}(x_j).$$ (8)

For $k_j=0$, as $H_0=1$ and ${\gamma }_1$ is a probability measure, (8) is true. Suppose that (8) is true for $\le k_j$. That is, for each $0\le h\le k_j$, $T_t{H}_h=e^{-ht}{H}_h$. For any $l$, because the Hermite polynomial $H_l$ is a polynomial of degree $l$, one checks that $T_t{H}_l(x_j)$ is a polynomial of degree $l$: using the binomial formula,

$${\int }_{ℝ}{(e^{-t}{x}_j+\sqrt{1-e^{-2t}}{y}_j)}^l\exp \left(-\frac{y_j^2}{2}\right)𝑑{\gamma }_1(y_j)$$

is a polynomial in $x_j$ of degree $l$. Hence $T_t{H}_l$ a linear combination of $H_0,H_1,\mathrm{\dots },H_l$. For $0\le h\le k_j$,

$${⟨T_t{H}_{k_j+1},H_h⟩}_{L^2({\gamma }_1)}={⟨H_{k_j+1},T_t{H}_h⟩}_{L^2({\gamma }_1)}={⟨H_{k_j+1},e^{-ht}{H}_h⟩}_{L^2({\gamma }_1)}=0.$$

Therefore there is some $c\in ℝ$ such that $T_t{H}_{k_j+1}=c{H}_{k_j+1}$. Then check that $c=e^{-(k_j+1)t}$. ∎

We now give an explicit expression for the domain $𝒟(L)$ of the Ornstein-Uhlenbeck operator $L$ and for $L$ applied to an element of its domain.

Theorem 17.

For $f\in 𝒟(L)$,

$$Lf=-\sum _{k=0}^{\mathrm{\infty }}k{I}_f(f).$$

Proof.

Let $f\in 𝒟(L)$, i.e. $\frac{T_{t}f-f}{t}\to Lf$ in $L^2({\gamma }_n)$ as $t\downarrow 0$. For any $k\ge 0$, using Theorem 16,

$I_{k}Lf$ $=I_k\left(\underset{t\downarrow 0}{lim}{\displaystyle \frac{T_{t}f-f}{t}}\right)$

$=\underset{t\downarrow 0}{lim}{\displaystyle \frac{I_k{T}_{t}f-I_{k}f}{t}}$

$=\underset{t\downarrow 0}{lim}{\displaystyle \frac{T_t{I}_{k}f-I_{k}f}{t}}$

$=\underset{t\downarrow 0}{lim}{\displaystyle \frac{e^{-kt}{I}_{k}f-I_{k}f}{t}}$

$=\left(\underset{t\downarrow 0}{lim}{\displaystyle \frac{e^{-kt}-1}{t}}\right)I_{k}f$

$={{\left(e^{-kt}\right)}^{\prime }|}_{t=0}{I}_{k}f$

$=-k{I}_{k}f.$

Using this,

${\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{k}^2{\parallel I_{k}f\parallel }_{L^2({\gamma }_n)}^2$ $={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\parallel I_{k}Lf\parallel }_{L^2({\gamma }_n)}^2$

$={\parallel {\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{I}_{k}Lf\parallel }_{L^2({\gamma }_n)}^2$

$={\parallel Lf\parallel }_{L^2({\gamma }_n)}^2$

Moreover,

$$Lf=L\left(\sum _{k=0}^{\mathrm{\infty }}{I}_{k}f\right)=\sum _{k=0}^{\mathrm{\infty }}L{I}_{k}f=\sum _{k=0}^{\mathrm{\infty }}{I}_{k}Lf=\sum _{k=0}^{\mathrm{\infty }}-k{I}_f.$$

Let $f\in L^2({\gamma }_n)$ satisfy

For $t>0$,

${\parallel {\displaystyle \frac{T_{t}f-f}{t}}+{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}k{I}_{k}f\parallel }_{L^2({\gamma }_n)}^2$ $={\parallel {\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left({\displaystyle \frac{e^{-kt}{I}_{k}f-I_{k}f}{t}}+k{I}_{k}f\right)\parallel }_{L^2({\gamma }_n)}^2$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\left|{\displaystyle \frac{e^{-kt}-1}{t}}+k\right|}^2{\parallel I_{k}f\parallel }_{L^2({\gamma }_n)}^2.$

For $t>0$ and $k\ge 0$,

$$|t^{-1}(e^{-kt}-1)|\le k,$$

and thus

For each $k\ge 0$, as $t\downarrow 0$,

$$\frac{e^{-kt}-1}{t}+k\to 0,$$

thus as $t\downarrow 0$,

$$\sum _{k=0}^{\mathrm{\infty }}{\left|\frac{e^{-kt}-1}{t}+k\right|}^2{\parallel I_{k}f\parallel }_{L^2({\gamma }_n)}^2\to 0$$

and hence

$${\parallel \frac{T_{t}f-f}{t}+\sum _{k=0}^{\mathrm{\infty }}k{I}_{k}f\parallel }_{L^2({\gamma }_n)}^2\to 0.$$

This means that $\frac{T_{t}f-f}{t}$ converges in $L^2({\gamma }_n)$ to $-{\sum }_{k=0}^{\mathrm{\infty }}k{I}_{k}f$ as $t\downarrow 0$, and since $\frac{T_{t}f-f}{t}$ converges, $f\in 𝒟(L)$. ∎