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

Jordan Bell
June 27, 2015

## 1 Definitions

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

We write $\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$. With the order topology, $\overline{\mathbb{R}}$ is a compact metrizable space, and $\mathbb{R}$ has the subspace topology inherited from $\overline{\mathbb{R}}$, namely the inclusion map is an embedding $\mathbb{R}\to\overline{\mathbb{R}}$. It follows that11 1 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 138, Lemma 4.20.

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

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

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

and

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

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

## 2 Kolmogorov’s inequality

Kolmogorov’s inequality is the following.22 2 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., p. 322, Theorem 10.11.

###### Theorem 1 (Kolmogorov’s inequality).

Suppose that $(\Omega,\mathscr{S},P)$ is a probability space, that $X_{1},\ldots,X_{n}\in L^{2}(P)$, that $E(X_{1})=0,\ldots,E(X_{n})=0$, and that $X_{1},\ldots,X_{n}$ are independent. Let

 $S_{k}(\omega)=\sum_{j=1}^{k}X_{j}(\omega),\qquad\omega\in\Omega,$

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

 $P\left(\left\{\omega\in\Omega:\bigvee_{k=1}^{n}|S_{k}(\omega)|\geq\lambda% \right\}\right)\leq\frac{1}{\lambda^{2}}\sum_{j=1}^{n}V(X_{j})=\frac{1}{% \lambda^{2}}V(S_{n}).$

## 3 Gaussian measures on R

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

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

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

 $\gamma=\delta_{a}$

for some $a\in\mathbb{R}$ or has density

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

for some $a\in\mathbb{R}$ and some $\sigma>0$, with respect to Lebesgue measure on $\mathbb{R}$, we say that $\gamma$ is a Gaussian measure. We say that $\delta_{a}$ is a Gaussian measure with mean $a$ and variance $0$, and that a Gaussian measure with density $p(\cdot,a,\sigma^{2})$ has mean $a$ and variance $\sigma^{2}$. 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

 $\widetilde{\gamma}(y)=\int_{\mathbb{R}}\exp(iyx)d\gamma(x)=\exp\left(iay-\frac% {1}{2}\sigma^{2}y^{2}\right),\qquad y\in\mathbb{R}.$ (2)

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

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

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

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

using (1).

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

 $\displaystyle 1-\Phi(t)$ $\displaystyle=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{s^{% 2}}{2}\right)ds-\int_{-\infty}^{t}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{s^{2}}% {2}\right)ds$ $\displaystyle=\int_{t}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{s^{2}}{2}% \right)ds.$

The following lemma gives an estimate for $1-\Phi(t)$ that tells us something substantial as $t\to+\infty$, beyond the immediate fact that $(1-\Phi)(\infty)=1-\Phi(\infty)=0$.33 3 Vladimir I. Bogachev, Gaussian Measures, p. 2, Lemma 1.1.3.

###### Lemma 2.

For $t>0$,

 $\frac{1}{\sqrt{2\pi}}\left(\frac{1}{t}-\frac{1}{t^{3}}\right)e^{-t^{2}/2}\leq 1% -\Phi(t)\leq\frac{1}{\sqrt{2\pi}}\frac{1}{t}e^{-t^{2}/2}.$
###### Proof.

Integrating by parts,

 $\displaystyle 1-\Phi(t)$ $\displaystyle=\int_{t}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{s^{2}}{2}% \right)ds$ $\displaystyle=\int_{t}^{\infty}\frac{1}{s\sqrt{2\pi}}\cdot s\exp\left(-\frac{s% ^{2}}{2}\right)ds$ $\displaystyle=-\frac{1}{s\sqrt{2\pi}}\exp\left(-\frac{s^{2}}{2}\right)\bigg{|}% _{t}^{\infty}-\int_{t}^{\infty}\frac{1}{s^{2}\sqrt{2\pi}}\exp\left(-\frac{s^{2% }}{2}\right)ds$ $\displaystyle\leq\frac{1}{t\sqrt{2\pi}}\exp\left(-\frac{t^{2}}{2}\right).$

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

 $\displaystyle 1-\Phi(t)$ $\displaystyle=\frac{1}{t\sqrt{2\pi}}\exp\left(-\frac{t^{2}}{2}\right)-\int_{t}% ^{\infty}\frac{1}{s^{3}\sqrt{2\pi}}\cdot s\exp\left(-\frac{s^{2}}{2}\right)ds$ $\displaystyle=\frac{1}{t\sqrt{2\pi}}\exp\left(-\frac{t^{2}}{2}\right)+\frac{1}% {s^{3}\sqrt{2\pi}}\exp\left(-\frac{s^{2}}{2}\right)\bigg{|}_{t}^{\infty}$ $\displaystyle+\int_{t}^{\infty}\frac{3}{s^{4}\sqrt{2\pi}}\exp\left(-\frac{s^{2% }}{2}\right)ds$ $\displaystyle\geq\frac{1}{t\sqrt{2\pi}}\exp\left(-\frac{t^{2}}{2}\right)-\frac% {1}{t^{3}\sqrt{2\pi}}\exp\left(-\frac{t^{2}}{2}\right).$

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.44 4 Karl R. Stromberg, Probability for Analysts, p. 58, Theorem 4.6.

###### Theorem 3.

Suppose that $\xi_{j}\in L^{2}(\Omega,\mathscr{S},P)$, $j\geq 1$, are independent random variables each with mean $0$. If $\sum_{j=1}^{\infty}V(\xi_{j})<\infty$, then $\sum_{j=1}^{\infty}\xi_{j}$ converges almost surely.

###### Proof.

Define $S_{n}:\Omega\to\mathbb{R}$ by

 $S_{n}(\omega)=\sum_{j=1}^{n}\xi_{j}(\omega),$

define $Z_{n}:\Omega\to[0,\infty]$ by

 $Z_{n}=\bigvee_{j=1}^{\infty}|S_{n+j}-S_{n}|,$

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

 $Z=\bigwedge_{n=1}^{\infty}Z_{n}.$

If $S_{n}(\omega)$ converges and $\epsilon>0$, there is some $n$ such that for all $j\geq 1$, $|S_{n+j}(\omega)-S_{n}(\omega)|<\epsilon$ and so $Z_{n}(\omega)\leq\epsilon$ and $Z(\omega)\leq\epsilon$. Therefore, if $S_{n}(\omega)$ converges then $Z(\omega)=0$. On the other hand, if $Z(\omega)=0$ and $\epsilon>0$, there is some $n$ such that $Z_{n}(\omega)<\epsilon$, hence $|S_{n+j}(\omega)-S_{n}(\omega)|<\epsilon$ for all $j\geq 1$. That is, $S_{n}(\omega)$ is a Cauchy sequence in $\mathbb{R}$, and hence converges. Therefore

 $\left\{\omega\in\Omega:\textrm{S_{n}(\omega) converges}\right\}=\{\omega\in% \Omega:Z(\omega)=0\}.$ (3)

Let $\epsilon>0$. For any $n$ and $k$, using Kolmogorov’s inequality with $X_{j}=\xi_{n+j}$ for $j=1,\ldots,k$,

 $P\left(\bigvee_{j=1}^{k}|S_{n+j}-S_{n}|\geq\epsilon\right)\leq\frac{1}{% \epsilon^{2}}\sum_{j=1}^{k}V(X_{j})\leq\frac{1}{\epsilon^{2}}\sum_{j=n+1}^{% \infty}V(\xi_{j}).$

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

 $P(Z_{n}\geq\epsilon)\leq\frac{1}{\epsilon^{2}}\sum_{j=n+1}^{\infty}V(\xi_{j}),$

hence, for each $n$,

 $P(Z\geq\epsilon)\leq P(Z_{n}\geq\epsilon)\leq\frac{1}{\epsilon^{2}}\sum_{j=n+1% }^{\infty}V(\xi_{j}).$

Because $\sum_{j=1}^{\infty}V(\xi_{j})<\infty$, $\sum_{j=n+1}^{\infty}V(\xi_{j})\to 0$ as $n\to\infty$, so

 $P(Z\geq\epsilon)=0.$

Because this is true for all $\epsilon>0$, 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.55 5 Karl R. Stromberg, Probability for Analysts, p. 59, Theorem 4.7.

###### Theorem 4.

Suppose that $\xi_{j}\in L^{2}(\Omega,\mathscr{S},P)$, $j\geq 1$, are independent random variables each with mean $0$, and let $S_{n}=\sum_{j=1}^{n}\xi_{j}$. If

 $P\left(\bigvee_{n=1}^{\infty}|S_{n}|<\infty\right)>0$ (4)

and there is some $\beta\in[0,\infty)$ such that $\bigvee_{j=1}^{\infty}|\xi_{j}|\leq\beta$ almost surely, then

 $\sum_{j=1}^{\infty}V(\xi_{j})<\infty.$
###### Proof.

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

 $A=\left\{\omega\in\Omega:\bigvee_{n=1}^{\infty}|S_{n}(\omega)|\leq\alpha\right\}.$

For $p\geq 1$, let

 $A_{p}=\left\{\omega\in\Omega:\bigvee_{n=1}^{p}|S_{n}(\omega)|\leq\alpha\right\},$

which satisfies $A_{p}\downarrow A$ as $p\to\infty$. 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

 $\displaystyle E(\chi_{A_{p}}S_{p+1}^{2})$ $\displaystyle=E(\chi_{A_{p}}(S_{p}+\xi_{p+1})(S_{p}+\xi_{p+1}))$ $\displaystyle=E(\chi_{A_{p}}S_{p}^{2}+2\chi_{A_{p}}S_{p}\xi_{p+1}+\chi_{A_{p}}% \xi_{p+1}^{2})$ $\displaystyle=E(\chi_{A_{p}}S_{p}^{2})+2E(\chi_{A_{p}}S_{p})E(\xi_{p+1})+E(% \chi_{A_{p}})E(\xi_{p+1}^{2})$ $\displaystyle=E(\chi_{A_{p}}S_{p}^{2})+P(A_{p})V(\xi_{p+1})$ $\displaystyle\geq E(\chi_{A_{p}}S_{p}^{2})+P(A)V(\xi_{p+1}).$

Set $B_{p}=A_{p}\setminus A_{p+1}$. For $\omega\in A_{p}$, $|S_{p}(\omega)|\leq\alpha$, and for almost all $\omega\in\Omega$, $|\xi_{p+1}(\omega)|\leq\beta$, so for almost all $\omega\in B_{p}$,

 $|S_{p+1}(\omega)|\leq|S_{p}(\omega)|+|\xi_{p+1}(\omega)\leq\alpha+\beta,$

hence

 $\displaystyle P(A)V(\xi_{p+1})$ $\displaystyle\leq E((\chi_{B_{p}}+\chi_{A_{p+1}})S_{p+1}^{2})-E(\chi_{A_{p}}S_% {p}^{2})$ $\displaystyle=E(\chi_{B_{p}}S_{p+1}^{2})+E(\chi_{A_{p+1}}S_{p+1}^{2})-E(\chi_{% A_{p}}S_{p}^{2})$ $\displaystyle\leq P(B_{p})(\alpha+\beta)^{2}+E(\chi_{A_{p+1}}S_{p+1}^{2})-E(% \chi_{A_{p}}S_{p}^{2}).$

Adding the inequalities for $p=1,2,\ldots,n-1$, because $B_{p}$ are pairwise disjoint,

 $\displaystyle P(A)\sum_{p=1}^{n-1}V(\xi_{p+1})$ $\displaystyle=(\alpha+\beta)^{2}\sum_{p=1}^{n-1}P(B_{p})+E(\chi_{A_{n}}S_{n}^{% 2})-E(\chi_{A_{1}}S_{1}^{2})$ $\displaystyle\leq(\alpha+\beta)^{2}+E(\chi_{A_{n}}S_{n}^{2})$ $\displaystyle\leq(\alpha+\beta)^{2}+\alpha^{2}.$

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

 $\sum_{p=1}^{\infty}V(\xi_{p+1})<\infty,$

and with $V(\xi_{1})<\infty$ this completes the proof. ∎

## 4 Rn

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

 $\widetilde{\mu}(y)=\int_{\mathbb{R}^{n}}e^{i\left\langle y,x\right\rangle}d\mu% (x),\qquad y\in\mathbb{R}^{n}.$

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

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

for $E$ a Borel set in $\mathbb{R}$.

We now give a characterization of Gaussian measures on $\mathbb{R}^{n}$ and their densities.66 6 Vladimir I. Bogachev, Gaussian Measures, p. 3, Proposition 1.2.2; Michel Simonnet, Measures and Probabilities, p. 303, Theorem 14.5. In the following theorem, the vector $a\in\mathbb{R}^{n}$ is called the mean of $\gamma$ and the linear transformation $K\in\mathscr{L}(\mathbb{R}^{n})$ is called the covariance operator of $\gamma$. When $a=0\in\mathbb{R}^{n}$ and $K=\mathrm{id}_{\mathbb{R}^{n}}$, we say that $\gamma$ is standard.

###### Theorem 5.

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

 $\widetilde{\gamma}(y)=\exp\left(i\left\langle y,a\right\rangle-\frac{1}{2}% \left\langle Ky,y\right\rangle\right),\qquad y\in\mathbb{R}^{n}.$ (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 $\mathbb{R}^{n}$ is

 $x\mapsto\frac{1}{\sqrt{(2\pi)^{n}\det K}}\exp\left(-\frac{1}{2}\left\langle K^% {-1}(x-a),x-a\right\rangle\right),\qquad x\in\mathbb{R}^{n}.$
###### Proof.

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

 $\displaystyle\widetilde{\nu}(t)=\int_{\mathbb{R}}e^{its}d\nu(s)=\int_{\mathbb{% R}^{n}}e^{itf(x)}d\gamma(x),\qquad t\in\mathbb{R}.$

Let $v$ be the unique element of $\mathbb{R}^{n}$ such that $f(x)=\left\langle v,x\right\rangle$ for all $x\in\mathbb{R}^{n}$. Then

 $\widetilde{\nu}(t)=\int_{\mathbb{R}^{n}}e^{i\left\langle tv,x\right\rangle}d% \gamma(x)=\widetilde{\gamma}(tv).$

so by (5),

 $\widetilde{\nu}(t)=\exp\left(i\left\langle tv,a\right\rangle-\frac{1}{2}\left% \langle Ktv,tv\right\rangle\right)=\exp\left(if(a)t-\frac{1}{2}\left\langle Kv% ,v\right\rangle t^{2}\right).$

This implies that $\nu$ is a Gaussian measure on $\mathbb{R}$ with mean $f(a)$ and variance $\left\langle Kv,v\right\rangle$: if $\left\langle Kv,v\right\rangle=0$ then $\nu=\delta_{f(a)}$, and if $\left\langle Kv,v\right\rangle>0$ then $\nu$ has density

 $\frac{1}{\sqrt{\left\langle Kv,v\right\rangle}\sqrt{2\pi}}\exp\left(-\frac{(s-% f(a))^{2}}{2\left\langle Kv,v\right\rangle}\right),\qquad s\in\mathbb{R},$

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

Suppose that $\gamma$ is Gaussian and let $f\in(\mathbb{R}^{n})^{*}$. Then the pushforward measure $f_{*}\gamma$ is a Gaussian measure on $\mathbb{R}$. 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 $\mathbb{R}^{n}$ such that $f(x)=\left\langle x,v_{f}\right\rangle$ for all $x\in\mathbb{R}^{n}$. Using the change of variables formula,

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

and

 $\displaystyle\sigma^{2}(f)$ $\displaystyle=\int_{\mathbb{R}}(t-a(f))^{2}d(f_{*}\gamma)(t)$ $\displaystyle=\int_{\mathbb{R}^{n}}(f(x)-a(f))^{2}d\gamma(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}(f(x)^{2}-2f(x)a(f)+a(f)^{2})d\gamma(x).$

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

 $a(f)=\left\langle v_{f},a\right\rangle,\qquad f\in(\mathbb{R}^{n})^{*}.$

For $f,g\in(\mathbb{R}^{n})^{*}$,

 $\displaystyle\sigma^{2}(f+g)$ $\displaystyle=\int_{\mathbb{R}^{n}}(f(x)^{2}+2f(x)g(x)+g(x)^{2}$ $\displaystyle-2f(x)a(f)-2f(x)a(g)-2g(x)a(f)-2g(x)a(g)$ $\displaystyle+a(f)^{2}+2a(f)a(g)+a(g)^{2})d\gamma(x),$

so

 $\displaystyle\sigma^{2}(f+g)-\sigma^{2}(f)-\sigma^{2}(g)$ $\displaystyle=\int_{\mathbb{R}^{n}}(2f(x)g(x)-2f(x)a(g)-2g(x)a(f)$ $\displaystyle+2a(f)a(g))d\gamma(x).$

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

 $B(f,f)=2\int_{\mathbb{R}^{n}}(f(x)-a(f))^{2}d\gamma(x)\geq 0,$

namely, $B$ is positive semidefinite. It follows that there is a unique positive semidefinite $K\in\mathscr{L}(\mathbb{R}^{n})$ such that $B(f,g)=\left\langle Kv_{f},v_{g}\right\rangle$ for all $f,g\in(\mathbb{R}^{n})^{*}$. For $y\in\mathbb{R}^{n}$ and for $v_{f}=y$, using the change of variables formula, using the fact that $f_{*}\gamma$ is a Gaussian measure on $\mathbb{R}$ with mean

 $a(f)=\left\langle v_{f},a\right\rangle=\left\langle y,a\right\rangle$

and variance

 $\sigma^{2}(f)=B(f,f)=\left\langle Kv_{f},v_{f}\right\rangle=\left\langle Ky,y\right\rangle$

and using (2),

 $\displaystyle\widetilde{\gamma}(y)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{if(x)}d\gamma(x)$ $\displaystyle=\int_{\mathbb{R}}e^{it}d(f_{*}\gamma)(t)$ $\displaystyle=\exp\left(i\left\langle y,a\right\rangle\cdot 1-\frac{1}{2}\left% \langle Ky,y\right\rangle\cdot 1^{2}\right)$ $\displaystyle=\exp\left(i\left\langle y,a\right\rangle-\frac{1}{2}\left\langle Ky% ,y\right\rangle\right),$

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 $\{e_{1},\ldots,e_{n}\}$ for $\mathbb{R}^{n}$ such that $\left\langle Ke_{j},e_{j}\right\rangle>0$ for each $1\leq j\leq n$. Write $\left\langle Ke_{j},e_{j}\right\rangle=\sigma_{j}^{2}$, and for $y\in\mathbb{R}^{n}$ set $y_{j}=\left\langle y,e_{j}\right\rangle$, with which $y=y_{1}e_{1}+\cdots+y_{n}e_{n}$ and then

 $\displaystyle\left\langle Ky,y\right\rangle$ $\displaystyle=\left\langle y_{1}Ke_{1}+\cdots+y_{n}Ke_{n},y_{1}e_{1}+\cdots+y_% {n}e_{n}\right\rangle$ $\displaystyle=\left\langle y_{1}\sigma_{1}^{2}e_{1}+\cdots+y_{n}\sigma_{n}^{2}% e_{n},y_{1}e_{1}+\cdots+y_{n}e_{n}\right\rangle$ $\displaystyle=\sigma_{1}^{2}y_{1}^{2}+\cdots+\sigma_{n}^{2}y_{n}^{2}.$

And

 $\left\langle y,a\right\rangle=\left\langle y_{1}e_{1}+\cdots+y_{n}e_{n},a_{1}e% _{1}+\cdots+a_{n}e_{n}\right\rangle=a_{1}y_{1}+\cdots+a_{n}y_{n}.$

Let $\gamma_{j}$ be the Gaussian measure on $\mathbb{R}$ 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 $\mathbb{R}$, and thus

 $\displaystyle\widetilde{\gamma}(y)$ $\displaystyle=\exp\left(i\left\langle y,a\right\rangle-\frac{1}{2}\left\langle Ky% ,y\right\rangle\right)$ $\displaystyle=\exp\left(i\sum_{j=1}^{n}a_{j}y_{j}-\frac{1}{2}\sum_{j=1}^{n}% \sigma_{j}^{2}y_{j}^{2}\right)$ $\displaystyle=\prod_{j=1}^{n}\exp\left(ia_{j}y_{j}-\frac{1}{2}\sigma_{j}^{2}y_% {j}^{2}\right)$ $\displaystyle=\prod_{j=1}^{n}\widetilde{\gamma_{j}}(y_{j})$ $\displaystyle=\prod_{j=1}^{n}\int_{\mathbb{R}}\exp(iy_{j}t)d\gamma_{j}(t)$ $\displaystyle=\prod_{j=1}^{n}\int_{\mathbb{R}}\exp(iy_{j}t)p(t,a_{j},\sigma_{j% }^{2})dt$ $\displaystyle=\int_{\mathbb{R}^{n}}\prod_{j=1}^{n}\exp(iy_{j}x_{j})p(x_{j},a_{% j},\sigma_{j}^{2})dx$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{i\left\langle y,x\right\rangle}\prod_{j=% 1}^{n}p(x_{j},a_{j},\sigma_{j}^{2})dx.$

This implies that $\gamma$ has density

 $x\mapsto\prod_{j=1}^{n}p(x_{j},a_{j},\sigma_{j}^{2}),\qquad x\in\mathbb{R}^{n},$

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

 $\displaystyle\left\langle K^{-1}(x-a),x-a\right\rangle$ $\displaystyle=\left\langle\sum_{j=1}^{n}\sigma_{j}^{-2}(x_{j}-a_{j})e_{j},\sum% _{j=1}^{n}(x_{j}-a_{j})e_{j}\right\rangle$ $\displaystyle=\sum_{j=1}^{n}\frac{(x_{j}-a_{j})^{2}}{\sigma_{j}^{2}},$

so we have, as $\det K=\prod_{j=1}^{n}\sigma_{j}^{2}$,

 $\displaystyle\prod_{j=1}^{n}p(x_{j},a_{j},\sigma_{j}^{2})$ $\displaystyle=\prod_{j=1}^{n}\frac{1}{\sigma_{j}\sqrt{2\pi}}\exp\left(-\frac{(% x_{j}-a_{j})^{2}}{2\sigma_{j}^{2}}\right)$ $\displaystyle=\frac{1}{\sqrt{(2\pi)^{n}\det K}}\exp\left(-\frac{1}{2}\left% \langle K^{-1}(x-a),x-a\right\rangle\right).$

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

 $x\mapsto\frac{1}{\sqrt{(2\pi)^{n}}}\exp\left(-\frac{1}{2}\left\langle x,x% \right\rangle\right),\qquad x\in\mathbb{R}^{n}.$

It follows that $\gamma_{n}$ is equal to the product measure $\prod_{j=1}^{n}\gamma_{1}$, and thus that the probability space $(\mathbb{R}^{n},\mathscr{B}_{\mathbb{R}^{n}},\gamma_{n})$ is equal to the product $\prod_{j=1}^{n}(\mathbb{R},\mathscr{B}_{\mathbb{R}},\gamma_{1})$.

For $f_{1},\ldots,f_{n}\in L^{2}(\gamma_{1})$, we define $f_{1}\otimes\cdots\otimes f_{n}\in L^{2}(\gamma_{n})$, called the tensor product of $f_{1},\ldots,f_{n}$, by

 $(f_{1}\otimes\cdots\otimes f_{n})(x)=\prod_{j=1}^{n}f_{j}(x_{j}),\qquad x=(x_{% 1},\ldots,x_{n})\in\mathbb{R}^{n}.$

It is straightforward to check that for $f_{1},\ldots,f_{n},g_{1},\ldots,g_{n}\in L^{2}(\gamma_{1})$,

 $\left\langle f_{1}\otimes\cdots\otimes f_{n},g_{1}\otimes\cdots\otimes g_{n}% \right\rangle_{L^{2}(\gamma_{n})}=\prod_{j=1}^{n}\left\langle f_{j},g_{j}% \right\rangle_{L^{2}(\gamma_{1})}.$

One proves that the linear span of the collection of all tensor products is dense in $L^{2}(\gamma_{n})$, and that $\{v_{k}:k\geq 0\}$ is an orthonormal basis for $L^{2}(\gamma_{1})$, then

 $\{v_{k_{1}}\otimes\cdots\otimes v_{k_{n}}:(k_{1},\ldots,k_{n})\in\mathbb{Z}_{% \geq 0}^{n}\}$ (6)

is an orthonormal basis for $L^{2}(\gamma_{n})$.

We will later use the following statement about centered Gaussian measures.77 7 Vladimir I. Bogachev, Gaussian Measures, p. 5, Lemma 1.2.5.

###### Theorem 6.

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

###### Proof.

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

 $\displaystyle\int_{\mathbb{R}^{n}}\exp\left(i\left\langle y,x\right\rangle% \right)d\mu(x)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\exp\left(i\left\langle y% ,u\sin\theta+v\cos\theta\right\rangle\right)d(\gamma\times\gamma)(u,v)$ $\displaystyle=\left(\int_{\mathbb{R}^{n}}\exp\left(i\left\langle y\sin\theta,u% \right\rangle\right)d\gamma(u)\right)$ $\displaystyle\cdot\left(\int_{\mathbb{R}^{n}}\exp\left(i\left\langle y\cos% \theta,v\right\rangle\right)d\gamma(v)\right)$ $\displaystyle=\widetilde{\gamma}(y\sin\theta)\widetilde{\gamma}(y\cos\theta).$

By Theorem 5,

 $\displaystyle\widetilde{\gamma}(y\sin\theta)\widetilde{\gamma}(y\cos\theta)$ $\displaystyle=\exp\left(-\frac{1}{2}\left\langle Ky\sin\theta,y\sin\theta% \right\rangle\right)\exp\left(-\frac{1}{2}\left\langle Ky\cos\theta,y\cos% \theta\right\rangle\right)$ $\displaystyle=\exp\left(-\frac{1}{2}\sin^{2}(\theta)\left\langle Ky,y\right% \rangle-\frac{1}{2}\cos^{2}(\theta)\left\langle Ky,y\right\rangle\right)$ $\displaystyle=\exp\left(-\frac{1}{2}\left\langle Ky,y\right\rangle\right).$

Thus, the characteristic function of $\mu$ is

 $\widetilde{\mu}(y)=\exp\left(-\frac{1}{2}\left\langle Ky,y\right\rangle\right)% ,\qquad y\in\mathbb{R}^{n},$

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 $k\geq 0$, we define the Hermite polynomial $H_{k}$ by

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

It is apparent that $H_{k}(t)$ is a polynomial of degree $k$.

###### Theorem 7.

For real $\lambda$ and $t$,

 $\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)=\sum_{k=0}^{\infty}\frac{1}{% \sqrt{k!}}H_{k}(t)\lambda^{k}.$
###### Proof.

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

 $\displaystyle g(u)$ $\displaystyle=\sum_{k=0}^{\infty}\frac{g^{(k)}(t)}{k!}(u-t)^{k}$ $\displaystyle=\sum_{k=0}^{\infty}\frac{\sqrt{k!}}{(-1)^{k}}\exp\left(-\frac{t^% {2}}{2}\right)H_{k}(t)\frac{1}{k!}(u-t)^{k}$ $\displaystyle=\exp\left(-\frac{t^{2}}{2}\right)\sum_{k=0}^{\infty}\frac{(-1)^{% k}}{\sqrt{k!}}H_{k}(t)(u-t)^{k}.$

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

 $\displaystyle\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)$ $\displaystyle=\exp\left(\frac{1}{2}t^{2}-\frac{1}{2}(\lambda-t)^{2}\right)$ $\displaystyle=\exp\left(\frac{1}{2}t^{2}\right)g(\lambda-t)$ $\displaystyle=\exp\left(\frac{1}{2}t^{2}\right)g(t-\lambda)$ $\displaystyle=\exp\left(\frac{1}{2}t^{2}\right)\exp\left(-\frac{t^{2}}{2}% \right)\sum_{k=0}^{\infty}\frac{(-1)^{k}}{\sqrt{k!}}H_{k}(t)(-\lambda)^{k}$ $\displaystyle=\sum_{k=0}^{\infty}\frac{1}{\sqrt{k!}}H_{k}(t)\lambda^{k}.$

###### Theorem 8.

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

 $\{H_{k}:k\geq 0\}$

is an orthonormal basis for $L^{2}(\gamma_{1})$.

###### Proof.

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

 $\begin{split}&\displaystyle\int_{\mathbb{R}}\exp\left(\lambda t-\frac{1}{2}% \lambda^{2}\right)\exp\left(\mu t-\frac{1}{2}\mu^{2}\right)d\gamma_{1}(t)\\ \displaystyle=&\displaystyle e^{\lambda\mu}\int_{\mathbb{R}}\frac{1}{\sqrt{2% \pi}}\exp\left(-\frac{1}{2}(t-(\lambda+\mu))^{2}\right)dt\\ \displaystyle=&\displaystyle e^{\lambda\mu}.\end{split}$

On the other hand, using Theorem 7,

 $\begin{split}&\displaystyle\int_{\mathbb{R}}\exp\left(\lambda t-\frac{1}{2}% \lambda^{2}\right)\exp\left(\mu t-\frac{1}{2}\mu^{2}\right)d\gamma_{1}(t)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}}\left(\sum_{k=0}^{\infty}\frac{1}% {\sqrt{k!}}H_{k}(t)\lambda^{k}\right)\left(\sum_{l=0}^{\infty}\frac{1}{\sqrt{l% !}}H_{l}(t)\mu^{l}\right)d\gamma_{1}(t)\\ &\displaystyle=\int_{\mathbb{R}}\sum_{k,l\geq 0}\frac{1}{\sqrt{k!l!}}\lambda^{% k}\mu^{l}H_{k}(t)H_{l}(t)d\gamma_{1}(t)\\ &\displaystyle=\sum_{k,l\geq 0}\frac{1}{\sqrt{k!l!}}\lambda^{k}\mu^{l}\left% \langle H_{k},H_{l}\right\rangle_{L^{2}(\gamma_{1})}.\end{split}$

Therefore

 $\sum_{k,l\geq 0}\frac{1}{\sqrt{k!l!}}\lambda^{k}\mu^{l}\left\langle H_{k},H_{l% }\right\rangle_{L^{2}(\gamma_{1})}=\sum_{k=0}^{\infty}\frac{1}{k!}\lambda^{k}% \mu^{k}.$

From this, we get that if $k\neq l$ then $\frac{1}{\sqrt{k!l!}}\left\langle H_{k},H_{l}\right\rangle_{L^{2}(\gamma_{1})}=0$, i.e.

 $\left\langle H_{k},H_{l}\right\rangle_{L^{2}(\gamma_{1})}=0.$

If $k=l$, then $\frac{1}{\sqrt{k!l!}}\left\langle H_{k},H_{l}\right\rangle_{L^{2}(\gamma_{1})}% =\frac{1}{k!}$, i.e.

 $\left\langle H_{k},H_{k}\right\rangle_{L^{2}(\gamma_{1})}=1.$

Therefore, $\{H_{k}:k\geq 0\}$ is an orthonormal set in $L^{2}(\gamma_{1})$.

Suppose that $f\in L^{2}(\gamma_{1})$ satisfies $\left\langle f,H_{k}\right\rangle_{L^{2}(\gamma_{1})}=0$ for each $k\geq 0$. Because $H_{k}(t)$ is a polynomial of degree $k$, for each $k\geq 0$ we have

 $\mathrm{span}\{H_{0},H_{1},H_{2},\ldots,H_{k}\}=\mathrm{span}\{1,t,t^{2},% \ldots,t^{k}\}.$

Hence for each $k\geq 0$, $\left\langle f,t^{k}\right\rangle_{L^{2}(\gamma_{1})}=0$. One then proves that $\mathrm{span}\{1,t,t^{2},\ldots\}$ 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 $k\geq 1$,

 $H_{k}^{\prime}(t)=\sqrt{k}H_{k-1}(t),\qquad H_{k}^{\prime}(t)=tH_{k}(t)-\sqrt{% k+1}H_{k+1}(t).$
###### Proof.

Theorem 7 says

 $\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)=\sum_{k=0}^{\infty}\frac{1}{% \sqrt{k!}}H_{k}(t)\lambda^{k}.$

On the one hand,

 $\displaystyle\frac{d}{dt}\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)$ $\displaystyle=\lambda\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)$ $\displaystyle=\sum_{k=0}^{\infty}\frac{1}{\sqrt{k!}}H_{k}(t)\lambda^{k+1}$ $\displaystyle=\sum_{k=1}^{\infty}\frac{1}{\sqrt{(k-1)!}}H_{k-1}(t)\lambda^{k}.$

On the other hand,

 $\frac{d}{dt}\exp\left(\lambda t-\frac{1}{2}\lambda^{2}\right)=\sum_{k=0}^{% \infty}\frac{1}{\sqrt{k!}}H_{k}^{\prime}(t)\lambda^{k}.$

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

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

i.e.,

 $H_{k}^{\prime}(t)=\sqrt{k}H_{k-1}(t).$

For $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{Z}_{\geq 0}^{n}$, we define the Hermite polynomial $H_{\alpha}$ by

 $H_{\alpha}(x)=H_{k_{1}}(x_{1})\cdots H_{k_{n}}(x_{n}),\qquad x=(x_{1},\ldots,x% _{n})\in\mathbb{R}^{n}.$

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 $\mathbb{R}^{n}$, with mean $0\in\mathbb{R}^{n}$ and covariance operator $\mathrm{id}_{\mathbb{R}^{n}}$, the collection

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

is an orthonormal basis for $L^{2}(\gamma_{n})$.

For $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{Z}_{\geq 0}^{n}$, write $|\alpha|=k_{1}+\cdots+k_{n}$. For $k\geq 0$, we define

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

which is a subspace of $L^{2}(\gamma_{n})$ of dimension

 $\binom{k+n-1}{k}.$

As $\mathcal{X}_{k}$ 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 $\mathcal{X}_{k}$:

 $L^{2}(\gamma_{n})=\bigoplus_{k=0}^{\infty}\mathcal{X}_{k}.$

Let

 $I_{k}:L^{2}(\gamma_{n})\to\mathcal{X}_{k}$

be the orthogonal projection onto $\mathcal{X}_{k}$.

## 6 Ornstein-Uhlenbeck semigroup

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

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

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

 $\int_{\mathbb{R}^{n}}|f(x)|^{p}d\gamma(x)=\int_{\mathbb{R}^{n}\times\mathbb{R}% ^{n}}|f(M_{t}(u,v))|^{p}d(\gamma\times\gamma)(u,v).$

Applying Fubini’s theorem, the function

 $u\mapsto\int_{\mathbb{R}^{n}}|f(M_{t}(u,v))|^{p}d\gamma(v)=\int_{\mathbb{R}^{n% }}|f(e^{-t}u+\sqrt{1-e^{-2t}}v)|^{p}d\gamma(v)$

belongs to $L^{1}(\gamma)$. We define the Ornstein-Uhlenbeck semigroup $\{T_{t}:t\geq 0\}$ on $L^{p}(\gamma)$, $p\geq 1$, by

 $T_{t}(f)(u)=\int_{\mathbb{R}^{n}}f(M_{t}(u,v))d\gamma(v)=\int_{\mathbb{R}^{n}}% f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)d\gamma(v),$

for $u\in\mathbb{R}^{n}$.

###### Theorem 11.

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

 $\int_{\mathbb{R}^{n}}(T_{t}f)(x)d\gamma(x)=\int_{\mathbb{R}^{n}}f(x)d\gamma(x).$
###### Proof.

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

 $\displaystyle\int_{\mathbb{R}^{n}}(T_{t}f)(u)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}f(M_{t}(u,v))d% \gamma(v)\right)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}f(M_{t}(u,v))d(\gamma% \times\gamma)(u,v)$ $\displaystyle=\int_{\mathbb{R}^{n}}f(x)d({M_{t}}_{*}(\gamma\times\gamma))(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}f(x)d\gamma(x).$

###### Theorem 12.

Let $\gamma$ be a Gaussian measure on $\mathbb{R}^{n}$ with mean $0$. For $p\geq 1$ and $t\geq 0$, $T_{t}$ 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,

 $\displaystyle\left\|T_{t}f\right\|_{L^{p}(\gamma)}^{p}$ $\displaystyle=\int_{\mathbb{R}^{n}}\left|\int_{\mathbb{R}^{n}}f(M_{t}(u,v))d% \gamma(v)\right|^{p}d\gamma(u)$ $\displaystyle\leq\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}|f(M_{t}(u,v)% )|^{p}d\gamma(v)\right)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}T_{t}(|f|^{p})(u)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}|f|^{p}(u)d\gamma(u)$ $\displaystyle=\left\|f\right\|_{L^{p}(\gamma)}^{p},$

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

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

so $T_{t}$ has operator norm $1$. ∎

For a Banach space $E$, we denote by $\mathscr{B}(E)$ 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 $\mathscr{B}(E)\to\mathbb{E}$. To say that a map $Q:[0,\infty)\to\mathscr{B}(E)$ is strongly continuous means that for each $t\in[0,\infty)$, $Q(s)\to Q_{t}$ 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 $\mathscr{B}(E)$ is a map $Q:[0,\infty)\to\mathscr{B}(E)$ such that (i) $Q(0)=\mathrm{id}_{E}$ and (ii) for $s,t\geq 0$, $Q(s+t)=Q(s)\circ Q(t)$. 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 $x\in E$, $Q(t)x\to x$.88 8 Walter Rudin, Functional Analysis, second ed., p. 376, Theorem 13.35.

We now establish that the $\{T_{t}:t\geq 0\}$ is indeed a one-parameter semigroup and that it is strongly continuous.99 9 Vladimir I. Bogachev, Gaussian Measures, p. 10, Theorem 1.4.1.

###### Theorem 13.

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

###### Proof.

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

 $T_{0}(f)(u)=\int_{\mathbb{R}^{n}}f(u)d\gamma(v)=f(u),$

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

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

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

 $\begin{split}&\displaystyle(T_{t}(T_{s}f))(x)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}(T_{s}f)\left(e^{-t}x+\sqrt{1% -e^{-2t}}y\right)d\gamma(y)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}f% \left(e^{-s}\left(e^{-t}x+\sqrt{1-e^{-2t}}y\right)+\sqrt{1-e^{-2s}}w\right)d% \gamma(w)\right)d\gamma(y)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}f\left(e^% {-s-t}x+\sqrt{1-e^{-2t-2s}}P(y,w)\right)d(\gamma\times\gamma)(y,w)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}(f\circ M% _{s+t})(x,P(y,w))d(\gamma\times\gamma)(y,w)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}(f\circ M_{s+t})(x,z)d\gamma(% z)\\ \displaystyle=&\displaystyle T_{s+t}(f)(x),\end{split}$

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

For $f\in C_{b}(\mathbb{R}^{n})$, $u\in\mathbb{R}^{n}$, and $v\in\mathbb{R}^{n}$, 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

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

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

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

and hence

 $\displaystyle(T_{t}f-T_{0}f)(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}f\left(e^{-t}u+\sqrt{1-e^{-2t}}v\right)d% \gamma(v)-\int_{\mathbb{R}^{n}}f(u)d\gamma(v)$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(f\left(e^{-t}u+\sqrt{1-e^{-2t}}v% \right)-f(u)\right)d\gamma(v)$ $\displaystyle\to 0.$

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

 $|(T_{t}f-T_{0}f)(u)|\leq\int_{\mathbb{R}^{n}}2\left\|f\right\|_{\infty}d\gamma% (v)=2\left\|f\right\|_{\infty},$

by the dominated convergence theorem we then have

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

Now let $f\in L^{p}(\gamma)$. There is a sequence $f_{j}\in C_{b}(\mathbb{R}^{n})$ satisfying $\left\|f_{j}-f\right\|_{L^{p}(\gamma)}\to 0$, with $\left\|f_{j}\right\|_{L^{p}(\gamma)}\leq 2\left\|f\right\|_{L^{p}(\gamma)}$ for all $j$. For any $t\geq 0$,

 $\displaystyle\left\|T_{t}f-T_{0}f\right\|_{L^{p}(\gamma)}$ $\displaystyle\leq\left\|T_{t}f-T_{t}f_{j}\right\|_{L^{p}(\gamma)}+\left\|T_{t}% f_{j}-T_{0}f_{j}\right\|_{L^{p}(\gamma)}+\left\|T_{0}f_{j}-T_{0}f\right\|_{L^{% p}(\gamma)}$ $\displaystyle=\left\|T_{t}(f-f_{j})\right\|_{L^{p}(\gamma)}+\left\|T_{t}f-T_{0% }f_{j}\right\|_{L^{p}(\gamma)}+\left\|f_{j}-f\right\|_{L^{p}(\gamma)}$ $\displaystyle\leq\left\|f-f_{j}\right\|_{L^{p}(\gamma)}+\left\|T_{t}f-T_{0}f_{% j}\right\|_{L^{p}(\gamma)}+\left\|f_{j}-f\right\|_{L^{p}(\gamma)}.$

Let $\epsilon>0$ and let $j$ be so large that $\left\|f-f_{j}\right\|_{L^{p}(\gamma)}<\epsilon$. Because $f_{j}\in C_{b}(\mathbb{R}^{n})$, by (7) there is some $\delta>0$ such that when $0, $\left\|T_{t}f_{j}-f_{j}\right\|_{L^{p}(\gamma)}<\epsilon$. Then when $0,

 $\left\|T_{t}f-T_{0}f\right\|_{L^{p}(\gamma)}\leq\epsilon+\epsilon+\epsilon,$

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

For $t>0$, we define $L_{t}\in\mathscr{B}(L^{p}(\gamma))$ by

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

We define $\mathscr{D}(L)$ 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 $L:\mathscr{D}(L)\to L^{p}(\gamma)$. This is the infinitesimal generator of the semigroup $\{T_{t}:t\geq 0\}$, 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.1010 10 Walter Rudin, Functional Analysis, second ed., p. 376, Theorem 13.35.

###### Theorem 14.

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

1. 1.

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

2. 2.

For each $f\in\mathscr{D}(L)$ and for each $t\geq 0$,

 $\frac{d}{dt}(T_{t}f)=(L\circ T_{t})f=(T_{t}\circ L)f.$
3. 3.

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

4. 4.

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

 $R(\lambda)f=\int_{0}^{\infty}e^{-\lambda t}T_{t}fdt,\qquad f\in L^{p}(\gamma),$

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

 $((\lambda I-L)\circ R(\lambda))f=f,\quad f\in L^{p}(\gamma),\qquad(R(\lambda)% \circ(\lambda I-L))f=\mathscr{D}(L),$

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

We remind ourselves that if $H$ is a Hilbert space with inner product $\left\langle\cdot,\cdot\right\rangle$, an element $A$ of $\mathscr{B}(H)$ is said to be a positive operator when $\left\langle Ax,x\right\rangle\geq 0$ for all $x\in H$. We prove that each $T_{t}$ is a positive operator on the Hilbert space $L^{2}(\gamma)$.1111 11 Vladimir I. Bogachev, Gaussian Measures, p. 10, Theorem 1.4.1.

###### Theorem 15.

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

###### Proof.

For $t\geq 0$, define $N_{t}:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}^{n}\times\mathbb{R}^{n}$ by

 $O_{t}(x,y)=\left(e^{-t}x+\sqrt{1-e^{-2t}}y,-\sqrt{1-e^{-2t}}x+e^{-t}y\right),% \qquad(x,y)\in\mathbb{R}^{n}\times\mathbb{R}^{n},$

whose transpose is the linear operator $N_{t}^{*}:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}^{n}\times\mathbb{R}^% {n}$ defined by

 $O_{t}^{*}(u,v)=\left(e^{-t}u-\sqrt{1-e^{-2t}}v,\sqrt{1-e^{-2t}}u+e^{-t}v\right% ),\qquad(u,v)\in\mathbb{R}^{n}\times\mathbb{R}^{n}.$

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

 $\begin{split}&\displaystyle\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}e^{i\left% \langle(x,y),(u,v)\right\rangle}d({O_{t}}_{*}(\gamma\times\gamma))(u,v)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}e^{i\left% \langle(x,y),O_{t}(u,v)\right\rangle}d(\gamma\times\gamma)(u,v)\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}e^{i\left% \langle O_{t}^{*}(x,y),(u,v)\right\rangle}d(\gamma\times\gamma)(u,v)\\ \displaystyle=&\displaystyle\widetilde{\gamma\times\gamma}(O_{t}^{*}(x,y))\\ \displaystyle=&\displaystyle\widetilde{\gamma\times\gamma}(e^{-t}x-\sqrt{1-e^{% -2t}}y,\sqrt{1-e^{-2t}}x+e^{-t}y)\\ \displaystyle=&\displaystyle\widetilde{\gamma}(e^{-t}x-\sqrt{1-e^{-2t}}y)% \widetilde{\gamma}(\sqrt{1-e^{-2t}}x+e^{-t}y)\\ \displaystyle=&\displaystyle\exp\left(-\frac{1}{2}\left\langle K(e^{-t}x-\sqrt% {1-e^{-2t}}y),e^{-t}x-\sqrt{1-e^{-2t}}y\right\rangle\right)\\ &\displaystyle\cdot\exp\left(-\frac{1}{2}\left\langle K(\sqrt{1-e^{-2t}}x+e^{-% t}y),\sqrt{1-e^{-2t}}x+e^{-t}y\right\rangle\right)\\ \displaystyle=&\displaystyle\exp\left(-\frac{1}{2}\left\langle Kx,x\right% \rangle-\frac{1}{2}\left\langle Ky,y\right\rangle\right)\\ \displaystyle=&\displaystyle\widetilde{\gamma}(x)\widetilde{\gamma}(y)\\ \displaystyle=&\displaystyle\widetilde{\gamma\times\gamma}(x,y),\end{split}$

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 $t\geq 0$,

 $\displaystyle\left\langle T_{t}f,g\right\rangle_{L^{2}(\gamma)}$ $\displaystyle=\int_{\mathbb{R}^{n}}(T_{t}f)(x)g(x)d\mu(x)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}f\left(e^{-t}x+\sqrt{1-% e^{-2t}}y\right)g(x)d(\gamma\times\gamma)(x,y)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}(f\circ\pi_{1}\circ O_{% t})(x,y)(g\circ\pi_{1}\circ O_{t}^{-1}\circ O_{t})(x,y)d(\gamma\times\gamma)(x% ,y)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}(f\circ\pi_{1})(u,v)(g% \circ\pi_{1}\circ O_{t}^{-1})(u,v)d({O_{t}}_{*}(\gamma\times\gamma))(u,v)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}(f\circ\pi_{1})(u,v)(g% \circ\pi_{1}\circ O_{t}^{-1})(u,v)d(\gamma\times\gamma)(u,v)$ $\displaystyle=\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}f(u)g\left(e^{-t}u-% \sqrt{1-e^{-2t}}v\right)d(\gamma\times\gamma)(u,v)$ $\displaystyle=\int_{\mathbb{R}^{n}}f(u)\left(\int_{\mathbb{R}^{n}}g(M_{t}(u,-v% ))d\gamma(v)\right)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}f(u)\left(\int_{\mathbb{R}^{n}}g(M_{t}(u,v)% )d\gamma(v)\right)d\gamma(u)$ $\displaystyle=\int_{\mathbb{R}^{n}}f(u)(T_{t}g)(u)d\gamma(u)$ $\displaystyle=\left\langle f,T_{t}g\right\rangle_{L^{2}(\gamma)},$

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,

 $\left\langle T_{t}f,f\right\rangle_{L^{2}(\gamma)}=\left\langle T_{t/2}T_{t/2}% f,f\right\rangle_{L^{2}(\gamma)}=\left\langle T_{t/2}f,T_{t/2}^{*}f\right% \rangle_{L^{2}(\gamma)}=\left\langle T_{t/2}f,T_{t/2}f\right\rangle_{L^{2}(% \gamma)},$

which is $\geq 0$, which establishes that $T_{t}$ is a positive operator on $L^{2}(\gamma)$. ∎

We now write the Ornstein-Uhlenbeck semigroup using the orthogonal projections $I_{k}:L^{2}(\gamma_{n})\to\mathcal{X}_{k}$, where $\gamma_{n}$ is the standard Gaussian measure on $\mathbb{R}^{n}$.1212 12 Vladimir I. Bogachev, Gaussian Measures, p. 11, Theorem 1.4.4.

###### Theorem 16.

For each $t\geq 0$ and $f\in L^{2}(\gamma_{n})$,

 $T_{t}f=\sum_{k=0}^{\infty}e^{-kt}I_{k}(f).$
###### Proof.

Define $S_{t}:L^{2}(\gamma_{n})\to L^{2}(\gamma_{n})$ by $S_{t}f=\sum_{k=0}^{\infty}e^{-kt}I_{k}(f)$, which satisfies, using that the subspaces $\mathcal{X}_{k}$ are pairwise orthogonal,

 $\left\|S_{t}f\right\|_{L^{2}(\gamma_{n})}^{2}=\sum_{k=0}^{\infty}e^{-kt}\left% \|I_{k}(f)\right\|_{L^{2}(\gamma_{n})}^{2}\leq\sum_{k=0}^{\infty}\left\|I_{k}(% f)\right\|_{L^{2}(\gamma_{n})}^{2}=\left\|f\right\|_{L^{2}(\gamma_{n})}^{2},$

so $S_{t}\in\mathscr{B}(L^{2}(\gamma_{n}))$. 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 $\alpha=(k_{1},\ldots,k_{n})$ with $k=|\alpha|=k_{1}+\cdots+k_{n}$,

 $S_{t}H_{\alpha}=e^{-kt}H_{\alpha},$

and

 $\displaystyle(T_{t}H_{\alpha})(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}H_{\alpha}\left(e^{-t}x+\sqrt{1-e^{-2t}}y% \right)d\gamma_{n}(y)$ $\displaystyle=\int_{\mathbb{R}^{n}}\prod_{j=1}^{n}H_{k_{j}}\left(e^{-t}x_{j}+% \sqrt{1-e^{-2t}}y_{j}\right)d\gamma_{n}(y)$ $\displaystyle=\prod_{j=1}^{n}\int_{\mathbb{R}}H_{k_{j}}\left(e^{-t}x_{j}+\sqrt% {1-e^{-2t}}y_{j}\right)d\gamma_{1}(y_{j}).$

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 $x_{j}$,

 $\int_{\mathbb{R}}H_{k_{j}}\left(e^{-t}x_{j}+\sqrt{1-e^{-2t}}y_{j}\right)d% \gamma_{1}(y_{j})=e^{-k_{j}t}H_{k_{j}}(x_{j}).$ (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 $\leq k_{j}$. That is, for each $0\leq h\leq k_{j}$, $T_{t}H_{h}=e^{-ht}H_{h}$. 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,

 $\int_{\mathbb{R}}(e^{-t}x_{j}+\sqrt{1-e^{-2t}}y_{j})^{l}\exp\left(-\frac{y_{j}% ^{2}}{2}\right)d\gamma_{1}(y_{j})$

is a polynomial in $x_{j}$ of degree $l$. Hence $T_{t}H_{l}$ a linear combination of $H_{0},H_{1},\ldots,H_{l}$. For $0\leq h\leq k_{j}$,

 $\left\langle T_{t}H_{k_{j}+1},H_{h}\right\rangle_{L^{2}(\gamma_{1})}=\left% \langle H_{k_{j}+1},T_{t}H_{h}\right\rangle_{L^{2}(\gamma_{1})}=\left\langle H% _{k_{j}+1},e^{-ht}H_{h}\right\rangle_{L^{2}(\gamma_{1})}=0.$

Therefore there is some $c\in\mathbb{R}$ such that $T_{t}H_{k_{j}+1}=cH_{k_{j}+1}$. Then check that $c=e^{-(k_{j}+1)t}$. ∎

We now give an explicit expression for the domain $\mathscr{D}(L)$ of the Ornstein-Uhlenbeck operator $L$ and for $L$ applied to an element of its domain.1313 13 Vladimir I. Bogachev, Gaussian Measures, p. 12, Proposition 1.4.5.

###### Theorem 17.
 $\mathscr{D}(L)=\left\{f\in L^{2}(\gamma_{n}):\sum_{k=0}^{\infty}k^{2}\left\|I_% {k}(f)\right\|_{L^{2}(\gamma_{n})}^{2}<\infty\right\}.$

For $f\in\mathscr{D}(L)$,

 $Lf=-\sum_{k=0}^{\infty}kI_{f}(f).$
###### Proof.

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

 $\displaystyle I_{k}Lf$ $\displaystyle=I_{k}\left(\lim_{t\downarrow 0}\frac{T_{t}f-f}{t}\right)$ $\displaystyle=\lim_{t\downarrow 0}\frac{I_{k}T_{t}f-I_{k}f}{t}$ $\displaystyle=\lim_{t\downarrow 0}\frac{T_{t}I_{k}f-I_{k}f}{t}$ $\displaystyle=\lim_{t\downarrow 0}\frac{e^{-kt}I_{k}f-I_{k}f}{t}$ $\displaystyle=\left(\lim_{t\downarrow 0}\frac{e^{-kt}-1}{t}\right)I_{k}f$ $\displaystyle=\left(e^{-kt}\right)^{\prime}\big{|}_{t=0}I_{k}f$ $\displaystyle=-kI_{k}f.$

Using this,

 $\displaystyle\sum_{k=0}^{\infty}k^{2}\left\|I_{k}f\right\|_{L^{2}(\gamma_{n})}% ^{2}$ $\displaystyle=\sum_{k=0}^{\infty}\left\|I_{k}Lf\right\|_{L^{2}(\gamma_{n})}^{2}$ $\displaystyle=\left\|\sum_{k=0}^{\infty}I_{k}Lf\right\|_{L^{2}(\gamma_{n})}^{2}$ $\displaystyle=\left\|Lf\right\|_{L^{2}(\gamma_{n})}^{2}$ $\displaystyle<\infty.$

Moreover,

 $Lf=L\left(\sum_{k=0}^{\infty}I_{k}f\right)=\sum_{k=0}^{\infty}LI_{k}f=\sum_{k=% 0}^{\infty}I_{k}Lf=\sum_{k=0}^{\infty}-kI_{f}.$

Let $f\in L^{2}(\gamma_{n})$ satisfy

 $\sum_{k=0}^{\infty}k^{2}\left\|I_{k}f\right\|_{L^{2}(\gamma_{n})}^{2}<\infty.$

For $t>0$,

 $\displaystyle\left\|\frac{T_{t}f-f}{t}+\sum_{k=0}^{\infty}kI_{k}f\right\|_{L^{% 2}(\gamma_{n})}^{2}$ $\displaystyle=\left\|\sum_{k=0}^{\infty}\left(\frac{e^{-kt}I_{k}f-I_{k}f}{t}+% kI_{k}f\right)\right\|_{L^{2}(\gamma_{n})}^{2}$ $\displaystyle=\sum_{k=0}^{\infty}\left|\frac{e^{-kt}-1}{t}+k\right|^{2}\left\|% I_{k}f\right\|_{L^{2}(\gamma_{n})}^{2}.$

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

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

and thus

 $\sum_{k=0}^{\infty}\left|\frac{e^{-kt}-1}{t}+k\right|^{2}\left\|I_{k}f\right\|% _{L^{2}(\gamma_{n})}^{2}\leq\sum_{k=0}^{\infty}(2k)^{2}\left\|I_{k}f\right\|_{% L^{2}(\gamma_{n})}^{2}<\infty.$

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

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

thus as $t\downarrow 0$,

 $\sum_{k=0}^{\infty}\left|\frac{e^{-kt}-1}{t}+k\right|^{2}\left\|I_{k}f\right\|% _{L^{2}(\gamma_{n})}^{2}\to 0$

and hence

 $\left\|\frac{T_{t}f-f}{t}+\sum_{k=0}^{\infty}kI_{k}f\right\|_{L^{2}(\gamma_{n}% )}^{2}\to 0.$

This means that $\frac{T_{t}f-f}{t}$ converges in $L^{2}(\gamma_{n})$ to $-\sum_{k=0}^{\infty}kI_{k}f$ as $t\downarrow 0$, and since $\frac{T_{t}f-f}{t}$ converges, $f\in\mathscr{D}(L)$. ∎