Let ?i=??xi . The operator ?i is unbounded on L2(γ) . We will explore its adjoint operator ??i   in L2(γ) . For this purpose, take f,g∈C∞0 , i.e., infinitely many times differentiable functions with compact support. Then

<?if,g>L2(γ)===1(2π)d2∫?if(x)g(x)e?∣∣x∣∣22dx1(2π)d2∫f(x)[xig(x)??ig(x)]e?∣∣x∣∣22dx<f,(xi??i)g>L2(γ).

We see that ??i=xi??i , where the first term is a multiplication operator. Define a second-order differential operator by

L=∑i=1d??i?i=x???Δ

It is positive and symmetric and plays the role of the Laplacian on L2(γ) . Symmetry is shown by

<Lf,g>=∑i=1d<??i?if,g>=∑i=1d<?if,?ig>=∑i=1d<f,??i?ig>=<f,Lg>

Positivity follows by setting f=g in the middle expression above.

     The operator L is called the Ornstein-Uhlenbeck operator.

Proposition The Hermite polynomials are eigenvectors for the Ornstein-Uhlenbeck operator. Moreover, for any multi-index α∈Nd ,

LHα=|α|Hα.

Proof. Again consider d=1 . We first explore the action of D? on Hn .

<D?Hn?1,Hj>=<Hn?1,DHj>=n<Hn?1,Hj?1>=0,j≠n.

So, D?Hn?1 is a multiple of Hn . Take j=n .

<D?Hn?1,Hn>=n<Hn?1,Hn?1>=n(n?1)!=n!=<Hn,Hn>.

Thus D?Hn?1=Hn and it follows that ??iHα?ei=Hα , for d≥1 , Where e1,…,en is the standard orthonormal

system. Hence

LHα=∑i=1d??i?iHα=∑i=1d??iαiHα?ei=∑i=1dαiHα=|α|Hα.

        We now turn to the Ornstein-Uhlenbeck semigroup, i.e., the semigroup generated by L . For this purpose we use our spectral decomposition of L2(γ) . Since {Hα,α∈N} form a orthonormal system of L2(γ) , for any f∈L2(γ) ,

f=∑α∈NaαHα.

Let (Tt)t≥0=(e?tL)t≥0 be the family of bounded linear operators acting on L2(γ) by

e?tLf=∑α∈Ndet∣∣α∣∣aαHα.

In particular

e?tLHα=e?t∣∣α∣∣Hα.

It follows that e?tL is a bounded operator on L2(α) for any t≥0$andthat$e?tLe?sL=e?(s+t)L,s,t≥0 . Since T0 is the identity, (Tt)t≥0 forms a semigroup.

Any Φ∈L2(γ×γ) defines a bounded linear operator on L2(γ) by

Tf(x)=∫Φ(x,y)f(y)dγ(y).

It is not essential here that we work in our Gaussian setting. Any L2 -space would do fine. We verify the boundedness. The Cauchy-Schwardz inequality gives that

(Tf(x))2≤∫|Φ(x,y)|2dγ(y)∫|f(y)|2dγ(y).

Integrating both sides in x leads to

\left|\left|Tf\right|\right|^{2}\le\left|\left|\Phi\right|\right|_{L^{2}(\gamma\times\gamma)}^{2}\left|\left|f\right|\right|^{2}.

||Tf||2≤||Φ||2L2(γ×γ)||f||2.

We now leave the general situation. The operator T_{t}Tt , for t>0t>0 , is given by a kernel in the sense that

T_{t}f(x)=\int_{\mathbb{R}^{d}}M_{t}^{\gamma}(x,y)f(y)d\gamma(y).

Ttf(x)=∫RdMγt(x,y)f(y)dγ(y).

The explicit expression for this kernel was found already in 1866 by Mehler. It is named the Mehler kernel. Using the normalized Hermite polynomials h_{\alpha}hα , we shall first verify that the kernel can be expressed in the form

M_{t}^{\gamma}(x,y)=\sum_{\alpha\in\mathbb{N}^{d}}e^{-t|\alpha|}h_{\alpha}(x)h_{\alpha}(y).

Mγt(x,y)=∑α∈Nde?t|α|hα(x)hα(y).

It is easy to check that this series converges in L^{2}(\gamma\times\gamma)L2(γ×γ) . Consider, for $N\in\mathbb{N}$, the truncated kernel

\sum_{|\alpha|<N}e^{-t|\alpha|}h_{\alpha}(x)h_{\alpha}(y).

∑|α|<Ne?t|α|hα(x)hα(y).

For |\beta|<N|β|<N , the corresponding operator acts on H_{\beta}Hβ as

\int\sum_{|\alpha|<N}e^{t|\alpha|}h_{\alpha}(x)h_{\alpha}(y)H_{\beta}(y)d\gamma(y)=e^{-t|\beta|}<h_{\beta},H_{\beta}h_{\beta}(x)=e^{-t|\beta|}\left|\left|H_{\beta}\right|\right|h_{\beta}(x)=e^{-t|\beta|}H_{\beta}=T_{t}H_{\beta}.

∫∑|α|<Net|α|hα(x)hα(y)Hβ(y)dγ(y)=e?t|β|<hβ,Hβhβ(x)=e?t|β|∣∣∣∣Hβ∣∣∣∣hβ(x)=e?t|β|Hβ=TtHβ.

Since the truncated kernels converge in L^{2}(\gamma\times\gamma)L2(γ×γ) , the corresponding operators converge in the operator norm. We conclude that T_{t}Tt can be epresented by Mehler kernel. We next want to compute a closed expression for M_{t}^{\gamma}Mγt . Let d=1d=1 . Since \mathcal{F}\left(e^{-\xi^{2}}\right)(x)=\sqrt{\pi}e^{-\frac{x^{2}}{4}}F(e?ξ2)(x)=π√e?x24 , where \mathcal{F} denotes the Fourier transform, H_{n} can be written

\begin{array}{rcl}H_{n}\left(y\right) & = & \left(-1\right)^{n}e^{\frac{y^{2}}{2}}\frac{d^{n}}{dy^{n}}e^{-\frac{y^{2}}{2}}=\left(-1\right)^{n}e^{\frac{y^{2}}{2}}\frac{d^{n}}{dy^{n}}\frac{1}{\sqrt{2\pi}}\int e^{iy\xi-\frac{^{\xi^{2}}}{2}}d\xi\\& = & \left(-1\right)^{n}e^{\frac{y^{2}}{2}}\frac{i^{n}}{\sqrt{2\pi}}\int\xi^{n}e^{iy\xi-\frac{\xi^{2}}{2}}d\xi.\end{array}

Assuming that the order of summation and integration can be switched. By using the generating function of Hermite polynomial, we get

\begin{array}{rcl}M_{t}^{\gamma} & = & \sum_{n=0}^{\infty}e^{-tn}h_{n}\left(x\right)h_{n}\left(y\right)\\& = & \sum_{n=0}^{\infty}e^{-tn}\frac{1}{n!}H_{n}\left(x\right)\left(-1\right)^{n}e^{\frac{y^{2}}{2}}\frac{i^{n}}{\sqrt{2\pi}}\int\xi^{n}e^{iy\xi-\frac{\xi^{2}}{2}}d\xi\\& = & \frac{1}{\sqrt{2\pi}}e^{\frac{y^{2}}{2}}int\sum_{n=0}^{\infty}\frac{1}{n!}\left(-i\xi e^{-t}\right)^{n}H_{n}\left(x\right)e^{iy\xi-\frac{\xi^{2}}{2}}d\xi\\& = & \frac{1}{\sqrt{2\pi}}e^{\frac{y^{2}}{2}}\int e^{i\xi\left(y-e^{t}x+\frac{\xi^{2}}{2}e^{-2t}\right)}d\xi\end{array}

Let \xi^{t}=\xi\sqrt{1-e^{-2t}} . Then, taking the inverse Fourier transform yields

M_{t}^{\gamma}\left(x,y\right)=\frac{e^{\frac{y^{2}}{2}}}{\sqrt{1-e^{-2t}}}e^{-\frac{\left(y-e^{-t}x\right)^{2}}{1-e^{-2t}}}.

This is a closed expression for the kernel, but it remains to verify the switch of order above. BY using dominated convergence theorem, it is ease to get the conclusion. Let d\ge1 . Then 

M_{t}^{\gamma}\left(x,y\right)=\frac{e^{\frac{\left|y\right|^{2}}{2}}}{\sqrt{\left(1-e^{-2t}\right)^{d}}}e^{-\frac{\left|y-e^{-t}x\right|^{2}}{1-e^{-2t}}}.

Making the change of variable z=\frac{y-e^{-t}x}{\sqrt{1-e^{-2t}}} , we get

T_{t}f\left(x\right)=\int M_{t}^{\gamma}\left(x,y\right)f\left(y\right)d\gamma\left(y\right)=\int f\left(e^{-t}x+z\sqrt{1-e^{-2t}}\right)d\gamma\left(z\right).

This is sometimes called Mehler‘s formula.

\(\S2. \)The Ornstein-Uhlenbeck operator and its semigroup