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

Let \(\partial_i =\frac{\partial}{\partial x_i}\). The operator \(\partial_i\) is unbounded on \(L^2(\gamma)\). We will explore its adjoint operator \(\partial^*_i\)  in \(L^2(\gamma)\). For this purpose, take \(f,g\in C_0^{\infty}\), i.e., infinitely many times differentiable functions with compact support. Then

\[\begin{array}{rcl}<\partial_{i}f,g>_{L^{2}\left(\gamma\right)} & = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\int\partial_{i}f\left(x\right)g\left(x\right)e^{-\frac{\left|x\right|^{2}}{2}}dx\\& = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\int f\left(x\right)\left[x_{i}g\left(x\right)-\partial_{i}g\left(x\right)\right]e^{-\frac{\left|x\right|^{2}}{2}}dx\\& = & <f,\left(x_{i}-\partial_{i}\right)g>_{L^{2}\left(\gamma\right)}.\end{array}\]

We see that \(\partial_{i}^{*}=x_{i}-\partial_{i}\), where the first term is a multiplication operator. Define a second-order differential operator by

\[L=\sum_{i=1}^{d}\partial_{i}^{*}\partial_{i}=x\cdot\nabla-\Delta\]

It is positive and symmetric and plays the role of the Laplacian on \(L^{2}(\gamma)\). Symmetry is shown by

\[<Lf,g>=\sum_{i=1}^{d}<\partial_{i}^{*}\partial_{i}f,g>=\sum_{i=1}^{d}<\partial_{i}f,\partial_{i}g>=\sum_{i=1}^{d}<f,\partial_{i}^{*}\partial_{i}g>=<f,Lg>\]

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 \(\alpha\in\mathbb{N}^{d}\),

\[LH_{\alpha}=\left|\alpha\right|H_{\alpha}.\]

Proof. Again consider \(d=1\). We first explore the action of \(D^{*}\) on \(H_{n}\).

\[<D^{*}H_{n-1},H_{j}>=<H_{n-1},DH_{j}>=n<H_{n-1},H_{j-1}>=0,j\ne n.\]

So, \(D^{*}H_{n-1}\) is a multiple of \(H_{n}\). Take \(j=n\).

\[<D^{*}H_{n-1},H_{n}>=n<H_{n-1},H_{n-1}>=n(n-1)!=n!=<H_{n},H_{n}>.\]

Thus \(D^{*}H_{n-1}=H_{n}\) and it follows that \(\partial_{i}^{*}H_{\alpha-e_{i}}=H_{\alpha}\), for \(d\ge1\), Where \(e_{1},\ldots,e_{n}\) is the standard orthonormal

system. Hence

\[LH_{\alpha}=\sum_{i=1}^{d}\partial_{i}^{*}\partial_{i}H_{\alpha}=\sum_{i=1}^{d}\partial_{i}^{*}\alpha_{i}H_{\alpha-e_{i}}=\sum_{i=1}^{d}\alpha_{i}H_{\alpha}=\left|\alpha\right|H_{\alpha}.\]

We now turn to the Ornstein-Uhlenbeck semigroup, i.e., the semigroup generated by \(L\). For this purpose we use our spectral decomposition of \(L^{2}(\gamma)\). Since \(\left\{ H_{\alpha},\alpha\in\mathbb{N}\right\}\) form a orthonormal system of \(L^{2}(\gamma)\), for any \(f\in L^{2}(\gamma)\),

\[f=\sum_{\alpha\in\mathbb{N}}a_{\alpha}H_{\alpha}.\]

Let \(\left(T_{t}\right)_{t\ge0}=\left(e^{-tL}\right)_{t\ge 0}\) be the family of bounded linear operators acting on \(L^{2}(\gamma)\) by

\[e^{-tL}f=\sum_{\alpha\in\mathbb{N}^{d}}e^{t\left|\alpha\right|}a_{\alpha}H_{\alpha}.\]

In particular

\[e^{-tL}H_{\alpha}=e^{-t\left|\alpha\right|}H_{\alpha}.\]

It follows that \(e^{-tL}\) is a bounded operator on \(L^{2}(\alpha)\) for any \(t\ge0\) and that \(e^{-tL}e^{-sL}=e^{-(s+t)L},s,t\ge0\). Since \(T_{0}\) is the identity, \(\left(T_{t}\right)_{t\ge0}\) forms a semigroup.

Any \(\Phi\in L^{2}(\gamma\times\gamma)\) defines a bounded linear operator on \(L^{2}(\gamma)\) by

\[Tf(x)=\int\Phi(x,y)f(y)d\gamma(y).\]

It is not essential here that we work in our Gaussian setting. Any \(L^{2}\)-space would do fine. We verify the boundedness. The Cauchy-Schwardz inequality gives that

\[\left(Tf(x)\right)^{2}\le\int|\Phi(x,y)|^{2}d\gamma(y)\int|f(y)|^{2}d\gamma(y).\]

Integrating both sides in \(x\) leads to

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

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

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

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

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

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

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

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

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

Since the truncated kernels converge in \(L^{2}(\gamma\times\gamma)\), the corresponding operators converge in the operator norm. We conclude that \(T_{t}\) can be epresented by Mehler kernel. We next want to compute a closed expression for \(M_{t}^{\gamma}\). Let \(d=1\). Since \(\mathcal{F}\left(e^{-\xi^{2}}\right)(x)=\sqrt{\pi}e^{-\frac{x^{2}}{4}}\), 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.