$\S2. $The Ornstein-Uhlenbeck operator and its semigroup

时间：2014-01-22 16:49:51 阅读：402 评论：0 收藏：0 [点我收藏+]

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

< ? i f, g > L 2 (γ) = = = 1 ( 2 π ) d 2 \int ? i f (x) g (x) e ? ∣ ∣ x ∣ ∣ 2 2 d x 1 ( 2 π ) d 2 \int f (x) [x i g (x) ? ? i g (x)] e ? ∣ ∣ x ∣ ∣ 2 2 d x < f, (x i ? ? i) g > L 2 (γ) .

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

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 ? ? i ? i = x ? ? ? Δ

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

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

< L f, g > = \sum i = 1 d < ? ? i ? i f, g > = \sum i = 1 d < ? i f, ? i g > = \sum i = 1 d < f, ? ? i ? i g > = < f, L g >

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

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 $\alpha\in\mathbb{N}^{d}$ ,

L H α = | α | H α .

$LH_{\alpha}=\left|\alpha\right|H_{\alpha}.$

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

< D ? H n ? 1, H j > = < H n ? 1, D H j > = n < H n ? 1, H j ? 1 > = 0, j \neq 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 > .

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

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

L H α = \sum i = 1 d ? ? i ? i H α = \sum i = 1 d ? ? i α i H α ? e i = \sum i = 1 d α i H α = | α | H α .

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

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 α \in N a α H α .

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

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 ? t L f = \sum α \in N d e t ∣ ∣ α ∣ ∣ a α H α .

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

In particular

e ? t L H α = e ? t ∣ ∣ α ∣ ∣ H α .

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

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

T f (x) = \int Φ (x, y) f (y) d γ (y) .

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

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

(T f (x)) 2 \leq \int | Φ (x, y) | 2 d γ (y) \int | f (y) | 2 d γ (y) .

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

Integrating both sides in $x$ leads to

| | T f | | 2 \leq | | Φ | | 2 L 2 (γ \times γ) | | f | | 2 .

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

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 R d M γ t (x, y) f (y) d γ (y) .

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

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 γ t (x, y) = \sum α \in N d e ? t | α | h α (x) h α (y) .

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

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 | α | < N e ? t | α | h α (x) h α (y) .

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

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

\int \sum | α | < N e t | α | h α (x) h α (y) H β (y) d γ (y) = e ? t | β | < h β, H β h β (x) = e ? t | β | ∣ ∣ ∣ ∣ H β ∣ ∣ ∣ ∣ h β (x) = e ? t | β | H β = T t H β .

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

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.

$\S2. $The Ornstein-Uhlenbeck operator and its semigroup

原文：http://www.cnblogs.com/levin2013/p/3529557.html

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