## Lecture 3. Semigroups generated by diffusion operators

In this lecture, we consider a diffusion operator L which is essentially self-adjoint. Its Friedrichs extension is still denoted by L.

The fact that we are now dealing with a non negative self-adjoint operator allows us to use spectral theory in order to define the semigroup generated by L. We recall the following so-called spectral theorem.

Theorem: Let $A$ be a non negative self-adjoint operator on a separable Hilbert space $\mathcal{H}$. There exist a measure space $(\Omega, \nu)$, a unitary map $U: \mathbf{L}_{\nu}^2 (\Omega,\mathbb{R}) \rightarrow \mathcal{H}$ and a non negative real valued measurable function $\lambda$ on $\Omega$ such that $U^{-1} A U f (x)=\lambda(x) f(x),$ for $x \in \Omega$, $Uf \in \mathcal{D}(A)$. Moreover, given $f \in \mathbf{L}_{\nu}^2 (\Omega,\mathbb{R})$, $Uf$ belongs to $\mathcal{D}(A)$ if only if $\int_{\Omega} \lambda^2 f^2 d\nu < +\infty$.

We may apply the spectral theorem to the self-adjoint operator $-L$ in order to define $e^{tL}$. More generally, given a Borel function $g :\mathbb{R}_{\ge 0} \to \mathbb{R}$ and the spectral decomposition of $-L$, $U^{-1} L U f (x)=-\lambda(x) f(x)$, we may always define an operator $g(-L)$ as being the unique operator that satisfies $U^{-1} g(-L) U f (x)= g\circ \lambda (x) f(x).$ We may observe that $g(-L)$ is a bounded operator if $g$ is a bounded function.

As a particular case, we define the diffusion semigroup $(\mathbf{P}_t)_{t \ge 0}$ on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ by the requirement $U^{-1} \mathbf{P}_t U f (x)=e^{-t \lambda (x)} f(x).$

This defines a family of bounded operators $\mathbf{P}_t : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ whose following properties are readily checked from the spectral decomposition:

• For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\| \mathbf{P}_t f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}.$
• $\mathbf{P}_0=\mathbf{Id}$ and for $s,t \ge 0$, $\mathbf{P}_s \mathbf{P}_t =\mathbf{P}_{s+t}$.
• For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, the map $t \to \mathbf{P}_t f$ is continuous in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.
• For $f,g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\int_{\mathbb{R}^n} (\mathbf{P}_t f) g d\mu= \int_{\mathbb{R}^n} f(\mathbf{P}_t g) d\mu$

We summarize the above properties by saying that $(\mathbf{P}_t)_{t \ge 0}$ is a self-adjoint strongly continuous contraction semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.

From the spectral decomposition, it is also easily checked that the operator $L$ is furthermore the generator of this semigroup, that is for $f \in \mathcal{D}(L)$, $\lim_{t \to 0} \left\| \frac{\mathbf{P}_t f -f}{t} -Lf \right\|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0.$ From the semigroup property, it implies that for $t \ge 0$, $\mathbf{P}_t \mathcal{D}(L) \subset \mathcal{D}(L)$, and that for $f \in \mathcal{D}(L)$, $\frac{d}{dt}\mathbf{P}_t f= \mathbf{P}_t Lf=L \mathbf{P}_t f,$ the derivative on the left hand side of the above equality being taken in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.

It is easily seen that the semigroup $(\mathbf{P}_t)_{t \ge 0}$ is actually unique in the followings sense:

Proposition: Let $\mathbf{Q}_t : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $t \ge 0$, be a family of bounded operators such that:

• For $s,t \ge 0$, $\mathbf{Q}_s \mathbf{Q}_t =\mathbf{Q}_{s+t}$.
• For $f \in \mathcal{D}(L)$, $\lim_{t \to 0} \left\| \frac{\mathbf{Q}_t f -f}{t} -Lf \right\|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0,$

then for every $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $t \ge 0$, $\mathbf{P}_t f=\mathbf{Q}_t f$.

Exercise: Show that if $L$ is the Laplace operator on $\mathbb{R}^n$, then for $t > 0$, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$. Show that if the constant function $1 \in \mathcal{D}(L)$ and if $L1=0$, then $\mathbf{P}_t 1=1.$

Exercise: Let $L$ be an essentially self-adjoint diffusion operator on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$

• Show that for every $\lambda > 0$, the range of the operator $\lambda \mathbf{Id}-L$ is dense in $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.
• By using the spectral theorem, show that the following limit holds for the operator norm on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}.$

Exercise: As usual, we denote by $\Delta$ the Laplace operator on $\mathbb{R}^n$. The Mac-Donald’s function with index $\nu \in \mathbb{R}$ is defined for $x \in \mathbb{R} \setminus \{ 0 \}$ by $K_\nu (x)=\frac{1}{2} \left( \frac{x}{2} \right)^\nu \int_0^{+\infty} \frac{e^{-\frac{x^2}{4t} -t}}{t^{1+\nu}} dt$.

• Show that for $\lambda \in \mathbb{R}^n$ and $\alpha > 0$, $\frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{i \langle \lambda , x \rangle} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ) dx=\frac{1}{\alpha +\| \lambda \|^2}.$
• Show that for $\nu \in \mathbb{R}$, $K_{-\nu}=K_\nu$.
• Show that $K_{1/2}(x)=\sqrt {\frac{\pi}{2x}} e^{-x}.$
• Prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$ and $\alpha > 0$, $(\alpha\mathbf{Id}-\Delta)^{-1} f (x)=\int_{\mathbb{R}^n} G_\alpha (x-y) f(y) dy,$ where $G_\alpha(x)=\frac{1}{(2\pi)^{n/2}} \left( \frac{\| x \| }{\sqrt{\alpha}} \right)^{1-\frac{n}{2}} K_{\frac{n}{2}-1} (\sqrt{\alpha} \| x \| ).$ (You may use Fourier transform to solve the partial differential equation $\alpha g -\Delta g=f$).

Exercise: By using the previous exercise, prove that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$, $\lim_{n \to + \infty} \left(\mathbf{Id} -\frac{t}{n} \Delta \right)^{-n} f =\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy,$ the limit being taken in $\mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$. Conclude that almost everywhere, $\mathbf{P}_t f (x)=\frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-\frac{\|x-y\|^2}{4t} } f(y)dy.$

Exercise:

• Show the subordination identity $e^{-y | \alpha | } =\frac{y}{2\sqrt{\pi}} \int_0^{+\infty} \frac{e^{-\frac{y^2}{4t}-t \alpha^2}}{t^{3/2}} dt, \quad y > 0, \alpha \in \mathbb{R}.$
• The Cauchy’s semigroup on $\mathbb{R}^n$ is defined as $\mathbf{Q}_t=e^{-t \sqrt{-\Delta}}$. By using the subordination identity and the heat semigroup on $\mathbb{R}^n$, show that for $f \in \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{Q}_tf (x)=\int_{\mathbb{R}^n} q(t,x-y) f(y) dy,$ where $q(t,x)=\frac{\Gamma\left( \frac{n+1}{2} \right)}{\pi^{\frac{n+1}{2}} } \frac{t}{(t^2+\| x \|^2 )^{\frac{n+1}{2}} }.$

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