## Lecture 3. Semigroup generated by a symmetric diffusion operator

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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### 3 Responses to Lecture 3. Semigroup generated by a symmetric diffusion operator

1. NH, A Japanese student says:

In the statement of the spectral theorem, is H a comlex Hilbert space, or real Hilbert space?
In lecture 6 you will use this theorem in the case of H=L^2_\mu(R^n,R), the real Hilbert space.
But In many books I have ever seen the famous theorem is treated only in complex Hilbert spaces, not real. Is it also true in the real case?

2. The theorem is also true in the real case. There is a discussion in the following thread on Math Stack Exchange: