Lecture 3. Semigroups generated by diffusion operators

Posted on September 6, 2016 by Fabrice Baudoin

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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