## Lecture 4. Diffusion semigroups as solutions of heat equations

In this lecture, we show that the diffusion semigroup that was constructed in the previous lectures appears as the solution of a parabolic Cauchy problem. Under an ellipticity and completeness assumption, it is moreover the unique square integrable solution.

Proposition: Let $L$ be an essentially self-adjoint diffusion operator and let $(\mathbf{P}_t)_{t \ge 0}$ be the corresponding diffusion semigroup. Let $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, and let $u (t,x)= \mathbf{P}_t f (x), \quad t \ge 0, x\in \mathbb{R}^n.$ Then $u$ is a weak solution of the Cauchy problem
$\frac{\partial u}{\partial t}= L u,\quad u (0,x)=f(x).$

Proof: For $\phi \in \mathcal{C}_c ((0,+\infty) \times \mathbb{R}^n,\mathbb{R})$, we have
$\int_{\mathbb{R}^{n+1}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) u(t,x) d\mu(x) dt$
$=\int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) \mathbf{P}_t f (x) d\mu(x) dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \mathbf{P}_t \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) f (x) d\mu(x) dt$
$= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} -\frac{\partial}{\partial t} \left( \mathbf{P}_t \phi (t,x) f(x) \right) d\mu(x) dt =0.$

If the operator $L$ is furthermore assumed to be elliptic, then as we have seen in the previous lecture, the map $(t,x) \to \mathbf{P}_t f(x)$ is smooth and therefore, the above solution is also strong.

We now address uniqueness questions. We need further assumptions that already have been met before. We consider an elliptic diffusion operator $L$ with smooth coefficients on $\mathbb{R}^n$ such that:

• There is a Borel measure $\mu$, symmetric and invariant for $L$ on $\mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$;
• There exists an increasing sequence $h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$, $0 \le h_n \le 1$, such that $h_n\nearrow 1$ on $\mathbb{R}^n$, and $||\Gamma (h_n,h_n)||_{\infty} \to 0$, as $n\to \infty$.

Under these assumptions we already know that $L$ is essentially self-adjoint. The next proposition implies that $(t,x) \to \mathbf{P}_t f(x)$ is the unique solution of the parabolic Cauchy problem.

Proposition Let $L$ be a diffusion operator that satisfies the above assumptions. Let $u(t,x)$ be a smooth solution of the Cauchy problem $\frac{\partial u}{\partial t}= L u,\quad u (0,x)=0,$ where $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. Assume that $\| u (t , \cdot) \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} < +\infty$. Then $u(t,x)=0$

Proof: Let $h_n$ be as above. On one hand, we have $\int_0^\tau \int_{\mathbb{R}^n} h_n^2 u Lu d\mu dt =\frac{1}{2} \int_0^\tau \frac{\partial}{\partial t} \int_{\mathbb{R}^n} h_n^2 u^2 (t,x) d\mu dt = \int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu$.
On the other hand, we have $\int_{\mathbb{R}^n} h_n^2 u Lu d\mu =-\int_{\mathbb{R}^n} \Gamma(h_n^2 u, u) d\mu =-\int_{\mathbb{R}^n} h_n^2 \Gamma(u) d\mu -2 \int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu$.
From Cauchy-Schwarz inequality, we now have $\int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu \ge - \left( \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu\right)^{1/2}\left( \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu\right)^{1/2}$
$\ge -\frac{1}{2} \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu-\frac{1}{2} \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu$.

We deduce that $\int_{\mathbb{R}^n} h_n^2 u Lu d\mu \le \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu$. As a conclusion we obtain that $\int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu \le \int_0^\tau \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu dt$. Letting $n \to \infty$, yields $\int_{\mathbb{R}^n} u^2 d\mu \le 0$ and thus $u=0$ $\square$

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