## Lecture 4. The heat kernel of a diffusion semigroup

The goal of this lecture is to prove that if a diffusion operator L is elliptic, then the semigroup it generates admits a smooth kernel. As a consequence, the semigroup generated by an elliptic diffusion operator is regularizing in the sense that it transforms any function into a smooth function. The key point is the following estimate that can be proved by using the theory of Sobolev spaces.

Proposition: Let $L$ be an elliptic diffusion operator with smooth coefficients on $\mathbb{R}^n$ which is symmetric with respect to a Borel measure $\mu$. Let $u \in \mathbf{L}_\mu^2 (\mathbb{R}^n,\mathbb{R})$ such that $Lu,L^2u,\cdots, L^ku \in \mathbf{L}_\mu^2 (\mathbb{R}^n,\mathbb{R})$, for some positive integer $k$. If $k > \frac{n}{4}$, then $u$ is a continuous function, moreover, for any bounded open set $\Omega \subset \mathbb{R}^n$, and any compact set $K \subset \Omega$, there exists a positive constant $C$ (independent of $u$) such that $\left(\sup_{x \in K} | u(x) | \right)^2 \le C \left( \sum_{j=0}^k \|L^j u \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ More generally, if $k > \frac{m}{2}+\frac{n}{4}$ for some non negative integer $m$, then $u \in \mathcal{C}^m(\mathbb{R}^n,\mathbb{R})$ and for any bounded open set $\Omega \subset \mathbb{R}^n$, and any compact set $K \subset \Omega$, there exists a positive constant $C$ (independent of $u$) such that $\left(\sup_{|\alpha| \le m} \sup_{x \in K} |\partial^\alpha u(x) | \right)^2 \le C \left( \sum_{j=0}^k \|L^j u \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$

We can explain by simple computations on the Laplace operator $\Delta$ how the $\frac{m}{2}+\frac{n}{4}$ comes into the play in the above proposition. Let $u$ be a smooth and rapidly decreasing function. By the inverse Fourier transform formula $\| \Delta^j u \|^2_{ \mathbf{L}^2 (\Omega,\mathbb{R})} =(2\pi)^{4j} \int_{\mathbb{R}^n} \| \lambda \|^{4j} | \hat{u} (\lambda) |^2 d\lambda,$ so that by Cauchy-Schwarz inequality we may bound
$\partial^\alpha u(x) = \int_{\mathbb{R}^n} (2i\pi\lambda)^{\alpha} e^{2i\pi \langle x, \lambda\rangle} \hat{u} (\lambda) d\lambda = \int_{\mathbb{R}^n} (2i\pi\lambda)^{\alpha} e^{2i\pi \langle x, \lambda\rangle} \sqrt{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j} } \frac{\hat{u} (\lambda)}{\sqrt{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j} } } d\lambda,$
by $\sum_{j=0}^k \|\Delta^j u \|^2_{ \mathbf{L}^2 (\mathbb{R}^n,\mathbb{R})}$ only when $\int_{\mathbb{R}^n} \frac{\|\lambda\|^{2\alpha} d\lambda}{ \sum_{j=0}^k (2\pi)^{4j} \| \lambda \|^{4j}} < \infty$ that is if $k > \frac{m}{2}+\frac{n}{4}$. We are now in position to prove the following regularization estimate for the semigroup associated with an elliptic operator.

Proposition: Let $L$ be an elliptic diffusion operator with smooth coefficients that is essentially self-adjoint with respect to a measure $\mu$. Denote by $(\mathbf{P}_t)_{t \ge 0}$ the corresponding semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$.

• If $K$ is a compact set of $\mathbb{R}^n$, there exists a positive constant $C$ such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})},$ where $\kappa$ is the smallest integer larger than $\frac{n}{4}$.
• For $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, the function $(t,x)\rightarrow \mathbf{P}_tf (x)$ is smooth on $(0,+\infty)\times \mathbb{R}^n$.

Proof: Let us first observe that from the spectral theorem that if $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ then for every $k \ge 0$, $L^k \mathbf{P}_t f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $\|L^k \mathbf{P}_t f \|_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \le \left(\sup_{\lambda \ge 0} \lambda^k e^{-\lambda t}\right) \|f \|_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})}.$ Now, let $K$ be a compact set of $\mathbb{R}^n$. From the previous proposition, there exists therefore a positive constant $C$ such that $\left(\sup_{x \in K} | \mathbf{P}_t f(x) | \right)^2 \le C \left( \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ Since it is immediately checked that $\sup_{\lambda \ge 0} \lambda^k e^{-\lambda t}=\left( \frac{k}{t}\right)^k e^{-k},$ the bound $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$ easily follows. We now turn to the second part. Let $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. First, we fix $t > 0$. As above, from the spectral theorem, for every $k \ge 0$, $L^k \mathbf{P}_t f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, for any bounded open set $\Omega$. By hypoellipticity of $L$, we deduce therefore that $\mathbf{P}_t f$ is a smooth function.

Next, we prove joint continuity in the variables $(t,x)\in (0,+\infty)\times \mathbb{R}^n$. It is enough to prove that if $t_0 >0$ and if $K$ is a compact set in $\mathbb{R}^n$, $\sup_{x \in K} | \mathbf{P}_{t} f(x) - \mathbf{P}_{t_0} f(x) | \rightarrow_{t \to t_0} 0.$ From the previous proposition, there exists a positive constant $C$ such that $\sup_{x \in K} | \mathbf{P}_{t} f(x) - \mathbf{P}_{t_0} f(x) | \le C \left( \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f-L^k \mathbf{P}_{t_0} f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})} \right).$ Now, again from the spectral theorem, it is checked that $\lim_{t \to t_0} \sum_{k=0}^{\kappa} \|L^k \mathbf{P}_t f-L^k \mathbf{P}_{t_0} f \|^2_{ \mathbf{L}_\mu^2 (\Omega,\mathbb{R})}=0.$ This gives the expected joint continuity in $(t,x)$. The joint smoothness in $(t,x)$ is a consequence of the second part of the previous proposition and the details are let to the reader $\square$

Remark: If the bound $\sup_{x \in K} |\mathbf{P}_t f(x)| \le C(t) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}$ uniformly holds on $\mathbb{R}^n$, that is if $\| \mathbf{P}_t \|_{ \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R})} < \infty,$ then the semigroup $(\mathbf{P}_t)_{t \ge 0}$ is said to be ultracontractive.

Exercise: Let $L$ be an elliptic diffusion operator with smooth coefficients that is essentially self-adjoint with respect to a measure $\mu$ . Let $\alpha$ be a multi-index. If $K$ is a compact set of $\mathbb{R}^n$, show that there exists a positive constant $C$ such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\sup_{x \in K} |\partial^{\alpha} \mathbf{P}_t f(x)| \le C \left( 1 +\frac{1}{t^{|\alpha|+\kappa}} \right) \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})},$ where $\kappa$ is the smallest integer larger than $\frac{n}{4}$.

We are now in position to prove the following fundamental theorem:

Theorem: Let $L$ be an elliptic and essentially self-adjoint diffusion operator. Denote by $(\mathbf{P}_t)_{t \ge 0}$ the corresponding semigroup on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. There is a smooth function $p(t,x,y)$, $t \in (0,+\infty), x,y \in \mathbb{R}^n$, such that for every $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ and $x \in \mathbb{R}^n$ , $\mathbf{P}_t f (x)=\int_{\mathbb{R}^n} p(t,x,y) f(y) d\mu (y).$ The function $p(t,x,y)$ is called the heat kernel associated to $(\mathbf{P}_t)_{t \ge 0}$. It satisfies furthermore:

• (Symmetry) $p(t,x,y)=p(t,y,x)$;
• (Chapman-Kolmogorov relation) $p(t+s,x,y)=\int_{\mathbb{R}^n} p(t,x,z)p(s,z,y)d\mu(z)$.

Proof: Let $x\in \mathbb{R}^n$ and $t > 0$. From the previous proposition, the linear form $f \rightarrow \mathbf{P}_t f (x)$ is continuous on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, therefore from the Riesz representation theorem, there is a function $p(t,x,\cdot)\in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, such that for $f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\mathbf{P}_t f (x)=\int_{\mathbb{R}^n} p(t,x,y) f(y) d\mu (y).$ From the fact that $\mathbf{P}_t$ is self-adjoint on $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$, $\int_{\mathbb{R}^n} (\mathbf{P}_t f) g d\mu=\int_{\mathbb{R}^n} f(\mathbf{P}_t g) d\mu,$ we easily deduce the symmetry property $p(t,x,y)=p(t,y,x).$ And the Chapman-Kolmogorov relation $p(t+s,x,y)=\int_{\mathbb{R}^n} p(t,x,z)p(s,z,y)d\mu(z)$ stems from the semigroup property $\mathbf{P}_{t+s} =\mathbf{P}_t \mathbf{P}_s$. Finally, from the previous proposition the map $(t,x) \rightarrow p(t,x,\cdot) \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$ is smooth on $(0,+\infty) \times \mathbb{R}^n$ for the weak topology of $\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})$. This implies that it is also smooth on $(0,+\infty) \times \mathbb{R}^n$ for the norm topology. Since, from the Chapman-Kolmogorov relation $p(t,x,y)=\langle p(t/2,x,\cdot), p(t/2,y.\cdot) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) },$ we conclude that $(t,x,y)\rightarrow p(t,x,y)$ is smooth on $(0,+\infty) \times \mathbb{R}^n \times \mathbb{R}^n$ $\square$

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