# Regularity of the heat kernel

Let $(M,g)$ be a compact Riemannian manifold. Let $H:M\times M\times\mathbb{R}_{>0}\to\mathbb{R}$ be the heat kernel. i.e. $H\in C^0(M\times M\times\mathbb{R}_{>0})$ is the unique continuous function such that for all $y\in M$,

(A) $H^y\in C^{2,1}(M\times\mathbb{R}_{>0})$

(B) $\left(\Delta^g-\dfrac{\partial}{\partial t}\right)H^y=0$

(C) $\displaystyle\lim_{t\to0}H^y_t=\delta_y$

, where $H^y(x,t):=H(x,y,t)$, $H^y_t(x):=H^y(x,t)$,

$C^{2,1}(M\times\mathbb{R}_{>0}):=\{\varphi:M\times\mathbb{R}_{>0}\to\mathbb{R}|\text{ For each chart }(U;x^1,\cdots,x^m)\subset M, \dfrac{\partial\varphi}{\partial t}, \dfrac{\partial\varphi}{\partial x^i},\text{ and }\dfrac{\partial^2\varphi}{\partial x^i\partial x^j}:U\times\mathbb{R}_{>0}\to\mathbb{R} \text{ are well defined and continuous.}\}$.

${\bf [Question 1]}$ From (A) and (B) above it is derived that $H^y\in C^\infty(M\times\mathbb{R}_{>0})$ for all $y\in M$. How about the regularity of H as a function on $M\times M\times\mathbb{R}_{>0}$? Does it hold that $H\in C^\infty(M\times M\times \mathbb{R}_{>0})$? If not, aren't there any regularity result of $H:M\times M\times\mathbb{R}_{>0}\to\mathbb{R}$ which is useful to exchange integrals and differentiation?

${\bf [Question 2]}$ Suppose that $F:M\times[0,T]\to \mathbb{R}$ is a continuous function. Then is it true that the function \begin{eqnarray} u(x,t):=-\int_0^t\int_M H(x,y,t-\tau)F(y,\tau)\mu_g(dy)d\tau \end{eqnarray} belongs to $C^{1,0}(M\times[0,T])\cap C^{2,1}(M\times(0,T))$?

${\bf [Question 3]}$ Suppose that $f:M\to\mathbb{R}$ be a $C^1$ function. Does the function \begin{eqnarray} v(x,t):=\int_M H(x,y,t)f(y)\mu_g dy \end{eqnarray} belong to $C^{1,0}(M\times[0,\infty))\cap C^{2,1}(M\times\mathbb{R}_{>0})$?

Please tell me also references. Thank you.

• Possible reference for you: Chavel's book Eigenvalues in Riemannian Geometry has a chapter on the heat kernel. – Neal Nov 7 '14 at 18:50

## 2 Answers

Smoothness in all the questions has a local character. And in local coordinates it's a parabolic equation $Lu=F$ with (smooth) variable coefficients. So the local theory in $\mathbb R^n$ will do.

On Question2 the answer is no. If function $F$ is continuous it does not follow that $u$ is locally from $C^{2,1}$. For the heat equation see Nonclassical solution to u_t-\Delta u=f in one space dimension? question here.

On Question3 the answer is yes. As it was mentioned it's a question of local regularity. And for $\mathbb R^n$ one can differentiate the equation $Lu=0$ with respect to the space variable $x_i$, transfer all the terms with $u$ to the rhs and obtain a Cauchy problem for $\partial_iu$ with a continuous initial condition $\partial_i f$ and a continuous rhs. It is included in the definition of a fundamental solution that $v(x,t)\to f(x)$ as $t\to0\!+$ for continuous $f$, see, for example, A. Friedman, Partial Differential Equations of Parabolic Type.

For question 1, see Theorem 5.2.1 of E. B. Davies's book Heat Kernels and Spectral Theory, which asserts that indeed the heat kernel is a $C^\infty$ function on $M \times M \times (0,\infty)$. (It also applies to Riemannian manifolds $M$ which are complete but not compact.)

For question 2, it suffices to assume $F$ is bounded and measurable. Differentiation under the integral sign will show that $u \in C^\infty(M \times (0,T))$. Indeed, for this it suffices to assume $F$ is bounded and measurable. To get continuity up to $t=T$, extend $F$ to $\tilde{F} : M \times [0,T+\epsilon]$ by $\tilde{F}(x,t) = F(x,t)$ for $t \le T$ and $\tilde{F}(x,t) = 0$ for $t > T$. Then the corresponding function $\tilde{u}$ is continuous (even smooth) on $M \times (0, T+\epsilon)$ and $\tilde{u}(x,t) = u(x,t)$ for $t \le T$.

To get continuity at $t=0$, simply note that $$|u(x,t)| \le t \cdot\left(\sup_{M \times [0,t]} |F|\right)\left( \sup_{\tau \in [0,t]} \int_M H(x,y,t-\tau) \mu_g(dy)\right)$$ But $\sup_{M \times [0,t]} |F|$ is finite if $F$ is bounded, and $\int_M H(x,y,t-\tau) \mu_g(dy) = 1$ for any $x, t, \tau$.

I guess you still want to show that the spatial derivatives of $u$ are continuous up to $t=0$. That shouldn't be hard but maybe takes a little more thought.

For question 3, as before, if $f$ is merely bounded and measurable then $u \in C^\infty(M \times (0,\infty))$ by differentiating under the integral sign.

Still working on continuity at 0. It's really just going to come from the continuity of $f$, the fact that $H(x,y,t) \to \delta_x$ and the triangle inequality.

• $M$ is compact. – stb2084 Nov 7 '14 at 16:56
• @stb2084: Thanks, I had overlooked that. – Nate Eldredge Nov 7 '14 at 17:44
• Thanks a lot. But are the spacial derivatives of $v$ are continuous at $t=0$ in the 3rd question? – stb2084 Nov 8 '14 at 9:42
• In my 2nd question, in my understanding, $u$ satisfies $\left(\Delta-\dfrac{\partial}{\partial t}\right)u=F$ in $M\times (0,T)$. So if $u\in C^\infty(M\times(0,T))$, $F$ must be smooth. – stb2084 Nov 8 '14 at 12:07