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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.

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

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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.

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    $\begingroup$ $M$ is compact. $\endgroup$ – stb2084 Nov 7 '14 at 16:56
  • $\begingroup$ @stb2084: Thanks, I had overlooked that. $\endgroup$ – Nate Eldredge Nov 7 '14 at 17:44
  • $\begingroup$ Thanks a lot. But are the spacial derivatives of $v$ are continuous at $t=0$ in the 3rd question? $\endgroup$ – stb2084 Nov 8 '14 at 9:42
  • $\begingroup$ 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. $\endgroup$ – stb2084 Nov 8 '14 at 12:07

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