# Are heat kernels on metric measure spaces continuous?

Let $(M,d)$ be a separable, complete, compact metric space and $\mu$ a Radon measure with full support on it. Let $\mathcal{E}$ be a regular strongly local Dirichlet form on $L^2(M)$. There exists an associated self-adjoint non-negative operator, a strongly continuous semigroup and a heat kernel (or transition function) $p(t,x,y)$.

What I want to know is whether the heat kernel is continuous as a function of $x$ and $y$?

If $M$ is a Riemannian manifold and the operator is the Laplace-Beltrami, the heat kernel is smooth in all three variables. For metric spaces there are general results for example by Sturm saying that if $(M,d)$ satisfies a volume doubling property and $\mathcal{E}$ satisfies the Sobolev inequality, then the heat kernel satisfies some exponential decay bounds. A result saying that the heat kernel is continuous under these or similar assumptions would be useful as well.

Edit: in response to Nate Eldridge's comment. In general one only gets the existence of a transition kernel $p(t,x,A)$ and one can then construct a heat kernel $p(t,x,y)$ as the density function of the kernel by the Radon-Nikodym theorem. This only works if the transition kernel is absolutely continuous with respect to $\mu$. I will assume here that this is the case.

• Keep in mind that the existence of a heat kernel which is a function $p(t,x,y)$ is already an assumption. In general you only have a transition kernel $p(t,x,B)$ and it need not be absolutely continuous to $\mu$. I think it can happen that the heat kernel is a function but not continuous, but I cannot think of a counterexample right now. Typically to get continuity of the heat kernel you look for something like a Harnack inequality. – Nate Eldredge Feb 8 '17 at 19:00