On-diagonal to off-diagonal heat kernel lower bounds, Davies' argument

Question

Theorem 3.3.4 in Davies' Heat Kernels and Spectral Theory begins with ``on-diagonal'' lower bounds for the heat kernel $K$ of $H$, (i.e. $K = e^{-Ht}$), where $H$ is a uniformly elliptic operator acting on $L^{2}(\mathbb{R}^{N})$. That is, Davies has already proved $$K(t,x,y) \geq C t^{-N/2} \qquad \text{ when } \qquad |x-y|^{2} \leq C t$$ Now for arbitrary $x,y \in \mathbb{R}^{N}$ and $t > 0$, he defines a sequence of points $$x_{r} = x + r(y-x)/M$$ where $0 \leq r \leq M$ and $M$ is the smallest integer such that $4(y-x)^{2}/Ct \leq M$. Then he uses the inequality

$$K(t,x, y) \geq \int\cdots\int K(t/M,x,y_{1})K(t/M,y_{1},y_{2})\cdots K(t/M, y_{M-1},y)\,dy_{1}\cdots dy_{M-1}$$

where $y_{r}$ is being integrated over the set $$\{|y_{r} - x_{r}| < 1/4\,C(t/M)^{1/2}\}$$ But this step I do not understand at all. Where does the iterated integral come from??

Answer 1

This is a classic trick for heat kernel proofs:

The idea is that if $f(t,x)$ is a solution to the heat equation with initial data $f_0(x)$, then $g(t,x) := f(t+s,x)$ is a solution to the heat equation with initial data $g_0(x) = f(s,x)$.

Rephrasing that $f(t,x)$ is a solution to the heat equation as: $$ f(t,x) = \int K(t,x,y) f_0(y) dy, $$ we have that $$ f(t+s,x) = \int K(t+s,x,y) f_0(y)dy. $$

On the other hand, we also have that $$ f(t+s,x) = \int K(t,x,y) f(s,y) dy $$ by our second observation. Now, combine the first and third lines to give $$ f(t+s,x) = \int \int K(t+s,x,y) K(s,y,z) f_0(z) dz dy. $$

By uniqueness of the heat kernel, this shows that for $0 < s < t$ $$ K(t,x,y) = \int K(t-s,x,w) K(s,w,y) dw. $$

To go from this to your desired inequality, iterate, and then shrink the domain of integration as desired (using positivity of the heat kernel).