TITLE: On-diagonal to off-diagonal heat kernel lower bounds, Davies' argument
QUESTION [4 upvotes]: 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??

REPLY [4 votes]: 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).
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