# Does generator of continuous time random walk map heat kernel from L^2 to L^2?

Let $\Gamma = (G,E)$ be an undirected, infinite, connected graph with no multiple edges or loops. We equip $\Gamma$ with a set of edge weights $\pi_{xy}$, where, given $e=\{x,y\}\in E$, we write $\pi_{\{x,y\}} = \pi_{xy}=\pi_{yx}>0$. If $\{x,y\}\not\in E$, we set $\pi_{xy}=0$. We write $\pi_x$ for $\sum_{y\sim x} \pi_{xy}$. We also assign a set of positive vertex weights $(\theta_x)_{x\in G}$.

We consider the continuous time random walk on $\Gamma$ with generator $\mathcal{L}_\theta$ given by

\begin{equation*} (\mathcal{L}_\theta f)(x) = \frac{1}{\theta_x}\sum_{y\sim x}\pi_{xy}(f(y)-f(x)), \end{equation*}

and denote the resulting Markov process by $(X^\theta_t)_{t\geq 0}$. Roughly speaking, at a vertex $x$, this process waits an exponential time with mean $\theta_x/\pi_x$, and then jumps to one of its neighbors $y$ with probability $\pi_{xy}/\pi_x$. It's not hard to see that if one chooses the $\theta$ s and $\pi$ s in certain ways, the process can behave quite pathologically and run away from its starting point very quickly (e.g., the process may have a finite lifetime).

The function $p_t(x,y) := \frac{\mathbb{P}^x(X^\theta_t=y)}{\theta_y}$ is the called the heat kernel of the random walk.

I'm interested in some analytic aspects of the generator $\mathcal{L}\_\theta$. For example, $\mathcal{L}\_\theta$ is a bounded operator from $L^2(\theta)$ to $L^2(\theta)$ if and only if

\begin{equation*} \sup_{x\in G} \frac{\pi_x}{\theta_x}< \infty. \end{equation*}

Set $u(x) := p_t(x_0,x)$. Note that $u$ is always in $L^2(\theta)$, as

\begin{equation*} \sum_{x\in G} u^2(x)\theta_x = \sum_{x\in G} p_t(x_0,x)p_t(x,x_0)\theta_x = p_{2t}(x_0,x_0) \leq \frac{1}{\theta_{x_0}} \end{equation*}

Is it the case that $\mathcal{L}_\theta u$ is always in $L^2(\theta)$ also?

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I was able to answer this question, although it didn't turn out to be useful in the way I thought it would be.

Let $P^\theta_t$ be the transition operator, $(P^\theta_tf)(x) = \sum_{y\in G} p_t(x,y)f(y)\theta_y$. Then a fairly easy computation shows that $P_t$ maps $L^2(\theta)$ to $L^2(\theta)$, and one can also show that if $v\in C_c(G)$, then $P_t(\mathcal{L}\_\theta v) = \mathcal{L}_\theta (P_tv)$, basically from a self-adjointness calculation. If one applies this to the function $u=\theta^{-1}_{x_0}\delta_{x_0}$, then we get $\mathcal{L}\_\theta u = \mathcal{L}_\theta (P_tv) = P_t(\mathcal{L}_\theta v)$, and the right-hand side of this is in $L^2(\theta)$.

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