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?



Does generator of continuous time random walk map heat kernel from L^2 to L^2?
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 $L^2(\theta)$. 


