# Continuity of the Green function with respect to the measure

Let $$G$$ be a finitely generated group and let $$\mu$$ be a finite measure on $$G$$. Define the Green function as $$G(\mu)=\sum_{n\geq 0}\mu^{*n}(e),$$ where $$\mu^{*n}$$ is the $$n$$th convolution power of $$\mu$$. Usually, the Green function is defined for probability measures $$\mu$$, but it can be defined in more general settings, a typical example begin the measure $$r\mu$$, where $$\mu$$ is a probability measure and $$r$$ is a positive number below the spectral radius of $$\mu$$.

Of course, $$G(\mu)$$ can be infinite. We are interested in the following on measures $$\mu$$ such that $$G(\mu)<+\infty$$. To simplify things, we also restrict ourselves to finitely supported measures. Precisely, let $$S$$ be a finite set generating $$G$$ as a semi-group. Then, let $$P(S)^+=\{\mu \text{ finite measure whose support is exactly }S, G(\mu)<+\infty\}.$$ Endow the set of finite measures with the topology of pointwise convergence, that is $$\mu_n$$ converges to $$\mu$$ if and only if for every $$g\in G$$, $$\mu_n(g)$$ converges to $$\mu(g)$$.

Is the function $$G(\mu)$$ continuous in $$\mu$$ on $$P(S)^+$$ ?

• Is $\mu$ a signed measure, or a non-negative measure? (Not sure if this affects the answer in any way, just asking.) – Mateusz Kwaśnicki Jul 12 at 10:57
• @MateuszKwaśnicki Well I had implicitly in mind a non-negative measure, but you're right it might not change the result nor the proof – M. Dus Jul 12 at 22:10
• Since this has not received any answers yet, let me spell it out that (a) I find the question quite interesting, (b) by Fatou's lemma, $G(\mu)(g)$ is clearly lower semi-continuous; (c) I fail to find a counter-example. – Mateusz Kwaśnicki Jul 18 at 17:42
• @MateuszKwaśnicki Thanks a lot. Discussing it with a friend, we've made some progress. Proposition 6.3 in Mathieu and Sisto's paper (arxiv.org/abs/1411.7865) shows Lipschitz continuity of the Green distance $d_G(e,x)$ with respect to both $\mu$ and the word distance $d(e,x)$. One might be able to adapt their proof to show Lipschitz continuity of $G(\mu)$ with respect to $\mu$. However, they assume that (a) the group is non-amenable and (b) $\mu$ is a probability measure. In particular, it seems it could not be applied to $r\mu$, where $r$ is below the spectral radius of $\mu$... – M. Dus Jul 24 at 12:16