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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 $P(S)^+$ ?

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    $\begingroup$ Is $\mu$ a signed measure, or a non-negative measure? (Not sure if this affects the answer in any way, just asking.) $\endgroup$ – Mateusz Kwaśnicki Jul 12 at 10:57
  • $\begingroup$ @MateuszKwaśnicki Well I had implicitly in mind a non-negative measure, but you're right it might not change the result nor the proof $\endgroup$ – M. Dus Jul 12 at 22:10
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    $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki Jul 18 at 17:42
  • $\begingroup$ @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 $\mu$... $\endgroup$ – M. Dus Jul 24 at 12:16

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