Given $f$ is a bounded continuous function on $\mathbf{R}$ and $\mu$ is a probability measure such that for all $x \in \mathbf{R}$ $ f(x) = \int_\mathbf{R}f(x+y)\mu(dy) $
Please help to show that $f(x+s) = f(x)$ for all $s$ in the support of $\mu$. (The support of $\mu$ is the smallest set $E$ s.t. $\mu(E)=1$.) The hint is that we can try to use martingales to show $E[f(x+X_0)-f(x)]^2=0$, where $X_0$ has distribution $\mu$.
So far I've constructed a martingale $M_n=f(x+S_n)$. Here $S_n = \sum_{j=1}^nX_j$ and $\{ X_j\}$ are i.i.d. with distribution $\mu$. But I've no idea on how to connect this martingale with the previous hint.
I came cross another post about the same question on the forum, but it ends with the martingale that I stated here. So I'm wondering if anyone can provide some further insights about how to finish the proof. Since this is a hw problem, I've been working on it for three days only to get to this point. I sincerely appreciate any inputs that people will give.
Thanks a ton!