Martingale theory to show f(x+s) = f(x)

Assume f is a bounded continuous function on $\mathbf{R}$ and $X$ is a random variable with distribution $F$. Assume for all $x \in \mathbf{R}$ that
f(x) = \int_\mathbf{R}f(x+y)F(dy)

Please help conclude that $f(x+s) = f(x)$ where $s$ is any value in the support of $F$. The hints that I have come across are to use Martingale theory and consider $\{ X_n\}$ to be i.i.d. with distribution $F$ and make a martingale with some function of $S_n = \sum_{j=1}^nX_j$.


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Hint: Let $M_n=f(x+S_n)$. Then $(M_n)$ is a martingale. To wit, $x+S_{n+1}=x+S_n+X_{n+1}$ where $x+S_n$ is measurable with respect to $\mathcal F_n$ and $X_{n+1}$ is independent of $\mathcal F_n$. By the standard properties of conditional expectation (integrate that which is independent, leave out that which is measurable),
\mathbb E(M_{n+1}\mid\mathcal F_n)=g(x+S_n),\qquad g:y\mapsto\mathbb E(f(y+X_{n+1})).
It happens that a hypothesis in your post implies that $g=f$… Hence $\mathbb E(M_{n+1}\mid\mathcal F_n)=M_n$, and $(M_n)$ is indeed a martingale with respect to the filtration $(\mathcal F_n)$. Furthermore, $f$ is bounded hence $(M_n)$ is bounded.
It also happens that every lecture/book/set of notes on the subject mentions proeminently a convergence theorem about bounded martingales, which you can apply here.

To say more would be to provide you a full solution which you could then copy verbatim and hand out as fulfillment of your homework (since this is homework, ain’t it?), without understanding anything in it–and we do not want this to happen, do we?