We assume that for all $d \geqslant 0$, $\mathbb{P}(X \in d\mathbb{Z}) < 1$. We consider a function $h$ uniformly continuous and bounded on $\mathbb{R}$ such that for all $x \in \mathbb{R}$, $h(x) \leqslant h(0)$ and $$h(x) = \sum_{i=0}^{+\infty} p_i h\left(x - x_i\right)$$ We recall that for all $x \in \mathbb{R}$ and $n \in \mathbb{N}$, $h(x) = \mathbb{E}\left(h\left(x - S_n\right)\right)$. Show that for all $n \in \mathbb{N}$ and $x \geqslant 0$ such that $\mathbb{P}\left(S_n = x\right) > 0$, we have $h(-x) = h(0)$.
We assume that for all $d \geqslant 0$, $\mathbb{P}(X \in d\mathbb{Z}) < 1$. We consider a function $h$ uniformly continuous and bounded on $\mathbb{R}$ such that for all $x \in \mathbb{R}$, $h(x) \leqslant h(0)$ and
$$h(x) = \sum_{i=0}^{+\infty} p_i h\left(x - x_i\right)$$
We recall that for all $x \in \mathbb{R}$ and $n \in \mathbb{N}$, $h(x) = \mathbb{E}\left(h\left(x - S_n\right)\right)$. Show that for all $n \in \mathbb{N}$ and $x \geqslant 0$ such that $\mathbb{P}\left(S_n = x\right) > 0$, we have $h(-x) = h(0)$.