# One step forward, two steps back

Let $f:\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function. Does there existe $g:\mathbb R\to\mathbb R$ also infinitely differentiable such that $f(x) = g(x+1)-g(x)$ for all $x\in\mathbb R$?

1. It is true for polynomials. Let $V=\{f:\mathbb{R}\to\mathbb{R}| \text{ there exist } g \in C^\infty(\mathbb{R},\mathbb{R}) \text{ such that } f(x)=g(x+1)-g(x) \text{ for every } x\in\mathbb{R}\}$.
Note that $V$ is a real vector subspace of $C^\infty(\mathbb{R},\mathbb{R})$ and $\mathbb{R}[X]\subset V$ because, for every natural number $n$, $f(x)=(x+1)^n-x^n \in V$ is a polynomial of degree $n-1$, so by induction, it is easy to prove that $x^n \in V$.