TITLE: $e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)\to f(x)$ for $f$ continuous and bounded
QUESTION [19 upvotes]: Let $f:\mathbb R\to\mathbb R$ be continuous and bounded. Prove that for each $x>0$ we have
$$f(x)=\lim_{n\to\infty}\left(e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)\right).$$

When $x$ is an integer we have
$$e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)=\frac{e^{-nx}}{(nx)!}\cdot\sum_{k=0}^\infty\binom{nx}{k}(nx-k)!(nx)^k.$$
I then substitute the gamma function integral formula for the factorial, swap the order of $\sum$ and $\int$ and apply the binomial theorem to get that this is equal to
$$\frac{e^{-nx}}{(nx)!}\int_0^\infty(nx+y)^{nx}e^{-y}\,dy.$$
This doesn't look easy to solve. When $x$ is not an integer, I am having even more difficulties.
I would appreciate any help.

REPLY [5 votes]: For a complex exponential $h(x)=e^{sx}$ it works:
$$e^{-nx}\sum_{k\ge 0} \frac{(nx)^k}{k!} h(\frac{k}n) = e^{-nx} e^{nxe^{s/n}}=
e^{sx}+O(1/n)$$
Then look at $g(y) = e^{-y} \sin(y)$. There is a unique $a > 0$ where $g(a) = \sup_{y \ge 0} |g(y)|$.
Letting $G(y)=g(\frac{ya}{x})$ we'll have $$
\lim_{n\to \infty} \sum_{k\ge 0}\frac{e^{-nx}(nx)^k}{k!} G(\frac{k}{n})=G(x)$$
$G$ attains its maximum only at $x$  and $\sum_{k\ge 0}\frac{e^{-nx}(nx)^k}{k!}=1$, which implies that for every $\epsilon >0$, $$\lim_{n\to \infty}\sum_{k=0, \ |\frac{k}n-x|<\epsilon }^\infty \frac{e^{-nx}(nx)^k}{k!}=1$$
From which it is easy to conclude.