$e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right)\to f(x)$ for $f$ continuous and bounded

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.

Solution 1:

Here's a fun probabilistic proof.

Let $X_1,X_2,\dots$ be iid Poisson random variables with parameter $x>0$. Then $S_n=X_1+\dots+X_n$ is Poisson with parameter $nx$.

By the weak law of large numbers, $S_n/n\stackrel{\mathbb P}\to\mathbb EX_1=x$, so $S_n/n\stackrel{d}\to x$. Hence for any continuous bounded $f$, we have $\mathbb E[f(S_n/n)]\to f(x)$ as $n\to\infty$, i.e. $$f(x)=\lim_{n\to\infty}\left(\sum_{k=0}^\infty f(k/n)\cdot\mathbb P(S_n=k)\right)=\lim_{n\to\infty}\left(\sum_{k=0}^\infty f(k/n)\cdot\frac{(nx)^k}{k!}e^{-nx}\right),$$ as desired!

Solution 2:

There's a nice one-liner proof using probability theory, since you effectively are looking at $E[f(X/n)]$ where $X$ is Poisson($nx$) distributed. A purely real analysis argument which is really based in the same ideas looks like this.

Fix $x>0$, let $\varepsilon > 0$, find $\delta > 0$ such that $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta$. Write the error as

$$\left | \sum_{k=0}^\infty e^{-nx} \frac{(nx)^k}{k!} (f(k/n)-f(x)) \right |$$

noting that in this step it is crucial that $\sum_{k=0}^\infty \frac{(nx)^k}{k!}=e^{nx}$. Now split the sum based on whether $|k/n-x|<\delta$ and use the triangle inequality. You will be able to use the continuity setup defined above on one piece and the boundedness of $f$ on the other piece. The part that gets slightly technical is finding a bound on $\sum_{k : |k/n-x| \geq \delta} e^{-nx} \frac{(nx)^k}{k!}$.

Solution 3:

Hint : Let $(X_i)$ be a family of i.i.d Poisson variables of parameter $x$.

1. Show that $$e^{-nx}\cdot\sum_{k=0}^\infty\frac{(nx)^k}{k!}f\left(\frac{k}{n}\right) =\mathbb{E} \left(f \left(\frac{X_1 + ... + X_n}{n} \right)\right)$$

2. Conclude using the Law of Large Numbers.