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

• Where did $f(k/n)$ go in your summation?
– Ian
Jan 1 at 18:16
• @Ian looks like I forgot to copy it over, which means I can't apply binomial theorem as I did... Jan 1 at 18:19

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!

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 $$\sum_{k : |k/n-x| \geq \delta} e^{-nx} \frac{(nx)^k}{k!}$$.

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.

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.