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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.

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

4 Answers 4

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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!

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

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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.

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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.

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