Let f:R→R be continuous and bounded. Prove that for each x&gt;0 we havee−nx⋅∑k=0∞(nx)kk!f(kn))

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Let f:R→R be continuous and bounded. Prove that for each x&amp;gt;0 we havee−nx⋅∑k=0∞(nx)kk!f(kn))

Rosalind Barker · Accepted Answer

Here&#039;s a fun probabilistic proof.Let X1,X2,… be iid Poisson random variables with parameter x&amp;gt;0. Then Sn=X1+…+Xn is Poisson with parameter nx.By the weak law of large numbers, Snn→EX1=x, so Snn→x. Hence for any continuous bounded f, we have E[f(Snn)]→f(x) as n→∞, i.ef(x)=limn→∞(∑k=0∞f(kn)⋅P(Sn=k))=limn→∞(∑k=0∞f(kn)⋅(nx)kk!e−nx),as desired!

Tate Puckett · Accepted Answer

For a complex exponential h(x)=esx it works:e−nx∑k≥0(nx)kk!h(kn)=e−nxenxe1/n=esx+O(1/n)Then look at g(y)=e−ysin⁡(y). There is a unique a&amp;gt;0 where g(a)=supy≥0|g(y)|.Letting G(y)=g(yax) we&#039;ll havelimn→∞∑k≥0e−nx(nx)kk!G(kn)=G(x)G attains its maximum only at x and ∑k≥0e−nx(nx)kk!=1, which implies that for every ϵ&amp;gt;0,limn→∞∑k=0,|kn−x|&amp;lt;ϵe−nx(nx)kk!=1From which it is easy to conclude.

Let f:\mathbb{R}\to\mathbb{R} be continuous and bounded. Prove that for each

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