= Solution
Let $\varepsilon_1,\ldots,\varepsilon_n$ be independent $\operatorname{Bernoulli}(\lambda/n)$ variables and $S_n=\sum_i\varepsilon_i$. Apply the <tensorization of entropy> to $f(S_n)$ and then apply the stated Bernoulli log-Sobolev inequality in each coordinate. If $S_n^{(i)}=S_n-\varepsilon_i$, this gives
$$
\operatorname{Ent}(f(S_n))
\leq n\frac{\lambda}{n}\left(1-\frac{\lambda}{n}\right)
\mathbb E\left[
\frac{|Df(S_n^{(i)})|^2}{f(S_n)}\right].
$$
For each fixed $i$, $(S_n^{(i)},\varepsilon_i)$ converges in distribution to $(X,0)$, where $X\sim\operatorname{Poisson}(\lambda)$. The <Poisson limit theorem> in fact gives convergence in total variation. The assumptions $K_1\leq f\leq K_2$ and $|Df|\leq K_3$ make all displayed integrands bounded, so expectations and entropy pass to the limit. Since $n(\lambda/n)(1-\lambda/n)\to\lambda$,
$$
\operatorname{Ent}(f(X))
\leq\lambda\mathbb E\left[\frac{|Df(X)|^2}{f(X)}\right].
$$
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