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The entropy method combines a coordinatewise certificate with tensorization of entropy to bound the moment-generating function of a certifiable random quantity and hence its tails.

Let $X_1,\dots ,X_n$ be independent random variables, let $f=f(X_1,\dots ,X_n)$, and let each $f_i$ depend on every coordinate except $X_i$. The modified logarithmic Sobolev inequality states that, for every real $\lambda$ for which the expectations exist,

To prove it, first apply tensorization of entropy:Condition on all coordinates except $X_i$ and use the stated variational formula with the admissible constant $u=e^{\lambda f_i}$. The $i$th conditional entropy is at mostSumming and taking the remaining expectations proves the inequality.

Define the tilted probability measure $Q$ by $dQ=ZdP$; this is normalized because $\mathbb{E}Z=1$. ThenLet ${\mathbb{E}}_i$ average only coordinate $i$, keeping $X^{(i)}$ fixed. The marginal density of $Q_{X^{(i)}}$ relative to $P_{X^{(i)}}$ is ${\mathbb{E}}_{i}Z$, and henceMoreover,Substituting Han's inequality for relative entropy and rearranging gives the tensorization of entropy

Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be independent $\mathrm{Bernoulli}(\lambda /n)$ variables and $S_n={\sum }_i{\epsilon }_i$. Apply the tensorization of entropy to $f(S_n)$ and then apply the stated Bernoulli log-Sobolev inequality in each coordinate. If $S_n^{(i)}=S_n-{\epsilon }_i$, this givesFor each fixed $i$, $(S_n^{(i)},{\epsilon }_i)$ converges in distribution to $(X,0)$, where $X\sim \mathrm{Poisson}(\lambda )$. The Poisson limit theorem in fact gives convergence in total variation. The assumptions $K_1\le f\le K_2$ and $|Df|\le K_3$ make all displayed integrands bounded, so expectations and entropy pass to the limit. Since $n(\lambda /n)(1-\lambda /n)\to \lambda$,

Let $d_i=f(X)-f_i(X^{(i)})$. The self-bounding function assumptions say $0\le d_i\le 1$ and ${\sum }_i{d}_i\le Z$. Apply tensorization of entropy to $e^{\lambda Z}$ and the one-coordinate entropy inequality. Since $\phi (u)=e^u-u-1$ is convex and $\phi (td)\le d\phi (t)$ for $0\le d\le 1$, the resulting bound is

Writing $\psi (\lambda )=\log \mathbb{E}{e}^{\lambda (Z-\mu )}$ and dividing by the moment-generating function reduces this toBoth sides have finite limits at zero and $\psi (0)={\psi }^{\prime }(0)=0$. Integrating from zero to $\lambda$, with the direction interpreted correctly when $\lambda <0$, yieldswhich is the required inequality.