= Solution
Put $Y=g(X)-\mathbb Eg(X)$ and $H(\lambda)=\log\mathbb Ee^{\lambda Y}$. Apply the assumed Poisson log-Sobolev inequality to $f=e^{\lambda g}$. Since $|Dg|\leq1$ and $|e^a-1|\leq\lambda e^\lambda$ for $|a|\leq\lambda$,
$$
\frac{|D(e^{\lambda g})(x)|^2}{e^{\lambda g(x)}}
=e^{\lambda g(x)}|e^{\lambda Dg(x)}-1|^2
\leq\lambda^2e^{2\lambda}e^{\lambda g(x)}.
$$
After division by $\mathbb Ee^{\lambda g}$, the entropy bound becomes
$$
\lambda H'(\lambda)-H(\lambda)\leq C\lambda^2e^{2\lambda}.
$$
Because $(H(\lambda)/\lambda)'=(\lambda H'-H)/\lambda^2$ and $H'(0)=0$, integration from $0$ to $\lambda$ gives
$$
H(\lambda)\leq\psi(\lambda)
:=\frac{C\lambda}{2}(e^{2\lambda}-1).
$$
The <Chernoff bound> and optimization over $\lambda\geq0$ therefore yield
$$
\mathbb P(Y\geq t)
\leq\exp\left\{-\sup_{\lambda\geq0}
[\lambda t-\psi(\lambda)]\right\}
=e^{-\psi^*(t)},
$$
where $\psi^*$ is the <Legendre transform of a cumulant-generating function>, also called the Chernoff-Cramér transform.
Back to article page
