$\newcommand{\Ent}{\mathrm{Ent}}$

• Taken from Thm 7.4.1 in the book “Sur les inégalités de Sobolev logarithmiques”.

We first assume that $F$ is a smooth and bounded function such that $\|F\|_{\mathrm{Lip}} \leq 1$. Consider the Laplace transform of $F$: $$ H(\lambda) := \mathbb{E}_\mu\left(e^{\lambda F}\right), \quad \lambda > 0. $$

Apply the logarithmic Sobolev inequality to the function $f = e^{\lambda F / 2}$. Then: $$ \Ent_\mu(e^{\lambda F}) \leq c \, \mathbb{E}_\mu\left( \left| \nabla e^{\lambda F/2} \right|^2 \right). $$ Since $\nabla e^{\lambda F / 2} = \frac{\lambda}{2} e^{\lambda F / 2} \nabla F$, we get: $$ \left| \nabla e^{\lambda F / 2} \right|^2 = \frac{\lambda^2}{4} e^{\lambda F} |\nabla F|^2. $$ Because $\|\nabla F\|_\infty \leq 1$, it follows that: $$ \Ent_\mu(e^{\lambda F}) \leq \frac{c \lambda^2}{4} H(\lambda). $$

Also, we have the identity: $$ \Ent_\mu(e^{\lambda F}) = \int e^{\lambda F} \log \frac{e^{\lambda F}}{\int e^{\lambda F} \, d\mu} \, d\mu = \lambda H'(\lambda) - H(\lambda) \log H(\lambda), $$ which gives: $$ \lambda H'(\lambda) - H(\lambda) \log H(\lambda) \leq \frac{c \lambda^2}{4} H(\lambda). $$

Dividing by $\lambda^2 H(\lambda)$ (which is valid since $H(\lambda) > 0$), we obtain: $$ \frac{H'(\lambda)}{\lambda H(\lambda)} - \frac{\log H(\lambda)}{\lambda^2} \leq \frac{c}{4}. $$

Define: $$ K(\lambda) := \frac{1}{\lambda} \log H(\lambda), \quad \lambda > 0. $$

Then we have: $$ K'(\lambda) \leq \frac{c}{4}. $$

The function $K$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Taking the limit as $\lambda \to 0^+$ gives: $$ \lim_{\lambda \to 0^+} K(\lambda) = \mathbb{E}_\mu(F). $$

Therefore, for all $\lambda > 0$: $$ K(\lambda) \leq \mathbb{E}_\mu(F) + \frac{c}{4} \lambda, $$ that is: $$ \log H(\lambda) \leq \lambda \mathbb{E}_\mu(F) + \frac{c}{4} \lambda^2, $$ and hence: $$ H(\lambda) \leq \exp\left( \lambda \mathbb{E}_\mu(F) + \frac{c}{4} \lambda^2 \right). $$

Now apply Markov’s inequality: $$ \mu(F \geq \mathbb{E}_\mu(F) + r) = \mu\left(e^{\lambda F} \geq e^{\lambda (\mathbb{E}_\mu(F) + r)} \right) \leq \frac{H(\lambda)}{e^{\lambda (\mathbb{E}_\mu(F) + r)}}. $$

Using the previous bound on $H(\lambda)$, we get: $$ \mu(F \geq \mathbb{E}_\mu(F) + r) \leq \exp\left( \frac{c}{4} \lambda^2 - \lambda r \right). $$

Optimizing this upper bound by choosing $\lambda = \frac{2r}{c}$ gives: $$ \mu(F \geq \mathbb{E}_\mu(F) + r) \leq \exp\left( -\frac{r^2}{c} \right). $$

This proves the first inequality. The second follows by applying the same bound to $-F$ and combining both tails: $$ \mu(|F - \mathbb{E}_\mu(F)| \geq r) \leq \mu(F \geq \mathbb{E}_\mu(F) + r) + \mu(F \leq \mathbb{E}_\mu(F) - r) \leq 2 e^{-r^2 / c}. $$

Finally, the general case where $F$ is only Lipschitz (not smooth and bounded) is obtained by standard mollification (i.e., convolution with a smooth kernel) and a limiting argument using the semicontinuity of the Lipschitz norm and uniform upper bounds.