Lipschitz concentration for standard Gaussian measure

First suppose $f$ is smooth and bounded with bounded derivatives. For $\lambda >0$, apply [F1] to $g=e^{\lambda f}$. With $\psi (\lambda )=\log E{e}^{\lambda (f(Z)-Ef(Z))}$, the entropy identity and $\mathrm{\mid }\mathrm{\nabla }f\mathrm{\mid }\le L$ give $\lambda {\psi }^{\mathrm{\prime }}(\lambda )-\psi (\lambda )\le {\lambda }^2{L}^2\mathrm{/}2$.

Assume $L>0$. The function $q(\lambda )=\psi (\lambda )\mathrm{/}\lambda$ extends continuously to $q(0)=0$ because $\psi (0)={\psi }^{\mathrm{\prime }}(0)=0$. Its derivative on $(0,\mathrm{\infty })$ is $q^{\mathrm{\prime }}(\lambda )=(\lambda {\psi }^{\mathrm{\prime }}(\lambda )-\psi (\lambda ))\mathrm{/}{\lambda }^2\le L^2\mathrm{/}2$ by step 1.1. The mean value theorem [F5] on $[0,\lambda ]$ gives $\psi (\lambda )\le {\lambda }^2{L}^2\mathrm{/}2$. Markov's inequality [F6] therefore gives $P(f(Z)-Ef(Z)\ge r)\le e^{-\lambda r+{\lambda }^2{L}^2\mathrm{/}2}$; choosing $\lambda =r\mathrm{/}{L}^2$ proves the displayed bound for this $f$.

Suppose $L>0$. For a general Lipschitz $f$, clip its values to $[-n,n]$ to obtain a bounded $L$-Lipschitz function $f_n$. For $s>0$ set $P_s{f}_n(x)=\int f_n(e^{-s}x+\sqrt{1-e^{-2s}}y)d{\gamma }_d(y)$. Equivalently, write this as an integral against the smooth Gaussian density ${\rho }_s(y-e^{-s}x)$. Every derivative of the kernel in $x$ is a polynomial times a Gaussian density; on compact $x$-sets it has an integrable Gaussian majorant, so [F4] permits differentiation under the integral to every order. The global $L^1$ norm of each kernel derivative is finite, hence the derivatives of $P_s{f}_n$ are bounded by $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}$ times finite constants. Using the same integration variable for two points shows $P_s{f}_n$ is $L$-Lipschitz, so step 2.1 applies. For each fixed $n$, $P_s{f}_n(x)\to f_n(x)$ as $s\downarrow 0$, and then $f_n(x)\to f(x)$ as $n\to \mathrm{\infty }$. Throughout, $\mathrm{\mid }{P}_s{f}_n(x)\mathrm{\mid }\le \mathrm{\mid }f(0)\mathrm{\mid }+L\mathrm{\mid }x\mathrm{\mid }+L\int \mathrm{\mid }y\mathrm{\mid }d{\gamma }_d(y)$. For each fixed $\lambda$, $\lambda L\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}4+(\lambda L{)}^2$ and the density in [F2] give an integrable Gaussian majorant for $e^{\lambda \mathrm{\mid }{P}_s{f}_n(x)\mathrm{\mid }}$. Dominated convergence [F3] passes the means and exponential moments first as $s\downarrow 0$ and then as $n\to \mathrm{\infty }$, proving the bound for $f$.

If $L=0$, then $f$ is constant on ${\mathbb{R}}^d$, so $P\{f(Z)\ge Ef(Z)+r\}=0$ for every $r>0$. This handles the zero-Lipschitz case and completes the theorem.