The similar and more general question is asked [here](https://mathoverflow.net/questions/334335/lipschitz-function-of-independent-subgaussian-random-variables?noredirect=1&lq=1), whose setting is random vectors.


Let $X$ be $\sigma$-sub-Gaussian and $f$ is a Lipschitz function w.r.t. constant $L$. How to prove can the subgaussian moment of $f(X)$ be bounded in terms of $\sigma$ and $L$?


It might not work on random vectors but I think it works on random variable.



Lipschitz function of subgaussian random variable
$\newcommand{\si}{\sigma}\newcommand\R{\mathbb R}$One of mutually equivalent definitions of a ($\si$)-sub-Gaussian random variable (r.v.) $X$ is as follows:
\begin{equation\*}
 Ee^{X^2/\si^2}\le2. \tag{1}\label{1}
\end{equation\*}


It is now clear that the answer to the question is no. Indeed, suppose that $X\sim N(0,1)$, so that $X$ is $\si$-sub-Gaussian for some universal positive real constant $\si$. For any real $a$ and $x$, let now $f\_a(x):=x+a$, so that $f\_a$ is $L$-Lipschitz for $L=1$. Then
for any real $b>0$ we have $Ee^{f\_a(X)^2/b^2}\to\infty$ as $a\to\infty$, so that the sub-Gaussian norm of $f\_a(X)$ cannot be bounded in terms of $\si$ and $L$.


However, the sub-Gaussian norm of $f(X)-Ef(X)$ **can** be bounded in terms of $\si$ and $L$, as follows. Let $Y$ be an independent copy of $X$. Then, for any real $c>0$, by Jensen's inequality for the convex function $u\mapsto\exp\frac{(t-u)^2}{c^2}$,
\begin{equation\*}
\begin{aligned}
 E\exp\frac{(f(X)-Ef(X))^2}{c^2}&=E\exp\frac{(f(X)-Ef(Y))^2}{c^2} \\ 
 &=\int\_\R P(X\in dx)\exp\frac{(f(x)-Ef(Y))^2}{c^2} \\ 
 &\le\int\_\R P(X\in dx)E\exp\frac{(f(x)-f(Y))^2}{c^2} \\ 
 &=E\exp\frac{(f(X)-f(Y))^2}{c^2} \\ 
 &\le E\exp\frac{L^2(X-Y)^2}{c^2} \\ 
 &\le E\exp\frac{2L^2X^2+2L^2Y^2}{c^2} \\ 
 &=\Big(E\exp\frac{2L^2X^2}{c^2}\Big)^2 \\ 
 &\le E\exp\frac{4L^2X^2}{c^2}\le2 
\end{aligned}
\end{equation\*}
by \eqref{1} if
\begin{equation\*}
 c=2L\si.
\end{equation\*}
So, the sub-Gaussian norm of $f(X)-Ef(X)$ is $\le2L\si$, if $\si$ is the sub-Gaussian norm of $X$. $\quad\Box$


