If $f: \mathbb{R}^n \to \mathbb{R}$ is a Lipschitz function and $X$ is a standard $n$-dimensional Gaussian vector with $\mathbb{E} f(X) = 0$, then $f(X)$ is subgaussian (in a way that does not depend on $n$). If $f$ is $\mathcal{C}^1$, this is equivalent to saying that $|\nabla f|$ bounded implies $f(X)$ is subgaussian.


There seem to be two natural generalizations of this. The first is to ask for weaker bounds on $|\nabla f|$. For example, if $|\nabla f|$ is subgaussian, then $f$ should be subexponential. The second generalization concerns functions $f: \mathbb{R}^n \to \mathbb{R}^k$. If I want to control $|f|$ independently of $k$, it is no longer enough to assume that $f$ is Lipschitz, since for the function $f(x) = (x\_1, \dots, x\_k)$, $|f|$ concentrates around $\sqrt k$. The natural condition seems to be a bound on the Frobenius norm of $D f$ (the matrix of partial derivatives).


The following statement contains both generalizations simultaneously (and is not hard to prove): If $f: \mathbb{R}^n \to \mathbb{R}^k$ is continuously differentiable and $\mathbb{E} f(X) = 0$ then
$$
\big(\mathbb{E} |f(X)|^p\big)^{1/p} \le c \sqrt p \big(\mathbb{E} \|Df\|\_F^p\big)^{1/p}.
$$


My question is whether a statement like this is known and (if so) where I can find a reference.



Concentration of Gaussian vectors
To answer my own question, this follows from a more general result that is mentioned in ["On measure concentration of vector valued maps"](http://webmail.impan.gov.pl/cgi-bin/ba/pdf?ba55-3-07) by Ledoux and Oleszkiewicz, Theorem 4: for any convex function $\Psi: \mathbb{R}^k \to \mathbb{R}$,
$$
\mathbb{E} \Psi(f(X)) \le \mathbb{E} \Psi(\frac{\pi}{2} Y \cdot Df(X))
$$
where $X$ and $Y$ are independent standard Gaussians. If you condition the right hand side on $X$ and integrate $Y$, a standard result on the moments of order-2 Gaussian chaos gives
$$
\mathbb{E} (\frac{\pi}{2} Y \cdot Df(X))^p \le (cp)^{p/2} \mathbb{E} \|Df\|\_F^p
$$
which is what I claimed above. (By following the references a little more carefully, you can even get the sharp constant.)


