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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.

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To answer my own question, this follows from a more general result that is mentioned in "On measure concentration of vector valued maps" 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.)

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It would be good of you to cite the source directly in the text of the answer. (The link does not work and if it did, it would become unhelpful if it ever broke.) –  cardinal Apr 15 '12 at 23:39
    
Thanks, link fixed (and I've also written the authors, etc.) –  Joe Neeman Apr 16 '12 at 5:18

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