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Let $X$ be a separable Banach space, and let $\mathbb P$ be a Radon probability measure on $X$ with zero mean and covariance operator $K : X^* \to X$. Let $x$ be an $X$-valued random variable with distribution $\mathbb P$.

I would like a simple upper bound on the size of $\mathbb E \|x\|^2$ in terms of the operator norm $\|K\|$. In his textbook The Concentration of Measure Phenomenon, Ledoux proves that $$\mathbb E\|x\|^2 \le 4 \|K\|$$ as a consequence of a concentration inequality for a simpler example (where the vectors are sums of vectors with i.i.d. random coefficients $\eta_i$ such that $|\eta_i| \le 1$ a.s.). Because of the boundedness assumption, the argument doesn't quite work in this setting, though I'm certain I could generalize it if I needed to.

Nonetheless, I'm certain that the estimate I'm looking for is buried somewhere in the literature on concentration of measure (and in fact is probably due to Talagrand). Could you please point me in the right direction?

Edit: The inequality as I previously wrote it is incorrect. The correct inequality should be $$\operatorname{Var}(\|x\|) \le 4\|K\|,$$ implying $$\mathbb E\|x\|^2 \le 4\|K\| + \left( \mathbb E\|x\| \right)^2.$$ That is, the size of typical random element could be quite large (i.e. $\mathbb E\|x\| \gg 1$, as in Mark Meckes's example in the comments), but the deviation is only of the order $\|K\|^{1/2}$.

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  • $\begingroup$ Either I'm misinterpreting or something is missing somewhere: if $x$ is, say, a Rademacher sequence in $\ell_2^n$, then the LHS is $n$ and the RHS is 4, right? $\endgroup$
    – Mark Meckes
    Commented Sep 23, 2010 at 23:58
  • $\begingroup$ Mark, I didn't have my copy of Ledoux's book handy as I wrote this post, and your example exactly illustrates the flaw in my statement. The result I was trying to quote is that $$\operatorname{Var}(\|x\|) \le 4\|K\|,$$ so that $$\mathbb E\|x\|^2 \le 4\|K\| + \mathbb E\|x\|.$$ So in words, there is sharp concentration of the norm $\|x\|$ around its mean value $\mathbb E\|x\|$, which may be quite large. $\endgroup$
    – Tom LaGatta
    Commented Sep 24, 2010 at 16:07
  • $\begingroup$ That should be $(\mathbb{E} \Vert x \Vert)^2$ at the end then, right? $\endgroup$
    – Mark Meckes
    Commented Sep 24, 2010 at 16:55
  • $\begingroup$ It absolutely should be. $\endgroup$
    – Tom LaGatta
    Commented Sep 24, 2010 at 17:00

1 Answer 1

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The inequality can't be true without additional assumptions. To see this, let $X = \ell_2^n$ and let $x$ have a spherically symmetric distribution and let $R = \Vert x \Vert$. Then $R$ is an essentially arbitrary nonnegative random variable; indeed we could start by picking $R$ and defining $x=R\Theta$, where $\Theta$ is a uniform random vector in $S^{n-1}$ independent of $R$.

Now $\mathrm{Var}(\Vert x \Vert) = \mathrm{Var} (R)$, and $K = \frac{1}{n} \mathbb{E}(R^2) I_n$, so $\Vert K \Vert = \frac{1}{n} \mathbb{E} (R^2)$. Thus for any fixed distribution of $R$, the inequality fails for sufficiently large $n$.

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  • $\begingroup$ Mark, your example illustrates exactly what was lacking with my question: an independence assumption. I think I see the point of this study now: infinite-dimensional random vectors are essentially sums of independent objects. Thus, functionals which depend similarly on all the components (e.g., the norm) satisfy very strong large deviations estimates. $\endgroup$
    – Tom LaGatta
    Commented Sep 27, 2010 at 16:37

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