# Quantitative Version of Jensen's Inequality?

Hi,

I've been looking at situations where Jensen's inequality is almost tight, and found myself proving a lemma that I'm nearly certain exists somewhere in the literature.

The specifics are as follows:

Suppose we have some convex, increasing function $f(x)$ and a set of $n$ real numbers $x_i$. Define $$\delta := \frac{\sum f(x_i)}{ n } -f\left(\frac{\sum x_i}{n}\right)$$ We know that $\delta$ is positive by Jensen's, and that it is zero when all the $x_i$'s are equal to the average. Let $\delta$ be positive, but small. We now fix an epsilon, and ask how many $x_i$'s are either greater than $\frac{\sum x_i}{n}(1 + \epsilon)$ or smaller than $\frac{\sum x_i}{n}(1 - \epsilon)$. If we call that set $I$, the lemma would state that $$|I| \leq g(\delta, \epsilon) n$$ for $g$ continuous, vanishing as delta goes to zero for any fixed $\epsilon$, and depending only on the choice of $f$. What this shows is that if the Jensen's "deficit" is small, then the number of entries that are "far away" from the average is $o(n)$.

Is this some well known (or even not well known, but existent...) lemma?

Thanks!

-Matan

EDIT: Made a silly mistake in defining $\delta$ - the body should now contain the correct normalization (Thanks Daniel!)

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You could try to look for things pertaining to strict convexity, which has applications in the Calculus of Variations. A useful inequality to this end is $f(\theta x+(1−\theta)y)+c|x-y|^2 \leq \theta f(x) +(1-\theta)f(y)$. From this perspective it makes more sense to look at $\delta$ as the difference of the average of the values and the value of the average. – Daniel Spector 0 secs ago – Daniel Spector Jan 25 '13 at 15:39
There appears to be some literature on defect versions of Jensen's inequality, e.g. papers.ssrn.com/sol3/papers.cfm?abstract_id=2027471 or ams.org/mathscinet-getitem?mr=2274968 – Terry Tao Jan 25 '13 at 19:22
Daniel, could you plz let me know where that inequality come from? any reference? – math-Student Jan 25 '14 at 18:18

This is not an answer but it may be helpful to know that there exists the notion of modulus of convexity of a convex function $f:X\to ]-\infty,\infty]$ defined on a Banach space $X$ which quantifies how convex a convex function is. It is defined as $$\delta_f(t) = \inf\{\tfrac{1}{2}f(x) + \tfrac12 f(y) - f(\tfrac{x+y}{2})\ :\ \|x-y\|=t \}.$$ If $\delta_f(t)>0$ for $t>0$ then $f$ is uniformly convex, if $\delta_f(t)>Ct^p$ for some $C>0$, then $f$ one says that $f$ has a modulus of convexity of power type $p$.