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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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    $\begingroup$ 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 $\endgroup$ Jan 25, 2013 at 15:39
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    $\begingroup$ 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 $\endgroup$
    – Terry Tao
    Jan 25, 2013 at 19:22
  • $\begingroup$ Daniel, could you plz let me know where that inequality come from? any reference? $\endgroup$ Jan 25, 2014 at 18:18

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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 $p$.

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See the paper of Hussain and Pečarić, "AN IMPROVEMENT OF JENSEN'S INEQUALITY WITH SOME APPLICATIONS",

https://www.worldscientific.com/doi/abs/10.1142/S179355710900008X

For any random variable $X$ and concave function $\phi$, $$ \phi(\mathbb{E}\left(X\right))-\mathbb{E}\left(\phi(X)\right) \geq \Biggl|\mathbb{E}\left(\Bigl|\phi(X)-\phi(\mathbb{E}\left(X\right))\Bigr|\right) -\Bigl|\phi_{+}^{'}(\mathbb{E}\left(X\right))\Bigr|\cdot\mathbb{E}\left(\Bigl|X-\mathbb{E}\left(X\right)\Bigr|\right)\Biggr|, $$ where $\phi_{+}^{'}$ denotes the right-hand derivative of $\phi$.

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