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Let $K \subseteq \mathbb R^n$ be a compact convex set with volume $V$, and let $f: K \to [0,1]$ be a convex function with domain $K$. Assume that $\min_{x \in K} f(x) = 0$. I claim that, for every $\epsilon \in [0,1]$, the volume of $\{x \in K : f(x) \leq \epsilon\}$ is at least $\epsilon V$. In other words, the function $f$ is small on a large proportion of its domain.

Is the above true? I think I have a very simple proof, but I'm skeptical of myself because I can't seem to find a reference, and it would simplify some proof of a well-known fact. If it is true, could anyone provide a reference?

UPDATE: HTFB exhibited a nice counter-example, with $K$ the unit ball and $f$ the distance from the origin. There, the volume of $\{x \in K : f(x) \leq \epsilon\}$ is $\epsilon^n V$. I wonder if that is the worst case? I.e. if we can guarantee that the volume of the region in question is at least $\epsilon^n V$.

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    $\begingroup$ I don't think this can be quite right. Take $K$ the closed unit sphere centred on the origin in $\mathbb{R}^2$, and $f$ the distance from the origin. Then the volume of $\{x \in K: f(x) \leq \epsilon\}$ is exactly $\epsilon^2V$. So the dimension of $K$ must feature. $\endgroup$ – HTFB Nov 6 '13 at 12:20
  • $\begingroup$ I see. Thanks. The natural followup question, I suppose, is whether the volume of the region in question is at least $\epsilon^n V$? $\endgroup$ – sd234 Nov 7 '13 at 2:53
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As pointed out by HTFB, the original conjecture is false. My revised conjecture is that the volume of $\{ x \in K : f(x) \leq \epsilon\}$ is at least $\epsilon^n V$, matching HTFB's example. Here is a proposed proof sketch, in which I am only mildly confident.

Assume without loss of generality that $K$ includes the origin, and moreover that the minimum of $f$ is attained there --- i.e. $f(0) = 0$. Let $\epsilon K$ be the result of scaling down $K$ by a factor of $\epsilon$ (i.e. apply the linear tranformation $\epsilon I$ to $K$). The volume of $\epsilon K$ is $\epsilon^n V$. I claim that $f(x) \leq \epsilon$ for every $x \in \epsilon K$. Take $x \in \epsilon K$ and $y \in K$ such that $x = \epsilon y$. Clearly, $x= \epsilon y + (1-\epsilon) 0$. By Jensen's inequality:

$f(x) \leq \epsilon f(y) + (1-\epsilon) f(0) = \epsilon f(y) \leq \epsilon$.

Assuming I'm not missing anything, this should complete the proof.

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