The isodiametric inequality tells us that, of all sets of diameter $r$, the one with the largest Lebesgue measure is the ball of radius $r/2$ - and this holds regardless of norm. Let $\tau(r)$ be the measure of that ball. I now have another set $S$ of diameter at most $r$, but its area is $\tau(r) (1- \epsilon)$, for some very small $\epsilon$.

I am interested in showing that this set is "almost" a ball, in the sense that there exists a ball $B$ such that the Lebesgue measure of the symmetric difference $B \triangle S$ is bounded above by some $f(\epsilon)$ that is also small.

I feel like this is some kind of application of Brunn-Minkowski or some other classical theorem, but I just can't seem to find it. Even if this can be done for the Euclidean norm on $\mathbb{R}^2$, it would be a great help [for the record, this is rather simple in the $\ell^\infty$ norm - just draw squares around the leftmost, rightmost, topmost, and bottommost points in $S$, and use the measure constraints. Dealing with a curved geometry requires a more subtle approach].

Thank you all very much!


EDITED TO ADD: I'm not sure if this is required, but I'll also be fine if $S$ were restricted with rather strong regularity conditions - for example, connected, convex, $C^1$ boundary, or other such conditions.

  • $\begingroup$ I would introduce in the discussion, besides the diameter $\delta(A)$ of $A$, the "radius" $\rho(A)$ defined as the infimum of radii of balls containing $A$ (I think it corresponds to a unique minimal closed ball containing $A$, in the case of strictly convex norms). Then one may try to prove this lemma: If $\delta(A)=2$ and $\rho(A)\ge r > 1$, then $|A| \le |B(0,1)|(1−\epsilon(r))$ for some $0 < \epsilon(r)=o(1)$ as $r\to 1$. $\endgroup$ – Pietro Majer Nov 24 '12 at 23:32
  • $\begingroup$ Therefore, as a consequence: if $\delta(A)=2$ and $|A| > |B(0,1)|(1-\epsilon(r))$, then $\rho(A) < r$, that is $A\subset B(x,r)$ for some point $x$. Thus $|A\Delta B(x,1)|\le |B(x,r)\setminus A|+|B(x,r)\setminus B(x,1)|\le |B(0,1)|(2r^2-2+\epsilon(r))$. $\endgroup$ – Pietro Majer Nov 24 '12 at 23:42
  • $\begingroup$ A question: the existence of such function $f(\epsilon)$ is clear and what you want is a concrete function $f$, isn't it? $\endgroup$ – Pietro Majer Nov 25 '12 at 0:04
  • $\begingroup$ Pietro, thanks! I didn't think of this formulation. The lemma you suggested will definitely do it, though I guess I traded one problem for another... To answer your question, existence is sufficient, though making sure it vanishes polynomially fast in $\epsilon$ would be a plus! -Matan $\endgroup$ – matan.harel Nov 25 '12 at 2:23
  • $\begingroup$ See also the work of Fusco, Maggi, Pratelli: annals.math.princeton.edu/2008/168-3/p06 annals.math.princeton.edu/wp-content/uploads/… $\endgroup$ – Daniel Spector Nov 25 '12 at 9:28

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