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Nov 19, 2015 at 21:46 comment added Bruno Le Floch @TOM I'm changing $A$ to be some union of intervals fattened by $\epsilon$. I should have called it $B$ probably to limit confusion. So $B$ is essentially the image of Anton Petrunin's $A$ under your operation. Acting a second time with your operation will not change the volume much.
Nov 19, 2015 at 21:44 comment added TOM Just to make thing clearer - I assumed that $\varepsilon$ is the volume of one little ball and not the radius.
Nov 19, 2015 at 21:41 comment added TOM @BrunoLeFloch : I do not quite agree, the unions of shifts of one ball by a line segment of length $l$ will have length $\approx l\varepsilon$ and the whole union of shifts $\approx 2ldN^2\varepsilon$, which is quite different from the measure of $A$, which is like $N^2\varepsilon$. Or am I saying something silly?
Nov 19, 2015 at 19:59 comment added Bruno Le Floch @TOM: still, take $A$ to be the union of $\epsilon$-balls whose centers have all coordinates integer except one, all coordinates between $\pm N$. In other words the centers are along edges between neighboring integer-coordinate points considered by Anton Petrunin. Then the shifted $A$ will essentially have the same form, but with $N\to N+1$, and again there is no minimizer.
Nov 19, 2015 at 19:01 comment added TOM Perhaps the "closer" way of modifying the question is the following. So by adding a ball to the set I extend it by a segment of a fixed length in every direction. What about changing the discrete shifts to shifts in the same 2d directions by a segment of fixed length? That seems to at least kill your couterexample for the discrete question.
Nov 19, 2015 at 18:37 comment added Anton Petrunin @TOM, things might change if you assume convexity.
Nov 19, 2015 at 16:48 comment added TOM Thank you! I guess then one cannot get away with finitely many directions for shifts and get a non-trivial answer.
Nov 19, 2015 at 16:46 vote accept TOM
Nov 19, 2015 at 15:58 history answered Anton Petrunin CC BY-SA 3.0