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Let me state a standard result first. Let a $A\subset \mathbb{R}^d$ be a set of fixed volume. Define $A_t$ to be the set of all points at distance at most $t$ from $A$. Then the volume of $A_t$ is minimal if $A$ is a ball of the prescribed volume.

Another way to define $A_t$ is by $A_t=A+B(0,t)$, where $B(0,t)$ is the centered ball of radius $t$. We shall think of it as the union of translates of $A$ by all vectors in $B(0,t)$.

I am interested in extending such a result to the discrete setting. Say, we translate $A$ only in the $d$ orthogonal directions. That is, we look at the union $U(A)=\cup_v (A+v)$, where $v$ is either the zero vector or $\pm e_i$, where $e_i$ is an element of the standard orthonormal basis.

Given that the volume of $A$ is fixed, which $A$ minimize the volume of $U(A)$?

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  • $\begingroup$ What do you mean by minimizing the volume? Minimize with respect to the volume of $A$? Or, you mean that $A_t$ is made minimal by a ball, for all sets $A$ with a fixed volume? I guess the latter makes the most sense. Some other comments: Start with a lattice in the plane, and explore. What if you take the triangular lattice instead, or the hexagonal? $\endgroup$
    – Per Alexandersson
    Commented Nov 19, 2015 at 15:40
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    $\begingroup$ What I meant by minimizing the volume of $A_t$ was indeed minimizing with respect to all sets $A$ of fixed given volume (this was said in the first paragraph). $\endgroup$
    – TOM
    Commented Nov 19, 2015 at 16:43

1 Answer 1

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Fix large $N$. Take $A$ to be union of $\varepsilon$-balls which centers have integer coordinates between $\pm N$. (You have to ajust $\varepsilon$ to get the needed volume.)

In this case $U(A)$ is a union $\varepsilon$-balls which centers have integer coordinates between $\pm (N+1)$. Therefore $\mathrm{vol}[U(A)]$ can be made arbitrary close to $\mathrm{vol}A$ and there is no minimizer.

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  • $\begingroup$ Thank you! I guess then one cannot get away with finitely many directions for shifts and get a non-trivial answer. $\endgroup$
    – TOM
    Commented Nov 19, 2015 at 16:48
  • $\begingroup$ @TOM, things might change if you assume convexity. $\endgroup$
    – Anton Petrunin
    Commented Nov 19, 2015 at 18:37
  • $\begingroup$ 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. $\endgroup$
    – TOM
    Commented Nov 19, 2015 at 19:01
  • $\begingroup$ @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. $\endgroup$
    – Bruno Le Floch
    Commented Nov 19, 2015 at 19:59
  • $\begingroup$ @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? $\endgroup$
    – TOM
    Commented Nov 19, 2015 at 21:41

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