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For $S \subseteq \mathbb{R}^n$, let $f(S) = \{p \in \mathbb{R}^n\; |\;\text{d}(p,q) \le 1 \text{ for some }q \text{ in }S\}$.

In the thread

$\qquad$Given a convex set in a normed vector space, take a neighbourhood of it. Is still convex?

it was shown that if $S$ is convex, then so is $f(S)$.

It's easy to show that the converse is false.

But what about this question ...

Does there exist a bounded, nonempty set $S \subset \mathbb{R}^n$ such that the sets $$S, f(S), f(f(S)),...$$ are all non-convex?

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quasi
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1 Answers1

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In $\mathbb{R}^n$, we find that $f^k(S) = \{p \in \mathbb{R}^n \mid \exists q \in S \ d(p,q) \leq k\}$. This is because we can always pick a point on the line between $p$ and $q$ to witness the iterative containment.

In then follows that in $\mathbb{R}$ (so the one-dimensional case), for bounded $S$, some $f^k(S)$ will be an interval, and hence convex (just make $k$ bigger than the radius of $S$).

In higher dimensions, just consider a set with the two points $(0,...,0)$ and $(1,...,0)$. Assuming Euclidean distance, we find that $(0,k,0...,0)$ and $(1,k,0...,0)$ are in $f^k(S)$, but $(0.5,k,0,...,0)$ is not.

Arno
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    Doesn't the maximum norm have the same problem if we take $(1,0)$ and $(0,1)$? Surely, then $(k+1,k)$ and $(k,k+1)$ are both in $f^k(S)$, but $(k+1/2,k+1/2)$ is not. – Milo Brandt Jan 11 '17 at 20:56
  • @Milo Brandt You are right, I hadn't thought that part through, and will delete the remark. – Arno Jan 11 '17 at 21:17