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Let $f: \mathbb{R}^n \to \mathbb{R}^n$.

What is a necessary and sufficient condition for the following? If $C$ is a convex subset of $\mathbb{R}^n$, then so is $f(C)$.

It's easy to find various sufficient conditions (I won't attempt an exhaustive list here), but I've been unable to find interesting necessary conditions.

If it helps, I'm happy to assume that $C$ is closed and/or bounded.

If a necessary and sufficient condition is known for $\mathbb{R}^n$, can it be extended to general vector spaces? Alternatively, if such a condition is unknown, could someone explain why the problem is difficult?

aduh
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  • Do you want a characterisation of functions $f$ that map all convex sets to convex sets? Because the way you phrased your question, it looks like you want a characterisation for a fixed $C$. But this doesn't make much sense. – TonyK Mar 10 '17 at 17:52
  • And if the answer to my question is yes, I think it is necessary and sufficient that $f$ maps every straight line or line segment to a straight line, a line segment, or a point. – TonyK Mar 10 '17 at 17:56
  • The answer to your question is yes. I will edit for clarity. – aduh Mar 10 '17 at 18:13
  • @TonyK I'm having trouble showing necessity. Let $C$ be a line (segment) such that $f(C)$ is not a line (segment, point). If $n=1$, then it follows that $f(C)$ is not convex. But how to conclude for $n \geq 2$? – aduh Mar 10 '17 at 18:51
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    In full generality this question might be difficult: consider the existence of a space-filling curve from $[0,1]$ to $[0,1]^2$. Certainly all linear (affine) functions satisfy the property, as does any continuous function from $\Bbb R$ to $\Bbb R$. Does anyone have a simple example of another function that satisfies the property? – Greg Martin Mar 10 '17 at 18:56
  • @GregMartin Projections satisfy it (simple, but just a special case of linear maps). Another example I found, not quite as simple, is the "normalization" map $$r \mapsto \frac{1}{\sum_{i \in I} r_i}r,$$ where $I$ is a fixed subset if ${1,...,n }$. – aduh Mar 10 '17 at 19:08
  • @GregMartin But general continuous functions do not satisfy it, convexity is a strong property. I am not sure whether there is a space filling curve such that image of any interval is convex, it may be a good question for the start. On the other hand, if we do not require continuity, every function which maps every open set to the whole $\mathbb R^n$ satisfies it. – Mirek Olšák Mar 10 '17 at 19:11
  • @MirekOlšák: Are you sure about that last sentence? A straight line segment is a convex set; are you claiming that every function which maps every open set to the whole of $\mathbb R^n$ must also map straight line segments to the whole of $\mathbb R^n$? – TonyK Mar 10 '17 at 19:34
  • @TonyK Oh, you are right. I have forgotten about these flat convex sets at the moment. But there are still crazy mappings which map every segment to the whole $\mathbb R^n$. – Mirek Olšák Mar 10 '17 at 21:08
  • Here is a counterexample: Let $f$ be a function that maps every open set to the whole of $\mathbb R^n$. Define $g(x_1,\ldots,x_n) = f(x_1,\ldots,x_n)$ except when $x_1=0$, in which case $g=(0,\lfloor x_2\rfloor,\ldots,0)$. Then $g$ also maps every open set to the whole of $\mathbb R^n$, but the image of the (convex) line $x_1=0$ looks like (non-convex) $\mathbb Z$. – TonyK Mar 10 '17 at 21:33
  • The question becomes interesting if you assume continuity and also that preimages of convex sets are convex. – Moishe Kohan Oct 22 '19 at 22:49

1 Answers1

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The reason there is no necessary condition is that there are quite pathological maps $f$ that satisfy the condition (as discussed in comments). As discussed in this answer, there exists a function $g:\mathbb{R}^n\to \mathbb{R}$ that maps every nontrivial line segment onto $\mathbb{R}$. This can be composed with a surjection $\mathbb{R}\to\mathbb{R}^n$ (or a surjection of $\mathbb{R}$ onto a closed ball $B\subset \mathbb{R}^n$) to obtain a map $f$ such that $f(C)=\mathbb{R}^n$ for every convex set $C$ with more than one point. (Alternatively, $f(C)=B$ for any such $C$.)

This $f$ is quite terrible. There are no sensible function properties that it satisfies; yet, it preserves convexity. This is why we can't come up with a necessary property for convexity-preservation.

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    Thanks, this is a really nice answer. I'll have to study the answer you linked to a bit before accepting, but this is exactly the sort of answer I was looking for. – aduh Mar 14 '17 at 18:00