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A function $f:\Bbb R^n\to\Bbb R^m$ is said that maps convex sets on convex sets if it satisfies: (1) the image $f(A)$ of any convex set $A$ is a convex set; (2) the preimage $f^{-1}(B)$ of any convex set $B$ is a convex set. All affine transformations are examples of such functions.

My question is the following: if $f:\Bbb R^n\to\Bbb R^m$ maps convex sets on convex sets and $f(\Bbb R^n)$ is a subspace of $\Bbb R^m$, is $f$ an affine transformation?

Thanks for your help.

Addendum: I'm interested in the case $n,m>1$. I apologize for not specifying this before.

Ángel Valencia
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  • This might be relevant. – Giuseppe Negro Mar 18 '16 at 16:21
  • Angel Valencia: I just added a higher dimensional example to my answer below, which could be generalized by taking maps which cube one coordinate and map the rest to $0.$ I haven't found any whose image is more than one dimensional however. – coffeemath Mar 23 '16 at 14:14

2 Answers2

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In the book Computational and Analytical Mathematics, Springer 2013, Knecht and Vanderwerff prove in Theorem 21.3 on page 458ff the following:

Let $X$ and $Y$ be any Banach spaces where $X$ contains two linearly independent vectors. Suppose $T:X\to Y$ is a continuous and one-to-one mapping such that $T$ maps convex sets on convex sets. Then $T$ is affine.

So the answer is YES.

max_zorn
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Take $m=n=1$ and $f(x)=x^3$ as a map from the reals to the reals. It maps convex sets (i.e. intervals in this case) to convex sets, and also its inverse function $f^{-1}$ does this. But it is not affine. [Note also the image of $f$ is all of $\mathbb{R}$ so is a subspace.]

Added: A (perhaps trivial) example in the case $n=m=2$ can be done with the map $F(x,y)=(x^3,0).$ If $P=(a,b),\ Q=(c,d)$ where $a \le c$ then $F$ maps the segment $PQ$ to the vertical segment $[a^3,b^3] \times \{0\},$ which is convex, while $F^{-1}$ maps segment $PQ$ to the strip $[a^{1/3},b^{1/3}] \times \mathbb{R},$ which is again convex. Note also that here $F(\mathbb{R}^2)$ is the $x$ axis, a subspace of $\mathbb{R}^2.$ This $F$ is not affine.

coffeemath
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  • It's true, I forgot this case. But I'm interested in higher dimensions; being more specifical, I'm interested in the case $n,m>1$. Thank you. – Ángel Valencia Mar 18 '16 at 16:21
  • @AngelValencia OK will try for that. Suggestion: add the requirement $n,m>1$ to the body of the question. – coffeemath Mar 18 '16 at 16:25