# What class of functions is characterized by the property $f[\operatorname{conv} A] \subseteq \operatorname{conv} f[A]$

It is well-known that the inclusion $f[\overline A] \subseteq \overline{f[A]}$ (for every subset $A$) characterizes continuous functions.1

Asking similar questions for other closure operators seems to be a rather natural question. So this leads me to the question:

Let $X$, $Y$ are vector spaces and $f\colon X\to Y$. Which functions have the property $$f[\operatorname{conv} A] \subseteq \operatorname{conv} f[A]\tag{1}$$ for each $A\subseteq X$, where $\operatorname{conv} A$ denotes the convex hull of the set $A$?

And what about the following property? $$f[\operatorname{conv} A] = \operatorname{conv} f[A]\tag{2}$$

It is relatively easy to see that affine maps fulfill both of them, simply because they preserve convex combinations.2 But it is not immediately clear whether the above properties characterize affine maps. EDIT: Now some counterexamples showing that (1) and (2) can hold for some maps which are not affine were posted in the comments. In fact, I'd say that those counterexamples make it seem less likely that there is a nice characterization of functions with these properties.

1This result can be found on many textbooks, but there are also some posts on this site with the proof of this fact. See Prove that the following statements are equivalent characterizations of continuity or $f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$.

2In fact, a transformation is affine if and only if it preserves convex combinations. See: Is every convex-linear map an affine map?

• Well, this obviously is false in one dimension, but the case seems more interesting in more than one dimension. Clearly $f$ would be a projective map since it preserves collinearity, and it's my intuition that a projective map that doesn't send any points off to infinity is probably affine. (This might break down in the infinite dimensional case, though) – Milo Brandt Feb 8 '16 at 16:48
• @MiloBrandt You have some kind of example like $f(x)= \begin{cases} 0 & x\le 0, \\ 1 & x>0 \end{cases}$ in mind? This function fulfills (1). And similar counterexample should work in higher dimensions, too. Or do you also have counterexample for (2)? – Martin Sleziak Feb 8 '16 at 17:01
• When $X = Y = \mathbb{R}$ any monotonic function satisfies 1), and any monotonic continuous function satisfies 2). – Qiaochu Yuan Feb 8 '16 at 17:21

Convexity-preserving mappings are not necessarily affine unless additional requirements are imposed. For example, $f : \mathbb R^n \to \mathbb R^n$ given by $f(x_1, x_2, ... x_n) = (x_1^3, 0, ... 0)$ is convexity-preserving but not affine.
On the other hand, it has been proved that one-to-one convexity-preserving mappings between real vector spaces $f : \mathbb R^n \to \mathbb R^m$ with $n \ge 2$ are affine transformations.