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This question in the title came to my mind while I was sitting with my granny in front of my house maybe about half an hour ago.

Although it looks innocent I do not know at the moment some simple argument that would prove or disprove it.

So, let us suppose that we have some function $f$ which is defined on the whole $\mathbb R^{n}$ and that if $S \subset \mathbb R^{n}$ is bounded convex set which is not the straight line segment (with or without boundary points) that then $f(S)$ is also bounded convex set.

Is every such $f$ continuous everywhere?

Be aware that we do not consider functions defined on $\mathbb R$ because bounded convex sets on $\mathbb R$ must be straight line segments (with or without boundary points) so we seek to find an answer for $n \geq 2$.

Tomasz Kania
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Farewell
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1 Answers1

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This is a counterexample in $\mathbb R^2$, where every convex body that does not lie on a line has nonempty interior. First, take a function $g:\mathbb{R}^2\to\mathbb{R}$ that maps every nonempty open set onto all of $\mathbb{R}$. The same construction as in this answer works: take a countable basis of topology, insert a Cantor-type set in each, keeping them disjoint; then put each set in bijection with $\mathbb{R}$.

Then compose $g$ with a map of $\mathbb{R}$ onto the unit disk of $\mathbb{R}^2$ to obtain $f$ that satisfies the stated condition, and is discontinuous everywhere.

In higher dimensions, one can use the idea of this answer (transfinite induction) to construct a function $f:\mathbb{R}^n\to\mathbb{R}$ that maps every nondegenerate line segment onto $\mathbb{R}$. As above, this leads to an everywhere discontinuous function that maps every convex set (other than the sets with $0$ or $1$ points) onto the unit ball of $\mathbb{R}^n$.

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