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Let $f : \mathbb{R} \longrightarrow \mathbb{R}^2$ be a continuous map which sends any interval $I \subseteq \mathbb{R}$ to a convex subset $f(I)$ of $\mathbb{R}^2$. Is it true that there must be a line in $\mathbb{R}^2$ which contains the image $f(\mathbb{R})$ of $f$?

Yes, this question seems rather elementary, but I have already spent (or lost?) too much time on this devilish problem, and I have communicated this question to sufficiently many people to know that it is far from trivial...

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    $\begingroup$ I'm far for being an expert so I'm not sure whether this makes sense, anyway did you try to build a counterexample by using some kind of plane-filling curve? $\endgroup$
    – Francesco Polizzi
    Commented Mar 20, 2015 at 15:09
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    $\begingroup$ Yes, I tried. If one aims to construct a counterexample, then one can discretized the whole situation, and try to draw a path on a $n \times n$ grid such that 1) the path fills the square and 2) satisfies the convexity condition in an appropriate approximate sense. The problem is that in order to be able to extract a limit, one has to ensure the equicontinuity of our family of paths. I've never succeeded in drawing an equicontinuous family of paths satisfying 1) and 2) above (but of course it doesn't mean that it is not possible). $\endgroup$
    – Abcd
    Commented Mar 20, 2015 at 15:48
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    $\begingroup$ This is the Mihalik-Wieczorek problem. It's indeed devilish. :) Let me remark that if you only ask for the convexity of $f(I)$ for the intervals of the form $I=[0,t]$ then it is possible to construct a space-filling curve: see arxiv.org/abs/1407.5204 and the references there. $\endgroup$
    – Jairo Bochi
    Commented Mar 24, 2015 at 19:48
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    $\begingroup$ Thank you, Jairo, that was exactly what I was looking for ! Your comment should be an answer, by the way. $\endgroup$
    – Abcd
    Commented Mar 25, 2015 at 16:22

2 Answers 2

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I asked [a very similar question] (convexity of images of space-filling curves) here once.

Suppose $f:[0,1]\to[0,1]^2$ is continuous and for each $t\in[0,1]$, the area of $\lbrace f(s) : 0\le s\le t \rbrace$ is $t$. For what sets of values of $t\in[0,1]$ can $\lbrace f(s) : 0\le s\le t \rbrace$ be convex? All $t$? Only countably many $t$? If so, which countable sets? Topologically discrete ones? Dense ones?

Perhaps Pietro Majer's answer to that question will shed some light on this one as well.

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See Theorem 2 in this paper by Kay and Meyer (J LMS, 1973).

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    $\begingroup$ The theorem reads "Let $V$ and $W$ be real vector spaces. Any one-to-one mapping $f:V\to W$ which preserves convexity is collinearity-preserving." The OP did not assume the function to be one-to-one, so this does not fully answer the question. $\endgroup$
    – Joonas Ilmavirta
    Commented Mar 20, 2015 at 18:12
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    $\begingroup$ Indeed, the interesting case is when the function is not one to one (space-filling curves). For a one-to-one function from $\mathbb{R}$ to $\mathbb{R}^2$, the image of each $[-n, n]$ must be a convex set homeomorphic to $[0,1]$, hence a line segment. $\endgroup$
    – Ramiro de la Vega
    Commented Mar 20, 2015 at 18:25
  • $\begingroup$ Ah, did not notice the 1-1 hypothesis... $\endgroup$
    – Igor Rivin
    Commented Mar 20, 2015 at 19:30

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