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Let $T:\mathbb R^2\to \mathbb R^2$ be a mapping such that $T(C)$ is a convex set in $\mathbb R^2$ whenever $C$ is convex set in $\mathbb R^2$ and $T(0,0)=(0,0).$ Is $T$ $\mathbb R$-linear?

We have to show here that $T(ax+by)=aT(x)+bT(y)$ for all $a,b\in \mathbb R$ and $x,y\in \mathbb R^2.$

Any help is appreciated. Thank you.

Andrés E. Caicedo
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Mini_me
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  • @CameronWilliams you mean lines passing through origin goes to line passing through origin by $T$? – Mini_me Jan 06 '18 at 15:27
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    You can start with an easier question, take $T:\mathbb{R}\rightarrow \mathbb{R}$ with $T(0)=0$ taking a convex set to a convex set, is it linear? (I don't know the solution but I think this baby case should be a good starting point) – Yanko Jan 06 '18 at 15:29
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    @yanko, that's not actually a great idea. By which I mean the answer for $T:\Bbb{R}\to\Bbb{R}$ is that it's false, consider $x^3$. However, you can't generalize this example to two dimensions, because a diagonal line gets warped if you use a two-dimensional variant of the cubing map. – jgon Jan 06 '18 at 15:32
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    @jgon can't we take $T(x,y)=(x^3,0)$? – Yanko Jan 06 '18 at 15:37
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    @yanko, fair point. The projection is linear, and therefore convex. Right. Nice. Consider me corrected. – jgon Jan 06 '18 at 15:39
  • Though I have to say, I'm now very curious about this same question with the condition that $T$ be surjective added. – jgon Jan 06 '18 at 15:40
  • @jgon Good intuition.. the answer says that if $T$ is one to one (or equivalently surjective) then the statement is true. – Yanko Jan 06 '18 at 15:44
  • @yanko If T is linear then we can say that one-to-one is equivalent to onto,right? – Mini_me Jan 06 '18 at 15:51
  • It follows from the rank nullity theorem. (It is only true if the vector spaces are of the same dimension, in our case $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ ,$\dim \mathbb{R}^2=2$) – Yanko Jan 06 '18 at 17:15

2 Answers2

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Such map $T$ is not necessarily $\mathbb{R}$-linear. To see that consider the following counterexample.

Let $\pi_x : \mathbb{R}^2 \to \mathbb{R} : (x, y) \mapsto x$ the $x$-coordinate projection in $\mathbb{R}$, $e_x : \mathbb{R} \to \mathbb{R}^2 : x \mapsto (x, 0) $ the "canonical embedding" of the $x$-coordinate in $\mathbb{R}^2$, $$ f : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} \sin(\frac{1}{x}) &,& x \neq 0 \\ \hfill 0 \hfil &,& x = 0 \end{cases} $$ and $$ T : \mathbb{R}^2 \to \mathbb{R}^2 : (x, y) \mapsto e_x \circ f \circ \pi_x(x,y) = (f(x), 0 )$$

By definition, $T(0,0) = (0,0)$.

Now let $C \subseteq \mathbb{R}^2$ convex. Then $\pi_x(C) \subseteq \mathbb{R}$ is convex.

Since convex sets of $\mathbb{R}$ are intervals and $f$ maps intervals to intervals, it follows that $f \circ \pi_x(C) \subseteq \mathbb{R}$ is convex.

Therefore $ T(C) = e_x \circ f \circ \pi_x(C) \subseteq \mathbb{R}^2 $ is convex.

But $T$ is clearly not $\mathbb{R}$-linear.

mucciolo
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  • The map $f$ can be chosen to be continuous and monotone, as the example given by @yanko in comments above shows. – hardmath Jan 07 '18 at 21:54
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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.

Now your $T$ is actually linear because $T(0)=0$. So the answer is YES provided that $T$ is one-to-one and continuous. Note that assuming one-to-one is reasonable, else you could just send everything to the zero vector.

max_zorn
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  • When was any of "continuous and one-to-one" assumed? – Kenny Lau Jan 06 '18 at 15:51
  • @KennyLau: If you can remove these assumptions, that would be great. Do you have a proof or pointer to a result? The proof is not trivial even with continuity and injectivity. I do think injectivity is a reasonable assumption in view of the zero map. – max_zorn Jan 06 '18 at 17:35
  • Why must the answer be "yes"? – Kenny Lau Jan 06 '18 at 18:45
  • added continuity, Kenny. – max_zorn Jan 06 '18 at 18:48
  • I upvoted because it's a nice find. However the last sentence is beside the point because "just send everything to the zero vector" actually does give us $T$ being $\mathbb R$-linear. The crux here is requiring $1-1$ together with dimension of the domain $\gt 1$. – hardmath Jan 07 '18 at 16:10