Is the function T $\mathbb R$-linear? 
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.
 A: 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.
A: 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. 
