TITLE: Generalizing the Mazur-Ulam theorem to convex sets with empty interior in Banach spaces
QUESTION [16 upvotes]: The Mazur-Ulam theorem (1932) states that any isometry of a normed linear space is affine. See Nica (Expo. Math. 30 (2012), 397-398; arXiv:1306.2380) for a very elegant proof.
Question: Let $M$ be a closed convex set with empty interior in a normed linear space and $T:M\to M$ be an isometry of $M$ onto itself. Does it follow that $T$ is affine (maps linear segments to linear segments)?
This question naturally arises in connection with the work of Bader, Furman, Gelander, Monod, see p. 88 of their paper `Property (T) and rigidity for actions on Banach spaces' (Acta Math. 198 (2007), 57-105; arXiv:math/0506361). 
Related information: 1. The question has already been answered in the positive  by Mankiewicz (On extension of isometries in normed linear spaces, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 20 (1972), 367-371) for closed convex sets having nonempty interior. 

There is an active related work on the problem suggested by Tingley
(Geometriae Dedicata, 1987, p.371): Suppose that $f: S_X\to S_Y$is an (onto) isometry between spheres of normed spaces. Is $f$ necessarily the restriction to $S_X$ of a linear, or affine, transformation? But in this work a starting point is an isometry of a non-convex set.

REPLY [14 votes]: I was referred to this question by Uri Bader. Thanks Uri.
Here are two counter examples.
Consider the set $K$ of all continuous strictly increasing functions $f$ from $[0,1]$ onto $[0,1]$. Let $T$ be the map from $K$ onto $K$ that sends $f$ to its inverse. Endowing $K$ with the $L_1$ norm, $T$ is an isometry (the $L_1$ distance between two functions is the area enclosed by their graphs and, By Fubini theorem, this is the same for the functions and their inverses).
The first example, $M_1$, is the closure of $K$ in the $L_1$ norm and the extension of $T$ to $M_1$. It is easy to see it is not affine (it is also easy to describe $M_1$ and the extension of $T$ explicitly).
The second example $M_2$, is all the continuous increasing functions from $[0,1]$ onto $[0,1]$ satisfying $(y-x)/2\le f(y)-f(x)\le 2(y-x)$ for all $0\le x<y\le1$, endowed with the $L_1$ norm.
$T$ is the same as before. 
It is easy to see that both $M_1$ and $M_2$ are norm compact.