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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.

  1. 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.
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  • $\begingroup$ If $M$ has nonempty interior in its affine hull, the result reduces to Mankiewicz's theorem. But $M$ does not necessarily have a nonempty interior in its affine hull: consider the unit ball of $\ell_1$ in $\ell_2$. (This is an answer to a comment which has been deleted.) $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 8, 2015 at 19:37
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    $\begingroup$ @Carlo Beenakker Thank you very much for adding arXiv references, this can be helpful for some readers. $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 8, 2015 at 22:39

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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.

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  • $\begingroup$ I am having trouble seeing why $M_1, M_2$ are compact. Can you elaborate? $\endgroup$
    – Nate Eldredge
    Commented May 2, 2016 at 13:58
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    $\begingroup$ Consider the subset of $M_1$ consisting of the (non decreasing) functions which take constant values in $\{0,1/n,2/n,\dots,1\}$ on each interval $((i-1)/n,i/n)$. There is a finite number of such functions and each function in $K$ and thus in $M_1$ is of distance at most $2/n$ from one of these functions. To prove the last claim note that each function in $K$ intersects at most $2/n$ of the $n^2$ $1/n\times 1/n$ squares of the natural $1/n$ grid. (I'm not very precise about the constant 2 above...) $\endgroup$
    – Gideon Schechtman
    Commented May 2, 2016 at 14:33
  • $\begingroup$ Beautiful (counter)-examples ! $\endgroup$
    – BS.
    Commented May 2, 2016 at 15:11
  • $\begingroup$ Nice example. As for the edit, I just could not resist. :) $\endgroup$
    – Bill Johnson
    Commented May 2, 2016 at 15:49
  • $\begingroup$ That's a beautiful, Gideon. Next challenge: a fixed point free example :) Uri. $\endgroup$
    – Uri Bader
    Commented May 2, 2016 at 19:05

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