8
$\begingroup$

Here is a problem that has been bugging me for a while.

Let $\| \|$ be a norm over $\mathbb{R}^n$, let $C$ be a convex subset of $\mathbb{R}^n$ with non-empty interior, and let $f: (C,\|\|) \rightarrow (\mathbb{R}^n,\|\|)$ be a distance-preserving map.

Is it true that there exists an isometry $g$ of $(\mathbb{R}^n,\|\|)$, such that $f = g|_C$?

A few notes:

-I don't require an isometry to be surjective, though it follows from the bounded-compactness of $(\mathbb{R}^n,\|\|)$.

-There is flexibility regarding the caracteristics of $C$, which may (to serve my purpose) be: $C$ is an open ball or closed ball with non-empty interior. In that case, $f$ is a bijection from $C = B(x,r)$ onto $B(f(x),r)$.

-Mazur-Ulam's theorem states that every surjective isometry between two real normed vector-spaces is affine.

-Though it may not seem so, I think the easiest way to prove this is to show that $f$ perserves the middles of any two points of $C$. Unfortunately, this isn't an easy task, I do not think I would have been able to prove it in the case when $C = \mathbb{R}^n$ if I hadn't come across PFDs about Mazur-Ulam. I have failed to adapt the proof(s) of Mazur-Ulam theorem to this case. I don't understand the "fundamental reason" why this theorem works, making it difficult to use. Maybe it's just magic.

-The norm may not be strictly convex, therefore, given two distinct points, there may be infinitely many points whose distance to each of the original points is half the distance between the original points.

-There might be a solution if we assume that $f$ is differentiable on the interior of $C$, but once again I don't think it would be easy to prove that $f$ is differentiable without having established results which directly imply that it is affine.

Any idea how to proceed?

$\endgroup$
3
  • $\begingroup$ In the special case when $C$ is bounded, and the norm is Euclidean, this is known as rigidity problem for convex hypersurfaces. If the boundary of $C$ is $C^1$ smooth, the answer is yes, and is apparently due to Shenkin, see the paper [C. Vilcu, On typical degenerate convex surfaces. Math. Ann., 2008]. I have not read Sen'kin's paper though, and it is rather short, so this makes me worried a bit. There is an alternative proof in the $C^2$ case, see arxiv.org/abs/1306.1581. EDIT: actually, your setting is a bit different. $\endgroup$ – Igor Belegradek Jan 4 '14 at 3:10
  • 1
    $\begingroup$ By the way, about your comment #4: the distance preserving map has to be volume preserving. Therefore if $C=\mathbb R^n$ then $f$ has to be surjective. $\endgroup$ – Anton Petrunin Jan 4 '14 at 4:06
  • $\begingroup$ @Igor Belegradek: Thanks for the info. Unfortunately the maths in your arxiv link go far beyond my level, to such extent that I couldn't even decide wether this proves my result in that peculiar case or not. Anton Petrunin: Same here, I am unfamiliar with the notion of volume so I'm going to look it up. Like I said, surjectivity can also be proven by studying $f - f(x)$ on every $\overline{B(x,r-\frac{1}{n})}$ and applying the result: Let $(X,d)$ be a metric space and let $g: X \rightarrow X$ be a distance-preserving map; if $X$ is compact then $g$ is surjective. $\endgroup$ – nombre Jan 4 '14 at 12:19
3
$\begingroup$

So the problem is that your map $f$ is not surjective.

However, $f$ is locally surjective in the following sence:

If $\Omega$ is the interior of $C$, $x\in \Omega$ and $\varepsilon>0$ is such that $B_\varepsilon(x)\subset \Omega$ then the restriction $f|_{B_\varepsilon(x)}$ is a bijection from $B_\varepsilon(x)\to B_\varepsilon(f(x))$. This is true since the distance preserving map has to be volume preserving and $\mathrm{vol}\,B_\varepsilon(x)=\mathrm{vol}\,B_\varepsilon(f(x))$.

This local surjectivity can exchange the surjectivity in the proof of Jväisälä given here (the only proof of Mazur–Ulam's theorem I know). It gives the following: assume $B_{100{\cdot}|x-y|}(x)\subset \Omega$ then $$f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}{2},$$ i.e., the map $f$ is a restriction of affine map to $C$. Hence everything follows.

$\endgroup$
5
  • $\begingroup$ I like the idea, but I am not sure it applies to any norm. $\endgroup$ – nombre Jan 4 '14 at 0:56
  • $\begingroup$ Sorry for the double post. The problem is that given two points $X$ and $Y$ in $C$, you cannot expect $f(X)$ and $f(Y)$ to be in such places that there exists $Z \in C$ and a radius $r$ satisfying both $X,Y \in B(Z,r)$, $f(X),f(Y) \in B(f(Z),r)$ and $[XY] \subset \overline{B(Z,r)}$ and $[f(X)f(Y)] \subset \overline{B(f(Z),r)}$. Then again I might have misunderstood your point. $\endgroup$ – nombre Jan 4 '14 at 1:10
  • $\begingroup$ @nombre Okey, I think I see now what is your problem. The answer is rewritten. $\endgroup$ – Anton Petrunin Jan 4 '14 at 6:00
  • $\begingroup$ post-edit: What I tried in this direction was this: I replaced $W$ with the set of applications $f$ whose domain contain a neighborhood of the segment $[xy]$ (and is a subset of $\mathbb{R}^n$), that are distance-preserving, and that admit $x$ and $y$ as fixed points. The first step of the demonstration would require $x+y -f(\frac{x+y}{2})$ to be in $f(V)$ for any $f \in E$ and for some neighborhood $V$ of $[xy]$. I feel your condition $B_{100 - \|x-y\|}(x) \subset \Omega$ has a purpose related to this requirement; but I fail to see how you accomodate with Jvaisala's proof to make this work. $\endgroup$ – nombre Jan 4 '14 at 11:23
  • $\begingroup$ In Jvaisala's proof you may think that $g$ is a bijection from $Q=B_{|x-y|}(x)\cup B_{|x-y|}(y)$ to itself. No further changes are required. $\endgroup$ – Anton Petrunin Jan 4 '14 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.