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I am looking for an elementary way to prove the following theorem.

Theorem. Let $\alpha$ and $\beta$ be two simple convex closed curves in $\mathbb R^2$. Assume $$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$ and there is a 1-Lipschitz bijecction $f\colon\alpha\to\beta$. Then $f$ is an isometry.

It would be better if the same proof would work for Lobachevsky plane and unit sphere (for the sphere one has to assume that the length of the curves is $<2{\cdot}\pi$).

The proof I know is simple, but it use Alexandrov geometry quite a bit: If we cut from the plane the region bounded by $\alpha$ and glue instead the region bounded by $\beta$ then the obtained space will have curvature $\ge0$ in the sence of Alexandrov and it is easy to show that it has to be isometric to the Euclidean plane. Hence the result.

P.S. This morning I realized that this also follows from the following continuos version of Cauchy's Arm Lemma:

Let $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ be closed convex curves with unit-speed parameter. Assume that for any $t$ in a subinterval $[a,b]\subset [0,\ell]$, the curvature of $\alpha$ at $\alpha(t)$ is at most the curvature $\beta$ at $\beta(t)$. Then $|\alpha(a)-\alpha(b)|\ge|\beta(a)-\beta(b)|$ and equality holds only if the resriction $\alpha|[a,b]$ is isometric to the resriction $\beta|[a,b]$.

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I'm rather confused. Aren't any two curves of the same length always isometric? Do you really mean that $f$ is a 1-Lipschitz bijection of $\mathbb{R}^2$ to itself and not just of $\alpha$ to $\beta$? Or something else? – Deane Yang Jul 28 '11 at 15:55
@Deane: they are isometric in the length-metric, but not in the original Euclidean metric of $\mathbb{R}^2$. Also, if I understand this correctly the formulation given by Anton is the correct one: the bijection is required between $\alpha$ and $\beta$. The convexity assumption is the key point, as otherwise the claim would be trivially false (you could for example "turn bumps from the outside to the inside" with a non-convex curve). – Tapio Rajala Jul 29 '11 at 6:56
Tapio, if the map $f$ is defined only as a map from $\alpha$ to $\beta$, then the theorem is trivially true using the length metric and false using the Euclidean metric because any two simple convex curves with the same length satisfy the assumptions of the theorem. So the theorem makes sense to me only if $f$ more than just a map between the curves. Or am I missing something here? – Deane Yang Jul 29 '11 at 12:02
Deane, when using the Euclidean metric for example an ellipse and a circle with the same length do not satisfy the assumptions because the diameter of the ellipse is larger than the diameter of the circle and hence there is no 1-Lipschitz bijection. – Tapio Rajala Jul 29 '11 at 12:35
So "1-Lipschitz" here means $d(f(x),f(y)) \le d(x,y)$ for any $x, y \in \alpha$, where $d$ is the Euclidean distance function? I always think of "Lipschitz" as a local property, but I guess in metric geometry it is really a global condition. – Deane Yang Jul 29 '11 at 12:49

3 Answers 3

It seems there should a proof along the following lines, but I did not took the time to check every detail.

Edit: there is a problem in what follows; the fact that $\beta$ must be contained in the interior of $\alpha$ is not true. It is possible that another normalization makes it hold, but I know feel that the postscriptum of the question is the good point of view.

First, the assumptions give that $f$ is an isometry with respect to the length metrics on $\alpha$ and $\beta$. Let $a$, $a'$ be two farthest points on $\alpha$, $b=f(a)$ and $b'=f(a')$. Without lost of generality, apply to $\beta$ an isometry of $\mathbb{R}^2$ so that $b=a$ and $\beta$ is contained in the half-plane delimited by the line orthogonal to $[aa')$ at $a$. Using that $f$ is $1$-Lipschitz and that $[aa']$ is a diameter, we get that $\beta$ must be contained in the interior of $\alpha$. Considering the projection to the domain delimited by $\beta$, we get a $1$-Lipschitz map $\tilde f:\alpha\to\beta$ that contracts strictly distances around any point $x\in\alpha$ such that $\alpha$ at $x$ and $\beta$ at $f(x)$ do not share a common supporting direction. Since $\alpha$ and $\beta$ have the same length, this never happens and $\beta$ must be an homothetic image of $\alpha$. Since they have the same length, the homothety constant must be $1$ and we are (hopefully) done.

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Using that f is 1-Lipschitz and that [aa′] is a diameter, we get that β must be contained in the interior of α Why? – akopyan Jul 28 '11 at 12:55
You are right akopyan, this part fails in general. I mixed up two arguments (if one could make $[aa']=[bb']$, then it would hold). – Benoît Kloeckner Jul 31 '11 at 16:33

The proof for polygonal paths is almost trivial. Suppose $\alpha=P_1P_2\cdots P_n$ and $\beta=Q_1Q_2\cdots Q_m$. Subdivide the edges of $\beta$ to include $f(P_i)$'s as fake vertices and denote this polygon $\beta'$. Do the same with $\alpha$ and $f^{-1}(Q_j)$'s and use the Lipschitz condition to prove that $\alpha'$ and $\beta'$ have corresponding edges of equal length. Now assume that $f(P_i)\in [Q_j,Q_{j+1}]$, then $$|Q_jQ_{j+1}|=|Q_jf(P_i)|+|Q_{j+1}f(P_i)|=|P_if^{-1}(Q_j)|+|P_if^{-1}Q_{j+1}|$$ $$\geq |f^{-1}(Q_j)f^{-1}(Q_{j+1})|$$ so $f(P_i)=Q_j$ for some $j$, for all $i$. A similar argument shows that $\angle Q_j\le \angle P_i$ and we conclude that $f$ is an isometry.

Now it seems to me that this argument can be modified to include an approximation argument to imply your statement. (Approximate the curves by polygons, use the fact that $f$ is almost Lipschitz to conclude that it is almost an ismoetry and take the limit.)

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Right, polygonal is easy. In fact the inequality for the angles follows immediately. But even for polygonal one has to work a bit for the case of sphere. On the other hand I see some technical difficulties in your approximation. It seems that the way indicated in the post scriptum is easier to write (but still unpleasant). – Anton Petrunin Jul 29 '11 at 22:43
When I have some free time, I will try to work out the details, but I do believe that such an elementary solution is possible. As for the sphere, I thought that assuming the polygonal edges are great circle arcs, one can go through with similar calculations. – Gjergji Zaimi Jul 30 '11 at 0:26
Well for Lobachevsky plane the same works, but positive curvature of sphere makes some trouble, but it is not important. If the proof takes more than 10 lines then forget about it. – Anton Petrunin Jul 31 '11 at 9:08

If α is a convex shape and f is a 1-lipshitz map then perimeter of convex hull of f(α) ⩽ length of α. (Similar statement for higher dimension has been proven by Alexander)

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Indeed, it is easy to prove that 1-Lipschitz map does not increase the perimeter of convex hull. Now one has to show that equality holds only for isometry... – Anton Petrunin Jul 29 '11 at 22:33
Here it is for high dimensional Euclidean case. – akopyan Jul 30 '11 at 13:18
The equality case for finite set is easy. But to do equality for infinite set of points one has to make effective estimates for finite case; this is pain to write. [At least I do not see an other way.] – Anton Petrunin Jul 30 '11 at 13:27

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