(1-Lipschitz) + (length-preserving) = isometry

Question

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 $\beta|[a,b]$.

Answer 1

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)

Answer 2

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.)

Answer 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.

Answer 4

$ \alpha,\ \beta$ is images in $\mathbb{R}^2$. Then they are metric space with a canonical metric $|\ |$ in $\mathbb{R}^2$. Define $f : \alpha\rightarrow \beta$ to be a $1$-Lipschitz bijection ($[xy]$ : arc and $\overline{xy}$ : segment)

(1) If $f(x)=X,\ f(y)=Y$, then assume that $x_1=x,\ x_n=y$, If $X_i=f(x_i)$, then $$ {\rm length}\ [XY] <\sum_i\ |X_i-X_{i+1}| + \epsilon \leq {\rm length}\ [xy] +\epsilon $$

That is, $f$ is 1-Lipschitz wrt intrinsic metrics. By an assumption, it is isometric.

(2) By translation, we assume that $f(x)=x$. Then $\beta$ is in ${\rm conv}\ \alpha$ (Hence we complete the proof)

Proof : In further, consider $x_0\in \alpha\bigcap \beta$. Hence ${\rm conv}\ [xx_0]\bigcup \overline{xx_0}$ contains arc $[xx_0]_\beta$ in $\beta$.

In further, define $[xx_0]_\beta\subset [xY]$ where $f(y)=Y$ and $y\in [xx_0]$.

Hence $$ |x-Y| >|x-x_0|>|x-y| $$ which is a contradiction.

Answer 5

I'll prove the result in the following equivalent form:

THEOREM 0   Let $\ \alpha\ \beta\subseteq\mathbb R^2\ $ be two closed convex curves of the same (positive) length. Then either $\ \alpha\ \beta\ $ are isometric as subsets of the Euclidean plane $\ \mathbb R^2\ $ or there does not exist a metric function (i.e. Lip$_1)\ $ of any of the two curves onto the other one.

To this end, let's apply two classic results about curves:

THEOREM 1 The integral of the curve curvature over an arbitrary closed convex curve is always $\ 2\!\cdot\!\pi.$

This theorem holds for all closed convex curves with the following understanding: the curves corner points serve as measure atoms, where the integral over a corner is the angle between the extreme tangents at the given corner (in the case of a smooth point this angle is $\ 0$ (zero). (There is an extra to it about the orientation, but never mind).

Now, another classical theorem (a bit harder?): a continuous bijection $\ f\ $ of a plane curve onto another is an isometry $\ \Leftarrow:\Rightarrow\ $ the curvature at $\ x $ and $\ f(x)\ $ is the same at every point of the domain of $\ f$.

Now, the proof proper of THEOREM 0:

If the curves $\ \alpha\ \beta\ $ are isometric then there is nothing to prove.

Otherwise, consider an arbitrary curve-length preserving bijection $\ f :\alpha\rightarrow\beta\ $ hence $\ f\ $ is not an isometry (of subsets of $\ \mathbb R^2).\ $ Thus there exist points $\ x\ y\in\alpha\ $ such that

$$ \kappa_{\alpha}(x)<\kappa_{\beta}(f(x))\qquad\mbox{and}\qquad \kappa_{\alpha}(y)>\kappa_{\beta}(f(y)) $$

(where $\ \kappa$'s are the respective curvatures). These inequalities may happen for the corner points too.

We see that $\ f^{-1}\ $ is not locally metric around point $\ f(x),\ $ and $\ f\ $ is not locally metric at point $\ y.\ $ This proves THEOREM 0.

An essential misgiving:   My proof works when all points of the considered curves admit curvature (class $C^2$) or are the corner points. However, there can be also points of the curve for which there is only one tangent (i.e. supporting) straight line but for which the curvature is not defined. My proof still works when the measure of such nasty points is zero. However, I will provide an example where the 1-measure of such nasty points of a closed convex curve is positive and the said measure can be arbitrarily close to the length of the curve. I tried some simple reductions of the general case to the one I have proved but I have not succeeded yet--perhaps you can do it more promptly than me. Actually, I believe in a direct proof which doesn't even mention any derivatives; however, using the integral operation would make it simpler than without it.