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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.

EDIT I'll post a similar but simpler proof of the full theorem (within the time required by $\\LaTeX$.)

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.

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.

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.

EDIT I'll post a similar but simpler proof of the full theorem (within the time required by $\\LaTeX$.)

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 when 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.

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.