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Suppose $\gamma:\mathbb{S}^1\to\mathbb{R}^2$ is a smooth origin symmetric strictly convex curve. Is there any special linear transformation $A\in SL(2,\mathbb{R})~$ such that the length of $A\gamma~$ is minimized.

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There is an $A\in SL(2,{\mathbb R})$ that minimizes the length of $A\gamma$.

Proof. You may assume that $\gamma$ has length $L$ and lies outside of the unit circle. Any admissible $A$ can be written in the form $A=T_\phi\ D \ T_\psi$ where $T_\phi$ and $T_\psi$ are rotations and $D={\rm diag}(\rho, {1\over \rho})$ with $\rho\geq 1$. This $A$ transforms the unit circle into an ellipse of circumference at least $4\rho$, and $A\gamma$ will have an even larger length. This implies that we can restrict the set of admissible $A$'s to $\rho\leq L/4$ without loosing a possible candidate. But now we are minimizing an obviously continuous function over the compact set $[0,\pi]\times[1,{L\over 4}]\times [0,\pi]$; therefore the minimum exists.

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Nice -- we posted almost the same solution within 2 minutes of each other. Yours is a bit more compact, but presumes a bit more knowledge about $SL(2,\mathbb{R})$. – joriki Jun 13 '11 at 10:13
Awesome, very nice. – timhortons Jun 13 '11 at 20:32

$SL(2,\mathbb{R})$ is the group of special linear transformations, not of special affine transformations. However, since translations don't change the length of the curve, the answer is the same for both groups.

The answer is yes. Let $L$ be the length of $\gamma$, and let $d$ be the smallest distance of an image point of $\gamma$ from the origin. This smallest distance exists since, as a continuous function on the compact set $S^1$, the distance attains its minimum; it is non-zero since the origin cannot lie on the curve. Consider the set of elements $A\in SL(2,\mathbb{R})$ with entries less than or equal to $a_0:=\frac{L}{4d}$. This set is compact in the topology on $SL(2,\mathbb{R})$ induced by the elementwise metric, since it is closed and bounded. The length of $A\gamma$ is a continuous function on this compact set, and hence attains its minimum $L_0$, with $L_0\le L$ (since $L\ge4d$, hence $a_0\ge1$ and hence the set contains the identity). Now consider elements of $SL(2,\mathbb{R})$ with at least one entry $a>a_0$. Such a transformation maps one of the intersections of the curve with the coordinate axes to a point at a distance greater or equal to $ad$ from the origin. Thus, the length of the transformed curve is at least $4ad>4a_0d=L\ge L_0$. Thus $L_0$ is the minimal length. It is attained at least by a one-parametric family of transformations which differ from each other by rotations.

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Awesome, very nice. – timhortons Jun 13 '11 at 20:32

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