Closed, convex curves with minimized lengths Let $\gamma$ be closed, smooth, strictly convex curve. It is known that we can find a special linear transformation $A\in SL(2)$ so that the length of $A\gamma$ is minimized. 
I am interested in knowing of a test/ criteria $(\ast)$ that if a given closed, smooth, strictly convex curve $\gamma$ satisfies $(\ast)$, then its length must be already minimized; that is, we do not need to apply any special linear transformation to $\gamma$ to reach the minimum length.
In terms of the support function, I have the following necessary conditions that $\gamma$ must satisfy. Before stating the conditions, let me first define the support function of $\gamma.$ The support function of a smooth, strictly convex, closed curve $\gamma$ is defined by $h(\theta)=\langle x, (\cos \theta, \sin \theta) \rangle,$ where $x$ is the point on $\gamma$ with the outer unit normal $(\cos \theta, \sin \theta)$ and $\theta\in[0,2\pi]$ (It then follows that $x=(h\cos\theta -h_{\theta}\sin\theta,h\sin\theta+h_{\theta}\cos\theta)$ and thus we have a parametrization of $\gamma$ over the unit sphere $\mathbb{S}^1$ identified with $[0,2\pi]$.). It is also known that the length of $\gamma$ can be expressed as follows
$$L(\gamma)=\int_{0}^{2\pi}h~d\theta.$$
Now the conditions: if $L(A\gamma),~A\in SL(2),$ is minimized for $A=I$, then
$$0=\int_{0}^{2\pi}D_{L_i}hd\theta,i=1,2~~~(\ast)$$
where $L_1=\left(
  \begin{array}{cc}
   1  &  0\\
    0 &  -1\\
  \end{array}
\right)
$ and $L_2=\left(
  \begin{array}{cc}
   0  &  1\\
    1 &  0\\
  \end{array}
\right)
$. Note $L_1,L_2$ are the generatores of the Lie group of $SL(2).$ Here, I should say
$$D_{L_i}h=\langle L_ix^T, (\cos \theta, \sin \theta)\rangle.$$
Doing some calculations $(\ast)$ implies that $\int_0^{2\pi}h\cos2\theta d\theta=\int_0^{2\pi}h\sin2\theta d\theta=0.$
Now my questions are:
1- given a curve $\gamma$ satisfying $(\ast)$, is it true that the length of $\gamma$ is already minimized?
2- generally, given $\gamma,$ can I explicitly find an $A\in SL(2)$ such that the length of $A\gamma$ is minimum?
3- what is the direct generalization to higher dimensions for hypersurfaces? (I know this last question is very vague)
 A: The paper by C.M. Petty "Surface area of a convex body under affine transformations" gives a necessary and sufficient conditions that a convex body has minimal surface area among its affine transforms of the same volume. This, in particular, answers my questions asked above. The paper explicitly states in section 3 that the condition $\ast$ given above is necessary and sufficient. Equivalently, if the second Fourier coefficient of the support function are zero then the length is minimized.
A: Your condtion $(*)$ is certainly not sufficient for the global minimum: There could be several local extrema of the length, and all of them satisfy $(*)$. In the following I look at the problem in my pedestrian way. It then becomes clear that the global minimum has to be found numerically.
Let the curve $\gamma$ be given by
$$\gamma:\quad z(\phi)=p(\phi)u(\phi)+\dot p(\phi)\dot u(\phi)$$
with $u(\phi):=(\cos\phi,\sin\phi)$. Then
$$\dot z=(p+\ddot p)\dot u=\rho\dot u\ ,$$
whereby $\rho:=p+\ddot p$ denotes the radius of curvature. We now introduce a linear map $A\in SL(2,{\mathbb R})$ that transforms the curve $\gamma$ into the curve
$$\gamma_A:\quad w(\phi):=Az(\phi)\ .$$
In order to compute the length of $\gamma_A$ we have to look at
$$|\dot w|=\rho \sqrt{\langle A\dot u,A\dot u\rangle}\ .\tag{1}$$
Now $\langle A\dot u,A\dot u\rangle=\langle A'A\dot u,\dot u\rangle$. The matrix $A'A$ is symmetric, whence can be written in the form $A'A=T^{-1}DT$, where $D={\rm diag}\bigl(\lambda,{1\over\lambda}\bigr)$ with $\lambda\geq1$, and $T$ is a rotation matrix. Instead of $(1)$ we now have
$$|\dot w|=\rho \sqrt{\langle DT\dot u,T\dot u\rangle}=\rho \sqrt{\lambda\cos^2(\phi+\alpha)+{1\over\lambda}\sin^2(\phi+\alpha)}\>d\phi\ .$$
The length of $\gamma_A$ is therefore given by
$$L(\gamma_A)=\int_0^{2\pi}\rho(\theta-\alpha)\sqrt{\lambda\cos^2\theta+{1\over\lambda}\sin^2\theta}\>d\theta\ .$$
Looking at this formula I get the impression that we can minimize over $\alpha$ once and for all, independently of $\lambda$. This has to do with the partcular symmetries of the large surd. In a second step one would then find the optimal $\lambda$.
