Given a centrally symmetric convex body $K$ in the plane (with smooth boundary), it is easy to see that there exists a norm function $g:\mathbb{R}^2\to \mathbb{R}_{\geq 0}$ for which $K$ is the unit ball.
For the polar body $K^{\circ}$ we also have a norm function $h:\mathbb{R}^2\to \mathbb{R}_{\geq 0}$ for which $K^{\circ}$ is the unit ball. I have the following question:
Is $x\times \nabla g(x)=y\times \nabla h(y)$ whenever $(x,y)$ is s.t. $g(x)=h(y)$? Where $\nabla$ denotes the gradient of the function (it makes sense outside the origin in this case) and $x\times y:=x_1y_2-x_2y_1$.
This is certainly true for the K the unit ball in the $\ell^2$-norm.
Oddly enough, if we change $\times$ by the usual inner product this assertion follows from Euler's theorem on homogeneous functions.
Thanks in advance.
Best Answer
I think you need to assume that $K$ has a smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.
Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^*$ is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$. The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so everything can be computed explicitly.
For every $x \in \mathbb{R}^2$, set $$x^{p-1}:=(|x_1|^{p-1}\mathrm{sign(x_1)},|x_2|^{p-1}\mathrm{sign(x_2)}).$$ Then $\nabla (g^p)(x) = (p|x_1|^{p-1}\mathrm{sign(x_1)},p|x_2|^{p-1}\mathrm{sign(x_2)}) = px^{p-1}$. By the chain rule, if $x \ne 0$, $$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$ $$x \times \nabla g(x) = \frac{x_1x_2^{p-1}-x_2x_1^{p-1}}{g(x)^{p-1}}.$$
My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.