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.



Norm functions induced by convex bodies
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.


