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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.

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  • $\begingroup$ Choose $g(x)=h(y)=1$, which means $x\in\partial K$ and $y\in\partial K^\circ$. But then you can choose $x$ so that $x\times \nabla g(x)=0$, and if $K$ is not the disc, you can choose $y$ so that $y\times\nabla h(y)\not=0$, can't you? In the case of the dot product I think you have $x\cdot\nabla g(x)=g(x)$. $\endgroup$
    – M. Winter
    Commented Jul 12, 2023 at 6:10
  • $\begingroup$ What holds is $g(\nabla h)=1$ and $h(\nabla g) =1$. You can also differentiate this with respect to each coordinate to get a second equation. I think that’s the closest you can get to your conjecture. $\endgroup$
    – Deane Yang
    Commented Jul 12, 2023 at 6:19
  • $\begingroup$ Can you elaborate on how you get this equation and such a second equation @DeaneYang? I'm interested in you answer $\endgroup$
    – kvicente
    Commented Jul 12, 2023 at 11:31
  • $\begingroup$ I’ll let you try first. Do you know how to define $h$ in terms of $g$? $\endgroup$
    – Deane Yang
    Commented Jul 12, 2023 at 15:32
  • $\begingroup$ Good question @DeaneYang. Would it be h(x)=g(x)||x||^2? $\endgroup$
    – kvicente
    Commented Jul 12, 2023 at 21:04

1 Answer 1

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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.

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