# Norm functions induced by convex bodies

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.

• 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 $x\cdot\nabla g(x)=g(x)$. Commented Jul 12, 2023 at 6:10
• 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. Commented Jul 12, 2023 at 6:19
• Can you elaborate on how you get this equation and such a second equation @DeaneYang? I'm interested in you answer Commented Jul 12, 2023 at 11:31
• I’ll let you try first. Do you know how to define $h$ in terms of $g$? Commented Jul 12, 2023 at 15:32
• Good question @DeaneYang. Would it be h(x)=g(x)||x||^2? Commented Jul 12, 2023 at 21:04

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.