Does it holds that the $L^{\infty}$ norm of the support function of a convex body is minimal on balls with the same volume?

137 Views Asked by At

I was wondering if the following inequality holds.

Let $K$ be a convex body of $\mathbb{R}^n$ and let us denote by $h_K$ its support function, defined as, for $x\in\mathbb{R}^n$ $$ h_K(x)={\max}\{x\cdot y\;| y\in K \}.$$

Let $B_R$ be the ball of $\mathbb{R}^n$ with the same volume of $K$ (in the sense of Lebesgue measure), where $R$ is its radius.

Is it true that $$ ||h_K ||_{L^\infty(\mathbb{S}^{n-1})}\geq ||h_{B_R}||_{L^\infty(\mathbb{S}^{n-1})}(=R)?$$

Do you have some reference for this inequality? Perhaps is an easy consequence of the Alexandrov-Fenchel inequalities?

Thank you in advance!

1

There are 1 best solutions below

0
On BEST ANSWER

Assume without loss of generality that $R = 1$. On the way to a contradiction, suppose that $h_K(\eta) < 1$ for all $\eta \in S^{n-1}$. By compactness of $S^{n-1}$ and continuity of $h_K$ there exists an $\epsilon > 0$ such that $h_K(\eta) < 1 - \epsilon$ for all $\eta \in S^{n-1}$.

Since $K$ is a convex body it may be written as the intersection of its supporting half-spaces: $K = \bigcap_{\eta \in S^{n-1}} H_\eta$ where $H_\eta = \{x : \langle x, \eta\rangle \leq h_K(\eta)\}$. Therefore $K \subset B_{1-\epsilon}$, which contradicts the assumption that $|K| = |B_1|$.