Let $K$ be a symmetric convex body in $\mathbb{R}^n$ (that is the unit ball of a norm). Let $h\_K$ be its support function, that is $h\_K(u) = \sup\_{x \in K}\langle x,u \rangle$. The quantity $w(K) = \int\_{S^{n-1}} h\_K(\theta)\,d\theta$ is called the mean width of $K$ and is well studied.


Assume that $\operatorname{vol}(K) = \operatorname{vol}(B\_n)$ where $B\_n$ is the unit euclidean ball. Since $K$ does not contain $B\_n$, its polar $K^\circ = \{ y \in \mathbb{R^n}, h\_K(y) \leq 1 \}$ is not contained in $B\_n$ (polarity exchange inclusions and $B\_n$ is its polar), so that there is $y$ in $\mathbb{R^n}$ such that $\vert y \vert\_2 \geq 1 $ and $h\_K(y) \leq 1$. Setting $\theta = \frac{y}{\vert y \vert\_2}$, we have $h\_K(\theta) \leq 1$.


What can be said about the level-sets of the restriction $h\_K : S^{n-1} \to \mathbb{R}^+$, that is the sets:


$$W\_c^K = \{ \theta \in S^{n-1}, h\_K(\theta) \leq c\} \quad ?$$


The argument above shows that for all $c\geq 1$, $W\_c^K$ is non-empty.


Can we say something for instance about the (Haar) measure of $W\_{100}^K$, where 100 is an example of a dimension independent constant?


In particular, it is known that in some cases, such as the cross-polytope, the average of $h\_K$ is large, that is if we normalize the cross-polytope $C\_n$ so that it has the same volume as the ball then $w(C\_n)$ is of order $\log(n)$. However can we still hope that for instance $\mu(W\_{100}^K) \geq \frac{1}{2}$ for every $K$ of volume $1$, where $\mu$ is the Haar measure?



Distribution of the support function of convex bodies: beyond mean width
Actually the statement is pretty strong, in particular it implies the Bourgain-Milman (reverse Blaschke Santalo inequality) theorem, which says that the Mahler conjecture is true up to constant, that is, there exists $c\_0>0$ such that for all convex body $K$ :


$$\operatorname{vol}(K)\operatorname{vol}(K^\circ) \geq c\_0^n w\_n^2 $$


where $w\_n = \operatorname{vol}(B\_n)$ is the volume of the euclidean unit ball. Note that the theorem is stated in an equivalent unusual way.


Now, let $K$ be a convex body with $\operatorname{vol}(K) = w\_n$ assume that there exists $c>0$ such that $\mu(W\_K^c) \geq \frac{1}{2}$, then by integrating in polar :


\begin{align\*}
\operatorname{vol}(K^\circ) &= nw\_n\int\_{S^{n-1}}\int\_{0}^{\vert\vert \theta \vert\vert\_{K^\circ}^{-1}}t^{n-1}\, dt\,d\theta \\
& = w\_n \int\_{S^{n-1}}\vert\vert \theta \vert\vert\_{K^\circ}^{-n}\,d\theta \\
&= w\_n \int\_{S^{n-1}} h\_K^{-n}(\theta) \\
& \geq w\_n \,\mu(W\_K^c)\frac{1}{c^n} \geq \left(\frac{1}{2c}\right)^nw\_n
\end{align\*}


So that we indeed get $$ \operatorname{vol}(K)\operatorname{vol}(K^\circ) \geq\left(\frac{1}{2c}\right)^n w\_n^2 $$
(Note that the volume product is affine invariant, hence the assumption about the volume of K does not lose generality).


However, one cannot (a priori) invert the implication, so that it would be a stronger statement than Bourgain-Milman.


