Distribution of the support function of convex bodies: beyond mean width

Question

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?

Answer 1

This fails already in dimension $2$ for very unbalanced convex bodies. Given $\alpha > 0$, consider the ellipse $K_\alpha$ obtained as the image of the unit disk by the area-preserving transformation $\begin{pmatrix} \alpha & 0 \\ 0 &\alpha^{-1} \end{pmatrix}$. You are asking lower bounds on the measure of the set $$ W_c^{K_\alpha} = \{ \theta \in S^1 , \alpha^2 \cos^2 \theta + \alpha^{-2} \sin^2 \theta \leq c^2 \}, $$ which goes to $0$ as $\alpha$ tends $+\infty$.

Answer 2

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.