Almost Euclidean balls and support functions from High-Dimensional Probability (Roman Vershynin)

Question

I tried to solve Exercise 11.3.2 (Almost Euclidean balls and support functions) from High-Dimensional Probability (Roman Vershynin) and find out that there is a mistake in first part (I think it's a mistake) and reformulate it

< (i) Let $V \subset R^m$ be a bounded set. Show that $B^m_2 \subset \conv (V) \subset \closure {B^m_2}$ for every if and only if $$ \sup_{x \in V} <x,y> = \lVert y \rVert_2 \text{ for all } y \in R^m. $$ (ii) Let $V \subset R^m$ be a bounded set and $r_-,r_+ \geq 0$. Show that the inclusion $$ r_- B^m_2 \subset \conv(V) \subset r_+ \closure{B^m_2} $$ holds if and only if $$ r_- \lVert y \rVert_2 \leq \sup_{x \in V} <x,y> \leq r_+\lVert y \rVert_2 \text{ for all } y \in R^m. $$

Where $B^m_2$ is open Euclidean ball with 1 radian and 0 center $(\{x \in R^m : \lVert x \rVert_2 < 1 \})$ and $\closure{B^m_2}$ is it's closure $(\{x \in R^m : \lVert x \rVert_2 \leq 1 \})$.

I solved (i) but not the (ii). I hope someone can help me with it.

Answer 1

I reformulated the exercise by adding closedness assumption to $V$ for both (a) and (b). In addition, for part (a) we should have $\text{conv}(V) = B_2^m$ instead of $V = B_2^m$. I'm assuming $B_2^m$ is closed, which I believe is a consistent notation in Vershynin's book.

That $r_- B_2^m \subset \text{conv}(V)\subset r_+ B_2^m$ implies the inequalities is trivial. So, let's show the converse. The inequalities are equivalent to

$$ r_- \leq \sup_{x\in V}\langle x,\theta \rangle \leq r_+ \text{ for all $y\in \mathcal{S}^{m-1}$} $$

Suppose $\text{conv}(V)\setminus r_+ B_2^m \neq \emptyset$. Then there exists $x_0 \in V \setminus r_+ B_2^m$ indicating

$$ r_+ < \langle x_0, \frac{x_0}{\|x_0\|}\rangle \leq \sup_{x\in V} \langle x, \frac{x_0}{\|x_0\|}\rangle \leq r_+ $$

This is a contradiction.

Next, we show $r_- B_2^m \subset \text{conv}(V)$.

Let $\lambda_\theta$ be $\sup \{\lambda \geq 0 \text{ such that } \lambda \theta \in \text{conv}(V)\}$. It suffices to prove $\lambda_\theta \geq r_-$ for all $\theta \in \mathcal{S}^{m-1}$.

Suppose there exists $\theta_0 \in \mathcal{S}^{m-1}$ such that $\lambda_{\theta_0}< r_-$. Take a hyperplane $H$ which is tangent to $\text{conv}(V)$ at the point $x_0 = \lambda_{\theta_0}\theta_0$. Then choose $\theta_1\in \mathcal{S}^{m-1}$ so that the hyperplane $H$ is expressed as $\{\langle x-x_0, \theta_1 \rangle=0\}$.

From our construction, we may assume $V \in \{\langle x-x_0, \theta_1 \rangle \leq 0\}$.

Then, we have $$ r_- \leq \sup_{x\in V}\langle x, \theta_1\rangle \leq \langle x_0, \theta_1\rangle \leq \|x_0\| = \langle x_0, \theta_0 \rangle < r_- $$ This is a contradiction.

The first inequality is due to the assumption and the second is due to $V \in \{\langle x-x_0, \theta_1 \rangle \leq 0\}$. For the third inequality, note that $\langle x_0, \theta_1\rangle$ is the distance from the origin to the hyperplane $H$, so it must be smaller than $\|x_0\|$.