Proof about vertices of a convex hull how to prove that, given a set defined as
$S_{k}$ = {y: y = Ax, $\|x\|_{\infty}\leq$ 1}
its convex hull conv($S_{k}$) has its vertices defined by those vectors $x$ such that $\|x\|_{\infty}$ = 1. How to do it? 
Thank you for your help!
 A: So, the set under study is a linear image $S := A\mathbb{B}_\infty$, of $\mathbb{B}_\infty := \{x | \|x\|_\infty \le 1\}$, the  unit ball for the $\ell_\infty$-norm . We're interested in the vertices of $S$, namely $V(S)$. Define
\begin{equation}V_0 := \{Ax | \|x\|_\infty = 1\}.
\end{equation}
The aim is to show that $V(S) \subseteq V_0$.
Suppose on the contrary that $S$ has a vertex $ v = Ax$, with $\|x\|_\infty < 1$. We'll produce a contradiction. Now, by definition of $v$ as a vertex of $S$, there exists a half-space $H^+ := \{y|\langle w, y\rangle \le \lambda\}$ which contains $S$ and such that the intersection of its boundary $\partial H^+ = \{y | \langle w, y\rangle = \lambda\}$ with $S$ is the singleton $\{v\}$. Now, by the Cauchy-Schwarz inequality,
\begin{equation}
Ax \in \partial H^+ \land \|x\|_\infty < 1 \implies \lambda = \langle w, Ax\rangle = \langle A^*w, x\rangle \le \|A^*w\|_1\|x\|_\infty < \|A^*w\|_1.
\end{equation}
Finally, invoking the above inequality, the compacity of $\mathbb{B}_\infty$, and using the fact that $H^+ \supseteq S$, one has $\langle w, Az\rangle \le \lambda < \|A^*w\|_1 $ for all $z \in \mathbb{B}_\infty$, and so $\max_{z \in \mathbb{B}_\infty} \langle A^*w, z\rangle < \|A^*w\|_1$, i.e $\|A^*w\|_1 < \|A^*w\|_1$, a contradiction.
