$\overline{conv}(C)$, where $C = \{ e _{1}, \cdots e _{n} \}$, $e _{i} \in \ell ^{p, \infty}$ is diametral

Question

Let $C = \{ e _{1}, \cdots e _{n} \}$, where each $e _{i}$ are unit vectors in $\ell ^{p, \infty}$, and $1 < p < \infty$. I want prove that $\overline{conv}(C)$ is diametral. My doubt is: $\Vert x - y \Vert _{p, \infty} \leqslant 1$ for all $x, y \in \overline{conv}(C)$, where \begin{align*} \Vert x \Vert _{p, \infty} = \max \{ \Vert x ^{+}\Vert _{p}, \Vert x ^{-} \Vert _{p} \} \end{align*} and \begin{align*} (x ^{+}) ^{i} = \max \{ x _{i}, 0 \} = \frac{\vert x _{i}\vert + x _{i}}{2} \quad \mbox{and} \quad (x ^{-}) ^{i} = \max \{ -x _{i}, 0 \} = \frac{\vert x _{i}\vert - x _{i}}{2} \end{align*} for all $x _{i} \in \ell ^{p, \infty}$?.

Answer 1

Note that $\mathrm{diam}( \overline {\mathrm{conv}}(C) ) = \mathrm{diam}(C)$, and note that $\|e_i-e_j\|_{p,\infty}=1$ for $i\neq j$. This proves your inequality $\|x-y\|_{p,\infty}\le 1$.

However, you won't get diametrality from this. On page 39 of Goebel and Kirk's "Topics in metric fixed point theory" you will find a proof of the fact, due to Brodskii and Milman, that compact convex sets have normal structure. Note that your set actually lives in a finite-dimensional space.