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 $x \_{i} \in \ell ^{p, \infty}$?.



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


