Consider$\newcommand\binorm[1]{\lVert#1\rVert}\newcommand\trinorm[1]{\lVert\lvert#1\rvert\rVert}$Consider the space $\ell ^{2}$ with the standard norm \begin{align*} \Vert x \Vert _{2} = \left( \sum _{i =1} ^{\infty} x _{i} ^{2} \right) ^{1/2} \end{align*}\begin{align*} \binorm x_{2} = \left( \sum _{i =1} ^{\infty} x _{i} ^{2} \right) ^{1/2} \end{align*} and we define the equivalent norm \begin{align*} \Vert\vert x \Vert\vert _{\sqrt{2}}= \max\{ \Vert x \Vert _{2}, \sqrt{2}\Vert x \Vert _{\infty} \} \mbox{.} \end{align*}\begin{align*} \trinorm x _{\sqrt{2}}= \max\{ \binorm x _{2}, \sqrt{2}\binorm x _{\infty} \} \mbox{.} \end{align*}
Let's define the positive part of the unit ball \begin{align*} B _{\ell ^{2}} ^{+} = \lbrace x \in \ell ^{2}: \; \Vert x \Vert _{2} \leqslant 1, \; x _{i} \geqslant 0 \rbrace \mbox{.} \end{align*}\begin{align*} B _{\ell ^{2}} ^{+} = \lbrace x \in \ell ^{2}: \; \binorm x _{2} \leqslant 1, \; x _{i} \geqslant 0 \rbrace \mbox{.} \end{align*} I want show that $B _{\ell ^{2}} ^{+}$ with the norm $\Vert\vert \cdot \Vert\vert _{\sqrt{2}}$ don't$\trinorm\cdot_{\sqrt{2}}$ doesn't have normal structure, and for. To show that, I should show that diam$(B _{\ell ^{2}} ^{+})$$\operatorname{diam}(B _{\ell ^{2}} ^{+})$ = $r _{x}(B _{\ell ^{2}} ^{+})$. I showed that diam$(B _{\ell ^{2}} ^{+}) = 1$$\operatorname{diam}(B _{\ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{\ell ^{2}} ^{+}) =1$. What element in $B _{\ell ^{2}} ^{+}$ can I take forto prove that $r _{x}(B _{\ell ^{2}} ^{+}) =1$?
