## $B _{\ell ^{2}} ^{+}$ with the norm $\Vert\vert \cdot \Vert\vert _{\sqrt{2}}$ don’t have normal structure

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*} 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*}

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*} I want show that $$B _{\ell ^{2}} ^{+}$$ with the norm $$\Vert\vert \cdot \Vert\vert _{\sqrt{2}}$$ don’t have normal structure, and for show that, I should show that diam$$(B _{\ell ^{2}} ^{+})$$ = $$r _{x}(B _{\ell ^{2}} ^{+})$$. I showed that diam$$(B _{\ell ^{2}} ^{+}) = 1$$, but I don’t know why $$r _{x}(B _{\ell ^{2}} ^{+}) =1$$. What element in $$B _{\ell ^{2}} ^{+}$$ can take for prove that $$r _{x}(B _{\ell ^{2}} ^{+}) =1$$?