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



$B _{\ell ^{2}} ^{+}$ with the norm $\lVert\lvert \cdot \rvert\rVert _{\sqrt{2}}$ doesn't have normal structure
For $x$ in $B^+\_{\ell\_2}$ one has $$\sup\_{y\in B^+\_{\ell\_2}}\|x-y\|\_2^2=\sup\_{y\in B^+\_{\ell\_2}}\Big( \|x\|\_2^2+\|y\|\_2^2-2(x\cdot y)\Big)\le \|x\|\_2^2+1,$$ 
and
$$\sup\_{y\in B^+\_{\ell\_2}}\sqrt{2}\|x-y\|\_\infty=\sqrt{2}\sup\_{y\in B^+\_{\ell\_2},\,n\in\mathbb{N}}\big| x\_n -y\_n \big|\le \sqrt{2},$$ 
because $0\le x\_n\le1$ and $0\le y\_n\le1$. 


On the other hand, looking at $y$ in the standard basis, $\{e\_n\}\_n$, $$\sup\_{y\in B^+\_{\ell\_2}}\|x-y\|\_2^2\ge \lim\_{n\to\infty}\|x-e\_n\|\_2^2=\lim\_{n\to\infty} \Big(\|x\|\_2^2+1-2 x\_n\Big)=\|x\|\_2^2+1,$$ 
and
$$\sup\_{y\in B^+\_{\ell\_2}}\sqrt{2}\|x-y\|\_\infty\ge \sqrt{2}\lim\_{n\to\infty}\|x-e\_n\|\_\infty=\sqrt{2}.$$


Therefore $$\sup\_{y\in B^+\_{\ell\_2}}\|x-y\|\_2=\sqrt{\|x\|\_2^2+1}\le\sqrt{2}$$$$\sup\_{y\in B^+\_{\ell\_2}}\sqrt{2}\|x-y\|\_\infty=\sqrt2,$$ and 
$$r\_x:=\sup\_{y\in B^+\_{\ell\_2}}\max\big\{\|x-y\|\_2, \sqrt{2}\|x-y\|\_\infty\big\}$$$$=\max\big\{\sup\_{y\in B^+\_{\ell\_2}}\|x-y\|\_2, \sup\_{y\in B^+\_{\ell\_2}}\sqrt{2}\|x-y\|\_\infty\big\}=\sqrt{2}$$ for all $x$ in $B^+\_{\ell\_2}$.


