# $B _{\ell ^{2}} ^{+}$ with the norm $\lVert\lvert \cdot \rvert\rVert _{\sqrt{2}}$ doesn't have normal structure

$$\newcommand\binorm[1]{\lVert#1\rVert}\newcommand\trinorm[1]{\lVert\lvert#1\rvert\rVert}$$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 $$r _{x}(B _{\ell ^{2}} ^{+}) =1$$?

• What does $r_x(B^+_{l^2})$ mean? – Nik Weaver Jan 5 at 3:38
• $r _{x} (B _{\ell ^{2}} ^{+}) = \sup \{ \Vert\vert x - y \Vert\vert _{\sqrt{2}}: \ y \in B _{\ell ^{2}} ^{+} \}$ – Amanda Gael Jan 5 at 4:32
• So if $x = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0, 0, 0, \ldots)$ and $y = (0, 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0, \ldots)$ then $\|x - y\|_{\sqrt{2}} = \sqrt{2}$? – Nik Weaver Jan 5 at 13:55
• But, I need prove that for all $x \in B ^{+} _{\ell ^{2}}$ exist $y \in B ^{+} _{\ell ^{2}}$ such that $r _{x} (B ^{+} _{\ell ^{2}}) = 1$. For prove this, I need prove that $\Vert\vert x - y \Vert\vert _{\sqrt{2}} = 1$ for all $x \in B ^{+} _{\ell ^{2}}$. – Amanda Gael Jan 7 at 16:10

## 1 Answer

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 $$B^+_{\ell_2}$$.