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

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    $\begingroup$ What does $r_x(B^+_{l^2})$ mean? $\endgroup$
    – Nik Weaver
    Commented Jan 5, 2019 at 3:38
  • $\begingroup$ $r _{x} (B _{\ell ^{2}} ^{+}) = \sup \{ \Vert\vert x - y \Vert\vert _{\sqrt{2}}: \ y \in B _{\ell ^{2}} ^{+} \}$ $\endgroup$
    – G. P
    Commented Jan 5, 2019 at 4:32
  • $\begingroup$ 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 $\|x - y\|_{\sqrt{2}} = \sqrt{2}$? $\endgroup$
    – Nik Weaver
    Commented Jan 5, 2019 at 13:55
  • $\begingroup$ 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 $x \in B ^{+} _{\ell ^{2}}$. $\endgroup$
    – G. P
    Commented Jan 7, 2019 at 16:10

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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 $B^+_{\ell_2}$.

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  • $\begingroup$ So in order to have $diam(B) = r_x(B)$ we need that $diam(B) = \sqrt{2}$ as well and not 1 as the OP states. Now Nick Weavers example in the comments to the OP is interesting in the context, because by exhibiting two elements $x, y$ that are $\sqrt{2}$ apart he shows that indeed we have $diam(B) \geq \sqrt{2}$. $\endgroup$
    – Vincent
    Commented Jan 29, 2019 at 8:59
  • $\begingroup$ thanks for help me :) $\endgroup$
    – G. P
    Commented Feb 13, 2019 at 4:46

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