Consider a norm on $\mathbb C^2$ as $\|(z\_1,z\_2)\|:=\max\{|z\_1|,|z\_2|,\frac{1}{\sqrt{2}}|z\_1+iz\_2|\}.$


*Question.* Is $(\mathbb C^2,\|.\|)$ linearly isometric to $(\mathbb C^2,\|.\|\_{\infty})$ where $\|(z\_1,z\_2)\|\_\infty:=\max\{|z\_1|,|z\_2|\}?$



A space isometric to $\ell_\infty^2$
There is no such map $f$. Let's try to map from the second space (with the funny norm, which I'll denote simply by $\|\cdot\|$) back to $(\mathbb C^2, \|\cdot \|\_{\infty})$. Let $u=f(e\_1)$, $v=f(e\_2)$, so $\|u\|\_{\infty}=\|v\|\_{\infty}=1$. Since $|1\pm i|^2=2$, so $\|(1,\pm 1)\|=1$, we also have $\|u\pm v\|\_{\infty}=1$. However, if $|z|=1$ and $w\not= 0$, then $|z\pm w|>1$ for one choice of sign. So if (say) $|u\_1|=1$, then $v\_1=0$. It follows that $u=e^{i\alpha}e\_1$, $v=e^{i\beta}e\_2$, or the other way around.


But then $\|f((1,-i)/\sqrt{2})\|\_{\infty}=1/\sqrt{2}$ even though $\|(1,-i)/\sqrt{2}\|=1$.


