Prove that Cartesian composition $c_0 \times c_0$ with rate $ \Vert (x_1 ,x_2)\Vert = \Vert x_1 \Vert_{c_0} + \Vert x_2 \Vert_{c_0} $ is not isometric isomorphic to space $c_0$.
closed as offtopic by JanChristoph SchlagePuchta, Yemon Choi, David Handelman, David Roberts, S. Carnahan♦ Dec 22 '18 at 4:56
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3$\begingroup$ what is the origin of this question and why you say "prove" (not "is it true?")? $\endgroup$ – Fedor Petrov Dec 20 '18 at 8:54

$\begingroup$ It seems that the answer depends on $c_0$. If $c_0=\mathbb{R}^n$, the statement is true. If $c_0=\ell^1$, it seems to be false. $\endgroup$ – JanChristoph SchlagePuchta Dec 20 '18 at 12:08

1$\begingroup$ @JanChristophSchlagePuchta I'm confused by your comment. $c_0$ usually denotes a specific Banach space, so that part at least is not ambiguous $\endgroup$ – Yemon Choi Dec 20 '18 at 16:26

1$\begingroup$ I assume that by Cartesian composition you mean the direct sum of vetor spaces, and I assume that by "rate" you mean "norm". Then there are two slightly different interpretations of your question. One is to ask whether the norm you have just defined on $c_0({\bf N)}) \oplus_1 c_0({\bf N})$ is equal to the usual norm on $c_0({\bf N} \sqcup {\bf N})$, and then the answer is "no" by a trivial calculation. A slightly more interesting question (still not too difficult) is to ask whether there is any isometric isomorphism between the Banach space $c_0\oplus_1 c_0$ and $c_0$. $\endgroup$ – Yemon Choi Dec 20 '18 at 16:29

1$\begingroup$ Gera, where does this problem originate? It looks to me like it could be part of an assignment $\endgroup$ – Yemon Choi Dec 21 '18 at 21:48
If $c_0$ were isometrically isomorphic to $c_0\oplus_1 c_0$, the dual $\ell_1$ would be isometrically isomorphic to $\ell_1\oplus_\infty \ell_1$. Such an isomorphism, $A$, would map extreme points of the unit ball of $\ell_1$ to extreme points of the unit ball of $\ell_1\oplus_\infty \ell_1$ (and vice versa). The latter have the form $(\pm f_n, \pm g_m)$ where $(f_n)$ and $(g_n)$ are the unit vector bases in the two copies of $\ell_1$. Let $(e_n)$ denote the unit vector basis in the original $\ell_1$. Then there are distinct integers $j,k,l$ and signs $\varepsilon_j, \varepsilon_k, \varepsilon_l$ such that $A(\varepsilon_j e_j)=(f_1,g_1)$, $A(\varepsilon_k e_k)= (f_1,g_1)$, $A(\varepsilon_l e_l)= (f_1,g_2)$. Now calculate the norms $\\varepsilon_j e_j + \varepsilon_k e_k + \varepsilon_l e_l\=3$, but $\A(\dots)\=1$.
Actually, the only (real) Banach space that admits a decomposition $X_1 \oplus_\infty X_2$ and a decomposition $Y_1 \oplus_1 Y_2$ is $\mathbb{R}^2$ with the sup or sumnorm, and there is no such complex Banach space. (See E. Behrends, Studia Math. 55, 7185 (1976) or P. Harmand et al., Mideals in Banach spaces and Banach algebras. Lecture Notes in Mathematics. 1547. Berlin: SpringerVerlag (1993).)