Prove that Cartesian composition $c_0 \times c_0$ is not isometric isomorphic [closed]

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 off-topic by Jan-Christoph Schlage-Puchta, Yemon Choi, David Handelman, David Roberts, S. Carnahan♦Dec 22 '18 at 4:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – David Handelman, S. Carnahan
If this question can be reworded to fit the rules in the help center, please edit the question.

• what is the origin of this question and why you say "prove" (not "is it true?")? – Fedor Petrov Dec 20 '18 at 8:54
• 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. – Jan-Christoph Schlage-Puchta Dec 20 '18 at 12:08
• @Jan-ChristophSchlage-Puchta I'm confused by your comment. $c_0$ usually denotes a specific Banach space, so that part at least is not ambiguous – Yemon Choi Dec 20 '18 at 16:26
• 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 $c_0$. – Yemon Choi Dec 20 '18 at 16:29
• Gera, where does this problem originate? It looks to me like it could be part of an assignment – 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 sum-norm, and there is no such complex Banach space. (See E. Behrends, Studia Math. 55, 71-85 (1976) or P. Harmand et al., M-ideals in Banach spaces and Banach algebras. Lecture Notes in Mathematics. 1547. Berlin: Springer-Verlag (1993).)