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 $c_0$.
6
-
3$\begingroup$ what is the origin of this question and why you say "prove" (not "is it true?")? $\endgroup$– Fedor PetrovCommented Dec 20, 2018 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. $\endgroup$– Jan-Christoph Schlage-PuchtaCommented Dec 20, 2018 at 12:08
-
1$\begingroup$ @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 $\endgroup$– Yemon ChoiCommented Dec 20, 2018 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 ChoiCommented Dec 20, 2018 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 ChoiCommented Dec 21, 2018 at 21:48
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
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).)
answered Dec 21, 2018 at 22:33