$c_0(C[0,1])$ is the $c_0$-direct sum of countably many $C[0,1]$.How to prove
$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$.
Here,Banach-space isomorphism means a bounded invertible operator from $C[0,1]$ onto $c_0(C[0,1])$.
5
-
$\begingroup$ Is Banach-space isomorphic the term you want? Ordinarily I think of "Banach-space isomorphic" as synonymous with isometrically isomorphic. $\endgroup$– Todd TrimbleCommented Feb 25, 2016 at 15:37
-
$\begingroup$ @ToddTrimble I think (but I don't claim to speak for everyone) that "Banach-space isomorphic" is in contrast to "Banach-algebraically isomorphic" (both objects are Cstar algebras, so if one just said "isomorphic", one might think this is isomorphism of Cstar algebras) $\endgroup$– Yemon ChoiCommented Feb 25, 2016 at 17:02
-
$\begingroup$ @ToddTrimble As I think was discussed on the nLab a while ago, "isometrically isomorphic" is not usually used as the default by people who work with Banach spaces - if we want isometric we usually feel obliged to explicitly mention it $\endgroup$– Yemon ChoiCommented Feb 25, 2016 at 17:04
-
$\begingroup$ @YemonChoi (Referring to your second comment) I defer to your judgment here, although I've seen "linearly homeomorphic" also used to indicate the weaker notion of isomorphism (and FWIW, I personally believe that would be preferable language). $\endgroup$– Todd TrimbleCommented Feb 25, 2016 at 17:22
-
4$\begingroup$ In Banach space theory, isomorphism means linear homeomorphism. This is completely standard. $\endgroup$– Bill JohnsonCommented Feb 26, 2016 at 4:29
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
8
There is a useful simple
Lemma. If $X\sim X\oplus X$, $Y\sim Y\oplus Y$, and each of $X,Y$ is isomorphic to a complemented subspace of another, then $X\sim Y$.
Proof. We have $X\sim Y\oplus A$, then $X\sim (Y\oplus Y)\oplus A=Y\oplus(Y\oplus A)=Y\oplus X$, analogously $Y\sim X\oplus Y$.
Now let $X=C([0,1])$, $Y=c_0(X)$. Property $Y\sim Y\oplus Y$ is clear, property $X\sim X\oplus X$ follows from $X\oplus X=C([0,1]\times \{0,1\})$ and Milyutin theorem. $X$ is clearly complemented in $Y$. At last, $Y$ is the space of functions on the compact space on the plane $K=\cup \{\frac1n \times [0,\frac1n]\}\cup (0,0)$, which are equal to 0 at $(0,0)$. Again by Milyutin theorem we see that $X\sim C(K)$, hence $Y$ is isomorphic to a hyperplane in the space isomorphic to $X$. Hyperplane is of course complemented.
This is maybe not a very good proof, since your claim may be used in the proof of Milyutin theorem (I do not remember). But if so, study the proof, it should contain this claim.
answered Feb 25, 2016 at 13:21
-
$\begingroup$ @FedorPetrov.What is "A" in the proof of the lemma above? $\endgroup$ Commented Feb 25, 2016 at 13:40
-
-
$\begingroup$ @FedorPetrov.Thank you for help. I am confused at the statement of your lemma. Do you mean that if $X\sim X\oplus X, Y\sim \oplus Y$, "Then" each of $X,Y$ is isomorphic to a complemented subspace of another? $\endgroup$ Commented Feb 25, 2016 at 14:04
-
$\begingroup$ No, it is another assumption. Sorry, I forgot to formulate the conclusion, fixed now. $\endgroup$ Commented Feb 25, 2016 at 14:10
-
$\begingroup$ oh, I see. thank you. $Y=\{f\in C(K)| f(0,0)=0\}$. I can chek this statemet by building an isometric isomorphism. But is there any easier way to see this? $\endgroup$ Commented Feb 25, 2016 at 14:14
$\begingroup$
$\endgroup$
0
If you mean $c_0(X)=\lbrace (x_n)_{n\in\mathbb N} \in X^{\mathbb N}: \|x_n\|_X\to 0\rbrace$ with norm $\|(x_n)_n\|=\sup\lbrace \|x_n\|_X: n\in\mathbb N\rbrace$, then $c_0(C[0,1])$ and $C[0,1]$ are isomorphic (as explained by Fedor) but not isometrically isomorphic: The unit ball of $C[0,1]$ has extreme points but that of $c_0(C[0,1])$ does not.
answered Feb 25, 2016 at 16:05