Timeline for Is it true that $c_0(X)^* = \ell_1(X^*)$ ?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2012 at 2:14 | vote | accept | Rafael | ||
| Mar 20, 2012 at 2:08 | comment | added | Rafael | u mean by rotating $f \circ \pi_j (x_j)$ ? sorry, I'm embarassed, but I barely work with complex numbers :) | |
| Mar 20, 2012 at 1:58 | comment | added | Rafael | sorry, i made some typos, and i can't fix the posts, so I delete and rewrite... | |
| Mar 20, 2012 at 1:57 | comment | added | Rafael | but I'm not supposing that the field is $\mathbb{R}$. So, instead of getting \begin{equation} \sum^n_{j=1} \lVert f \circ \pi_j \rVert \le \frac{\sum^n_{j=1} f \circ \pi_j (x_j) }{1-\varepsilon} = \frac { f (x_1,\ldots,x_n,0,0,\ldots)} {1-\varepsilon}\le \frac {\lVert f \rVert}{1-\varepsilon} \end{equation} I'm gettin \begin{equation} \sum^n_{j=1} \lVert f \circ \pi_j \rVert \le \frac{\sum^n_{j=1} \lvert f \circ \pi_j (x_j) \rvert}{1-\varepsilon} \le \text{?} \end{equation} | |
| Mar 20, 2012 at 1:53 | comment | added | Bill Johnson | I don't understand your problem, Rafael. You just multiply $x_n$ by a complex number of modulus one to make $f\circ \pi_n(x_n)$ non negative. | |
| Mar 20, 2012 at 1:50 | comment | added | Rafael | see, I don't have $f \circ \pi_n(x_n) \ge (1- \varepsilon) \lVert f \circ \pi_n \rVert$, but I have $\lvert f \circ \pi_n(x_n) \rvert \ge (1- \varepsilon) \lVert f \circ \pi_n \rVert$ | |
| Mar 20, 2012 at 1:36 | comment | added | Rafael | From Hahn-Banach we know that, given $x \neq 0$, there's a functional $f$ such that $\lVert f \rVert = 1$ and $f(x) = \lVert x \rVert$. It follows that, if $X$ is reflexive, given $f \neq 0$, there's an $x \in X$ such that $\lVert x \rVert = 1$ and $f(x) = \lVert f \rVert$. Let $n \in \mathbb{N}$. For each $1 \le j \le n$, choose $x_j$ such that $\lVert x_j \rVert = 1$ and $f \circ \pi_j(x_j) = \lVert f \circ \pi_j \rVert$. Then \begin{equation} \sum^n_{j=1} \lVert f \circ \pi_j \rVert = \lvert f (x_1,\ldots,x_n,0,0,\ldots) \rvert \le \lVert f \rVert. \end{equation} | |
| Mar 20, 2012 at 1:28 | comment | added | Bill Johnson | Then, of course, you look at, for each n, the vector $(x_1,\dots,x_n,0,0,0,\dots)$. | |
| Mar 20, 2012 at 1:26 | comment | added | Bill Johnson | Yes, sure; but whether or not $X$ is reflexive is irrelevant, and the same problem occurs in proving the things you said you did prove. Are you hung up by the fact that $F\circ \pi_n$ need not achieve its norm? Then just choose a unit vector $x_n$ s.t. $F\circ \pi_n (x_n) \ge (1-\epsilon) \|F\circ \pi_n\|$. | |
| Mar 20, 2012 at 1:21 | comment | added | Rafael | Hi Bill, here is what I tried. Given a functional $f$ in $c_0(X)^*$, I'll try to define an element of $\ell_1(X^*)$ that depends on $f$. Define $\pi_n\colon: X \to c_0 (X)$ by $\pi_n(x) = (0,\ldots,0,x,0,\ldots)$. The natural idea would be \begin{equation} f \mapsto (f \circ \pi_1, f \circ \pi_2, \ldots), \end{equation} but first I need to prove that $(f \circ \pi_1, f \circ \pi_2, \ldots) \in \ell_1(X^*)$, that is, I must show that \begin{equation} \sum^\infty_{n=1} \lVert f \circ \pi_n \rVert < \infty. \end{equation} | |
| Mar 20, 2012 at 1:21 | comment | added | Bill Johnson | Yes, it is true, and the proof is essentially the same as what you said you did prove. I cannot imagine how you can prove this for reflexive $X$ but not for general $X$. | |
| Mar 20, 2012 at 1:17 | answer | added | Pietro Majer | timeline score: 6 | |
| Mar 20, 2012 at 1:08 | comment | added | Rafael | I'm studying an article an the author seems to use this argument. See, he writes \begin{equation} c_0 (\ell_2)^*** \approx \ell_\infy(\ell_2)^*. \end{equation} I've already proven that $\ell_p(X)^* \approx \ell_q(X^*)$, where $1 < p < \infty$ and $q$ is the conjugate of $p$. I also proved that $\ell_1(X)^* \approx \ell_\infty(X^*)$. I think I can prove that $c_0(X)^*=\ell_1(X^*)$ if I suppose that $X$ is reflexive -- and that's the case in the article. But I'm hoping this result is valid for a general $X$ normed space. | |
| Mar 20, 2012 at 0:45 | comment | added | Bill Johnson | This is an easy exercise. I cannot imagine what your difficulty might be. If you describe what you tried and where you got stuck perhaps someone can help. | |
| Mar 20, 2012 at 0:40 | comment | added | Yemon Choi | Also, how did the question arise? (I am not sure whether a hint or explanation would be more useful than a mere reference.) | |
| Mar 20, 2012 at 0:38 | comment | added | Yemon Choi | I think this is true, as a special case of results concerning the dual of an injective tensor product being sometimes equal to the projective tensor product of the duals. However, this is something I "know of" rather than "know", if you see what I mean, so I'll just stop here and wait for true experts to weigh in. | |
| Mar 20, 2012 at 0:35 | comment | added | Owen Sizemore | Do you mean just $l_1(x)$ not $l_1(X^*)$? | |
| Mar 19, 2012 at 23:53 | history | asked | Rafael | CC BY-SA 3.0 |