I'm trying to prove this that but I can't . Any help/reference ?
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$\begingroup$ Do you mean just $l_1(x)$ not $l_1(X^*)$? $\endgroup$– Owen SizemoreCommented Mar 20, 2012 at 0:35
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$\begingroup$ 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. $\endgroup$– Yemon ChoiCommented Mar 20, 2012 at 0:38
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$\begingroup$ Also, how did the question arise? (I am not sure whether a hint or explanation would be more useful than a mere reference.) $\endgroup$– Yemon ChoiCommented Mar 20, 2012 at 0:40
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4$\begingroup$ 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. $\endgroup$– Bill JohnsonCommented Mar 20, 2012 at 0:45
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$\begingroup$ 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. $\endgroup$– RafaelCommented Mar 20, 2012 at 1:08
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True. For any $n\in \mathbb{N}$ consider the inclusion to the $n$-th coordinate $j _ n : X\to c _ 0(X)$ which is right inverse to the evaluation at $n$, so that $(j _ n x)(n)= x$, for any $x\in X$. Let $j _ n ^ T : c _ 0(X) ^ * \to X^*$ its transpose operator. Any $\eta \in c _ 0(X)^ * $ defines a sequence $y:\mathbb{N}\to X ^ *$ such that $y(n) := j _ n ^T \eta $. The $\ell _ 1(X^*)$- norm of $y$ is $$\|y\|_ { \ell _ 1 (X^*)}=\sum _{n\in\mathbb{N}}\, \|y(n)\| _ {X ^ *} = \sum _{n\in\mathbb{N}}\, \, \sup _ {\|x\| _ X \le 1} \langle y(n), x \rangle=$$$$ = \sup _ {m\in\mathbb{N}}\, \, \sup _ {\|\xi\| _ { c _ {0} (X)} \le 1} \, \sum _{n\le m}\, \langle y(n), \xi(n) \rangle =$$$$= \sup _ {\|\xi\| _ { c _ {0} (X)} \le 1} \, \langle \eta, \xi \rangle = \|\eta\| _ {c _ 0(X)^*}\, \, .$$ This shows that the inclusion $\ell _ 1(X^*)\to c _ 0(X)^ * $ is actually a linear (surjective) isometry.
answered Mar 20, 2012 at 1:17