Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to prove this that but I can't . Any help/reference ?

share|improve this question
    
Do you mean just $l_1(x)$ not $l_1(X^*)$? –  Owen Sizemore Mar 20 '12 at 0:35
    
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. –  Yemon Choi Mar 20 '12 at 0:38
    
Also, how did the question arise? (I am not sure whether a hint or explanation would be more useful than a mere reference.) –  Yemon Choi Mar 20 '12 at 0:40
4  
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. –  Bill Johnson Mar 20 '12 at 0:45
    
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. –  Rafael Mar 20 '12 at 1:08

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
thanks a lot :) –  Rafael Mar 20 '12 at 2:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.