1
$\begingroup$

Let $X$ be a Banach space. Define $c_0(X)$ to be the vector space of all sequence $(x_n)$ of $X$ such that $\lim |x_n|=0$; and $\ell_1(X^*)$ to be the vector space of all sequence $(y_n)$ of $X^*$ such that $\sum_{n=1}^\infty |y_n|<\infty$. If $c_0(X)$ is equipped with the norm $$|(x_n)|_0 = \max |x_n|$$ and $\ell_1(X^*)$ is equipped with the norm $$|(y_n)|_1= \sum_{n=1}^\infty |y_n|$$ is it true that $c_0(X)^*$ is isomorphic to $\ell_1(X^*)$?

$\endgroup$
  • $\begingroup$ Well i think not in general! Take $X=\mathbb{R}$ and $\left{\frac{1}{n}\right\}\in c_{0}$ but certainly not in $\mathscr{l}^{1}$, since the harmonic sum diverges $\endgroup$ – TheOscillator Nov 18 '15 at 23:34
  • 1
    $\begingroup$ @TheOscillator I was going to say the same, but the OP was edited. $\endgroup$ – Ian Nov 18 '15 at 23:35
  • 1
    $\begingroup$ Ah, he sure did. Well at least i tried answering some question $\endgroup$ – TheOscillator Nov 18 '15 at 23:36
  • $\begingroup$ When $X$ is Hilbert, given $y_n \in \ell^1$ you have the functional $f_{y_n}(x_n)=\sum_{n=1}^\infty (x_n,y_n)$. This is definitely bounded on $c_0$ with the norm you've given, indeed it is bounded on all of $\ell^\infty$. So at least in the Hilbert case, the question is whether there are any other bounded linear functionals on $c_0$. I sort of expect the answer is yes, because (assuming the axiom of choice) the dual of $L^\infty([0,1])$ is not isomorphic to $L^1([0,1])$. $\endgroup$ – Ian Nov 18 '15 at 23:37
  • 1
    $\begingroup$ @Ian But the dual of the standard $c_0$ is $\ell^1$. $\endgroup$ – David C. Ullrich Nov 19 '15 at 0:28
3
$\begingroup$

There was a major misconception here in the original version; the obvious guess for the dual of $c_0(X)$ is $\ell^1(X^*)$, not $\ell^1(X)$.

In fact the dual is $\ell^1(X^*)$. Leaving the easier inclusion to you:

Say $L$ is a bounded linear functional on $c_0(X)$. Let $e_j$ be the vector consisting of all zeroes except for one $1$, in the $j$-th place. Now $$x\mapsto L(xe_j)$$ defines a bounded linear functional on $X$, so there exists $y_j\in X^*$ such that $$L(xe_j)=y_j(x)\quad(x\in X).$$

Let $c_{00}(X)$ denote the subspace of sequences that vanish except for finitely many terms. By linearity it follows that $$Lx=\sum_jy_j(x_j)\quad(x\in c_{00}(X)).$$It follows easily from this that $$\sum_{j=1}^N||y_j||\le||L||.$$ So $\sum||y_j||<\infty$, so $y=(y_1,y_2,\dots)\in\ell^1(X^*)$.

Now $y$ defines an element of the dual of $c_0(X)$ that agress with $L$ on $c_{00}(X)$. Since $c_{00}(X)$ is dense in $c_0(X)$ it follows that $y=L$, or less informally that $$Lx=\sum y_j(x_j)\quad(x\in X).$$

$\endgroup$
  • $\begingroup$ I'm sorry for the error, you are right, I'll edit it. Thanks. $\endgroup$ – user34870 Nov 19 '15 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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