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 $\ell_1(X^*)$?

  • $\begingroup$ 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
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    $\begingroup$ @TheOscillator I was going to say the same, but the OP was edited. $\endgroup$ – Ian Nov 18 '15 at 23:35
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    $\begingroup$ 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
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    $\begingroup$ @Ian But the dual of the standard $c_0$ is $\ell^1$. $\endgroup$ – David C. Ullrich Nov 19 '15 at 0:28

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 $$Lx=\sum y_j(x_j)\quad(x\in X).$$

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

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