# a reference for the isomorphism $\ell_1(E')\cong [c_0(E)]'$

Let $$E$$ be a Banach space, and let $$E'$$ denote its topological dual. Let us consider the spaces $$\ell_1(E')$$ and $$c_0(E)$$ defined by

$$\ell_1(E')=\{(x_n^{'})_{n=1}^\infty\subset E': \sum_{n=1}^\infty||x_n^{'}||<\infty\}$$, and

$$c_0(E)=\{(x_n)_{n=1}^\infty\subset E: x_n\longrightarrow 0 \ {\rm{in}} \ E\}$$.

Can anybody give a reference containing an elementary proof of the topological isomorphism $$\ell_1(E')\cong [c_0(E)]'$$ ?

With the same idea as when $$E=\mathbb C$$, define $$\gamma:\ell_1(E')\to c_0(E)'$$ by $$\tag1 \gamma(x')x=\sum_n x'_n(x_n),\ \ \ \ x\in c_0(E).$$ The conditions on $$\ell_1(E')$$ and $$c_0(E)$$ guarantee that $$\gamma$$ is well-defined (that is, the series makes sense). Moreover, $$\gamma$$ is isometric; indeed, from $$(1)$$ we get $$\|\gamma (x')\|\leq\sum_n\|x'_n\|=\|x'\|$$. Given $$\varepsilon>0$$ and $$m\in\mathbb N$$, for each $$n\leq m$$ choose $$x_n\in E$$ with $$\|x_n\|=1$$ and $$x'_n (x_n)>\|x'_n\|-\varepsilon/2^n$$, and put $$x_n=0$$ for $$n>m$$; then $$x=\{x_n\} \in c_0 (E)$$ and $$\gamma(x')x=\sum_nx'_n (x_n)=\sum_{n=1}^mx'_n (x_n)>-\varepsilon+\sum_{n=1}^m\|x'_n\|.$$ As we can do this for all $$\varepsilon$$ and $$m$$, we get $$\|\gamma (x')\|\geq\sum_n\|x'_n\|=\|x'\|$$, and thus $$\gamma$$ is isometric.

It is trivial that $$\gamma$$ is linear. If $$\gamma(x)=0$$, evaluating on sequences that have a single nonzero element we get that $$x'_n=0$$ for all $$n$$, so $$x'=0$$ and $$\gamma$$ is injective. Now let $$\phi\in c_0(E)'$$. Any element in $$c_0(E)$$ is of the form $$x=\sum_n x_n\,e_n$$, where $$x_n\in E$$ and $$e_n\in c_0(\mathbb N)$$ is the sequence with the $$n^{\rm th}$$ entry equal to 1 and zeroes elsewhere. From $$x_n\to0$$, we get that the series for $$x$$ converges (since the norm is the supremum norm). As $$\phi$$ is continuous, $$\phi(x)=\sum_n \phi(x_ne_n)=\sum_n\phi_n(x_n),$$ where $$\phi_n\in E'$$ is defined by $$\phi_n(x)=\phi(xe_n)$$ (by $$xe_n$$ we mean the sequence with $$x$$ in the $$n^{\rm th}$$ position and zeroes elsewhere). Thus $$\phi=\gamma(\{\phi_n\})$$, if we show that $$\{\phi_n\}\in\ell_1(E')$$.

Fix $$\varepsilon>0$$, $$m\in\mathbb N$$. For each $$n\leq m$$, choose $$x_n\in E$$ with $$\|x_n\|=1$$, $$\phi_n(x_n)>\|\phi_n\|-\varepsilon/2^n$$, and $$x_n=0$$ if $$n>m$$. Then $$x=\{x_n\}\in c_0(E)$$, and $$\|\phi\|\geq|\phi(x)|=\sum_n\phi_n(x_n)=\sum_{n=1}^m\phi_n(x_n)\geq-\varepsilon+\sum_{n=1}^m \|\phi_n\|.$$ As we can do this for all $$\varepsilon>0$$ and all $$m\in\mathbb N$$, we get that $$\sum_n\|\phi_n\|\leq\|\phi\|$$ and so $$\{\phi_n\}\in\ell_1(E')$$. So $$\gamma$$ is surjective.

By the Inverse Mapping Theorem, $$\gamma$$ is an isomorphism.

• Thank for the very nice proof. Just one more question. It is easy to see that $||\gamma(x')||\leq||x'||_1$. Could you add the proof of the fact that the reverse implication is true also? – serenus Jan 10 at 7:53
• Done. You'll notice that the argument is familiar. – Martin Argerami Jan 10 at 10:06
• In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $\gamma (x')x$, isn't it? – serenus Jan 10 at 10:33
• Yes. $\ \ \ \ \$ – Martin Argerami Jan 10 at 15:35
• Thanks for the proof. – serenus Jan 10 at 15:43