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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 $\ell_1(E')\cong [c_0(E)]'$ ?

Thanks in advance.

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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.

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  • $\begingroup$ 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? $\endgroup$ – serenus Jan 10 at 7:53
  • $\begingroup$ Done. You'll notice that the argument is familiar. $\endgroup$ – Martin Argerami Jan 10 at 10:06
  • $\begingroup$ In the isometric part, in the beginning of the inequality, I think $x'(x)$ should be $\gamma (x')x$, isn't it? $\endgroup$ – serenus Jan 10 at 10:33
  • $\begingroup$ Yes. $\ \ \ \ \ $ $\endgroup$ – Martin Argerami Jan 10 at 15:35
  • $\begingroup$ Thanks for the proof. $\endgroup$ – serenus Jan 10 at 15:43

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