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Consider $\ell_1$ as a subspace of $(\ell_\infty)^*$ this is: $$\ell_1 \cong S:=\left\lbrace T \in (\ell_\infty)^* \colon \exists x\in \ell^1 \, \forall y \in \ell_\infty \ \ T(y)=\sum_{i=1}^\infty x_i y_i\right\rbrace$$ Then I need to calculate $$(\ell_1)^\perp = \{x^{**} \in (\ell_\infty)^{**} \colon x^{**}(T)= 0 \ \ \forall T \in S \}$$ Is $(\ell_1)^\perp = \{0\}$?

I don't know how to proceed, any help will be appreciated.

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    $\begingroup$ It's not $\{0\}$. $S$ is a proper closed subspace of $(\ell_\infty)^{\ast}$. Have you a description of $(\ell_\infty)^{\ast\ast}$? $\endgroup$ – Daniel Fischer Oct 6 '15 at 15:01
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Let $\Lambda = \{ x \in l_\infty | x_k \text{ has a limit} \}$. Define $\lambda(x) = \lim_k x$ for $x \in \Lambda$. It is easy to check that $\|\lambda\| = 1$. The Hahn Banach theorem allows us to extend the domain of $\lambda$ to $l_\infty$ without increasing the norm, we abuse notation slightly by letting $\lambda$ denote this extension.

Let $e_n=(0,...0,1,0,...)$, with the one being in the $n$th place. Note that $e_n \in \Lambda$ and $\lambda(e_n) = 0$ for all $n$.

Define $\phi$ by $\phi(f)(x) = f(x)-\sum_n f(e_n) x_n$. Note that $|\phi(f)(x)| \le |f(x)| + \sum_n |f(e_n)| \|x\| \le (\|f\|+\sum_n |f(e_n)|)\|x\|$, hence $\|\phi(f)\| \le \|f\|+\sum_n |f(e_n)|$. A little more work shows that $\sum_n |f(e_n)| \le \|f\|$, hence $\|\phi\| \le 2$ and so $\phi: l_\infty^{*} \to l_\infty^{*} $ is continuous.

Now show that $S = \ker \phi$, this shows that $S$ is closed. Now evaluate $\phi(\lambda)$ at some point to show that $\phi(\lambda) \neq 0$ and so $S^\bot $ is not trivial.

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