Let $e_n = (0,0,\dots,1,0,0,\dots)$,(i.e. the n-th component is 1). Show that $e_n\rightharpoonup 0$ in $\sigma(\ell^{\infty},(\ell^{\infty})')$. I'm having trouble because the dual of $\ell^{\infty}$ is not $\ell^1$. Could someone give me some ideas? Thanks!
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You're right that you can't use the direct argument for $1 \le p < \infty$. Have you tried contradiction? This gives you an $x' \in (\ell^\infty)'$ and $\varepsilon > 0$ with $| x'(e_{n_k}) | \ge \varepsilon$ for some subsequence. Can you use this to construct $x \in \ell^\infty$ that exploits the $e_n$?
jubumbo
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Yeah, you are right! Thank you so much! Shouldn't have forgotten contradiction. – QD666 Oct 14 '18 at 04:09
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You can use a direct argument. – copper.hat Oct 14 '18 at 04:17
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Well, in some sense, $l_1$ is 'enough' of a dual for the purposes of this problem!
Suppose $f$ is a linear functional, let $g_k = f(e_k)$.
Choose $x_k$ such that $|x_k| = 1$ and $f(x_k) = |f(e_k)| = |g_k|$.
By choosing $y_n = (x_1,...,x_n,0,...)$ we see that $\|y_n\| = 1$ and $|f(y_n)| = \sum_{k=1}^n |g_k| \le \|f\|$ and so $g \in l_1$ and hence $f(e_k) = g_k \to 0$.
Aside: Not that it helps here, a little more work shows that any element of the dual $f$ can be written as $f=g+h$ where $g \in l_1$ and $h = 0$ on $c_0$.
copper.hat
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@daw: My answer was incorrect, thanks for pointing that out. I have repaired it. – copper.hat Oct 14 '18 at 19:16
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I would see this answer from the following perspective: Since $e_n \in c_0$ (space of null sequences with $\ell^\infty$ norm), it is sufficient to consider the restriction $f_{|c_0} \in (c_0)^* = \ell^1$ of $f \in (\ell^\infty)^*$. And to show $e_n \rightharpoonup 0$ in $c_0$ uses the 'classical argument'. – gerw Oct 16 '18 at 06:15
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@copper.hat How do you get $x_k$ such that $|x_k|=1$ and $f(x_k)=|f(e_k)|$? – Math Feb 23 '22 at 11:59
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