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there is a well known fact that $l^1$ is not the dual of $l^\infty$. An exercise Folland's Real analysis serves as an example for this.(Page 192 ex 19)

Define $\phi_n \in (l^\infty)^*$ by $\phi_n(f)=n^{-1}\sum_1^nf(j)$, Then the sequence ${\phi_n}$ has a weak* cluster point $\phi$, and $\phi$ is an element of $(l^\infty)^*$ that does not arise from an element of $l^1$.

By the knowledge of elementary calculus, it's not difficult to show that for each $f\in l^\infty$, the sequence $\phi_n(f)$ has cluster points, but I don't know how to determine an element $\phi$ of $(l^\infty)^*$ via this, because there can be millions of cluster points. How can I make $\phi$ linear and bounded?

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  • You don't need to find $\phi$ explicitly. If $e_m$ is the $m$'th standard unit vector, what is $\phi(e_m)$? (Note if $\phi$ arose from an element of $\ell_1$, $\phi(e_m)$ would be its $m$'th coordinate.) – David Mitra Jan 12 '15 at 21:58

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it's not difficult to show that for each $f\in l^\infty$, the sequence $\phi_n(f)$ has cluster points

You are right, this elementary calculus argument (looking at the values $\phi_n(f)$ for each $f$ individually) is not enough. Instead, you should use Alaoglu's theorem (5.18 in the book), which provides a weak* cluster point $\phi$.

There can be no explicit form for $\phi$, because the Axiom of Choice is an essential tool behind this argument (precise, technical facts here). But the fact that $\phi$ is a cluster point of that sequence tells you that

  1. $\phi$ takes value $0$ on each $e_m$
  2. $\phi$ takes value $1$ on the constant sequence $1,1,\dots$.

Hence $\phi$ does not arise from $\ell^1$.

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