How does one show that $L^1\subsetneq (L^\infty)^*$? I am having trouble in this. Any help would be appreciated.
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Look here http://math.stackexchange.com/questions/711489/showthatlinftyisstrictlycontainedinl1?rq=1 (see answer, don't look in title of question) – Cortizol Mar 18 '14 at 17:02

@Cortizol Thanks! – Mar 18 '14 at 17:05

No problem. But that is also your question, so I am little bit confused. But if this is helpful I am glad to hear that. – Cortizol Mar 18 '14 at 17:08

very sorry, misread the question! :( – Frank Mar 18 '14 at 19:08
1 Answers
This is a very standard and yet elegant proof.
We shall create a bounded linear functional on $\ell^\infty(\mathbb N)$, which is not realized by any element of $\ell^1(\mathbb N) $, using the HahnBanach Theorem.
Let first $c(\mathbb N)$ be the set of converging sequences. Clearly $c(\mathbb N)$ is a closed subspace of $\ell^\infty(\mathbb N)$. Let $u=\{u_n\}_{n\in\mathbb N}\in c(\mathbb N)$. It is readily seen that the functional $$ \varphi(u)=\lim_{n\to\infty}u_n, \quad \varphi : c(\mathbb N)\to \mathbb R, $$ is a bounded linear functional of $c(\mathbb N)$, and in fact $\\varphi\=1$.
Using HahnBanach we can extend it to a bounded linear functional $\tilde\varphi$ on $\ell^\infty(\mathbb N)$, and with the same norm.
We shall show that $\tilde\varphi$ can not be realized by any element of $\ell^1(\mathbb N)$. Assume there is an $v=\{v_n\}_{n\in\mathbb N}$, such that $$ \tilde\varphi(u)=\langle u,v\rangle=\sum_{n=1}^\infty u_nv_n, $$ Setting $$u^n=(\underbrace{0,0,\ldots,0}_{n\,\,\text{zeros}},1,1,\ldots), $$ then $$ \langle u^n,v\rangle=\sum_{k=n}^\infty v_k, \quad\text{and hence}\quad \lim_{n\to\infty}\langle u^n,v\rangle=0. $$ However, $\tilde\varphi(u^n)=1$, since the limit of each sequence $u^n$ is equal to 1. Thus $\tilde\varphi$ is not representable by any element of $\ell^1(\mathbb N)$.
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It's something wrong with site I would say. I am on question "Show that $L^1\subsetneq (L^\infty)^*$", but in link it is `.....showthatl1subsetneqlinfty`. I really don't know. I am confused. – Cortizol Mar 18 '14 at 20:02

I was looking at site and I think this is answer for this question http://math.stackexchange.com/questions/717827/showthatmathcallinftyisnothomeomorphictomathcall1?lq=1. Again, maybe this is only for me, but I see your answer on question about $L^1$ etc and not $\ell^1$ – Cortizol Mar 20 '14 at 00:20