Show that $L^1\subsetneq (L^\infty)^*$

Question

How does one show that $L^1\subsetneq (L^\infty)^*$? I am having trouble in this. Any help would be appreciated.

Answer 1

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 Hahn-Banach 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 Hahn-Banach 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 $\ell^1(\mathbb N)$.