Does this hold for $p=\infty $, i.e., is it true that $(l^{\infty})'= l^1? $

Question

Let $E=l^p$ where $1 \le p < \infty $ we know $E'=l^q$ Where $q$ is the dual exponent of $p$, i.e. $q$ is such that $\frac{1}{p}+\frac{1}{q}=1$

Does this hold for $p=\infty $, i.e., is it true that $(l^{\infty})'= l^1? $

And what is the $(l^{\infty})'= ?$

Answer 1

This is not true because of the following Theorem

$\textbf{Theorem:}$ If $X$ is a normed space such that its dual $X'$ is separable, then $X$ itself is separable.

So, if $(l^{\infty})' =l_1$, then it will follow that $l^{\infty}$ is separable which is not true.

Answer 2

Another proof. More elementary than Urban's proof. (But less general.)

Consider the subspace $c \subseteq l^\infty$ consisting of the real sequences that converge. So $c$ is a closed linear subspace of $l^\infty$. Let $L \;:\; c \to \mathbb R$ be the "limit" linear functional. That is, for $x = (x_1,x_2,\cdots)$, define $$ L(x) := \lim_{n\to\infty} x_n. $$
Then $L$ is a bounded linear functional on $c$, $L \in c'$. By the Hahn-Banach theorem, there is an extension $\widetilde{L}$ of $L$ with $\widetilde{L} \in (l^\infty)'$.

But we claim $\widetilde{L}$ is not the linear functional arising from any element of $l^1$. Indeed, if $y=(y_1,y_2,\dots) \in l^1$ what would it mean for $\langle y, x\rangle = \widetilde{L}(x)$ for all $x \in l^\infty$? Well, fix an $n$ and let $x=e_n$, the sequence with $1$ in the $n$th spot, zero elsewhere. We have $e_n \in c$ and $L(e_n)=0$, so $$ y_n = \langle y, e_n\rangle = \widetilde{L}(x) = L(x) = 0 $$ This holds for all $n$, so $y$ is the zero element of $l^1$.

On the other hand, for $x=u$ the sequence of all ones, $u=(1,1,1,\cdots)$, we have $u \in c$ and $L(u)=1$, so $$ 1 = L(u) = \widetilde{L}(u) = \langle y,u \rangle =\sum_{n=1}^\infty y_n \cdot 1 = 0 $$ Impossible.