Functional Analysis – Dual of l^\infty is Not l^1

Question

I know that the dual space of $l^\infty$ is not $l^1$, but I didn't understand the reason.
Could you give me a example of an $x \in l^1$ such that if $y \in l^\infty$, then $ f_x(y) = \sum_{k=1}^{\infty} x_ky_k$ is not a linear bounded functional on $l^\infty$, or maybe an example of a $x \notin l^1$ such that if $y \in l^\infty$, then $ f_x(y) = \sum_{k=1}^{\infty} x_ky_k$ is a linear bounded functional on $l^\infty$?

Answer 1

The point is the following: There are bounded functionals on $\ell^\infty$, which are not of the form $$ f(y) = \sum_k x_k y_k $$ for some $x$. I do not know if such a functional can be given explicitly, but they do exist. Let $f \colon c \to \mathbb R$ (where $c \subseteq \ell^\infty$ denotes the set of convergent sequences) be given by $f(x) = \lim_n x_n$. Then $f$ is bounded, as $|\lim_n x_n| \le \sup_n |x_n| = \|x\|$. Let $g \colon \ell^\infty \to \mathbb R$ be a Hahn-Banach extension. If $g$ where of the above mentioned form, we would have (with $e_n$ the $n$-th unit sequence) $$ x_n = g(e_n) = f(e_n) = 0 $$ hence $g = 0$. But $g \ne 0$, as for example $g(1,1,\ldots) = 1$.