Dual of $l^\infty$ is not $l^1$
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 $g(1,1,\ldots) = 1$.
For any ultrafilter $\mathscr U$ the function $$\newcommand{\Ulim}{\operatorname{{\mathscr U}-lim}}f \colon x = (x_n) \mapsto \Ulim x_n$$ is a bounded linear function from $\ell_\infty$ to $\mathbb R$.
Since $f(e^{i})=0$, this function is not from $\ell_1$.
The limit of a sequence $(x_n)$ along an ultrafilter $\mathscr U$ or ultralimit is defined as:
$$\Ulim x_n = a \qquad\Leftrightarrow\qquad (\forall \varepsilon>0) \{n\in\mathbb N; |x_n-a|<\varepsilon\}\in\mathscr U.$$
To prove that the function $f$ defined above has the required properties we can use the following facts:
- The $\mathscr U$-limit $\Ulim x_n$ exists for every bounded sequence $(x_n)$.
- If $(x_n)$ is a convergent sequence, then $\Ulim x_n = \lim\limits_{n\to\infty} x_n$.
- If $\Ulim x_n$ and $\Ulim y_n$ exist, then \begin{gather*} \Ulim (x_n+y_n) = \Ulim x_n + \Ulim y_n\\ \Ulim (x_n \cdot y_n) = \Ulim x_n \cdot \Ulim y_n \end{gather*}
- If $x_n\le y_n$ for each $n\in\mathbb N$, then $\Ulim x_n \le \Ulim y_n$.
For some basic facts and references about $\mathscr U$-limits, see:
- Basic facts about ultrafilters and convergence of a sequence along an ultrafilter
- Applications of ultrafilters
- Where has this common generalization of nets and filters been written down?
We can show actually more that $\ell_1$ and $\ell_\infty^*$ are not Banach-space isomorphic. (There are non-reflexive Banach spaces isometrically isomorphic to their second duals.)
If you accept the fact that $\ell_\infty \cong C(\beta \mathbb{N})$ (which follows from the very definition of the Stone–Čech compactification applied to the discrete space of natural numbers), we can prove more. Once you see this, the dual of $C(\beta \mathbb{N})$ is non-separable as it contains an uncountable discrete set $\{\delta_x\colon x\in \beta\mathbb{N}\}$ (here $\delta_x$ stands for the Dirac delta measure supported on $x$). Of course, $\ell_1$ is separable so it cannot be Banach-space isomorphic to $\ell_\infty^*$.