Dual of $l^\infty$ is not $l^1$

Solution 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 $g(1,1,\ldots) = 1$.

Solution 2:

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?

Solution 3:

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 $\ell_\infty^*$.

Solution 4:

Another argument:

Let $e_n$ be the usual "basis" of $\ell^\infty$, i.e. $e_n$ is the sequence with a 1 in the $n$th position and 0 elsewhere, and let $e_n^* \in (\ell^\infty)^*$ be the "dual basis", i.e. $e_n^*(x) = x(n)$.

In particular, all the $e_n^*$ are in the unit ball of $(\ell^\infty)^*$, which by the Banach-Alaoglu theorem is compact in the weak-* topology. So the sequence $(e_n^*)$ must have at least one weak-* cluster point; let $f$ be one of them. Then for any $x \in \ell^\infty$, the number $f(x)$ must be a cluster point of the sequence of numbers $(e_n^*(x)) = (x(n))$. In particular, if $x = e_n$, then $f(e_n)$ is a cluster point of the sequence $(0,\dots, 0, 1, 0,0 ,\dots)$, so that $f(e_n) = 0$.

Thus if $f$ were a functional coming from $\ell^1$, it would have to be the zero functional. However, taking $x = 1$ to be the sequence of all $1$s, $f(1)$ is a cluster point of $(1,1,1,\dots)$, so $f(1)=1$. This is a contradiction.

The functional $f$ is rather interesting; it is somewhat similar yet different from a Banach limit. It has the interesting property that for any $x$, $f(x)$ is a cluster point of the sequence $x$; since $\mathbb{R}$ is first countable, that means $f(x)$ picks out some subsequential limit of $x$. So if $x = (1,0,1,0,1,0,\dots)$, then $f(x)$ will be either 0 or 1. However, this means that, unlike a Banach limit, it cannot be shift invariant.

Another interesting note is that the sequence $e_n^*$ does not have any weak-* convergent subsequence in $(\ell^\infty)^*$. For if there were some subsequence $e_{n_k}^*$ converging weak-* to some $g \in (\ell^\infty)^*$, we could consider an element $x \in \ell^\infty$ with $x(n_k) = 0$ for odd $k$ and $x(n_k) = 1$ for even $k$. Then the sequence $e_{n_k}^*(x)$, which is $(0,1,0,1,0,\dots)$, would have to converge to $g(x)$, which is absurd. This is no contradiction of the weak-* compactness of the ball, since the weak-* topology on the ball is not guaranteed to be first countable (indeed, it would only be first countable if $\ell^\infty$ were separable). So there will exist a weak-* cluster point, but it need not be a subsequential limit. This illustrates that metric space intuition is unhelpful for thinking about the weak-* topology in general.