Dual of $l^\infty$ is not $l^1$ 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 $l^\infty$?
 A: 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 $\ell_\infty^*$.
A: 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?
A: 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 $g(1,1,\ldots) = 1$.
A: 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.
A: This isn't the usual answer (the usual answer is the one given by Martini) but I believe it works. Consider $X$ the subspace of $\ell^\infty$ given as the span of $e_1,e_2,\ldots$, and $(1,1,1,\ldots)$. (Here, $e_i:=(0,\ldots,0,1,0,\ldots)$, with the $1$ in the $i$-th slot). Define $$ f: X \to \mathbb{C} $$ via $f(e_i)=\frac{1}{2^{i}}$ and $f(1,1,\ldots)=1.1$, and extend it to the whole of $X$ using linearity. In particular, $f(1,1,\ldots) \neq \lim_{i\to\infty} f(e_1+\ldots+e_i)$. We claim $|f(x)| \le 20||x||_\infty$ for any $x\in X$. This is because, if $x=\sum_{i=1}^\infty z_ie_i + z(1,1,\ldots)$, with only finitely many of the $\{z_i\}$ nonzero, then $||x||_\infty = \sup_N\{|z+z_N|\}$, and $|f(x)| = |1.1z+\sum_{i=1}^\infty \frac{z_i}{2^{i}} |$. WLOG $z_i=0$ for all $i>M$, where $M$ is a big number. We claim there must exist an $N$ with $$ 20|z+z_N| \ge |1.1z + \sum_{i=1}^M \frac{z_i}{2^{i}}|.$$ To see this, we have that the RHS is at most $1.1|z| + \sup_i\{|z_i|\}$. Assume such an $N$ doesn't exist. Then $N=M+1$ fails so $20|z| < 1.1|z| + \sup|z_i|$ and so $|z_n| > 18|z|$, for the $n$ with $\sup|z_i|=|z_n|$. Then $N=n$ must also fail so $20|z+z_n| < 1.1|z|+ |z_n|$. But 
$$ 20|z+z_n| \ge 20*\frac{17}{18}|z_n| \ge 2.1|z_n| \ge 1.1|z|+\sup|z_i|,$$ so in fact $n=N$ works! A contradiction, so $|f(x)|\le 20||x||_\infty$ is verified.
Since $f$ is bounded, Hahn-Banach now implies we can extend $f$ to a bounded linear functional on $\ell_\infty$ to obtain some $F\in \ell_\infty^*$, which satisies $|F(x)| \le 20||x||_\infty\forall x\in\ell_\infty$. This $F$ will satisfy $F(1,1,\ldots)=1.1 \neq \lim_{i\to\infty} \sum_{n=1}^i F(e_n)$, so it can't come from an element of $\ell_1$. (If it did, this element of $\ell_1$ would necessarily be $(\frac{1}{2}, \frac{1}{4},\ldots)$.) Thus, $F\in\ell^*_\infty$ but $F\not\in\ell_1$, finishing the problem.
A: Counterexample: 
Consider the linear functional $\phi$, defined on $l_\infty$ and given by  $\phi(x) = \lim_{N\rightarrow\infty} \frac{1}{N}\sum_{n=1}^N x_n$. Now, $\phi$ (which you might call the average functional) is bounded since $|\phi(x)|\le \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N ||x||_\infty = ||x||_\infty$.
Assume $\phi(x) = \sum_{n=1}^\infty x_n y_n$, for some $y\in l_1$.
Then, for $\delta = \{1, 0, 0, ... \}\in l_\infty$, you have $\phi(\delta)=0=y_1$. Similarly, for each $e^n = \{0, ... 0, 1, 0, ...\}$  with a "single $1$" at position $n$ (thus $\delta=e^1$), you have that $\phi(e^n)=0=y_n$. Then $y_n = 0_n$ is the zero sequence. Then $\phi$ is the zero functional. But, since, e.g. for $x_n = (-1)^n$ you have $\phi(x)=0.5\ne 0$, you have a contradiction. 
