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 $l^\infty$?

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    $\begingroup$ $X^{**}$ always contains a copy of $X$ via the canonical embedding, so you cannot find an $\ell ^1$ element which is not in $(\ell^\infty)^*$. $\endgroup$ – Adam Hughes Jul 16 '14 at 8:53
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    $\begingroup$ It's hard to explicitly write down a functional like that, because the axiom of choice is necessary for producing such functional. $\endgroup$ – Asaf Karagila Jul 16 '14 at 8:59

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

  • $\begingroup$ Maybe I'm getting confused by something trivial but if $z$ is an element of $(l^\infty)^\ast$, i.e. a bounded linear functional on $l^\infty$ (maybe not in the form of a sum like in my post), then shouldn't be also a bounded linear functional on the subspace $c_0$ of $l^\infty$? But the dual of $c_0$ is isomorphic with $l^1$... $\endgroup$ – Benzio Jul 16 '14 at 9:32
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    $\begingroup$ Yes. But the restriction map $(\ell^\infty)^* \to c_0^*$ is not one-to-one, note for example that the functional $g$ given above restricts to $0$. $\endgroup$ – martini Jul 16 '14 at 9:35
  • $\begingroup$ Thank you, now is much clearer :-) $\endgroup$ – Benzio Jul 16 '14 at 9:42
  • $\begingroup$ Such functionals cannot be given explicitly. This fact was mentioned in several posts on this site, for example, math.stackexchange.com/questions/103476/…, math.stackexchange.com/questions/55651/… and the posts listed there among linked questions. $\endgroup$ – Martin Sleziak Jul 17 '14 at 11:50
  • $\begingroup$ @MartinSleziak Thanks. ${}$ $\endgroup$ – martini Jul 17 '14 at 12:15

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:

  • $\begingroup$ "Since f(ei)=0, this function is not from ℓ1"why this function is not from ℓ1? $\endgroup$ – mathrookie Nov 7 '18 at 19:38
  • $\begingroup$ @mathrookie If a function $f\in\ell_\infty^*$ is represented by $(c_n)\in\ell_1$, this means that $f(x)=\sum c_nx_n$. In particular we get $c_n=f(e^i)$. So the only possible representation would be using the zero sequence. However, zero sequence gives use the function $f=0$, which is not the case here. (BTW this is basically the argument explained in detail in martini's answer.) $\endgroup$ – Martin Sleziak Nov 7 '18 at 19:48

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 $\ell_\infty^*$.

  • $\begingroup$ BTW the isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$ is described in the Wikipedia article on Stone–Čech compactification. $\endgroup$ – Martin Sleziak Jul 17 '14 at 11:54
  • $\begingroup$ Since it was mentioned in several other posts that $\ell_1\ne\ell_\infty^*$ cannot be shown in ZF, this nice argument also must use some form of AC somewhere. I suppose that the place, where Axiom of Choice is used, is the existence of uncountably many elements in $\beta\mathbb N$. $\endgroup$ – Martin Sleziak Jul 17 '14 at 11:57
  • $\begingroup$ Oh, yes, but functional analysis without choice is a nightmare! $\endgroup$ – Tomek Kania Jul 17 '14 at 13:10
  • $\begingroup$ I certainly agree with that. But still the information that this cannot be shown in ZF is useful, since this tells us, that there must be some kind of non-constructive step in the proof of this. $\endgroup$ – Martin Sleziak Jul 20 '14 at 12:54

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.

  • $\begingroup$ Your functional is not well defined as limit might not exists. $\endgroup$ – user99914 Sep 14 '18 at 16:28
  • $\begingroup$ You are right. I found $x = \{x_n\} = \{1, -1, -1, -1, 1, ...\}$ with $x_n=\pm1$, with changes of sign at 1, 4, 12, 36 , 108... ($n_{k+1} = 3n_k$). Thanks. $\endgroup$ – A.Restrepo Sep 16 '18 at 3:12

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