# 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 $l^\infty$?

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

• 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 $l^1$... – Benzio Jul 16 '14 at 9:32
• 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 $0$. – martini Jul 16 '14 at 9:35
• Thank you, now is much clearer :-) – Benzio Jul 16 '14 at 9:42
• 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. – Martin Sleziak Jul 17 '14 at 11:50
• @MartinSleziak Thanks. ${}$ – 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:

• "Since f(ei)=0, this function is not from ℓ1"why this function is not from ℓ1? – math112358 Nov 7 '18 at 19:38
• @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.) – 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 $\ell_\infty^*$.

• BTW the isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$ is described in the Wikipedia article on Stone–Čech compactification. – Martin Sleziak Jul 17 '14 at 11:54
• 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 $\beta\mathbb N$. – Martin Sleziak Jul 17 '14 at 11:57
• Oh, yes, but functional analysis without choice is a nightmare! – Tomasz Kania Jul 17 '14 at 13:10
• 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. – Martin Sleziak Jul 20 '14 at 12:54

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.

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.

• Alas, if you use $f(e_i)=0$ and $f(1,1,\ldots,1)=1$ then the same proof works and it's much easier. Moreover it gives essentially the same construction as the "standard" one above. – Juan Carlos Ortiz Jan 27 at 5:01

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.

• Your functional is not well defined as limit might not exists. – user99914 Sep 14 '18 at 16:28
• 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. – A.Restrepo Sep 16 '18 at 3:12