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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 $(\ell^\infty)^*$. $\endgroup$ Commented Jul 16, 2014 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
    Commented Jul 16, 2014 at 8:59

6 Answers 6

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

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    $\begingroup$ 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
    Commented Jul 16, 2014 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
    Commented Jul 16, 2014 at 9:35
  • $\begingroup$ Thank you, now is much clearer :-) $\endgroup$
    – Benzio
    Commented Jul 16, 2014 at 9:42
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    $\begingroup$ 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. $\endgroup$ Commented Jul 17, 2014 at 11:50
  • $\begingroup$ @MartinSleziak Thanks. ${}$ $\endgroup$
    – martini
    Commented Jul 17, 2014 at 12:15
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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 $\ell_\infty^*$.

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  • $\begingroup$ BTW the isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$ is described in the Wikipedia article on Stone–Čech compactification. $\endgroup$ Commented Jul 17, 2014 at 11:54
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    $\begingroup$ 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 $\beta\mathbb N$. $\endgroup$ Commented Jul 17, 2014 at 11:57
  • $\begingroup$ Oh, yes, but functional analysis without choice is a nightmare! $\endgroup$ Commented Jul 17, 2014 at 13:10
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    $\begingroup$ 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. $\endgroup$ Commented Jul 20, 2014 at 12:54
  • $\begingroup$ Are the $\delta_x$ really in $C(\beta \mathbb{N})$? After all, each $x \in \beta\mathbb{N} \setminus \mathbb{N}$ is a limit of elements of $\mathbb{N}$, and since $\delta_x(y) = 0$ for $y \in \mathbb{N}$, we should have $\delta_x(x) = 0$, which is a contradiction. $\endgroup$
    – Smiley1000
    Commented Nov 21, 2023 at 14:01
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For any free 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 $\mathscr U$ is a free ultrafilter and $(x_n)$ 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:

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  • $\begingroup$ "Since f(ei)=0, this function is not from ℓ1"why this function is not from ℓ1? $\endgroup$
    – math112358
    Commented Nov 7, 2018 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.) $\endgroup$ Commented Nov 7, 2018 at 19:48
  • $\begingroup$ This requires the ultrafilter to be free, right? $\endgroup$
    – Smiley1000
    Commented Nov 21, 2023 at 11:28
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    $\begingroup$ @Smiley1000 Thanks for the correction, I have added the condition that $\mathscr U$ has to be a free ultrafilter. (This is needed to show that the $\mathscr U$-limit extends the usual limit.) $\endgroup$ Commented Nov 21, 2023 at 14:03
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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jan 27, 2020 at 5:01
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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.

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  • $\begingroup$ Your functional is not well defined as limit might not exists. $\endgroup$
    – user99914
    Commented Sep 14, 2018 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
    Commented Sep 16, 2018 at 3:12

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