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In Functional Analysis we treated the Hahn-Banach theorem, and if I understood correctly, the dual space of $l^1$ (space of all absolutely summable sequences) is isomorphic to the space of all bounded sequences, $l^\infty$. However, we also saw that the dual of $l^\infty$ is NOT isomorphic to $l^1$, but is bigger. In fact, we proved the existence of a nonzero functional $f:l^\infty \to \mathbb{R}$, which satisfies $f(e_n) = 0$ for all $n \in \mathbb{N}$, where $e_n$ is the sequence $(e_n)_k = \delta_{nk}$. My question is, how is this possible? Every bounded sequence is a linear combination of $e_n$'s, so how can $f$ be nonzero?

Thanks in advance.

user337331
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  • Let $u_N = \sum_{n =1}^N (u)n e_n$. In $l^\infty$ it is weird because If you only know that $f$ is continuous $l^\infty \to \mathbb{R}$, you can't write $f(u) = \lim{N \to \infty} f(u_N)$ since $|u - u_N|$ doesn't always $\to 0$ as $N \to \infty$, whereas it does in $l^p$ for every $p \in [1, \infty[$ (in $l^p$ we can write that $f(u+\epsilon v) \to f(u)$ as $\epsilon \to 0$ for any $u,v \in l^p$, hence $f(u_N) = f(u + (u_N-u)) \to f(u)$ as $N \to \infty$). Now I admit that I don't know how to define the general form of $f \in (l^\infty)^*$, did you ? (we'll probably need measures) – reuns May 05 '16 at 21:14
  • So in $l^\infty$ there are $c$ and $c_0$ the (closed, Banach) subspaces of convergent and converging to $0$ sequences. In class you proved that there is $f \in (l^\infty)^*$ such that $f(u) = 0$ for any $u \in c_0$, but $f(v) \ne 0$ for some $v$ ? Did you say that $f(u) = \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N (u)_n$ for every $u \in c$ is such an example, and use Hahn-Banach for extending it to any $u \in l^\infty$ ? – reuns May 05 '16 at 21:29
  • We used the result that if $S$ is a linear subspace of $X$ which is not dense in $X$, then there exists a nonzero functional $f$ such that $S\subset ker f$. In this case we chose $S={(x)_n \in l^\infty | (x)_n \neq 0 \text{ for finitely many } n}$ – user337331 May 05 '16 at 21:38
  • Your subspace $S$ is not dense in $\ell^{\infty}$ because every element of $S$ is at least 1 unit away from the constant sequence ${ 1,1,1,1,\cdots }\in \ell^{\infty}$ – Disintegrating By Parts May 05 '16 at 21:46
  • @TrialAndError : if I didn't write too many mistakes, do you see any way for visualizing how $f(u) = \lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N (u)_n$ for $u \in c$ can be extended to $l^\infty$ ? – reuns May 05 '16 at 21:51
  • Take a look at this question: http://math.stackexchange.com/questions/47395/the-duals-of-l-infty-and-l-infty – Disintegrating By Parts May 05 '16 at 22:19
  • @TrialAndError : I saw that, but probably as the OP, I understand nothing of $\beta \mathbb{N}$ – reuns May 05 '16 at 22:30
  • @user1952009 : The accepted answer to that question does not use the Stone-Czech compactification. – Disintegrating By Parts May 05 '16 at 22:32
  • @TrialAndError : the link in the 1st comment is useful, but it doesn't say how we define such measures $\lambda$ in general, and how $\lambda(E)$ probably isn't always definable for any weird subset $E \subset \mathbb{N}$. – reuns May 05 '16 at 23:01
  • @user1952009 : If you started with $\Phi \in (\ell^{\infty})^{*}$, how do you think you might define a finitely-additive set function $\mu E$ for any $E\subseteq\mathbb{N}$? – Disintegrating By Parts May 05 '16 at 23:13
  • @TrialAndError : $\mu(E) = \Phi(I_E)$ (with $I_E$ the sequence of $0,1$ indicating $E$), but what I ask is how defining those for some weird sets $E$, I think that in general it stays undefinable, and we can just say "the function extended to the whole $l^\infty$ exists by Hahn-Banach" but not define it. – reuns May 05 '16 at 23:25
  • @user1952009 : That function you stated would be defined for all subsets $E$ of $\mathbb{N}$. It would be finitely-additive. Not a problem. And you can see that the variation of $\mu$ would be finite. – Disintegrating By Parts May 05 '16 at 23:43
  • @TrialAndError : when you say it would be defined, do you need the axiom of choice, or can you really define a function $\Phi : l^\infty \to \mathbb{R}$ such that for any definable subset $E \subset \mathbb{N}$ you can define the value of $\Phi(I_E)$ ? (and say for which $\Phi(u) = \lim_{n \to \infty} u_n$ for any $u \in c$) – reuns May 05 '16 at 23:54
  • @user1952009 : At this point, I'm not sure what you're asking. – Disintegrating By Parts May 06 '16 at 00:02

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we can take $Y$ to be the linear span of all $e_{n}$ and $x=\{ 1,1,1,....\}$. Since $x$ is not in $\bar{Y}$, by Hahn-Banach theorem there is a bounded functional $f$ on $l^{\infty }$ such that vanish on $Y$ and $f(x)=1$.

M. H. Sattari
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