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On this link http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2004;task=show_msg;msg=1414.0001 is the argument that a linear subspace in a normed space is closed w.r.t. norm iff it is weakly closed.

On the other hand, $c_0$ (sequences convergent to $0$) is a norm-closed linear subspace of $l_\infty$ (bounded sequences), but it is not weakly closed, since the base vectors $e_i$ are weakly dense in $l_\infty$.

Since I studied func.an. quite a while ago, my question is - what am I missing?

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How do you prove that the base vectors are weakly dense in $\ell_\infty$? –  Davide Giraudo Jul 22 '13 at 8:32
    
It is a direct consequence of Hahn-Banach (geometric) that the norm closure of a convex set in a normed vector space (or the closure in a locally convex topological vector space) is equal to its weak closure. In particular, $c_0$ is indeed weakly closed in $\ell^\infty$. –  1015 Jul 22 '13 at 12:13
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the (span of the) base vectors is weak* dense, but not weakly dense. –  GEdgar Jul 22 '13 at 12:51

2 Answers 2

up vote 4 down vote accepted

Edit: the previous answer was wrong. Below is a new version.

Weak convergence of a net $\{x_n\}$ to $1\in\ell^\infty$ means that, for any functional $\varphi$ in the dual of $\ell^\infty$, $\varphi(x_n-1)\to0$.

We can see $\ell^\infty$ as $C(\beta\mathbb N)$, the continuous functions on the Stone-Čech compactification of $\mathbb N$. Let $\omega\in\beta\mathbb N\setminus\mathbb N$ (in other words, $\omega$ is a free ultrafilter). Then $1(\omega)=1$ and $x(\omega)=0$ for all $x\in c_0$. So we have, letting $\varphi_\omega$ be the point-evaluation at $\omega$, $$ \varphi_\omega(x-1)=\varphi_\omega(x)-\varphi_\omega(1)=0-1=-1, $$ and so no net in $c_0$ will make the limit go to zero. This means that $c_0$ is not weakly dense in $\ell^\infty$. Weak-star density works because one has to deal with less functionals.

As was mentioned, the Hahn-Banach theorem guarantees that $c_0$ (being convex) is both norm and weakly closed.

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Thank you, I can see that I have confused weak with *-weak topology. Let me make sure if I understand the details correctly: we can embed $l^\infty$ (with weak topology) into a direct product of copies of $\mathbb{R}$, indexed by elements of $l^1$. Your argument shows that for each $x\in c_0$ there exists an element $z\in l^1$, such that neighborhoods of '1' in that product, with values in z-coordinate <1/2, do not contain $x$. I am still a little confused - why does it imply 1 is not in the weak closure of $c_0$? We cannot intersect all such neighborhoods, since it will not be an open set. –  user35953 Jul 23 '13 at 3:30
    
It is simpler than that. Weak convergence of a net $\{x_n\}$ in $c_0$ (each $x_n$ is an element of $c_0$) to $1$ means that for any $f\in (\ell^\infty)^*$, $f(x_n-1)\to0$. In my answer I'm showing that I can find an $f$ (many, actually) such that the limit does not go to zero; so $x_n$ does not converge weakly to $1$. The trick is that I'm only using those functionals coming from $\ell^1$. –  Martin Argerami Jul 23 '13 at 3:41
    
@David: you are completely right, my answer is way off the mark. I'll change it completely. –  Martin Argerami Jul 23 '13 at 15:38

$c_0$ is not weakly dense in $\ell_\infty$. Indeed, let $u:=\sum_{k=0}^{+\infty}e^{2k}$, where $e^k_j:=\delta_{kj}$. Take a Banach limit $L$. Then $$U:=\{x\in\ell_\infty, |L(x-u)|<1/3\}$$ is a neighborhood of $u$ in the weak topology. If $x\in U\cap c_0$, then $L(x)=0$, hence $|L(u)|<1/3$. It's a contradiction, since $L(u)=1/2$.

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