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?
Best Answer
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.