Example of $N\subseteq E^*$ such that $(N^\perp)^\perp \neq \overline{N}$

Question

Let $E$ be a normed space, $E^*$ its dual space and $F\subseteq E$, $G\subseteq E^*$ subspaces. As usual, we define $$F^\perp:=\{f\in E^*: f(F)=0\}$$ and $$G^\perp:=\{x\in E: g(x)=0, \ \forall \ g \in G\}.$$ I'd like some help with this exercise from Brezis's Functional Analysis, Sobolev Spaces and PDE:

Let $E=l_1$, so that $E^*=l_\infty$. Consider $N=c_0$ as a closed subspace of $E^*$. Check that $(N^\perp)^\perp \neq N$.

Here, $c_0$ is the set of bounded sequences with limit $0$.

My attempt. Observing the way $l_\infty$ is associated with $(l_1)^*$ I think that, in this case, $$N^\perp=\{(x_n)\in l_1:\Sigma x_ny_n=0, \ \forall \ (y_n)\in c_0\}$$ and $$(N^\perp)^\perp = \{(z_n)\in l_\infty:\Sigma z_nx_n=0, \ \forall \ (x_n)\in N^\perp\}.$$ So, since $(N^\perp)^\perp \supseteq N$, I have to find a sequence in $(N^\perp)^\perp \backslash N$. I've been trying to get it, but I couldn't. I'd be grateful for any hints or comments. Thanks in advance!

Answer 1

I'll provide some hints. If you require clarification, let me know and I'll expand.

You do have the correct forms for $N^\perp$ and $(N^\perp)^\perp$ (I find the notation slightly confusing, and much prefer Rudin's $N_\perp$ for the pre-annihilator). Now just identify what $N^\perp$ and $(N^\perp)^\perp$ are.

One can easily check that $N^\perp=0$, that is, for any $x=(x_n)\in\ell_1$ non-zero, we can find some $y=(y_n)\in c_0$ such that $\sum_nx_ny_n\neq0$. Can you now identify $(N^\perp)^\perp$?

Answer 2

As already noted, it's easy to see that $N^\perp=0$, hence $N^{\perp\perp}=\ell_\infty$. Someone should add a comment that the reason this is possible is that $N$ in this example is not weak-* closed; in fact the weak-* closure of $c_0$ is all of $\ell_\infty$. (In general, if $E$ is a Banach space and $N$ is a subspace of $E^*$ then $N^{\perp\perp}$ is the weak-* closure of $N$.)