This is a question from an old exam:
Let $\ell_0$ be a linear functional defined on $c \subset \ell^{\infty}$ (on the set of all convergent sequences) by $$\ell_0(x)=\lim_{n\to\infty}x_n.$$ Show that the set $\mathcal{L}$ of all continuous linear extensions of $\ell_0$ to $\ell^{\infty}$ is closed in the weak-$*$ topology on $(\ell^{\infty})'$."
I think that a a sequence $ \{ l_n \} $ is sequentially weak-$*$ convergent in $(\ell^{\infty})'$ if for every $x \in \ell^{\infty}$, $\lim_{n\to\infty} \ell_n (x)=\ell(x)$, but is this the same thing as being convergent in the weak-$*$ topology here?
If somebody could help me solve this exercise I would be very grateful.
Let $f\notin\mathcal L$, then there exists an $x\in c$ so that $f(x)\neq\lim_n x_n$. Denote $|f(x)-\lim_n x_n|$ as $r$. Consider the set: $$B_r(f;x)=\{g\in\mathscr (l^\infty)^*\mid |f(x)-g(x)|<r\}$$ For any $g\in B_r(f;x)$ you have $g(x)\neq\lim_n x_n$ simply from the lower triangle inequality, thus $g\notin\mathcal L$. On the other hand $B_r(f;x)$ is an open set (it is the pre-image of $(-r,r)$ under the map $g\mapsto g(x)-f(x)$, which is continuous on the weak-$*$ topology).
So for every point not in $\mathcal L$ you have an open set that does not intersect $\mathcal L$ and that contains the point. So the complement of $\mathcal L$ is open and $\mathcal L$ is closed.
It is not enough to show that any weak-$*$ limit of elements in $\mathcal L$ lies in $\mathcal L$, because $(\ell^\infty)^*$ is not a sequential space, meaning the sequential closure is not the same as the topological closure. For a counter example see here.