Weak star closedness and norm closedness I wondering about the following: Suppose we have a set $C\subset L^\infty$ which is closed in the weak-star topology $\sigma(L^\infty,L^1)$. Therefore we view $L^\infty$ as dual of $L^1$. Is this set $C$ then also norm closed on $L^\infty$? 
 A: One of the interests of weak, weak-star topologies on normed spaces is that they are contained in the norm topology (therefore, we can hope to have more compact sets).
Here, the complement of $C$ is open in the topology $\sigma(E^*,E)$ (nothing specific about $L^p$ spaces), hence in the strong topology, hence $C$ is closed in this topology.
A: It can be easy to get confused by notation like $\sigma(E^*,E)$, so here's a direct argument.
Let $E$ be any Banach space (in your example, take $E = L^1$).  Suppose $C$ is  a weak-* closed subset of $E^*$; we wish to show it is also norm closed.  So suppose $f$ is in the norm closure of $C$.  Then there is a sequence $f_n \in C$ converging in norm to $f$ (the norm topology is a metric space, so we can reason about it using sequences).  In particular, for every $x \in E$, we have $|f_n(x) - f(x)| \le \|f_n - f\| \|x\| \to 0$, so $f_n(x) \to f(x)$.  But this is precisely what it means for $f_n$ to converge to $f$ in the weak-* topology.  $C$ is weak-* closed, so $f \in C$.  Thus we have shown that $C$ contains its norm closure, which is to say it is norm closed.
