Let $E$ be a normed vector space. For $S \subseteq E$ let $S_L$ denote $\{\phi \in E^*: \phi(x)=0 \forall x\in S\}$ and for $T\subseteq E^*$ denote $T^L=\{x\in E| \phi(x)=0 \forall \phi \in T$} Show $(S_L)^L$ is the closed linear span of $S$ and prove or disprove that $(T^L)_L$ is the closed linear span of $T$
First note that $S_L=\{\phi| \overline{SpanS}\subseteq \ker(\phi)\}$ and $T^L=\cap_{\phi \in T} \ker\phi$. First i show that $(S_L)^L=\overline{SpanS}$. It is clear that $\overline{SpanS}\subseteq (S_L)^L$ using the noted "new" definitions. Now let $x \not \in \overline{SpanS}$ then take the subspace generated by $\overline{SpanS}$ and $x$ and by hanh banach we can define a linear functional that has $\overline{SpanS}$ as kernel and $x$ is not in the kernel. Thus $x \not \in (S_L)^L$ as desired.
Trouble comes with the second statement. I think I showed that $\overline{SpanT}\subset (T^L)_L$ but I could not show the reverse direction. It is clear $Span(T)\subseteq (T^L)_L$ we also know that $(T^L)_L$ is closed since if $f_n \to f$ in norm, then also they converge pointwise, but if $f_n(x)=0 \forall n$ then $f(x)=0$. Is the reverse inclusion true or what is the counterexample?
It is enough to prove that $(T^L)_L$ is the weak$^*$-closure of $T$ in $E^*$. If $x^*\in T$ then $x^*(x)=0$ for every $x\in T^L$ so $T\subseteq (T^L)_L$. On the other hand, $(T^L)_L$ is weak$^*$ closed in $E^*$ so $\overline T^w\subseteq (T^L)_L$. Now, if $x^*\neq \overline T^w$, then the Hahn-Banach theorem gives us an $x\in T^L$ such that $x^*(x)\neq 0.$ That is, $x^*\neq (T^L)_L$.