Let $X$ be a nonreflexive banach space, $X^\ast$ its dual and $U$ some closed subset of $X$. Denote with $U^\perp$ the annihilator of $U$ given by $$ U^\perp = \{ x^\ast \in X^\ast \;|\; \forall x\in U: x^\ast(x)=0 \} $$ I am interested in the dual of $X^\ast/U^\perp$. What is $(X^\ast/U^\perp)^\ast$?
My attempt:
It is known that $$ (X/U)^\ast \cong U^\perp \\ X^\ast/U^\perp \cong U^\ast \\ U^\perp \subseteq X^\ast \\ U^\perp \text{ closed} $$ Hence $$ (X^\ast/U^\perp)^\ast = (U^\perp)^\perp $$ Based on the defintion of the annihilator above $$ (U^\perp)^\perp = \{ x^{\ast\ast} \in (X^\ast)^\ast \;|\; \forall x^\ast\in U^\perp: x^{\ast\ast}(x^\ast)=0 \} $$ If $X$ was reflexive, then $(X^\ast)^\ast=X$ and this would mean that $$ (U^\perp)^\perp = \{ x \in X \;|\; \forall x^\ast\in U^\perp: x^\ast(x)=0 \} $$ If $X$ is not reflexive, like I have, then $(U^\perp)^\perp$ can also be written as $$ (U^\perp)^\perp = \overline{span(U)} $$ where $\overline{span(U)}$ is the weak$^\ast$ closed span of $U$.
Is this correct?
Sources:
I think the result is correct. However, I have some remarks:
First, the $\operatorname{span}$ is not needed, because $U$ is already a subspace. Some people denote the weak$*$-closure by $\overline{U}^{w*}$.
Second, your second source defines the annihilator in slightly different way: With their definition, one would have $(U^\perp)^\perp \subset X$, but with your definition it is $(U^\perp)^\perp \subset X^{**}$.