annihilator of an intersection in infinite dimension Given two subspaces of an infinite dimensional Banach space, is the sum of their annihilators dense in the annihilator of their intersection?
 A: Without further constraints the answer is: In general, no.
Let $E$ be the Banach space, and $T \subsetneq E$ a dense subspace. Let $S \neq \{0\}$ any finite dimensional subspace having trivial intersection with $T$.
Then $(S \cap T)^\perp = \{0\}^\perp = E'$, but, since $T$ is dense, $S^\perp + T^\perp = S^\perp$ is a weak$^\ast$-closed, hence norm-closed proper subspace of $E'$.
Generally, $F^\perp = \overline{F}^\perp$, and $\overline{S \cap T} \subset \overline{S} \cap \overline{T}$, hence $(\overline{S} \cap \overline{T})^\perp \subset \overline{S \cap T}^\perp$, and the inclusion is proper if $\overline{S \cap T} \neq \overline{S} \cap \overline{T}$. Furthermore $S^\perp + T^\perp \subset (\overline{S} \cap \overline{T})^\perp$, so a necessary condition is that $\overline{S \cap T} = \overline{S} \cap \overline{T}$.
That condition is certainly fulfilled if $S$ and $T$ are closed subspaces, and often not when at least one of them is not closed. Therefore let us demand that the two subspaces are closed.
By factoring out $S \cap T$, we may also assume that $S \cap T = \{0\}$.
Since $^\perp\bigl(S^\perp + T^\perp\bigr) \subset \vphantom{i}^\perp\bigl(S^\perp\bigr) = \overline{S} = S$, and similarly for $T$, we have $^\perp\bigl(S^\perp + T^\perp\bigr) \subset S \cap T = \{0\}$, i.e. $S^\perp + T^\perp$ is weak$^\ast$-dense.
Now if $E$ is reflexive, then the weak$^\ast$ topology on $E'$ is the weak topology, and that means that in this case $S^\perp + T^\perp$ is dense (a subspace is a convex subset, hence its weak closure and its norm closure coincide).
If $E$ is not reflexive, then I think it can happen that $S^\perp + T^\perp$ is not dense, but I have neither found an example nor a proof that it is always dense even in that case yet.
