annihilator of an intersection in infinite dimension

Question

Given two subspaces of an infinite dimensional Banach space, is the sum of their annihilators dense in the annihilator of their intersection?

Answer 1

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.