I have the following problem (p.73 in Marsden's 'Introduction to Mechanics and Symmetry' Book):
Given is an infinite-dimensional Banach space $Z$ and on it a weakly non-degenerate, symplectic form $\Omega$, namely a continuous, bilinear, skew-symmetric mapping $$\Omega : Z \times Z \to \mathbb R,$$ s.t. if $z \in Z$ and $\Omega(z, v) = 0 \; \forall v \in Z$, then $z = 0$.
Furthermore the symplectic complement of a subspace $E$ is defined as the set $E^{\Omega} = \{z \in Z \mid \Omega(z, v) = 0 \; \forall v \in E\}$.
Now I have to show that if $E, F \subset Z$ are closed subspaces then $$ (E \cap F)^{\Omega} = E^{\Omega} + F^{\Omega}.$$
Using the already proven results $$ E^{\Omega} \cap F^{\Omega} = (E + F)^{\Omega}, $$ and $$ E = E^{\Omega\Omega}, \mbox{ for E closed, } $$ (a proof can be found here) I already managed to obtain the inclusion $$(E \cap F)^{\Omega} \supset E^{\Omega} + F^{\Omega}.$$
To other inclusion follows immediately if I can show that $E^{\Omega} + F^{\Omega}$ is closed. I know that by construction both summands are closed, but that's about it.
My solution attempt
Let $(w_i) \subset E^{\Omega} + F^{\Omega}$, s.t. $w_i \rightarrow w.$ Then $w_i = e_i + f_i, e_i \in E^{\Omega}, f_i \in F^{\Omega}.$
If I could somehow show that there exist $e, f \in Z$ s.t. $w = e + f$ and that $e_i \rightarrow e, f_i \rightarrow f$ (or maybe using subsequences?) then since the summands are closed one has $e \in E^{\Omega}, f \in F^{\Omega}$ which immediately implies $w \in E^{\Omega} + F^{\Omega}$. But as I have no experience with Banach spaces or topological vector spaces in general, I'm at a loss here. Where could I get the necessary theory to solve this problem? Am I on the right track?
In reaction to Franklin Pezzuti Dyer's Answer:
Thanks a lot for your answer. And here my answer to your questions:
Yes I already did that, but just didn't bother to mention the details in my question. But here it goes: As $\Omega$ is continuous the restrictions $\Omega_z := v \mapsto \Omega(z, v)$ are continuous for any $z \in Z$ ($\Omega_z = \Omega^\flat z$ in Marsden notation). Now $\{0\}$ is closed in $\mathbb R$ and hence so are the preimages $\Omega_z^{-1}(\{0\})$. Let $E \subset Z$ then $$ E^{\Omega} = \bigcap_{z \in E} \Omega_z^{-1}(\{0\}), $$which is closed as an arbitrary intersection of closed sets.
and 3. Yes: Maybe I misunderstood what you are trying to do but I don't see how this gets me any further: The only unary operation here is $(\cdot)^{\Omega}$, and it is only involutive on the set of all closed subsets of $Z$. Therefore I have $$ \begin{align} A &= \{X \subset Z \mid X \mbox{ closed }\},\\ f&: A \to A, X \mapsto X^{\Omega},\\ \blacksquare&: A^2 \to A, (X, Y) \mapsto X \cap Y,\\ \blacktriangle&: A^2 \to 2^Z, (X, Y) \mapsto X + Y \end{align} $$
My problem specifically is that I don't know if $A$ is closed under $\blacktriangle$, resp. $+$. So the equality I actually get is $(E \cap F)^{\Omega} = \mathrm{cl}(E^{\Omega} + F^{\Omega}) \supset E^{\Omega} + F^{\Omega}$.
And there seem to be examples where the sum of two closed subspaces in a banach space is not closed: see here (actually this seems to be for a hilbert space so I'm not completely sure how relevant this is here) and here (this needs a bounded linear operator. I don't know if $\Omega^{\flat}$ satisfies that). I would suspect that here the continuity and weak/strong non-degeneracy properties of $\Omega$ play a huge role which makes the closed sets $E^{\Omega}, F^{\Omega}$ have additional properties which need to be used somehow?
Edit: I think I finally found an answer:
It is as I suspected: The statement $$ (E \cap F)^{\Omega} = E^{\Omega} + F^{\Omega} $$ is (even for closed $E$ and $F$) not true in general. And the situation is actually even worse: While rereading the proof of the $$ E \subset Z, E \mbox{ closed } \Rightarrow E^{\Omega\Omega} = E $$statement I realised, that it is only true for strongly non-degenerate symplectic forms, but I assumed only weak symplecticity.
This made a case distinction on non-degeneracy necessary:
For in the strongly non-degenerate case I can show the following: Since a Hilbert space is also a Banach space any of the Hilbert space counterexample pairs of closed sets given here, here or here can be used as follows:
Let $A, B \subset Z$ such a pair of closed sets with $A + B$ not closed. Then define $E = A^{\Omega}, F = B^{\Omega}$. Since $\Omega$ is strongly non-degenerate one has $A^{\Omega\Omega} = A, A^{\Omega\Omega} = B.$ Thus $E^{\Omega} + F^{\Omega} = A^{\Omega\Omega} + B^{\Omega\Omega} = A + B \subsetneq \mathrm{cl}(A + B) = \mathrm{cl}(E^{\Omega} + E^{\Omega}) = (E \cap F)^{\Omega}.$ Thus the valid statement has to be$$(E\cap F)^{\Omega} = \mathrm{cl}(E^{\Omega} + F^{\Omega}).$$
If $\Omega$ is only weakly non-degenerate I get the slightly weaker statement:$$(E\cap F)^{\Omega} \supset \mathrm{cl}(E^{\Omega} + F^{\Omega}).$$
I believe that the question is now answered. Is there way to mark this question as closed?
