Let $W$ be a infinite dimensional Banach space equipped with a symplectic form $\Omega$. Let $E\subset W$ be a closed subspace and $E^\Omega$ the symplectic orthogonal complement of E w.r.t. $\Omega$.
I learned from a book that $(E^\Omega)^\Omega=E$. But I cannot find a way to prove it. Can someone give me a proof? Thanks!
This claim is not true for not strongly symplectic vector spaces, as is implied by the corollary at the bottom of this answer. (It is true for strongly symplectic ones, as expressed in the comments to the question.)
I recall that a (real) topological vector space $(W, \Omega)$ is presymplectic if it is equipped with a continuous antisymmetric bilinear form $\Omega : W \times W \to \mathbb{R}$. This form induces a linear map $\Omega^{\flat} : W \to W^{\ast} : v \mapsto \Omega(v, -)$. $\Omega$ is weakly symplectic if $\Omega^{\flat}$ is injective and strongly symplectic if it is bijective. Both notions coincide in finite dimension, but they are strictly distinct as claimed by the first proposition below.
Proposition 1. There exists a weakly but not strongly symplectic (Hilbert) space $(W, \Omega)$.
Proof: This follows for instance from the theorem in this article of Marsden and from a theorem of Weinstein referenced therein. $\square$
Proposition 2. Given a symplectic vector space $(W, \Omega)$, the kernel $E$ of a nonzero map $\alpha \in W^{\ast}$ is a closed proper subspace such that $E^{\Omega} \subset E$. Moreover, $\mathrm{dim} \, E^{\Omega} \le 1$ with equality only if $\alpha \in \mathrm{Im} \, \Omega^{\flat}$.
Proof: Given a nonzero element $\alpha \in W^{\ast}$, that is a nonzero continuous map $\alpha : W \to \mathbb{R}$, its kernel $E$ is clearly closed and a proper subspace of $E$. Set $F := E^{\Omega}$.
Observe that for any $w \in W \setminus E$ (such an element exists), we have $W = E + \mathbb{R}w \cong E \oplus \mathbb{R}w$. Given any nonzero element $f \in F$ (if it exists), we have $\Omega(f, w) \neq 0$; Indeed, the nondegeneracy of $\Omega$ implies that there exists $e \in E$ and $r \in \mathbb{R}$ such that
$$ 0 \neq \Omega(f, e + rw) = r \Omega(f,w) $$
with the equality due to the fact that $f \in F := E^{\Omega}$.
This implies $F \subset E$. Indeed, if it were otherwise, there would exists $f \in F \setminus E$ and we would have $W = E + \mathbb{R}f \cong E \oplus \mathbb{R}f$. Taking $w = f$, the above inequality and antisymmetry would imply $0 \neq \Omega(w,f) = \Omega(f,f) = 0$, a contradiction.
This implies also $\mathrm{dim} \, F \le 1$. Indeed, otherwise, there would exist two independent vectors $f, g \in F$ normalised such that $\Omega(w, f) = 1 = \Omega(w, g)$, and we would have that the nonzero vector $h := f-g \in F$ satisfies $\Omega(w, h) = 0$, contradicting the above.
Suppose that $\mathrm{dim} F = 1$; We claim that this implies $\alpha \in \mathrm{Im} \, \Omega^{\flat}$. Indeed, let $f \in F$ be the unique element such that $\Omega(f, w) = \alpha(w) \neq 0$. It readily follows that $\alpha = \Omega^{\flat}(f)$. $\square$
Corollary. If $(W, \Omega)$ is a weakly but not strongly symplectic topological vector space, then there exists a closed proper subspace $E \subsetneq W$ such that $(E^{\Omega})^{\Omega} = W$.
Proof: By hypothesis, there exists $\alpha \in W^{\ast} \setminus \mathrm{Im} \, \Omega^{\flat}$. Set $E = \mathrm{Ker} \, \alpha$ and $F = E^{\Omega}$. From the previous proposition, $F = \{0\}$, hence $F^{\Omega} = W \neq E$. $\square$
This means in particular that not strongly symplectic vector spaces admit codimension one symplectic subspaces!
