Every closed subspace of a Hilbert space has a topological complement, namely its orthogonal complement. I'm wondering if every algebraic complement of a closed subspace of a Hilbert space is automatically a topological complement.
In other words, if $C$ is closed, $V$ and $C$ have trivial intersection, and $V+C$ is the whole space, then is $V$ automatically closed?
Thank you for your help!
Not in infinite dimension. Yes, of course, in finite dimension since every subspace is closed.
Take a nonzero, unbounded linear functional $\phi$ on your infinite-dimensional Hilbert space $H$. Take $x_0\in H$ such that $\phi(x_0)=1$. Then $$ H=\ker \phi \oplus \mathbb{C}x_0\qquad x=(x-\phi(x)x_0)+\phi(x)x_0. $$ Since $\mathbb{C}x_0$ is finite-dimensional, it is closed. And since $\phi$ is unbounded, $\ker \phi$ is not closed.
Notes:
To "construct" such a $\phi$, take $\{e_n\}$ an infinite orthonormal set in $H$ and set $\phi(e_n):=n$. Extend by linearity to $\mbox{span}\{e_n\}$. Then extend to any algebraic complement of $\mbox{span}\{e_n\}$ by $0$.
If $H=F\oplus G$ is an orthogonal direct sum of any two subspaces of $H$, though, then both $F$ and $G$ must be closed and we have $F=G^\perp$ and $G=F^\perp$.
