I know that if $H$ is a Hilbert space and $C$ a closed subspace one can define the orthogonal projection onto $C$ as the map $x \oplus y \in H = C \oplus C^\bot \mapsto x$.
I am wondering:
Is it valid, for an open subspace $O$, to define a (possibly non-orthogonal) projection onto $O$?
And if the answer is affirmative, how to do it?
If $P$ is a bounded idempotent, i.e., $P^2=P$ (hence it is a projection which is maybe not orthogonal) and its range is $O\subseteq H$, then $O$ is closed. Namely, if $\{ x_n\}_{n=1}^{\infty}$ is a sequence in $O$ which converges to $x\in H$, then $$ \| Px-x\|\leq \| Px-Px_n\|+\|x_n-x\|\leq (1+\| P\|)\| x-x_n\|\to 0$$ which means $Px=x$, i.e., $x\in O$.
