Nevertheless, I wonder whether there are conditions for existence of a complement $M$ to the kernel $N$ of a bounded linear operator $T:V\to Q$. That is, under which conditions there is a closed subspace $M\subset V$ such that $N \oplus M = V$?
In my particular case, the operator $T\colon V \to Q$ fulfills these equivalent properties:
- $T'\colon Q' \to V'$ is an homeomorphism on its range
- $T'$ is injective and has a closed range
- $T$ is surjective
$T$ has a bounded right inverse(I was wrong here, see comments below)
Disclaimer: This relates to the problem I have posted the day before.
EDIT: I additionally assume that $V$ and $Q$ are reflexive and separable.
UPDATE: I have answered the questions, based on the comments.