Conditions for a kernel of a bounded operator to be complemented

Question

I am well aware of the problem of complementing subspaces in Banach spaces as it was discussed here and here .

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)

Any ideas?

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.

Answer 1

Let me summarize the comments and my findings to answer this question:

Having read through this survey and some earlier papers, I have come the conclusion that the condition $T\colon V\to Q$ being bounded and surjective does not imply that $\ker T$ is complementable. Assume the contrary, then

• For any $N$ subspace of $V$, one can take $Q:= V/N$ and the natural surjection $T\colon V \to Q=V/N$ having $N$ as the kernel
• By boundedness of $T$ one has $V$ separable, then $Q$ separable
• As $Q$ is a quotient space of a reflexive Banach space, $Q$ is reflexive, cf. Palmer's book (pp. 79)
• and thus, one would have the contradictory outcome that every closed subset of a separable and reflexive B-space has a complement

This is equivalent to $V$ being isomorph to a Hilbert-space, cf. the discussion here. And this is known to be not true for $L^p$ spaces, $p\neq2$, see this paper.