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.

  • $\begingroup$ The fourth bullet does not seem to be equivalent to the other ones (or do you talk about a bounded right inverse that is allowed to be non-linear?). $\endgroup$ – Martin May 16 '13 at 19:00
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    $\begingroup$ Notice that every closed subspace is the kernel of some map to another space, so our restriction to kernels is not very operative :-) $\endgroup$ – Mariano Suárez-Álvarez May 17 '13 at 9:40
  • $\begingroup$ @Mariano Right. But, this means that if there were conditions for existence of complements, than they would be expressible in terms of properties of the operator. $\endgroup$ – Jan May 17 '13 at 10:06
  • $\begingroup$ @Martin . Not sure. I was using the map $T_f:V/N \to Q$ that is a bijection and concluded that, choosing the right representative of $[v]\in V/N$, $T_f^{-1}$ is the 'right' inverse to $T$. However, this 'right' is maybe not safe. $\endgroup$ – Jan May 17 '13 at 10:11
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    $\begingroup$ The problem is that $T_{f}^{-1} \colon Q \to V/N$, but you don't get from $V/N$ back up to $V$ with a linear map unless $N$ is complemented in $V$. In fact, if $T$ has a bounded and linear right inverse $R$ then $M = \ker{(1_V - RT)} = \operatorname{im} R$ is a complement of $N$. $\endgroup$ – Martin May 17 '13 at 10:16

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.

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