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The following question came to me while working on a technical matter about transversality in infinite dimension, and I'm really curious to know whether it has an affirmative answer at least under extra hypotheses.

Let A be a bounded linear operator on a Banach space E. Does it exist a bounded linear operator S such that 0 and 1 do not belong to the same connected component of the spectrum of the operator AS:= A + A(I-A)S?

That is, S is OK if either 0 or 1 is not in the spectrum of AS, or if they both are in the spectrum, they should belong to different connected component of it. Thus it may be assumed that 0 and 1 belong to the same component of spec(A), otherwise S=0 trivially solves the problem.

The first idea is to look for S of the form f(A), but this can't work if f is continuous, since then spec(AS) is the continuous image of spec(A) with a map that fixes 0 and 1. However, if A admits a discontinuous functional calculus (e.g. a normal operator on Hilbert space), the trick does work.

I do not know the answer to the question even on Hilbert spaces. In a general Banach space the problem seems even harder, due the difficulty of building operators.

I'd very grateful of any suggestion! (Pietro Majer).

edit (17/11/2011).

Here are a few more or less trivial facts that I know.

  • For a Banach space $X$, the set $\mathcal{A}$ of all $A\in L(X)$ such that there exists $S\in L(X)$ such that no connected component of $\operatorname{spec}(A_S)$ contains both $0$ and $1$, is an open set;

  • If $A\in \mathcal{A}$, then there is $S$ such that $A_S$ is even a linear projector (thus satifying the condition on the spectrum ad abundantiam);

  • If $A\in L(X)$ and $A_S\in \mathcal{A}$ for some $S\in L(X)$, then $A$ itself is in $\mathcal{A}$;

  • $A\in \mathcal{A}$ if and only if there are closed subspaces $V$ and $W$ of $X$ such that $V\times W\ni (v,w) \mapsto Av + (I-A)w \in X$ is bijective;

  • if $AX$ is a closed subspace and $(I-A)^{-1}(AX)$ is a complemented subspace of $X$, then $A\in \mathcal{A}$.

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    $\begingroup$ On the HI spaces of Gowers-Maurey every operator has countable spectrum, so at least this is true on some infinite dimensional spaces. $\endgroup$
    – Bill Johnson
    Commented May 15, 2010 at 1:34
  • $\begingroup$ Not sure if it helps but I'll mention it just in case: There is a classical theorem by Weyl-von Neumann that says: If A is a selfadjoint operator on a separable Hilbert space, then for any $\epsilon \gt 0$ there is a Hilbert-Schmidt operator S with Hilbert-Schmidt norm less than $\epsilon$ such that A+S has a pure point spectrum. (Kato: "Perturbation Theory for Linear Operators", chapter 10, paragraph 2.1). Unless I'm mistaken that answers the question for separable Hilbert spaces, no? $\endgroup$
    – Tim van Beek
    Commented May 15, 2010 at 6:38
  • $\begingroup$ Tim: the result you quote is only for self-adjoint operators, which is already covered by the original poster's comment about normal operators. $\endgroup$
    – Yemon Choi
    Commented May 15, 2010 at 7:13
  • $\begingroup$ Slight improvement : The conjecture is also true, if A is a compact perturbation of a normal operator. $\endgroup$
    – jjcale
    Commented Aug 28, 2011 at 18:16
  • $\begingroup$ For me, only your first fact is "more or less trivial", Pietro. $\endgroup$
    – Bill Johnson
    Commented Nov 18, 2011 at 19:20

1 Answer 1

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For Hilbert spaces, the conjecture follows from fact 4 and the answer to question Complement of a subspace which is a cartesian product applied to the kernel of the map $H\times H\ni (v,w) \mapsto Av + (I-A)w \in X$ .

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  • $\begingroup$ that's true ! $\endgroup$
    – Pietro Majer
    Commented Nov 24, 2011 at 1:02
  • $\begingroup$ I'm glad to accept this answer, as the case of Hilbert spaces is already quite interesting to me. $\endgroup$
    – Pietro Majer
    Commented Nov 24, 2011 at 7:09

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