sp.spectral theory - Perturbations of an operator that disconnect the spectrum

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 A_S:= A + A(I-A)S?

That is, S is OK if either 0 or 1 is not in the spectrum of A_S, 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(A_S) 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 $Ain L(X)$ such that there exists $Sin L(X)$ such that no connected component of $operatorname{spec}(A_S)$ contains both $0$ and $1$, is an open set;




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




• If $Ain L(X)$ and $A_Sin mathcal{A}$ for some $Sin L(X)$, then $A$ itself is in $mathcal{A}$;




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




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