Every operator has almost-invariant subspaces (Q2637098)

The following question, which is closely related to the Invariant Subspace Problem, is examined: Given a bounded linear operator \(T\) acting on a complex Banach space \(X\), can we perturb it by a finite-rank operator \(F\) such that \(T-F\) has an invariant subspace of infinite dimension and codimension in \(X\)? A subspace \(Y\) of a Banach space \(X\) is called an almost-invariant subspace for a bounded operator \(T\) if there exists a finite-dimensional subspace \(M\subseteq X\) such that \(T Y \subseteq Y + M\). The minimal dimension of such a subspace \(M\) is said to be the defect of \(Y\) for \(T\). It is obvious that every finite-dimensional or finite-codimensional \(Y\subseteq X\) is almost invariant under \(T\). Thus, the question about almost-invariant subspaces is non-trivial only if we restrict our search to subspaces that are of infinite dimension and codimension in \(X\); such subspaces are called half-spaces. One of the main results of the paper says that every operator which has a point in the boundary of the spectrum which is not an eigenvalue admits an almost-invariant half-space with defect one. If it is assumed that \(X\) is reflexive, then the previous result holds for every operator. Equivalently, any bounded operator on a separable, reflexive, infinite-dimensional Banach space admits a rank-one perturbation which has an invariant subspace of infinite dimension and codimension. In the context of Hilbert spaces, it is possible to get perturbations that are also small in norm.