for every rational function ƒ with poles off K. In this note it is shown that any operator for which the spectrum is a spectral set has a nontrivial invariant subspace. In [6] von Neumann introduced the notion of spectral set and showed that if T has II T\\ = 1 then the closed unit disc, D"~, is a spectral set for T. For this reason any operator whose spectrum is a spectral set is called a von ...

Notes for my lectures in the PSU Analysis Seminar during the winter and spring terms 2013-14, with special emphasis on the 1972 results of Victor Lomonosov.

The notion of an invariant subspace is fundamental to the subject of operator theory. Given a linear operator T on a Banach space X, a closed subspace M of X is said to be a non-trivial invariant subspace for T if T (M) ⊆M and M 6= {0}, X. This generalizes the idea of eigenspaces of n×n matrices. A famous unsolved problem, called the “invariant subspace problem,” asks whether every bounded line...

A number of recent results in the geometric framework have been obtained for nonsingular systems, using the notion of Self Bounded Controlled Invariant Subspace, and of Self Hidden Conditioned Invariant Subspace. The aim of this note is to extend the above mentioned notion of Self Bounded Controlled Invariant Subspace to singular systems, to investigate its dynamical properties and to study its...

The n-variable PARITY function is computable (by a well-known recursive construction) by AC formulas of depth d+ 1 and leafsize n·2dn1/d . These formulas are seen to possess a certain symmetry: they are syntactically invariant under the subspace P of even-weight elements in {0, 1}, which acts (as a group) on formulas by toggling negations on input literals. In this paper, we prove a 2 −1) lower...

Heeft elke begrensde lineaire operator, werkend op een Hilbert ruimte, een niet-triviale invariante deelruimte? Het antwoord is positief voor zowel eindig-dimensionale ruimtes als voor niet-separabele ruimtes. Het onopgeloste probleem voor het geval daar tussenin, dus voor separabele Hilbert ruimtes staat bekend als het invariante deelruimte probleem. Professor B.S. Yadav van de Indian Society ...

Proof. Suppose that U1, . . . , Um are invariant subspaces of T ∈ L(V ). We wish to show that U1 + · · ·+ Um is also an invariant subspace. Consider any ~v ∈ U1 + · · ·+ Um. By definition, we can write ~v = ~u1 + · · ·+ ~um, for some ~u1 ∈ U1, . . . , ~um ∈ Um. Since U1, . . . , Um are invariant subspaces, by definition, we know T~u1 ∈ U1 ⊂ U1 + · · ·+ Um, ..., T~um ∈ Um ⊂ U1 + · · ·+ Um. Hence...