Von Neumann operators are reflexive.

Abstract

Von Neumann operators are reflexive Dedicated to Heran S. Collins, the thesis advisor oft he first author and the mathematical grandfather ofthe second, on the occasion ofhis retirement from Louisiana State UniversityBy John B. Conway at Bloomington and James J. Dudziak*) at Lewisburg A compact subset K of the complex plane is a spectral set for an operator T on a Hubert space Jif if the spectrum of T, σ (T), is contained in K and for every rational function / with poles off K, \\f(T)\\^\\f\\ K = sup{\f(z)\:zeK}. The operator T is called a von Neumann operator if the spectrum of T is itself a spectral set for T. It has been known for some time that a von Neumann operator has a non-trivial invariant subspace [1]. In this paper it will be shown that every von Neumann operator is reflexive. That is, if A is any operator such that AJi^M for every subspace M of 3f that is invariant for Γ, then A belongs to the weakly closed operator algebra generated by T and the identity.For any operator T on J^9 let H^(T] be the closure of {p(T) : p is a polynomial} in the weak operator topology (WOT) of @(Μ ) and let P 00 (T) denote the closure of {p(T):p is a polynomial} in the natural weak* topology that &(3if) has s the dual space of the trace class operators on J^. In the process of proving the main result of this paper it will also be shown that for a von Neumann operator T, ^(T) = P°°(T) and if £:Ρ°°(Τ) -> c is a weak* continuous linear functional, then there are vectors g and h in tf such that L(X) = <Xg, fc> for all X in P 00 (T) (Corollary 3. 2).Actually the main accomplishment of this work is probably not the result proclaimed in the title, but rather the development of a functional calculus for von Neumann operators. Let Q denote the non-peak points of the function algebra R (σ (T)) and let /1 Q denote the restriction of planar Lebesgue measure to Q.