Lattice Trace Operators

Brian Jefferies

doi:10.1155/2014/629502

Abstract

A bounded linear operator T on a Hilbert space ℋ is trace class if its singular values are summable. The trace class operators on ℋ form an operator ideal and in the case that ℋ is finite-dimensional, the trace tr(T) of T is given by ∑jajj for any matrix representation {aij} of T. In applications of trace class operators to scattering theory and representation theory, the subject is complicated by the fact that if k is an integral kernel of the operator T on the Hilbert space L2(μ) with μ a σ-finite measure, then k(x,x) may not be defined, because the diagonal {(x,x)} may be a set of (μ⊗μ)-measure zero. The present note describes a class of linear operators acting on a Banach function space X which forms a lattice ideal of operators on X, rather than an operator ideal, but coincides with the collection of hermitian positive trace class operators in the case of X=L2(μ).

Highlights

A trace class operator A on a separable Hilbert space H is a compact operator whose singular values λj(A), j = 1, 2, . . ., satisfy ∞‖A‖1 = ∑λj (A) < ∞. (1) j=1The decreasing sequence {λj(A)}∞ j=1 consists of eigenvalues of (A∗A)1/2
The number tr (A) = ∑ (Ahj, hj) j=1 is called the trace of A and is independent of the orthonormal basis {hj}∞ j=1 of H
The difficulty is addressed by Brislawn [4, 5], [1, Appendix D] who shows that, for a trace class operator Tk : L2(μ) → L2(μ) with integral kernel k, the equality tr (Tk) = ∫ ̃k (x, x) dμ (x) holds

Summary

Introduction

The difficulty is addressed by Brislawn [4, 5], [1, Appendix D] who shows that, for a trace class operator Tk : L2(μ) → L2(μ) with integral kernel k, the equality tr (Tk) = ∫ ̃k (x, x) dμ (x). Brislawn [5, Theorem 4.3] shows that a hermitian positive Hilbert-Schmidt operator Tk is a trace class operator if The present panapdeornelxyaimf ∫inΣẽks(xth, ex)spdaμc(exC) 1

Banach Function Spaces and Regular Operators

Martingale Regularisation

Trace Class Operators

Lattice Properties