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DOMINATION BY POSITIVE STRICTLY SINGULAR OPERATORS

Published online by Cambridge University Press:  24 March 2003

JULIO FLORES
Affiliation:
Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense, 28040 Madrid, Spainjflores@sunam1.mat.ucm.espacoh@mat.ucm.es
FRANCISCO L. HERNÁNDEZ
Affiliation:
Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad Complutense, 28040 Madrid, Spainjflores@sunam1.mat.ucm.espacoh@mat.ucm.es
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Abstract

The problem of domination for positive compact operators between Banach lattices was solved by Dodds and Fremlin in [6]: given a Banach lattice $E$ with order continuous dual norm, an order continuous Banach lattice $F$ and two positive operators $0 \leqslant S \leqslant T : E \longrightarrow F$ , the operator $S$ is compact if $T$ is. A similar problem has been considered in the class of weakly compact operators by Abramovich [1] and in a general form by Wickstead [26]. Precisely, Wickstead's result shows that the operator $S$ is weakly compact if $T$ is whenever one of the following two conditions holds: either $E^{\prime}$ is order continuous or $F$ is order continuous. When it comes to Dunford–Pettis operators, Kalton and Saab [15] have proved that the operator $S$ is Dunford–Pettis if $T$ is, provided that the Banach lattice $F$ is order continuous. On the other hand, Aliprantis and Burkinshaw settled the problem of domination for compact [2] and weakly compact [3] endomorphisms (that is, the case when $E = F$ ). For example, they proved that if either the norm on $E$ or the norm on $E^{\prime}$ is order continuous, then the compactness of $T$ is inherited by the power operator $S^2$ . Also, they showed that, for $E$ an arbitrary Banach lattice, $T$ being compact always implies that $S^3$ is compact. More recently, Wickstead studied converses for the Dodds–Fremlin and Kalton–Saab theorems in [27] and for the Aliprantis–Burkinshaw theorems in [28].

Type
Notes and Papers
Copyright
The London Mathematical Society, 2002

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