1996
DOI: 10.1017/s0305004100074752
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Converses for the Dodds–Fremlin and Kalton–Saab theorems

Abstract: The two theorems in the title give conditions on Banach lattices E and F under which a positive operator from E into F, dominated by another positive operator with some property, must also have that property. The Dodds-Fremlin theorem says that this is true for compactness provided both E′ and F have order continuous norms, whilst the Kalton–Saab theorem establishes such a result for Dunford–Pettis operators provided F has an order continuous norm. These results were originally provided, in their full generali… Show more

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Cited by 83 publications
(77 citation statements)
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References 7 publications
(11 reference statements)
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“…If the norm of E is not order continuous, then it follows from the proof of Theorem 1 of Wickstead [9], that E contains a sublattice that is isomorphic to l 1 and there exists a positive projection P from E into l 1 . Consider the operator product…”
Section: Each Positive Dunford-pettis Operator From E Into F Is Semi-mentioning
confidence: 99%
See 1 more Smart Citation
“…If the norm of E is not order continuous, then it follows from the proof of Theorem 1 of Wickstead [9], that E contains a sublattice that is isomorphic to l 1 and there exists a positive projection P from E into l 1 . Consider the operator product…”
Section: Each Positive Dunford-pettis Operator From E Into F Is Semi-mentioning
confidence: 99%
“…DunfordPettis) operators does not satisfy the analogue of Schauder's Theorem. However, the class of semi-compact operators satisfies the domination problem (Theorem 18.20 of [3]), but the class of Dunford-Pettis operators fails to satisfy this property, as was proved in [1], [7] and [9].…”
Section: Introduction and Notationmentioning
confidence: 96%
“…Supposons que la norme du treillis de Banach E n'est pas continue pour l'ordre. Alors il résulte de la démonstration du Théorème 1 de [10] que E contient un sous-treillis vectoriel isomorphe (en ordre) au treillis de Banach l 1 , et il existe aussi une projection positive P : E −→ l 1 . Il est clair que l'opérateur P est de Dunford-Pettis, mais l'opérateur P 2 = P n'est pas compact.…”
Section: Le Resultat Principalunclassified
“…Assume that the norms of E and F are not order continuous. Then, it follows from the proof of Theorem 1 of Wickstead [14] that E (resp. F) contains a sublattice isomorphic to l ∞ and there exists a positive projection P 1 : E −→ l ∞ (resp.…”
Section: ) the Operator T Is Order Weakly Compact Whenever Its Adjoimentioning
confidence: 95%