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On the domination of limited and order Dunford-Pettis operators. (English. French summary) Zbl 1350.46018

A bounded subset \(A\) of a Banach space \(X\) is called limited in \(X\) if, for every weak\(^*\) null sequence \((x_n')\) in \(X'\), we have \(x_n'(x)\to 0\) uniformly for \(x\in A\). An operator \(T:X\to Y\) between Banach spaces is said to be limited whenever the set \(T(B_X)\) is limited in \(Y\). A bounded subset \(A\) of a Banach space is said to be a Dunford-Pettis set whenever every weakly compact operator from \(X\) to another Banach space carries \(A\) to a norm totally bounded set. An operator \(T:E\to X\) is said to be an order Dunford-Pettis operator whenever \(T\) carries order bounded subsets of \(E\) to Dunford-Pettis sets in \(X\).
In this paper, the authors consider the so-called “domination problem” for limited (resp.order limited, order Dunford-Pettis) operators: suppose that positive operators \(S,T:E\to F\) between Banach lattices \(E\) and \(F\) satisfy \(0\leq S\leq T\). If \(T\) is limited (resp.order limited, order Dunford-Pettis), does it follow that \(S\) is also limited (resp.order limited, order Dunford-Pettis)? The authors prove that \(S\) is an order Dunford-Pettis operator whenever \(T\) is. However, they provide an example of a rank-one operator \(T:\ell_1\to c\) which dominates a nonlimited operator. Under additional assumptions, when \(F\) is Dedekind \(\sigma\)-complete and \(E'\) is order continuous, the authors prove that \(S\) is limited whenever \(T\) is limited. They also prove that whenever \(F\) is \(\sigma\)-Dedekind complete and \(T\) is order limited, then \(S\) is order limited.

MSC:

46B42 Banach lattices
47B60 Linear operators on ordered spaces
47B65 Positive linear operators and order-bounded operators
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