نتایج جستجو برای: pettis integral
تعداد نتایج: 115289 فیلتر نتایج به سال:
Abstract In this paper convergence theorems for sequences of scalar, vector and multivalued Pettis integrable functions on a topological measure space are proved varying measures vaguely convergent.
There are several generalizations of the space L1(R) of Lebesgue integrable functions taking values in the real numbers R (and defined on the usual Lebesgue measure space (Ω,Σ, μ) on [0, 1] ) to a space of strongly-measurable “integrable” (suitably formulated) functions taking values in a Banach space X. The most common generalization is the space L1(X) of Bochner-Lebesgue integrable functions....
Di Piazza and Preiss asked whether every Pettis integrable function defined on [0, 1] and taking values in a weakly compactly generated Ba-nach space is McShane integrable. In this paper we answer this question in the negative.
If all bounded linear operators from L 1 into a Banach space X are Dunford-Pettis (i.e. carry weakly convergent sequences onto norm convergent sequences), then we say that X has the complete continuity property (CCP). The CCP is a weakening of the Radon-Nikod ym property (RNP). Basic results of Bourgain and Talagrand began to suggest the possibility that the CCP, like the RNP, can be realized a...
A Banach space X is said to have the alternative Dunford-Pettis property if, whenever a sequence xn → x weakly in X with ‖xn‖ → ‖x‖, we have ρn(xn) → 0 for each weakly null sequence (ρn) in X∗. We show that a C∗-algebra has the alternative Dunford-Pettis property if and only if every one of its irreducible representations is finite dimensional so that, for C∗-algebras, the alternative and the u...
This paper serves as a corrigendum to the paper titled Application of Pettis integration to delay second order differential inclusions appearing in EJQTDE no. 88, 2012. We present here a corrected version of Theorem 3.1, because Proposition 2.2 is not true. 1 Correction In the above article, Proposition 2.2 is not true since the normed space P 1 E ([0, 1]) is not complete. Consequently, to corr...
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