On completely continuous integration operators of a vector measure (Q2922476)
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scientific article; zbMATH DE number 6353712
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | On completely continuous integration operators of a vector measure |
scientific article; zbMATH DE number 6353712 |
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10 October 2014
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completely continuous operators
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vector measures
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space of integrable functions
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0.9277539
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0.9250931
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0.9236964
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0.92317915
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0.92129266
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0.9209039
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0.91511136
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On completely continuous integration operators of a vector measure (English)
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Let \(X\) be a Banach space and \(m\) be an \(X\)-valued countably additive measure. Let \(L^1(m)\) denote the space of equivalence classes of scalar-valued functions \(f\) that are integrable w.r.t.\ \(x^\ast \circ m\) for all \(x^\ast \in X^\ast\) and such that for any measurable set \(A\), there is a vector \(\int_A f dm\) in \(X\) such that \(x^\ast(\int_A f dm) = \int_A f (d x^\ast \circ m)\) for all \(x^\ast \in X^\ast\). This space is a Banach lattice w.r.t.\ the norm \(\sup \{\int |f| d|x^\ast \circ m| : \|x^\ast \| \leq 1 \}\) and point-wise-\(\|m\|\)-a.e. order. Consider the operator \(I_m:L^1(m) \rightarrow X\) defined by \(I_m(f) = \int f dm\). In this paper, the authors study the implications of this operator being completely continuous. They show that, if \(X\) is an Asplund space, then this condition implies that \(m\) has finite variation and \(L^1(m) = L^1(|m|)\).
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