Discrepancy norms on the space \(M[0,1]\) of Radon measures (Q1280724)

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scientific article; zbMATH DE number 1262678
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Discrepancy norms on the space \(M[0,1]\) of Radon measures
scientific article; zbMATH DE number 1262678

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    Discrepancy norms on the space \(M[0,1]\) of Radon measures (English)
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    13 December 1999
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    If \(X\) is a Banach space with symmetric basis, \((e_n)_{n\in\mathbb{N}}\), see the book of \textit{J. Lindenstrauss} and \textit{L. Tzafriri} [``Classical Banach spaces'', Vol I (1996; Zbl 0852.46015)], which satisfies: \[ \Biggl\| \sum^\infty_{n= 1}a_n\sigma_n e_{\pi(n)}\Biggr\|= \Biggl\| \sum^\infty_{n= 1}a_n e_n\Biggr\| \] for every permutation \(\pi\) of the integers and every choice of \(\text{signs }\sigma_n= \pm 1\), the space \(M[X]\) is the space \(M[0,1]\) of all Radon measures \(\mu\) on the unit interval \([0,1]\), with the norm \(\|\mu\|= \sup\{\| \sum^d_{i=1} \mu(I_i)e_i\|\); \((I_i)\), \(i\leq d\) disjoint subintervals of \([0,1]\), \(d\in N\}\). The paper studies the structure of the space \(M[X]\). One of the results proved in this direction is: Theorem. If \(X\) is a Banach space with symmetric basis and \(X\) contains no copy of \(\ell_1\) then a measure \(\mu\in M[0,1]\) belongs to the closure of \(L_1[0,1]\) under the \(M[X]\) norm if and only if \(\mu\) is a diffuse measure. Also, in the paper are defined \(M[X]\) continuous operators, showing that each \(M[X]\) continuous operator is nearly representable, see [\textit{R. Kaufmann}, \textit{M. Petrakis}, \textit{L. H. Riddle} and \textit{J. J. Uhl jun.}, Trans. Am. Math. Soc. 312, No. 1, 315-333 (1989; Zbl 0673.47016)], if \(X\) has a symmetric basis and contains no copy of \(\ell_1\). Examples of linear functionals on \(L_1[0,1]\) that are not \(M[\ell_p]\) continuous for any \(p>1\) are given. In the paper is proved also the following: Theorem. Suppose \(X\) is a Banach space with symmetric basis and \(X\) contains no copy of \(\ell_1\). Suppose that \(Z\) is a Banach space and \(T: L_1[0,1]\to Z\) is a non-compact operator. If \(T\) is also \(M[X]\) continuous, then there exists a non-compact bounded operator \(S:X\to Z\). This extends an earlier result obtained by the author of the paper and others, see the second paper cited above.
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    spaces of measures
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    Radon measures
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    nearly representable
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    non-compact operator
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