Stability of vector measures and twisted sums of Banach spaces (Q387246)

Let \(\mathcal{F}\) be an algebra of subsets of a set \(\Omega\) (a set-algebra) and \(X\) a Banach space. A set function \(\nu:\mathcal{F}\to X\) is called \(\varepsilon\)-additive if \(\|\nu(A\cup B)-(\nu(A)+\nu(B))\|\leq\varepsilon\) whenever \(A\) and \(B\) are disjoint members of \(\mathcal F\). If \(\varepsilon\) can be taken to be \(0\), we have an additive set function, or a vector measure. The paper under review departs from the following result of \textit{N. J. Kalton} and \textit{J. W. Roberts} [Trans. Am. Math. Soc. 278, 803--816 (1983; Zbl 0524.28008)]:NEWLINENEWLINEThere is an absolute constant \(K<45\) such that if \(\nu:\mathcal{F}\to\mathbb{R}\) is 1-additive, then there exists an additive \(\mu:\mathcal{F}\to\mathbb{R}\) with \(|\nu(A)-\mu(A)|\leq K\) for all \(A\in\mathcal{F}\).NEWLINENEWLINEIt is known that \(K\geq 3/2\), but the exact value of \(K\) is not known. A Banach space for which the Kalton-Roberts theorem is true is said to have the SVM (stability of vector measures) property. So \(\mathbb{R}\) enjoys the SVM property. If (like \(c_0\)) a Banach space does not have the SVM, it still makes sense to ask: ``For what set-algebras \(\mathcal{F}\) does it have the SVM?'' A little more precisely, \(X\) has the \(\kappa\)-SVM if it has the SVM for all \(\mathcal{F}\) with \(\text{card}(\mathcal{F})<\kappa\). The minimal cardinal number \(\kappa\) so that \(X\) lacks the \(\kappa\)-SVM is denoted \(\tau(X)\) and called the SVM-character of \(X\).NEWLINENEWLINEAlready in Section 2 we find enlightening observations. For instance, every Banach space has \(\tau(X)\geq\text{card}(\mathbb{N})=\omega\), \(\tau(c_0(\Gamma)),\tau(C(K))>\omega\), and the \(\omega\)-SVM implies the SVM if \(X\) is complemented in its bidual. Easy, but important, is the observation that the \(\kappa\)-SVM passes down to complemented subspaces, which in turn implies that all injective Banach spaces have the SVM-property since all \(\ell_\infty(\Gamma)\)-spaces enjoy SVM.NEWLINENEWLINESection 3 contains necessary background on twisted sums and the three-space problem.NEWLINENEWLINEThe results in the first part of Section 4 can be illustrated by Theorem 4.2 (ii): If \(X\) has the \(\Gamma^+\)-SVM property, then the pair \((c_0(\Gamma),X)\) splits. Here, \(\Gamma^+\) is the cardinal successor of \(\Gamma\). In the second part of Section 4, it is then concluded from the first part that, e.g., \(L_p\), \(1\leq p<\infty\), has \(\tau=\omega\).NEWLINENEWLINESection 5 is an application of the machinery of twisted sums, the main result is probably that \(\kappa\)-injectivity implies \(\kappa\)-SVM when \(\kappa\) has uncountable cofinality.NEWLINENEWLINEIn Section 6, the theme of Section 5 is continued and some rather heavy constructions involving the Johnson-Lindenstrauss space \(JL_\infty\) are made to conclude that \(\tau(c_0(\Gamma))=\omega_2=\tau(JL_\infty)\).NEWLINENEWLINEThe main result in Section 7 is that \(\kappa\)-SVM indeed is a three-space property. The proof is by contradiction and is very involved.NEWLINENEWLINEFrom Section 2, \(\omega\)-SVM implies the SVM if \(X\) is complemented in its bidual. The author now, in Section 8, goes on to characterize those spaces complemented in its bidual that enjoy SVM and obtains, e.g., that such a space enjoys the SVM if and only if \(\text{Ext}(X^\ast,\ell_1)=0\).NEWLINENEWLINEThe final Section 9 is a list of reflections on 6 open problems, two of the problems being: \newline\noindent (a) Does the SVM-property of \(X\) imply that \(X\) (or \(X^{\ast\ast}\)) is injective? \newline\noindent (b) Does the SVM-property of \(X\) imply that \(X\) contains a copy of \(c_0\)?