Boolean independent sequences and Bourgain-Rosenthal's theorem (Q402865)

Let \(S\) be a set. A sequence of pairs \((A_n, B_n)\), \(A_n, B_n \subset S\), \(n = 1, 2, \dots\), is called Boolean independent if, for arbitrary finite not empty disjoint sets \(I, J \subset \mathbb N\), the intersection \(\left(\bigcap_{i \in I}A_i\right) \cap \left(\bigcap_{j \in J}B_j\right)\) is not empty.NEWLINENEWLINEThe author proves that, if for a uniformly bounded sequence of functions \(f_n: S \to \mathbb R\) and \(0 < a < b < \infty\) one defines \(A_n = \{s \in S: |f_n(s)| < a\}\), \(B_n = \{s \in S: |f_n(s)| > b\}\), \(n = 1, 2, \dots\), and the sequence of pairs \((A_n, B_n)\) happens to be Boolean independent, then there is a subsequence \((g_n) \subset (f_n)\) such that, for every finite collection \((c_k)_{k=1}^n \subset \mathbb R\), NEWLINE\[NEWLINE \sup_{s \in S}\left|\sum_{k=1}^nc_kg_k(s)\right| \geq \frac{b-a}{2} \sum_{k=1}^n|c_k|. NEWLINE\]NEWLINE In other words, the author eliminates the positivity condition of the functions \(f_n\) condition in Rosenthal's famous lemma [\textit{H. P. Rosenthal}, Bull. Am. Math. Soc. 84, 803--831 (1978; Zbl 0391.46016)]. This fixes a gap in the proof of an important Bourgain-Rosenthal theorem [\textit{J. Bourgain} and \textit{H. P. Rosenthal}, J. Funct. Anal. 52, 149--188 (1983; Zbl 0541.46020)]. In a modern reformulation, the aforementioned Bourgain-Rosenthal theorem states that an operator \(T: L_1 \to X\) which does not fix a copy of \(\ell_1\) is necessarily narrow in the sense of Plichko and Popov. More about this class of operators can be found in the recent book by \textit{M. Popov} and \textit{B. Randrianantoanina} [Narrow operators on function spaces and vector lattices. Berlin: de Gruyter (2013; Zbl 1258.47002)].