Sums of SCD sets and their applications to SCD operators and narrow operators (Q2379266)

This paper deals with SCD sets and SCD operators. These concepts were introduced recently [\textit{A. Aviles} et al, C. R., Math., Acad. Sci. Paris 347, No. 21-22, 1277--1280 (2009; Zbl 1187.46007)] as follows. A bounded convex subset \(A\) of a Banach space \(X\) is called \textit{SCD} if there is a sequence of slices \((S_n)\) such that each slice of \(A\) contains one of the \(S_n\) and \(A\) is said to be \textit{hereditarily SCD} when every convex subset of \(A\) is SCD. A (separable) Banach space \(X\) is called \textit{SCD} if every bounded convex subset of \(X\) is SCD. Finally, an operator \(T:X\longrightarrow Y\) is called \textit{(hereditarily) SCD} if the image by \(T\) of the unit ball of \(X\) is (hereditarily) SCD. Spaces not containing \(\ell_1\) and spaces with the Radon-Nikodým property are SCD; operators not fixing copies of \(\ell_1\) and strong Radon-Nikodým operators are hereditarily SCD. In this paper it is shown that the direct sum of two hereditarily SCD sets is SCD and this result is used to give an interesting result about narrow operators. Also, the questions 7.7 of the authors' paper with \textit{A. Avilés}, \textit{M. Martín} and \textit{J. Merí} [Trans. Am. Math. Soc. 362, 4871--4900 (2010; Zbl 1214.46004)] are answered in the negative: a sum of two (hereditarily) SCD operators need not be an SCD operator. Editorial remark: The authors have pointed out to us that there is a misprint in Corollary~2.2 which should read ``A sum of two hereditarily SCD sets need not be hereditarily SCD'' instead of ``\dots need not be an SCD set.''