Spaces of operators, \(c_0\) and \(l^1\). (English) Zbl 1073.46010

Bator, Elizabeth M.; Lewis, Paul W.

Rocky Mt. J. Math. 34, No. 3, 829-832 (2004).

A classical result of Pelczynski says that if the dual \(X^\ast\) of a Banach space \(X\) contains a copy of \(c_0\), then \(X^\ast\) already contains a (1-complemented) copy of \(\ell_\infty\) and \(X\) contains a complemented copy of \(\ell_1\). Thus, for Banach spaces \(X\) and \(Y\), \(L(X,Y^{\ast\ast})\) contains a copy of \(c_0\) iff it contains a copy of \(\ell_\infty\) iff \(X\otimes_\pi Y^\ast\) contains a complemented copy of \(\ell_1\). The authors observe that the embedding of \(\ell_1\) into \(X\otimes_\pi Y^\ast\) maps each unit vector in \(\ell_1\) to a finite tensor.
The important result of this short note is, however, that we cannot substitute \(Y^{\ast\ast}\) by \(Y\) in the last equivalence of the above theorem. More precisely, whenever \(Y\) is such that \(Y^\ast\) is separable, \(\ell_1\) embeds complementably into \(Y^\ast\) but \(c_0\) does not embed as a subspace of \(Y\), then there exist a Banach space \(X\) and an isomorphic embedding \(J\) of \(\ell_1\) into \(X\otimes_\pi Y^\ast\) such that its image is a complemented subspace and each unit vector is taken to a finite rank tensor, but \(L(X,Y)\) does not contain a copy of \(c_0\). This completes the results from P. Lewis [Stud. Math. 145, 213–218 (2001; Zbl 0986.46011)].

Reviewer: Olav Nygaard (Kristiansand)

MSC:

46B28	Spaces of operators; tensor products; approximation properties
46B20	Geometry and structure of normed linear spaces
46B25	Classical Banach spaces in the general theory

Keywords:

copy of \(c_0\) in \(L(X,Y)\)

Citations:

Zbl 0986.46011

References:

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