8

A compact operator between Banach spaces is an operator that maps bounded sets into relatively compact sets, while a completely continuous operator maps all weakly convergent sequences into convergent sequences.

Compact operators are always completely continuous, but completely continuous operators may be non-compact: the identity operator in the Schur space ${\rm l}_1$ is an example. In reflexive spaces, completely continuous operators are compact, so the two classes of operators are the same.

Here are my questions:

  1. If the classes of compact operators and completely continuous operators are the same, what can we say about the space? Must it be reflexive?

  2. What can we say about the spectrum of completely continuous operators?

t.b.
  • 78,116
Strongart
  • 4,767
  • I made some changes in your question, hoping it is easier to understand now. Please check I didn't distort your intentions. A small remark: "totally continuous" operators are usually called completely continuous. – t.b. Sep 04 '12 at 12:02

2 Answers2

9

I do not know off the top of my head a characterisation of when the completely continuous operators coincide with the compact operators, but certainly such a space need not be reflexive; for example, consider the James space. In particular it is shown in Proposition 4.9 of Niels Laustsen's paper Maximal ideals in the algebra of operators on certain Banach spaces, Proceedings of the Edinburgh Mathematical Society 45 (2002), 523–546, that the compact operators coincide with the completely continuous operators on the James space.

Another example of a non-reflexive space where every completely continuous operator is compact is the space $C(K)$ of all continuous scalar-valued functions on $K$ (with the sup norm), where $K$ is a scattered, compact Hausdorff space. In fact, the converse is true also: a compact Hausdorff space is scattered if and only if every completely continuous operator on $C(K)$ is compact.

Edit: Of course, an easy counterexample (implicitly contained in my $C(K)$ counterexample above) to the question of whether reflexivity is necessary for every completely continuous operator to be compact is the sequence space $c_0$. The standard unit vector basis of $c_0$ is weakly null but not norm null, so the ideal of completely continuous operators on $c_0$ is properly contained in the algebra of all bounded linear operators on $c_0$. On the other hand it is a classical result that the algebra of all bounded linear operators on $c_0$ contains only one non-trivial closed two-sided ideal, namely the compact operators. It follows then that the ideal of compact operators on $c_0$ and the ideal of completely continuous operators $c_0$ must be the same.

Philip Brooker
  • 1,806
  • Thanks, but I need more details. 1) I have not heard about the scattered space before, can tell me anything more or which books discuss it? 2)For the c$_0$ part, if I change the c$_0$ by l$_1$, which step will break? – Strongart Sep 07 '12 at 11:44
  • The definition of scattered space (also known as a dispersed space) is one where every subset contain an isolated point (with respect to the subspace topology). There are probably better places to learn about them, but I learnt about them from Semadeni's book Banach spaces of continuous functions; this book is over 40 years old, but it is still very good and widely used as a reference. 2) The argument for $c_0$ doesn't work for $\ell_1$ because $\ell_1$ has the Schur property (that is, every weakly convergent sequence in $\ell_1$ is norm convergent); in particular, every bounded...
    • – Philip Brooker Sep 24 '12 at 20:03
  • linear operator on $\ell_1$ is completely continuous, and so the ideal of completely continuous operators is not proper. On a related point, notice that the standard unit vector basis of $\ell_1$ is not weakly convergent. – Philip Brooker Sep 24 '12 at 20:09