# Completely-continuous operator

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Completely-Continuous Operator

A bounded linear operator \$f\$, acting from a Banach space \$X\$ into another space \$Y\$, that transforms weakly-convergent sequences in \$X\$ to norm-convergent sequences in \$Y\$. Equivalently, an operator \$f\$ is completely-continuous if it maps every relatively weakly compact subset of \$X\$ into a relatively compact subset of \$Y\$. It is easy to see that every compact operator is completely continuous, however the converse is false. For example, recall that the Banach space \$X=l_1\$ has the Schur Property, that is weak sequential and norm sequential convergence coincide. It follows that the identity operator from \$X\$ to \$X\$ is completely-continuous, but it is not compact since \$X\$ is infinite-dimensional. If \$X\$ is reflexive, then every completely-continuous operator is compact, so the two classes of operators do coincide in that case. The term "completely-continuous operator" originally meant what we now call "compact operator", which has sometimes resulted in confusion.

It can be assumed that the space \$X\$ is separable (for \$Y\$ this is not a necessary condition; however, the image of a completely-continuous operator is always separable).

The class of compact operators is the most important class of the set of completely-continuous operators (cf. Compact operator).

How to Cite This Entry:
Completely-continuous operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Completely-continuous_operator&oldid=29590
This article was adapted from an original article by M.I. Voitsekhovskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article