# 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. In the past, the term "completely-continuous operator" was often used to mean 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).