# Difference between revisions of "Completely-continuous operator"

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''Completely-Continuous Operator'' | ''Completely-Continuous Operator'' | ||

− | A bounded linear operator | + | A bounded linear operator $f$, acting from a [[Banach space|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 | + | 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). |

## Revision as of 06:18, 1 April 2013

*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).

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

#### References

[1] | D. Hilbert, "Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen" , Chelsea, reprint (1953) |

[2] | F. Riesz, "Sur les opérations fonctionelles linéaires" C.R. Acad. Sci. Paris Sér. I Math. , 149 (1909) pp. 974–977 |

[3] | S.S. Banach, "Théorie des opérations linéaires" , Hafner (1932) |

#### Comments

#### References

[a1] | N. Dunford, J.T. Schwartz, "Linear operators. General theory" , 1 , Interscience (1958) |

[a2] | A.E. Taylor, D.C. Lay, "Introduction to functional analysis" , Wiley (1980) |

**How to Cite This Entry:**

Completely-continuous operator.

*Encyclopedia of Mathematics.*URL: http://www.encyclopediaofmath.org/index.php?title=Completely-continuous_operator&oldid=29569