# Difference between revisions of "Completely-continuous operator"

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

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+ | A bounded linear operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239501.png" />, acting from a [[Banach space|Banach space]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239502.png" /> into another space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239503.png" />, that transforms weakly-convergent sequences in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239504.png" /> to norm-convergent sequences in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239505.png" />. Equivalently, an operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239508.png" /> is completely-continuous if it maps every relatively weakly compact subset of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239509.png" /> into a relatively compact subset of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c02395010.png" />. 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''<sub>1</sub> 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. | ||

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+ | It can be assumed that the space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239506.png" /> is separable (for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c023/c023950/c0239507.png" /> this is not a necessary condition; however, the image of a completely-continuous operator is always separable). | ||

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

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====References==== | ====References==== | ||

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====Comments==== | ====Comments==== | ||

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====References==== | ====References==== | ||

<table><TR><TD valign="top">[a1]</TD> <TD valign="top"> N. Dunford, J.T. Schwartz, "Linear operators. General theory" , '''1''' , Interscience (1958)</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top"> A.E. Taylor, D.C. Lay, "Introduction to functional analysis" , Wiley (1980)</TD></TR></table> | <table><TR><TD valign="top">[a1]</TD> <TD valign="top"> N. Dunford, J.T. Schwartz, "Linear operators. General theory" , '''1''' , Interscience (1958)</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top"> A.E. Taylor, D.C. Lay, "Introduction to functional analysis" , Wiley (1980)</TD></TR></table> |

## Revision as of 04:14, 31 March 2013

*Completely-Continuous Operator*

A bounded linear operator , acting from a Banach space into another space , that transforms weakly-convergent sequences in to norm-convergent sequences in . Equivalently, an operator is completely-continuous if it maps every relatively weakly compact subset of into a relatively compact subset of . 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 is separable (for 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=29567