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Difference between revisions of "Completely-continuous operator"

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''Completely-Continuous Operator''
 
''Completely-Continuous 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.
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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.  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).  
 
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).  
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====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  D. Hilbert,  "Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen" , Chelsea, reprint  (1953)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  F. Riesz,  "Sur les opérations fonctionelles linéaires"  ''C.R. Acad. Sci. Paris Sér. I Math.'' , '''149'''  (1909)  pp. 974–977</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  S.S. Banach,  "Théorie des opérations linéaires" , Hafner  (1932)</TD></TR><TD valign="top">[4]</TD> <TD valign="top"> R. E. Megginson,  "An Introduction to Banach Space Theory" , Springer (1998)p.\ 336  </table>
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<table><TR><TD  
 
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valign="top">[1]</TD> <TD valign="top">  D. Hilbert,  "Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen" , Chelsea, reprint  (1953)</TD></TR><TR><TD  
 
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valign="top">[2]</TD> <TD valign="top">  F. Riesz,  "Sur les opérations fonctionelles linéaires"  ''C.R. Acad. Sci. Paris Sér. I Math.'' , '''149'''  (1909)  pp. 974–977</TD></TR><TR><TD  
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valign="top">[3]</TD> <TD valign="top">  S.S. Banach,  "Théorie des opérations linéaires" , Hafner  (1932)</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top"> R. E. Megginson,  "An Introduction to Banach Space Theory" , Springer (1998) pp. 336-339 </TD></TR><TR><TD
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valign="top">[5]</TD> <TD  valign="top"> A. Pietsch,  "History of Banach Spaces and Linear Operators" , Birkhauser (2007)  pp. 49-50 </table>
  
 
====Comments====
 
====Comments====

Latest revision as of 20:40, 2 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. 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).


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)
[4] R. E. Megginson, "An Introduction to Banach Space Theory" , Springer (1998) pp. 336-339
[5] A. Pietsch, "History of Banach Spaces and Linear Operators" , Birkhauser (2007) pp. 49-50

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://encyclopediaofmath.org/index.php?title=Completely-continuous_operator&oldid=29586
This article was adapted from an original article by M.I. Voitsekhovskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article