Namespaces
Variants
Actions

Difference between revisions of "Completely-continuous operator"

From Encyclopedia of Mathematics
Jump to: navigation, search
m (tex encoding is done)
Line 1: Line 1:
 
+
{{TEX|done}}
  
  
 
''Completely-Continuous Operator''
 
''Completely-Continuous Operator''
  
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.
+
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 <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).  
+
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
This article was adapted from an original article by M.I. Voitsekhovskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article