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<record version="5" id="5966">
 <title>compact operator</title>
 <name>CompactOperator</name>
 <created>2004-06-26 14:23:15</created>
 <modified>2008-06-29 14:09:28</modified>
 <type>Definition</type>
 <creator id="572" name="mhale"/>
 <author id="3771" name="CWoo"/>
 <author id="572" name="mhale"/>
 <classification>
	<category scheme="msc" code="46B99"/>
 </classification>
 <synonyms>
	<synonym concept="compact operator" alias="completely continuous"/>
 </synonyms>
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 <content>Let $X$ and $Y$ be two Banach spaces.
A \textbf{compact operator} (completely continuous operator) is a linear operator $T \colon X \to Y$
that maps the unit ball in $X$ to a set in $Y$ with compact closure.  It can be shown that a compact operator is necessarily a bounded operator.

The set of all compact operators on $X$, commonly denoted by $\Kset(X)$,
is a closed two-sided ideal of the set of all bounded operators on $X$, $\Bset(X)$.

Any bounded operator which is the norm limit of a sequence of finite rank operators is compact.
In the case of Hilbert spaces, the converse is also true.
That is, any compact operator on a Hilbert space is a norm limit of finite rank operators.

\begin{example}[Integral operators]
Let $k(x,y)$, with $x,y \in [0,1]$, be a continuous function.
The operator defined by $$(T\psi)(x) = \int_0^1 k(x,y) \psi(y) \,\d y, \qquad \psi \in C([0,1])$$ is compact.
\end{example}</content>
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