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<record version="3" id="4434">
 <title>$T_f$ is a distribution of zeroth order</title>
 <name>T_fIsADistributionOfZerothOrder</name>
 <created>2003-07-09 05:18:27</created>
 <modified>2003-12-20 05:24:38</modified>
 <type>Proof</type>
<parent id="4433">every locally integrable function is a distribution</parent>
 <selfproof>0</selfproof>
 <creator id="127" name="Koro"/>
 <author id="1858" name="matte"/>
 <classification>
	<category scheme="msc" code="46F05"/>
	<category scheme="msc" code="46-00"/>
 </classification>
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To check that $T_f$ is a \PMlinkname{distribution of zeroth order}{Distribution4},
we shall use
condition (3) on \PMlinkname{this page}{Distribution4}.
First, it is clear that $T_f$ is a linear mapping.
To see that $T_f$ is continuous, suppose $K$ is a compact set in
$U$ and $u\in \cD_K$, i.e., $u$ is a smooth function with support in $K$.
We then have
\begin{eqnarray*}
|T_f(u)| &amp;=&amp; |\int_K f(x) u(x) dx |\\
   &amp;\le&amp; \int_K |f(x)| \,\, |u(x)| dx \\
   &amp;\le&amp; \int_K |f(x)| dx \, ||u||_\infty.
\end{eqnarray*}
Since $f$ is locally integrable, it follows that $C=\int_K |f(x)| dx$
is finite, so
$$ |T_f(u)| \le C ||u||_\infty.$$
Thus $f$ is a distribution of zeroth order (\cite{lang}, pp. 381). $\Box$

\begin{thebibliography}{9}
 \bibitem{lang}
 S. Lang, \emph{Analysis II},
 Addison-Wesley Publishing Company Inc., 1969.
 \end{thebibliography}</content>
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