54A20-ContractiveMapsAreUniformlyContinuous.tex

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{\bf Theorem} A contraction mapping is uniformly continuous.

{\bf Proof} Let $T:X\to X$ be a contraction mapping in a metric space
$X$ with metric $d$. Thus, for some $q\in [0,1)$, we have
for all $x,y\in X$,
$$ d(Tx,Ty)\le q d(x,y).$$
To prove that $T$ is uniformly continuous, let $\varepsilon>0$ be given.
There are two cases.
If $q=0$, our claim is trivial, since then for all $x,y\in X$,
 $$ d(Tx,Ty)=0<\varepsilon.$$
On the other hand, suppose $q\in(0,1)$. Then for all $x,y\in X$ with
$d(x,y)<\varepsilon/q$, we have
$$ d(Tx,Ty) \le q d(x,y) < \varepsilon.$$
In conclusion, $T$ is uniformly continuous. $\Box$

The result is stated without proof in \cite{rudin}, pp. 221.

\begin{thebibliography}{9}
 \bibitem{rudin}
 W. Rudin, \emph{Principles of Mathematical Analysis}, McGraw-Hill Inc., 1976.
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