# contractive maps are uniformly continuous

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 $x,y\in X$,

 $d(Tx,Ty)\leq qd(x,y).$

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

 $d(Tx,Ty)\leq qd(x,y)<\varepsilon.$

In conclusion, $T$ is uniformly continuous. $\Box$

The result is stated without proof in [1], pp. 221.

## References

Title contractive maps are uniformly continuous ContractiveMapsAreUniformlyContinuous 2013-03-22 13:46:28 2013-03-22 13:46:28 mathcam (2727) mathcam (2727) 6 mathcam (2727) Theorem msc 54A20