contractive maps are uniformly continuous

Theorem A contraction mapping is 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 $x,y\in X$,

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

To prove that $T$ is uniformly continuous, let $\epsilon >0$ be given. There are two cases. If $q=0$, our claim is trivial, since then for all $x,y\in X$,

On the other hand, suppose $q\in (0,1)$. Then for all $x,y\in X$ with , we have

In conclusion, $T$ is uniformly continuous. $\mathrm{\square }$

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

Title	contractive maps are uniformly continuous
Canonical name	ContractiveMapsAreUniformlyContinuous
Date of creation	2013-03-22 13:46:28
Last modified on	2013-03-22 13:46:28
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	6
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 54A20