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 $x,y\in X$ , $$ d(Tx,Ty)\le q d(x,y).$$ 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.

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

Bibliography

1

W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Inc., 1976.