contractive maps are uniformly continuous

Proof Let T:XX be a contraction mapping in a metric space X with metric d. Thus, for some q[0,1), we have for all x,yX,

d(Tx,Ty)qd(x,y).

To prove that T is uniformly continuous, let ε>0 be given. There are two cases. If q=0, our claim is trivial, since then for all x,yX,

d(Tx,Ty)=0<ε.

On the other hand, suppose q(0,1). Then for all x,yX with d(x,y)<ε/q, we have

d(Tx,Ty)qd(x,y)<ε.

In conclusionMathworldPlanetmath, T is uniformly continuous.

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

References

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