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 $x,y\in X$,
$$d(Tx,Ty)\le qd(x,y).$$ |
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.
References
- 1 W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Inc., 1976.
Title | contractive maps are uniformly continuous |
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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 |