Math 2.0

Real functions Proposition

Lipschitz maps are uniformly continuous

Builds on Lipschitz mapEpsilon-delta uniform continuity

Statement

Let XX and YY be metric spaces and let f ⁣:X→Yf \colon X \to Y be a Lipschitz map. Then ff is uniformly continuous.

Remarks

Proof: let K>0K > 0 be a Lipschitz constant of ff. For ε>0\varepsilon > 0, take δ=ε/K\delta = \varepsilon / K. If d(x,y)<δd(x, y) < \delta, then d(f(x),f(y))≤Kd(x,y)<εd(f(x), f(y)) \le K d(x, y) < \varepsilon. If ff is Lipschitz with K=0K = 0, then ff is constant. The converse fails: x↦xx \mapsto \sqrt{x} on [0,1][0, 1] is uniformly continuous, but it is not Lipschitz. In particular, each Lipschitz map is continuous.

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Where this leads

4 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Lipschitz_continuity, Properties
About this page
Tag07DCKDXRBP
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Kindstatement, proposition
WritingW0 stub
MSC26A16
StatusThis copy of the site has no build data. Run m2 status 07DCKDXRBP.
UsesLipschitz map, Epsilon-delta uniform continuity

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