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uniform continuity of Lipschitz functions
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(Theorem)
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Proof. Let $f\colon X\to Y$ be a mapping such that for some $C>0$ and $\alpha$ with $0 < \alpha \le 1$ one has $$ d_Y(f(p),f(q)) \le C d_X(p,q)^\alpha. $$ For every given $\epsilon > 0$ , choose $\delta=\left(\epsilon/(C+1)\right)^{\frac 1 \alpha}$ . If $p,q\in X$ are given points satisfying $$ d_X(p,q)<\delta $$ then $$ d_Y(f(p),f(q))\leq C \delta^\alpha \le C\frac{\epsilon}{C+1} < \epsilon, $$ as desired. 
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"uniform continuity of Lipschitz functions" is owned by paolini.
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Cross-references: points, mapping, Lipschitz continuous, uniformly continuous, continuous mapping
This is version 5 of uniform continuity of Lipschitz functions, born on 2005-02-28, modified 2009-06-11.
Object id is 6834, canonical name is UniformContinuityOfLipschitzFunctions.
Accessed 5628 times total.
Classification:
| AMS MSC: | 26A16 (Real functions :: Functions of one variable :: Lipschitz classes) |
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Pending Errata and Addenda
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