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Math Help - proof involving lipschitz and uniformly continuity. is it correct?

  1. #1
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    proof involving lipschitz and uniformly continuity. is it correct?

    Let f: D-R satisfy the Lipschitz condition: there exists an L such that for all x,y \inD |f(x)-f(y)|<=L|x-y|

    Prove that f is uniformly continuous on D

    Here is what I have, is this correct?

    Let \epsilon>0. Choose \delta= \epsilon/L. Then by the definition of uniform continuity, if |x-y|< \delta= \epsilon/L, then |f(x)-f(y)|<L \epsilon/L= \epsilon.
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  2. #2
    Senior Member
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    Feb 2008
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    Your logic is correct but your writing is misleading. When you say "by the definition of uniform continuity" you are signalling that you're about to use uniform continuity as a hypothesis. But instead you use Lipschitz continuity---which is what you should do!

    A proper write-up would look something like this...

    Let \to R" alt="f\to R" /> satisfy a Lipschitz condition with constant L, and let \epsilon>0. Then there is \delta:=\epsilon/L>0 such that for any x,y\in D, if |x-y|<\delta then |f(x)-f(y)|<L|x-y|=L\epsilon/L=\epsilon. Hence f is uniformly continuous.
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