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Math Help - Lipschitz Continuous

  1. #1
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    Lipschitz Continuous

    I know this proof should be short and sweet, but maybe there is something that I'm missing. I need to prove that every Lipschitz mapping is uniformly continuous. The map f: A c R --->R. I know the definition of Lipschitz is C=>0 ||f(x)-f(y)||<= C||x-y||

    I assumed that the function is Lipschitz, but I don't know what to do next. Should I let E>0 exist? I am not sure if I can create an epsilon..

    Thanks!
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  2. #2
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    Quote Originally Posted by EricaMae View Post
    to prove that every Lipschitz mapping is uniformly continuous. The map f: A c R --->R. I know the definition of Lipschitz is C=>0 ||f(x)-f(y)||<= C||x-y||
    Is this something new? I have seen it three times now.
    The historical Lipschitz condition requires C > 0.
    If \varepsilon  > 0 the let \delta  = \frac{\varepsilon }{C}.
    Then it is simple.
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