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Math Help - Holder Continuity

  1. #1
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    Holder Continuity

    A function f: (a,b) ->R is said to be Holder continuous of order alpha>0, if there exists M>0 such that |f(x)-f(y)|<=M|x-y|^alpha for all x,y elements of (a,b).

    a) Prove that if alpha>0 and f is Holder continuous of order alpha, then f is uniformly continuous on (a,b).
    b) Prove that if alpha>1 and f is Holder continuous of order alpha, then f is constant on (a,b).
    c) Prove that if f is differentiable and Holder continuous of order alpha=1, then f'(x) is a bound function on (a,b).

    My work:
    a) Take any E>0 and choose delta=(E/M)^(1/alpha)>0. Then for all x,y elements of (a,b) such that |x-y|<delta, we have |f(x)-f(y)|<=M|x-y|^alpha<M*delta^alpha=E. Is that all I need to prove part (a)?

    I don't know where to begin for (b) and (c). Any help?
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  2. #2
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    Re: Holder Continuity

    Hint for b): under these assumptions, f is differentiable and its derivative equals 0.
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