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Math Help - Lipschitz continuity proof given boundedness of a derivate (not a derivative)

  1. #1
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    Lipschitz continuity proof given boundedness of a derivate (not a derivative)

    Hey guys. Tough problem here (tough for me anyway...)

    Suppose the derivate D^+ of a function f on [a,b]\subseteq\overline{\mathbb{R}} is bounded, where

    D^+=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}.

    Show that f is Lipschitz continuous.
    I can show that f is of bounded variation, and therefore is differentiable almost everywhere. I don't know if that's a promising approach, however, nor even if it is, how to finish the proof.

    One other possible avenue is this: The proof that a function is Lipschitz if its derivative is bounded uses the mean value theorem. Is there maybe some variation of the mean value theorem for D^+ which I could use in this case?

    Any help would be much appreciated!
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  2. #2
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    That result looks false to me. Suppose that f is the function

    f(x) = \begin{cases}0&(x<0),\\1&(x\geqslant0).\end{cases}

    Then f has D^+ equal to 0 everywhere (in the interval [1,1) say), but it isn't even continuous, let alone Lipschitz continuous.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    That result looks false to me. Suppose that f is the function

    f(x) = \begin{cases}0&(x<0),\\1&(x\geqslant0).\end{cases}

    Then f has D^+ equal to 0 everywhere (in the interval [–1,1) say), but it isn't even continuous, let alone Lipschitz continuous.
    Yeah, that's what I was thinking, too, but I didn't have the confidence to say so. Okay, then. Thanks!
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  4. #4
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    It occurs to me that you may have got the definition of D^+ a bit wrong. Suppose that it should be D^+ = \limsup_{h\to0}\frac{f(x+h)-f(x)}h. In other words, the "+" in D^+ indicates that the lim (in the definition of a derivative) has become a limsup, but it is taken in both directions, not just from the right.

    With that definition, it seems plausible that if D^+ is bounded then f should be Lipschitz continuous. But I don't offhand see how to prove that.
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