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Math Help - [SOLVED] Prove differentiability from Lipschitz condition

  1. #1
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    [SOLVED] Prove differentiability from Lipschitz condition

    A function a,b)\to R" alt="fa,b)\to R" /> satisfies a Lipschitz condition at x\in(a,b) iff there is M>0 and \epsilon>0 such that |x-y|<\epsilon and y\in(a,b) imply that |f(x)-f(y)|\leq M|x-y|. If f is differentiable at x, prove that f satisfies a Lipschitz condition at x.
    The similarities between this condition and the definition of the derivative are obvious, but I can't seem to find an approach whereby I can actually prove a derivative exists in this case.

    Any ideas?

    Thanks!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    The similarities between this condition and the definition of the derivative are obvious, but I can't seem to find an approach whereby I can actually prove a derivative exists in this case.

    Any ideas?

    Thanks!
    The question asks to show that "differentiable" implies "Lipschitz", not the converse (which is false) (however, a Lipschitz function is differentiable almost-everywhere, but this is a hard theorem).

    To solve the question, you just have to write the \varepsilon-\delta definition of the differentiability at x for a fixed \delta, for instance \delta=1, and to use a triangle inequality: there is \ell(=f'(x)) and \varepsilon>0 such that for y such that |x-y|<\varepsilon, |\frac{f(y)-f(x)}{y-x}-\ell|\leq 1, hence |\frac{f(y)-f(x)}{y-x}|\leq \ell + 1.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    The question asks to show that "differentiable" implies "Lipschitz", not the converse (which is false) (however, a Lipschitz function is differentiable almost-everywhere, but this is a hard theorem).

    To solve the question, you just have to write the \varepsilon-\delta definition of the differentiability at x for a fixed \delta, for instance \delta=1, and to use a triangle inequality: there is \ell(=f'(x)) and \varepsilon>0 such that for y such that |x-y|<\varepsilon, |\frac{f(y)-f(x)}{y-x}-\ell|\leq 1, hence |\frac{f(y)-f(x)}{y-x}|\leq \ell + 1.
    Oh, oops. Here I am worrying about my math abilities, and the problem is in my comprehension of English. No wonder I couldn't prove it!

    Thanks!
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