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Math Help - uniform continuity

  1. #1
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    uniform continuity

    Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.
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    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by friday616 View Post
    Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.
    Three words: MEAN VALUE THEOREM

    \frac{|f(x)-f(y)|}{|x-y|}=|f'(t)|\leq M\implies |f(x)-f(y)|\leq M|x-y|.

    Let \delta=\frac{\epsilon}{M} and you're done.

    If d(f(x),f(y))\leq Md(x,y), then f is said to be Lipschitz continuous, with the Lipschitz constant being the smallest such M that satisfies that inequality. Lipschitz continuity always implies uniform continuity, but not necessarily the other way around.
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  3. #3
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    Quote Originally Posted by friday616 View Post
    Suppose that f:R--> R is differentiable on R and that the derivative of f is bounded on R. Prove that f is uniformly continuous on R.
    If the derivative is continous you could use the MVT \vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vertand so f is Lipschitz and thus uniformly continous. Without continous derivative though...
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    If the derivative is continous you could use the MVT \vert f(x)-f(y) \vert = \vert f(c)(x-y) \vert \leq M\vert x-y\vertand so f is Lipschitz and thus uniformly continous. Without continous derivative though...
    MVT only requires the derivative to exist, not for it to be continuous.
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