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Math Help - Proof Uniformly Continous

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    Proof Uniformly Continous

    Need help with this proof. Thanks so much!

    Let j be an interval and f a differentiable function on j whose derivative is bounded. Show that f is uniformly continuous (use the mean value theorem).
    Observe that boundedness of f ' is not necessary for uniform continuity of f. Indeed, the function x sqrtx (x > 0) is uniformly continuous and differentiable, but its derivative is not bounded.
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    Quote Originally Posted by ml692787 View Post
    Need help with this proof. Thanks so much!

    Let j be an interval and f a differentiable function on j whose derivative is bounded. Show that f is uniformly continuous (use the mean value theorem).
    Let J be an open inverval. Say f: J\mapsto \mathbb{R} is a differenciable function. And |f'|\leq C is bounded by positive constant. Let x,y\in J be two distinct points. Then |f(x)-f(y)|\leq C|x-y| so if we choose |x-y| < \delta where \delta = \frac{\epsilon}{C} this will complete the proof.
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    Quote Originally Posted by ml692787 View Post
    Observe that boundedness of f ' is not necessary for uniform continuity of f. Indeed, the function x sqrtx (x > 0) is uniformly continuous and differentiable, but its derivative is not bounded.
    For any \epsilon >0 choose \delta = \epsilon^2 then |f(x)-f(y)| = |\sqrt{x}-\sqrt{y}| say x\not = y because otherwise this difference is zero which is certainly less than \epsilon. In that case we get |\sqrt{x}-\sqrt{y}| = \frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{|x-y|}{\sqrt{|x-y|} }= \sqrt{|x-y|} < \sqrt{\delta} = \epsilon.
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