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Thread: Proof Uniformly Continous

  1. #1
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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 $\displaystyle J$ be an open inverval. Say $\displaystyle f: J\mapsto \mathbb{R}$ is a differenciable function. And $\displaystyle |f'|\leq C$ is bounded by positive constant. Let $\displaystyle x,y\in J$ be two distinct points. Then $\displaystyle |f(x)-f(y)|\leq C|x-y|$ so if we choose $\displaystyle |x-y| < \delta$ where $\displaystyle \delta = \frac{\epsilon}{C}$ 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 $\displaystyle \epsilon >0$ choose $\displaystyle \delta = \epsilon^2$ then $\displaystyle |f(x)-f(y)| = |\sqrt{x}-\sqrt{y}|$ say $\displaystyle x\not = y$ because otherwise this difference is zero which is certainly less than $\displaystyle \epsilon$. In that case we get $\displaystyle |\sqrt{x}-\sqrt{y}| = \frac{|x-y|}{\sqrt{x}+\sqrt{y}} \leq \frac{|x-y|}{\sqrt{|x-y|} }= \sqrt{|x-y|} < \sqrt{\delta} = \epsilon$.
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