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Math Help - Prove it's not uniformly continous

  1. #1
    Member thaopanda's Avatar
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    Prove it's not uniformly continous

    Examine the uniform continuity of f : [0,\infty) \rightarrow R given by f(x) := e^{-x} on [0,\infty).

    I know it's not uniformly continuous but how do I prove that?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    Examine the uniform continuity of f : [0,\infty) \rightarrow R given by f(x) := e^{-x} on [0,\infty).

    I know it's not uniformly continuous but how do I prove that?
    Derivative of e^{-x} is -e^{-x}, so |f'(x)|\leq1 on [0,\infty). This means that it is uniformly continuous.
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  3. #3
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    It is uniformly continous:

    By the mean value theorem we have \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } = \vert f'(c)\vert where c\in [x,y]. Taking the supremum over all x,y\in [0, \infty ) and noting that \vert f'(x) \vert \leq 1 we have that \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq 1 for all x\neq y and so \vert f(x)-f(y) \vert \leq \vert x-y\vert. Now taking \delta < \epsilon in the definition of unif. cont. the result follows.

    Actually you can prove that for a continously diff. function it's equivalent:

    1) f' is bounded
    2) There exists m>0 such that \vert f(x)-f(y)\vert \leq m\vert x-y\vert
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