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Thread: 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 : $\displaystyle [0,\infty) \rightarrow $ R given by f(x) := $\displaystyle e^{-x}$ on $\displaystyle [0,\infty)$.

    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 : $\displaystyle [0,\infty) \rightarrow $ R given by f(x) := $\displaystyle e^{-x}$ on $\displaystyle [0,\infty)$.

    I know it's not uniformly continuous but how do I prove that?
    Derivative of $\displaystyle e^{-x}$ is $\displaystyle -e^{-x}$, so $\displaystyle |f'(x)|\leq1$ on $\displaystyle [0,\infty)$. 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 $\displaystyle \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } = \vert f'(c)\vert$ where $\displaystyle c\in [x,y]$. Taking the supremum over all $\displaystyle x,y\in [0, \infty )$ and noting that $\displaystyle \vert f'(x) \vert \leq 1$ we have that $\displaystyle \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq 1$ for all $\displaystyle x\neq y$ and so $\displaystyle \vert f(x)-f(y) \vert \leq \vert x-y\vert$. Now taking $\displaystyle \delta < \epsilon$ in the definition of unif. cont. the result follows.

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

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