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Math Help - Uniform Continuity

  1. #1
    Senior Member I-Think's Avatar
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    Uniform Continuity

    There are 2 ways of saying a function,  f, is uniformly continuous

    A. For every \epsilon>0, there exists a \delta>0 such that
    |f(x)-f(p)|<\epsilon whenever  x,p\in{I} and |x-p|<\delta

    B. For every \epsilon>0, there exists a \delta>0 such that
    |f(x)-f(p)|\leq\epsilon whenever  x,p\in{I} and |x-p|\leq\delta

    Prove f satisfies A if and only if it satisfies B

    This is a confusing proposition. I think if f satisfies B, then it obviously satisfies A, but I could be mistaken.
    And if it satisfies A, how do I go about demonstrating B?

    Thanks in advance for the help
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    There are 2 ways of saying a function,  f, is uniformly continuous

    A. For every \epsilon>0, there exists a \delta>0 such that
    |f(x)-f(p)|<\epsilon whenever  x,p\in{I} and |x-p|<\delta

    B. For every \epsilon>0, there exists a \delta>0 such that
    |f(x)-f(p)|\leq\epsilon whenever  x,p\in{I} and |x-p|\leq\delta

    Prove f satisfies A if and only if it satisfies B

    This is a confusing proposition. I think if f satisfies B, then it obviously satisfies A, but I could be mistaken.
    And if it satisfies A, how do I go about demonstrating B?

    Thanks in advance for the help
    As a heuristics less than and less than or equal to are indistinguishable in hard analysis. If f satisfies A. then it clearly satisfies B. Conversely, try for a given \varepsilon using \frac{\varepsilon}{2} in B. and taking half of the guaranteed \delta
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    As a curiosity:

    We say that A\leq 0 is obtained by relaxing the strict inequalyty A<0 .


    Fernando Revilla
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