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Thread: Uniform Continuity

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

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

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

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

    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,$\displaystyle f$, is uniformly continuous

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

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

    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 $\displaystyle f$ satisfies A. then it clearly satisfies B. Conversely, try for a given $\displaystyle \varepsilon$ using $\displaystyle \frac{\varepsilon}{2}$ in B. and taking half of the guaranteed $\displaystyle \delta$
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    As a curiosity:

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


    Fernando Revilla
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