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Math Help - Uniform Continuity on an Interval

  1. #1
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    Uniform Continuity on an Interval

    Show f(x)=\frac{1}{x^2} is not uniformly continuous on I=(0,\infty) .
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  2. #2
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    Basically, it can't be uniformly continuous in that interval because the slope goes to (negative) infinity as x goes to zero. To prove this rigorously, remember the definition of uniform continuity: for any \epsilon>0, there is a fixed \delta such that |x-y|<\delta implies |\frac{1}{x^2}-\frac{1}{y^2}|< \epsilon.

    To show that it's not uniformly continuous, choose any epsilon, and show that for any fixed delta, |\frac{1}{x^2}-\frac{1}{(x+\delta)^2}| goes to infinity as x goes to 0.
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  3. #3
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    so to rigorously prove it can I just choose a delta like 0?
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  4. #4
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    oh ok never mind that was a stupid question lol

    the only thing I dont see is how to replace y^2 with (x+ \delta)^2
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  5. #5
    Senior Member Pinkk's Avatar
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    You can also use the fact that you can find a Cauchy sequence (s_{n})\in (0,\infty ) where (f(s_{n})) is NOT Cauchy.
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    Quote Originally Posted by frenchguy87 View Post
    Show f(x)=\frac{1}{x^2} is not uniformly continuous on I=(0,\infty) .
    There is a theorem concerning uniform continuity that says:

    f is uniform continuous on its domain ,and in our case (0, \infty) iff for any pair of sequences on (0, \infty { x_{n}},{ y_{n}} we have:

    lim_{n\to\infty} |x_{n}-y_{n}| = 0\Longrightarrow lim_{n\to\infty}|f(x_{n})-f(y_{n})| =0

    Hence to prove that : \frac{1}{x^2} is not uniform continuous on (0, \infty) we must find a pair of sequences that:

    lim_{n\to\infty} |x_{n}-y_{n}| = 0 ,and

    lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0

    So if we put :

    x_{n}=\frac{1}{\sqrt{n}} and

     y_{n} =\frac{2}{\sqrt{n}},then


    lim_{n\to\infty} |x_{n}-y_{n}| = 0 and



    lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0
    Last edited by xalk; March 27th 2010 at 07:00 AM.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by frenchguy87 View Post
    Show f(x)=\frac{1}{x^2} is not uniformly continuous on I=(0,\infty) .
    Notice that uniformly continuous functions preserve Cauchy sequences. Is \frac{1}{\sqrt{n}} Cauchy? Is n?
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