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Math Help - Continuity Proof

  1. #1
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    Continuity Proof

    Question:

    Show that the function: f(x):=\frac{1}{x^2} is uniformly continous on A:=[1,\infty), but that it is not uniformly continuous on B := (0,\infty).

    My thoughts:

    So far I've gotten to |f(x)-f(u)|\leq|x-u||x+u| when in A.
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  2. #2
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    Quote Originally Posted by Luxury View Post
    Question:

    Show that the function: f(x):=\frac{1}{x^2} is uniformly continous on A:=[1,\infty), but that it is not uniformly continuous on B := (0,\infty).

    My thoughts:

    So far I've gotten to |f(x)-f(u)|\leq|x-u||x+u| when in A.
    You may want to browse through the forum because I could swear I saw it recently

    EDIT: http://www.mathhelpforum.com/math-he...ontinuity.html OK not exactly the same, but close enough/
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  3. #3
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    I actually did see that, and I already knew how to do that proof. I'm still lost on this one thought. It might be just because I've been staring at this stuff for too long.
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  4. #4
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    Quote Originally Posted by Luxury View Post
    Question:

    Show that the function: f(x):=\frac{1}{x^2} is uniformly continous on A:=[1,\infty), but that it is not uniformly continuous on B := (0,\infty).

    My thoughts:

    So far I've gotten to |f(x)-f(u)|\leq|x-u||x+u| when in A.
    Here's one way to show its not uniformly continuous on B := (0,\infty)

    Proof by contradiction

    Suppose f(x) is uniformly continuous on B := (0,\infty). Fix e > 0 Thus, there is a d such that:

    lx-yl < d --> l f(x) - f(y) l < e

    Choose an n so large such that n > max{1/ \sqrt{d}, 7e/3}.

    Let x = 1/ \sqrt{n}, y = x + d/2

    Clearly, lx-yl < d

    But, lf(x) - f(y)l = l n - 1/(1/ \sqrt{n}+d/2)^2 l = l n + 4n/(4+n*d^2+2d* \sqrt{n}) l = n* l 1 - 4/(4+n*d^2+2d* \sqrt{n})l

    but we said n > 1/ \sqrt{d}, thus

    1 - 4/(4+n*d^2+2d* \sqrt{n}) > 3/7 and
    n*l(1 - 4/(4+n*d^2+2d* \sqrt{n})l > n*3/7 > e

    contradiction

    Note, there are likely mistakes here, I rushed it a bit.
    Last edited by southprkfan1; April 1st 2010 at 05:42 AM.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Also note that a unif. cont. function preserves Cauchy sequences. But, \frac{1}{\sqrt{n}}\to 0 but f\left(\frac{1}{\sqrt{n}}\right)=n is not Cauchy.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Also note that a unif. cont. function preserves Cauchy sequences. But, \frac{1}{\sqrt{n}}\to 0 but f\left(\frac{1}{\sqrt{n}}\right)=n is not Cauchy.
    Interesting, never knew that.
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  7. #7
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    South, correct me if I'm wrong, but I think you did it for f(x) = 1/x not f(x)=1/x^2.
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  8. #8
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    Quote Originally Posted by Luxury View Post
    South, correct me if I'm wrong, but I think you did it for f(x) = 1/x not f(x)=1/x^2.
    Correct

    Fixed (I hope)
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