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Math Help - Uniform Continuity

  1. #1
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    Uniform Continuity

    f(x)=1/x how would you show this is continuous on A=[1,\infty) but not on B=(0,\infty)?
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    Senior Member Tinyboss's Avatar
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    The first by exhibiting a function \varepsilon(\delta) valid on that interval, and the second by contradiction.
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    Quote Originally Posted by CrazyCat87 View Post
    f(x)=1/x how would you show this is continuous on A=[1,\infty) but not on B=(0,\infty)?
    I assume you mean uniformly continuous

    First note that 1/x - 1/y = y-x/xy

    For the first part:

    For x, y in B, we have x,y>1 so 1/xy < 1

    Thus, for all x,y in B lf(x) - f(y)l = l1/x - 1/yl = lx-yl/lxyl < lx-yl


    and uniform continuity is not hard to show (let delta = epsilon).

    For the second part, you have to show that there exists an epsilon > 0, you can find an x, y small enough such that their difference is greater than epsilon.
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    I'm still having trouble with proving uniform continuity on B=(0,\infty).

    How would you show that g(x)=\frac{1}{x^2} is uniformly continuous on B?
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    Proving Non-Uniform Continuity

    I'd like to see that one also...
    Last edited by frenchguy87; March 25th 2010 at 02:37 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    f(x)=1/x how would you show this is continuous on A=[1,\infty) but not on B=(0,\infty)?
    To prove that it's uniformly convergent on [1,\infty) merely note that it's differentiable and \left|\frac{-2}{x^3}\right|\leqslant 2. Thus, the function is Lipschitz and trivially unfiromly continuous.
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    Quote Originally Posted by CrazyCat87 View Post
    f(x)=1/x how would you show this is continuous on A=[1,\infty) but not on B=(0,\infty)?
    To prove that 1/x is not uniformly continuous on (0, \infty) we must show that:

    There exists ε>0 such that for all δ>0 there exists x,wε(0, \infty) such that:

    |x-w|<δ and |f(x)-f(w)| \geq\epsilon.

    Let ε=2.

    Take 0<δ<1 .

    Choose x= δ and w= δ/4 then we have:

    |x-w| = |δ- \frac{\delta}{4}|= \frac{3\delta}{4}<\delta

    AND

    |f(x)-f(w)| = |\frac{1}{\delta}-\frac{4}{\delta}|=\frac{3}{\delta}\geq 3\geq\epsilon.


    Hence 1/x is not uniformly continuous on (0, \infty)
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  8. #8
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    as aside note, f(x)=\frac1x is uniformly continuous for each [\alpha,\infty) where \alpha>0.

    the proof follows easily since given \epsilon>0 and by taking \delta=\alpha^2\epsilon we have \left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    as aside note, f(x)=\frac1x is uniformly continuous for each [\alpha,\infty) where \alpha>0.

    the proof follows easily since given \epsilon>0 and by taking \delta=\alpha^2\epsilon we have \left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare
    Or that |f'(x)|\leqslant\frac{2}{\alpha^2}
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  10. #10
    Math Engineering Student
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    i know, the problem is that bounded derivative implies Lipschitz is not something that many people learn.
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