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Thread: Uniform Continuity

  1. #1
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    Uniform Continuity

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

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

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

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

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

    Let ε=2.

    Take 0<δ<1 .

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

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

    AND

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


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

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

    the proof follows easily since given $\displaystyle \epsilon>0$ and by taking $\displaystyle \delta=\alpha^2\epsilon$ we have $\displaystyle \left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare$
    Or that $\displaystyle |f'(x)|\leqslant\frac{2}{\alpha^2}$
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  10. #10
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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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