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Math Help - Uniform Convergence

  1. #1
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    Uniform Convergence

    Hello my lecturer gave a proof that f_n(x)=x^n is not uniformly convergent to f(x)=0\  \mbox{if} \ x<1\  \mbox{and}\  1\  \mbox{if}\  x=1 on [0,1]. Which went as follows:

    Given \epsilon\in(0,\frac{1}{3}) and suppose we have uniform convergence then:

    \exists N s.t. \forall n\geq N\forall x\in[0,1]:\ |f_n(x)-f(x)|<\epsilon

    Given such an N take n=N x=(1-\epsilon)^{(\frac{1}{n})})<1

    then |f_n(x)-f(x)|=|((1-\epsilon)^{(\frac{1}{n})^n)}=1-\epsilon>epsilon which is a contradiction and we are done.

    However I am confused why this breaks down if the function is only defined on [0,a) where a\in(0,1) in which case it is uniformly convergent?

    Thanks for any help
    Last edited by hmmmm; October 15th 2011 at 07:33 AM.
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  2. #2
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    Re: Uniform Convergence

    In LaTeX, if the exponent consists of more than one token, it has to be surrounded by { }. E.g., (1-\epsilon)^{1/n} gives (1-\epsilon)^{1/n}.

    Quote Originally Posted by hmmmm View Post
    However I am confused why this breaks down if the function is only defined on (0,1) in which case it is uniformly convergent?
    This sequence is still not uniformly convergent on (0, 1), but it is uniformly convergent on [0, 1 - a] for any 0 < a < 1.
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  3. #3
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    Re: Uniform Convergence

    Ah yes sorry I miss read my notes, however I am still unsure as to why this is?

    Thanks for the help (sorry about the LaTex-I will edit it)

    thanks for any help
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  4. #4
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    Re: Uniform Convergence

    Quote Originally Posted by hmmmm View Post
    I am still unsure as to why this is?
    Why what is? The sequence converges uniformly on [0, 1 - a]? Because for each \epsilon we can choose N to be the smallest such that (1-a)^N<\epsilon. Then for any n\ge N we have |f_n(x)|=x^n\le(1-a)^n\le(1-a)^N<\epsilon.

    The fact that the sequence does not converge uniformly on (0, 1) is shown using your lecturer's proof, as you said.
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  5. #5
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    Re: Uniform Convergence

    Thanks I understand that but I was wondering where the proof that the function does not uniformly converge on [0,1] breaks down if we have that the function is defined on [0,a)\ \mbox{where}\ a\in(0,1) instead?

    Thanks for the help
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  6. #6
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    Re: Uniform Convergence

    Quote Originally Posted by hmmmm View Post
    I was wondering where the proof that the function does not uniformly converge on [0,1] breaks down if we have that the function is defined on [0,a)\ \mbox{where}\ a\in(0,1) instead?
    Quote Originally Posted by hmmmm View Post
    Given \epsilon\in(0,\frac{1}{3}) and suppose we have uniform convergence then:

    \exists N s.t. \forall n\geq N\forall x\in[0,1]:\ |f_n(x)-f(x)|<\epsilon

    Given such an N take n=N x=(1-\epsilon)^{(\frac{1}{n})})<1
    This x does not have to belong to [0, a).
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  7. #7
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    Re: Uniform Convergence

    ah of course, sorry about that.

    Thanks for all the help there.
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