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Thread: Uniform Convergence

  1. #1
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    Uniform Convergence

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

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

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

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

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

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

    Thanks for any help
    Last edited by hmmmm; Oct 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 $\displaystyle (1-\epsilon)^{1/n}$.

    Quote Originally Posted by hmmmm View Post
    However I am confused why this breaks down if the function is only defined on $\displaystyle (0,1)$ 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 $\displaystyle \epsilon$ we can choose N to be the smallest such that $\displaystyle (1-a)^N<\epsilon$. Then for any $\displaystyle n\ge N$ we have $\displaystyle |f_n(x)|=x^n\le(1-a)^n\le(1-a)^N<\epsilon$.

    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 $\displaystyle [0,1]$ breaks down if we have that the function is defined on $\displaystyle [0,a)\ \mbox{where}\ a\in(0,1)$ 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 $\displaystyle [0,1]$ breaks down if we have that the function is defined on $\displaystyle [0,a)\ \mbox{where}\ a\in(0,1)$ instead?
    Quote Originally Posted by hmmmm View Post
    Given $\displaystyle \epsilon\in(0,\frac{1}{3})$ and suppose we have uniform convergence then:

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

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