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Math Help - Pointwise Convergence vs. Uniform Convergence

  1. #1
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    Pointwise Convergence vs. Uniform Convergence

    Hi,
    Let fn(x)=(x-(1/n))^2 for x ele [0,1].
    1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
    2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.

    So, f(x) = x^2 pointwise on [0,1].
    Then I wrote a proof but i'm not sure if it is for proving it is pointwise or uniform convergence:
    Let eps>0. Choose N = 1/(sqrt(eps)). Then n> N => n > 1/(sqrt(eps)) => 1/(n^2)< eps => |-2x/n +1/(n^2)|< eps => |(x-1/n)^2-x^2|<eps.

    Now i'm wondering how to do the other one. I don't really see also the difference between both definitions:
    Pointwise:
    for each esp > 0 and x ele S there exists N such that |fn(x)-f(x)|< eps for n>N.

    Uniform Convergence:
    for each eps > 0 there exists a number N such that |fn(x)-f(x)|<eps for all x ele S and all n > N.

    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    Let fn(x)=(x-(1/n))^2 for x ele [0,1].
    1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
    2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.

    So, f(x) = x^2 pointwise on [0,1].
    Then I wrote a proof but i'm not sure if it is for proving it is pointwise or uniform convergence:
    Let eps>0. Choose N = 1/(sqrt(eps)). Then n> N => n > 1/(sqrt(eps)) => 1/(n^2)< eps => |-2x/n +1/(n^2)|< eps => |(x-1/n)^2-x^2|<eps.

    Now i'm wondering how to do the other one. I don't really see also the difference between both definitions:
    Pointwise:
    for each esp > 0 and x ele S there exists N such that |fn(x)-f(x)|< eps for n>N.

    Uniform Convergence:
    for each eps > 0 there exists a number N such that |fn(x)-f(x)|<eps for all x ele S and all n > N.

    Thanks
    to (a): f_n \to f point-wise on [a,b] if \lim_{n \to \infty} f_n = f on [a,b]

    to (b): you went about this backwards. i think it would be neater to use a different kind of proof here. you need to show that:

    \lim_{n \to \infty} \left[ \sup_{x \in [0,1]} |f_n - f| \right] = 0
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  3. #3
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    So to prove it is uniform convergence on [0,1]:
    lim [|sup(x^2 - (x-1/n)^2)|:x ele [0,1]] = lim[|sup(2x/n - 1/(n^2))|: x ele [0,1]] = lim[2n/(n^2) - 1/(n^2)] = 0. Thus uniform convergence by theorem or remark as they call it in the book.

    So to prove it is pointwise convergence:
    Do I just need to state lim as n-> inf (x-1/n)^2 = x^2 for all x ele [0,1]? Or is there anything more rigorous I need to do?

    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    So to prove it is uniform convergence on [0,1]:
    lim [|sup(x^2 - (x-1/n)^2)|:x ele [0,1]] = lim[|sup(2x/n - 1/(n^2))|: x ele [0,1]] = lim[2n/(n^2) - 1/(n^2)] = 0. Thus uniform convergence by theorem or remark as they call it in the book.
    yes. but the sup is outside of the absolute values


    So to prove it is pointwise convergence:
    Do I just need to state lim as n-> inf (x-1/n)^2 = x^2 for all x ele [0,1]? Or is there anything more rigorous I need to do?

    Thanks
    nope, that's it. just take the limit. if you want to make it rigorous, you can use the definition of a limit to prove that the limit is as you say, but that's completely unnecessary as far as i'm concerned--and i think your professor would feel the same way. if you're uncomfortable though, you can prove that the limit is in fact x^2 by the definition of the limit
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  5. #5
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    Let fn(x)=(x-(1/n))^2 for x ele [0,1].
    1) Show that the sequence of functions {fn} converges to some function pointwise on [0,1] and write the limit function.
    2) Does {fn} converge to f uniformly on [0,1]? Prove your idea.
    I remember having the same problem for homework. In fact, I explained this problem to Jhevon.

    f_n(x) = x^2 - \frac{2x}{n}+\frac{1}{n^2} so f_n(x) \to x^2.
    For uniform convergence you need to prove that,
    |f(x)-f_n(x)|
    Can be made small.
    Thus,
    \left| x^2 - \frac{2x}{n}+\frac{1}{n^2} -x^2 \right| \leq \frac{2x}{n}+\frac{1}{n^2} \leq \frac{2}{n} + \frac{1}{n} = \frac{3}{n}
    Thus, choose N = \frac{3}{\epsilon}.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I remember having the same problem for homework. In fact, I explained this problem to Jhevon.

    f_n(x) = x^2 - \frac{2x}{n}+\frac{1}{n^2} so f_n(x) \to x^2.
    For uniform convergence you need to prove that,
    |f(x)-f_n(x)|
    Can be made small.
    Thus,
    \left| x^2 - \frac{2x}{n}+\frac{1}{n^2} -x^2 \right| \leq \frac{2x}{n}+\frac{1}{n^2} \leq \frac{2}{n} + \frac{1}{n} = \frac{3}{n}
    Thus, choose N = \frac{3}{\epsilon}.
    indeed! in fact, he uses the same text that we did. he is doing homework from section 24 i believe

    the method i instructed him to use was the one our professor used in class
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    Quote Originally Posted by Jhevon View Post
    the method i instructed him to use was the one our professor used in class
    I do not like that method. It does not always works. For example, if you have unbounded functions then the supremum is not a real number so the entire limit just does not make any sense.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I do not like that method. It does not always works. For example, if you have unbounded functions then the supremum is not a real number so the entire limit just does not make any sense.
    yes, i agree. he asked another question in which that was the case and i instructed him not to use this method. but i think it's a neater way to use when we are bounded
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  9. #9
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    I have a feeling you go to the same school as me and we are in the same class. Thanks for posting this problem, i really needed the help.
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