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Math Help - Pointwise and uniform convergence

  1. #1
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    Pointwise and uniform convergence

    I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
    1) Let \{a_n\} be a sequence that contains each rational in [0,1] precisely once. For each n\in N, define
    f_n(x)= 0 if x is irrational
    f_n(a_i)=0 if i > n
    f_n(a_i)=1 if i \leq n
    Prove that \{f_n\} converges pointwise on [0,1] to a function that is not R-integrable.
    My attempt:
    I guess that this sequence of functions converges to the rational characteristic function, but I can't really show lim f_n=f as n approaches infinity.
    I tried this again and I got this far: f(x)=0 if x is irrational and f(x)=1 if x is rational.
    Consider: absolute value of f_n(x)-f(x)= 0-0=0< \epsilon if x is irrational. Choose \epsilon=1/2
    If x is rational, f_n(x)-f(x)=1-1=0< \epsilon. I think f(x)=1 because eventually i \leq n as n approaches infinity.

    2) For n \in N define f_n: [0,1] \rightarrow R by f_n(x)=1/x if x \in [1/n,1] and f_n(x)=n^2x if x \in [0,1/n). Prove that \{f_n\} does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

    I don't know what the pointwise limit of f_n, so I can't go any anywhere. It would be really helpful if someone helps me how to find the pointwise limit
    Last edited by jackie; May 11th 2010 at 08:19 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by jackie View Post
    I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
    1) Let \{a_n\} be a sequence that contains each rational in [0,1] precisely once. For each n\in N, define
    f_n(x)= 0 if x is irrational
    f_n(a_i)=0 if i > n
    f_n(a_i)=1 if i \leq n
    Prove that \{f_n\} converges pointwise on [0,1] to a function that is not R-integrable.
    My attempt:
    I guess that this sequence of functions converges to the rational characteristic function, but I can't really show lim f_n=f as n approaches infinity.
    I tried this again and I got this far: f(x)=0 if x is irrational and f(x)=1 if x is rational.
    Consider: absolute value of f_n(x)-f(x)= 0-0=0< \epsilon if x is irrational. Choose \epsilon=1/2
    If x is rational, f_n(x)-f(x)=1-1=0< \epsilon. I think f(x)=1 because eventually i \leq n as n approaches infinity.
    The part for x rational sounds a litte sloppy to me, apart from that I don't quite see where your problem is. To fix that sloppiness you should argue that if x\in [0;1]\cap\mathbb{Q}, then there exists a unique n_0 such that x=a_{n_0} and therefore for all n\geq n_0 we have that |f_n(x)-1|=|f_n(a_{n_0})-1|=|1-1|=0<\epsilon

    f is not integrable because all the lower sums are 0 and all the upper sums are 1.

    2) For n \in N define f_n: [0,1] \rightarrow R by f_n(x)=1/x if x \in [1/n,1] and f_n(x)=n^2x if x \in [0,1/n). Prove that \{f_n\} does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

    I don't know what the pointwise limit of f_n, so I can't go any anywhere. It would be really helpful if someone helps me how to find the pointwise limit
    What about f(x) := \begin{cases}0, &\text{if } x=0\\\tfrac{1}{x}, &\text{otherwise}\end{cases}
    Last edited by Failure; May 12th 2010 at 08:02 AM.
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