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Math Help - Convergence of a sequence of functions

  1. #1
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    Convergence of a sequence of functions

    Suppose \{f_n\} is a sequence of functions converging uniformly to  f. Define g_n(x)=f_n\left(x+\frac{1}{n}\right) . Show that the sequence \{g_n\} converges pointwise to f.

    So I need to show that f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)

    Thanks in advance
    Last edited by akolman; November 30th 2008 at 04:21 PM. Reason: misspelling, clumsy fingers
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  2. #2
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    I am pretty confused, does pointwise convergence means
    f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right) ???

    Also, I noticed that if the all {f_n} are continuous, then the proof wouldn't be hard. However, they are not mentioning anything, is continuity necessary? How can I prove that?
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  3. #3
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    Quote Originally Posted by akolman View Post
    I am pretty confused, does pointwise convergence means
    f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right) ???

    Also, I noticed that if the all {f_n} are continuous, then the proof wouldn't be hard. However, they are not mentioning anything, is continuity necessary? How can I prove that?
    No it means:

    f(x)=\lim \limits_{n \to \infty} g_n(x)=\lim \limits_{n \to \infty} f_n\left(x+\frac{1}{n}\right)

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    No it means:

    f(x)=\lim \limits_{n \to \infty} g_n(x)=\lim \limits_{n \to \infty} f_n\left(x+\frac{1}{n}\right)

    CB
    How different is that from uniform continuity??
    Last edited by akolman; December 1st 2008 at 01:23 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by akolman View Post
    How different is that from uniform continuity??
    Where did contiuity enter the question?

    The difference between uniform and pointwise convergence is that:

    Pointwise: For all \varepsilon>0 there exists an N_x such that:

     <br />
|f_n(x)-f(x)|<\varepsilon ,\ \ \forall n>N_x<br />

    but N_x may depend on x

    Uniform:For all \varepsilon>0 there exists an N independednt of x such that:

    |f_n(x)-f(x)|<\varepsilon ,\ \ \forall n>N.

    CB
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  6. #6
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    I get the difference now, so how do you do this proof?

    Thanks in advance.
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  7. #7
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    If I let x_n=x+\frac{1}{n}, then g_n(x)=f_n(x_n). I know that x_n converges uniformly to x

    \exists N_1 such that for n>N_1 , |x_n-x|<\varepsilon

    Since f_n converges uniformly to f(x),
    \exists N_2 such that for n>N_2, |f(x)-f_n(x)|<\varepsilon,

    Is this right?
    If I let N=max(N_1,N_2) ,
    n>N \Rightarrow |x_n-x| < \varepsilon
    \Rightarrow |f_n(x_n)-f(x_n)| <|f_n(x)-f(x)|<\varepsilon




    I am sorry if I pushing this question too much, but I am still clueless.
    Last edited by akolman; December 2nd 2008 at 10:55 PM.
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