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Thread: Convergence of a sequence of functions

  1. #1
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    Convergence of a sequence of functions

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

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

    Thanks in advance
    Last edited by akolman; Nov 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
    $\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$ ???

    Also, I noticed that if the all $\displaystyle {f_n}$ 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
    $\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$ ???

    Also, I noticed that if the all $\displaystyle {f_n}$ 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:

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

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

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

    CB
    How different is that from uniform continuity??
    Last edited by akolman; Dec 1st 2008 at 01:23 AM.
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  5. #5
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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 $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_x$ such that:

    $\displaystyle
    |f_n(x)-f(x)|<\varepsilon ,\ \ \forall n>N_x
    $

    but $\displaystyle N_x$ may depend on $\displaystyle x$

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

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

    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 $\displaystyle x_n=x+\frac{1}{n}$, then $\displaystyle g_n(x)=f_n(x_n)$. I know that $\displaystyle x_n$ converges uniformly to $\displaystyle x$

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

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

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




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