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Math Help - Real Analysis - Uniform Convergence

  1. #1
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    Real Analysis - Uniform Convergence

    Let  f_n(x):=\frac{1}{n\sqrt{2\pi}}\exp{{\frac{-x^2}{2n^2}}}
    and  f(x):=0

     f_n(x) \rightarrow f(x) pointwise.

     \forall n\in\omega\int_{-\infty}^{\infty}f_n(x)dx = 1

    and \int_{-\infty}^{\infty}f(x)dx=0

    By The Uniform Limit of L.I. functions is L.I. and the limits are the same, we can conclude
    f_n(x)\not\rightarrow f(x) uniformly.

    Although that's a straight forward proof, I'd like to prove this from the definition. That is
    \exists \varepsilon >0 \,\forall \delta >0 \,\exists x,y\in\mathbb{R}(|x-y|<\delta\wedge|f(x)-f(y)|\geq\varepsilon)

    But this is turning out to be more difficult. Can someone help?
    Last edited by blmalikov; May 11th 2013 at 05:04 PM.
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  2. #2
    Super Member girdav's Avatar
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    Re: Real Analysis - Uniform Convergence

    No, there is uniform convergence on the real line (because |f_n(x)|\leqslant \frac 1{n\sqrt{2\pi}}). The integral of the limit is not the limit of the integral in this case. In general, considering unbounded domains, uniform convergence is not enough in order to switch the limit and the integral.
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