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Math Help - Uniform Convergence of (x^n)f(x)

  1. #1
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    Uniform Convergence of (x^n)f(x)

    I'm having difficulty showing that if f(1)=0 and f:[0,1] to R is continuous then (x^n)f(x) is uniformly convergent.
    I get that I'll probably need the fact f is continuous on a closed bounded interval so is bounded but this seems to result in requiring x^n to be uniformly convergent on (0,1) which I don't think it is (since it's pointwise convergent to 0, but take 0<e<1 and then given N take x bigger than the (N+1)th [positive] root of e. Then there is an n>N [namely N+1] for which |x^n| > e)

    Am I wrong in thinking x^n is not uniformly convergent?

    Can anyone give me a hint?

    Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Uniform Convergence of (x^n)f(x)

    Quote Originally Posted by eigensheep View Post
    if f(1)=0 and f:[0,1] to R is continuous then (x^n)f(x) is uniformly convergent.
    We have g(x)=\lim_{n\to +\infty}x^nf(x)=0 for all x in the compact [0,1] i.e. the limit is a continuous function. Besides g_n(x)=x^nf(x) is a decreasing monotonous sequence. According to the Dini's Theorem, g_n\to 0 uniformly in [0,1] .
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