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Uniform convergence question

  1. Mar 26, 2007 #1
    Suppose f(x) is continuous on [0,1], and that f(1)=0. Prove that [itex]x^n f(x)[/itex] converges uniformly on [0,1] as [itex]n \rightarrow \infty[/itex]

    By continuity, if [itex]|x-1|< \delta [/itex] then [itex]|f(x)|< \epsilon[/itex] for [itex]x \in [x_0 ,1][/itex] for some [itex]x_0 \in [0,1][/itex].

    And there is some N such that if n>N, then [itex]|x^n|<\epsilon[/itex] since [itex]x^n \rightarrow 0[/itex] for [itex]x \in [0,1][/itex].

    The endpoints work since x^nf(x) is 0 there. So I have an N that works for [itex] \{ 0 \} \cup [x_0, 1][/itex].

    I'm having trouble getting the rest of the interval. I thought about covering the set and using compactness, but was wondering if there was a better way.
    Last edited: Mar 26, 2007
  2. jcsd
  3. Mar 26, 2007 #2
    You get uniform continuity on compact sets, my friend.
  4. Mar 26, 2007 #3
    I'm asking about uniform convergence.

    ETA: But the function would be uniformly continuous, giving me delta that works for all x in the interval.

  5. Mar 27, 2007 #4


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    What's the point of this?

    "If [itex]|x-1|< \delta [/itex] then [itex]|f(x)|< \epsilon[/itex]"

    means that for [itex]x \in (1-\delta ,1][/itex], [itex]|f(x)|<\epsilon[/itex]. Why add the part about "for [itex]x \in [x_0, 1][/itex] for some [itex]x_0 \in [0,1][/itex]."?
    This isn't true for x=1.
    What is the definition of uniform convergence?
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