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Pointwise & uniform boundednes

  1. Jan 26, 2012 #1
    In Baby Rudin, Theorem 7.25 states:
    If [itex]K[/itex] is compact, [itex]f_n \in C(K)[/itex] for [itex]n=1,2,3,...[/itex] and if [itex]{f_n}[/itex] is pointwise bounded and equicontinuous on [itex]K[/itex], then
    (a) [itex]{f_n}[/itex] is uniformly bounded on [itex]K[/itex]

    The theorem continues with point (b), which I understand.

    My question is, whether point (a) needs the equicontinuity, since:
    * [itex]C(K)[/itex] is a set of continuous bounded functions on a compact space, i.e. uniformly continuous, hence there is no function [itex]g[/itex] s.t. [itex]g(x) \to \infty[/itex].
    * if the previous holds, then by the hypothesis of pointwise boundedness of [itex]\{f_n\}[/itex], there must exist a number [itex]\sup_{x \in K, n=1,2,3,...} |f_n(x)| = M<\infty[/itex], [itex]x\in K[/itex] arbitrary and hence the uniform boundedness should hold.

    Am I wrong?
     
    Last edited: Jan 27, 2012
  2. jcsd
  3. Jan 26, 2012 #2

    micromass

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    The sequence

    [itex]f_n:[0,1]\rightarrow \mathbb{R}:x\rightarrow \frac{nx}{x^2+(1-nx)^2}[/itex]

    seems to be a counterexample to your claim. See example 7.21 of Rudin, I modified it a bit.
     
  4. Jan 27, 2012 #3
    Thanks a lot! I assume, that the reason is that [itex]f_n[/itex] converges pointwise for every [itex]x\in K[/itex], but due to the points [itex]1/n[/itex] in your example, it is impossible to find a global finite [itex]M[/itex] s.t. [itex]|f_n(x)| \leq M[/itex] for all [itex]x \in K[/itex]. Right?

    I really do not understand why usually things go quite well, but at some moments I get stuck with (obvious?) things :(
     
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