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Math Help - Uniform Convergence Nonproof?

  1. #1
    Guy
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    Uniform Convergence Nonproof?

    Hey guys. I think this is a nonproof since the proof given in baby Rudin is a bit longer, but I'm not sure where it breaks down. The well-known theorem is:

    Suppose {f_n} converges uniformly to f on a set E in \mathbb{C}. Let x be a limit point of E and suppose that \lim_{t \to x} f_n (t) = A_n.

    Then \{A_n\} converges, and \lim_{t \to x} f(t) = \lim_{n \to \infty} A_n.

    Attempted proof:

    Denote \lim_{t \to x} f(t) by L. For arbitrary n, note that |A_n - L| = \lim_{t \to x} |f_n (t) - f(t)| by continuity of | \cdot |. Given \epsilon, choose N so that for n \ge N, |f_n (t) - f(t)| \le \epsilon, for all t \in E. Since this holds for all t \in E, it holds for the limit as well, i.e. \lim_{t \to x} |f_n (t) - f(t)| \le \epsilon so that if n \ge N, |A_n - L| \le \epsilon. QED.

    If that's wrong, I need to know since I have the feeling it would indicate some big flaw in my understanding. If it's okay, I think the issue might be that the proof doesn't generalize to arbitrary metric spaces perhaps?
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  2. #2
    Senior Member roninpro's Avatar
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    Quote Originally Posted by Guy View Post
    Denote \lim_{t \to x} f(t) by L.
    I think that you already have a problem here. You do not know that \lim_{t \to x} f(t) exists.

    Can you find a way to prove that it does?
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  3. #3
    Guy
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    Oh, wow, you're right. For some reason I thought it was a given.

    The proof given makes a lot more sense now. The strategy seems to be to show that {A_n} converges first, then use that to show that the limit exists, and that they are equal.

    Heh, that's actually kind of embarrassing.
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