Results 1 to 3 of 3

Thread: Uniform Convergence Nonproof?

  1. #1
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98

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

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

    Attempted proof:

    Denote $\displaystyle \lim_{t \to x} f(t)$ by $\displaystyle L$. For arbitrary n, note that $\displaystyle |A_n - L| = \lim_{t \to x} |f_n (t) - f(t)| $ by continuity of $\displaystyle | \cdot |$. Given $\displaystyle \epsilon$, choose $\displaystyle N$ so that for $\displaystyle n \ge N, |f_n (t) - f(t)| \le \epsilon$, for all $\displaystyle t \in E$. Since this holds for all $\displaystyle t \in E$, it holds for the limit as well, i.e. $\displaystyle \lim_{t \to x} |f_n (t) - f(t)| \le \epsilon$ so that if $\displaystyle n \ge N, |A_n - L| \le \epsilon$. 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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    Quote Originally Posted by Guy View Post
    Denote $\displaystyle \lim_{t \to x} f(t)$ by $\displaystyle L$.
    I think that you already have a problem here. You do not know that $\displaystyle \lim_{t \to x} f(t)$ exists.

    Can you find a way to prove that it does?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Guy
    Guy is offline
    Member
    Joined
    Mar 2010
    Posts
    98
    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 $\displaystyle {A_n}$ converges first, then use that to show that the limit exists, and that they are equal.

    Heh, that's actually kind of embarrassing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Uniform convergence vs pointwise convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 15th 2012, 11:03 PM
  2. Replies: 1
    Last Post: Oct 31st 2010, 07:09 PM
  3. Pointwise convergence to uniform convergence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: Nov 29th 2009, 08:25 AM
  4. Pointwise Convergence vs. Uniform Convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Oct 31st 2007, 05:47 PM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 28th 2007, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum