Uniform Convergence Nonproof?

**Guy** · June 18th 2010, 05:09 PM

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?

**roninpro** · June 18th 2010, 07:54 PM

Originally Posted by Guy

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?

**Guy** · June 18th 2010, 07:58 PM

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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