# Thread: Uniform Convergence Nonproof?

1. ## 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?

2. Originally Posted by Guy
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?

3. 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.