# Rudin 7.11 uniform convergence alternate proof using the limit inequality theorem

I came up with a simpler proof of theorem 7.11 from Rudin PMA but I feel unsure that my reasoning is sound. Am I allowed to apply the limit inequality theorem as I have on the line marked (1)?

My concern is that I do not know if the limits exist which is necessary to use this theorem.

Theorem 7.11: Suppose $$f_n \to f$$ uniformly on a set $$E$$ in a metric space. Let $$x$$ be a limit point of $$E$$, and suppose that $$\lim_{t\to x} f_n(t) = A_n \qquad (n \in N).$$ Then $$\{A_n\}$$ converges, and $$\lim_{t\to x} f(t) = \lim_{n\to \infty} A_n.$$ In other words, the conclusion is that

$$\lim_{n\to\infty}\lim_{t\to x}f_n(t)=\lim_{t\to x}\lim_{n\to \infty}f_n(t)$$ My proof:

Fix $$\epsilon > 0$$. Then by uniform convergence, there exists some $$N\in \mathbb{N}$$ such that $$\forall n>N$$ $$|f_n(t)-\lim_{n\to \infty}f_n(t)|<\epsilon \qquad \forall t\in E$$ Now we may write this inequality as

$$\lim_{n\to \infty}f_n(t)-\epsilon< f_n(t)<\lim_{n\to \infty}f_n(t)+\epsilon$$

Now since this holds $$\forall t \in E$$, we have by the limit inequality theorem that

$$\lim_{t\to x}(\lim_{n\to \infty}f_n(t)-\epsilon)< \lim_{t\to x}f_n(t)<\lim_{t\to x}(\lim_{n\to \infty}f_n(t)+\epsilon) \qquad (1)$$

which implies that

$$|\lim_{t\to x}f_n(t)-\lim_{t\to x}\lim_{n\to \infty}f_n(t)|<\epsilon$$

Therefore, we have shown that $$\lim_{n\to\infty}\lim_{t\to x}f_n(t)=\lim_{t\to x}\lim_{n\to \infty}f_n(t)$$

• How do you know that the first and third limits as $t \to x$ in $(1)$ exist? May 29, 2021 at 21:49

I think the answer is no. As you note to apply the limit inequality theorem we need to know that $$\{A_n\}$$ will converge which we do not know. Thus, the first step of Rudin's proof is to show that $$\{A_n\}$$ does in fact converge using the fact that it is Cauchy.

So why do we need this? Well if we look at the proof of the limit inequality theorem we essentially want to bound by $$\epsilon$$ something that looks like,

$$|\lim_{t\to x}\lim_{n\to\infty}f_n(t)-\lim_{n\to\infty}f_n(t) - (f(x)-f(t))|<\epsilon$$

But in order for this bound to be true, we do actually need to know that these objects are converging to our supposed limits.

I would probably change the following:

$$\lim_{n\to \infty}f_n(t)-\epsilon< f_n(t)<\lim_{n\to \infty}f_n(t)+\epsilon$$

Now since this holds $$\forall t \in E$$, we have by the limit inequality theorem that

$$\lim_{t\to x}(\lim_{n\to \infty}f_n(t)-\epsilon)< \lim_{t\to x}f_n(t)<\lim_{t\to x}(\lim_{n\to \infty}f_n(t)+\epsilon) \qquad (1)$$

and change this into:

$$f_n(t)-\epsilon< \lim_{n\to \infty}f_n(t)

Now since this holds $$\forall t \in E$$, we have by the limit inequality theorem that

$$\lim_{t\to x}f_n(t)-\epsilon< \limsup_{t\to x}\lim_{n\to \infty}f_n(t)<\lim_{t\to x}f_n(t)+\epsilon \qquad (1)$$ same applies to $$\liminf$$

P.S. also need to show $$\{A_n\}$$, whish is $$\{\lim_{t\to x}f_n(t)\}$$, converge.