I am trying to prove Theorem 7.13 in baby Rudin and find that my proof goes in a totally different direction than the proof in the book. I briefly sketch what I did.
Since $f_n(x)$ is continuous, then $f_n(x)$ is bounded in compact domain $K$ (theorem 4.15). Then $$M_n = \sup_{x\in E} |f_n(x)-f(x)|$$ is well defined and is non-increasing because $f_n(x)\ge f_{n+1}(x) \ge f(x)$. Moreover, $lim_{n\to \infty}M_{n} = 0$, then from Theorem 7.9, we conclude that $f_n\to f$ uniformly.
Could someone tell me whether my proof is wrong ? Or whether it is possible to prove this theorem in this way?
Theorem 7.13 Suppose $K$ is compact and
(a) $\{f_n\}$ is a sequence of continuous functions on $K$
(b) $\{f_n\}$ converges pointwise to a continuous function on $K$
(c) $f_n(x) \ge f_{n+1}(x)$ for all $x \in K$, $n=1,2,3,\dots$ Then $f_n \to f$ uniformly on $K$.Rudin's Proof of Theorem 7.13 Put $g_n = f_n -f$. Then $g_n$ is continuous, $g_n \to 0$ pointwise, and $g_n \ge g_{n+1}$. We have to prove that $g_n \to 0$ uniformly on $K$.
Let $\epsilon > 0$ be given. Let $K_n$ be the set of all $x \in K$ with $g_n(x) \ge \epsilon$. Since $g_n$ is continuous, $K_n$ is closed (Theorem 4.8) hence compact (Theorem 2.35). Since $g_n \ge g_{n+1}$ we have $K_{n+1} \subset K_{n}$. Fix $x \in K$. Since $g_n(x) \to 0$, we see that $x \not\in K_n$ if $n$ is sufficiently large. Thus $x \not\in \bigcap K_n$. In other words, $\bigcap K_n$ is empty.Hence $K_N$ is empty for some $N$ (Theorem 2.36). It follows that $0 \le g_n(x) < \epsilon$ for all $x \in K$ and for all $n \ge N$. This proves the theorem.
Here is also the crucial Theorem 7.9:
Theorem 7.9 Suppose $$\underset{n \to \infty}{\lim} f_n(x) = f(x) \quad (x \in E).$$
Put $$M_n = \underset{x \in E}{\sup}|f_n(x)-f(x)|.$$ Then $f_n \to f$ uniformly on $E$ if and only if $M_n \to 0$ as $n \to \infty$.
