Rudin PMA p.150

This is the theorem;


(i)Let $X$ be a metric space and $K$ be a compact subset.

(ii)Let $\{f_n\}$ be a sequence of continuous functions such that $f_n:K\rightarrow \mathbb{R}$

(iii)$\forall x\in K, f_n(x)≧f_{n+1}(x)$

(iv)$f_n \rightarrow f$ pointwise on $K$.

(v)$f$ is continuous.

Then $f_n\rightarrow f$ uniformly on $K$.


I think the condition (iii) is not essential.

I think (iii) can be replaced to "$\{f_n(x)\}_{n\in \mathbb{N}}$ is monotonic for all $x\in K$" and i actually proved it.

Here's what i tried;

Define $g_n=f_n - f$.

Let $A=\{x\in K|\{g_n(x)\}_{n\in\mathbb{N}} \text{ is monotonically decreasing}\}$ and $B=\{x\in K|\{g_n(x)\}_{n\in\mathbb{N}} \text{ is monotonically increasing}\}$

Let $K_n=\{x\in K|g_n(x)≧\epsilon \bigvee g_n(x)≦-\epsilon\}$.

Then $K_n$ is compact.

Also, it can be shown that$\forall x\in A \forall n\in \mathbb{N}, g_n(x)≧0$ and $\forall x\in B \forall n\in \mathbb{N} g_n(x)≦0$.

Thus $K_{n+1} \subset K_n$.

Since $g_n\rightarrow 0$ pointwise on $K$, $K\cap (\bigcap K_n)=\emptyset$.

Thus, $\exists N\in \mathbb{N}$ such that $n≧N \Rightarrow K_n=\emptyset$. Q.E.D.

Is my argument correct?


1 Answer 1


I think your argument is perfect. Here is just a comment. Since $g_n(x)$ monotonically converges to $0$ for every $x\in K$, $|g_n(x)|$ decreases to $0$ for every $x\in K$, so your replacement of (iii) could be reduced to (iii).

  • $\begingroup$ what if K is not compact, what need to be done in order to get the same conclusion? $\endgroup$
    – math101
    Mar 5, 2015 at 13:54

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