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On an exam question (Question 21H), it is claimed that if $K$ is compact and $f_n : K \to \mathbb{R}$ are continuous functions increasing pointwise to a continuous function $f : K \to \mathbb{R}$, then $f_n$ converges to $f$ uniformly. I have tried proving this claim for the better part of an hour but I keep coming short. I suspect a hypothesis on equicontinuity has been omitted — partly because the first half of the question is about the Arzelà–Ascoli theorem — but I don't have access to the errata for the exam so I can't be sure.

Here is my attempted proof: let $g_n = f - f_n$, so that $(g_n)$ is a sequence of continuous functions decreasing pointwise to $0$. Clearly, $0 \le \cdots \le \| g_n \| \le \| g_{n-1} \| \le \cdots \le \| g_1 \|$, so we must have $\| g_n \| \longrightarrow L$ for some constant $L$. $K$ is compact, so for each $g_n$, there is an $x_n \in K$ such that $g_n(x_n) = \| g_n \|$, and there is a convergent subsequence with $x_{n_k} \longrightarrow x$ for some $x \in K$. By hypothesis, $g_n(x) \longrightarrow 0$, and by construction $g_{n_k} (x_{n_k}) \longrightarrow L$. I'd like to conclude that $L = 0$, but to do this I would need to know that the two sequences have the same limit. This is true if, say, $\{ g_n \}$ is an equicontinuous family, but this isn't one of the hypotheses, so I'm stuck.

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You may want to see this:… – user9413 May 20 '11 at 13:47
The link is broken now. – Jack Feb 19 at 23:18
up vote 10 down vote accepted

The result is true as stated and is called Dini's Theorem. A proof can be found in Chapter III, Section 1.4 of these notes or indeed in this wikipedia article. (My notes are for a sophomore-junior level undergraduate class, so it is stated in the case in which $K$ is a closed interval. But it is clear that the argument works for any compact topological space.)

(For some reason this is one of those named theorems that tends to be assigned as an exercise or come up on exams...)

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Suppose this was not the case. Then there would be an $\epsilon_0 >0$ such that for each $n \in \mathbb N$ there exists $x_n \in K$ such that

$$ f(x_n) - f_n(x_n) > \epsilon_0$$

(note that $f \ge f_n$ by assumption on monotonicity)

Now fix $m$ for the moment. Because of monotonicity we must have $f(x_n) - f_m(x_n) \ge f(x_n) - f_n(x_n) > \epsilon_0$ for all $n \ge m$.

Now note that the sequence $(x_n)$ has a limit point $x$. Passing to this limit in the above, we get

$$f(x) - f_m(x) \ge \epsilon_0$$

But this is of course true of every $m$. Hence letting $m \to \infty$ we obtain

$$|f(x) - \lim_{m \to \infty} f_m(x)| \ge \epsilon_0$$

a contradiction.

Please note also, that continuity of all functions involved is crucial!

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I was of course assuming that $K$ be sequentially compact... Which you might not have intended. – Sam May 20 '11 at 11:21
I think the proof doesn't work because the inequality $f(x_n) - f_m(x_n) \ge f(x_n) - f_n(x_n) $ should be the other way around. – tst Nov 6 '15 at 0:39

Your argument is very close.

Fix $\epsilon > 0$. Since $g_n(x) \downarrow 0$, there exists $m$ such that $g_m(x) \le \epsilon$. Now, for all $n_k \ge m$ we have $g_{n_k}(x_{n_k}) \le g_m(x_{n_k})$. As $k \to \infty$, the left side converges to $L$ by your construction. And by the continuity of $g_m$, the right side converges to $g_m(x) \le \epsilon$. So $L \le \epsilon$. Since $\epsilon$ was arbitrary, the proof is complete.

Edit: Incidentally, a uniformly convergent sequence is always equicontinuous, so equicontinuity does in fact hold. One could also prove it directly from the assumptions given (try it!).

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$\textbf{Dini's Theorem (with some extra conditions):}$ Let $K$ be a compact subset of $\mathbb{R}$ and suppose $f:K \to \mathbb{R}$ is continuous. Let $\{f_n\}$ be a monotonically decreasing sequence of continuous functions such that $f_n(x) \to f(x)$ pointwise. Prove that $f_n(x) \to f(x)$ uniformly.

$\textbf{Proof:}$ We want to show that $\forall \epsilon >0$ $\exists N$ such that $\forall n \ge N$ and $\forall x \in K$, $$|f_n(x) - f(x) | < \epsilon$$

Let $g_n(x) = f_n(x) - f(x)$. Then since $f_n(x) \searrow f(x)$ point wise we have that $g_n(x)$ is a monotonically decreasing sequence of continuous function. Now let $\epsilon > 0$ be fixed but arbitrary. We want to show the existence $N$ to satisfy the conditions of uniform convergence. Let, $$ E_n := \{ x \in K : g_n(x) = f_n(x) - f(x) < \epsilon \}$$ Then $E_n$ is open because it is the pre image of the continuous functions $g_n(x)$. Notice that $g_n(x)$ is continuous as it is the difference of continuous functions. We claim that $E_n$ is an ascending sequence of open sets. That is, $$ E_0 \subset E_1 \subset \dots \subset E_n \subset \dots$$ The sequence is ascending because $g_n(x)$ is decreasing. That is if $x \in E_n$, then $g_n(x) < \epsilon$, but this implies that $g_{n+1} (x) < \epsilon$, and so $x \in E_n$ implies that $x \in E_{n+1}$. Now as $n \to \infty$, we have that $g_n(x) \to 0$. Since $g_n(x) = 0 < \epsilon$, we have that the sets of $x \in K$ such that $g_n(x) < \epsilon$ will eventually be all of $K$. Thus, $$ \bigcup_{n=1}^{\infty} E_n \supset K$$ and so the above union is an open cover of $K$. Since $K$ is compact, there exists a finite subcover, such that, $$ \bigcup_{n=1}^N E_n \supset K$$ At this point, we should be happy as we have an appearance of $N$. Remember we are looking for $N$. $$K \subset \bigcup_{n=1}^N E_n$$ implies that $\forall x \in K$, we have that $|g_n(x)|= | f_n(x) - f(x)| < \epsilon$ Since $g_n(x)$ is decreasing, this immediately gives us that $\forall x \in K$, $\forall n > N$, $|f_n(x) - f(x)| < \epsilon$. Thus we have show $f_n(x)$ converge to $f(x)$ for arbitrary $\epsilon$ and so we are finished.

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