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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: mathoverflow.net/questions/45784/… –  user9413 May 20 '11 at 13:47

3 Answers 3

up vote 6 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

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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