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I was looking at this question, and trying to come up with a counterexample. After thinking about it, I thought the following might be true:

Claim: let $\{f_n\}$ be a sequence of continuous, monotonically increasing functions on a compact $K \subset \Bbb R$. That is, each $f_n : K \to \mathbb{R}$ satisfies $x < y \implies f(x) \leq f(y)$.

Suppose $f_n \to f$ pointwise, where $f$ is a continuous function. Then the family $\{f_n\}$ is equicontinuous.

Is it true? If so, I'd like to prove it, and if not, I'd like a counterexample.

My thoughts so far: I think it's true, but perhaps this is due to a lack of imagination. It's clear that $f$ must be monotonically increasing. Also, $f_n - f$ is a sequence of continuous functions that converges to $0$ pointwise.

I was also thinking of perhaps starting with a lack of continuity and proving the existence of a jump discontinuity in $f$, though it's not clear this would work.

Any input is appreciated!

EDIT: This is much healthier's answer in the original question suggests a certain working $\epsilon$-$\delta$ proof, so if anyone wants to work through that I'd appreciate it. Alternate approaches would be even more interesting to me.

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You say f is increasing. Is it the sequence that is increasing or the functions themselves? –  mathematician Jul 2 at 19:18
    
The functions themselves. I'll clarify. –  Omnomnomnom Jul 2 at 19:20
    
That is, we do not (necessarily) have $f_n(x) \leq f_{n+1}(x)$ for all $x \in K$. –  Omnomnomnom Jul 2 at 19:22

1 Answer 1

Such a family is equicontinuous, and hence uniformly equicontinuous, since $K$ is compact.

Fix $x\in K$ arbitrarily, and let $\varepsilon > 0$. Since $f$ is continuous and monotonically increasing (not necessarily strictly), there is a $\delta > 0$ such that for every $y \in [x-\delta,x+\delta] \cap K$ we have $f(x) - \varepsilon/4 \leqslant f(y) \leqslant f(x) + \varepsilon/4$.

Let $a = \min [x-\delta,x]\cap K$ and $b = \max [x,x+\delta]\cap K$. Since $f_n \to f$ pointwise, there is an $n_0$ such that $\lvert f_n(a) - f(a)\rvert \leqslant \varepsilon/4$ and $\lvert f_n(b) - f(b)\rvert \leqslant \varepsilon/4$ for all $n \geqslant n_0$.

For $n \geqslant n_0$, we then have $\lvert f_n(y) - f_n(x)\rvert \leqslant f_n(b) - f_n(a) \leqslant (f(x)+2\varepsilon/4) - (f(x) - 2\varepsilon/4) = \varepsilon$ for all $y \in [x-\delta,x+\delta]\cap K$.

For each of the finitely many $n < n_0$, there is a $\delta_n > 0$ such that $\lvert f_n(y) - f_n(x)\rvert \leqslant \varepsilon$ for $y \in [x-\delta_n,x+\delta_n]\cap K$.

Then we have $\lvert f_n(y) - f_n(x)\rvert \leqslant \varepsilon$ for all $y\in [x-\eta,x+\eta]\cap K$ and all $n$ if we choose

$$\eta = \min \{\delta\}\cup \{\delta_n : n < n_0\}.$$

Thus $\{ f_n\}$ is equicontinuous in $x$. Since $x$ was arbitrary, $\{f_n\}$ is equicontinuous on $K$. Since $K$ is compact, $\{f_n\}$ is uniformly equicontinuous on $K$, and therefore the topologies of pointwise convergence and uniform convergence on $K$ coincide on $\{f_n\}$.

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