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I'm working on a question on uniform convergence. After spending an hour, I'm still trying to write a formal proof for it. The question seems obvious to me, but I cannot prove it formally.

Consider $f_n$ as a sequence of continuos functions pn [a,b]. Suppose for any $x \in [a,b]$ $f_n(x)$ is non-increasing. I need to show that if $f_n \rightarrow 0$ pointwise pn [a,b], then $f_n \rightarrow 0$ uniformly on [a,b].

Thank you very much for your help.

This is what I've done so far:

I tried to prove it with contradiction. Let's suppose that $f_n$ does not converge to 0 uniformly, so then there exists a sequence $x_n \in [a,b]$ that converges to $x_0$ which is for sure in [a,b] and for that series, we have an epsilon for which $f_n(x_n) \ge 0$. I don't know how to continue from here. Can I say since f converges point wise, so then for every element of $x_n$ (let's say $x_k$), then $f_n(x_k)$ converges to 0 as n goes to $\infty$ and therefore the limit should converge to?

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Wow, after an hour of work you must have quite a bit to say about what you attempted. How about you include it in your post? –  rschwieb Feb 1 '13 at 18:44
    
@rschwieb, Sure. –  Sepehr Feb 1 '13 at 18:49
    
This is called Dini's theorem, by the by. –  Chris Eagle Feb 1 '13 at 19:00

2 Answers 2

up vote 3 down vote accepted

Since, for each $x$, the sequence $f_n(x)$ is nonincreasing with $n$ and converges to $0$, it must have nonnegative terms, so $f_n(x)\ge 0$ for all $n$ and $x$.

Pick any $\epsilon>0$. Then, pick a point $x\in[a,b]$. Since $f_n\to 0$ pointwise, there is some $n$ such that $f_n(x)<\epsilon$, and, since $f_n$ is continuous, there is some $\delta>0$ such that $f_n(y)<2\epsilon$ whenever $|y-x|<\delta$. Then, since $f_n(y)$ is nonincreasing with $n$, you must also have $f_N(y)\le f_n(y)<2\epsilon$ for all $N\ge n$ and $|y-x|<\delta$, i.e., you have converged to within $2\epsilon$ in a neighborhood of $x$. Now use compactness to cover $[a,b]$ with a finite number of these neighborhoods.

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Hint: Suppose for the sake of argument that $f_n \geq 0$ for all $n$. Then the worst $|f_n(x)|$ can be for all $x \in [a,b]$ is $f_n(a)$ by your nonincreasing assumption. Because $f_n$ converges pointwise to zero, how do you conclude uniform convergence in this case?

Moreover how do you generalize to $f_n$ nonnegative? (Hint: think about $\{f_n(b)\}$).

EDIT: I misread the question. The above is not pertinent.

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The hypothesis is that, for each $x$, the sequence $(f_n(x))_n$ is nonincreasing, not that for each $n$, the function $f_n$ is nondecreasing over $[a,b]$ (yes, this is not so clear in the post but otherwise the question is absurd). –  Did Feb 1 '13 at 19:20

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