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If sequence of functions {$f_n$} converges uniformly, then {$f_n$} is a cauchy sequence.
That is, it satisfies $|f_n(x)-f_m(x)| \le \epsilon$.

Then if {$f_n$} is a cauchy sequence, $|f_n(x)-f_m(x)| \le \epsilon$,
this sequence of functions {$f_n$} converges uniformly?
Or only we can conclude is that it converges?

It would be better to write a whole question.
In Rudin's book, theorem 7.8 says
The sequence of functions {$f_n$} defined on E converges uniformly on E if and only if for every $\epsilon>0$ there exists an integer N such that $N \le m,n$ , x belongs to E implies
$$|f_n(x)-f_m(x)| \le \epsilon$$.

My question is that: we have $|A_n-A_m| \le \epsilon$ for $N \le m,n$, so that {$A_n$} is a cauchy sequence converges to A. Therefore $|A_n-A| < 3/\epsilon$.
In here, that inequality is from convergence or uniform convergence?
I wonder whether I can use Theorem 7.8 in here or not. (because it says if and only if)

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How do you define "Cauchy sequence" (of functions)? If you consider it pointwise (i.e. that for each $x$, $\{f_n(x)\}$ is Cauchy) I don't think the result will follow. A possible counterexample over some domain might be $f_n(x)=x^n$. – Pedro Tamaroff Nov 30 '12 at 3:12
up vote 2 down vote accepted

It depends on what you mean by saying "Cauchy." There are uniformly Cauchy sequences (which will converge uniformly) and pointiwse Cauchy sequences (which are only guaranteed to converge pointwise). Both are described in the linked article.

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In Rudin's book, theorem 7.8 says "{$f_n$} converges uniformly iff {$f_n$} is a cauchy sequence". Then, in here, I can say that it converges uniformly by this theorem? – niagara Nov 30 '12 at 3:14
Well, to be precise though, it says The sequence of functions $\{f_n\}$, defined on $E$ converges uniformly on $E$ if and only if for every $\epsilon>0$ there exists $N$ such that $m,n\geq N$ and $x\in E$ implies $\vert f_n(x)-f_m(x)\vert\leq\epsilon$. This is a uniform statement - it is saying $\{f_n\}$ converges uniformly iff $\{f_n\}$ is uniformly Cauchy. – icurays1 Nov 30 '12 at 3:25
Oh, I see. I added my whole question, but I think your comment give me a clue. Maybe in here, I can't say "{$A_n$} converges uniformly" carelessly. So that inequality $|A_n-A|<3/\epsilon$ comes from convergence not uniformly convergence... – niagara Nov 30 '12 at 3:31
I'm not sure what your sequence $A_n$ is, but yes - you have to be very careful about claiming uniform convergence. – icurays1 Nov 30 '12 at 3:34
Actually, it is a part of proof theorem 7.11 in Rudin. If you have a book you can check. – niagara Nov 30 '12 at 3:36

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