# Uniform convergence and cauchy sequence

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)

-
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

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