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?
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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)
Best Answer
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.