# If $\sum f_k$ converges uniformly in $X$, then $f_k \to 0$ uniformly in $X$

This problem must be very easy but I find it complicated, and I want some help here.

If $\sum f_k$ converges uniformly in $X$, then prove that $f_k \to 0$ uniformly in $X$.

It's clear that converges pointwise to that function, but how can I prove that the convergence is uniform?

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How would you prove that the convergence of the numerical series $\sum a_n$ implies $a_n \to 0$? Try to extend the same technique to the case at hand. –  Giuseppe Negro Nov 17 '11 at 16:47

First, show more generally if $\{a_n(x)\}$ converges uniformly to $a(x)$ and $\{b_n(x)\}$ converges uniformly to $b(x)$, then $\{a_n(x)-b_n(x)\}$ converges uniformly to $a(x)-b(x)$.

Then let $b_n(x)=\sum_{i=1}^n f_i(x)$ and $a_n(x)=\sum_{i=1}^{n+1} f_i(x) = b_{n+1}(x)$.

Then $a_n(x)$ and $b_n(x)$ both converge uniformly to the sum $\sum_{i=1}^\infty f_i(x)$

But then $a_n(x)-b_n(x)=f_{n+1}(x)$ converges uniformly to $0$.

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Since $\sum f_k$ converges uniformly in $X$, the sequence $S_n=\sum_{m=1}^n f_m$ is uniformly Cauchy. That is, for every $\epsilon>0$, there is an integer $N$ such that $|S_n(x)-S_m(x)|<\epsilon$ for all admissible $x$ and all $n,m\ge N$.

But then, for every $\epsilon>0$, there would be an integer $N$ such that $|f_n(x)| =| S_{n+1}(x)-S_n(x)|<\epsilon$ for all admissible $x$ and all $n \ge N$; which shows that $\{f_n\}$ converges uniformly to 0.

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