How can I prove that this sequence converges uniformly?

Question

let $\sum_{n=1} ^{\infty} f_{n}(x) $ be an uniformly convergent series of functions in a subset $A \subseteq \mathbb{R}$.prove that the sequence $f_{n}$ converges uniformly to $0$ in $A$.

I tried to prove it like this:

let $(S_{m})_{m \geq 1}$ be the sequence of the partial sums of $(f_{n})_{n \geq 1}$ s.t:

$(I)$ $S_{m}= f_{1}(x)+f_{2}(x)+...+f_{m}(x)$

By the Cauchy Criterion for Uniform Convergence we have that:

$\forall \:\epsilon\gt0 , \:\exists \:N \geq 1 :$

$\forall\:$ $n$>$m$>N $\implies |S_{n}-S_{m}|$<$\epsilon$; $\forall $ $x$ in $A$

chose $n_{0}$ such that $(n_{0}-1)$> $N$:

$|S_{n_{0}}-S_{n_0-1}|$<$\epsilon$

$(I) \implies$ $|f_{n}(x)|$<$\epsilon$

which gives that $\forall \:\epsilon\gt0 , \:\exists \:N \geq 1 :$

$\forall \:\epsilon\gt0 , \:\exists \:N \geq 1 :$

$(n_{0}-1)$> $N$ $ \implies$ $|f_{n}(x)-0|$<$\epsilon$; $\forall $ $x$ in $A$

$\therefore f_n(x)^\rightarrow_\rightarrow 0$ in A

Answer 1

The clearest way to prove uniform convergence is to prove that

$sup_{X}|S_{n}(x)-S(x)|$ tends to zero!!

Answer 2

Let $f(x)=\sum_{n=1}^{\infty}f_n(x).$ Let $F_n(x)=\sum_{j=1}^n f_j(x).$

For $\epsilon >0,$ take $n_{\epsilon}\in \Bbb N$ such that $$\forall m\ge n_{\epsilon}\,(\,\sup_{x\in A}|f(x)-F_m(x)|<\epsilon /2\,).$$ Then for all $n> n_{\epsilon}$ we have $$\sup_{x\in A}|f_n(x)|=\sup_{x\in A}|(f(x)-F_{n-1}(x))-(f(x)-F_n(x)|\le$$ $$\le\sup_{x\in A}|f(x)-F_{n-1}(x)|+|f(x)-F_n(x)|<$$ $$<\epsilon /2+\epsilon /2=\epsilon.$$