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The question is:

Let $\displaystyle \sum_{n=1}^{\infty}f_n(x)$ a series converges uniformly on $I$ to $S(x)$. Prove that $\lim\limits_{n\to\infty}f_n(x)=0$.

My try: $\forall N\in\mathbb N$, we have $f_N(x)=\sum\limits_{n=1}^{N}f_n(x)-\sum\limits_{n=1}^{N-1}f_n(x)$, so the statement is proved if I prove that

$$f_n(x),g_n(x)\to f(x),g(x)\text{ uniformly on }J\implies\\ f_n(x)+g_n(x)\to f(x)+g(x) \text{ uniformly on }J.$$ On the one hand: $$\begin{align*} \lim_{n\to\infty}\sup_{x\in J}|f_n(x)-f(x)+g_n(x)-g(x)|\leq&\quad\text{(triangle inequality) }\\ \limsup\limits_{x\in J}{|f_n(x)-f(x)|+|g_n(x)-g(x)|}=0+0=0. \end{align*}$$

Now I'm in trouble proving that $\limsup\limits_{x\in J}{|f_n(x)+g_n(x)-f(x)-g(x)|}\geq 0$. How can I do that ?

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We know:
$\forall x\in J:|f_n(x)+g_n(x)-f(x)-g(x)|\ge0 $
then:
$\limsup\limits_{x\in J}{|f_n(x)+g_n(x)-f(x)-g(x)|}\ge0$

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It easier to work with the remainder $R_n(x) = \sum_{k=n}^\infty f_k(x)$. By hypothesis, we know that $$ \lim_{n\to\infty} \sup_{x} |R_n(x)| = 0. $$

Since $f_n(x) = R_n(x) - R_{n+1}(x)$, we deduce $$ \sup_x |f_n(x)| \leq \sup_x |R_n(x)| + \sup_x |R_{n+1}(x)| \xrightarrow[n\to\infty]{} 0. $$

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  • $\begingroup$ Sorry for illteracy but how $f_n(x)\to 0$ uniformly in your solution? $\endgroup$ – user65985 Jun 8 '13 at 15:15
  • $\begingroup$ What do you mean? I proved that $\lim_{n\to\infty} \sup_x |f_n(x)| = 0$. $\endgroup$ – Siméon Jun 8 '13 at 15:35
  • $\begingroup$ @PeterTamaroff: it is not clear. Check the title. $\endgroup$ – Siméon Jun 8 '13 at 16:01
  • $\begingroup$ @Ju'x you showed that $\sup|f_n(x)|$ is less than something tends to zero. How it proved the limit is exactly zero? $\endgroup$ – user65985 Jun 8 '13 at 16:14
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    $\begingroup$ @CoarguAliquis: $\sup_x |f_n(x)|$ is non-negative. Do you know the squeeze theorem? $\endgroup$ – Siméon Jun 8 '13 at 16:23

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