Assume $f_n,g_n : [0,1] \longrightarrow \mathbb{R} $ are continuous, $f_n\geq 0$ for $n \geq 1$ and for all $x \in [0,1]$ both series $\sum_{n=1}^{+\infty}f_n(x),\sum_{n=1}^{+\infty}g_n(x)$ are convergent as well as $\sum_{n=1}^{+\infty}f_n(x)=\sum_{n=1}^{+\infty}g_n(x)$.
Show that if $\sum_{n=1}^{+\infty}g_n(x)$ uniformly convergent so is $\sum_{n=1}^{+\infty}f_n(x)$.
I feel it should be obvious, but I look at the definition of series uniform convergence and can't think of anything constructive.
From comments by Daniel Fischer
The only thing the uniform convergence of $\sum g_n$ is needed for is the continuity of the limit function. Forget about $g_n$, and prove that $\sum f_n$ converges uniformly under the condition that $$S(x) = \sum_{n=1}^\infty f_n(x)$$ is a continuous function (plus your conditions on the $f_n$ of course). Dini's theorem may help.
Proof of the theorem: for all $n$ (and $\varepsilon > 0$), the set $U_n(\varepsilon) = \{ x : \lvert S(x) - S_n(x)\rvert < \varepsilon\}$ is open (by continuity), $U_n(\varepsilon) \subset U_{n+1}(\varepsilon)$ (by monotonicity), and $\bigcup U_n(\varepsilon) = [0,1]$ (by pointwise convergence); since $[0,1]$ is compact, $\{ U_n(\varepsilon) : n \in\mathbb{N}\}$ has a finite subcover, by nestedness $U_n(\varepsilon) = [0,1]$ for all $n\geqslant n_0$. The crucial point is that $U_n\subset U_{n+1}$, which follows by monotonicity.