(b) Prove that if $\sum |g_n|$ converges uniformly on $X$ and if there exists $K \gt 0$ such that $f_n(x) \le K$ for all $n \in\mathbb N$ e $x \in X$, then $\sum |f_n||g_n|$ also converges uniformly on $X$.
My attempt
Let's use Cauchy's criterion for uniform convergence $f_n(x) \le K$ for all $n \in\mathbb N$ and
$x \in X$ so $ k \gt n$ $\sum f_n \le K $ large enough, we can take, $ |g_{n + 1} (x)| \lt \frac {\epsilon}{3K}, |g_{k + 1}(x )| \lt \frac{\epsilon}{3K}, g_{n + 1}(x) - g_ {k + 1} (x) \ge \frac {\epsilon}{3K} $, all valid for $\forall x \in X $, we now apply the soma technique by parts
$$|\sum_{m=n+1}^k g_m(x)f_m(x) |=| \sum_{m=n+1}^k g_m(x)\Delta s_{m-1}(x)| = |s_{m-1}(x)g_m(x)]_{n+1}^{k+1} + \sum_{m=n+1}^k s_m(x) \Delta - g_m(x)|$$ $$\le |s_k(x) g_{k+1} - s_n(x)g_{n+1}(x)|+ \sum_{m=n+1}^k |s_m(x)| \Delta - g_m(x)$$ $$\le K|g_{k+1}(x)| + K|g_{n+1}|+ K(g_{n+1}(x) - g_{k+1}(x)) \le \epsilon \forall X$$
This was the only thing I was able to do, how bad is it?
Thanks in advance for any help.
As stated this is false. You have to change $f_n (x) \leq K$ to $|f_n(x)| \leq K$.
$ \sum\limits_{m=n+1}^{k}|f_m(x)g_m(x)| \leq K \sum\limits_{m=n+1}^{k}|g_m(x)|$ and uniform convergence of $\sum |g_n|$ implies that given $\epsilon >0$ we can find $N$ such that $\sum\limits_{m=n+1}^{k}|g_m(x)|<\frac {\epsilon} K$ for $k>n>N$ for all $x$. Hence $ \sum\limits_{m=n+1}^{k}|f_m(x)g_m(x)| <\epsilon$ for $k>n>N$ for all $x$ and this finishes the proof.