Prove that if the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent over set E and the function $g(x)$ is bounded over E that the series $\sum_{n=1}^\infty g(x)f_n(x)$ is also uniformly convergent over the set E.
What I thought of: If $g(x)$ is bounded then $ \exists \sup{g(x)} = M $ and $\exists \inf{g(x)} = m$. Then if $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent then there exist $S(x)$ which is the sum of the series and $S(x)M$ $< \infty$ and also we have $S(x)m$ $< \infty$. However, I don't know how to continue from here.
The answer is not different between a function series and a function sequence (since $g$ does not depend on $n$, you can place it in front of the series). So let $f_n$ be a function sequence that converges uniformly.
Let $f$ be the uniform limit of the sequence, then:
$$ (1):\forall \epsilon >0, \exists N_\epsilon \in \mathbb{N}, \forall n\in \mathbb{N} \text{ s.t. } n\geqslant N_\epsilon, \forall x\in A, ||f_n(x)-f(x)|| \leqslant \epsilon $$
Then, $\forall x\in A, ||g(x)f_n(x)-g(x)f(x)||\leqslant |g(x)|\times ||f_n(x)-f(x)|| \leqslant \sup |g| \times \epsilon$
Thus, applying (1) for $\epsilon' = \frac{\epsilon}{\sup |g|}$ is enough to see that the definition of uniform convergence holds for $(g\times f_n)$.