Uniform convergence of a seriesobtained by dominating it with another series

Question

Let $\{f_n\}$ be a sequence of complex valued functions on the real line and $n$ be integers. If there is a sequence of nonnegative numbers $\{a_n\}$ such that $\sum _{n} \mid f_n(x) \mid \leq \sum_{n} a_n < \infty$ for all $x$ in the domain, then can I conclude that $\sum_{n} f_n$ converges uniformly on the domain? If so, how do I prove it? I first tried the Weierstrass test but it does not seem to fit in this situation. Could anyone please help me?

Answer 1

Your conjecture is wrong in general: consider $f_n: [0,1] \to \mathbb R: x \mapsto x^n \cdot (1-x)$.

Then for $x\neq 1$, $\sum_n \vert f_n(x) \vert = 1 = \sum_n \frac1{2^n}<\infty$ (see this question), but $\sum_n \vert f_n(1) \vert=0$. Thus the pointwise limit is not continuous and thus the convergence is not uniform since each $f_n$ is continuous.

Answer 2

$\sum_n f_n(x)$ is uniformly Cauchy since $\sum_{k=n}^m |f_k(x)| \leq \sum_{k=n}^m a_n \to 0$ as $m,n \to \infty$

Hence $\sum_n f_n(x)$ converges uniformly.