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?
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2I suppose that you forgot to add that $\sum_na_n$ converges. – José Carlos Santos May 09 '19 at 17:30
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Oh I forgot. Yes it must converges. Then could you explain how to prove the question? – Keith May 09 '19 at 17:37
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No, but I'll think about it. – José Carlos Santos May 09 '19 at 17:39
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You can't. Take $f_n=\chi_{[n,n+1)}$, $a_1=1$, and $a_n=0$ for $n>1$. – David Mitra May 09 '19 at 17:42
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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.
Maximilian Janisch
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OK, I forgot to add that the inequality also holds for every subset of $\mathbb{Z}$... Then I think what I claim is true. – Keith May 09 '19 at 17:47
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$\sum_n f_n(x)$ 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.
Mustafa Said
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Something is wrong here, since my sequence also seems to be uniformly Cauchy – Maximilian Janisch May 10 '19 at 05:53
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The inequality is incorrect because it does not hold term-wise; it only applies to the partial sums. – Pion Chan May 02 '25 at 16:05