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Let $X$ be a compact metric space and let $f_n$ and $g_n$ be sequences of continuous functions from the metric space into the complex plane. If $\Sigma g_n$ converges uniformly and absolutely on $X$ and the absolute value of $f_n$ is equal or smaller than the absolute value of $g_n$ for each $n$, then, it is clear that $\Sigma f_n$ converges absolutely on $X$. However how can I show that $\Sigma f_n$ converges uniformly on $X$? I tried the Weierstrass M-test but failed... Could anyone please help me?

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    $\begingroup$ Use Cauchy convergence criterion. $\endgroup$ – C.Ding Nov 14 '17 at 8:25
  • $\begingroup$ What is Cauchy convergence criterion? Could you explain in more detail? $\endgroup$ – Keith Nov 14 '17 at 8:27
  • $\begingroup$ If $\forall \varepsilon >0$, $\exists N\in\mathbb{N}$, $\forall m\geq n>N$, $|\sum_{k=n}^m f_n |<\varepsilon,$ then $ f_n$ converges uniformly. The google translation tells me that it is called Cauchy convergence criterion, which I'm not sure about. $\endgroup$ – C.Ding Nov 14 '17 at 8:33
  • $\begingroup$ I tried that too but cannot apply it to $f_n$. I cannot bound the partial sum of $f_n$ by $g_n$. Could you show me your way? $\endgroup$ – Keith Nov 14 '17 at 8:36
  • $\begingroup$ It is sufficient and necessary to show that $\sum |g_n|$ conveges uniformly. And by Dini's theorem, it is equivalent to show $\sum|g_n|$ is continuous. $\endgroup$ – C.Ding Nov 14 '17 at 11:55
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This is wrong. It assumes that $\sum |g_n|$ converges uniformly, which does not follow from the assumptions.

Let $\varepsilon > 0$.

Since $\sum_{n=1}^\infty |g_n|$ converges uniformly, there exists $n_0 \in \mathbb{N}$ such that

$$n \ge n_0 \implies \forall x\in X \text{ holds }\sum_{k=n+1}^\infty |g_k(x)| = \left|\sum_{k=1}^\infty |g_k(x)|- \sum_{k=1}^n|g_k(x)|\right| < \varepsilon$$

Now for $M, N \ge n_0$ and $x \in X$ we have:

$$\left|\sum_{k=1}^M f_k(x)- \sum_{k=1}^N f_k(x)\right|= \left|\sum_{k=N+1}^M f_k(x)\right| \le \sum_{k=N+1}^M |f_k(x)| \le \sum_{k=N+1}^M |g_k(x)| \le \sum_{k=N+1}^\infty |g_k(x)| < \varepsilon$$

Therefore, $\sum_{k=1}^n f_k$ is uniformly Cauchy. This means that $\sum_{k=1}^n f_k$ is Cauchy in $C(X)$, the space of all continuous functions from $X$ to $\mathbb{C}$, equipped with the uniform metric: $$d_\infty(f_1, f_2) = \sup_{x \in X} |f_1(x) - f_2(x)|$$ Since $(C(X), d_\infty)$ is a complete metric space, the series $\sum_{k=1}^n f_k$ converges with respect to $d_\infty$, which in turn implies that $\sum_{k=1}^n f_k$ converges absolutely.

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  • $\begingroup$ Only the sum of $g_n$ converges uniformly. It is not that the sum of absolute value of $g_n$ converges uniformly. How can I prove that the sum of absolute value of $g_n$ converges uniformly? $\endgroup$ – Keith Nov 14 '17 at 8:41
  • $\begingroup$ @Keith Ah, I see. It does not have to follow from uniform and absolute convergence of $\sum g_n$, see here. $\endgroup$ – mechanodroid Nov 14 '17 at 8:45
  • $\begingroup$ In the link there is the "distinction" between two concepts. What I intend is the uniform 'and' absolute convergence of $g_n$ not uniform absolute convergence. $\endgroup$ – Keith Nov 14 '17 at 8:53
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    $\begingroup$ @Keith I know, I'm trying to fix it. $\endgroup$ – mechanodroid Nov 14 '17 at 8:54
  • $\begingroup$ @Keith Wait, but your problem states that $|f_n(x)| \le g_n(x)$. This automatically implies $g(x) \ge 0$ so $|g(x)| = g(x)$. $\endgroup$ – mechanodroid Nov 14 '17 at 9:05

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