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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.