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Let us start with a given function $f_n$ defined on a certain domain $I$ and s.t. $f_n:I\mapsto\text{ some normed vector space}$. We know that: $$\sum_{n=0}^\infty \|f_n\| := \sum_{n=0}^\infty \sup_I |f_n(x)| < \infty\implies \sum_{n=0}^\infty f_n(x)\text{ is normally convergent}\tag{1}$$ I was thinking that, at this point, it could have been enough to apply the so called Weierstrass M-test to show that $\sum_{n=0}^{\infty}f_n$ is uniformly and absolutely convergent as well.
However, Weierstrass M-test first requires absolute convergence of the series, that is: $$\sum_{n=0}^\infty |f_n(x)|\text{ convergent}$$ and I know that in general it holds true that: $$\sum_{n=0}^\infty f_n(x)\text{ convergent}\nRightarrow \sum_{n=0}^\infty |f_n(x)|\text{ convergent} $$

Hence, starting from normal convergence of a series, how is it possible to use Weierstrass M-test so as to show that the series converges absolutely and uniformly as well? Which is the missing part above?

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    $\begingroup$ You wrote $\implies$ but isn't that just the definition of normal convergence? $\endgroup$
    – zhw.
    Sep 20, 2020 at 17:12
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    $\begingroup$ Saying $\sum f_n$ is normally convergent is the same as saying the Weierstrass M test applies to $\sum f_n.$ $\endgroup$
    – zhw.
    Sep 20, 2020 at 17:20
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    $\begingroup$ No, not at all. What I'm saying is simple: The hypothesis $\sum M_n<\infty$ in the Weierstrass M test holds iff $\sum f_n$ converges normally. $\endgroup$
    – zhw.
    Sep 20, 2020 at 18:14
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    $\begingroup$ Why not just say "by definition ....." $\endgroup$
    – zhw.
    Sep 20, 2020 at 18:26
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    $\begingroup$ I suppose so. But using $\implies$ for a definition can be confusing. In fact, I thought you were confused about it, which is why I wrote my first comment. $\endgroup$
    – zhw.
    Sep 21, 2020 at 19:09

1 Answer 1

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For any $x \in I$, $$\sum_{n=0}^\infty |f_n(x)| < \sum_{n=0}^\infty \sup_{x \in I}|f_n(x)| = \sum_{n=0}^\infty \|f_n\| < \infty$$ Then by the $M$-test the series is uniformly and absolutely convergent if the normed vector space is a Banach space.

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  • $\begingroup$ Pardon, just one question. In your answer you are considering $(1)$ as an $\iff$ relation, if I am not wrong, aren't you? Why that? According to $(1)$ (i.e. normal convergence definition), if I know that $\sum_{n=0}^{\infty}f_n$ is normally convergent, I cannot be certain that $\sum_{n=0}^\infty \|f_n\| := \sum_{n=0}^\infty \sup_I |f_n(x)| < \infty$ @abhi01nat $\endgroup$
    – Strictly_increasing
    Sep 20, 2020 at 17:58
  • $\begingroup$ Normal convergence of the series is precisely the convergence of the sum of uniform norms, i.e., the series is defined to be normally convergent if the sum $\sum_{n=0}^\infty \|f_n\|$ converges. This is what the definition in the wiki page you linked to says. $\endgroup$
    – abhi01nat
    Sep 20, 2020 at 18:00
  • $\begingroup$ Aren't "is precisely" and "if..." contradictory? I interpret "is" as an "iff" relation, i.e. $\iff$ ,and not as an "if" relation @abhi01nat $\endgroup$
    – Strictly_increasing
    Sep 20, 2020 at 18:02
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    $\begingroup$ Convergent means there exists a limit point in the space. The Cauchy condition is necessary for convergence in a normed vector space but is not sufficient if is not a Banach space. $\endgroup$
    – DanielWainfleet
    Sep 20, 2020 at 18:40
  • $\begingroup$ @DanielWainfleet Good point. $\endgroup$
    – zhw.
    Sep 21, 2020 at 1:37

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