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I want to show that all absolute convergent series in $L^\infty$ with $||.||_\infty$ converges, directly using definition. But I am confused on what exactly do I need to show. My goal is to use this to show that $L^\infty$ is complete.

Definitions in the book I'm reading,

$X$ be a normed vector space, a series in $X$, $\sum x_n$ is said to converge to $x$ if $\sum^N x_n\to x$ as $N\to \infty$ and it is absolutely convergent if $\sum ||x_n||<\infty$

If I supposed that $\{f_n\}$ is a sequence in $L^\infty$ and the series $\sum ||f_n||_\infty<\infty$, do I need to show that for all $\epsilon>0$ there exists $K$ such that $||\sum^N f_n -f||_\infty<\epsilon$ for all $N>K$?

It would be greatly appreciated, if you could provide some insight and logic of how to approach this. Many thanks!

Remu X
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    What do you mean by $| g - f | < \epsilon$ (with $g = \sum^N f_n$)? – cs89 Apr 28 '23 at 07:02
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    In any Banach space $X$, $\sum |x_n|<\infty$ implies $\sum x_n$ converges because the partial sums form a Cauchy sequence. See: https://math.stackexchange.com/questions/2889441/normed-space-x-is-complete-iff-every-absolutely-convergent-series-in-x-conve?rq=1 – Kavi Rama Murthy Apr 28 '23 at 07:25
  • @geetha290krm Thanks for the reply! I forgot to mention, but I am trying to show it is a Banach space, however I want to use the fact that a normed vector space is complete iff every abs convergent series in the space converges. (Which is from the link you attached) – Remu X Apr 28 '23 at 07:29
  • @cs89 ah you are right, that doesnt make any sense, since $g,f $ are functions not real number. thanks! – Remu X Apr 28 '23 at 07:32

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Given $j$ there exists $N$ such that $ \|\sum\limits_{k=n}^{m}f_n\|<\frac 1 j$ whenever $m>n>N$. This implies that $ |\sum\limits_{k=n}^{m}f_n(x)|<\frac 1 j$ almost everywhere whenever $m>n>N$. [Note that there are only a coutable number of null sets involved here and their union is a null set]. Thus, for almost all $x$, $(\sum\limits_{k=1}^{m}f_k(x))$ is a Cauchy sequence of real numbers and $f(x)=\lim \sum\limits_{k=1}^{m}f_n(x)$ exists. To finish the proof observe that $ |\sum\limits_{k=1}^{m}f_n(x)-f(x)|\leq\sum\limits_{k=m+1}^{\infty}\|f_n(x)\|\leq \frac 1 j$ almost everywhere whenever $m>N$. Thus, $ \|\sum\limits_{k=1}^{m}f_n(x)-f(x)\|\leq \frac 1 j$ whenever $m>N$ proving that $\sum_n f_n=f$ in the norm of $L^{\infty}$.

Kavi Rama Murthy
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  • Hi thanks! But I do still have a confusion, since I am only assuming that the series $\sum ||f_n||$ is absolutely convergent (which by definition only means the sum is finite) and I do not know the space is complete, why can we argue that there exists a $N$ such that the property in ur first line holds? – Remu X Apr 28 '23 at 18:07
  • Ok I think I understand, if such condition is not satisfied then the sum can not be finite. Thanks – Remu X Apr 28 '23 at 18:18