cdsommegolfzp

2022-07-01

My instructor briefly discussed a result in lecture that I need help with. Here was the set up.

We assumed that $\{{f}_{n}\}$ is an orthonormal system in ${L}^{2}(0,1)$ and we supposed that there exists an $M>0$ such that $|{f}_{n}(x)|\le M$ a.e for all $n\in \mathbb{N}$. Furthermore, we let $\{{c}_{n}\}$ be a sequence of real numbers such that $\sum _{n=1}^{\mathrm{\infty}}{c}_{n}{f}_{n}$cnfn converges a.e.

They said that it was "clear" that

$\underset{n\to \mathrm{\infty}}{lim}{c}_{n}=0.$

This is not clear or intuitive for me.

After reading through the page, I am still a bit confused. Can someone put together a working proof to clear things up?

We assumed that $\{{f}_{n}\}$ is an orthonormal system in ${L}^{2}(0,1)$ and we supposed that there exists an $M>0$ such that $|{f}_{n}(x)|\le M$ a.e for all $n\in \mathbb{N}$. Furthermore, we let $\{{c}_{n}\}$ be a sequence of real numbers such that $\sum _{n=1}^{\mathrm{\infty}}{c}_{n}{f}_{n}$cnfn converges a.e.

They said that it was "clear" that

$\underset{n\to \mathrm{\infty}}{lim}{c}_{n}=0.$

This is not clear or intuitive for me.

After reading through the page, I am still a bit confused. Can someone put together a working proof to clear things up?

Shawn Castaneda

Beginner2022-07-02Added 17 answers

Suppose the sum of an infinite series is $\phantom{\rule{thickmathspace}{0ex}}S\phantom{\rule{thickmathspace}{0ex}}$ ,and we denote by $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\phantom{\rule{thickmathspace}{0ex}}$ the $\phantom{\rule{thickmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}-$−th element of the sequence of partial sums of the series. By definition, the series converges iff the limit of $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\phantom{\rule{thickmathspace}{0ex}}$ exists finitely ( and in this case $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\to S\phantom{\rule{thickmathspace}{0ex}}$) , so:

$S=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{S}_{n}:=\sum _{k=1}^{n}{a}_{k}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{n}={S}_{n}-{S}_{n-1}\underset{n\to \mathrm{\infty}}{\overset{}{\to}}\stackrel{\text{(arith. of limits)}}{S}-S=0$

and we have the basic necessary condition for convergence of infinite series.

In your case you have $\phantom{\rule{thickmathspace}{0ex}}|{c}_{n}{f}_{n}|\underset{n\to \mathrm{\infty}}{\overset{}{\to}}0\phantom{\rule{thickmathspace}{0ex}}$ (no matter whether real of complex numbers), but $\phantom{\rule{thickmathspace}{0ex}}\{{f}_{n}\}\phantom{\rule{thickmathspace}{0ex}}$ is a bounded sequence...! Finish the argument.

$S=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{S}_{n}:=\sum _{k=1}^{n}{a}_{k}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}_{n}={S}_{n}-{S}_{n-1}\underset{n\to \mathrm{\infty}}{\overset{}{\to}}\stackrel{\text{(arith. of limits)}}{S}-S=0$

and we have the basic necessary condition for convergence of infinite series.

In your case you have $\phantom{\rule{thickmathspace}{0ex}}|{c}_{n}{f}_{n}|\underset{n\to \mathrm{\infty}}{\overset{}{\to}}0\phantom{\rule{thickmathspace}{0ex}}$ (no matter whether real of complex numbers), but $\phantom{\rule{thickmathspace}{0ex}}\{{f}_{n}\}\phantom{\rule{thickmathspace}{0ex}}$ is a bounded sequence...! Finish the argument.

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