Is $\int_{\Omega} \bigg( \sum_{n=1}^{\infty} |f_n| \bigg)^p d \mu$ really in $L^p$?

Question

What confuses me that I think that $|f_n|$ should have some power of $p$.

$f_n$ are elements of $L^p$. $\sum_n f_n$ is an absolutely convergent sequence in $L^p$.

Def. of $\| \cdot \|_p$ of $L^p$:

https://en.wikipedia.org/wiki/Lp_space#Lp_spaces

So I think the form I give, doesn't look like that. Yet my notes claim it's that (in order for the series to be in $L^p$?).

By absolutely convergent, do you mean $\sum_{n=1}^\infty ||f_n||_p <\infty$ ? — Danny Pak-Keung Chan, May 09 '19 at 19:37
Are you asking whether $$ \int_\Omega \left( \sum_n |f_n| \right)^p ,\mathrm d\mu < +\infty, $$ i.e. whether the function $$ \sum_n |f_n| $$ is in $L^p$? — MSDG, May 09 '19 at 19:45
@MisterRiemann That's part of the proof I'm looking at, but I don't understand how it concludes that $\sum_{n=1}^{\infty} |f_n|$ $\in L^p$. Because I think $|f_n|$ should have some power of $p$. Since the formulation given here doesn't look like the one for $L^p$. — mavavilj, May 09 '19 at 19:46
A similar proof is here: "12 proposition". https://folk.ntnu.no/hanche/notes/lp/lp-a4.pdf — mavavilj, May 09 '19 at 19:50

MSDG · Answer 1 · 2019-05-09T20:10:09.727

2

By Minkowski, $$ \left\Vert \sum_{n=1}^N |f_n| \right\Vert_p \leq \sum_{n=1}^N \Vert f_n\Vert_p, $$ and so by Fatou, $$ \left\Vert \sum_{n=1}^\infty |f_n| \right\Vert_p = \left\Vert \lim_N\sum_{n=1}^N |f_n| \right\Vert_p \leq \liminf_N \left\Vert \sum_{n=1}^N |f_n| \right\Vert_p \leq \liminf_N \sum_{n=1}^N \Vert f_n\Vert_p = \sum_{n=1}^\infty \Vert f_n\Vert_p < +\infty. $$ EDIT: By request, the line above (skipping a few steps) could also be written in the form \begin{align} \left(\int_\Omega \left(\sum_{n=1}^\infty |f_n|\right)^p \,\mathrm d\mu\right)^{1/p} \leq \liminf_N \left(\int_\Omega \left(\sum_{n=1}^N |f_n|\right)^p \,\mathrm d\mu\right)^{1/p} \leq \liminf_N \sum_{n=1}^N \left( \int_{\Omega} |f_n|^p \,\mathrm d\mu \right)^{1/p}. \end{align}

edited May 09 '19 at 20:10

answered May 09 '19 at 19:49

MSDG

7,143

So are you referring to the $| \cdot |_p$ of $L^p$? How does one apply that "integral form" to an infinite series? – mavavilj May 09 '19 at 19:53
Yes, of course. – MSDG May 09 '19 at 19:53
Shouldn't $\sum_{n=1}^{\infty} |f_n|$ have power $p$? Since in the $| \cdot |_p$ of $L^p$ there's $|f|^p$. – mavavilj May 09 '19 at 19:54
I don't understand why this would affect the argument. – MSDG May 09 '19 at 19:56
It doesn't, but I'm asking about this notational confusement. See the question and the title. I'm asking why is that form I give in $L^p$, because it doesn't look the same as the $|\cdot|_p$ of $L^p$. – mavavilj May 09 '19 at 19:56
It would be the same if you wrote it as an integral, then you would see your power of $p$. I'll edit – MSDG May 09 '19 at 19:57
Does the edit help out? – MSDG May 09 '19 at 20:01
Yes, but what do you do at the very last inequality? – mavavilj May 09 '19 at 20:02
Minkowski's inequality, as I wrote in the very first line of my answer. – MSDG May 09 '19 at 20:03
How does one apply it? I don't see. – mavavilj May 09 '19 at 20:04
Just like the usual triangle inequality. That's why I wrote everything in norm notation instead of integral notation since it's easier to see what's going on. – MSDG May 09 '19 at 20:06
And do you use $1/p < 1$ for the outer $1/p$? – mavavilj May 09 '19 at 20:06
Btw, it still doesn't look the same as the form I gave. Your $p$ is inside the sum, mine is out of an infinite sum. – mavavilj May 09 '19 at 20:07
No, note that the norm is defined with the $1/p$, i.e. $$ \Vert f \Vert_p = \left(\int_\Omega |f|^p,\mathrm d\mu\right)^{1/p}. $$ Otherwise it is not a norm. EDIT: I forgot to write $1/p$ in the last inequality of the answer, sorry. – MSDG May 09 '19 at 20:08

Is $\int_{\Omega} \bigg( \sum_{n=1}^{\infty} |f_n| \bigg)^p d \mu$ really in $L^p$?

1 Answers1