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I have been struggling to prove the following:

Let $ \{ f_n \}$ be a sequence in $ L^p(E) $ for some $ p \geq 1 $. Then,

$$ \left( \sum_{n=1}^\infty | \int_E f_n \mathrm{d}\mu |^p \right)^{ \frac{1}{p}} \leq \int_E \left( \sum_{n=1}^\infty |f_n|^p \right)^{\frac{1}{p}} \mathrm{d} \mu $$.

I'm not sure if any of my attempts have been promising enough to include. Any hints would be greatly appreciated. :)

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  • $\begingroup$ Have you considered the fact that on $\mathbb{R}^\infty$, $\|x\|_p=(\sum_{i}^\infty |x|^p)^{1/p}$ is a norm? $\endgroup$ – Alex R. Mar 7 at 21:59
  • $\begingroup$ Do you mean something like rewriting the inequality as, $ || \{ \int_E f_n d \mu \} ||_p \leq \int_E || \{ f_n(x) \} ||_p d \mu $? $\endgroup$ – user38770 Mar 7 at 22:11
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Let $p >1$. We have $\sum |a_n|^{p} =\sup \{ |\sum a_n b_n|: \sum |b_n|^{q} \leq 1\}$ where $q$ is the index conjugate to $p$. Hence it is enough to show that $|\sum b_n \int_E f_n \,d\mu| \leq $ RHS whenever $\sum |b_n|^{q} \leq 1$. This is very easy since $|\sum b_n f_n| \leq (\sum |f_n|^{p})^{1/p} (\sum |b_n|^{q})^{1/q}$. The case $p=1$ is trivial.

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  • $\begingroup$ Thank you very much for your insightful answer. $\endgroup$ – user38770 Mar 8 at 4:39

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