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I am a little bit stuck on solving the following problem:

Let $f_n \in L_1$ for all $n \in \mathbb{N}$. Which is always larger?

$$ \left(\sum_{n = 1} ^\infty \left\lvert\int f_n \,d\mu\right\rvert^2\right)^\frac{1}{2} $$ or $$ \int\left(\sum_{n = 1} ^\infty \lvert f_n\rvert^2\right)^\frac{1}{2} \,d\mu? $$

My guess is that the second expression is always greater. The reason for the guess is if we consider this sum to be just one term, then we simply just have the triangle inequality for integrals. However, when I try to show this result to the finite case, I run into problems: \begin{align} \left(\sum_{n = 1} ^k \left\lvert\int f_n \,d\mu\right\rvert^2\right)^{\frac{1}{2}} \leq \left(\sum_{n = 1} ^k \left(\int \lvert f_n\rvert \,d\mu\right)^2\right)^{\frac{1}{2}} \leq \left(\int \sum_{n = 1} ^k \lvert f_n\rvert \,d\mu\right). \end{align} How do I incorporate the square and square root into the expression? If I could show this, the result should then be true through Monotone Convergence.

Lorago
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Partial T
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    Interesting: you've actually already proved it if the supports of the $f_n$ are disjoint, while the two expressions are equal if their supports are all a single point. How to interpolate between these two extremes.... – Greg Martin Aug 28 '22 at 21:01
  • @GregMartin I am not sure if I get the hint. Is there an obvious way to interpolate between the situations? – Partial T Aug 28 '22 at 21:59
  • No :D That wasn't a rhetorical device, that really was me wondering how – Greg Martin Aug 28 '22 at 23:36

1 Answers1

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One way that I sometimes look at $L^p$ inequalities is by linearizing against the dual space (e.g. a related result). Here, we are dealing with $\ell^2$, which is self dual.

Consider the set of sequences of complex numbers $$ {\scr A}=\left\{\{a_n\}:\sum_{n=1}^\infty|a_n|^2=1\right\}\tag1 $$ and the set of sequences of complex valued functions $$ {\scr B}=\left\{\{b_n\}:\forall x\in X,\sum_{n=1}^\infty|b_n(x)|^2=1\right\}\tag2 $$ Then we have $$ \left(\sum_{n=1}^\infty\left|\int_Xf_n(x)\,\mathrm{d}\mu\,\right|^2\right)^{1/2}=\sup_{a\in{\scr A}}\sum_{n=1}^\infty\int_Xa_nf_n(x)\,\mathrm{d}\mu\tag3 $$ and $$ \int_X\left(\sum_{n=1}^\infty|f_n(x)|^2\right)^{1/2}\,\mathrm{d}\mu=\sup_{b\in{\scr B}}\int_X\sum_{n=1}^\infty b_n(x)f_n(x)\,\mathrm{d}\mu\tag4 $$ We can view ${\scr A}$ as the subset of constant functions in ${\scr B}$. Therefore, because the sup over a set is at least as great as the sup over a subset, we have that the sup over ${\scr B}$ must be at least as big as the sup over ${\scr A}$. That is, $$ \left(\sum_{n=1}^\infty\left|\int_Xf_n(x)\,\mathrm{d}\mu\,\right|^2\right)^{1/2}\le\int_X\left(\sum_{n=1}^\infty|f_n(x)|^2\right)^{1/2}\,\mathrm{d}\mu\tag5 $$

robjohn
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