Comparing the quantity of two Lebesgue integrals of infinite sums

Question

This is question 2 from here:

Let $f_n \in L_1$ for all $n \in \mathbb{N}$. Which is larger, $$\left(\sum_{n=1}^\infty \left|\int f_n d\mu\right|^2\right)^{1/2}\label{1}$$ or $$\int \left(\sum_{n=1}^\infty |f_n|^2\right)^{1/2} d\mu$$

I’ve not made much progress on this question, and I’d like to ask for some help.

Taking the first expression, we have \begin{align} \left(\sum_{n=1}^\infty \left|\int f_n d\mu\right|^2\right)^{1/2} & \leq \left(\sum_{n=1}^\infty\left(\int |f_n| d\mu\right)^2\right)^{1/2}\\ & \stackrel{\text{Holder}}{\leq}\left(\sum_{n=1}^\infty\left(\int 1 d\mu\right)\left(\int |f_n|^2 d\mu\right)\right)^{1/2}\\ & = \left(\left(\int 1 d\mu\right) \left(\sum_{n=1}^\infty\int |f_n|^2 d\mu\right)\right)^{1/2}\\ & \stackrel{\text{LMCT}}{=} \left(\left(\int 1 d\mu\right) \left(\int \sum_{n=1}^\infty |f_n|^2 d\mu\right)\right)^{1/2}\\ & = \left(\int 1 d\mu\right)^{1/2} \left(\int \sum_{n=1}^\infty |f_n|^2 d\mu\right)^{1/2} \end{align} I don’t think what I have is correct. Since there is now an extra $\int 1$ term, and I have no good way of bringing the $\frac{1}{2}$ into the integral. I think at some point I need to make use of the Minkowski inequality.

There is a proof here making use of the $\ell_p$ spaces, is there any way to avoid it and show this inequality within the $L_p$ spaces?

Answer 1

In this particular case, an elementary proof is available. For each $N$, consider the vector

$$ \mathbf{f}_N = (f_1, \ldots, f_N) \in (L_1)^N, $$

and let $\| (x_1, \ldots, x_N) \|_{\mathbb{R}^N} = \sqrt{x_1^2 + \cdots + x_N^2} $ denote the Euclidean norm on $\mathbb{R}^N$. Then

$$ \left(\sum_{n=1}^{N} \left|\int f_n \, \mathrm{d}\mu\right|^2\right)^{1/2} = \left\| \int \mathbf{f}_N \, \mathrm{d}\mu \right\|_{\mathbb{R}^N},$$

whereas

$$ \int \left(\sum_{n=1}^{N} |f_n|^2\right)^{1/2} \, \mathrm{d}\mu = \int \|\mathbf{f}_N\|_{\mathbb{R}^N} \, \mathrm{d}\mu. $$

Then a standard argument shows that

$$ \left\| \int \mathbf{f}_N \, \mathrm{d}\mu \right\|_{\mathbb{R}^N} \leq \int \|\mathbf{f}_N\|_{\mathbb{R}^N} \, \mathrm{d}\mu. $$

Proof. Choose a unit vector $\mathbf{u} \in \mathbb{R}^N$ pointing to the same direction of $\int \mathbf{f}_N \, \mathrm{d}\mu$:

$$ \int \mathbf{f}_N \, \mathrm{d}\mu = \left\| \int \mathbf{f}_N \, \mathrm{d}\mu \right\|_{\mathbb{R}^N} \mathbf{u} $$ Then

\begin{align*} \left\| \int \mathbf{f}_N \, \mathrm{d}\mu \right\|_{\mathbb{R}^N} &= \left( \int \mathbf{f}_N \, \mathrm{d}\mu \right) \cdot \mathbf{u} = \int (\mathbf{f}_N \cdot \mathbf{u}) \, \mathrm{d}\mu \leq \int \|\mathbf{f}_N\|_{\mathbb{R}^N} \, \mathrm{d}\mu. \end{align*}

Now letting $N \to \infty$ proves the desired inequality.

---

Remark. This approach is not far from the one using Minkowski integral inequality, because the latter can be thought of as a special case of the inequality

$$ \left\| \int_{\mathcal{X}} f(x) \, \mu(\mathrm{d}x) \right\| \leq \int_{\mathcal{X}} \|f(x)\| \, \mu(\mathrm{d}x), $$

where $f : \mathcal{X} \to E$ is an integrable function taking values in a Banach space $E$ and $\int_{\mathcal{X}} f(x) \, \mu(\mathrm{d}x)$ is the corresponding Bochner integral. Then the Minkowski inequality corresponds to the case where $E$ is an $L^p$-space on another measure space.

Answer 2

By the general Minkowski inequality

$$\Big(\int_{\mathbb{N}}\Big|\int_Xf_n(x)\,\mu(dx)\Big|^2\,\#(dn)\Big)^{1/2}\leq\int_X\Big(\int_{\mathbb{N}}|f_n(x)|^2\,\#(dn)\Big)^{1/2}\,\mu(dx)$$

where $\#$ is the counting measure on $(\mathbb{N},2^{\mathbb{N}})$ which acts on sequences as $$\int_{\mathbb{N}}g(n)\,\#(dn)=\sum_n g(n)$$

The assumption that each $f_n\in L_1(\mu)$ implies that $n\mapsto\int_X |f_n(x)|\,\mu(dx)$ is finite.

---

A more pedestrian way to obtain inequality above is by using the finite dimensional version of the problem (see here) which yields, for any $n\in\mathbb{N}$ \begin{align} \Big(\sum^n_{k=1}\Big|\int_Xf_n\,d\mu\Big|^2\Big)^{1/2} \leq \int_X\Big(\sum^n_{k=1}|f_n|^2\Big)^{1/2}\,d\mu\tag{1}\label{one} \end{align} Then pass to the limit as $n\rightarrow\infty$ and applying monotone convergence on the right-hand-side of \eqref{one}