TITLE: Showing a series is finite a.e. given an $L^p$ function.
QUESTION [7 upvotes]: Q: Suppose $f \in L^2 (\mathbb{R})$. Show that $\sum_{n = 1}^\infty \frac{f(x + n)}{n^{3/4}}$ is finite a.e. $x \in \mathbb{R}$.
It seems that we have to show the sum is in $L^p (\mathbb{R})$ for some $p \geq 1$. Let's try $p = 2$, applying Cauchy-Schwarz inequality, we have
$$\int \left|\sum_{n = 1}^\infty \frac{f(x + n)}{n^{3/4}}\right|^2 dx \leq \left(\sum_{n = 1}^\infty \frac{1}{n^{3/2}}\right) \int \left(\sum_{n = 1}^\infty (f(x + n))^2\right) dx = C \sum_{n = 1}^\infty \|f\|_2^2 = \infty .$$
It seems that this estimate is too large.
For $p = 1$, it seems that a similar situation occurs when CS-inequality is applied.
I am thinking if we can show
$$\int \left|\sum_{n = 1}^\infty \frac{f(x + n)}{n^{3/4}}\right| dx \leq \sum_{n = 1}^\infty \int \frac{[f(x + n)]^2}{n^{3/2}} dx,$$
then we are done.
However I have no idea how to get this (or a similar) estimate.

REPLY [6 votes]: Define $G : \Bbb{R} \to [0,\infty], x \mapsto \sum_{n \in \Bbb{Z}} |f(x+n)|^2$. I first claim that $G(x) < \infty$ almost everywhere. Since $G$ is $1$-periodic, it suffices to show this for almost all $x \in [0,1)$. Now, apply the periodization trick to see
$$
\infty
> \int_{\Bbb{R}} |f(y)|^2 \, d y
= \sum_{n \in \Bbb{Z}} \int_{n + [0,1)} |f(y)|^2 d y
= \sum_{n \in \Bbb{Z}} \int_{[0,1)} |f(x+n)|^2 \, d x
= \int_{[0,1)} G(x) \, d x,
$$
where the interchange of integration and summation is justified by the monotone convergence theorem.
The preceding calculation easily implies that indeed $G(x) < \infty$ for almost all $x \in [0,1)$ and hence for almost all $x \in \Bbb{R}$.
Now, finally note by the Cauchy-Schwarz inequality that if $G(x) < \infty$, then
$$
\sum_{n=1}^\infty |f(x+n)| n^{-3/4}
\leq \sqrt{\sum_{n=1}^\infty |f(x+n)|^2} \sqrt{\sum_{n=1}^\infty n^{-3/2}}
\leq \sqrt{G(x)} \cdot \sqrt{\sum_{n=1}^\infty n^{-3/2}}
< \infty.
$$
This easily implies the claim.