The problem goes as follows:

Let $r_{n}$ be an enumeration of the rationals. Using Fubini-Tonelli, show that:

$$ F(x) = \sum_{n \geq 1} \frac{1}{n^{2}} \frac{1}{|r_{n} - x |^{1/2}}$$

is finite almost everywhere.

Attempt at solution:

My idea is that I want to show the $F(x)$ is integrable. Then using Fubini-Tonelli, taking the sum outside of the integral:

$$ \int_{\mathbb{R}}F(x) dx = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{|r_{n} - x |^{1/2}} dx$$ $$ = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{|x |^{1/2}}dx $$by traslation invariance.

Now I know that $\frac{1}{|x |^{1/2}}$ locally integrable, but not integrable on all of $\mathbb{R}$.

I was also thinking of defining $f_{n}(x) = \frac{1}{|x |^{1/2}}$ when $ x$ in $(-n,n)$ and $0$ elsewhere. Then let $$F_{N}(x) = \sum_{n=1}^{N} \frac{1}{n^{2}} f(x-r_{n}) $$

My reasoning starts to fall apart here and I can't seem to make it work.

Any hints or solutions are greatly appreciated!

  • 2
    $\begingroup$ maybe try to integrate over any $(a,b)$ instead of over $\mathbb{R}$. That is, $$\int_a^b F(x) dx\quad \text{ instead of }\quad \int_\mathbb{R} F(x) dx.$$ If you can show that $\int_a^b F $ is finite for any $(a,b)$, it implies $F$ is finite a.e. $\endgroup$ – Xiao Oct 29 '16 at 3:12

Set $f(x)=\frac{1}{|x|^{1/2}}$. Then $$F(x)=\sum_n \frac{1}{n^2} f(x-q_n).$$ You can decompose $F$ in two parts: one summing the tails of $f$, and one summing the spikes. Namely set $g(x)=f(x)\chi_{[-1,1]}$ and $h(x)=f(x)-g(x)=f(x)\chi_{[-1,1]^c}$. Then $$F(x)=\sum_n \frac{1}{n^2}g(x-q_n)+\sum_n \frac{1}{n^2}h(x-q_n)$$ and to the first sum you can apply your argument, while the second sum is uniformly bounded because $\|h\|_\infty=1$.

  • $\begingroup$ Thank you for this answer. I understand that $h$ is uniformly bounded, but isn't the integral of $h$ still infinite? $\endgroup$ – blueyesister Oct 29 '16 at 20:22
  • 1
    $\begingroup$ @blueyesister Yes it is, but don't take the integral of it. To prove that $F$ is finite a.e. it suffices to show that this is the case for both sums. As for the first sum, take the integral and apply the reasoning you cited in the question. As for the second sum, it is finite everywhere since $\|h\|_\infty=1$ and $\sum_n \frac{1}{n^2}<\infty$. Therefore their sum is finite a.e. $\endgroup$ – Del Oct 29 '16 at 20:35
  • $\begingroup$ Ok, I see... But I don't see how the second sum is finite everywhere. Doesn't the function $h(x-q_{n})$ have spikes at every rational outside $[-1,1]$? $\endgroup$ – blueyesister Oct 29 '16 at 21:04
  • 1
    $\begingroup$ @blueyesister No, $h$ is the function $\frac{1}{|x|^{1/2}}\chi{[-1,1]^c}$, which is everywhere between $0$ and $1$. Then we sum the pieces $\frac{1}{n^2}h(x-q_n)$, which are just translated and rescaled versions of $h$, thus they don't have spikes. Maybe you were thinking of summing among all the rationals outside $[-1,1]$, but this is not what we are doing here $\endgroup$ – Del Oct 29 '16 at 21:42
  • $\begingroup$ Ok, I understand now. Thanks so much. $\endgroup$ – blueyesister Oct 29 '16 at 21:58

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