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I would like to show that $$\sum_{n\ge0}\left\vert \frac{1}{\sqrt n} f \left(x-\sqrt n \right)\right\vert \tag{$*$}$$ converges for almost every (a.e.) $x$.

The only technique I have is based on the answer to If $f\in L^1(\mathbb{R})$, then $\sum_{n\ge 1}f(x+n)$ Converges for a.e. $x$., but I think some additional technique is needed. In that question I was hoping for a technique general enough to answer both questions, but the $\sqrt \cdot$ is throwing me off. Here's what I have:

Given an integer $k$, $\int_k^{k+1}\sum_{n\ge0}\vert \frac{1}{n}f(x-\sqrt n)\vert=\sum_{n\ge 0}\int_k^{k+1}\frac{1}{\sqrt n}\vert f(x-\sqrt n)\vert$. The idea is to show that this integral is finite (implying the sum $(*)$ converges on $(k,k+1)$) and use the fact that $k$ is arbitrary to conclude that the sum converges a.e. on $\mathbb{R}$. A naive attempt would be to bound the RHS by $\sum_{n\ge 0}\int_k^{k+1}\vert f(x-\sqrt n)=\sum_{n\ge 0}\int_{-\sqrt n - k}^{-\sqrt n - k +1}\vert f(x) \vert$, but the latest expression is larger than $\int_{-\infty }^\infty\vert f\vert$.

I don't have a nice way to (a) deal with the $1/\sqrt n$ and (b) deal with the limits of integration, but they can probably be dealt with simultaneously I just don't see how.

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    $\begingroup$ Hint: Without loss of generality, note that $$\sum_{n\geqslant1}\int_0^1\frac{1}{\sqrt n}g(\sqrt n-x) dx=\int_{-\infty}^\infty S(t)g(t) dt$$ where $$S(t)=\sum_{n\geqslant1}\frac1{\sqrt{n}}\,[t^2\lt n\lt (t+1)^2],$$ and show that $S(t)$ is uniformly bounded (by $2$, give or take). $\endgroup$ – Did Feb 8 '15 at 9:24
  • $\begingroup$ I can see that $S(t)<2$ for all $t$, but can you expand on the first equality? $\endgroup$ – The Substitute Feb 8 '15 at 11:09
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    $\begingroup$ Just exchange the sum and the integral and perform the change of variable $t=\sqrt{n}-x$ in each integral on $(0,1)$. (But, in the RHS, the integral over $(-\infty,\infty)$ should be over $(0,\infty)$, sorry about that. $\endgroup$ – Did Feb 8 '15 at 11:35

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