I'm studying for a qualifying exam and having difficulty showing the following:
Let $f\in L^1(\mathbb{R})$. Show that $\sum_{n=1}^\infty n^{-1/2} f(x-\sqrt{n})$ converges absolutely for almost every $x\in\mathbb{R}$.
I understand that it suffices to show that $\int_\mathbb{R} n^{-1/2}f(x-\sqrt{n})\,dx$ is summable, but I can't seem to show this. Is this even the correct approach?
Assume wlog that $f\ge0$. It's enough to show that $$\sum n^{-1/2}\int_a^bf(x-\sqrt n)\,dx<\infty.$$ By monotone convergence (and translation-invariance of Lebesgue measure) that's the same as $$\int_{-\infty}^\infty f(x)\phi(x)\,dx,$$where $$\phi=\sum n^{-1/2}\chi_{[a-\sqrt n,b-\sqrt n]}.$$So you need only show that $\phi\in L^\infty$. Which it is.
I'll leave the details of that to you; special case this negative that positive something else and you find $A(x)$ and $B(x)$ such that $$\phi(x)=\sum_{n=A(x)}^{B(x)}n^{-1/2},$$you estimate that sum by an integral and you're done.