Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue integrable function. Prove that $$\sum_{n=1}^{\infty} \frac{f(x-\sqrt{n})}{\sqrt{n}}$$ converges almost for every $x \in \mathbb{R}$.

My tactic here (which of course might lead me to nowhere) is to express this series as an integral or sum of integrals and use in some way the fact that $f$ is integrable along with a convergence theorem of course.

From integrability of $f$ we know that $f$ is finite almost everywhere which I believe would help me in my proof.

But I don't know exactly how to start with this.

Any useful hint would be appreciated. I don't want a full solution to this.

Thank you in advance.

  • $\begingroup$ We can prove that the above trigonometric series converges for all $s\neq 0$ using Dirichlet's test. $\endgroup$
    – Michael L.
    Aug 23, 2017 at 2:39
  • $\begingroup$ @JackD'Aurizio You're right, I worked through the unboundedness of the partial sums as done here. However, I don't see why this shouldn't contradict the convergence of $\sum_{n\geq 1} e^{is\sqrt{n}}/\sqrt{n}$ at all $s$. $\endgroup$
    – Michael L.
    Aug 23, 2017 at 3:01
  • $\begingroup$ As best as I can tell, it isn't convergent at any $s\in \mathbb{R}$, as $\sum_{n\geq 1} \cos(s\sqrt{n})/\sqrt{n}$ diverges everywhere. $\endgroup$
    – Michael L.
    Aug 23, 2017 at 3:04
  • $\begingroup$ Mumble mumble... so the Fourier transform is probably not the best approach, or a working approach at all... $\endgroup$ Aug 23, 2017 at 3:06
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    $\begingroup$ By a change of variable $u=\sqrt t$ we have that $$\int_0^{\infty} \frac{|f(x-\sqrt t)|}{\sqrt t} dt = 2 \int_0^{\infty} |f(x-u)|du \leq 2 \|f\|_{L^1}$$ Hence the associated integral is easily shown to be absolutely convergent. If we could somehow relate this integral to the associated sum, then life would be great. @JackD'Aurizio Do you know if there is there a rigorous way to do this? For instance, this would also give a straightforward way to show finiteness of sums of the form $$\sum \frac{f(x-n^{1/3})}{n^{2/3}}$$ $\endgroup$
    – shalop
    Aug 23, 2017 at 10:51

1 Answer 1


A brute-force method: Define $g(x):= \sum_0^{\infty} \frac{|f(x-\sqrt n)|}{\sqrt n}$, which a priori might be infinite at many points. Fix some $j \in \Bbb Z$. By Tonelli's theorem we may interchange sum and integral to compute \begin{align*}\int_j^{j+1} g(x)dx &=\sum_{n=1}^{\infty} \int_j^{j+1}\frac{|f(x-\sqrt n)|}{\sqrt n} dx \\ &=\sum_{n=1}^{\infty} n^{-1/2}\int_{j-\sqrt n}^{j+1-\sqrt n} |f(x)|dx \\ &=\sum_{k=1}^{\infty} \sum_{n=k^2}^{(k+1)^2-1} n^{-1/2}\int_{j-\sqrt n}^{j+1-\sqrt n} |f(x)|dx\\&= \sum_{k=1}^{\infty} \int_{\Bbb R}\bigg(\sum_{n=k^2}^{(k+1)^2-1}n^{-1/2}\cdot1_{[j-\sqrt n,\;j+1-\sqrt n)}(x) \bigg)|f(x)|dx\end{align*} Now if $k^2 \leq n <(k+1)^2$, then $n^{-1/2} \leq k^{-1}$ and also $[j-\sqrt n,j+1-\sqrt n) \subset [j-k-1,j-k+1)$.

Therefore we find that \begin{align*}\sum_{n=k^2}^{(k+1)^2-1}n^{-1/2}\cdot 1_{[j-\sqrt n,\;j+1-\sqrt n)}(x) &\leq \big[(k+1)^2-k^2\big]\cdot k^{-1} \cdot 1_{[j-k-1,j-k+1)}(x) \\ &=(2k+1) \cdot k^{-1} \cdot 1_{[j-k-1,j-k+1)}(x) \\ & \leq 3 \cdot 1_{[j-k-1,j-k+1)}(x)\end{align*} Consequently, \begin{align*}\int_j^{j+1} g(x)dx & \leq 3\sum_{k=1}^{\infty} \int_{j-k-1}^{j-k+1}|f(x)|dx \\ & = 3\sum_{k=1}^{\infty} \int_{j-k-1}^{j-k}|f(x)|dx+3\sum_{k=1}^{\infty} \int_{j-k}^{j-k+1}|f(x)|dx \\ &= 3\int_{-\infty}^{j-1} |f(x)|dx + 3\int_{-\infty}^j |f(x)|dx \\ &\leq 6 \|f\|_{L^1}<\infty\end{align*} We conclude that $g(x)<\infty$ for a.e. $x \in [j,j+1]$. But $j\in \Bbb Z$ was arbitrary, so we find that $g(x)<\infty$ for a.e. $x \in \Bbb R$. With a little more effort, this argument may be generalized to show that $\sum_n n^{\alpha-1}|f(x-n^{\alpha})|<\infty$ for any $\alpha \in (0,1]$. This was the special case $\alpha=\frac{1}{2}$.

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    $\begingroup$ @MariosGretsas Added a spoiler. $\endgroup$
    – shalop
    Aug 23, 2017 at 10:45
  • $\begingroup$ @MariosGretsas Then read just the beginning! $\endgroup$
    – zhw.
    Aug 24, 2017 at 15:50

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