Question: Let $f\in L^1(\Bbb R)$. For which increasing sequences $\{a_k\}_{k=1}^\infty$ of positive real numbers does $$\sum_{k=1}^\infty \frac{1}{a_k} f(x+a_k)$$ converge absolutely for almost every $x\in \Bbb R$?
I believe this might be true for many such sequences, with possible restrictions on how fast $a_k$ grows to infinity. I have sketched the proof for $a_k = \sqrt{k}$ below, which I shall try to generalize. The key step seems to involve the "inverse" sequence. Let $a_k = \phi(k)$ for $k\in \Bbb N$, where $\phi:[1,\infty) \to [1,\infty)$ is a strictly increasing surjective function. Then, $\phi^{-1}: [1,\infty) \to [1,\infty)$ exists. I refer to $b_k = \phi^{-1}(k)$ as the "inverse" sequence of $a_k.$
Toy Case: One can do the following if $a_k = \sqrt k$. It is enough to show that the series converges a.e. on $[n,n+1]$, for all $n\in \Bbb Z$. WLOG, let $n=0$. Now, it is enough to show $$\int_0^1 \sum_k \frac{1}{\sqrt k} |f(x+\sqrt k)|\,dx < \infty$$ as $\int_0^1 |g| < \infty \implies g$ is finite a.e. on $[0,1]$. Using the Monotone Convergence Theorem, it is enough to show $$ \sum_k \frac{1}{\sqrt k} \int_0^1|f(x+\sqrt k)|\,dx < \infty.$$ For fixed $m \in \Bbb N$ and $m^2 \le k < (m+1)^2$, $$\frac{1}{\sqrt k} \int_0^1|f(x+\sqrt k)|\,dx \le \frac{1}{m} \sup_{y\in [m, m+1)} \int_0^1 |f(x+y)|\, dx \le \frac1m \int_m^{m+2} |f(x)|\, dx$$ so that $$ \begin{align*} \sum_k \frac{1}{\sqrt k} \int_0^1|f(x+\sqrt k)|\,dx &\le \sum_{m=1}^\infty \sum_{k=m^2}^{(m+1)^2-1} \frac1m \int_m^{m+2} |f(x)|\, dx\\ &= 2 \sum_{m=1}^\infty \int_m^{m+2} |f(x)|\, dx + \sum_{m=1}^\infty \frac 1 m \int_m^{m+2} |f(x)|\, dx\\ &\le 3 \sum_{m=1}^\infty \int_m^{m+2} |f(x)|\, dx\\ &\le 6\|f\|_1 < \infty. \end{align*}$$
Generalization: I propose the following hypotheses on $\{a_k\}_{k=1}^\infty$. Let $\phi:[1,\infty) \to [1,\infty)$ be a strictly increasing surjective function, and $a_k := \phi(k)$ for all $k\in \Bbb N$. Let $\{b_k\}$ be the inverse sequence of $a_k$, defined using $\phi^{-1}:[1,\infty) \to [1,\infty)$. Lastly, assume there exists $\beta > 0$ such that $b_{k+1} - b_k \le \beta k$ for all $k\in \mathbb N$.
Once again, it is enough to show that the series converges a.e. on $[n,n+1]$, for all $n\in \Bbb Z$. WLOG, let $n=0$. Now, it is enough to show $$\int_0^1 \sum_k \frac{|f(x+a_k)|}{a_k} \,dx < \infty$$ as $\int_0^1 |g| < \infty \implies g$ is finite a.e. on $[0,1]$. Using the Monotone Convergence Theorem, it is enough to show $$ \sum_k \frac{1}{a_k} \int_0^1|f(x+a_k)|\,dx < \infty.$$ For fixed $m \in \Bbb N$ and $b_m \le k < b_{m+1}$, we have $m = \phi(b_m) \le a_k < \phi(b_{m+1}) = m +1$. $$\frac{1}{a_k} \int_0^1|f(x+a_k)|\,dx \le \frac{1}{m} \sup_{y\in [m, m+1)} \int_0^1 |f(x+y)|\, dx \le \frac1m \int_m^{m+2} |f(x)|\, dx$$ so that $$\sum_k \frac{1}{a_k} \int_0^1|f(x+a_k)|\,dx \le \sum_{m=1}^\infty \sum_{k=b_m}^{b_{m+1}-1} \frac1m \int_m^{m+2} |f(x)|\, dx = \sum_{m=1}^\infty \frac{b_{m+1}-b_m}{m} \int_m^{m+2} |f(x)|\, dx .$$ As $b_{m+1} - b_m \le \beta m$ for each $m\ge 1$, we have $$\sum_k \frac{1}{a_k} \int_0^1|f(x+a_k)|\,dx \le \beta \sum_{m=1}^\infty\int_m^{m+2} |f(x)|\, dx \le 2\beta \|f\|_1 < \infty.$$
Other Thoughts:
- In an ideal world, I'd like to know necessary and sufficient conditions on $\{a_k\}_{k=1}^\infty$ so that $\sum_{k=1}^\infty \frac{1}{a_k} f(x+a_k)$ converges absolutely for almost every $x\in \Bbb R$.
- For sequences growing slower than $\sqrt k$, e.g. $a_k = k^{1/3}$, the "inverse" sequences grow faster. In this case, $b_k = k^3$, and there is no constant $\beta > 0$ such that $b_{k+1} - b_k \le \beta k$ for all $k \in \Bbb N$. As a result, I'd expect the series $$\sum_{k=1}^\infty \frac{1}{k^{1/3}} f(x+k^{1/3})$$ to not converge absolutely on a set of positive measure. I do not know if this is true.
Thank you!