Suppose that $f \in L^2(\mathbb R)$ is an arbitrary square integrable function and $(a_n)_{n \in \mathbb Z}$ is a sequence which is square summable, i.e. $(a_n)_{n \in \mathbb Z} \in \ell^2(\mathbb Z)$. Does the series of translates of $f$, $$ \sum_{n=\infty}^\infty a_n f(\cdot  n) $$ always converges in $L^2(\mathbb R)$ to a square integrable function? If yes, does it converge unconditionally? If no, can we make additional properties on the sequence $(a_n)_n$ so that the series converges? References to papers are very welcome

9$\begingroup$ Go to the Fourier side and you'll see everything that there is to see about this question there. $\endgroup$– fedjaOct 25, 2021 at 12:21

$\begingroup$ Thank you for the comment. Am I right to assume that the answer to the above question (regarding the convergence assuming that the sequence is square summable) is "yes"? $\endgroup$– J. SwailOct 27, 2021 at 11:26
1 Answer
Let $\xi_n=\pm 1$ be arbitrary signs. You want to show that $$\forall\epsilon>0\hspace{3mm}\exists N\in\mathbb{N}\hspace{3mm}\forall m\geq n\geq N\hspace{3mm}\\sum_{n\leqk\leq m}\xi_ka_kf(\cdotk)\_{L^2}<\epsilon. $$ By Parseval theorem, $$\\sum_{n\leqk\leq m}\xi_na_nf(\cdotn)\_{L^2}^2 = \\widehat{f}(\cdot)\sum_{n\leqk\leq m}\xi_ka_ke^{2\pi ik\cdot}\_{L^2}^2 $$ At this point, you may consider cases. For example, if the function $F\in L^1([0,1])$ defined by $F(\gamma):=\sum_{n\in\mathbb{Z}}\widehat{f}(\gamman)^2$ is bounded, then the RHS of the above $(=)$ is equal to $$\int_0^1 F(\gamma)\left\sum_{n\leqk\leq m}\xi_na_ne^{2\pi ik\gamma}\right^2 d\gamma \leq \F\_{L^{\infty}([0,1])}\hspace{2mm} \sum_{n\leqk\leq m}\xi_ka_k^2.$$ Thus, the series of translates converges unconditionally. Generally, if the Fourier series $$\sum_{k\in\mathbb{Z}}\xi_ka_ke^{2\pi ik\cdot}$$ converges in the weighted $L^2$ space with the norm $$\g\=\left(\int_0^1 g(x)^2F(x)\ dx\right)^{1/2}$$ for all possible choice of signs $\xi_k=\pm1$, then the series of translates converges unconditionally.
On the contrary, let $f\in L^2$ such that $$\int_0^1 \widehat{f}^4 = \infty.$$ Expand the Fourier series of the restriction of $\widehat{f}$ to the interval $[0,1]$, and let $(a_n)$ be its Fourier coefficients. Then, the series of translates diverges for these choices of $(a_n)$ and $f$.