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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

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    $\begingroup$ Go to the Fourier side and you'll see everything that there is to see about this question there. $\endgroup$
    – fedja
    Oct 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. Swail
    Oct 27, 2021 at 11:26

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Let $\xi_n=\pm 1$ be arbitrary signs. You want to show that $$\forall\epsilon>0\hspace{3mm}\exists N\in\mathbb{N}\hspace{3mm}\forall m\geq n\geq N\hspace{3mm}\|\sum_{n\leq|k|\leq m}\xi_ka_kf(\cdot-k)\|_{L^2}<\epsilon. $$ By Parseval theorem, $$\|\sum_{n\leq|k|\leq m}\xi_na_nf(\cdot-n)\|_{L^2}^2 = \|\widehat{f}(\cdot)\sum_{n\leq|k|\leq m}\xi_ka_ke^{2\pi ik\cdot}\|_{L^2}^2 $$ At this point, you may consider cases. For example, if the function $F\in L^1([0,1])$ defined by $F(\gamma):=\sum_{n\in\mathbb{Z}}|\widehat{f}(\gamma-n)|^2$ is bounded, then the RHS of the above $(=)$ is equal to $$\int_0^1 F(\gamma)\left|\sum_{n\leq|k|\leq m}\xi_na_ne^{2\pi ik\gamma}\right|^2 d\gamma \leq \|F\|_{L^{\infty}([0,1])}\hspace{2mm} \sum_{n\leq|k|\leq m}|\xi_ka_k|^2.$$ 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 $f$.

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