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. $\endgroup$
    – 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

1 Answer 1


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 $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.