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Let $f : [0,2 \pi)^d \rightarrow \mathbb{R}$ be a square-integrable periodic function in $L^2( [0,2 \pi)^d )$ with $d \geq 1$. We assume moreover that the square-summable Fourier coefficients of $f$, denoted by $c_n(f)$ for $n \in \mathbb{Z}^d$, are such that:

  1. $c_n(f)$ is real, strictly positive, with $c_{n}(f) = c_m(f)$ for any $n,m \in \mathbb{Z}^d$ with equal Euclidian norms $\lVert n \rVert = \lVert m \rVert$,
  2. the sum $\sum_{n \in\mathbb{Z}^d} c_n(f) = \infty$, and
  3. the $c_n(f)$ are decreasing in the sense that $c_n(f) \leq c_m(f)$ as soon as $\lVert n \rVert \geq \lVert m \rVert$.

Can we prove that $f$ is necessarily discontinuous and even unbounded at $0$ with these assumptions?

Context: I am especially interested by the situation where $c_n(f)$ behaves asymptotically like (or even is equal to for $n \neq 0$) $\lVert n \rVert^{- \alpha}$ for some $\alpha \in (d/2, d]$. The case $d = 1$ with $2\pi$ periodic functions is already interesting and may be quite simpler.

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    $\begingroup$ At least in the context situation, this sounds right, since at least in the continuous case $|t|^{-\alpha}$ has FT $c|x|^{\alpha -d}$, and the perturbation has a smooth FT, so won't affect the behavior at $x=0$. $\endgroup$ Jun 1, 2020 at 14:14
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    $\begingroup$ I am not sure to understand: what is the perturbation you are talking about? Do you know if there is anything like the result $\mathcal{F} \{ \lvert t \rvert^{-\alpha} \}(x) = c \lVert x \rvert^{\alpha - d}$ for the periodic setting? (I realized there was a mistake in the context for the parameter $\alpha$, I corrected it.) $\endgroup$
    – Goulifet
    Jun 1, 2020 at 14:20
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    $\begingroup$ The perturbation in my comment refers to $d_n$, if you write $c_n=|n|^{-\alpha}+d_n$. $\endgroup$ Jun 1, 2020 at 15:08
  • $\begingroup$ I see, the comparison with the continuous case is effectively interesting. $\endgroup$
    – Goulifet
    Jun 1, 2020 at 15:10

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