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Let $f : \mathbb{S}_1 \rightarrow \mathbb{C}$ be a square-integrable fnction and let $(\widehat{f}_k)_{k \in \mathbb{Z}}$ be its Fourier-coefficients. It is very well known, that the condition $\sum\limits_{k\in \mathbb{Z}} |\widehat{f}_k| < \infty$ is sufficient for $f$ to be continuous and thus of course bounded.

My Question is whether or not the slightly weaker condition $$ \sum\limits_{k\in \mathbb{Z}} |\widehat{f}_k|^2 \, |k| < \infty $$ already suffices for $f$ to be bounded. (Maybe even continuous, but this is doubtful.)

Sorry, if I'm not seeing something obvious. Any help is much appreciated.

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  • $\begingroup$ PS. Instead of calling $f$ bounded, it would be more rigorous to say $f \in L^2(\mathbb{S}_1)$ admits a bounded representative. $\endgroup$
    – Ben Deitmar
    Commented Feb 16, 2022 at 11:10
  • $\begingroup$ The condition is close to implying $f$ is bounded. By fourier inversion, $|f(\theta)| = \left|\sum_k \widehat{f_k}e(k\theta)\right| \le \left(\sum_k |\widehat{f_k}|^2 |k|\right)^{1/2}\left(\sum_k \frac{1}{|k|}\right)^{1/2}$ by Cauchy-Schwarz. So if you change the condition to, say, $\sum_k |\widehat{f_k}|^2|k|^{1.01} < \infty$, then $f$ has to be bounded. $\endgroup$
    – mathworker21
    Commented Feb 16, 2022 at 11:40
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    $\begingroup$ Your condition means that $f$ is in the Sobolev space $H^{1/2}$, which contains unbounded functions $\endgroup$
    – Piero D'Ancona
    Commented Feb 16, 2022 at 14:52
  • $\begingroup$ Thank you, I wasn't aware Sobolev spaces can be characterized by the Fourier coefficients. $\endgroup$
    – Ben Deitmar
    Commented Feb 16, 2022 at 15:00

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