I have the following "argument" about Fourier series, which I know is wrong because it yields a ridiculous conclusion. However, I don't know where the mistake is, and need to know which step is the problem.

We consider $f \in L^2[0,2 \pi]$ and $R_f (\theta) := \int_0^{2 \pi} f(u) f(u +\theta) \, d \theta $. Here, we define $f(u+\theta) := f(u+\theta-2 \pi)$ whenever $u+\theta > 2 \pi$ so that $R_f$ is a function on the unit circle.

We rewrite $R_f (\theta)$ in terms of a convolution in the following way: $$R_f (\theta) = [f \star f(- \,\cdot) ]( -\theta).$$ Using the properties of convolutions, this shows that $R_f \in L^\infty[0,2 \pi]$ In particular, $R_f$ is $L^2$. Taking the Fourier transform of $R_f$ and using the convolution theorem, we find the following: $$ \mathcal{F}[R_f](n) = \mathcal{F}[f](n)\cdot \overline{ \mathcal{F}[f](n)} = |\mathcal{F}[f](n)|^2 $$

However, by the Plancherel theorem, $\|R_f \|_2 = \| \mathcal{F}[R_f] \|_2$. Therefore, this implies that $\sum_n |\mathcal{F}[R_f](n)|^2 < \infty $ and hence that $\sum_n |\mathcal{F}[f](n)|^4 < \infty$.

However, an arbitrary square integrable sequence is not necessarily $L^4$, so clearly something is wrong here.I'm not sure what the fishy part is, and I was wondering if anyone had some pointers.

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    $\begingroup$ @AlexandreEremenko: On a compact group, I think it does...? The assertion that $f \ast f \in L^\infty$ follows from Cauchy-Schwartz (or Young's inequality), and $L^\infty \subset L^2$. $\endgroup$ – Nate Eldredge Oct 1 '18 at 14:21

No contradiction; an arbitrary square integrable sequence is $\ell^4$. The sequence is decaying to zero, so its fourth powers decay faster. Indeed, we have $\ell^p \subset \ell^q$ for any $q > p$.

  • $\begingroup$ Ah, right. I was getting the inclusion backwards for sequences. I'm so used to functions on compact spaces where $L^p$ norms get more restrictive as $p$ increases. Thanks. $\endgroup$ – Gabe K Oct 1 '18 at 17:18

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