# A question about the convolution theorem

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.

• @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 $L^\infty \subset L^2$. – Nate Eldredge Oct 1 '18 at 14:21

## 1 Answer

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 $$q > p$$.

• 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. – Gabe K Oct 1 '18 at 17:18