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If $ \sum_{n=-\infty} ^{\infty} c_n e^{i n t}$ is the Fourier series of $f$, does it always exist $g(t) \in \mathbb{L}^1(\mathbb{T})$, with $g(t) = \sum_{n=-\infty} ^{\infty} c_n^2 e^{i n t}$

I know that $$c_n = \frac{1}{2 \pi} \int_{-\pi} ^{\pi} f(t) e^{-int}dt$$ therefore one has to show that $$ \sum_{n=-\infty} ^{\infty} \left(\frac{1}{2 \pi} \int_{-\pi} ^{\pi} f(t) e^{-int}dt\right)^2e^{-int}$$ is a Fourier series. Don't have a clue on how to move forward.

Ayoub
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pendermath
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1 Answers1

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I assume we know something about the regularity $f$. Like, it's continuous or at least in $L^2(T)$ ?

It this is the case the result follows from Parseval's identity. Since $\sum_{n\in\mathbb{Z}} c_ne^{int}$ is the Fourier series of $f$, we know that $\|f\|_2= \sum_{n\in\mathbb{Z}} |c_n|^2$. In particular, $(c_n)\in L^2(\mathbb{Z})$.

But now, we're done because then $\sum c_n^2e^{int}$ is normally convergent. Its sum, $g$, is therefore continuous and therefore in $L^1(T)$.

Ayoub
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    I guess the OP meant $f\in L^1$ in which case $g=f\ast f$ is $L^1$ as well. With $f = pv(1/\sin(t))$ which is not $L^1$ then $g=\delta$ is not a function. – reuns Apr 13 '21 at 11:34
  • @reuns, How does it follow that $g$ is $\delta$ in your example ? – Ayoub Apr 13 '21 at 11:37
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    The Fourier coefficients of $pv(1/(e^{it}−1))$ are a constant times $sign(n)$, it follows from expansion of $\frac{1}{re^{it}-1}+\frac1{r^{-1} e^{it}-1}$ in Fourier (geometric series) and letting $r\to 1^-$. – reuns Apr 13 '21 at 12:41