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.
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 $L^1(T)$.