(Tao Analysis 2, P.125, Q.5.5.3) If $f \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$ and $P$ is a trigonometric polynomial, show that $$\widehat{f*P}(n) = \hat{f}(n)c_n = \hat{f}(n)\hat{P}(n)$$ for all integers $n$. More generally, if $f, g \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$, show that $$\widehat{f*g}(n) = \hat{f}(n)\hat{g}(n)$$ for all integers $n$.
$\hat{f}: \mathbb{Z} \to \mathbb{C}$ is the Fourier transform of $f$ defined by $\hat{f}(n) : = \langle f, e_n \rangle$, where $e_n(x) : = e^{2\pi i n x}$. I denote the convolution operation by $*$.
Let $P(x) = \sum_{m= -M}^M c_m e_m(x)$ for some $c_m \in \mathbb{C}$.Then, $$\widehat{f*P}(n) =\langle f*P, e_n \rangle = \sum_{m= -M}^M c_m \langle f * e_m, e_n \rangle = \sum_{m= -M}^M c_m \hat{f}(m) \langle e_m, e_n \rangle = c_n \hat{f}(n).$$
Similarly, I can show that $c_n = \hat{P} (n)$. I am unsure about the second part. I know that by Weierstrass approximation theorem, for any $g \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$, there exists a trigonometric polynomial $P$ such that for every $\epsilon$, $||g - P||_{\infty} \le \epsilon$. Therefore, intuitively, by the first part of the exercise, the second part also holds. But, I don't know how to prove it rigorously. I would appreciate if you give some help.