I need to show that a continuous function $f :\mathbb{S^1} \rightarrow \mathbb{C} $ is smooth if and only if the sequence of its Fourier coefficients $\{ \hat {f(n)}\}_{-\infty}^{+\infty}$ is a sequence of rapid decay.
Now a sequence of complex numbers $\{ a_n\}_{n=-\infty}^{\infty}$ is a sequence of rapid decay if $$ \sum_{-\infty}^{\infty} |n|^k |a_n| < \infty $$ for any $k \in \mathbb{N}\cup\{0\}$.
I don't know how to proceed with this definition. Any kind of help is appreciated!
Let $c_n(f)= \int_0^1 f(x) e^{-2i\pi n x}dx$. Clearly $|c_n(f)| \le \|f\|_\infty$
if $f$ is periodic with its derivative continuous then $f(1) = f(0)$ and integrating by parts $$c_n(f') = \int_0^{2\pi} f'(x) e^{-2i\pi n x}dx = \left.f(x) e^{-2i \pi nx}\right|_0^1+ 2i \pi n \int_0^1 f(x) e^{-2i \pi n x}dx = 2i \pi n \ c_n(f)$$ Hence if $f$ is smooth (on $S^1$) then for every $k$ : $f^{(k)}$ is continuous and (for $n \ne 0$) : $$c_n(f) = \frac{c_n(f^{(k)})}{(2i \pi n)^k} = \mathcal{O}(n^{-k})$$ i.e. $\sum_n |c_n(f) n^{k-2}|$ converges and $c_n(f)$ is rapidly decreasing.
Conversely if $c_n$ is rapidly decreasing, then for every $k$ : $f_k(x) = \sum_{n=-\infty}^\infty c_n (2i \pi n)^k e^{2 i \pi n x} $ converges uniformly, so that $f_k = f_0^{(k)}$ and it is continuous (on $S^1$), i.e. $f_0$ is smooth (on $S^1$).