Is it possible to build a 1-priodic smooth function from a rapidly decreasing sequence such that the sequence be the Fourier coefficients of the function? More precisely:
Let $\lbrace c_k\rbrace _{k \in \mathbb{Z}}$ be rapidly decreasing sequence. Is it possible to show that there is a 1-priodic smooth function $F: \mathbb{R}\rightarrow \mathbb{C}$ such that $\lbrace c_k\rbrace$ is the sequence of the Fourier coefficients of $f = F|_{[0,1]}$
Yes. (Assuming that that rapidly decreasing means that $\sup_{k\in \mathbb Z}\vert k^m c_k \vert <\infty$ for all $m=0,1,2,\dots$.)
The function must be $f(t)=\sum_{k\in \mathbb Z} c_k \exp(2\pi i k t)$.
Just write $f_N(t)=\sum_{\vert k \vert \le N} c_k \exp(2\pi i k t)$. As $\vert c_k\vert = O(k^{-2})$, this the sequence $(f_N)$ is Cauchy in $L^2([0,1])$ and thus $f=\lim_{N\rightarrow \infty}f_N$ exists in $L^2([0,1]) $. Clearly $f$ is $1$-periodic and has the correct Fourier coefficients. I claim that it is also smooth.
Taking derivatives of $f_N$ produces powers of $k$, but they do not affect summability, as $c_k$ is rapidly decreasing. Hence the $k$th derivative $f_N^{(k)}$ is also Cauchy in $L^2$. Thus $(f_N)$ is Cauchy in the Sobolev space $H^k([0,1])$. This implies that $f\in H^k$ for every $k\in \mathbb N$ and hence it must be smooth.
