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 $f = F|\_{[0,1]}$



Building a smooth function from a rapidly decreasing sequence
**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.


