Let $c_0$ be the Banach space of doubly infinite sequences $$\lbrace
a_n: -\infty\lt n\lt \infty, \lim_{|n|\to \infty} a_n=0 \rbrace.$$ Let $T$ be the space of $2\pi$ periodic functions integrable on $[0,2\pi]$.
Let $$S=\lbrace \lbrace a_n\rbrace \in c_0: a_n=\hat{f}(n) \forall n \mbox{ for some function } f\in T\rbrace,$$
where $\hat{f}(n)$ denotes the $n$-th Fourier coefficient of $f$, i.e. $$\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,dx.$$
When I was a graduate student, I was told that no known characterizations of $S$ were known. Is this still true?
Best Answer
This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.
The linear subspace $S$ of $c_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.
More precisely, $\lbrace a_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b_n \rbrace$ and $\lbrace c_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a_n=\sum_{k=-\infty}^\infty b_k c_{n-k} $$ for all $n$.
This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has, by Parseval identity, $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k)$$ where $h(x)=\overline{g(x)}$.
(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence in $\ell^2(\mathbb{Z})$ is a surjective isomorphic isometry.)