## Return to Answer

3 minor changes.

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, $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(-x)}e^{-inx}dx=\sum_{k=-\infty}^\infty f(x)\overline{h(x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{h}(n-k)$$ hat{g}(n-k) where $h(x)=\overline{g(-x)}$. 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 an a surjective isomorphic isometry.)

2 added 78 characters in body

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 $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{g(-x)}e^{-inx}\,dx=\sum_{k=-\infty}^\infty f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(-x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k).$$ hat{h}(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 is an isomorphic isometry.)

1

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 $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{g(-x)}e^{-inx}\,dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k).$$

(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence is an isomorphic isometry.)
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