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?

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    $\begingroup$ Well, what does "characterisation" count as? S is just the Fourier algebra of $\mathbb Z$, which is the convolution of two $\ell^2(\mathbb Z)$ functions. But that follows rather immediately from it also being the Fourier transform of $L^1(\mathbb T)$. $\endgroup$ Commented Nov 19, 2010 at 13:39
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    $\begingroup$ The Fourier transform takes $L^2(\mathbb T) \rightarrow \ell_2(\mathbb Z)$ unitary (if you normalise the measure on $[0,2\pi]$ correctly). Every $L^1(\mathbb T)$ function is the pointwise product of two $L^2(\mathbb T)$ functions, and the fourier transform converts pointwise product to convolution, so that's the proof! (Maybe a few $\epsilons$s and $\delta$s are needed to make this 100% rigorous). $\endgroup$ Commented Nov 19, 2010 at 14:22
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    $\begingroup$ The product of two (periodic) $L^2$ functions is an $L^1$ function, and conversely every $L^1$ function is the square of a (complex) $L^2$ function. And multiplying functions corresponds to convolving Fourier coefficients. Is that it?. $\endgroup$ Commented Nov 19, 2010 at 14:27
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    $\begingroup$ The statement that the Fourier transform of $L^1(T)$ cannot be characterized as a subspace of $c_0(Z)$ has always struck me as something of a folk principle rather than a precise theorem. I think there is a theorem in this direction which says that there exists a sequence $(a_n)$ which is the Fourier series of an integrable function, but where $(|a_n|)$ is not the Fourier series of any integrable function -- however I'd have to check this as I've forgotten the example. $\endgroup$
    – Yemon Choi
    Commented Nov 19, 2010 at 15:16
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    $\begingroup$ Provided $f,g$ are in $L^2$, not in $L^1$, that first expression is correct. You can even see it as a consequence of Parseval's lemma, if you group the integrand as $f(x)$ and $g(x)e^{-inx}$ and use that phase rotation in physical space is the same as translation in frequency space... $\endgroup$ Commented Nov 19, 2010 at 16:40

1 Answer 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, 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.)


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