Given a finite Borel measure $\mu$ on $\mathbb{S}^1 = \mathbb{R}/\mathbb{Z}$, define its Fourier coefficients by

$$ \hat\mu(n) = \int e^{2i\pi nx} d\mu(x) \qquad\forall n\in \mathbb{Z}.$$

Clearly, $(\hat\mu(n))_n$ is bounded.

  1. What sufficient conditions on a bounded sequence $(a_n)_n$ are known that ensure that there is a finite measure with $\hat\mu(n)=a_n$ ?
  2. same with probability measures instead of finite ones;
  3. same with necessary conditions.

I would guess that characterizations are out of reach, but maybe I am wrong?

Added in Edit: Yemon Choi rightfully asks what kind of sufficient or necessary condition I am after. Any is good for my culture, but I am especially interested in sufficient condition that enable one to construct measure satisfying constraints on Fourier coefficient. To be honest, one of my goals was to understand why it is not easy to disprove Furstenberg's $\times 2$, $\times 3$ conjecture by simply picking Fourier coefficients $(c_n)_n$ such that $c_{2^p3^qm}=c_m$ (and $c_0=1$) inside the set of Fourier series of probability measures. I think I am starting to get the point.

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    $\begingroup$ Bochner's theorem seems relevant. The difference between probability measures versus finite measures is essentially that $\hat{\mu}(0)$ is the total measure. $\endgroup$ Oct 3, 2018 at 16:05
  • $\begingroup$ @WillieWong "Finite measures" are not necessarily positive measures. $\endgroup$ Oct 3, 2018 at 16:41
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    $\begingroup$ For finite signed measures, you could combine Bochner's (Herglotz's) theorem with the Jordan decomposition: $a_n$ is the Fourier transform of a signed measure iff it is the difference of two positive definite sequences. Maybe there's a simpler way to express that condition? $\endgroup$ Oct 3, 2018 at 17:40
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    $\begingroup$ My impression is that it is folklore among those working in Fourier analysis that there is no good set of nec+suff conditions, unless one allows essentially tautological reformuations. It might help if you can restrict the kinds of nec or suff conditions you want. E.g. do you want a condition on the sequence which remains invariant when the sequence $(a_n)$ is replaced by $|a_n|$? $\endgroup$
    – Yemon Choi
    Oct 3, 2018 at 23:36
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    $\begingroup$ If $\widehat{\mu}(n)=0$ for $n<0$ (say), then $\mu$ is absolutely continuous (Riesz+Riesz), and of course this makes the other half of the coefficients go to zero. This already pretty much rules out characterizations in terms of the asymptotic size of the $\widehat{\mu}(n)$ as $|n|\to\infty$. $\endgroup$ Oct 4, 2018 at 0:08

3 Answers 3


The most elegant solution exists for problem 2: the necessary and sufficient condition for $a_n=\hat{\mu}(n)$ for a positive measure is that he sequence $(a_n)$ is non-negative semi-definite, which means that all Toplitz forms $$\sum_{i,j=0}^na_{i-j}z_i\overline{z}_j\geq 0$$ for all integers $n$ and complex $z_i$. This is a theorem of Caratheodory and Toplitz (see, for example N. Akhiezer, Classical moment problem, Ch V, section 1).

For probability measures one has to add to this $a_0=1$.

For 1, the best necessary and sufficient condition is that $a_n$ is a difference of two on-negative semi-definite sequences, which is not very effective, of course. Of course these conditions are not always easy to check, but one can derive many simple necessary conditions.

Bochner's theorem is a continuous analog of this (for Fourier transforms of non-periodic measures).


This is the content of the following theorem (from Hoffman's Banach spaces of analytic functions, 1962).

Hoffman, K., Banach spaces of analytic functions, Prentice-Hall Series in Modern Analysis. Englewood Cliffs, N.J.: Prentice-Hall, Inc. XIII, 217 p. (1962). ZBL0117.34001.

Notice that $f$ is defined in the unit disk as

$$f(r e^{2\pi i\theta}) = \sum c_n r^{|n|} e^{2\pi i n\theta},$$ and the last eventuality (iv) is the one relevant.

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I'm not sure if the following are really what you are after, but I hope they might be useful as a first step in working out what kinds of condition are or aren't good enough for your purposes.

Here is a "nice" or "easy" sufficient condition, in the sense that it is stated in terms of "intrinsic" properties of the sequence: if $(a_n)\in \ell^2({\bf Z})$ then by the classical Parseval/Plancherel theorem $(a_n)$ is the Fourier series of an $L^2$-function $f$ on the circle; and any finite measure whose Fourier series is $(a_n)$ must be $f$ times the usual uniform measure.

Here is a necessary condition which, in a similar flavour, is stated in terms of the sequence itself without invoking the Fourier transform. Suppose ${\bf a}=(a_n)$ is the Fourier series of a finite measure on ${\bf T}$. Then the sequence $(a_n)$ is weakly almost periodic as a function on ${\bf Z}$; that is, if $L_k :\ell^\infty({\bf Z}) \to \ell^\infty({\bf Z})$ is translation by a fixed integer $k$, then the set $\{ L_k{\bf a} \colon k\in {\bf Z}\}$ of all translates of ${\bf a}$ is relatively weakly compact as a subset of the Banach space $\ell^\infty({\bf Z})$. (N.B. this really is the weak topology not the weak-star topology.)

I am not sure exactly where this was first proved, but it might be in work of W. Eberlein; subsequently W. Rudin produced examples to show that not every element of $\ell^\infty({\bf Z})$ with this sequence with this property arises as the Fourier series of a finite measure. In fact, dropping into more modern terminology/jargon, ${\rm WAP}({\bf Z})$ is much much bigger than the norm-closure of $B({\bf Z})$ inside $\ell^\infty({\bf Z})$.

(Weak compactness is something I always find a bit mysterious but often one can decide whether or not a given element of $\ell^\infty({\bf Z})$ is w.a.p. by using something like "Grothendieck's double-limit criterion".)

I should add that the conditions above can be generalized with ${\bf Z}$ replaced by any discrete group $G$, provided that one accepts $B({\bf Z})$ as the object of study rather than $M({\bf T})$, and accepts the Fourier algebra $A({\bf Z})$ as the replacement for $L^1({\bf T})$. That is, when $G$ is discrete we have

$$\ell^2(G) \subseteq A(G) \subseteq B(G)\cap C_0(G) \subseteq B(G) \subseteq {\rm WAP}(G) \subseteq \ell^\infty(G)$$ and indeed all inclusions except the first one remain valid for any locally compact $G$, not just the discrete groups


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