TITLE: Which $f \in L^\infty$ are the Fourier transform of a bounded complex measure?
QUESTION [6 upvotes]: A measure on $\mathbb R$ is a set function $\mu,$ defined for all Borel sets of $\mathbb R,$ which is countably additive(that is, $\mu(E)=\sum \mu(E_{i})$ if $E$ is the union of the countable family $\{E_{i}\}$ of pairwise disjoint Borel sets of $\mathbb R$), and for which $\mu(E)$ is finite if the closure of $E$ is compact.
With each $\mu$ on $\mathbb R$ there is associated a set function $|\mu|,$ the total variation of $\mu,$ defined by,
$$|\mu|(E)= \sup \sum |\mu(E_{i})|,$$
the supreme being taken over all finite collections of pairwise disjoint  Borel sets $E_{i}$ whose union is $E.$ Then $|\mu|$ is also a measure on $\mathbb R.$
If $\|\mu\|= |\mu|(\mathbb R)< \infty,$ we say $\mu$ is a bounded complex Borel measure on $\mathbb R.$
Put, $M(\mathbb R)=$ The space of all bounded complex Borel measures on $\mathbb R.$
Put, $ L^{\infty}(\mathbb R)=$ The space of essentially bounded functions on $\mathbb R.$

My Question is: Let $f\in L^{\infty}(\mathbb R).$ Can we expect to find bounded complex Borel measure $\mu$ on $\mathbb R$ such that its Fourier–Stieltjes transform is $f$, that is,
  $$f(\xi)=\hat{\mu}(\xi)=\int_{\mathbb R}  e^{-2\pi i x\cdot \xi} d\mu(x); (\xi\in \mathbb R) ?$$

If answer is negative,then what extra condition one should think of to  impose on the given member of $L^{\infty}$, to solve the above integral  equation ?
Thanks,
Edit: We note that, Fourerie-Steltije transform maps, $M(\mathbb R) \to L^{\infty}(\mathbb R): \mu \mapsto \hat{u}$ with inequality, $\|\hat{\mu}\|_{L^{\infty}} \leq \|\mu\|.$

REPLY [5 votes]: This may or may not be useful, but Bochner's theorem says that a function $f$ is the Fourier transform of a positive bounded measure $\mu$ if and only if $f$ is continuous and positive definite.  (A simple way to describe a positive definite function is this: given any $x_1, \dots, x_n \in \mathbb{R}$, the $n \times n$ symmetric matrix whose $ij$ entry is $f(x_i - x_j)$ should be a positive definite matrix.)
So thanks to the Jordan decomposition, we could say:

$f$ is the Fourier transform of a bounded complex measure if and only if $f$ can be written as $f = f_1 - f_2 + if_3 - if_4$ where all the functions $f_j$ are continuous and positive definite.

In practice, however, it tends to be difficult to verify directly that a function is positive definite.