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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 $\mathbb R.$

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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, $\|\hat{\mu}\|_{L^{\infty}} \leq \|\mu\|.$

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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.

Corrected answer: Regarding $f$ and $\mu$ as tempered distributions, you are looking to solve $\check{f} = \mu$. A bounded linear functional on $C_c(\mathbb R)$ (which is included the Schwarz class) may be represented by a bounded complex Borel measure. This is the Riesz Representation Theorem. In the present context, it suffices for there to exist a constant $C$ satisfying $|\check{f}(\phi)| \le C \|\phi\|_\infty$ for all $\phi \in C_c(\mathbb R)$. As below, you have the identity $$\check{f}(\phi) = \int_{\mathbb R} f \mathcal F^3\phi \, dx,$$ so a sufficient condition is that $$\left| \int_{\mathbb R} f \mathcal F^3\phi \, dx \right| \le C \|\phi\|_\infty$$ for some constant $C$ and all $\phi \in C_c(\mathbb R)$.

This solution makes use (or at least mention) of tempered distributions.

Regarding $f$ and $\mu$ as tempered distributions, you are looking to solve $\check{f} = \mu$. A tempered distribution is a Radon measure exactly when it is positive: that is, whenever $\check{f}(\phi) \ge 0$ for every nonnegative Schwarz function $\phi$. Since the Fourier transform has period 4 on the space of tempered distributions you can use the Plancherel formula to obtain (I’ll pass on the hat notation to make typesetting easier) $$\mathcal F^{-1} f (\phi) = \mathcal F^3 f(\phi) = f(\mathcal F^3\phi).$$ Since $f$ is a function you require $$\int_{\mathbb R} f \mathcal F^3 \phi \, dx \ge 0$$ whenever $\phi$ is a nonnegative Schwarz function.

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