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For a bounded complex Borel measure $\mu$ on $\mathbb R$, we define, its Fourier-Stieltjes transform, $\hat{\mu}(y)= \int_{\mathbb R} e^{-2\pi ix\cdot y} d\mu(x); (y\in \mathbb R).$

Let $1\leq p \leq \infty $ and we put, $$X_{p}= \{f\in L^{p}(\mathbb R)\cap L^{\infty}(\mathbb R) :\hat{f}\in L^{p}(\mathbb R)\cap L^{\infty}(\mathbb R)\};$$ and we consider the algebra of Fourier-Stieltjes transforms(functions of Fourier-Stieltjes transforms), namely,

$$B(\mathbb R) = \{f:\mathbb R \to \mathbb C : \exists \ \text{bounded complex Borel measure} \ \mu \ \text{on} \ \mathbb R \ni \ \hat{\mu}= f \}.$$

For $p=1,$ clearly, by inversion formula, $X_{p} \subset B(\mathbb R).$

My Question: Can we expect, $X_{p}\subset B(\mathbb R)$ for $1<p\leq \infty $ ? At least for some values of $p;$ or we get a counter examples for some values of $p$ ?

Thanks,

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    $\begingroup$ My first instinct is to try and disprove this indirectly by using a Closed Graph Theorem / Open Mapping Theorem argument, similar to the known standard proof that the Fourier transform is not a surjection from $L^1({\bf R})$ onto $C_0({\bf R})$. In other words, rather than look for an explicit counter-example, try to prove that the norms don't match $\endgroup$
    – Yemon Choi
    Commented Apr 15, 2014 at 13:41
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    $\begingroup$ So to rephrase, is every $f \in L^p \cap L^\infty$ the Fourier transform of some complex Borel measure? $\endgroup$
    – Nate Eldredge
    Commented Apr 15, 2014 at 18:01
  • $\begingroup$ If $X_p\subset B$, then, in particular, any $f$ with $f,\widehat{f}\in L^p\cap L^{\infty}$ would have to satisfy $f\in L^1$ also. I don't see why this would be true (start with a smooth $f\in L^p\cap L^{\infty}$ that is not in $L^1$). $\endgroup$
    – Christian Remling
    Commented Apr 15, 2014 at 19:19

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