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,

known standardproof 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 – Yemon Choi Apr 15 '14 at 13:41