$1\leq p \leq \infty $. 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 transforms(functions of Fourier transforms), namely, $$A(\mathbb R) =\{f:\mathbb R \to \mathbb C : \exists \ g\in L^{1} (\mathbb R) \ \text{such that } \ \hat{g}=f \}. $$
For $p=1,$ clearly, by inversion formula, $X_{p} \subset A(\mathbb R).$
My Question: Can we expect, $X_{p} \subset A(\mathbb R)$; for $1<p \leq \infty$? At least for some values of $p$; Or we get a counter example for some values of $p$ ?
Thanks,
We will answer negatively to this question. Note that the elements in $A(\Bbb{R})$ are all uniformly continuous functions. So, to find a counter example, we only need to consider some discontinuous function $f$.
Let $f(x)={\bf 1}_{[-1,1]}(x)$ then it is easy to see that $f\in X_p$ for every $p\in(1,+\infty]$ but $f\notin A(R)$. So $X_p\not\subset A(\Bbb{R})$ for every $1<p<+\infty$.