$\{f\in L^{1} : \hat{f} \in L^{p} \}$ closed under convolution?

Question

Let $f, g\in L^{1}(\mathbb R),$ we may define the convolution of $f$ and $g$ as follows: $f\ast g(x)= \int_{\mathbb R} f(x-y)g(y) dy, (x\in \mathbb R).$

We note that $L^{1}(\mathbb R) \ast L^{1}(\mathbb R) \subset L^{1}(\mathbb R)$ and $L^{p}(\mathbb R)$ is not closed under convolution for $p>1.$

We put, $A^{p}(\mathbb R)=\{f\in L^{1}(\mathbb R) : \hat{f} \in L^{p}(\mathbb R) \}.$

My Question is: Can we expect $A^{p}(\mathbb R) \ast A^{p}(\mathbb R) \subset A^{p}(\mathbb R)$ for $1\leq p \leq \infty$?

Edit: Take $1<p\leq \infty$

Answer 1

For $f\in A^p$, you know $\widehat{f} \in L^p \cap L^\infty$, since $\mathcal{F} L^1 \subset C_0 \subset L^\infty$ (Riemann Lebesgue Lemma).

Now, for $f,g \in A^p$, you get $f\ast g \in L^1$ (classical result), as well as

$$ \widehat{f\ast g} = \widehat{f} \cdot \widehat{g} \in L^p \cdot L^\infty \subset L^p. $$

Hence, $f\ast g \in A^p$.