Question about Fourier transform and its itegrability Let $n\geq 1$ an integer. Let $f\in L^1(\mathbb{R}^n)$ be an integrable function. Then its Fourier transform $\widehat{f}$ is defined as
$$\widehat{f}(\xi)= \int_{\mathbb{R}^n} f(x) e^{-2\pi i x\cdot \xi} dx,$$
where $\cdot$ denotes the scalar product.
The Fourier transform has many properties and there is an extensive theory about it. It is known that if $\widehat{f}$ integrates against polynomials then this implies that $f$ is differentiable up to some order.
My question is the following:
If $f$ is such that $\widehat{f}\in L^p(\mathbb{R}^n)$ for some $p>1$. Does this tell us anything about $f$ itself? Like for instance if $f$ is bounded, continuous or any other thing?
Thanks a lot! :)

From the Hausdorff-Young inequality, we know that $\mathcal {F} : L^p \to L^{p'} $ is bounded for $1\leq p \leq 2$. It is easy to see that the same holds for the inverse Fourier transform.
Hence, if $\hat {f} \in L^p $ for some $1\leq p \leq 2$, then $f =\mathcal {F}^{-1}\hat {f } \in L^{p'} $.
