$f,\hat f\in L^1(\mathbb R)$ implies $f\in L^p(\mathbb R)$ for $1\leq p\leq\infty$

Question

Could someone please hint how I may show $f,\hat f\in L^1(\mathbb R)$ implies $f\in L^p(\mathbb R)$ for all $p\in[1,\infty]$?

This is what I tried:

$$\begin{align*}\int\left|f\left(x\right)\right|^pdx&=\int\left|\int\widehat{f}\left(\xi\right)e^{2\pi i\xi}d\xi\right|^pdx\\&\leq\int\left(\int\left|\widehat{f}\left(\xi\right)\right|\left|e^{2\pi i\xi}\right|d\xi\right)^pdx\\&\leq\int\left(\int\left|\widehat{f}\left(\xi\right)\right|d\xi\right)^pdx\\&=\int\left\|\widehat{f}\right\|_1^pdx.\end{align*}$$

However, the integral does not converge if $\left\|\widehat{f}\right\|_1\neq0$. Moreover, I am not sure how to use the assumption that $f\in L^1\left(\mathbb{R}\right)$.

Answer 1

By the Fourier inversion, under the assumption $\hat f\in L^1$ we get $f$ back as the inverse Fourier transform of $\hat f$, and estimating the integral defining this transform yields $\|f\|_\infty \le C\|\hat f\|_1$ (the constant $C$ is subject to normalization choices for $\hat f$). So we know $f\in L^1\cap L^\infty$. For any $1<p<\infty$, $$ \int |f|^p \le \|f\|_\infty^{p-1} \int|f| = \|f\|_\infty^{p-1}\|f\|_1 <\infty $$ (One could say this is a kind of interpolation, but certainly not anything deep like Riesz-Thorin.)

Answer 2

if f and $f$ and $\hat f$ in $L^1$, then $f$ is in $L^\infty $ and $f$ is in $L^2$ (Plancherel), then by Riesz-Thorin interpolation theorem, we can extend the continuité of the Fourier operator to $L^p$ space.