I am looking for the following questions:


(1) **True or false?** for every $p<q$, one may find a function $f\in L^1(\mathbb{R})$ such that $\hat{f}\in L^q (\mathbb{R})$ but $\hat{f}\notin L^p (\mathbb{R})$.


(2) **True or false?** There exists a function $f\in L^1(\mathbb{R})$ such that $\hat{f}\notin L^p (\mathbb{R})$ for all $1\leq p<\infty$.


P.s. $\hat{f}$ is the Fourier transform of $f$.



Two classic problems concerning Fourier transform of an integrable function
Here is a proof that (2) is true.


It suffices to find $f\in L^1(\mathbf{R})$ such that for any integer $n\in\mathbf{N}^\star$, one has $f^{\star n}\notin L^2(\mathbf{R})$, where
\begin{align\*}
f^{\star n}:=\stackrel{n\text{ times}}{\overbrace{f\star f\star \cdots \star f}.}
\end{align\*}
Indeed, using the formula $\widehat{f\star g}=\hat{f}\hat{g}$ and Plancherel's Theorem we see that \begin{align\*}
f^{\star n}\notin L^2(\mathbf{R})\Longleftrightarrow \hat{f}\notin L^{2n}(\mathbf{R}).\end{align\*}
But by interpolation, since $\hat{f}\in L^\infty(\mathbf{R})$, one has \begin{align\*}\hat{f}\notin \bigcup\_{1\leq p<\infty} L^p(\mathbf{R})\Longleftrightarrow \hat{f}\notin \bigcup\_{n\in\mathbf{N}^\star}L^{2n}(\mathbf{R}),
\end{align\*}
Now consider $\text{B}$ the closed unit ball of $L^1(\mathbf{R})$. This is a complete metric space. If for all $f\in L^1(\mathbf{R})$ there exists $n\in\mathbf{N}^\star$ such that $f^{\star n}\in L^2(\mathbf{R})$, this means in particular that
\begin{align\*}
\text{B} = \bigcup\_{k\in\mathbf{N}^\star} F\_k,
\end{align\*}
where
\begin{align\*}
F\_k:=\Big\{f\in \text{B}\,:\, f^{\star k}\in L^2(\mathbf{R})\text{ and }\|f^{\star k}\|\_2\leq k\Big\}.
\end{align\*}
Indeed, if $f\in \text{B}$ is such that $f^{\star n}\in L^2(\mathbf{R})$ for some integer $n$, then Young's inequality ensures that $f^{\star k}\in L^2(\mathbf{R})$ for all integer $k\geq n$, with the estimate $\|f^{\star k}\|\_2\leq \|f^{\star n}\|\_2$.


On the other hand, all the $F\_k$ are closed sets for the $L^1(\mathbf{R})$ topology : if $(f\_p)\_p\in F\_k^{\mathbf{N}}$ converges to $f$, we have obviously $f\in \text{B}$. The sequence $(f\_p^{\star k})\_p$ is bounded in $L^2(\mathbf{R})$ by $k$ so it has a weak cluster point $g$ in this space which satisfies also $\|g\|\_2\leq k$. On the other hand Young's inequality ensures that $(f^{\star k}\_p)\_p$ converges to $f^{\star k}$ in $L^1(\mathbf{R})$, since $(f\_p)\_p$ converges to $f$ in this space. This is enough to identify the limits (working locally, for instance) and ensures that $g=f^{\star k}\in L^2(\mathbf{R})$ with the bound $\|f^{\star k}\|\_2 \leq k$ : we have proven the claimed closedness of $F\_k$.


The complete metric space $\text{B}$ has been covered by a countable family of closed sets : Baire's lemma applies to get the existence of $k\_0\in\mathbf{N}^\star$ such that $F\_{k\_0}$ has a non-empty interior.


Now, note that all the elements of $F\_{k\_0}$ have their Fourier transform in $L^{2k\_0}(\mathbf{R})$ (again Plancherel's theorem and the convolution formula). In particular, denoting by $\mathscr{F}:L^1(\mathbf{R})\rightarrow L^\infty(\mathbf{R})$ the Fourier map, we have thus found a non trivial open ball inside $\mathscr{F}^{-1}(L^{2k\_0}(\mathbf{R}))$, which is a vector subspace of $L^1(\mathbf{R})$ : by translation and homogeneity this would imply $\mathscr{F}(L^1(\mathbf{R}))\subset L^{2k\_0}(\mathbf{R})$. By the closed graph theorem this would lead to an estimate of the form $\|\hat{\varphi}\|\_{2k\_0} \lesssim \|\varphi\|\_1$ which is just not reasonnable and we have a contradiction.


