Two classic problems concerning Fourier transform of an integrable function 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 $f$.
 A: 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.
A: For (1): there exists a non-trivial Schwartz function $\psi$ whose Fourier transform is supported in $[-2,-1] \cup [1,2]$.
Let $\phi_k = 4^k \psi(4^k \xi)$, then we have $\hat{\phi_k} = \hat{\psi}(4^{-k}\xi)$
Choose $r\in (p,q)$. Let
$$ f = \sum 4^{-k/r} \phi_k $$
Since all of the $\|\phi_k\|_{L^1}$ are the same, this is a geometric series and converges in $L^1$.
You have that using the disjointness of the Fourier support, that
$$ \| \hat{f}\|_{L^p}^p = \sum 4^{k(1-p/r)} \|\hat{\psi}\|_{L^p}^p $$
which fail to converge, yet
$$ \| \hat{f}\|_{L^q}^q = \sum 4^{k(1-q/r)} \|\hat{\psi}\|_{L^q}^q $$
converges as a geometric series.
