\begin{document}
\begin{frontmatter}


\title{On the unboundedness in $L^{q}$ of the Fourier transform of $L^{p}$ functions}
\author{Ahmed A. Abdelhakim}
\address{Mathematics Department, Faculty of Science, Assiut University, Assiut 71516 - Egypt\\
Email: ahmed.abdelhakim@aun.edu.eg}
\begin{abstract}
The Hausdorff-Young inequality confirms the continuity of the Fourier transform $\mathcal{F}$ from
$L^{p}(\mathbb{R}^{d})$
to $L^{q}(\mathbb{R}^{d})$ exclusively when
$1\leq p\leq 2$ and $q$ is its conjugate Lebesgue
exponent. Let $\Omega$ be a Lebesgue measurable bounded subset of $\mathbb{R}^{d}$.
We show that $\mathcal{F}$ is bounded
from $L^{p}(\mathbb{R}^{d})$
to $L^{q}(\Omega)$ precisely when
$\frac{1}{p}+\frac{1}{q}\geq 1$
and $1\leq p\leq 2$. Let $N>>1$ be arbitrarily large.
We construct concrete counterexamples
that prove the unboundedness of $\mathcal{F}$ from $L^{p}(\mathbb{R}^{d})$
to $L^{q}\left(\left\{|x|>>N\right\}\right)$
outside the admissible range of
the Hausdorff-Young inequality.
\end{abstract}
\begin{keyword}
Fourier transform\sep Hausdorff-Young inequality\sep
frequency domain \sep $L^{p}$-$L^{q}$ boundedness
\MSC[2010] 42A38\sep 42B10
\end{keyword}
\end{frontmatter}
\section{Introduction}
If $f\in L^{1}(\mathbb{R}^{d})$
then its Fourier transform \cite{rmander,Lieb,Rudin}
\begin{equation*}
\mathcal{F}{f}(\xi):=
\int_{\mathbb{R}^{d}}
f(x)\,e^{-\dot{\imath} x\cdot\xi}\,dx
\end{equation*}
is a continuous bounded function
with
\begin{equation}\label{qi1}
\|\mathcal{F}{f}\|_{L^{\infty}(\mathbb{R}^{d})}\leq
\|f\|_{L^{1}(\mathbb{R}^{d})}.
\end{equation}
Moreover, if $f$ is an $L^{2}(\mathbb{R}^{d})$
function then so is $\hat{f}$
that now satisfies the Plancherel's identity
\begin{equation}\label{qi2}
\|\mathcal{F}{f}\|_{L^{2}(\mathbb{R}^{d})}=
\|f\|_{L^{2}(\mathbb{R}^{d})}.
\end{equation}
Riesz-Thorin interpolation \cite{Reed} between
(\ref{qi1}) and (\ref{qi2}) gives the Hausdorff-young
inequality
\begin{equation}\label{hy}
\|\mathcal{F}{f}\|_{L^{q}(\mathbb{R}^{d})}\leq
\|f\|_{L^{p}(\mathbb{R}^{d})}
\end{equation}
for all $p$, $q$ such that
\begin{equation}\label{hy1}
\frac{1}{p}+\frac{1}{q}=1,\;1\leq p\leq 2.
\end{equation}
\indent Suppose $\Omega$ is a Lebesgue measurable bounded subset of $\mathbb{R}^{d}$. First we consider the boundedness
of $\mathcal{F}:L^{p}(\mathbb{R}^{d})
\longrightarrow L^{q}(\Omega)$. This is a weaker problem
since $ L^{q}(\mathbb{R}^{d}) \subset
L^{q}(\Omega)$. So we expect a range
of admissible exponents larger than
the range (\ref{hy1}) of the Hausdorff-Young inequality. We find the optimal such range.\\
\indent
Second, let $1\leq p\leq \infty$. Then $\mathcal{F}f\in L^{\infty}_{\text{loc}}$
for every $f\in L^{p}$ with compact support.
The classical assertion
that (\ref{hy}) need not hold outside the range
(\ref{hy1}) then shows that the integrability
issues of $|\mathcal{F}f|^{q}$ appear near infinity.
We provide concrete counterexamples
that show the unboundedness of
$\mathcal{F}:L^{p}(\mathbb{R}^{d})
\longrightarrow L^{q}\left(\left\{|x|>>N\right\}\right)$ outside the range (\ref{hy1}) for arbitrarily large $N$.\\
\indent
To our best knowledge,
the sharpness of the range (\ref{hy1})
for (\ref{hy}) is handled in the literature
by a scaling argument that imposes
the necessary condition $1/p +1/q=1$,
while the restriction $1\leq p \leq 2\,$
is inferred by proving
that $\mathcal{F}{f}$ can be a distribution
that is not a function for $f\in L^{p}(\mathbb{R}^{d})$ with $2<p\leq \infty$ (see e.g. \cite{rmander}).
Our counterexamples are concrete
$L^{p}$ functions that have $L^{p}$ Fourier transforms. They take advantage of
the uncertainty principle
to forbid the inequality (\ref{hy})
with any constant independent of $f$
where $\|\mathcal{F}{f}\|_{L^{q}}$
is taken over arbitrarily high frequencies.
The idea is to increase accordingly the frequency
in the oscillatory input functions.
\section{Bounded frequency domain}
\begin{thm}
The Fourier integral operator
$\mathcal{F}:L^{p}(\mathbb{R}^{d})
\longrightarrow L^{q}(\Omega)$
is continuous precisely if
${1}/{p}+{1}/{q}\geq 1$
and $1\leq p \leq 2$.
\end{thm}
\begin{proof}
The Hausdorff-Young inequality (\ref{hy}) asserts
\begin{equation}\label{p11}
\|\mathcal{F}{f}\|_{L^{q}(\Omega)}\leq
\|f\|_{L^{p}(\mathbb{R}^{d})},\quad
\frac{1}{p}+\frac{1}{q}=1,\;1\leq \frac{1}{p}\leq \frac{1}{2}.
\end{equation}
But H\"{o}lder's inequality implies
\begin{equation}\label{p12}
\|\mathcal{F}{f}\|_{L^{\widetilde{q}}(\Omega)}\leq
|\Omega|^{\frac{1}{\widetilde{q}}-\frac{1}{q}}
\|\mathcal{F}{f}\|_{L^{q}(\Omega)}
\end{equation}
whenever $q\geq \widetilde{q}$. Using
(\ref{p12}) in (\ref{p11}) yields
the estimate
\begin{equation}\label{p13}
\|\mathcal{F}{f}\|_{L^{q}(\Omega)}\leq
|\Omega|^{\frac{1}{p}+\frac{1}{q}-1}
\|f\|_{L^{p}(\mathbb{R}^{d})}
\lesssim_{d,p,q}
\|f\|_{L^{p}\left(\mathbb{R}^{d}\right)}
\end{equation}
when
\begin{equation}\label{range1}
\frac{1}{q}\geq 1-\frac{1}{p},\;1\leq \frac{1}{p}\leq \frac{1}{2}.
\end{equation}
\indent It remains to prove that (\ref{p13})
is false with any constant
independent of $f$ outside the range
(\ref{range1}).\\
\indent Let $B(1)$ be the unit ball and let
$0<\lambda<1$. Suppose $f$
is a Schwartz function and define
$f_{\lambda}(x):=f(\lambda x)$. Then
\begin{equation}\label{ee1}
\|f_{\lambda}\|_{L^{p}(\mathbb{R}^{d})}
=\lambda^{-\frac{d}{p}}\|f\|_{L^{p}(\mathbb{R}^{d})}.
\end{equation}
We also have
\begin{equation*}
\mathcal{F}{f_{\lambda}}(\xi)=\lambda^{-d}
\mathcal{F}{f}(\frac{\xi}{\lambda}).
\end{equation*}
Therefore, rescaling $\xi/\lambda\rightarrow \xi$,
we find
\begin{equation*}
\int_{B(1)}|\mathcal{F}{f_{\lambda}}(\xi)|^{q}d\xi=
\lambda^{d(1-q)}\,
\int_{B(1/\lambda)} |\mathcal{F}{f}(\xi)|^{q}d\xi
\geq \lambda^{d(1-q)}\,
\int_{B(1)} |\mathcal{F}{f}(\xi)|^{q}d\xi
\end{equation*}
which implies
\begin{equation}\label{ee2}
\|\mathcal{F}{f}_{\lambda}
\|_{L^{q}(B(1))}
\geq\lambda^{d(\frac{1}{q}-1)}
\|\mathcal{F}{f}\|_{L^{q}(B(1))}.
\end{equation}
Assume (\ref{p13}) holds. Then $f_{\lambda}$ must satisfy it with a constant independent of
$\lambda$, and by (\ref{ee1}) together with  (\ref{ee2}) we should get
\begin{equation*}
\|\mathcal{F}{f}\|_{L^{q}(\Omega)}\lesssim_{d,p,q}
\lambda^{d(1-\frac{1}{p}-\frac{1}{q})}
\|f\|_{L^{p}(\mathbb{R}^{d})},\quad 0< \lambda<1.
\end{equation*}
The latter estimate is a contradiction unless
$\frac{1}{p}+\frac{1}{q}\geq 1$. What is left then
is the region ${1}/{p}+{1}/{q}\geq 1$,
$2<p\leq \infty$.
\begin{exm}\label{stphs}
Let $f$ be such that
\begin{equation*}
\mathcal{F}{f}(\xi)=e^{-\frac{\dot{\imath}}{2} N |\xi|^{2}}
\chi_{\raisebox{-.5ex}{$\scriptstyle B(1)$}}
{(\xi)},\;
N>>1.
\end{equation*}
Obviously $\|\mathcal{F}{f}\|_{L^{q}
\left(B(1)\right)}\sim_{d,q} 1$,
$1\leq q\leq \infty$.
Furthermore $f=\mathcal{F}^{-1}\left({\mathcal{F}{f}}\right)$
is well defined and has the oscillatory integral
representation
\begin{equation*}
f(x)=\int_{B(1)}e^{
\dot{\imath} N \psi_{N}(\xi;x)}\,d \xi
\end{equation*}
with the phase
\begin{equation*}
\psi_{N}(\xi;x):=\frac{x}{N}\cdot\xi-\frac{1}{2}|\xi|^{2}.
\end{equation*}
Since
\begin{eqnarray*}
  \nabla_{\xi}\psi_{N}(\xi;x) &=& \frac{x}{N}-\xi,\;
  \xi\neq 0, \\
  \left(D_{\xi}^2\psi_{N}\right)(\xi)&:=&\left(\frac{\partial \psi_{N}(\xi;x) }{\partial\xi_{i}\partial\xi_{j}}\right)
 = -(\delta_{ij})
\end{eqnarray*}
then
$\xi\mapsto\psi_{N}(\xi;x)$
has a unique nondegenerate critical point at
$\xi={x}/{N}$ that lies inside the spherical shell $B(\frac{3}{4})\setminus B(\frac{1}{4})$
when $x\in \mathcal{S}:=\left\{{N}/{4}<|x|\leq{3N}/{4}\right\}$. Thus
$f$ has the stationary phase asymptotic
(see \cite{Stein})
\begin{eqnarray*}
f(x)&=&\left(\frac{2\pi}{N}\right)^{\frac{d}{2}}
e^{\dot{\imath}\frac{\pi}{4}\text{sgn}\left(\left(D_{\xi}^2
\psi_{N}\right)(x/N)\right)}
\frac{e^{\dot{\imath} N \psi_{N}(\frac{x}{N};x)}}{|\text{det}\left(\left(D_{\xi}^2
\psi_{N}\right)(x/N)\right)|^{\frac{1}{2}}}\left(1+O
\left(\frac{1}{N}\right)
\right) \\
   &=&\left(\frac{2\pi}{N}\right)^{\frac{d}{2}}
   e^{-\dot{\imath}\frac{\pi}{4}n}
e^{\dot{\imath}\frac{|x|^2}{2N}}
\left(1+O
\left(\frac{1}{N}\right)\right),\quad x\in \mathcal{S}.
 \end{eqnarray*}
And since
\begin{equation*}
\int_{\mathcal{S}}|f(x)|^{p}dx \gtrsim
\frac{1}{N^{\frac{d}{2}p}}|\mathcal{S}|\sim
N^{\frac{d}{2}(2-p)}
\end{equation*}
then we have
\begin{equation*}
\frac{\|\mathcal{F}{f}\|_{L^{q}
\left(B(1)\right)}}{\|f\|_{L^{p}
\left(\mathbb{R}^{d}\right)}}\gtrsim N^{-\frac{d}{2p}(2-p)}
\end{equation*}
which blows up when $N\rightarrow +\infty$
unless $1\leq p\leq 2$.
\end{exm}
\end{proof}
\begin{center}
\begin{tikzpicture}[scale=.92]
\fill[fill = black!10] (4,0)--(4,4)--(2,4)--(2,2)--cycle;
\draw[thin,->] (0.0, 0)-- (4.5,0) node[black,right] {$\frac{1}{p}$};
\draw[thin,->] (0.0, 0)-- (0,4.5) node[black,above left] {$\frac{1}{q}$};
\draw[thick,black] (2, 4)-- (4,4);
\draw[thick,black] (4, 0)-- (4,4);
\draw[thick,black] (2, 2)-- (2,4);

\draw[very thin,dotted] (2, 0)-- (2,2);
\draw[very thin,dotted] (2, 4)-- (0,4);
\draw (2, 0) node[below] {$\frac{1}{2}$};
\draw (4, 0) node[below] {$1$};
\draw (0, 4) node[left] {$1$};
\draw[very thin,dotted] (2, 2)-- (0,4);
\draw[thick,black] (4, 0)-- (2,2);


 \end{tikzpicture}\\
The region on and inside the dark right trapezoid represents the optimal range
for the estimate  $\|\mathcal{F}f\|_{L^{q}\left(\Omega\right)}
\lesssim_{d,p,q}
\|f\|_{L^{p}\left(\mathbb{R}^{d}\right)}$.
\end{center}
\section{Unboundedness in high frequencies}
\begin{thm}\label{thm2}
Let $\Omega$ be any bounded measurable subset
of $\mathbb{R}^{d}$. Then there exists
$f\in L^{p}(\mathbb{R}^{d})$ with compact support
such that the estimate
\begin{equation}\label{thmest}
\|\mathcal{F}{f}\|_{L^{q}\left(\mathbb{R}^{d}\setminus \Omega\right)}
\lesssim_{d,p,q}
\|f\|_{L^{p}(\mathbb{R}^{d})}
\end{equation}
fails outside the admissibility range (\ref{hy1})
of the Hausdorff-Young inequality.
\end{thm}
\begin{proof}
Losing no generality,
assume $\Omega\subset B(N)$
for some $N>1$. The following counterexamples prove
Theorem \ref{thm2}:
\begin{exm}\label{he1}
Fix $\:0<\delta<<1$.
Inspired by the uncertainty principle,
let $f=e^{\dot{\imath}N^{2} x\cdot u}\chi_{\raisebox{-.5ex}{$\scriptstyle [0,\delta/N]^{d}$}}$,
$u=(1,...,1)$. Then, when
$\,\xi\in \mathcal{R}:=
[N^{2}+N,N^{2}+2N]^{d}$ we have
\begin{equation*}
e^{\dot{\imath} x\cdot\left(N^{2}u -\xi\right)}=
e^{O(\delta)}=1+O(\delta).
\end{equation*}
Hence we get
\begin{equation*}
\mathcal{F}{f}(\xi)=
\int_{[0,\delta/N]^{d}}
\left(1+O(\delta)\right)\,dx\gtrsim_{\delta}N^{-d},
\;\xi\in\mathcal{R}.
\end{equation*}
Since $\:\mathcal{R}\subset
\mathbb{R}^{d}\setminus \Omega,\,$
and $\,|\mathcal{R}|\sim N^{d}\:$ then
\begin{equation*}
\|\mathcal{F}{f}\|_{L^{q}\left(\mathbb{R}^{d}
\setminus \Omega\right)}\geq
\|\mathcal{F}{f}\|_{L^{q}\left(\mathcal{R}\right)}
\gtrsim_{\delta}N^{d\left(\frac{1}{p}+\frac{1}{q}-1\right)}
\|f\|_{L^{p}(\mathbb{R}^{d})}.
\end{equation*}
This proves the restriction
${1}/{p}+{1}/{q}\leq1$.
\end{exm}
\begin{exm}\label{rem1}
Suppose
\begin{equation*}
{f}(x)=e^{{\dot{\imath}} N^{2} x\cdot u}
\chi_{\raisebox{-.5ex}{$\scriptstyle [N,2N]^{d}$}}{(x)}.
\end{equation*}
Then, for the frequencies
$\xi\in \mathcal{T}:=[N^{2}+\delta/N,N^{2}+2\delta/N]^{d}$ we have
\begin{equation*}
\mathcal{F}{f}(\xi)=
\prod_{k=1}^{d}
\int_{N}^{2N}
e^{\dot{\imath}x_{k}(N^{2}-\xi_{k})}\,dx_{k}\gtrsim_{\delta}
N^{d}.
\end{equation*}
And since $\mathcal{T}\subset \Omega^{\text{\textbf{c}}}$
and $\,|\mathcal{T}|\sim N^{-d}$ then
\begin{equation*}
\|\mathcal{F}{f}\|_{L^{q}\left(\mathbb{R}^{d}
\setminus \Omega\right)}
\geq \|\mathcal{F}{f}\|_{L^{q}\left(\mathcal{T}\right)}
\gtrsim_{\delta}N^{d\left(1-\frac{1}{p}-\frac{1}{q}\right)}
\|f\|_{L^{p}(\mathbb{R}^{d})}
\end{equation*}
which yields the restriction
${1}/{p}+{1}/{q}\geq1$.
\end{exm}
Counterexamples \ref{he1} and \ref{rem1}
put us on the
line ${1}/{p}+{1}/{q}=1$
in the $\frac{1}{p}$-$\frac{1}{q}$ plane
(see the figures below).
\begin{exm}\label{ce3}
Let
\begin{equation*}
f(x)=e^{\frac{\dot{\imath}}{2}N^{2}|x|^{2}}
\chi_{\raisebox{-.5ex}{$\scriptstyle
\left\{2<|x|<5 \right\}
$}}{(x)}
\end{equation*}
 Then
\begin{equation*}
\mathcal{F}{f}(\xi)=\int_{\left\{2<|x|<5 \right\}}e^{
-\dot{\imath} N^{2} \psi_{N^{2}}(x;\xi)}\,d x.
\end{equation*}
Proceeding with the stationary phase method as in Counterexample \ref{stphs} gives
\begin{equation*}
|\mathcal{F}{f}(\xi)|\gtrsim
N^{-{d}},\; 3N^{2}<|\xi|<4N^{2}.
\end{equation*}
Observe that, when $3N^{2}<|\xi|<4N^{2}$,
the only (nondegenerate) critical point
$x=\xi/N^{2}$ of $x\mapsto \psi_{N^{2}}(x;\xi)$ lives
in $\left\{3<|x|<4 \right\}$,
away from the boundary of
$\left\{2<|x|<5 \right\}$. So we obtain
\begin{equation*}
{\|\mathcal{F}{f}\|_{L^{q}
\left(\left\{3N^{2}<|x|<4N^{2} \right\}\right)}}\gtrsim N^{-d+\frac{2d}{q}}{\|f\|_{L^{p}
\left(\mathbb{R}^{d}\right)}}
\end{equation*}
which imposes
the necessary condition
$2\leq q \leq \infty$ for the estimate (\ref{thmest}).
\end{exm}
\begin{rem}
Note in Counterexamples \ref{he1} through \ref{ce3}
that we can concentrate
the highest contribution of
$\mathcal{F}f$ at very high frequencies,
as far from $supp f \cup \Omega$
as we wish, by increasing the frequency
in the input $f$.
\end{rem}
\begin{exm}
Another counterexample that
forces the restriction
$q\geq p$, and consequently
could replace Counterexample \ref{ce3},  is
\begin{equation*}
f(x)=e^{\frac{\dot{\imath}}{2}|x|^{2}}
\chi_{\raisebox{-.5ex}{$\scriptstyle
\left\{2N<|x|<5N \right\}$}}{(x)}.
\end{equation*}
Rescaling $x/N\rightarrow x$ we may write
\begin{equation*}
\mathcal{F}{f}=
N^{d}
\int_{\left\{2<|x|<5\right\}}
e^{-{\dot{\imath}}N^2
\psi_{N}(x;\xi)}\,dx.
\end{equation*}
Again, the stationary phase technique gives
$|\mathcal{F}{f}|\gtrsim 1$ in the
spherical shell
$\left\{ 3N<|\xi|<4N\right\}$, hence the estimate
\begin{equation*}
{\|\mathcal{F}{f}\|_{L^{q}
\left(\left\{3N<|x|<4N \right\}\right)}}\gtrsim N^{\frac{d}{q}-\frac{d}{p}}{\|f\|_{L^{p}
\left(\mathbb{R}^{d}\right)}}.
\end{equation*}
\end{exm}
\begin{center}
\begin{tikzpicture}[scale=0.92]
\fill[fill = black!10] (4,0)--(4,4)--(0,4)--cycle;
\draw[very thin,->] (0.0, 0)-- (4.5,0) node[black,right] {$\frac{1}{p}$};
\draw[very thin,->] (0.0, 0)-- (0,4.5) node[black,above left] {$\frac{1}{q}$};
\draw (2.5, 3.2) node {$
e^{{\dot{\imath}} N^{2} x\cdot u}
\chi_{\raisebox{-.5ex}{$\scriptstyle [0,\delta/N]^{d}$}}$};
\draw[thick,black] (4,0.03)-- (4,4);
\draw[thick,black] (4, 4)-- (0.03,4);
\draw (4, 0) node[below] {$1$};
\draw (0, 4) node[left] {$1$};
\draw[thin,dashed] (4, 0)-- (0.02,3.98);
\draw [black, thin] (0,4) circle [radius=0.04];
\draw [black, thin] (4,0) circle [radius=0.04];
 \end{tikzpicture}
\begin{tikzpicture}[scale=0.92]
\fill[fill = black!10] (4,0)--(0,0)--(0,4)--cycle;
\draw[very thin,->] (0.0, 0)-- (4.5,0) node[black,right] {$\frac{1}{p}$};
\draw[very thin,->] (0.0, 0)-- (0,4.5) node[black,above left] {$\frac{1}{q}$};
\draw (1.5, 1) node {$
e^{{\dot{\imath}} N^{2} x\cdot u}
\chi_{\raisebox{-.5ex}{$\scriptstyle [N,2N]^{d}$}}$};
\draw[ thin,dotted] (4, 0)-- (4,4);
\draw[ thin,dotted] (4, 4)-- (0,4);
\draw[thick,black] (0, 0)-- (3.97,0);
\draw[thick,black] (0, 0)-- (0,3.97);
\draw (4, 0) node[below] {$1$};
\draw (0, 4) node[left] {$1$};
\draw[thin,dashed] (4, 0)-- (0.02,3.98);
\draw [black, thin] (0,4) circle [radius=0.04];
\draw [black, thin] (4,0) circle [radius=0.04];
 \end{tikzpicture}
 \begin{tikzpicture}[scale=0.92]
\fill[fill = black!10] (4,2)--(4,4)--(0,4)--(0,2)--cycle;
\draw[thin,->] (0.0, 0)-- (4.5,0) node[black,right] {$\frac{1}{p}$};
\draw[thin,->] (0.0, 0)-- (0,4.5) node[black, above left] {$\frac{1}{q}$};
\draw[thick,black] (4, 2.03)-- (4,4);
\draw[thick,black] (4, 4)-- (0,4);
\draw[thick,black] (0, 4)-- (0,2.03);
\draw[thin,dashed] (0.02, 2)-- (4,2);
\draw[very thin,dotted] (4, 0)-- (4,2);
\draw (0, 2) node[left] {$\frac{1}{2}$};
\draw (4, 0) node[below] {$1$};
\draw (0, 4) node[left] {$1$};
\draw [black, thin] (0,2) circle [radius=0.04];
\draw [black, thin] (4,2) circle [radius=0.04];
\draw (0.3, 3) node[right] {$e^{\frac{\dot{\imath}}{2}N^{2}|x|^{2}}
\chi_{\raisebox{-.5ex}{$\scriptstyle
\left\{2<|x|<5 \right\}
$}}$};
 \end{tikzpicture}
  \begin{tikzpicture}[scale=0.92]
\fill[fill = black!10] (0,0)--(0,4)--(4,4)--cycle;
\draw[thin,->] (0.0, 0)-- (4.5,0) node[black,right] {$\frac{1}{p}$};
\draw[thin,->] (0.0, 0)-- (0,4.5) node[black, above left] {$\frac{1}{q}$};
\draw[thick,black] (0, 4)-- (3.97,4);
\draw[thick,black] (0, 0.03)-- (0,4);
\draw[thin,dashed] (0.03, 0.03)-- (3.97,3.97);
\draw[very thin,dotted] (4, 0)-- (4,3.97);
\draw (4, 0) node[below] {$1$};
\draw (0, 4) node[left] {$1$};
\draw [black, thin] (0,0) circle [radius=0.04];
\draw [black, thin] (4,4) circle [radius=0.04];
\draw (0.0, 3.5) node[right] {$e^{\frac{\dot{\imath}}{2}|x|^{2}}
\chi_{\raisebox{-0.5ex}{$\scriptstyle
\left\{2<\frac{|x|}{N}<5\right\}
$}}$};
 \end{tikzpicture}\\
Compactly supported inputs for which the estimate
$\|\mathcal{F}{f}\|_{L^{q}\left(\mathbb{R}^{d}\setminus B(N)\right)}
\lesssim_{d,p,q}
\|f\|_{L^{p}(\mathbb{R}^{d})}$\\
fails in the corresponding shadowed region.
\end{center}
\end{proof}


\begin{small}
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