\section{The boundary with Fourier quasicrystals}\label{sec:fq}

\subsection{Background}
Recall that a measure $\nu$ is \emph{strongly tempered} if $\int(1+|x|)^{-N}\,d|\nu|(x)<\infty$
for some $N$; a crystalline measure $\mu$ is a Fourier quasicrystal if $\mu$ and
$\widehat\mu$ are both strongly tempered. The difference between temperedness of
$\mu$ and of $|\mu|$ is discussed in general by Baake and Strungaru \cite{BS23}. For
crystalline measures, Favorov \cite[Theorem~1]{Fav24a} (see also \cite{Fav23})
proved that a tempered measure with polynomially separated support,
$|x-x'|\ge c\min\{1,|x|^{-h},|x'|^{-h}\}$, has tempered total variation; in
particular a crystalline measure with polynomially separated support and spectrum
is a Fourier quasicrystal. He constructed crystalline measures that are not
Fourier quasicrystals \cite{Fav24a,Fav24b}, and Maz\'a\v{c}, Richard and Strungaru
\cite{MRS26} studied almost periodicity of crystalline measures in several
distributional categories, constructing in particular a crystalline Fourier
eigenmeasure that is not strongly tempered, hence not strongly tempered on either
side \cite[Theorem~4.11, Remark~5.2]{MRS26}. These works answer, for several
notions of almost periodicity, the question raised by Meyer \cite{Mey17b,Mey21}
whether every crystalline measure is an almost periodic distribution. Boyvalenkov
and Favorov \cite{BF26} studied measures of unbounded total variation that are
positive or translation bounded and have pure point Fourier transform; they showed
that the squared masses of $\widehat\mu$ form a translation bounded measure, and that
$\widehat\mu$ itself is translation bounded if the spectrum is locally linearly
independent over $\Z$, which gives conditions for a crystalline measure to be a
Fourier quasicrystal.

Gon\c{c}alves \cite{Gon23} classified the Fourier summation pairs $(\mu,a)$,
$\int\widehat\varphi\,d\mu=\sum_\lambda a(\lambda)\varphi(\lambda)$, in which the
left-hand side is an absolutely convergent integral, under the hypotheses
$\int(1+t^2)^{-1}\,d|\mu|(t)<\infty$ and
$\sum_\lambda|a(\lambda)|e^{-c|\lambda|}<\infty$, in terms of almost periodic
holomorphic functions in the upper half-plane, Hermite--Biehler functions and de
Branges spaces. Gon\c{c}alves and Vedana \cite{GV25} removed the restriction on the
polynomial degree, requiring
$\int(1+t^2)^{-c_1}\,d|\mu|(t)+\sum_\lambda|a(\lambda)|e^{-c_2|\lambda|}<\infty$ for some
$c_1,c_2>0$, and remark that they are unaware of a summation formula valid for all
compactly supported smooth $\varphi$ that violates this condition. A crystalline
measure $\nu$ gives such a formula in two ways, with $(\mu,a)=(\nu,\widehat\nu)$ or,
by Fourier inversion, with the roles of $\nu$ and $\widehat\nu$ exchanged; if $\nu$
is not strongly tempered, the left-hand side has to be read as the distributional
pairing \eqref{eq:summation}. In Favorov's example \cite{Fav24b} of a crystalline
measure that is not a Fourier quasicrystal, the Fourier transform appears to be
strongly tempered, so the exchanged formula may well satisfy the hypotheses of
\cite{GV25}. Alfes, Kiefer and Maz\'a\v{c} \cite{AKM25} relate Fourier eigenmeasures to
modular-type transformation laws.

\subsection{No element is strongly tempered}
On our sets the failure of strong temperedness is a property of every nonzero
element of an infinite-dimensional space.

\begin{theorem}\label{thm:nonstrong}
For $\Lambda$ as in Theorem~\ref{thm:A} or~\ref{thm:B}, no nonzero
$\mu\in\M_\Lambda$ is strongly tempered, and neither is its Fourier transform. In
particular, for $A',B'\subseteq\Lambda$, no nonzero element of $\Ncal(A',B')$ is a
Fourier quasicrystal.
\end{theorem}

Consequently every nonzero element of the spaces of Theorems~\ref{thm:A}
and~\ref{thm:B} and Corollary~\ref{thm:C} gives a summation formula in which neither
$\nu$ nor $\widehat\nu$ is strongly tempered, so that it violates the hypotheses of
\cite{Gon23,GV25} in either reading. The super-polynomial clustering of our sets is
necessary for this, by Favorov's theorem quoted above.

The theorem follows from an exclusion statement for a single half-endpoint
carrier, in which strong temperedness is assumed on \emph{both} sides; the
notation is that of Section~\ref{sec:blocks} and Appendix~\ref{app:rapid}.

\begin{lemma}\label{lem:twosided}
Let $(\eps_j)$ and $(\delta_j)$ be strictly decreasing sequences of positive numbers
tending to zero, with $64\eps_1\le10^{-12}$ and $64\delta_1\le10^{-12}$, and put $\alpha_j=\frac12-\eps_j$ and
$\alpha'_j=\frac12-\delta_j$. If $T\in\Ncal(L_{\alpha'},L_\alpha)$ and both $T$ and $\F T$
are strongly tempered, then $T=0$.
\end{lemma}

\emph{Idea of the proof.} In the Zak coordinates of Appendix~\ref{app:rapid}, the two
support conditions say that the moment array of $T$ vanishes on one side of the
diagonal and that of $\F T$ on the other, and the two arrays are related by the
gauge $E$. Strong temperedness on both sides allows us to use moments whose
weights absorb the products of the small distances $\eps_j$ and $\delta_j$; in these
weights the weighted contraction of Proposition~\ref{prop:contraction} applies, and
its only solution is zero. For merely tempered $T$ these weights are not available,
in accordance with the existence of the nonzero block measure of
Theorem~\ref{thm:rapid}(a).

\begin{lemma}\label{lem:mixed}
In the situation of Lemma~\ref{lem:twosided}, let $V$ be as in Lemma~\ref{lem:zak},
with $\eta=\eps$ and $\eta'=\delta$, and let $\phi_{i,e}$ and $\phi'_{j,f}$ be built
from $\eps$ and $\delta$ as in \eqref{eq:newton}. Let $K=[-10^{-2},10^{-2}]^2$. There
are $C$ and $r$ such that for all $f\in C^\infty(\R^2)$, $i,j\ge0$ and
$e,f'\in\{0,1\}$,
\[
 |V(f\,\phi_{i,e}\otimes\phi'_{j,f'})|\le C\norm f_{C^r(K)}\,32^{i+j}(1+i+j)^r
 \min\Bigl\{\prod_{l\le i}\eps_l^2,\ \prod_{l\le j}\delta_l^2\Bigr\}.
\]
\end{lemma}

\begin{proof}
Write $T=\sum_\lambda c_\lambda\delta_\lambda$, with $\sum_\lambda|c_\lambda|(1+|\lambda|)^{-N}<\infty$,
and $\lambda=m_\lambda+\frac12+x_\lambda$ with $m_\lambda\in\Z$ and $x_\lambda\in\Xi_\eps$.
Since $T$ is strongly tempered, $T(\psi)=\sum_\lambda c_\lambda\psi(\lambda)$ for all
$\psi\in\Sc$, with absolute convergence, and the definition of $V$ gives
\[
 V(F)=\sum_\lambda c_\lambda(-1)^{m_\lambda}\chi(x_\lambda)\int\chi(y)F(x_\lambda,y)\,
 e^{-2\pi\ii(m_\lambda+\frac12)y}\,dy\qquad(F\in C^\infty(\R^2)).
\]
Since $V$ is supported in $[-10^{-12},10^{-12}]^2$, we may replace $F$ by
$(\chi_0\otimes\chi_0)F$, with $\chi_0$ as in Section~\ref{subsec:A-stage1}, so that only
the values of $F$ on $K$ matter. Integrating by parts $r\ge N$ times in $y$ gives
\[
 |V(F)|\le C\sum_\lambda|c_\lambda|(1+|m_\lambda|)^{-r}\max_{\nu\le r}\sup_y|\partial_y^\nu F(x_\lambda,y)|,
\]
an estimate in which $F$ is evaluated only at the atoms $x_\lambda$ and no
derivative in $x$ occurs. For $F=f\,\phi_{i,e}\otimes\phi'_{j,f'}$, we have
$\phi_{i,e}(x_\lambda)=0$ unless $x_\lambda=\pm\eps_l$ with $l>i$, and then
\[
 |\phi_{i,e}(x_\lambda)|\le|D_i(x_\lambda)|\le\prod_{l'\le i}a_{l'}\le20^i\prod_{l\le i}\eps_l^2,
\]
since $0\le z(x_\lambda)\le a_{l'}\le2\pi^2\eps_{l'}^2$ for $l'\le i$. The function
$\phi'_{j,f'}$ is a trigonometric polynomial whose coefficients have $\ell^1$ norm at
most $2^j$ and whose frequencies are at most $j+\frac12$ in modulus, so its
derivatives of order $\nu\le r$ are bounded by $2^j(2\pi(j+1))^r$. This gives the
bound with $\prod_{l\le i}\eps_l^2$. For the other bound we write
$V(F)=(gV)(\bar gF)$ with $\bar g=e^{2\pi\ii xy}$ and use the analogous representation
of $gV$ through the atoms of $\F T$, which follows from \eqref{eq:zakfourier} in
the same way; it involves derivatives in $x$ only, and evaluations at the
recentred atoms $y_\xi\in\Xi_\delta$ of $\F T$. The derivatives of $\bar g$ are
bounded on the chart.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:twosided}]
Use the moments \eqref{eq:moments} with $\kappa=10^{-3}$ and the weights
$\omega_i=\prod_{l\le i}64\eps_l$ and $\omega'_j=\prod_{l\le j}64\delta_l$, which are of
the form considered in Section~\ref{subsec:A-stage2} with $\rho=10^{-12}$. Since
$\min\{P^2,Q^2\}\le PQ$, Lemma~\ref{lem:mixed} shows that
$|M(fV)_{ie,jf'}|\le C\kappa^{-2}\norm f_{C^r(K)}2^{-i-j}(1+i+j)^r$. Hence the arrays
$M(fV)$ lie in $\mathbb H$, continuously in $f\in C^r(K)$, and Lemma~\ref{lem:gauge}
gives $M(gV)=E\,M(V)$. By Lemma~\ref{lem:flags}, $\mathbf A=M(V)$ and
$\mathbf B=M(gV)$ satisfy \eqref{eq:flags} and \eqref{eq:diagrel}, so
Proposition~\ref{prop:contraction} gives $\mathbf A=0$. By Lemma~\ref{lem:faithful},
$V=0$, and by \eqref{eq:zakphys} all clusters of $T$ vanish; hence $T=0$.
\end{proof}

\begin{proposition}\label{prop:blocknonstrong}
For all positive integers $P,R$, no nonzero element of $\M_{C(P,R)}$ is strongly
tempered, and no nonzero element has strongly tempered Fourier transform.
\end{proposition}

\begin{proof}
Let $\mu$ span $\M_{C(P,R)}$. Since $\F\mu=\zeta\mu$, strong temperedness of $\mu$ or
of $\F\mu$ implies strong temperedness of both. The inverse of the hole exchange
of Lemma~\ref{lem:holes} subtracts an element of $W_k$, a finite combination of
modulated translated lattice combs, which is strongly tempered on both sides; it
therefore produces an element of $\M_{L_{\gamma'}}$ that is strongly tempered on both
sides. The map $\mathcal R$ of Theorem~\ref{thm:rapid}(d) is built from finitely many
integer translations and multiplications by a trigonometric polynomial, so it maps
this element to an element of $\M_{L_\gamma}$ that is strongly tempered on both sides.
The sequence $(\eps_j)$ of \eqref{eq:blockparams} satisfies
$64\eps_1\le2^{-446}<10^{-12}$, so this element vanishes by
Lemma~\ref{lem:twosided} (with $\delta=\eps$). Since both maps are injective,
$\mu=0$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:nonstrong}]
Let $0\ne T\in\M_\Lambda$, so that $T=\sum_\ell c_\ell e_\ell$ is a finite sum with some
$c_\ell\ne0$ (Corollary~\ref{cor:exhaust}). Each $e_\ell$ is a dilate of a block
measure and hence, by Proposition~\ref{prop:blocknonstrong}, not strongly tempered.
The supports of the $e_\ell$ are pairwise disjoint, so
$|T|=\sum_\ell|c_\ell|\,|e_\ell|\ge|c_\ell|\,|e_\ell|$, and $|T|$ is not tempered. Since
$\F T=\sum_\ell c_\ell\zeta_{i(\ell)}e_{\ell^*}$, where $i(\ell)$ is the index of $\ell$,
is of the same form, $|\F T|$ is not tempered
either. The statement for subpairs follows from $\Ncal(A',B')\subseteq\M_\Lambda$.
\end{proof}

\begin{remark}\label{rem:nonstrong}
Since every element of $\M_\Lambda$ is a \emph{finite} combination of the $e_\ell$, no
infinite combination can produce a strongly tempered element; cf.\
Corollary~\ref{cor:nosums}. Theorem~\ref{thm:nonstrong} concerns the sets
constructed in Sections~\ref{sec:blocks} and~\ref{sec:exhaustion}. It does not
mean that countable dimension requires masses that grow faster than every
polynomial: on the different set of Theorem~\ref{thm:D} (Section~\ref{sec:fqcount}),
the Fourier quasicrystals form a space of dimension $\aleph_0$.
\end{remark}