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\title[Triangles in sets of positive upper density in the plane]{Triangles in sets of positive upper density in the plane}
\author{Generated by GPT-6 Astra Pro}
\date{November 6, 2026}
\hypersetup{pdftitle={Triangles in sets of positive upper density in the plane},
 pdfauthor={Generated by GPT-6 Astra Pro},
 pdfsubject={Triangle configurations and uniform smoothing}}
\begin{document}
\begin{abstract}
This note proves that every measurable subset of the plane with positive upper
Banach density contains a translate of a rotation of every sufficiently large
dilate of any fixed noncollinear triangle.
\end{abstract}
\maketitle

\section{Triangles and the counting estimate}\label{sec:intro}
The upper Banach density of a Lebesgue measurable set $E\subset\R^2$ is
\[
 d^*(E)=\limsup_{L\to\infty}\sup_{z\in\R^2}
       \frac{|E\cap(z+[0,L]^2)|}{L^2}.
\]
\begin{theorem}\label{main:density}
Let $\Delta\subset\R^2$ consist of three noncollinear points, and let
$E\subset\R^2$ be measurable with $d^*(E)>0$. There exists
$\lambda_0>0$ such that for every $\lambda\ge\lambda_0$ the set $E$
contains a translate of a rotation of $\lambda\Delta$.
\end{theorem}
Furstenberg, Katznelson, and Weiss \cite[Theorem~B]{FKW} proved that,
for all sufficiently large $\lambda$, every positive-radius neighborhood
of $E$ contains a congruent copy of $\lambda\Delta$. Ziegler \cite{Ziegler}
extended this neighborhood result to arbitrary finite configurations.
Bourgain's theorem for nondegenerate simplices \cite{Bourgain} gives
exact copies of large triangles in ambient dimension at least three.
Falconer, Kova\v{c}, and Yavicoli \cite[Theorem~1.4]{FKY} obtained
the planar conclusion when $d^*(E)>1/2$. Castro-Silva
\cite[Conjecture~1]{CS} conjectured the conclusion up to congruence
for sets of positive upper density in squares centered at the origin.
Theorem~\ref{main:density} gives orientation-preserving copies under
the upper Banach density hypothesis.

The proof combines the density Hales--Jewett theorem with an operator-norm
continuity estimate for the triangle-counting form, derived from the
smoothing theorem (Theorem~\ref{main:smoothing}). If there were an unbounded
sequence of missed scales, sufficiently many of them could be chosen
far apart. The density Hales--Jewett theorem would then give a positive
lower bound for a weighted sum of triangle counts at their nonempty
subsums. Each subsum is close to its largest constituent scale relative
to that scale, so continuity in operator norm makes the corresponding
count close to zero, contradicting the lower bound.

Write $J(x_1,x_2)=(-x_2,x_1)$. After translation, rotation, and a
fixed change of scale, a noncollinear triangle has vertex set
$\Delta_M=\{0,e_1,Me_1\}$ for $M=\alpha I+\beta J$ with $\beta\ne0$. Since $M$ commutes with rotations, a copy at
scale $s$ has vertices $x,x+s u_\theta,x+s Mu_\theta$, where
$u_\theta=(\cos\theta,\sin\theta)$, with $\theta\in\R/(2\pi\mathbb Z)$.
Put $v_\theta=Ju_\theta$ and let
$\sigma$ be the probability measure $\dd\theta/(2\pi)$ on the unit circle.
Integrable densities are identified with their measures. For a finite
measure $\nu$ on $\R^2$ and $s>0$, let $\nu_s$ denote its pushforward
by $u\mapsto su$, and put
\begin{equation}\label{pre:form}
 \Lambda_\nu(f_0,f_1,f_2)=\iint f_0(x)f_1(x+u)f_2(x+Mu)\dd\nu(u)\dd x.
\end{equation}
Denote its $L^2\times L^2\times L^\infty$ norm by $\|\Lambda_\nu\|$.
Cauchy--Schwarz gives $\|\Lambda_\nu\|\le\|\nu\|_{\mathrm{TV}}$,
and a change of variables gives $\|\Lambda_{\nu_s}\|=\|\Lambda_\nu\|$.

\begin{theorem}\label{main:smoothing}
For every probability density $\rho$ on $\R^2$,
\[
 \lim_{\varepsilon\downarrow0}
 \sup_{s>0}\big\|\Lambda_{(\sigma*\rho_\varepsilon-\sigma)_s}\big\|=0,
 \qquad \rho_\varepsilon(u)=\varepsilon^{-2}\rho(u/\varepsilon).
\]
This includes the Gaussian $\rho(u)=\exp(-\pi|u|^2)$.
\end{theorem}
Let $\xi_j$ denote the frequency at vertex $j$. The frequency dual to
the displacement $u$ in~\eqref{pre:form} is $b=\xi_1+M^T\xi_2$.
Stationary equations play a similar role in the trilinear estimates of
Christ \cite{Christ,ChristSub} and Christ, Durcik, and Roos \cite{CDR}.
Here their exact form forces $b=0$ almost everywhere on a set where
measurable frequency fields satisfy them (Section~\ref{sec:rig}).
Next, consider translations of the inputs on the scale $R^{-1}$. Only
stationary frequency triples contribute to the limiting correlations,
which gives cancellation on $|b|\asymp R$ (Section~\ref{sec:profile}). Finally, after fixed linear
changes of coordinates, the three vertex frequencies are pairwise
separated by distances comparable to $|b|$. A telescoping estimate
then sums the annuli without a rate of decay (Section~\ref{tree:sec}).

\subsection{From smoothing to triangles}

Theorem~\ref{main:smoothing} gives
\begin{equation}\label{den:new:cont}
 \lim_{r\to1}\|\Lambda_{\sigma_r-\sigma}\|=0 .
\end{equation}
Indeed, for fixed $t>0$ the norm is at most
$2\|\Lambda_{\sigma*\rho_t-\sigma}\|+\|(\sigma*\rho_t)_r-\sigma*\rho_t\|_1$.
The last term tends to zero by continuity of dilation in $L^1$.
Let $t\downarrow0$.

\begin{proof}[Proof of Theorem~\ref{main:density}]
It suffices to consider $\Delta_M$. Write
$D_s(f)=\Lambda_{\sigma_s}(f,f,f)$ and fix $0<\delta<d^*(E)$.
The density Hales--Jewett theorem \cite{FK} gives $n$ such that
every subset of $\{0,1,2\}^n$ of density at least $\delta/4$
contains a combinatorial line. There are $4^n-3^n$ lines, each with a nonempty set $J$ of
variable coordinates. The words of a line
agree off $J$ and take the values $0,1,2$, respectively, throughout $J$.

Suppose the missed scales are unbounded. Choose missed scales
$\lambda_1<\cdots<\lambda_n$ so rapidly increasing that for every
nonempty $J$, with $s_J=\sum_{j\in J}\lambda_j$ and $k=\max J$,
\begin{equation}\label{den:new:close}
 \|\Lambda_{\sigma_{s_J}-\sigma_{\lambda_k}}\|
 =\|\Lambda_{\sigma_{s_J/\lambda_k}-\sigma}\|
 < \delta/(4\cdot4^n).
\end{equation}
This is possible by \eqref{den:new:cont}, since the relative error
is at most $(\lambda_1+\cdots+\lambda_{k-1})/\lambda_k$.
Put $S=\max(1,\|M\|)\sum_{j=1}^n\lambda_j$. Choose a square $Q$
with $|E\cap Q|\ge\delta|Q|$ and with side so large that
$Q'=Q+[-S,S]^2$ satisfies $|Q'|\le2|Q|$.
Take $f=\one_{E\cap Q}$, replacing $E\cap Q$ by a Borel subset of
the same measure if necessary.

For fixed $\theta$ put $d_0=0$, $d_1=u_\theta$, $d_2=Mu_\theta$.
At base point $x$ mark every word $w$ satisfying
$f(x+\sum_j\lambda_j d_{w_j})=1$. The marked fraction integrates
to $|E\cap Q|$ over $Q'$, by translation in $x$. Consequently it
is at least $\delta/4$ on a set of area at least $\delta|Q|/2$.
At every such $x$ at least one whole line is marked.
For a line with variable coordinates $J$, translating $x$ by the contribution of
its fixed letters yields the three shifts $0,s_Ju_\theta,s_JMu_\theta$.
Integrating over all $x$ and then over $\theta$ therefore gives
\[
 \sum_{\varnothing\ne J\subset\{1,\ldots,n\}}
 3^{n-|J|}D_{s_J}(f)\ge \delta|Q|/2 .
\]
But $D_{\lambda_k}(f)=0$ at every chosen scale, so
\eqref{den:new:close} bounds every term $D_{s_J}(f)$ by
$\delta|Q|/(4\cdot4^n)$. Summing over the lines gives a contradiction.
\end{proof}

\section{Measurable stationarity}\label{sec:rig}

Write $p=(x,\theta)$, $\dd\mu(p)=\dd x\dd\sigma(\theta)$,
$L_0=0$, $L_1=I$, $L_2=M$, and $\pi_jp=x+L_ju_\theta$.
For constants $a_j=\xi_j$, differentiating the phase $\sum_j\xi_j\cdot\pi_jp$
in $x$ and $\theta$ gives its stationary equations. The same equations
for measurable planar vector fields $a_j$ are
\begin{equation}\label{rig:station}
 \sum_{j=0}^2a_j(\pi_jp)=0,\qquad
 v_\theta\cdot\big(a_1(\pi_1p)+M^Ta_2(\pi_2p)\big)=0 .
\end{equation}
The first equation is applied separately to each scalar coordinate. Its
finite-dimensional kernel permits passage to analytic fields.

\subsection{Circle operators}
Use the Fourier convention $\widehat f(\xi)=\int f(x)\e(-x\cdot\xi)\dd x$,
where $\e(t)=\exp(2\pi it)$, and write $\check f$ for the inverse
Fourier transform. For functions on $\R^2$, unadorned $L^p$ norms are
with respect to Lebesgue measure.

\begin{lemma}\label{rig:compact}
Let $N$ be an invertible linear similarity of $\R^2$ and $W$ a measurable weight with
$|W|\le1$. For
\[
 Tf(z)=\int W(z,\theta)f(z+Nu_\theta)\dd\sigma(\theta)
\]
there is a nondecreasing function $\omega_N(s)\to0$ as $s\downarrow0$ such that
$\|\sqrt e\,T\|_{2\to2}\le\omega_N(\|e\|_1)$ whenever $e\in L^1$ and $0\le e\le1$.
If $W$ vanishes for $z$ outside a bounded set, then $T$ is compact on $L^2$.
\end{lemma}
\begin{proof}
The difference of two independent uniform points on the circle of radius
$|Ne_1|$ has a density $w\in L^1$, since the map
$(\theta,\varphi)\mapsto N(u_\theta-u_\varphi)$ has nonzero Jacobian
except when the two tangents are parallel. Expanding $TT^*$ shows that
its kernel is bounded in absolute value by $w(y-z)$. Where $w\le H$,
the kernel of $\sqrt e\,TT^*\sqrt e$ is dominated by the rank-one kernel
$H\sqrt{e(z)e(y)}$. The remainder has norm at most
$\|w\one_{\{w>H\}}\|_1$ by Young's inequality. Hence
\[
 \|\sqrt e\,T\|_{2\to2}^2\le H\|e\|_1+\|w\one_{\{w>H\}}\|_1,
\]
which gives the claimed modulus by first choosing $H$ large. For bounded
$z$-support of $W$, the part of the $TT^*$ kernel with $w\le H$ is
Hilbert--Schmidt. Thus $TT^*$, and hence $T$, is compact.
\end{proof}

When $M$ is a rotation, Greenleaf, Iosevich, Krause, and Liu \cite{GIKL}
proved a restricted strong-type estimate at the $L^2\times L^2\to L^2$
endpoint for the corresponding bilinear average.

\begin{lemma}\label{pre:circleimprove}
The triangle form satisfies
\begin{equation}\label{pre:triple}
 |\Lambda_\sigma(f_0,f_1,f_2)|\le C_M\prod_{j=0}^2\|f_j\|_2 .
\end{equation}
On any permutation of $L^2\times L^2\times L^\infty$, its norm is at
most $\omega_M(s)\to0$ when one of the $L^2$ inputs is supported on a
set of area at most $s$. For every finite measure $\kappa$, both estimates hold for $\sigma*\kappa$ with
an extra factor $\|\kappa\|_{\mathrm{TV}}$.
\end{lemma}
\begin{proof}
For \eqref{pre:triple}, partition the circle into small arcs $I$.
Cauchy--Schwarz in the angle gives
\begin{align*}
 &\int\left|\int_I f_1(x+u_\theta)f_2(x+Mu_\theta)\dd\sigma(\theta)\right|^2\dd x\\
 &\qquad\le\int_{I^2}\int |f_1(y)|^2
          |f_2(y+Mu_\varphi-u_\theta)|^2\dd y\dd\sigma(\theta)\dd\sigma(\varphi).
\end{align*}
On the diagonal the absolute Jacobian of
$(\theta,\varphi)\mapsto Mu_\varphi-u_\theta$ is $|\beta|$.
Choose the arcs so that on each $I^2$ this map is one-to-one with Jacobian
bounded below. Changing variables bounds the last integral by
$C_M\|f_1\|_2^2\|f_2\|_2^2$. Sum over $I$ and pair with $f_0$.
For the second assertion, bound one input by its supremum. On the
opposite edge use Lemma~\ref{rig:compact} with $W=1$ and $e$ the indicator
of the small support. Take $\omega_M$ to be the maximum of the moduli for
the finitely many edge maps $L_j-L_i$. Convolution by $\kappa$ integrates
translations of $f_1,f_2$ by $w,Mw$ against $\dd\kappa(w)$.
\end{proof}

\subsection{The scalar kernel}

For a set $E$ of configurations with vertices in bounded measurable
sets $X_j$, the degree at $z\in X_j$ is
\[
 d_{j,E}(z)=\int\one_E(z-L_ju_\theta,\theta)\dd\sigma(\theta).
\]
A \emph{core of degree $c$} has $d_{j,E}\ge c>0$ almost everywhere
on each of its vertex sets. Every $E$ of positive measure contains a
positive-measure core. Indeed, fix $0<c<\mu(E)/\sum_j|X_j|$.
At each step simultaneously delete all vertices of current degree less than $c$
and their incident configurations. The total loss is at most
$c\sum_j|X_j|$ since each vertex is deleted once. On the decreasing
intersection all remaining degrees are at least $c$ by monotone
convergence.

On a core $E$ set $\mathcal H_E=\bigoplus_jL^2(X_j;\R)$. Define
$D_E:\mathcal H_E\to L^2(E,\mu)$ by $D_E f=\sum_j f_j\circ\pi_j$,
and put $K_E=\ker D_E$.

\begin{lemma}\label{rig:kernel}
The space $K_E$ is finite dimensional and
$\operatorname{dist}(f,K_E)\le C_E\|D_Ef\|_{L^2(E,\mu)}$.
For each fixed $t>0$, if $F\subset E$ is a core of degree $t$
on vertex sets $Y_j\subset X_j$ and $\mu(E\setminus F)$ is
sufficiently small depending on $E,t$,
restriction is an isomorphism from $K_E$ onto $K_F$.
\end{lemma}
\begin{proof}
The diagonal blocks of $D_E^*D_E$ are multiplication by $d_{j,E}\ge c$.
Each off-diagonal block is a circle operator with weight bounded by one
and shift $(L_i-L_j)u_\theta$. These differences are nonzero similarities,
so the blocks are compact by Lemma~\ref{rig:compact}. Conjugating
by the inverse square root of the diagonal gives the identity plus
a compact self-adjoint operator. Its positive spectrum is bounded
away from zero. Conjugating back gives finite dimensionality and
the distance bound.

If $f$ belongs to the kernel on a core of degree $t$, the block
equations and Lemma~\ref{rig:compact} give
\[
 \|\sqrt e\,f_j\|_2\le C_M t^{-1}\omega_M(\|e\|_1)\|f\|_2
 \qquad (0\le e\le1).
\]
Indeed, the $j$th block equation expresses the degree times $f_j$ as a sum
of two weighted circle operators applied to the other components, up to sign. With $\eta=\mu(E\setminus F)$,
each omitted vertex loses all its incidences, so $|X_j\setminus Y_j|\le\eta/c$.
The displayed estimate with $e=\one_{X_j\setminus Y_j}$ makes restriction
injective on $K_E$ for small $\eta$. Conversely extend $f\in K_F$ by zero.
The degrees $e_j$ of $E\setminus F$ are bounded by one and have integral
$\eta$. Consequently
\[
 \|D_Ef\|_2^2\le3\sum_j\|\sqrt{e_j}\,f_j\|_2^2
       \le C_Mt^{-2}\omega_M(\eta)^2\|f\|_2^2 .
\]
The distance bound makes projection onto $K_E$ injective on this copy
of $K_F$ for small $\eta$.
The two injections give equality of dimensions and hence surjectivity
of restriction.
\end{proof}

\subsection{Rigidity of the stationary equations}

\begin{lemma}\label{rig:pieces}
Every positive-measure core $E$ has a positive-measure subset where
each component of a basis of $K_E$ agrees, at its vertex, with a real
analytic function defined on all of $\R^2$.
\end{lemma}
\begin{proof}
Use an orthonormal basis, write $\Phi_j(x)$ for the row of its values
at vertex $j$, and extend by zero outside $X_j$. Its three Gram matrices
$\int\Phi_j^T\Phi_j$ sum to the identity. Let $E_h$ be $E$ translated
by $-h$ in the base variable, with basis $\Phi_j(\cdot+h)$. Then
$\mu(E\triangle E_h)\to0$.

There is a common core $F_h\subset E\cap E_h$ of degree $c/4$ with
$\mu(E\setminus F_h)\to0$. Indeed, start by deleting
$\eta=\mu(E\setminus E_h)$ incidences from $E$, then prune at $c/4$.
If $V$ is the sum of the measures of the pruned vertex sets, at most
$cV/4$ further incidences are lost. Their original degrees give
$cV\le3(\eta+cV/4)$, so the total loss is at most $4\eta$.
Since $\mu(E_h)=\mu(E)$ the same bound holds for $E_h$.
Lemma~\ref{rig:kernel} applies to both cores for small $h$, with the
same threshold by translation. The surviving vertex sets $Y_{j,h}$
lie in $X_j\cap(X_j-h)$ up to null sets. Since the two bases restrict
to bases of $K_{F_h}$, there is a unique invertible matrix $C(h)$ with
\begin{equation}\label{rig:trans}
 \Phi_j(x+h)=\Phi_j(x)C(h),\qquad |X_j\setminus Y_{j,h}|\to0 .
\end{equation}

By absolute continuity the Gram matrices on $Y_{j,h}$ sum to $I+o(1)$.
Multiply \eqref{rig:trans} by $\Phi_j^T$, integrate, and use $L^2$
translation continuity to get $C(h)\to I$. To compose \eqref{rig:trans}
at $h,k,h+k$, restrict to
$Y_{j,h}\cap(Y_{j,k}-h)\cap Y_{j,h+k}$. The complements in $X_j$
again tend to zero in measure, since $|X_j\triangle(X_j-h)|\to0$.
The summed Gram matrix on the intersections is invertible for small
$h,k$, so $C(h+k)=C(h)C(k)$. By the continuous one-parameter group
theorem on the coordinate axes, $C(h)=P(h):=\exp(h_1T_1+h_2T_2)$ near
zero for commuting real matrices $T_1,T_2$.

Write $z_j=\pi_jp$. For every triple of sufficiently small $h_j$,
the equalities $\Phi_j(z_j)=\Phi_j(z_j-h_j)P(h_j)$ hold simultaneously
on a subset of $E$ of measure at least $\mu(E)/2$. Indeed, the loss is
at most $\sum_j|X_j\setminus(Y_{j,h_j}+h_j)|=o(1)$ since the vertex
degrees are at most one. The equality event is jointly measurable in
$p,h_0,h_1,h_2$ by taking Borel versions of the $\Phi_j$. Integrate
over the three small shifts, then change variables $b_j=z_j-h_j$ at
fixed $p$. Fubini gives fixed $b_0,b_1,b_2$ and a positive-measure
subset of $E$ where $\Phi_j(z_j)=\Phi_j(b_j)P(z_j-b_j)$ for all $j$.
These expressions are real analytic functions of their vertex variables.
\end{proof}

\begin{lemma}\label{rig:exact}
On a core $E$ of positive degree, every square-integrable measurable
solution of \eqref{rig:station} satisfies
$a_1(\pi_1p)+M^Ta_2(\pi_2p)=0$ for almost every $p\in E$.
\end{lemma}
\begin{proof}
If the asserted equality fails on a set of positive measure, prune that
set to a positive-measure core and use this new core in what follows.
The coordinates of the vector triple belong to its scalar kernel.
On the positive-measure subset given by Lemma~\ref{rig:pieces} the
three fields therefore agree with global real analytic functions.
An analytic function on a connected real analytic manifold that vanishes
on a set of positive volume vanishes identically. Thus these functions satisfy
\eqref{rig:station} on all of $\R^2\times S^1$.

Write $A=a_0(x)$, $B=a_1(x+u)$, $C=a_2(x+Mu)$ for these analytic fields,
where $u=u_\theta$ and $v=Ju$. The equations give $B+M^TC=tu$ for a
smooth scalar $t=t(x,\theta)$, and
\[
 (M^T-I)C=A+tu,\qquad -(M^T-I)B=M^TA+tu .
\]
Put $K=Da_0(x)$, $X_1=\partial_\theta-v\cdot\nabla_x$, and
$X_2=\partial_\theta-Mv\cdot\nabla_x$. Since $X_1B=X_2C=0$,
\[
 (X_1t)u+tv=M^TKv,\qquad (X_2t)u+tv=KMv .
\]
The $v$-component of the second identity gives $t=v^TKMv$, which
is even under $(u,v)\mapsto(-u,-v)$. The spatial transport terms
$(L_jv\cdot\nabla_x t)u$ are therefore even. Taking the odd part
of both identities gives
$\partial_\theta(tu)=M^TKv=KMv$. Hence $M^TK=KM$, and integration
in $\theta$ gives $tu=KMu+c(x)$. Both $tu$ and $KMu$ are odd, so
$c=0$. Thus $KM$ preserves every line and is scalar. In particular $K$ commutes
with $M$. It follows that $(M^T-M)K=0$. As $M^T-M=-2\beta J$ is
invertible, $K=0$ and $t=0$. This contradicts the choice of the core.
\end{proof}

\section{Cancellation on one frequency annulus}\label{sec:profile}

\begin{proposition}\label{loc:cor:annulus}
Let $\Psi\in L^1(\R^2)$ have integral zero and put
$\Psi_R(u)=R^2\Psi(Ru)$. For every bounded box $Q$ and $C<\infty$,
\[
 \sup_{|f_j|\le\one_Q,\ \operatorname{Lip}f_j\le CR}
 |\Lambda_{\sigma*\Psi_R}(f_0,f_1,f_2)|\longrightarrow0
 \qquad(R\to\infty).
\]
\end{proposition}
\begin{proof}
Take any sequence $R_n\to\infty$ and admissible inputs $f_{j,n}$.
Extract local correlations of the translates
$f_{j,n}(z+s/R_n)$. All weak-* limits below are in $L^\infty$ of
the vertex variable $z$ or the configuration variable $p=(x,\theta)$.
First extract a common subsequence for the two-point products at each
vertex and the three-point products on configuration space, with all
shifts rational. Their uniform Lipschitz bounds in the shift variables
give versions of the limits that extend continuously to all real shifts
off a single null set in each space. The same bounds give weak-* convergence
to these extensions at every real shift.

At each vertex, the weak-* limit of
$f_{j,n}(z+s/R_n)\overline{f_{j,n}(z+t/R_n)}$ depends only on
$s-t$, by translating $z$ by $t/R_n$ in a compactly supported test.
The limits of squared absolute values of finite linear combinations of
translates show that for almost every $z$ this continuous function of
$s-t$ is positive definite.
By Bochner's theorem it equals
$\int \e((s-t)\cdot\xi)\dd\nu_{j,z}(\xi)$ for a positive measure of
mass at most one. The measures form measurable kernels, as follows
by testing against smooth functions and Fourier inversion.

For $e_s(\xi)=\e(s\cdot\xi)$ and a trigonometric polynomial
$a=\sum_s c_s e_s$, set $F_{j,n}[a](z)=\sum_s c_s f_{j,n}(z+s/R_n)$.
Write $\nu_j=\nu_{j,\pi_jp}$. For $G=\sum_d b_d\otimes c_d$, with
$b_d,c_d$ trigonometric polynomials, put
\[
 G_n(p)=\sum_d F_{1,n}[b_d](\pi_1p)F_{2,n}[c_d](\pi_2p).
\]
Denote the weak-* limit of $F_{0,n}[a](\pi_0p)G_n(p)$ by $\ell_p(a,G)$.
This defines a bilinear form satisfying
\begin{equation}\label{corr:CS}
 |\ell_p(a,G)|\le\|a\|_{L^2(\nu_0)}\|G\|_{L^2(\nu_1\otimes\nu_2)} .
\end{equation}
Indeed, the weak-* limit of $|G_n|^2$ is the square of the second norm in
\eqref{corr:CS}. To see this, expand the square and test against a
compactly supported function $\varphi$ of $p$ with $|\varphi|\le1$. In the coordinates
$z=x+u_\theta$, each term pairs a product of two translates at vertex
$1$ with such a product at vertex $2$ through the operator of
Lemma~\ref{rig:compact}, with shift $(M-I)u_\theta$ and weight
$\varphi(z-u_\theta,\theta)$. This operator is compact. The
one-vertex correlation formula gives the respective weak $L^2$
limits on bounded vertex boxes. Likewise $|F_{0,n}[a](\pi_0 p)|^2$
has weak-* limit $\|a\|_{L^2(\nu_0)}^2$. Cauchy--Schwarz on shrinking
configuration balls gives \eqref{corr:CS} by Lebesgue differentiation.
Rational shifts and rational complex coefficients give a single
exceptional null set. Continuity then extends the assertions to real
shifts and all coefficients. Characters span a dense subspace of $L^2$
of each finite measure by uniqueness of Fourier transforms. Thus
\eqref{corr:CS} extends $\ell_p$ to the two indicated $L^2$ spaces.

For $s\in\R^2$ and $h\in\R$, changing variables by
$(x,\theta)\mapsto(x+s/R_n,\theta+h/R_n)$ in a continuous test gives
\begin{equation}\label{corr:inv}
 \ell_p(a,G)
 =\ell_p\big(e_s a,(e_{s+h v_\theta}\otimes e_{s+hMv_\theta})G\big).
\end{equation}
Indeed the displacement at vertex $j$ differs from
$(s+h L_jv_\theta)/R_n$ by $O_M(h^2/R_n^2)$, so the input error
tends to zero. The weak-* limit at these angle-dependent shifts
follows by approximating them by simple functions, using the uniform
Lipschitz bound. Rational parameters and continuity again give
simultaneous invariance for almost every $p$. At each such $p$, density
and \eqref{corr:CS} extend this identity to arbitrary $a$ and $G$, since
multiplication by each character preserves the $L^2$ norm.

In \eqref{corr:inv} set $s=0$ and average over $0\le h\le H$.
Since $H^{-1}\int_0^H\e(hc)\dd h\to\one_{\{0\}}(c)$, dominated convergence
in $L^2(\nu_1\otimes\nu_2)$ and \eqref{corr:CS} restrict $G$ to the hyperplane
\[
 v_\theta\cdot\xi_1+Mv_\theta\cdot\xi_2=0.
\]
For almost every $p$ the product measure on this hyperplane is
concentrated on pairs of atoms. Indeed, for $j=1,2$ put
$d_j(\theta)=L_jv_\theta$. With $z=\pi_jp$ fixed, Fubini gives
\[
 \int\!\iint \one_{\{\xi\ne\xi',\ d_j(\theta)\cdot(\xi-\xi')=0\}}
     \dd\nu_{j,z}(\xi)\dd\nu_{j,z}(\xi')\dd\sigma(\theta)=0,
\]
since distinct points contribute at most two angles.
As $(x,\theta)\mapsto(\pi_jp,\theta)$ preserves measure, for almost
every $p$ each positive-mass fiber of
$\xi\mapsto d_j(\theta)\cdot\xi$ is concentrated at a single point.
Write $\eta_j$ for the pushforward of $\nu_j$ by this projection.
By Fubini the product mass of the hyperplane is
$\sum_{t:\,\eta_1(\{t\})>0}\eta_1(\{t\})\eta_2(\{-t\})$.
Each positive term therefore comes from a pair of point atoms.

For a pair of atoms $(\xi_1,\xi_2)$ on the hyperplane, set
$G=\one_{\{(\xi_1,\xi_2)\}}$. With $h=0$, average \eqref{corr:inv}
over $s\in[0,H]^2$ and let $H\to\infty$. The characters in $s$
restrict $a$ in $L^2(\nu_0)$ to the single point
$\xi_0=-\xi_1-\xi_2$, again by \eqref{corr:CS}. This gives zero
unless $\xi_0$ is an atom. Thus the only contributing triples of
atoms satisfy both equations
\[
 \xi_0+\xi_1+\xi_2=0,\qquad
 v_\theta\cdot(\xi_1+M^T\xi_2)=0 .
\]

For almost every $p$, all such atom triples satisfy
$\xi_1+M^T\xi_2=0$. To cover all atoms by countably many measurable
fields, normalize each nonzero $\nu_{j,z}$ to a probability measure and
take all pairs of quantiles of its two coordinate marginals at levels
in $\mathbb Q\cap(0,1)$. The quantiles are measurable in $z$ by the distribution
function formula. Each coordinate of an atom is an atom of the
corresponding marginal and hence is attained on an interval of quantile
levels of positive length. Set the fields to zero where $\nu_{j,z}=0$.
If the two displayed equations and $\xi_1+M^T\xi_2\ne0$ held on a
positive-measure set for any triple of these fields, restrict to a
positive-measure subset on which the vertices and the fields are bounded,
then prune to a positive-measure core.
Lemma~\ref{rig:exact} gives a contradiction.

For $|G|\le1$, truncation on the hyperplane to finitely many atom pairs
has $L^2(\nu_1\otimes\nu_2)$ error at most the square root of the omitted
mass. On each contributing pair $\e(y\cdot(\xi_1+M^T\xi_2))=1$.
Thus \eqref{corr:CS} gives $\ell_p(1,e_y\otimes e_{My})=\ell_p(1,1)$
for every $y\in\R^2$. Take $w\in C_c(\R^2)$ equal to one on $Q$.
Every starting sequence has a subsequence along which dominated
convergence and $\int\Psi=0$ give
\[
 \lim_{n\to\infty}\Lambda_{\sigma*\Psi_{R_n}}(f_{0,n},f_{1,n},f_{2,n})
 =\int\Psi(y)\int w(x)\ell_p(1,e_y\otimes e_{My})\dd\mu(p)\dd y=0.
 \qedhere
\]
\end{proof}

\subsection{Localization of the inputs}
The following observation will also give the endpoint in Theorem~\ref{main:smoothing}.
Let $\mu_n=\sigma*\kappa_n$, where $\sup_n\|\kappa_n\|_{\mathrm{TV}}<\infty$
and $|\mu_n|(\{|u|>U\})\to0$ for some fixed $U$. Let $P_n$ be convolution
by measures supported in the unit ball and uniformly bounded in total variation. This
includes the identity operator.
\begin{lemma}\label{localize}
Suppose that for each fixed triple of smooth compactly supported cutoffs
$\chi_l$,
\[
 \sup_{\|f_l\|_\infty\le1}
 |\Lambda_{\mu_n}(\chi_0P_nf_0,\chi_1P_nf_1,\chi_2P_nf_2)|\to0.
\]
Then $\Lambda_{\mu_n}(P_nf_0,P_nf_1,P_nf_2)$ has norm tending to zero
on every permutation of $L^2\times L^2\times L^\infty$.
\end{lemma}
\begin{proof}
Take a smooth nonnegative partition $\sum_q\chi_{0,q}(x)=1$ of the base
plane by translates of one cutoff on the unit lattice. For $l=1,2$ take
translated cutoffs $0\le\chi_{l,q}\le1$ equal to one at $x+L_lu$ whenever
$x\in\operatorname{supp}\chi_{0,q}$ and $|u|\le U$. The sum of the products
lies in $[0,1]$ and equals one for $|u|\le U$. Thus localization costs
$o(1)$ in the desired norm by the tail assumption and Cauchy--Schwarz in $x$.

For a localized form restrict the two $L^2$ inputs to a fixed box
translated by $q$, containing the unit neighborhoods of the cutoff
supports. This is exact since $P_n$ uses only that neighborhood.
Divide by their local $L^2$ norms, omitting zero terms. Truncate the normalized
inputs at magnitude $T\ge1$. Each discarded part has $L^2$ norm at
most one and is supported on a set of area at most $T^{-2}$. Its form
error is $O(\omega_M(T^{-2}))$ for a unit bounded input. Indeed,
write its convolution under $P_n$ as an integral of translates and
apply Lemma~\ref{pre:circleimprove} to each translate after multiplication
by the cutoff. For the truncated inputs the
hypothesis gives $o_n(1)T^2$, uniformly in $q$ by translation invariance.
Restore the local norms and sum their products by Cauchy--Schwarz and
bounded overlap. First let $n\to\infty$ and then $T\to\infty$.
\end{proof}

\begin{corollary}\label{ann:cor:global}
Let $\psi$ be an even smooth function supported in a compact annulus
away from zero and set $K_R=\sigma*(R^2\check\psi(R\,\cdot))$.
For inputs with Fourier supports in $B(0,C_0R)$, where $C_0$ is fixed
and support for a bounded input is understood distributionally,
\[
 |\Lambda_{K_R}(f_0,f_1,f_2)|\le a(R)\|f_i\|_\infty\prod_{j\ne i}\|f_j\|_2,
 \qquad i=0,1,2,
\]
where $a(R)\to0$ as $R\to\infty$.
\end{corollary}
\begin{proof}
Choose $p$ Schwartz with $\widehat p=1$ on a neighborhood of $B(0,C_0)$, so
$p_R=R^2p(R\,\cdot)$ reproduces the given inputs. Multiply $p_R$ by a
fixed smooth cutoff supported in the unit ball and equal to one near zero,
obtaining a kernel $q_R$ with $\|q_R-p_R\|_1=o(1)$ and
$\|\nabla q_R\|_1=O(R)$. Apply Lemma~\ref{localize} along any sequence
$R\to\infty$ with $P_nf=q_R*f$. The total variation of $K_R$ outside
$B(0,2)$ tends to zero because $\check\psi$ is Schwartz. For bounded
inputs the localized projections have fixed supports, bounded magnitudes,
and Lipschitz constants $O(R)$, so Proposition~\ref{loc:cor:annulus}
gives the hypothesis. Replacing $q_R$ by $p_R$ costs $o(1)$ in form norm
by Young's inequality and the uniform total variation bound for $K_R$.
\end{proof}

\section{Summing the frequency annuli}\label{tree:sec}
Choose a smooth radial $\chi$ equal to one on $B(0,1)$ and supported
in $B(0,2)$. Put $\psi(b)=\chi(b/2)-\chi(b)$ and write $\Lambda_r$
for the form with kernel $\sigma*(r^2\check\psi(r\,\cdot))$. For Schwartz
inputs it has the Fourier representation
\[
 \iint\widehat f_0(-\xi_1-\xi_2)\widehat f_1(\xi_1)\widehat f_2(\xi_2)
       \widehat\sigma(b)\psi(b/r)\dd\xi_1\dd\xi_2,
 \qquad b=\xi_1+M^T\xi_2 .
\]
\begin{lemma}\label{tree:lem:tail}
For dyadic $R\ge1$ and $\|f_j\|_2,\|f_j\|_\infty\le B$, one has
\[
 \sum_{r=2^k\ge R}|\Lambda_r(f_0,f_1,f_2)|
       \le C_{M,B}\big(\sup_{r\ge R} a(r)\big)^{1/2},
\]
with $a(r)$ from Corollary~\ref{ann:cor:global} for a sufficiently large
fixed $C_0=C_0(M)$.
\end{lemma}
\begin{proof}
Set $A=M^T$ and $\xi_0=-\xi_1-\xi_2$. In the coordinates
\[
 z_0=(A-I)^{-1}\xi_0,\qquad z_1=-A^{-1}\xi_1,\qquad z_2=\xi_2
\]
the three differences are
\[
 z_0-z_2=-(A-I)^{-1}b,\quad z_1-z_2=-A^{-1}b,\quad
 z_1-z_0=A^{-1}(A-I)^{-1}b.
\]
Thus all pairwise distances are comparable to $r$ on the support of
$\psi(b/r)$. Translating all three frequency supports by $w$ in these coordinates
amounts to modulating the inputs by $((A-I)w,-Aw,w)$. The factors
cancel in~\eqref{pre:form}, so the form is unchanged.

Let $\mu$ be the sum of the three pushforwards of
$|\widehat f_j(\xi_j)|^2\dd\xi_j$ to the $z$-plane, so that
$\mu(\R^2)\le3B^2$. At each $r$, partition each input smoothly in
$z$-frequency on a square lattice of mesh $hr$, using translates and
dilates of a single bump. The projections have uniformly bounded
convolution $L^1$ norms and their support squares have side at most $3hr$.
Choose $h>0$ so small that the distances between squares in every
contributing triple are comparable to $r$. Each square has only
$O_M(1)$ possible partners. The expansion converges absolutely by
\eqref{pre:triple}, Cauchy--Schwarz on two families of projections,
and bounded overlap. The identity follows by taking finite projection
sums and then limits in $L^2$.

For a contributing triple of squares $Q_0,Q_1,Q_2$, write $m\le n\le p$
for their $\mu$-masses in increasing order. The convolved version of
\eqref{pre:triple}, with convolution factor of $L^1$ norm $\|\check\psi\|_1$,
bounds its contribution by $C(mnp)^{1/2}$. For a second bound translate
one square to the origin in the common frequency coordinates. All three
supports now lie in $B(0,C_Mr)$ in their original frequency planes.
Apply Corollary~\ref{ann:cor:global} with the largest-mass input in the
bounded slot to obtain $C_B a(r)(mn)^{1/2}$. Taking the geometric mean
bounds the contribution by $C_B a(r)^{1/2} n p^{1/4}$.

To sum this bound, put
\[
 H(t)=\int\mu(B(z,t))^{1/4}\dd\mu(z),
 \qquad 0\le H(t)\le\mu(\R^2)^{5/4}.
\]
Fix $c>0$ small enough that the $cr$-neighborhoods of $Q_0,Q_1,Q_2$
are disjoint. They lie in a ball $B_*$ which in turn lies in $B(z,Cr)$
for every $z\in\bigcup_j Q_j$, with $C$ depending only on $M$.
At least two of the neighborhoods have mass at most $\mu(B_*)/2$.
For $z$ in either corresponding square,
\[
 \mu(B(z,Cr))^{1/4}-\mu(B(z,cr))^{1/4}
   \ge(1-2^{-1/4})\mu(B_*)^{1/4}\ge(1-2^{-1/4})p^{1/4}.
\]
Those two squares have total mass at least $n$. Integrate the difference
over them and sum over all triples. Bounded overlap and the bounded
number of partners give
\[
 |\Lambda_r(f_0,f_1,f_2)|\le C_{M,B}a(r)^{1/2}\big(H(Cr)-H(cr)\big).
\]
Choose an integer $D$ with $2^Dc\ge C$. Since $H$ is increasing,
the sum of the differences over dyadic $r\ge R$ telescopes to at most
$D\mu(\R^2)^{5/4}$. This proves the lemma.
\end{proof}

\begin{corollary}\label{tree:cor:smoothing}
Uniformly for $\|f_j\|_2,\|f_j\|_\infty\le B$,
\[
 \Lambda_{\sigma*\rho_\varepsilon-\sigma}(f_0,f_1,f_2)\to0
 \quad(\varepsilon\downarrow0).
\]
\end{corollary}
\begin{proof}
For a finite measure $\kappa$, smoothing its $u$ variable by
$\rho_\varepsilon$ averages translates of the two inputs at vertices
$1,2$, by $v,Mv$ respectively. These translations preserve both
norms. Thus Lemma~\ref{tree:lem:tail} bounds the tails $r\ge R$
for both $\sigma$ and $\sigma*\rho_\varepsilon$ by the same quantity.
The remaining kernel of the difference is
$\sigma*(\rho_\varepsilon*\check\chi_R-\check\chi_R)$,
where $\check\chi_R(u)=R^2\check\chi(Ru)$. For fixed $R$ its
total variation tends to zero by translation continuity in $L^1$.
The low and high parts exhaust the form, since translation continuity
in $L^2$ and \eqref{pre:triple} give
$\Lambda_{\sigma*\check\chi_S}(f_0,f_1,f_2)\to
\Lambda_{\sigma}(f_0,f_1,f_2)$ as $S\to\infty$, and likewise
after convolution with $\rho_\varepsilon$.
First fix $R$ large and then let $\varepsilon\downarrow0$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{main:smoothing}]
Take $P_n$ to be the identity in Lemma~\ref{localize} and apply it
to $\mu_n=\sigma*\rho_{\varepsilon_n}-\sigma$ for an arbitrary
$\varepsilon_n\downarrow0$. The hypothesis follows from
Corollary~\ref{tree:cor:smoothing}, since bounded inputs times fixed
cutoffs have bounded $L^2$ norms. The tail outside $B(0,2)$ is at most
$\int_{|w|>1}\rho_{\varepsilon_n}(w)\dd w\to0$.
This proves the estimate at unit scale. Dilation invariance gives
uniformity for all scales.
\end{proof}

\begin{remark*}
One also recovers the Fourier inverse estimate. For $g\in L^2\cap L^\infty$
with bounded Fourier transform, fix $t>0$ and put $k=\sigma*\rho_t$ with $\rho$ a smooth,
even, compactly supported probability density. Then
\[
 \sup_{\|f_0\|_2,\|f_1\|_2\le1}|\Lambda_\sigma(f_0,f_1,g)|
 \le\|\Lambda_{\sigma-k}\|\,\|g\|_\infty +C_M\|\widehat k\|_1\|\widehat g\|_\infty.
\]
For Schwartz $f_0,f_1$, the function paired with $g$ in $\Lambda_k$ is
$z\mapsto\int k(u)f_0(z-Mu)f_1(z+(I-M)u)\dd u$, which is Schwartz
since $k\in C_c^\infty$. Fourier pairing is thus valid for $g\in L^2$.
At fixed $b=\xi_1+M^T\xi_2$, apply Cauchy--Schwarz in $\xi_2$ to
$\widehat f_0(-b+(M^T-I)\xi_2)$ and $\widehat f_1(b-M^T\xi_2)$.
Both linear parts are invertible. Integration against $|\widehat k(b)|$
gives the second term, and density extends the bound to $L^2$ tests.
Consequently the left side tends to zero for any sequence
$g_n\in L^2\cap L^\infty$ with $\sup_n\|g_n\|_\infty<\infty$ and
$\|\widehat g_n\|_\infty\to0$, by first taking the upper limit in $n$
and then letting $t\downarrow0$.
\end{remark*}

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\end{document}