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+% ============================================================
+% ANNULAR VARIATION OF THE TRIANGULAR HILBERT TRANSFORM
+% AT THE SYMMETRIC POINT
+%
+% An expository rewrite. Written so that a determined
+% high-school student can read it from page one, and so that a
+% graduate student can find every technical step.
+%
+% HOW TO READ THIS FILE
+%
+% Every line that starts with % is a remark for you, the reader.
+% The mathematics comes afterward. The order is always:
+% story -> statement -> proof.
+% Whenever a formula appears, the next sentence says what it
+% means in English. Whenever a symbol is new, its first use is
+% defined right there. Skim the technical sections the first
+% time through; come back to them when a proof points you there.
+%
+% A formula is bookkeeping. The sentence next to it is the idea.
+% ============================================================
+
+\documentclass[11pt]{article}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{microtype}
+\usepackage[colorlinks=true,linkcolor=blue!45!black,urlcolor=blue!45!black]{hyperref}
+\newcommand{\R}{\mathbb R}
+\newcommand{\C}{\mathbb C}
+\newcommand{\dd}{\,\mathrm d}
+\DeclareMathOperator{\tr}{tr}
+\DeclareMathOperator{\Dil}{Dil}
+\newcommand{\HS}{\mathrm{HS}}
+\newcommand{\op}{\mathrm{op}}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{definition}
+\newtheorem{definition}[theorem]{Definition}
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+\numberwithin{equation}{section}
+
+\title{Annular variation of the triangular Hilbert transform\\at the symmetric point\\[4pt]
+ \large a rewrite for students}
+\author{after the research paper}
+\date{}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+% ============================================================
+\section{The big picture}\label{sec:picture}
+% ============================================================
+
+% This section assumes only that you know what an integral and a
+% limit are. Everything else is built here. Read it twice.
+%
+% The whole paper is about ONE operator. It takes two functions
+% F and G of two variables (x,y) and returns a third function of
+% (x,y). The recipe: step from (x,y) in the horizontal direction
+% through F, step from (x,y) in the vertical direction through G,
+% multiply, and weight the step length t by 1/t. Then integrate
+% over t. The question: is this operator stable under EVERY way
+% of truncating the integration in t, including a way that changes
+% from point (x,y) to point (x,y)? The answer is yes, at a cost
+% that grows like the square root of the number of truncations.
+
+\subsection{The Hilbert transform in one dimension}\label{subsec:hilbert-1d}
+
+% Start in one variable. For a smooth function f that vanishes
+% far away, the Hilbert transform of f is
+\begin{equation}\label{eq:HT-1d}
+ Hf(x)=\frac1\pi\,\operatorname{p.v.}\!\int_{-\infty}^{\infty}
+ \frac{f(x+t)}{t}\,\dd t .
+\end{equation}
+% Read it out loud. At the point x, look at f at the point x+t,
+% stepping t to the right. Weight the value by 1/t. Positive t
+% contributes f(x+t)/t; negative t contributes the SAME sign in
+% front, and 1/t already has the opposite sign for negative t.
+% So the transform measures how lopsided f is around x: it is a
+% machine for detecting antisymmetry. A constant has zero Hilbert
+% transform. A symmetric bump has zero Hilbert transform.
+
+% Two warnings about (1.1).
+%
+% (i) The integral is singular. Near t=0, f(x+t)/t looks like
+% f(x)/t, which has infinite integral. The letters "p.v."
+% stand for principal value: integrate the truncated version
+% over eps<|t|<R and only THEN let eps go to zero and R go
+% to infinity. For smooth compactly supported f the limit
+% exists at every x.
+%
+% (ii) The transform is not bounded on L^1. It IS bounded on
+% every L^p with 1<p<infinity: a constant C_p satisfies
+% ||Hf||_p <= C_p ||f||_p.
+% "L^p" means the function has finite p-th power integral,
+% ||f||_p = (integral of |f|^p)^(1/p),
+% which you may think of as a measure of size with a knob p.
+% Large p emphasises peaks. Small p emphasises mass.
+
+% The TRUNCATIONS are the objects we are going to "vary":
+\begin{equation}\label{eq:HT-trunc}
+ H_{\varepsilon,R}f(x)
+ =\frac1\pi\int_{\varepsilon<|t|<R}\frac{f(x+t)}{t}\,\dd t ,
+ \qquad 0<\varepsilon<R<\infty .
+\end{equation}
+% Even though Hf exists as a principal value, the truncated
+% operators H_{eps,R} are far from innocent. As eps decreases to
+% zero and R grows to infinity, the functions H_{eps,R}f do
+% converge pointwise almost everywhere -- but proving this
+% UNIFORMLY in x is a genuine theorem (a one-sided version of
+% the classical Carleson theorem for the Hilbert transform).
+% "Almost everywhere" means: for every x outside a set of
+% measure zero.
+%
+% The present paper proves the analogue of that statement in TWO
+% dimensions, for a bilinear operator, in the strongest possible
+% form: not just pointwise convergence, but control of the total
+% amount of oscillation across all truncations at once. That is
+% the meaning of "variation".
+
+
+\subsection{The bilinear Hilbert transform}\label{subsec:BHT}
+
+% Before the two-dimensional version, there is a one-dimensional
+% bilayer. Take two functions f and g of ONE variable. Send one
+% input to the right and the other to the LEFT by the same amount
+% t. Then weight by 1/t and integrate.
+\begin{equation}\label{eq:BHT}
+ \mathrm{BH}(f,g)(x)
+ =\operatorname{p.v.}\!\int_{-\infty}^{\infty}
+ f(x+t)\,g(x-t)\,\frac{\dd t}{t}.
+\end{equation}
+% This is the bilinear Hilbert transform. "Bilinear" means linear
+% in f alone and linear in g alone, but not in f+g together. It
+% was open for decades whether it is bounded on any nontrivial
+% range of L^p spaces; Lacey and Thiele settled this in 1997.
+% The relevant point for us: the two inputs move in OPPOSITE
+% directions along the SAME axis. That opposition is what makes
+% the analysis tractable.
+
+\subsection{The triangular Hilbert transform}
+\label{subsec:triangular}
+
+% Now come to the operator of the paper. It is
+% two-dimensional: F and G are functions of (x,y). Instead of
+% moving both inputs along the same x-axis, we move them along
+% ORTHOGONAL directions.
+\begin{equation}\label{eq:triangular-def}
+ B_{\varepsilon,R}(F,G)(x,y)
+ =\int_{\varepsilon<|t|<R}
+ F(x+t,y)\,G(x,y+t)\,\frac{\dd t}{t},
+ \qquad 0<\varepsilon<R<\infty .
+\end{equation}
+% Read it out loud. Fix an output point (x,y). Step from (x,y)
+% to (x+t,y): that is a HORIZONTAL move, and we read F there.
+% Step from (x,y) to (x,y+t): that is a VERTICAL move, and we
+% read G there. Multiply the two values, weight by 1/t, sum
+% over t.
+%
+% Three points participate:
+% (x,y) the corner,
+% (x+t,y) the horizontal vertex,
+% (x,y+t) the vertical vertex.
+% They form a right triangle. The two inputs sit on the two legs
+% that leave the corner. This is why the operator is called
+% TRIANGULAR.
+%
+% Compare (1.3) to the bilinear Hilbert transform (1.2). In (1.2)
+% both inputs lie on the SAME line. In (1.3) they lie on
+% different lines. The one-dimensional opposition in (1.2) was a
+% gift; the two-dimensional orthogonality in (1.3) does not give
+% the same gift, and the operator is harder. This paper is the
+% first proof that it behaves well at the symmetric endpoint
+% (defined in the next subsection).
+
+% There is a subtlety in the word "coherent". F(x+t,y) and
+% G(x,y+t) can be highly aligned with the kernel 1/t for large
+% stretches of t; only cancellations across changes of scale keep
+% the operator bounded. In the classical Hilbert transform, those
+% cancellations are packaged by orthogonality of frequencies.
+% Here the two different directions break that orthogonality and
+% a new mechanism is required. That mechanism is what the paper
+% discovers.
+
+
+\subsection{The scalar simplex form and the symmetric point}
+\label{subsec:simplex}
+
+% Dualise the bilinear operator into a TRILINEAR form: multiply
+% the output of B by a third function G_0 and integrate.
+% Holder's inequality, the standard pairing tool, says
+% ||B(F,G)||_p <= C ||F||_3 ||G||_3,
+% || G_0 B(F,G)||_1 <= ||G_0||_q ||B(F,G)||_p
+% provided 1/q + 1/p = 1. If p=3/2 then q=3. So for a L^3
+% basis, the natural companion G_0 is also in L^3. The scalar
+% form we want is exactly
+\begin{equation}\label{eq:scalar-annular}
+ \Lambda_{\varepsilon,R}(G_0,G_1,G_2)
+ =\iiint G_0(a,b)\,G_1(b,c)\,G_2(c,a)\,
+ \mathbf 1_{\varepsilon<a+b+c<R}\,
+ \frac{\dd a\,\dd b\,\dd c}{a+b+c}.
+\end{equation}
+% Where does (1.4) come from, and why does the kernel become
+% 1/(a+b+c) instead of 1/t? Use the determinant-one substitution
+% a=-x, b=-y, t=a+b+c,
+% whose inverse is x=-a, y=-b, t=a+b+c.
+% The three variables (a,b,c) satisfy c=t+x+y. Then
+% F(x+t,y) becomes G_1(b,c)=F(b+c,-b),
+% G(x,y+t) becomes G_2(c,a)=G(-a,c+a),
+% G_0(x,y) becomes G_0(a,b).
+% Each of these maps (F,G) -> (G_1,G_2) and back has matrix
+% determinant one, and preserves the L^3 norm. Computing from
+% (1.3) yields (1.4): the compactly supported part of
+% the kernel is 1/t=1/(a+b+c).
+%
+% The three inputs in (1.4) each see a DIFFERENT PAIR of the
+% three scalar variables (a,b,c):
+% G_0 sees (a,b),
+% G_1 sees (b,c),
+% G_2 sees (c,a).
+% The pairs form a CYCLE b->c->a->b. This is the "simplex":
+% the graph with vertices a,b,c and edges the three pairs. No
+% one input sees all three variables. That is what makes the
+% operator "entangled" from the point of view of real-interpolation.
+
+% Now for the exponents. Holder in three variables says:
+% the bound
+% |\Lambda| <= C ||G_0||_{p_0} ||G_1||_{p_1} ||G_2||_{p_2}
+% applies exactly when
+% 1/p_0 + 1/p_1 + 1/p_2 = 1.
+% That is a whole SURFACE of choices, not a single point.
+%
+% The SYMMETRIC POINT of that surface is the choice
+% p_0 = p_1 = p_2 = 3.
+% It is the choice in which all three inputs are treated equally,
+% with exponent 3 for each. Symmetric-point variation and
+% symmetric-point principal values are the endpoint of the whole
+% theory: they are what we would want if we did not already
+% know a comfortable exponent tuple.
+
+% Which of these was open in 2025?
+% - The lower exponents, p_0,p_1,p_2 close to 2, are easier
+% because there is more room in the Holder relation.
+% - The symmetric point p=3,3,3 is the HARDEST nontrivial
+% choice on the boundary: each input is exactly at the
+% point where the coarea/decomposition tools start to lose.
+% - Even harder is the VARIATION version of the estimate:
+% not just a single annulus eps<|t|<R, but a weighted sum
+% of many disjoint annuli (equation (1.7) below).
+% This paper proves the hardest version.
+
+
+\subsection{Variation}\label{subsec:variation-intro}
+
+% The naive object is the single truncated operator B_{eps,R}.
+% But we care about how far the family B_{eps,R} can wander as
+% the truncation moves. For a real number r>2 (the "variation
+% exponent"), define
+\begin{equation}\label{eq:variation-def}
+ V_r(F,G)(x,y)
+ =\sup_{\substack{J\ge1\\0<t_0<\cdots<t_J\\t_j\in\mathbb Q}}
+ \Big(\sum_{j=1}^{J}
+ |B_{t_{j-1},t_j}(F,G)(x,y)|^r\Big)^{1/r}.
+\end{equation}
+% Read the outer object first. For EACH point (x,y), pick any
+% partition 0<t_0<t_1<...<t_J of the positive real line, form
+% the truncated operators on the slices (t_{j-1},t_j), take the
+% absolute value of the scalar at (x,y), raise it to the r-th
+% power, sum over the slices, take the r-th root, and THEN take
+% the supremum over all partitions. The answer is a function of
+% (x,y). The map F,G |-> V_r(F,G) is the r-th variation of the
+% family B_{eps,R}.
+%
+% Two features matter.
+%
+% (i) The partition is allowed to depend on (x,y). Different
+% points (x,y),(x',y') may use different partitions; the
+% supremum is taken pointwise FIRST and only afterwards
+% are the L^{3/2} norms taken.
+%
+% (ii) There is no limit on the number of intervals J. The
+% supremum is over every finite partition. We do not
+% prefer dyadic endpoints.
+%
+% These features are what make variation strictly stronger than
+% pointwise convergence and stronger than the maximal estimate.
+
+\subsection{The count estimate}\label{subsec:count-intro}
+
+% Theorem A below follows from a quantitative estimate for
+% finitely many annuli. For an integer n>=1, define
+\begin{equation}\label{eq:count-def}
+ \mathcal S_n(F,G)(x,y)
+ =\sup\Big\{\sum_{j=1}^{m}|B_{\varepsilon_j,R_j}(F,G)(x,y)|
+ : 0\le m\le n,\ 0<\varepsilon_j<R_j\in\mathbb Q,\
+ (\varepsilon_j,R_j)\text{ pairwise disjoint}\Big\}.
+\end{equation}
+% Same story. For each point (x,y): pick at most n annuli
+% eps_j<|t|<R_j that are pairwise disjoint (endpoints may touch;
+% "disjoint" is about the interiors), form the truncated operator
+% on each, take absolute values, sum. Take the supremum over all
+% choices. The result is \mathcal S_n(F,G), a function of (x,y).
+%
+% The main theorem is the size bound on \mathcal S_n. It is a
+% count estimate: it costs the number of annuli n but nothing
+% else. From it, variation follows by a short combinatorial
+% step (Section 1.7 below).
+
+% Here is why the count estimate is the right reduction.
+% From n annuli we lose sqrt(n). From the count estimate to
+% VARIATION at exponent r>2, we sort the increments by size
+% and bound the resulting series:
+%
+% V_r(F,G) <= sum_j 2^{j(1/r-1/2)} \mathcal S_{2^j}(F,G),
+%
+% and the series on the right converges whenever r>2. In
+% particular, the count estimate on the right directly implies
+% the full-variation estimate. The series converges because
+% (1/r - 1/2) < 0.
+
+
+\subsection{From count to variation}\label{subsec:count-to-variation}
+
+% Why does the count estimate imply the variation estimate?
+% Take any partition t_0<t_1<...<t_J at a point (x,y). Let
+% d_j = |B_{t_{j-1},t_j}(F,G)(x,y)|
+% be the increment sizes. Sort them in decreasing order:
+% d_{(1)} >= d_{(2)} >= ... >= d_{(J)}.
+% (The (j) is rank of the increment, not index of the interval;
+% the procedure is a pointwise rearrangement.)
+%
+% For every rank 2^k at which the list still has an entry,
+% the first 2^k increments together are a valid choice of 2^k
+% pairwise disjoint intervals in the count supremum. Hence
+% d_{(2^k)} <= 2^{-k} * \mathcal S_{2^k}(F,G).
+%
+% The increments of ranks 2^k <= l < 2^{k+1} form a block of
+% size 2^k, and each is at most d_{(2^k)}, so this block's
+% l^r sum is at most 2^k * d_{(2^k)}^r, i.e. its l^r norm is at
+% most 2^{k/r} d_{(2^k)}. Summing over k gives
+\begin{equation}\label{eq:count-to-variation}
+ V_r(F,G)
+ \le \sum_{k\ge0} 2^{k(1/r-1/2)}\, \mathcal S_{2^k}(F,G)
+ \quad\text{pointwise},
+\end{equation}
+% The coefficient 2^{k(1/r-1/2)} is positive and decreases
+% geometrically when r>2, which is exactly the range we need
+% for the series to converge. Applying the L^{3/2} norm,
+% then Minkowski, and then the count estimate,
+% ||V_r(F,G)||_{3/2}
+% <= C ||F||_3 ||G||_3 \sum_k (1+k)2^{k(1/r-1/2)},
+% and the geometric series is finite because 1/r-1/2<0. That is
+% the whole content of Theorem B below, aside from measurability.
+
+% ============================================================
+\section{Statements of the main theorems}\label{sec:statements}
+% ============================================================
+
+% Here is the exact content of the paper. Everything else is
+% preparation or consequences. We use \R for the real numbers,
+% \C for the complex numbers, L^p(\R^2) as usual, and
+% ||f||_p for the L^p norm. All functions are complex-valued.
+
+\subsection{The count estimate}
+
+\begin{theorem}[Disjoint-annulus count estimate]\label{thm:count}
+There is an absolute constant $C$ such that, for every integer
+$n\ge1$ and every complex $F,G\in L^3(\R^2)$,
+\begin{equation}\label{eq:count-bound}
+ \|\mathcal S_n(F,G)\|_{3/2}
+ \le C\sqrt n\,\log(2+n)\,\|F\|_3\,\|G\|_3 .
+\end{equation}
+\end{theorem}
+
+% Read the right side. There are three factors.
+%
+% ||F||_3 ||G||_3 is the size of the inputs. If F=0 or
+% G=0 then the left side is zero.
+%
+% sqrt(n) is the price for the number of annuli. Doubling
+% the number of annuli costs at most sqrt(2) more.
+%
+% log(2+n) is a negligible extra cost; it appears because we
+% allow arbitrary disjoint annuli, not just dyadic ones.
+%
+% The constant C is absolute: it does not depend on n, on F, G,
+% on the annuli, on the point (x,y), or on anything else.
+%
+% The answer to "why sqrt(n)" is a rescaling trick. We
+% normalize the third input G_0 to unit size, and we can split
+% the whole energy of the operator across n annuli. Each annulus
+% contributes a bounded amount per switch, but we can choose the
+% switch thresholds adaptively. A duality rescaling by a factor
+% n^{1/2} absorbs the linear count of switches into the trace
+% of the energy (see the smooth-annuli section later).
+
+\subsection{The variation estimate}
+
+\begin{theorem}[Full annular variation]\label{thm:variation}
+For every real $r>2$ there is a finite constant $C_r$ such that
+\[
+ \|V_r(F,G)\|_{3/2}
+ \le C_r\,\|F\|_3\,\|G\|_3
+\]
+for all complex $F,G\in L^3(\R^2)$.
+\end{theorem}
+
+% Theorem~\ref{thm:variation} is the headline. It says the
+% family B_{eps,R} has bounded r-variation for every r>2,
+% uniformly. In particular (take r->infinity formally) it
+% implies a maximal estimate, and it implies joint convergence.
+% Both consequences are corollaries below.
+
+
+\subsection{The maximal and principal-value corollaries}
+\label{subsec:corollaries}
+
+% Take a single increment in the partition: two endpoints. Then
+% variation dominates the maximal operator over all endpoints:
+\begin{corollary}[Maximal operator]\label{cor:maximal}
+For complex $F,G\in L^3(\R^2)$, define
+\[
+ B_*(F,G)(x,y)
+ =\sup_{0<\varepsilon<R<\infty}|B_{\varepsilon,R}(F,G)(x,y)|
+\]
+on a common full-measure set (on which every truncated operator
+is defined), and set it to zero elsewhere. Then $B_*$ is
+measurable and
+\begin{equation}\label{eq:hard-maximal-bound}
+ \|B_*(F,G)\|_{3/2}\le C_M\|F\|_3\|G\|_3
+\end{equation}
+with $C_M=C_2$ from Theorem~\ref{thm:variation}.
+\end{corollary}
+
+% The supremum is over BOTH hard truncation endpoints, not just
+% the upper one. The constant C_M is absolute. From the maximal
+% estimate the principal-value theorem follows in one more step.
+\begin{corollary}[Joint principal value]\label{cor:bilinear-pv}
+For every complex $F,G\in L^3(\R^2)$, the joint limit
+\[
+ B(F,G)=\lim_{\substack{\varepsilon\downarrow0\\R\uparrow\infty}}
+ B_{\varepsilon,R}(F,G)
+\]
+exists almost everywhere and in $L^{3/2}$, and
+$\|B(F,G)\|_{3/2}\le C_M\|F\|_3\|G\|_3$.
+Moreover, the entire tail converges in the maximal sense:
+\begin{equation}\label{eq:maximal-tail-convergence}
+ \Big\|\sup_{\substack{0<\varepsilon<1/n\\R>n}}
+ |B_{\varepsilon,R}(F,G)-B(F,G)|\Big\|_{3/2}
+ \longrightarrow 0 .
+\end{equation}
+\end{corollary}
+
+% The maximal operator controls the two-parameter (eps,R)
+% truncations simultaneously: the SUPREMUM inside the norm on
+% the left of (2.4) shrinks to zero as n grows. So the
+% principle-value limit is not merely pointwise; it is
+% dominated in L^{3/2} by an integrable envelope.
+%
+% Section 6 proves all of the above from the count estimate.
+% Here we only stated them for context.
+
+% ============================================================
+\section{Strategy: how a single-annulus bound becomes a count}
+\label{sec:strategy}
+% ============================================================
+
+% The proof of Theorem~\ref{thm:count} has three layers, each
+% built to solve a problem the one beneath it could not.
+%
+% 1. HEAT FLOW. Replace the hard annulus indicator by a smooth,
+% Gaussian-derived window. The hard truncation is recovered as
+% the endpoint of a Gaussian scale integral. Smooth windows
+% have the miraculous property that their second derivatives
+% are TRACES OF MATRICES with positive spectrum.
+%
+% 2. MATRIX ENERGY. Sample the trilinear form on a fine grid.
+% It becomes a trace of a product of three matrices times
+% diagonal insertions. The second derivative of the trace
+% trace(R^*R)^{3/2} in the insertion directions is a sum of
+% squares of entries (a Hessian form). A "mixed trace
+% inequality" bounds the cyclic product by those costs.
+%
+% 3. ROUGH KERNELS. Coming back to the actual annulus
+% indicator, the endpoint errors are bounded using the
+% frequency-block estimate, which exploits disjointness of
+% the annuli. Counting disjointness is exactly what the
+% sqrt(n) loss comes from.
+%
+% Here is the logic in one sentence: SMOOTHING COSTS A TRACE,
+% DISJOINTNESS BOUNDS THE TRACE, AND RANK SORTING
+% CONVERTS A COUNT INTO A VARIATION.
+
+
+\subsection{The two difficulties, and where sqrt(n) comes from}
+\label{subsec:difficulties}
+
+% The proof has to handle TWO kinds of error, and each looks at
+% first like it costs a factor of n.
+%
+% difficulty (S) -- SMOOTH SWITCHES. The smooth annuli
+% \int_{eps^2}^{R^2} (k_s * k_s * k_s)(t) ds/s have a
+% beginning and an end at each truncation endpoint. Summing
+% over every interval, the number of switches is 2n at each
+% output point. Naively the bound is n times something, not
+% sqrt(n). Section~\ref{sec:smooth} shows how to rescale the
+% third input to absorb this linear cost into a square root.
+%
+% difficulty (H) -- HARD ENDPOINT ERRORS. The gap between the
+% smooth annulus and the hard Hilbert annulus is a sum of
+% "endpoint error" kernels E_eps, one per endpoint. Naively
+% there are 2n of them, which again looks linear. Section
+% \ref{sec:rough} groups endpoints at comparable scales. Within
+% a dyadic group, disjointness gives a uniform pointwise bound
+% on the total, so each group behaves like ONE kernel. The
+% frequency-block estimate (Section~\ref{sec:frequency}) divides
+% the groups into two classes: those with many jumps (there are
+% at most 2 sqrt(n) of them) and those with few (which are
+% handled by the frequency estimate). The sqrt(n) is the price
+% of the changeover.
+
+% The point is that both difficulties are paid for ONCE by
+% sqrt(n), and not by n. This is the content of the paper.
+
+% ============================================================
+\section{Preliminaries}\label{sec:preliminaries}
+% ============================================================
+
+% This section fixes three things: the representatives of
+% truncated operators, the change of variables to simplex
+% coordinates, and the step-function structure that makes the
+% analysis measurable. Then it records the one-dimensional
+% maximal estimates that will pay for the switch costs.
+
+\subsection{Finite truncations}\label{subsec:truncations}
+
+% The truncated operator B_{eps,R}(F,G) is a two-dimensional
+% integral in t of a product, and it is not obvious whether it
+% even makes sense at a given (x,y). The following lemma
+% handles this. We call a "null set" a set of measure zero on
+% R^2.
+\begin{lemma}[Finite truncations]\label{lem:finite-truncations}
+For $F,G\in L^3(\R^2)$, all finite annular integrals defining
+$B_{\varepsilon,R}(F,G)$ are absolutely convergent outside one
+common null set. On its complement they depend continuously on
+$(\varepsilon,R)$ in $0<\varepsilon<R<\infty$. Changing the
+representatives of $F$ or $G$ changes this family only on a
+common null set. Moreover,
+\begin{equation}\label{eq:finite-annulus-bound}
+ \Big\|\int_{\varepsilon<|t|<R}
+ |F(x+t,y)G(x,y+t)|\,\frac{\dd t}{|t|}\Big\|_{3/2}
+ \le 2\log(R/\varepsilon)\|F\|_3\|G\|_3 .
+\end{equation}
+\end{lemma}
+\begin{proof}
+Fix $t$. Holder's inequality in $(x,y)$ and translation
+invariance give
+$\|F(\cdot+t,\cdot)G(\cdot,\cdot+t)\|_{3/2}\le\|F\|_3\|G\|_3$.
+Minkowski's integral inequality yields \eqref{eq:finite-annulus-bound}.
+Apply it to the countably many annuli $1/m<|t|<m$, $m\ge2$;
+their union of exceptional null sets works for every finite
+annulus. Absolute continuity of the $t$-integral gives endpoint
+continuity. A zero-$L^3$-difference estimate on each of those
+annuli proves the assertion about representatives.
+\end{proof}
+
+% Read (3.1) as a bookkeeping bound: on any finite annulus the
+% truncated operator is dominated by $\log(R/\eps)$ times the
+% product of input norms. The $\log$ is essential: without it the
+% principal value would not be an L^{3/2}-bounded operator on the
+% hard annulus. The lemma lets us use any representative we like;
+% the truncation acts on L^3 as a well-defined continuous family
+% of operators.
+%
+% In (1.7) the count supremum uses rational endpoints. By
+% continuity in $(\varepsilon,R)$, rational endpoints exhaust all
+% endpoints; we may (and will) replace rational by arbitrary
+% throughout. Similarly, measurability of $B_*$ and $V_r$ comes
+% from this exhaustion.
+
+
+\subsection{Simplex coordinates}\label{subsec:coordinates}
+
+% Most of the analysis is done in simplex coordinates. The
+% change of variables below has absolute determinant one, so it
+% preserves every L^p norm and commutes with the operator up to
+% name changes.
+\begin{equation}\label{eq:coords-change}
+ x=-a,\quad y=-b,\quad t=a+b+c,
+ \quad\text{with inverse}\quad
+ a=-x,\quad b=-y,\quad c=t+x+y .
+\end{equation}
+% Pulled back along this map, the two inputs F and G become
+\begin{equation}\label{eq:pair-maps}
+ G_1(b,c)=F(b+c,-b),\qquad G_2(c,a)=G(-a,c+a).
+\end{equation}
+% Each of these two maps is a norm-preserving bijection of
+% L^3(\R^2). Its companion change of the dual input is the
+% identity: G_0(a,b) is just G_0(a,b). Combining (3.2) with
+% (1.3) yields the simplex form (1.4). Thus the bilinear
+% theorem on B_{eps,R} and the trilinear theorem on
+% \Lambda_{eps,R} are equivalent, at matching constants.
+%
+% In particular, (1.4) is not merely a formal change of
+% variables. It has the SAME optimal constant as the scalar
+% flat form (Section 6). We mark this now because it justifies
+% working only with the simplex coordinates from here on.
+
+\subsection{Step choices}\label{subsec:step-choices}
+
+% The count and variation suprema are taken pointwise. But the
+% operators appearing under the supremum are not arbitrary
+% functions of (a,b); they are built from a FINITE list of
+% choices. To keep the analysis measurable and Riemann-summable
+% we restrict to choices that are locally constant on a
+% partition of \R^2 by finitely many horizontal and vertical
+% lines.
+
+\begin{definition}[Step choice]\label{def:step-choice}
+A \emph{step choice} for the annulus is a choice of data
+\[
+ \text{ menu }=\big\{\,(\varepsilon_j^{(m)},R_j^{(m)},
+ \beta_j^{(m)})_{j=1,\ldots,n}
+ \big\}_{m=1,\ldots,M}
+\]
+together with a partition of $\R^2$ into finitely many open
+rectangles by horizontal and vertical lines, and a map from
+rectangles to $m$. The bounded rectangle is where the operator
+is genuinely varying; the values on the measure-zero boundaries
+of the rectangles are immaterial.
+\end{definition}
+
+% In plain words: we pre-commit to a finite list of annuli
+% choices, one per row of a big table. The row is chosen
+% independently in each rectangle. In the interior of a rectangle
+% the annuli and coefficients do not change as (a,b) varies.
+%
+% Why step choices? Three reasons.
+%
+% (1) Measurability. A measurable assignment of annuli to (a,b)
+% can be approximated in measure by a step choice. So proving
+% the bound for step choices with uniform constants gives the
+% bound for all measurable assignments by exhaustion.
+%
+% (2) Riemann sums. Fixing a step choice means we are reduced to
+% finitely many kernels. Riemann-summing against a smooth
+% companion amounts to a finite calculation.
+%
+% (3) Endpoints at cell boundaries are measure zero, and they
+% contribute nothing to the trilinear form.
+
+% We will be careful about two points.
+% - The cell choice can depend on (a,b) and on the specific
+% rectangular partition (of the OUTPUT point), but the finitely
+% many kernels under the sum are fixed once the partition is
+% fixed.
+% - Nothing about the step choice needs to be the same at
+% different points (a,b), nor consistent across dyadic scales.
+
+
+\subsection{Line maximal estimates}\label{subsec:line-maximal}
+
+% Every jump in a matrix energy is controlled by a maximal
+% average along a line. We record the two forms below. The
+% constants are absolute.
+
+% Let M_R and M_Z denote the CENTERED Hardy--Littlewood
+% maximal operators on the real line and on the integer lattice
+% respectively: for functions on R,
+% M_R f(x) = sup_{r>0} (2r)^{-1}\int_{x-r}^{x+r}|f(y)|dy,
+% and for sequences on Z,
+% M_Z f(j) = sup_{m>=0} (2m+1)^{-1}\sum_{|l-j|<=m}|f(l)|.
+
+\begin{lemma}[Line maximal estimates]\label{lem:line-maximal}
+On either $\R$ or $\mathbb Z$, for $1<p<\infty$,
+\begin{equation}\label{eq:line-maximal}
+ \mu\{M f>\rho\}\le\frac3\rho\|f\|_1,
+ \qquad
+ \|Mf\|_p^p\le\frac{6p\,2^{p-1}}{p-1}\|f\|_p^p .
+\end{equation}
+Let $g_\sigma(t)=(2\pi\sigma)^{-1/2}e^{-t^2/(2\sigma)}$. There is
+an absolute constant $C$ such that, uniformly in $\sigma$,
+\begin{align}
+ \sum_m h g_\sigma((m-j)h)|f(m)|
+ &\le C M_{\mathbb Z}f(j), &&\sigma\ge h^2,
+ \label{eq:discrete-gaussian-maximal}\\
+ \int g_\sigma(t)|f(x-t)|\,\dd t
+ &\le C M_{\R}f(x), &&\sigma>0 .
+ \label{eq:continuous-gaussian-maximal}
+\end{align}
+\end{lemma}
+\begin{proof}
+The weak-type estimate: from any finite family of intervals,
+greedily select the widest interval not already overlapping
+another, and discard any covered by their triple enlargements;
+this yields a disjoint sub-covered set of at most 3x the original
+measure (also for integer intervals). Combining with the
+truncation argument for strong $L^p$ gives
+\eqref{eq:line-maximal}.
+
+The Gaussian bounds follow from splitting the sum or integral
+into shells $2^{k-1}a<|m-j|\le 2^ka$ with $a=\sqrt\sigma/h$.
+Each shell contributes at most $C2^k a M f$, and the resulting
+geometric series converges.
+\end{proof}
+
+% ============================================================
+\section{Dimension-free matrix inequalities}\label{sec:matrix}
+% ============================================================
+
+% This is the linear-algebra engine of the proof. Everything
+% below is finite-dimensional. The point of this section is to
+% introduce a QUADRATIC COST attached to each possible direction
+% of variation of a matrix trace, and to prove that this cost
+% controls the trilinear form we want to estimate.
+
+% Setup. Spaces $H_0,H_1,H_2$ are finite-dimensional complex
+% Hilbert spaces. Maps $W_v:H_{v+1}\to H_v$ are linear (indices
+% mod 3). The ROW STAR at vertex $v$ is
+\begin{equation}\label{eq:row-star}
+ R_v=[\,W_v\quad W_{v-1}^*\,]:H_{v+1}\oplus H_{v-1}\to H_v,
+\end{equation}
+% with two Gram matrices
+\begin{equation}\label{eq:star-grams}
+ T_v=R_v^*R_v,\qquad S_v=R_vR_v^* .
+\end{equation}
+% The two Grams have equal nonzero spectra, so their energies
+% coincide:
+\[
+ \Phi(T_v)=\Phi(S_v),\qquad \Phi(P):=\tr(P^{3/2}) .
+\]
+
+% The energy objective is a smooth function of a positive
+% semidefinite matrix $P\ge0$, with a natural Hessian cost.
+% We first record the Hessian formula.
+\begin{lemma}[Supported Hessian]\label{lem:hessian}
+On the cone of matrices positive definite on a fixed support,
+$\Phi(P)=\tr(P^{3/2})$ is smooth, and its first and second
+derivatives at a supported direction $K$ (and at supported
+directions $K,L$) are
+\begin{equation}\label{eq:hessian}
+ d\Phi_P(K)=\frac32\tr(\sqrt P\,K),\qquad
+ d^2\Phi_P(K,L)=\mathcal B_P(K,L),
+\end{equation}
+where the Hessian form is
+\begin{equation}\label{eq:hessian-form}
+ \mathcal B_P(K,L)
+ =\frac32\operatorname{Re}\big\langle K,
+ (\mathsf L_{\sqrt P}+\mathsf R_{\sqrt P})^{-1}L\big\rangle_{\HS} .
+\end{equation}
+% Here \mathsf L_A(X)=AX and \mathsf R_A(X)=XA$ are the left and
+% right multiplication operators. In a diagonal basis for $P$
+% with positive eigenvalues $p_i$, we have
+\begin{equation}\label{eq:hessian-metric}
+ b_P(K)=\mathcal B_P(K,K)
+ =\frac32\sum_{i,j}\frac{|K_{ij}|^2}{\sqrt{p_i}+\sqrt{p_j}} .
+\end{equation}
+% The form $b_P$ is symmetric positive-definite on supported
+% Hermitian directions, and it satisfies
+% $|\mathcal B_P(K,L)|^2\le b_P(K)b_P(L)$ (after Cauchy--Schwarz).
+\end{lemma}
+\begin{proof}
+Restrict to the support $\supp P$ (where $P$ is positive
+definite) and set $A=\sqrt P$. Differentiating the squaring map
+$A\mapsto A^2$ at $A$ multiplies the $(i,j)$ entry by
+$\sqrt{p_i}+\sqrt{p_j}>0$, so the inverse function theorem
+gives a smooth positive square root. Then
+$d\Phi_P(K)=\frac32\tr(AK)$ by the chain rule, and one more
+differentiation gives (4.4)-(4.5). Diagonalising $A$ yields
+(4.6). Cauchy--Schwarz on the Hilbert-Schmidt inner product
+along eigenvectors of $\mathsf L_A+\mathsf R_A$ gives the
+bound $|\mathcal B_P(K,L)|^2\le b_P(K)b_P(L)$ after noting that
+$(\mathsf L_A+\mathsf R_A)^{-1}$ is positive definite on
+supported Hermitian matrices.
+\end{proof}
+
+% The key use is a COMPARISON: if we enlarge the Gram, the
+% cost shrinks. This is a monotonicity of the Hessian form,
+% analogous to Loewner--Heinz monotonicity of the square root.
+
+
+% The following lemma is the monotonicity of the Hessian cost
+% under a Loewner-type order on the Gram.
+\begin{lemma}[Order comparison]\label{lem:order}
+If $P,Q\ge0$ on the same space, $a\ge1$, and $P\le aQ$, then
+every Hermitian $K$ supported on $P$ is supported on $Q$, and
+\begin{equation}\label{eq:matrix-order}
+ b_P(K)\ge a^{-1/2}\,b_Q(K).
+\end{equation}
+\end{lemma}
+\begin{proof}
+$\ker Q\subseteq\ker P$ follows directly from $0\le P\le aQ$,
+which gives the support containment. The order facts: spectrum
+calculus and the integral representation
+$\sqrt X=\pi^{-1}\int_0^\infty X(X+tI)^{-1}t^{-1/2}\dd t$
+combined with inversion order
+($0<X\le Y\Rightarrow Y^{-1}\le X^{-1}$) and monotonicity of
+the multiplication operator $\mathsf L_A+\mathsf R_A$ in $A$
+give $\sqrt{P+\eta I}\le \sqrt a\sqrt{Q+\eta I}$ for all $\eta>0$.
+Then
+$(\mathsf L_{\sqrt{P+\eta I}}+\mathsf R_{\sqrt{P+\eta I}})^{-1}$,
+being the inverse of a positive operator on Hilbert--Schmidt
+space, decreases as $\eta\downarrow0$ by monotonicity of
+inversion. Taking $\eta\downarrow0$ proves (4.7).
+\end{proof}
+
+% The following lemma bounds a matrix cost of the form
+% $R^*LR$ (Gram of $R$, insertion $L$) by the size of $L$
+% times $\Phi(R^*R)$. It is the single most-used inequality in
+% the paper.
+\begin{lemma}[Inserted Gram bound]\label{lem:insertion}
+For a linear map $R:E\to F$ and a Hermitian map $L:F\to F$,
+the matrix $R^*LR$ is supported on $R^*R$, and
+\begin{equation}\label{eq:insertion-bound}
+ b_{R^*R}(R^*LR)
+ \le \frac34\|L\|_{\op}^2\,\Phi(R^*R) .
+\end{equation}
+\end{lemma}
+\begin{proof}
+Let $\rho_i>0$ be the positive singular values of $R$ and
+$C$ the compression of $L$ to the left singular subspace.
+Then
+$b_{R^*R}(R^*LR)
+=\frac32\sum_{i,j}\frac{\rho_i^2\rho_j^2}{\rho_i+\rho_j}
+ |C_{ij}|^2$.
+The pointwise inequality
+$x^2y^2/(x+y)\le (x^3+y^3)/4$ for $x,y>0$ and the fact that
+each row (and column) of $C$ has squared $\ell^2$ norm at most
+$\|L\|_{\op}^2$ yield
+$\le\frac34\|L\|_{\op}^2\sum_i\rho_i^3$, which equals
+the right side of (4.8).
+\end{proof}
+
+% Finally, if we form a PROBABILITY AVERAGE of Grams and their
+% insertions, the cost can only decrease. This is what lets
+% us average over Gaussian center shifts.
+\begin{lemma}[Joint convexity]\label{lem:joint-convexity}
+Let $E$ be a fixed subspace and $T_\omega\ge0$, $K_\omega$
+measurable, $\supp T_\omega=E$, $K_\omega$ supported on
+$T_\omega$, and $\int(\|T_\omega\|_{\op}+\|K_\omega\|_{\op})\dd\nu<\infty$.
+Then $\overline T=\int T_\omega\dd\nu$ has support $E$,
+$\overline K=\int K_\omega\dd\nu$ is supported on $\overline T$,
+and
+\begin{equation}\label{eq:joint-convexity}
+ b_{\overline T}(\overline K)
+ \le\int b_{T_\omega}(K_\omega)\,\dd\nu(\omega) .
+\end{equation}
+\end{lemma}
+\begin{proof}
+For any $x\in E$, $\langle\overline T x,x\rangle>0$, so
+$\overline T$ is positive definite on $E$, and $K$ is supported
+on it. Put $A_\omega=\sqrt{T_\omega}$ and $A=\int A_\omega\dd\nu$.
+Cauchy--Schwarz in $\omega$ gives $A^2\le\overline T$. By
+$\sqrt X$ monotonicity and
+$\langle K,\mathsf M^{-1}K\rangle
+=\sup_X\{2\langle K,X\rangle-\langle X,\mathsf M X\rangle\}$
+(varational formula for a positive selfadjoint operator), the
+cost at $(\overline T,\overline K)$ is controlled by the average
+of the costs at $(T_\omega,K_\omega)$.
+\end{proof}
+
+
+% The Hessian cost is defined on a fixed support. But our
+% matrices CHANGE rank: when entries are switched on or off
+% they go from zero to nonzero. The following lemma covers
+% this case: it is a first-derivative formula valid across
+% rank changes.
+
+\begin{lemma}[Rank changes and entrywise derivatives]\label{lem:rank-change}
+For rectangular $R_-,R_+$ of the same size, set
+$R(z)=(1-z)R_-+zR_+$ and $D=R_+-R_-$. Then
+\begin{equation}\label{eq:rank-change-derivative}
+ \Phi(R_+^*R_+)-\Phi(R_-^*R_-)
+ =3\int_0^1\operatorname{Re}\langle R(z)\sqrt{R(z)^*R(z)},D\rangle_{\HS}\,\dd z .
+\end{equation}
+For every $(i,j)$,
+\begin{equation}\label{eq:entry-gradient}
+ |(R\sqrt{R^*R})_{ij}|
+ \le\|\operatorname{row}_iR\|_2\|\operatorname{col}_jR\|_2 .
+\end{equation}
+\end{lemma}
+\begin{proof}
+Approximate by $\sqrt{R(z)^*R(z)+\eta I}$ and let $\eta\downarrow0$.
+The main claim is the spectral-calculus continuity
+$\|\sqrt{R^*R+\eta I}-\sqrt{R^*R}\|_{\op}\le\sqrt\eta$.
+Then (4.9) follows from the fundamental theorem of calculus.
+(4.10) is Cauchy--Schwarz in the matrix product.
+\end{proof}
+
+% The centrepiece of the section: a cyclic product of three
+% matrices, each with a diagonal insertion, is dominated by
+% the three Hessian costs.
+
+\begin{lemma}[Mixed trace inequality]\label{lem:mixed-trace}
+For Hermitian maps $D_v:H_v\to H_v$ put
+$D_*=\max_v\|D_v\|_{\op}$. Then
+\begin{equation}\label{eq:mixed-trace}
+ |\tr(D_0W_0D_1W_1D_2W_2)|
+ \le\frac53 D_*\sum_{v=0}^2 b_{T_v}(R_v^*D_vR_v) .
+\end{equation}
+The constant is independent of dimensions, and no Gram is
+assumed invertible.
+\end{lemma}
+\begin{proof}
+The proof proceeds in three stages.
+%
+First, package the three costs into one quadratic form.
+Let $H=H_0\oplus H_1\oplus H_2$ and set
+\[
+ M=\begin{pmatrix}0&W_0&W_2^*\\W_0^*&0&W_1\\W_2&W_1^*&0\end{pmatrix},
+ \quad D=\operatorname{diag}(D_0,D_1,D_2).
+\]
+If $P_v$ projects onto $H_v$ and $A_v=MP_vM$, $K_v=MP_vDM$,
+then after permuting blocks these equal $T_v$ and $R_v^*D_vR_v$,
+and $K_v$ is supported on $A_v\le M^2$. If $b$ denotes $b_{M^2}$
+on $H$, then
+\begin{equation}\label{eq:trace-common-metric}
+ \mathcal E:=\sum_v b_{T_v}(R_v^*D_vR_v)
+ \ge\sum_v b_{M^2}(K_v)\ge\tfrac13 b_{M^2}(MDM).
+\end{equation}
+%
+Second, bound a cubic trace using the eigenvalue structure
+of $M$: in a diagonal eigenbasis $m_i$ of $M$ with $q_i=|m_i|$
+nonincreasing, formula (4.6) yields
+$\mathcal E\ge\frac12 Q$ with
+\[
+ Q:=\sum_{i,j:q_i+q_j>0}\frac{q_i^2q_j^2}{q_i+q_j}|D_{ij}|^2 .
+\]
+Set
+$A_i=q_i^3|D_{ii}|^2$, $E_i=q_i\sum_{j>i}q_j^2|D_{ij}|^2$,
+$A=\sum A_i$, $E=\sum E_i$. Then $Q\ge A/2+E$ (Hermitian
+symmetry with $q_j\le q_i$ for $j>i$).
+Now group the cubic trace $\tr((DM)^3)$ by the smallest index
+of each triple:
+%
+(case 1) smallest index once: the contribution is at most
+$3D_*E_i$ by Cauchy--Schwarz in the off-diagonal direction.
+% (case 2) smallest index twice: the contribution is at most
+$3D_*\sqrt{A_iE_i}$ (Cauchy--Schwarz with
+$\sum_k|D_{ik}|^2\le D_*^2$).
+% (case 3) all three indices equal: modulus at most $D_*A_i$.
+%
+Summing and using Cauchy--Schwarz,
+\[
+ |\tr((DM)^3)|
+ \le D_*(A+3E+3\sqrt{AE})
+ \le D_*(\tfrac52A+\tfrac92E)
+ \le 5D_*(\tfrac A2+E)
+ \le 5D_*Q .
+\]
+%
+Third, recover the complex phase. Set
+$\tau=\tr(D_0W_0D_1W_1D_2W_2)$. The matrix $M$ has zero
+diagonal blocks, so each closed three-step block product visits
+all three vertices. The three starting points in one
+orientation give $\tau$, in the other give $\overline\tau$, so
+$\tr((DM)^3)=6\operatorname{Re}\tau$. If $\tau\ne0$ choose
+$|z|=1$ with $z\tau=|\tau|$ and replace $W_0\to zW_0$; this
+multiplies $R_0,R_1$ on the right by unitary phase diagonals,
+preserving every cost. Applying the previous bound gives
+$6|\tau|\le5D_*Q\le10D_*\mathcal E$, which is (4.11).
+\end{proof}
+
+% For complex $D_v$, split $D_v=D_v^{(0)}+iD_v^{(1)}$ into
+% real and imaginary parts, both Hermitian, of operator norm at
+% most $\|D_v\|_{\op}$. Expanding gives eight products with
+% Hermitian diagonals. Each is bounded by (4.11) with $D_*$
+% unchanged; each cost occurs four times across the eight.
+% The result is
+\begin{equation}\label{eq:complex-mixed-trace}
+ |\tr(D_0W_0D_1W_1D_2W_2)|
+ \le\frac{20}{3}D_*\sum_{v=0}^2\sum_{\epsilon=0}^1
+ b_{T_v}(R_v^*D_v^{(\epsilon)}R_v) .
+\end{equation}
+% All directions on the right remain Hermitian.
+
+% That concludes the dimension-free matrix engine. What matters
+% for the sequel: the cyclic product is controlled by a sum of
+% three positive quadratic costs, one for each vertex.
+
+
+% ============================================================
+\section{Gaussian heat flow}\label{sec:heat}
+% ============================================================
+
+% The matrix costs from Section~\ref{sec:matrix} are second
+% derivatives. The heat flow turns those second derivatives
+% into derivatives in a scale parameter s, so that integrating
+% over s controls the accumulated costs by initial and final
+% energies. This is the "energy method": we pay the quadratic
+% costs by heat dissipation.
+
+% Setup. Fix a finite collection of vertices u_i=ih with
+% -N<=i<=N and h>0. Vertex indices v range over Z/3. Edge
+% arrays A_v(i,j) are finite, and we set n_v = (sum h^2
+% |A_v(i,j)|^3)^{1/3}. Centers p_v are real; variances sigma_v
+% positive. Weights:
+\begin{equation}\label{eq:weights}
+ g_\sigma(t)=(2\pi\sigma)^{-1/2}e^{-t^2/(2\sigma)},\qquad
+ w_v(i)=h g_{\sigma_v}(u_i-p_v),\qquad
+ W_v(i,j)=A_v(i,j)\sqrt{w_v(i)w_{v+1}(j)} .
+\end{equation}
+% The star at vertex v has two blocks
+\begin{equation}\label{eq:heat-blocks}
+ C_{v,v+1}=W_v,\quad C_{v,v-1}=W_{v-1}^*,\quad
+ R_v=[C_{v,v+1}\ C_{v,v-1}] ,
+\end{equation}
+% and the weights enter two matrix cubes through
+$e_v=\Phi(T_v)=\Phi(S_v)$ on the corresponding Gram.
+% The Gaussian scores and their insertions are
+\begin{equation}\label{eq:heat-scores}
+ N_v=\operatorname{diag}_i\frac{u_i-p_v}{\sigma_v},\quad
+ U_v=R_v^*N_vR_v,\quad
+ P_{v,w}=C_{v,w}N_wC_{v,w}^* .
+\end{equation}
+% The QUADRATIC COSTS we want to dissipate are
+\begin{equation}\label{eq:heat-costs}
+ V_v=b_{T_v}(U_v),\quad
+ Y_{v,w}=b_{S_v}(P_{v,w}),\quad
+ Z_v=\mathcal B_{S_v}(P_{v,v+1},P_{v,v-1}) .
+\end{equation}
+% All insertions are supported on their own Gram, since a
+% sandwiched matrix $B^*DB$ annihilates $\ker B$ both ways.
+
+% The plane over which we integrate is $\Pi_d=\{p_0+p_1+p_2=d\}$,
+% with measure inherited from any pair of centers (both changed
+% to be the free coordinates; the Jacobian is always one).
+
+% We first record the heat identities. These relate the
+% derivative in s to center derivatives and the costs.
+\begin{lemma}[Heat identities]\label{lem:heat-identities}
+For fixed finite arrays and any positive rates $\alpha_v>0$,
+evaluated at $\sigma_i=\alpha_is$, $e_v$ is smooth in the
+centers and in $s$. On $\Pi_d$,
+\begin{equation}\label{eq:heat-identity}
+ 2\partial_se_v=\sum_i\alpha_i\partial_{p_i}^2e_v
+ -\alpha_vV_v-\sum_{w\ne v}\alpha_wY_{v,w} ,
+\end{equation}
+and
+\begin{equation}\label{eq:heat-mixed}
+ \partial_{p_{v+1}}\partial_{p_{v-1}}e_v=Z_v .
+\end{equation}
+The plane integrals $J_v(d,s)=\int_{\Pi_d}e_v\,\dd\pi_d$ are
+finite, smooth, and satisfy
+\begin{equation}\label{eq:heat-plane-identities}
+ \begin{aligned}
+ 2\partial_sJ_v
+ &=\Sigma_\alpha\partial_d^2J_v
+ -\int_{\Pi_d}\Big(\alpha_vV_v+\sum_{w\ne v}\alpha_wY_{v,w}\Big)\dd\pi_d,\\
+ \partial_d^2J_v&=\int_{\Pi_d}Z_v\,\dd\pi_d ,
+ \end{aligned}
+\end{equation}
+with $\Sigma_\alpha=\sum\alpha_i$. On any compact
+$s$-interval $[s_0,s_1]\subset(0,\infty)$, every such plane
+integral grows at most polynomially in $|d|$.
+\end{lemma}
+\begin{proof}
+The support issue: each star factors as
+$R_v=D_{\rm row}R_v^0D_{\rm col}$ with positive invertible
+diagonal multipliers, so $\ker R_v$, $\supp T_v$,
+$\ran R_v$, $\supp S_v$ are locally fixed. Smoothness reduces
+to smoothness of the positive eigenvalues of a product of two
+positive matrices, which is standard. The rank-zero case is
+trivial.
+
+The Gaussian identities
+$\partial_pw=Nw$, $\partial_p^2w=(N^2-\sigma^{-1})w$,
+$2\partial_\sigma w=\partial_p^2 w$ combined with the chain
+rule on $e_v$, restricted to its support, give (5.6) and (5.7).
+Dominating by polynomial times $e_v$ and using
+$\int_{\Pi_d}e_v\,\dd\pi_d\le$ polynomial in $|d|$, we can
+differentiate under the integral. Choosing $p_i$ as dependent
+coordinate yields $\partial_d^2J_v=\int\partial_{p_i}^2e_v\dd\pi_d$;
+choosing $p_{v+1}$ and then $p_{v-1}$ yields
+$\partial_d^2J_v=\int\partial_{p_{v+1}}\partial_{p_{v-1}}e_v\dd\pi_d$.
+Then (5.6) and (5.7) give (5.8).
+\end{proof}
+
+
+\subsection{Strict dissipation at comparable variances}
+\label{subsec:dissipation}
+
+% The mixed term $Z_v$ does not have a sign. We trade it for
+% column costs and then compare them to the row costs at the
+% neighboring vertex. The loss is $\sqrt2$, small enough that
+% strict dissipation survives when the three variance rates
+% are comparable.
+
+\begin{lemma}[Plane dissipation]\label{lem:plane-dissipation}
+For finite arrays and any positive rates,
+\begin{equation}\label{eq:heat-plane-dissipation}
+ 2\partial_sJ(d,s)\le\int_{\Pi_d}\sum_w
+ \Big[-\alpha_w+\sqrt2\max\{0,\Sigma_\alpha/2-\alpha_w\}\Big]
+ V_w\,\dd\pi_d .
+\end{equation}
+In particular, put $\lambda=11/10$. For each permutation of
+$(1,\lambda,\lambda)$,
+\begin{equation}\label{eq:heat-strict}
+ -\partial_sJ(d,s)\ge c_*\int_{\Pi_d}\sum_wV_w\,\dd\pi_d,
+ \qquad c_*=\tfrac12\min\{1-\tfrac35\sqrt2,\tfrac{11}{10}-\tfrac12\sqrt2\}>0 .
+\end{equation}
+\end{lemma}
+\begin{proof}
+The core inequality is the neighbor comparison
+\begin{equation}\label{eq:heat-neighbors}
+ Y_{w+1,w}+Y_{w-1,w}\le\sqrt2\,V_w .
+\end{equation}
+Let $A=W_w$, $B=W_{w-1}^*$, $N=N_w$. The elementary inequality
+$\|Az+Bz'\|^2\le 2\|Az\|^2+2\|Bz'\|^2$ implies
+$T_w\le2Q$ with $Q=\operatorname{diag}(A^*A,B^*B)$. By
+Lemma~\ref{lem:order},
+$V_w\ge2^{-1/2}b_Q([A\ B]^*N[A\ B])$. In a block-preserving
+eigenbasis of $Q$, discarding off-diagonal blocks gives
+$b_Q([A\ B]^*N[A\ B])\ge b_{A^*A}(A^*NA)+b_{B^*B}(B^*NB)$.
+Then $A^*A\le S_{w+1}$ with insertion $P_{w+1,w}$, and
+$B^*B\le S_{w-1}$ with insertion $P_{w-1,w}$. A second use of
+Lemma~\ref{lem:order} with comparison factor one proves
+(5.10). Summing (5.8) over $v$ and using
+$2Z_v\le Y_{v,v+1}+Y_{v,v-1}$ gives (5.9); the two column
+costs carrying the score at vertex $w$ have common coefficient
+$\Sigma_\alpha/2-\alpha_w$. For the stated rates $\Sigma=16/5$;
+the two coefficients are $-1+(3/5)\sqrt2$ and $-11/10+(1/2)\sqrt2$,
+both negative. This gives (5.10).
+\end{proof}
+
+\subsection{Energy bounds and single-edge dissipation}
+\label{subsec:energy-bounds}
+
+% The initial energy is bounded by the discrete $L^3$ norms
+% once the Gaussian variance is at least the mesh square.
+\begin{lemma}[Initial energy]\label{lem:initial-energy}
+For all $d\in\R$ and all variances $\sigma_v\ge h^2$,
+\begin{equation}\label{eq:initial-energy}
+ \begin{gathered}
+ \int_{\Pi_d}\|W_v\|_{\HS}^3\,\dd\pi_d\le M n_v^3 ,
+ \qquad M=1+\sqrt{2/\pi},\\
+ \int_{\Pi_d}\sum_ve_v\,\dd\pi_d\le2\sqrt2\,M\sum_v n_v^3 .
+ \end{gathered}
+\end{equation}
+The bounds also hold after replacing entries by smaller ones
+in absolute value.
+\end{lemma}
+\begin{proof}
+The Gaussian grid mass obeys
+\begin{equation}\label{eq:heat-mesh-mass}
+ \sum_i h g_\sigma(u_i-p)\le 1+\frac{2h}{\sqrt{2\pi\sigma}} .
+\end{equation}
+For $\sigma\ge h^2$, the right side is at most $M$. Weighted
+Holder gives
+$\|W_v\|_{\HS}^3\le M\sum|A_v|^3w_vw_{v+1}$.
+Using $p_v,p_{v+1}$ as free coordinates on $\Pi_d$ and
+integrating gives the first estimate; summing over $v$ and
+double-counting each edge once in each star gives the second.
+\end{proof}
+
+% When variance rates become very unequal, summed star
+% dissipation no longer holds. But a SINGLE EDGE still
+% dissipates exactly, because its star has no mixed term.
+\begin{lemma}[Single-edge energy]\label{lem:single-edge-energy}
+Fix $v\ne w$ and retain only the weighted block $C=C_{v,w}$
+at vertex $v$. For arbitrary positive rates, set
+\[
+ e^e_{v,w}=\Phi(CC^*), \quad
+ V^e_{v,w}=b_{C^*C}(C^*N_vC), \quad
+ Y^e_{v,w}=b_{CC^*}(CN_wC^*),
+\]
+\[
+ J^e_{v,w}(s)=\iint_{\R^2}e^e_{v,w}\,\dd p_v\dd p_w .
+\]
+The plane integrals of $e^e,V^e,Y^e$ are independent of $d$,
+and
+\begin{equation}\label{eq:heat-single-edge}
+ -2\partial_sJ^e_{v,w}=\iint_{\R^2}
+ (\alpha_vV^e_{v,w}+\alpha_wY^e_{v,w})\,\dd p_v\dd p_w .
+\end{equation}
+With $n_e$ the edge array's $L^3$ norm, whenever
+$\alpha_vs_0,\alpha_ws_0\ge h^2$ and $s_1>s_0$,
+\begin{equation}\label{eq:heat-single-edge-integral}
+ \int_{s_0}^{s_1}\iint_{\R^2}
+ (\alpha_vV^e_{v,w}+\alpha_wY^e_{v,w})\,\dd p_v\dd p_w\dd s
+ \le2M n_e^3 .
+\end{equation}
+For the full two-block star, $Y_{v,w}\le Y^e_{v,w}$.
+\end{lemma}
+\begin{proof}
+The integrands depend only on the two endpoint centers; use
+those as free coordinates. In the one-edge star the other
+column block is zero, so $Z_v=0$. After permuting blocks the
+matrices are
+$R=[C\ 0]$, $T=\operatorname{diag}(C^*C,0)$,
+$U=\operatorname{diag}(C^*N_vC,0)$, $S=CC^*$. The Hessian
+costs are exactly $V^e_{v,w}$ and $Y^e_{v,w}$; (5.8) gives
+(5.12). Since $e^e_{v,w}\le\|C\|_{\HS}^3$, the initial-energy
+lemma bounds $J^e_{v,w}(s_0)\le Mn_e^3$; integrate (5.12) and
+use that the terminal energy is nonnegative. Finally $CC^*\le S_v$
+and $CN_wC^*$ is supported on $CC^*$, so Lemma~\ref{lem:order}
+with factor one bounds $Y_{v,w}\le Y^e_{v,w}$.
+\end{proof}
+
+% With the two dissipation lemmas in hand, the heat machinery
+% is complete. We can now turn to the main smooth-annulus
+% argument.
+
+
+% ============================================================
+\section{Smooth annuli and repeated masks}\label{sec:smooth}
+% ============================================================
+
+% We prove the smooth-annulus version of the count estimate.
+% Same structure: at most n disjoint scale intervals at each
+% output point, complex coefficients $|\beta|\le1$, smooth
+% derivative windows. The result costs $\sqrt n$ but no more.
+
+% Define the smooth kernels and their scale-integrals by
+\begin{equation}\label{eq:smooth-kernels}
+ k_s(t)=\frac{t}{\sqrt s}g_s(t),\qquad
+ I_{\varepsilon,R}(t)
+ =\int_{\varepsilon^2}^{R^2}(k_s*k_s*k_s)(t)\,\frac{\dd s}{s} .
+\end{equation}
+
+\begin{proposition}[Smooth annuli with repeated choices]
+\label{prop:smooth-switches}
+Let $G_v\in C_c^\infty(\R^2)$ be complex. At each output point
+$(a,b)$, make a step choice of at most $n\ge1$ disjoint radius
+intervals $(\varepsilon_j,R_j)$ and coefficients $|\beta_j|\le1$.
+Then for $\mathcal K_{a,b}(t)=\sum_j\beta_jI_{\varepsilon_j,R_j}(t)$,
+\[
+ |\Lambda_{\mathcal K}(G_0,G_1,G_2)|
+ \le C\sqrt n\prod_{v=0}^2\|G_v\|_3 .
+\]
+The constant is absolute and independent of the step partition,
+the menu, and all endpoints.
+\end{proposition}
+
+\subsection{The cost of the switches}
+\label{subsec:switch-cost}
+
+% Work on the finite grid of the heat section. The masked dual
+% array (choosing which annuli are active at each (i,j))
+% is
+\begin{equation}\label{eq:repeated-mask}
+ A_0^s(i,j)=A_0(i,j)\sum_\ell\beta_\ell
+ \mathbf 1_{\{\varepsilon_\ell^2<s<R_\ell^2\}} .
+\end{equation}
+% The sum runs through the list chosen at THIS index pair.
+% Keep $A_1,A_2$ fixed.
+
+% The key inputs are that the mask stays bounded by $|A_0|$ and
+% that at every switch, the total variation is linear in n.
+\begin{equation}\label{eq:mask-switch-count}
+ |A_0^s(i,j)|\le|A_0(i,j)|,\qquad
+ \sum_\tau|A_0^{\tau+}(i,j)-A_0^{\tau-}(i,j)|\le2n|A_0(i,j)|.
+\end{equation}
+
+\begin{lemma}[Repeated-mask dissipation]\label{lem:repeated-dissipation}
+With masked arrays above and variances $\sigma_v=\alpha_vs$ where
+$\boldsymbol\alpha$ is a permutation of $(1,11/10,11/10)$,
+\begin{equation}\label{eq:repeated-dissipation}
+ \int_{s_-}^{s_+}\int_{\Pi_0}\sum_vV_v\dd\pi_0\dd s
+ \le C\Big(\sum_vn_v^3+n(n_0^3+n_0^2(n_1+n_2))\Big).
+\end{equation}
+\end{lemma}
+\begin{proof}
+Between consecutive switch scales the arrays are fixed, so
+\eqref{eq:heat-strict} applies. The only new work is bounding
+the total absolute energy jump.
+
+At a switch $\tau$, freeze the weights and set
+$\delta A(i,j)=A_0^{\tau+}(i,j)-A_0^{\tau-}(i,j)$.
+Interpolate all changed entries linearly; convexity keeps
+moduli bounded by $|A_0|$. For fixed centers and index, define
+the abbreviations
+$X^2=\sum_m|A_0(i,m)|^2w_1(m)$,
+$Y^2=\sum_m|A_0(m,j)|^2w_0(m)$,
+$H^2=\sum_m|A_2(m,i)|^2w_2(m)$,
+$Q^2=\sum_m|A_1(j,m)|^2w_2(m)$.
+In $R_0=[W_0\ W_2^*]$, the changing entry has row norm at most
+$\sqrt{w_0(i)}(X+H)$, column norm $\sqrt{w_1(j)}Y$,
+derivative $|\delta A(i,j)|\sqrt{w_0(i)w_1(j)}$; the adjoint
+in $R_1=[W_1\ W_0^*]$ has row and column bounds
+$\sqrt{w_1(j)}(Y+Q)$ and $\sqrt{w_0(i)}X$. The star at vertex 2
+is untouched. Lemma~\ref{lem:rank-change} gives
+\[
+ |\Delta e_0|+|\Delta e_1|\le C\sum_{i,j}w_0(i)w_1(j)|\delta A(i,j)|
+ (2XY+HY+QX).
+\]
+The measure $h^{-2}w_0(i)w_1(j)\dd\pi_0$ is a probability measure,
+under which $p_0,p_1$ are independent normals with means
+$u_i,u_j$ and variances $\sigma_0,\sigma_1$, and $p_2=-p_0-p_1$.
+Adding Gaussian variances gives
+$\mathbb E w_0(m)=hg_{2\sigma_0}(u_m-u_i)$,
+$\mathbb E w_1(m)=hg_{2\sigma_1}(u_m-u_j)$,
+$\mathbb E w_2(m)=hg_{\sigma_0+\sigma_1+\sigma_2}(u_m+u_i+u_j)$, and
+Lemma~\ref{lem:line-maximal} bounds the $\mathbb E$ of
+$X^2,Y^2,H^2,Q^2$ by constant multiples of
+$X_*^2=M_{\mathbb Z}(|A_0(i,\cdot)|^2)(j)$,
+$Y_*^2=M_{\mathbb Z}(|A_0(\cdot,j)|^2)(i)$,
+$H_*^2=M_{\mathbb Z}(|A_2(\cdot,i)|^2)(-i-j)$,
+$Q_*^2=M_{\mathbb Z}(|A_1(j,\cdot)|^2)(-i-j)$,
+all independent of the switch scale. Cauchy--Schwarz under the
+probability measure controls the three products. Applying the
+$L^{3/2}$ line maximal inequality on slices gives (via
+bijection $j\mapsto-i-j$ on relevant slices) that the three
+$L^3$ norms of the majorants are at most $Cn_0,Cn_2,Cn_1$.
+Integrating, summing the switches using
+\eqref{eq:mask-switch-count}, and H\"older on the grid gives
+the total jump cost
+$\sum_\tau|\Delta\sum_vJ_v(0,\tau)|\le Cn(n_0^3+n_0^2(n_1+n_2))$.
+Discarding the terminal energy and adding the initial-energy
+bound from Lemma~\ref{lem:initial-energy} finishes the proof.
+\end{proof}
+
+
+\subsection{From the energy to the smooth kernel}
+\label{subsec:energy-to-kernel}
+
+\begin{proof}[Proof of Proposition~\ref{prop:smooth-switches}]
+Put $\lambda=11/10$ and use base variances $\lambda s$ at every
+vertex. In Lemma~\ref{lem:mixed-trace}, insert the diagonal matrix
+with entries $k_s(u_i-p_v)/g_{\lambda s}(u_i-p_v)$ at vertex $v$.
+It is Hermitian and uniformly bounded: the polynomial factor in
+$k_s$ is absorbed by the extra Gaussian decay of $g_{\lambda s}$
+relative to $k_s$.
+
+To handle the cost at vertex $v$, narrow just its row variance to
+$s$. Write $T_v^{\rm b}$ for the base Gram and $T_v^{\rm n}$ for
+the narrowed-row Gram. Then
+$(R_v^{\rm b})^*D_vR_v^{\rm b}=\sqrt s\,U_v^{\rm n}$ and
+$T_v^{\rm n}\le\sqrt\lambda\,T_v^{\rm b}$. Row scaling preserves
+support, so Lemma~\ref{lem:order} bounds the base cost by
+$CsV_v^{\rm n}$. The narrowed variances are one of the
+permutations allowed in Lemma~\ref{lem:repeated-dissipation}.
+
+Integrate the mixed trace inequality over base centers in $\Pi_0$
+and over $s$ with measure $\dd s/s$. Its right side is controlled
+by \eqref{eq:repeated-dissipation}. On the left the density at
+each vertex is $hk_s(u_i-p_v)$; integration over the center plane
+convolves the three windows, giving
+\[
+ h^3\sum_{i,j,l}A_0(i,j)A_1(j,l)A_2(l,i)
+ \sum_{\text{chosen at }(i,j)}\beta\,
+ I_{\varepsilon,R}(u_i+u_j+u_l).
+\]
+If the original norms are nonzero, apply this to arrays normalized
+to have norms $n^{-1/2},1,1$. The right side is bounded by an
+absolute constant because $n(n^{-3/2}+2n^{-1})$ is bounded.
+Rescaling gives $C\sqrt n\prod_vn_v$. The Riemann passage of the
+preliminaries gives the proposition for smooth compactly supported
+inputs.
+\end{proof}
+
+\subsection{The hard endpoint identity}
+\label{subsec:hard-endpoint}
+
+% Smooth annuli differ from hard Hilbert annuli by a
+% difference of two odd endpoints. We record the exact
+% identity now so that the next two sections can estimate
+% the endpoint errors together.
+%
+% Put
+% psi = k_1*k_1*k_1 = -g_3''',
+% P(v) = 2int_0^v psi(z)dz (v>=0),
+% c_0 = P(inf) = 2g_3''(0) != 0.
+% Then P has a smooth even extension vanishing to second order
+% at zero. For rho>0 define
+\begin{equation}\label{eq:endpoint-kernel}
+ E_\rho(t)=\frac{P(|t|/\rho)-c_0\mathbf 1_{\{|t|>\rho\}}}{t},
+\end{equation}
+% with the continuous value zero at $t=0$. Gaussian scaling and
+% the substitution $z=|t|/\sqrt s$ in (6.1) give, away from
+% the endpoint values,
+\begin{equation}\label{eq:hard-smooth}
+ I_{\varepsilon,R}(t)
+ =\frac{c_0}{t}\mathbf 1_{\{\varepsilon<|t|<R\}}
+ +E_\varepsilon(t)-E_R(t) .
+\end{equation}
+% This is the BRIDGE between smooth and hard annuli. The hard
+% kernel is $c_0/t$ on the annulus $(\varepsilon,R)$; the
+% two endpoint errors $E_\varepsilon$ and $E_R$ account for the
+% Gaussian tail. Each $E_\rho$ is an odd dilate of one fixed
+% kernel with two jumps at $\pm\rho$.
+
+% So the hard count estimate reduces to two problems.
+% (i) The smooth-annuli estimate of Proposition~\ref{prop:smooth-switches}
+% bounds the correct $c_0/t$ piece after adding the
+% endpoint errors.
+% (ii) A frequency-block estimate, developed next, controls
+% the endpoint errors with square-root loss.
+
+
+% ============================================================
+\section{Uniform frequency-block estimate}\label{sec:frequency}
+% ============================================================
+
+% The second main new ingredient: a Fourier estimate for
+% oscillatory Gaussian windows, with coefficients that can
+% depend jointly on scale, frequency, and output point. Its
+% decisive hypothesis is a POINTWISE square-integral bound over
+% scales and frequencies. No sum of frequency suprema is
+% required.
+
+% Setup. Fix q>0, D>=1, and let Xi_D subset [-2D,2D] be
+% measurable with measure dxi/D, so mu(Xi_D)<=4. The frequency
+% windows and their convolutions are
+\begin{equation}\label{eq:frequency-windows}
+ \ell_\xi(z)=D^{-1}\frac{\dd}{\dd z}
+ \bigl(g_q(z)e^{i\xi z}\bigr),\qquad
+ H_\xi=\ell_\xi*\ell_\xi*\ell_\xi .
+\end{equation}
+% The parameters q,D are suppressed in this notation.
+
+\begin{proposition}[Frequency-block estimate]\label{prop:frequency-block}
+Let $\mathcal I\subset\mathbb Z$ be finite and nonempty, and
+$L_k=2^k$. Let $A_0,A_1,A_2$ be finite arrays on the lattice
+$u_i=ih$ with the $L^3$ norms $n_v$ defined as in
+Section~\ref{sec:heat}. Suppose measurable coefficients
+$\mu_{ij,k}:\Xi_D\to\C$ satisfy
+\begin{equation}\label{eq:frequency-coefficient-condition}
+ |\mu_{ij,k}(\xi)|\le C_1,\qquad
+ \sum_{k\in\mathcal I}\int_{\Xi_D}
+ |\mu_{ij,k}(\xi)|^2\,\dd\gamma(\xi)\le C_1^2,
+ \quad \dd\gamma=\dd\xi/D,
+\end{equation}
+for every index pair $(i,j)$, the first inequality holding
+almost everywhere. If
+\begin{equation}\label{eq:frequency-mesh}
+ h^2\le\frac{q\min_{k\in\mathcal I}L_k^2}{16D^2},
+\end{equation}
+then
+\begin{equation}\label{eq:frequency-finite-estimate}
+ \Big|h^3\int_{\Xi_D}\sum_{k\in\mathcal I}\sum_{i,j,l}
+ A_0(i,j)\mu_{ij,k}(\xi)A_1(j,l)A_2(l,i)
+ \mathrm{Dil}_{L_k}H_\xi(u_i+u_j+u_l)\,\dd\gamma(\xi)\Big|
+ \le C(q,C_1)\prod_{v=0}^2n_v .
+\end{equation}
+\end{proposition}
+
+% There are two reasons for refining one Gaussian variance in the
+% proof.
+% (1) A window oscillating at frequency D becomes an average
+% of score insertions at variance comparable to D^{-2}s.
+% (2) The boundary energy from changing the output array can
+% still be estimated uniformly at this small variance.
+% The refined row scores are controlled by the heat identity
+% through single-edge energies and changes of star energy at
+% slab boundaries. Such a change contains the square of the
+% changing edge norm times the norm of the unchanged edge; we
+% isolate a uniform estimate for that mixed term first.
+
+\begin{lemma}[Uniform mixed boundary estimate]
+\label{lem:frequency-boundary}
+Fix distinct $v,w\in\{0,1\}$, so $\{v,w\}=\{0,1\}$. The edge
+$v-w$ has array $B$, indexed in original order $(0,1)$, and the
+edge $v-2$ has array $E$, indexed in the order $(v,2)$. Put
+variances $(\theta s,s,s)$ with $0<\theta\le1/16$ and
+$h^2\le\theta s$. Let $\dd\nu_s(d)=g_{(1-\theta)s}(d)\dd d$.
+There is a nonnegative array $\mathcal H_E(i,j)$, independent of
+$(s,\theta)$, such that
+\begin{align}
+ &\int_{\R}\int_{\Pi_d}
+ \|C_{v,w}\|_{\HS}^2\|C_{v,2}\|_{\HS}\,\dd\pi_d\dd\nu_s(d)
+ \le C\sum h^2|B(i,j)|^2\mathcal H_E(i,j),\label{eq:frequency-boundary-mixed}\\
+ &\Big(\sum h^2\mathcal H_E(i,j)^3\Big)^{1/3}
+ \le C\Big(\sum h^2|E(m,l)|^3\Big)^{1/3}.\label{eq:frequency-boundary-norm}
+\end{align}
+Here $(i,j)$ are in the original endpoint order of edge $0$;
+$i_v,i_w$ denote the same two indices in the order $(v,w)$. The
+arrays $B,E$ are extended by zero on $\mathbb Z^2$.
+\end{lemma}
+\begin{proof}
+Expand the first Hilbert--Schmidt squared. For each $(i,j)$,
+$h^{-2}w_v(i_v)w_w(i_w)\dd\pi_d\dd\nu_s(d)$ is a probability
+measure: with $p_v,p_w$ the free coordinates, it makes them
+independent normals with means $u_{i_v},u_{i_w}$ and variances
+$\theta s,s$, while $d\sim g_{(1-\theta)s}$ and $p_2=d-p_v-p_w$.
+
+Under this measure, $\mathbb E h^{-2}w_v(m)w_2(l)$ is a bivariate
+Gaussian density at $(u_m,u_l)$ with mean $(u_{i_v},-u_{i_v}-u_{i_w})$
+and covariance
+$\Sigma=s\big(\!\begin{smallmatrix}2\theta&-\theta\\-\theta&3\end{smallmatrix}\!\big)$.
+The bounds $\Sigma\le4s\operatorname{diag}(\theta,1)$,
+$\det\Sigma=\theta(6-\theta)s^2$ and the Gaussian density
+formula give
+\begin{equation}\label{eq:frequency-covariance-domination}
+ \mathbb E[w_v(m)w_2(l)]
+ \le Ch^2g_{C'\theta s}(u_m-u_{i_v})
+ g_{C's}(u_l+u_{i_v}+u_{i_w})
+\end{equation}
+with $C'=4$ absolute.
+
+Let $M_{\mathbb Z}^{(1)},M_{\mathbb Z}^{(2)}$ be line maximal
+operators in the two coordinates. By Lemma~\ref{lem:line-maximal},
+take
+\begin{equation}\label{eq:frequency-boundary-majorant}
+ \mathcal H_E(i,j)^2
+ =C\bigl(M_{\mathbb Z}^{(1)}M_{\mathbb Z}^{(2)}|E|^2\bigr)
+ (i_v,-i_v-i_w).
+\end{equation}
+Then $\mathbb E\|C_{v,2}\|_{\HS}^2\le\mathcal H_E(i,j)^2$ by
+(7.4) and $h^2\le\theta s$. Cauchy--Schwarz under the
+probability measure gives (7.2). The map
+$(i,j)\mapsto(i_v,-i_v-i_w)$ is a bijection $\mathbb Z^2$;
+the $\ell^{3/2}$ bounds for the two line maximal operators
+applied to $|E|^2$ give (7.3).
+\end{proof}
+
+
+\subsection{Proof of the frequency-block estimate}
+\label{subsec:freq-proof}
+
+\begin{proof}[Proof of Proposition~\ref{prop:frequency-block}]
+
+\paragraph{Reduction to pointwise bounds.}
+Discard a common null set of frequencies so that both bounds in
+\eqref{eq:frequency-coefficient-condition} hold pointwise.
+Set
+\[
+ B_{0,k}^\xi(i,j)=A_0(i,j)\mu_{ij,k}(\xi),\quad
+ a_k=qL_k^2,\quad \theta=\frac1{16D^2}.
+\]
+For each $\xi$, use $B_{0,k}^\xi$ on the open slab
+$2a_k<s<3a_k$, and zero outside; keep $A_1,A_2$ throughout.
+Slabs are disjoint because $a_{k+1}=4a_k$. The mesh hypothesis
+$h^2\le\theta s$ holds on the whole scale interval
+$[s_-,s_+]=[\min a_k,4\max a_k]$.
+
+\paragraph{Refining a row variance.}
+Fix a slab, put $L=L_k$, $a=a_k$, $\omega=\xi/L$. The dilated
+window is
+\[
+ f_{\xi,k}(u)=\Dil_L\ell_\xi(u)
+ =\frac LD\frac{\dd}{\dd u}\bigl(g_a(u)e^{i\omega u}\bigr).
+\]
+Initially all three vertices have variance $s$. The diagonal
+insertion with entries $f_{\xi,k}(u_i-p_v)/g_s(u_i-p_v)$ has
+operator norm at most $C(q)$: $|L\omega/D|\le2$, and Gaussian
+decay in $g_a/g_s$ kills the rest. Splitting into real and
+imaginary parts gives eight terms, each a Hermitian insertion of
+uniformly bounded norm (Lemma~\ref{lem:mixed-trace}).
+
+Consider the row cost at $v$. Replace row variance $s$ by
+$b'=\theta s$, keep column variances at $s$, and shift the row
+center $p_v\to p_v+d$. Denote the refined Gram and score insertion
+by $T_v^d,U_v^d$. Gaussian variance addition gives
+\[
+ T_v=\int_\R T_v^d\dd\nu_s(d),\qquad
+ \dd\nu_s(d)=g_{s-b'}(d)\dd d .
+\]
+All these Grams share a common support (positive row scaling on
+the kernel preserves the kernel of the row matrix).
+
+To represent the inserted Gram in the same way, put
+$\omega'=a\omega/(a-b')$. Since $b'<a/2$, completing the square
+gives the exact identity
+\[
+ g_a(u)e^{i\omega u}
+ =\exp\!\Big(\frac{\omega^2ab'}{2(a-b')}\Big)
+ \int_\R g_{b'}(u-d)e^{i\omega'd}g_{a-b'}(d)\dd d .
+\]
+Differentiating in $u$, the derivative of $g_{b'}(u-d)$
+supplies the negative refined row score $-(u-d)/b'$. The
+prefactor is at most $C(q)$; also
+$g_{a-b'}(d)\le Cg_{s-b'}(d)$ and $L/D\le C(q)\sqrt{\theta s}$.
+Applying Lemma~\ref{lem:joint-convexity} and quadratic
+homogeneity to the resulting average gives
+\begin{equation}\label{eq:frequency-score-refinement}
+ b_{T_v}(K_v)\le C(q)\theta s\int_\R
+ V_v(p_0,\ldots,p_v+d,\ldots,p_2;s)\dd\nu_s(d).
+\end{equation}
+The refined variances are now $(\sigma_v,\sigma_w,\sigma_{v'})=(b',s,s)$
+with $b'<a/2$.
+
+\paragraph{Integrated refined scores.}
+Fix $v$; all stars, energies, costs refer to the slab edge $0$.
+For a center-dependent $Q$, abbreviate
+$\widetilde Q(s)=\int_\R\int_{\Pi_d}Q(p,s)\dd\pi_d\dd\nu_s(d)$.
+On any open scale interval with fixed arrays, the heat identities
+give
+\begin{equation}\label{eq:frequency-averaged-heat}
+ 2\partial_s\widetilde J_v=3\widetilde Z_v-\theta\widetilde V_v
+ -\sum_{w\ne v}\widetilde Y_{v,w}.
+\end{equation}
+The coefficient $3$ comes from the original second derivative
+$2+\theta$ in the plane, plus $1-\theta$ from differentiating
+$\nu_s$ and integrating by parts. Dominating by polynomials
+times $e_v$ and using Gaussian decay of $\nu_s$ justifies the
+$s$-differentiation and the two integrations by parts, and gives
+one-sided limits at slab boundaries.
+
+The metric Cauchy--Schwarz inequality $2Z_v\le\sum_{w\ne v}Y_{v,w}$
+gives
+\begin{equation}\label{eq:frequency-score-differential}
+ \theta\widetilde V_v
+ \le-2\partial_s\widetilde J_v+\tfrac12\sum_{w\ne v}\widetilde Y_{v,w}.
+\end{equation}
+Pay column costs by single edges: $S_v\ge C_{v,w}C_{v,w}^*$, so
+Lemma~\ref{lem:order} yields $Y_{v,w}\le Y^e_{v,w}$. The integrated
+single-edge bound (5.13) controls the total of $Y^e_{v,w}$. If the
+edge is a fixed one (say at vertices $1,2$), it costs at most
+$C(n_1^3+n_2^3)$; if it is the slab edge $0$, it costs at most
+$C\sum_k\sum_{i,j}h^2|B_{0,k}^\xi(i,j)|^3$, which after frequency
+integration by \eqref{eq:frequency-coefficient-condition} is at
+most $C(C_1)n_0^3$. All column terms are thus controlled, and the
+coefficient $1$ of the column cost makes this independent of the
+small variance ratio $\theta$.
+
+\paragraph{Summing slab boundaries.}
+Before the first slab, edge $0$ is zero, so the initial energy is
+at most $C(n_1^3+n_2^3)$ by Lemma~\ref{lem:initial-energy}. At a
+boundary of each slab, the change in the pointwise star energy at
+$v\in\{0,1\}$ is at most
+\[
+ C\bigl(\|C_{\rm sw}\|_{\HS}^3+\|C_{\rm sw}\|_{\HS}^2\|C_{\rm fix}\|_{\HS}\bigr).
+\]
+The cubic term integrates by Lemma~\ref{lem:initial-energy} to at
+most $C\sum_{i,j}k\sum h^2|B_{0,k}^\xi|^3$, i.e.\ at most
+$C(C_1)n_0^3$ after frequency integration.
+
+For the mixed term, let $A^{(v,2)}$ be the fixed unweighted array
+on edge $v-2$, indexed in the order $(v,2)$. Apply
+Lemma~\ref{lem:frequency-boundary} with this array; its majorant
+$\mathcal H_v$ is independent of $(k,\xi,s)$, and its cube norm
+is at most $C(n_1+n_2)$. Both boundaries together contribute at
+most a constant times
+\[
+ \sum_{i,j}h^2|A_0(i,j)|^2\mathcal H_v(i,j)
+ \sum_k\int_{\Xi_D}|\mu_{ij,k}|^2\dd\gamma
+ \le C(C_1)n_0^2(n_1+n_2)
+\]
+by \eqref{eq:frequency-coefficient-condition} and H\"older. This
+step is exactly where averaging coefficient squares over
+frequency matters; taking a separate frequency supremum per scale
+would not give the bound.
+
+\paragraph{Recovering the frequency-block form.}
+Integrate (7.7) on intervals between consecutive boundaries and
+telescope. Discard the terminal energy (nonnegative). The
+initial energy, the absolute jumps just estimated, and the
+single-edge costs give
+\begin{equation}\label{eq:frequency-integrated-score}
+ \int_{\Xi_D}\int_{s_-}^{s_+}
+ \theta\widetilde V_v\,\dd s\,\dd\gamma
+ \le C(C_1)\sum_{j=0}^2n_j^3 .
+\end{equation}
+Values at the finitely many slab boundaries are immaterial.
+Measurability in $\xi$ follows from measurability of the arrays
+and Borel dependence of finite-matrix energies; Tonelli's theorem
+justifies every nonnegative frequency integration above.
+
+Return to equal-variance stars and apply Lemma~\ref{lem:mixed-trace}
+to the real and imaginary parts of the three window insertions.
+Integrate over base centers on $\Pi_0$. Shifting the row center in
+(7.6) maps this plane to $\Pi_d$; the two column centers are free
+coordinates, so its Jacobian is one. The cost at vertex $v$ is at
+most $C(q)\theta s\widetilde V_v$; integrate over every slab with
+$\dd s/s$ and over $\Xi_D$ with $\dd\gamma$; (7.8) bounds the sum.
+
+On the other side, multiplying the three weighted windows in the
+trace and integrating the centers gives exactly
+\[
+ h^3\sum_{i,j,l}B_{0,k}^\xi(i,j)A_1(j,l)A_2(l,i)
+ \Dil_{L_k}H_\xi(u_i+u_j+u_l),
+\]
+the convolution identity on $\Pi_0$: the three window arguments
+sum to $u_i+u_j+u_l$. The expression is independent of $s$
+within its slab, whose logarithmic length is $\log(3/2)$.
+
+This proves \eqref{eq:frequency-finite-estimate} with the right
+side replaced by $C(q,C_1)\sum_vn_v^3$. If all $n_v>0$, apply
+this to $A_v/n_v$; coefficients and mesh hypothesis are unchanged,
+and trilinearity gives the product bound. If any norm is zero,
+the form is zero.
+\end{proof}
+
+
+\subsection{Continuous version}
+
+\begin{corollary}[Continuous frequency-block estimate]
+\label{cor:frequency-continuum}
+Let $q,D,\Xi_D,\mathcal I$ be as in Proposition~\ref{prop:frequency-block}.
+Suppose $\mu_{a,b,k}(\xi)$ is constant in $(a,b)$ on every cell
+of one finite rectangular partition, independent of $\xi$ and
+common to all $k$. Suppose its values on cells are bounded
+measurable functions of $\xi$ and that
+\begin{equation}\label{eq:frequency-continuous-condition}
+ |\mu_{a,b,k}(\xi)|\le C_1,\qquad
+ \sum_k\int_{\Xi_D}|\mu_{a,b,k}(\xi)|^2\dd\gamma\le C_1^2 .
+\end{equation}
+Define
+\begin{equation}\label{eq:frequency-continuous-kernel}
+ \mathcal K_{a,b}(t)=\int_{\Xi_D}\sum_{k\in\mathcal I}
+ \mu_{a,b,k}(\xi)\Dil_{L_k}H_\xi(t)\dd\gamma(\xi).
+\end{equation}
+Then for complex $G_v\in L^3(\R^2)$,
+\begin{equation}\label{eq:frequency-continuous-estimate}
+ |\Lambda_{\mathcal K}(G_0,G_1,G_2)|\le C(q,C_1)\prod_{v=0}^2\|G_v\|_3 .
+\end{equation}
+\end{corollary}
+\begin{proof}
+Take $G_v$ smooth compactly supported. On each step choice the
+kernel (7.10) is continuous in $t$ (bounded block, bounded
+measurable coefficients, Gaussian common majorant). After the
+frequency integration, the trilinear integral is the limit of
+its lattice Riemann sums. The finitely many rectangular step
+boundaries have measure zero. The mesh condition (7.2) holds
+for all sufficiently fine lattices, and the discrete norms
+converge to $\|G_v\|_3$. Proposition~\ref{prop:frequency-block}
+proves (7.11) for these inputs.
+
+For general $L^3$ inputs, the fixed finite block and scale set
+admit an integrable majorant $k_*(t)\ge|\mathcal K_{a,b}(t)|$,
+uniform in $(a,b)$, obtained as a finite sum of
+$\sup_{\xi\in\Xi_D}|H_\xi|$ dilates. For any integrable $k\ge0$,
+\[
+ \iiint|G_0(a,b)G_1(b,c)G_2(c,a)|k(a+b+c)\dd a\dd b\dd c
+ \le\|k\|_1\prod_v\|G_v\|_3 .
+\]
+Set $t=a+b+c$, apply Holder in $(a,b)$ for fixed $t$, then use
+the determinant-one changes of variables for the other two edge
+functions. Thus the defining integral is absolutely convergent,
+is continuous in the three $L^3$ inputs, and smooth
+approximation finishes the proof. The final constant is
+independent of $D$, of the number of scales, and of the step
+choice.
+\end{proof}
+
+% ============================================================
+\section{Rough kernels and endpoint errors}\label{sec:rough}
+% ============================================================
+
+% The comparison (6.3) leaves one error kernel at each annular
+% endpoint. Naively estimating them separately would cost the
+% number of endpoints. We instead collect endpoints at
+% comparable scales, so disjointness of the annuli gives a
+% Gaussian bound independent of the number of discontinuities.
+% The frequency-block estimate from the previous section
+% controls the groups with few discontinuities; the remaining
+% groups are few enough to estimate pointwise.
+
+\subsection{A pointwise Gaussian estimate}
+\label{subsec:gaussian-envelope}
+
+% For a locally integrable function on R^2, let M_x and M_y
+% denote the centered HL maximal operators in the two
+% coordinates. For F,G in L^3, put
+\begin{equation}\label{eq:rough-maximal-envelope}
+ \mathcal M_{F,G}(x,y)
+ =\bigl(M_x(|F|^{5/2})(x,y)\bigr)^{2/5}
+ \bigl(M_y(|G|^{5/2})(x,y)\bigr)^{2/5}.
+\end{equation}
+% The one-dimensional maximal inequality on L^{6/5}, then
+% Hölder, gives
+\begin{equation}\label{eq:rough-maximal-norm}
+ \|\mathcal M_{F,G}\|_{3/2}\le C\|F\|_3\|G\|_3 .
+\end{equation}
+
+\begin{lemma}[Gaussian envelope]\label{lem:gaussian-envelope}
+Suppose $c>0$ and a measurable kernel $f$ satisfies
+$|f(z)|\le e^{-cz^2}w(z)$ with $w\ge0$ and $\|w\|_5\le\delta$.
+Then for every $L>0$,
+\begin{equation}\label{eq:gaussian-envelope}
+ \int_\R|F(x+t,y)G(x,y+t)|\,|\Dil_Lf(t)|\dd t
+ \le C_c\delta\,\mathcal M_{F,G}(x,y)
+\end{equation}
+at every point where the right side is finite.
+\end{lemma}
+\begin{proof}
+After setting $t=Lz$, apply Holder with exponents
+$(5/2,5/2,5)$ to
+$|F(x+Lz,y)|e^{-cz^2/2}$,
+$|G(x,y+Lz)|e^{-cz^2/2}$, $w(z)$.
+The first two resulting integrals are controlled by
+$M_x(|F|^{5/2})$ and $M_y(|G|^{5/2})$, uniformly in $L$, by
+splitting $\R$ into $|z|\le1$ and shells $2^j<|z|\le2^{j+1}$.
+The last is $\|w\|_5\le\delta$.
+\end{proof}
+
+% In the triangular form, $G_1(b,c)=F(b+c,-b)$ and
+% $G_2(c,a)=G(-a,c+a)$, both norm-preserving. With
+% $t=a+b+c$, the c-integral is bounded by
+% $C_c\delta\mathcal M_{F,G}(-a,-b)$, so integrating
+% against |G_0(a,b)| gives a contribution at most
+% $C_c\delta\prod_v\|G_v\|_3$. Everything can depend on (a,b)
+% uniformly, so this statement holds for any kernel family
+% with a uniform closed-form Gaussian envelope.
+
+
+\subsection{Rough kernels with a controlled number of jumps}
+\label{subsec:rough-family}
+
+% The following proposition turns the frequency estimate into a
+% bound for piecewise smooth kernels. The POINTWISE bound on
+% the number of jumps at one scale and the bound on their TOTAL
+% number play different roles: they supply the uniform bound and
+% the square-integral bound for the frequency coefficients.
+
+\begin{proposition}[Rough kernel families]\label{prop:rough-family}
+Fix $A\ge1$, $c>0$. Let $n\ge1$, $\mathcal I\subset\mathbb Z$
+finite. For $k\in\mathcal I$, let $K_{a,b,k}:\R\to\C$ and
+$m_{a,b,k}\in\{0,1,2,\ldots\}$ be finite-valued step choices in
+$(a,b)$ with $K_{a,b,k}=0$ whenever $m_{a,b,k}=0$. When
+$m_{a,b,k}\ge1$, suppose $K_{a,b,k}$ is $C^1$ off a set of at
+most $Am_{a,b,k}$ points, and that outside this set
+\begin{equation}\label{eq:rough-family-decay}
+ |K_{a,b,k}(z)|+|K_{a,b,k}'(z)|\le Ae^{-cz^2} .
+\end{equation}
+Suppose also that
+\begin{equation}\label{eq:rough-family-hypotheses}
+ \int_\R z^\ell K_{a,b,k}(z)\dd z=0\quad(\ell=0,1,2),
+ \quad\sum_k m_{a,b,k}\le An,
+ \quad m_{a,b,k}\le A\sqrt n .
+\end{equation}
+For $\mathcal K_{a,b}(t)=\sum_k\Dil_{2^k}K_{a,b,k}(t)$ and
+$G_v\in C_c^\infty$, one has
+\begin{equation}\label{eq:rough-family-bound}
+ |\Lambda_{\mathcal K}(G_0,G_1,G_2)|
+ \le C_{A,c}\sqrt n\log(2+n)\prod_v\|G_v\|_3 .
+\end{equation}
+\end{proposition}
+\begin{proof}
+
+\paragraph{A Gaussian-weighted primitive.}
+Seek a representation $K=(g_{3q}f)'''$. Define
+\[
+ U(z)=\frac12\int_{-\infty}^z(z-u)^2K(u)\dd u .
+\]
+The three moment conditions imply, for $j=0,1,2$,
+\[
+ U^{(j)}(z)=-\frac1{(2-j)!}\int_z^\infty(z-u)^{2-j}K(u)\dd u .
+\]
+Use the defining integral when $z\le0$, the tail integral when
+$z\ge0$; the Gaussian tail estimates give
+$|U^{(j)}(z)|\le Ce^{-c_1 z^2}$ for $j=0,1,2$, with $c_1>0$ fixed.
+Moreover $U'''=K$ as a weak derivative. Choose $q>0$ large enough,
+depending on $(A,c)$, and put $f=U/g_{3q}$. Then
+\begin{equation}\label{eq:rough-f-decay}
+ |f^{(j)}(z)|\le Ce^{-c_2z^2}\quad(0\le j\le3).
+\end{equation}
+The functions $f,f',f''$ are continuous. Using $K'$ between
+exceptional points bounds the classical derivative of $f'''$ by
+$Ce^{-c_2z^2}$; the jump at each exceptional point is bounded by
+$C$, and there are at most $Am$ of them. Thus
+\begin{equation}\label{eq:rough-f-bv}
+ \|f^{(j)}\|_\infty\le C,\quad
+ \operatorname{Var}(f^{(j)})\le Cm\quad(0\le j\le3).
+\end{equation}
+
+\paragraph{Fourier bounds.}
+Normalize $\widehat f(\xi)=(2\pi)^{-1}\int f(z)e^{-i\xi z}\dd z$.
+Let $\Xi_1=\{|\xi|<2\}$ and $\Xi_D=\{D\le|\xi|<2D\}$ for dyadic
+$D\ge2$. For $D\ge2$, the distributional derivative of $f'''$ is
+a finite measure of total variation at most $Cm$, so
+$|\xi|^4|\widehat f(\xi)|\le Cm$ gives the first bound below. For
+the second, the translation inequality for bounded variation
+gives
+$\|f'''(\cdot+D^{-1})-f'''\|_1\le Cm/D$; applying the supremum
+bound \eqref{eq:rough-f-bv} to bound its $L^2$ norm squared by
+$Cm/D$, and using Plancherel
+$\widehat{f'''}(\xi)=(i\xi)^3\widehat f(\xi)$ together with
+$|e^{i\xi/D}-1|\ge2\sin(1/2)$ on $\Xi_D$, yields
+\begin{equation}\label{eq:rough-fourier-bounds}
+ \sup_{\xi\in\Xi_D}|D^3\widehat f(\xi)|\le\frac{Cm}{D},\quad
+ \int_{\Xi_D}|D^3\widehat f(\xi)|^2\dd\xi\le\frac{Cm}{D}.
+\end{equation}
+For $D=1$, both bounds follow from \eqref{eq:rough-f-decay} and
+$m\ge1$.
+
+\paragraph{A finite frequency cutoff.}
+Choose $\chi\in C_c^\infty$ supported on $[-2,2]$ equal to one on
+$[-1,1]$, set $P_*=(2+n)^8$, and let
+$\widehat{f_*}(\xi)=\widehat f(\xi)\chi(\xi/P_*)$,
+$K_*=(g_{3q}f_*)'''$. Let $\eta$ be the Schwartz kernel of the
+multiplier $\chi$, $\eta_{P_*}(z)=P_*\eta(P_*z)$; then
+$f_*^{(j)}=f^{(j)}*\eta_{P_*}$ for $0\le j\le3$. The
+translation inequality gives
+$\|f^{(j)}-f_*^{(j)}\|_1\le Cm/P_*$. The supremum norms are
+uniformly bounded. Interpolation between $L^1$ and $L^\infty$
+yields $\|f^{(j)}-f_*^{(j)}\|_5\le C(m/P_*)^{1/5}$.
+Since every derivative of $g_{3q}$ is bounded by a constant
+multiple of a slightly wider Gaussian, the product rule shows
+\begin{equation}\label{eq:rough-cutoff-error}
+ |K(z)-K_*(z)|\le e^{-c_3z^2}w(z),\quad
+ \|w\|_5\le C(m/P_*)^{1/5}
+\end{equation}
+for a nonnegative $w$. Apply this separately to each
+$K_{a,b,k}$. By \eqref{eq:rough-family-hypotheses},
+$(m_{a,b,k}/P_*)^{1/5}\le C/n$ whenever $m_{a,b,k}\ne0$, and at
+each $(a,b)$ at most $An$ such indices are nonzero.
+Lemma~\ref{lem:gaussian-envelope} bounds the sum of all
+replacement errors by $C\mathcal M_{F,G}(-a,-b)$ after
+integration in $c$, hence their contribution to
+$\Lambda_{\mathcal K}$ is at most $C\prod_v\|G_v\|_3$.
+
+\paragraph{Applying the frequency estimate.}
+For each dyadic $D$ use the kernels of
+Corollary~\ref{cor:frequency-continuum}:
+$\ell_\xi(z)=D^{-1}d/dz(g_q(z)e^{i\xi z})$ and
+$H_\xi=\ell_\xi*\ell_\xi*\ell_\xi$. Gaussian convolution and
+modulation give $(g_{3q}(z)e^{i\xi z})'''=D^3H_\xi(z)$. For block
+$\Xi_D$, the contribution to $K_{a,b,k,*}$ is
+\[
+ \int_{\Xi_D}c_{a,b,k}(\xi)H_\xi(z)\dd\xi,\quad
+ c_{a,b,k}(\xi)=D^3\widehat f_{a,b,k}(\xi)\chi(\xi/P_*).
+\]
+Set $\mu_{a,b,k}(\xi)=Dc_{a,b,k}(\xi)/\sqrt n$. The two estimates
+in \eqref{eq:rough-fourier-bounds} give
+\[
+ |\mu_{a,b,k}(\xi)|\le\frac{Cm_{a,b,k}}{\sqrt n}\le C,\quad
+ \sum_k\int_{\Xi_D}|\mu_{a,b,k}|^2\frac{\dd\xi}{D}
+ \le C\sum_k\frac{m_{a,b,k}}{n}\le C .
+\]
+Every step choice in $(a,b)$ has been preserved; different $k$
+share a common finite rectangular partition. The block kernel is
+$\sqrt n\int_{\Xi_D}\mu_{a,b,k}(\xi)H_\xi\,\dd\xi/D$. By
+Corollary~\ref{cor:frequency-continuum} (with measure $\dd\xi/D$),
+this block contributes at most $C\sqrt n\prod_v\|G_v\|_3$. Only
+$O(\log(2+n))$ blocks meet the cutoff's support; summing their
+bounds and adding the replacement error proves
+\eqref{eq:rough-family-bound}.
+\end{proof}
+
+
+\subsection{Removing the first moment of an odd kernel}
+\label{subsec:odd-family}
+
+% Endpoint kernels are odd but need not have vanishing first
+% moment. A dilation identity reduces them to the previous
+% proposition.
+
+\begin{corollary}[Odd kernel families]\label{cor:odd-family}
+Proposition~\ref{prop:rough-family} remains valid if the
+three moment conditions in \eqref{eq:rough-family-hypotheses}
+are replaced by the assumption that every $K_{a,b,k}$ is odd
+almost everywhere.
+\end{corollary}
+\begin{proof}
+Choose a real, smooth, compactly supported odd $\varphi$ with
+$\int z\varphi(z)\dd z=1$. Put $u_{a,b,k}=\int zK_{a,b,k}\dd z$
+(uniformly bounded, vanishing when $m_{a,b,k}=0$). The kernels
+$K_{a,b,k}-u_{a,b,k}\varphi$ satisfy all three moment conditions
+and the other hypotheses of Proposition~\ref{prop:rough-family},
+with enlarged fixed constants and unchanged counts. It remains
+to bound the subtracted terms.
+
+Define $\Psi=\varphi-2^{-1}\Dil_2\varphi$; oddness gives its
+zeroth and second moments, and its first moment is
+$1-2^{-1}\cdot2=0$. For any integer $J\ge1$,
+\[
+ \Dil_{2^k}\varphi
+ =\sum_{j=0}^{J-1}2^{-j}\Dil_{2^{k+j}}\Psi
+ +2^{-J}\Dil_{2^{k+J}}\varphi.
+\]
+For each fixed $j$, apply Proposition~\ref{prop:rough-family} to
+shifted scales $k+j$ with kernels $u_{a,b,k}\Psi$ and shifted
+counts $m_{a,b,k}$; the constants are independent of $j$ and the
+outer coefficients $2^{-j}$ are summable. The remainder
+contributes at most $C2^{-J}n\prod_v\|G_v\|_3$ by
+Lemma~\ref{lem:gaussian-envelope}, since there are at most $An$
+nonzero counts at any output point. Letting $J\to\infty$ proves
+the claim.
+\end{proof}
+
+\subsection{Grouping the endpoint errors}
+\label{subsec:endpoint-errors}
+
+\begin{proposition}[Endpoint errors]\label{prop:endpoint-errors}
+At each $(a,b)$, choose at most $n$ pairwise disjoint open radius
+intervals $(\varepsilon_j,R_j)$ with $0<\varepsilon_j<R_j<\infty$
+and complex coefficients $\beta_j$, $|\beta_j|\le1$. Suppose
+these are a finite-valued step function of $(a,b)$. Then the
+kernel $\mathcal E_{a,b}(t)=\sum_j\beta_j(E_{\varepsilon_j}-E_{R_j})(t)$
+satisfies
+\begin{equation}\label{eq:endpoint-error-bound}
+ |\Lambda_{\mathcal E}(G_0,G_1,G_2)|
+ \le C\sqrt n\log(2+n)\prod_{v=0}^2\|G_v\|_3
+\end{equation}
+for $G_v\in C_c^\infty$.
+\end{proposition}
+\begin{proof}
+Place each endpoint $\rho$ in the unique half-open dyadic
+interval $[L_k,2L_k)$, $L_k=2^k$. Let $m_{a,b,k}$ count
+endpoints with multiplicity, collect their contributions into
+$\Dil_{L_k}K_{a,b,k}$. Thus $K_{a,b,k}$ is a linear combination
+of the dimensionless kernels
+\begin{equation}\label{eq:dimensionless-endpoint}
+ e_\alpha(z)=\frac{P(|z|/\alpha)-c_0\mathbf 1_{\{|z|>\alpha\}}}{z},
+ \qquad 1\le\alpha<2,
+\end{equation}
+with coefficient $\beta_j$ at a lower endpoint and $-\beta_j$ at
+its upper endpoint. Every kernel is odd, its possible jump
+points are $\pm\alpha$, and
+\begin{equation}\label{eq:endpoint-count}
+ \sum_k m_{a,b,k}\le2n .
+\end{equation}
+
+The important assertion is the uniform bound, with constants
+independent of $m_{a,b,k}$,
+\begin{equation}\label{eq:paired-endpoint-bound}
+ |K_{a,b,k}(z)|+|K_{a,b,k}'(z)|\le Ce^{-cz^2}
+\end{equation}
+off at most $2m_{a,b,k}$ points. Fix $(a,b)$ and $k$. Pair the
+two endpoints of each annulus whose endpoints both lie in
+$[L_k,2L_k)$. After scaling, its contribution is
+$\beta_j(e_{\alpha_j}-e_{\gamma_j})$ with
+$1\le\alpha_j<\gamma_j<2$. The intervals $(\alpha_j,\gamma_j)$
+are disjoint, so $\sum_j(\gamma_j-\alpha_j)\le1$. At most two
+endpoints are unpaired, corresponding to intervals that straddle
+$L_k$ and $2L_k$ respectively (an interval that strictly
+contains $[L_k,2L_k)$ has no endpoint in this group).
+
+For $|z|\le3$, the smooth term $s_\alpha(z)=P(|z|/\alpha)/z$ has
+a smooth even extension, and
+$\sup_{|z|\le3,1\le\alpha\le2}(|\partial_\alpha s_\alpha|+
+|\partial_z\partial_\alpha s_\alpha|)<\infty$. The fundamental
+theorem of calculus in $\alpha$, followed by the length bound,
+bounds the sum of the paired smooth terms independently of their
+number. For the paired hard terms, at most one difference is
+nonzero at any $|z|$, and on its support $|z|\ge1$, so the size
+and classical derivative are uniformly bounded. The two unpaired
+terms obey the same bounds on this compact range.
+
+For $|z|>3$, all indicators equal one and
+$P(|z|/\alpha)-c_0=-2g_3''(|z|/\alpha)$. Consequently
+$e_\alpha$, $\partial_ze_\alpha$, $\partial_\alpha e_\alpha$,
+and $\partial_z\partial_\alpha e_\alpha$ are bounded by
+$Ce^{-cz^2}$, uniformly for $1\le\alpha\le2$. The length bound
+and the two unpaired terms give the same bound for their sum.
+Combining the compact and tail estimates proves
+\eqref{eq:paired-endpoint-bound}.
+
+Finish by separating groups by their count. At each output
+point, \eqref{eq:endpoint-count} implies at most $2\sqrt n$
+indices have $m_{a,b,k}>\sqrt n$. The uniform bound
+\eqref{eq:paired-endpoint-bound} and
+Lemma~\ref{lem:gaussian-envelope} bound all these groups
+together, after $c$-integration, by $C\sqrt n\mathcal M_{F,G}$;
+their trilinear contribution is at most
+$C\sqrt n\prod_v\|G_v\|_3$.
+
+Set the kernels and counts of those groups to zero. The
+remaining family satisfies \eqref{eq:rough-family-decay}, has
+odd kernels, and has counts satisfying
+$\sum_km_{a,b,k}\le2n$ and $m_{a,b,k}\le\sqrt n$.
+Corollary~\ref{cor:odd-family} applies and proves
+\eqref{eq:endpoint-error-bound}. Every grouping and deletion is
+made at the output point; no common choice of endpoints or
+dyadic groups is imposed at different points.
+\end{proof}
+