==== TABLE OF CONTENTS ====
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/masks.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/triangle.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//main.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//preamble.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/consequences.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/heat.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/introduction.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/jumps.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/matrix.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/maximal.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/trace.tex
The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//references.bib
===========================
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/masks.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[x=1cm,y=1cm,font=\small,
active/.style={line width=2pt,draw=blue!65!black},
event/.style={densely dotted,draw=black!24},
endpoint/.style={circle,draw=blue!65!black,fill=white,
line width=.65pt,inner sep=0pt,minimum size=4pt},
endpoint label/.style={font=\footnotesize,fill=white,inner sep=2pt},
energy/.style={line width=1pt,draw=black},
jump/.style={-{Stealth[length=3pt]},densely dashed,
line width=.8pt,draw=red!65!black}]
\foreach \t in {2.3,3.8,6.4,9.0,11.6}
\draw[event] (\t,.1)--(\t,6.15);
\node[anchor=west] at (1.9,6.65) {Output-dependent scale intervals};
\foreach \yy in {5.8,4.8,3.8}
\draw[black!22] (1.95,\yy)--(12.25,\yy);
\node[anchor=east] at (1.7,5.8) {$(i_1,j_1)$};
\node[anchor=east] at (1.7,4.8) {$(i_2,j_2)$};
\node[anchor=east] at (1.7,3.8) {$(i_3,j_3)$};
\draw[active] (2.3,5.8)--(6.4,5.8);
\draw[active] (3.8,4.8)--(9.0,4.8);
\draw[active] (6.4,3.8)--(11.6,3.8);
\foreach \xx/\yy in {2.3/5.8,6.4/5.8,3.8/4.8,9.0/4.8,6.4/3.8,11.6/3.8}
\node[endpoint] at (\xx,\yy) {};
\node[endpoint label,above] at (2.3,5.91) {$\varepsilon_{i_1j_1}^{\,2}$};
\node[endpoint label,above] at (6.4,5.91) {$\rho_{i_1j_1}^{\,2}$};
\node[endpoint label,above] at (3.8,4.91) {$\varepsilon_{i_2j_2}^{\,2}$};
\node[endpoint label,above] at (9.0,4.91) {$\rho_{i_2j_2}^{\,2}$};
\node[endpoint label,above] at (6.4,3.91) {$\varepsilon_{i_3j_3}^{\,2}$};
\node[endpoint label,above] at (11.6,3.91) {$\rho_{i_3j_3}^{\,2}$};
\draw[-{Stealth[length=4pt]},black!65] (1.95,3.35)--(12.6,3.35)
node[right] {$s$};
\node[anchor=west,fill=white,inner sep=2pt] at (1.9,2.9) {Integrated energy};
\draw[-{Stealth[length=4pt]},black!65] (1.95,.1)--(12.6,.1)
node[right] {$s$};
\node[anchor=east] at (1.7,1.7) {$J(s)$};
\draw[energy] (2.3,2.2) .. controls (2.9,2.05) and (3.3,1.91) .. (3.8,1.82);
\draw[jump] (3.8,1.82)--(3.8,2.38);
\draw[energy] (3.8,2.38) .. controls (4.7,2.13) and (5.4,1.85) .. (6.4,1.64);
\draw[jump] (6.4,1.64)--(6.4,1.97);
\draw[energy] (6.4,1.97) .. controls (7.2,1.75) and (8.1,1.37) .. (9.0,1.18);
\draw[jump] (9.0,1.18)--(9.0,.78);
\draw[energy] (9.0,.78) .. controls (9.9,.61) and (10.7,.43) .. (11.6,.32);
\node[below] at (2.3,.02) {$s_-$};
\node[below] at (11.6,.02) {$s_+$};
\end{tikzpicture}
\caption{Three selected output pairs and a schematic integrated energy,
not an exact heat-flow solution. The entry $A_0^s(i,j)$ is retained only
on $(\varepsilon_{ij}^2,\rho_{ij}^2)$. Between successive event scales,
$J(s)$ is nonincreasing by the fixed-array dissipation inequality.
Dashed arrows connect its one-sided limits at events; the values at the
events themselves are not shown. Jumps may have either sign, and
endpoints may coincide, as illustrated
by $\rho_{i_1j_1}^2=\varepsilon_{i_3j_3}^2$. The bound
$\sum_\tau |\Delta J(\tau)|\le C\mathcal S(A)$, together with the
initial-energy bound, controls the accumulated dissipation independently
of the number of event scales.}
\label{fig:mask-energy}
\end{figure}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/masks.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/triangle.tex ====
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[>=Stealth, every node/.style={font=\small},
space/.style={circle,draw=black!60,minimum size=9mm,inner sep=2pt},
edge/.style={->,thick,blue!65!black}]
\node[space] (h0) at (0,1.9) {$H_0$};
\node[space] (h1) at (-1.8,-.8) {$H_1$};
\node[space] (h2) at (1.8,-.8) {$H_2$};
\draw[edge] (h1) -- node[left=5pt] {$W_0$} (h0);
\draw[edge] (h2) -- node[below=5pt] {$W_1$} (h1);
\draw[edge] (h0) -- node[right=5pt] {$W_2$} (h2);
\node[below=18pt] at (0,-.8) {the cyclic product $W_0W_1W_2$};
\node[space] (r1) at (5,1.3) {$H_1$};
\node[space] (r2) at (5,-.8) {$H_2$};
\node[space] (r0) at (7.5,.25) {$H_0$};
\draw[edge] (r1) -- node[above=5pt] {$W_0$} (r0);
\draw[edge] (r2) -- node[below=5pt] {$W_2^*$} (r0);
\node[below=18pt] at (6.25,-.8) {$R_0=[\,W_0\quad W_2^*\,]$};
\end{tikzpicture}
\caption{The three edge maps and the row star at vertex $0$.
Taking the adjoint of $W_2$ makes both blocks of $R_0$
have target $H_0$. The two Grams $R_0^*R_0$ and $R_0R_0^*$
have the same positive eigenvalues on different spaces.}
\label{fig:triangle}
\end{figure}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//figures/triangle.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//main.tex ====
\input{preamble}
\title{The maximal triangular Hilbert transform\\at the symmetric point}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
For arbitrary complex inputs in $L^3(\R^2)$, we prove the pointwise
maximal $L^3\times L^3\to L^{3/2}$ estimate for the triangular Hilbert
transform, with the supremum over both hard truncation endpoints.
The estimate yields joint almost-everywhere and $L^{3/2}$ convergence
as the lower endpoint tends to zero and the upper endpoint tends to infinity.
This also proves the conjectured scalar estimate at the symmetric point.
\end{abstract}
\tableofcontents
\input{sections/introduction}
\input{sections/matrix}
\input{sections/trace}
\input{sections/heat}
\input{sections/jumps}
\input{sections/maximal}
\input{sections/consequences}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//main.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//preamble.tex ====
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==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//preamble.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/consequences.tex ====
\section{General inputs and principal values}\label{sec:consequences}
We now pass the hard maximal estimate from compactly supported smooth
inputs to all complex $L^3$ inputs. The main point is to control all
truncations on one set of full measure. We then prove joint principal-value
convergence and recover the scalar simplex formulation.
Write $p=3/2$ and
\[
B_{\varepsilon,R}(F,G)(x,y)
=\int_{\varepsilon<|t|n}}
|B_{\varepsilon,R}(F,G)-B(F,G)|\right\|_{3/2}
\longrightarrow0.
\end{equation}
The limit is the unique bounded bilinear extension of the
principal-value operator on compactly supported smooth inputs.
\end{theorem}
\begin{proof}
For compactly supported smooth $f,g$ and a fixed $(x,y)$, the function
$h(t)=f(x+t,y)g(x,y+t)$ is smooth and compactly supported. The truncated
integral is
\[
\int_\varepsilon^R\frac{h(t)-h(-t)}t\,\dd t.
\]
Its numerator is $O(t)$ at zero and vanishes for large $t$, so it has a
joint limit at every point.
For general inputs, let
\[
\Omega_n(F,G)=
\sup_{\substack{0<\varepsilon,\varepsilon'<1/n\\R,R'>n}}
|B_{\varepsilon,R}(F,G)-B_{\varepsilon',R'}(F,G)|.
\]
These are measurable: by endpoint continuity the supremum may be taken
over rational parameters. They decrease with $n$ and are bounded by
$2B_*(F,G)$. Fix smooth approximants $f_m,g_m$ as above. Outside a
single null set for this countable collection of input pairs,
\[
\Omega_n(F,G)\le\Omega_n(f_m,g_m)
+2B_*(F-f_m,G)+2B_*(f_m,G-g_m).
\]
Letting $n\to\infty$, the smooth joint convergence and
Theorem~\ref{thm:main} give
\[
\left\|\lim_{n\to\infty}\Omega_n(F,G)\right\|_p
\le 2C_M\bigl(\|F-f_m\|_3\|G\|_3
+\|f_m\|_3\|G-g_m\|_3\bigr).
\]
The right side tends to zero with $m$. Hence the limiting oscillation
is zero almost everywhere. This is the Cauchy criterion for all pairs
of endpoints in the indicated tails, so the joint complex limit exists.
It is measurable by taking a fixed cofinal sequence of endpoint pairs,
and its magnitude is at most $B_*(F,G)$.
Taking $(\varepsilon',R')$ to the limit in the definition of
$\Omega_n$ bounds the supremum in
\eqref{eq:maximal-tail-convergence} by $\Omega_n(F,G)$.
Dominated convergence applies because
$0\le\Omega_n(F,G)\le2B_*(F,G)\in L^p$.
This proves \eqref{eq:maximal-tail-convergence}, and hence joint norm
convergence. Passing to the norm limit in the bilinear identities for
the truncations proves bilinearity of $B$. Its bound and uniqueness
follow from Theorem~\ref{thm:main} and density.
\end{proof}
\subsection{Scalar forms and the simplex coordinates}
For complex $F_0,F_1,F_2\in L^3(\R^2)$, define the flat scalar form
\[
\mathcal L_{\varepsilon,R}(F_0,F_1,F_2)
=\iiint_{\varepsilon<|t|0$, an integer $N\ge0$, and nodes $u_i=ih$, $-N\le i\le N$.
At vertex $v\in\mathbb Z/3\mathbb Z$ take $H_v=\C^{2N+1}$, and prescribe
arbitrary complex edge values $A_v(i,j)$. For centers $p_v\in\R$ and
variances $\sigma_v>0$, define
\begin{gather*}
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),\\
W_v(i,j)=A_v(i,j)\sqrt{w_v(i)w_{v+1}(j)}.
\end{gather*}
Use the stars $R_v=[W_v\ W_{v-1}^*]$ and Grams
$T_v=R_v^*R_v$, $S_v=R_vR_v^*$ from Section~\ref{sec:trace}, and set
\[
e_v=\Phi(T_v)=\Phi(S_v).
\]
Define the diagonal score and its insertions into the incident stars by
\begin{align*}
L_w&=\diag_i\frac{u_i-p_w}{\sigma_w},
&U_v&=R_v^*L_vR_v,\\
U_{v,v+1}&=W_vL_{v+1}W_v^*,
&U_{v,v-1}&=W_{v-1}^*L_{v-1}W_{v-1}.
\end{align*}
Define their quadratic costs and the mixed term by
\begin{equation}\label{eq:heat-costs}
V_v=b_{T_v}(U_v),\qquad
Y_{v,w}=b_{S_v}(U_{v,w})\quad(w\ne v),\qquad
Z_v=\mathcal B_{S_v}(U_{v,v+1},U_{v,v-1}).
\end{equation}
Every insertion is supported on the Gram used in its definition: an
inserted Gram $A^*BA$ annihilates $\ker A$ on both sides.
For $d\in\R$, let $\Pi_d=\{p_0+p_1+p_2=d\}$, with the measure
\begin{equation}\label{eq:heat-plane-measure}
\int_{\Pi_d}H\,\dd\pi_d
=\int_{\R^2}H(p_0,p_1,d-p_0-p_1)\,\dd p_0\,\dd p_1.
\end{equation}
Any other pair of free coordinates gives the same measure, since the
change of coordinates has absolute determinant one. Put $\lambda=11/10$.
We also write
\[
\mathcal S(A)=\sum_{v=0}^2\sum_{i,j=-N}^Nh^2|A_v(i,j)|^3.
\]
Our goal in this section is the fixed-array inequality
\[
-\partial_sJ(d,s)\ge c_*\int_{\Pi_d}\sum_vV_v\,\dd\pi_d,
\qquad J(d,s)=\int_{\Pi_d}\sum_ve_v\,\dd\pi_d,
\]
when the variances are $\sigma_i=\alpha_i s$ and
$\boldsymbol\alpha$ is a permutation of $(1,\lambda,\lambda)$.
The odd-window estimate in Section~\ref{sec:maximal} uses wider Gaussian
weights for uniformly bounded diagonal factors, but compares each score cost
after narrowing only its own vertex. The fixed enlargement
$\lambda=11/10$ retains strict dissipation for all three comparisons.
Lemma~\ref{lem:plane-dissipation} gives an absolute $c_*>0$;
Lemma~\ref{lem:initial-energy} bounds $J$ by $C\mathcal S(A)$ at scales
$s\ge h^2$. Later, when entries change at prescribed scales, the same
differential inequality will apply between successive changes.
We first differentiate the star energies, choosing the Gram orientation
whose support stays fixed. We then integrate over the center plane and
compare neighboring stars. These two steps produce the strict decrease
that controls the score costs.
\subsection{Differentiating the star energies}
\begin{lemma}\label{lem:star-smoothness}
For fixed finite edge arrays, each $e_v$ is smooth in all centers and
positive variances. If only the parameters at vertex $v$ vary, the support
of $T_v$ is fixed. If the parameters at either or both other vertices
vary while those at $v$ are fixed, the support of $S_v$ is fixed.
\end{lemma}
\begin{proof}
The weighted star has the form $R_v=D_{\rm row}R_v^0D_{\rm col}$, where
$R_v^0=[A_v\ A_{v-1}^*]$ is fixed and both diagonal multipliers are strictly
positive; the column multiplier has one block for each neighbor.
Invertible row multiplication preserves the kernel, and invertible column
multiplication preserves the range. This proves the two support
assertions, since
\[
\ker R_v=\ker(R_v^0D_{\rm col}),\qquad
\operatorname{ran}R_v=D_{\rm row}\operatorname{ran}R_v^0
\]
when the respective other multiplier is held fixed.
For joint smoothness, the rank-zero case is immediate. Otherwise factor
$R_v^0=XY^*$ with $X,Y$ of full column rank. Write
$X'=D_{\rm row}X$, $Y'=D_{\rm col}Y$ and
$P'=X'^*X'$, $Q'=Y'^*Y'$. These last two matrices are positive definite
and depend smoothly on the parameters. The nonzero eigenvalues of
$R_vR_v^*=X'Q'X'^*$ are those of $Q'^{1/2}P'Q'^{1/2}$, by comparing
$AA^*$ and $A^*A$ for $A=X'Q'^{1/2}$. Therefore
\[
e_v=\Phi(Q'^{1/2}P'Q'^{1/2}),
\]
which is smooth on the positive definite cone.
\end{proof}
\begin{lemma}\label{lem:heat-identities}
Fix a positive vector $\boldsymbol\alpha$ and set $\sigma_i=\alpha_i s$.
For all centers and all $s>0$,
\begin{align}
2\partial_s e_v
&=\sum_i\alpha_i\partial_{p_i}^2e_v
-\alpha_vV_v-\sum_{w\ne v}\alpha_wY_{v,w},
\label{eq:heat-identity}\\
\partial_{p_{v+1}}\partial_{p_{v-1}}e_v&=Z_v.
\label{eq:heat-mixed}
\end{align}
\end{lemma}
\begin{proof}
For $w=h g_\sigma(u-p)$ and $L=(u-p)/\sigma$,
\[
\partial_pw=Lw,\qquad
\partial_p^2w=(L^2-\sigma^{-1})w,\qquad
\partial_\sigma w=\tfrac12\partial_p^2w.
\]
With the columns fixed, $T_v$ is linear in the row density, so
\[
\partial_{p_v}T_v=U_v,\qquad
\partial_{\sigma_v}T_v=\tfrac12\partial_{p_v}^2T_v.
\]
Its support is fixed by Lemma~\ref{lem:star-smoothness}. The chain rule
and Lemma~\ref{lem:hessian} thus give
\[
\partial_{\sigma_v}e_v
=\tfrac12\partial_{p_v}^2e_v
-\tfrac12\mathcal B_{T_v}(U_v,U_v)
=\tfrac12\partial_{p_v}^2e_v-\tfrac12V_v.
\]
For a column vertex $w\ne v$, use instead $S_v$, which is linear in that
column density and has fixed support. This gives
\[
\partial_{p_w}S_v=U_{v,w},\qquad
\partial_{\sigma_w}e_v=\tfrac12\partial_{p_w}^2e_v-\tfrac12Y_{v,w}.
\]
Summing the three variance derivatives with coefficients $\alpha_i$
proves \eqref{eq:heat-identity}.
The two summands in $S_v=W_vW_v^*+W_{v-1}^*W_{v-1}$ depend on different
column densities. Its mixed derivative in the two column centers is
therefore zero. Only the Hessian term remains in the mixed derivative of
$e_v$, giving \eqref{eq:heat-mixed}.
\end{proof}
\subsection{Comparing neighbors and integrating centers}
\begin{lemma}\label{lem:neighbors}
For every choice of the weighted matrices and every vertex $w$,
\begin{equation}\label{eq:heat-neighbors}
Y_{w+1,w}+Y_{w-1,w}\le\sqrt2\,V_w.
\end{equation}
\end{lemma}
\begin{proof}
Abbreviate $A=W_w$, $B=W_{w-1}^*$, and $L=L_w$. The inequality
$\lVert Az+Bz'\rVert^2\le2\lVert Az\rVert^2+2\lVert Bz'\rVert^2$
implies
\[
T_w\le2Q,\qquad Q=\diag(A^*A,B^*B).
\]
Since $U_w=[A\ B]^*L[A\ B]$ is supported on $T_w$,
Lemma~\ref{lem:order} gives $b_{T_w}(U_w)\ge2^{-1/2}b_Q(U_w)$.
In a block-preserving eigenbasis of $Q$, the formula for $b_Q$ is a sum
of nonnegative squared entries. Discarding the off-diagonal blocks yields
\[
b_Q(U_w)\ge b_{A^*A}(A^*LA)+b_{B^*B}(B^*LB).
\]
Here $A^*A\le S_{w+1}$ and $B^*B\le S_{w-1}$, and the respective
insertions are $U_{w+1,w}$ and $U_{w-1,w}$. Each is supported on its
smaller edge Gram. Apply Lemma~\ref{lem:order} again, now with comparison
factor one, to obtain \eqref{eq:heat-neighbors}.
\end{proof}
\begin{lemma}\label{lem:plane-dissipation}
Fix finite edge arrays and a positive vector $\boldsymbol\alpha$, and
evaluate the energies at $\sigma_i=\alpha_i s$. Set
\[
J_v(d,s)=\int_{\Pi_d}e_v\,\dd\pi_d,
\qquad J(d,s)=\sum_vJ_v(d,s),
\qquad \Sigma_\alpha=\sum_i\alpha_i.
\]
These integrals are finite, continuously differentiable in $s>0$, and
twice continuously differentiable in $d$. Moreover,
\begin{equation}\label{eq:heat-plane-dissipation}
2\partial_sJ(d,s)\le\int_{\Pi_d}\sum_w
\left[-\alpha_w+
\sqrt2\max\{0,\Sigma_\alpha/2-\alpha_w\}\right]V_w\,\dd\pi_d.
\end{equation}
In particular, 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_* = \frac12\min\{1-\tfrac35\sqrt2,\tfrac{11}{10}-\tfrac12\sqrt2\}>0.
\end{equation}
\end{lemma}
\begin{proof}
We first justify differentiation at the fixed mesh. On compact positive
$s$-intervals the nodes are bounded, the scores grow at most linearly in
the centers, and the first and second density ratios grow at most
quadratically. Lemma~\ref{lem:insertion} gives
\[
V_v\le\tfrac34\lVert L_v\rVert_{\op}^2e_v,
\qquad Y_{v,w}\le\tfrac34\lVert L_w\rVert_{\op}^2e_v.
\]
For the second inequality, pad $L_w$ by zero on the other column block,
obtaining $B_w$ with $U_{v,w}=R_vB_wR_v^*$, and apply the lemma to
$R_v^*$. Also $2|Z_v|\le Y_{v,v+1}+Y_{v,v-1}$.
For a Hermitian matrix $B$,
\[
|\tr(\sqrt{T_v}R_v^*BR_v)|\le\lVert B\rVert_{\op}e_v,
\]
since $R_v\sqrt{T_v}R_v^*$ is positive with trace $e_v$. The corresponding
column estimate follows in the same way with $S_v$ and a padded column
insertion. Together with the density-ratio formulas in
Lemma~\ref{lem:heat-identities}, these bounds control every first center
derivative, pure second center derivative, mixed column derivative, and
first $s$-derivative by a polynomial in the centers times $e_v$.
To dominate the energy itself, use
\begin{equation}\label{eq:heat-energy-bound}
e_v\le(\lVert W_v\rVert_{\HS}^2+
\lVert W_{v-1}\rVert_{\HS}^2)^{3/2}
\le\sqrt2(\lVert W_v\rVert_{\HS}^3+
\lVert W_{v-1}\rVert_{\HS}^3).
\end{equation}
Each edge norm cubed is bounded by a Gaussian in its two endpoint
centers, locally uniformly for $s>0$. Indeed, the finite set of nodes is
bounded and $(u-p)^2\ge p^2/2-u^2$. On $\Pi_d$ with $|d|\le D$, any
pair of centers satisfies
\[
p_0^2+p_1^2+p_2^2\le4(p_i^2+p_j^2)+3D^2.
\]
Thus an endpoint-pair Gaussian, multiplied by any of the preceding
polynomials, has an integrable majorant in any fixed plane chart, locally
uniformly in $(d,s)$. This proves all needed differentiation statements
by dominated convergence. The constants in this justification may
depend on the finite mesh and the compact parameter intervals; they will
not enter the uniform estimate.
We can now differentiate the plane integrals. It remains to express
their second derivatives in $d$ using both pure and mixed center
derivatives, then apply Lemma~\ref{lem:neighbors} to the resulting costs.
Choosing $p_i$ as the dependent coordinate in
\eqref{eq:heat-plane-measure} gives
\[
\partial_d^2J_v=\int_{\Pi_d}\partial_{p_i}^2e_v\,\dd\pi_d
\qquad\text{for every }i.
\]
For the column indices $j=v+1$, $k=v-1$, first choose $p_j$ dependent to
write $\partial_dJ_v=\int_{\Pi_d}\partial_{p_j}e_v\,\dd\pi_d$.
Reparametrize this last integral with $p_k$ dependent, then differentiate
again. It follows from \eqref{eq:heat-mixed} that
\begin{equation}\label{eq:heat-plane-mixed}
\partial_d^2J_v
=\int_{\Pi_d}\partial_{p_k}\partial_{p_j}e_v\,\dd\pi_d
=\int_{\Pi_d}Z_v\,\dd\pi_d.
\end{equation}
These are identities of integrated derivatives; no pointwise equality
between pure and mixed derivatives is required.
Integrating \eqref{eq:heat-identity} and using
\eqref{eq:heat-plane-mixed}, we obtain
\[
2\partial_sJ_v=\int_{\Pi_d}
\left[\Sigma_\alpha Z_v-\alpha_vV_v
-\sum_{w\ne v}\alpha_wY_{v,w}\right]\dd\pi_d.
\]
Cauchy--Schwarz gives $2Z_v\le Y_{v,v+1}+Y_{v,v-1}$. Sum over $v$ and
collect the two neighbor costs carrying the score at $w$. Their common
coefficient is $\Sigma_\alpha/2-\alpha_w$. If it is negative, discard
those terms; otherwise apply Lemma~\ref{lem:neighbors}. This proves
\eqref{eq:heat-plane-dissipation}.
For the indicated permutations, $\Sigma_\alpha=16/5$. The two possible
coefficients in \eqref{eq:heat-plane-dissipation} are
$-1+(3/5)\sqrt2$ and $-11/10+(1/2)\sqrt2$, both strictly negative. Dividing by two gives
\eqref{eq:heat-strict}.
\end{proof}
\subsection{The initial energy}
The previous lemma controls the accumulated costs by the energy lost along
the heat flow. We now bound the energy itself at scales no smaller than the
mesh. This estimate will also apply immediately after entries are switched
on or off, since it requires only a bound on their absolute values.
\begin{lemma}[Initial energy]\label{lem:initial-energy}
Let $h>0$, let $N\ge0$ be an integer, and let $A_v(i,j)$ be the finite complex arrays above.
For any $d\in\R$ and variances $\sigma_v\ge h^2$,
\begin{equation}\label{eq:initial-energy}
\int_{\Pi_d}\sum_ve_v\,\dd\pi_d
\le C_0\mathcal S(A),\qquad
C_0=2\sqrt2\left(1+\sqrt{2/\pi}\right).
\end{equation}
The same bound holds if any entries are replaced by smaller entries in
absolute value before forming the weighted matrices.
\end{lemma}
\begin{proof}
For every center $p$ and variance $\sigma>0$,
\begin{equation}\label{eq:heat-mesh-mass}
\sum_{i=-N}^Nh g_\sigma(u_i-p)
\le1+\frac{2h}{\sqrt{2\pi\sigma}}.
\end{equation}
To see this, compare a grid point $u_i\ge p+h$ with the integral over its
preceding interval of length $h$, where the Gaussian is at least its
value at $u_i$. For $u_i\le p-h$, use the succeeding interval. These
intervals are disjoint, and the at most two remaining grid points each
contribute at most $h\sup g_\sigma$.
Since each $\sigma_v\ge h^2$, every density has total mass at most
$M=1+\sqrt{2/\pi}$. Weighted H\"older gives
\begin{align*}
\lVert W_v\rVert_{\HS}^3
&=\left(\sum_{i,j}|A_v(i,j)|^2w_v(i)w_{v+1}(j)\right)^{3/2}\\
&\le M\sum_{i,j}|A_v(i,j)|^3w_v(i)w_{v+1}(j).
\end{align*}
Integrate with $p_v,p_{v+1}$ as free coordinates on $\Pi_d$. Each density
has integral $h$, so
\[
\int_{\Pi_d}\lVert W_v\rVert_{\HS}^3\,\dd\pi_d
\le M\sum_{i,j}h^2|A_v(i,j)|^3.
\]
Each edge occurs in exactly two stars. Equation~\eqref{eq:heat-energy-bound}
therefore proves \eqref{eq:initial-energy}. Replacing entries by smaller
ones in absolute value only reduces the last upper bounds, proving the
final assertion.
\end{proof}
For fixed arrays and a permutation $\boldsymbol\alpha$ of
$(1,\lambda,\lambda)$, Lemmas~\ref{lem:plane-dissipation} and
\ref{lem:initial-energy} now give, whenever $h^2\le s_00}\frac1{2r}\int_{x-r}^{x+r}|f(t)|\,\dd t,
\qquad
M_{\mathbb Z}f(n)=\sup_{r\in\mathbb Z_{\ge0}}
\frac1{2r+1}\sum_{|m-n|\le r}|f(m)|.
\]
Thus discrete averages include singletons. We use Lebesgue measure on
$\R$ and counting measure on $\mathbb Z$.
The estimates below are versions of the classical Hardy--Littlewood
maximal inequalities~\cite[Section~IV]{HardyLittlewood1930}; we include
the centered covering argument and the Gaussian comparison with the
constants needed here.
\Needspace{3\baselineskip}
\begin{lemma}[One-dimensional maximal estimates]\label{lem:line-maximal}
On either of these measure spaces, write $M$ for the corresponding
centered maximal operator. For $f\in L^1$ and $\rho>0$,
\[
\mu\{Mf>\rho\}\le\frac3\rho\|f\|_1.
\]
For every $10$, $\sigma\ge h^2$ and $n\in\mathbb Z$,
\begin{equation}\label{eq:discrete-gaussian-maximal}
\sum_{m\in\mathbb Z}h g_\sigma((m-n)h)|f(m)|
\le C_g M_{\mathbb Z}f(n).
\end{equation}
\end{lemma}
\begin{proof}
For a finite collection of intervals, repeatedly choose an interval of
largest radius and discard all intervals meeting it. The chosen
intervals are disjoint. Every discarded interval has radius no larger
than that of the interval which removed it and is contained in its
triple-radius enlargement. This enlargement has three times the
original length on $\R$. On $\mathbb Z$, its cardinality is
$6r+1\le3(2r+1)$, including $r=0$.
For each point in a finite subset of $\{M_{\mathbb Z}f>\rho\}$, choose
a witnessing interval whose average exceeds $\rho$. The preceding
selection bounds the size of that subset by three times the sum of
the cardinalities of the chosen intervals, and hence by
$3\|f\|_1/\rho$. Taking the supremum over finite subsets proves the
discrete weak estimate. On $\R$, fixed-radius averages of an
$L^1$ function are continuous, so $\{M_{\R}f>\rho\}$ is open.
Apply the same selection to a finite witnessing cover of any compact
subset and then use inner regularity to obtain the continuous estimate.
For $f\in L^q$, split $f$ into its parts on $\{|f|\le\rho/2\}$ and
$\{|f|>\rho/2\}$. The first part has maximal function at most
$\rho/2$, and the second is integrable. The weak estimate gives
\[
\mu\{Mf>\rho\}
\le\frac6\rho\int_{|f|>\rho/2}|f|\,\dd\mu.
\]
The distribution formula and Tonelli's theorem now yield
\[
\begin{aligned}
\|Mf\|_q^q
&\le6q\int |f(x)|
\int_0^{2|f(x)|}\rho^{q-2}\,\dd\rho\,\dd\mu(x)\\
&=\frac{6q\,2^{q-1}}{q-1}\|f\|_q^q.
\end{aligned}
\]
These formulas are valid first with a bounded range of distribution
parameters and then by monotone convergence; finiteness of $Mf$ is
not assumed in advance. Both maximal functions are measurable: the
discrete supremum is countable, and continuous averages are continuous
in their radius, so positive rational radii suffice on $\R$.
Finally put $a=\sqrt\sigma/h\ge1$. In the sum in
\eqref{eq:discrete-gaussian-maximal}, the kernel is
\[
h g_\sigma((m-n)h)
=\frac1{\sqrt{2\pi}a}
\exp\left(-\frac{(m-n)^2}{2a^2}\right).
\]
Split the integers into $|m-n|\le a$ and the annuli
$2^{j-1}a<|m-n|\le2^ja$, $j\ge1$. The sum of $|f|$ over
$|m-n|\le2^ja$ is at most $3\cdot2^ja\,M_{\mathbb Z}f(n)$.
Bounding the Gaussian by its maximum on each region proves the claim
with, for example,
\[
C_g=\frac3{\sqrt{2\pi}}
\left(1+\sum_{j\ge1}2^j e^{-4^{j-1}/2}\right)<\infty.
\]
All constants are independent of the mesh and of the support of $f$.
\end{proof}
\subsection{Changing a matrix entry}
At a truncation endpoint the rank of a star can change. We therefore
need a first-derivative formula that does not require fixed support.
For the sum of cubes of the singular values, this formula is a case
of the general gradient formula for unitarily invariant matrix
functions~\cite[Theorem~3.1]{Lewis1995}. We give a direct regularization
proof and then establish the entrywise bound used in the jump estimate.
\begin{lemma}[A derivative through rank changes]\label{lem:rank-change}
For finite rectangular complex matrices $R_-,R_+$ of the same size,
put $R(z)=(1-z)R_-+zR_+$ and $D=R_+-R_-$. Then
\begin{equation}\label{eq:rank-change-derivative}
\begin{aligned}
&\Phi(R_+^*R_+)-\Phi(R_-^*R_-)\\
&\quad=3\int_0^1\operatorname{Re}
\langle R(z)\sqrt{R(z)^*R(z)},D\rangle_{\HS}\,\dd z.
\end{aligned}
\end{equation}
For every rectangular matrix $R$ and every entry $(i,j)$,
\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}
For $\eta>0$, differentiate
$\tr((R(z)^*R(z)+\eta I)^{3/2})$ using
Lemma~\ref{lem:hessian}. Its derivative is
\[
3\operatorname{Re}
\langle R(z)\sqrt{R(z)^*R(z)+\eta I},D\rangle_{\HS}.
\]
Spectral calculus gives
\[
\|\sqrt{R^*R+\eta I}-\sqrt{R^*R}\|_{\op}\le\sqrt\eta.
\]
Since $R(z)$ stays in a bounded set, the derivatives converge uniformly
in $z$, and the endpoint energies converge. Integrating the
regularized identity before letting $\eta\downarrow0$ proves
\eqref{eq:rank-change-derivative}.
Cauchy--Schwarz in the matrix-product formula bounds the left side of
\eqref{eq:entry-gradient} by the norm of row $i$ of $R$ times the norm
of column $j$ of $\sqrt{R^*R}$. The squared norm of that column is
$(R^*R)_{jj}$, which is also the squared norm of column $j$ of $R$.
This proves \eqref{eq:entry-gradient}, including zero columns and
singular matrices.
\end{proof}
\subsection{The total jump bound}
Return to the finite arrays and Gaussian weights of
Section~\ref{sec:heat}, and write $\dd\pi=\dd\pi_0$ on $\Pi_0$.
Independently for every pair $(i,j)$, choose
finite endpoints
\[
h\le\varepsilon_{ij}<\rho_{ij},
\qquad
A_0^s(i,j)=A_0(i,j)
\mathbf1_{\{\varepsilon_{ij}^20$, $N\ge0$, complex arrays $A_v$ on
$\{-N,\ldots,N\}^2$, and endpoint choices as above, let
$\alpha$ be any permutation of $(1,11/10,11/10)$ and use variances
$\sigma_v=\alpha_vs$. The corresponding masked costs satisfy
\begin{equation}\label{eq:masked-dissipation}
\int_{s_-}^{s_+}\int_{\Pi_0}
\sum_{v=0}^2 V_v(p,s)\,\dd\pi\,\dd s
\le C\mathcal S(A).
\end{equation}
The constant is independent of the arrays, the mesh, the endpoints,
and the number of distinct endpoint values.
\end{proposition}
\begin{proof}
Write $J(s)=\int_{\Pi_0}\sum_ve_v(p,s)\,\dd\pi$ for the masked
energy, and put $\Delta J(\tau)=J(\tau+)-J(\tau-)$ at an endpoint.
We first estimate its jump at a single scale, then sum those
estimates and telescope the heat inequality.
\paragraph{The two stars at a jump.}
At an endpoint scale $\tau$, hold the weights at $s=\tau$ and let
\[
\delta A(i,j)=A_0^{\tau+}(i,j)-A_0^{\tau-}(i,j).
\]
The superscripts denote one-sided masks. Interpolate all changed
entries at once between these two arrays. Each interpolated entry
has absolute value at most $|A_0(i,j)|$, even when some entries open
and others close at the same scale.
For fixed $i,j,p$, define nonnegative numbers using the original,
unmasked arrays:
\begin{equation}\label{eq:jump-slice-quantities}
\begin{aligned}
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),&
K^2&=\sum_m|A_1(j,m)|^2w_2(m).
\end{aligned}
\end{equation}
In $R_0=[W_0\;W_2^*]$, a changing entry has row norm at most
$\sqrt{w_0(i)}\sqrt{X^2+H^2}$, column norm at most
$\sqrt{w_1(j)}Y$, and derivative modulus
$\sqrt{w_0(i)w_1(j)}|\delta A(i,j)|$.
In $R_1=[W_1\;W_0^*]$, its adjoint entry has row norm at most
$\sqrt{w_1(j)}\sqrt{Y^2+K^2}$ and column norm at most
$\sqrt{w_0(i)}X$, with the same derivative modulus.
Star $R_2=[W_2\;W_1^*]$ is unchanged. Lemma~\ref{lem:rank-change}
and $\sqrt{x^2+y^2}\le x+y$ therefore give
\begin{equation}\label{eq:jump-entry-bound}
\begin{aligned}
|\Delta e_0|+|\Delta e_1|
\le3\sum_{i,j}|\delta A(i,j)|w_0(i)w_1(j)
\bigl[(X+H)Y+(Y+K)X\bigr].
\end{aligned}
\end{equation}
This estimate requires no assumption on the ranks during interpolation.
\paragraph{A probability measure on the center plane.}
For each fixed $i,j$, the measure
\[
\dd\nu_{ij}=h^{-2}w_0(i)w_1(j)\,\dd\pi
\]
is a probability measure. Indeed, in the free coordinates $(p_0,p_1)$,
its density is
$g_{\sigma_0}(u_i-p_0)g_{\sigma_1}(u_j-p_1)$.
Thus $p_0,p_1$ are independent Gaussian variables with means
$u_i,u_j$ and variances $\sigma_0,\sigma_1$, while
$p_2=-p_0-p_1$. Adding Gaussian variances gives
\begin{equation}\label{eq:jump-gaussian-expectations}
\begin{aligned}
\mathbb E_{\nu_{ij}}w_0(m)
&=h g_{2\sigma_0}(u_m-u_i),\\
\mathbb E_{\nu_{ij}}w_1(m)
&=h g_{2\sigma_1}(u_m-u_j),\\
\mathbb E_{\nu_{ij}}w_2(m)
&=h g_{\sigma_0+\sigma_1+\sigma_2}(u_m+u_i+u_j).
\end{aligned}
\end{equation}
For the last identity, $p_0+p_1$ has mean $u_i+u_j$ and variance
$\sigma_0+\sigma_1$, and averaging $g_{\sigma_2}(u_m+p_0+p_1)$
adds the remaining variance. Each identity also follows directly
by completing the square in its Gaussian integral.
Extend the arrays by zero to $\mathbb Z^2$ and define
\begin{equation}\label{eq:jump-starred-majorants}
\begin{aligned}
X_*^2(i,j)&=M_{\mathbb Z}(|A_0(i,\cdot)|^2)(j),\\
Y_*^2(i,j)&=M_{\mathbb Z}(|A_0(\cdot,j)|^2)(i),\\
H_*^2(i,j)&=M_{\mathbb Z}(|A_2(\cdot,i)|^2)(-i-j),\\
K_*^2(i,j)&=M_{\mathbb Z}(|A_1(j,\cdot)|^2)(-i-j).
\end{aligned}
\end{equation}
Every variance in \eqref{eq:jump-gaussian-expectations} is at least
$h^2$. Lemma~\ref{lem:line-maximal} consequently bounds
$\mathbb E X^2,\mathbb E Y^2,\mathbb E H^2,\mathbb E K^2$ by
$C_g$ times the corresponding starred squares. Cauchy--Schwarz
under the probability $\nu_{ij}$ gives, in particular,
\[
\mathbb E(XY)\le C_gX_*Y_*,\qquad
\mathbb E(HY)\le C_gH_*Y_*,\qquad
\mathbb E(KX)\le C_gK_*X_*.
\]
No independence of these four random variables is needed.
The starred quantities depend only on the original arrays and the
output indices, not on the scale $\tau$.
\paragraph{Summing the jumps.}
Integrating \eqref{eq:jump-entry-bound} over the centers and using
$w_0(i)w_1(j)\,\dd\pi=h^2\,\dd\nu_{ij}$ yields
\[
|\Delta J(\tau)|\le3C_g\sum_{i,j}h^2|\delta A(i,j)|
(2X_*Y_*+H_*Y_*+K_*X_*).
\]
Every entry opens once and closes once. Therefore its total absolute
change, summed over any subset of endpoint scales, is at most
$2|A_0(i,j)|$. We obtain
\begin{equation}\label{eq:total-jump-majorant}
\sum_\tau|\Delta J(\tau)|
\le6C_g\sum_{i,j}h^2|A_0(i,j)|
(2X_*Y_*+H_*Y_*+K_*X_*).
\end{equation}
The sum may include all endpoints or just the interior ones.
To bound this expression, set
\[
n_v=\left(\sum_{i,j}h^2|A_v(i,j)|^3\right)^{1/3},
\qquad \|f\|_{3,h}^3=\sum_{i,j\in\mathbb Z}h^2|f(i,j)|^3.
\]
The $q=3/2$ case of \eqref{eq:line-maximal-strong}, applied on slices,
gives
\begin{equation}\label{eq:jump-slice-norms}
\|X_*\|_{3,h},\ \|Y_*\|_{3,h}\le Cn_0,
\qquad \|H_*\|_{3,h}\le Cn_2,
\qquad \|K_*\|_{3,h}\le Cn_1.
\end{equation}
For $H_*$, fix $i$ and change the summation index from $j$ to $-i-j$;
this is a bijection of $\mathbb Z$. For $K_*$, fix $j$ and change
$i$ to $-i-j$. Thus both estimates are ordinary slice estimates,
with no multiplicity or mesh factor. Restricting the sums back to
the original finite square can only decrease them.
H\"older's inequality in \eqref{eq:total-jump-majorant}, followed by
\eqref{eq:jump-slice-norms}, proves
\begin{equation}\label{eq:total-jump-bound}
\sum_\tau|\Delta J(\tau)|
\le C(2n_0^3+n_0^2n_2+n_0^2n_1)
\le C\mathcal S(A).
\end{equation}
The last step uses $a^2b\le(2a^3+b^3)/3$.
In particular, the estimate has no dependence on how many distinct
scales occur.
\paragraph{Telescoping the heat estimate.}
List the distinct squared endpoints as
$t_0=s_-0$ are the eigenvalues on the support, then in an orthonormal
eigenbasis
\begin{equation}\label{eq:hessian-metric}
b_P(K)=\frac32\sum_{i,j}
\frac{|K_{ij}|^2}{\sqrt{p_i}+\sqrt{p_j}}.
\end{equation}
In particular, $\mathcal B_P$ is a positive definite symmetric bilinear
form on the real space of supported Hermitian directions, and
\[
|\mathcal B_P(K,L)|\le b_P(K)^{1/2}b_P(L)^{1/2}.
\]
\end{lemma}
\begin{proof}
Restrict to the support and put $A=\sqrt P$.
The derivative of the squaring map at $A$ is $X\mapsto AX+XA$.
In an eigenbasis of $A$, this map multiplies its $(i,j)$ entry by
$\sqrt{p_i}+\sqrt{p_j}>0$. The inverse function theorem therefore
gives a smooth positive square root. Its derivative $X_K$ in
direction $K$ satisfies
\[
AX_K+X_KA=K.
\]
Taking a trace after multiplication by $A$ yields
$\tr(PX_K)=\tfrac12\tr(AK)$. Differentiation of
$\Phi(P)=\tr(P\sqrt P)$ now gives
\[
d\Phi_P(K)=\tr(KA)+\tr(PX_K)=\frac32\tr(AK).
\]
Differentiating this identity in direction $L$ proves the Hessian
formula. The entrywise formula follows by diagonalizing $A$.
It also proves symmetry, positivity, and Cauchy--Schwarz on the
real space of Hermitian directions.
\end{proof}
The energy and its Hessian are unchanged by simultaneous unitary
conjugation of their arguments. They are also unchanged if all
arguments are extended by zero to a larger Hilbert space: the
positive eigenvalues and the supported entries in
\eqref{eq:hessian-metric} stay the same. We will use these
observations to compare Grams acting on different subspaces.
\Needspace{3\baselineskip}
\begin{lemma}[Order comparison]\label{lem:order}
Let $P,Q$ be positive semidefinite matrices on the same Hilbert space,
let $a\ge1$, and suppose $P\le aQ$. 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}
If $x\in\ker Q$, then $0\le\langle Px,x\rangle
\le a\langle Qx,x\rangle=0$, so $Px=0$.
Thus $\ker Q\subseteq\ker P$, proving the support assertion.
We recall the two operator-order facts needed for the comparison.
Inversion reverses order on positive definite matrices: for
$00$, the assumptions give
\[
P+\eta I\le a(Q+\eta I),
\qquad
\sqrt{P+\eta I}\le\sqrt a\,\sqrt{Q+\eta I}.
\]
If $A\le B$, then
$\mathsf L_A+\mathsf R_A\le\mathsf L_B+\mathsf R_B$ as
selfadjoint operators on Hilbert--Schmidt space. Indeed, left and
right multiplication by the positive matrix $B-A$ have the
nonnegative quadratic forms
$\tr(X^*(B-A)X)$ and $\tr(X^*X(B-A))$.
Inverting the resulting positive definite operators gives
\[
b_{P+\eta I}(K)\ge a^{-1/2}b_{Q+\eta I}(K).
\]
Finally let $\eta\downarrow0$.
Formula~\eqref{eq:hessian-metric}, applied in a full eigenbasis,
shows that both limits are the asserted supported costs: every
entry of $K$ meeting the kernel of the respective matrix is zero.
\end{proof}
We will apply the Hessian to Gram matrices with an insertion between
the two factors. The following bound controls such derivatives by
the energy itself.
\begin{lemma}[Inserted Gram bound]\label{lem:insertion}
Let $R:E\to F$ be a linear map between finite-dimensional complex
Hilbert spaces, and let $L:F\to F$ be Hermitian. The matrix $R^*LR$
is supported on $R^*R$, and
\begin{equation}\label{eq:insertion-bound}
b_{R^*R}(R^*LR)
\le\frac34\lVert L\rVert_{\op}^2\Phi(R^*R).
\end{equation}
\end{lemma}
\begin{proof}
The insertion annihilates $\ker R=\ker(R^*R)$ and is Hermitian,
so it is supported on the Gram. The assertion is immediate if
$R=0$. Otherwise take a singular-value decomposition, with positive
singular values $\rho_i$, and let $C$ be the compression of $L$ to
the corresponding left singular vectors. Then
$\lVert C\rVert_{\op}\le\lVert L\rVert_{\op}$ and
\[
b_{R^*R}(R^*LR)
=\frac32\sum_{i,j}
\frac{\rho_i^2\rho_j^2}{\rho_i+\rho_j}|C_{ij}|^2.
\]
For $x,y>0$, the arithmetic--geometric mean inequality gives
\[
\frac{x^2y^2}{x+y}
\le\frac{(xy)^{3/2}}2
\le\frac{x^3+y^3}4.
\]
Each row and each column of $C$ has squared Euclidean norm at most
$\lVert L\rVert_{\op}^2$. Consequently
\[
b_{R^*R}(R^*LR)
\le\frac38\sum_{i,j}(\rho_i^3+\rho_j^3)|C_{ij}|^2
\le\frac34\lVert L\rVert_{\op}^2\sum_i\rho_i^3,
\]
which is \eqref{eq:insertion-bound}.
\end{proof}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/matrix.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/maximal.tex ====
\section{From masked energies to the maximal transform}\label{sec:maximal}
The estimate of Section~\ref{sec:jumps} allows different truncation
endpoints at every output position. We first use this freedom on a finite
mesh, choosing the endpoints that maximize each output. We then pass to
functions and compare the resulting smooth kernels with hard truncations.
The comparison will be pointwise and uniform in both endpoints.
\subsection{Odd windows and selected endpoints}
For $s>0$ and $0<\varepsilon0$, let $N\ge0$ be an integer, set $u_i=ih$, and let
$A_v(i,j)$, $v=0,1,2$, be complex
arrays indexed by $-N\le i,j\le N$. Assign an arbitrary endpoint pair
$h\le\varepsilon_{ij}<\rho_{ij}<\infty$ to each $(i,j)$. Then
\begin{equation}\label{eq:selected-kernel}
\left|h^3\sum_{i,j,l}A_0(i,j)A_1(j,l)A_2(l,i)
I_{\varepsilon_{ij},\rho_{ij}}(u_i+u_j+u_l)\right|
\le C\prod_{v=0}^2\left(\sum_{i,j}h^2|A_v(i,j)|^3\right)^{1/3}.
\end{equation}
The constant is independent of the arrays, mesh and selected endpoints.
\end{proposition}
\begin{proof}
We apply the mixed trace inequality with Gaussian weights of larger
variance than the windows $k_s$, so that the diagonal insertions are
bounded. We then compare each quadratic cost with a variance choice
covered by Proposition~\ref{prop:masked-dissipation}.
Use the masked array
$A_0^s(i,j)=A_0(i,j)\mathbf1_{\{\varepsilon_{ij}^20$ so that both supports lie in the interior of $[-B,B]^2$.
Take $h=B/N$, $-N\le i,j,l\le N$, and sample
\[
A_1(j,l)=G_1(jh,lh),\qquad A_2(l,i)=G_2(lh,ih).
\]
The corresponding continuum outputs are
\[
\mathcal U_k(a,b)=\int_\R G_1(b,c)G_2(c,a)
I_{\varepsilon_k,R_k}(a+b+c)\,\dd c.
\]
They vanish outside $[-B,B]^2$: $G_2(c,a)$ restricts $a$ and $G_1(b,c)$
restricts $b$. A nonzero integrand also has $c\in[-B,B]$.
For our fixed finite endpoint family the integrands and their $c$
derivatives are uniformly bounded on this compact cube. Riemann sums
in $c$ consequently converge uniformly for $(a,b)\in[-B,B]^2$ and
$1\le k\le m$. In particular,
\[
\max_{i,j}\max_{1\le k\le m}
|U_k(i,j)-\mathcal U_k(ih,jh)|\longrightarrow0.
\]
The finite maximum of magnitudes preserves this uniform convergence.
Since $\max_k|\mathcal U_k|$ is continuous and compactly supported,
the two-dimensional Riemann sums on the left of
\eqref{eq:discrete-finite-maximum} converge to its $L^{3/2}$ norm.
The cube roots of the input sums converge to $\|G_1\|_3$ and
$\|G_2\|_3$.
The condition on $h$ holds for all sufficiently large $N$.
We obtain
\[
\|\max_k|\mathcal U_k|\|_{3/2}\le C\|G_1\|_3\|G_2\|_3.
\]
No convergence or regularity of the maximizing indices is needed: we
passed to the limit in the finite maximum itself.
With $x=-a$, $y=-b$, and $t=a+b+c$, the last integral is
$B^I_{\varepsilon_k,R_k}(F,G)(x,y)$. The output coordinate change
preserves Lebesgue measure, giving the asserted estimate for every
finite endpoint family.
Finally enumerate all rational pairs $0<\varepsilon0$, such that,
for $t\ne0$, apart from endpoint values,
\begin{equation}\label{eq:maximal-hard-decomposition}
I_{\varepsilon,R}(t)
=\frac{c_0}{t}\mathbf1_{\{\varepsilon<|t|0$, followed by oddness for
$t<0$, yields
\[
I_{\varepsilon,R}(t)
=\frac{P(|t|/\varepsilon)-P(|t|/R)}{t}.
\]
Define $Q(v)=P(v)-c_0\mathbf1_{\{v>1\}}$ for $v\ge0$.
Smooth oddness of $\psi$ gives $P(v)=O(v^2)$ at zero. For $v>1$,
\[
Q(v)=-2g_3''(v)
=-2\left(\frac{v^2}{9}-\frac13\right)g_3(v).
\]
Thus $Q$ has Gaussian decay at infinity and bounded one-sided values
at its sole jump, $v=1$. Enlarging an absolute constant on a compact
interval, we have
\[
\frac{|Q(v)|}{v}\le C(1+v)^{-2}\qquad(v>0).
\]
Set $E_r(t)=Q(|t|/r)/t$ for $t\ne0$ and set $E_r(0)=0$.
Substituting the definition of $Q$ proves
\eqref{eq:maximal-hard-decomposition} and
\eqref{eq:maximal-error-envelope}. Values at the four hard endpoints
have no effect on any integral.
\end{proof}
Let $M_x$ and $M_y$ denote the continuous centered interval maximal
operators in the first and second variables, respectively. The next
estimate follows from their one-dimensional bounds in
Lemma~\ref{lem:line-maximal}.
\begin{lemma}\label{lem:pointwise-endpoint-error}
For complex $F,G\in C_c^\infty(\R^2)$ and every $r>0$,
\[
\left|\int_\R F(x+t,y)G(x,y+t)E_r(t)\,\dd t\right|
\le C\bigl[M_x(|F|^2)(x,y)M_y(|G|^2)(x,y)\bigr]^{1/2}.
\]
The right side has $L^{3/2}(\R^2)$ norm at most $C\|F\|_3\|G\|_3$.
\end{lemma}
\begin{proof}
Put $q_r(t)=r^{-1}(1+|t|/r)^{-2}$. For nonnegative locally integrable
$f$ on $\R$, the part of $\int q_r(t)f(x+t)\,\dd t$ with $|t|\le r$
is at most $2M_{\R}f(x)$. On the shell
$2^kr<|t|\le2^{k+1}r$, $k\ge0$, its contribution is at most
\[
r^{-1}2^{-2k}\int_{|t|\le2^{k+1}r}f(x+t)\,\dd t
\le 4\,2^{-k}M_{\R}f(x).
\]
Summing the geometric series proves a bound independent of $r$.
By \eqref{eq:maximal-error-envelope} and weighted Cauchy--Schwarz,
\begin{align*}
\left|\int F(x+t,y)G(x,y+t)E_r(t)\,\dd t\right|
&\le C\left(\int |F(x+t,y)|^2q_r(t)\,\dd t\right)^{1/2}\\
&\hspace{12mm}\cdot
\left(\int |G(x,y+t)|^2q_r(t)\,\dd t\right)^{1/2}.
\end{align*}
Apply the preceding one-dimensional bound on the respective coordinate
lines. This gives the pointwise assertion.
H\"older's inequality and Lemma~\ref{lem:line-maximal}, applied on slices
and then integrated in the other coordinate, give
\begin{align*}
\|[M_x(|F|^2)M_y(|G|^2)]^{1/2}\|_{3/2}
&\le\|M_x(|F|^2)\|_{3/2}^{1/2}
\|M_y(|G|^2)\|_{3/2}^{1/2}\\
&\le C\||F|^2\|_{3/2}^{1/2}\||G|^2\|_{3/2}^{1/2}
=C\|F\|_3\|G\|_3.
\end{align*}
\end{proof}
\begin{proposition}[Compact smooth inputs]\label{prop:compact-maximal}
For all complex $F,G\in C_c^\infty(\R^2)$,
\[
\|B_*(F,G)\|_{L^{3/2}(\R^2)}
\le C\|F\|_{L^3(\R^2)}\|G\|_{L^3(\R^2)},
\]
with an absolute constant. The supremum defining $B_*$ is over both
finite hard truncation endpoints, inside the spatial norm.
\end{proposition}
\begin{proof}
Lemma~\ref{lem:maximal-hard-kernel} and
Lemma~\ref{lem:pointwise-endpoint-error} give, at each spatial point,
\[
|c_0|\sup_{0<\varepsilon0$.
\end{proof}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/maximal.tex ====
==== BEGIN The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/trace.tex ====
\section{A mixed trace inequality}\label{sec:trace}
The triangular form is naturally represented by a cyclic product of
operators. Such a representation also appears in the dyadic model of
Kova\v{c}, Thiele and Zorin-Kranich~\cite[Equation~(1.5)]{KovacThieleZorinKranich2015}.
We prove a finite-dimensional estimate that controls this product by
three of the quadratic costs from Section~\ref{sec:matrix}.
Let $H_v$, $v\in\mathbb Z/3\mathbb Z$, be finite-dimensional complex
Hilbert spaces, and let $W_v:H_{v+1}\to H_v$ be arbitrary linear maps.
Combine the two edges incident to $v$ into the \emph{row star}
\begin{equation}\label{eq:row-star}
R_v=[\,W_v\quad W_{v-1}^*\,]:
H_{v+1}\oplus H_{v-1}\longrightarrow H_v.
\end{equation}
Its two Gram matrices are
\begin{equation}\label{eq:star-grams}
T_v=R_v^*R_v,\qquad S_v=R_vR_v^*.
\end{equation}
Their positive eigenvalues are the squares of the nonzero singular
values of $R_v$, with the same multiplicities. In particular,
$\Phi(T_v)=\Phi(S_v)$.
Figure~\ref{fig:triangle} shows the cyclic edges and the row star at
vertex $0$.
\input{figures/triangle}
\begin{lemma}[Mixed trace estimate]\label{lem:mixed-trace}
Let $W_v:H_{v+1}\to H_v$ be linear maps between finite-dimensional
complex Hilbert spaces, with row stars $R_v$ and Gram matrices $T_v$
as in \eqref{eq:row-star}--\eqref{eq:star-grams}.
For Hermitian maps $D_v:H_v\to H_v$, put
$D_* =\max_v\lVert D_v\rVert_{\op}$. Then
\begin{equation}\label{eq:mixed-trace}
\left|\tr(D_0W_0D_1W_1D_2W_2)\right|
\le \frac53D_*\sum_{v=0}^2
b_{T_v}(R_v^*D_vR_v).
\end{equation}
The constant is independent of the dimensions of the spaces.
\end{lemma}
\begin{proof}
We first put the three quadratic costs in a common metric. We then
estimate a cubic trace in an eigenbasis and recover the absolute value
of the original cyclic product by changing the phase of one edge.
\paragraph{A common quadratic form.}
On $H=H_0\oplus H_1\oplus H_2$, define the Hermitian matrices
\[
M=\begin{pmatrix}
0&W_0&W_2^*\\
W_0^*&0&W_1\\
W_2&W_1^*&0
\end{pmatrix},
\qquad D=\diag(D_0,D_1,D_2).
\]
Thus $\lVert D\rVert_{\op}=D_*$. Let $P_v$ be the orthogonal
projection onto $H_v$ and set
\[
A_v=MP_vM,\qquad K_v=MP_vDM.
\]
Only the $v$-th block row of $P_vM$ can be nonzero; it consists of
$R_v$ with its columns placed in the corresponding vertex spaces.
Consequently $A_v$ and $K_v$ are,
respectively, $T_v$ and $R_v^*D_vR_v$, after a permutation of blocks
and extension by zero. These operations preserve the quadratic cost.
Since $P_v$ commutes with $D$, the matrix $K_v$ is Hermitian. Moreover,
\[
\ker A_v=\ker(P_vM)\subseteq\ker K_v,
\qquad 0\le A_v\le M^2.
\]
The first identity follows from
$\langle A_vx,x\rangle=\lVert P_vMx\rVert^2$.
Thus $K_v$ is supported on $A_v$, and hence also on $M^2$.
Lemma~\ref{lem:order}, followed by the quadratic-form inequality for
three vectors, yields
\begin{equation}\label{eq:trace-common-metric}
\begin{aligned}
\mathcal E
&:=\sum_v b_{T_v}(R_v^*D_vR_v)\\
&\ge\sum_v b_{M^2}(K_v)
\ge\frac13 b_{M^2}\left(\sum_vK_v\right)
=\frac13 b_{M^2}(MDM).
\end{aligned}
\end{equation}
All these quadratic forms are taken on their supports; no
invertibility of $M$ is required.
\paragraph{The cubic trace.}
Choose an orthonormal eigenbasis $(e_i)$ of $M$, write the real
eigenvalues as $m_i$, and arrange the indices so that $q_i:=|m_i|$ is
nonincreasing. Equal values may be ordered arbitrarily. Write $D_{ij}$
for the entries of $D$ in this basis. The spectral formula for
$b_{M^2}$ gives
\begin{equation}\label{eq:trace-Q}
\mathcal E\ge\frac12Q,
\qquad
Q:=\sum_{i,j:q_i+q_j>0}
\frac{q_i^2q_j^2}{q_i+q_j}|D_{ij}|^2.
\end{equation}
If either eigenvalue is zero, its summand is zero; pairs of zero
eigenvalues are omitted. Define
\[
A_i=q_i^3|D_{ii}|^2,\qquad
E_i=q_i\sum_{j>i}q_j^2|D_{ij}|^2,
\qquad A=\sum_iA_i,\quad E=\sum_iE_i.
\]
Hermitian symmetry and the ordering of the $q_i$ imply
\begin{equation}\label{eq:trace-AE}
\begin{aligned}
Q
&=\frac A2+
2\sum_{\substack{i0}}
\frac{q_i^2q_j^2}{q_i+q_j}|D_{ij}|^2
\ge\frac A2+E.
\end{aligned}
\end{equation}
Indeed, for $ii}m_jm_kD_{ij}D_{jk}D_{ki}.
\]
Here $j=k$ is allowed. For the vector $x=(m_jD_{ji})_{j>i}$, the inner
sum is $x^*D_{>i}x$, where $D_{>i}$ is the compression of $D$ to the
indices greater than $i$. Since
$\lVert D_{>i}\rVert_{\op}\le D_*$, the modulus of this contribution
is at most $3D_*E_i$. This argument retains the signs of the $m_j$
and uses $D_{ji}=\overline{D_{ij}}$.
If the smallest index $i$ occurs twice, the contribution is
\[
3m_i^2D_{ii}\sum_{k>i}m_k|D_{ik}|^2.
\]
Cauchy--Schwarz and
$\sum_k|D_{ik}|^2=\lVert De_i\rVert^2\le D_*^2$ give
\[
\left|\sum_{k>i}m_k|D_{ik}|^2\right|
\le D_*\left(\sum_{k>i}q_k^2|D_{ik}|^2\right)^{1/2}.
\]
Its contribution therefore has modulus at most
$3D_*\sqrt{A_iE_i}$. This also holds when $q_i=0$, without any
division by $q_i$.
Finally, when all three indices equal $i$, the modulus is at most
$D_*A_i$. These three cases exhaust all triples, including repeated
indices. Summing them and applying Cauchy--Schwarz once more, we obtain
\begin{equation}\label{eq:trace-cubic}
\begin{aligned}
|\tr((DM)^3)|
&\le D_*\bigl(A+3E+3\sqrt{AE}\bigr)\\
&\le D_*\left(\frac52A+\frac92E\right)
\le 5D_*\left(\frac A2+E\right)
\le5D_*Q.
\end{aligned}
\end{equation}
\paragraph{Recovering the complex cyclic product.}
We have bounded the cubic trace by the star costs. It remains to
extract the modulus of the desired triangular trace. Put
\[
\tau=\tr(D_0W_0D_1W_1D_2W_2).
\]
Because $M$ has zero diagonal blocks, a nonzero closed three-step
block product must visit all three vertices. The three starting
points in one orientation each give $\tau$, by cyclicity of the
trace. Reversing the orientation gives $\overline\tau$, since the
$D_v$ are Hermitian. Hence
\[
\tr((DM)^3)=6\operatorname{Re}\tau.
\]
If $\tau\ne0$, choose $z\in\mathbb C$ with $|z|=1$ and
$z\tau=|\tau|$, and replace $W_0$ by $zW_0$. The row stars change by
\[
\begin{aligned}
R_0&\longmapsto
R_0\diag(zI_{H_1},I_{H_2}),\\
R_1&\longmapsto
R_1\diag(I_{H_2},\overline zI_{H_0}),
\end{aligned}
\]
while $R_2$ is unchanged. Each displayed right multiplier is unitary
and conjugates the corresponding Gram and inserted Gram by the same
unitary. Thus $\mathcal E$ is unchanged, as is $D_*$.
Applying \eqref{eq:trace-Q} and \eqref{eq:trace-cubic} to the modified
matrices, with $Q$ computed for the new $M$, gives
\[
6|\tau|\le5D_*Q\le10D_*\mathcal E,
\]
which is \eqref{eq:mixed-trace}. The case $\tau=0$ is immediate.
\end{proof}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//sections/trace.tex ====
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author = {Tao, Terence},
title = {Cancellation for the multilinear {Hilbert} transform},
journal = {Collectanea Mathematica},
volume = {67},
number = {2},
pages = {191--206},
year = {2016},
eprint = {1505.06479},
archivePrefix = {arXiv},
primaryClass = {math.CA},
url = {https://arxiv.org/abs/1505.06479v3},
note = {Version consulted: arXiv:1505.06479v3}
}
@article{Daletskii1957,
author = {Daletskii, Yu. L.},
title = {Integration and differentiation of functions of {Hermitian} operators depending on a parameter},
journal = {Uspekhi Matematicheskikh Nauk},
volume = {12}, number = {1(73)}, pages = {182--186}, year = {1957},
url = {https://www.mathnet.ru/eng/rm7543},
note = {In Russian; Section 1, formula (1)}
}
@article{Kwong1975,
author = {Kwong, Man Kam},
title = {Inequalities for the powers of nonnegative {Hermitian} operators},
journal = {Proceedings of the American Mathematical Society},
volume = {51},
number = {2},
pages = {401--406},
year = {1975},
doi = {10.1090/S0002-9939-1975-0374970-X},
url = {https://doi.org/10.1090/S0002-9939-1975-0374970-X}
}
==== END The-maximal-triangular-Hilbert-transform-at-the-symmetric-point//references.bib ====