==== TABLE OF CONTENTS ==== Endpoint-convergence-for-the-planar-Schrodinger-equation//01-introduction.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//02-preliminaries.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//03-transverse.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//04-sparse.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//05-frame.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//06-flat.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//07-geometry.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//08-dyadic.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//09-endpoint.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//bibliography.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//dependencies.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//endpoint-tree.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//geometry.tex Endpoint-convergence-for-the-planar-Schrodinger-equation//main.tex =========================== ==== BEGIN Endpoint-convergence-for-the-planar-Schrodinger-equation//01-introduction.tex ==== \section{Introduction}\label{sec:introduction} For $f\in L^2(\R^2)$ we use the Fourier transform \[ \widehat f(\xi)=\int_{\R^2}e^{-ix\cdot\xi}f(x)\,dx, \qquad \|f\|_{H^{1/3}}^2=(2\pi)^{-2}\int_{\R^2} (1+|\xi|^2)^{1/3}|\widehat f(\xi)|^2\,d\xi. \] The first formula is initially defined on integrable functions and then extended by continuity to $L^2$. For $a>0$ put \begin{equation}\label{eq:intro-abel} U_af(x,t)=(2\pi)^{-2}\int_{\R^2} e^{ix\cdot\xi-it|\xi|^2-a|\xi|^2}\widehat f(\xi)\,d\xi. \end{equation} This integral is absolutely convergent by Cauchy--Schwarz. At a Lebesgue point of $f$, write \[ f^*(x)=\lim_{r\downarrow0}\frac1{\pi r^2} \int_{|y-x|1/2$ in higher dimensions. In the plane, Bourgain and Moyua--Vargas--Vega \cite{Bourgain95,MVV96} developed the connection with Fourier restriction. The bilinear approach of Tao and Vargas \cite{TV00} gave $s>15/32$; Tao's sharp bilinear restriction theorem \cite{Tao03} improved this to $s>2/5$, and Lee \cite{Lee06} obtained $s>3/8$. In dimensions $n\ge3$, Bourgain \cite{Bourgain13} obtained the sufficient range $s>1/2-1/(4n)$. Bourgain's later counterexample \cite[Proposition~1]{Bourgain16} forces $s\ge n/[2(n+1)]$ when $n\ge2$, but does not rule out equality. Du, Guth and Li \cite[Theorems~1.1--1.2]{DGL17} proved planar convergence and a local $L^3$ maximal estimate for $s>1/3$. Their proof combined polynomial partitioning with wave packets and refined Strichartz estimates, using $\ell^2$ decoupling to control packets tangent to the partitioning surface. Du, Guth, Li and Zhang \cite[Theorems~1.1 and~4.2]{DGLZ18} then obtained higher-dimensional convergence for $s>(n+1)/[2(n+2)]$ and developed multilinear refined Strichartz estimates. Du and Zhang \cite[Theorem~1.1]{DZ19} established convergence for $s>n/[2(n+1)]$ when $n\ge3$. A recent Bessel-potential extension still states its positive $p=2$ range with the strict inequality $s>n/[2(n+1)]$ \cite[Theorems~1.3--1.4]{PanSunTan26}. The choice of pointwise representative also has a history: Sj\"ogren and Sj\"olin \cite[Theorem~2]{SjogrenSjolin89} constructed, for $s>1/2$, representatives continuous on the entire time line for almost every spatial point, using spatial or space-time averages. Theorem~\ref{thm:main} resolves positively the planar Sobolev endpoint of Carleson's convergence problem: it supplies the equality case $s=1/3$, with the order of limits and the all-real-times representative specified above. Its maximal estimate has local $L^1$ output. The stronger question of an endpoint local $L^3$ maximal estimate is not asserted here. \subsection{Three ingredients of the proof} The main analytic conclusion is a local maximal estimate at the exact Sobolev exponent. Write $S(t)f=e^{it\Delta}f$ for the $L^2$ evolution with Fourier multiplier $e^{-it|\xi|^2}$. For bounded-frequency data, the pointwise maximal estimate takes the form \begin{equation}\label{eq:intro-maximal} \int_{B(x_0,1)}\sup_{0\le t<1}|S(t)f(x)|\,dx \le C\|f\|_{H^{1/3}}, \end{equation} with a constant independent of the center $x_0$ and the frequency bound. Theorem~\ref{thm:maximal} proves this estimate. The final section then constructs the representative in Theorem~\ref{thm:main}; it does not infer an all-times assertion by intersecting time-dependent exceptional sets. We work first at unit frequency, with packets of spatial width $L$ and propagation time $L^2$. A measure satisfying a two-dimensional ball bound models the spatial area on a measurable time graph. The proof has three main components. \paragraph{A transverse power saving.} Multilinear restriction \cite{BCT06} and the refined fractal restriction estimate of Du and Zhang \cite[Theorem~1.6]{DZ19} reduce a hypothetical saturated transverse configuration to weighted incidences between points and tubes. Conditional information across scales has additive profiles whose extremal relations come from three transverse directions. A robust projection theorem \cite[Theorem~1]{He20} restricts the derivatives of these profiles to zero and one. A further argument rules out arbitrarily many alternations; polynomial partitioning \cite[Theorem~4.1]{GK15} then excludes the remaining jump. This gives a fixed power saving, which is propagated to the sparse packet estimate in Proposition~\ref{prop:sparse}. \paragraph{A bilinear estimate for smooth packet arrays.} Theorem~\ref{thm:frame} treats packets with general smooth profiles and separated frequency groups. Positive masks remove large pieces of a tight frequency frame, charging them to the lost energy. A quartic potential, together with the curvature identity for the parabola, controls changes in the spatial smoothing scale. The resulting estimate is uniform in the number of scales. This uniformity is needed for the long, nearly flat frequency configurations that arise in the proof of Theorem~\ref{thm:distribution}. \paragraph{Square summation at the endpoint.} Theorem~\ref{thm:distribution} yields a short-time weak estimate with some fixed output exponent $p>2$, with no growing frequency loss. To combine frequencies, we linearize all of them on the same time graph. A binary time tree separates heavier children from their lighter siblings. Spatial localization supplies decay in the relative depth of this tree, while $p>2$ supplies a positive convexity deficit that sums powers of the lighter-child masses across frequency scales. Together these estimates retain the Sobolev square norm and prove \eqref{eq:intro-maximal}. The packet and measure formulations allow arbitrary bounded phase-space counts and sufficiently smooth compact profiles. In particular, the bilinear estimate applies to the composite packets created by localization; no exact evolution equation is imposed on those profiles. \subsection{Organization and dependencies} Section~\ref{sec:preliminaries} fixes the packet conventions and the restriction estimates used with small losses. The transverse and sparse arguments follow in Sections~\ref{sec:transverse} and~\ref{sec:sparse}. The independent frame estimate is proved before its applications, in Section~\ref{sec:frame}. Sections~\ref{sec:flat} and~\ref{sec:geometry} provide the flat-frequency and geometric reductions, and Section~\ref{sec:dyadic} proves the uniform dyadic distribution estimate. Section~\ref{sec:endpoint} performs the endpoint summation and proves Theorem~\ref{thm:main}. Figure~\ref{fig:dependencies} records the logical order. The distinction between a fixed power saving and an estimate uniform in the packet scale is essential throughout. \input{figures/dependencies} ==== END Endpoint-convergence-for-the-planar-Schrodinger-equation//01-introduction.tex ==== ==== BEGIN Endpoint-convergence-for-the-planar-Schrodinger-equation//02-preliminaries.tex ==== \section{Packets, localization, and fractal estimates} \label{sec:preliminaries} We first fix the packet conventions and the analytic estimates used in the transverse argument. There are two packet classes: compactly supported smooth atoms, which are convenient for the later geometric arguments, and exact Schr\"odinger packets, which permit the use of restriction estimates. The latter have rapidly decreasing spatial tails. Keeping these two classes distinct will also make the localization errors explicit. \subsection{Conventions and the compact atom class} For frequency data $v\in L^2(\R^2)$ supported in a fixed bounded set, put \begin{equation} Ev(x,t)=\int_{\R^2}e^{i(x\cdot\xi-t|\xi|^2)}v(\xi)\,d\xi. \label{eq:prelim-extension} \end{equation} Thus extension input norms are frequency-space $L^2$ norms; the Fourier normalization in Section~\ref{sec:introduction} differs only by the fixed factor $(2\pi)^{-2}$. In particular, with that Fourier convention, $\|f\|_2^2=(2\pi)^{-2}\|\widehat f\|_2^2$. All frequency sets below lie in fixed bounded boxes unless another scale is specified. The direction associated with $\xi$ is the line parallel to $(2\xi,1)$, or its unit representative with positive last coordinate. On a fixed bounded frequency set, distances between frequencies and between these unit representatives are comparable. Unless stated otherwise, all measures are nonnegative Borel measures on space-time $\R^3$. A ball bound is required for balls with arbitrary centers. Its normalizing parameter may be assumed positive, since the zero case is immediate. Lattice partitions are half-open; a supremum on a cell may be taken on its closure. Fixed enlargements, boundedly overlapping grids, and fixed changes of frequency boxes cost constants independent of the large scales and of the number of packets. All distribution thresholds are positive. An estimate with a free factor $S^\epsilon$ is asserted for each fixed $\epsilon>0$, with a constant that may depend on $\epsilon$. We can therefore use a smaller exponent to absorb a fixed number of logarithms. On a countersequence, $o(1)$ denotes a quantity tending to zero; simultaneous statements at countably many fixed parameters are obtained by subsequences and diagonal choices. Parameters reused in different arguments have only their local meaning. \begin{definition}[Compact atoms] \label{def:prelim-atoms} Let $L\geq1$, and set $R=L^2$. An atom with label $\alpha$ is the field \begin{equation} A_\alpha(x,t)=L^{-1} e^{i(x\cdot\omega_\alpha-t|\omega_\alpha|^2)} H_\alpha\left(\frac{x-b_\alpha-2\omega_\alpha t}{L}, \frac{t}{L^2}\right). \label{eq:src5} \end{equation} The frequencies $\omega_\alpha$ lie in a fixed square. The labels obey a joint phase-space bound: every unit ball in $\R^2\times\R^2$ contains at most a fixed number of the pairs \[ \left(\frac{b_\alpha}{L},L\omega_\alpha\right). \] The profiles $H_\alpha(y,\tau)$ are supported in a fixed bounded box, and have uniform bounds for sufficiently many derivatives. More precisely, each application requires a finite integer $J$ and fixed bounds $\|\partial^\beta H_\alpha\|_\infty\leq C_\beta$ for $|\beta|\leq J$. The integer $J$ is chosen after the fixed parameters of that application; it does not grow with $L$ or with the array cardinality. A packet field is a finite sum $\sum_\alpha c_\alpha A_\alpha$, and its coefficient energy is $\sum_\alpha|c_\alpha|^2$, without weights. \end{definition} The joint phase-space condition does not require bounded multiplicity of frequencies alone. In particular, arbitrarily many spatial translates of one frequency are allowed. Constants in packet estimates may depend on the fixed support, multiplicity, and derivative bounds in Definition~\ref{def:prelim-atoms}. Whenever a later estimate permits an arbitrary fixed decay order, sufficiently many of these finitely many derivatives are to be fixed first. Suppose that $\mu$ is restricted to $|t|\leq C L^2$ and satisfies \begin{equation} \mu(B(z,r))\leq m r^2 \qquad(z\in\R^3,\ r\geq1). \label{eq:prelim-ball-bound} \end{equation} The additional sparsity assumption, with the fixed choice $M=100$, is \begin{equation} \int_{\R^3} \left(1+\frac{|x-b_\alpha-2\omega_\alpha t|}{L}\right)^{-M} \,d\mu(x,t) \leq\frac{mL^3}{D} \quad\text{for every label used.} \label{eq:src6} \end{equation} Other sufficiently large fixed values of $M$ would work. The ball bound alone implies \eqref{eq:src6} with $D=1$ up to a fixed constant. Indeed, the $\rho$-neighborhood of one trajectory on this time slab is covered by $O(1+L^2/\rho)$ balls of radius $C\rho$. For $\rho=2^kL$, its mass is at most \[ Cm\bigl(2^kL^3+2^{2k}L^2\bigr). \] Summing these bounds against $2^{-Mk}$, $k\geq0$, proves the assertion. The normalization in \eqref{eq:src5} is chosen to be stable under cap rescaling. For labels in a square of side $h^{-1}$ centered at $v$, where $1\leq h\leq L$, set \begin{equation} (x',t')=\left(\frac{x-2vt}{h},\frac{t}{h^2}\right),\qquad L'=\frac Lh,\quad b'_\alpha=\frac{b_\alpha}{h},\quad \omega'_\alpha=h(\omega_\alpha-v). \label{eq:prelim-atom-rescaling} \end{equation} An exact substitution gives \begin{equation} A_\alpha(x,t)=h^{-1}e^{i(x\cdot v-t|v|^2)}A'_\alpha(x',t'), \label{eq:prelim-atom-covariance} \end{equation} where $A'_\alpha$ has the same profile and the parameters in \eqref{eq:prelim-atom-rescaling}. The coefficients are unchanged, and \[ \left(\frac{b'_\alpha}{L'},L'\omega'_\alpha\right) =\left(\frac{b_\alpha}{L},L\omega_\alpha-Lv\right), \] so the phase-space ball counts are unchanged as well. If $\mu'$ is the pushforward of $\mu$, an inverse image of an $r$-ball is covered by $O(h)$ balls of radius $Chr$. Hence $\mu'$ has ball parameter $m'=Cmh^3$. The integrand in \eqref{eq:src6} is invariant under this change, and $m'(L')^3=CmL^3$, so its sparsity parameter is preserved up to the allowed fixed constants. On a time slab of length $O((L')^2)$ centered at zero, a compact packet meets only $O(1)$ spatial squares of side $(L')^2$. Thus the energies assigned to these squares sum with bounded overlap. \subsection{Exact packets and spatial localization} The corresponding energy statement for exact fields requires controlling the tails rather than declaring the packets compactly supported. \begin{lemma}[Exact packet localization] \label{lem:packets} Let $s\geq1$, let $v$ have frequency support in a fixed bounded set, and consider $|t|\leq Cs^2$. There is a decomposition $v=\sum_{\omega,n}a_{\omega,n}v_{\omega,n}$, with $\omega\in s^{-1}\mathbb{Z}^2$ in a fixed bounded enlargement and $b_n=sn$, $n\in\mathbb{Z}^2$, such that \begin{align} \sum_{\omega,n}|a_{\omega,n}|^2&\leq C\|v\|_2^2, \label{eq:prelim-packet-energy}\\ \left\|\sum_{(\omega,n)\in\mathcal I} a_{\omega,n}v_{\omega,n}\right\|_2^2 &\leq C\sum_{(\omega,n)\in\mathcal I}|a_{\omega,n}|^2 \quad\text{for every subfamily }\mathcal I, \label{eq:prelim-subseries-energy}\\ |Ev_{\omega,n}(x,t)|&\leq C_Ns^{-1} \left(1+\frac{|x-b_n-2t\omega|}{s}\right)^{-N} \quad(N\geq1). \label{eq:prelim-packet-decay} \end{align} The constants are independent of the number of terms retained. The phase-space label counts are bounded, and every packet has frequency support in an $O(s^{-1})$ neighborhood of $\omega$. Fix $0<\eta<1$. On a set $\Omega$ in this time slab, retain the packets whose trajectories come within $s^{1+\eta}$ of $\Omega$ at the same time. For every $A>0$, the sum of the omitted packets is bounded pointwise on $\Omega$ by $C_{A,\eta}s^{-A}\|v\|_2$. For a partition into spatial squares of side $s^2$, extended over the common time slab, the retained data $v_Q$ can consequently be chosen so that \begin{equation} \sum_Q\|v_Q\|_2^2\leq C\|v\|_2^2, \qquad |Ev-Ev_Q|\leq C_{A,\eta}s^{-A}\|v\|_2 \quad\text{on }Q\times[-Cs^2,Cs^2]. \label{eq:prelim-box-localization} \end{equation} The same conclusions hold on translated time slabs after modulating the initial datum and writing the trajectories in the recentered time $t-t_0$. \end{lemma} \begin{proof} Take a smooth bounded-overlap partition $\{\chi_\omega\}$ of frequency space at resolution $s^{-1}$. We can arrange that $\chi_\omega(\omega+\eta/s)$ is supported in $[-1,1]^2$. Put \[ g_\omega(\eta)=s^{-1}(\chi_\omega v)(\omega+\eta/s). \] Then $\|g_\omega\|_2=\|\chi_\omega v\|_2$. On $[-\pi,\pi]^2$, expand $g_\omega$ in the orthonormal Fourier basis: \[ g_\omega(\eta)=\sum_{n\in\mathbb{Z}^2} a_{\omega,n}(2\pi)^{-1}e^{-in\cdot\eta}. \] Choose one smooth cutoff $\psi$ supported in $(-\pi,\pi)^2$ and equal to one on $[-1,1]^2$. Multiplying the series by $\psi$ leaves $g_\omega$ unchanged, and gives the packet data \[ v_{\omega,n}(\xi)=\frac{s}{2\pi} \psi\bigl(s(\xi-\omega)\bigr) e^{-isn\cdot(\xi-\omega)}. \] Parseval, followed by the bounded overlap of the frequency partition, gives \eqref{eq:prelim-packet-energy}. Multiplication by $\psi$ is bounded on $L^2$, and its translated frequency supports have bounded overlap. Applying these two facts to any selected Fourier coefficients proves \eqref{eq:prelim-subseries-energy}. This also proves convergence of the decomposition in $L^2$. Since all frequency supports remain in a fixed bounded set, the corresponding extensions converge uniformly in $(x,t)$. Changing variables $\xi=\omega+\eta/s$ gives \begin{equation} Ev_{\omega,n}(x,t)=\frac{s^{-1}}{2\pi} e^{i(x\cdot\omega-t|\omega|^2)} \int \psi(\eta) e^{i((x-sn-2t\omega)/s)\cdot\eta-i(t/s^2)|\eta|^2}\,d\eta. \label{eq:prelim-exact-profile} \end{equation} For $|t|/s^2\leq C$, all derivatives of $\psi(\eta)e^{-i(t/s^2)|\eta|^2}$ are bounded. Repeated integration by parts proves \eqref{eq:prelim-packet-decay}. In particular the profile in \eqref{eq:prelim-exact-profile} is Schwartz in its first argument, uniformly on the prescribed time interval. For clarity, the tail bound is uniform even for infinite packet expansions. At a fixed point, there are $O(s^2)$ frequency centers. For each one the normalized spatial labels form a translate of $\mathbb{Z}^2$, and, for $N>1$, \[ \sum_{n:\ |x-sn-2t\omega|>s^{1+\eta}} \left(1+\frac{|x-sn-2t\omega|}{s}\right)^{-2N} \leq C_Ns^{-\eta(2N-2)}. \] The factor $s^{-2}$ from the squared packet bound cancels the $O(s^2)$ frequency count. Cauchy--Schwarz and \eqref{eq:prelim-packet-energy} therefore bound the discarded field by $C_Ns^{-\eta(N-1)}\|v\|_2$. Choose $N$ sufficiently large in terms of $A$ and $\eta$. Finally, a bounded-velocity trajectory traverses distance $O(s^2)$ in the time slab. Since $s^{1+\eta}\leq s^2$, its padded trajectory meets only $O(1)$ squares of side $s^2$. Each coefficient is consequently assigned to only $O(1)$ of the subfamilies defining $v_Q$. Equation~\eqref{eq:prelim-subseries-energy} proves their summed energy bound, and the preceding tail estimate proves the second part of \eqref{eq:prelim-box-localization}. To center the time interval at $t_0$, replace $v(\xi)$ by $e^{-it_0|\xi|^2}v(\xi)$ before decomposing it. \end{proof} In applications involving polynomially many boxes or a measure of polynomial total mass, the exponent $A$ is chosen large enough that all these tail errors remain smaller than any prescribed negative power. If only a subpower padding is desired on a countersequence, first fix a positive padding exponent for each required estimate and then let it decrease through a diagonal choice. This does not assert a uniform tail constant as $\eta\downarrow0$. \subsection{The refined fractal estimate and unit-cube suprema} We use the following specialization of the refined fractal estimate of Du--Zhang. We include the reductions from its uniform occupancy formulation, since both the weights and the suprema will matter later. \begin{lemma}[Refined fractal estimate] \label{lem:fractal} Let $S\geq1$. Let $\mathcal Q$ be a collection of distinct unit lattice cubes in a ball of radius $CS$ in $\R^3$. Suppose that \begin{equation} \#\{Q\in\mathcal Q:Q\cap B(z,r)\ne\varnothing\} \leq\gamma r^2\quad(1\leq r\leq CS), \qquad \#\{Q\in\mathcal Q:Q\cap B(z,\sqrt S)\ne\varnothing\} \leq\lambda_0 \label{eq:prelim-fractal-counts} \end{equation} for all centers $z$, with fixed enlargements of the radius range allowed. If $v$ has frequency support in a fixed bounded set, then \begin{align} \int_{\bigcup\mathcal Q}|Ev|^2\,dx\,dt &\leq C_\epsilon S^\epsilon (\gamma\lambda_0S)^{1/3}\|v\|_2^2, \label{eq:prelim-fractal-integral}\\ \sum_{Q\in\mathcal Q}\sup_Q|Ev|^2 &\leq C_\epsilon S^\epsilon (\gamma\lambda_0S)^{1/3}\|v\|_2^2. \label{eq:prelim-fractal-sup} \end{align} These estimates are uniform under translations of the collection of cubes. \end{lemma} \begin{proof} For unit-ball frequency support, Theorem~1.6 of \cite{DZ19}, specialized to spatial dimension $2$ and counting exponent $2$, gives \[ \|Ev\|_{L^2(X)} \leq C_\epsilon S^\epsilon \gamma^{1/6}\lambda^{1/6}S^{1/6}\|v\|_2 \] when every occupied square-root-scale lattice cube contains a comparable number $\lambda$ of the unit cubes of $X$. Its time sign is changed by reflection, and its physical-space input norm is converted to the frequency-space norm by Plancherel. A bounded frequency enlargement is reduced to unit frequency by a fixed parabolic dilation and bounded cube coverings. None of these operations changes a large-scale exponent. For the asserted upper occupancy version, take an ambient radius comparable to $S$ whose square root is an integer, large enough to contain the given ball. Its square-root grid then contains whole unit lattice cubes. The occupancy of each parent is at most $C\lambda_0$, since such a parent is covered by a fixed number of $\sqrt S$-balls. Sort the parents by their dyadic occupancy $\lambda$. Every resulting union retains the counting bound $C\gamma r^2$. Apply the quoted theorem to each union, square it, and sum. The geometric sum of $\lambda^{1/3}$ is at most $C\lambda_0^{1/3}$; using a logarithmic bound would also suffice after decreasing the chosen small exponent. Renaming that exponent proves \eqref{eq:prelim-fractal-integral}. Ball centers can be translated at the cost of replacing $v(\xi)$ by $e^{i(y\cdot\xi-u|\xi|^2)}v(\xi)$, which preserves its norm and support. To obtain suprema, let $F=Ev$. Its space-time Fourier transform is supported on a fixed compact portion of the paraboloid. Choose a Schwartz function $\rho$ whose Fourier transform is one on that compact set, so that $F=F*\rho$. For every unit cube $Q$ and every $z\in Q$, \begin{equation} \sup_{y\in Q}|F(y)| \leq C_N\int_{\R^3}(1+|w|)^{-N}|F(z+w)|\,dw. \label{eq:prelim-reproducing} \end{equation} Indeed, taking the supremum over $y-z$ in a fixed bounded set in the convolution formula leaves a rapidly decreasing majorant for the kernel. Cauchy--Schwarz with this integrable weight, followed by integration over $z\in Q$ and summation in $Q$, gives \[ \sum_{Q\in\mathcal Q}\sup_Q|F|^2 \leq C_N\int_{\R^3}(1+|w|)^{-N} \int_{\bigcup\mathcal Q}|Ev(z+w)|^2\,dz\,dw. \] For each fixed $w=(y,u)$ the inner field is the extension of the modulated datum above. Apply \eqref{eq:prelim-fractal-integral} on the unchanged domain and integrate the weight. This proves \eqref{eq:prelim-fractal-sup} without enlarging the ambient scale according to the translation. \end{proof} The same reproducing argument applies separately to factors in a product. For example, if $F_i=Ev_i$, $i=1,2,3$, then \begin{equation} \sum_{Q\in\mathcal Q}\prod_{i=1}^3\sup_Q|F_i| \leq C_N\int_{(\R^3)^3}\prod_{i=1}^3(1+|w_i|)^{-N} \int_{\bigcup\mathcal Q}\prod_{i=1}^3|Ev_i(z+w_i)|\,dz \,dw_1\,dw_2\,dw_3. \label{eq:prelim-product-suprema} \end{equation} To see this, multiply \eqref{eq:prelim-reproducing} for the three fields, integrate over each unit cube, and use Tonelli. For each fixed triple of shifts, the three exact fields are tested at the same integration point on the same domain. Their data have independent unimodular modulations, which preserve supports, norms, and frequency transversality. This is the version used when applying multilinear restriction to products of cube suprema. \subsection{Weights and cap rescaling} \begin{lemma}[Weighted refined estimate] \label{lem:prelim-weighted} Let $\nu$ be supported in a ball of radius $CS$, $S\geq1$, and suppose \begin{equation} \nu(B(z,r))\leq mr^2\quad(r\geq1),\qquad \nu(B(z,\sqrt S))\leq mSd \quad(z\in\supp\nu), \label{eq:prelim-weighted-hypotheses} \end{equation} where $m,d>0$. For bounded frequency support, \begin{equation} \int |Ev|^2\,d\nu \leq C_\epsilon S^\epsilon mS^{2/3}d^{1/3}\|v\|_2^2. \label{eq:prelim-weighted-fractal} \end{equation} The same bound holds for $\sum_Q\nu(Q)\sup_Q|Ev|^2$, summed over unit cubes. The ball bound alone gives this conclusion with $d=1$, up to a fixed constant. \end{lemma} \begin{proof} Fixed-radius enlargements of the second bound in \eqref{eq:prelim-weighted-hypotheses} follow by a bounded covering with balls of radius $\sqrt S$ centered on the support. This also extends it to arbitrary centers, up to a fixed constant. Every unit cube has mass at most $Cm$. Split the positive-mass cubes into classes \[ m\chi<\nu(Q)\leq2m\chi, \qquad \chi\in\{C_02^{-k}:k\geq0\}, \] with harmless endpoint adjustments and fixed $C_0$. Cubes in one class have counting parameter $C\chi^{-1}$: the cubes meeting an $r$-ball have disjoint masses inside a fixed enlargement of that ball. Their square-root-scale occupancy is at most $CSd\chi^{-1}$ by the same argument. Lemma~\ref{lem:fractal} therefore gives \begin{align*} \sum_{Q\text{ in the }\chi\text{ class}}\nu(Q)\sup_Q|Ev|^2 &\leq C_\epsilon m\chi S^\epsilon \bigl(\chi^{-1}\cdot Sd\chi^{-1}\cdot S\bigr)^{1/3} \|v\|_2^2\\ &=C_\epsilon S^\epsilon mS^{2/3}d^{1/3} \chi^{1/3}\|v\|_2^2. \end{align*} The series $\sum_\chi\chi^{1/3}$ converges. Summing proves both assertions, including measures with arbitrarily small positive cube masses; no lower mass cutoff is needed here. Finally the first ball bound, evaluated at $r=\sqrt S$, supplies $d=1$. \end{proof} \begin{lemma}[Weighted cap estimate] \label{lem:cap} Fix $\alpha_0<1/2$. Suppose that $S\geq1$, that $\nu$ is supported in a ball of radius $CS$ and satisfies $\nu(B(z,r))\leq mr^2$ for $r\geq1$, and that $v$ is supported in a frequency ball $B(v_0,q)$, with $v_0$ in a fixed bounded set and \[ S^{-\alpha_0}\leq q\leq1. \] Then, for every $\epsilon>0$, \begin{equation} \int |Ev|^2\,d\nu \leq C_{\epsilon,\alpha_0}S^\epsilon mS^{2/3}q^{1/3}\|v\|_2^2. \label{eq:prelim-cap-estimate} \end{equation} The estimate also holds with the integral replaced by $\sum_Q\nu(Q)\sup_Q|Ev|^2$ over unit cubes. Bounded translations, and averages of translations with a fixed integrable Schwartz weight, obey the same bounds. \end{lemma} \begin{proof} Translate the space-time support ball to the origin, absorbing the translation in a unimodular factor on $v$. Write \[ g(\eta)=q\,v(v_0+q\eta),\qquad (y,\tau)=\bigl(q(x-2v_0t),q^2t\bigr),\qquad S'=Sq^2. \] Then $\|g\|_2=\|v\|_2$, $\supp g\subset B(0,1)$, and \begin{equation} Ev(x,t)=q\,e^{i(x\cdot v_0-t|v_0|^2)}Eg(y,\tau). \label{eq:prelim-cap-identity} \end{equation} Let $\nu'$ be the pushforward of $\nu$. The inverse map is \[ t=q^{-2}\tau,\qquad x=q^{-1}y+2v_0q^{-2}\tau. \] Consequently, the inverse image of a ball of radius $r\geq1$ is a bounded shear of a box with side lengths $r/q,r/q,r/q^2$. It is covered by $O(q^{-1})$ balls of radius $Cr/q$. The original ball bound applies at these radii and gives \begin{equation} \nu'(B(z,r))\leq Cm q^{-3}r^2\qquad(r\geq1). \label{eq:prelim-rescaled-density} \end{equation} There is no Jacobian multiplier on this pushforward measure. The rescaled time interval has length $O(S')$, but the spatial projection has diameter $O(Sq)=O(S'/q)$. Partition the latter into squares $Q$ of side $S'$. On the common time slab, use Lemma~\ref{lem:packets} at packet width $s=\sqrt{S'}$, with any fixed padding exponent in $(0,1)$. Let $g_Q$ denote its localized data. Then \[ \sum_Q\|g_Q\|_2^2\leq C\|g\|_2^2, \qquad |Eg-Eg_Q|\leq C_A(S')^{-A}\|g\|_2 \quad\text{on the corresponding prism}, \] where $A$ can be made arbitrarily large. Each restricted measure lies in a ball of radius $CS'$, so Lemma~\ref{lem:prelim-weighted} with $d=1$ and \eqref{eq:prelim-rescaled-density} gives \[ \sum_Q\int_{Q\times[-CS',CS']}|Eg_Q|^2\,d\nu' \leq C_\epsilon(S')^\epsilon m q^{-3}(S')^{2/3}\|g\|_2^2. \] The sum uses the localized energies, rather than the number of spatial squares. This distinction is essential for the cap power. For completeness, $\nu'(\R^3)=\nu(\R^3)\leq CmS^2$, whereas $S'\geq S^{1-2\alpha_0}$. Thus the integrated squared error, after multiplication by the $q^2$ in \eqref{eq:prelim-cap-identity}, is at most \[ C_A m q^2 S^2(S')^{-2A}\|g\|_2^2. \] Choose $A$ in terms of $\alpha_0$ so that this is bounded by the desired right side; bounded $S$ is absorbed in the constant. For the principal term, \begin{equation} q^2\,m q^{-3}(Sq^2)^{2/3} =mS^{2/3}q^{1/3}. \label{eq:prelim-cap-power} \end{equation} Since $(S')^\epsilon\leq S^\epsilon$, this proves \eqref{eq:prelim-cap-estimate}. Finally, square \eqref{eq:prelim-reproducing} by weighted Cauchy--Schwarz, integrate in $z\in Q$ against $d\nu(z)$, and sum the original unit cubes. This bounds the sum of weighted suprema by \[ C_N\int_{\R^3}(1+|w|)^{-N} \int_{\R^3}|Ev(z+w)|^2\,d\nu(z)\,dw. \] For every shift the field is an exact extension of modulated data with the same cap support and norm. Apply the already proved cap estimate and integrate the rapidly decreasing weight. This proves the supremum assertion and the asserted translated versions. \end{proof} \begin{remark}[Enlarged cubes in the inductive formulation] \label{rem:prelim-enlarged-cubes} The proof of Lemma~\ref{lem:fractal} is sufficient for all later applications. For comparison, we record an alternative reduction of \eqref{eq:prelim-fractal-integral} to the inductive Proposition~3.1 of \cite{DZ19}. Choose $K=S^\delta$ with $0<\delta<1/4$ sufficiently small relative to the desired loss, and enlarge occupied unit cubes to occupied $K^2$-cubes. Selecting one original unit cube in each occupied large cube shows, by ball covering, that the large cubes have counting parameter $O(\gamma)$ for $r\geq K^2$, and square-root-scale occupancy $O(\lambda_0)$. Normalize $\|v\|_2=1$ and discard large cubes with $L^6$ norm below $S^{-A}$. There are polynomially many cubes, so their total $L^2$ contribution is arbitrarily small when $A$ is sufficiently large. The remaining $L^6$ norms have only $O(\log S)$ dyadic classes, since $|Ev|\leq C$ and each large cube has volume $K^6$. After fixing a norm class, recompute its parent occupancies and sort them dyadically. On each resulting union $Y$ of $M_Y$ large cubes, Proposition~3.1 gives \[ \|Ev\|_{L^6(Y)} \leq C_\epsilon M_Y^{-1/3} (\gamma\lambda_0S)^{1/6}S^\epsilon\|v\|_2. \] H\"older contributes $|Y|^{1/3}=M_Y^{1/3}K^2$ on passing to $L^2$, cancelling $M_Y^{-1/3}$. After squaring, the additional loss is $K^4$; the finitely many logarithmic losses and this factor are absorbed by the choice of $\delta$ and a smaller initial $\epsilon$. The grids in this deduction can be made compatible as follows. Increase the ambient scale by a bounded factor so that $S^{1/2}/S^{2\delta}$ is an integer. Such scales have successive ratios tending to one. Use nested grids at these two scales and a bounded family of translated parent grids. For sufficiently large $S$, each original unit cube can be assigned to a parent grid in which it is farther than $2\sqrt3 K^2$ from the parent boundary; keep only the occupied large cubes within that parent. Split the original collection according to these finitely many assignments. The same original ball counts control each part, while translation is absorbed into the datum as in Lemma~\ref{lem:fractal}. Bounded scales are covered by a fixed constant. Thus neither boundary cubes nor rounding changes the resulting estimate. \end{remark} ==== END Endpoint-convergence-for-the-planar-Schrodinger-equation//02-preliminaries.tex ==== ==== BEGIN Endpoint-convergence-for-the-planar-Schrodinger-equation//03-transverse.tex ==== \section{A power saving for transverse inputs} \label{sec:transverse} The first step is a rigidity statement for almost extremizers of the fractal estimate. A hypothetical extremizing configuration produces a probability distribution on points and incident tubes. Information at successive spatial scales first forces three identical one-dimensional profiles, then forces those profiles to alternate only finitely often between slopes zero and one. Polynomial partitioning rules out the remaining transition. \begin{proposition}[Transverse power saving]\label{prop:transverse} Fix a bounded frequency region. There are $c>0$ and $L_0<\infty$, depending only on that region, for which the following configuration cannot exist when $L\ge L_0$. Let $\mathcal Q$ be unit lattice cubes inside a ball of radius $L^2$ in $\R^3$, with \[ \#\mathcal Q\ge L^{4-c},\qquad \#\{Q\in\mathcal Q:Q\cap B(z,r)\ne\varnothing\} \le L^c r^2\quad(r\ge1), \] where cube counts may equivalently use their centers, up to fixed constants. Suppose $v_1,v_2,v_3$ have frequency support in the fixed region, $\|v_i\|_2\le1$, and \[ \sup_Q|Ev_i|\ge L^{-4/3-c} \qquad(Q\in\mathcal Q,\ i=1,2,3). \] Then it is impossible that every triple of unit frequency directions, one from each support, has absolute determinant at least $L^{-c}$. \end{proposition} We argue by contradiction along $L\to\infty$ and $c\downarrow0$. Every $o(1)$ in this section refers to this sequence or a subsequence. A diagonal choice converts estimates valid with every fixed positive power loss into $L^{o(1)}$ estimates; constants are fixed before making that choice. All rounding, support enlargements, and packet padding are included in these subpower errors. The same convention applies when a scale is a fixed positive power of $L$. \subsection{Uniform multilinear estimates and the tube model} We need the multilinear estimates when transversality itself deteriorates by a subpower. An unspecified transversality-dependent constant in the restriction theorem would not justify this use. The following finite induction supplies the required uniformity. \begin{lemma}[Multilinear estimates with subpower transversality] \label{lem:transverse-product} Suppose three exact extension inputs have bounded frequency supports, and every triple of their unit directions has determinant at least $\nu$, where $0<\nu\le1$. For each $\epsilon>0$ there are $C_\epsilon,M_\epsilon<\infty$ such that, on every ball of radius $S\ge2$, \begin{equation}\label{eq:transverse-product} \int_{B_S}\prod_{i=1}^3|Ev_i(z)|\,dz \le C_\epsilon\nu^{-M_\epsilon}S^\epsilon \prod_{i=1}^3\|v_i\|_2. \end{equation} The same conclusion holds for the sum over unit lattice cubes in the ball of $\prod_i\sup_Q|Ev_i|$. In particular, when $\nu=S^{-o(1)}$ both bounds are $S^{o(1)}\prod_i\|v_i\|_2$. For weighted tubes of width $\Delta$ in a bounded region, the corresponding multilinear Kakeya bound is \begin{equation}\label{eq:transverse-kakeya} \int\prod_{i=1}^3 \left(\sum_T w_i(T)\mathbf1_T\right)^{1/2} \le C_\epsilon\nu^{-M_\epsilon}\Delta^{3-\epsilon} \prod_{i=1}^3\left(\sum_Tw_i(T)\right)^{1/2}. \end{equation} Here fixed enlargements of tubes and of the ambient region are allowed; if those enlargements are subpower on a sequence, they cost a subpower. \end{lemma} \begin{proof} Partition each direction support into neighborhoods of radius a sufficiently small fixed power of $\nu$. There are at most $\nu^{-O(1)}$ such neighborhoods. For each triple choose representative directions and send them to the coordinate axes by a linear map. Its norm, inverse norm, and volume distortion are bounded by fixed powers of $\nu^{-1}$. The transformed direction neighborhoods lie in fixed small neighborhoods of the axes. The weighted multilinear Kakeya inequality of \cite[Theorem~1.15 and Remark~1.10]{BCT06}, applied there and summed over triples, proves \eqref{eq:transverse-kakeya}. Weights follow first by repetitions and then by approximation. Widths larger than a fixed constant can instead be treated by the trivial bound, after increasing the inverse-power constant. For completeness, use this Kakeya inequality to obtain restriction uniformity. Let $A_B(S,\nu)$ be the least normalized product-integral constant for inputs supported in $[-B,B]^2$. Fix padding exponent $\theta>0$ and Kakeya loss $\sigma>0$. Cover a ball of radius $S$ by cubes of side comparable to $\sqrt S$. The exact packet decomposition restricts each input on a smaller cube to packets incident to its $S^\theta\sqrt S$ enlargement. Its retained exact input has squared norm at most a fixed constant times the incident coefficient energy. Its support is within $O_B(S^{-1/2})$ of the original support: only caps intersecting that support occur, and every enlarged cap lies within this distance of such an intersection. Thus determinant continuity preserves $\nu/2$ transversality when $S^{-1/2}\le c_B\nu$, without any openness or connectedness assumption on the supports. Apply the smaller-scale product estimate in every cube. Its geometric coefficient is the product of the three square roots of incident energies. After division of coordinates by $S$, the padded tubes have width $\Delta=S^{-1/2+\theta}$. Enlarge them by a fixed factor, so an incident tube contains the whole smaller grid cube. Equation~\eqref{eq:transverse-kakeya}, divided by that cube's volume $S^{-3/2}$, gives the bound \[ C_{B,\theta,\sigma}\nu^{-M_\sigma} S^{3/2}(S^{-1/2+\theta})^{3-\sigma} \prod_i\|v_i\|_2 \le C_{B,\theta,\sigma}\nu^{-M_\sigma} S^\beta\prod_i\|v_i\|_2,\qquad \beta=3\theta+\sigma/2, \] for the sum of these geometric coefficients. To bound discarded-field errors, normalize each parent input, use rapid tails for a discarded factor and the fixed-frequency $L^\infty$ bound for the others, and expand the three-factor product. The seven error terms together contribute at most $C_{B,\theta,N}S^{3-N}\prod_i\|v_i\|_2$, for every prescribed $N$. This is relative to the current \emph{parent} norms; no lower bound for a retained local norm is required. We obtain the recurrence \begin{equation}\label{eq:transverse-square-root} A_B(S,\nu)\le C_{B,\theta,\sigma}\nu^{-M_\sigma}S^\beta A_{B+1}(C_B\sqrt S,\nu/2) +C_{B,\theta,N}S^{3-N}, \qquad S^{-1/2}\le c_B\nu. \end{equation} Spatial translations only modulate the exact inputs. Taking $N>3$ and setting $\widetilde A_B=1+A_B$ absorbs the additive error in the same multiplicative recurrence for $\widetilde A_B$. Fix the target $\epsilon>0$. Choose a finite $J$ with $3\cdot2^{-J}<\epsilon/2$, then choose $\theta,\sigma$ with $2\beta<\epsilon/2$. The finitely many radii defined by $S_0=S$, $S_{j+1}=C_{B+j}\sqrt{S_j}$ are comparable to $S^{2^{-j}}$, with constants depending only on $B,J$. If $S\ge(C_{B,J}/\nu)^{2^J}$ for a sufficiently large constant, all $J$ support-enlargement conditions hold. Iterating \eqref{eq:transverse-square-root} and using the terminal trivial bound $\widetilde A_{B+J}(R,\nu')\lesssim_{B,J}R^3$ gives \[ \widetilde A_B(S,\nu) \le C_{B,\epsilon}\nu^{-M_{B,\epsilon}} S^{\beta\sum_{j=0}^{J-1}2^{-j}+3\cdot2^{-J}} \le C_{B,\epsilon}\nu^{-M_{B,\epsilon}}S^\epsilon. \] In the complementary range, the trivial bound $A_B(S,\nu)\lesssim_B S^3$ is absorbed by increasing the fixed inverse-transversality exponent to at least $3\cdot2^J$. This proves \eqref{eq:transverse-product} for all $S,\nu$. On a subpower-transverse sequence the threshold is eventually satisfied for each fixed $J$, and every intermediate enlarged support remains subpower transverse. This is a finite version of the restriction bootstrapping in \cite[Proposition~2.1]{BCT06}; polynomial dependence is supplied by this argument, not by the statement of their restriction theorem alone. Finally, fixed-frequency reproducing kernels give, for $z\in Q$, \[ \sup_Q|Ev_i|\lesssim_N \int(1+|w|)^{-N}|Ev_i(z+w)|\,dw. \] Integrate the product over $Q$ and sum. For each triple of shifts the three shifted fields are exact extensions on the same domain, with independently modulated inputs of unchanged norms and supports. Apply \eqref{eq:transverse-product} and integrate the Schwartz weights. \end{proof} \begin{lemma}[Reduction to weighted incident tubes] \label{lem:transverse-tubes} A countersequence to Proposition~\ref{prop:transverse} yields a set $P$ in a fixed bounded region and three weighted tube families satisfying \begin{equation}\label{eq:transverse-point-counts} |P|=L^{2+o(1)},\qquad |P\cap B(z,r)|\le L^{2+o(1)}r^2 \quad(L^{-1}\le r\le1). \end{equation} The points are centers of distinct cubes of side comparable to $L^{-1}$. The tubes have width $L^{-1+o(1)}$, their coefficient weights obey $\sum_Tw_i(T)\lesssim1$, and every cross-family direction triple has determinant at least $L^{-o(1)}$. At each $p\in P$ there is a retained incident subfamily of total weight $e_i^*(p)=L^{-4/3+o(1)}$ such that its normalized weight law satisfies, for fixed $\kappa>0$, $u_0>0$, \begin{equation}\label{eq:src1} \Prob\bigl(\operatorname{dir}T_i\in B(v,L^{-u})\mid X=p\bigr) \le L^{-\kappa u+o(1)} \qquad(00$, instead has squared sum at most $L^{4/3-\delta/3+o(1)}$. Thus some retained class has $L^{2-o(1)}$ cubes in each occupied box. The upper counting bound gives occupancy at most $L^{2+o(1)}$. Divide coordinates by $L^2$ and take the centers of these boxes. Counting cubes in an enlarged ball and dividing by the minimum occupancy proves \eqref{eq:transverse-point-counts}. Translation of the original ambient ball merely modulates the exact data. Apply Lemma~\ref{lem:packets} at packet width $L$ on the time interval of length $O(L^2)$. After division of coordinates by $L^2$, the resulting tubes have width $L^{-1+o(1)}$. Give each tube the squared coefficient of its packet as weight. On each retained box, packets reaching a padded enlargement give an exact local input and an error $O(L^{-100})$. The padding exponent tends sufficiently slowly to zero; the rapid-tail estimates permit this by diagonalization. By \eqref{eq:prelim-subseries-energy}, the squared norm of this exact local input is bounded by the incident coefficient energy $e_i(p)=\sum_{T\ni p}w_i(T)$ up to a constant. On the at least $L^{2-o(1)}$ selected unit cubes in this box the required squared supremum sum is $L^{-2/3-o(1)}$. The weighted fractal estimate at radius $L$ bounds it above by $L^{2/3+o(1)}e_i(p)$, so \[ e_i(p)\ge L^{-4/3-o(1)}. \] Lemma~\ref{lem:transverse-product}, in its Kakeya form and with $\Delta=L^{-1+o(1)}$, yields \[ \sum_{p\in P}\prod_{i=1}^3e_i(p)^{1/2}\le L^{o(1)}. \] For fixed $\delta>0$, points at which some $e_i(p)\ge L^{-4/3+\delta}$ therefore number at most $L^{2-\delta/2+o(1)}$. Discarding these points by a diagonal choice leaves \eqref{eq:transverse-point-counts} and makes all three energies $L^{-4/3+o(1)}$ uniformly. It remains to remove concentration in direction without losing the local lower bound. Choose $0<\kappa<1/3$ and $00$ this is negligible relative to $L^{-2/3-o(1)}$. Delete successively for any fixed finite list of rational $u$'s, assigning a packet deleted more than once to just one bin. The triangle inequality in the square-sum norm shows that the surviving field retains its lower local norm. The unrestricted local estimate then gives surviving weight $e_i^*(p)=L^{-4/3+o(1)}$. Every surviving bin has original weight at most $q^\kappa e_i(p)$; normalizing by $e_i^*(p)$ costs only $L^{o(1)}$. Boundedly many neighboring bins cover a ball, and a diagonal choice proves \eqref{eq:src1} at all fixed rational scales. All estimates were uniform in $p$, so this point-dependent pruning supplies legitimate conditional laws without requiring a single globally pruned family. \end{proof} \subsection{Information at spatial scales} From now on only the abstract configuration of Lemma~\ref{lem:transverse-tubes} is used. Draw $X$ uniformly from $P$. Given $X$, draw six tubes $T_i,T_i'$ independently, two from each retained family. A tube label includes its exact direction. For rational $0C+\eta\}\le L^{-\eta+o(1)}. \] The same counting argument with a varying threshold proves uniform integrability. For the conditioning assertion put $p=\Prob(Y\mid V)$ and $q=\Prob(Y\mid V,W)$ at the sampled labels. Conditional on $(V,W)$, the expectation of $p/q$ is at most one. Thus \[ \Prob\{I_L(Y\mid V)-I_L(Y\mid V,W)<-\eta\} \le L^{-\eta}. \] This proves the almost-sure sign in the limit; no finite-scale pointwise monotonicity is asserted. The chain rule is pointwise already at finite scale. The determination rule follows by summing over the stated small set of possible labels. This argument also works on restricted events, since it bounds the total mass of the bad low-probability labels in the event's permitted set. If the event indicator is included in the joint limit, apply weak convergence to the open set where it is one and the information is strictly above $\gamma+\eta$, and then let $\eta\downarrow0$. For events defined through other limiting variables, the same argument uses strict inequalities and slack, as detailed in the zero-ratio application below. Tightness and diagonal weak convergence on a countable product give the joint law. Include in this product all rational block and prefix parameters, all finitely many choices of reference and replacement tubes, their direction bins, coordinate grids, projections, the information differences used below, and the angular quantities to be introduced. Direction grids use deterministic representatives of occupied bins. Realized bases and inverse bases have norms $L^{o(1)}$, so every coordinate partition has at most $L^{C+o(1)}$ labels. Standard open-set inequalities for weak convergence justify the assertion about finite systems after enlarging their bounds by slack. Uniform integrability permits passage of expectations in the cost estimates below. \end{proof} \begin{lemma}[The spatial profiles]\label{lem:transverse-profiles} In the limiting law set \[ F(s)=I(X_s),\qquad G_i(s)=I(X_s\mid T_i),\qquad G_i'(s)=I(X_s\mid T_i'). \] These functions extend from rational scales to nondecreasing Lipschitz functions on $[0,1]$, vanishing at zero. Their Lipschitz constants are respectively $3$ and $1$. Almost surely, \begin{equation}\label{eq:transverse-endpoint-information} F(s)\ge2s,\qquad F(1)=2,\qquad G_i(1),G_i'(1)\ge\tfrac23. \end{equation} \end{lemma} \begin{proof} The exact chain increment of $G_i$ between $s$ and $s+t$ is $I(X_{s+t}\mid X_s,T_i)$. Inside a parent, the tube meets at most $L^{t+o(1)}$ child cubes, including its padding. The determination rule bounds this increment between zero and $t$; the unrestricted spatial count gives the bounds zero and $3t$ for $F$. Rational comparisons simultaneously yield the Lipschitz extensions. The ball count in \eqref{eq:transverse-point-counts} gives every $X_s$-cell mass at most $L^{-2s+o(1)}$, proving $F(s)\ge2s$. An $X_1$-cell contains boundedly many points and $X$ is uniform, so $F(1)=2$. Finally the retained weights give \[ \Prob(X=p,T_i=T) =|P|^{-1}\frac{w_i(T)\mathbf1_{\{T\text{ retained at }p\}}} {e_i^*(p)} \le L^{-2/3+o(1)}w_i(T). \] The probability of sampling a tube whose marginal is smaller than $L^{-\eta}w_i(T)$ is at most $L^{-\eta}\sum_Tw_i(T)=O(L^{-\eta})$. Outside this event, every conditional point mass is at most $L^{-2/3+\eta+o(1)}$. Letting $\eta\downarrow0$ proves the conditional endpoint bounds, also for all primed tubes. This uses only summability of weights, so exact tube labels need not have bounded cardinality. \end{proof} \subsection{Direction costs and equality of the profiles} To compare the three tube-conditional profiles, we use coordinates adapted to one reference tube from each family. Specifying those coordinates requires revealing coarse direction bins. This can lower the spatial information; the costs below measure that loss, and their telescoping bound will let the loss tend to zero with the block size. Fix a rational block $[s,s+b]\subset[0,1]$ and rational $b_*\ge b$. Let $C=X_s$ and let $D$ record direction bins of precision $L^{-b_*}$ for a fixed sublist of the six tubes, including one reference tube from each family. Within the parent subtract its center, multiply coordinates by $L^s$, and use the approximate basis supplied by the three reference bins. Its norms and inverse norms are $L^{o(1)}$. For rational $0L^{-\eta/2}$. The approximate basis fixed by $D$ differs from the exact basis by $L^{-b_*+o(1)}$, including the subpower condition numbers. Its corresponding coefficient thus has magnitude at least $L^{-\eta}$ for large $L$. Given $(T_i',Z_j,C,D)$ alone, possible child positions on this event number at most $L^{O(\eta)+o(1)}$: the approximate $j$th coordinate parametrizes the tube with inverse slope at most $L^{\eta+o(1)}$. This permitted set depends only on those displayed labels. Exact reference directions need not be added to the conditioning. Apply the restricted determination bound and the information conditioning rule. The probability that $r_{j,L}<\eta/2$ and \[ I_L(X_{s+t}\mid C,D,T_i')-I_L(Z_j\mid C,D)>C\eta+\eta \] tends to zero, with a fixed absolute $C$. The joint limit therefore gives the corresponding comparison on the open event $\{r_j<\eta/2\}$ with error at most $(C+1)\eta$; one can first use any strict upper error bound and then decrease it. This is an implication about the support of the joint limiting law and requires no coupling of finite-scale and limiting events. Let rational $\eta\downarrow0$. On $\{r_j=0\}$ we obtain \[ G_i'(s+t)-G_i'(s)-\ell_i' \le I(Z_j\mid C,D), \] where $D$ includes the alternative bin and $\ell_i'$ is its analogous cost. Replacement invariance and \eqref{eq:src3} now compare the $i$ and $j$ profile increments up to the costs. Take dyadic partitions of the exponent interval, with block size $b=b_*$. On each block distribute its cost uniformly over its length. The integral of this density is the cost; hence its expected integral is $O(b)$ by Lemma~\ref{lem:transverse-costs}. Summability over dyadic $b$ shows that all normalized costs tend to zero for almost every pair consisting of a path and an exponent. At simultaneous differentiability points it follows that $\dot G_i\le\dot G_j$ whenever the corresponding limiting ratio information is zero. The restricted-event determination just used is applied before differentiation, so no uniform positive lower bound for the coefficient is required. For almost every path each vertex $i$ therefore has an edge $i\to j$ for which $\dot G_i\le\dot G_j$ almost everywhere. Since both integrals are $2/3$, this forces $G_i=G_j$. Every equivalence class of equal profiles has at least two members. With three families there is only one such class, proving \eqref{eq:src4}. \end{proof} \subsection{Projection growth on a block} The next input is the robust form of the discretized projection theorem: one direction improves the projection of every sufficiently large subset. That quantifier is essential because the subset we construct will depend on the direction. \begin{lemma}[Robust planar discretized projection]\label{lem:transverse-projection} Fix $00$. There are $\varepsilon_{\rm pr}>0$ and $\eta_{\rm pr}>0$ with the following property for all sufficiently small $\Delta$. Suppose $A\subset B(0,1)\subset\R^2$, with $\mathcal N_\Delta$ denoting covering number by radius-$\Delta$ balls, satisfies \[ \Delta^{-2d+\eta_{\rm pr}}\le\mathcal N_\Delta(A) \le\Delta^{-2d-\eta_{\rm pr}},\qquad \mathcal N_\Delta(A\cap B(z,r)) \le\Delta^{-\eta_{\rm pr}}r^{\rho_0} \mathcal N_\Delta(A) \] for $\Delta\le r\le1$. Let $\sigma$ be a probability on projection lines obeying $\sigma(B(V,r))\le\Delta^{-\eta_{\rm pr}}r^{\rho_0}$ on the same range. Some $V\in\supp\sigma$ then satisfies \[ \mathcal N_\Delta(\pi_V A')\ge\Delta^{-d-\varepsilon_{\rm pr}} \quad\hbox{for every }A'\subset A\hbox{ with } \mathcal N_\Delta(A')\ge\Delta^{\eta_{\rm pr}} \mathcal N_\Delta(A). \] One may decrease both the tolerance and the gain. \end{lemma} \begin{proof} This is the $n=2$, $m=1$, $\alpha=2d$ case of \cite[Theorem~1]{He20}, a general form of the planar Bourgain discretized projection theorem. In dimension two its Grassmannian nonconcentration condition is the displayed angular-ball condition. That theorem gives the conclusion simultaneously for every large subset, for a direction set of probability at least one minus a positive power of $\Delta$. Decreasing the tolerance relative to its exponent gives the stated formulation. \end{proof} Take $T_1,T_2,T_3$ as reference tubes and write the direction of $T_1'$ in their exact basis as $(1,a_2,a_3)$. Given $X$ and these three exact references, redraw $T_1'$ from its original conditional law. Let $h_j(u)$, for rational $0L^\eta p_z}p_z \le L^{-\eta}\sum_zq_z\lesssim L^{-\eta}. \] A boundedly finer grid and the determination rule handle balls not contained in one grid cell. The argument is identical for joint and marginal grids. The chain rule and \eqref{eq:transverse-conditioning} bound joint cell information by the sum of marginal informations. This proves \eqref{eq:transverse-angular-sum}; monotonicity follows directly from nesting of centered balls. \end{proof} \begin{lemma}[The projection block test]\label{lem:transverse-block-test} Fix $00$, and $\kappa_0>0$. There is $\zeta>0$, depending only on these parameters, for which the following conditions cannot hold on an event of positive limiting probability. Fix rational $[s,s+b]\subset[0,1]$, with $b