Proof. For each fixed \(x\), the map \(\varphi_{x,t}(y)=y-tu(x,y)\) is increasing, onto, and satisfies \[(1-\left|t\right|\,C_{\rm Lip})\left|y-y'\right| \le\left|\varphi_{x,t}(y)-\varphi_{x,t}(y')\right| \le(1+\left|t\right|\,C_{\rm Lip})\left|y-y'\right|.\] The change-of-variables formula for a one-dimensional bi-Lipschitz map, followed by translation in \(x\), proves (3). A difference of endpoint truncations is pointwise dominated by \[\int_{c' r<|t|<C' r}\left|A_t h\right|\frac{\,\mathrm dt}{|t|}\] with \(c'=\min(1,c)\) and \(C'=\max(1,C)\). Minkowski and (3) bound this by \(C_p\log(C'/c')\left\|h\right\|_p\). The same proof applies to inner and outer endpoints, and permits endpoints depending on \((x,y)\). ◻

Proof. Use disjoint amplitude sets \(F_i=\{2^{i-1}<|f|\le2^i\}\) and write \(f=\sum_i2^i f_i\), where \(|f_i|\le\mathbf 1_{F_i}\). For the moment all sums are finite. Put \(a_i=2^i\sqrt{|F_i|}\), so \(\sum_i a_i^2\le4\left\|f\right\|_2^2\). Group the indices into measure classes \[\mathcal I_k=\{i:2^k\le|F_i|<2^{k+1}\},\qquad A_k=\left(\sum_{i\in\mathcal I_k}a_i^2\right)^{1/2}.\] If \(i_0\) is the largest index in a nonempty class, then \[\sum_{i\in\mathcal I_k}a_i \le 2^{(k+1)/2}\sum_{i\le i_0}2^i \le 2\sqrt2\,a_{i_0}\le2\sqrt2\,A_k.\] Make the analogous decomposition \(g=\sum_j2^j g_j\), with measure classes \(\mathcal J_l\) and square sums \(B_l\). For \(i\in\mathcal I_k\), \(j\in\mathcal J_l\), \[(1+|\log(|F_i|/|E_j|)|)^{-3} \le C(1+|k-l|)^{-3}.\] The assumed bound, followed by the within-class estimates, gives \[\left|\mathcal B(f,g)\right| \le C C_0\sum_{k,l}(1+|k-l|)^{-3}A_kB_l \le C C_0\left\|A\right\|_{\ell^2}\left\|B\right\|_{\ell^2}.\] The final step is Young’s inequality for the summable sequence \((1+|k|)^{-3}\). This proves the assertion. Truncating the amplitude decompositions and using the asserted continuity or completion gives the last statement, with no dependence on the number of classes. ◻

Proof. For a dyadic interval \(J\) of length \(2^{-j}\), split \(q=q\mathbf 1_{3J}+q\mathbf 1_{(3J)^c}\). By (5), the first part satisfies \[\sum_{k\ge j}\int_J |\Delta_k(q\mathbf 1_{3J})|^2 \le C\left\|q\right\|_\infty^2|J|.\] The kernels of \(\Delta_k\) have Schwartz decay at scale \(2^{-k}\). For \(x\in J\) and \(k\ge j\), this gives \[|\Delta_k(q\mathbf 1_{(3J)^c})(x)| \le C\left\|q\right\|_\infty 2^{j-k}.\] After summing the squares, we obtain \[ \sum_{k\ge j}\int_J|\Delta_kq|^2 \le C\left\|q\right\|_\infty^2|J|. \tag{7}\] For each dyadic interval \(I\) of length \(2^{-k}\), set \(\alpha_I=\int_I|\Delta_kq|^2\). Then (7) says \(\sum_{I\subset J}\alpha_I\le C\left\|q\right\|_\infty^2|J|\).

Write \(M\) for the Hardy–Littlewood maximal operator. Kernel decay, summed on dyadic annuli about a point of \(I\), gives \[\sup_{x\in I}|S_{k+c}h(x)| \le C_c\operatorname*{ess\,inf}_{y\in I}Mh(y).\] The shift \(c\) is fixed, so moving the kernel center across \(I\) changes the bound by only a fixed factor. If \(F\ge0\) belongs to \(L^2\), the dyadic Carleson condition implies \[\sum_I\alpha_I \bigl(\operatorname*{ess\,inf}_I F\bigr)^2 \le C\left\|q\right\|_\infty^2\left\|F\right\|_2^2.\] Indeed, for each \(t>0\), sum over the maximal dyadic intervals on which \(\operatorname*{ess\,inf}_I F>t\), apply the Carleson condition on each such interval, and integrate against \(2t\,\mathrm dt\). Their union lies, up to null sets, in \(\{F>t\}\). Apply this assertion with \(F=Mh\), and use the \(L^2\) maximal theorem. This proves (6). This is also the usual continuous Littlewood–Paley Carleson embedding; compare the proof of (Tao 2007, Proposition 4.1). ◻

Proof. Put \(L_b=\left\|b'\right\|_\infty\) and \(b_k=\Delta_kb\). Cancellation and the first absolute moment of the smooth kernels give \[ \left\|b_k\right\|_\infty\le C2^{-k}L_b, \qquad \left\|\partial S_kb\right\|_\infty\le C L_b, \qquad b_k'=\Delta_kb'. \tag{9}\] Also \(\left\|S_Nb-b\right\|_\infty\le C2^{-N}L_b\). Consequently, for each fixed \(j\), the exact identity \[ b=S_{j-4}b+\sum_{k\ge j-3}b_k \tag{10}\] has a uniformly convergent high-frequency sum. This identity retains the constant and low-frequency part of \(b\); no homogeneous reconstruction of a bounded function at frequency zero is needed.

For any fixed smooth annular cutoff \(\rho\), let \(K_j\) be the kernel of \(\mathfrak m(D)\rho(2^{-j}D)\). Rescaling and three integrations by parts on its fixed compact frequency support show \[ \int_\mathbb R|z|\,|K_j(z)|\,\mathrm dz \le C A(\mathfrak m)2^{-j}. \tag{11}\] In fact the rescaled kernel is bounded by \(C A(\mathfrak m)(1+|z|)^{-3}\), which proves the assertion.

First suppose that \(\widehat h\) is smooth and compactly supported away from zero. Put \(h_j=\Delta_jh\); only finitely many \(h_j\) are nonzero. Apply (10) to each of them. There are three interactions.

For the low frequencies of \(b\), put \(B_j=S_{j-4}b\). The supports of both \(h_j\) and \(B_jh_j\) lie in a fixed annulus of radius \(2^j\). A slightly enlarged annular multiplier \(T_j\) therefore satisfies \([B_j,T]\partial h_j=[B_j,T_j]\partial h_j\). The kernel formula and (11) give \[\left\|[B_j,T_j]\partial h_j\right\|_2 \le \mathop{\mathrm{Lip}}(B_j)\int |z||K_j(z)|\,\mathrm dz\, \left\|\partial h_j\right\|_2 \le C A(\mathfrak m)L_b\left\|h_j\right\|_2.\] The outputs remain in fixed annuli of radius \(2^j\). Finite frequency overlap and (5) bound their sum by \(C A(\mathfrak m)L_b\left\|h\right\|_2\).

For \(k\ge j+4\), the output of \([b_k,T]\partial h_j\) lies in a fixed annulus of radius \(2^k\). Plancherel, Bernstein’s inequality, and (9) yield \[\left\|[b_k,T]\partial h_j\right\|_2 \le C A(\mathfrak m)L_b 2^{j-k}\left\|h_j\right\|_2.\] Group by \(k\), use frequency overlap, and apply Young’s inequality on \(\ell^2(\mathbb Z)\) to the summable sequence \(2^{-n}\mathbf 1_{n\ge4}\). This bounds the high–low sum by the same quantity and proves its convergence in \(L^2\).

It remains to consider \(|k-j|\le3\). Fix one such difference and write \(H_k=h_{k+r}\), \(|r|\le3\). The exact identity \[ [b_k,T]\partial H_k =\partial([b_k,T]H_k)-b_k'TH_k+T(b_k'H_k) \tag{12}\] treats the possible low output frequencies. Indeed, \(\partial([b_k,T]H_k)\) has spectrum in a ball of radius \(C2^k\). Its \(\Delta_i\) projection vanishes for \(i>k+c_0\), with \(c_0\) fixed, and otherwise \[\left\|\Delta_i\partial([b_k,T]H_k)\right\|_2 \le C A(\mathfrak m)L_b2^{i-k}\left\|H_k\right\|_2.\] Another application of (5) and Young’s inequality to \(2^{-n}\mathbf 1_{n\ge-c_0}\) bounds the sum of these derivative terms. Thus arbitrarily low output frequencies cause no logarithmic loss.

Choose a fixed \(c_1\) so that \(S_{k+c_1}\) equals one on the spectrum of every product \(b_k'H_k\). For \(g\in L^2\), \[\begin{split} \left|\left\langle\sum_k b_k'H_k,g\right\rangle\right| &\le \left(\sum_k\left\|H_k\right\|_2^2\right)^{1/2} \left(\sum_k\left\|(\Delta_kb')S_{k+c_1}g\right\|_2^2\right)^{1/2}\\ &\le C L_b \left(\sum_k\left\|H_k\right\|_2^2\right)^{1/2}\left\|g\right\|_2, \end{split}\] by 6. Since \(\left\|T\right\|_{2\to2}\le A(\mathfrak m)\), this bounds \(T\sum_k b_k'H_k\). Replacing \(H_k\) by \(TH_k\) in the same calculation bounds \(\sum_k b_k'TH_k\): the frequency supports are unchanged and \(\sum_k\left\|TH_k\right\|_2^2\le A(\mathfrak m)^2 \sum_k\left\|H_k\right\|_2^2\). Equation (12), summed over its seven possible differences, completes the proof on the chosen dense class.

For \(h\in W^{1,2}\), approximate by \((S_N-S_{-N})h\), followed if necessary by smooth compact Fourier approximations away from zero. The approximation converges in \(W^{1,2}\). Since \(b\) is bounded and \(T\) is \(L^2\)-bounded, the defining commutators converge in \(L^2\). The estimate passes to the limit, and then defines a bounded operator on all of \(L^2\). ◻

Proof. Work first with a smooth field and on one fixed \(x\)-fiber. Let \(\varphi_s(y)=y-su(x,y)\). Direct differentiation gives \[C_s^{-1}C_s'=b_s\partial_y, \qquad b_s=-u(x,\cdot)\circ\varphi_s^{-1},\] and hence, since \(T\) commutes with \(\partial_y\), \[ \frac{\,\mathrm d}{\,\mathrm ds}(C_sTC_s^{-1}) =C_s[b_s,T]\partial_y C_s^{-1}. \tag{14}\] The bi-Lipschitz estimates in 3 imply \[\left\|C_s\right\|_{2\to2}+\left\|C_s^{-1}\right\|_{2\to2}\le C, \qquad \mathop{\mathrm{Lip}}(b_s)\le \frac{C_{\rm Lip}}{1-|s|C_{\rm Lip}}\le C.\] Apply 7 inside (14), integrate from \(0\) to \(t\), and multiply by \(C_t\). The estimate is uniform in the fiber, so integration in \(x\) proves (13).

A bounded Lipschitz field is approximated by convolutions with smooth approximate identities, preserving the bounds on its size and Lipschitz seminorm. For each fixed small \(s\), the corresponding composition operators converge strongly on \(L^2\): this follows first for continuous compactly supported inputs by uniform convergence of the maps, and then by density and their uniform operator bounds. Passing to the limit proves the assertion for the original field. ◻

Proof. Apply 5 to the localized band form. It gives a uniform local \(L^2\) bound for \(H_{u,a}^{\epsilon}\widetilde P_w\), where \(\widetilde P_w\) is an enlarged cutoff equal to one on the spectrum of \(P_w\). Since \(P_w f\) retains the same horizontal support, \[\left\|\mathbf 1_{I\times\mathbb R}H_{u,a}^{\epsilon}P_w f\right\|_2 =\left\|\mathbf 1_{I\times\mathbb R}H_{u,a}^{\epsilon} \widetilde P_w(P_w f)\right\|_2 \le C\left\|P_w f\right\|_2.\] Insert this in (16) and let \(\mathcal W\) increase to all dyadic scales. Plancherel and the square partition give \(\left\|\mathbf 1_{I\times\mathbb R}H_{u,a}^{\epsilon}f\right\|_2 \le C\left\|f\right\|_2\), uniformly in \(\epsilon\). The localization already proved above gives the asserted global bound for the unit field. Finally, for a Schwartz input the symmetric principal value exists pointwise; Fatou gives its \(L^2\) bound, and Chebyshev gives the weak estimate in 1. ◻

Proof. The first assertion follows from 3, the uniform \(L^1\) norm of the convolution kernel of \(P_w\), and \[\int_{\tau<|t|<a}\frac{\,\mathrm dt}{|t|} =2\log(a/\tau)\le C(1+L^{0.97}).\] For the second, the fundamental theorem of calculus gives \[\begin{split} &(P_wh)(x-t,y-tu(x,y))-(P_wh)(x-t,y)\\ &\qquad=-t u(x,y)\int_0^1 (\partial_yP_wh)(x-t,y-\theta t u(x,y))\,\mathrm d\theta. \end{split}\] The intermediate maps satisfy 3, and the kernel of \(\partial_yP_w\) has \(L^1\) norm \(O(w^{-1})\). Consequently, \[ \left\|(A_t-S_t^x)P_wh\right\|_p \le C_p |t|w^{-1}\left\|h\right\|_p. \tag{18}\] Integrating over any subinterval of \(0<|t|<w\) against \(\,\mathrm dt/|t|\) gives a constant. Equation (1) handles the horizontal part, uniformly in its inner and outer endpoints. ◻

Proof. For \(r<|t|<R\), the omitted scales below \(r\) have sum bounded by \(C r/|t|^2\), using \(\left|k_\ell(t)\right|\le C\ell/|t|^2\) and a geometric sum. The omitted scales above \(R\) have sum at most \(C/R\). The integrals of these bounds over the middle interval are bounded. For \(|t|<r\), the retained sum is at most \(C\sum_{\ell\ge r}\ell^{-1}\le C/r\); for \(|t|>R\), it is at most \(C R/|t|^2\). These are integrable on their respective regions and prove (20). For the moment bound, use \[\int |t||k_\ell(t)|\,\mathrm dt =\ell\int |s||k(s)|\,\mathrm ds\] and \(\sum_{\ell\le R}\ell\le2R\). Finally, changing an endpoint by a fixed factor changes the spatial kernel only on an annulus having bounded \(\,\mathrm dt/|t|\) measure. ◻

Proof. The amplitude upper bound implies \(2^i\le2Q\) on every nonempty level. If \(|A_i|<e^{-L/4}M_0\), then \(a_i\le2^i e^{-L/8}\), and summing the geometric series in \(i\) gives \(C Qe^{-L/8}\). If \(|A_i|>e^{L/4}M_0\), the \(L^1\) bound gives \[2^i\le2Qe^{-L/4},\qquad a_i^2=4^i|A_i|/M_0\le2Q\,2^i.\] Summing the square-root geometric series over the allowed amplitudes again gives \(C Qe^{-L/8}\). Since \(\log Q=O(L^{0.9})\), both tails are \(O(e^{-cL})\).

The remaining level sets occupy at most \(C(1+L)\) dyadic measure classes. Within a class, the geometric estimate in the proof of 5 bounds the sum of \(a_i\) by a constant times its square sum. Cauchy–Schwarz over the classes and \(\sum_i a_i^2\le4C_0\) give the first bound in (28). The already bounded tails can be included in that bound. ◻

Proof. We give the construction, including the normalization and the separation of the different series forms. At each length \(\ell\), partition the frequency slope \(-\xi/\eta\) smoothly into intervals of width \(c\,d\), \(d=w/\ell\), with centers \(\theta\) on a lattice of spacing a fixed small multiple of \(d\). Take \(c\) so small that on every such frequency piece \(\left|\ell(\xi+\theta\eta)\right|\le c_*/4\). Introduce normalized coordinates \[ X'=\ell(\xi+\theta\eta),\qquad Y'=W(\eta-\eta_j),\qquad U'=(u-\theta)/d. \tag{34}\] Then \[ \ell(\xi+u\eta)=X'+(w\eta)U',\qquad w\eta=w\eta_j+Y'/m. \tag{35}\] The vertical cutoff restricts \(Y'\) to a fixed compact interval and \(w\eta\) to a fixed positive annulus. On the support of \(\psi\), its argument is bounded away from zero. In view of \(\left|X'\right|\le c_*/4\), this forces \(c_b'\le|U'|\le C_b'\), with fixed positive bounds. Enlarge these bounds slightly, retaining separation from zero. Since \(d\le1\) and \(\left|u\right|\le1\), only bounded values of \(\theta\) can contribute.

In the coordinates (34), the product of the annular multiplier, the slope partition, and the vertical cutoff is smooth on a fixed compact coordinate region, with all fixed derivatives bounded uniformly. For example the slope partition is a smooth function of \(-X'/(w\eta)\), and (35) shows that its derivatives are uniform because \(w\eta\) stays away from zero and \(m\ge1\). Choose slightly enlarged smooth plateaus in \(X',Y',U'\), with the \(X'\)-plateau supported in \(|X'|<c_*\), the \(U'\)-plateau supported away from zero, and an additional vertical plateau keeping \(w\eta\) in a positive annulus. Fourier series on a larger fixed coordinate box now give \[\sum_{\nu\in\mathbb Z^3} a_{\nu,\ell,\theta,j} A_{\nu,\ell,\theta,j}(\xi,\eta) B_{\nu,\ell,\theta,j}((u-\theta)/d),\] where \(A_\nu\) is a localized exponential in \((X',Y')\), \(B_\nu\) is a localized exponential in \(U'\), and for every fixed \(N\), \[ \sup_{\ell,\theta,j,w,m}|a_{\nu,\ell,\theta,j}| \le C_N(1+|\nu|)^{-N}. \tag{36}\] This follows by integration by parts on the fixed coordinate box. Multiplication by the plateaus restores the original symbol exactly; their supports give the strict spectral and selector restrictions in 13.

Factor the frequency function \(A_\nu\) into two smooth frequency functions by inserting a plateau equal to one on its support. Both factors retain \(|X'|<c_*\), a vertical interval of length \(C/W\), and a positive annular vertical support. Their inverse Fourier transforms are modulated Schwartz kernels at the scales \(\ell,W\). Their fixed seminorms, and those of the selector factor, grow at most polynomially in \(|\nu|\). The coefficients in (36) absorb these losses uniformly. One can do this simultaneously for every fixed seminorm as follows. Let \(A_\nu^*=\sup_{\ell,\theta,j,w,m}|a_{\nu,\ell,\theta,j}|\) and omit zero values. Extract the common coefficient \(c_\nu=(A_\nu^*)^{1/4}\), multiply each of the two frequency factors by \((A_\nu^*)^{1/4}\), and multiply the selector factor by \(a_{\nu,\ell,\theta,j}/(A_\nu^*)^{3/4}\). The product is unchanged. Each of the three factors now has uniform fixed seminorms, by (36), and \(\sum_\nu c_\nu<\infty\). In particular, no derivative order in this construction grows with \(L\) or a truncation parameter.

For clarity, consider one such factored term \(b(u)A(D)B(D)\). Write \(K_A,K_B\) for its two convolution kernels and \(|R|=\ell W\). At an intermediate convolution center \(z\), put \[\Psi_z(x)=|R|K_A(x-z),\qquad \Phi_z(y)=|R|\overline{K_B(z-y)}.\] Then the contribution to the bilinear form is exactly \[ \frac1{|R|^2}\int_{\mathbb R^2} \langle f,\Phi_z\rangle \langle b(u)\Psi_z,g\rangle\,\mathrm dz. \tag{37}\] The conjugation in \(\Phi_z\) keeps its Fourier support in the same positive-frequency region as that of \(\Psi_z\). Both waves have the carrier \(\eta_j(-\theta,1)\) and satisfy (30)–(31).

Let \(\mathcal Q\) be a fundamental cell of the box lattice, of area \(|R|\). For any integrable function \(F\), \[\int_{\mathbb R^2}F(z)\,\mathrm dz =\int_{\mathcal Q}\sum_{n,k\in\mathbb Z} F\bigl(z+(n\ell,\theta n\ell+kW)\bigr)\,\mathrm dz.\] Applying this identity to (37) and normalizing the average over \(\mathcal Q\) leaves precisely the factor \(|R|^{-1}\) in (33). Use the same translation for every \(j\) at this scale and slope. There are only finitely many lengths and relevant slope labels, so a finite product of these normalized translation measures supplies a common average for their complete sum.

We also need one angular grid per form. The union of the wave slope supports and all selector supports for a label \(\theta\) is contained in \([\theta-Cd,\theta+Cd]\), with \(C\) fixed and independent of \(j\). Choose a sufficiently large fixed dyadic multiple \(H d\). Among three adjacent dyadic grids, this interval is contained in a cell of length \(H d\) with a margin comparable to \(d\). One may see this directly from the three grids \[\bigl\{2^{-k}([0,1)+n+(-1)^k\alpha):n,k\in\mathbb Z\bigr\}, \qquad \alpha\in\{0,1/3,2/3\}:\] they are nested, and their boundary positions at any fixed scale are spaced by one third of that scale. Taking \(H\) large gives the claimed containment and margin. This is the usual adjacent-grid principle; see (Lerner and Nazarov 2019, Theorem 3.1 and Remark 3.2). Assign each label to one suitable grid and split the sum into the three resulting forms. Since \(d=w/\ell\) is dyadic, the chosen cell level depends only on \(\ell\).

At a fixed length and label, the lattice sums converge absolutely in a bilinear pairing of \(L^2\) functions. Indeed, envelope decay and Cauchy–Schwarz give \[\sum_z |R|^{-1}|\langle f,\Phi_z\rangle|^2 \le C\left\|f\right\|_2^2, \qquad \sum_z |R|^{-1}|\langle b_z(u)\Psi_z,g\rangle|^2 \le C\left\|g\right\|_2^2.\] For example, bound each squared pairing by \(C|R|\) times the integral of \(|f|^2\) against a decaying box envelope, and then sum those envelopes over the lattice. Their sum is uniformly bounded. The same argument works with the bounded selector and with vector components, by a further Cauchy–Schwarz inequality in \(j\). Since the length and label sets are finite, the continuous representation is justified by these bounds and the summability in \(\nu\), first on regular inputs and then on \(L^2\). Exhausting each lattice by finite subsets gives the claimed finite packet approximations.

For each fixed series index and angular grid, all scales and labels are kept in one form before its estimate is applied. Different series indices remain separate forms and are summed only using \(\sum_\nu c_\nu<\infty\). In particular, their copies of the same spatial lattice are never combined into a collection with unbounded multiplicity. This completes the representation. ◻

Proof. We apply the assumed bounds to the local field and sets of 4, for which \(\left|u\right|\le1\), \(\mathop{\mathrm{Lip}}(u)\le C_{\rm Lip}\), and \(F,E\subset I\times\mathbb R\) with \(\left|I\right|\le b_0\). Suppose first that \(L=\log\sqrt{|E|/|F|}\ge L_0\). The time and scale reductions give exponentially small normalized errors by (19). The bin error is smaller than any fixed power of \(L^{-1}\), by (26). It remains to estimate the diagonal form in (24).

Apply the amplitude decompositions from 12 to \(F^0,G^0\). The projections defining these vectors preserve horizontal support. After omitting levels of measure zero, their level sets have positive finite measure by (27) and lie in \(I\times\mathbb R\); the normalized slices satisfy the stated indicator bounds. First use finite decompositions. Each pair of slices has a packet representation by 14; its cutoff is \(\widetilde Q_j\), so the slice itself need not have narrow spectrum. Summing the common series coefficients and averaging the translations preserve each assumed model bound. Pairs for which both measures are retained satisfy (29); apply (39) to them. Their total contribution, divided by \(\sqrt{|F||E|}\), is at most \[C L^{-20}(1+L)\le C L^{-19}.\] If either measure is discarded, use (38). The sum of those normalized weights is at most \(C\sqrt{1+L}\,e^{-cL}+Ce^{-2cL}\); the factor \((1+\log(2m))^{C_1}\) is a fixed power of \(L\). These terms are therefore exponentially small. The estimates are uniform for the finite amplitude sums. Their tails converge absolutely by the same bounds, so they apply to the full diagonal form. All errors and the main term are bounded by \(C(1+L)^{-3}\sqrt{|F||E|}\), as required.

If \(L\) ranges over a fixed bounded interval, use \(m=1\). No bin projections or their amplitude losses are needed. The time and scale errors have bounded operator norms, and (38), with the scalar packet representation, bounds the remaining form. The factor \((1+|\log(|F|/|E|)|)^{-3}\) is bounded below by a positive absolute constant on this interval.

Finally suppose that \(L\) is large and negative. Again use \(m=1\), retain the original scalar input and test, and apply (40). It gives a positive power of \(|E|/|F|\). The errors give the power \((|E|/|F|)^{1/4}\) by the fixed choice \(p=4\) in (19). Every fixed positive power of this ratio is bounded by a constant times \((1+|\log(|F|/|E|)|)^{-3}\). This proves the remaining case.

All arguments include an arbitrary positive inner cutoff. If \(\epsilon>w\), the lower retained scale is raised to a dyadic scale comparable to \(\epsilon\), which simply selects a finite subcollection of the allowed scales. If it exceeds \(\tau\), the middle form is empty and the error estimates suffice. If \(w\ge\tau\), the short-time and horizontal bounds already cover the whole remaining integral. None of the bounds used here depends on how many scales lie between the endpoints. ◻

Proof. From the packet support and \(t\in\Omega_s\), \[\xi+t\eta=(\xi+\theta_s\eta)+(t-\theta_s)\eta, \qquad \frac{a_0}{\ell_s}\leq|\xi+t\eta| \leq\frac{a_1}{\ell_s},\] where \(0<a_0<a_1<\infty\) are fixed. Positivity of \(a_0\) uses \(c_*\ll c_1\) and the two fixed bounds on \(w\eta\). For envelopes, replacing slope \(\theta_s\) by \(t\) changes the transverse coordinate over one tile length by at most \(C d_s\ell_s=Cw\leq CW\). Thus every fixed Schwartz envelope remains a uniform envelope in the \(t\)-coordinates.

At one length there are boundedly many labels, since all labels of the tree lie in an interval of length \(O(d_s)\). Within each label the centers lie on a translated box lattice. The inner product of two normalized packets at this length is bounded by a fixed rapidly decreasing function of their normalized center separation. Its row and column sums are bounded. The elementary Schur inequality, obtained by applying \(2|a_sa_q|\leq |a_s|^2+|a_q|^2\) to these row and column sums, proves the fixed-length synthesis bound.

This argument also applies after multiplying packets by \(\mathfrak p_{T,J_0}\). Their centers belong to \(R_T\), their lengths are at most \(l\), and their widths are \(W\). Consequently the polynomial can be absorbed into a fixed larger number of packet decay powers. Multiplication by a polynomial differentiates the Fourier transform and does not enlarge its support.

Choose a fixed integer \(q_0\) such that \(2^{q_0}a_0>4a_1\), and split dyadic lengths according to their exponents modulo \(q_0\). Within each resulting list, the displayed annuli are disjoint and have gaps. Orthogonality in the first \(t\)-coordinate therefore proves (49), also for the polynomially weighted packets.

On one such list, the sum over all lengths above a given endpoint is exactly a smooth low-pass convolution in that coordinate applied to the complete sum: choose the cutoff to equal one on the lower-frequency annuli retained at that endpoint and zero on all higher-frequency annuli. The gap permits dilates of one fixed smooth cutoff. Their absolute kernels are dominated by the one-dimensional maximal operator. An interval of lengths is a difference of two endpoint sums. Apply this observation to the full weighted sum, use its just-proved \(L^2\) bound, and sum over the fixed number of lists. This gives (50).

For Hilbert-valued sums, the fixed-length and orthogonality bounds follow by summing the scalar squared bounds over components. The same low-pass convolution acts on every component, since \(a_0,a_1\) are common. Its norm is pointwise bounded by the positive kernel acting on the Hilbert norm of the full sum. The maximal-function proof is therefore dimension free. Alternatively one can sum the squared scalar maximal estimates. ◻

Proof. First consider positive separated tests, with \(t-\theta_s\geq c_1d_s\). While a test has energy larger than \(\sigma^2|R_T|/2\), choose one whose direction is within \(\varepsilon\) of the infimum of the directions of all such tests. Here \(\varepsilon>0\) is a sufficiently small fixed multiple of \(\min_{s\in\mathcal S}d_s\). Delete all remaining tiles with \(\ell_s\leq l_T\), \(t_T\in\Omega_s\), and \(z_s\in A_0R_T\), including nonseparated tiles. The fixed absolute dilation \(A_0\) is chosen below. Record the positive separated test tiles present at selection as \(\mathcal Q_T\). The procedure terminates because every selection deletes at least one tile. Thereafter perform the negative separated selection, using directions within \(\varepsilon\) of the supremum and reversing the order. At the end, each signed energy is at most \(\sigma^2|R_T|/2\); hence the size is at most \(\sigma\).

We prove the area bound for the positive selection. The \(\mathcal Q_T\) are disjoint and \[ B:=\sum_T\sum_{s\in\mathcal Q_T}|v_s|^2|R_s| \asymp \sigma^2\sum_T|R_T|. \tag{54}\] The lower comparison is the selection threshold and the upper one follows from the initial size bound, which persists on every remaining subcollection. We claim \[ \left\|\sum_T\sum_{s\in\mathcal Q_T}v_s\Phi_s\right\|_2^2 \leq CB. \tag{55}\]

Expand the squared norm. Pairs with lengths in a fixed ratio have total contribution \(O(B)\). Indeed, spectral overlap forces their labels to lie within \(O(d_s)\), so there are boundedly many neighboring labels. Their box shapes are comparable because the transverse discrepancy over either length is \(O(w)\leq O(W)\). Packet decay and the spatial lattices then give the bounded row and column sums for the normalized Gram matrix, exactly as in the fixed-length proof above.

For the other pairs fix the top \(T\) of the longer member \(q\in\mathcal Q_T\), and write \(t=t_T\). Let \(\mathcal A_T\) consist of the selected tiles \(s\) for which \(\ell_q>C_0\ell_s\) and \(\langle\Phi_s,\Phi_q\rangle\ne0\) for at least one such \(q\). Choose the fixed \(C_0\) sufficiently large. Spectral overlap and the common vertical frequency range give \[|\theta_s-\theta_q|\leq Cc_*(d_s+d_q),\qquad |\theta_q-t|\leq C_\Omega d_q.\] Thus, by the choices of \(c_*\) and \(C_0\), \[ |\theta_s-t|\leq \tfrac14 c_1d_s, \qquad t\in\Omega_s. \tag{56}\] The second assertion uses the fixed margin about the wave slope support inside \(\Omega_s\).

We next verify a bounded overlap fact for the boxes of \(\mathcal A_T\) in the coordinates of slope \(t\). These boxes can all be replaced by fixed comparable \(\ell_s\times W\) boxes in those coordinates. Suppose two of the latter overlap, with \(\ell_s\) much smaller than \(\ell_{s'}\). Let \(t_s,t_{s'}\) be the directions of their respective selecting tests. Because \(s\) was in a positive separated test, \[t_s\geq\theta_s+c_1d_s>t+\tfrac12c_1d_s, \qquad |t_{s'}-t|\leq C d_{s'}.\] After separating the lengths by another fixed factor, \(t_{s'}<t+\tfrac14c_1d_s\). If the test selecting \(s'\) had not yet been selected when \(s\) was selected, all its eventual test tiles would still have been present at that time. It would therefore already have been failing, since deleting tiles only decreases each fixed test energy. The near-infimum rule, with \(\varepsilon\ll c_1\min d_s\), therefore forces the test of \(s'\) to have been selected first.

Its length is at least \(\ell_{s'}\). Overlap of the comparable \(t\)-boxes bounds the horizontal center separation by \(C\ell_{s'}\). In the coordinates of \(t_{s'}\), the transverse center separation is \(O(W+w)=O(W)\): use \(|t_{s'}-t|\ell_{s'}=O(w)\), or equivalently compare each original tile over its own length. Also \(t_{s'}\in\Omega_s\), by the margin in (56). For an absolute \(A_0\) this places \(s\) in the enlarged deletion tree of \(s'\), a contradiction. At comparable lengths, bounded label and lattice multiplicity give bounded overlap directly. Splitting lengths into a fixed number of lists proves the asserted overlap bound for all of \(\mathcal A_T\). The same reasoning applies to any fixed enlargement of these boxes by increasing \(A_0\) once.

Let \(B_s^*\) be sufficiently rapidly decreasing uniform tile envelopes in the \(t\)-coordinates, and put \(D_T=\mathfrak p_{T,J_0}^{-1}\), with \(J_0\) a fixed large integer. Then \[ \left\|D_T\sum_{s\in\mathcal A_T}B_s^*\right\|_2 \leq C|R_T|^{1/2}. \tag{57}\] Here centers of \(\mathcal A_T\) need not belong to \(R_T\). To prove the estimate, test against \(\left\|h\right\|_2\leq1\). Since \(\ell_s\leq l_T\), the top weight can be passed across a tile envelope by spending fixed decay powers. Decomposition into dilated tile boxes, followed by the overlap just proved, gives \[\begin{align*} \sum_s\int |h|D_T B_s^* &\leq C\sum_s |R_s| \inf_{R_s^t}D_T\inf_{R_s^t}\mathcal M_t h\\ &\leq C\int D_T\mathcal M_t h \leq C\left\|D_T\right\|_2\left\|h\right\|_2 \leq C|R_T|^{1/2}. \end{align*}\] The notation \(R_s^t\) denotes the fixed comparable \(t\)-box; changing it by an absolute dilation is harmless. This proves (57).

For each \(s\in\mathcal A_T\), the sum of its cross terms with \(\mathcal Q_T\) can include all \(q\in\mathcal Q_T\) with \(\ell_q>C_0\ell_s\), since the added inner products are zero. Therefore its absolute value is bounded using \[U_T(z)=\sup_{r>0} \left|\sum_{\substack{q\in\mathcal Q_T\\\ell_q>r}} v_q\Phi_q(z)\right|.\] The coefficient bound \(|v_s|\leq C\sigma\), (57), and 17 show that all such cross terms for this top have sum at most \[C\sigma\int\left(\sum_{s\in\mathcal A_T}B_s^*\right)U_T \leq C\sigma |R_T|^{1/2}\left\|D_T^{-1}U_T\right\|_2 \leq C\sigma^2|R_T|.\] Summing and using (54) proves (55). Pairing its synthesis with \(f\) gives \(B\leq\left\|f\right\|_2(CB)^{1/2}\), hence \(B\leq C\left\|f\right\|_2^2\). The negative selection has the identical proof with order reversed.

Finally, partition each enlarged deletion box \(A_0R_T\) into a fixed number of translated ordinary top boxes of length \(l_T\) and width \(W\), assigning tiles by their centers. Their direction and length conditions are unchanged, and their total area is \(O(|R_T|)\). This proves the last assertion and (53). ◻

Proof. For each tile to be removed select one box in (42) witnessing an average greater than \(\eta\). Since \[\int_{\mathbb R^2\setminus KR}B_R\leq C K^{2-D_0}|R|,\] choosing the fixed \(D_0\) sufficiently large and then \(K=K_\eta\leq C(1+\eta^{-1/100})\) retains at least \(\eta|R|/2\) of the weighted test inside \(KR\). Its unweighted tested set has measure at least \(\eta|R|/2\), because \(B_R\leq1\).

Process these finitely many witnesses in decreasing length order. Keep a witness unless both its \(K\)-dilated spatial box and its dyadic direction interval intersect the corresponding objects of an earlier keeper. Assign it to any such keeper when it is rejected. The tested sets of the keepers are disjoint: a common tested point would force both intersections and preclude keeping the later one. This proves (58) and (59).

Let a rejected witness have length \(l'\), direction \(t'\), and original associated tile \(s\), and let its keeper have length \(l_T\geq l'\) and direction \(t_T\). Dyadic nesting gives \[\Omega(t_T,l_T)\subset\Omega(t',l')\subset\Omega_s, \qquad |t_T-t'|\leq Cw/l'.\] Choose a point common to the two dilated spatial boxes. Moving from it to the rejected box center, or to \(z_s\), costs horizontally \(O(Kl')\) and transversely in the \(t_T\)-coordinates \(O(KW+K l'|t_T-t'|)=O(KW)\). This proves (60). A keeper’s own associated tiles satisfy the same conclusion.

Partition \(CKR_T\) into \(O(K^2)\) translated boxes of length \(l_T\), width \(W\), and slope \(t_T\), assigning tiles by their centers. The length and angular requirements for a test tree persist. Their total area is at most \(C(1+\eta^{-1/50})\eta^{-1}|E|\), which is bounded by (61) after changing the absolute constant. ◻

Proof. The assertion is immediate for an empty tree, so assume the tree is nonempty. Work in centered coordinates \((x,y-tx)\) of the top, so its coordinate extents are \(l\times W\). For clarity write the transverse coordinate as \(y'\), with top center \((0,0)\). All tile centers satisfy \(|x_s|\leq l/2\) and \(|y_s'|\leq W/2\). The packet envelopes are uniform in these coordinates.

Partition the horizontal line into maximal dyadic intervals \(I\) for which there is no tile with \(x_s\in3I\) and \(\ell_s\leq |I|\). The tiles are finite and have positive minimum length, so these intervals cover the line up to endpoints and are pairwise disjoint. Each parent fails the condition, furnishing a tile \(q_I\) with \[ \ell_{q_I}\leq2|I|,\qquad \mathop{\mathrm{dist}}(x_{q_I},I)\leq C|I|. \tag{63}\] Fix a sufficiently large absolute dyadic constant \(C_{\mathrm W}\). We treat lengths \(\ell_s\leq C_{\mathrm W}|I|\) and longer lengths separately on the strip above \(I\).

For a short tile, density tested at its own box and wave decay give, for every fixed large \(N_0\), \[ \int_{E\cap\{x\in I\}} |v_s b_s(u)\Psi_s| \leq C\sigma\eta|R_s| \left(1+\frac{\mathop{\mathrm{dist}}(x_s,I)}{\ell_s}\right)^{-N_0}. \tag{64}\] Indeed the selector support lies in \(\Omega_s\), the density definition includes \(R_s\) with direction \(\theta_s\), and a fixed number of additional wave-decay powers extracts the last factor while retaining the density weight.

Here is an explicit summation of (64). At a fixed length there are \(O(1)\) labels, and each horizontal column contains \(O(1)\) centers within the transverse top strip. If \(\ell_s\leq r:=|I|\), the defining property of \(I\) gives \(\mathop{\mathrm{dist}}(x_s,I)\geq r\). Summing over the one-dimensional center lattice then bounds its contribution, divided by \(C\sigma\eta W\), by \(C r(\ell_s/r)^{N_0-1}\). For \(r<\ell_s\leq C_{\mathrm W}r\), the same lattice sum is \(O(r)\) after multiplication by \(\ell_s\), and there are only a fixed number of such lengths. Thus the total is \(O(\sigma\eta Wr)\). All intervals with \(r\leq l\) lie within a fixed enlargement of the top’s horizontal interval, by (63); their disjoint lengths sum to \(O(l)\). If \(r>l\), all tile lengths are at most \(l\), all centers are at distance at least \(r\) from \(I\), and their restriction \(|x_s|\leq l/2\) improves the preceding sum to \(C\sigma\eta Wl(l/r)^{N_0-1}\). For each such dyadic \(r\), only \(O(1)\) intervals can occur, again by the parent witness. Summing proves a total short-tile contribution \(C\sigma\eta lW\).

Now let \(I\) admit longer tiles. Choose a dyadic \(l'\asymp |I|\), greater than \(\ell_{q_I}\) and small enough that \(l'<\ell_s\) for every longer tile. Increasing \(C_{\mathrm W}\) once ensures this. It also ensures \(l'\leq\tau\). By (63), a box of slope \(t\), length \(l'\), and width \(W\) can contain \(z_{q_I}\) and the portion \(I\times[-W/2,W/2]\), with a fixed enlargement absorbed into the choice of \(l'\). Dyadic nesting puts every longer selector support inside \(\Omega(t,l')\). Consequently the original longer-tile sum vanishes outside \(A_I=\{u\in\Omega(t,l')\}\). We perform all following freezing and error decompositions after multiplying that sum by \(\mathbf 1_{A_I}\), and retain this indicator on both the main term and the error. Individually the frozen terms need not vanish outside \(A_I\). For the cells \[Q_{I,k}=I\times[(k-\tfrac12)W,(k+\tfrac12)W], \qquad k\in\mathbb Z,\] the density test at \(q_I\) therefore yields \[ \mu_{I,k}:= |Q_{I,k}\cap E\cap\{u\in\Omega(t,l')\}| \leq C\eta(1+|k|)^{D_1}|I|W, \tag{65}\] where the fixed \(D_1\) can be taken to be \(D_0\). The density weight on the cell is bounded below by a constant times \((1+|k|)^{-D_0}\).

Put \(a=|u-t|\) at the point of evaluation. For nonseparated tiles, selector support implies \(a\asymp d_s\), so only a fixed number of dyadic lengths occur. Their pointwise sum is bounded by \(C\sigma(1+|k|)^{-N_1}\) on \(Q_{I,k}\), for any fixed \(N_1\). For separated tiles replace \(b_s(u)\) by \(\mathbf 1_{\{d_s\geq a\}}b_s(t)\). If \(d_s\geq a\), the error is at most \(Ca/d_s\), using the derivative bound on \(b_s\). If \(d_s<a\), the actual selector vanishes unless \(d_s\geq a/C\), since \(|t-\theta_s|\leq C_\Omega d_s\). There are only a fixed number of these remaining scales. Summing the geometric series in \(a/d_s\), and using lattice overlap at each scale, bounds the entire error by \(C\sigma(1+|k|)^{-N_1}\). This argument also covers \(a=0\), when the separated error is zero.

The resulting main sum, at each point, is an interval sum of the separated packets with coefficients \(v_sb_s(t)\): its lengths obey \(C_{\mathrm W}|I|<\ell_s\leq w/a\). Let \(M_{I,k}\) be the supremum on \(Q_{I,k}\) of the absolute values of all such interval sums with arbitrary upper endpoint. We claim that, for every fixed sufficiently large \(N_2\), \[ \sum_{I,k} M_{I,k}^2(1+|k|)^{N_2}|I|W \leq C\sigma^2|R_T|, \tag{66}\] where only the \(I\) admitting longer tiles occur. To see this, let \(U\) be the maximal interval sum over all separated lengths of the tree. For each fixed \(I\), every interval sum under consideration has horizontal \(t\)-frequency at most \(C/|I|\) and vertical frequency, after the common carrier is removed, at most \(C/W\). The same is true after multiplication by any top polynomial. Apply (46) to the weighted sums and dominate their absolute values by \(\mathfrak p_{T,J_0}U\). Since the cells are disjoint, the squared cell suprema are bounded in sum by \(C\left\|\mathcal M_t (\mathfrak p_{T,J_0}U)\right\|_2^2\). For all the present cells \(|x|\leq Cl\), and \(\mathfrak p_{T,J_0}\asymp (1+|k|)^{2J_0}\) there. Taking \(4J_0\geq N_2\), (45) and 17 prove (66).

The intervals admitting longer tiles have total length \(O(l)\), by their parent witnesses and \(|I|<l/C_{\mathrm W}\). Choose \(N_2>2D_1+2\). Factoring the full density bound before Cauchy–Schwarz, \[\begin{align*} \sum_{I,k}M_{I,k}\mu_{I,k} &\leq C\eta \left(\sum_{I,k}M_{I,k}^2 (1+|k|)^{N_2}|I|W\right)^{1/2}\\ &\hspace{1.2cm}\cdot \left(\sum_{I,k}(1+|k|)^{2D_1-N_2}|I|W\right)^{1/2} \leq C\sigma\eta|R_T|. \end{align*}\] The previously bounded pointwise errors have the same total bound by (65), taking \(N_1>D_1+2\). Adding the short contribution proves (62). ◻

Proof of 16. We first prove the scalar part of (43). If \(\left\|f\right\|_2=0\) or \(|E|=0\), the form is zero; assume otherwise. Process thresholds \(\eta=2^{-n}\) in decreasing order, starting high enough that no density or size condition fails. At each step first remove densities exceeding \(\eta\) by 19, then lower size to \[\sigma_\eta=\sqrt\eta\, \left\|f\right\|_2/|E|^{1/2}\] by 18. At the start of the step, the remaining collection has density at most \(2\eta\), as well as the original bound \(\gamma\), and size at most \(\sqrt2\sigma_\eta\). These bounds persist under all removals within the step.

The density removals have total ordinary top area at most \(C(1+\eta^{-1/20})\eta^{-1}|E|\), and the size removals at most \(C\sigma_\eta^{-2}\left\|f\right\|_2^2 =C\eta^{-1}|E|\). Applying 20 to every resulting ordinary tree and summing bounds the form by \[ C\left\|f\right\|_2|E|^{1/2} \sum_{\eta\ \text{dyadic}} \min(\gamma,2\eta)\sqrt\eta\, (1+\eta^{-1/20})\eta^{-1}. \tag{67}\] There is no limiting remainder to estimate for a finite collection. Every positive-density tile is eventually removed. If \(D_E(s)=0\), testing at \(R_s\) shows that \(b_s(u)g=0\) almost everywhere, since the weight is strictly positive; such a term is zero. Zero-size terms are zero as well.

For \(0<\gamma\leq1\), the terms with \(\eta\leq\gamma\) sum to \(C\gamma^{9/20}\), and the terms with \(\eta\geq\gamma\) do also: the relevant powers are \(\eta^{1/2},\eta^{9/20}\) below the crossover and \(\gamma\eta^{-1/2}, \gamma\eta^{-11/20}\) above it. For bounded larger \(\gamma\) the sum is uniformly bounded. This proves the asserted scalar \(\gamma^{1/4}\) estimate, with room in its exponent.

For (44), put \(r=|E|/|F|<1\). In addition to the threshold size estimate, every tree has size at most a fixed constant by (52). Its actual size at step \(\eta\) is therefore at most \(C\min(1,\sqrt{\eta/r})\). No change in the removal algorithm is necessary. With the universal density bound, the sum of the tree contributions at that step is at most \[C|E|\min(1,\sqrt{\eta/r}) \min(1,\eta)\eta^{-1} (1+\eta^{-1/20}).\] For \(r\leq\eta\leq1\), this sums to \(C|E|r^{-1/20}\); the scales \(\eta>1\) cost \(C|E|\). For \(\eta\leq r\), the sum is at most \[C|E|r^{-1/2} \sum_{\eta\leq r} (\eta^{1/2}+\eta^{9/20}) \leq C|E|r^{-1/20} =C\sqrt{|F||E|}\,r^{9/20}.\] This proves the reverse-disparity claim, including the density-area inflation from the selection.

Finally consider \(\mathcal H=\mathbb C^J\). For a scalar measurable \(h\) supported in \(E\), write \[\left\|h\right\|_{2,1}^{\#} =\int_0^\infty |\{|h|>a\}|^{1/2}\,\,\mathrm da.\] Layer-cake decomposition, retaining the complex phase of \(h\), extends the scalar estimate to \[|\Lambda_j(f_j,g_j)| \leq C\gamma^{1/4}\left\|f_j\right\|_2 \left\|g_j\right\|_{2,1}^{\#}.\] Every sliced support lies in \(E\), so its density still has upper bound \(\gamma\). As \(|g_j|\leq1\), split the last integral at \(a_0=(2+m)^{-2}\). The portion below \(a_0\) is at most \(a_0|E|^{1/2}\); for the rest, Cauchy–Schwarz gives \[\int_{a_0}^1|\{|g_j|>a\}|^{1/2}\,\,\mathrm da \leq \left(\int_{a_0}^1 a|\{|g_j|>a\}|\,\,\mathrm da\right)^{1/2} \left(\int_{a_0}^1\frac{\,\mathrm da}{a}\right)^{1/2} \leq C\sqrt{1+\log(2m)}\left\|g_j\right\|_2.\] Since \(J\leq C_{\rm bin}m\) and \(\sum_j\left\|g_j\right\|_2^2\leq|E|\), \[\sum_j(\left\|g_j\right\|_{2,1}^{\#})^2 \leq C(1+\log(2m))|E|.\] Cauchy–Schwarz in \(j\) proves (43), even with a square root of this logarithmic factor. All arguments used only the stated packet hypotheses, which persist on every subcollection. Every choice of decay or derivative order above is fixed before the parameters and the finite tile collection are chosen. This completes the proof of 16. ◻

Proof. If \(S=0\), the assertion is immediate, so assume \(S>0\). We first state the maximal-function input in the form used here. For a Lipschitz unit vector field \(v_0\), let \(M_{v_0,\delta}\) be the supremum of averages over Euclidean rectangles containing the evaluation point, with long side at most \(c_{\mathrm{LL}}/\mathop{\mathrm{Lip}}(v_0)\), for which the field belongs to the rectangle’s direction uncertainty arc on at least a \(\delta\) fraction of the rectangle. The arc is centered at the long-side direction and has total length equal to the short-to-long side ratio, as in 2. 2, namely (M. Lacey and Li 2006, Theorem 1.4), gives \[ \left\|M_{v_0,\delta}a\right\|_{2,\infty} \le C\delta^{-1/2}\left\|a\right\|_2, \qquad 0<\delta<1. \tag{79}\] The same estimate holds if the admissible rectangles are required to satisfy a smaller fixed uncertainty interval, since this restricts the family of rectangles. When \(\mathop{\mathrm{Lip}}(v_0)=0\), the length restriction is understood as absent. We apply this result to \(v_0=(1,u)/\sqrt{1+u^2}\), whose Lipschitz constant is bounded by an absolute constant. We also use the ordinary uncentered strong maximal operator \(M_{\mathrm{str}}\) over axis-parallel rectangles, which is bounded on \(L^2(\mathbb R^2)\).

Let \(Q\subset\mathcal D\) have finite measure, let \(D\ge D_*\), and put \[a_T(D)=\frac{|Q\cap DR_T|}{|DR_T|} =\frac{|Q\cap DR_T|}{D^2|R_T|}.\] Here and below dilation is about the box center in its slope coordinates. We prove, for \(0<\beta\le1\), that \[ \sum_{T:a_T(D)>\beta}|R_T| \le \min\left(S,\, C L^{2C_{\mathrm{as}}}D^2\beta^{-2}|Q|\right). \tag{80}\]

Fix a top occurring on the left-hand side. Set \[J_T=I\cap[x_T-Dl_T/2,x_T+Dl_T/2],\qquad a=|J_T|, \qquad b=C_{\mathrm{w}}DW,\] where \(C_{\mathrm{w}}\ge1\) is a fixed absolute constant. The witness set is contained in \(DR_T\cap\mathcal D\), so \(a>0\). The slope parallelogram \[\widetilde R_T=\{(x,y):x\in J_T,\quad |(y-y_T)-s_T(x-x_T)|\le b/2\}\] contains both \(E_T\) and \(Q\cap DR_T\). It satisfies \[ a\le\min(|I|,Dl_T),\qquad ab\le C_{\mathrm{w}}D^2|R_T|,\qquad \frac ba\ge C_{\mathrm{w}}\frac W{l_T} =C_{\mathrm{w}}m d_T. \tag{81}\] Thus the possibly very long horizontal dilation has been clipped before any Euclidean enclosing rectangle is constructed.

Suppose first that \(b/a\le\varepsilon_0\), for a fixed small absolute \(\varepsilon_0\). Projection onto the orthonormal directions \[e_T=\frac{(1,s_T)}{\sqrt{1+s_T^2}},\qquad e_T^\perp=\frac{(-s_T,1)}{\sqrt{1+s_T^2}}\] encloses \(\widetilde R_T\) in a Euclidean rectangle \(U_T\) whose side lengths are \[L_T=\sqrt{1+s_T^2}\,a +\frac{|s_T|}{\sqrt{1+s_T^2}}b, \qquad B_T=\frac{b}{\sqrt{1+s_T^2}}.\] The slopes are absolutely bounded. Consequently \(L_T\asymp a\), \(B_T\asymp b\), \(|U_T|\le C D^2|R_T|\), and \[L_T\le C|I|,\qquad \frac{B_T}{L_T}\ge c m d_T.\] On the witness set, normalization of the direction gives \(\left\|v_0-e_T\right\|\le C d_T\). Since \(m\to\infty\), this error is smaller than any required fixed small multiple of \(B_T/L_T\), for sufficiently large \(L\). The fixed widening constant \(C_{\mathrm{w}}\) can absorb the absolute aperture constants as well. Choose the horizontal localization constant so that \(C|I|\) meets the length restriction in (79). The rectangle \(U_T\) is then admissible with the common popularity parameter \[\delta_D=c L^{-C_{\mathrm{as}}}D^{-2},\] where \(c>0\) is sufficiently small and absolute. Indeed, \(|E_T|\ge L^{-C_{\mathrm{as}}}|R_T|\), and \(|U_T|\le CD^2|R_T|\). Moreover, \[\frac1{|U_T|}\int_{U_T}\mathbf 1_Q \ge\frac{|Q\cap DR_T|}{CD^2|R_T|}>c\beta.\] Every point of \(E_T\) therefore belongs to \(\{M_{v_0,\delta_D}\mathbf 1_Q>c\beta\}\).

If \(b/a>\varepsilon_0\), enclose \(\widetilde R_T\) in the axis-parallel rectangle with horizontal side \(J_T\) and vertical side of length \(b+|s_T|a\). Its area is at most \(C ab\le CD^2|R_T|\). It contains \(E_T\) and \(Q\cap DR_T\), so every point of \(E_T\) belongs to \(\{M_{\mathrm{str}}\mathbf 1_Q>c\beta\}\). This argument also covers boxes whose transverse side exceeds the horizontal side.

The sets \(E_T\) are pairwise disjoint. Summing their lower measure bounds and using the two maximal estimates yields \[\begin{align*} \sum_{T:a_T(D)>\beta}|R_T| &\le L^{C_{\mathrm{as}}} \left|\bigcup_{T:a_T(D)>\beta}E_T\right|\\ &\le C L^{C_{\mathrm{as}}} (\delta_D^{-1}+1)\beta^{-2}|Q|\\ &\le C L^{2C_{\mathrm{as}}}D^2\beta^{-2}|Q|. \end{align*}\] The bound by \(S\) is immediate, proving (80).

For positive numbers \(S_0,A_0\), direct integration, split at \(\min(1,\sqrt{A_0/S_0})\), gives \[\int_0^1\min(S_0,A_0\beta^{-2})\,\mathrm d\beta \le2\sqrt{S_0A_0}.\] Layer cake in (80) therefore gives \[\begin{align*} \sum_T|R_T|a_T(D) &\le C L^{C_{\mathrm{as}}}D\sqrt{S|Q|}, \tag{82}\\ \sum_T|Q\cap DR_T| &\le C L^{C_{\mathrm{as}}}D^3\sqrt{S|Q|}. \tag{83}\end{align*}\] In the second line the additional factor \(D^2\) is the dilation-area ratio.

Set \(D_k=2^kD_*\). The explicit formula for \(h_T\) implies the pointwise bound \[h_T(z)\le C_{D_2}\sum_{k\ge0}2^{-kD_2}\mathbf 1_{D_kR_T}(z).\] To verify it outside \(D_*R_T\), choose the least \(k\ge1\) such that \(z\in D_kR_T\). Then \(\mathop{\mathrm{dist}}_T(z)>2^{k-2}D_*\), while \(D_*\ge2K_0\), so \(h_T(z)\le C_{D_2}2^{-kD_2}\). Inside \(D_*R_T\), use \(h_T\le C_3\). Applying (83), \[ \int_Q n \le C L^{C_{\mathrm{as}}}D_*^3\sqrt{S|Q|} \sum_{k\ge0}2^{-k(D_2-3)} \le C L^{C_{\mathrm{as}}}D_*^3\sqrt{S|Q|}. \tag{84}\] The sum converges because \(D_2>4\). Finally take \(Q=\{z\in\mathcal D:n(z)>\lambda\}\). This set has finite measure, since the collection is finite and \(D_2>2\). The inequality \(\lambda|Q|\le\int_Q n\) proves the asserted weak \(L^2\) bound. All exponents in this argument are fixed: in particular the dilation cost before the final summation is exactly \(D^3\), independently of the number of tops and angular scales. ◻

Proof. For the Fourier convention \(e^{ix\xi}\), the transform of \(\operatorname{sinc}(x)\) is supported in \([-1,1]\). Taking its \(r_L\)-th power convolves that transform \(r_L\) times, so the transform of \(\operatorname{sinc}(X/L^A)^{r_L}\) is supported in \([-r_L/L^A,r_L/L^A]\). The product of the two one-dimensional functions has this support bound in each normalized frequency coordinate. The affine coordinate change (75) sends these coordinates to \(l_T(\xi+s_T\eta)\) and \(W\eta\), proving (88). The center translation contributes only a Fourier phase. The calculation can equivalently be performed first for compactly supported Fourier distributions and then their convolutions; the resulting mask is integrable for \(r_L>1\).

Write \(d=\mathop{\mathrm{dist}}_T(z)\). Since at least one of \(|X_T(z)|,|Y_T(z)|\) is at least \(d/2\), the elementary inequality \(|\operatorname{sinc}(a)|\le\min(1,|a|^{-1})\) gives, for \(d>0\), \[ |q_T(z)|\le\min\left(1,\left(\frac{2L^A}{d}\right)^{r_L}\right). \tag{92}\] If \(d>K_0=L^{A+2}\), this is at most \((2L^{-2})^{r_L}\le\exp(-L\log L)\), proving (89). To include arbitrary distances in the weighted bound, write \(d=tK_0\), with \(t>1\). Once \(r_L\ge D_2\), \[|q_T(z)|h_T(z)^{-1} \le 2^{-D_2}(1+t)^{D_2} (2L^{-2})^{r_L}t^{-r_L} \le (2L^{-2})^{r_L}.\] For \(d\le K_0\), use \(q_T\le1\) and \(h_T\ge1\). This proves (90), including the region arbitrarily far from the top.

Finally, if \(d\le L^{A/2}\), both arguments of the sinc functions have absolute value at most \(L^{-A/2}\). They lie in \([0,1]\) after evaluation by sinc, and \[0\le1- \operatorname{sinc}(X_T/L^A) \operatorname{sinc}(Y_T/L^A) \le C\frac{X_T^2+Y_T^2}{L^{2A}} \le C L^{-A}.\] The inequality \(1-a^{r_L}\le r_L(1-a)\), for \(0\le a\le1\), proves (91). Since \(r_L/L^A=O(L^{1-A})\), the bandwidth assertion follows from (88) and addition of Fourier supports under multiplication. ◻

Proof. Minkowski’s inequality bounds the norm of an analyzing coefficient by the integral of its scalar analyzing envelope times the largest modulus of its multiplier among the components, divided by \(\ell W\); the factor \(\left\|g\right\|_{\mathcal H}\) can be replaced by one. At a fixed source point the support condition for \(b_{s,j}(u)\) allows only a bounded number of labels at this scale. This remains true after taking the maximum over components, since the support condition is the common condition on the label slope at precision \(d\).

Choose the available analyzing decay order \(N\geq2n\), with \(n,n',N\) all fixed. For one label, the lattice gives the unequal-exponent bound \[\sum_{s\text{ in that lattice}} \frac{\mathcal B_{s,N}^{t}(z')\mathcal B_{s,n}^{t}(z)}{\ell W} \leq C_{n,N,n'} k_{\ell,n'}(x-x') k_{W,n'}(Y_t(z)-Y_t(z')).\] Indeed, in each normalized coordinate, \(1+\left|a-b\right|^2\leq2(1+\left|a-c\right|^2)(1+\left|b-c\right|^2)\) extracts the relative kernel with exponent \(n'\) from the two endpoint envelopes. The remaining exponents are \(N-n'\) at the source and \(n-n'\) at the output. Drop the source factor, which is at most one, and sum the output envelope. Its lattice sum is uniformly bounded since \(n-n'>1/2\): first sum the vertical centers in each horizontal row, then the horizontal rows. The transverse translations of the rows do not affect these one-dimensional bounds. The comparison with the reference coordinates already gives the required analyzing envelope. Summing the bounded number of labels active at the source and integrating proves (96).

For the error, if \(\left|u-t\right|\leq cd\) with \(c\) sufficiently small, the replacement indicator is one and smoothness gives \(\left|b_{s,j}(u)-b_{s,j}(t)\right|\leq C\left|u-t\right|/d\). The union of labels on which either multiplier is nonzero is bounded. Outside this region, any nonzero original multiplier requires \(\left|u-t\right|\leq C'B^4d\) by (93), whereas a nonzero replacement requires \(\left|u-t\right|\leq d\). The multipliers are bounded there. Again, only boundedly many labels are active at the source, counting both families. Applying the same lattice calculation with the common factor (97) proves (98). ◻

Proof. Let \(\mathcal S_Q\) be the finite set in the sum and dualize its weighted coefficient norm against vectors \(c_s\) satisfying \(\sum_{s\in\mathcal S_Q}\left\|c_s\right\|^2\left|R_s\right|=1\). Using (95), the dual pairing is bounded by the integral of the norm of the synthesis \[\sum_{s\in\mathcal S_Q} \mathbf 1_{\{\left|u(z)-t\right|\leq d_s\}} \big(c_{s,j}b_{s,j}(t)\Psi_{s,j}(z)\big)_j.\] At each \(z\), the indicators choose a partial interval of the dyadic lengths: \(\left|u(z)-t\right|\leq w/\ell_s\) is monotone in \(\ell_s\). The norm of this synthesis is therefore at most its maximal partial length sum. All its nonzero component frequencies satisfy (99).

Write \(r=\mathrm{length}(Q)>0\). Apply the proof of the Hilbert-valued synthesis estimate in 17, including a fixed polynomial growth weight \(P_Q\) about \(Q\). This proof applies to the present reference box even if \(r\) is not dyadic or \(r>\tau\). For the weighted packet bounds it uses only center containment in \(Q\) and \(\ell_s\leq r\); multiplication by \(P_Q\) preserves frequency support and gives tile bounds with slower fixed decay. The annular separation and maximal partial sums depend on the dyadic tile lengths, not on the reference length \(r\). For example, a fixed sufficiently large integer \(q\) gives the admissible polynomial \[P_Q(z)=\left(1+\frac{(x-x_Q)^2}{r^2}\right)^q \left(1+\frac{(Y_t(z)-Y_t(z_Q))^2}{W^2}\right)^q, \qquad \int P_Q^{-2}\leq C_q rW=C_q\left|Q\right|.\] The main label count is bounded, so the same synthesis proof yields \[\int\left\|\text{selected synthesis}\right\| \leq \left(\int P_Q^{-2}\right)^{1/2} \left\|P_Q\,\text{maximal synthesis}\right\|_2 \leq C\left|Q\right|^{1/2}.\] Taking the supremum over \(c_s\) proves (102). Deleting packets preserves both the synthesis hypotheses and this proof. ◻

Proof. We give the smooth-convolution versus dyadic-expectation comparison explicitly; compare (Jones et al. 2008, Lemma 3.2 and Section 4). Split the integers into a fixed number \(q\) of residue classes, large enough that the annuli in one class have gaps of ratio greater than four. The variation of the sum is at most the sum of the variations of these classes, so it suffices to treat one class. Its supports are disjoint, and \(F=\sum_kF_k\) satisfies \(\left\|F\right\|_2^2=\sum_k\left\|F_k\right\|_2^2\).

Choose a smooth function \(\chi\) equal to one on \([-1,1]\) and supported in \([-2,2]\), and let \(P_i\) have multiplier \(\chi(2^{-i}\nu)\). At a sequence of integer indices lying in the annular gaps, \(P_iF\) is exactly the partial sum of all the preceding bands. Include an index below the first band, where this partial sum is zero.

For each integer \(i\), let \(\mathcal F_i\) be the sigma algebra generated by the intervals \([a2^{-i},(a+1)2^{-i})\), \(a\in\mathbb Z\), and let \(E_i\) be averaging on its atoms. The sigma algebras increase with \(i\) and \((E_iF)_i\) is a martingale. We prove the comparison bound \[ \sum_i\left\|(P_i-E_i)F\right\|_2^2\leq C\left\|F\right\|_2^2. \tag{105}\] Take a smooth dyadic partition \(F=\sum_jf_j\) with \(f_j\) supported in a fixed annulus of size \(2^j\) and \(\sum_j\left\|f_j\right\|_2^2\leq C\left\|F\right\|_2^2\). If \(j\leq i-C\), then \(P_if_j=f_j\), and Poincaré’s inequality on each interval of length \(2^{-i}\), followed by Bernstein’s inequality, gives \[\left\|(P_i-E_i)f_j\right\|_2\leq C2^{j-i}\left\|f_j\right\|_2.\] If \(j\geq i+C\), then \(P_if_j=0\). Define its band primitive \(A_j\) by \(\widehat A_j(\nu)=\widehat f_j(\nu)/(i\nu)\). Thus \(A_j'=f_j\) and \(\left\|A_j\right\|_2\leq C2^{-j}\left\|f_j\right\|_2\). With \(h=2^{-i}\), averaging on a cell gives \(E_if_j=(A_j((a+1)h)-A_j(ah))/h\) there. The sampling estimate \[ \sum_{a\in\mathbb Z}\left\|A_j(ah)\right\|_{\mathcal H}^{2} \leq C2^j\left\|A_j\right\|_2^2,\qquad h\geq c2^{-j}, \tag{106}\] follows directly from smooth reproduction. Namely, take a Schwartz kernel \(K_j\) whose Fourier transform equals one on the spectrum of \(A_j\) and which satisfies \(\left|K_j(x)\right|\leq C_N2^j(1+2^j\left|x\right|)^{-N}\). Cauchy–Schwarz in \(A_j=K_j*A_j\) gives \(\left\|A_j(ah)\right\|^2\leq C\int\left|K_j(ah-x)\right|\left\|A_j(x)\right\|^2\,\mathrm dx\); summing uses \(\sup_x\sum_a\left|K_j(ah-x)\right|\leq C2^j\). Consequently \[\left\|E_if_j\right\|_2^2 \leq C h^{-1}\sum_a\left\|A_j(ah)\right\|^2 \leq C2^{i-j}\left\|f_j\right\|_2^2.\] For \(\left|i-j\right|\leq C\) use the bounded \(L^2\) norms of \(P_i,E_i\). The resulting bounds have the form \(\left\|(P_i-E_i)f_j\right\|_2\leq a_{i-j}\left\|f_j\right\|_2\) with a fixed \(\ell^1(\mathbb Z)\) sequence \(a\). The triangle inequality followed by the \(\ell^1*\ell^2\) convolution inequality proves (105).

For any sequence \((D_i)\), its usual \(3\)-variation is bounded pointwise by \(2(\sum_i\left\|D_i\right\|^2)^{1/2}\): each endpoint of an increasing sequence is used at most twice, and \(\ell^3\) is bounded by \(\ell^2\). Thus (105) controls the variation of \((P_i-E_i)F\).

Lépingle’s martingale variation inequality (Lépingle 1976), in the unweighted \(L^2\) case and at variation exponent \(3\), gives \(\left\|V^3(E_iF)\right\|_2\leq C\left\|F\right\|_2\); see (Zorin-Kranich 2020, Theorem 1.1, with \(p=2\), \(r=3\), and weight one). For a finite index range on \(\mathbb R\), apply its probability-space version on every atom of the coarsest sigma algebra, normalizing Lebesgue measure there, and sum the squared inequalities. This proves the same statement on the line without a finite-measure assumption. For Hilbert-valued \(F\), apply the scalar inequality to its real and imaginary components. Minkowski’s inequality, in the form \(\ell^3(\ell^2)\leq\ell^2(\ell^3)\), and the outer \(L^2\) norm give the same constant independently of the number of components. Combining with the comparison and then taking the selected low-pass indices proves the estimate for one spaced class. Summing the fixed number of classes proves (104). ◻

Proof. If \(A=0\), dyadic differentiation makes \(f\) constant almost everywhere, and the assertion is immediate. Otherwise take the maximal proper dyadic subintervals \(J\) of \(I\) on which the average of \(\left|f-f_I\right|\) exceeds \(2A\). Their total length is at most \(\left|I\right|/2\), the function has \(\left|f-f_I\right|\leq2A\) almost everywhere off their union, and dyadic parent maximality gives \(\left|f_J-f_I\right|\leq4A\) on each selected interval. Repeat the same selection inside each \(J\), now subtracting its own mean. At generation \(k\) the selected union has measure at most \(2^{-k}\left|I\right|\). Outside it, the successive differences of means and the final differentiation bound show \(\left|f-f_I\right|\leq(4k+2)A\). This proves an exponential distributional bound for \(\left|f-f_I\right|/A\). Integrating that bound by layer cake gives \(\left\|f-f_I\right\|_{L^b(I,\,\mathrm dx/\left|I\right|)}\leq CbA\); adding the mean proves the lemma. Finite stopping generations and then monotone convergence justify the argument for any locally integrable \(f\). ◻

Proof. We give the local mean and oscillation bounds, including their behavior on intervals shorter than \(l\). All convolutions in what follows are in the \((x,Y_t)\) coordinates.

First, (97) satisfies \[ 0\leq f_d\leq C,\qquad \sum_{d\text{ dyadic}}f_d\leq C(1+\log B). \tag{107}\] The first indicators are nonzero for only \(O(1+\log B)\) dyadic \(d\) at a fixed point, and the second terms sum geometrically. Average first with the common vertical kernel \(k_{W,n'}\). Its \(L^1\) norm is fixed, so (107) persists on each reference line. The right side of (98), summed over scales, therefore defines a nonnegative majorant \(\mathcal A_t\) for \(\mathcal E_t\).

Let \(J\) be any horizontal interval of length \(r\leq l\). If \(\ell\leq r\), \[\frac1r\int_J k_{\ell,n'}(x-x')\,\mathrm dx \leq \frac Cr\left(1+\frac{\mathop{\mathrm{dist}}(x',J)}r\right)^{-2}.\] Hence the local mean of the short-scale part of \(\mathcal A_t\) is at most \(C(1+\log B)\), by summing the source functions before integration. For \(\ell>r\), the derivative of its convolution with a bounded source has supremum at most \(C/\ell\). The total oscillation of these terms on \(J\) is at most \(Cr\sum_{\ell>r}\ell^{-1}\leq C\). It follows that \(\mathcal A_t\) has local BMO norm \(O(1+\log B)\). Its mean on an interval of length \(l\) has the same bound, since all scales then belong to the short-scale part. John–Nirenberg on that interval gives \(\left\|\mathcal E_t\right\|_{L^b(I,\,\mathrm dx/l)}\leq Cb(1+\log B)\). We only use domination by the BMO majorant here; no assertion that an arbitrary function below a BMO function is BMO is required.

For the square function, fixed-scale lattice overlap gives \[ U_\ell(z)^2 \leq C\sum_{\ell_s=\ell} \left\|a_s^{\rm main}\right\|^2\mathcal B_{s,n}^{t}(z). \tag{108}\] For the same interval \(J\), partition center space into \(t\)-boxes \(Q_{ab}\) of dimensions \(r\times W\), with horizontal offset \(a\) from \(J\) and transverse offset \(b\) from the reference line. For \(z_s\in Q_{ab}\) and \(\ell_s\leq r\), the envelope bound implies \[\int_J\mathcal B_{s,n}^{t}(x,y_0+tx)\,\mathrm dx \leq C\ell_s(1+\left|a\right|+\left|b\right|)^{-n_1}\] for a fixed \(n_1>4\), after choosing \(n\) sufficiently large. Indeed the transverse decay gives the \(b\) factor; for a distant horizontal box the integral is bounded by \(Cr(\ell_s/(r\left|a\right|))^{n_1}\), which is at most \(C\ell_s(1+\left|a\right|)^{-n_1}\) because \(\ell_s\leq r\). Equation (102) gives exactly \[\sum_{\substack{z_s\in Q_{ab}\\\ell_s\leq r}} \left\|a_s^{\rm main}\right\|^2\ell_s =W^{-1}\sum_{\substack{z_s\in Q_{ab}\\\ell_s\leq r}} \left\|a_s^{\rm main}\right\|^2\left|R_s\right|\leq Cr.\] Summing the offsets in (108) proves \[ \frac1r\int_J\sum_{\ell\leq r}U_\ell^2\,\mathrm dx\leq C. \tag{109}\] This argument applies on each line directly; it is not a restriction of an almost-everywhere plane estimate to a line. Notice also that the factor \(W\) has canceled, so the estimate has no dependence on \(m\). Since main coefficients are bounded, fixed-scale lattice bounds give \(U_\ell\leq C\) and \(\left|\partial_x U_\ell\right|\leq C/\ell\) on the line. The long-scale part of \(\mathcal S_t^2\) has oscillation at most \(C\) on \(J\). By (109), its local BMO norm is bounded; its mean on every length-\(l\) interval is bounded as well. John–Nirenberg applied to \(\mathcal S_t^2\) gives \(\left\|\mathcal S_t\right\|_{L^b(I,\,\mathrm dx/l)}\leq C\sqrt b\).

It remains to prove the main-term variation bound, with or without the mask. Fix \(J\) and its center boxes \(Q_{ab}\) as above, and first restrict to scales \(\ell_s\leq r\). Group the synthesis by those boxes. For a fixed box multiply its synthesis by \[P_{ab}(x)=\left(1+\frac{(x-x_{Q_{ab}})^2}{r^2}\right)^{q_0},\] where \(q_0\) is a sufficiently large fixed integer. Because centers lie in \(Q_{ab}\) and \(\ell_s\leq r\), these multiplied waves still have fixed tile envelope bounds. Their Fourier supports are preserved by polynomial multiplication. Using (108) and integrating on the entire line therefore gives \[\sum_\ell\left\|P_{ab} \sum_{\substack{\ell_s=\ell\\z_s\in Q_{ab}}} a_s^{\rm main}\Phi_s\right\|_{L^2(\mathbb R;\mathcal H)}^2 \leq Cr(1+\left|b\right|)^{-n_1},\] and the same bound holds with \(q_{T(s)}\Phi_s\) in place of \(\Phi_s\). Here and elsewhere \(a_s^{\rm main}\Phi_s\) denotes componentwise multiplication. The bound follows from (102) as above, retaining transverse decay while integrating the horizontal envelope. Apply 25 with the length index reversed; this preserves the interval-sum variation. The support hypotheses hold by (99) and (100). On \(J\), \(P_{ab}\geq c(1+\left|a\right|)^{2q_0}\), and multiplication by this one scalar commutes with every interval sum and with its norm. Thus the \(L^2(J)\) variation of the synthesis from \(Q_{ab}\) is at most \[Cr^{1/2}(1+\left|a\right|)^{-2q_0}(1+\left|b\right|)^{-n_1/2}.\] Minkowski and summation over \(a,b\) give a bound \(Cr^{1/2}\) for the short-scale main variation.

For each longer scale the derivative of main synthesis on the line is bounded by \(C/\ell\), also after masking by (101). The variation of the long-scale part therefore changes by at most \(Cr\sum_{\ell>r}\ell^{-1}\leq C\) on \(J\). The full main variation differs from the long-scale variation by at most the short-scale variation pointwise, by the triangle inequality for (103). This proves its bounded local BMO norm. Taking \(r=l\) proves its bounded mean on \(I\). John–Nirenberg gives normalized \(L^b\) norm at most \(Cb\). The variation of the error synthesis is bounded by the positive sum \(C\mathcal E_t\), since disjoint intervals use each scale at most once. For the masked error the same is true because \(\left|q_T\right|\leq1\). The already proved error bound completes the proposition. ◻

Proof. Use 27 on assigned top tiles, with \(t=s_T\) and \(l=l_T\). These satisfy (93) with a fixed angular constant. Write \(X=(x-x_T)/l_T\) and \(Y=(Y_{s_T}(z)-Y_{s_T}(z_T))/W\). For a sufficiently large fixed integer \(q\), let \[P_T(z)=\left(1+\frac{X^2+Y^2}{K_1^2}\right)^q.\] Its product with any assigned wave is bounded by a slower fixed tile envelope with uniform constants. Indeed at the centers \(X,Y\) are bounded by a fixed fraction of \(K_1\), and in moving away from a center the top scales are no smaller than the corresponding tile scales. The same statement holds for the fixed derivatives in use. Multiplication by \(P_T\) preserves spectrum. Applying the last observation after 27 gives a uniform polynomial moment bound on every length-\(l_T\) interval of every parallel line for \(\sup_{\mathcal J}\left\|P_TF_{T,\mathcal J}\right\|\) and \(\sup_{\mathcal J}\left\|P_T\mathcal D F_{T,\mathcal J}\right\|\).

We justify the required vertical neighborhood supremum. Demodulate component \(j\) by \(e^{-i\eta_jY_{s_T}}\). Each of the just described weighted sums has vertical Fourier support in \([-C/W,C/W]\). Choose one Schwartz reproduction kernel \(K_W\) with Fourier transform one on that interval. Its absolute values at all arguments shifted by at most \(W\) are bounded by \(C_nW^{-1}(1+\left|y\right|/W)^{-n}\). Reproduction, the Hilbert norm inequality, and the supremum over \(\mathcal J\) show that its vertical neighborhood supremum is bounded by convolution of the corresponding unshifted supremum with this positive integrable kernel. Minkowski in normalized \(L^b\) on horizontal intervals preserves the uniform moment estimate. Finally \(P_T(z+(0,v))\asymp P_T(z)\) for \(\left|v\right|\leq W\), uniformly in \(z\), because \(K_1\geq1\). We have therefore proved, on every top-scale box, polynomial moments for \(P_T\) times the entire left side of (110).

For clarity, this yields a global exceptional-area bound, not merely one inside the top box. Partition the plane into boxes of dimensions \(l_T\times W\), indexed by \((a,b)\in\mathbb Z^2\). On each such box, \(1+\mathop{\mathrm{dist}}_T/K_1\asymp1+(\left|a\right|+\left|b\right|)/K_1\). Choose \(q\) with \(2q>8D_2+4\). If the left side of (110) exceeds its right side in that box, its product with \(P_T\) exceeds \[c L^{C_{\mathrm{top}}} \big(1+(\left|a\right|+\left|b\right|)/K_1\big)^{2q-8D_2}.\] Take moment order \(b_L=\lceil L\rceil\). The available moment bound is at most a fixed power of \(L\), since \(1+\log B\) is such a power. Choose \(C_{\mathrm{top}}\) larger by a fixed amount. Markov’s inequality and summation over \((a,b)\) then give \[\left|\mathcal X_T\right| \leq C\left|R_T\right|K_1^2 e^{-(J_*+1)L} \leq \left|R_T\right|e^{-J_*L}\] for sufficiently large \(L\), after increasing the threshold exponent or the fixed lower bound on \(L\). Equivalently one can first arrange \(e^{-(J_*+2)L}\) to absorb the polynomial \(K_1^2\). Taking the union over tops proves the asserted bound by \(S=\sum_T\left|R_T\right|\). This argument uses fixed \(q\) and fixed packet orders, even though the moment order grows with \(L\).

For (111), coefficient norms are bounded, there are boundedly many assigned labels per scale, and each differentiated wave has a bound \(C\ell_s^{-1}\) times a fixed tile envelope in slope \(s_T\). At a fixed scale, summing this envelope over assigned centers is at most \(C\ell_s^{-1}(1+\mathop{\mathrm{dist}}_T/K_1)^{-8D_2}\). To see the decay, multiply by the inverse top factor, absorb it into a slower tile envelope as for \(P_T\), and use bounded lattice overlap. Summing over \(\ell_s\geq l_*\) costs \(\sum_{\ell\geq l_*}\ell^{-1}\leq C/l_*\). The same bound applies to any interval of these lengths by the triangle inequality. This proves the unconditional derivative estimate and completes the corollary. ◻

Proof. First fix, for example, \(J_*=10\) in 28, and fix the resulting \(C_{\mathrm{top}}\) before making any choice of \(A\). Its exceptional area is smaller than any prescribed fixed inverse power of \(L\) times \(\left|F\right|\), since \(S/\left|F\right|\leq L^C N^2\leq L^C e^{4L}\). On \(\mathop{\mathrm{dist}}_T\leq L^{A/2}\), the elementary estimate for the mask is \(\left|1-q_T\right|\leq C L^{1-A}\): use \(\left|1-\operatorname{sinc}v\right|\leq Cv^2\) near zero and the integer power \(2\lceil L\rceil\) in (87). Outside this region, \(\left|1-q_T\right|\leq2\), whereas \[\big(h_T^{(1)}\big)^7 \leq C L^{-7D_2(A/2-A_1)}.\] Equation (110), with the full length interval, therefore implies outside \(\mathcal X_{\rm top}\) that, for any prescribed fixed \(H_1\), \[ \left\|P-P_q\right\|\leq L^{-H_1}\sum_T h_T^{(1)} \tag{113}\] once \(A\) is chosen large enough. In the near region the factor \(L^{C_{\mathrm{top}}+1-A}\) suffices; in the far region use the displayed seventh power to reduce \((h_T^{(1)})^8\) to \(h_T^{(1)}\).

Apply (77) with dilation \(K_1\) to \(n_1=\sum_Th_T^{(1)}\). For any finite-measure set \(A_0\) in the testing domain, weak \(L^2\) and layer cake imply \[\int_{A_0}n_1\leq2\left\|n_1\right\|_{2,\infty}\left|A_0\right|^{1/2} \leq L^C S^{1/2}\left|A_0\right|^{1/2}.\] Since \(S\leq L^C\left|E\right|\) and \(\left|E\right|=N^2\left|F\right|\), choosing \(H_1\) larger than \(H\) by the fixed powers in these inequalities proves the claimed mask integral even when \(A_0=F\setminus\mathcal X_{\rm top}\).

After \(A\) has been fixed, (77) with dilation \(K_0\) gives \[\left|\{z\in F:n(z)>NL^{C_n}\}\right| \leq L^{C-2C_n} S/N^2 \leq L^{C-2C_n}\left|F\right|.\] Take \(C_n\) sufficiently large and define \(F_{\rm conc}\) by deleting this set and \(\mathcal X_{\rm top}\) from \(F\). Increasing the fixed thresholds slightly makes the sum of their areas at most \(L^{-H}\left|F\right|\). This proves all assertions. ◻

Proof. For \(t\in\{0,1/3,2/3\}\) use the dyadic grid whose intervals of length \(2^k\) are \[\mathcal D_k^t =\big\{2^k[j+(-1)^kt,j+1+(-1)^kt):j\in\mathbb Z\big\}.\] These grids are nested: a boundary at scale \(k+1\) is a boundary at scale \(k\) because \(3t\) is an integer. At a fixed scale their combined boundary points have spacing \(2^k/3\). Given an interval \(I\) of positive finite length, choose the least integer \(k\) with \(2^k>3|I|\). Its closure meets boundary points from at most one of the three grids. Hence \(I\) is contained in an interval \(Q\) from one of the other grids, and \(|Q|\leq6|I|\).

Choose an integer \(k_H\) with \(6H\leq2^{k_H}<12H\), and in each grid retain only scales \(k\leq k_H\). Put \(\beta=\vartheta/6\). For each \(b\) and each grid, call \(Q\) bad if \[|Q|^{-1}\int_Q f_b\geq\beta.\] Let \(\mathcal J_b^t\) be the intervals maximal under inclusion among these bad intervals. Every bad interval is contained in a maximal one, since it has only finitely many ancestors up to level \(k_H\). For fixed \(b,t\), the members of \(\mathcal J_b^t\) are pairwise disjoint. Define \[\mathcal N_t(x)=\sum_{b\in\mathcal B} \sum_{Q\in\mathcal J_b^t}\mathbf 1_Q(x), \qquad \mathcal N(x)=\sum_{t\in\{0,1/3,2/3\}}\mathcal N_t(x).\] Each interval in (118) produces a bad dyadic interval containing \(x\), at an allowed scale. Thus each distinct index in \(\mathcal B_x\) contributes at least once to \(\mathcal N(x)\).

For every dyadic interval \(P\) at an allowed scale, disjointness within each bin gives the Carleson estimate \[ \begin{split} \sum_b\sum_{\substack{Q\in\mathcal J_b^t\\Q\subseteq P}}|Q| &\leq\beta^{-1}\sum_b \sum_{\substack{Q\in\mathcal J_b^t\\Q\subseteq P}} \int_Q f_b\\ &\leq\beta^{-1}\int_P\sum_b f_b \leq\mathcal A|P|,\qquad \mathcal A=C_0/\beta. \end{split} \tag{120}\] All sums here are of nonnegative terms, so this argument also applies to countably many bins and intervals.

For completeness, (120) has an elementary exponential-tail consequence. Fix a root interval \(P_0\) of length \(2^{k_H}\). First take any finite subfamily of the selected interval occurrences inside \(P_0\), retaining multiplicities when different bins select the same interval, and let \(V\) count that subfamily. For an integer \(r\geq1\), expansion of \(V^r\) and the nesting of intersecting dyadic intervals give \[ \int_{P_0} V^r \leq r!\sum_{Q_1\supseteq\cdots\supseteq Q_r}|Q_r| \leq r!\mathcal A^r|P_0|. \tag{121}\] The chain sum counts occurrences with their multiplicities. To obtain the second inequality, sum first over \(Q_r\subseteq Q_{r-1}\) and apply (120); repeat until the remaining sum is bounded by \(\mathcal A|P_0|\). Repeated intervals are allowed throughout. Ordering the \(r\) occurrences accounts for the factor \(r!\) and can only overcount.

Enumerate all selected occurrences inside \(P_0\). Their finite initial sums increase to \(\mathcal N_t\) on \(P_0\), so monotone convergence and (121) show \[ \int_{P_0}\exp(\mathcal N_t/(2\mathcal A)) \leq \sum_{r=0}^{\infty}2^{-r}|P_0|=2|P_0|. \tag{122}\] In particular the count is finite almost everywhere. Markov’s inequality bounds its superlevel measure in \(P_0\) by \(2|P_0|\exp(-\lambda/(2\mathcal A))\). The root intervals of one grid meeting \(U\) have total length at most \(|U|+2^{k_H+1}\). Sum the root estimates and observe that \(\mathcal N>\lambda\) implies \(\mathcal N_t>\lambda/3\) for at least one grid. Since \(\mathcal A=6C_0/\vartheta\) and \(2^{k_H}<12H\), this proves (119).

There are countably many bins and dyadic intervals. For each fixed interval its badness is determined by a measurable integral. Its maximality adds only finitely many conditions on its ancestors before level \(k_H\). Consequently the displayed count is measurable. The same argument, using measurability of integrals with a parameter, proves the final assertion. No maximality selection or exceptional set depending on an uncountable family of intervals is required. ◻

Proof. Fix a shift \(\sigma\). For the transitions \(i\equiv\sigma\pmod K\) with \(i<k\), (154) gives \[R_i\le\log\frac{n_{a(i)}}{n_{i+1}}.\] For this fixed shift, the intervals of levels \([a(i),i+1]\) are successive blocks starting at zero. Their logarithms telescope. The endpoint of the last completed block is at most \(k\), so the sum is at most \(\log(n_0/n_k)\), by monotonicity of the masses. Summing over the \(K\) shifts proves the first inequality in (167). The second is an ordinary telescope. This is a full-root bound. In particular, it remains applicable to weights occurring on a later segment even when another shift’s block starts before that segment.

Set \(a=\mu/K\), solely for this proof, and use the specified logarithm \[L_k=-a\log n_k-\sum_{i=h}^{k-1}\zeta_{i\bmod K}R_i, \qquad \omega_{T,h,k}=e^{L_k}.\] Its increment is \[ L_{k+1}-L_k=aP_k-\zeta_{k\bmod K}R_k. \tag{169}\] Since \(n_k\ge h_T\) and all real transition exponents are nonnegative, \(|e^{L_k}|\le h_T^{-a}\). For any complex \(A,B\), \[|e^B-e^A| \le |B-A|\int_0^1|e^{(1-t)A+tB}|\,\,\mathrm dt \le |B-A|\max(|e^A|,|e^B|).\] Using this inequality for adjacent logarithms and then (167) proves (168). Notice that the exponential of the sum of the absolute logarithmic increments has not been used. Such an exponential would be unsuitable when \(h_T\) is very small. ◻

Proof. For vectors \(f_h,\ldots,f_k\) and scalars \(\omega_h,\ldots,\omega_k\), summation by parts gives \[ \left\|\sum_{i=h}^k\omega_i f_i\right\| \le \left(|\omega_k|+\sum_{i=h}^{k-1}|\omega_{i+1}-\omega_i|\right) \max_{h\le j\le k}\left\|\sum_{i=h}^j f_i\right\|. \tag{173}\] Group the packets assigned to \(T\) according to their stopping level. The resulting initial sums, or the sums obtained after restricting to an interval of packet lengths, are interval partial sums of the original top’s length sequence. By (110) and the fact that the \(K_0\)-dilated weight is at least the corresponding \(K_1\)-dilated weight, their norm at \(z\in Z_0\) is at most \(L^C h_T(z)^8\). The mask \(q_T(z)\) is common to this whole top. Apply (173) and (168).

To see that the logarithmic factor can be absorbed uniformly, use (112) and \(\log N\le2L\). Since \(n_{v_0(T)}\le n\) and \(h_T\le C_3\), \[0\le\log\frac{n_{v_0(T)}}{h_T} \le C L+\log^+\frac1{h_T}.\] Here and below \(\log^+x=\max(0,\log x)\). Also \(\mu/K<1\). The elementary boundedness of \(x^\delta\log^q(1/x)\) on \((0,1]\), for fixed \(\delta>0\) and fixed \(q\), shows that \[h_T^{8-\mu/K} \left[1+K(1+\left\|\zeta\right\|_2) \log\frac{n_{v_0(T)}}{h_T}\right] \le L^C(1+\left\|\zeta\right\|_2)h_T^4.\] This proves (171). Summing over the eligible tops and using \(h_T^4\le C_3^3h_T\) proves (170). For insignificant tops, the mask estimate after (87) gives \(|q_T|\le e^{-L\log L}\). Applying the same summation proves (172). ◻

Proof. All descendants of \(J\) have slopes inside \(J\), so their top slopes are within \(\rho_j\) of \(s_R\). Their eligibility cutoffs are at most \(\rho_j/B^2\). A mass in an additional boundary term (153) at a transition \(i\ge i_j\) can come from an interval outside \(J\), because the block associated to another shift can start before \(i_j\). However, that interval is within \(G^{-1/4}d(i)\) of the relevant child and has width \(d(i+1)\). The relevant child is a descendant of \(J\), and \(d(i)\le\rho_j\). Hence every added slope is within \(C\rho_j\) of \(s_R\). The tops in an added interval have precision at most \(d(i+1)/B^2\); those in the immediate-parent mass have precision at most \(d(i)/B^2\). This proves (175), including for weights below the block. Taking a difference of the relevant nested top sets preserves the same property. This argument uses the neighbor radius at the immediate transition, rather than the possibly larger width of the other shift’s block ancestor.

In normalized coordinates of a top \(T\), horizontal motion along \(s_R\) has velocities \(1/l_T\) and \((s_R-s_T)/W\). Differentiating (76), using the Lipschitz property of the absolute value at almost every point, gives \[|\partial_x\log h_T| \le C\left(\frac1{l_T}+\frac{|s_R-s_T|}{W}\right) \le C\left(\frac{\rho_j}{B^2w}+\frac{\rho_j}{mw}\right).\] The factor \(K_0^{-1}\) available from the dilation is not needed. Multiplying by \(H_j\) and using (158) yields \[H_j|\partial_x\log h_T| \le\frac C{\sqrt G} \left(\frac{B}{B^2}+\frac Bm\right) \le C G^{-1/2}.\] For a positive sum \(a=\sum_T h_T\), its logarithmic derivative is the weighted average of the logarithmic derivatives of the summands. This proves (176), also for positive differences, which are such sums by their set interpretation.

If \(a,b>0\), direct differentiation gives \[\partial_x\log\frac a{a+b} =\frac b{a+b}\bigl(\partial_x\log a-\partial_x\log b\bigr).\] This proves (177); the inequality \(1-r\le-\log r\) holds for \(0<r\le1\). If the complementary top set is empty, \(r=1\) everywhere. In particular, writing \(q_r=-\log r\ge0\), \[ |\partial_x q_r|\le\frac{2\varepsilon_{\mathrm{mov}}}{H_j}q_r, \qquad e^{-2\varepsilon_{\mathrm{mov}}}q_r(x_0) \le q_r(x)\le e^{2\varepsilon_{\mathrm{mov}}}q_r(x_0) \quad(x,x_0\in I). \tag{179}\] The inequalities remain valid for an identically zero loss.

Finally, a block factor is a product of at most \(K\) transition factors. The simpler bound \(|\partial_x\log r|\le2\varepsilon_{\mathrm{mov}}/H_j\) shows that its specified complex logarithm changes by at most \(C K(1+\left\|\zeta\right\|_2)G^{-1/2}\). For (157) this is \(L^C G^{-1/2}\) and tends to zero faster than any negative fixed power of \(L\). Exponentiating proves (178) after decreasing \(c_0\). This last argument does not require any bound on the size of a logarithmic loss itself. ◻

Proof. Parametrize the interval by its horizontal coordinate \(x\), and write \(h=i_{j+1}\) and \(a=\mu/K\) in this proof. Use \(P_i,R_i\) from (166) for \(h\le i<k_T\). By (179), each of these nonnegative losses is comparable throughout \(I\) to its value at \(x_0\), with factor \(e^{O(\varepsilon_{\mathrm{mov}})}\), and its absolute derivative is bounded by \(C\varepsilon_{\mathrm{mov}}/H_j\) times itself. Define \[\mathcal B(x)=\sum_{i=h}^{k_T-1} \bigl(aP_i(x)+|\zeta_{i\bmod K}|R_i(x)\bigr).\] It follows that \(\mathcal B(x)\le e^{C\varepsilon_{\mathrm{mov}}}\mathcal B(x_0)\). By the full-root bound, \[ \mathcal B(x_0) \le C K(1+\left\|\zeta\right\|_2) \log\frac{n_{v_0(T)}(z_0)}{h_T(z_0)}. \tag{182}\] The full-root mass is used only at the snapshot in this inequality; its spatial derivative is not needed.

Let \(L_k=\log\omega_k\) be the specified logarithm in the proof of 34, and put \(q_k=\partial_x L_k\). Equation (169) and the mass derivative bound imply \[\begin{align*} \sup_k|q_k| &\le\frac{C\varepsilon_{\mathrm{mov}}}{H_j} \bigl(a+\mathcal B(x)\bigr),\\ \sum_k|q_{k+1}-q_k| &\le\frac{C\varepsilon_{\mathrm{mov}}}{H_j}\mathcal B(x). \end{align*}\] For example, the initial term is \(q_h=-a\partial_x\log n_h\), and subsequent differences are the derivatives of \(aP_i-\zeta_{i\bmod K}R_i\). At every point of \(I\), \[\sup_k|\omega_k(x)|\le h_T(x)^{-a} \le e^{a\varepsilon_{\mathrm{mov}}}h_T(z_0)^{-a}.\] The complex-exponential estimate used for (168) therefore gives \[\left\|\omega(x)\right\|_{\mathrm{BV}} \le C h_T(z_0)^{-a}\bigl(1+\mathcal B(x_0)\bigr).\]

For finite sequences the product rule for differences gives \[\left\|ab\right\|_{\mathrm{BV}} \le \sup_k|a_k|\left\|b\right\|_{\mathrm{BV}} +\sup_k|b_k|\sum_k|a_{k+1}-a_k|.\] Using \(\partial_x\omega_k=\omega_k q_k\) and the preceding estimates yields \[ \left\|\partial_x\omega(x)\right\|_{\mathrm{BV}} \le\frac{C\varepsilon_{\mathrm{mov}}}{H_j} h_T(z_0)^{-a}\bigl(1+\mathcal B(x_0)\bigr)^2. \tag{183}\] The functions involved are absolutely continuous on the line. Integrate (183) between \(x_0\) and \(x'\), over distance at most \(H_j\), and use the triangle inequality for the finite-sequence norm. This proves the difference bound with right side \(C\varepsilon_{\mathrm{mov}} h_T(z_0)^{-a}(1+\mathcal B(x_0))^2\).

At the snapshot, (112), (157), and (182) bound the last expression by \(G^{-1/2}L^C h_T(z_0)^{-4}\). Indeed the only possible nonpolynomial quantity is \(\log^+(1/h_T(z_0))\), whose square is absorbed by the difference between exponents \(a<1\) and \(4\). The same argument proves (180). There is no assumption that \(G^{-1/2}\mathcal B(x_0)\) is small: very small \(h_T\) is allowed, and only a polynomial in its logarithm has appeared. ◻

Proof. Fix one top \(T\) contributing to \(A_c\). Its contributing packets have \(k_s\ge i_{j+1}\), hence \[ \ell_s\ge l_*:=\frac{B^2w}{\rho_{j+1}}, \qquad l_T\ge l_*. \tag{185}\] Also \(s_T\in c\), so \(|s_T-t_c|\le\rho_{j+1}/2\), while \(|s_T-s_R|\le C\rho_j\).

Consider any interval partial sum \(S\) of these unweighted packets on \(T\). The good-snapshot estimates (110), including their demodulated vertical-derivative and vertical-neighborhood versions, bound \(S\) and \(W(\partial_y-i\eta_\nu)S_\nu\) by \(L^Ch_T(z_0)^8\) throughout the vertical neighborhood of radius \(W\) at \(z_0\). Put \(v=(s_R-s_T)b\) and first move to \(z_1=z_0+(0,v)\). Since \[|v|\le C\rho_jH_j\le\frac{CBw}{\sqrt G}\ll W,\] the fundamental theorem of calculus after componentwise demodulation gives \[\left\|S(z_1)-\operatorname{diag}(e^{i\eta_\nu v})S(z_0)\right\| \le L^Ch_T(z_0)^8\frac{|v|}{W} \le G^{-1/2}L^Ch_T(z_0)^8.\] Next move from \(z_1\) to \(z'\) along slope \(s_T\) by horizontal distance \(b\). The nonexceptional horizontal derivative estimate in 28, applied to lengths at least \(l_*\), bounds this change by \[L^Ch_T(z_0)^8\frac{|b|}{l_*} \le G^{-1/2}L^Ch_T(z_0)^8.\] To justify its spatial weight here, the two displacements change the normalized coordinates of \(T\) by at most \(C/(B^2\sqrt G)+CB/(m\sqrt G)\). Its weight at every intermediate point is therefore comparable to \(h_T(z_0)\). If the derivative estimate is expressed using the smaller dilation \(K_1\), that weight is bounded above by the \(K_0\)-dilated weight used here. No good-point estimate at \(z_1\) or \(z'\) has been invoked for this horizontal move.

The replacement of \(s_T\) by \(t_c\) in the diagonal phase costs at most \[C|s_T-t_c|\frac{|b|}{w}\left\|S(z_0)\right\| \le G^{-1/2}L^Ch_T(z_0)^8,\] because \(\eta_\nu\asymp w^{-1}\) uniformly in \(\nu\). This proves, uniformly over the interval partial sum, \[ \left\|S(z')-\mathcal D_c(b)S(z_0)\right\| \le G^{-1/2}L^Ch_T(z_0)^8. \tag{186}\]

The same bound holds with the mask \(q_T\) multiplying \(S\) at its own evaluation point. Indeed \(|q_T|\le1\), and its formula (87), together with boundedness of the derivative of sinc, gives \[|q_T(z')-q_T(z_0)| \le L^C\left(\frac{|b|}{l_T} +\frac{|s_R-s_T|\,|b|}{W}\right) \le G^{-1/2}L^C.\] Combining this with the snapshot bound for \(S\) proves the masked version of (186). This estimate uses absolute motion of the mask and does not divide by \(q_T\), which may vanish.

Group the masked packets by stopping level and denote their sums by \(f_k(z)\). Their weighted contribution to the difference in (184) is \[\begin{align*} &\sum_k\omega_k(z') \bigl(f_k(z')-\mathcal D_c(b)f_k(z_0)\bigr)\\ &\hspace{15mm} +\sum_k\bigl(\omega_k(z')-\omega_k(z_0)\bigr) \mathcal D_c(b)f_k(z_0). \end{align*}\] All interval partial sums of the first parenthesis have norm at most \(G^{-1/2}L^Ch_T(z_0)^8\), by the masked version of (186). The snapshot partial sums in the second line have norm at most \(L^Ch_T(z_0)^8\). Apply (173) to both lines, then (180) and (181). The result for this top is at most \(G^{-1/2}L^Ch_T(z_0)^4\). Sum over the tops eligible in \(c\). Since \(h_T^4\le C_3^3h_T\), this is at most \(G^{-1/2}L^Cn_c(z_0)\). Finally \(\log G\asymp L^{0.7}\) absorbs every fixed power of \(L\) into a smaller positive power of \(G\), proving (184). ◻

Proof. Fix \(R\), a line of slope \(s_R\), and any horizontal interval \(I\) of length \(l_R\) on that line. By 27, the normalized \(L^b(I)\) norm of each indicated quantity is at most \(C b^C(1+\log B)^C\). The same bound holds for \(\mathcal V_{R,\sigma}\) because (188) is a fixed subset satisfying (93). As \(\log B=K\log G\) is bounded by a fixed power of \(L\), this bound is polynomial in \(L\) when \(b\) is a fixed positive multiple of \(L\).

For clarity, fix an arbitrarily large absolute number \(J_0\). Choose \(b=\lceil L\rceil\) and then a fixed exponent \(C_{\mathrm{path}}\) sufficiently larger than those in the moment bound. Markov’s inequality gives, for all large \(L\), \[\begin{aligned} &|\{x\in I:\mathcal E_R(x)>L^{C_{\mathrm{path}}}\}| +|\{x\in I:\mathcal S_R(x)>L^{C_{\mathrm{path}}}\}|\\ &\qquad +\sum_{\sigma=0}^{K-1} |\{x\in I:\mathcal V_{R,\sigma}(x)>L^{C_{\mathrm{path}}}\}| \le e^{-J_0L}|I|. \end{aligned}\] The factor \(K+2\) is polynomial and is absorbed by the same choice. Alternatively one may increase the fixed multiple in the choice of \(b\); neither choice depends on the number of scales or tops.

Partition the whole reference line into intervals of length \(l_R\). Along this line \(h_R\) changes by at most an absolute factor on each such interval, directly from (76). Hence the preceding estimate implies \[\int_I h_R\mathbf 1_{\{R\text{ bad}\}}\,\,\mathrm dx \le C e^{-J_0L}\int_I h_R\,\,\mathrm dx.\] Sum over the intervals and integrate in the transverse coordinate \(y-s_Rx\); this change of coordinates has determinant one. Using (147) gives \[\int_{\mathbb R^2}h_R\mathbf 1_{\{R\text{ bad}\}}\,\,\mathrm dz \le L^C e^{-J_0L}|R_R|.\] Sum over \(R\) and take \(J_0\) large enough to obtain (190). This establishes the estimate with its full spatial weight, not only on the central part of \(R\).

Markov’s inequality applied to the nonnegative sum defining (191) gives its first measure bound. Since \(S\le L^C|E|=L^CN^2|F|\) and \(N\le e^{2L}\), the second follows. For example, Cauchy–Schwarz and the previously proved adjoint estimate give \[\int_{Z_0\setminus Z}\left\|P\right\| \le |Z_0\setminus Z|^{1/2}\left\|P\right\|_2 \le L^Ce^{-5L/2}N|F|,\] which is smaller than any required fixed negative power of \(L\) times \(N|F|\). The estimate is applied to the original sum \(P\), as required for exceptional discards.

Finally, all packet collections used here depend only on the finite geometric data, the angular translation, and the shift. They do not depend on \(z\), \(\zeta\), significance, or a mass cutoff. Their finite length variations and the other thresholded quantities are measurable in \(z\). They are measurable jointly with \(\vartheta\) as well: at the finitely many relevant angular levels, membership of each of the finitely many top slopes in its interval changes at finitely many translation endpoints. On the complementary parameter intervals all the relevant membership data are fixed. This also proves the asserted independence from \(\zeta\). ◻

Proof. Identity (196) follows directly from (160), because \(V_c-n_c=n_c^{\alpha-p}\left\|A_c\right\|^p\). The term \(n_J-\sum_c n_c\) is a sum of top weights by (150), now between the endpoints of the whole block. Every top either remains eligible in exactly one block child or loses eligibility and terminates at \(J\). The construction therefore branches by disjoint top sets, assigning each top exactly one path. It terminates after finitely many blocks even if no mass stop occurs, since no top is eligible beyond the final level. At such a final expanding node the block-child sum is empty and all tops terminate directly.

If \(R\) terminates directly at an expanding node, its terminal amount is \(h_R\). If it reaches a mass-stopped node \(c\), the terminal amount is \((V_c/n_c)h_R\), which is at most \(C_Mh_R\) by (195). Repeated use of (196) proves the equality in (197).

To verify the claim about bad paths, it is useful to normalize the expansion by actual probabilities. At an active node define \[D_J=n_J-\sum_c n_c+\sum_cV_c =n_J+\sum_c(V_c-n_c)>M.\] Choose a block child \(c\) with probability \(V_c/D_J\), or a directly terminating top \(R\) with probability \(h_R/D_J\). These probabilities sum to one. At a mass-stopped node choose an eligible top with probability \(h_R/n_c\). The probability of a given complete terminal path is exactly its summand in (197) divided by \(V_{\mathrm{root}}\). In fact, for an active edge \(J\to c\), \(V_c/D_J=\beta_JV_c/V_J\), and at a direct terminal edge \(h_R/D_J=\beta_Jh_R/V_J\). The successive \(V_c\) factors cancel; at a mass stop the last factor \(h_R/n_c\) leaves the amount \((V_c/n_c)h_R\). This proves the asserted probability identity without an independence assumption on the amplitudes.

For a directly terminating top, its last conditional probability is \(h_R/D_J\le h_R/M\le h_R\). If a top terminates in the next node by a mass stop, the combined last choice of that child and then that top has conditional probability \[\frac{V_c}{D_J}\frac{h_R}{n_c} \le C_M\frac{h_R}{M}\le C_Mh_R.\] Earlier conditional probabilities are at most one. Thus the unconditional probability of any specified terminal top is at most \(C_Mh_R\). This estimate also covers tops which terminate after many active ancestors; it uses only the last active step. By (191), the total probability of bad terminal tops is at most \[C_M\sum_{R\text{ bad at }z}h_R(z)\le C_Me^{-L}<\frac12\] for all sufficiently large \(L\). Here \(\log C_M=O(p\log L+\log(1+M))=o(L)\) for each fixed \(A_M\). The probability identity now implies that good tops carry at least half of the total amount, and the bound on \(e_R\) proves (198). ◻

Proof. If there are no active indices there is nothing to prove in (201). Otherwise let \(q\) be the last active index. For \(j<q\), the next mass is also greater than \(M\), and \[u_j=1-\frac{n_{j+1}}{n_j} \le\log\frac{n_j}{n_{j+1}}.\] The last term \(u_q\) is at most one, whether its next mass is positive or zero. Therefore the sum is at most \(1+\log(n_0/n_q)\le1+\log(n_0/M)\). This argument deliberately does not telescope through a possibly tiny mass below the stopping cutoff. Equation (112) gives the last inequality.

Both \(n_j\) and \(h_j^*\) are positive sums of weights of tops eligible at \(J_j\), except that the difference can be identically zero. In the last possible block it is simply \(n_j\). Thus 36 gives relative variation \(1+O(G^{-1/2})\) across the whole \(H_j\) cell. A marked cell has positive supremum of \(h_j^*\), so this difference is not identically zero there, and its infimum is at least \(\tfrac12 e^{-CG^{-1/2}}\ge\tfrac14\) for large \(L\). The first marking condition supplies activity everywhere on the cell.

Suppose instead that \(z\) is active, its cell is unmarked, and \(h_j^*(z)\ge1\). The second marking condition then holds, so the first one fails: the cell has infimum mass at most \(M\). The relative-motion bound implies \[M<n_j(z)\le e^{CG^{-1/2}}M<M+1,\] where the last inequality holds because \(M\) is a fixed polynomial in \(L\). Its drop of at least one gives \(n_{j+1}(z)<M\) (or zero at the last possible block), and all subsequent masses are at most that mass. This is consequently the last active index and can occur at most once. At a point of \(Z_0\), since the denominator of (160) is at least \(n_j\), (170) gives \[\beta_{J_j} \le\frac{V_{J_j}}{n_j} \le1+L^{Cp}n_j^{\alpha-1} \le L^{Cp}(1+M)^C,\] proving (203). If \(n_j(z)\le M\), its cell cannot have infimum greater than \(M\), proving the assertion about inactive indices. This observation will justify assigning the value one to the auxiliary phase factor on every unmarked cell.

Finally, eligibility at the starting level gives \(d_R\le\rho_j/B^2\), equivalently \(l_R\ge B^2w/\rho_j\). Together with (158), this yields \[\frac{H_j}{l_R} \le\frac{\rho_j}{B^2\rho_{j+1}\sqrt G} \le\frac1{B\sqrt G}.\] This only uses eligibility at the start of the block, and therefore applies to the final possible block as well. ◻

Proof. Use only children for which \(\widetilde X_c\ne0\); an empty collection gives the assertion immediately. Apply the clock selection from (163) to this finite subfamily at the snapshot, with \(Q\) as in (214). The clock rates are fixed as \(b\) varies. Each retained set has pairwise interval distance at least \(Q\), and the retention probabilities satisfy \(\pi_c\ge n_c/m_c\).

Every used child other than the possible unshrunk path child contains a significant eligible top at \(z_0\). From each retained child other than that possible exception, select one such top. Their slopes are \(Q\)-separated, their precisions satisfy \[d_T\le\rho_{j+1}/B^2\le Q/B^2,\] and their diameter is at most \(\rho_j\le B\rho_{j+1}\le BQ\). The simultaneous count in 30, with \(D\) as chosen in [count:choice-D], bounds this list by \(D/100\). Adding the possible retained path child leaves at most \(D\) members. Clock exclusion separates that extra child too; it does not require a significant representative for that one child.

In each Hilbert component, the retained carrier frequencies are separated by at least \(c Q/w\), with \(c>0\) absolute, since \(\eta_\xi\asymp w^{-1}\). Choose \(\epsilon'\le\epsilon_0c\). For a retained set \(\mathcal S\), let \[F_{\mathcal S}(b)= \sum_{c\in\mathcal S}\pi_c^{-1} l_c^{1-\mu}\lambda_c\mathcal D_c(b)\widetilde X_c.\] Expectation over the clocks gives \(\mathbb E_{\mathcal S}F_{\mathcal S}=F\); no independence of retention indicators is asserted or needed. Put \(r=r_*\). Minkowski in that expectation, (139), and Jensen for the concave function \(t^{1/r}\) give \[\begin{align*} \left\|F\right\|_{L^p(\nu;\mathcal H)} &\le \mathbb E_{\mathcal S} \left(\sum_{c\in\mathcal S} \pi_c^{-r}l_c\left|\lambda_c\right|^r \left\|\widetilde X_c\right\|^r\right)^{1/r} \\ &\le\left(\sum_c\pi_c^{1-r}l_c\left|\lambda_c\right|^r \left\|\widetilde X_c\right\|^r\right)^{1/r} \\ &\le\left(\sum_c l_c\left\|\widetilde X_c\right\|^r\right)^{1/r} \le\left(\sum_c l_c\left\|\widetilde X_c\right\|^p\right)^{1/p}. \tag{215}\end{align*}\] The power of \(l_c\) uses \(r(1-\mu)=1\). The third line uses the retention cancellation (164) for the present child subfamily. The last inequality in (215) is Hölder with weights of total mass at most one and \(r\le p\); the missing mass can be placed at a zero vector. Raise to power \(p\), add one, and divide by the constant denominator in (213). ◻

Proof of 42. Perform the construction on each marked cell meeting \(Z\). Equations (224), (229), and (230) give \[\beta_J^0(z)\le e^{a_L\upsilon}\mathcal B(b) \le e^{a_L\upsilon}\mathcal B^\sharp(b) =e^{a_L\upsilon}m_I\Gamma_j(b).\] By (231), the exponent is at most \[\bigl(G^{-\gamma/3}+CpK_*G^{-1/2} +CpK_*G^{-1/16}\bigr)\upsilon.\] Equations (222) and (206) bound this by \(G^{-c_{\rm av}}u_j(z)\), for a fixed positive exponent: choose \(c_{\rm av}<\min(\gamma/4,1/64)\) and then take \(L\) sufficiently large. This proves (205) with no error independent of the mass drop.

There are countably many cells on each fixed line and finitely many levels in the current truncation. Choose one snapshot in every relevant cell, use (230) there, and set \(\Gamma_j=1\) elsewhere. This is a measurable step function of the line parameter. The proof of 46 integrates the original integrand on lines where its slice is measurable, as holds for almost every transverse parameter. It therefore needs no jointly measurable snapshot choices across parallel lines. ◻

Proof. A residual packet in block \(j\) has \[ \frac{\rho_{j+1}}{B^2}<d_s\le\frac{\rho_j}{B^2}, \qquad i_j\le k_s<i_{j+1}. \tag{236}\] Because \(\rho_j/\rho_{j+1}\le B\), this block contains at most \(N_{\mathrm{len}}:=C(1+\log B)\) dyadic packet lengths. Also \(s_{T(s)},s_R\in J_j\), so the top assignment and (236) imply \[\left|\theta_s-s_R\right| \le C d_s+\rho_j \le C B^3 d_s\le C B^4d_s.\] Thus the reference-line estimates apply whenever \(\ell_s\le l_R\).

The coefficient multiplying \(q_{T(s)}\phi_s\) in \(B_{J_j}\) is a product of transition weights of modulus at most one, followed by \(n_{v_{k_s}(T(s))}^{-\mu/K}\). The ending mass is at least \(h_{T(s)}\). Consequently the modulus of the coefficient multiplying \(\phi_s\) is at most \(|q_{T(s)}|h_{T(s)}^{-\mu/K}\le C\). This last bound follows from the mask estimate (87): on the significant region \(h_T\ge1\), while off that region the mask has faster decay than this fixed small negative power of \(h_T\).

At \(z\) let \(e_\ell\) and \(a_\ell\) be the sums of the absolute error and main envelopes, respectively, at length \(\ell\) in the fixed family \(\mathcal P_R\) of (187). The preceding bounds place every residual packet with \(\ell_s\le l_R\) in this family, so its absolute envelopes are included in these positive sums. Goodness of \(R\), as specified in 39, gives \[\sum_{\ell\le l_R}e_\ell\le L^C, \qquad \left(\sum_{\ell\le l_R}a_\ell^2\right)^{1/2}\le L^C.\] The contribution of such lengths to \(\left\|B_{J_j}\right\|\) is bounded by \(C\sum_{\ell\in\mathcal I_j}(e_\ell+a_\ell)\), where \(\mathcal I_j\) is the set of residual lengths in that block. The sets \(\mathcal I_j\) are disjoint. Positivity and Cauchy–Schwarz within each block therefore give \[\begin{align*} \left\{\sum_j \left(\sum_{\ell\in\mathcal I_j,\,\ell\le l_R} e_\ell\right)^2\right\}^{1/2} &\le\sum_{\ell\le l_R}e_\ell,\\ \sum_j\left(\sum_{\ell\in\mathcal I_j,\,\ell\le l_R} a_\ell\right)^2 &\le N_{\mathrm{len}}\sum_{\ell\le l_R}a_\ell^2. \end{align*}\] The factor \(N_{\mathrm{len}}\) is polynomial in \(L\).

Before the last possible block, \(R\) remains eligible at the child level. Hence \(d_R\le\rho_{j+1}/B^2<d_s\), and consequently \(\ell_s<l_R\). Lengths larger than \(l_R\) can occur only in the last possible block. At each of its at most \(N_{\mathrm{len}}\) lengths, the total absolute coefficient envelope is bounded by a constant, by 23, applied with reference length equal to that scale. To recall the reason this bound has no factor counting the available labels, the condition \(|\theta_s-s_R|\le CB^4d_s\) makes the analyzing and synthesizing box envelopes comparable to common \(s_R\)-coordinate envelopes, since \(W\gg B^4w\). Summing centers first produces a common positive convolution kernel of integral \(O(1)\). At each source point only boundedly many selector labels are nonzero. Summing these labels then uses \(\left\|g\right\|_{\mathcal H}\le1\), and costs \(O(1)\). The uniformly bounded multipliers just considered preserve this absolute estimate. The last block therefore contributes at most \(C N_{\mathrm{len}}\) in norm. Combining the estimates proves (235). ◻

Proof. Every ending mass in this sequence is at least \(a_{i_f}=n_f>M-2\). Since all real parts of \(\zeta\) are nonnegative, \(|\omega_j(k)|\le a_k^{-\mu/K}\le1\) when \(M\ge3\). Using the real logarithms of positive masses, a logarithmic increment of the coefficient is \[\log\omega_j(k+1)-\log\omega_j(k) =\frac{\mu}{K}\log\frac{a_k}{a_{k+1}} -\zeta_{k\bmod K}\log\frac{m_{v_{k+1}}}{a_{k+1}}.\] The logarithm here is the specified linear expression defining the coefficient, so no branch of a complex logarithm is being chosen. The first sum of absolute increments telescopes. For the second use 34 on the entire root-to-endpoint path. All its summands are nonnegative, whence \[\begin{align*} \sum_{k=i_j}^{i_f-2} |\log\omega_j(k+1)-\log\omega_j(k)| &\le \left(\frac{\mu}{K} +K\max_\sigma|\zeta_\sigma|\right) \log\frac{n_{\mathrm{root}}}{n_f} \\ &\le L^{C_w}. \end{align*}\] The last step uses \(n_{\mathrm{root}}\le NL^C\) on \(Z\), \(\log N\le2L\), \(n_f> M-2\ge1\), and \(\max_\sigma|\zeta_\sigma|\le\mu+L^2\). The full-root bound is needed: a denominator \(m_v\) belonging to a different shift may involve a block whose starting level precedes the segment. No estimate using only \(\log(n_e/n_f)\) is asserted.

For two successive logarithmic values \(z_0,z_1\) in this sequence, both \(|e^{z_0}|\) and \(|e^{z_1}|\) are at most one. Along the straight segment between them, \(|e^{(1-t)z_0+tz_1}|\le1\). Integration of the derivative of the exponential consequently yields \(|e^{z_1}-e^{z_0}|\le|z_1-z_0|\). The logarithmic-increment bound proves (247). ◻

Proof. Fix a line of slope \(s_R\), and on that line use the functions \(\Gamma_j\) constructed in 42. At a point of \(Z\) where \(R\) is good, the active indices fall into the marked indices, the unary indices, and at most one active unmarked index with \(h_j^*\ge1\). For the marked indices, combine (205), (239), and (233). Their phase-domination losses have sum at most \(CLG^{-c}\), which is bounded. The unary factors have bounded log product by (251) and (252).

At the possible remaining index, 41 gives \(M<n_j<M+1\). From (170) and the baseline \(n_j\) in the denominator of \(\beta_{J_j}\), \[\beta_{J_j}\le 1+L^{Cp}n_j^{\alpha-1} \le L^{Cp}(1+M)^C.\] Here \(C\) is fixed, since \(\alpha\) stays in a fixed compact interval below three. We have therefore proved the pointwise domination \[ \prod_{j\text{ active at }z}\beta_{J_j}(z) \le L^{Cp}(1+M)^C\prod_{j=0}^{J_R}\Gamma_j(z), \qquad z\in Z\cap\{R\text{ good}\}, \tag{253}\] where \(J_R\) is the last possible block index for \(R\). It is legitimate to include every possible index on the right: if \(n_j(z)\le M\), its mesh cell cannot satisfy \(\inf n_j>M\), so it is unmarked and \(\Gamma_j(z)=1\). Unmarked active indices likewise have \(\Gamma_j=1\).

Recall that \(\Gamma_j\) is constant on the mesh of length \(H_{j-1}\) and has average at most one on every cell of length \(H_j\). The meshes are nested. On any cell \(I\) of the largest mesh \(H_{J_R}\), successive averaging from the smallest mesh upward therefore gives \[ \int_I\prod_{j=0}^{J_R}\Gamma_j(x)\,\,\mathrm dx\le |I|. \tag{254}\] More explicitly, all later factors \(\Gamma_k\), \(k>j\), are constant on each \(H_j\) cell. Integrating \(\Gamma_j\) on those cells can thus only decrease the integral of the nonnegative product. Iterate this observation for \(j=0,1,\ldots,J_R\).

The weight \(h_R\) is uniformly comparable on \(I\). This remains true for a terminal block and for the short first block: eligibility of \(R\) at \(i_j\) gives \(\rho_j\ge B^2d_R\), and every block has \(\rho_j/\rho_{j+1}\le B\). Thus \[ H_j=\frac{w}{\rho_{j+1}\sqrt G} \le\frac{l_R}{B\sqrt G}\le l_R. \tag{255}\] Along the line of slope \(s_R\), the transverse normalized coordinate of \(R\) is fixed and the horizontal normalized coordinate changes by at most \(H_j/l_R\). The definition (76) consequently gives \(\sup_I h_R\le C\inf_Ih_R\), with an absolute constant depending only on the fixed decay exponent. By (254), \[\int_I h_R\prod_{j=0}^{J_R}\Gamma_j\,\,\mathrm dx \le (\sup_Ih_R)|I|\le C\int_Ih_R\,\,\mathrm dx.\] Sum over the largest-mesh cells and use (253). Whenever the original integrand in (199) has a measurable slice on this line, we obtain \[\int_{Z\cap\{R\text{ good}\}\text{ on the line}} h_R\prod_{j\text{ active}}\beta_{J_j}\,\,\mathrm dx \le L^{Cp}(1+M)^C \int_{\text{line}}h_R\,\,\mathrm dx.\]

The snapshots used to construct the \(\Gamma_j\) need only be chosen on this one line. There are countably many mesh cells, so these choices give measurable step functions on that line. No jointly measurable choice of snapshots across different lines is required. The original integrand in (199), expressed entirely through \(h_R\), the good sets, and the finite hierarchy factors \(\beta_{J_j}\), is nonnegative and Lebesgue measurable. In the coordinates \((x,y-s_Rx)\), it has a measurable slice for almost every transverse parameter. The preceding line estimate therefore holds for almost every such parameter. Integrating it and applying Tonelli, using the Jacobian-one coordinate change, proves (199). ◻

Proof. Recall the nonnegative amount \[V_v=n_v+n_v^{\alpha-p}\|A_v\|^p.\] On the set where \(n_v\le M\), Equation (170) gives \[V_v\le L^{Cp}(1+M)^{\alpha-1}n_v.\] Here and below the bound on the imaginary parts is absorbed into \(L^{Cp}\). On the remaining set, apply Lemma 40. At least half of the root amount is carried by good paths, and the terminal amount of a path ending at \(R\) is at most \(L^{Cp}(1+M)^{\alpha-1}h_R\). Consequently, \[\begin{align*} \int_Z V_v &\le L^{Cp}(1+M)^{\alpha-1}\int n_v\\ &\quad+2L^{Cp}(1+M)^{\alpha-1} \sum_{R:\,s_R\in v} \int_{Z\cap\{\text{good }R\}} h_R\prod_{\text{active path of }R}\beta_J. \end{align*}\] The root and its descendants contain only their eligible tops; sums over tops here and subsequently are understood with that convention. Theorem 46, namely Equation (199), bounds each integral in the second line by \(L^{Cp}(1+M)^C\int h_R\). Moreover, \[\sum_R\int h_R\lesssim L^C S,\] by the definition of the dilated decaying weights. We conclude that \[\int_Z V_v\le L^{Cp}(1+M)^C S.\] Since \(M\) is a fixed power of \(L\), taking a \(p\)-th root and using \(n_v^{\alpha-p}\|A_v\|^p\le V_v\) proves the result. ◻

Proof. Let \[\mathfrak S_\mu =\{a\in[0,\infty)^K:\ \textstyle\sum_\sigma a_\sigma=\mu\}, \qquad \mathcal F_v(\zeta,z)=n_v(z)^{\alpha/p-1}A_v(\zeta,z).\] Every nonempty mass is positive. For each fixed \(z\), the defining sum is a finite sum of exponentials of linear functions of \(\zeta\), with real logarithms of positive ratios as coefficients. In particular it is entire in \(\zeta\).

We first give a uniform bound on the tube over \(\mathfrak S_\mu\), without using the vertex estimate. From Equations (170) and (112), and from \(\int n_v\le\int n\lesssim L^C S\), we have, for \(a\in\mathfrak S_\mu\) and \(\eta\in\mathbb R^K\), \[\begin{align*} \|\mathcal F_v(a+i\eta)\|_{L^p(Z)} &\le L^C(1+\mu+\|\eta\|_2) \left(\int_Z n_v^\alpha\right)^{1/p}\\ &\le L^C(1+\|\eta\|_2) (NL^C)^{(\alpha-1)/p}S^{1/p}. \tag{257}\end{align*}\] The factor involving \(N\) is at most \(\exp(C\sqrt L)\), since \(\log N\le2L\), \(p=L^{1/2}\), and \(2<\alpha<3\) for sufficiently large \(L\).

Multiply the analytic family by \[\mathcal D(\zeta)=\exp\left(\sum_\sigma\zeta_\sigma^2\right).\] On the tube under consideration, \[|\mathcal D(a+i\eta)| =\exp(\|a\|_2^2-\|\eta\|_2^2) \le\exp(\mu^2-\|\eta\|_2^2).\] At a vertex, Proposition 47 controls \(\mathcal D\mathcal F_v\) whenever \(\|\eta\|_2\le L^2\). For \(\|\eta\|_2>L^2\), Equation (257) gives the same bound: indeed \[(1+\|\eta\|_2) \exp(\mu^2-\|\eta\|_2^2+C\sqrt L)\] is bounded by an absolute constant in this range, after increasing the fixed lower threshold for \(L\). Thus the damped family has norm at most \(L^C S^{1/p}\) at every vertex, uniformly in all imaginary parts.

Here are details that justify applying scalar three-lines to this family. Test against any element \(b\) of the dual unit ball \(L^{p'}(Z;\mathcal H)\). One may first restrict the pairing to a bounded set on which every nonempty mass used in the finite family is bounded below by a positive constant. Such sets exhaust \(Z\), and the restriction does not increase the norm of \(b\). On each such set, integration defines an entire scalar function; logarithms of the finitely many masses are bounded there. The crude bound above and the damping make this scalar function bounded on every interpolating strip. If two real vectors \(a_0,a_1\) have established uniform bounds for all imaginary parts, apply the bounded scalar three-lines lemma (Hajłasz 2012, Lemma 3.3, p. 57) to \[z\longmapsto \int_Z \mathcal D\big((1-z)a_0+za_1+i\eta\big) \big\langle \mathcal F_v\big((1-z)a_0+za_1+i\eta,\cdot\big),b \big\rangle\,\,\mathrm dx,\] with the indicated preliminary restriction understood. The boundary lines have the required bounds because their imaginary vectors are still arbitrary. The real parts throughout the strip remain in the simplex. Starting with its vertices and successively taking convex combinations proves the same bound at their barycenter, with no factor depending on \(K\). Exhaustion of \(Z\), followed by duality, removes the preliminary restriction. Finally \(\mathcal D(\zeta^{\mathrm c})=\exp(\mu^2/K)\) is an absolute factor. This proves the lemma. ◻

Proof. Fix \(z\in Z_0\) and a top \(T\), and suppress \(z\) from the masses. For \(i<k_T\) let \[r_i(\vartheta) =\log\frac{m_{v_{i+1}(T)}}{n_{v_i(T)}}\ge0, \qquad M_i(b)=\sum_{\substack{U:\,|s_U-s_T|\le b\,d(i)\\ d_U\le d(i)/B^2}}h_U.\] Unlike \(r_i\), the latter mass does not depend on the grid translation. Each mass in this display is at least \(h_T\) when \(i<k_T\), and is at most \(n\).

Let \(b_i(\vartheta)\in[0,1/2]\) be the distance from \(s_T\) to the nearer boundary of its level-\(i\) interval, divided by \(d(i)\). An extra top counted in \(m_{v_{i+1}(T)}-n_{v_i(T)}\) has slope within \(\epsilon d(i)\) of \(s_T\). Indeed the neighbor distance in Equation (153) is \(G^{-1/4}d(i)\), and the two level-\((i+1)\) interval lengths add at most \(2d(i)/G\). Such a top is eligible at level \(i+1\), hence also at level \(i\). It lies outside the immediate parent. It follows that \(r_i=0\) if \(b_i>\epsilon\). If \(0<b_i\le\epsilon\), the immediate parent contains every top counted by \(M_i(b_i)\), whereas its extra mass is disjoint from these inner tops and is at most \(M_i(\epsilon)-M_i(b_i)\). Consequently \[ r_i(\vartheta) \le\log\frac{M_i(\epsilon)}{M_i(b_i(\vartheta))}. \tag{261}\] Endpoint coincidences can be omitted: for the finite set of slopes and levels they form a null set of translations.

For a fixed number \(0<b<\epsilon\), set \(r=\lceil\log_G(\epsilon/b)\rceil\). The inclusion \[ M_{i+r}(\epsilon)\le M_i(b) \tag{262}\] uses both conditions in the definition of \(M_i\): the radius satisfies \(\epsilon d(i+r)\le b d(i)\), and the precision threshold \(d(i+r)/B^2\) is no larger than \(d(i)/B^2\). To handle the end of the path without losing a term, define \[a_i=\begin{cases} \log(M_i(\epsilon)/h_T),&0\le i<k_T,\\ 0,&i\ge k_T. \end{cases}\] If \(i+r<k_T\), Equation (262) applies. If \(i+r\ge k_T\), use \(M_i(b)\ge h_T\) instead. In both cases \[\log\frac{M_i(\epsilon)}{M_i(b)}\le a_i-a_{i+r}.\] Summing and cancelling gives \[ \sum_{i<k_T}\log\frac{M_i(\epsilon)}{M_i(b)} \le\sum_{0\le i<\min(r,k_T)}a_i \le r\log\frac n{h_T}. \tag{263}\] This argument also covers \(k_T=0\), with empty sums.

Because \(d(0)/d(i)=G^i\) is an integer, the common uniform translation is uniform modulo every \(d(i)\). Thus each \(b_i\) has density \(2\) on \([0,1/2]\). Linearity of expectation, Equation (261), and Equation (263) imply \[\begin{align*} \mathbb E_\vartheta\sum_{i<k_T}r_i(\vartheta) &\le2\int_0^\epsilon \sum_{i<k_T}\log\frac{M_i(\epsilon)}{M_i(b)}\,\,\mathrm db\\ &\le2\log(n/h_T) \int_0^\epsilon\lceil\log_G(\epsilon/b)\rceil\,\,\mathrm db \le C\epsilon\log(n/h_T). \tag{264}\end{align*}\] For completeness the integral in the last line equals \(\epsilon/(1-G^{-1})\): express the ceiling as a sum of indicators of \(b<\epsilon G^{-j}\), \(j\ge0\), and integrate. The dependence of the different \(b_i\)’s is absent from this calculation.

On the fixed top \(T\), the value of \(\kappa_s\) depends only on the stopping level \(k_s\). As packet length increases, \(k_s\) increases; Equation (165) shows that these values are nonincreasing. Their deficits from one are nondecreasing and satisfy \[0\le1-\kappa_s \le\frac\mu K\sum_{i<k_T}r_i(\vartheta),\] using \(k_s\le k_T\) and \(1-e^{-x}\le x\). Group the packets by stopping level. Every interval of these levels is an interval of packet lengths, so Equation (110) bounds its unweighted sum by \(L^C h_T^8\). Summation by parts against the monotone deficits, and \(|q_T|\le1\), therefore give \[\left\|\sum_{s:T(s)=T}(1-\kappa_s)q_T\phi_s\right\| \le L^C h_T^8\frac\mu K\sum_{i<k_T}r_i(\vartheta).\] For example, write the deficits as sums of their nonnegative increments and use the bound on the corresponding suffix sums; the sum of these increments is their largest value. Taking expectations, using \(\mu/K\le1\), and summing over tops proves \[\mathbb E_\vartheta\|P_q-\mathcal W_\vartheta\| \le L^C\epsilon\sum_T h_T^8\log(n/h_T).\] The weights have a fixed absolute upper bound. For every such weight, \[h_T^8\log(n/h_T) \le C h_T(1+\log_+ n),\] because \(h^7\log_+(1/h)\) is bounded on a fixed bounded interval. Equation (112) gives \(1+\log_+n\lesssim L\) on \(Z_0\). This proves Equation (260). ◻

Proof. We make the exceptional-set comparison explicit because the good set depends on the grid. Write \(\|P\|_2\le L^{C_P}\sqrt{|E|}\), as supplied by Equation (43). Proposition 29 and Proposition 30 allow the fixed polynomial thresholds to be chosen so that \[ \int_{F\setminus Z_0}\|P\|\le L^{-30}N|F|, \qquad \int_{Z_0}\|P-P_q\|\le L^{-30}N|F|. \tag{265}\] Indeed a discarded area at most \(L^{-2C_P-60}|F|\) has the stated first bound by Cauchy–Schwarz. The exponentially small concentration and count exceptions meet this area bound for sufficiently large \(L\).

For each fixed translation the good-path construction gives \[|Z_0\setminus Z_\vartheta|\le S e^{-9L}.\] This is Markov’s inequality applied to the weighted bad-path mass: its integral is at most \(S e^{-10L}\), and the deletion threshold is \(e^{-L}\). Since \(S\lesssim L^C N^2|F|\) and \(N^2\le e^{4L}\), \[ \int_{Z_0\setminus Z_\vartheta}\|P\| \le L^{C_P}\sqrt{|E|}\,(S e^{-9L})^{1/2} \le L^C e^{-5L/2}N|F| \le L^{-30}N|F|. \tag{266}\] These are bounds for each translation; no union over translations is taken.

For every \(\vartheta\), the triangle inequality gives the following decomposition, whose left side is independent of \(\vartheta\): \[\begin{align*} \int_F\|P\| &\le \int_{F\setminus Z_\vartheta}\|P\| +\int_{Z_\vartheta}\|P-P_q\| +\int_{Z_\vartheta}\|P_q-\mathcal W_\vartheta\| +\int_{Z_\vartheta}\|\mathcal W_\vartheta\|. \tag{267}\end{align*}\] The first two terms are controlled by Equations (265) and (266). The final term is controlled by Equation (259) with \(Q=30\). Average the remaining term over translations. It is nonnegative and \(Z_\vartheta\subset Z_0\), so Tonelli’s theorem and Lemma 49 yield \[\begin{align*} \mathbb E_\vartheta\int_{Z_\vartheta} \|P_q-\mathcal W_\vartheta\| &\le\int_{Z_0}\mathbb E_\vartheta \|P_q-\mathcal W_\vartheta\|\\ &\le L^C\epsilon\int_{Z_0}n \le L^C\epsilon N|F| \le L^{-30}N|F|. \end{align*}\] The penultimate inequality uses Equation (112); the last uses \(\epsilon=4G^{-1/4}\) and \(\log G\asymp L^{.7}\). All functions being averaged are measurable: for finitely many tops, packets, and levels, the angular assignments change only at finitely many translation values, and the good sets were defined by measurable quantities. Thus this averaging is legitimate even though \(Z_\vartheta\) and \(\mathcal W_\vartheta\) use the same translation. Averaging Equation (267) now proves the asserted estimate. In particular, on every discarded set we have estimated the original sum \(P\). ◻

Proof. The first and third assertions are precisely the ordinary estimates proved with (43). For the second, the fixed deletion of tiles with \(D_E(s)\le L^{-C_2}\) in 2.7 contributes at most \(CL^{-40}\sqrt{|F||E|}\), by (68). Assign the remaining tiles to the popular tops as already constructed. Their adjoint sum is \(P\), and the coefficient convention in Equation (33) gives \[|\Lambda_{\mathrm{popular}}(f,g)| \le\int_F\|P\|.\] Theorem 50 bounds this by \(L^{-20}N|F|=L^{-20}\sqrt{|F||E|}\). Adding the deleted contribution proves the second assertion. All preceding arguments apply to arbitrary finite tile sets and thus give the claimed uniformity. ◻

Completion of the proofs of [thm:band-restricted,thm:main]. Theorem 51 supplies exactly the three hypotheses of Proposition 15. That proposition therefore proves the band restricted estimate (4). The dependence here is forward: the reductions established the implication from the model estimates, which have now been proved without using the band restricted theorem.

All estimates were obtained first for finite packet collections and positive inner truncations, with constants independent of both. The summable smooth expansions and translated-lattice averages in the packet reduction consequently pass to their limits as specified in Proposition 15. The amplitude summation, vertical Littlewood–Paley reassembly, and spatial localization proved in the reductions give a uniform strong \(L^2\) estimate for the original fixed-field transform with a positive inner cutoff. For a Schwartz input its symmetric principal value exists pointwise; Fatou’s lemma gives the same strong bound for the principal-value operator, which extends by density. For the already chosen outer length \(a=a_*\), Chebyshev’s inequality gives the weak \((2,2)\) assertion of Theorem 1. The constants are independent of the vector field, the input, and the level in that inequality. The dilation calculation following 1 gives the outer length \(a_*/\mathop{\mathrm{Lip}}(v)\) for positive Lipschitz seminorm; constant directions have already been handled by the one-dimensional theorem. ◻