Diagonal Fourier extension for positively curved surfaces in three dimensions

Proof. Choose a smooth cutoff \zeta_0 equal to one on B(0,\rho), supported in B(0,2\rho), and identically 1/2 on a neighborhood V of a point \xi_1 with \rho<|\xi_1|<2\rho. It can be chosen by rescaling a fixed cutoff with an annular plateau. The phase

has Hessian I+O(\rho^2): the remainder q_s-Q_s and its derivatives of orders zero, one, and two are respectively O(\rho^4), O(\rho^3), and O(\rho^2) on the support of the cutoff. Fix \rho small enough that q'' is positive definite everywhere.

After translating the target base point, let its height function be h_0 near zero and put H=h_0''(0)>0 and B=q''(\xi_1)>0. Choose an invertible real matrix A with A^{\mathsf T}HA=B; for example, A=H^{-1/2}B^{1/2}. Define, near \xi_1,

Thus \psi and q have the same value, gradient, and Hessian at \xi_1. This change of graph is induced by an invertible affine map of ambient frequency space: its base part is x\mapsto \xi_1+A^{-1}x, and its height part is addition of an affine function of the new base coordinates.

Let \vartheta\in C_c^\infty(B(0,2)) equal one on B(0,1) and take 0\le\vartheta\le1. For a small fixed \epsilon>0 put

Choose \epsilon so that its support is contained in V and the target graph is defined there. Taylor’s theorem and the agreement of the two-jets give

Consequently \phi=q+R_\epsilon remains uniformly convex for sufficiently small \epsilon and agrees with \psi on B(\xi_1,\epsilon). It equals Q_s outside a compact set, so all of its derivatives of order at least two are bounded.

To verify the required continuation form, observe that D=q_s-Q_s is strictly positive for 0<|\xi|<1. Shrink V, if necessary, so that D\ge d_0>0 there. The function R_\epsilon/D, extended by zero outside V, is smooth. Set

On the perturbation support, \zeta_0=1/2 and |R_\epsilon/D|\le C\epsilon^3/d_0. Choosing \epsilon smaller makes this last quantity at most 1/4. Hence 0\le\widetilde\zeta\le1, with the same inner and outer support properties as \zeta_0, and \phi=Q_s+\widetilde\zeta(q_s-Q_s). Every choice was made using only the fixed target graph.□

Proof. Since N^2w\theta=1, the vectors N^2z run through (D_{\mathrm f}\theta)^{-1}\mathbb Z^2. For each fixed \nu, Equation (3.3) is the Fourier coefficient array of f\chi_\theta^\nu e(N^2t_I\phi) on a square of side D_{\mathrm f}\theta. Parseval and Fourier inversion give

and, after synthesis, f(\chi_\theta^\nu)^2. Sum over \nu and use \sum_\nu(\chi_\theta^\nu)^2=1. The norm of synthesis follows from \mathsf S_{m,I}=(D_{\mathrm f}\theta)^{-2}\mathsf P_{m,I}^*. These identities first hold for finite partial Fourier sums in L^2 and then for their limits.□

Proof. The disk of radius \theta, with its two translations chosen independently, contains any prescribed index in both tests. Its weight is \theta^2, so K_{\theta,r}(c)\ge\|c\|_\infty^2 and

For every atomic domination of c, the triangle inequality in \ell^3 now gives

Taking the infimum and summing its cube over bins with weight r proves Equation (3.8). A singleton has cost w^{2/3}|c_{\nu,z}|, so the infimum is finite; the displayed lower bound makes it positive away from zero. Homogeneity, subadditivity, and absolute monotonicity follow directly from the definition, as does contraction under masks.□

Specialization of the cited theorem. Set m=N. The scales in Equation (3.2) become

Thus the companion’s synthesis and analysis operators are exactly the displayed operators, with its window constant L_{\mathrm f} equal to our D_{\mathrm f}. The initial gauge is exactly the last expression in Equation (3.13), and Equation (3.8) gives

All bounded label sets fit one fixed admissible complexity range for sufficiently large N, since m=N; the finitely many smaller dyadic scales may be included by enlarging the constant. The bound at any such fixed scale follows from finite-dimensional boundedness and \mathcal K_\kappa(d)\ge\|d\|_\infty^2. Applying Equation (3.10) proves Equation (3.13). This argument uses arbitrary input arrays, so it also applies after any restriction of their support.□

Proof. By Equation (2.4), the velocity map is bi-Lipschitz on the fixed frequency-label range, where it is bounded. The map U\mapsto (U,1)/\sqrt{1+|U|^2} is bi-Lipschitz on this bounded range: its inverse is \omega\mapsto\omega'/\omega_3, whose derivative is bounded where \omega_3 is bounded away from zero. Thus the directions

are separated by a fixed positive multiple of \delta. Choose spherical caps \Omega_\nu centered at \omega_\nu with radius c_1\delta, where the fixed c_1>0 is small enough that these caps are disjoint. Their surface areas satisfy c_2\delta^2\le |\Omega_\nu|\le C_2\delta^2.

The center segment of T_{\nu,z} has Euclidean length \sqrt{1+|U_\nu|^2}, bounded by a fixed constant. For each \omega\in\Omega_\nu, its distance from the line through (z,0) in direction \omega is O(\delta). Adding the horizontal cross-section of radius \delta preserves this bound. The projection of the entire tube onto that line is an interval of bounded length. Cover that interval by a bounded number of intervals of length one. It follows that T_{\nu,z} is contained in at most C_3 cylinders \mathcal T_{C_4\delta}(a,\omega), where C_3 and C_4\ge1 are fixed. In particular, for every g\in L^3(\mathbb R^3),

Averaging over \Omega_\nu, multiplying by |d_{\nu,z}|^2, and summing gives

Set B_d=\sum_\nu b_\nu(d)\mathbf1_{\Omega_\nu}. Disjointness of the caps gives

Choose N_0 so that C_4\delta<1. Hölder’s inequality on S^2 and Equation (4.3) now yield

We used (C_4\delta)^{-s}\le\delta^{-s}, so the displayed constant is independent of s. Duality between L^{3/2} and L^3, followed by taking the power 3/2, proves Equation (4.4). The sum over positions causes no loss because K_\rho already takes the supremum over cylinder locations.□

Proof. Write D=\max_{\nu,z}|d_{\nu,z}|^2>0. A disk of radius \delta, independently translated to a prescribed velocity and position, is an admissible test with weight \delta^2. On the other hand, every test has W_\kappa(E)\ge\delta^2. Hence

Capacity is monotone under restriction to a subset of coordinates.

Put A=D/n^2. The cost of any remaining array with capacity at most A is controlled by the contribution of a largest original coefficient to the tube integral. Indeed, any coordinate subarray r with \mathcal K_\kappa(r)\le A satisfies

An original entry whose squared modulus is D contributes D\mathbf1_{T_{\nu,z}} to H_d, whence \int H_d^{3/2}\ge\pi\delta^2D^{3/2}. Thus the cost of any such remainder is controlled by \pi^{-1}\int H_d^{3/2}, independently of which entries remain.

Start with remainder r_1=d. Whenever \mathcal K_\kappa(r_j)>A, choose an ellipse test whose density is at least half of that supremum. Let q_j contain all nonzero coordinates of r_j selected by that test, write E_j for its ellipse, and put

Such a test exists without assuming that the supremum is attained. It selects at least one nonzero entry, so the process terminates after at most n removals. Denote the terminal remainder by r_*; then \mathcal K_\kappa(r_*)\le A, including the possibility r_*=0.

We group the removed bundles by the capacity of the remainder from which they were selected. This controls the capacity of each group by its first remainder, while leaving only logarithmically many groups. For each integer k\ge0, set \lambda_k=2^kA, and group together all q_j for which

Let d_k be the union array at this level, omitting empty levels. The array d_k is a coordinate restriction of the remainder at the first removal in that level. Therefore \mathcal K_\kappa(d_k)<2\lambda_k. Every bundle at this level has \rho_j\ge\lambda_k/2, so additivity of m gives

Use Equation (4.5) for each selected test. Since the q_j are disjoint coordinate subarrays, their tube functions sum to a function bounded by H_d. For nonnegative numbers, \sum_j x_j^{3/2}\le(\sum_jx_j)^{3/2}. Consequently

The stopping estimate applies to r_*. Adding its cost to Equation (4.8) proves Equation (4.6).

Finally, at every removal the capacity lies in (A,nD], by monotonicity and Equation (4.7). The ratio of these endpoints is n^3, so there are at most 1+\lfloor3\log_2 n\rfloor occupied dyadic levels. Including the nonzero remainder gives J\le C\log(2+n). If n=1, then \mathcal K_\kappa(d)=D=A; the process makes no removal and the remainder alone gives the conclusion. No lower bound on the smallest coefficient has been used.□

Proof. There is nothing to prove if d=0. Otherwise use the decomposition in Lemma 4.2, and let c^{(i)} be the terminal array of d^{(i)} with the same output mask. The sampling map is linear, so c=\sum_i c^{(i)}. The triangle inequality in \ell^3, Hölder’s inequality for the finite sum over i, and Theorem 3.4 give

There are O(N^2) initial frequency labels and O(N^2) initial position labels in the fixed bounded ranges, so n\le C N^4. It follows that

Insert Lemma 4.1 and use \delta^{-3s/2}=N^{3s/2} to obtain Equation (4.9). Every coordinate restriction used above obeys the same label bounds as d, so the same packet constant P(\kappa,\eta) applies to every piece.□

Proof. Write \delta=N^{-1} and D=D_{\mathrm f}. Every window \chi_\delta^\nu meeting Q_0 lies in a fixed compact neighborhood Q_1 of Q_0, since \delta\leq1. Fix

The initial spatial labels will satisfy |z|\leq Z and the terminal spatial labels |z'|\leq1. Thus all label bounds used in Proposition 4.3 are fixed independently of the small exponents. Throughout the proof we decrease \mu_0, if necessary, only in terms of this fixed setup. In particular, we may require N to exceed any fixed lower threshold needed below. The hypotheses give

whenever H is the separation of two distinct centers.

Removing resonant frequencies. Set a_B=x_B-(0,0,N^2/2). For each pair of distinct balls write a_B-a_{\widetilde B}=(q,s_0)\in\mathbb R^2\times\mathbb R and H=|(q,s_0)|. Its resonant set is

If this set is nonempty, then

so |s_0|\geq H/[2(G+1)] for sufficiently small \mu_0. For two points of \Omega_{B,\widetilde B}, Equation (2.4) gives

Consequently each resonant set has area at most C\mu^{40}. The union \Omega over all pairs therefore has area at most C\mu^{24}. Define h_*=h\mathbf1_{Q_0\setminus\Omega}, and let F_* be the extension defining F with h replaced by h_*. By Hölder’s inequality,

The powers 20, 40, and 100 provide ample room both for this discarded mass and for the phase estimates below; their precise values are not essential.

A joint bound for the initial coefficients. Let

For u\in\mathcal U and B\in\mathcal B define

Only angular windows meeting Q_0 need be retained. We claim that, uniformly in u,

If h_*\chi_\delta^\nu=0 almost everywhere, the claim is immediate. Otherwise choose a point \xi_g\in\Gamma_\nu\cap(Q_0\setminus\Omega). For this \nu, define the operator on unrestricted L^2(\mathbb R^2) by

We will apply it to f=h_*\mathbf1_{\Gamma_\nu}. This enlargement of the operator domain is useful: the kernel of T_\nu T_\nu^* has the smooth amplitude (\chi_\delta^\nu)^2, with no indicator of the excised set.

For a single ball block, Parseval’s identity on the square of side D\delta containing \Gamma_\nu gives

Here the full spatial sum is understood before its restriction to |z|\leq Z; its Fourier frequencies are N^2z\in(D\delta)^{-1}\mathbb Z^2. It follows that each diagonal ball block of T_\nu T_\nu^* has operator norm at most C\delta^2.

For distinct B,\widetilde B, an entry of this Gram operator is

To estimate it, put \xi=\nu+\delta v and denote the resulting phase by \Psi(v). Let

The direction v_0 is defined because \xi_g avoids every resonant set. On the entire support of the smooth amplitude \chi(v)^2,

For Equation (5.6), the two errors relative to H\mu^{20} are at most C\mu^{30} and 32Z\mu^{20} by Equation (5.2); their sum is at most 1/2 after fixing \mu_0. Equation (5.7) follows from the bounded higher derivatives of \phi and \delta\mu^{-20}\leq\mu^{30}/2.

Repeated integration by parts in the fixed direction v_0 now gives, for each fixed integer k\geq1,

For completeness, use the operator (2\pi iD_{v_0}\Psi)^{-1}D_{v_0} on the exponential. The lower bound for D_{v_0}\Psi and the higher derivative bounds imply that each fixed directional derivative of (D_{v_0}\Psi)^{-1} is O_k(\Lambda^{-1}). Applying the adjoint k times to the fixed smooth bump therefore costs O_k(\Lambda^{-k}). No upper bound on the first derivative is required. In particular, these constants remain uniform as H becomes arbitrarily large: the lower bound and the higher derivative bounds carry the same factor H.

There are at most C_ZN^2 retained spatial labels in each ball block. Thus every row and column of the off-diagonal block matrix has at most C_Z\mu^{-8}N^2\leq C_ZN^3 entries. Taking k=4 in Equation (5.8) and applying the row and column sum bound for the \ell^2 operator norm shows that its norm is at most C\delta^2. Together with the diagonal block estimate, this gives \|T_\nu\|_{2\to2}^2\leq C\delta^2. Applying it to h_*\mathbf1_{\Gamma_\nu} proves Equation (5.5).

Since the windows have bounded overlap and |\Gamma_\nu|\leq C\delta^2, Hölder’s inequality gives the precise power needed for Proposition 4.3:

Reconstruction and the discarded spatial tail. Let d^{B,u} now denote the finite array in Equation (5.4) restricted to |z|\leq Z. Apply the synthesis and sampling map to this array, retaining only \nu'=\nu_*, |z'|\leq1, and 0\leq j<N^2. Write the resulting array as c^{B,u}_{z',j}. We next justify, uniformly in B,u,z',j,

To do so, write \widehat d^{B,u} for the initial analysis array with the spatial grid untruncated. Lemma 3.2 at the initial scale gives

The same lemma gives the exact squared-mass identity

These statements require only h\in L^2(Q_0), which follows from the assumed L^3 bound and compact support. In particular, the possibly very large center modulation does not affect the bound.

A matrix entry from an omitted |z|>Z to a retained output is

After the substitution \xi=\nu+\delta v, the prefactor is D^{-2}, the amplitude has uniformly bounded smooth derivatives, and the derivative of the phase in the direction -z/|z| is at least

All phase derivatives of order at least two are bounded uniformly, because j\delta^2\leq1. The same integration by parts argument therefore gives

There are O(N^2) angular labels, and on the spatial mesh D^{-1}N^{-1}\mathbb Z^2,

Consequently each discarded part of a matrix row has squared \ell^2 norm at most C_kN^{4-2k}. By Equation (5.12), its pairing with \widehat d^{B,u} is at most C_kN^{1-k}. Taking k=4 proves Equation (5.10). One may justify all infinite sums by the L^2 convergence in Equation (5.11); the tail pairing is also absolutely convergent by Cauchy–Schwarz. Thus neither the data nor the sharp excision boundary has been differentiated.

Summing samples and filling the balls. Apply Proposition 4.3 to each retained array and sum over B. Equation (5.9) cancels its factor N^6 exactly and gives

There are O(N^4) terminal spatial labels and N^2 times per ball, so the total number of samples is O(\mu^{-8}N^6)=O(N^7). The cubed errors from Equation (5.10) sum to at most CN^7N^{-9}=CN^{-2}. Using |a+b|^3\leq4(|a|^3+|b|^3) and P,M\geq1, we obtain

uniformly for u\in\mathcal U.

Integrating in the common offset u turns the sum for each ball into an integral over the disjoint half-open cells

These cells cover B. Indeed, for y\in B set r=y-a_B; then

The lower corner \ell' of its horizontal cell in D^{-1}\mathbb Z^2 satisfies |\ell'|\leq N^2/4+\sqrt2/D<N^2 for the fixed large N under consideration. Hence z'=\ell'/N^2 is a retained label, and j=\lfloor r_3\rfloor is one of the retained times. The change of variables has Jacobian one; the offset domain has fixed volume D^{-2}. Equation (5.14) therefore implies

On \{|F|>\mu\}, Equation (5.3) gives |F_*|>\mu/2. Chebyshev’s inequality now proves Equation (5.1).

All choices made in this proof involve only fixed geometric constants and fixed orders of differentiation. Thus \mu_0 is independent of \kappa,\eta,s, as asserted.□

Proof. Let \lambda be the finite complex measure on the graph of \phi whose density in base coordinates is g. For v\in C_c^\infty(\mathbb R^3), Cauchy–Schwarz gives

Put m=\xi+\xi' and w=(\xi-\xi')/2; this linear change has absolute Jacobian one. Write w=r\omega, where r>0 and \omega\in S^1, and set

Uniform convexity implies

Thus H_{m,\omega} is strictly increasing for r>0, and the substitution s=H_{m,\omega}(r) yields

The omitted set r=0 has measure zero. Consequently \lambda*\lambda has an L^2 density with norm at most (\pi/c)^{1/2}\|g\|_2^2, by the L^2 representation theorem. Its Fourier transform is G(-\,\cdot)^2 in the sense of tempered distributions. Plancherel therefore gives Equation (6.1); this reasoning applies to the stated rough data and does not assume the convolution has a density in advance.

For g=h\chi_1^{\nu_*}, compact support and Hölder’s inequality bound \|g\|_2 by a fixed multiple of \|h\|_3. Furthermore

Differentiation under the integral is valid because the integrands are supported in a fixed compact set. Another application of Hölder’s inequality proves Equation (6.2).□

Proof. First take any finite collection \mathcal Q of occupied cubes and choose x_Q\in Q\cap A_\mu for each of them. By Equation (6.2), on B(x_Q,r_\mu), where r_\mu=\mu/(2C_{\mathrm b})\leq1/2, one has |F|>\mu/2. These balls have overlap at most 125: if a point belongs to one of them, its center lies within distance one, and at most 5^3 unit lattice cubes can contain such centers. Separation of the selected points is not required. Therefore

For example, one may take C_X=\max\{1,12000C_{\mathrm b}^7/\pi\}. This bound holds for every finite collection of occupied cubes. An infinite collection would have finite subsets of arbitrarily large cardinality, a contradiction. Thus the full collection is finite with the same bound. Finally C_X\mu^{-7}\leq\mu^{-8} when \mu\leq C_X^{-1}.□

Proof. If X is empty there is nothing to cover. Otherwise, using open balls throughout, set

For each x there is an index 0\leq j<J with

Indeed, failure at every index would imply n_J(x)>\mu^{-\beta J}n_0(x)\geq\mu^{-\beta J}>\mu^{-8}, since \beta J>8, contrary to \#X\leq\mu^{-8}. Assign each x one such index j and the unique integer k\geq0 for which

There are at most J(1+8\log_2(1/\mu)) nonempty groups with a common pair (j,k).

In one group, choose a maximal pairwise disjoint collection of the open balls B(x,R_j), and denote its centers by Y. The collection is finite. Each unselected ball intersects a selected one, so every center in the group is within distance 2R_j of a point of Y. A point of the unit cube containing that center is within a further distance \sqrt3. Since R_j\geq\mu^{-100}, the balls \{B(y,4R_j):y\in Y\} cover all cubes assigned to the group.

Make a finite graph on Y, joining distinct x,y when |x-y|<R_{j+1}/2. For a fixed vertex x, the original balls B(y,R_j) of all its neighbors are disjoint and lie in B(x,R_{j+1}): Equation (6.3) gives R_j<R_{j+1}/2. Each neighbor ball contains at least 2^k points of X, whereas Equation (6.7) gives

The degree of x is therefore less than 2\mu^{-\beta}. A greedy coloring uses at most 3\mu^{-\beta} colors. Centers of the same color are separated by at least R_{j+1}/2, and

by Equation (6.3). Each color thus supplies a family satisfying the separation condition, with at most \#X\leq\mu^{-8} balls and R=4R_j\geq\mu^{-100}.

Multiplying the group and color counts proves Equation (6.5). A dyadic N with 4R\leq N^2<16R exists, and satisfies

The argument uses neither the absolute locations of the points nor the diameter of X. In particular, different clusters may be arbitrarily far apart. The use of open balls also resolves boundary cases: unselected balls meet selected balls by maximality, and equality in the graph’s distance threshold leaves the desired weak separation inequality intact.□

Proof. By homogeneity it suffices to take \|h\|_3\leq1. In Proposition 5.1 choose \kappa=\eta=s=\theta. These choices lie in the permitted parameter ranges, and

For J in Equation (6.4), we have

Consequently the scale in every family furnished by Lemma 6.3 satisfies

Applying Proposition 5.1 to each family and summing over the cover yields, for 0<\mu\leq\mu_0,

Here the colors cost \mu^{-\beta}=\mu^{-\tau/4}, while the scale factor costs at most \mu^{-\tau/4}. The group count contributes one factor of J(1+\log(1/\mu)), and the logarithm squared in Proposition 5.1 contributes the other two. All constants and \mu_0 are independent of \tau.

The layer-cake identity, valid by Tonelli even before finiteness of the norm is known, is

For its portion below \mu_0, Equation (6.11) and p=3+\tau reduce the needed integral to

For example, the substitution u=\log(1/\mu) turns the integral into \int_0^\infty e^{-\tau u/2}(1+u)^3\,du, which gives the displayed identity by integrating each monomial. Since p\leq4 and J^3\leq34^3\tau^{-3}, the small-level contribution is at most C P(\theta,\theta)M(\theta)^{3/2}\tau^{-7}.

For the remaining levels, again by Tonelli,

This is uniform for 3<p\leq4 and is absorbed into the previous bound because P,M\geq1 and \tau\leq1. Taking the pth root and restoring homogeneity proves Equation (6.9). If p>4, directly use

□

Proof. Choose a nonempty interior chart; a nonempty smooth surface with boundary also has such a chart. Write it in the affine graph form

where L is invertible, and write d\sigma=J_\Sigma(\xi)\,d\xi on this chart. Take a fixed nonzero g\in C_c^\infty(D) and define f(\Psi(\xi))=g(\xi)/J_\Sigma(\xi) on the chart, with f=0 elsewhere. This is a smooth surface datum with

Let

Horizontal Plancherel and its derivative identity give, for t\geq0,

where one may take B_g=2(2\pi)^{-2}\|\nabla g\|_2^2+2\|g\nabla\phi\|_2^2>0. Set K=\max\{1,(2B_g/A)^{1/2}\}. The integral outside the disk |y'|<K(1+t) is at most B_g/K^2\leq A/2, so at least A/2 of the squared norm lies inside that disk. Hölder’s inequality on the disk implies

uniformly for 3\leq p\leq4, with the explicit positive choice

Integrating Equation (6.14) over t\geq0 yields \|G\|_p^p\geq c_g/(p-3).

The affine coordinates introduce exactly the factors

Hence, on setting d_g=|\det L|^{-1}c_g>0, we obtain

This proves Equation (6.13), with every choice of chart and datum fixed independently of p.□

Proof of Corollary 1.2. On the compact paraboloid \Sigma=\{(y,|y|^2):|y|\le1\}, write \dd\sigma=w(y)\dd y, where w(y)=\sqrt{1+4|y|^2}. Apply Theorem 1.1 to the datum g(y)/w(y) on \Sigma. A fixed change of output variables then gives

This is the compact-frequency extension estimate in Lee–Rogers–Seeger [10, Equation (1.3)] with both exponents equal to p. Their Theorem 1.1, with spatial dimension two and Besov summation exponent two, yields

For an inhomogeneous Littlewood–Paley partition (P_k)_{k\ge0}, the uniform annular multiplier bound \|P_kf\|_p\le C_s2^{-ks}\|(I-\Delta)^{s/2}f\|_p gives

The series converges for s>\alpha. This proves Equation (1.2) on Schwartz functions, and density gives the stated bounded extension.□

Proof. Shrink the product neighborhood so that its input domain is a ball. Since M is a Hessian in y, commuting its derivatives in \partial_tM=cM gives

Multiplication by (M^{-1})_{j\ell}, summed over j, yields (\partial_{y_k}c)\delta_{i\ell} =(\partial_{y_i}c)\delta_{k\ell}. For each k, choose \ell=i\ne k; hence \nabla_yc=0. The input ball is connected, so c=c(t). Set

The equation for M now implies D_y^2(\partial_t\psi-\rho h)=0. Thus \partial_t\psi=\rho(t)h(y)+b(t)\cdot y+d(t) for smooth functions b,d. Integrating and using \psi(0;y)=0 gives

Since S'=\rho>0, it has a smooth local inverse t=t(s). The output change \kappa(u,s)=(u-B(t(s)),t(s)) therefore satisfies

Here D_y^2h=M(0,y) is definite. No input coordinate change is needed.

Put z=(u,s) and A(z,y)=a(\kappa(z);y). Extend this fixed amplitude smoothly by zero, and periodically in z across a fixed cube Q of side L whose interior contains its output support. On Q,

Indeed, integration by parts in z gives \|A_k\|_\infty\le C_m(1+|k|)^{-m} for any fixed m>3. All coefficients have a common compact frequency support, and the constants are independent of N. Choose a compact graph patch \Sigma, parametrized by \omega(y)=(y,h(y)), with smooth boundary and parameter domain containing this support, and write \dd\sigma=w(y)\dd y, where w(y)=\sqrt{1+|\nabla h(y)|^2}. For

we have

The factor e^{iNr(s)} is unimodular, and each remaining Fourier term is e^{2\pi i k\cdot z/L}E_\Sigma g_k(Nz/(2\pi)). Minkowski’s inequality and Theorem 1.1, applied globally to each term, bound their sum on Q by

Changing back by \kappa contributes only the bounded factor \sup|\det D\kappa|^{1/p} on the fixed output support. Absolute coefficient summability and the local L^1 integrability of f justify the series and its interchange with the integral. This proves Equation (7.1), with no power loss in N.□