Annular variation of the triangular Hilbert transform at the symmetric point

Proof. For each fixed t, Hölder’s inequality and translation invariance bound the L^{3/2} norm of the product by \|F\|_3\|G\|_3. Minkowski’s integral inequality gives (2.1). Apply this estimate to the countably many annuli 1/m<|t|<m, m\ge2. Their union of exceptional null sets works for every finite annulus. Absolute continuity of the integral gives endpoint continuity. The same estimate applied to a zero L^3 difference proves the assertion about representatives, first on those countably many annuli and hence on all finite annuli.□

Proof. We recall the covering proof of the Hardy–Littlewood estimates [10]. From any finite family of intervals, repeatedly select one of largest radius and discard those meeting it. The selected intervals are disjoint, and the union of the family lies in their triple-radius enlargements. Each enlargement has at most three times the original measure, also for integer intervals. Apply this to witnessing intervals centered at a finite subset of a discrete level set, or to a finite witnessing cover of a compact subset of a continuous level set. The latter level set is open because fixed-radius averages are continuous. Taking suprema over the finite or compact subsets proves the weak bound.

For f\in L^p, split at height \rho/2. The low part has maximal function at most \rho/2, while the high part is integrable. Thus

The distribution formula and Tonelli’s theorem give the strong bound after integrating p\rho^{p-1}\dd\rho. Truncation of the distribution integral first makes this argument independent of any a priori finiteness assertion.

For (2.4), put a=\sqrt\sigma/h\ge1. The kernel is (\sqrt{2\pi}a)^{-1}e^{-(m-j)^2/(2a^2)}. Split the sum into |m-j|\le a and the annuli 2^{l-1}a<|m-j|\le2^la, l\ge1. The sum of |f| in the corresponding ball is at most 3\cdot2^la M_{\mathbb Z}f(j). The resulting series is bounded by a constant times 1+\sum_{l\ge1}2^l e^{-4^{l-1}/2}. The same shell argument with integrals proves (2.5).□

Proof. Restrict to the support and put A=\sqrt P. The derivative of the squaring map at A is X\mapsto AX+XA; in an eigenbasis, it multiplies entry (i,j) by \sqrt{p_i}+\sqrt{p_j}>0. The inverse function theorem gives a smooth positive square root, whose derivative X_K solves

Multiplying by A and taking traces gives \tr(PX_K)=\tfrac12\tr(AK). Hence differentiation of \Phi(P)=\tr(P\sqrt P) yields d\Phi_P(K)=\tfrac32\tr(AK). Differentiating once more proves the Hessian formula. Formula (3.4) follows by diagonalizing A, and proves the remaining assertions. This is the square-root case of the classical divided-difference calculus [2, Section 1].□

Proof. For x\in\ker Q we have 0\le\langle Px,x\rangle \le a\langle Qx,x\rangle=0, so \ker Q\subseteq\ker P. This proves the support assertion.

We recall the order facts used in the proof. Inversion reverses order for positive definite matrices, as follows by conjugating 0<X\le Y by X^{-1/2}, inverting, and conjugating back. Also

Spectral calculus reduces this norm-convergent formula to the scalar integral. Since X(X+tI)^{-1}=I-t(X+tI)^{-1}, inversion order shows that square root preserves order. This is the classical Loewner–Heinz square-root monotonicity; see, for example, Kwong [16, Theorem 1 and Section 2]. The supported Hessian comparison below is derived from these order facts here.

For \eta>0,

If A\le B, then \mathsf L_A+\mathsf R_A\le\mathsf L_B+\mathsf R_B on Hilbert–Schmidt space: the difference has quadratic form \tr(X^*(B-A)X)+\tr(X^*X(B-A))\ge0. Invert these positive operators to obtain

Let \eta\downarrow0. In an eigenbasis, every entry of K meeting the kernel of the corresponding limiting matrix is zero. Thus Formula (3.4) gives precisely the supported costs on both sides of (3.5).□

Proof. The insertion is Hermitian and annihilates \ker R=\ker(R^*R). If R=0, there is nothing to prove. Otherwise let \rho_i>0 be the positive singular values of R, and let C be the compression of L to the corresponding left singular vectors. Then \lVert C\rVert_{\op}\le\lVert L\rVert_{\op} and

For x,y>0, x^2y^2/(x+y)\le (xy)^{3/2}/2\le(x^3+y^3)/4. Every row and column of C has squared Euclidean norm at most \lVert L\rVert_{\op}^2. Therefore

as required.□

Proof. If E=\{0\}, all matrices and costs vanish. Otherwise restrict every matrix to E. For every nonzero x\in E, the positive quantity \langle T_\omega x,x\rangle has positive integral, so \overline T is positive definite on E. All averages vanish on E^\perp, proving the support assertions.

Put A_\omega=\sqrt{T_\omega} and A=\int A_\omega\,\dd\nu. The integral exists by the stated integrability and Cauchy–Schwarz. For every vector x, the vector-valued Cauchy–Schwarz inequality gives

Consequently A^2\le\overline T, and square-root order, proved in Lemma 3.2, implies

For a positive definite selfadjoint operator \mathsf M on a real Hilbert space, completing the square proves

Apply this identity on Hermitian matrices with real Hilbert–Schmidt inner product. The operators \mathsf L_{\sqrt T}+\mathsf R_{\sqrt T} preserve that real space. By (3.8), for each Hermitian X the expression inside the supremum for (\overline T,\overline K) is at most the average of the corresponding expressions for (T_\omega,K_\omega). Taking the supremum and then bounding the supremum of an integral by the integral of the pointwise suprema gives

Every inverse in this argument is taken on matrices acting on E; uniform invertibility over \omega was never assumed.□

Proof. For \eta>0, Lemma 3.1 gives

Spectral calculus implies \lVert\sqrt{R^*R+\eta I}-\sqrt{R^*R}\rVert_{\op}\le\sqrt\eta. The matrices R(z) stay bounded for 0\le z\le1, so the derivatives converge uniformly as \eta\downarrow0, and the endpoint energies converge. Integrating first and passing to the limit proves (3.10).

For (3.11), Cauchy–Schwarz in the matrix product bounds the entry by the norm of row i of R times the norm of column j of \sqrt{R^*R}. The latter column has squared norm (R^*R)_{jj}, also the squared norm of column j of R.□

Proof. We put the three costs in one quadratic form, bound a cubic trace, and then recover the complex phase of the cyclic product.

One quadratic form. On H=H_0\oplus H_1\oplus H_2, define

Both are Hermitian and \lVert D\rVert_{\op}=D_*. If P_v is the orthogonal projection onto H_v, put A_v=MP_vM and K_v=MP_vDM. After permuting blocks and extending by zero, these are respectively T_v and R_v^*D_vR_v. The matrix K_v is Hermitian because P_v commutes with D. Moreover,

Thus K_v is supported on A_v and on M^2. Lemma 3.2 and the inequality \sum_{v=0}^2 b(X_v)\ge\tfrac13 b(\sum_vX_v) for a positive quadratic form give

A cubic trace estimate. Choose an orthonormal eigenbasis of M, with real eigenvalues m_i ordered so that q_i=|m_i| is nonincreasing. Ties are arbitrary. In that basis Formula (3.4) yields

Any term with one zero q_i is zero; pairs with both zero are omitted. Write

Hermitian symmetry and q_j\le q_i for j>i imply

Expand

and group triples by the multiplicity of their smallest index i. If it occurs once, its three cyclic positions give

The inner sum is x^*D_{>i}x for x=(m_jD_{ji})_{j>i}, where D_{>i} is the corresponding compression of D. Since \lVert D_{>i}\rVert_{\op}\le D_*, this contribution has modulus at most 3D_*E_i. The argument allows j=k and retains the signs of the m_j.

If the smallest index occurs twice, its contribution is

By Cauchy–Schwarz and \sum_k|D_{ik}|^2=\lVert De_i\rVert^2\le D_*^2,

The contribution is therefore at most 3D_*\sqrt{A_iE_i}, also when q_i=0. Finally, if all three indices equal i, the modulus is at most D_*A_i. These cases exhaust all triples. Summing and using Cauchy–Schwarz gives

Recovering the complex phase. Set \tau=\tr(D_0W_0D_1W_1D_2W_2). Since M has zero diagonal blocks, every nonzero closed three-step block product visits all three vertices. The three starting vertices in one orientation give \tau, by cyclicity of trace; the reverse orientation gives \overline\tau, because the D_v are Hermitian. Consequently

When \tau\ne0, choose |z|=1 so that z\tau=|\tau| and replace W_0 by zW_0. This changes the stars by

and leaves R_2 unchanged. These right multipliers are unitary; each conjugates the Gram and its insertion by the same unitary. Thus \mathcal E and D_* are unchanged. Apply (3.14) and (3.16) to the modified matrices, with their own value of Q, to obtain

This is (3.12). The case \tau=0 is immediate.□

Proof. We first check that the possible kernels of the Grams cause no differentiability problem. Each star has the form

where both diagonal multipliers are positive and invertible. With the column parameters fixed, \ker R_v is fixed, so T_v has fixed support as the row parameters vary. With the row parameters fixed, \operatorname{ran}R_v is fixed, so S_v has fixed support as either or both column parameters vary. For joint smoothness, factor a nonzero R_v^0 as XY^* with X,Y of full column rank. Set X'=D_{\rm row}X, Y'=D_{\rm col}Y, P'=X'^*X', and Q'=Y'^*Y'. These last two matrices are positive definite and smooth, and the nonzero eigenvalues of R_vR_v^* are those of Q'^{1/2}P'Q'^{1/2}. Consequently

is smooth. The rank-zero case is immediate.

For one Gaussian density w=h g_\sigma(u-p) and its score N=(u-p)/\sigma,

With the columns fixed, T_v is linear in the row density, and hence

The chain rule on its fixed support and Lemma 3.1 give

For a column vertex w\ne v, use S_v instead. It is linear in that column density, has fixed support, and satisfies \partial_{p_w}S_v=P_{v,w}. Thus 2\partial_{\sigma_w}e_v=\partial_{p_w}^2e_v-Y_{v,w}. Taking the variance derivative along \sigma_i=\alpha_i s proves (4.4). The two summands of S_v=C_{v,v+1}C_{v,v+1}^*+C_{v,v-1}C_{v,v-1}^* depend on different column densities. Its mixed derivative in those two centers is zero; the Hessian term in the chain rule is therefore exactly Z_v, proving (4.5).

Before integrating these identities, we give the domination needed for differentiation under the integral. Lemma 3.3 yields

For the second inequality, apply the insertion bound to R_v^*, padding N_w by zero on the other column block. For any Hermitian B on the row space,

because R_v\sqrt{T_v}R_v^* is positive and has trace e_v. The corresponding estimate for a column insertion follows by using S_v. The Gaussian derivative formulas and the Hessian chain rule therefore bound every first center derivative, pure second center derivative, mixed column derivative, and first scale derivative used above by a polynomial in the centers times e_v, locally uniformly in s>0.

The energy itself satisfies

Since the node set is finite, each edge norm cubed is bounded by a Gaussian in its two endpoint centers, locally uniformly for positive variances. On \Pi_d, with |d|\le D, each pair of distinct indices i,j satisfies

Thus the preceding bounds have integrable majorants in every plane chart, locally uniformly in (d,s). Dominated convergence justifies all the asserted derivatives. To obtain polynomial growth in d, use the two endpoints of each bounding Gaussian as free coordinates; the third center is d minus their sum. Integrating a polynomial times that Gaussian leaves at most polynomial growth in |d|. Constants in these differentiability arguments may depend on the fixed arrays, mesh, rates, and compact scale interval, but do not enter the estimates below.

Choosing p_i as the dependent coordinate gives, for each i,

To obtain the mixed derivative, first choose p_{v+1} dependent and differentiate once. Reparametrize the resulting integral with p_{v-1} dependent and differentiate once more. This gives

Integrating (4.4) proves the first identity in (4.6). These comparisons concern integrated derivatives; no pointwise equality of pure and mixed center derivatives is asserted.□

Proof. We first prove the neighbor comparison

Write A=W_w, B=W_{w-1}^*, and N=N_w. The elementary inequality \|Az+Bz'\|^2\le2\|Az\|^2+2\|Bz'\|^2 gives

By Lemma 3.2, V_w\ge2^{-1/2}b_Q([A\ B]^*N[A\ B]). In a block-preserving eigenbasis of Q, the Hessian cost is a sum of nonnegative squared entries. Discarding the off-diagonal blocks gives

Now A^*A\le S_{w+1} and B^*B\le S_{w-1}; the respective insertions are P_{w+1,w} and P_{w-1,w}. A second use of Lemma 3.2, with comparison factor one, proves (4.10).

In (4.6), replace 2Z_v by its upper bound Y_{v,v+1}+Y_{v,v-1} and sum over v. The two column costs carrying the score at vertex w have common coefficient \Sigma_\alpha/2-\alpha_w. If this coefficient is negative, discard those terms; otherwise apply (4.10). This proves (4.8). For the specified rates, \Sigma_\alpha=16/5, and the two possible coefficients are -1+(3/5)\sqrt2 and -11/10+(1/2)\sqrt2. Both are negative, giving (4.9).□

Proof. The Gaussian grid mass obeys

Indeed, for u_i\ge p+h compare the summand with the integral over the preceding interval of length h; for u_i\le p-h, use the succeeding interval. These intervals are disjoint, and the at most two remaining nodes contribute at most 2h\sup g_\sigma. For \sigma\ge h^2, the right side is at most M. Weighted Hölder therefore gives

Use p_v,p_{v+1} as free coordinates on \Pi_d. Each density integrates to h, proving the first estimate in (4.11). Each edge occurs in two stars, so (4.7) proves the second. The same upper bounds decrease when the entrywise absolute values decrease.□

Proof. The three integrands depend only on the two endpoint centers. Using these as free coordinates proves independence of d. In the one-edge star the other column insertion is zero, so Z_v=0. After permuting column blocks, its matrices are

The Hessian cost is unchanged by zero extension, so its row and column costs are exactly V^e_{v,w} and Y^e_{v,w}. Equation (4.6) now proves (4.14). Also e^e_{v,w}\le\|C\|_{\HS}^3, so Lemma 4.3 gives J^e_{v,w}(s_0)\le M n_e^3. Integrate (4.14) and discard the nonnegative terminal energy to obtain (4.15). Finally CC^*\le S_v, while CN_wC^* is supported on CC^*. Lemma 3.2 with factor one gives Y_{v,w}\le Y^e_{v,w}.□

Proof. Between switch scales the arrays are fixed, so (4.9) applies. It remains to bound the energy jumps. At a switch \tau, freeze the weights and set \delta A(i,j)=A_0^{\tau+}(i,j)-A_0^{\tau-}(i,j). Interpolate all changed entries linearly. Their moduli stay bounded by |A_0| by convexity. For fixed centers and (i,j) let

These quantities use the original arrays, and square roots are nonnegative. In R_0=[W_0\ W_2^*], the changing entry has row norm at most \sqrt{w_0(i)}(X+H), column norm at most \sqrt{w_1(j)}Y, and derivative modulus \sqrt{w_0(i)w_1(j)}|\delta A(i,j)|. The adjoint entry in R_1=[W_1\ W_0^*] instead has row and column bounds \sqrt{w_1(j)}(Y+Q) and \sqrt{w_0(i)}X. The star R_2 is unchanged. Lemma 3.5 therefore gives

The measure h^{-2}w_0(i)w_1(j)\dd\pi_0 is a probability measure. Under it, p_0,p_1 are independent normals with means u_i,u_j and variances \sigma_0,\sigma_1, while p_2=-p_0-p_1. Adding Gaussian variances gives

Extend all arrays by zero to \mathbb Z^2. Lemma 2.2 bounds \mathbb E X^2,\mathbb E Y^2,\mathbb E H^2,\mathbb E Q^2 by constant multiples of, respectively,

All these majorants are independent of the switch scale. Cauchy–Schwarz under the probability measure controls the three products in (5.5). Their square-root majorants have L^3(\mathbb Z^2,h^2) norms at most Cn_0,Cn_0,Cn_2,Cn_1. This follows from the L^{3/2} line maximal inequality, applied on slices; the changes j\mapsto-i-j or i\mapsto-i-j are bijections on the relevant slices.

Integrate (5.5), sum the jumps using (5.3), and apply Hölder on the grid. The result is

Finally integrate (4.9) between consecutive switches. The resulting sum is bounded by the initial energy plus the sum of absolute jumps, after discarding the nonnegative terminal energy. Lemma 4.3 bounds the initial energy by C\sum_vn_v^3. This proves (5.4).□

Proof of Proposition 5.1. Put \lambda=11/10 and use base variances \lambda s at all vertices. In Lemma 3.6, insert the diagonal matrix with entries k_s(u_i-p_v)/g_{\lambda s}(u_i-p_v) at vertex v. It is Hermitian and uniformly bounded: the polynomial factor in k_s is absorbed by the extra Gaussian decay relative to g_{\lambda s}.

For the cost at vertex v, narrow just its row variance to s. Write T_v^{\rm b} for the base Gram and T_v^{\rm n} for the narrowed-row Gram. Directly from the densities,

Invertible row scaling preserves the support. Lemma 3.2 therefore bounds the base cost by C sV_v^{\rm n}. The narrowed variances are one of the permutations allowed in Lemma 5.2.

Integrate the mixed trace inequality over base centers in \Pi_0 and over s with measure \dd s/s. Its right side is bounded by (5.4). On the left, the density at each vertex is now h k_s(u_i-p_v). Integration over the center plane convolves the three windows, giving

All signed integrals here are absolutely convergent for fixed arrays on the compact positive scale interval.

If the original norms are nonzero, apply this bound to arrays normalized to have norms n^{-1/2},1,1. Its right side is bounded by an absolute constant, since n(n^{-3/2}+2n^{-1}) is bounded. Rescaling gives C\sqrt n\prod_vn_v. The zero-norm case is immediate. The Riemann passage described in Section 2 now proves the proposition. The finitely many endpoints are fixed before the mesh tends to zero, so its scale restriction causes no loss.□

Proof. Expand the square of the first Hilbert–Schmidt norm. For each pair (i,j), the measure

is a probability measure. Indeed, with p_v,p_w as the free coordinates on \Pi_d, it makes p_v,p_w,d independent normal variables with respective means u_{i_v},u_{i_w},0 and variances \theta s,s,(1-\theta)s. The remaining center is p_2=d-p_v-p_w.

Under this probability measure, the expectation of h^{-2}w_v(m)w_2(l) is a bivariate Gaussian density, evaluated at (u_m,u_l). Its mean is (u_{i_v},-u_{i_v}-u_{i_w}), and its covariance matrix is

Call this covariance matrix \Sigma. The bounds \Sigma\leq4s\operatorname{diag}(\theta,1) and \det\Sigma=\theta(6-\theta)s^2 show, directly from the Gaussian density formula, that

where C is absolute and one may take C'=4.

Let M_{\mathbb Z}^{(1)} and M_{\mathbb Z}^{(2)} be the discrete line maximal operators in the two coordinates. By Lemma 2.2, we can take

In fact, (6.7) and h^2\leq\theta s show that \mathbb E\|C_{v,2}\|_{\HS}^2\leq\mathcal H_E(i,j)^2. Cauchy–Schwarz under this probability measure now proves (6.5). The map (i,j)\mapsto(i_v,-i_v-i_w) is a bijection of \mathbb Z^2. The \ell^{3/2} bounds for the two line maximal operators, applied to |E|^2, prove (6.6).□

Proof of Proposition 6.1. We may discard a common null set of frequencies so that all pointwise bounds in (6.2) hold. Set

For each \xi, use B_{0,k}^\xi on the open slab 2a_k<s<3a_k, and use the zero array on edge 0 outside these slabs. The slabs are pairwise disjoint because a_{k+1}=4a_k. The other two arrays remain A_1,A_2 throughout. All scale integrals below are over

The mesh hypothesis gives h^2\leq\theta s on this entire interval.

We first turn the oscillatory windows into refined row scores. We then bound the integrated row scores by telescoping the star energies, using Lemma 6.2 at every slab boundary.

Refining a row variance. Fix a slab, put L=L_k, a=a_k, and \omega=\xi/L. The window at each vertex is

Initially give all three vertices the variance s. The diagonal insertion with entries f_{\xi,k}(u_i-p_v)/g_s(u_i-p_v) has operator norm at most C(q), uniformly on 2a<s<3a. Indeed, |L\omega/D|\leq2, and the remaining factor is bounded using Gaussian decay in g_a/g_s. Split this diagonal into its real and imaginary parts before applying Lemma 3.6. The resulting eight terms involve only Hermitian insertions of uniformly bounded norm.

Consider the cost belonging to the row star R_v. Replace its row variance s by b'=\theta s, leave both column variances equal to s, and shift the row center from p_v to p_v+d. Denote the refined Gram and score insertion by T_v^d and U_v^d. Addition of Gaussian variances gives

All these Grams have a common support: changing a strictly positive row density amounts to invertible row scaling and leaves the kernel of the row matrix unchanged.

To represent the inserted Gram in the same way, put \omega'=a\omega/(a-b'). Since b'<a/2, completing the square gives the exact identity

The exponential prefactor is at most C(q), because \omega^2b'\leq C(q). Also

Differentiate (6.12) in u. The derivative of the first Gaussian supplies the negative refined row score -(u-d)/b'. Consequently, for either the real or the imaginary diagonal insertion, its inserted Gram K_v has the form

where the scalar coefficient is explicitly

The preceding bounds give |\alpha(d)|\leq C(q)\sqrt{\theta s}. The averaged matrices are integrable, since the finite-array Gaussian weights dominate the polynomial score factors. By (6.11), Lemma 3.4, and the quadratic homogeneity of b_T in its second argument,

The cost on the right uses variances

Thus the factor \theta in the refined score is gained at exactly the scale needed to accommodate the oscillation.

Integrated refined scores. Fix v and use (6.14) for the rest of this part of the proof. All stars, energies, and costs refer to the slab-dependent edge 0 just defined. Dependence on \xi is suppressed until frequency integration is needed. For a center-dependent cost Q, write

On each open interval where the arrays are fixed, the heat identities (4.6) give

Here the coefficient 3 has a useful interpretation. The original plane second derivative has coefficient 2+\theta. Differentiating \nu_s, applying its heat equation, and integrating twice by parts adds 1-\theta. The second heat identity identifies the averaged plane second derivative with \widetilde Z_v.

For fixed arrays on a compact positive scale interval, Lemma 4.1 bounds the plane integrals of the energies, costs, and relevant absolute derivatives by polynomials in |d|. The density \nu_s and its derivatives have Gaussian decay. These bounds justify differentiation under the integral and the two integrations by parts in d, and give the required one-sided limits at slab boundaries. Their constants may depend on the mesh and \theta; they justify the identities and do not enter the estimates.

The metric Cauchy–Schwarz inequality gives 2Z_v\leq\sum_{w\ne v}Y_{v,w}. Hence

The column costs in this inequality can be paid for by individual edges. Indeed, S_v\geq C_{v,w}C_{v,w}^*, so the order comparison of Lemma 3.2 yields

The inserted matrix is supported on the single-edge Gram, so this comparison also applies if the larger star Gram has a larger support. Define the integrated single-edge energy

The integral of Y^e_{v,w} over \Pi_d is independent of d, because only the two endpoint centers occur. With just this edge present, the heat identities give

The coefficient of this column cost is 1; the remaining row cost is nonnegative. Thus the estimate is independent of the small variance ratio \theta.

If the edge is unchanged throughout [s_-,s_+], integration of (6.18) and Lemma 4.4 bound its total cost by C(n_1^3+n_2^3). If it is edge 0, apply the same argument separately on its slabs. The result is bounded by

After integration in \xi, this is at most C(C_1)n_0^3, since

We have therefore controlled all column terms in (6.16). The remaining issue is the boundary energy of the stars.

Summing the slab boundaries. Before the first slab, edge 0 is zero. By Lemma 4.3, \widetilde J_v(s_-)\leq C(n_1^3+n_2^3). The star at vertex 2 has no jumps. For v\in\{0,1\}, let w be the other vertex of edge 0. At either boundary of a slab, write C_{\rm sw}=C_{v,w} for the slab-side weighted block and C_{\rm fix}=C_{v,2} for the unchanged weighted block. The absolute change in the pointwise star energy is at most

To verify this bound, integrate the derivative of \Phi(C_{\rm fix}C_{\rm fix}^*+tC_{\rm sw}C_{\rm sw}^*) for 0<t\leq1. Its support is fixed there, and the derivative is

Continuity gives the endpoint at t=0. Removing the edge has the same absolute change as adding it.

The integrated cubic term in (6.20) is bounded by the cube of the norm of B_{0,k}^\xi, by Lemma 4.3. Summing both boundaries of every slab and using (6.19) gives C(C_1)n_0^3 after frequency integration.

For the mixed term, let A^{(v,2)} be the fixed unweighted array on the edge joining v,2, indexed in that order. Apply Lemma 6.2 with this array and denote its majorant by \mathcal H_v. It is independent of k,\xi,s, and its cube norm is at most C(n_1+n_2). Both boundaries together consequently contribute at most a constant times

by (6.2) and Hölder’s inequality. This is the point at which averaging coefficient squares over frequency is essential. Taking a separate supremum in frequency for each scale would not give this bound.

Integrate (6.16) on the intervals between consecutive boundaries and telescope. The final energy is nonnegative and may be discarded. The initial energy and the absolute jumps just estimated, together with the single-edge costs, give

Here we absorbed n_0^2(n_1+n_2) by Young’s inequality. Values at the finitely many boundaries are immaterial. All integrands are measurable in \xi: the arrays are measurable, and the finite-matrix energies and supported quadratic forms are Borel functions of their entries. The latter can also be obtained as limits of positive-definite regularizations. Thus every nonnegative frequency integration above is justified by Tonelli’s Theorem, without any regularity assumption on the coefficient functions beyond measurability.

Recovering the frequency-block form. Return to the equal-variance stars and apply Lemma 3.6 to the real and imaginary parts of the three window insertions. Integrate the bound over base centers on \Pi_0. Shifting the row center in (6.13) maps this plane to \Pi_d; using the two column centers as free coordinates shows that its Jacobian is 1. The integrated cost at vertex v is therefore at most C(q)\theta s\widetilde V_v(s). Integrate over every slab with \dd s/s, and then over \Xi_D with \dd\gamma. The sum of the resulting bounds is controlled by (6.21).

On the other side, multiplying the three weighted windows in the trace and integrating the centers gives exactly

This is the convolution identity on \Pi_0: the three arguments of the windows sum to u_i+u_j+u_l. The expression is independent of s within its slab, whose logarithmic length is \int_{2a_k}^{3a_k}\dd s/s=\log(3/2). All signed center integrations are absolutely convergent by Gaussian decay. We have proved (6.4) with its right side replaced by C(q,C_1)\sum_v n_v^3.

Finally, if all n_v are nonzero, apply this estimate to A_v/n_v. The coefficients and mesh hypothesis are unchanged, and trilinearity gives the product bound claimed in the Proposition. If any norm is zero, the form is zero.□

Proof. First take smooth compactly supported G_v. Within any fixed step choice, the kernel in (6.23) is continuous in t: the frequency block is bounded, the coefficients are bounded and measurable, and the Gaussian formulas supply a common majorant for dominated convergence. After performing the frequency integration, the trilinear integral is therefore the limit of its lattice Riemann sums. The finitely many rectangular step boundaries have measure zero. The mesh condition (6.3) holds for all sufficiently fine lattices, and the discrete norms converge to \|G_v\|_3. Proposition 6.1 proves (6.24) for these inputs.

To pass to general L^3 inputs, note that for the fixed finite block and scale set there is an integrable majorant k_*(t) with |\mathcal K_{a,b}(t)|\leq k_*(t), uniformly in (a,b). One may take a constant times the finite sum of the dilates of \sup_{\xi\in\Xi_D}|H_\xi|, which has Gaussian decay. For every integrable nonnegative k,

Indeed, set t=a+b+c, apply Hölder’s inequality in (a,b) for fixed t, and use the determinant-one changes of variables for the other two edge functions. Thus the defining integral is absolutely convergent and continuous in the three L^3 inputs. Smooth approximation completes the proof. The final constant comes from the Proposition and remains independent of D, the number of scales, and the particular step choices.□

Proof. After setting t=Lz, apply Hölder’s inequality with exponents 5/2,5/2,5 to

The first resulting integral is \int |F(x+Lz,y)|^{5/2}e^{-5cz^2/4}\,\dd z. Splitting \R into |z|\le1 and 2^j<|z|\le2^{j+1} bounds it by C_cM_x(|F|^{5/2})(x,y), uniformly in L. The second integral has the corresponding bound with G and M_y. This proves the first assertion. For the last assertion, retain half the Gaussian decay in w, whose L^5 norm is then at most C_cA.□

Proof. We first obtain Fourier estimates for a single kernel K=K_{a,b,k}; write m=m_{a,b,k}. All constants below depend only on A,c. There is nothing to prove for m=0, so suppose that m\ge1.

A Gaussian-weighted primitive. We seek a representation K=(g_{3q}f)''' for a suitable fixed q: the Fourier modes of f then produce exactly the differentiated Gaussian windows controlled by Corollary 6.3. Define

The three moment conditions imply, for j=0,1,2,

Use the defining integral when z\le0 and this tail integral when z\ge0. Gaussian tail estimates give |U^{(j)}(z)|\le C e^{-c_1z^2} for j=0,1,2, with c_1>0 fixed. Moreover, U'''=K as a weak derivative. Choose q>0 sufficiently large, depending only on A,c, and put

The quotient and product rules, with a slight reduction in the Gaussian exponent, give

Here and below derivatives of order three are weak derivatives, or their almost-everywhere representatives. The functions f,f',f'' are continuous. Between the exceptional points, the same calculation using K' bounds the classical derivative of f''' by Ce^{-c_2z^2}. Its jumps are bounded by C, and there are at most Am of them. Thus, for 0\le j\le3,

The same bounds hold with zero right sides when m=0.

Fourier bounds. Use the normalization \widehat f(\xi)=(2\pi)^{-1}\int f(z)e^{-i\xi z}\,\dd z. Let

We claim that

For D\ge2, the distributional derivative of f''' is a finite measure of total variation at most Cm. Taking its Fourier transform gives |\xi|^4|\widehat f(\xi)|\le Cm, proving the first bound. For the second, the translation inequality for functions of bounded variation gives

Indeed, integrate the absolute derivative measure over each traversed interval and apply Fubini’s theorem. By the uniform supremum bound in (7.8), the squared L^2 norm of this difference is also at most Cm/D. Plancherel’s identity and \widehat{f'''}(\xi)=(i\xi)^3\widehat f(\xi) now yield the second bound, since |e^{i\xi/D}-1|\ge2\sin(1/2) on \Xi_D. When D=1, both assertions follow from (7.7) and m\ge1.

A finite frequency cutoff. Choose \chi\in C_c^\infty(\R) supported on [-2,2] and equal to one on [-1,1], and set

We show that this replacement has a summable pointwise error. Let \eta be the Schwartz convolution kernel of the multiplier \chi, normalized so that \int\eta=1, and let \eta_{P_*}(z)=P_*\eta(P_*z). For 0\le j\le3, f_*^{(j)}=f^{(j)}*\eta_{P_*}. The translation inequality gives

Their supremum norms are uniformly bounded, by (7.8) and the L^1 norm of \eta. Interpolation therefore gives

Since every derivative of g_{3q} is bounded by a constant times a slightly wider Gaussian, the product rule shows that

for a nonnegative w. This calculation also holds for weak derivatives, so no additional terms occur at the jumps of K.

Apply this construction separately to every K_{a,b,k}. By (7.5), (m_{a,b,k}/P_*)^{1/5}\le C/n whenever m_{a,b,k}\ne0, and at each (a,b) there are at most An such indices. Lemma 7.1 bounds the sum of all replacement errors pointwise by C\mathcal M_{F,G}(-a,-b) after integration in c. Their contribution to \Lambda_{\mathcal K} is consequently at most C\prod_v\|G_v\|_3.

Applying the frequency estimate. It remains to estimate the kernels K_*. For each dyadic D use the kernels from Corollary 6.3, namely

Gaussian convolution and modulation give

For the block \Xi_D, the contribution to K_{a,b,k,*} is therefore

Set \mu_{a,b,k}(\xi)=D c_{a,b,k}(\xi)/\sqrt n. The two estimates in (7.9) give precisely

All transformations were applied separately to finitely many kernel values. The coefficients \mu_{a,b,k} therefore remain step choices on a common finite rectangular refinement of the original partitions. The block kernel is \sqrt n\int_{\Xi_D}\mu_{a,b,k}(\xi) H_\xi\,\dd\xi/D. Thus Corollary 6.3, with measure \dd\xi/D, bounds this block by C\sqrt n\prod_v\|G_v\|_3. Only O(\log(2+n)) blocks meet the support of the cutoff. Summing their bounds and adding the replacement error proves (7.6).□

Proof. Choose a real, smooth, compactly supported odd function \varphi with \int z\varphi(z)\,\dd z=1. Put

These coefficients are uniformly bounded and vanish when m_{a,b,k}=0. The kernels K_{a,b,k}-u_{a,b,k}\varphi satisfy all three moment conditions and the other hypotheses of Proposition 7.2, with enlarged fixed constants and unchanged counts. It remains to bound the subtracted terms.

Define \Psi=\varphi-2^{-1}\Dil_2\varphi. Oddness gives its zeroth and second moments, and its first moment is 1-2^{-1}\cdot2=0. For every integer J\ge1,

For each fixed j, apply Proposition 7.2 to the shifted scales k+j with kernels u_{a,b,k}\Psi and the same shifted counts m_{a,b,k}. The constants are independent of j, and the outer coefficients 2^{-j} are summable. The remainder contributes at most C2^{-J}n\prod_v\|G_v\|_3 by Lemma 7.1, since there are at most An nonzero counts at any output point. Letting J\to\infty proves the claim.□

Proof. Place each endpoint \rho in the unique half-open dyadic interval [L_k,2L_k), where L_k=2^k. Let m_{a,b,k} count these endpoints, with multiplicity, and collect their contributions into \Dil_{L_k}K_{a,b,k}. Thus K_{a,b,k} is a linear combination of

with coefficient \beta_j at a lower endpoint and -\beta_j at its upper endpoint. Every kernel is odd, its possible jump points are the \pm\alpha, and

The important assertion is the uniform bound

away from at most 2m_{a,b,k} points, with constants independent of m_{a,b,k}. Fix (a,b) and k. Pair the two endpoints of every annulus whose endpoints both lie in [L_k,2L_k). After scaling, its contribution is \beta_j(e_{\alpha_j}-e_{\gamma_j}) with 1\le\alpha_j<\gamma_j<2. The intervals (\alpha_j,\gamma_j) are disjoint, so

There are at most two unpaired endpoint contributions. Their intervals must satisfy, respectively,

Disjointness allows at most one interval of each type. An interval with \varepsilon_j<L_k and R_j\ge2L_k has no endpoint in this group. Figure 1 illustrates the decomposition.

For |z|\le3, first consider the smooth term s_\alpha(z)=P(|z|/\alpha)/z in (7.12). The function P has a smooth even extension and vanishes quadratically at zero. Hence s_\alpha extends smoothly across z=0, and

The fundamental theorem of calculus in \alpha, followed by (7.15), bounds the sum of the paired smooth terms and their z-derivatives independently of their number. For the hard terms, the paired difference is

Away from endpoints, at most one such difference is nonzero at any fixed |z|, because the intervals (\alpha_j,\gamma_j) are disjoint. On its support |z|\ge1, so its size and its classical derivative are uniformly bounded. The at most two unpaired terms have the same bounds on this compact range.

For |z|>3, all indicators in (7.12) equal one, and

Consequently e_\alpha, \partial_z e_\alpha, \partial_\alpha e_\alpha, and \partial_z\partial_\alpha e_\alpha are bounded by Ce^{-cz^2}, uniformly for 1\le\alpha\le2. The paired-length estimate and the two unpaired contributions prove the same bound for their sum. Combining the compact and tail estimates proves (7.14).

We finish by separating the groups according to their counts. At each output point, (7.13) implies that there are at most 2\sqrt n indices with m_{a,b,k}>\sqrt n. The uniform Gaussian bound (7.14) and Lemma 7.1 bound all these groups together, after integration in c, by C\sqrt n\mathcal M_{F,G}(-a,-b). Their trilinear contribution is at most C\sqrt n\prod_v\|G_v\|_3.

Set the kernels and counts of those groups to zero. The remaining family satisfies (7.4), has odd kernels, and has counts satisfying \sum_km_{a,b,k}\le2n and m_{a,b,k}\le\sqrt n. Corollary 7.3 applies and proves (7.11). Every grouping and deletion is made at the output point, so the proof imposes no common choice of endpoints or dyadic groups at different output points.□

Proof of Theorem 1.2. First let F,G\in C_c^\infty(\R^2), use the pair maps (2.2), and take G_0\in C_c^\infty(\R^2). For step choices of disjoint annuli and coefficients, the identity (5.7) writes the hard linearization, multiplied by c_0\ne0, as its smooth linearization minus its endpoint errors. Propositions 5.1 and 7.4 therefore give

The interval list and coefficients in the sum may vary with (a,b).

Fix a finite menu of such choices. Its measurable choice sets can be approximated in measure on a compact set by finite unions of rectangles. To see this, approximate each of the finitely many sets using Lebesgue regularity and then refine all rectangle boundaries to one coordinate partition. Resolve overlaps by a fixed ordering of the menu; the total mismeasured set remains bounded by the sum of the approximation errors. Assign any menu item on the remaining cells. For the fixed menu, the annuli are uniformly bounded away from zero, and the integrals against the smooth inputs are bounded on the compact support needed for G_0. Consequently the corresponding linearized integrals converge. Thus (8.1) holds for arbitrary measurable choices from a fixed finite menu.

Now fix a finite collection of lists of disjoint rational annuli, each of length at most n. The maximum of their sums of absolute increments is measurable. Choose a maximizing list measurably by resolving ties in a fixed order. For each of its increments choose a coefficient from \{1,-1,i,-i\} whose product with that increment has real part at least half its modulus. These choices are again measurable and range over a finite menu. Testing (8.1) against nonnegative smooth compactly supported G_0 shows, by L^3 duality, that this finite maximum has L^{3/2} norm at most C\sqrt n\log(2+n)\|F\|_3\|G\|_3. The reflection (x,y)=(-a,-b) preserves that norm. Exhausting the countable family of all rational lists and applying monotone convergence proves the count bound for smooth inputs.

For general complex F,G\in L^3, choose smooth compactly supported approximants F_m,G_m. For every fixed annulus, bilinearity and (2.1) give convergence of B_{\varepsilon,R}(F_m,G_m) in L^{3/2}. For a fixed finite collection of lists, the difference of the two finite maxima is bounded by a finite sum of such truncation differences. It therefore converges in L^{3/2}. Pass the estimate to the limit for this finite maximum, then exhaust the rational lists once more. Lemma 2.1 supplies the common representatives required in the definition of \mathcal S_n.□

Proof of Theorem 1.1. For any finite rational partition, arrange the absolute annular increments in nonincreasing order, denoting them by d_1\ge d_2\ge\cdots\ge0. Whenever the index 2^j exists, the corresponding largest increments are a permissible list in \mathcal S_{2^j}, so

The group of indices 2^j\le l<2^{j+1} has at most 2^j members. Bounding the full \ell^r norm by the sum of the group norms yields, uniformly in the chosen partition,

Theorem 1.2 and Minkowski’s inequality now imply

The series converges for every r>2. In particular, neither a bound on the number of increments nor a restriction to dyadic endpoints survives in the conclusion.□

Proof. Endpoint continuity in Lemma 2.1 identifies the supremum with its restriction to rational endpoint pairs. Consequently it is measurable and equals \mathcal S_1(F,G). Theorem 1.2 gives (9.1), with its absolute factor \log 3 absorbed into C_M.

For two pairs of inputs, bilinearity of each finite truncation gives, almost everywhere,

Taking L^p norms and using (9.1) proves continuity. Density of compactly supported smooth functions gives the asserted uniqueness.□

Proof. For compactly supported smooth f,g and fixed (x,y), put h(t)=f(x+t,y)g(x,y+t). Then

The numerator is O(t) at zero and vanishes for large t, so the joint limit exists at every point.

For general F,G, define the tail oscillation

These functions are measurable by endpoint continuity, decrease with n, and satisfy 0\le\Omega_n(F,G)\le2B_*(F,G). Choose compactly supported smooth f_m\to F and g_m\to G in L^3. Take a common full-measure set for the pairs (F,G), (f_m,g_m), (F-f_m,G), and (f_m,G-g_m), for all m. On this set, bilinearity gives

The smooth joint convergence implies \Omega_n(f_m,g_m)\to0 pointwise. Thus, writing \Omega_\infty=\lim_n\Omega_n(F,G), we obtain

Letting m\to\infty shows that \Omega_\infty=0 almost everywhere. This is the joint Cauchy criterion for the two endpoints, and hence defines B(F,G) almost everywhere. Taking, for example, the cofinal sequence (\varepsilon,R)=(1/(2n),2n) shows that the limit is measurable; its magnitude is at most B_*(F,G). Define it to be zero on the remaining null set.

Letting (\varepsilon',R') tend jointly to (0,\infty) in the definition of \Omega_n gives

The supremum is measurable, again by endpoint continuity. Dominated convergence applies because \Omega_n(F,G)\le2B_*(F,G)\in L^p, proving (9.3) and therefore joint norm convergence. The norm bound follows from |B(F,G)|\le B_*(F,G). Passing to the L^p limit in the bilinear identities for finite truncations proves complex bilinearity. The bound and density then prove uniqueness.□

Proof. Hölder’s inequality in (x,y) bounds the integral of the absolute value by 2\log(R/\varepsilon)\prod_v\|F_v\|_3. Fubini’s Theorem therefore identifies the finite form with \int F_0 B_{\varepsilon,R}(F_1,F_2). Corollary 9.1 gives its uniform bound, and pairing the joint L^p limit in Corollary 9.2 with F_0 gives the scalar principal value. The limiting form is bounded and complex trilinear; density proves uniqueness.□

Proof. Use the determinant-one change of variables

For flat inputs set

Each pair map has determinant one. The inverse formulas are

These maps preserve L^3 norms, compact smoothness, and the Schwartz class. Direct substitution identifies both the integrands and the finite truncation regions. Absolute integrability, the uniform bound, and joint scalar convergence therefore follow from Corollary 9.3. The inverse maps realize every simplex triple, so the optimal constants are equal. Density proves uniqueness of the bounded trilinear extension.□