Proof. We recall the covering proof of the Hardy–Littlewood estimates (Hardy and Littlewood 1930). 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 \[\mu\{Mf>\rho\} \le\frac6\rho\int_{|f|>\rho/2}|f|\,\mathrm d\mu.\] The distribution formula and Tonelli’s theorem give the strong bound after integrating \(p\rho^{p-1}\,\mathrm d\rho\). Truncation of the distribution integral first makes this argument independent of any a priori finiteness assertion.

For (9), 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 (10). ◻

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 \[AX_K+X_KA=K.\] Multiplying by \(A\) and taking traces gives \(\mathop{\mathrm{tr}}(PX_K)=\tfrac12\mathop{\mathrm{tr}}(AK)\). Hence differentiation of \(\Phi(P)=\mathop{\mathrm{tr}}(P\sqrt P)\) yields \(d\Phi_P(K)=\tfrac32\mathop{\mathrm{tr}}(AK)\). Differentiating once more proves the Hessian formula. Formula (14) follows by diagonalizing \(A\), and proves the remaining assertions. This is the square-root case of the classical divided-difference calculus (Daletskii 1957, sec. 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 \[\sqrt X=\frac1\pi\int_0^\infty X(X+tI)^{-1}\,\frac{\,\mathrm dt}{\sqrt t}\qquad(X\ge0).\] 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 (Kwong 1975, Theorem 1 and Section 2). The supported Hessian comparison below is derived from these order facts here.

For \(\eta>0\), \[P+\eta I\le a(Q+\eta I),\qquad \sqrt{P+\eta I}\le\sqrt a\sqrt{Q+\eta I}.\] If \(A\le B\), then \(\mathsf L_A+\mathsf R_A\le\mathsf L_B+\mathsf R_B\) on Hilbert–Schmidt space: the difference has quadratic form \(\mathop{\mathrm{tr}}(X^*(B-A)X)+\mathop{\mathrm{tr}}(X^*X(B-A))\ge0\). Invert these positive operators to obtain \[b_{P+\eta I}(K)\ge a^{-1/2}b_{Q+\eta I}(K).\] Let \(\eta\downarrow0\). In an eigenbasis, every entry of \(K\) meeting the kernel of the corresponding limiting matrix is zero. Thus Formula (14) gives precisely the supported costs on both sides of (15). ◻

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 \(\lVert C\rVert_{\mathrm{op}}\le\lVert L\rVert_{\mathrm{op}}\) and \[b_{R^*R}(R^*LR) =\frac32\sum_{i,j} \frac{\rho_i^2\rho_j^2}{\rho_i+\rho_j}|C_{ij}|^2.\] For \(x,y>0\), \(x^2y^2/(x+y)\le (xy)^{3/2}/2\le(x^3+y^3)/4\). Every row and column of \(C\) has squared Euclidean norm at most \(\lVert L\rVert_{\mathrm{op}}^2\). Therefore \[b_{R^*R}(R^*LR) \le\frac38\sum_{i,j}(\rho_i^3+\rho_j^3)|C_{ij}|^2 \le\frac34\lVert L\rVert_{\mathrm{op}}^2\sum_i\rho_i^3,\] 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 \(A=\int A_\omega\,\,\mathrm d\nu\). The integral exists by the stated integrability and Cauchy–Schwarz. For every vector \(x\), the vector-valued Cauchy–Schwarz inequality gives \[\lVert Ax\rVert^2 \le\int\lVert A_\omega x\rVert^2\,\,\mathrm d\nu =\langle\overline T x,x\rangle.\] Consequently \(A^2\le\overline T\), and square-root order, proved in Lemma 6, implies \[ \sqrt{\overline T}\ge\int\sqrt{T_\omega}\,\,\mathrm d\nu. \tag{18}\] For a positive definite selfadjoint operator \(\mathsf M\) on a real Hilbert space, completing the square proves \[ \langle K,\mathsf M^{-1}K\rangle =\sup_X\{2\langle K,X\rangle-\langle X,\mathsf M X\rangle\}. \tag{19}\] 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 (18), 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 \[b_{\overline T}(\overline K) \le\int b_{T_\omega}(K_\omega)\,\,\mathrm d\nu.\] Every inverse in this argument is taken on matrices acting on \(E\); uniform invertibility over \(\omega\) was never assumed. ◻

Proof. For \(\eta>0\), Lemma 5 gives \[\frac{\,\mathrm d}{\,\mathrm dz}\mathop{\mathrm{tr}}((R(z)^*R(z)+\eta I)^{3/2}) =3\operatorname{Re} \langle R(z)\sqrt{R(z)^*R(z)+\eta I},D\rangle_{\mathrm{HS}}.\] Spectral calculus implies \(\lVert\sqrt{R^*R+\eta I}-\sqrt{R^*R}\rVert_{\mathrm{op}}\le\sqrt\eta\). 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 (20).

For (21), 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.

Proof. We first check that the possible kernels of the Grams cause no differentiability problem. Each star has the form \[R_v=D_{\rm row}R_v^0D_{\rm col},\qquad R_v^0=[A_v\ A_{v-1}^*],\] 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 \[e_v=\Phi(Q'^{1/2}P'Q'^{1/2})\] is smooth. The rank-zero case is immediate.

For one Gaussian density \(w=h g_\sigma(u-p)\) and its score \(N=(u-p)/\sigma\), \[\partial_pw=Nw,\qquad \partial_p^2w=(N^2-\sigma^{-1})w,\qquad 2\partial_\sigma w=\partial_p^2w.\] With the columns fixed, \(T_v\) is linear in the row density, and hence \[\partial_{p_v}T_v=U_v,\qquad 2\partial_{\sigma_v}T_v=\partial_{p_v}^2T_v.\] The chain rule on its fixed support and Lemma 5 give \[2\partial_{\sigma_v}e_v=\partial_{p_v}^2e_v-V_v.\] 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 (31). 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 (32).

Before integrating these identities, we give the domination needed for differentiation under the integral. Lemma 7 yields \[V_v\le\tfrac34\|N_v\|_{\mathrm{op}}^2e_v, \qquad Y_{v,w}\le\tfrac34\|N_w\|_{\mathrm{op}}^2e_v, \qquad 2|Z_v|\le Y_{v,v+1}+Y_{v,v-1}.\] 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, \[\big|\mathop{\mathrm{tr}}(\sqrt{T_v}R_v^*BR_v)\big| \le\|B\|_{\mathrm{op}}e_v,\] 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 \[ e_v\le\left(\|W_v\|_{\mathrm{HS}}^2+ \|W_{v-1}\|_{\mathrm{HS}}^2\right)^{3/2} \le\sqrt2\left(\|W_v\|_{\mathrm{HS}}^3+ \|W_{v-1}\|_{\mathrm{HS}}^3\right). \tag{34}\] 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 \[p_0^2+p_1^2+p_2^2\le4(p_i^2+p_j^2)+3D^2.\] 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 \(i\), \[\partial_d^2J_v=\int_{\Pi_d}\partial_{p_i}^2e_v\,\,\mathrm d\pi_d.\] 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 \[\partial_d^2J_v =\int_{\Pi_d}\partial_{p_{v-1}}\partial_{p_{v+1}}e_v\,\,\mathrm d\pi_d =\int_{\Pi_d}Z_v\,\,\mathrm d\pi_d.\] Integrating (31) proves the first identity in (33). These comparisons concern integrated derivatives; no pointwise equality of pure and mixed center derivatives is asserted. ◻

Proof. We first prove the neighbor comparison \[ Y_{w+1,w}+Y_{w-1,w}\le\sqrt2\,V_w. \tag{37}\] 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 \[T_w\le2Q,\qquad Q=\operatorname{diag}(A^*A,B^*B).\] By Lemma 6, \(V_w\ge2^{-1/2}b_Q([A\ B]^*N[A\ B])\). In a block-preserving eigenbasis of \(Q\), the Hessian cost is a sum of nonnegative squared entries. Discarding the off-diagonal blocks gives \[b_Q([A\ B]^*N[A\ B]) \ge b_{A^*A}(A^*NA)+b_{B^*B}(B^*NB).\] 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 6, with comparison factor one, proves (37).

In (33), 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 (37). This proves (35). 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 (36). ◻

Proof. The Gaussian grid mass obeys \[ \sum_{i=-N}^Nh g_\sigma(u_i-p) \le1+\frac{2h}{\sqrt{2\pi\sigma}}. \tag{39}\] 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 \[\|W_v\|_{\mathrm{HS}}^3 =\left(\sum_{i,j}|A_v(i,j)|^2w_v(i)w_{v+1}(j)\right)^{3/2} \le M\sum_{i,j}|A_v(i,j)|^3w_v(i)w_{v+1}(j).\] Use \(p_v,p_{v+1}\) as free coordinates on \(\Pi_d\). Each density integrates to \(h\), proving the first estimate in (38). Each edge occurs in two stars, so (34) 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 \[R=[C\ 0],\qquad T=\operatorname{diag}(C^*C,0),\qquad U=\operatorname{diag}(C^*N_vC,0),\qquad S=CC^*.\] 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 (33) now proves (41). Also \(e^e_{v,w}\le\|C\|_{\mathrm{HS}}^3\), so Lemma 13 gives \(J^e_{v,w}(s_0)\le M n_e^3\). Integrate (41) and discard the nonnegative terminal energy to obtain (42). Finally \(CC^*\le S_v\), while \(CN_wC^*\) is supported on \(CC^*\). Lemma 6 with factor one gives \(Y_{v,w}\le Y^e_{v,w}\). ◻

Proof. Between switch scales the arrays are fixed, so (36) 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 \[\begin{align*} X^2&=\sum_m|A_0(i,m)|^2w_1(m),& Y^2&=\sum_m|A_0(m,j)|^2w_0(m),\\ H^2&=\sum_m|A_2(m,i)|^2w_2(m),& Q^2&=\sum_m|A_1(j,m)|^2w_2(m). \end{align*}\] 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 9 therefore gives \[ |\Delta e_0|+|\Delta e_1| \le C\sum_{i,j}w_0(i)w_1(j)|\delta A(i,j)| (2XY+HY+QX). \tag{47}\]

The measure \(h^{-2}w_0(i)w_1(j)\,\mathrm d\pi_0\) 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 \[\begin{align*} \mathbb E w_0(m)&=h g_{2\sigma_0}(u_m-u_i),\\ \mathbb E w_1(m)&=h g_{2\sigma_1}(u_m-u_j),\\ \mathbb E w_2(m)&=h g_{\sigma_0+\sigma_1+\sigma_2}(u_m+u_i+u_j). \end{align*}\] Extend all arrays by zero to \(\mathbb Z^2\). Lemma 4 bounds \(\mathbb E X^2,\mathbb E Y^2,\mathbb E H^2,\mathbb E Q^2\) by constant multiples of, respectively, \[\begin{align*} X_*^2(i,j)&=M_{\mathbb Z}(|A_0(i,\cdot)|^2)(j),& Y_*^2(i,j)&=M_{\mathbb Z}(|A_0(\cdot,j)|^2)(i),\\ H_*^2(i,j)&=M_{\mathbb Z}(|A_2(\cdot,i)|^2)(-i-j),& Q_*^2(i,j)&=M_{\mathbb Z}(|A_1(j,\cdot)|^2)(-i-j). \end{align*}\] All these majorants are independent of the switch scale. Cauchy–Schwarz under the probability measure controls the three products in (47). 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 (47), sum the jumps using (45), and apply Hölder on the grid. The result is \[\sum_\tau\left|\Delta\sum_vJ_v(0,\tau)\right| \le Cn\bigl(n_0^3+n_0^2(n_1+n_2)\bigr).\] Finally integrate (36) 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 13 bounds the initial energy by \(C\sum_vn_v^3\). This proves (46). ◻

Proof of Proposition 15. Put \(\lambda=11/10\) and use base variances \(\lambda s\) at all vertices. In Lemma 10, 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, \[(R_v^{\rm b})^*D_vR_v^{\rm b}=\sqrt s\,U_v^{\rm n}, \qquad T_v^{\rm n}\le\sqrt\lambda\,T_v^{\rm b}.\] Invertible row scaling preserves the support. Lemma 6 therefore bounds the base cost by \(C sV_v^{\rm n}\). The narrowed variances are one of the permutations allowed in Lemma 16.

Integrate the mixed trace inequality over base centers in \(\Pi_0\) and over \(s\) with measure \(\,\mathrm ds/s\). Its right side is bounded by (46). 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 \[h^3\sum_{i,j,l} A_0(i,j)A_1(j,l)A_2(l,i) \sum_{\text{chosen at }(i,j)}\beta\, I_{\varepsilon,R}(u_i+u_j+u_l).\] 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. After setting \(t=Lz\), apply Hölder’s inequality with exponents \(5/2,5/2,5\) to \[|F(x+Lz,y)|e^{-cz^2/2},\qquad |G(x,y+Lz)|e^{-cz^2/2},\qquad w(z).\] The first resulting integral is \(\int |F(x+Lz,y)|^{5/2}e^{-5cz^2/4}\,\,\mathrm dz\). Splitting \(\mathbb R\) 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 \(m\ge1\).

Proof. Choose a real, smooth, compactly supported odd function \(\varphi\) with \(\int z\varphi(z)\,\,\mathrm dz=1\). Put \[u_{a,b,k}=\int zK_{a,b,k}(z)\,\,\mathrm dz.\] 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 21, with enlarged fixed constants and unchanged counts. It remains to bound the subtracted terms.

Define \(\Psi=\varphi-2^{-1}\mathop{\mathrm{Dil}}_2\varphi\). Oddness gives its zeroth and second moments, and its first moment is \(1-2^{-1}\cdot2=0\). For every integer \(J\ge1\), \[\mathop{\mathrm{Dil}}_{2^k}\varphi =\sum_{j=0}^{J-1}2^{-j}\mathop{\mathrm{Dil}}_{2^{k+j}}\Psi +2^{-J}\mathop{\mathrm{Dil}}_{2^{k+J}}\varphi.\] For each fixed \(j\), apply Proposition 21 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 20, 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 \(\mathop{\mathrm{Dil}}_{L_k}K_{a,b,k}\). Thus \(K_{a,b,k}\) is a linear combination of \[ e_\alpha(z) =\frac{P(|z|/\alpha)-c_0\mathbf1_{\{|z|>\alpha\}}}{z}, \qquad 1\le\alpha<2, \tag{85}\] 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 \[ \sum_km_{a,b,k}\le2n. \tag{86}\]

The important assertion is the uniform bound \[ |K_{a,b,k}(z)|+|K_{a,b,k}'(z)|\le Ce^{-cz^2} \tag{87}\] 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 \[ \sum_j(\gamma_j-\alpha_j)\le1. \tag{88}\] There are at most two unpaired endpoint contributions. Their intervals must satisfy, respectively, \[\varepsilon_j<L_k\le R_j<2L_k, \qquad\text{or}\qquad L_k\le\varepsilon_j<2L_k\le R_j.\] 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 (85). The function \(P\) has a smooth even extension and vanishes quadratically at zero. Hence \(s_\alpha\) extends smoothly across \(z=0\), and \[\sup_{\substack{|z|\le3\\1\le\alpha\le2}} \bigl(|\partial_\alpha s_\alpha(z)| +|\partial_z\partial_\alpha s_\alpha(z)|\bigr)<\infty.\] The fundamental theorem of calculus in \(\alpha\), followed by (88), bounds the sum of the paired smooth terms and their \(z\)-derivatives independently of their number. For the hard terms, the paired difference is \[-\frac{c_0}{z} \bigl(\mathbf1_{\{|z|>\alpha_j\}} -\mathbf1_{\{|z|>\gamma_j\}}\bigr).\] 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 (85) equal one, and \[P(|z|/\alpha)-c_0=-2g_3''(|z|/\alpha).\] 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 (87).

We finish by separating the groups according to their counts. At each output point, (86) implies that there are at most \(2\sqrt n\) indices with \(m_{a,b,k}>\sqrt n\). The uniform Gaussian bound (87) and Lemma 20 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 (77), has odd kernels, and has counts satisfying \(\sum_km_{a,b,k}\le2n\) and \(m_{a,b,k}\le\sqrt n\). Corollary 22 applies and proves (84). 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. 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 24 gives its uniform bound, and pairing the joint \(L^p\) limit in Corollary 25 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 \[x=-a,\qquad y=-b,\qquad t=a+b+c, \qquad a=-x,\quad b=-y,\quad c=t+x+y.\] For flat inputs set \[\begin{aligned} G_0(u,v)&=F_0(-u,-v),\\ G_1(u,v)&=F_1(u+v,-u),\\ G_2(u,v)&=F_2(-v,u+v). \end{aligned}\] Each pair map has determinant one. The inverse formulas are \[\begin{aligned} F_0(r,s)&=G_0(-r,-s),\\ F_1(r,s)&=G_1(-s,r+s),\\ F_2(r,s)&=G_2(r+s,-r). \end{aligned}\] 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 26. The inverse maps realize every simplex triple, so the optimal constants are equal. Density proves uniqueness of the bounded trilinear extension. ◻