The maximal triangular Hilbert transform at the symmetric point

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 of A, this map multiplies its (i,j) entry by \sqrt{p_i}+\sqrt{p_j}>0. The inverse function theorem therefore gives a smooth positive square root. Its derivative X_K in direction K satisfies

Taking a trace after multiplication by A yields \tr(PX_K)=\tfrac12\tr(AK). Differentiation of \Phi(P)=\tr(P\sqrt P) now gives

Differentiating this identity in direction L proves the Hessian formula. The entrywise formula follows by diagonalizing A. It also proves symmetry, positivity, and Cauchy–Schwarz on the real space of Hermitian directions.□

Proof. If x\in\ker Q, then 0\le\langle Px,x\rangle \le a\langle Qx,x\rangle=0, so Px=0. Thus \ker Q\subseteq\ker P, proving the support assertion.

We recall the two operator-order facts needed for the comparison. Inversion reverses order on positive definite matrices: for 0<X\le Y, conjugate by X^{-1/2}, invert the resulting matrix at least I, and conjugate back. Moreover,

This norm-convergent formula follows from the scalar integral by spectral calculus. Since X(X+tI)^{-1}=I-t(X+tI)^{-1}, inversion order proves that the positive square root preserves order. This is the classical Loewner–Heinz square-root monotonicity; see, for example, Kwong [14, Theorem 1 and Section 2]. The supported Hessian comparison below is derived from these order facts here.

For \eta>0, the assumptions give

If A\le B, then \mathsf L_A+\mathsf R_A\le\mathsf L_B+\mathsf R_B as selfadjoint operators on Hilbert–Schmidt space. Indeed, left and right multiplication by the positive matrix B-A have the nonnegative quadratic forms \tr(X^*(B-A)X) and \tr(X^*X(B-A)). Inverting the resulting positive definite operators gives

Finally let \eta\downarrow0. Formula (2.1), applied in a full eigenbasis, shows that both limits are the asserted supported costs: every entry of K meeting the kernel of the respective matrix is zero.□

Proof. The insertion annihilates \ker R=\ker(R^*R) and is Hermitian, so it is supported on the Gram. The assertion is immediate if R=0. Otherwise take a singular-value decomposition, with positive singular values \rho_i, 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, the arithmetic–geometric mean inequality gives

Each row and each column of C has squared Euclidean norm at most \lVert L\rVert_{\op}^2. Consequently

which is (2.3).□

Proof. We first put the three quadratic costs in a common metric. We then estimate a cubic trace in an eigenbasis and recover the absolute value of the original cyclic product by changing the phase of one edge.

A common quadratic form. On H=H_0\oplus H_1\oplus H_2, define the Hermitian matrices

Thus \lVert D\rVert_{\op}=D_*. Let P_v be the orthogonal projection onto H_v and set

Only the v-th block row of P_vM can be nonzero; it consists of R_v with its columns placed in the corresponding vertex spaces. Consequently A_v and K_v are, respectively, T_v and R_v^*D_vR_v, after a permutation of blocks and extension by zero. These operations preserve the quadratic cost. Since P_v commutes with D, the matrix K_v is Hermitian. Moreover,

The first identity follows from \langle A_vx,x\rangle=\lVert P_vMx\rVert^2. Thus K_v is supported on A_v, and hence also on M^2. Lemma 2.2, followed by the quadratic-form inequality for three vectors, yields

All these quadratic forms are taken on their supports; no invertibility of M is required.

The cubic trace. Choose an orthonormal eigenbasis (e_i) of M, write the real eigenvalues as m_i, and arrange the indices so that q_i:=|m_i| is nonincreasing. Equal values may be ordered arbitrarily. Write D_{ij} for the entries of D in this basis. The spectral formula for b_{M^2} gives

If either eigenvalue is zero, its summand is zero; pairs of zero eigenvalues are omitted. Define

Hermitian symmetry and the ordering of the q_i imply

Indeed, for i<j we have q_j\le q_i, so 2q_i/(q_i+q_j)\ge1 whenever the denominator is positive.

We now bound

by grouping the triples according to how often their smallest index occurs. This is the index of a largest absolute eigenvalue in the triple.

If the smallest index i occurs once, its three cyclic positions give

Here j=k is allowed. For the vector x=(m_jD_{ji})_{j>i}, the inner sum is x^*D_{>i}x, where D_{>i} is the compression of D to the indices greater than i. Since \lVert D_{>i}\rVert_{\op}\le D_*, the modulus of this contribution is at most 3D_*E_i. This argument retains the signs of the m_j and uses D_{ji}=\overline{D_{ij}}.

If the smallest index i occurs twice, the contribution is

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

Its contribution therefore has modulus at most 3D_*\sqrt{A_iE_i}. This also holds when q_i=0, without any division by q_i.

Finally, when all three indices equal i, the modulus is at most D_*A_i. These three cases exhaust all triples, including repeated indices. Summing them and applying Cauchy–Schwarz once more, we obtain

Recovering the complex cyclic product. We have bounded the cubic trace by the star costs. It remains to extract the modulus of the desired triangular trace. Put

Because M has zero diagonal blocks, a nonzero closed three-step block product must visit all three vertices. The three starting points in one orientation each give \tau, by cyclicity of the trace. Reversing the orientation gives \overline\tau, since the D_v are Hermitian. Hence

If \tau\ne0, choose z\in\mathbb C with |z|=1 and z\tau=|\tau|, and replace W_0 by zW_0. The row stars change by

while R_2 is unchanged. Each displayed right multiplier is unitary and conjugates the corresponding Gram and inserted Gram by the same unitary. Thus \mathcal E is unchanged, as is D_*. Applying (3.5) and (3.7) to the modified matrices, with Q computed for the new M, gives

which is (3.3). The case \tau=0 is immediate.□

Proof. The weighted star has the form R_v=D_{\rm row}R_v^0D_{\rm col}, where R_v^0=[A_v\ A_{v-1}^*] is fixed and both diagonal multipliers are strictly positive; the column multiplier has one block for each neighbor. Invertible row multiplication preserves the kernel, and invertible column multiplication preserves the range. This proves the two support assertions, since

when the respective other multiplier is held fixed.

For joint smoothness, the rank-zero case is immediate. Otherwise factor R_v^0=XY^* with X,Y of full column rank. Write X'=D_{\rm row}X, Y'=D_{\rm col}Y and P'=X'^*X', Q'=Y'^*Y'. These last two matrices are positive definite and depend smoothly on the parameters. The nonzero eigenvalues of R_vR_v^*=X'Q'X'^* are those of Q'^{1/2}P'Q'^{1/2}, by comparing AA^* and A^*A for A=X'Q'^{1/2}. Therefore

which is smooth on the positive definite cone.□

Proof. For w=h g_\sigma(u-p) and L=(u-p)/\sigma,

With the columns fixed, T_v is linear in the row density, so

Its support is fixed by Lemma 4.1. The chain rule and Lemma 2.1 thus give

For a column vertex w\ne v, use instead S_v, which is linear in that column density and has fixed support. This gives

Summing the three variance derivatives with coefficients \alpha_i proves (4.3).

The two summands in S_v=W_vW_v^*+W_{v-1}^*W_{v-1} depend on different column densities. Its mixed derivative in the two column centers is therefore zero. Only the Hessian term remains in the mixed derivative of e_v, giving (4.4).□

Proof. Abbreviate A=W_w, B=W_{w-1}^*, and L=L_w. The inequality \lVert Az+Bz'\rVert^2\le2\lVert Az\rVert^2+2\lVert Bz'\rVert^2 implies

Since U_w=[A\ B]^*L[A\ B] is supported on T_w, Lemma 2.2 gives b_{T_w}(U_w)\ge2^{-1/2}b_Q(U_w). In a block-preserving eigenbasis of Q, the formula for b_Q is a sum of nonnegative squared entries. Discarding the off-diagonal blocks yields

Here A^*A\le S_{w+1} and B^*B\le S_{w-1}, and the respective insertions are U_{w+1,w} and U_{w-1,w}. Each is supported on its smaller edge Gram. Apply Lemma 2.2 again, now with comparison factor one, to obtain (4.5).□

Proof. We first justify differentiation at the fixed mesh. On compact positive s-intervals the nodes are bounded, the scores grow at most linearly in the centers, and the first and second density ratios grow at most quadratically. Lemma 2.3 gives

For the second inequality, pad L_w by zero on the other column block, obtaining B_w with U_{v,w}=R_vB_wR_v^*, and apply the lemma to R_v^*. Also 2|Z_v|\le Y_{v,v+1}+Y_{v,v-1}. For a Hermitian matrix B,

since R_v\sqrt{T_v}R_v^* is positive with trace e_v. The corresponding column estimate follows in the same way with S_v and a padded column insertion. Together with the density-ratio formulas in Lemma 4.2, these bounds control every first center derivative, pure second center derivative, mixed column derivative, and first s-derivative by a polynomial in the centers times e_v.

To dominate the energy itself, use

Each edge norm cubed is bounded by a Gaussian in its two endpoint centers, locally uniformly for s>0. Indeed, the finite set of nodes is bounded and (u-p)^2\ge p^2/2-u^2. On \Pi_d with |d|\le D, any pair of centers satisfies

Thus an endpoint-pair Gaussian, multiplied by any of the preceding polynomials, has an integrable majorant in any fixed plane chart, locally uniformly in (d,s). This proves all needed differentiation statements by dominated convergence. The constants in this justification may depend on the finite mesh and the compact parameter intervals; they will not enter the uniform estimate.

We can now differentiate the plane integrals. It remains to express their second derivatives in d using both pure and mixed center derivatives, then apply Lemma 4.3 to the resulting costs.

Choosing p_i as the dependent coordinate in (4.2) gives

For the column indices j=v+1, k=v-1, first choose p_j dependent to write \partial_dJ_v=\int_{\Pi_d}\partial_{p_j}e_v\,\dd\pi_d. Reparametrize this last integral with p_k dependent, then differentiate again. It follows from (4.4) that

These are identities of integrated derivatives; no pointwise equality between pure and mixed derivatives is required.

Integrating (4.3) and using (4.9), we obtain

Cauchy–Schwarz gives 2Z_v\le Y_{v,v+1}+Y_{v,v-1}. Sum over v and collect the two neighbor costs carrying the score at w. Their common coefficient is \Sigma_\alpha/2-\alpha_w. If it is negative, discard those terms; otherwise apply Lemma 4.3. This proves (4.6).

For the indicated permutations, \Sigma_\alpha=16/5. The two possible coefficients in (4.6) are -1+(3/5)\sqrt2 and -11/10+(1/2)\sqrt2, both strictly negative. Dividing by two gives (4.7).□

Proof. For every center p and variance \sigma>0,

To see this, compare a grid point u_i\ge p+h with the integral over its preceding interval of length h, where the Gaussian is at least its value at u_i. For u_i\le p-h, use the succeeding interval. These intervals are disjoint, and the at most two remaining grid points each contribute at most h\sup g_\sigma.

Since each \sigma_v\ge h^2, every density has total mass at most M=1+\sqrt{2/\pi}. Weighted Hölder gives

Integrate with p_v,p_{v+1} as free coordinates on \Pi_d. Each density has integral h, so

Each edge occurs in exactly two stars. Equation (4.8) therefore proves (4.10). Replacing entries by smaller ones in absolute value only reduces the last upper bounds, proving the final assertion.□

Proof. For a finite collection of intervals, repeatedly choose an interval of largest radius and discard all intervals meeting it. The chosen intervals are disjoint. Every discarded interval has radius no larger than that of the interval which removed it and is contained in its triple-radius enlargement. This enlargement has three times the original length on \R. On \mathbb Z, its cardinality is 6r+1\le3(2r+1), including r=0.

For each point in a finite subset of \{M_{\mathbb Z}f>\rho\}, choose a witnessing interval whose average exceeds \rho. The preceding selection bounds the size of that subset by three times the sum of the cardinalities of the chosen intervals, and hence by 3\|f\|_1/\rho. Taking the supremum over finite subsets proves the discrete weak estimate. On \R, fixed-radius averages of an L^1 function are continuous, so \{M_{\R}f>\rho\} is open. Apply the same selection to a finite witnessing cover of any compact subset and then use inner regularity to obtain the continuous estimate.

For f\in L^q, split f into its parts on \{|f|\le\rho/2\} and \{|f|>\rho/2\}. The first part has maximal function at most \rho/2, and the second is integrable. The weak estimate gives

The distribution formula and Tonelli’s theorem now yield

These formulas are valid first with a bounded range of distribution parameters and then by monotone convergence; finiteness of Mf is not assumed in advance. Both maximal functions are measurable: the discrete supremum is countable, and continuous averages are continuous in their radius, so positive rational radii suffice on \R.

Finally put a=\sqrt\sigma/h\ge1. In the sum in (5.2), the kernel is

Split the integers into |m-n|\le a and the annuli 2^{j-1}a<|m-n|\le2^ja, j\ge1. The sum of |f| over |m-n|\le2^ja is at most 3\cdot2^ja\,M_{\mathbb Z}f(n). Bounding the Gaussian by its maximum on each region proves the claim with, for example,

All constants are independent of the mesh and of the support of f.□

Proof. For \eta>0, differentiate \tr((R(z)^*R(z)+\eta I)^{3/2}) using Lemma 2.1. Its derivative is

Spectral calculus gives

Since R(z) stays in a bounded set, the derivatives converge uniformly in z, and the endpoint energies converge. Integrating the regularized identity before letting \eta\downarrow0 proves (5.3).

Cauchy–Schwarz in the matrix-product formula bounds the left side of (5.4) by the norm of row i of R times the norm of column j of \sqrt{R^*R}. The squared norm of that column is (R^*R)_{jj}, which is also the squared norm of column j of R. This proves (5.4), including zero columns and singular matrices.□

Proof. Write J(s)=\int_{\Pi_0}\sum_ve_v(p,s)\,\dd\pi for the masked energy, and put \Delta J(\tau)=J(\tau+)-J(\tau-) at an endpoint. We first estimate its jump at a single scale, then sum those estimates and telescope the heat inequality.

The two stars at a jump. At an endpoint scale \tau, hold the weights at s=\tau and let

The superscripts denote one-sided masks. Interpolate all changed entries at once between these two arrays. Each interpolated entry has absolute value at most |A_0(i,j)|, even when some entries open and others close at the same scale.

For fixed i,j,p, define nonnegative numbers using the original, unmasked arrays:

In R_0=[W_0\;W_2^*], a changing entry has row norm at most \sqrt{w_0(i)}\sqrt{X^2+H^2}, column norm at most \sqrt{w_1(j)}Y, and derivative modulus \sqrt{w_0(i)w_1(j)}|\delta A(i,j)|. In R_1=[W_1\;W_0^*], its adjoint entry has row norm at most \sqrt{w_1(j)}\sqrt{Y^2+K^2} and column norm at most \sqrt{w_0(i)}X, with the same derivative modulus. Star R_2=[W_2\;W_1^*] is unchanged. Lemma 5.2 and \sqrt{x^2+y^2}\le x+y therefore give

This estimate requires no assumption on the ranks during interpolation.

A probability measure on the center plane. For each fixed i,j, the measure

is a probability measure. Indeed, in the free coordinates (p_0,p_1), its density is g_{\sigma_0}(u_i-p_0)g_{\sigma_1}(u_j-p_1). Thus p_0,p_1 are independent Gaussian variables with means u_i,u_j and variances \sigma_0,\sigma_1, while p_2=-p_0-p_1. Adding Gaussian variances gives

For the last identity, p_0+p_1 has mean u_i+u_j and variance \sigma_0+\sigma_1, and averaging g_{\sigma_2}(u_m+p_0+p_1) adds the remaining variance. Each identity also follows directly by completing the square in its Gaussian integral.

Extend the arrays by zero to \mathbb Z^2 and define

Every variance in (5.8) is at least h^2. Lemma 5.1 consequently bounds \mathbb E X^2,\mathbb E Y^2,\mathbb E H^2,\mathbb E K^2 by C_g times the corresponding starred squares. Cauchy–Schwarz under the probability \nu_{ij} gives, in particular,

No independence of these four random variables is needed. The starred quantities depend only on the original arrays and the output indices, not on the scale \tau.

Summing the jumps. Integrating (5.7) over the centers and using w_0(i)w_1(j)\,\dd\pi=h^2\,\dd\nu_{ij} yields

Every entry opens once and closes once. Therefore its total absolute change, summed over any subset of endpoint scales, is at most 2|A_0(i,j)|. We obtain

The sum may include all endpoints or just the interior ones.

To bound this expression, set

The q=3/2 case of (5.1), applied on slices, gives

For H_*, fix i and change the summation index from j to -i-j; this is a bijection of \mathbb Z. For K_*, fix j and change i to -i-j. Thus both estimates are ordinary slice estimates, with no multiplicity or mesh factor. Restricting the sums back to the original finite square can only decrease them.

Hölder’s inequality in (5.10), followed by (5.11), proves

The last step uses a^2b\le(2a^3+b^3)/3. In particular, the estimate has no dependence on how many distinct scales occur.

Telescoping the heat estimate. List the distinct squared endpoints as t_0=s_-<t_1<\cdots<t_m=s_+. On every open interval (t_k,t_{k+1}) the array is fixed, so Lemma 4.4 applies. Its fixed-array continuity and Gaussian domination also justify the one-sided endpoint limits. Integrating on these intervals and summing gives

Here the terminal energy was discarded by nonnegativity. The initial right-hand mask has entries bounded by those of the original array, and all its variances are at least h^2. Lemma 4.5 therefore gives J(s_-+)\le C\mathcal S(A). Together with (5.12), this proves (5.5). The initial opening is accounted for by the right-hand initial energy; a final closing jump is unnecessary. Simultaneous endpoints have already been combined in the list of distinct scales, and their values at the single endpoint do not affect the scale integral.□

Proof. We apply the mixed trace inequality with Gaussian weights of larger variance than the windows k_s, so that the diagonal insertions are bounded. We then compare each quadratic cost with a variance choice covered by Proposition 5.3.

Use the masked array A_0^s(i,j)=A_0(i,j)\mathbf1_{\{\varepsilon_{ij}^2<s<\rho_{ij}^2\}} and the scale interval [s_-,s_+] from Proposition 5.3. All weighted matrices below use this same mask. Let superscript \mathrm b denote the base variances (\lambda s,\lambda s,\lambda s), where \lambda=11/10, and put

These matrices are real diagonal. With z=(u_i-p_v)/\sqrt{s},

Thus their operator norms have an absolute bound.

For each vertex v, let \boldsymbol\alpha(v) equal 1 at v and \lambda at the other two vertices. In the star centered at v, changing from the base variances to \boldsymbol\alpha(v)s changes only its row weight. In particular,

The last inequality uses g_s\le\sqrt\lambda\,g_{\lambda s}. The score L_v^{\boldsymbol\alpha(v)} equals \diag_i((u_i-p_v)/s), so the row densities also give the exact identity

Indeed, both inserted row densities are h k_s(u_i-p_v), and all column weights agree. The insertion on the left is therefore supported on T_v^{\boldsymbol\alpha(v)}. Lemma 2.2 and quadratic homogeneity give

Apply Lemma 3.1 to the base matrices. Integrate its three nonnegative terms separately, using Proposition 5.3 with the corresponding \boldsymbol\alpha(v). The factor s cancels \dd s/s, and we obtain

We identify the signed integral in this estimate. Expanding the trace leaves one signed Gaussian weight at each vertex:

On \Pi_0, substitute r=u_i-p_0 and q=u_j-p_1. The Jacobian is one and the third argument becomes u_i+u_j+u_l-r-q. Consequently the center integral is (k_s*k_s*k_s)(u_i+u_j+u_l). Since k_s(t)=s^{-1/2}k_1(t/\sqrt{s}), this convolution equals

The sums are finite and the Gaussian products are absolutely integrable on the center plane and the compact positive scale interval. Thus all these integrations can be performed term by term. The mask restricts the scale integral of the (i,j,l) term to (\varepsilon_{ij}^2,\rho_{ij}^2). Taking the absolute value after integration in (6.3) proves (6.1) with its right side initially replaced by C\mathcal S(A).

Put n_v=(\sum h^2|A_v|^3)^{1/3}. If any n_v is zero the signed sum vanishes. Otherwise apply the last estimate to A_v/n_v for all three vertices, keeping the selected endpoints fixed. The normalized value of \mathcal S is 3. Scaling the signed trilinear expression back proves the proposition.□

Proof. First fix a nonempty finite family of endpoint pairs (\varepsilon_k,R_k), 1\le k\le m, and take h\le\min_k\varepsilon_k. For finite arrays A_1,A_2, define

At each output pair choose the least index attaining this maximum, and write U(i,j) for its output. These choices provide admissible \varepsilon_{ij},\rho_{ij} in Proposition 6.1. They are fixed while the testing array A_0 varies.

Set H=\sum_{i,j}h^2 M(i,j)^{3/2}. If H=0 the desired bound is immediate. Otherwise choose

Then \sum h^2|A_0|^3=1 and h^2\sum A_0U=H^{2/3}. Proposition 6.1 therefore gives

This estimate has no dependence on the number of endpoint pairs. The complex phase of A_0 accounts for the absolute value at each output.

To pass to functions, define the compactly supported smooth inputs

Both pair maps have determinant one, so \|G_1\|_3=\|F\|_3 and \|G_2\|_3=\|G\|_3. Choose B>0 so that both supports lie in the interior of [-B,B]^2. Take h=B/N, -N\le i,j,l\le N, and sample

The corresponding continuum outputs are

They vanish outside [-B,B]^2: G_2(c,a) restricts a and G_1(b,c) restricts b. A nonzero integrand also has c\in[-B,B]. For our fixed finite endpoint family the integrands and their c derivatives are uniformly bounded on this compact cube. Riemann sums in c consequently converge uniformly for (a,b)\in[-B,B]^2 and 1\le k\le m. In particular,

The finite maximum of magnitudes preserves this uniform convergence. Since \max_k|\mathcal U_k| is continuous and compactly supported, the two-dimensional Riemann sums on the left of (6.4) converge to its L^{3/2} norm. The cube roots of the input sums converge to \|G_1\|_3 and \|G_2\|_3. The condition on h holds for all sufficiently large N. We obtain

No convergence or regularity of the maximizing indices is needed: we passed to the limit in the finite maximum itself.

With x=-a, y=-b, and t=a+b+c, the last integral is B^I_{\varepsilon_k,R_k}(F,G)(x,y). The output coordinate change preserves Lebesgue measure, giving the asserted estimate for every finite endpoint family.

Finally enumerate all rational pairs 0<\varepsilon<R and take increasing finite subfamilies. At each spatial point the output is continuous in the two positive endpoints, by its defining integrals on compact scale and spatial intervals. Thus the supremum over these rational pairs equals the supremum over all real pairs. It is measurable as a countable supremum of continuous spatial functions. Monotone convergence of the 3/2 powers of the finite maxima proves the result.□

Proof. Since k_1=-g_1' and Gaussian variances add under convolution,

The convolution identity follows by completing the square; differentiation under convolution is justified by Schwartz decay. For v\ge0, put

The substitution z=t/\sqrt{s} for t>0, followed by oddness for t<0, yields

Define Q(v)=P(v)-c_0\mathbf1_{\{v>1\}} for v\ge0. Smooth oddness of \psi gives P(v)=O(v^2) at zero. For v>1,

Thus Q has Gaussian decay at infinity and bounded one-sided values at its sole jump, v=1. Enlarging an absolute constant on a compact interval, we have

Set E_r(t)=Q(|t|/r)/t for t\ne0 and set E_r(0)=0. Substituting the definition of Q proves (6.5) and (6.6). Values at the four hard endpoints have no effect on any integral.□

Proof. Put q_r(t)=r^{-1}(1+|t|/r)^{-2}. For nonnegative locally integrable f on \R, the part of \int q_r(t)f(x+t)\,\dd t with |t|\le r is at most 2M_{\R}f(x). On the shell 2^kr<|t|\le2^{k+1}r, k\ge0, its contribution is at most

Summing the geometric series proves a bound independent of r. By (6.6) and weighted Cauchy–Schwarz,

Apply the preceding one-dimensional bound on the respective coordinate lines. This gives the pointwise assertion.

Hölder’s inequality and Lemma 5.1, applied on slices and then integrated in the other coordinate, give

□

Proof. Lemma 6.3 and Lemma 6.4 give, at each spatial point,

Both endpoint errors have the same majorant, independently of their radii. The hard outputs are continuous in both positive endpoints, so their supremum also equals the measurable supremum over rational pairs. Take L^{3/2} norms, apply Proposition 6.2 and Lemma 6.4, and divide by |c_0|>0.□

Proof. For each fixed t, Hölder’s inequality and translation invariance give \|F(\cdot+t,\cdot)G(\cdot,\cdot+t)\|_p\le\|F\|_3\|G\|_3. Minkowski’s integral inequality proves (7.1). Apply it to the countable annuli 1/n<|t|<n, n\ge2. Outside their combined exceptional null set, the integrand is absolutely integrable on every compact annulus away from zero. Absolute continuity of its integral gives the asserted endpoint continuity. If either input changes on a null set, apply the same bound to the zero L^3 difference, and use the same countable annuli. This proves independence simultaneously at every endpoint pair.□

Proof of Theorem 1.1. Choose compactly supported smooth f_m\to F and g_m\to G in L^3. For a fixed endpoint pair, bilinearity and (7.1) show that B_{\varepsilon,R}(f_m,g_m)\to B_{\varepsilon,R}(F,G) in L^p. Fix a finite family of rational endpoint pairs. The difference of the two finite maxima of magnitudes is bounded by the sum of the corresponding truncation differences. Thus these maxima also converge in L^p. Passing the compact smooth maximal estimate to the limit bounds this finite maximum by C_M\|F\|_3\|G\|_3. The constant is independent of the family. Exhaust all rational endpoint pairs and use monotone convergence and Lemma 7.1 to obtain Theorem 1.1.

We also obtain a useful continuity statement. For two pairs of inputs, bilinearity of every truncation gives, almost everywhere,

The bound just established makes the maximal map continuous from L^3\times L^3 to L^{3/2}. Density shows that it is the unique continuous extension of its compact smooth values.□

Proof. For compactly supported smooth f,g and a fixed (x,y), the function h(t)=f(x+t,y)g(x,y+t) is smooth and compactly supported. The truncated integral is

Its numerator is O(t) at zero and vanishes for large t, so it has a joint limit at every point.

For general inputs, let

These are measurable: by endpoint continuity the supremum may be taken over rational parameters. They decrease with n and are bounded by 2B_*(F,G). Fix smooth approximants f_m,g_m as above. Outside a single null set for this countable collection of input pairs,

Letting n\to\infty, the smooth joint convergence and Theorem 1.1 give

The right side tends to zero with m. Hence the limiting oscillation is zero almost everywhere. This is the Cauchy criterion for all pairs of endpoints in the indicated tails, so the joint complex limit exists. It is measurable by taking a fixed cofinal sequence of endpoint pairs, and its magnitude is at most B_*(F,G).

Taking (\varepsilon',R') to the limit in the definition of \Omega_n bounds the supremum in (7.3) by \Omega_n(F,G). Dominated convergence applies because 0\le\Omega_n(F,G)\le2B_*(F,G)\in L^p. This proves (7.3), and hence joint norm convergence. Passing to the norm limit in the bilinear identities for the truncations proves bilinearity of B. Its bound and uniqueness follow from Theorem 1.1 and density.□

Proof. Hölder’s inequality in (x,y) bounds the absolute integral by 2\log(R/\varepsilon)\prod_v\|F_v\|_3. Thus Fubini identifies the truncated form with \int F_0 B_{\varepsilon,R}(F_1,F_2). Theorem 1.1 gives its endpoint-uniform bound, and Theorem 7.2 gives joint scalar convergence by pairing the L^{3/2} limit with F_0. The limiting form is bounded and complex trilinear. Density of compact smooth triples proves uniqueness.□

Proof. Use the determinant-one change of variables

For flat inputs define

Each pair map has determinant one. Their inverses are

They preserve L^3 norms, compact smoothness, and Schwartz class. Direct substitution identifies the flat and simplex integrands and their truncation regions. The absolute integrability, uniform bound, and joint scalar convergence therefore follow from the flat conclusions. The inverse maps realize every simplex triple, which proves equality of the two optimal scalar constants. Density gives uniqueness of the bounded trilinear extension.□