Proof. A point \((w,t)\) lies in \(D\) exactly when \(t\ge\sup_{x\in K-z_0}\langle-w,x\rangle\). Since zero is interior to \(K-z_0\), this forces \(t\ge0\), and the slice at height \(t\) is \(-t(K-z_0)^\circ\). Integrating the slices gives \[\chi_C((0,1))=n!|K|,\qquad \chi_D((0,m))=n!m^{-m}|(K-z_0)^\circ|.\] For general cone data, interiority of \(V\) gives \(\langle V,x\rangle\ge c|x|\) on \(C\), and interiority of \(U\) gives the analogous estimate on \(D\). These inequalities prove finiteness. Under the displayed transformation, the two change-of-variables factors are \(|\det B|\) and \(|\det B|^{-1}\), and the inner products in the exponential factors are unchanged. ◻

Proof. We have \(X'=1+xX\), \(Y'=xY-1\), \(f'=-\phi(1-p)/p\), and \(j'=\phi p/(1-p)\). Direct substitution gives \[X^2f'+Y^2j'=0,\qquad XX'f'+YY'j'=-1,\qquad (X')^2f'+(Y')^2j'=-v'.\] Also \(X''=xX'+X\) and \(Y''=xY'+Y\). Thus \(g'=-(XX'f+YY'j)\), and differentiating once more proves \((\mathcal N+2)g=d\). Differentiating \(d=(X')^2f+(Y')^2j\) proves its stated first-order identity. Finally, differentiating the first identity twice gives \(d''=(\mathcal N+4)g''\); combining this with the derivative of the second identity proves the third. ◻

Proof. Moreau’s cone decomposition, with the positive-dual convention, gives \(\Pi_C(w)-\Pi_D(-w)=w\) and orthogonality of the two terms (Moreau 1962; Soltan 2019). Here is the sign convention directly. Writing \(x=\Pi_C(w)\), the projection variational inequality is \(\langle w-x,c-x\rangle\le0\) for \(c\in C\). Taking \(c=0,2x\) gives \(\langle w-x,x\rangle=0\); the same inequality then gives \(y=x-w\in D\). For \(d\in D\), \(\langle-w-y,d-y\rangle=-\langle x,d\rangle\le0\), so \(y=\Pi_D(-w)\). Applying the projection inequality to two inputs also gives \(|\Pi_C(w)-\Pi_C(w')|^2\le\langle\Pi_C(w)-\Pi_C(w'),w-w'\rangle\), hence the projection is one-Lipschitz. It is the gradient of the convex function \[\Phi(w)=\tfrac12|\Pi_C(w)|^2 =\sup_{x\in C}\bigl(\langle w,x\rangle-\tfrac12|x|^2\bigr).\] Rademacher’s theorem gives its a.e. differentiability (Rademacher 1919; Nekvinda and Zajíček 1988). A Lipschitz function is absolutely continuous on coordinate lines; one-dimensional integration by parts and Fubini’s theorem identify these derivatives with the weak derivatives. The convex-gradient representation therefore makes the a.e. derivative symmetric and positive semidefinite, and the Lipschitz bound makes it at most \(I\). Symmetry can equivalently be seen from commutation of weak mixed derivatives.

For fixed \(a>0\), minimize \(\Psi_a(\xi)=\mathbb E\Phi(Z+\xi)-aU\cdot\xi\). Write the orthogonal cone decomposition of \(\xi\) as \(\xi=x-y\). Since \(U\) is interior to \(C\), there is \(c>0\) with \(U\cdot y\ge c|y|\) for \(y\in D\). Jensen’s inequality gives \[\Psi_a(\xi)\ge\tfrac12|x|^2-a|U||x|+ac|y|.\] This lower bound is coercive. Gaussian convolution makes \(\Psi_a\) smooth, with Hessian \(\mathbb E D\Pi_C(Z+\xi)\). This Hessian is strictly between \(0\) and \(I\): the nondegenerate Gaussian assigns positive probability to open subsets of \(\operatorname{int}C\), where the derivative is \(I\), and of \(-\operatorname{int}D\), where it is zero. Thus the minimizer is unique, and its stationarity condition is \(\mathbb EX=aU\). The implicit function theorem proves smooth dependence on \(z\) and the displayed formula for \(\xi_z'\).

Choose a measurable bounded symmetric-contraction representative for the derivative where it is undefined. For each layer these exceptional inputs are Gaussian-null; Fubini’s theorem makes the choice immaterial to all layer integrals. Such a representative can be chosen jointly measurably by using difference quotients for the projection.

Gaussian integration by parts gives \(\mathbb E(X_z\cdot Z)=\mathbb E\operatorname{tr}(\Sigma P_z)\). Since \(\mathbb EY_z=aU-\xi_z\), orthogonality gives \[\mathbb E(X_z\cdot Z)=\mathbb E|X_z|^2-aU\cdot\xi_z.\] Subtracting \(|\mathbb EX_z|^2\) proves the variance expression in Equation [eq:6]. Moreover, \(\nabla|\Pi_C|^2=2\Pi_C\), so \[\frac d{dz}\mathbb E|X_z|^2=2aU\cdot\xi_z',\qquad h'=\frac1m(aU\cdot\xi_z'-pU\cdot\xi_z)=pB+ar_1.\] Finally, \[r_1=\frac p m U\cdot\bigl((\mathbb EP_z)^{-1}-I\bigr)U>0.\] No commutation of \(\Sigma\) with \(\mathbb EP_z\) is used. ◻

Proof. Interiority and Equation [eq:5] give \[\mathbb E|X_z|\le Ca(z),\qquad \mathbb E|Y_z|\le CB(z).\] For any one-Lipschitz vector function \(F\), \(|F(Z)|\le\mathbb E|F(Z)|+\mathbb E|Z|+|Z|\); apply this to both projections. For \(s\ge0\), \[a(-s)=\phi(s)\int_0^\infty t e^{-st-t^2/2}\,dt \asymp\frac{\phi(s)}{(1+s)^2},\qquad p(-s)\le C\frac{\phi(s)}{1+s}.\] The lower bound follows by restricting the integral to \([0,(1+s)^{-1}]\); for the upper bound omit one or the other exponential factor. On \([0,\infty)\), \(a(z)\asymp1+z\).

For negative \(z\), split \(\mathbb E(X_z\cdot Z)\) at \(|Z|=C_0(1+|z|)\). The first part is bounded by \(C(1+|z|)a(z)\). In the second part the pointwise Lipschitz bound and Gaussian tails give the same bound when \(C_0\) is sufficiently large. Hence \[0\le h(z)\le C(1+|z|)a(z),\qquad B(z)\le h(z)/a(z)\le C(1+|z|).\] Since \(0\le P_z\le I\) and \(\Sigma\ge.978I\), this gives Gaussian decay of every finite moment of \(P_z\). The first moment of \(X_z\) is Gaussian-small, while the Lipschitz bound gives bounded moments of arbitrarily high order on the negative half-line. Interpolation gives the asserted decay of its \(L^r\) norms.

For positive \(z\), Equation [eq:6] gives \(aB\le h\le s_0\), and thus \[|\mathbb EY_z|\le Cs_0/a(z),\qquad \xi_z=a(z)U-\mathbb EY_z.\] A fixed interior ball about \(U\) shows that \(\xi_z\) is a distance at least \(cz\) from the complement of \(C\) for all sufficiently large \(z\). On \(|Z|<cz\) the projection is the identity, so \(Y_z=0\) and \(P_z=I\) almost surely. Gaussian tails and the pointwise Lipschitz bound prove the remaining estimates.

The monotone function \(h\) tends to \(0\) and \(s_0\) at the two ends, with Gaussian tails. Integrating these tail functions proves the mass and moment assertions for \(\mu\). Finally \(ar_1\le h'\), and \(p/a\) has polynomial growth, proving the corresponding assertions for the other two densities. ◻

Proof. At negative layers the integrand tends to zero in \(L^2\) at a Gaussian rate. At positive layers use \[X_z-\mathbb EX_z=Z+Y_z-\mathbb EY_z\] to write it as \((p(z)-1)Z-Y_z+\mathbb EY_z\), with the same decay. The weak Jacobian of the integrand is \((p(z)I-P_z)\Sigma^{1/2}\), whose \(L^2\) norm has integrable Gaussian tails as well. Thus the truncated integrals and their weak derivatives are Cauchy in \(L^2\); closedness of the weak derivative gives the claim. Each integrand is centered. This argument differentiates the Lipschitz projections, not their derivative fields \(P_z\). ◻

Proof. We use the scalar bounds in Proposition 5. For a symmetric matrix \(Q\), transform the original data by \[C_Q=e^QC_{\rm in},\qquad D_Q=e^{-Q}D_{\rm in},\qquad U_Q=e^QU_{\rm in},\qquad V_Q=e^{-Q}V_{\rm in}.\] Lemma 3 preserves the Laplace product and \(U_Q\cdot V_Q=m\). Let \(T\) vary over its spectral box. For each pair \((Q,T)\), construct the biases and fields as above; we suppress their dependence on \(Q,T\) in the notation. We first establish continuity, then construct a continuous self-map and exclude its fixed points on a sufficiently large \(Q\)-sphere.

Proof. We first establish integrability and an exact formula for the contribution of \(\mathcal J_X\). Set \[J(z)=\int_{-\infty}^zW_X(l)P_l\,dl,\qquad S_z=J(z)/c_X(z),\qquad F_X(z)=\tau\log(c_X(z)I+J(z)).\] For finite \(z\) the matrix inside the logarithm is bounded below by \(c_X(z)I\), and its expected trace is finite. Hence its log determinant is integrable. Its derivative is \(-W_X(z)(I-P_z)\), so the matrix and \(F_X\) decrease with \(z\). The trace differentiation formula gives, almost everywhere, \[F_X'(z)=-j_X(z)(1-h_1(z))+\beta_X(z),\qquad h_1(z)=\tau P_z,\] where \[\beta_X(z)=j_X(z)\tau\bigl[S_z(I+S_z)^{-1}(I-P_z)\bigr]\ge0.\] The derivative is locally bounded uniformly in the Gaussian input after normalization, which justifies differentiation under expectation on compact intervals. The nonnegativity uses positivity under the trace and requires no commutation.

Compare \(F_X\) with \[F_0(z)=\int\log c_X(\min(z,x))\,d\gamma(x),\qquad F_0'=-j_X(1-p).\] Both functions equal \(\log c_X(z)+o(1)\) at negative infinity. For \(F_X\), this follows from \(0\le\tau\log(I+S_z)\le\tau S_z=o(1)\), using the layer tails; for \(F_0\), it follows from the Gaussian tail and the polynomial growth of \(\log c_X\). Thus \[(F_X-F_0)'=j_X(h_1-p)+\beta_X,\] and the signed first term is absolutely integrable. Integrating from a finite lower endpoint and letting it tend to negative infinity proves that \(\beta_X\) is integrable on every left half-line. At positive infinity \(F_0\) tends to the finite number \(\gamma\log c_X\). If the remaining integral of \(\beta_X\) were infinite, the last identity would force \(F_X\) to tend to positive infinity, contrary to its decrease. The same identity bounds \(F_X\) below. Consequently the limiting expectation is finite and \[\lim_{z\to\infty}F_X(z) =\gamma\log c_X+\int j_X(h_1-p)\,dz+\int\beta_X(z)\,dz.\] The matrices decrease to \(\mathcal J_X\). Subtracting the integrable log determinant at any finite layer permits monotone convergence to their limit, even if a zero eigenvalue is initially allowed there. The finite limiting expectation excludes that possibility almost surely. The finite expected trace controls the positive part of the limiting log determinant, so its negative part is integrable as well. We have therefore proved \[\tau\log\mathcal J_X =\gamma\log c_X+\int j_X(h_1-p)\,dz+\int\beta_X(z)\,dz.\] Reflection, replacing \(P_z\) by \(I-P_{-z}\), gives the corresponding identity for \(\mathcal J_Y\).

It remains to estimate the nonnegative surplus. For \(l\le z\), set \(D_{l,z}=c_X(z)^{-1}\int_l^zW_X(y)P_y\,dy\). Then \[0\le D_{l,z}\le S_z,\qquad D_{l,z}\le\bigl(c_X(l)/c_X(z)-1\bigr)I.\] The matrix function \(D\mapsto D(I+D)^{-1}=I-(I+D)^{-1}\) is order preserving. Applying this fact and then scalar spectral calculus to \(D_{l,z}\) yields \[S_z(I+S_z)^{-1} \ge D_{l,z}(I+D_{l,z})^{-1} \ge\frac1{c_X(l)}\int_l^zW_X(y)P_y\,dy.\]

We will average this inequality over \(l\). First, \(W_X\) is decreasing. To prove this, let \(g_1\) be standard normal and condition on \(g_1\le z\). The nonnegative random variable \(z-g_1\) has mean \(j_X(z)\) and variance \(j_X'(z)=1-c_X(z)j_X(z)>0\). Its survival function \[H_*(s)=p(z-s)/p(z),\qquad s\ge0,\] is log-concave, because \((\log p)''=-j_Xc_X\). Since \(H_*(0)=1\), concavity of its logarithm implies \(H_*(s+t)\le H_*(s)H_*(t)\). Integrating over the positive quadrant gives \[\tfrac12\mathbb E[(z-g_1)^2\mid g_1\le z] =\int_0^\infty\!\int_0^\infty H_*(s+t)\,ds\,dt \le j_X(z)^2.\] Thus \(j_X'\le j_X^2\), and \(W_X'=c_X(j_X'-j_X^2)\le0\).

For \(z\ge0\), choose the positive measure \(\omega_z\) on \([-z,z]\) specified by \[\int_{[-z,y]}\frac{d\omega_z(l)}{c_X(l)} =\frac1{4j_X(z)W_X(y)},\qquad -z\le y\le z.\] Its existence follows from the monotonicity of \(1/W_X\), with the indicated initial atom at \(-z\). Integration by parts gives \[\omega_z([-z,z]) =\frac1{4j_X(z)^2}+\frac{z}{2j_X(z)} \le\frac\pi8+\frac12<1,\] since \(j_X\) is increasing, \(j_X(0)=\sqrt{2/\pi}\), and \(j_X(z)\ge z\). Integrating the matrix inequality against this subprobability measure and using positivity gives \[\beta_X(z)\ge\frac14\int_{-z}^z\operatorname{nt}(y,z)\,dy\qquad(z\ge0).\] The reflected argument gives the corresponding estimate on the other half of the ordered region \(y\le z\): the two regions are distinguished by \(y+z\ge0\) and \(y+z\le0\). Their common boundary has measure zero. Since \(N\) is the integral of a symmetric kernel over the whole plane, the two surplus integrals sum to at least \(N/8\).

Finally, the two scalar comparison terms sum to \(-\gamma v\), because \(c_Xc_Y=1/(XY)\). The linear terms sum to \[\int\bigl(j_X(z)-j_X(-z)\bigr)(h_1(z)-p(z))\,dz=-\tau H_v,\] using \(j_X(z)-j_X(-z)=v'(z)\). This proves Equation [eq:11]. ◻

Proof. Let \(Z'\) be an independent copy of \(Z\), fix \(0<\zeta<1\), and put \(\widetilde Z=\zeta Z+\sqrt{1-\zeta^2}Z'\). The law of \(\widetilde Z\) is the same as that of \(Z\). Temporarily write \(X_z(h),Y_z(h),P_z(h)\) for the fields evaluated with \(Z=h\). For integers \(j\ge1\), define maps of \(G\) by \[\begin{aligned} \mathcal T_{X,j}(G)&=\int_{-j}^jW_X(z)\mathbb E_{Z'}X_z(\widetilde Z)\,dz,\\ \mathcal T_{Y,j}(G)&=\int_{-j}^jW_Y(z)\mathbb E_{Z'}Y_z(\widetilde Z)\,dz. \end{aligned}\] Their Jacobians are \(J_{X,j}\zeta\Sigma^{1/2}\) and \(-J_{Y,j}\zeta\Sigma^{1/2}\), where \[\begin{aligned} J_{X,j}&=\int_{-j}^jW_X(z)\mathbb E_{Z'}P_z(\widetilde Z)\,dz,\\ J_{Y,j}&=\int_{-j}^jW_Y(z)\mathbb E_{Z'}(I-P_z(\widetilde Z))\,dz. \end{aligned}\] Gaussian smoothing makes the maps smooth. Both factors are positive definite, because the conditional Gaussian sees an open set where the relevant projection derivative is the identity. Viewed as functions of \(Z\), the two maps have symmetric positive and negative definite derivatives, respectively. Integration along line segments proves strict monotonicity up to sign, hence injectivity. Their values belong to \(C\) and \(D\); since their derivatives are nonsingular, their images are open and lie in the interiors of these cones. The inverse function theorem gives a smooth inverse on each image.

For fixed \(j,\zeta\), the log determinants of the factors are integrable. Indeed, restrict the layer integral to a fixed compact subinterval of \([-1,1]\). Its biases are bounded. A fixed ball in \(\operatorname{int}C\), or in \(-\operatorname{int}D\), has conditional Gaussian probability at least \(c\exp[-C(1+|Z|)^2]\) after the relevant translations. Thus each factor is bounded below by this scalar times \(I\). The upper bound is deterministic, since \(0\le P_z\le I\) and the layer interval is finite. The map values also have finite first moments by the Lipschitz bounds.

For completeness, let \(\rho\) be the standard Gaussian density on \(\mathbb R^m\). Change of variables over the image of \(\mathcal T_{X,j}\), followed by Jensen’s inequality, gives \[\begin{aligned} \log\chi_C(V) &\ge\log\mathbb E\frac{e^{-V\cdot\mathcal T_{X,j}(G)} |\det D_G\mathcal T_{X,j}(G)|}{\rho(G)}\\ &\ge\tfrac m2\log(2\pi e) +\mathbb E\log|\det D_G\mathcal T_{X,j}|-\mathbb E[V\cdot\mathcal T_{X,j}]. \end{aligned}\] The integrability just proved justifies this step. The same argument applies to \(\mathcal T_{Y,j}\) and \(\chi_D(U)\). It uses only containment of the images in the cones, not surjectivity.

As \(j\to\infty\), the factors increase in matrix order to \(\mathbb E_{Z'}\mathcal J_X(\widetilde Z)\) and \(\mathbb E_{Z'}\mathcal J_Y(\widetilde Z)\). Their expected traces are finite by equality of the marginal laws. Subtracting the integrable log determinant for \(j=1\) makes monotone convergence applicable; the positive parts are controlled by the expected traces. Conditional concavity of log determinant (Boyd and Vandenberghe 2004) then gives \[\mathbb E\log\det\mathbb E_{Z'}\mathcal J_X(\widetilde Z) \ge\mathbb E\log\det\mathcal J_X(Z),\] and similarly for \(Y\). Lemma 10 supplies the two-sided integrability needed for this conditional Jensen inequality.

After dividing the sum of the two bounds by \(m\), the two right Jacobian factors contribute \(2\log\zeta+\tau\log\Sigma\). By the marginal law of \(\widetilde Z\), Equation [eq:5], and \(U\cdot V=m\), the cost terms converge to \(\int W_Xa\) and \(\int W_YB\). The layer tails justify both limits. Integration by parts gives \[\int W_Xa=\int c_Xp=\int\phi=1,\qquad \int W_YB=\int c_Yr_1=C_Y.\] The boundary products vanish by the same tails. Letting \(\zeta\uparrow1\) removes the term \(2\log\zeta\) and completes the proof. No limiting change of variables for the unsmoothed maps is needed. ◻

Proof of Proposition 12. Integration by parts gives \[\tau_\Sigma H_v=-\int h_\delta v'\,dz=\Delta_\mu v.\] We will prove \(C_Y-s_0+\Delta_\mu v=\delta[d]\). First, \(\int p\Delta B=\int a\Delta r\), by integrating the derivative of \(a\Delta B\); their sum is the zero total mass of \(\Delta_\mu\). Thus both integrals vanish. Since \(\int c_Yq_0=1\) and \(c_Y=2a-pv'\), \[C_Y-s_0+\Delta_\mu v=-\int(p\Delta r+h_\delta)v'\,dz.\]

For any smooth polynomial-growth test \(w_0\), we have \[\int\bigl[p\Delta B(2+\mathcal N)w_0 +(p\Delta r+h_\delta)w_0'\bigr]\,dz=0. \tag{12}\] Indeed, integrate the \(w_0''\) term once by parts and use \((\Delta B)'=-\Delta r\), \(h_\delta'=p\Delta B+a\Delta r\), and \(a-\phi=zp\). The expression becomes \[\int\bigl[2p\Delta B\,w_0+(a\Delta B+h_\delta)w_0'\bigr]\,dz,\] which is zero because \((a\Delta B+h_\delta)'=2p\Delta B\). The boundary products vanish: \(\Delta B=O(1+|z|)\) at the negative end, where \(a,p\) have Gaussian decay, and \(B,h-s_0\) have Gaussian decay at the positive end. The densities involving \(\Delta r\) have all needed polynomial moments by Lemma 7.

Apply Equation [eq:12] to \(w_0=g\). Equation [eq:2] gives \[\int\bigl[p\Delta B\,d+(p\Delta r+h_\delta)g'\bigr]\,dz=0.\] Substituting \(2zd-d'=v'+2g'\) in the definition of \(\delta[d]\) now proves the desired scalar identity. Finally, \[\tau H_v=\Delta_\mu v-\tau_T H_v,\qquad s_0=1+\tau T.\] Inserting these identities in the entropy bound yields exactly Equation [eq:10]. ◻

Proof. Use the orthonormal multivariate Hermite basis, indexed by \(\mathbf a=(a_1,\ldots,a_m)\in\{0,1,2,\ldots\}^m\); see, for example, (Bell 2015, secs. 5–6). The number operator has eigenvalue \(|\mathbf a|=\sum_i a_i\), and differentiation in coordinate \(i\) lowers the index by \(e_i\) with factor \(\sqrt{a_i}\). Thus, on a nonconstant mode, differentiation and application of \(\mathcal K_G\) give the multiplier \[\sum_{i:a_i>0}\frac{a_i}{1+|\mathbf a|-1}=1.\] This proves the covariance formula for finite Hermite sums and then for \(W^{1,2}\) fields by convergence of the coefficients and their weak derivatives.

For a matrix field with coefficient \(E_{\mathbf a}\), the contribution of \(\mathcal K_GE\) to its squared form norm is \[\frac{\langle E_{\mathbf a},E_{\mathbf a}\rangle_\Sigma} {1+|\mathbf a|}.\] It is summable whenever \(E\) is square-integrable. The deterministic weight \(\Sigma\) acts only on matrix coefficients, so it commutes with the Hermite degree decomposition. The same expansion gives the stated form norm and its Dirichlet expression. ◻

Proof. Consider the centered vector fields \[\mathcal W_f =\int_{\mathbb R} \bigl(p(z)Z-X_z+\mathbb E X_z\bigr)f'(z)\,dz\] and \(\mathcal W_j\). By Lemma 8, they belong to Gaussian \(W^{1,2}\), with Jacobians \(H_f\Sigma^{1/2}\) and \(H_j\Sigma^{1/2}\). Applying Lemma 13 and dividing by \(m\) gives \(\langle H_f,\mathcal K_GH_j\rangle_\Sigma\).

We also compute this covariance from the layers. If \(x\leq z\), the identities \(X_z=Z+\xi_z+Y_z\) and \(\xi_z=a(z)U-\mathbb E Y_z\) imply \[\frac1m\mathbb E \bigl[(X_x-\mathbb E X_x)\cdot(X_z-\mathbb E X_z)\bigr] =h(x)-a(x)B(z)+\operatorname{ct}(x,z).\] Indeed, the terms involving \(a(x)a(z)|U|^2\) cancel after subtracting the product of the means. For the centered differences defining \(\mathcal W_f\), the remaining covariance kernel, apart from \(\operatorname{ct}(x,z)-a(\min(x,z))B(\max(x,z))\), is \[s_0p(x)p(z)-p(x)h(z)-h(x)p(z)+h(\min(x,z)).\] Since \(\mu\) has mass \(s_0\) and cumulative function \(h\), this equals \[\int \bigl(p(x)-\mathbf1_{\{y\leq x\}}\bigr) \bigl(p(z)-\mathbf1_{\{y\leq z\}}\bigr)\,d\mu(y).\] The centered-test identity \[\int\bigl(p(z)-\mathbf1_{\{y\leq z\}}\bigr)f'(z)\,dz=f(y)\] therefore turns its integral against \(f'(x)j'(z)\) into \(\mu[fj]\).

The mixed distributional derivative of \(a(\min(x,z))B(\max(x,z))\) is \[d\mu(x)\,\delta_x(dz) -p(\min(x,z))r_1(\max(x,z))\,dx\,dz.\] Off the diagonal this follows from \(a'=p\) and \(B'=-r_1\). On the diagonal, the jump is \(pB+a r_1=h'\). Integrating by parts in both variables thus gives \(\mu[fj]-\mathcal Q[f,j]\) for this kernel. Subtraction proves [eq:13].

For noncompact tests, insert smooth cutoffs and pass to the limit. The tail estimates imply integrability with polynomial weights of \(a(\min(x,z))B(\max(x,z))\): one separates the ordered region into \(x\leq z\leq0\), \(x\leq0\leq z\), and \(0\leq x\leq z\). The diagonal measure and the density of \(\mathcal Q\) have all polynomial moments, since their marginals are \(\mu\). These bounds justify the integrations by parts and the limit. ◻

Proof. Write \(b_1=h_f^*\), \(b_2=h_j^*\). Direct differentiation gives \[\int_{-\infty}^z p f=a b_1-p b_1',\qquad \int_z^\infty q_0 f=B_0 b_1+q_0 b_1',\] and the analogous formulas for \(j\). Boundary terms vanish by polynomial growth and the Gaussian tails. As \(q_0a+pB_0=\phi\), insertion into \(\mathcal Q_0\) gives \(\mathcal Q_0[f,j]=\gamma[jb_1]=\gamma[fb_2]\), using self-adjointness of \(1+\mathcal N\) under \(\gamma\).

Put \[H_1=fb_2+jb_1,\qquad J_1=fb_2'+jb_1',\qquad w'=H_1'-J_1.\] Split the kernel variation into its two ordered half-planes. Symmetry and the preceding primitive formulas give \[\begin{aligned} (\mathcal Q-s_0\mathcal Q_0)[f,j] &=\int\Delta r(z)\left[ j(z)\int_{-\infty}^z p(x)f(x)\,dx +f(z)\int_{-\infty}^z p(x)j(x)\,dx\right]dz\\ &=\int\Delta r(aH_1-pJ_1). \end{aligned}\] Use \(\Delta r=-(\Delta B)'\), \(\Delta_\mu w=-\int h_\delta w'\), and \((a\Delta B+h_\delta)'=2p\Delta B\). After integration by parts, subtracting \(\Delta_\mu w\) leaves \[-\int\bigl[p\Delta B\,H_1+(p\Delta r+h_\delta)J_1\bigr].\] The product rule for \(\mathcal N\) gives \[\begin{aligned} H_1&=(2+\mathcal N)(b_1b_2)+2b_1'b_2',\\ J_1&=(b_1b_2)'+2z\,b_1'b_2'-(b_1'b_2')'. \end{aligned}\] Equation [eq:12] cancels the terms involving \(b_1b_2\). The remaining expression is \(\delta[b_1'b_2']\) by [eq:9], proving [eq:14]. All boundary terms contain the weighted differences controlled by the preceding tail estimates. Finally \(\Delta_\mu\) has total mass zero, so constants in \(w\) do not affect the formula. ◻

Proof. Abbreviate \(\Theta_y=\Theta_y(B_*)\). For \(x<z\), \(\Theta_x\Theta_z=\Theta_x\), and expansion gives \[E_x^{B_*}\circ E_z^{B_*}+P_x\circ(I-P_z) =P_x\circ(I-\Theta_z)+(I-P_z)\circ\Theta_x.\] Multiply by \(f'(x)f'(z)\), integrate over \(x<z\), and include the opposite order. The first two terms give the left side of [eq:15] and \(\operatorname{Nt}_\Lambda(f,f)\). The terms on the right give \(2e_\Lambda^{B_*}(f,f)\), by the spectral identity \[f(B_*)-f(y)I =\int\bigl(\mathbf1_{\{y\leq l\}}I-\Theta_l\bigr)f'(l)\,dl.\] All rearrangements are justified by the polynomially weighted integrability above. In particular, no nesting of \(P_z\), or commutation of \(P_z\) with a threshold or with \(\Lambda\), has been assumed. ◻

Proof. The first identity follows from \(R=-\int E_z^L\,dz\): \[\langle R,j'(L)\rangle_\Lambda+e_\Lambda(\iota,j') =-\int j'(z)\tau_\Lambda E_z^L\,dz.\] For the second, integration by parts in \(\tau_\Sigma H_j=\int(s_0p-h)j'\) gives \(\Delta_\mu j\). For the third, subtract \(f(L)\) from \(H_f\) in the pairing and write \(j(L)=(j(L)-j(z)I)+j(z)I\). The first part gives \(-e_\Sigma(f,j)\). For the scalar part use \(f'j=(fj-w)'\) and the second identity. Combining with \(\tau_\Sigma(fj)(L)\) gives the last line of [eq:16]. ◻

Proof. For the linear identity, evaluate \(\mu[\iota f_l]-2\mathcal Q[\iota,f_l]\) in two ways. Since \((1+\mathcal N)^{-1}\iota=\iota/2\), Equation [eq:14] gives \[\mu[\iota f_l]-2\mathcal Q[\iota,f_l] =\Delta_\mu\psi_l-\delta[l].\] The reference Gaussian term vanishes by self-adjointness of \(1+\mathcal N\), and the derivative of the remaining primitive is \(f_l-2h_l=\psi_l'\).

For the second evaluation, replace \(\mathcal Q[\iota,f_l]\) using Equation [eq:13] and replace \(\mu[\iota f_l]\) using the last line of Equation [eq:16]. Since \(H_\iota=A=L+R\) and \(\mathcal K_GL=L/2\), the same quantity \(\mu[\iota f_l]-2\mathcal Q[\iota,f_l]\) equals \[-(\Delta_\Sigma-\Delta_\mu)v_0+e_\Sigma(f_l,\iota) -2\langle R,\mathcal K_GH_{f_l}\rangle_\Sigma +2\operatorname{Ct}(\iota,f_l),\qquad v_0'=f_l.\] Equate these two evaluations. Equation [eq:16], applied to \(\psi_l-v_0\), whose derivative is \(-2h_l\), rewrites the remaining difference between \(\Delta_\mu\) and \(\Delta_\Sigma\). Its pairing with \(R\) combines with \(2\langle R,\mathcal K_GH_{f_l}\rangle_\Sigma\) to give \(2\langle R,d_l^*\rangle_\Sigma=\sigma_\Sigma(l)\). The other terms are precisely the first line of Equation [eq:17].

For the quadratic identity write \(b_1=h_l\), \(f=f_l\). Expand the squared form norm of \(d_l^*=\mathcal K_GH_f-b_1(L)\). In Hermite degree \(j\), the form contributes a factor \(1+j\) and \(\mathcal K_G\) contributes its reciprocal. Thus the cross term is \(-2\langle H_f,b_1(L)\rangle_\Sigma\), and the square of \(\mathcal K_GH_f\) is \(\langle H_f,\mathcal K_GH_f\rangle_\Sigma\). This uses the form regularity already proved, not a derivative of \(H_f\). Consequently \[\begin{split} D_\Sigma(l)^2 ={}&\langle H_f,\mathcal K_GH_f\rangle_\Sigma -2\langle H_f,b_1(L)\rangle_\Sigma +\tau_\Sigma b_1(L)^2\\ &+\sum_i\tau_\Sigma \bigl(\partial_{G_i}b_1(L)\bigr)^2 . \end{split}\] Adding \(J_\Sigma(l)\) cancels exactly the same weighted Dirichlet term. By [eq:13], the remaining expression is \[\mathcal Q[f,f]+\operatorname{Ct}(f,f) -2\langle H_f,b_1(L)\rangle_\Sigma +\tau_\Sigma\bigl(b_1(L)^2+Lh_{l^2}(L)\bigr).\] Insert [eq:14] for \(\mathcal Q[f,f]\) and [eq:16] for the pairing. The reference terms cancel because \[\gamma[xh_{l^2}]=\gamma[l^2] =\gamma[(f-b_1)b_1].\] The \(\Delta_\mu\) terms have the primitive \[2\int^x f'b_1-2fb_1+2\int^x fb_1',\] whose derivative is zero. Besides \(\delta[l^2]+\operatorname{Ct}(f,f)+2e_\Sigma(f,b_1)\), what remains is \(\Delta_\Sigma\) applied to \[b_1^2+xh_{l^2}-2\int^x fb_1'.\] Its derivative is \[h_{l^2}-x l^2+2ll' =-(\mathcal N-1)h_{l^2}=-\psi_{l^2}'.\] Since \(\Delta_\Sigma\) annihilates constants, this proves the second identity. All form expansions used a deterministic weight; none requires it to commute with the matrix fields. ◻

Proof. Use the first line of [eq:17] for \(F\), subtract the second line for each component of \(u\), and use \(\delta[d]=\delta[F]-\delta[|u|^2]\). The combined layer profile for the weight \(\Sigma\) is \[2F(x)+S_F(z)-2S_u(z)\cdot u(x) =4c+4kq(x)+2(k-q(x))S_q(z)-2S_\pi(z)\cdot\pi(x) =H(z,x).\] The contact integral being subtracted is \[\iint \operatorname{ct}(x,z) \bigl(S_F(x)+S_F(z)-S_u(x)\cdot S_u(z)\bigr)\,dx\,dz.\] It is nonnegative: \(\operatorname{ct}\geq0\), and its scalar factor is the second positive expression in [eq:3]. Discarding this integral preserves the upper bound.

Finally, subtract \(\tau_TH_v\) using the first line of [eq:16]. Its derivative correction is \(-\langle R,v'(L)\rangle_T\), and its layer correction is \(-e_T(\iota,v')=-\mathcal L_T[v''(x)]\). Splitting \(\Sigma=I+T\) gives the two layer terms in [eq:18], while \(\Delta_\Sigma\psi_d-\Delta_Tv=\Delta_{\rm p}\). ◻

Proof. The definitions of \(h_l,f_l,d_l^*\) depend linearly on \(l\). For the constant profile \(l=1\), its centered primitive is \(x\) and its image under \(1+\mathcal N\) is \(2x\). Hence \[d_1^*=2\mathcal K_G A-L=2\mathcal K_G R,\] because \(\mathcal K_G L=L/2\). Since \(F=c+2kq\), it follows that \[\sigma_\Sigma(F) =4c\langle R,\mathcal K_G R\rangle_\Sigma +4k\langle R,d_q^*\rangle_\Sigma.\] The vector \(u=(\pi,q)\) gives \(D_\Sigma(u)^2\ge\|d_q^*\|_{1,\Sigma}^2\). We estimate the resulting quadratic expression separately on odd and even Hermite degrees.

On the odd degrees, completion of the square in the \((1+\mathcal N_G)\)-form norm gives \[4k\langle R_o,(d_q^*)_o\rangle_\Sigma -\|(d_q^*)_o\|_{1,\Sigma}^2 \le 4k^2\langle R_o,\mathcal K_G R_o\rangle_\Sigma.\] Here the square is centered at \(2k\mathcal K_G R_o\), so the calculation remains valid in the form domain. Since \(c+k^2=b\) and \(\mathcal K_G\le1/4\) on the degrees of \(R_o\), the odd contribution is at most \(b\tau_\Sigma R_o^2\).

For the even degrees, the evenness of \(q\) implies that its Gaussian-centered primitive \(h_q\) is odd, and hence \(f_q\) is odd. The field \(L\) is odd in \(G\), so \(h_q(L)\) and \(f_q(L)\) are odd matrix fields. Thus \[(d_q^*)_e=\mathcal K_G(M_{f_q}^L)_e.\] Put \(\mathfrak b=b+\eta\) and \[b_m'=\frac{b_m}{1+t_+} =\frac{b-3\eta}{3(3\mathfrak b-4c)}=\frac{134}{309}>0.\] On a positive even degree, the eigenvalue \(\rho\) of \(\mathcal K_G\) lies in \((0,1/3]\). Writing \(R_j,M_j\) for the components of \(R,M_{f_q}^L\) in that degree, the quadratic form to estimate is \[4c\rho\|R_j\|_\Sigma^2 +4k\rho\langle R_j,M_j\rangle_\Sigma-\rho\|M_j\|_\Sigma^2.\] The scalar matrix inequality \[\begin{pmatrix}4c\rho&2k\rho\\2k\rho&-\rho\end{pmatrix} \le \begin{pmatrix}\mathfrak b&0\\0&b_m'\end{pmatrix}\] holds for \(0\le\rho\le1/3\). Indeed, the first diagonal entry of the difference is at least \[\mathfrak b-\frac{4c}{3}=\frac{103}{750}>0,\] and its determinant is \[(\mathfrak b-4c\rho)(b_m'+\rho)-4k^2\rho^2 =(1-3\rho)\left(\mathfrak b\,b_m'+\frac{4b}{3}\rho\right)\ge0.\] The constant degree has zero \(R\)-coefficient and contributes a nonpositive term. Summing the even degrees therefore gives at most \[\mathfrak b\tau_\Sigma R_e^2 +b_m'\|(M_{f_q}^L)_e\|_\Sigma^2 \le \mathfrak b\tau_\Sigma R_e^2+b_m\|M_{f_q}^L\|_2^2,\] where we used \(\Sigma\le(1+t_+)I\). Together with the odd estimate, this proves Equation [eq:19].

Finally, Equation [eq:15] yields \[\|M_{f_q}^L\|_2^2 =2e(f_q,f_q)-\operatorname{Nt}(f_q,f_q).\] The kernel defining \(\operatorname{Nt}\) is nonnegative, and \(f_q'=S_q>0\); hence its last term is nonnegative. The identity \(e(f_q,f_q)=\mathcal L_I[S_q(z)S_q(x)]\) proves the remaining assertion. ◻

Proof. Write \(Q=1_{(-\infty,z]}(B_*)\). Since \(Q\) is diagonal in the chosen basis and \(P^2\le P\), \[(E^2)_{ii}=(P^2)_{ii}-2Q_{ii}P_{ii}+Q_{ii} \le P_{ii}-2Q_{ii}P_{ii}+Q_{ii} =\epsilon_iE_{ii}.\] On the block \(x_i>z\), the compression of \(E\) is the compression of \(P\), and is positive semidefinite. On the block \(x_i<z\), the compression of \(-E\) is the compression of \(I-P\), and has the same property. Diagonal congruence by the factors \(\sqrt{a_i}\) proves the assertion about \(S\). Finally, symmetry of \(E\) gives \[\sum_{i,j}\frac{a_iw_i+a_jw_j}{2}|E_{ij}|^2 =\sum_i a_iw_i(E^2)_{ii} \le\sum_i a_iw_i\epsilon_iE_{ii}.\] ◻

Proof. Equation [eq:15], applied with threshold \(A\), gives \(N_T=2e_T^A(\iota,\iota)\) and \(N=2\mathcal B^A\). In a spectral basis of \(A\), the scalar kernel multiplying \((E_z^A)_{ij}T_{ji}\) in \(2e_T^A(\iota,\iota)\) is \(x_i+x_j-2z\). Separate this kernel into a comparison term and a remainder. The comparison term is \(2\epsilon\sqrt{a_i a_j}\) when both endpoints have sign \(\epsilon\), and zero when they have opposite signs. Its pairing is \(2\operatorname{tr}(TS_z)\), where \(S_z\ge0\) is supplied by Lemma 21. Thus the original pairing equals \(2\operatorname{tr}(TS_z)\) plus the pairing with the remaining kernel. Since \(T\ge t_-I\), the normalized expected integral of the comparison term is at least \(2t_-\mathcal B^A=t_-N\).

Let \(k_z(x_i,x_j)\) be the remaining kernel, and put \(\omega_z=(a_i+a_j)/2\). On a common-sign block it is \(\epsilon(\sqrt{a_i}-\sqrt{a_j})^2\); between the endpoints it remains \(x_i+x_j-2z\). We claim that \[\int_{\mathbb R}\frac{k_z(x_i,x_j)^2}{\omega_z}\,dz =c_g(x_i-x_j)^2.\] If the endpoints coincide, the remaining kernel vanishes; the quotient is assigned value zero wherever its denominator vanishes. Otherwise, translation and scaling reduce the calculation to endpoints \(0,1\). The interval between them contributes \[2\int_0^1(1-2z)^2\,dz=\frac23.\] The two exterior intervals contribute \[4\int_0^\infty \frac{(\sqrt{1+t}-\sqrt t)^4}{1+2t}\,dt=2\log2-1.\] For completeness, use the scalar substitution \[u_0=(\sqrt{1+t}-\sqrt t)^2,\qquad t=\frac{(1-u_0)^2}{4u_0},\qquad |dt|=\frac{1-u_0^2}{4u_0^2}\,du_0.\] The last integral becomes \[2\int_0^1\frac{u_0(1-u_0^2)}{1+u_0^2}\,du_0 =2\log2-1.\] This proves the claim.

Cauchy–Schwarz in the variables \(G,z,i,j\), with weight \(\omega_z\), bounds the absolute value of the remaining pairing by \[\left(\frac1m\mathbb E\int \sum_{i,j}\omega_z|(E_z^A)_{ij}|^2\,dz\right)^{1/2} \left(\frac1m\mathbb E\sum_{i,j}|T_{ji}|^2 \int\frac{k_z(x_i,x_j)^2}{\omega_z}\,dz\right)^{1/2}.\] Lemma 21 bounds the first factor by \(\sqrt{\mathcal B^A}\). The second factor is \(\sqrt{c_g}\|[A,T]\|_2\), since the commutator has entries \((x_i-x_j)T_{ij}\). This proves Equation [eq:20]. ◻

Proof. Work in a spectral basis of \(L\), and use \(a_i,\epsilon_i,E_z^L,S_z\) as in Lemma 21. Write \(g_i=g_z(x_i)\). The kernel multiplying \(T_{ji}(E_z^L)_{ij}\) in \(\mathcal L_T[H_0]\) is \[\frac{(x_i-z)g_i+(x_j-z)g_j}{2}.\] Separate this kernel into the comparison term \(\epsilon_i\sqrt{a_i a_j}(g_i+g_j)/2\) on a common-sign block, zero on the other blocks, and a remainder \(k_z(x_i,x_j)\). The original pairing therefore equals \(\operatorname{tr}(S_z(T\circ g_z(L)))\) plus the pairing with \(k_z\). We will bound the comparison term from below and the absolute value of the remainder from above.

We first bound this pairing in matrix order. For any real symmetric matrix \(V_0\), the spectral range of \(T\) implies \[T\circ V_0\ge t_cV_0-\frac{t_r}{2}\left(h_sI+\frac{V_0^2}{h_s}\right).\] Indeed, writing \(D=T-t_cI\), we have \(\|D\|\le t_r\), and for every vector \(v_0\), \[|\langle v_0,DV_0v_0\rangle| \le t_r|v_0|\,|V_0v_0| \le\frac{t_r}{2} \left(h_s|v_0|^2+h_s^{-1}|V_0v_0|^2\right).\] This proves the matrix inequality without a commutation assumption or a sign restriction on \(V_0\). Apply it with \(V_0=g_z(L)\), and pair against \(S_z\ge0\). After integration, the comparison part of \(\mathcal L_T[H_0]\) is bounded below by \[\int\left[t_cg_z-\frac{t_r}{2} (h_s+g_z^2/h_s)\right]\,d\beta.\] Together with \(\mathcal L_I[H-P_0]=\int\overline{H-P_0}\,d\beta\), this accounts for the term \(-\int W\,d\beta\).

It remains to estimate the pairing with the remainder \(k_z(x_i,x_j)\). Use the nonnegative scalar weight \[D_z=\frac{a_iW_0(z,x_i)+a_jW_0(z,x_j)}2.\] Lemma 21 gives \[\frac1m\mathbb E\int\sum_{i,j} D_z|(E_z^L)_{ij}|^2\,dz \le\int W_0\,d\beta.\] We will prove the scalar estimate \[\int_{\mathbb R}\frac{k_z(x_i,x_j)^2}{D_z}\,dz \le\alpha(\log2+.25)(x_i-x_j)^2.\] Whenever \(D_z=0\), the defining integrals of \(H_0(z,\cdot)^2\) on both relevant segments vanish. The corresponding integrals of \(H_0\), and hence \(k_z\), vanish as well; we set the quotient equal to zero there. For coincident endpoints, \(k_z\) vanishes identically. We therefore assume below that \(\ell=|x_i-x_j|>0\).

If \(z\) lies between the endpoints, put \(a=|x_i-z|\), \(b=|x_j-z|\), so \(a+b=\ell\). Let \(Q_z\) be the integral of \(H_0(z,y)^2\) over the interval between \(x_i,x_j\). Then \(D_z=Q_z/(2\alpha)\). Twice \(k_z\) is a signed sum of the integrals of \(H_0\) over the two subintervals, so Cauchy–Schwarz gives \[k_z^2\le\frac{\ell Q_z}{4},\qquad \frac{k_z^2}{D_z}\le\frac{\alpha\ell}{2}.\] The integral over this middle interval is therefore at most \(\alpha\ell^2/2\).

If \(z\) lies outside the interval, order the distances so that \(a\ge b>0\), and let \(\epsilon\) be their common sign. Writing \(g_a,g_b\) for the two segment averages, we have \[k_z^2 =\frac{(\sqrt a-\sqrt b)^2}{4} (\sqrt a\,g_a-\sqrt b\,g_b)^2.\] Set \(h_z(t)=H_0(z,z+\epsilon t)\), for \(0\le t\le a\). Then \[\sqrt a\,g_a-\sqrt b\,g_b =\frac1{\sqrt a}\int_b^a h_z(t)\,dt +\left(\frac1{\sqrt a}-\frac1{\sqrt b}\right) \int_0^b h_z(t)\,dt.\] Furthermore, \[2\alpha D_z =\int_b^a h_z(t)^2\,dt+2\int_0^b h_z(t)^2\,dt.\] Cauchy–Schwarz, with multiplicity \(1\) on \((b,a]\) and multiplicity \(2\) on \([0,b]\), consequently gives \[(\sqrt a\,g_a-\sqrt b\,g_b)^2 \le 2\alpha D_z \left[\frac{a-b}{a} +\frac12(1-\sqrt{b/a})^2\right].\] In either exterior interval, write \(a=\ell(1+t)\), \(b=\ell t\). Integrating the last estimate over both exterior intervals gives at most \[\alpha\ell^2\int_0^\infty (\sqrt{1+t}-\sqrt t)^2 \left[\frac1{1+t} +\frac12\left(1-\sqrt{\frac{t}{1+t}}\right)^2\right]dt.\] Use the substitution \(u_0=(\sqrt{1+t}-\sqrt t)^2\) from the proof of Lemma 22. The integral becomes \[\int_0^1\left(\frac1{1+u_0}-\frac{u_0}{2}\right)du_0 =\log2-\frac14.\] Adding the middle-interval contribution proves the scalar estimate with coefficient \(\alpha(\log2-\tfrac14+\tfrac12) =\alpha(\log2+\tfrac14)\).

Put \(\mathcal V=\int W_0\,d\beta\). Weighted Cauchy–Schwarz in \(G,z,i,j\), followed by the scalar estimate, bounds the absolute value of the remaining pairing by \[\sqrt{\mathcal V}\, \sqrt{\alpha(\log2+.25)}\,\|[L,T]\|_2.\] The convention for zero weights already covers \(\mathcal V=0\). The comparison term has already contributed \(-\int W\,d\beta\) to the upper bound for \(-\mathcal L_I[H-P_0]-\mathcal L_T[H_0]\). The remainder can increase this bound by at most the absolute value just estimated. Writing \(W=(W-W_0)+W_0\), we obtain \[-\int(W-W_0)\,d\beta -\mathcal V+\sqrt{\alpha(\log2+.25)\mathcal V}\,\|[L,T]\|_2.\] Completion of the square in \(\sqrt{\mathcal V}\) proves Equation [eq:22]. ◻

Proof. For a smooth scalar function \(h\), its matrix derivative at a real symmetric matrix satisfies \[\bigl(Dh(L)[M]\bigr)_{ij} =\left(\int_0^1 h'((1-t)x_i+tx_j)\,dt\right)M_{ij}.\] This is the Daleckiı̆–Kreı̆n divided-difference derivative formula; see (Noferini 2017, Theorem 2.11). To justify it directly here, first use polynomials and then approximate \(h\) in \(C^1\) on a compact interval containing the spectrum. The divided differences converge uniformly, and the derivative maps converge in Hilbert–Schmidt norm. This gives the formula locally, including coincident eigenvalues, without differentiating a choice of eigenvectors.

Apply this formula to \(h=F_*'\). In the fixed coordinates, \[\partial_{G_\ell}F_*'(L)=DF_*'(L)[M_\ell] =\bar f\odot M_\ell.\] The last expression is interpreted by the preceding spectral convention. Its Hilbert–Schmidt norm is at most \(\sup_{|x|\le\|L\|_{\rm op}}|f(x)|\,\|M_\ell\|_{\rm HS}\). Since \(\|L\|_{\rm op}\le C|G|\), polynomial growth permits Gaussian integration by parts, for example by inserting smooth cutoffs in \(G\) and then letting their supports increase. Thus the local polynomial approximation is used only to establish the derivative formula; no global interchange of polynomial approximants and Gaussian expectations is needed.

Finally, \(L\) commutes with \(F_*'(L)\), so \[\begin{aligned} \mathbb E[(\mathcal N F_*)(L)] &=\mathbb E[L\circ F_*'(L)]-\mathbb E f(L)\\ &=\sum_\ell M_\ell\circ \mathbb E[\bar f\odot M_\ell]-\mathbb E f(L). \end{aligned}\] Adding and subtracting \(\mathbf S\circ\mathbb E f(L)\) proves [eq:24]. ◻

Proof. The scalar functional. Since \(d\) is even, \(\psi_d\) is even, and the differential identities [eq:2] give \[\psi_d''=\mathcal N d=(\mathcal N+2)\mathsf K, \qquad (\psi_d-v)''=(\mathcal N+2)(\mathsf C-a_R g'').\] For any smooth \(F_*\), one has \((\mathcal N F_*)''=(\mathcal N+2)F_*''\). Define the even second primitives \[A_K(x)=\int_0^x(x-y)\mathsf K(y)\,dy,\qquad A_C(x)=\int_0^x(x-y)\mathsf C(y)\,dy.\] They satisfy \(A_K''=\mathsf K\) and \(A_C''=\mathsf C\), and their derivatives have polynomial growth. The preceding identities show that \(\psi_d-\mathcal N A_K\) and \(\psi_d-v-\mathcal N(A_C-a_Rg)\) have zero second derivative. Since \(v\) and \(g\) are also even, both differences are even and therefore constant. Thus, for constants \(c_K,c_C\), \[\psi_d=\mathcal N A_K+c_K,\qquad \psi_d-v=\mathcal N A_C-a_R\mathcal N g+c_C.\] The reference-mean difference \(\Delta_\Lambda j=\tau_\Lambda(j(L)-(\gamma j)I)\) annihilates constants. Scalar Gaussian integration by parts gives \(\gamma(\mathcal N A_K)=\gamma(\mathcal N A_C) =\gamma(\mathcal N g)=0\), justified by polynomial growth. We can therefore apply Lemma 24 to \(A_K\) and \(A_C\). The remaining term is evaluated directly: \(\mathcal N g=\mathsf K\), so its matrix expectation is \(-a_R\mathbb E\mathsf K(L)=-a_R\mathbf K\).

The unweighted trace of the first remainder is \(\tau\mathfrak R(\mathsf K)=\nu[e_K]\). Applying Lemma 24 and using trace cyclicity therefore gives \[\begin{aligned} \Delta_{\rm p} &=\nu[e_K]+\tau\bigl[(\mathbf S-I) (\mathbf K+T\circ\mathbf C)\bigr] -a_R\tau(T\mathbf K)+\tau_T\mathfrak R(\mathsf C)\\ &=\nu[e_K]+\tau_T\mathfrak R(\mathsf C) -\lambda\tau(\mathbf S T^2) -\tau\bigl[T^2(-a_R\mathbf C-a_R\lambda T-\lambda I)\bigr]\\ &\le\nu[e_K]+\tau_T\mathfrak R(\mathsf C) -\lambda\tau(\mathbf S T^2)-1.94\tau T^2. \end{aligned} \tag{25}\] For the second line use \(\mathbf K+T\circ\mathbf C=-\lambda T^2\) from [eq:8]. For the last line, the same normalization and \(\mathbf C\le-.341I\) give \[-\mathbf C-\lambda T =\frac{\mathbf H-\mathbf C}{2} \ge\frac{m_*+.341}{2}I, \qquad a_R\frac{m_*+.341}{2}-\lambda=1.94875>1.94.\] Pairing this operator inequality with \(T^2\ge0\) is legitimate without commutation. In particular, the term involving \(\mathbf S-I\) has been eliminated without imposing a sign on that matrix.

The weighted Dirichlet term. By linearity in the trace weight, \(J_\Sigma=J_I+J_T\). Applying Gaussian integration by parts to the term containing \(L\), and the derivative formula above to the Dirichlet term, yields \[J_I(u)=\nu\bigl[\overline{|u|^2}-|\bar u|^2\bigr].\] For the weighted part, fix a Gaussian input and express \(T\) and every \(M_\ell\) in the chosen eigenbasis of \(L\); they need not be diagonal. Expand the products in \(J_T(u)\) and average the summands with outer indices \(i,k\) interchanged. The scalar kernel multiplying \((M_\ell)_{ij}(M_\ell)_{jk}T_{ki}\) is \[\frac12\bigl(\overline{|u|^2}_{ij} +\overline{|u|^2}_{kj}\bigr) -\bar u_{ij}\cdot\bar u_{kj}.\] Set \(u_j=u(x_j)\). Replacing both copies of the profile by \(u-u_j\) leaves this kernel unchanged: its constant and linear terms cancel. The symmetrized kernel of \(\tau_T\mathfrak R(\mathsf C)\) is \[\frac12\bigl(\overline{\mathsf C}_{ij}-\mathsf C_i +\overline{\mathsf C}_{kj}-\mathsf C_k\bigr), \qquad \mathsf C_i=\mathsf C(x_i).\] Swapping \(i,k\) identifies its two contributions after summation; thus its contraction can also use the single term \(\overline{\mathsf C}_{ij}-\mathsf C_i\). Define \[b_{ij}^*=\overline{\mathsf C}_{ij}-\mathsf C_i -\overline{|u-u_j|^2}_{ij}.\] Subtracting the shifted kernel of \(J_T(u)\) now gives the kernel of \(\tau_T\mathfrak R(\mathsf C)-J_T(u)\): \[\frac12(b_{ij}^*+b_{kj}^*) +(\bar u_{ij}-u_j)\cdot(\bar u_{kj}-u_j).\] For fixed \(\ell,j\), this is multiplied by \(T_{ki}(M_\ell)_{ij}(M_\ell)_{kj}\) and summed over \(i,k\); we have used symmetry to replace \((M_\ell)_{jk}\) by \((M_\ell)_{kj}\). Swapping \(i,k\) in half of the first term yields \[\sum_{i,k}T_{ki}(M_\ell)_{ij}(M_\ell)_{kj} \frac{b_{ij}^*+b_{kj}^*}{2} =\sum_i(TM_\ell)_{ij}(M_\ell)_{ij}b_{ij}^*.\] The second term is \(\operatorname{tr}(TQ)\), where \[Q_{ik}=(M_\ell)_{ij}(M_\ell)_{kj} (\bar u_{ij}-u_j)\cdot(\bar u_{kj}-u_j).\] This is the Gram matrix of the vectors \((M_\ell)_{ij}(\bar u_{ij}-u_j)\), so \(Q\ge0\) and \(\operatorname{tr}(TQ)\le t_+\operatorname{tr}Q\). Sum over \(\ell,j\), take expectation, and divide by \(m\). The definition of \(\nu\) then gives \[\begin{aligned} \tau_T\mathfrak R(\mathsf C)-J_T(u) &\le\frac1m\mathbb E\sum_{\ell,i,j} (TM_\ell)_{ij}(M_\ell)_{ij}b_{ij}^* +t_+\nu[|\bar u_{ij}-u_j|^2]\\ &\le\lambda\tau(\mathbf S T^2) +\frac1{4\lambda}\nu[(b_{ij}^*)^2] +t_+\nu[|\bar u_{ij}-u_j|^2]. \end{aligned} \tag{26}\] The second step is scalar Young’s inequality entry by entry, followed by \(\sum_\ell\|TM_\ell\|_{\rm HS}^2 =\operatorname{tr}(\mathbf S T^2)\). The first term in this bound is exactly the one retained with a minus sign in [eq:25].

The commutator term. For any fixed \(M=M_\ell\), commute \(M\) with the quadratic matrix equation in [eq:8]. Expanding the Jordan products gives \[\mathbf H\circ[M,T]=[M,\mathbf K]+T\circ[M,\mathbf C].\] The matrix \([M,T]\) is skew-symmetric, but the lower norm estimate still holds. In an orthonormal basis diagonalizing \(\mathbf H\), the map \(B\mapsto\mathbf H\circ B\) multiplies entry \(ij\) by \((h_i+h_j)/2\ge m_*\); hence it increases the Hilbert–Schmidt norm by at least the factor \(m_*\) on all matrices. Likewise, \[\|(T-t_cI)\circ B\|_2\le t_r\|B\|_2,\] because the spectrum of \(T-t_cI\) lies in \([-t_r,t_r]\). Splitting the right side at \(t_cI\) and using \(\|A+B\|_2^2\le1.25\|A\|_2^2+5\|B\|_2^2\) gives \[m_*^2\|[M,T]\|_2^2 \le1.25\|[M,\mathbf K+t_c\mathbf C]\|_2^2 +5t_r^2\|[M,\mathbf C]\|_2^2.\] All matrices \(M_\ell\) are deterministic in this step. Thus, in fixed coordinates, \[[M,\mathbb E f(L)]=\mathbb E[M,f(L)], \qquad \|\mathbb E[M,f(L)]\|_2^2 \le\mathbb E\frac1m\|[M,f(L)]\|_{\rm HS}^2.\] Only after Jensen’s inequality do we pass to an eigenbasis of \(L\). There the commutator has entries \((M_\ell)_{ij}(f(x_j)-f(x_i))\). The resulting endpoint weights are therefore precisely those defining \(\nu\). Gaussian orthogonality also gives \(\|[L,T]\|_2^2=\sum_\ell\|[M_\ell,T]\|_2^2\). With the definition of \(\Gamma\) in [eq:4], these observations yield \[.80\|[L,T]\|_2^2 =.80\sum_\ell\|[M_\ell,T]\|_2^2\le\nu[\Gamma]. \tag{27}\]

Assembly and strictness. Combine [eq:25]–[eq:27]. The two terms involving \(\lambda\tau(\mathbf S T^2)\) cancel, giving \[\begin{aligned} &\Delta_{\rm p}-J_\Sigma(u)+.80\|[L,T]\|_2^2\\ &\quad\le\nu\biggl[e_K+\Gamma +\frac{(b_{ij}^*)^2}{4\lambda} +t_+|\bar u_{ij}-u_j|^2\\ &\hspace{20mm} -\bigl(\overline{|u|^2}_{ij}-|\bar u_{ij}|^2\bigr)\biggr] -1.94\tau T^2. \end{aligned}\] The measure \(\nu\) is symmetric. Averaging the integrand with the one obtained by swapping \(i,j\) changes its square term to \(\bigl((b_{ij}^*)^2+(b_{ji}^*)^2\bigr)/(8\lambda)\), and its mean-displacement term to \(t_+(|\bar u_{ij}-u_j|^2+|\bar u_{ij}-u_i|^2)/2\). In the first square, \(j\) is the endpoint \(\sigma\) of [eq:4] and \(i\) is the opposite endpoint. The symmetrized integrand is therefore strictly negative when \(x_i\ne x_j\), by [eq:4], and is zero when the endpoints coincide. All endpoint terms are integrable by polynomial growth, so this proves the linear-field bound in Proposition 25.

Substituting the displayed estimate into Equation [eq:23] bounds \(\mathcal E\) above by the same nonpositive \(\nu\)-integral, together with \(-.14\tau T^2-\int(W-W_0)\,d\beta\). The layer integral is finite by the projection-layer tails and the Gaussian tails of \(L\), and \(W-W_0>0\) by [eq:3] and the averaging argument preceding [eq:22]. This proves the asserted bound on \(\mathcal E\), in particular \(\mathcal E\le0\). If \(\mathcal E=0\), these three nonpositive contributions must all vanish. It follows that \(T=0\), \(\nu\) assigns zero mass to distinct endpoints, and \(\beta=0\). No uniform strict margin is required: a nonnegative measurable function that is positive on a set can have integral zero only if that set has measure zero. ◻

Proof of Theorem 1. In dimension one, write \(K=[a,b]\) with \(a<b\). For \(a<z<b\), \[|K|\,|(K-z)^\circ| =(b-a)\left(\frac1{z-a}+\frac1{b-z}\right)\ge4,\] with equality exactly at \(z=(a+b)/2\). Every one-dimensional convex body is a simplex, so this proves the theorem for \(n=1\). For the remaining argument let \(n\ge2\) and \(m=n+1\).

The Laplace bound [eq:1] follows from Equation [eq:10] and Proposition 25. Through the cone-volume identity this gives the required lower bound at every translation point \(z_0\in\operatorname{int}K\).

Suppose first that equality holds in the Laplace bound. Equation [eq:10] gives \(0\ge-\mathcal E\ge0\), so \(\mathcal E=0\). Hence \(\nu\) is supported on the diagonal and \(\beta=0\). The diagonal support has the following basis-independent consequence: \[\nu[(x-y)^2] =\frac1m\sum_\ell\mathbb E\|[L,M_\ell]\|_{\rm HS}^2 =\frac1m\sum_{\ell,k}\|[M_k,M_\ell]\|_{\rm HS}^2.\] The second identity uses \(L=\sum_kG_kM_k\) and Gaussian orthogonality. Its left side is zero, so all the deterministic real symmetric matrices \(M_\ell\) commute. Choose a single orthonormal basis that diagonalizes them simultaneously; in this basis \(L\) is diagonal for every \(G\). Repeated eigenvalues do not affect this conclusion.

In this fixed basis, the vanishing of \(\beta\) says that the nonnegative integrand \[\sum_i\bigl((P_z)_{ii}-\mathbf1_{\{x_i\le z\}}\bigr)(x_i-z)\] vanishes for almost every \((G,z)\). Away from the spectral endpoints, each diagonal entry of \(P_z\) is therefore the corresponding zero or one of \(\Theta_z(L)\). A positive semidefinite matrix with a zero diagonal entry has a zero row and column there. Apply this fact to \(P_z\) at a zero entry and to \(I-P_z\) at an entry equal to one. Since \(0\le P_z\le I\), it follows that \(P_z=\Theta_z(L)\) for almost every \((G,z)\).

For every \(G\) the spectral endpoint set contains only finitely many values of \(z\), so it has zero product measure. Fubini’s theorem now provides one fixed layer \(z\) at which \(D\Pi_C(Z+\xi_z)\) is diagonal almost surely. The covariance \(\Sigma\) is positive definite, so \(Z+\xi_z\) has an everywhere positive Gaussian density. Thus \(D\Pi_C\) is diagonal Lebesgue-almost everywhere in the same fixed basis.

The projection \(\Pi_C\) is globally Lipschitz. Its cross weak derivatives vanish, and consequently its \(i\)th component depends only on coordinate \(i\). For completeness, mollifying gives smooth maps with the same vanishing cross derivatives; each mollified component is constant along the other coordinate directions, and local uniform convergence gives the assertion for \(\Pi_C\). Write \[\Pi_C(x_1,\ldots,x_m)=(f_1(x_1),\ldots,f_m(x_m)).\] Its image is both \(C\) and the Cartesian product of the ranges of the \(f_i\). Since \(\Pi_C(0)=0\), every \(f_j(0)=0\), and evaluation at \(t e_i\) gives \(\Pi_C(t e_i)=f_i(t)e_i\in C\). Conversely, if \(t e_i\in C\), the projection fixes that point, so \(f_i(t)=t\). The \(i\)th range is therefore exactly the axis section \(\{t\in\mathbb R:t e_i\in C\}\), a closed convex cone. Solidity of \(C\) excludes the factor \(\{0\}\), and pointedness excludes the factor \(\mathbb R\). Each factor is therefore a half-line, so the normalized cone is an orthant up to signs.

Undoing the invertible transformation shows that the original cone has \(m\) linearly independent generating rays. Every nonzero point of the original cone has strictly positive height, so each ray meets the height-one hyperplane. Rescale the generators to these intersection points. In a nonnegative linear combination of the rescaled generators, height one means that the coefficients sum to one. The section is therefore their convex hull. An affine relation among the intersection points would be a linear relation among the independent generators, so these \(m=n+1\) points are affinely independent. Thus \(K\) is a simplex. If equality holds for the infimum defining \(P(K)\), apply this argument at its interior attaining point.

Conversely, the translated volume product is unchanged by applying an invertible affine map to both the body and its translation point: the linear part on the body is paired with its inverse transpose on the polar. It suffices to take \[K=\operatorname{conv}\{0,e_1,\ldots,e_n\}, \qquad z_0=\frac{\mathbf1}{m},\] where \(\mathbf1\) is the all-ones vector. The defining polar inequalities give \[(K-z_0)^\circ =\operatorname{conv}\{m\mathbf1,-m e_1,\ldots,-m e_n\}.\] Indeed, writing \(s=\sum_{i=1}^n y_i\), the polar inequalities are \(s\ge-m\) and \(y_i\le1+s/m\). Set \[\alpha_0=\frac{s+m}{m^2},\qquad \alpha_i=\alpha_0-\frac{y_i}{m}\quad(1\le i\le n).\] These coefficients sum to one and satisfy \(y=\alpha_0(m\mathbf1)+\sum_{i=1}^n\alpha_i(-m e_i)\). They are nonnegative exactly when the polar inequalities hold, proving the displayed convex-hull formula. Its volume is \[|(K-z_0)^\circ| =\frac{m^n\det(I+\mathbf1\mathbf1^{\mathsf T})}{n!} =\frac{m^m}{n!}.\] Since \(|K|=1/n!\), this gives equality in the lower bound. The scalar estimates used in the argument are established in Appendices 9 and 10. ◻

Proof of Corollary 2. Set \(f=e^{-\varphi}\), a nonzero integrable log-concave function. For \(z\in\mathbb R^n\), define \[f^z(y)=\inf_{\{x:f(x)>0\}} \frac{e^{-\langle x-z,y-z\rangle}}{f(x)}, \qquad \mathcal P(f)= \left(\int_{\mathbb R^n}f(x)\,dx\right) \inf_{z\in\mathbb R^n}\int_{\mathbb R^n}f^z(y)\,dy.\] Fradelizi and Meyer proved that the general geometric Mahler inequality in every dimension implies \(\mathcal P(f)\ge e^n\) for every log-concave function on \(\mathbb R^n\) with \(0<\int_{\mathbb R^n}f<\infty\) (Fradelizi and Meyer 2008, Proposition 2(a)). Theorem 1 supplies precisely this all-dimensional hypothesis. At \(z=0\) the defining infimum gives \(f^0=e^{-\varphi^*}\), so the product in Corollary 2 is at least \(\mathcal P(f)\). ◻

Proof. Gozlan’s equivalence between the functional inverse Santaló and entropy–transport inequalities, as stated in (Fradelizi et al. 2021, Theorem 1.3), applies to Corollary 2 with \(c=e\) and gives the constant \(-n\log(e\,e^2)=-3n\). ◻

Proof. At each node, we first enclose \(x_i=s\sinh(i/40)\), \(a_{t_i}\), \(\phi(x_i)\), and \(Y(x_i)\) by the rules below. These give \(u,p,h,j\), and hence \(d,g',g'',\mathsf K,R_v,V_2\) by the stable formulas. Next compute \(D_1,D_2,G_3,G_4\), and use \(P_F,Q_F\) to obtain the dotted derivatives and \(n_F\). The enclosure \(d_i\le.5\) at each node gives \(b-r^2-d_i\ge.0536\); only then do we evaluate \(q_i\) and its derivative formulas. This is a nodewise check, not yet a positivity assertion between nodes.

Taking interval hulls of these samples gives the value columns. Subtraction gives the first and second differences on the index ranges specified above. For the second difference at zero, use the odd reflection for dotted first derivatives and \(R_v\), and the even reflection for the other rows. Evaluating the four expressions in the second table directly at the same nodes gives their simultaneous bounds. The same arithmetic evaluates the two finite sums in the statement. Interpolation will later turn them into integral bounds; here they are only finite expressions in the enclosed node data. The primitive enclosures are as follows.

Proof. First suppose \(\Re t\ge0\), and write \(x=A+iy=s\sinh t\). Then \(A\ge0\), and the elementary bounds \(\sin(.45)<.435\), \(\tan(.45)<.5\) give \[y^2\le2.5+.25A^2.\] The Laplace integrals for \(Y\) and \(B=-Y'\) give \[|Y|\le Y(A)\le\min(1.254,1/A),\qquad |B|\le\min(1,1/A^2),\] where expressions containing \(1/A\) are omitted at \(A=0\). The corresponding tail integral for \(u=1-p\) gives \[|u|\le e^{y^2/2}u(A)\le1.75,\] since \(u(A)e^{A^2/8}\) decreases on \([0,\infty)\). Indeed \(Au(A)\le\phi(A)\) follows by integrating the Gaussian tail, and differentiating the product proves this monotonicity.

If \(A\le.9\), then \(A|y|<\pi/2\). Horizontal integration of \(\phi\) from \(iy\) to \(A+iy\), starting with \(\Re p(iy)=1/2\), gives \(\Re p\ge1/2\). If \(A\ge.9\), then \[|u|\le u(.9)e^{1.25+.125\cdot.9^2}<.80, \qquad u(.9)\le\tfrac12-\frac{.9-.9^3/6}{\sqrt{2\pi}}.\] Thus \(\Re p>0\) throughout the right-half image. The identity \[h(u)=\int_0^1\frac{v}{1-uv}\,dv\] shows analyticity and \(|h|\le2.5\): the denominator has real part at least \(1/2\) in the first region and modulus at least \(.2\) in the second.

We also need a nonvanishing bound for \(Y\). Its Laplace transform and the characteristic function \(e^{-A|s'|}\) of a Cauchy variable \(T_A\) of scale \(A\) give \[\Re Y(A+iy)=\sqrt{\pi/2}\, \mathbb E e^{-(y+T_A)^2/2}>0.\] At \(A=0\), take \(T_A=0\). If \(A\le2\), with probability at least \(1/4\) one has \(|y+T_A|\le\max(|y|,A)\le2\). For instance, when \(y\ge0\) the permitted interval contains \([-A,0]\); reflection handles the other sign. Hence \(\Re Y\ge\sqrt{\pi/2}e^{-2}/4>.02\). If \(A\ge2\), the mean of the integration variable under the Laplace weight is at most \(1/A\), because the Gaussian factor is a decreasing tilt of an exponential density. Consequently \[\left|\frac{Y(A+iy)}{Y(A)}-1\right|\le\frac{|y|}{A} \le\sqrt{7/8}<.936, \qquad Y(A)\ge\frac{e^{-1-1/(2A^2)}}{A}>\frac{.324}{A}.\] In both regions, \[|Y|\ge\frac{.02}{1+A}.\]

The principal logarithms of \(p\) and \(Y\) are therefore analytic here. The analytic logarithm of \(1-p=\phi Y\) is \(\log Y-x^2/2-\log\sqrt{2\pi}\), which agrees with its real value on the positive axis. This specifies the branch in \(j\). The preceding bounds yield \[|j|\le .625A^2+\log(1+A)+10.5,\qquad |Bj|\le12,\qquad |Y^2j|\le19.\] For example, use \(|\log Y|\le\log50+\log(1+A)+\pi/2\), \(|p|\le2.75\), and then split at \(A=1\) for the last two bounds. The stable formulas, together with \(|1-pB|\le3.75\) and \(|xY|=|1-B|\le2\), now imply \[|d|<48,\qquad |\mathsf K|<99,\qquad |xg'|<76, \qquad |\mathsf C|\le48+6.5(76+99)<1300.\] Also \[|\phi/p|\le7e^{-.375A^2},\qquad |B/Y|\le100,\qquad |B/Y^2|\le10000.\] Thus \(|R_v|\le107\). Since \(|x|\le1.25A+1.6\) and \(Ae^{-.375A^2}\le1\), \(|x\phi/p|<20\), giving \(|V_2|\le2+20+49+10000<11000\).

Reflection gives the left-half bounds. On the imaginary segment in question, both \(p\) and \(1-p\) have real part \(1/2\), and \(\Re Y>0\). The formulas are therefore regular on a neighborhood of that segment. The right-half expressions and their parity reflections agree near zero and hence by analytic continuation across the segment. This proves the asserted full-strip analyticity and bounds. ◻

Proof. Fix one of the functions in the statement and denote it by \(\Psi\). Use the 16 nodes \(i-7,\ldots,i+8\) for the cell \([t_i,t_{i+1}]\), reflecting by parity when necessary. Let \(\mathcal I_8\) be its interpolating polynomial on this stencil, of degree at most 15. We first bound \(\Psi-\mathcal I_8\), and then compare \(\mathcal I_8\) with the central chord. The first error is at most \[4\cdot10^5 \prod_{j=0}^7((j+1/2)/(40\cdot .36))^2 < .00001\] in the cell. Indeed Lemma 30 and Cauchy’s estimates from the strip of half-width \(.45\) to that of half-width \(.36\) bound, across the gap \(.09\), \(\dot F\) by \(1300/.09\), \(\ddot F\) by \(2\cdot1300/.09^2\), and \(n_F\) by their sum, which is less than \(4\cdot10^5\). Here \(|\tanh t|\le1\) on \(|\Im t|\le.36\). Apply the real interpolation remainder and Cauchy’s estimate of order 16 on the inner strip. The paired nodal factors attain their maximum at the midpoint of the cell; the displayed bound is less than \(.000007336661\).

For the comparison with the chord, normalize the cell by \(t=t_i+\delta\theta\), where \(\delta=1/40\) and \(0\le\theta\le1\), and put \(\Psi_k=\Psi(t_i+k\delta)\). For \(1\le l\le8\), let \(\mathcal I_l\) interpolate these values at \(k=1-l,\ldots,l\); thus \(\mathcal I_1\) is the central chord. For \(2\le l\le8\), let \(\mathcal J_l^-\) omit the rightmost node \(l\), and let \(\mathcal J_l^+\) omit the leftmost node \(1-l\). The interpolation identity \[\mathcal I_l(\theta) =\frac{l-\theta}{2l-1}\mathcal J_l^-(\theta) +\frac{\theta+l-1}{2l-1}\mathcal J_l^+(\theta)\] follows by checking the nodal values and the degree. Its weights are nonnegative and sum to one on the central cell.

The common nodes of \(\mathcal J_l^-\) and \(\mathcal J_l^+\) are exactly those of \(\mathcal I_{l-1}\). With \(\Delta\Psi_k=\Psi_{k+1}-\Psi_k\), Newton interpolation therefore gives \[\begin{aligned} \mathcal J_l^--\mathcal I_{l-1} &=\frac{\Delta^{2l-2}\Psi_{1-l}}{(2l-2)!} \prod_{k=2-l}^{l-1}(\theta-k),\\ \mathcal J_l^+-\mathcal I_{l-1} &=\frac{\Delta^{2l-2}\Psi_{2-l}}{(2l-2)!} \prod_{k=2-l}^{l-1}(\theta-k). \end{aligned}\] There is no mesh factor in these formulas because the nodes in the \(\theta\) coordinate have spacing one. Equivalently, the powers of \(\delta\) in the nodal product cancel those in the divided difference. Expanding \(\Delta^{2l-2}=\Delta^{2l-4}\Delta^2\) and using the second-difference bound gives \[|\Delta^{2l-2}\Psi_k|\le 2^{2l-4}M.\] Pair the nodal factors about \(1/2\). On \(0\le\theta\le1\), \[\left|\prod_{k=2-l}^{l-1}(\theta-k)\right| =\prod_{j=0}^{l-2}\big((j+\tfrac12)^2-(\theta-\tfrac12)^2\big) \le\prod_{j=0}^{l-2}(j+\tfrac12)^2.\] The convex-blend identity bounds \(|\mathcal I_l-\mathcal I_{l-1}|\) by the product of these two bounds divided by \((2l-2)!\). Summing over the seven successive pairs of added nodes yields \[|\mathcal I_8-\mathcal I_1| \le M\sum_{l=2}^8 \frac{2^{2l-4}}{(2l-2)!}\prod_{j=0}^{l-2}(j+1/2)^2 =\frac{4387}{8192}M<.536M.\] Adding the analytic remainder proves the claimed chord error. For the final cell \(i=143\), the largest stencil index is \(151\); near zero, parity supplies the negative indices. These stencils cover the entire interval \([0,3.6]\). ◻

Proof. It suffices to work on \(x\ge64\): the endpoint of the finite region is \(x_{144}>65.8>64\), so the two arguments overlap. Put \(D=x\partial_x,\ \xi=x^{-2},\ \xi_0=1/4096,\ m=xY=1-\xi n\), with \[n=\int_0^\infty y e^{-y-\xi y^2/2}\,dy,\qquad L=\log x+\tfrac12\log(2\pi)-\tfrac12,\qquad Z=\log x+\tfrac12\log(2\pi)-\log m-p+p^2h.\] Useful exact expressions are \[\begin{aligned} d&=h+\xi(\tfrac12 n^2-2p h n)+\xi^2 n^2 Z,\qquad g=\tfrac14+\tfrac14(1-m^2)-\tfrac12 \xi m^2 Z,\\ \mathsf K&=(h-\tfrac12)+\xi\big(\mathcal A+(m^2+\xi n^2)Z\big),\qquad \mathcal A=\tfrac12 n^2-2p h n-n+\tfrac12\xi n^2 . \end{aligned}\] We prove the three rows by keeping each derivative bound and its prefactor explicit. For a nonnegative numerical vector \(V=(V_j)_{j=0}^4\), write \[F\preccurlyeq aV \quad\hbox{to mean}\quad |D^jF(x)|\le a(x)V_j \quad(0\le j\le4,\ x\ge64).\] The prefactor \(a\) is a pointwise bound; the notation does not differentiate \(a\). If \(F\preccurlyeq aA\) and \(G\preccurlyeq bB\), the product rule gives \(FG\preccurlyeq ab(AB)\), where juxtaposition now denotes binomial convolution: \[(AB)_j=\sum_{i=0}^j\binom ji A_iB_{j-i}.\] Use the numerical vectors \(P_j=2^j\), \(I_j={\bf1}_{j=0}\), and \(J_j={\bf1}_{j=1}\). In particular \(\xi\preccurlyeq\xi P\), since \(D\xi=-2\xi\). Thus multiplying a function by \(\xi\) introduces both that prefactor and a convolution by \(P\). All vector inequalities below are componentwise.

Proof of Lemma 28. We apply the real bounds separately to \(t_x=\operatorname{arsinh}(x/s)\) and \(t_z=\operatorname{arsinh}(z/s)\). The following table encloses the values of the even profiles \(q,S_q,V_2,\rho\), not merely their samples. Each row specifies an interval of \(|t|\) in units of \(1/40\), and every displayed bound is multiplied by 1000.

For \(q,S_q,V_2\), take the sample minima and maxima on each closed row through index \(144\), and enlarge by the errors \(.000042,.003,.00101\), respectively. The last row also includes the tail bounds of Lemma 32. For \(\rho\), the exact formula makes an endpoint bound possible. Put \(y=\tanh^2t\) and \(v_1=(1-y)/s^2\). Differentiating gives \[\frac{d}{dy}\frac{\rho^2}{r^2w^2} =(1+v_1)^2-\frac{2(2+w^2v_1)}{s^2} -\frac{2y(1+v_1)}{s^2} \ge1-\frac6{s^2}-\frac{2(w^2+1)}{s^4}>.27.\] Here \(0\le y\le1\) and \(0\le v_1\le1/s^2\). Thus \(\rho\) increases for \(t\ge0\), giving its upper endpoint bound in each row, including the limiting bound in the last row. \[\begin{array}{c|cc cc cc c} & q_{\min}&q_{\max}& S_{q,\min}& S_{q,\max}& V_{2,\min}& V_{2,\max}& \rho_{\max}\\\hline 0{:}4&77&81&113&132&725&743&804\\ 4{:}8&80&88&125&166&740&785&814\\ 8{:}12&87&100&159&220&782&848&831\\ 12{:}16&99&116&213&288&845&920&853\\ 16{:}20&115&134&281&359&917&986&877\\ 20{:}28&133&171&352&464&983&1055&927\\ 28{:}40&170&211&457&504&1041&1059&992\\ 40{:}68&210&245&497&513&1009&1044&1072\\ 68{+}&244&256&506&514&999&1012&1104 \end{array}\] In particular, the table proves \(.113\le S_q\le.514\) and \(\rho\le1.104\). Number its rows from 1 to 9, and write \(q_i^\pm,S_i^\pm,V_i^\pm,\rho_i^+\) for their endpoints after division by 1000. Let \(i_x,i_z\) be the rows containing \(|t_x|,|t_z|\); these two row choices are independent.

We first enclose \(H\) on this pair of rows. Its scalar part \(4c+4kq+2(k-q)S\) increases in both \(q\) and \(S\), because its partial derivatives are \(4k-2S>0\) and \(2(k-q)>0\). The inequalities \(q<k\) and \(S_q<2k\) used here follow from the table. Also \(|\pi(x)|=r\), so \(|2S_\pi(z)\cdot\pi(x)|\le2r\rho_{i_z}^+\). Consequently \(H\in[H_-,H_+]\), where \[\begin{aligned} H_-&=4c+4kq_{i_x}^-+ 2(k-q_{i_x}^-)S_{i_z}^- -2r\rho_{i_z}^+,\\ H_+&=4c+4kq_{i_x}^++ 2(k-q_{i_x}^+)S_{i_z}^+ +2r\rho_{i_z}^+. \end{aligned}\] At the same time \(V_2(x)\in[V_{i_x}^-,V_{i_x}^+]\).

To verify the first layer inequality, subtract its right side from its left side and use \(J=H+V_2(x)-2\kappa\). Since \(S_q(x),S_q(z)\) are positive, replacing their product by \(S_{i_x}^+S_{i_z}^+\) gives the lower bound \[\begin{aligned} \mathscr D_{i_x,i_z}(H,V) ={}&H+t_c(H+V-2\kappa) -\left(\frac{t_r}{2h_s}+\frac1\alpha\right) (H+V-2\kappa)^2\\ &-2(b+\eta+(b+\eta-\kappa)t_-) -2b_mS_{i_z}^+S_{i_x}^+ -2e_0(2t_r)^2-\frac{t_rh_s}{2}. \end{aligned}\] This is a concave function of the two coordinates \(H,V\). Every point of \([H_-,H_+]\times[V_{i_x}^-,V_{i_x}^+]\) is a convex combination of its four vertices, so its value is at least the smallest vertex value. There are \(9\cdot9=81\) pairs of rows and four vertex evaluations for each pair. The resulting lower bounds, multiplied by 1000 and minimized over all rows of \(x\), are \[11,\ 14,\ 33,\ 64,\ 63,\ 39,\ 54,\ 45,\ 40\] in the order of the rows of \(z\). Each is positive, proving the strict inequality on every rectangle.

For the last inequality in the lemma, \(4c+2k(S_q(x)+S_q(z))-S_q(x)S_q(z)\) increases in each \(S_q\) argument, again because the other argument is less than \(2k\). The vector product is at most \(\rho(x)\rho(z)\le1.104^2\). Thus its left side is bounded below by \[4c+4k\cdot.113-.113^2-1.104^2>.25.\]

Finally the auxiliary estimates follow directly from rational arithmetic and the logarithm enclosure in Lemma 29: \[b+\eta+\kappa t_-+.125>0,\qquad \frac{1.053\kappa^2}{8(b+\eta+\kappa t_-+.125)}<.256, \qquad .256+\frac{(\log2+.25)2.26}{4}<.80.\] All rectangle margins are strictly positive. This proves the pointwise assertions for every pair of real arguments and completes the scalar estimates. ◻

Proof. The uniform trapezoidal error in the \(x\) coordinate is \(-(x_+-x_-)^2\int_0^1\theta(1-\theta)F''(x_-+\theta(x_+-x_-))\,d\theta/2\). Substitute \(x_+-x_-=2s\cosh m\sinh h\) and \(d\theta=h\cosh t\,d\beta/(2\cosh m\sinh h)\) to obtain the stated identity. For the first upper bound on \(W\), rearranging its definition reduces the claim to \((\sinh t-\sinh m\cosh h)^2\ge \cosh m(\cosh m-\cosh t)\sinh^2h\). To check this inequality, regard \(m,t\) as fixed and put \(\zeta=\cosh h-1\ge0\). The difference of its two sides is \[(\sinh t-\sinh m)^2 +2(\cosh(m-t)-1)\zeta +(\cosh t\cosh m-1)\zeta^2.\] Every coefficient is nonnegative. This proves \(W\le3S/2\).

The sharper bound near the endpoints comes from factoring the two differences of hyperbolic sines in \(\theta(1-\theta)\). With \(u_\pm=(1\pm\beta)/2\), the factorization gives \[\theta(1-\theta)\frac{\cosh m}{\cosh t} =\frac{\sinh(u_+h)\sinh(u_-h)}{\sinh^2h} \frac{\cosh(2m+\beta h)+\cosh h} {\cosh(2m+\beta h)+\cosh(\beta h)}.\] Since \(0\le u_\pm\le1\), convexity gives \(\sinh(u_\pm h)\le u_\pm\sinh h\). Thus the product on the left is bounded by \[\frac{1-\beta^2}{4}\frac{\cosh(2m+\beta h)+\cosh h}{\cosh(2m+\beta h)+\cosh(\beta h)}\] The quotient in this display is at most \((1+\cosh h)/2\): subtract one and use \(\cosh(2m+\beta h)+\cosh(\beta h)\ge2\). This proves the pointwise bound involving \(g_h\). Its integral uses \(\int_{-1}^1(1-\beta^2)\,d\beta/2=2/3\) and gives \(\int W\,d\beta/2\le S(1+(\cosh h-1)/2)\), as required. ◻

Proof. We establish the concentration, interpolation, and primitive assertions in turn.

Concentration from finite threshold sums. For the concentration assertion, compare with a threshold \(v\): the total on any given set is at most \(2Hv+I_j(v)\), where \(I_j(v)=\int_0^{3.6}(A_j-v)_+ dt\) and we use \(v\ge0\). The same threshold bound holds for a measurable density \(0\le\vartheta\le1\) on \([0,3.6]\) with \(\int_0^{3.6}\vartheta=2H\): \[\int_0^{3.6}\vartheta(t)(A_j(t))_+\,dt \le 2Hv+I_j(v),\qquad v\ge0.\] This follows by multiplying the pointwise threshold bound by \(\vartheta\).

Here is a small table for this comparison on successive intervals of \(1000H\) with endpoints \[0,75,150,200,250,300,350,400,540.\] Row entries give \(1000v\), followed below by upper bounds for \(10^5 I_j(v)\) in the same order. \[\begin{array}{c|rrrrrrrr} D&61&54&44&37&29&21&14&0\\ E&247&224&192&167&141&116&90&42\\ N&216&178&143&115&77&42&17&0\\ C&664&545&444&347&234&152&97&21\\\hline D&0&105&419&716&1124&1598&2069&3167\\ E&0&351&1326&2363&3673&5133&6885&10779\\ N&0&577&1576&2624&4378&6352&8026&9323\\ C&0&1776&4528&8066&13250&17831&21389&27670 \end{array}\] At both ends of each interval these upper bounds divided by \(10^5\) are \(\le 2H(\alpha_j-\beta_j H-v)\), which suffices by concavity. For the integral entries use exactly \[\sum_{i=0}^{144}\frac{\delta_i}{80}(A_j(t_i)+\varepsilon_j-v)_+,\qquad \varepsilon_j=.00040,\ .00120,\ .00217,\ .00741\] respectively, with \(t_i=i/40\), \(\delta_i=2\) except \(\delta_0=\delta_{144}=1\), using the values and interval rules of the one-variable estimates. Indeed by the chord errors there, each \(A_j\) has upper error from its chord at most \(\varepsilon_j\), so the formula works by convexity. The sample values are enclosed by the finite rational rules of Lemma 29; the interpolation estimates in Lemma 31 justify the passage between nodes. For example, with \(\mathcal R(M)=.536M+.00001\), the required errors are \[\begin{aligned} \mathcal R(.000673)+.021\mathcal R(.00222)&=.00039592632<.00040,\\ \mathcal R(.00222)&=.00119992<.00120,\\ \mathcal R(.00402)&=.00216472<.00217,\\ \mathcal R(.0138)&=.0074068<.00741. \end{aligned}\] Thus these sums are rigorous upper bounds for the integrals; no additional derivative evaluation between nodes is needed.

Averaging the interpolated profiles. For the remaining assertions put \[P_F(t)=\int_0^1 F(ts')\cosh(ts')\,ds'\ \Big/\int_0^1\cosh(ts')\,ds', \qquad P=P_{\mathsf K}.\] Use as approximate samples (starred) at \(t_i\), with \(F_i=F(t_i)\), \[P_{F,i}^*=F_i-40\sum_{k=1}^i\frac{\cosh t_k-\cosh t_{k-1}}{\sinh t_i}(F_k-F_{k-1}), \qquad P_{F,0}^*=F_0.\] This is an exact integration formula for the piecewise affine interpolant of \(F\) in the \(t\) coordinate, averaged against physical length. Indeed, integration by parts gives \(P_F(t)=F(t)-\int_0^t\dot F(v)\sinh v\,dv/\sinh t\); on each mesh cell the interpolant’s derivative is the constant \(40(F_k-F_{k-1})\). Errors for \(P_F\) from the chord through its starred samples are at most \[\begin{array}{c|ccc} F&\mathsf K&q&q^2\\\hline \text{error}&.000051&.000128&.000076 \end{array}\] Indeed sample errors are bounded by \(G_F/12800\) for \(G_F=\sup |\ddot F|\); also \(|\ddot P_F|\le G_F+2L_F+o_F\) where \(L_F=\sup|\dot F|,\ o_F=\operatorname{osc} F\). Differentiating the quotient, the terms besides the weighted mean of \((s')^2\ddot F\) are \(2\operatorname{Cov}(s'\dot F,s'\tanh(ts'))+\operatorname{Cov}(F,(s')^2) -2\mathbb E_*[s'\tanh(ts')]\operatorname{Cov}(F,s'\tanh(ts'))\) in its weighted expectation \(\mathbb E_*\). Bounding covariances by one quarter of the products of their ranges establishes this derivative estimate. Adding the error at the starred nodes and the chord error of the exact average gives the total bound \((2G_F+2L_F+o_F)/12800\). For \((F,G_F,L_F,o_F)=(\mathsf K,.238,.066,.0345)\) and \((q,.53,.193,.1784)\) this is below the first two stated errors. For \(q^2\) we can use \(G_F=.347,\ L_F=.099,\ o_F=.060\). Chord errors for \(\mathsf K,\mathsf D,\mathsf C,q\) are bounded by \(.000019,.000020,.000062,.000042\) using the one-variable estimates.

Here are the sample bounds for this part. In the formulas inside the table use sampled values at \(t_i\), including the starred samples for \(P_F\). Put \(q_M=.2559\), and define \[\begin{aligned} B_0(t)&=P(t)-(\mathsf K(t)+\mathsf K(0))/2+ d_*^K(\mathsf D(t)-\mathsf D(0))^2+d_*^C(\mathsf C(t)-\mathsf C(0))^2,\\ B_1(t)&=-\tfrac12(\mathsf K(t)-\mathsf K(0)) +d_*^K(\mathsf D(t)-\mathsf D(0))_+(\mathsf D(t)+\mathsf D(0)+.0214)_+,\\ Z_1(t)&=\tanh t\,(.0178-P(t)). \end{aligned}\] \[\begin{array}{c|l|r} i&\text{expression}&\text{bound}\\\hline 0{:}144&\mathsf D-\mathsf D(0)&[-10^{-8},.0314]\\ &P&\le .01767\\ &B_0&\le .01773\\ &B_0-.0146 t_i^2&\le .00001\\ &\dot{\mathsf D}&\ge -.0213\\\hline 0{:}56&B_1+\tfrac12(P-.0049)_+&\le .000001\\ &Z_1 &[0,.00742]\\ &B_1+78Z_1^2 &\le .000001\\ &P_{q^2}-(q+q_M)P_q+(q^2+q_M^2)/2&\le .01587\\\hline 0{:}144& -\mathsf C,\quad q,\quad \min(\mathsf D,-.01065),\quad \min(P,.0049) &\text{nondecreasing (samples)} \end{array}\] The arithmetic for these tables proceeds in the following order. At each node \(t_i=i/40\), first use the one-variable sample rules to enclose \(\mathsf K_i\), \(q_i\) and \(\mathsf C_i=-d_i+6.5g''(s\sinh t_i)\), together with their indicated derivatives. Form \(\mathsf D_i=\mathsf K_i+.021\mathsf C_i\). For the threshold table the four columns are \(\dot{\mathsf K}_i+.021\dot{\mathsf C}_i\), \(-\dot{\mathsf C}_i\), \(-n_{\mathsf K,i}\) and \(n_{\mathsf C,i}\).

The starred primitive samples are not point evaluations. For each of them, start the numerator sum in its defining formula at zero and add the profile differences times the corresponding differences of \(\cosh t_i\). Equivalently, use \(a_{t_i}=s\cosh t_i\) and \(x_i=s\sinh t_i\); the common factor \(s\) cancels in the quotient. With these starred samples and the point-value columns in hand, substitute into \(B_0,B_1,Z_1\) and the endpoint-distance expression. Squares are the full ranges of squares of the interval arguments. At \(t=0\), take \(\tanh t=0\) exactly.

For the last table row, compare successive nodes \(i,i+1\) for \(0\le i<144\). The differences for \(-\mathsf C\) and \(q\) are bounded below by \(.000005\) and \(.000012\), respectively. For the two clipped columns the lower bound is zero; clipping is applied to the interval endpoints before taking these differences. These operations, with the finite enclosure rules from the one-variable table, specify both tables here.

The monotonicity line and the chord errors prove the two claimed almost-monotonicity bounds, and for \(0\le t\le y\le3.6\), \[\begin{aligned} P(y)&\ge\min(P(t),.0049)-2\cdot.000051,\qquad P\le.0178,\\ \mathsf D(y)&\ge\min(\mathsf D(t),-.01065)-.000040 . \end{aligned}\] To justify this implication, let \(J_F\) be the piecewise affine interpolant of the relevant samples. If \(\min(F_i,c)\) is nondecreasing, then \(\min(J_F,c)\) is nondecreasing as well: each cell connects its two clipped endpoint values monotonically, and a cell with both endpoints above \(c\) remains above \(c\). Since clipping is 1-Lipschitz, \(|F-J_F|\le\epsilon\) implies, for \(t\le y\), \[F(y)\ge\min(F(y),c) \ge\min(J_F(y),c)-\epsilon \ge\min(F(t),c)-2\epsilon.\] For starred P samples the same argument uses their total chord error. Saturated interval values in the finite calculation are the exact constant \(c\); uncertain overlapping intervals are not used to infer monotonicity. Also \(B_0(t)\le\min(.018,.0146t^2+.00010)\) by convexity in its profile arguments. Here the added errors are \(<.000075\) (use the sample ranges for \(\mathsf D-\mathsf D(0),\mathsf C-\mathsf C(0)\)), and \(.0146t^2\) has chord error less than .000003.

The primitive budget. Take \(t=|l|\). Since \(\mathsf K\) is even, its primitive in physical length is represented by \(s\sinh(t)P(t)\) on the positive half-axis. Subtracting the primitives at the two endpoints gives \(\overline{\mathsf K}=P(y)+\theta(P(y)-P(t))\), where \(\theta=\sinh l/(\sinh y-\sinh l)\). After subtracting \(B_0(y)\) from \(e_K+\Gamma\), the remaining terms are \(\theta(P(y)-P(t))\) and \[-\tfrac12(\mathsf K(t)-\mathsf K(0))+ d_*^K (\mathsf D(t)-\mathsf D(0))(\mathsf D(t)+\mathsf D(0)-2\mathsf D(y)) +d_*^C (\mathsf C(t)-\mathsf C(0))(\mathsf C(t)+\mathsf C(0)-2\mathsf C(y)).\] We first bound this common correction, before distinguishing the sign of \(l\). Put \(X=\mathsf D(t)-\mathsf D(0)\) and \(\varepsilon=.000040\). The product with coefficient \(d_*^K\) is \[X\bigl(\mathsf D(t)+\mathsf D(0)-2\mathsf D(y)\bigr) =X\bigl(2[\mathsf D(t)-\mathsf D(y)]-X\bigr).\] The node and chord bounds give \(-.000021\le X\le.032\). If \(X<0\), use \(\mathsf D(y)-\mathsf D(0)\le.032\) to bound this product above by \(.000021(2\cdot.032+.000021)\). If \(X\ge0\) and \(\mathsf D(t)\le-.01065\), clipped monotonicity gives \(\mathsf D(y)\ge\mathsf D(t)-\varepsilon\). Hence \[X\bigl(2[\mathsf D(t)-\mathsf D(y)]-X\bigr) \le -X^2+2\varepsilon X\le\varepsilon^2.\] In the other case, \(\mathsf D(t)>-.01065\), the same clipped bound gives \(\mathsf D(y)\ge-.01065-.000040>-.0107\). The product is then bounded by \((\mathsf D(t)-\mathsf D(0))_+ (\mathsf D(t)+\mathsf D(0)+.0214)_+\), the product already present in \(B_1(t)\).

For the \(\mathsf C\) product, suppose first \(\mathsf C(t)>\mathsf C(0)\). Almost-monotonicity bounds the first factor by \(.000124\), while the global range bounds the second factor above by \(.320\). If \(\mathsf C(t)\le\mathsf C(0)\), set \(a=\mathsf C(0)-\mathsf C(t)\ge0\). The bound \(\mathsf C(y)\le\mathsf C(t)+.000124\) makes the product at most \(-a^2+2(.000124)a\le.000124^2\). After multiplying these errors by \(d_*^K\) and \(d_*^C\), their sum is less than \(.000014\). Thus the common correction is at most \(B_1(t)+.000014\).

We next pass from the samples of \(B_1\) to its true values. The product \((x-b)_+(x-b')_+\) is zero up to the larger threshold and is a nondecreasing convex quadratic thereafter. Thus the product in \(B_1\) is convex in \(\mathsf D(t)\), with \(\mathsf D(0)\) fixed. Its two clipped factors sum to at most \(.058\), both for the chords and for the true arguments. Here we use \(\mathsf D(0)<-.0140\); the exact formula is \((1-6.5t_c)((2+\pi)c_*-1)-2t_cc_*\), with \(c_*=\log2-1/2\). Consequently the upper error of \(B_1\) from the chord through its sample values is at most \(\tfrac12\cdot.000019+d_*^K\cdot.058\cdot.000020<.000020\).

If \(l<0\), then \(\theta\in[-1/2,0]\). The lower bound \(P(y)\ge\min(P(t),.0049)-2\cdot.000051\) therefore gives \(\theta(P(y)-P(t))\le.5(P(t)-.0049)_++ .000051\). Use the table row for \(B_1+\tfrac12(P-.0049)_+\). Convexity of the positive part and the chord error of \(P\) add at most \(.000051/2\) to the \(B_1\) chord error. The total correction is bounded by \[.000014+.000020+.000001+.000051/2+.000051<.00015.\] Combining this with \(B_0(y)\le\min(.018,.0146y^2+.00010)\) and \(y\le2h\) proves the \(l<0\) line of Equation [eq:s1].

For \(l\ge0\), \(t=l\) and \(0\le\theta\le\tanh l/\sinh(2h)\). Since \(P(y)\le.0178\) and \(P(t)\le.0178\), the nonnegative function \(Z_1(t)=\tanh t(.0178-P(t))\) satisfies \(\theta(P(y)-P(t))\le Z_1(t)/\sinh(2h)\). This time we use the table row for \(B_1+78Z_1^2\).

The error of \(Z_1\) from its starred chord is at most \(.000066\). To see this, combine the separate errors in the two factors with \(\tfrac14|\Delta P_i^*||\Delta\tanh t_i|\), the difference between the product of their chords and the chord of their products. The derivative estimate \(|\dot P|\le L_{\mathsf K}+o_{\mathsf K}/4\) follows by the same weighted differentiation used above. Explicitly, \[.000051+\frac{.0178+.007+.238/12800}{12800} +\frac1{4\cdot40}\left((.066+.0345/4)/40+2\cdot.238/12800\right)<.000066 .\] The starred samples lie in \([0,.00742]\). Convexity of the square therefore bounds the error in \(78Z_1^2\) by \(78\cdot.000066(2\cdot.00742+.000066)\). Including the common correction, the \(B_1\) chord error and the table margin gives \(.000014+.000020+.000001+ 78\cdot.000066(2\cdot.00742+.000066)<.00015\). It follows that \[e_K+\Gamma-B_0(y)\le .00015+Z_1(t)/\sinh(2h)-78 Z_1(t)^2 ,\] Maximizing \(Z/\sinh(2h)-78Z^2\) over real \(Z\) gives \(1/(312\sinh^2(2h))\). Together with the bound on \(B_0\), this proves [eq:s1] when \(l\ge0\), completing the primitive estimates.

The q endpoint-distance bound. Finally, on the part \(|l|\le t'\le y\), \(q(t')\) lies between the two endpoint values up to .000084, so the half sum of squared differences pointwise is at most \(.1784^2/2+.000084( .1784+.000084)<.0163\). If \(l<0\), reflect the extra part \([l,|l|]\) to \([0,t]\), where \(t=|l|\). To bound its mean endpoint-distance, maximize the convex quadratic in the other endpoint value \(q(y)\in[q(t)-.000084,q_M]\). The upper endpoint \(q_M\) suffices: throughout \(0\le t\le1.4\), the \(q_{\max}\) column in the proof of Lemma 28 (rows through \(40{:}68\)) gives \(q(t)\le.245\), whereas \(q(v)\le q(t)+.000084\) on \([0,t]\). Hence \(q_M-q(t)>3(.000084)\), so the squared distance to \(q_M\) is at least the squared distance to \(q(t)-.000084\) at every such \(q(v)\). After averaging, we obtain the following conservative bound: \[P_{q^2}(t)-(q(t)+q_M)P_q(t)+(q(t)^2+q_M^2)/2+.000084^2/2 \qquad (t=|l|).\] The table applies with upper error at most .00018 on its profile expression, proving \(b_a\le.0163\). Indeed for the product use the given chord errors plus \(\tfrac14|\Delta P_{q,i}^*||\Delta q_i|\), with \(|\dot P_q|\le L_q+o_q/4\); altogether the upper error is at most \[\begin{aligned} &.000076+2q_M\cdot.000128+(q_M+.53/12800)\cdot.000042\\ &\quad+\frac{.193}{4\cdot40} \left((.193+.1784/4)/40+2\cdot.53/12800\right)\\ &\quad+\tfrac12\cdot.000042(2q_M+.000042)<.00018, \end{aligned}\] using also convexity for the \(q(t)^2/2\) term. This completes the proof. ◻