The Mahler Conjecture for General Convex Bodies

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

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

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 [20, 27]. 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

Rademacher’s theorem gives its a.e. differentiability [22, 25]. 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

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

Subtracting |\mathbb EX_z|^2 proves the variance expression in Equation (6). Moreover, \nabla|\Pi_C|^2=2\Pi_C, so

Finally,

No commutation of \Sigma with \mathbb EP_z is used.□

Proof. Interiority and Equation (5) give

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,

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

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 (6) gives aB\le h\le s_0, and thus

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

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 2.3. For a symmetric matrix Q, transform the original data by

Lemma 2.1 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.

Continuity. Under convergent invertible transformations the cone projections converge uniformly on compact sets. To see this, projected points are bounded by the norm of the input; subsequential limits belong to the limiting cone and minimize the limiting distance, so uniqueness identifies the limit. Equicontinuity upgrades the conclusion to compact-uniform convergence. The coercive lower bound in Lemma 3.1 is locally uniform, so the biases also converge at every fixed layer.

There is no need to assert pointwise convergence of the derivatives of these projections. Gaussian integration by parts instead gives

where e_i is the i-th coordinate vector. The pointwise Lipschitz bounds dominate these expectations locally in the parameters. Lemma 3.2, also with a Gaussian factor G_i by Hölder’s inequality, permits integration over layers. Hence \mathbb EA and all M_i are continuous. Their local boundedness and the polynomial growth of the profiles then give continuity of \mathbf C,\mathbf K.

The covariance update. Put \Delta=\mathbf C^2-4\lambda\mathbf K, and set

The scalar inequalities, spectral calculus, and expectation imply

For the first inequality use (\mathbf C+.37I)^2\ge0. For the other two, expand the squares and use the corresponding endpoint inequalities in Proposition 2.3. Moreover \mathbf C+2\lambda t_\pm I<0, since \mathbf C\le-.341I and 2\lambda t_+=.0832.

The Löwner–Heinz inequality states that positive square root is order preserving, including for noncommuting matrices [5, Theorem 2.3]. Indeed its integral representation uses the order-preserving matrices B(B+tI)^{-1}. Taking square roots in the preceding sandwich therefore gives

so t_-I\le T_{\rm new}\le t_+I. The positive discriminant gap also gives continuity of this update.

A uniform inward estimate. We claim that \operatorname{tr}(Q\mathbb EA)<0 whenever \sigma=\|Q\|_{\rm Frob} is sufficiently large, uniformly over its eigenbasis and over T in the box. All constants in this paragraph depend only on the original cone data and the fixed dimension, not on Q,T.

Diagonalize Q, write its eigenvalues as d_j, and put h_j(z)=\mathbb E(P_z)_{jj}. Applying the original interior-vector bounds to e^{-Q}X_z and e^QY_z gives

Gaussian integration by parts in this basis gives h_j=\mathbb E[(X_z)_j(\Sigma^{-1}Z)_j], and the analogous identity for 1-h_j uses -Y_z. The covariance bounds and the one-Lipschitz coordinate bounds imply, for every cutoff J\ge1,

For example, bound |(X_z)_j| by its absolute mean plus \mathbb E|Z|+|Z|, and split the Gaussian integral at |Z|=J. A tail term containing the absolute mean is absorbed by the first term. These bounds require no commutation of Q and \Sigma.

Set A_j=\int_{\mathbb R}(p-h_j)\,dz and \varepsilon=1/\sqrt m. The quantity to be made negative is \operatorname{tr}(Q\mathbb EA)=\sum_jd_jA_j. Thus a negative eigenvalue requires a lower bound on A_j, whereas a positive eigenvalue requires an upper bound. We obtain a uniform bound for each sign and then improve it when |d_j|\ge\varepsilon\sigma.

If d_j\le0, use J=1+|z| on the negative half-line in the estimate for h_j. The bound \mathbb E|(X_z)_j|\le Ca(z)e^{d_j}\le Ca(z) makes its integral bounded independently of Q,T. On the positive half-line, p-h_j\ge p-1, whose integral is finite. Together these give A_j\ge-C.

If in addition d_j\le-\varepsilon\sigma, the same estimate with a sufficiently large fixed cutoff gives h_j\le1/4 on [0,e^{\varepsilon\sigma/2}] for large \sigma: here a(z)e^{d_j}\le C(1+e^{\varepsilon\sigma/2})e^{-\varepsilon\sigma} tends to zero. On this interval p-h_j\ge1/4. The complementary integrals retain their uniform lower bound, so

For positive eigenvalues we first note the uniform integral bound

On [0,e^{2\sigma}], use J=C_1\sqrt\sigma, with C_1 large enough that the integrated Gaussian error is negligible. This bounds the integral of 1-h_j there by C\sqrt\sigma e^{-d_j}+o(1).

The remaining tail must be controlled despite the changing cone. The transformed and scaled image of a fixed interior ball about the original U contains a ball about aU_Q of radius at least cae^{-\sigma}. On the other hand,

For z\ge e^{2\sigma}, this error is negligible relative to the radius. Thus \xi_z=aU_Q-\mathbb EY_z has interior distance at least cze^{-\sigma}, and

Combining these bounds with \int_{-\infty}^0p<\infty, for every d_j\ge0 we have

If d_j\ge\varepsilon\sigma, choose \rho>0 with \rho^2/2<\varepsilon. The lower bound on a gives

For every real z, the bound aB\le h\le s_0 gives B(z)\le s_0/a(z). A large fixed cutoff in the estimate for 1-h_j consequently gives h_j\ge3/4 throughout this interval. Its contribution to A_j is at most -\rho\sqrt\sigma/4, since p\le1/2 there. On the rest of the negative half-line use p-h_j\le p; its integral is bounded by \int_{-\infty}^0p. On the positive half-line use p-h_j\le1-h_j and the preceding bound C\sqrt\sigma e^{-d_j}+o(1), which is bounded when d_j\ge\varepsilon\sigma. Hence A_j\le-c\sqrt\sigma+C.

At least one eigenvalue has |d_j|\ge\varepsilon\sigma. A positive such eigenvalue contributes at most -c\sigma^{3/2}+O(\sigma) to \sum_jd_jA_j; a negative such eigenvalue contributes a still larger negative amount in magnitude. Every remaining negative eigenvalue contributes at most C|d_j|, and every remaining positive eigenvalue contributes at most Cd_j+C\sqrt\sigma d_je^{-d_j}=O(\sigma). There are only m terms. Hence \operatorname{tr}(Q\mathbb EA)=\sum_jd_jA_j<0 for all sufficiently large \sigma, as claimed.

The fixed point. Take a closed Frobenius ball of such a radius and the closed spectral box for T. On their product update T as above and update Q by nearest-point projection of Q+\mathbb EA back to the ball. This is a continuous self-map of a finite-dimensional compact convex set. Brouwer’s fixed point theorem [4] applies. A fixed point on the Q-sphere would require \mathbb EA to be a nonnegative multiple of Q, contradicting the inward estimate. At an interior fixed point, \mathbb EA=0.

Finally write S=\sqrt\Delta. At the fixed point T=(-\mathbf C-S)/(2\lambda), and direct expansion gives

The mixed products cancel without any commutation assumption. Also \mathbf H=S\ge m_*I, proving Equation (8).□

Proof. We first establish integrability and an exact formula for the contribution of \mathcal J_X. Set

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,

where

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

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

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

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

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

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

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

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

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

Its existence follows from the monotonicity of 1/W_X, with the indicated initial atom at -z. Integration by parts gives

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

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

using j_X(z)-j_X(-z)=v'(z). This proves Equation (9).□

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

Their Jacobians are J_{X,j}\zeta\Sigma^{1/2} and -J_{Y,j}\zeta\Sigma^{1/2}, where

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

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 [3] then gives

and similarly for Y. Lemma 4.1 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 (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

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 4.3. Integration by parts gives

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',

For any smooth polynomial-growth test w_0, we have

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

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 3.2.

Apply Equation (12) to w_0=g. Equation (2) gives

Substituting 2zd-d'=v'+2g' in the definition of \delta[d] now proves the desired scalar identity. Finally,

Inserting these identities in the entropy bound yields exactly Equation (11).□

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, [1, Sections 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

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

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

and \mathcal W_j. By Lemma 3.3, they belong to Gaussian W^{1,2}, with Jacobians H_f\Sigma^{1/2} and H_j\Sigma^{1/2}. Applying Lemma 5.1 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

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

Since \mu has mass s_0 and cumulative function h, this equals

The centered-test identity

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

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 (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

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

Split the kernel variation into its two ordered half-planes. Symmetry and the preceding primitive formulas give

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

The product rule for \mathcal N gives

Equation (12) cancels the terms involving b_1b_2. The remaining expression is \delta[b_1'b_2'] by (10), proving (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

Multiply by f'(x)f'(z), integrate over x<z, and include the opposite order. The first two terms give the left side of (15) and \operatorname{Nt}_\Lambda(f,f). The terms on the right give 2e_\Lambda^{B_*}(f,f), by the spectral identity

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:

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 (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 (14) gives

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 (13) and replace \mu[\iota f_l] using the last line of Equation (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

Equate these two evaluations. Equation (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 (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

Adding J_\Sigma(l) cancels exactly the same weighted Dirichlet term. By (13), the remaining expression is

Insert (14) for \mathcal Q[f,f] and (16) for the pairing. The reference terms cancel because

The \Delta_\mu terms have the primitive

whose derivative is zero. Besides \delta[l^2]+\operatorname{Ct}(f,f)+2e_\Sigma(f,b_1), what remains is \Delta_\Sigma applied to

Its derivative is

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 (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

The contact integral being subtracted is

It is nonnegative: \operatorname{ct}\geq0, and its scalar factor is the second positive expression in (3). Discarding this integral preserves the upper bound.

Finally, subtract \tau_TH_v using the first line of (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 (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

because \mathcal K_G L=L/2. Since F=c+2kq, it follows that

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

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

Put \mathfrak b=b+\eta and

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

The scalar matrix inequality

holds for 0\le\rho\le1/3. Indeed, the first diagonal entry of the difference is at least

and its determinant is

The constant degree has zero R-coefficient and contributes a nonpositive term. Summing the even degrees therefore gives at most

where we used \Sigma\le(1+t_+)I. Together with the odd estimate, this proves Equation (19).

Finally, Equation (15) yields

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,

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

□

Proof. Equation (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 6.2. 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

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

The two exterior intervals contribute

For completeness, use the scalar substitution

The last integral becomes

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

Lemma 6.2 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 (20).□

Proof. Work in a spectral basis of L, and use a_i,\epsilon_i,E_z^L,S_z as in Lemma 6.2. Write g_i=g_z(x_i). The kernel multiplying T_{ji}(E_z^L)_{ij} in \mathcal L_T[H_0] is

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

Indeed, writing D=T-t_cI, we have \|D\|\le t_r, and for every vector v_0,

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

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

Lemma 6.2 gives

We will prove the scalar estimate

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

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

Set h_z(t)=H_0(z,z+\epsilon t), for 0\le t\le a. Then

Furthermore,

Cauchy–Schwarz, with multiplicity 1 on (b,a] and multiplicity 2 on [0,b], consequently gives

In either exterior interval, write a=\ell(1+t), b=\ell t. Integrating the last estimate over both exterior intervals gives at most

Use the substitution u_0=(\sqrt{1+t}-\sqrt t)^2 from the proof of Lemma 6.3. The integral becomes

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

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

Completion of the square in \sqrt{\mathcal V} proves Equation (22).□

Proof. For a smooth scalar function h, its matrix derivative at a real symmetric matrix satisfies

This is the Daleckiĭ–Kreĭn divided-difference derivative formula; see [23, 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,

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

Adding and subtracting \mathbf S\circ\mathbb E f(L) proves (24).□

Proof. The scalar functional. Since d is even, \psi_d is even, and the differential identities (2) give

For any smooth F_*, one has (\mathcal N F_*)''=(\mathcal N+2)F_*''. Define the even second primitives

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,

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 7.1 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 7.1 and using trace cyclicity therefore gives

For the second line use \mathbf K+T\circ\mathbf C=-\lambda T^2 from (8). For the last line, the same normalization and \mathbf C\le-.341I give

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

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

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

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

Subtracting the shifted kernel of J_T(u) now gives the kernel of \tau_T\mathfrak R(\mathsf C)-J_T(u):

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

The second term is \operatorname{tr}(TQ), where

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

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 (25).

The commutator term. For any fixed M=M_\ell, commute M with the quadratic matrix equation in (8). Expanding the Jordan products gives

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,

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

All matrices M_\ell are deterministic in this step. Thus, in fixed coordinates,

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 (4), these observations yield

Assembly and strictness. Combine (25)–(27). The two terms involving \lambda\tau(\mathbf S T^2) cancel, giving

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 (4) and i is the opposite endpoint. The symmetrized integrand is therefore strictly negative when x_i\ne x_j, by (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 7.2.

Substituting the displayed estimate into Equation (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 (3) and the averaging argument preceding (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 1.1. In dimension one, write K=[a,b] with a<b. For a<z<b,

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 (1) follows from Equation (11) and Proposition 7.2. 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 (11) 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:

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

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 ith 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

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 ith 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

where \mathbf1 is the all-ones vector. The defining polar inequalities give

Indeed, writing s=\sum_{i=1}^n y_i, the polar inequalities are s\ge-m and y_i\le1+s/m. Set

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

Since |K|=1/n!, this gives equality in the lower bound. The scalar estimates used in the argument are established in Appendices A and B.□

Proof of Corollary 1.2. Set f=e^{-\varphi}, a nonzero integrable log-concave function. For z\in\R^n, define

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 \R^n with 0<\int_{\R^n}f<\infty [9, Proposition 2(a)]. Theorem 1.1 supplies precisely this all-dimensional hypothesis. At z=0 the defining infimum gives f^0=e^{-\varphi^*}, so the product in Corollary 1.2 is at least \mathcal P(f).□

Proof. Gozlan’s equivalence between the functional inverse Santaló and entropy–transport inequalities, as stated in [7, Theorem 1.3], applies to Corollary 1.2 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.

Node coordinates and Gaussian density. Take E=\big(\sum_{m=0}^{26}(i/160)^m/m!+[-10^{-27},10^{-27}]\big)^4. This encloses e^{i/40}, so s(E-1/E)/2 and s(E+1/E)/2 enclose x_i and a_{t_i}, respectively; intersect the former with [0,90]. For i\le73, enclose \phi by the same exponential sum with argument -x_i^2/128 and the same error, raised to the 64th power and divided by \sqrt{2\pi}. For i\ge74, use [0,10^{-26}]. Indeed x_{73}<11<x_{74}, and e^{11^2/2}>10^{26}, for instance from e^{1/2}>1.645. Every Taylor sum just used has a true argument of absolute value at most 1. Thus e^1/27!<3/27!<10^{-27} bounds its error before taking powers.

Gaussian tails. For i\le30, take y_m=x_i(-x_i^2/2)^m/m! and L=\sum_{m=0}^{39}y_m/(2m+1)+[0,y_{40}/81], then use Y=(1/2-L/\sqrt{2\pi})/\phi. This integrates the Taylor bounds for a negative exponential. The terms can be formed recursively as y_{m+1}=-y_m x_i^2/(2m+2).

For i\ge31, division by the small Gaussian density is avoided. Start with T_{34}=[0,34/x_i], put T_m=m/(x_i+T_{m+1}) for m=33,\ldots,1, and use Y=1/(x_i+T_1). To justify this enclosure, put I_m(x)=\int_0^\infty y^m e^{-xy-y^2/2}\,dy. Integration by parts gives mI_{m-1}=xI_m+I_{m+1}, so T_m=I_m/I_{m-1}=m/(x+T_{m+1}) and 0<T_{34}<34/x. Also xI_0+I_1=1, giving the formula for Y=I_0.

Logarithmic profiles and square roots. The tail enclosure gives u=(Y\phi)\cap[0,1/2] and p=1-u. For h, use \sum_{m=0}^{44}u^m/(m+2)+[0,2u^{45}/47]; the last interval bounds the remaining positive series because u\le1/2. To evaluate the logarithms in j, scale each argument by a power of 2 into [1,2.01]. For a scaled argument z, including z=2, take 2\sum_{m=0}^{24}y^{2m+1}/(2m+1), where y=(z-1)/(z+1), with absolute error 2|y|^{51}/(51(1-y^2)). Correct for the scaling by the corresponding multiple of \log2. Use \log\sqrt{2\pi}=\frac12\log(2\pi) and 3.14159265358979323846264338\le\pi\le3.14159265358979323846264339. These endpoints follow, for example, from the arctangent series at 1/5,1/239 in \pi=16\arctan(1/5)-4\arctan(1/239). For the square roots in the profiles, consecutive multiples of 10^{-26} enclosing each endpoint suffice.

Every addition, multiplication, reciprocal, and power is evaluated by the full range of that operation on its interval arguments; endpoints are rounded outwards. In particular the square of an interval crossing zero has lower endpoint zero. The intersections and square-root rules above preserve inclusion. They therefore give rational certificates, without an assumption about the accuracy of numerical special functions. For example, the two displayed finite integral sums lie respectively in [.1250504493,.1250504494] and [.3344848649,.3344848650], and the four simultaneous bounds have strict slack before rounding to the displayed table.

The supplementary programs scalar_intervals.py and scalar_algebra.py implement these rules with integer endpoints on a 10^{-70} grid and the specified 10^{-26} square-root grid. They also record each resulting interval. The formulas and remainder bounds here specify the certificates independently of those programs. A program-to-claim index and reproduction instructions for both appendices are supplied in verification/README.md.□

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

The Laplace integrals for Y and B=-Y' give

where expressions containing 1/A are omitted at A=0. The corresponding tail integral for u=1-p gives

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

Thus \Re p>0 throughout the right-half image. The identity

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

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

In both regions,

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

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

Also

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

in the cell. Indeed Lemma A.4 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

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

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

Pair the nodal factors about 1/2. On 0\le\theta\le1,

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

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

Useful exact expressions are

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

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:

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.

The bounds for n and m. Differentiating the integral for n, with D=-2\xi\partial_\xi, gives |n^{(i)}_\xi|\le(2i+1)!/2^i and |1-n|\le3\xi. For 1\le j\le4, expand (\xi\partial_\xi)^j. After division by 2^j\xi, each resulting bound is at most

Consequently n-1\preccurlyeq3.1\xi P. Set N=I+3.1\xi_0P, so n\preccurlyeq N. The identity m-1=-\xi n then gives

With M=I+1.02\xi_0P, we also have m\preccurlyeq M.

The logarithmic term. The series for -\log m is a series in 1-m. For a power of order k, convolution satisfies (P^k)_j=k^jP_j. Including the factor 1/k from the logarithm and using j\le4, its derivative bounds are therefore controlled by

The vector on the right is at most 1.04\xi P, using \sum_{k\ge1}k^3z^{k-1}=(1+4z+z^2)/(1-z)^4; the series and its derivatives converge absolutely in this tail.

The remaining corrections in Z are Gaussian. The product rule applied to u=\phi m/x gives |D^ju|\le10^{-18}\xi^2 for j\le4, using x^{20}e^{-x^2/2}<10^{-200}. Substitution into the series for h gives

Since Z-L=-\log m-p+p^2h+\tfrac12, these bounds imply Z-L\preccurlyeq1.05\xi P. We also need a bound for Z itself, whose logarithmic growth prevents a uniform absolute bound on the tail. Put H_0=4.59I+J+1.05\xi_0P. Since DL=1, D^jL=0 for j\ge2, and \xi L decreases on x\ge64, the precise bounds represented by these vectors are

Thus Z\preccurlyeq(\xi_0/\xi)H_0. The last prefactor will cancel one power of \xi when we bound the exact formulas above.

The algebraic corrections. The two corrections needed for \mathsf K are \mathcal A+1.5 and m^2+\xi n^2-1. To bound the first, use the identity

For the second, write m^2-1=(m-1)(m+1). The product rule then bounds their derivative vectors, after division by \xi, by the first and second lines below, respectively:

The comparisons on the right hold through order 4, as follows by expanding N,M and using (P^k)_j=k^jP_j. The bound 10\xi P also applies to \tfrac12n^2-2phn+\tfrac12. Its expansion replaces -2(n-1) by -(n-1) and omits \tfrac12\xi n^2, so the same nonnegative bound suffices.

The remainder in \mathsf K. Subtract the main term from its exact expression:

The first two terms inside brackets have combined bound 11.05\xi P; the last has bound 3.2\xi_0PH_0. Multiplication by the outside factor \xi, followed by \xi\le\xi_0, therefore bounds the derivative vector of this remainder by \xi times

Its first three components are at most (.0064,.027,.11), as in the first table row.

The remainder in d. Isolate -\xi/2 in the formula for d-\tfrac12:

The four terms give d-\tfrac12\preccurlyeq\xi E_d, where

The last term uses \xi^2\preccurlyeq\xi^2P^2 and Z\preccurlyeq(\xi_0/\xi)H_0. The first three components of E_d are at most (.51,1.02,2.1), proving the second row.

From g to the remainder in \mathsf C. Using m=1-\xi n, rewrite

The product bounds give the absolute estimate g-\tfrac14\preccurlyeq\xi_0E_g, with the constant prefactor \xi_0, where

This estimate is not in units of the varying \xi. Now g''=\xi(D^2-D)g, so two more D-derivatives and one factor \xi are required. Explicitly, for 0\le j\le2,

Since \mathsf C+\tfrac12=-(d-\tfrac12)+6.5g'', its derivative bounds in units of \xi, through order 2, are

Their components are at most (.55,1.16,2.61), proving the third row. This completes the derivative table.

Conversion to the real derivative bounds. The D-bounds give the derivatives used in Lemma A.1 through

In particular, (D^2-D)(\xi(L-1.5))=\xi(6(L-1.5)-5). The first row of the table therefore gives

because L\ge L(64)>4.5. Hence n_{\mathsf K}>0. For upper bounds, use L(64)<4.59 and the fact that \xi(L-1.5) and \xi(2L-4) decrease on this tail. The table and the conversion formulas give

These imply the global and restricted derivative bounds for \mathsf K,\mathsf C on the tail.

The q and layer-profile ranges. The second row of the table first gives

This tail bound on the radicand is independent of the estimates for q, and permits division by \ell in its derivative formulas. Substitution of the value and derivative bounds for d gives

Thus q\in[.255,.2556]. The identity S_q=2q+\tanh(t)\dot q-a_t^{-2}n_q then yields S_q\in[.509,.512]. The bounds for \dot q,n_q also give the required bound for \ddot q=n_q+\tanh(t)\dot q.

For the two derivatives of v, the integral for n gives 1-3\xi\le n\le1, while m=1-\xi n gives 1-\xi\le m\le1. Consequently

The Gaussian term in V_2 is smaller than .0001. The same Gaussian estimate and m\ge1-\xi_0 bound |R_v| by .319, as required.

The covariance inequalities and U_*. The value estimates just obtained imply 0\le\mathsf K\le.0008 and -.501\le\mathsf C\le-.499. Inserting these intervals in \lambda t_\pm^2+t_\pm\mathsf C+\mathsf K gives the lower bound .010 for t_- and the upper bound -.027 for t_+. For the third covariance inequality, the same intervals give

Finally n_{\mathsf K}>0 removes the positive-part term in U_*. Using |\dot{\mathsf K}|<.002 and |\dot{\mathsf C}|<.001 in its two squares gives U_*<.001.

The positive-part integral. The first table row and \log\sqrt{2\pi}-2+.0064<0 give \mathsf K\le(\log x)x^{-2}. Hence

Combining this with the finite upper bound .335 gives a total less than .422. All derivatives needed in the finite argument are also bounded by the displayed tail estimates. Reflection completes the proof of Lemma A.1.□

Proof of Lemma A.2. 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 A.6. For \rho, the exact formula makes an endpoint bound possible. Put y=\tanh^2t and v_1=(1-y)/s^2. Differentiating gives

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.

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

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

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

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

Finally the auxiliary estimates follow directly from rational arithmetic and the logarithm enclosure in Lemma A.3:

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

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

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

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:

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

Row entries give 1000v, followed below by upper bounds for 10^5 I_j(v) in the same order.

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

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 A.3; the interpolation estimates in Lemma A.5 justify the passage between nodes. For example, with \mathcal R(M)=.536M+.00001, the required errors are

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

Use as approximate samples (starred) at t_i, with F_i=F(t_i),

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

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

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,

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,

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

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

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

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

Combining this with B_0(y)\le\min(.018,.0146y^2+.00010) and y\le2h proves the l<0 line of Equation (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,

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

Maximizing Z/\sinh(2h)-78Z^2 over real Z gives 1/(312\sinh^2(2h)). Together with the bound on B_0, this proves (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 A.2 (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:

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

using also convexity for the q(t)^2/2 term. This completes the proof.□