A sharp entropy bound and the simplex inequality for isotropic constants

Proof. Let X_a have the conditional law of G-a given G>a, and write \mathbb E_a for expectation under this law. Gaussian integration by parts gives

Since X_a>0 almost surely, d>0 and hence q'=qd>0. Its strictly positive variance gives q'<1.

For completeness, this variance is at most the square of its mean. The survival function

satisfies S_a(u+v)\le S_a(u)S_a(v), since its hazard rate q(a+u) is increasing. Integrating over u,v\ge0 yields

Consequently 1-qd=\operatorname{Var}(X_a)\le d^2. Differentiating q'=qd now gives q''=q(d^2+qd-1)\ge0. Finally, q(a)\to0 as a\to-\infty, whereas q(a)>a for a>0. These limits prove surjectivity.□

Proof. We use a homotopy from the Gaussian and show that its zeros stay in a compact subset of the positive definite cone. For 0\le\theta\le1, define

where Z_\theta is the normalizing integral. There are constants a_-,a_+>0, independent of \theta, with a_-I\le D^2W_\theta\le a_+I. These densities have uniformly Gaussian tails. Their covariance matrices C_\theta depend continuously on \theta, are positive definite, and therefore satisfy

for fixed c_0,C_0>0.

For a positive definite symmetric P, let J_{\theta,P} be the Jacobian of the transport to P_\#\mu_\theta. Define a map with values in the vector space of symmetric matrices by

Contraction in both directions gives

We first verify continuity of F. Restrict P to any compact subset of the positive definite cone. The transports have common Lipschitz constants by (2.12); their values at zero are bounded because their means and second moments are bounded. Every convergent sequence (\theta_j,P_j) thus has locally uniformly convergent subsequences of transport maps. Their limits push forward \gamma_m to the limiting target and remain gradients of convex functions, so Brenier uniqueness identifies the limit. Write J_j,J for the corresponding Jacobians. They converge weak-* locally and obey common positive spectral bounds.

The change-of-variables equation expresses \log\det J_j in terms of the target potential composed with its transport map. It therefore converges locally uniformly to \log\det J. On a fixed compact spectral interval, strict concavity of \log\det gives a constant c>0 such that

Integration on any ball, followed by weak-* convergence, proves J_j\to J in local L^2. Functional calculus then gives local L^2 convergence of q^{-1}(J_j). Their common spectral bounds control the Gaussian tails, proving continuity of F.

We next bound all zeros, uniformly in \theta. Suppose \mathbb EA=0, where A=q^{-1}(J_{\theta,P}). Convexity of q and Jensen’s inequality on the expected spectral measure of each unit vector give

Gaussian integration by parts gives \mathbb E(x_r\partial_i\phi)=(\mathbb E J_{\theta,P})_{ir}. Orthogonal projection of the centered components of \nabla\phi onto the span of x_1,\ldots,x_m consequently yields

Together with (2.11), this bounds P^{-1} uniformly. The lower estimate in (2.12) now gives a uniform lower bound A\ge a_*I, where we may take a_*\le0. Since q'\le1,

On the other hand, Gaussian Poincaré and (2.12) imply

This also bounds P uniformly.

Choose a bounded open set in the space of symmetric matrices whose closure lies in the positive definite cone and whose interior contains every zero just bounded. The map F(\theta,\cdot) has no zeros on its boundary. At \theta=0, the target is N(0,P^2), so J_{0,P}=P and F(0,P)=q^{-1}(P). Its unique zero is q(0)I; its derivative there is multiplication by the positive scalar 1/q'(0). Its Brouwer degree is therefore nonzero. Homotopy invariance of degree gives a zero at \theta=1, as required.□

Proof. The density equation is

Differentiating twice gives (2.14). The divergence of the cofactor matrix of a Hessian vanishes. Applying this identity to J, and then differentiating (2.15) once, shows that

Thus, for smooth u and compactly supported smooth v,

Choose 0<\lambda\le\Lambda<\infty such that \lambda I\le J\le\Lambda I. Taking the trace in (2.14), and using D^2V\ge0, gives

For a compactly supported cutoff 0\le\chi\le1, integration by parts and |\nabla\operatorname{tr}J|\le\sqrt m\,|\partial J| imply

Let \chi tend to one through standard radial cutoffs with gradients bounded by a constant divided by their radii. Fatou’s lemma proves \|\partial J\|<\infty.

All eigenvalues of J lie in [\lambda,\Lambda]. The Fréchet derivatives of q^{-1} and of k\circ q^{-1} are bounded on that spectral interval, so A,k(A)\in H^1(\gamma_m). The same applies to any smooth scalar function on this interval. For such a matrix multiplier, insert a cutoff into (2.16) and let it tend to one. The terms containing its gradient vanish by Cauchy–Schwarz and the energy bound; the remaining differentiated terms are bounded by a constant times 1+|\partial J|^2. This justifies the asserted integrations by parts.□

Proof. The proof has two parts. Symmetry of the third derivatives of \phi first replaces the transport equation by averages of the scalar function M. Comparing arithmetic and logarithmic averages then produces the term \mathcal C.

The identity qk=t will match the entropy term in the covariance completion of Section 3; we first estimate the derivative cost of k(A). Put F(\lambda)=k(q^{-1}(\lambda))^2. Multiply (2.14) by F(J) in trace and integrate by parts. Lemma 2.4 justifies this operation. At a fixed point choose an orthonormal eigenbasis of J and rotate all three indices of the symmetric tensor \Theta_{ijr}=\partial_rJ_{ij}=\partial_i\partial_j\partial_r\phi into that basis. The quadratic term on the right of the Jacobian equation has trace contraction

where we used the symmetry of \Theta. The term obtained by integration by parts is

The remaining term is \operatorname{tr}(F(J)JD^2VJ)\ge0. Discarding it gives

Only pointwise changes of basis are made here; no eigenvectors are differentiated.

Regard M as a function of \lambda=q(a). The identities (2.3) give

For two positive numbers u,v, let M^{\mathrm{ar}}_{uv} be the uniform average of M on the interval between u and v, with its continuous interpretation when u=v. With the weight \Theta_{ijr}^2/(\lambda_i\lambda_j\lambda_r), symmetrization in i,j changes the coefficient in (2.20) to

In the full sum we can replace this coefficient by \lambda_rM^{\mathrm{ar}}_{\lambda_i\lambda_j}, decreasing its value. To see this, order three eigenvalues as u\le v\le w. Lemma 2.2 makes M increasing, so

The ordered coefficients u+v-2w, u+w-2v, and v+w-2u sum to zero. Their scalar product with these three ordered averages is nonnegative, by the rearrangement inequality. Full symmetry of \Theta permits precisely this averaging over permutations of the three indices. We have proved

We now fix a pair of eigenvalues. Bars in the rest of the proof denote uniform averages in \log\lambda over their interval. Set

with w_0=1 at x=0. Since da/d\log\lambda=D_0, the divided-difference formula shows that the coefficients of (Z_r)_{ij}^2 and (\partial_r k(A))_{ij}^2, respectively, when written with the same weight as in (2.22), are

Both D_0(1+b) and D_0(1-b)=k are increasing: d'=qd-1<0, and b'=l>0, k'=b>0. The covariance inequality for two increasing functions on an interval therefore yields

We also need a quantitative comparison of the two averaging measures. On the logarithmic interval, (2.6) says that M(\lambda)\lambda^{-m_*} is increasing. Exponential weighting and the same covariance inequality give

Indeed, after centering the logarithmic interval, the left side is \mathbb E_0(e^sM)/[\mathbb E_0e^s\,\mathbb E_0M], where \mathbb E_0 is its uniform average. Under the measure proportional to e^{m_*s}, the increasing factor Me^{-m_*s} can only increase the mean of e^s. Taking M=e^{m_*s} gives the displayed ratio.

Here is a convenient explicit lower bound for this ratio. Define A_0(x)=x\coth x-1, with A_0(0)=0. The identity

implies both A_0(x)\ge1-w_0^2 and that A_0(x)/x^2 is nonincreasing for x>0. To verify the identity, clear the denominator and expand \tfrac{x}{2}\sinh(2x)+x^2-\cosh(2x)+1 in powers of x; the coefficients vanish through degree four and are nonnegative thereafter. Since 0\le m_*\le1, we obtain A_0(m_*x)\ge m_*^2A_0(x). The addition formula for \sinh now gives

Every formula here extends continuously to m_*=0 or x=0.

Combining (2.24)–(2.26), the amount by which the coefficient in (2.22) exceeds the difference of the two coefficients in (2.23) is at least

Relative to the coefficient of (Z_r)_{ij}^2, this is at least

The last inequality is the Taylor expansion of (\sinh x/x)^2, whose omitted coefficients are positive. Substituting in (2.22) and summing proves (2.19).□

Proof. Put F=\nabla\phi. For two degree-d components of F, Gaussian integration by parts gives

Differentiation lowers the degree by one, and R therefore cancels the factor d. Summing over d\ge1 proves C_{ij}=\sum_\ell\E[J_{i\ell}R J_{j\ell}]. Symmetry of J gives the first matrix identity in (3.4). Expanding the row Gram matrix of q-(\Id+N)k gives cross terms -2\E(qk)=-2\E t. The remaining term is

by entrywise integration by parts. This proves the second identity. The calculation is valid by the H^1–H^{-1} pairing, even if Nk is not square-integrable.

For a test field F, the pairing with u-\partial^*v is \langle u,F\rangle-\langle v,\partial F\rangle. Cauchy–Schwarz bounds it by (\|u\|^2+\|v\|^2)^{1/2}\|F\|_{H^1}. Taking the dual norm proves (3.5) for S=\Id, and applying this argument to S^{1/2}U proves the general case.□

Proof. Let F_0=Y(b-\beta)+\sum_rX_rL_r. The relations (3.10) give

For the cross term, the order of the factors follows from \tr(F_0^tTl)=\tr(TF_0l), by transposition and cyclicity. Integration by parts on the linear part of A yields

Together with af-f'=2t-k^2-b^2-1, this accounts for the terms involving f' and f.

To expand the energy term, use both versions of K_r=X_rb+D_r=bX_r+D_r^t:

The two cross terms agree under transposition inside the trace. Moving b past T, rather than past X_r, gives

The terms containing B-\Id now combine as

In the other terms, substituting (3.12) produces G_re+Z_r\circ\psi and the Y multiplier f-2(b-\beta)l/\alpha=h_0+ck.

It remains to handle c\tau(TYk). Since \partial^*g=Y,

Here integration by parts first gives \tau(Tg_rK_r); transposition makes the two expressions equal. Substituting K_r=bX_r+D_r^t and moving b past T introduces cg_rb in M_r and replaces D_r by D_r-\gamma g_r in the commutator term. The remaining square is completed by

This proves (3.19).□

Proof. Work pointwise and diagonalize S. The relation B=\alpha^{-1}(S^{-1}-\Id/s) gives

Indeed, the contribution of b_{ij}^2 for i<j on either side is (S_i-S_j)^2/(\alpha S_iS_j). The norm in (3.22) includes expectation. Set F=\sum_rX_rw_r and let \operatorname{skew}M=(M-M^t)/2. Completing the square in the skew-symmetric matrix S^{-1/2}[T,b]S^{-1/2} bounds the first two terms on the left of (3.21) by

For completeness, we estimate this square together with the last term. Write S_i=su_i, where 0<u_i\le1, so \alpha S_iB_i=1-u_i. The rows of the block matrix X=(X_1,\ldots,X_m) are orthogonal, since XX^t=B is diagonal. For column j of the vertically stacked matrix (w_r)_r, let z_{ij} be its component along the normalized ith row of X. Zero rows can be completed to orthonormal directions, and cause no contribution to F. Then F_{ij}=\sqrt{B_i}\,z_{ij} and the remaining column components are orthogonal to these directions.

For i<j, the quadratic form in (z_{ij},z_{ji}) has matrix

Its largest eigenvalue is at most 1+s(z-2z^2)\le1+s/8, where z=\sqrt{u_iu_j}; here (1-u_i)(1-u_j)\le(1-z)^2. Diagonal and orthogonal column components have coefficient at most 1. Summation and expectation prove (3.21).□

Proof of Theorem 3.1. Apply Lemma 3.3 in (3.11). Bound its last line by Lemma 3.4, its first square by (3.23), and its linear T term by (3.25). The H terms combine to -\tau(H\rho p). Since Y=Y^t and h_0,p,b are commuting functions of A, the two terms quadratic in Y have exactly the form appearing in (3.26), by cyclicity of the trace; no commutation with Y is required. Set \Omega_B=\sum_{i=1}^m\omega(\lambda_i). Consequently,

Use S\le s\Id and T\le(s-1)\Id. In an eigenbasis of A, the entries of G_r,g_r,Z_r are symmetric in i,j. Pairing the entries (i,j) and (j,i) therefore gives precisely the endpoint average in (3.27); right multiplication by p^{-1/2} gives the denominator p_j for entry (i,j). Proposition 3.6 now bounds the last five terms of (3.28) by

The separation surplus is the same \mathcal C as in Proposition 2.5: its coefficient at (i,j) is independent of the derivative index r, so summing over r is unchanged by rotating that index. All changes of eigenbasis are pointwise; no eigenvectors are differentiated. The separation surplus cancels. Because Q has constant coefficients, its summed quadratic form is invariant under orthonormal changes of basis. We may therefore return to fixed coordinates and apply the Gaussian chaos decomposition.

For a degree-d component of Y, where d\ge2, the associated components of g and G satisfy g=G/d, and \|G\|^2=d\|Y\|^2. Different degrees remain orthogonal after differentiation. The required bound follows, degree by degree, from

Indeed, the right side minus the left side factors as \frac1{50}(1-2u)(1+68u)\ge0. Set u=1/d, multiply by \|G\|^2, and sum over the chaoses; the sums converge because Y\in H^1(\gamma_m). The summed Q term is at most V_0-\frac7{50}\|Y\|^2. Substitution into (3.28) therefore gives

which is (3.2).□

Proof of Proposition 3.5. By 0<b<1/2 and (4.2),

It remains to bound the terms involving T-\log S while retaining a negative multiple of the weight \omega in (3.1).

Fix a unit eigenvector v of B with eigenvalue \lambda\ge0. The corresponding eigenvalues of S,T,H,\rho are

Let \nu_v be the probability measure obtained by averaging the spectral measure of A at v. The decomposition of A into its first chaos and Y gives

Indeed, orthogonality of the Gaussian chaoses gives \E A^2=B+\E Y^2, including the vanishing of both matrix cross terms.

For 0<u\le.13, define

We claim the following quadratic minorant:

For a\le0, put x=-\sqrt u\,a. If x\ge2, the right side is nonpositive. If 0\le x\le2, it is enough to use p(a)\ge u/(x^2+7u) and

For 0\le a\le4, the right side of (4.9) is at most .13+2(.13)^{3/2}<.25. For a\ge4, its maximum over all real a is u+n_u^2/(4d_u)=.46. Thus (4.9) follows from (4.2) in all three ranges.

Choose

with u=.13 when \lambda=0. Since \lambda u^2\le1.28^2, integration of the minorant against (4.8) yields

The second inequality uses 1-1.28^2/[16(.46-.13)]>.689. We also have

To check this, note that \rho\le.64 and that H\le\lambda for \lambda\ge1, whereas H\le\alpha s for \lambda\le1. In the former case,

in the latter, \alpha s\cdot.64\cdot .13^3/[16(.46-.13)]<.027.

For completeness, we next verify the elementary logarithmic bound needed to compare T-\log S with the positive term in (4.10). Set

For 0<S\le s and \lambda=(1/S-1/s)/\alpha, we claim

where the second bound in the minimum is interpreted as +\infty at \lambda=0. First,

is concave. Its tangent at S=.28 has intercept below .16 and slope below .52, so F_0(S)\le .16+.52S. These two strict inequalities can, for example, be checked from

using twelve terms and the geometric bound for the remainder. Coefficient comparison now gives

which proves the first bound in the minimum.

For the second bound we use

Here is a direct verification, including the point S=1 by continuity. For x>0, let

Then

Descartes’ rule of signs gives at most three positive roots of P, counting multiplicity. The signs at .24,.26,.6, together with P(1)=0 and P'(1)>0, give exactly three: one in (.24,.26), one in (.26,.6), and 1. Thus the only local minima of J occur at the first root and at 1. We have J(1)=0 and J(.25)>.14; the latter follows already from \log2<.694. On [.24,.26], |P(.25)|<.01 and |P'|<4, hence |J'|<2. The first minimum is therefore positive. Also J(x)\to+\infty at both ends of (0,\infty). This proves J\ge0, which is equivalent to (4.13) away from x=1.

Using \alpha\lambda=1/S-1/s, inequality (4.13) gives

The last ratio decreases in S because 1.4\cdot.15<.478\cdot.45325. Finally, .089<.688\cdot.13 and .88<.688\cdot1.28 prove the last inequality in (4.12).

For \lambda\ne1 we have

Combining (4.10)–(4.12), and treating \lambda=1 by continuity, gives

The definitions give H\rho u=\omega(\lambda). Sum over an orthonormal eigenbasis of the matrix B, and add (4.7). This summation evaluates \tau(H\rho p(A)) without requiring p(A) to commute with B. Likewise, (4.7) may be integrated against the positive semidefinite matrix Y^2 without a commutation assumption. Since \sum_v r=\norm{Y}^2, the result is

as asserted.□

Proof of Proposition 3.6. Fix two endpoints a_-\le a_+ and write

When the endpoints coincide, all secant increments vanish, and (4.4) immediately proves the assertion. We henceforth suppose \delta>0.

We first isolate the cost of the secant increments at one endpoint a_j. Use coordinates (G,h) with h=g-G/2. With Z=0, the left side of the endpoint quadratic form in Proposition 3.6 is exactly the two matrices subtracted from Q in (4.4). For an increment vector (x,y,w), define

Completing squares against the surplus \ell_{1j}G^2+.75h^2 in (4.4) shows that the full endpoint form is bounded above by

Indeed, the two coefficients of 2GZ and 2hZ are precisely the numerators of the two squared terms in (4.14). The first line of that formula is the coefficient of Z^2 before completion of squares.

The next step estimates these increment norms by integrating derivatives between the endpoints. Recall that a secant slope is the uniform average of the derivative over [a_-,a_+]. With \xi=\log q(a), differentiation therefore gives the vector

More explicitly, (\Delta k,\Delta b,\psi_j) is the uniform average in a of \int_{\log q(a_j)}^{\log q(a)}v_j(q^{-1}(e^\xi))\,d\xi. This follows from b'=l, l'=zl, d(\log q)/da=d, and f''=l(1+3b+kz)=lh_*. Put

We record the bounds furnished by Lemma 4.1. If the endpoint a_j belongs to R_1,U_1,V_1,T_1, respectively, then

For a\le-1, the R_1 bound improves to .20. For a_j\in P_1, the bounds are .42^2 when a\le-1 and .61^2 when a\ge-1. If a_j\in O, write t_j=-a_j; then

For clarity, the complete finite arithmetic behind these bounds is given by the following table. Its three entries bound, in order,

To verify it, substitute the intervals in (4.3) and (4.6); each expression is affine in each of its variables separately, so its extrema occur at corners of the corresponding box. For the middle expression, first write it as

Substituting these three bounds into (4.14), together with the bounds for \mathsf D_j,\ell_{1j},z,\chi, proves (4.17)–(4.18). All these are finite rational comparisons; the exact verifier accompanying Appendix A also checks them.

It remains to compare the integrated derivative cost with the logarithmic separation surplus. Let u_- be the mean of (\log q(a)-\log q(a_-))/\delta under the uniform probability measure on [a_-,a_+], and put u_+=1-u_-. In the \xi coordinate this measure has density proportional to D_0=1/d. The excess-variance bound in Lemma 2.1 gives

The density D_0 is increasing, whereas D_0e^{-\xi} is decreasing. The covariance of an increasing function with an increasing function is nonnegative, and with a decreasing function is nonpositive. Applying this observation to the coordinate \xi compares its mean first with constant density and then with density proportional to e^\xi. The latter comparison gives u_-\le\tfrac12+\tfrac12\coth(\delta/2)-1/\delta, and hence

For the upper bound one uses \coth x-1/x\le x/3, obtained by comparing the power series of x\cosh x and (1+x^2/3)\sinh x.

Suppose P_- and P_+ have the following property: on every partial \xi-interval starting at the indicated endpoint, the mean of \mathcal N_j(v_j) is at most \sqrt{P_j}. Pointwise bounds for F_j are one way to ensure this property. Minkowski’s inequality, followed by the uniform average in a, gives

Since u_-\ge u_+\ge0,

Averaging (4.15) and applying (4.20) therefore proves the desired estimate whenever

Indeed, the resulting coefficient of Z^2 is at most \delta^2(1+\delta^2/36)\Theta_{m_*}(\delta)/4 =r_{-+}\delta^2.

We verify (4.21) for all endpoint positions. The function \Theta_m increases in \delta\ge0: differentiation with respect to y=\delta^2 leaves a positive numerator 1/180+2y/1680+y^2/(1680\cdot36). We will also use the following observation about partial intervals: if a smaller bound holds on an initial segment of logarithmic length L, then its fraction in a partial interval of length h\le\delta is at least \min(1,L/h)\ge L/\delta, provided \delta\ge L. If a_-\in R_1, take P_-\le.295 and P_+\le.39; then (P_-+P_+)/2\le.3425<\Theta_{.38}(0)=.345. For a_-\in U_1 or V_1, use P_-\le.23, P_+\le.39 and \Theta_{.68}(0)=.42. For a_-\in T_1, both bounds are .39<.42.

Now suppose a_-\in P_1. If a_+\le-1, both endpoint costs are at most .20<\Theta_{.19}(0), using the R_1 row at a_+=-1. Otherwise, the part of the logarithmic interval below -1 has length at least 1.62. Hence every partial interval from the lower endpoint has mean norm at most

The length assertion follows from q(-1)>.287, q(-2)<.056. If the upper endpoint lies in T_1, then \delta>4 because q(4)>4. For \delta\le4, therefore, P_+\le.23 and P_-\le.284143<\Theta_{.19}(0)=.2975. For 4\le\delta\le5, use P_-<.300787, P_+\le.39, and

For \delta\ge5, use P_-\le.61^2=.3721 and

This covers P_1.

Finally suppose a_-\in O. For any endpoint a_j\in O and any argument a\le-1 between the endpoints,

The first follows by integrating d\ge-a on the negative half-line; the second follows from l\le .075\cdot1.19<.09 and d\ge|a|. Using \chi\le.07 in (4.18) gives

If the whole interval lies at or below -1, take P_-=.144+.011\delta and P_+\le\max(P_-,.20); the latter bound also covers upper endpoints in P_1 and R_1. For 0\le\delta\le4, both are below .25. For \delta\ge4, use

This follows on clearing denominators and using \delta^2\ge16. The quadratic .106-.011\delta+.00260\delta^2 is strictly positive, so (4.21) follows in this case too.

If a_+>-1, the first part of the logarithmic interval has length at least 4, since q(-3)<.005 and q(-1)>.287. For a\ge-1, the same integration gives t_-^2\le1+2\delta; using \chi\le.097 in (4.18) yields F_-(a)\le .236+.031\delta. On every partial interval from the lower endpoint, Jensen’s inequality therefore permits the uniform bound

For partial intervals contained below -1, the same bound follows from (4.22), because \delta\ge4. By (4.23),

the quadratic has negative discriminant and positive leading coefficient. If a_+\notin T_1, then P_+\le.23<\Theta_0(\delta). If a_+\in T_1, then \delta>6.5 from q(4)>4 and q(-3)<.005, and P_+\le.39<\Theta_0(6.5)<\Theta_0(\delta). This proves (4.21) in every case and completes the proof.□

Proof. The convex set \{V<\infty\} has positive measure and hence nonempty interior: otherwise its affine hull would be a proper affine subspace. Choose a simplex with all vertices in this set and with nonempty interior. Convexity bounds V above by the largest of its values at the vertices throughout the simplex. Consequently, some ball B(x_0,r) lies in a sublevel set \{V\leq M\}, where M\geq V(x_0).

The closed convex set A=\{V\leq M+1\} has finite volume, since

It is bounded. Indeed, the convex hull of B(x_0,r) and a point z\in A contains a cone with base an (m-1)-dimensional disk of radius r through x_0, perpendicular to z-x_0, and height |z-x_0|. Its volume tends to infinity with |z-x_0|. In dimension one the same assertion follows from the length of the interval joining x_0 to z.

Choose R>r so large that A\subset B(x_0,R). If d=|x-x_0|\geq R and V(x)<\infty, put y=x_0+(R/d)(x-x_0). Then y\notin A, and convexity gives

Thus V(x)>V(x_0)+d/R. The inequality is automatic if V(x)=+\infty. On the compact ball \overline B(x_0,R), lower semicontinuity and properness give a finite lower bound: a sequence with values tending to -\infty would have a convergent subsequence and violate lower semicontinuity. Combining these two bounds proves (5.1), after enlarging b.

The moment assertion follows from e^{-V(x)}\leq e^b e^{-a|x|}. For the entropy assertion, the elementary bound

holds because |u|e^{-u/2} is bounded on [-b,\infty). Substitute u=V(x) and use (5.1).□

Proof. Fix a,b from Lemma 5.1. For each positive integer j, define the Moreau envelope [24, Section 7]

The expression in braces is lower semicontinuous, strictly convex on its effective domain, and tends to infinity as |y|\to\infty. It therefore has a unique minimizer p_j(x), and Q_j is finite everywhere. Moreover,

Indeed, V(y)\geq a|x|-a|x-y|-b, and minimizing jr^2/2-ar over r\geq0 proves the bound.

For completeness, the needed regularity of Q_j follows directly from the minimization. The optimality condition is j(x-p_j(x))\in\partial V(p_j(x)), where \partial V denotes the convex subdifferential. To obtain it, compare the minimum at p=p_j(x) with p+s(z-p), use convexity to bound V(p+s(z-p)) above, divide by s>0, and let s\downarrow0. The resulting inequality is V(z)\geq V(p)+j\langle x-p,z-p\rangle. Adding the two subgradient inequalities at p_j(x) and p_j(x') gives

Consequently both p_j and x\mapsto x-p_j(x) are 1-Lipschitz. Evaluating the infimum at p_j(x) and, in the reverse direction, at p_j(x+h) shows that

For example, the first comparison gives an upper error j|h|^2/2 after subtracting j\langle x-p_j(x),h\rangle, and the second gives a lower error of order j|h|^2 by the Lipschitz estimate just proved. Thus Q_j is continuously differentiable with j-Lipschitz gradient. It is convex, since the function minimized in (5.5) is jointly convex in (x,y).

We next check convergence before smoothing. The functions Q_j increase with j, and Q_j(x)\leq V(x) when V(x) is finite. If Q_j(x) stays bounded above as j\to\infty, the identity at the minimizer and V\geq-b imply

Lower semicontinuity then gives V(x)\leq\liminf_j V(p_j(x))\leq\lim_jQ_j(x). This proves

including the value +\infty.

Choose a nonnegative even smooth function \eta supported in the unit ball with integral one, and set \eta_\varepsilon(x)=\varepsilon^{-m} \eta(x/\varepsilon). Define

Convolution preserves convexity and the Lipschitz bound on the gradient; hence (5.3) holds. Evenness gives \int z\eta(z)\,\dd z=0, so Jensen’s inequality and (5.6) yield the common bound

Smoothing at this scale does not affect the limit. The j-Lipschitz bound on the gradient gives

Integrating against \eta_{1/j} cancels the linear term, so

Together with (5.7), this proves V_j(x)\to V(x) everywhere.

Now (5.9) dominates (1+|x|^2)e^{-V_j(x)} by an integrable function independent of j. Applying (5.2) with B in place of b also gives

At points where V=+\infty, both e^{-V_j} and V_je^{-V_j} tend to zero. Dominated convergence therefore proves convergence of the normalizing constants, first moments, second moments, and the integrals \int V_je^{-V_j}. Finally,

which proves all of (5.4).□

Proof of the inequality in Theorem 1.2. Apply Theorem 3.1 to the normalized potentials V_j+\log Z_j of Lemma 5.2. Their Hessians satisfy (5.3), so

The covariance of f is positive definite: zero variance in a nonzero direction would concentrate this Lebesgue density on an affine hyperplane. Thus (5.4) permits passage to the limit in both sides and proves the inequality.□

Proof. Write e_1,\ldots,e_m for the standard basis and J=q(A). Commutation of distributional derivatives of F gives

The scalar function q is real analytic near zero. Since A(0)=0, its convergent power series, interpreted by matrix multiplication, gives

near zero; the differentiated remainder is O(|x|^2). The identity q'(a)=q(a)(q(a)-a) yields

Thus (6.2), which is a pointwise identity between analytic functions near zero, has constant term

Equivalently, A(x)z=A(z)x for all x,z\in\R^m; in particular, A(x)e_s=X_sx. Its degree-one term is

The terms with leftmost factor A(x) cancel by (6.3). The remaining equality is (X_rX_s-X_sX_r)x=0 near zero, and hence

Commuting real symmetric matrices admit a common orthonormal eigenbasis. Choose the orthogonal matrix U with those basis vectors as columns, and rotate both the input and the output. Then

has diagonal coefficients. Orthogonality of U gives \sum_i\widetilde X_i^2=I, and the identity A(x)z=A(z)x gives \widetilde X_i e_j=\widetilde X_j e_i. If i\ne j, these two vectors are multiples of different basis vectors, so both are zero. Consequently, only the ith diagonal entry of \widetilde X_i can be nonzero. The sum-of-squares identity makes that entry a sign:

It follows that

Since t'=q and \sigma_i^2=1, the right side is also the Jacobian of y\mapsto(\sigma_i t(\sigma_i y_i))_{i=1}^m. Their difference has zero weak gradient on the connected set \R^m and is therefore almost everywhere constant. This proves (6.1).

For independent standard Gaussian coordinates G_i, the variables t(\sigma_iG_i) are independent mean-one exponentials. Their variances are one, and the signs and the orthogonal matrix preserve covariance I. This proves the final assertion.□

Proof. Let (u_i) and (v_k) be orthonormal eigenbases of M and N, with eigenvalues \lambda_i and \mu_k. Then

whereas the corresponding entry of M-N is (\lambda_i-\mu_k)u_i^{\mathsf T}v_k. The scalar bound 0<q'<1 therefore bounds the absolute value of each former entry by the absolute value of the latter. Squaring and summing over i,k proves (6.5), because the Frobenius norm is unchanged by orthogonal changes of the row and column bases separately.□

Proof of the equality statement in Theorem 1.2. Let f be a log-concave probability density on \R^m with \Dgap(f)=0. Choose the approximating densities f_j from Lemma 5.2. Write a_j,a for their means and the mean of f, and C_j,C for their covariance matrices. That lemma gives

Here C is positive definite, as recalled in the introduction, so the covariance convergence also implies convergence of the log determinants.

For each j, apply Proposition 2.3 to f_j. Let P_j be its positive definite normalizing matrix and let F_j transport \gamma_m to (P_j)_\#(f_j(x)\,\dd x). In the notation of Theorem 3.1, put

The entropy gap is affine invariant. Theorem 3.1 and (6.6) therefore imply

The function \omega is continuous and nonnegative on [0,\infty), vanishes only at 1, and tends to infinity at infinity. For every \varepsilon>0, it consequently has a positive infimum on \{\lambda\ge0:|\lambda-1|\ge\varepsilon\}. Thus every eigenvalue of B_j tends to 1, and B_j\to I in operator norm.

Since \sum_r|X_{j,r}|_{\mathrm F}^2=\tr B_j, the coefficients form a bounded sequence in a finite-dimensional space. Pass to a subsequence such that X_{j,r}\to X_r for every r. Each X_r is symmetric, and Gaussian orthogonality together with (6.7) gives

Lemma 6.2 now gives

To obtain a limiting map, remove the irrelevant translations: \widehat F_j=F_j-\E F_j. Each \widehat F_j lies in H^1(\gamma_m;\R^m): its law has finite second moments and, for this fixed j, its Jacobian has the bounds (2.9) applied to the transformed target. Gaussian Poincaré, applied componentwise to the centered difference, gives

Consequently \widehat F_j converges strongly in Gaussian H^1 to a centered map \widehat F with weak Jacobian D\widehat F=J. Indeed, (6.10) and (6.9) give convergence of both maps and first derivatives; passage to weak derivatives is also valid on every bounded set, where the Gaussian density has a positive lower bound. In particular, \widehat F\in W^{1,2}_{\mathrm{loc}}(\R^m;\R^m). Lemma 6.1 applies to (6.8) and shows that \widehat F_\#\gamma_m is an affine exponential product with covariance I.

It remains to return from the normalized laws to f. By Cauchy–Schwarz, the strong L^2(\gamma_m) convergence of the centered maps gives

We can now bound the normalizing matrices uniformly, without any uniform Hessian bounds on the approximating potentials. Choose positive constants c,C_0,d,D such that, for all sufficiently large j,

Congruence by P_j then yields

Hence P_j and P_j^{-1} are bounded. Pass to a further subsequence with P_j\to P; the lower bound in (6.12) makes P invertible.

The law of \widehat F_j is the image of f_j(x)\,\dd x under x\mapsto P_j(x-a_j). For a bounded continuous function \psi,

The first term tends to zero by (6.6); the second does so by dominated convergence. Strong convergence of \widehat F_j also identifies the weak limit of these laws as \widehat F_\#\gamma_m. It follows that the image of f(x)\,\dd x under x\mapsto P(x-a) is the affine exponential product already determined above. Since P is invertible, f itself has the form asserted in Theorem 1.2, up to equality almost everywhere.

Conversely, the density of a product of m independent mean-one exponentials is log-concave, has entropy m, and has covariance I. Every invertible affine image remains log-concave and has the same entropy gap by affine invariance. Thus every density in the stated class satisfies \Dgap=0.□

Proof of Theorem 1.1. Let K\subset\R^n be a convex body. Translation preserves both L_K and the property of being a simplex, so assume that K has centroid zero. Write \Sigma_K=\Cov(X) for X uniform on K, and define

Convexity of K makes \mathcal C_K convex. It is closed because K is compact: a convergent sequence (t_jx_j,t_j) with t_j\to0 has t_jx_j\to0, and when the limiting height is positive one can divide by that height. The cone has nonempty interior, and its only point at height zero is the origin.

Define the log-concave density

The change of variables y=tx, with Jacobian t^n for t>0, gives

so f_K is a probability density. The same change of variables shows that it is the density of (TX,T), where T is independent of X and has density t^ne^{-t}/n! on (0,\infty). Integration by parts therefore gives

Since \E X=0, independence makes the cross-covariance of TX and T vanish. Hence

Applying the entropy inequality of Theorem 1.2 in dimension n+1 yields

This is precisely

Every step from the entropy inequality to (7.6) is reversible. Thus equality in this bound holds if and only if f_K has equality in Theorem 1.2.

Suppose equality holds. By that theorem, the law with density f_K is an invertible affine image of the product of n+1 mean-one exponential laws. The closed support of f_K is exactly \mathcal C_K: it has positive density on the interior of the cone, and this interior is dense in the cone. Taking closed supports gives

for an invertible matrix M and a vector b.

The origin is the unique extreme point of \mathcal C_K. Indeed, every nonzero z\in\mathcal C_K is the midpoint of 0 and 2z, whereas 0=(u+v)/2 with u,v\in\mathcal C_K forces the nonnegative heights of u and v to vanish, and hence u=v=0. The orthant also has the origin as its unique extreme point. Invertible affine maps preserve extreme points, so (7.7) forces b=0.

Choose positive multiples w_0,\ldots,w_n of the respective columns of M, and let a_i be the last coordinate of w_i. Each w_i is a nonzero point of \mathcal C_K, so a_i>0. Put p_i=w_i/a_i. The section at height one is

The vectors p_i are linearly independent, since the vectors w_i are. They are consequently affinely independent in the height-one hyperplane. Thus K is a simplex. Figure 1 illustrates the normalization of the generating rays in this step.

Conversely, let K=\operatorname{conv}\{v_0,\ldots,v_n\} be a simplex, and let M have columns (v_i,1). These columns are linearly independent and

For the determinant identity, subtract the first column from the others and expand along the last row; the remaining determinant is n! times the volume of K, up to sign. If E_0,\ldots,E_n are independent mean-one exponential variables, the density of M(E_0,\ldots,E_n)^{\mathsf T} at (y,t)\in\mathcal C_K is

because the last row of M consists of ones. This is exactly f_K, up to immaterial boundary values. Theorem 1.2 therefore gives equality in (7.6).□

Proof of Lemma 4.1 and Lemma 2.2. We use the definitions of q,k,d,b,l from the transport section and of p,h_0,\mathsf D,\mathsf L,\mathsf E,K_0,J_0,A_*,\eta_*,\nu_* from the scalar section. The differential identities needed below are

In particular,

We verify the estimates separately on a\le-6, a\ge10, and [-6,10]. The central calculation is also specified in verification/verify_scalar.py; it uses rational arithmetic only.

The negative tail

Put r=-a\ge6, u=r^{-2}, and

Writing \varphi for the standard Gaussian density, the lower Gaussian tail is P=\Phi(-r)=\varphi(r)I_0(r). Thus

Integration by parts gives

The inequality 0\le1-(1-P)(-\log(1-P))/P\le2P gives |k-I_0|\le2P/r. Since P\le\varphi(r)/r and \varphi(6)<7\cdot10^{-9}, we have q\le1.01\varphi(r). Applying (A.1) successively therefore gives

Each Gaussian error decreases faster than its required inverse power for r\ge6: indeed r^N\varphi(r) is decreasing there for N\le7. Checking at r=6 yields, with g_0=k,g_1=b,g_2=l,g_3=l',

The elementary bounds 1-v^2/2\le e^{-v^2/2}\le1 also give

For \zeta=1+.01u, consequences we shall use are

In particular 0<b<1/2 and l>0. Moreover,

The last two inequalities follow by inserting u\le1/36; after dividing by positive factors they reduce to 3.01+4\zeta^2u/\alpha<4.54 and (1-4.54u)(1+7u)\ge1. The two summands of h_0 in (A.2) have opposite signs and magnitudes at most .501/r and .43/r. Hence

This establishes the required bounds on p,h_0 in this tail.

For clarity, the remaining substitutions can be made with the following ratios, avoiding cancellation in p:

They imply the O row of the coefficient table in Lemma 4.1. For example, \mathsf D\le D_*(7+r^2)<.295(7+r^2), \mathsf L\le2\zeta/(\alpha r^3)<.60,

The bounds for z=l'/l, h_*=1+3b+kz, and \chi=l/d follow from

where d=r+q\ge r. Since d'<0, both d^{-1}(1+b) and d^{-1}(1-b)=k are increasing. Their product is M, proving the required logarithmic slope bound m_0=0.

It remains to check the matrix inequality in this tail. Write N for the matrix subtracted from Q in that inequality, so that

Equations (A.5) give \mathsf E<.05072. Using b<.028, K_0<.56, and J_0<.72 in (A.6) gives

Since Q_{11}=.43, Q_{12}=.13, and Q_{22}=2.72, the leading minor of Q-N-\operatorname{diag}(.13,.75) is positive, and its determinant is at least

This proves the local matrix bound on a\le-6.

The positive tail

For r=a\ge10 put u=r^{-2} and \theta=\log(r\sqrt{2\pi}). Gaussian tail integration gives

where

Let \mathcal U=u\partial_u, with \theta held fixed, and let \mathcal D=r\,d/dr=\partial_\theta-2\mathcal U. Define

Here B,J,R are auxiliary scalar functions used only in this appendix. Differentiating k gives

We first establish uniform bounds for these three derivatives:

Differentiation under the integral gives, for 0\le u\le.01,

Indeed the jth derivative is bounded in absolute value by (2j)!/2^j. To control the apparent quotient in G, use (F(u)-1)/u=\int_0^1F'(tu)\,dt. Since F\ge.99 and |1+\log F|\le1, differentiating (A.7) yields

The values F(0)=1, F'(0)=-1, G(0)=-1/2, and G'(0)=5/2 therefore give

with

The same bounds apply to \partial_\theta R_0 with g=3,c_0=90; all higher \theta derivatives vanish. For example, \mathcal U^2R_0=uR_0'+u^2R_0'' and \mathcal U^3R_0=uR_0'+3u^2R_0''+u^3R_0''', so these statements follow directly from Taylor’s theorem and the derivative bounds just given.

Expanding the three operators in (A.8), the linear-in-u errors are respectively

The errors contributed by R_0 are bounded respectively by

For example the second operator is 2-3\mathcal D+\mathcal D^2 and the third is 6-11\mathcal D+6\mathcal D^2-\mathcal D^3; substituting \mathcal D=\partial_\theta-2\mathcal U into (A.10) gives exactly the displayed constants.

Here is a rational way to check uniformity on the unbounded interval. Write x=\log(r/10)\ge0, so u=.01e^{-2x} and \theta=\theta_0+x, where 3.221<\theta_0<3.222. The terms in (A.11) decrease with x. For the linear-in-u terms, maximize a linear function times e^{-2x} at its endpoint or stationary point, using e^{-t}\le1/(1+t) for t\ge0. More explicitly, set

Upper bounds for the total errors are

These prove (A.9). The bound on \theta_0 itself can be checked with the logarithm series

applied to \theta_0=\tfrac12\log(200\pi). For a completely specified enclosure, use Machin’s identity \pi=16\arctan(1/5)-4\arctan(1/239) and the alternating arctangent sums through degrees 47 and 49. In the logarithm series first divide the argument by a power of two to place it in [1,2), and take N=35 for that argument and for \log2. These rational operations give 3.221523626198<\theta_0<3.221523626199.

Combining (A.9) with the preceding Taylor bounds on C gives the following estimates needed in the lemmas:

To see the ratio bounds directly, J>2\theta-4.13>0, R-1.7J>2.6\theta-8.171>0, and 3J-R>.86. For the bounds involving k and p, we have C\ge .99\theta-.5>0 by (A.7), whereas its Taylor expansion gives

Thus 0<C<\theta-.5<\theta-.475, and 0<uCJ<.075. In particular,

which proves .493<p<.5. The inverse-power majorants used for these consequences take their maxima at r=10, as follows by differentiating the linear or quadratic polynomials in \log(r/10) times the indicated inverse power of r. For example, u(\theta-.475)(2\theta-3.87) decreases because 1/(\theta-.475)+2/(2\theta-3.87)<2 at \theta=3.221 and hence for every larger \theta. Similarly B/r<.18, and consequently

These bounds prove the claimed estimates on p,h_0 and the T_1 row of the coefficient table; for example

They also give -.3<z<0 and

To check the remaining estimates involving d, let V have density F(u)^{-1}e^{-v-uv^2/2} on v>0. The Gaussian excess above r is V/r, so

Taylor’s lower bound in the numerator and F\le1 give \mathbb EV\ge1-3u and \mathbb EV^2\ge2(1-6u). Monotone weighting of the exponential density gives \mathbb EV\le1. Hence d\ge.97/r, \operatorname{Var}V\ge.88, and

Equation (A.2) now implies

This verifies the logarithmic slope required by Lemma 2.2.

For the matrix inequality, use \mathsf E\le p+.482(.75-.482) in (A.6). The first part of N_{11} increases with b on [.482,.5], and (p+v)^2/p increases with p when 0<v<p. Therefore

For the off-diagonal estimate use N_{12}=\nu_*/2+b(cK_0-\eta_*) and .367<K_0<.4. Thus Q-N-\operatorname{diag}(.022,.75) has positive leading minor and determinant at least

This finishes the positive tail.

Polynomial enclosures on the central interval

We next provide all data and rules for the finite certificate on [-6,10]. On each row of Table 1, write a=w+rx, -1\le x\le1. The corresponding closed intervals are consecutive and cover [-6,10]. The tabulated constants are rational seeds; we do not assume that they are Gaussian function values. The differential residuals and the two analytic anchors q(10) and k(-6) will certify the entire piecewise polynomial family, including all its initial coefficients. Construct polynomials P,\widehat k of degree 14 from their tabulated constant terms by the recurrences

The subscript denotes a coefficient, so the constant 1 contributes only when i=0. At each step only coefficients already computed occur on the right. The letters P,\widehat k,D denote polynomial approximations to q,k,d, respectively, throughout this central calculation.

For a polynomial A, let [A] denote the sum of the absolute values of its coefficients. Direct use of (A.13) gives

For each adjacent pair, the values of both P and \widehat k at their common endpoint differ by less than 2\cdot10^{-10}. Also

and the first-row value \widehat k(-1) differs from .16237766 by less than 3\cdot10^{-9}, while the last-row value P(1) differs from 10.098093234 by less than 2\cdot10^{-10}. These are rational coefficient checks. For example the first residual is bounded by

with missing coefficients taken as zero. The bound on D follows from the quartic lower-bound rule (A.21) below, after charging discarded coefficients to the remainder.

Here are analytic anchors for those coefficient checks. For any r>0, Taylor’s formula under the integral gives alternating upper and lower bounds for I_0(r) by

where (-1)!!=1. At r=10, use S_{11},S_{12} and q(10)=1/I_0(10); at r=6, use S_{13},S_{14} and |k(-6)-I_0(6)|\le2\varphi(6)/36. Exact substitution gives

The rough bound \varphi(6)<7\cdot10^{-9} used here follows, for example, from \sqrt{2\pi}>12/5 and e^{18}>\sum_{i=0}^{59}18^i/i!: division gives (5/12)/\sum_{i=0}^{59}18^i/i!<7\cdot10^{-9}.

These anchors and residuals imply the following uniform, rather than just endpoint, errors:

Indeed the equation for q-P has multiplier r(q+P-a)>0, so backward integration from a=10 is stable. Its absolute error increases by at most twice each bound in (A.14) and by each joining error. The resulting bound is

The equation for k-\widehat k has multiplier -rd<0, so forward integration from a=-6 is stable. The additional forcing is bounded by r[\widehat k]e_q. The portions with [\widehat k]<2 and [\widehat k]<5.6 have total lengths 6.5 and 9.5, respectively. Thus an upper bound for its accumulated error is

Construct the following full polynomials before truncating any of them:

They approximate b,l,l'. Their error bounds, obtained directly from (A.1) and (A.17), are

The order in (A.18) matters for the strength of the certificate: form B,J,Y with the full degree-14 inputs, and only then apply the quartic arithmetic described next.

The rational positivity certificate

An enclosure (A,e) means that the function in question differs from A(x) by at most e on [-1,1]. Replace a polynomial by its terms of degree at most four and add the sum of the absolute values of all discarded coefficients to e. Addition and subtraction add error bounds. For a product use the coefficient convolution and the error

then truncate and charge the discarded coefficients as before. Scalar multiplication scales the error by the absolute value of the scalar. Initialize this arithmetic with

using the full polynomials in (A.18). Truncate each of these six initial enclosures to degree four, charging discarded coefficients to its error, before evaluating the groups below. For a resulting quartic A(x)=\sum_{j=0}^4A_jx^j, a rational lower bound is

This is the Bernstein coefficient bound on each of [0,1] and [-1,0]: on x=\sigma t the displayed coefficients are precisely the coefficients in the degree-four Bernstein basis, whose basis functions are nonnegative and sum to one.

We list every polynomial to which this rule is applied. In a given row, use the corresponding bin in Lemma 4.1 to define the bounds

The constant m_0 is from Lemma 2.2. The triples (z_{\min},z_{\max},\chi_{\max}) are (-.08,.70,.070) for rows to the left of -1, and (-.475,0,.097) for rows to the right. Within the quartic arithmetic set

The four groups are as follows:

Products are evaluated from left to right, with repeated multiplication for powers. In the last expression the product in the middle is 2\mathsf E\,o\,f_1. The bounds below are the minimum over all members of a group, so positivity proves every inequality listed in it.

Table 2 is obtained by the rational operations (A.13), (A.18), (A.19), (A.20), and (A.21), with no additional function evaluations. The accompanying checker reproduces all 44 bounds, all 22 differential residual bounds, the 20 joining bounds, the anchor checks, and the coefficient norm bounds used to obtain (A.17). The same endpoint enclosures give q(-1)>.287, q(-2)<.056, q(-3)<.005, and q(4)>4, as used in the secant argument. The logarithm series above also certifies

We explain why these finite checks finish the proof. Group \mathcal G_1 gives 0<b<1/2, l>0, the bounds for p,h_0, and the bounds for b,\mathsf L. Its strict positive margins give p>0. Group \mathcal G_2 gives the bounds for z,h_*,\chi after division by l or d, and the logarithmic slope bound by division by d^2(1-b^2). Group \mathcal G_3 gives the remaining coefficient bounds after division by p; the O bound for \mathsf D instead follows from D_*<.295 and the lower bound on p. Finally

Its leading minor and determinant, multiplied by p, are exactly the two expressions in \mathcal G_4. Their positivity proves the matrix inequality. Continuity supplies every bin endpoint, including the stronger of the two adjacent logarithmic slope bounds. Together with the two tails, this proves both lemmas on the whole real line.□