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 \[\mathbb E_a X_a=d(a),\qquad \operatorname{Var}(X_a)=1-q(a)d(a).\] 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 \[S_a(u)=\frac{\Phi(-a-u)}{\Phi(-a)},\qquad u\ge0,\] 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 \[\frac12\mathbb E_a X_a^2 =\int_0^\infty uS_a(u)\,du \le\left(\int_0^\infty S_a(u)\,du\right)^2=d(a)^2.\] 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 \[W_\theta(y)=(1-\theta)|y|^2/2+\theta V(y),\qquad d\mu_\theta=Z_\theta^{-1}e^{-W_\theta(y)}\,dy,\] 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 \[ c_0I\le C_\theta\le C_0I \tag{15}\] 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 \[F(\theta,P)=\mathbb E q^{-1}(J_{\theta,P}).\] Contraction in both directions gives \[ \frac{\lambda_{\min}(P)}{\sqrt{a_+}}I \le J_{\theta,P}\le\frac{\|P\|_{\mathrm{op}}}{\sqrt{a_-}}I. \tag{16}\]

We first verify continuity of \(F\). Restrict \(P\) to any compact subset of the positive definite cone. The transports have common Lipschitz constants by (16); 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 \[\log\det J_j\le\log\det J+ \operatorname{tr}\bigl(J^{-1}(J_j-J)\bigr)-c|J_j-J|_{\mathrm F}^2.\] 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 \[\mathbb E J_{\theta,P}\ge q(0)I.\] 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 \[P C_\theta P\ge(\mathbb E J_{\theta,P})^2\ge q(0)^2I.\] Together with (15), this bounds \(P^{-1}\) uniformly. The lower estimate in (16) now gives a uniform lower bound \(A\ge a_*I\), where we may take \(a_*\le0\). Since \(q'\le1\), \[q(a)\le q(a_*)+a-a_*\quad(a\ge a_*), \qquad \mathbb E\operatorname{tr}J_{\theta,P} \le m\bigl(q(a_*)-a_*\bigr).\] On the other hand, Gaussian Poincaré and (16) imply \[c_0\|P\|_{\mathrm{op}}^2 \le\operatorname{tr}(P C_\theta P) \le\mathbb E\operatorname{tr}J_{\theta,P}^2 \le\frac{\|P\|_{\mathrm{op}}}{\sqrt{a_-}} \mathbb E\operatorname{tr}J_{\theta,P}.\] 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 \[ \log\det J=V(\nabla\phi)-|x|^2/2-(m/2)\log(2\pi). \tag{19}\] Differentiating twice gives (18). The divergence of the cofactor matrix of a Hessian vanishes. Applying this identity to \(J\), and then differentiating (19) once, shows that \[\mathcal L u=e^{|x|^2/2}\operatorname{div} \bigl(e^{-|x|^2/2}J^{-1}\nabla u\bigr).\] Thus, for smooth \(u\) and compactly supported smooth \(v\), \[ \mathbb E(v\mathcal Lu) =-\mathbb E\langle\nabla v,J^{-1}\nabla u\rangle. \tag{20}\]

Choose \(0<\lambda\le\Lambda<\infty\) such that \(\lambda I\le J\le\Lambda I\). Taking the trace in (18), and using \(D^2V\ge0\), gives \[\mathcal L(\operatorname{tr}J) \ge-m+\Lambda^{-2}\sum_r|J_r|_{\mathrm F}^2.\] For a compactly supported cutoff \(0\le\chi\le1\), integration by parts and \(|\nabla\operatorname{tr}J|\le\sqrt m\,|\partial J|\) imply \[\begin{align*} \Lambda^{-2}\mathbb E\chi^2|\partial J|^2 &\le m+2\lambda^{-1}\sqrt m\, \mathbb E\bigl(\chi|\nabla\chi|\,|\partial J|\bigr)\\ &\le m+\tfrac12\Lambda^{-2}\mathbb E\chi^2|\partial J|^2 +2m\Lambda^2\lambda^{-2}\mathbb E|\nabla\chi|^2. \end{align*}\] 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 (20) 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 (18) by \(F(J)\) in trace and integrate by parts. Lemma 6 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 \[\sum_i F(\lambda_i)\operatorname{tr}(J^{-1}J_iJ^{-1}J_i) =\sum_{i,j,r}\frac{F(\lambda_i)}{\lambda_j\lambda_r}\Theta_{ijr}^2,\] where we used the symmetry of \(\Theta\). The term obtained by integration by parts is \[\sum_r\lambda_r^{-1}\operatorname{tr}\bigl((\partial_rF(J))J_r\bigr) =\sum_{i,j,r}\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\Theta_{ijr}^2.\] The remaining term is \(\operatorname{tr}(F(J)JD^2VJ)\ge0\). Discarding it gives \[ \mathbb E\operatorname{tr}k(A)^2 \ge\mathbb E\sum_{i,j,r}\Theta_{ijr}^2 \left(\frac{F(\lambda_i)}{\lambda_j\lambda_r} +\frac{F[\lambda_i,\lambda_j]}{\lambda_r}\right). \tag{24}\] Only pointwise changes of basis are made here; no eigenvectors are differentiated.

Regard \(M\) as a function of \(\lambda=q(a)\). The identities (7) give \[ \frac{d}{d\lambda}(\lambda F(\lambda)) =k^2+\frac{2kb}{d} =\frac{1-b^2}{d^2}=M. \tag{25}\] 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 (24) to \[\frac{\lambda_i+\lambda_j}{2}\, M^{\mathrm{ar}}_{\lambda_i\lambda_j}.\] 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 4 makes \(M\) increasing, so \[M^{\mathrm{ar}}_{uv}\le M^{\mathrm{ar}}_{uw} \le M^{\mathrm{ar}}_{vw}.\] 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 \[ \mathbb E\operatorname{tr}k(A)^2 \ge\mathbb E\sum_{i,j,r} \frac{\Theta_{ijr}^2}{\lambda_i\lambda_j\lambda_r}\, \lambda_r M^{\mathrm{ar}}_{\lambda_i\lambda_j}. \tag{26}\]

We now fix a pair of eigenvalues. Bars in the rest of the proof denote uniform averages in \(\log\lambda\) over their interval. Set \[x=\delta_{ij}/2,\qquad w_0=\frac{x}{\sinh x},\qquad m_*=m_*(a_i,a_j),\] 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 (26), are \[ \lambda_r w_0^2\overline{D_0}^{\,2},\qquad \lambda_r w_0^2\overline{bD_0}^{\,2}. \tag{27}\] 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 \[ \overline M\ge M_1:=\overline{D_0}^{\,2}-\overline{bD_0}^{\,2} \ge\tfrac34\overline{D_0}^{\,2}. \tag{28}\]

We also need a quantitative comparison of the two averaging measures. On the logarithmic interval, (10) says that \(M(\lambda)\lambda^{-m_*}\) is increasing. Exponential weighting and the same covariance inequality give \[ \frac{M^{\mathrm{ar}}_{\lambda_i\lambda_j}}{\overline M} \ge\frac{\operatorname{shc}((1+m_*)x)} {\operatorname{shc}(x)\operatorname{shc}(m_*x)}, \qquad \operatorname{shc}(x)=\frac{\sinh x}{x}. \tag{29}\] 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 \[x\coth x+\frac{x^2}{\sinh^2x}\ge2\] 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 \[\begin{align*} \frac{\operatorname{shc}((1+m_*)x)} {\operatorname{shc}(x)\operatorname{shc}(m_*x)} &=1+\frac{m_*A_0(x)+A_0(m_*x)}{1+m_*}\\ &\ge1+m_*(1-w_0^2). \tag{30}\end{align*}\] Every formula here extends continuously to \(m_*=0\) or \(x=0\).

Combining (28)–(30), the amount by which the coefficient in (26) exceeds the difference of the two coefficients in (27) is at least \[\lambda_r(1+m_*)(1-w_0^2)M_1.\] Relative to the coefficient of \((Z_r)_{ij}^2\), this is at least \[\frac34(1+m_*)(w_0^{-2}-1) \ge\frac{1+m_*}{16} \left(\delta_{ij}^2+\frac{\delta_{ij}^4}{30} +\frac{\delta_{ij}^6}{1680}\right).\] The last inequality is the Taylor expansion of \((\sinh x/x)^2\), whose omitted coefficients are positive. Substituting in (26) and summing proves (23). ◻

Proof. Put \(F=\nabla\phi\). For two degree-\(d\) components of \(F\), Gaussian integration by parts gives \[\sum_\ell\mathbb E[(\partial_\ell F_{i,d})(\partial_\ell F_{j,d})] =d\,\mathbb E[F_{i,d}F_{j,d}].\] Differentiation lowers the degree by one, and \(R\) therefore cancels the factor \(d\). Summing over \(d\ge1\) proves \(C_{ij}=\sum_\ell\mathbb E[J_{i\ell}R J_{j\ell}]\). Symmetry of \(J\) gives the first matrix identity in (34). Expanding the row Gram matrix of \(q-(I+N)k\) gives cross terms \(-2\mathbb E(qk)=-2\mathbb Et\). The remaining term is \[\mathbb E\bigl[k(I+N)k\bigr] =\mathbb E\left(k^2+\sum_r K_r^2\right),\] 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 (35) for \(S=I\), 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 (40) give \[\|F_0+(B-I)l\|_S^2 =\|F_0\|_S^2+\frac1\alpha\tau(Hl^2) -\frac2\alpha\tau(TF_0l).\] For the cross term, the order of the factors follows from \(\mathop{\mathrm{tr}}(F_0^tTl)=\mathop{\mathrm{tr}}(TF_0l)\), by transposition and cyclicity. Integration by parts on the linear part of \(A\) yields \[\tau(TAf)=\tau\left[T\left(Yf+\sum_rX_r\partial_rf\right)\right], \qquad \partial_rf=X_rf'+G_rf'+Z_r\circ\Delta f.\] 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 \(K_r=X_rb+D_r=bX_r+D_r^t\): \[ \tau\left(T\sum_rK_r^2\right) =\tau(TbBb)+2\tau\left(Tb\sum_rX_rD_r\right) +\tau\left(T\sum_rD_r^tD_r\right). \tag{50}\] The two cross terms agree under transposition inside the trace. Moving \(b\) past \(T\), rather than past \(X_r\), gives \[\tau(TbX_rD_r)=\tau(TX_rD_rb)+\tau([T,b]X_rD_r).\] The terms containing \(B-I\) now combine as \[\tau\bigl(T(B-I)f'\bigr)+\tau(Tb^2-TbBb) +\frac1\alpha\tau(Hl^2) =-\tau(Hp)-P_T.\] In the other terms, substituting (42) 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 \(\partial^*g=Y\), \[\tau(TYk)=\sum_r\tau(TK_rg_r).\] 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 \[-\tau(TD_r^tD_r)+c\tau(TD_r^tg_r) =-\tau(Tw_r^tw_r)+\gamma^2\tau(Tg_r^2).\] This proves (49). ◻

Proof. Work pointwise and diagonalize \(S\). The relation \(B=\alpha^{-1}(S^{-1}-I/s)\) gives \[ P_T=\frac1{2\alpha} \bigl\|S^{-1/2}[T,b]S^{-1/2}\bigr\|^2. \tag{52}\] 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 (52) 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 (51) by \[2\alpha\bigl\|\operatorname{skew}(S^{1/2}FS^{1/2})\bigr\|^2.\]

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 \(i\)th 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 \[\begin{pmatrix} 1-su_iu_j&-s\sqrt{u_iu_j(1-u_i)(1-u_j)}\\ -s\sqrt{u_iu_j(1-u_i)(1-u_j)}&1-su_iu_j \end{pmatrix}.\] 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 (51). ◻

Proof of Theorem 8. Apply Lemma 10 in (41). Bound its last line by Lemma 11, its first square by (53), and its linear \(T\) term by (55). 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 (56), by cyclicity of the trace; no commutation with \(Y\) is required. Set \(\Omega_B=\sum_{i=1}^m\omega(\lambda_i)\). Consequently, \[\begin{align*} \mathcal{D}\le {}&-V_0-\mathcal C+\frac{69}{500}\|Y\|^2 -\frac{\Omega_B}{1000} +\|D-\beta g\|_S^2 +\left(1+\frac s8\right)\|D-\gamma g\|^2 +\frac1\alpha\|L\|^2\\ &\hspace{22mm}+\gamma^2\tau\left(T\sum_rg_r^2\right) +\frac{\|Mp^{-1/2}\|^2}{4(1-\epsilon)}. \tag{58}\end{align*}\] Use \(S\le sI\) and \(T\le(s-1)I\). 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 (57); right multiplication by \(p^{-1/2}\) gives the denominator \(p_j\) for entry \((i,j)\). Proposition 13 now bounds the last five terms of (58) by \[\mathbb E\sum_{i,j,r}Q\bigl((G_r)_{ij},(g_r)_{ij}-(G_r)_{ij}/2\bigr) +\mathcal C.\] The separation surplus is the same \(\mathcal C\) as in Proposition 7: 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 \[ Q(1,u-1/2)+\frac7{50}u\le1-u, \qquad 0\le u\le\frac12. \tag{59}\] 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 (58) therefore gives \[\mathcal{D}\le-\frac1{500}\|Y\|^2-\frac{\Omega_B}{1000},\] which is (32). ◻

Proof of Proposition 12. By \(0<b<1/2\) and (61), \[ s\left((b-\beta)^2+ \frac{\alpha h_0^2}{4(1-\epsilon-\kappa)p}\right) \le s\left(.28^2+\frac{.075\cdot .67^2}{4\cdot .36}\right)<.111. \tag{66}\] It remains to bound the terms involving \(T-\log S\) while retaining a negative multiple of the weight \(\omega\) in (31).

Fix a unit eigenvector \(v\) of \(B\) with eigenvalue \(\lambda\ge0\). The corresponding eigenvalues of \(S,T,H,\rho\) are \[S=(1/s+\alpha\lambda)^{-1},\quad T=S-1,\quad H=\alpha S(\lambda-1)^2,\quad \rho=\epsilon+\kappa S/s.\] 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 \[ \int a\,d\nu_v(a)=0,\qquad \int a^2\,d\nu_v(a)=\lambda+r, \qquad r=v^{\mathsf T}\mathbb E(Y^2)v\ge0. \tag{67}\] Indeed, orthogonality of the Gaussian chaoses gives \(\mathbb EA^2=B+\mathbb EY^2\), including the vanishing of both matrix cross terms.

For \(0<u\le.13\), define \[n_u=\frac{u^{3/2}}{2},\qquad d_u=\frac{u^3}{16(.46-u)}.\] We claim the following quadratic minorant: \[ p(a)\ge u+n_ua-d_ua^2\qquad(a\in\mathbb R). \tag{68}\] 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 \[1-(1-x/2)(x^2+7u) \ge \frac{x(x-1)^2}{2}+.045(2-x)\ge0.\] For \(0\le a\le4\), the right side of (68) 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 (68) follows from (61) in all three ranges.

Choose \[u=\min(.13,1.28/\sqrt\lambda),\] with \(u=.13\) when \(\lambda=0\). Since \(\lambda u^2\le1.28^2\), integration of the minorant against (67) yields \[ \int p\,d\nu_v\ge u-d_u(\lambda+r) \ge .689u-d_ur. \tag{69}\] The second inequality uses \(1-1.28^2/[16(.46-.13)]>.689\). We also have \[ H\rho d_u\le .027. \tag{70}\] 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, \[H\rho d_u\le \frac{.64\cdot1.28^2\cdot.13}{16(.46-.13)}<.02582;\] 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 (69). Set \[F_0(S)=\frac{S(S-1-\log S)}{(1-S)^2},\qquad F_0(1)=\tfrac12.\] For \(0<S\le s\) and \(\lambda=(1/S-1/s)/\alpha\), we claim \[ \frac{\alpha F_0(S)}{\epsilon+\kappa S/s} \le\min(.089,.88/\sqrt\lambda)\le .688u, \tag{71}\] where the second bound in the minimum is interpreted as \(+\infty\) at \(\lambda=0\). First, \[F_0(S)=\int_0^1 \frac{Sx}{1-x+xS}\,dx\] 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 \[-\log(.28)=2\sum_{j=0}^{\infty}\frac{(9/16)^{2j+1}}{2j+1},\] using twelve terms and the geometric bound for the remainder. Coefficient comparison now gives \[.075(.16+.52S)<.089(.15+.45325S),\] which proves the first bound in the minimum.

For the second bound we use \[ \frac{F_0(S)}{\sqrt S}\le .478+1.4S\qquad(S>0). \tag{72}\] Here is a direct verification, including the point \(S=1\) by continuity. For \(x>0\), let \[J(x)=\frac{(.478+1.4x^2)(1-x^2)^2}{x} -(x^2-1-2\log x).\] Then \[x^2J'(x)=P(x):=-.478+2x+.444x^2-2x^3-6.966x^4+7x^6.\] 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 (72) away from \(x=1\).

Using \(\alpha\lambda=1/S-1/s\), inequality (72) gives \[\frac{\alpha F_0(S)\sqrt\lambda}{\rho} \le \sqrt\alpha\sqrt{1-S/s}\, \frac{.478+1.4S}{.15+.45325S} \le \frac{\sqrt{.075}\cdot .478}{.15}<.88.\] 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 (71).

For \(\lambda\ne1\) we have \[\frac{T-\log S}{H\rho}=\frac{\alpha F_0(S)}{\rho}.\] Combining (69)–(71), and treating \(\lambda=1\) by continuity, gives \[T-\log S-H\rho\int p\,d\nu_v \le -.001H\rho u+.027r.\] The definitions give \(H\rho u=\omega(\lambda)\). Sum over an orthonormal eigenbasis of the matrix \(B\), and add (66). This summation evaluates \(\tau(H\rho p(A))\) without requiring \(p(A)\) to commute with \(B\). Likewise, (66) may be integrated against the positive semidefinite matrix \(Y^2\) without a commutation assumption. Since \(\sum_v r=\lVert Y\rVert^2\), the result is \[(.027+.111)\lVert Y\rVert^2-.001\sum_i\omega(\lambda_i) =\frac{69}{500}\lVert Y\rVert^2-\frac1{1000}\sum_i\omega(\lambda_i),\] as asserted. ◻

Proof of Proposition 13. Fix two endpoints \(a_-\le a_+\) and write \[\delta=\log q(a_+)-\log q(a_-),\qquad m_*=m_0(a_-).\] When the endpoints coincide, all secant increments vanish, and (63) 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 13 is exactly the two matrices subtracted from \(Q\) in (63). For an increment vector \((x,y,w)\), define \[\begin{align*} \mathcal N_j(x,y,w)^2={}&A_*x^2+y^2/\alpha+\mathsf D_jw^2\\ &+\frac{((A_*b_j-\eta_*/2)x+\mathsf L_jy+K_{0j}w)^2}{\ell_{1j}} +\frac{(-\eta_*x+J_{0j}w)^2}{.75}. \tag{73}\end{align*}\] Completing squares against the surplus \(\ell_{1j}G^2+.75h^2\) in (63) shows that the full endpoint form is bounded above by \[ Q(G,h)+Z^2\mathcal N_j(\Delta k,\Delta b,\psi_j)^2. \tag{74}\] Indeed, the two coefficients of \(2GZ\) and \(2hZ\) are precisely the numerators of the two squared terms in (73). 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 \[ v_j(a)=\chi(a)(1,z(a),W_j(a)),\qquad W_j(a)=h_*(a)-2b_j-2\mathsf L_jz(a). \tag{75}\] 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 \[F_j(a)=\mathcal N_j(v_j(a))^2.\]

We record the bounds furnished by Lemma 14. If the endpoint \(a_j\) belongs to \(R_1,U_1,V_1,T_1\), respectively, then \[ F_j(a)\le .295,\quad .23,\quad .23,\quad .39 \qquad\hbox{for all }a. \tag{76}\] 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 \[ \frac{F_j(a)}{\chi(a)^2}\le \begin{cases} 20+1.04(7+t_j^2),&a\le-1,\\ 12+1.63(7+t_j^2),&a\ge-1. \end{cases} \tag{77}\] For clarity, the complete finite arithmetic behind these bounds is given by the following table. Its three entries bound, in order, \[|W_j|,\qquad |A_*b_j-\eta_*/2+\mathsf L_jz+K_{0j}W_j|,\qquad |-\eta_*+J_{0j}W_j|.\] \[ \begin{array}{c|cc} &a\le-1&a\ge-1\\\hline O&(1.867,.52,1.314)&(2.34,.55,1.155)\\ P_1&(1.766,.49,1.813)&(2.564,.433,1.155)\\ R_1&(1.681,.664,2.057)&(2.621,.420,1.155)\\ U_1&(1.55,.847,2.101)&(2.461,.615,1.155)\\ V_1&(1.263,.912,1.787)&(1.885,.770,1.205)\\ T_1&(.913,.869,1.386)&(1.071,.840,1.251) \end{array} \tag{78}\] To verify it, substitute the intervals in (62) and (65); 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 \[(A_*-2K_{0j})b_j-\eta_*/2+K_{0j}h_*+ (1-2K_{0j})\mathsf L_jz.\] Substituting these three bounds into (73), together with the bounds for \(\mathsf D_j,\ell_{1j},z,\chi\), proves (76)–(77). All these are finite rational comparisons; the exact verifier accompanying Appendix 8 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 3 gives \[0\le \frac{d\log D_0}{d\xi}=\frac{1-qd}{d^2}\le1.\] 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 \[ \tfrac12\le u_-\le\tfrac12+\delta/12, \qquad u_-^2+u_+^2\le\tfrac12(1+\delta^2/36). \tag{79}\] 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 \[\mathcal N_-(\Delta k,\Delta b,\psi_-)\le\sqrt{P_-}\,\delta u_-, \qquad \mathcal N_+(\Delta k,\Delta b,\psi_+)\le\sqrt{P_+}\,\delta u_+.\] Since \(u_-\ge u_+\ge0\), \[P_-u_-^2+P_+u_+^2 \le\max\left(P_-,\frac{P_-+P_+}{2}\right)(u_-^2+u_+^2).\] Averaging (74) and applying (79) therefore proves the desired estimate whenever \[ \max\left(P_-,\frac{P_-+P_+}{2}\right)\le \Theta_{m_*}(\delta):= \frac{1+m_*}{4}\, \frac{1+\delta^2/30+\delta^4/1680}{1+\delta^2/36}. \tag{80}\] 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 (80) 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 \[.61-\frac{(.61-.42)1.62}{\delta},\qquad P_-=(.61-.3078/\delta)^2.\] 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 \[\max(.300787,(.300787+.39)/2)<.345394 <\Theta_{.19}(4).\] For \(\delta\ge5\), use \(P_-\le.61^2=.3721\) and \[\max(.3721,(.3721+.39)/2)=.38105<\Theta_{.19}(5).\] 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, \[t_j^2\le a^2+2\delta,\qquad |a|\chi(a)\le .09.\] 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 (77) gives \[ F_j(a)\le .144+.011\delta. \tag{81}\] 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 \[ \Theta_0(\delta)\ge .25+.00260\delta^2. \tag{82}\] This follows on clearing denominators and using \(\delta^2\ge16\). The quadratic \(.106-.011\delta+.00260\delta^2\) is strictly positive, so (80) 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 (77) yields \(F_-(a)\le .236+.031\delta\). On every partial interval from the lower endpoint, Jensen’s inequality therefore permits the uniform bound \[P_-=.236+.031\delta-\frac{4}{\delta}(.092+.020\delta) =.156+.031\delta-.368/\delta.\] For partial intervals contained below \(-1\), the same bound follows from (81), because \(\delta\ge4\). By (82), \[\Theta_0(\delta)-P_- \ge .094-.031\delta+.00260\delta^2+.368/\delta>0;\] 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 (80) 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 \[|A|\leq e^{M+1}\int e^{-V}.\] 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 \[M+1<V(y)\leq \left(1-\frac Rd\right)V(x_0)+\frac RdV(x).\] 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 (83), after enlarging \(b\).

The moment assertion follows from \(e^{-V(x)}\leq e^b e^{-a|x|}\). For the entropy assertion, the elementary bound \[ |u|e^{-u}\leq C_b e^{-u/2}\qquad(u\geq-b) \tag{84}\] holds because \(|u|e^{-u/2}\) is bounded on \([-b,\infty)\). Substitute \(u=V(x)\) and use (83). ◻

Proof. Fix \(a,b\) from Lemma 15. For each positive integer \(j\), define the Moreau envelope (Moreau 1965, sec. 7) \[ Q_j(x)=\inf_{y\in\mathbb R^m}\left\{V(y)+\frac j2|x-y|^2\right\}. \tag{87}\] 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, \[ Q_j(x)\geq a|x|-b-\frac{a^2}{2j}. \tag{88}\] 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 \[\langle x-x',p_j(x)-p_j(x')\rangle \geq |p_j(x)-p_j(x')|^2.\] 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 \[\nabla Q_j(x)=j(x-p_j(x)).\] 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 (87) 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 \[|x-p_j(x)|^2\leq\frac{2(Q_j(x)+b)}j\longrightarrow0.\] Lower semicontinuity then gives \(V(x)\leq\liminf_j V(p_j(x))\leq\lim_jQ_j(x)\). This proves \[ Q_j(x)\longrightarrow V(x)\quad\hbox{for every }x, \tag{89}\] 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 \[ V_j=Q_j*\eta_{1/j}+\frac{|x|^2}j. \tag{90}\] Convolution preserves convexity and the Lipschitz bound on the gradient; hence (85) holds. Evenness gives \(\int z\eta(z)\,\,\mathrm dz=0\), so Jensen’s inequality and (88) yield the common bound \[ V_j(x)\geq Q_j(x)\geq a|x|-B, \qquad B=b+\frac{a^2}2. \tag{91}\]

Smoothing at this scale does not affect the limit. The \(j\)-Lipschitz bound on the gradient gives \[Q_j(x-z)\leq Q_j(x)-\langle\nabla Q_j(x),z\rangle+\frac j2|z|^2.\] Integrating against \(\eta_{1/j}\) cancels the linear term, so \[0\leq (Q_j*\eta_{1/j})(x)-Q_j(x)\leq\frac1{2j} \qquad(x\in\mathbb R^m).\] Together with (89), this proves \(V_j(x)\to V(x)\) everywhere.

Now (91) dominates \((1+|x|^2)e^{-V_j(x)}\) by an integrable function independent of \(j\). Applying (84) with \(B\) in place of \(b\) also gives \[|V_j(x)|e^{-V_j(x)}\leq C e^{-a|x|/2}.\] 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, \[h(f_j)=\frac1{Z_j}\int V_je^{-V_j}+\log Z_j,\] which proves all of (86). ◻

Proof of the inequality in Theorem 2. Apply Theorem 8 to the normalized potentials \(V_j+\log Z_j\) of Lemma 16. Their Hessians satisfy (85), so \[h(f_j)\geq m+\frac12\log\det\mathop{\mathrm{Cov}}(f_j).\] The covariance of \(f\) is positive definite: zero variance in a nonzero direction would concentrate this Lebesgue density on an affine hyperplane. Thus (86) 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 \[ (\partial_rJ)e_s=(\partial_sJ)e_r \qquad(1\le r,s\le m). \tag{93}\] The scalar function \(q\) is real analytic near zero. Since \(A(0)=0\), its convergent power series, interpreted by matrix multiplication, gives \[J(x)=q(0)I+q'(0)A(x)+\tfrac12q''(0)A(x)^2+O(|x|^3)\] near zero; the differentiated remainder is \(O(|x|^2)\). The identity \(q'(a)=q(a)(q(a)-a)\) yields \[q(0)=\sqrt{\frac2\pi},\qquad q'(0)=\frac2\pi>0,\qquad q''(0)=\sqrt{\frac2\pi}\left(\frac4\pi-1\right)>0.\] Thus (93), which is a pointwise identity between analytic functions near zero, has constant term \[ X_re_s=X_se_r. \tag{94}\] Equivalently, \(A(x)z=A(z)x\) for all \(x,z\in\mathbb R^m\); in particular, \(A(x)e_s=X_sx\). Its degree-one term is \[(X_rA(x)+A(x)X_r)e_s =(X_sA(x)+A(x)X_s)e_r.\] The terms with leftmost factor \(A(x)\) cancel by (94). The remaining equality is \((X_rX_s-X_sX_r)x=0\) near zero, and hence \[ X_rX_s=X_sX_r\qquad(1\le r,s\le m). \tag{95}\]

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 \[\widetilde A(y)=U^{\mathsf T}A(Uy)U =\sum_{i=1}^m y_i\widetilde X_i, \qquad \widetilde X_i=U^{\mathsf T} \left(\sum_{r=1}^m U_{ri}X_r\right)U\] 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 \(i\)th diagonal entry of \(\widetilde X_i\) can be nonzero. The sum-of-squares identity makes that entry a sign: \[\widetilde X_i=\sigma_i e_ie_i^{\mathsf T}, \qquad \widetilde A(y)=\operatorname{diag}(\sigma_1y_1,\ldots,\sigma_my_m).\] It follows that \[D_y\bigl(U^{\mathsf T}F(Uy)\bigr) =\operatorname{diag}\bigl(q(\sigma_1y_1),\ldots,q(\sigma_my_m)\bigr).\] 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 \(\mathbb R^m\) and is therefore almost everywhere constant. This proves (92).

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 \[u_i^{\mathsf T}(q(M)-q(N))v_k =(q(\lambda_i)-q(\mu_k))\,u_i^{\mathsf T}v_k,\] 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 (96), because the Frobenius norm is unchanged by orthogonal changes of the row and column bases separately. ◻

Proof of the equality statement in Theorem 2. Let \(f\) be a log-concave probability density on \(\mathbb R^m\) with \(\delta(f)=0\). Choose the approximating densities \(f_j\) from Lemma 16. Write \(a_j,a\) for their means and the mean of \(f\), and \(C_j,C\) for their covariance matrices. That lemma gives \[ \lVert f_j-f\rVert_{L^1(\mathbb R^m)}\longrightarrow0,\qquad a_j\longrightarrow a,\qquad C_j\longrightarrow C,\qquad \delta(f_j)\longrightarrow0. \tag{97}\] 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 5 to \(f_j\). Let \(P_j\) be its positive definite normalizing matrix and let \(F_j\) transport \(\gamma_m\) to \((P_j)_\#(f_j(x)\,\,\mathrm dx)\). In the notation of Theorem 8, put \[\begin{gathered} J_j=DF_j,\qquad A_j=q^{-1}(J_j),\qquad X_{j,r}=\mathbb E[x_rA_j(x)],\\ A_j(x)=\sum_{r=1}^m x_rX_{j,r}+Y_j(x),\qquad B_j=\sum_{r=1}^m X_{j,r}^2. \end{gathered}\] The entropy gap is affine invariant. Theorem 8 and (97) therefore imply \[ \lVert Y_j\rVert_{L^2(\gamma_m)}\longrightarrow0, \qquad \sum_{i=1}^m \omega(\lambda_i(B_j))\longrightarrow0. \tag{98}\] 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 \(\sum_r|X_{j,r}|_{\mathrm F}^2=\mathop{\mathrm{tr}}B_j\), 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 (98) gives \[ \sum_{r=1}^mX_r^2=I,\qquad A_j\longrightarrow A(x):=\sum_{r=1}^m x_rX_r \quad\text{in }L^2(\gamma_m). \tag{99}\] Lemma 18 now gives \[ J_j=q(A_j)\longrightarrow J:=q(A) \quad\text{in }L^2(\gamma_m). \tag{100}\]

To obtain a limiting map, remove the irrelevant translations: \(\widehat F_j=F_j-\mathbb EF_j\). Each \(\widehat F_j\) lies in \(H^1(\gamma_m;\mathbb R^m)\): its law has finite second moments and, for this fixed \(j\), its Jacobian has the bounds (13) applied to the transformed target. Gaussian Poincaré, applied componentwise to the centered difference, gives \[ \lVert \widehat F_j-\widehat F_k\rVert_{L^2(\gamma_m)}^2 \le \lVert J_j-J_k\rVert_{L^2(\gamma_m)}^2. \tag{101}\] Consequently \(\widehat F_j\) converges strongly in Gaussian \(H^1\) to a centered map \(\widehat F\) with weak Jacobian \(D\widehat F=J\). Indeed, (101) and (100) 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, \(\widehat F\in W^{1,2}_{\mathrm{loc}}(\mathbb R^m;\mathbb R^m)\). Lemma 17 applies to (99) 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 \[ D_j:=\mathop{\mathrm{Cov}}((\widehat F_j)_\#\gamma_m) =P_jC_jP_j^{\mathsf T}\longrightarrow I. \tag{102}\] 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 \(j\), \[cI\le C_j\le C_0I,\qquad dI\le D_j\le DI.\] Congruence by \(P_j\) then yields \[ \frac d{C_0}I\le P_jP_j^{\mathsf T}\le\frac Dc I. \tag{103}\] Hence \(P_j\) and \(P_j^{-1}\) are bounded. Pass to a further subsequence with \(P_j\to P\); the lower bound in (103) makes \(P\) invertible.

The law of \(\widehat F_j\) is the image of \(f_j(x)\,\,\mathrm dx\) under \(x\mapsto P_j(x-a_j)\). For a bounded continuous function \(\psi\), \[\begin{align*} &\left|\int\psi(P_j(x-a_j))f_j(x)\,\,\mathrm dx -\int\psi(P(x-a))f(x)\,\,\mathrm dx\right|\\ &\quad\le\lVert \psi\rVert_{\infty}\lVert f_j-f\rVert_{L^1} +\int\left|\psi(P_j(x-a_j))-\psi(P(x-a))\right|f(x)\,\,\mathrm dx \longrightarrow0. \end{align*}\] The first term tends to zero by (97); 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 \(f(x)\,\,\mathrm dx\) 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 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 \(\delta=0\). ◻