The Mahler Conjecture for General Convex Bodies

2. Theorem 1.1 — Cone-product bound and equality

Proof target：Theorem 1.1 · p. 1

Proof route. Set $m=n+1$ and transfer the volume product to Laplace integrals on positively dual cones. Simultaneously normalize coordinates and Gaussian covariance, then apply change of variables to maps obtained by integrating projections over layers. Split the remaining error into a linear Gaussian matrix field and a positive layer defect. A strict interval inequality between eigenvalues absorbs the error. Equality eliminates the defects and separates the projection into individual coordinates in one fixed basis.

1. Fix the center and convert volume to Laplace integrals

Fix any $z_0\in\operatorname{int}K$. Let $C=\{(tx,t):t\ge0,\ x\in K-z_0\}$ and $D=C^*=\{y:\langle y,x\rangle\ge0\ (x\in C)\}$. The cone star denotes positive duality, unlike body polarity $\circ$. For $U=(0,m)$ and $V=(0,1)$, $$ \chi_C(V)\chi_D(U)=\frac{(n!)^2}{m^m}|K|\,|(K-z_0)^\circ|, \qquad \chi_C(V)=\int_Ce^{-\langle V,x\rangle}dx. $$ The height-$t$ slice of $D$ is $-t(K-z_0)^\circ$. This sign and $\int_0^\infty t^ne^{-at}dt=n!a^{-m}$ fix the constant. It suffices to prove that the Laplace product is at least $1$ for closed solid pointed convex cones, with interior $U,V$ and $\langle U,V\rangle=m$. The transformation $(C,D,U,V)\mapsto(BC,B^{-T}D,BU,B^{-T}V)$ preserves both the product and the pairing. Klartag states the correspondence using negative duality; the manuscript recalculates it with positive duality.

Lemma 2.1 · pp. 6–7 [Kla18, Lemma 2.1; Corollary 4.1, pp. 6–7, 15]

2. Normalize the projection means and covariance together

Let $G$ be standard Gaussian, $\Sigma=I+T$, $Z=\Sigma^{1/2}G$, and $a(z)=\phi(z)+z\Phi(z)$, where $\phi,\Phi$ are the one-dimensional Gaussian density and distribution function. Choose a bias $\xi_z$ at every layer $z\in\mathbb R$ such that $$ X_z=\Pi_C(Z+\xi_z),\quad Y_z=\Pi_D(-Z-\xi_z),\quad \mathbb EX_z=a(z)U $$ Moreau’s decomposition gives $X_z-Y_z=Z+\xi_z$ and $\langle X_z,Y_z\rangle=0$. The bias exists by coercivity and strict convexity of $\mathbb E\frac12|\Pi_C(Z+\xi)|^2-a(z)U\cdot\xi$. The weak projection derivative $P_z=D\Pi_C(Z+\xi_z)$ satisfies $0\le P_z\le I$, but the layers need not commute or be nested.

Lemma 3.1 · pp. 9–10 [Sol19, Theorem 1.1, p. 1]

Put $A=\int(\Phi(z)I-P_z)dz$, $M_i=\mathbb E(G_iA)$, and $L=\sum_iG_iM_i$. Denote the manuscript’s fixed scalar profiles by $\mathsf C,\mathsf K$, and distinguish their matrix averages $\mathbf C=\mathbb E\mathsf C(L)$ and $\mathbf K=\mathbb E\mathsf K(L)$. Proposition 3.4 simultaneously chooses linear coordinates and $T$ so that $$ \mathbb EA=0,\qquad \lambda T^2+\mathbf C\circ T+\mathbf K=0, \qquad -\mathbf C-2\lambda T\ge m_*I, $$ Here $\circ$ is the Jordan product, $\lambda=.65$, $m_*=.352$, and $-.022I\le T\le.064I$. Operator monotonicity of the square root keeps the covariance update inside this box; an inward coordinate estimate and Brouwer’s fixed point theorem also enforce mean balance. Fixing the covariance to $I$ beforehand would lose the matrix cancellation needed later.

Equation (7) · p. 11 Proposition 3.4 · pp. 12–14 [Cha13, Theorem 2.3, p. 4]

3. Split the entropy error into a linear field and layer defects

Integrating $X_z,Y_z$ against positive weights and smoothing produces injective cone-valued maps. Change of variables and Jensen’s inequality compare their log Jacobians and linear costs to the Laplace integrals. Proposition 4.3 yields $$ \frac1m\log(\chi_C(V)\chi_D(U))\ge-\mathcal E $$ where $\mathcal E$ is the manuscript’s entropy error functional. This estimate alone does not yet prove the product bound.

Proposition 4.3; Equation (11) · pp. 19–20

The Gaussian degrees of $R=A-L$ start at $2$, and those of its odd part start at $3$. The covariance formula, expressed through $(1+N_G)^{-1}$, controls the coefficients in each degree. It is applied to Gaussian Sobolev vector fields, without differentiating the measurable $P_z$ again. Lemma 5.1 reproves the Houdré–Privault semigroup identity by Hermite expansion.

Lemma 5.1 · pp. 20–21 [HP02, Proposition 2.1, Equation (2.4), p. 3]

Comparison with $\Theta_z(L)=1_{(-\infty,z]}(L)$ defines a positive measure $\beta$ through the nonnegative quantities $((P_z)_{ii}-1_{x_i\le z})(x_i-z)$ in a spectral basis of $L$. This retains the non-nesting and noncommutation of the layers. Proposition 5.7 and the estimates of Section 6 give $$ \mathcal E\le\Delta_p-J_\Sigma(u)+.80\|[L,T]\|_2^2 +1.8\tau T^2-\int(W-W_0)\,d\beta, \qquad W-W_0>0. $$ Here $\tau B=m^{-1}\mathbb E\operatorname{tr}B$ and $\|B\|_2^2=\tau(B^TB)$. The terms $\Delta_p,J_\Sigma(u)$ are a profile mean discrepancy and a Dirichlet term evaluated on the linear field. Apart from the final integral, the right side now depends only on $L,T$.

Section 5.3; Lemma 5.4 · pp. 23–24 Proposition 5.7 · pp. 26–27 Section 6.5; Equation (23) · p. 34

4. Close the bound with strict spectral interval estimates

Divided-difference differentiation of matrix functions produces uniform averages between eigenvalues $x_i,x_j$. The Daleckiĭ–Kreĭn formula needs no differentiation of eigenvectors; symmetric matrices and smooth profiles suffice. Gaussian integration by parts and the normalization equation cancel the term involving $S-I$, where $S=\sum_iM_i^2$, despite its lack of sign. The strict interval inequality in Proposition 2.3 absorbs the remaining error for distinct endpoints into the interval variance of the profile.

Lemma 7.1 · pp. 35–36 [Nof17, Theorem 2.11, print p. 6 / PDF p. 7] The profile is $u=(\pi,q)$: $\pi$ is a circular component with angle proportional to $\operatorname{arsinh}(x/s)$, and $q=k-\sqrt{b-r^2-d}$. The identity $d+|u|^2=c+2kq$ reduces the quadratic correction to one component, while variation of the circular component supplies strict variance on every nontrivial interval. That variance simultaneously controls divided-difference errors and the covariance commutator cost.

Proposition 2.3 · p. 8

Proposition 7.2 gives $\Delta_p-J_\Sigma(u)+.80\|[L,T]\|_2^2\le-1.94\tau T^2$, hence $$ \mathcal E\le-.14\tau T^2-\int(W-W_0)\,d\beta\le0. $$ Returning to the entropy estimate proves that the cone product is at least $1$. The slice formula gives the lower bound at every $z_0$, and choosing $z_0=s(K)$ proves the inequality in Theorem 1.1. Appendices A/B supply rational interval certificates at nodes, analytic interpolation between nodes and estimates on the unbounded tails. This article has not rerun all certificates or independently checked every estimate; Proposition 2.3 is a substantial internal proof input, not a verified numerical experiment.

Proposition 7.2 · pp. 36–39 Appendix A.2; Proposition B.1 · pp. 43–45, 54–55

5. Recover the simplex from zero defects and compute the converse

3. Corollary 1.2 — From all dimensions to the functional bound

Proof target：Corollary 1.2 · p. 4

Proof route. Use the Fradelizi–Meyer implication through auxiliary bodies in increasing dimensions. The translation-minimized functional product controls the origin-centered Legendre product, and a separate calculation supplies an extremizer.

4. Corollary 8.1 — From functional products to entropy–transport

Proof target：Corollary 8.1 · p. 41

Proof route. Read Corollary 1.2 as functional inverse Santaló $IS_n(c)$ with $c=e$, then apply Gozlan’s equivalence to essentially continuous densities.