The symmetric Mahler conjecture and its equality cases

1. From integrated simplex volume to arbitrary convex bodies

Start with a finite strip body $A=\{X:|b_i\cdot X|\le1\}$. Using the conformal lens width $\lambda$, Proposition 6.1 constructs a union of feasible simplices $\Sigma_X\subset A^\circ$ such that

The first inequality comes from holomorphic mass, the identity from the uniform boundary law, and the last inequality from containment in the polar. For an arbitrary $K$, approximate $Q=K^\circ$ by inner symmetric polytopes $Q_j$. The inclusions $(1-\delta_j)Q\subset Q_j\subset Q$ give $K\subset Q_j^\circ\subset(1-\delta_j)^{-1}K$ by polarity. Taking the limits of both volumes proves the bound, without smoothness or general-position assumptions.

Proposition 6.1 · pp. 15–16 Corollary 6.2 · pp. 16–17

2. From equality to the three-ball property of the polar

If $P(K)=4^n/n!$, then $B=K\oplus_1[-1,1]$ also attains equality in dimension $d=n+1$. The integrated missing volume for approximations to $B^\circ=Q\times[-1,1]$ tends to zero. Proposition 7.1 thereby obtains, at every $X\in\operatorname{int}B$ and for every $\xi\ne0$, one feasibility witness that compares with all competing representations. Proposition 8.1 lets $X$ approach the apex $(0,1)$ and uses the cost comparison to construct a median for each triple in $(\mathbb R^n,p_Q)$.

Lemma 8.2 converts medians to the three-ball property for arbitrary radii. For pairwise-intersecting balls with centers $x_i$ and radii $r_i$, their distances $d_i$ to a median $m$ satisfy $d_i+d_j\le r_i+r_j$. If $m$ is outside one ball, moving it toward that center by the excess distance places it in all three balls. Radius zero is included. The normed space is finite-dimensional, real and complete, so Hansen–Lima's classification applies and gives $Q=TH$. Polarity yields $K=T^{-T}H^\circ$; closure of the Hanner class under polarity finishes the necessary classification.

Proposition 7.1 · pp. 18–20 Proposition 8.1 · pp. 21–22 Lemma 8.2 · p. 23 [HL81, Corollary 7.4, p. 15] Theorem 8.3; Proposition 8.4 · p. 23

3. Every linear Hanner body attains equality

The converse does not require the classification theorem. An interval has volume product 4, and for bodies $A,A'$ of dimensions $k,l$ and $s\in\{1,\infty\}$,

The recursive construction thus preserves $4^{k+l}/(k+l)!$. Polarity exchanges the $\ell_1$ and $\ell_\infty$ sums, and the two determinant factors of an invertible linear map cancel in the volume product. This proves attainment for the whole equality family, including the base dimension $n=1$.

Lemmas 2.1–2.2 · p. 6

1. High powers at an isolated zero produce boundary probabilities

The conformal map of Lemma 3.1 is $F(z)=\frac8{\pi^2}\sum_{j\ge0}(-1)^jz^{2j+1}/(2j+1)^2$. Its boundary law is $\epsilon\lambda(T)+iT$, where $T$ is uniform on $[-1,1]$ and $\epsilon$ is an independent fair sign. Set $g=F^{-1}$ and $f_k=(g(b_jz)^k)_j$ on $\Omega=\{z:b_jz\in D\ \forall j\}$. Spanning ensures the zero is unique and the degree-$k$ homogeneous leading term is nonzero away from the origin. A sublevel $|f_k|^2\le s<1$ confines every row coordinate to a fixed compact subset of the lens, also verifying the compact-sublevel hypothesis of Lemma 4.1.

With $d^c=\frac i4(\bar\partial-\partial)$, Lemma 4.1 bounds the integrated sum of squared Jacobian minors below by $(\pi k)^d/d!$. Its proof compares sublevel mass by Stokes using the positivity of $dd^c\log|f_k|^2$, and calculates the homogeneous leading term's sphere flux as $(\pi k)^d$. Invariance under common phase rotation makes the correction to radial homogeneous flux vanish; this is the normalization step.

For each basis $I$, change variables to $u_i=g(b_iz)$, then set $u_i=\rho_i^{1/k}e^{i\theta_i}$. The factor $k^d$ cancels. On the fixed region $\sum_i\rho_i^2<1$, the inferred coordinates approach the boundary samples. Upper Fatou on a finite measure space yields $\limsup M_{I,k}/k^d\le(\pi^d/d!)P_I$. Summing over finitely many bases gives $S\ge1$. Additional boundary equalities do not have to be excluded at this stage.

Lemma 3.1 · pp. 7–8 Lemma 4.1 · pp. 10–12 Lemma 5.1 · pp. 12–14

2. Avoid double counting and retain the deficit

For each fixed sign choice the boundary law has measure $4^{-d}dt_I$. With $t_I=B_IX$ and $D_I=|\det B_I|$, it gives $P_I=4^{-d}D_I\int_A\sum_\epsilon\mathbf1_{(I,\epsilon)\text{ feasible}}\,dX$. By strict curvature of the lens boundary and nonproportionality of the rows, Lemma 5.2 makes the angles at which an extra row also lies on the boundary a null set. Thus, for almost every $X$, different feasible pairs give different vertices of $L_X$.

A vector in the interior of a feasible simplex defines a linear functional uniquely maximized at the corresponding vertex. Hence distinct simplices have disjoint interiors, allowing their volumes $D_I/d!$ to be summed. This proves the central identity. Containment $\Sigma_X\subset C$ and $S\ge1$ then give

and its nonnegative upper bound. The output for the equality argument is this integrated deficit; it does not assert $\Sigma_X=C$ at each $X$.

Lemma 5.2 · p. 14 Proposition 6.1 · pp. 15–16

1. Product neighborhoods capture every point and direction

We have $|B||C|=\frac4{n+1}P(K)=4^d/d!$. Choose symmetric inner polytopes $t_jQ\subset Q_j\subset Q$ with $t_j\to1$, and put $C_j=Q_j\times[-1,1]$, $B_j=C_j^\circ$. Proposition 6.1 gives $D_j=\int_{B_j}(|C_j|-|\Sigma_{j,X}|)dX\to0$.

For prescribed $X\in\operatorname{int}B$ and $e\in\partial C$, choose small positive-volume balls $U_k\subset\operatorname{int}B$ and $V_k\subset\operatorname{int}C$ close to $X$ and $e$. Choose $\overline{V_k}\subset\operatorname{int}C$; then $t_jC\subset C_j$ and $t_j\to1$ let us take $j_k$ large enough to ensure both $V_k\subset C_{j_k}$ and $D_{j_k}<|U_k||V_k|$. If every $\Sigma_{j_k,Z}$ with $Z\in U_k$ missed $V_k$, the deficit would be at least that product. Therefore there are $X_k\to X$, $e_k\to e$ with $e_k\in\Sigma_{j_k,X_k}$. This uses product neighborhoods, not a claim of pointwise convergence of the deficit at the prescribed point.

§7.1, (36)–(38) · pp. 18–19 §7.2, (39)–(41) · pp. 19–20

2. Preserve the witness even when bases become singular

A simplex containing $e_k$ supplies $d$ rows $c_{ik}$, weights $\theta_{ik}\ge0$ with $\sum_i\theta_{ik}\le1$, and a common witness $Y_k$. Concavity of $\lambda$ extends feasibility from vertices to all of $C_{j_k}$. Since $\lambda\le\lambda(0)$, we have $Y_k\in\lambda(0)B_{j_k}$; the polar sandwich makes these witnesses uniformly bounded. No uniform bound on the inverse basis matrices is needed, and the limiting rows may be dependent.

Take a simultaneous convergent subsequence of $Y_k,c_{ik},\theta_{ik}$. For any $h\in C$, use $t_{j_k}h\in C_{j_k}$ to pass to all feasibility inequalities with the same limit $Y$. Since $p_C(e)=1$, we get $1\le\sum_i\theta_i\le1$, so no mass remains at the simplex vertex 0. Setting $e=\xi/T$ and $\alpha_i=T\theta_i$ gives the desired representation. Finally,

proves optimality against every competitor using that common witness.

§7.2, (39)–(41) · pp. 19–20 Proposition 7.1 · pp. 18–20

1. Compare the two sign costs at the apex

Translate the triple to $0,x,y$. Write $\|\cdot\|=p_Q$, $p_1=\|x\|$, $p_2=\|y\|$, $p=\|x-y\|$, $s=p_1+p_2$, $a=p_1-p_2$, $A_\pm=(p\pm a)/2$ and $c=(s-p)/2$. Repeated points are immediate, so consider distinct points. Fix $u\in\operatorname{int}K$ and apply Proposition 7.1 at $X_\delta=(\delta u,1-\delta)\in\operatorname{int}B$ to $\xi=(x-y,a)$. Since $p\ge|a|$, the total weight is $p$ and the two sign weights are $A_\pm$.

The aggregates $r_\pm=\sum_{\sigma_i=\pm1}\alpha_i b_i$ satisfy $r_++r_-=x-y$ and $\|r_\pm\|\le A_\pm$. Their possible differences form the compact convex set

Lemma 3.2 gives $\lambda(1-\eta)=\frac2\pi\eta\log(1/\eta)+O(\eta)$, hence $\lambda(1-\delta t)/\lambda(1-\delta)\to t$ uniformly on compact positive intervals. Here $t=1-\sigma\langle b,u\rangle$ lies in $[1-p_K(u),1+p_K(u)]\subset(0,\infty)$. This uniformity gives the lower bound $p-h_{D_0}(u)$ for the limiting normalized optimal cost, even though the chosen representation varies with $\delta$.

Compare with $(x-y,a)=p_1(x/p_1,1)+p_2(-y/p_2,-1)$. Its normalized cost tends to $s-\langle x+y,u\rangle$, so optimality yields

No uniform selection of representations across $u$ is needed.

Lemma 3.2 · pp. 8–9 §8, (42)–(46) · pp. 21–22

2. Convex separation forces additive distances

Extend the inequality to $K$ by continuity and use $p_K=h_Q$ to homogenize it. For every direction $v$ we obtain $h_{\{x+y\}}(v)\le h_{D_0}(v)+(s-p)h_Q(v)$. Convex separation gives $x+y\in D_0+(s-p)Q$. Choose the corresponding $r_\pm$ and let $m=x-r_+=y+r_-$. Then

Because $c+A_+=p_1$, $c+A_-=p_2$ and $A_++A_-=p$, all three triangle inequalities become equalities. The same argument permits $c=0$ or $A_\pm=0$. This is the median used in the main theorem's classification.

Proposition 8.1 · pp. 21–22

1. Return from translation minimization to the origin

Set $f=e^{-\varphi}$. Fradelizi–Meyer's product is $\mathcal P(f)=(\int f)\inf_z\int f^z$, where $f^z(y)=\inf_{f(x)>0}e^{-\langle x-z,y-z\rangle}/f(x)$. Their Proposition 1(a) assumes the symmetric geometric Mahler inequality in every dimension and concludes $\mathcal P(f)\ge4^n$ for even log-concave $f$. Theorem 1.1 supplies this assumption. Since $f^0=e^{-\varphi^*}$, the product at the origin is at least $\mathcal P(f)$. If the second integral is infinite, the required bound is immediate.

The external implication uses auxiliary bodies in dimension $n+m$ and lets $m\to\infty$; the geometric inequality in dimension $n$ alone is insufficient. For $\varphi=\|x\|_1$, the conjugate is zero on the cube and $+\infty$ outside, proving sharpness. This limiting argument does not classify equality for general functions.

[FM08, Proposition 1(a), print p. 1433 / PDF p. 5] [FM08, Proof of Proposition 1(a), print pp. 1434–1435 / PDF pp. 6–7] [FM08, Remark 2, print p. 1437 / PDF p. 9] Corollary 1.2 · pp. 4–5

1. Establish the domains of the cost and entropy

For $\rho_i=e^{-V_i}$, the identity $|\nabla\rho_i|=|\nabla V_i|e^{-V_i}$ holds almost everywhere in the support's interior. Cordero-Erausquin–Klartag's Lemma 4 gives $\int|\nabla\rho_i|<\infty$, so the pushforward definition implies $\int|y|\,d\nu_i(y)<\infty$. The proof of their Proposition 7 also gives $V_i\in L^1(\eta_i)$, making $H(\eta_i)=-\int V_i\,d\eta_i$ finite. The statement of the external Proposition 7 concerns integrability of $V_i^*$ against the moment measure; the integrability of $V_i$ used here is obtained from equations (14)–(15) in its proof.

[CEK15, Lemma 4, print p. 3840 / PDF p. 7] [CEK15, Proof of Proposition 7, (14)–(15), print p. 3844 / PDF p. 11]

2. Insert the constant 4 in the symmetric equivalence

Fradelizi–Gozlan–Zugmeyer Theorem 1.3 states, with credit to Gozlan, the equivalence of $\mathrm{IS}_{n,s}(c)$ and $\mathrm{ET}_{n,s}(c)$. Their Definitions 1.1–1.2 match the manuscript's essential continuity, entropy sign and cost $\mathcal T$. Apply Corollary 1.2 with $c=4$ to obtain $-n\log(4e^2)$. Origin symmetry places the potentials and moment measures in the even/symmetric setting, while full dimensionality supplies the setting of log-concave probabilities with densities. The conclusion remains meaningful if $\mathcal T=+\infty$.

[FGZ21, Definitions 1.1–1.2, pp. 2–3] [FGZ21, Theorem 1.3, p. 3] Corollary 1.3 · p. 5