A counterexample to the Monge ansatz for the three-marginal Coulomb cost

Proof. For each j, define the Borel set

Its image H_j(S_j) is Borel. Pairwise disjointness of the component images and injectivity of H_j give

Since S_j\subset T^{-1}(H_j(S_j)), preservation of \mu would imply

Thus \nu(S_j)=0 for all four indices, contradicting (3).□

Proof. For a parameter K>0, use the potentials in the statement and set

By (4), its (x,z)-gradient vanishes at the k-th center triple. We choose K to obtain a unique minimizing pair on a small product neighborhood for each fixed y, and to make each coordinate of the resulting minimizing pair an invertible function of y.

The Coulomb cost is smooth near both center triples. At the k-th center, write

The (x,z)-Hessian of F_k there is M_k+KI_6, while its mixed derivative in y is N_k. For h(b)=|b|^{-1}, b\ne0, differentiation gives

This matrix has eigenvalues 2|b|^{-3},-|b|^{-3},-|b|^{-3} and is therefore invertible. The two 3\times3 blocks of N_k are

and hence are both invertible. The matrices M_k are fixed symmetric 6\times6 matrices. For all sufficiently large K, both M_k+KI_6 are positive definite. Moreover,

Invertibility is an open condition on 3\times3 matrices, so we may choose a single K for which both blocks of (M_k+KI_6)^{-1}N_k are invertible for each k.

Fix such a K. The implicit function theorem applied to \nabla_{(x,z)}F_k=0 gives a smooth stationary branch

with the prescribed center values. Differentiating this equation at y=a_k gives

The two block invertibilities allow the inverse function theorem to be applied separately to X_k and Z_k. After restricting to one sufficiently small neighborhood of a_k, both maps are diffeomorphisms onto their open images.

By continuity, the (x,z)-Hessian of F_k remains positive definite on a product of sufficiently small open balls around the three centers. Shrink the y-ball so that the stationary pair (X_k(y),Z_k(y)) belongs to the product of the other two balls. For fixed y, this latter product is convex. The positive definite Hessian makes F_k(\cdot,y,\cdot) strictly convex on it, so its stationary pair is its unique minimum there.

Define its minimum value by

The preceding minimum property gives (5) with exactly the stated equality set. The center values in (4) give w_k(a_k)=0. All neighborhoods can be restricted further as needed to retain the separate diffeomorphism properties.□

Proof. Use the functions and neighborhoods of Lemma 3.1. Choose small balls contained in these neighborhoods, taking U_0 inside both central neighborhoods. On the respective components define

The balls will be shrunk below. Since the functions are smooth near their centers, we can first restrict to fixed preliminary balls on which their absolute values are bounded by one constant M>0. Take the final balls sufficiently small that their closures are pairwise disjoint.

At the five centers the values of u are 5/2,0,0,0,0. For a triple of distinct centers other than a permutation of (0,a_k,-a_k), there are two possibilities. If zero is absent, its potential sum is zero and its cost is positive. If zero is present, the other two centers come from different opposite pairs. Their cost is

so again there is a strict gap. There are finitely many such triples, and the cost and local potentials are continuous near each one. All these strict inequalities therefore persist on the corresponding products after sufficiently small restrictions.

For a repeated-center type, two coordinates belong to the same ball. If all radii are at most r, the distance between those two coordinates is at most 2r, and their pair cost is at least 1/(2r), with the usual interpretation at coincidence. Taking r small enough that 1/(2r)>3M makes the cost strictly larger than the potential sum on every such product.

The inequalities of Lemma 3.1 remain valid when their domains are restricted, with the same equality characterization. Permuting a triple leaves both c and the sum of the three values of u unchanged. Thus the two prescribed component types and their permutations satisfy (7), with equality exactly as in (8); every other type has strict inequality. Finally, keep U_0,U_k^- fixed and shrink each U_k^+ once more. Since X_k(a_k)=0 and Z_k(a_k)=-a_k, continuity ensures

All inequalities and diffeomorphism properties survive this restriction. The function u is bounded and continuous on the final disjoint union U.□

Proof. Both X_k(U_k^+) are open neighborhoods of zero. Choose a positive-radius ball B about zero inside their intersection and U_0. The inverse maps then take B into U_k^+, and Proposition 4.1 gives Z_k(Y_k(B))\subset U_k^-. The inverse and composition of the relevant diffeomorphisms are diffeomorphisms onto open images. The identities X_k(Y_k(x))=x and W_k=Z_k\circ Y_k give equality by Proposition 4.1. Disjointness follows from the five component balls.□

Proof of Proposition 6.1. Write K_0=\supp g, and let K_1,\ldots,K_4 be its images under the four maps in \calH. These are nonempty compact sets in the five disjoint components, and \mu(K)=1 for K=\bigcup_{a=0}^4K_a. In particular,

The optimal coupling \pi_0 constructed in Section 5 is supported on products of three distinct component sets.

Fix \delta>0. Partition each K_a into finitely many Borel cells of diameter at most \delta, for example by a finite grid of half-open cubes. Discard the cells of zero \mu-mass and denote the remaining cells by A_1,\ldots,A_N. They partition a set of full \mu-measure and each belongs to a single K_a. Put

The marginal constraints on \pi_0 read

For each fixed i, split A_i into Borel sets B_{ijk} of masses q_{ijk}, ignoring zero-mass pieces. Such a split can be made by successive first-coordinate slabs: the first-coordinate marginal of \mu|_{A_i} is absolutely continuous, so its cumulative distribution is continuous and takes every value between zero and \mu(A_i). Endpoints of the slabs have zero mass; any unused null set is included in a fixed piece.

For each q_{ijk}>0, Lemma 6.2 supplies maps on B_{ijk} sending the probability q_{ijk}^{-1}\mu|_{B_{ijk}}, respectively, to

Reset any exceptional values outside the indicated target cell to a fixed point in that cell. These changes are Borel and occur on null sets. Glue the maps over the finitely many pieces to obtain T_2,T_3, and give them a fixed value on the remaining Borel null set in \R^3. Different source pieces may use the same normalized target law. Their weights q_{ijk} add to the full mass of the target cell, so no target partition is needed. Indeed, for any Borel set D,

and the same calculation proves (T_3)_\#\mu=\mu. The graph coupling \pi_\delta=(\Id,T_2,T_3)_\#\mu gives each product A_i\times A_j\times A_k exactly the mass q_{ijk}.

Whenever q_{ijk}>0, the three cells lie in three distinct component sets. Thus all pairwise distances in their product are at least \eta, including after the redistribution within the cells. If x and y are two triples in the same such product, then |x_\ell-y_\ell|\le\delta for each coordinate. Consequently

Both costs are finite, bounded by 3/\eta on all active products. Since \pi_\delta and \pi_0 have identical masses on these products, summing their oscillation bounds gives

Letting \delta\downarrow0 proves the required upper bound on the Monge infimum. The reverse bound follows because every admissible graph plan is a coupling. Strict nonattainment is Theorem 1.1.□

Proof of Lemma 6.2. We construct a Borel map from each such probability to uniform measure on (0,1)^d, and a Borel map in the reverse direction. Let f be a nonnegative Borel probability density. The forward map is Rosenblatt’s successive conditional-distribution transform [10]; we also give the inverse quantile construction and make the Borel choices explicit in both directions.

The forward map. We successively transform each coordinate by its conditional distribution function given the preceding coordinates. Write x_{1:j}=(x_1,\ldots,x_j), with x_{1:0} the empty prefix, and set

The prefix densities are Borel, have integral one, and are finite almost everywhere. Tonelli’s theorem gives \int_\R f_{1:j}(a,r)\dd r=f_{1:j-1}(a) for every prefix a, including the empty prefix when j=1. For j=1,\ldots,d, define

This formula is used where its denominator belongs to (0,\infty); on the remaining fibers use instead the fixed distribution function t\mapsto(1+e^{-t})^{-1}. Thus every fiber is a continuous distribution function with limits zero and one, and all F_j are jointly Borel. For j\ge2, the exceptional prefix fibers have zero probability under f_{1:j-1}(a)\dd a; there are no exceptional fibers when j=1.

For a random variable with a continuous distribution function F, the variable F(X) is uniform on (0,1). Indeed, continuity and the limits of F allow a cut at each probability level u\in(0,1); any interval on which F=u has probability zero, and hence \mathbb P(F(X)\le u)=u. At stage j, this identity and Fubini show that, conditionally on the preceding original prefix, the new transformed coordinate is uniform. It is therefore independent of that prefix, and hence of all preceding transformed coordinates. A finite induction over j=1,\ldots,d shows that the Borel map

sends f\dd x to product uniform measure on (0,1)^d. Its values lie in [0,1]^d, including on exceptional fibers. This argument does not require a density to be positive everywhere or a distribution function to be strictly increasing.

The reverse map. Given independent uniform coordinates, we successively recover the original conditional laws by quantiles. For each j and 0<u<1, define

This value is finite, and

proves joint Borel measurability on 0<u<1. At u=0,1, define q_j(a,u)=0. For 0<u<1, continuity of F_j(a,\cdot) gives

Consequently q_j(a,U) has distribution function F_j(a,\cdot) for uniform U. For u\in[0,1]^d, recursively set

starting with the empty prefix. This finite recursion defines a Borel map. For independent uniform inputs, induction shows that after step j the output prefix has density f_{1:j}: the preceding prefix has density f_{1:j-1}, its exceptional fibers have probability zero, and on every other fiber the quantile has the conditional distribution just defined. Thus S_f has output law f\dd x. If f and g are densities of \alpha and \beta, respectively, then S_g\circ R_f has the required property.□

Proof. Fix d\ge2 and s>0, and put a_1=e_1 and a_2=e_2 in \R^d. The measure argument already applies in this dimension. We verify the three properties of the cost that it needs: local equality families given by diffeomorphisms, strict supporting inequalities on all other component products, and a uniform cost-oscillation bound on separated products.

Local branches. For h_s(b)=|b|^{-s}, b\ne0, the center jets are

As before, we keep the positive outer point y free and minimize in the other two coordinates. The potentials in Lemma 3.1 are replaced by

The central potential is common to the two branches, and F_k=c_s-u_0-v_k has vanishing (x,z)-gradient at each center. Direct differentiation gives

Its radial eigenvalue is s(s+1)|b|^{-s-2}, and its tangential eigenvalues are -s|b|^{-s-2}, so it is invertible.

At the k-th center, define M_k and N_k as in Section 3, with c_s in place of c. The two d\times d blocks of N_k are -D^2h_s(-a_k) and -D^2h_s(-2a_k), hence are invertible, while M_k is a symmetric 2d\times2d matrix. For sufficiently large K, both M_k+KI_{2d} are positive definite and both d\times d blocks of (M_k+KI_{2d})^{-1}N_k are invertible, because

One K works for the two branches. The implicit and inverse function arguments of Lemma 3.1 now give separate local diffeomorphisms X_k,Z_k on d-dimensional neighborhoods. The cost is smooth near the collision-free center triples; after shrinking the product balls, the (x,z)-Hessian of F_k remains positive definite. Strict convexity on the product of the x- and z-balls therefore gives the same local supporting inequality and exact equality set, with w_k(a_k)=0.

The supporting inequality and the marginal. The five center values of the potential are now 2+2^{-s},0,0,0,0. A distinct-center triple containing zero but not an opposite pair has cost

A distinct-center triple without zero has positive cost and potential sum zero. These finitely many strict gaps persist on small products by continuity. If two points lie in one ball of radius at most r, their pair cost is at least (2r)^{-s}. This exceeds 3M for sufficiently small r, where M bounds the absolute values of the preliminary potentials. Positivity of the other pair terms and permutation symmetry of c_s therefore give the full supporting inequality and exact contact set described in Proposition 4.1, now in \R^d. The proof of Lemma 4.2 then also applies in \R^d.

On that common ball B, choose a nonnegative g\in C_c^\infty(B) with \int_B g^2\dd x=1, and form \nu and \mu by (9) using the four resulting diffeomorphisms. The change-of-variables construction in Section 5, now with d-dimensional Jacobians, gives the same smooth compactly supported density and square root: each image function is supported compactly inside its open component, and the five supports are disjoint. The same random choice of a branch and permutation gives an optimal coupling \pi_0 with the three marginals equal to \mu. The bounded supporting potential proves its cost is finite and forces every optimal coupling onto the contact set. Applying Lemma 2.1 to the four branch maps then excludes every marginal-preserving Borel pair, exactly as in Section 5.

Approximation by preserving maps. The five compact component sets have a positive mutual separation \eta, and \pi_0 uses only products of three distinct components. Repeat the cell partition and redistribution from the proof of Proposition 6.1 in \R^d, using Lemma 6.2. The first-coordinate marginals used to split the cells are absolutely continuous by Fubini in every such dimension. The redistribution gives a graph coupling \pi_\delta of a preserving Borel pair with the same mass as \pi_0 on each product of cells of diameter at most \delta. Every active product still has all pair distances at least \eta. For two triples x,y in one such product, the scalar mean value theorem gives

Both costs are bounded by 3\eta^{-s} on the active products. Summing their oscillations against the common product masses yields

The reverse inequality for the infima holds because every graph plan is a coupling. This proves the final assertion. In particular, no restriction relating s to d is needed: the optimizer and these approximating plans stay away from all collisions.□