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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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An attaining coupling and potential on a three-point metric space

Example

Let X={x1,x2,x3} have the metric with zero diagonal and d(x1,x2)=d(x2,x3)=1, d(x1,x3)=2, with symmetric values. For p=(1/2,0,1/2) and q=(0,1,0), the coupling

π=(01/2000001/20)

and normalized potential f=(0,−1,0) both attain value 1.

The straight segments show the two distance-one edges, the node labels give the source and target masses, and the curved arrows represent the two positive coupling entries.

x1:(p1;q1)=(1=2;0)x2:(p2;q2)=(0;1)x3:(p3;q3)=(1=2;0)d(x1;x2)=1d(x2;x3)=1¼12=1=2¼32=1=2

Facts & Assumptions

Given: The specified three-point distance matrix and the probability vectors in the Example.

[F1]

The finite-metric theorem equates the minimum coupling cost and maximum normalized 1-Lipschitz objective and gives complementary slackness on positive coupling entries (Finite-metric transport duality and complementary slackness).

[F2]

A coupling is a nonnegative matrix with prescribed row and column sums, and its cost is the finite sum of distance times transported mass (Couplings of finite probability vectors, The finite-metric 1-Wasserstein transport cost).

[F3]

A probability vector has nonnegative coordinates summing to one (Finite probability vectors).

Verification

1.1givenF2F3algebra

The specified distances are symmetric, vanish exactly on equal points, and are positive for distinct points. For the three distinct points, the triangle inequalities are 2≤1+1, 1≤2+1, and 1≤1+2; when a point repeats, each inequality is 0≤d+d or d≤d+0. Thus this is a metric. The displayed matrix is nonnegative, has row sums (1/2,0,1/2) and column sums (0,1,0), and has cost (1/2)⋅1+(1/2)⋅1=1.

1.2givenF3algebra

The vector f=(0,−1,0) has f1=0. Its pairwise differences have magnitudes 1,1,0, bounded respectively by distances 1,1,2. Its objective is (1/2)⋅0−1⋅(−1)+(1/2)⋅0=1.

2.1step 1.1step 1.2F1algebra∎

The feasible plan and potential in steps 1.1 and 1.2 have the same value. By [F1], the primal optimum is at most the plan cost, the dual optimum is at least the potential value, and the two optima are equal; hence both are attained with value 1. The positive entries are π12=π32=1/2, and they satisfy f1−f2=d(x1,x2)=1 and f3−f2=d(x3,x2)=1, as complementary slackness requires.

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