Couplings, tightness, and existence
Prove the direct-method existence argument for transport plans and see why lower semicontinuity of the cost matters.
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The question for today
What you’ll understand
- Show that fixed marginals make the family of couplings tight.
- Pass marginal constraints and lower-semicontinuous costs through weak limits.
- Construct a nonattainment example when the cost loses lower semicontinuity.
A numerical sequence of increasingly cheap transport plans does not yet prove that a cheapest plan exists. The limit might lose mass, violate a marginal constraint, or have a suddenly larger cost. The direct method handles these three possibilities separately. Fixed probability marginals provide tightness, bounded continuous tests preserve the marginals, and a lower-semicontinuous nonnegative cost prevents a bad upward jump at the limit. This chain of reasoning is more informative than saying that transport is convex, since convexity alone does not establish attainment.
Fixed marginals prevent escape of mass
Work on Polish source and target spaces, meaning complete separable metric spaces, with probability marginals μ and ν. Each probability is tight: for any positive tolerance there is a compact set carrying all but that tolerance of its mass. Choose compact K and L for the two marginals. For every coupling, mass outside K × L is at most the source mass outside K plus the target mass outside L, by the union bound.
The same compact product works for every coupling because the marginals are fixed, giving uniform tightness of the feasible family. Prokhorov’s theorem then gives weakly convergent subsequences of any sequence of couplings. Compact support of the original distributions is unnecessary. On Euclidean space, finite moments can help estimate tails, but tightness of the fixed marginals already supplies the compactness step for a general nonnegative cost.
Closure and lower semicontinuity finish the argument
If couplings πₙ converge weakly to π, test the first coordinate with a bounded continuous function f(x). Its integral against every πₙ is the fixed integral of f against μ, so the same equality holds for π. Repeat for the second coordinate. The limiting measure is therefore still a coupling. This proves weak closure of the marginal constraints; weak convergence here is convergence against bounded continuous test functions.
Suppose the cost is nonnegative and lower semicontinuous on the product space. The Portmanteau inequality gives that the limit plan’s cost is at most the liminf of the sequence’s costs. Apply this to a minimizing sequence to obtain an optimizer. If at least one feasible coupling has finite cost, the optimum is finite. For squared Euclidean cost, finite second moments of both marginals make the independent coupling finite, using |x − y|² ≤ 2|x|² + 2|y|².
Existence is weaker than uniqueness or smoothness
In a finite transport matrix, the feasible set is a closed bounded polytope and the linear objective attains its minimum. There can still be many minimizers, especially when some costs are equal. In continuous problems an optimal plan may likewise split mass or be singular. The direct method neither gives a unique optimizer nor produces a differentiable transport map. Those questions require extra geometry or strict structural assumptions, rather than more compactness.
Lower semicontinuity cannot be dropped casually. Raising a cost at the diagonal can make arbitrarily small shifts cheap while making their weak limit expensive. The practice problem turns this into a concrete nonattainment example on a compact square. This illustrates the roles of the assumptions cleanly: even compact feasible couplings cannot rescue an objective that jumps upward at its limiting candidate. Negative unbounded costs introduce further integrability issues, which are excluded by the nonnegative-cost theorem used here.
Work it through
A finite compact family with an attained optimum
Both source and target put mass one-half at zero and one-half at one. Parameterize all couplings and minimize squared Euclidean cost.
Let the top-left matrix entry be t. The row and column constraints force the full matrix [[t, 1/2 − t], [1/2 − t, t]], with 0 ≤ t ≤ 1/2 from nonnegativity.
Diagonal entries have cost zero and off-diagonal entries have cost one. The objective is therefore 1 − 2t, a continuous affine function on the closed interval [0, 1/2].
The minimum occurs at t = 1/2, giving the diagonal coupling and value zero. The endpoint belongs to the feasible interval, so the infimum is actually attained rather than merely approached.
The unique squared-cost optimizer in this example is the diagonal plan. Compactness proves attainment; the particular objective establishes uniqueness here.
The living notebook
Explore the model
Interactive lab · probability
A fixed finite support cannot lose tail mass
Pause and Reset, then set success probability p = 0.25 and trials n = 1. Predict masses 0.75 and 0.25 at the fixed positions zero and one, with total one. The finite support makes tightness immediate, independently of the probability value. This lab displays one marginal family, not joint couplings or an existence test for arbitrary costs; the closure and lower-semicontinuity proof must still be supplied.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forCompactness of measures and good behavior of the objective are separate requirements for an attained optimum.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let both marginals be uniform on [0, 1]. Define c(x, y) = |x − y|² off the diagonal and c(x, x) = 1. Show that the infimum cost is zero but no plan attains it.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the measure-preserving shift Tε(x) = x + ε modulo one, with 0 < ε < 1, and split its integral into the wrapped and unwrapped parts.
Compare with the solution
The shift preserves the uniform law and never fixes a point. On a set of length 1 − ε its displacement is ε; on the wrapped set of length ε the displacement magnitude is 1 − ε. Its cost is (1 − ε)ε² + ε(1 − ε)² = ε(1 − ε), tending to zero. Yet c is strictly positive at every pair, so its integral against any probability coupling is strictly positive. Hence no plan has cost zero. The diagonal cost jump violates lower semicontinuity, exactly where the existence proof needs it.
Keep the idea
Take it with you
- Fixed tight marginals make the coupling family uniformly tight.
- Weak tests preserve the marginal constraints, and Portmanteau controls a nonnegative lower-semicontinuous cost.
- Attainment alone does not imply uniqueness, a map, or regularity.
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