Brenier maps and convex potentials
Identify when quadratic optimal transport is a unique gradient map and separate that structural theorem from smooth PDE regularity.
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The question for today
What you’ll understand
- State the quadratic Brenier theorem with finite moments and source absolute continuity.
- Relate quadratic dual contact to a convex gradient map.
- Compute simple optimal maps and explain why arbitrary shears are different.
Allowing a transport plan to split mass is essential in general, but quadratic Euclidean transport has a remarkable deterministic structure when the source is sufficiently diffuse. Brenier’s theorem gives a unique optimal plan induced by the gradient of a convex function. It does not say that every cost, every source measure, or every invertible deformation has such an optimizer. Its strength lies in a precise set of assumptions, and its uniqueness concerns the map almost everywhere rather than one globally determined formula for the potential.
State the theorem before using its geometry
Let μ and ν be probability measures on real d-space with finite second moments, and suppose μ is absolutely continuous with respect to d-dimensional Lebesgue measure. For cost c(x, y) = one-half times |x − y|², Brenier’s theorem gives a unique optimal coupling, of the form (id, T)#μ, with T the gradient of a convex potential ψ at μ-almost every source point. The target need not have a density; it may be atomic.
Absolute continuity of the source is a sufficient hypothesis, not a requirement imposed symmetrically on both marginals. Without a diffuse-source condition, splitting may be unavoidable, as a single source atom sent to two target atoms shows. The map is unique only up to μ-null sets. A potential can be altered by an additive constant, and its behavior away from the source’s relevant region need not be uniquely fixed. None of these qualifications weakens uniqueness of the actual optimal joint distribution.
Quadratic duality explains the gradient form
At a smooth source point where the dual constraint is tight for a destination T(x), differentiating the contact relation in x yields gradient f(x) = x − T(x). Thus T is the gradient of |x|²/2 − f(x). The c-concavity structure of an optimal quadratic potential makes this new function convex. This differentiable calculation is motivation; the full theorem uses almost-everywhere differentiability and convex analysis rather than assuming all data are smooth.
Convex gradients are monotone: their differences have nonnegative inner product with differences of the inputs. In one dimension this is the familiar nondecreasing rearrangement. In higher dimensions, ordinary two-point monotonicity is not sufficient to characterize all convex gradients; cyclic monotonicity supplies the stronger structure. A nonsymmetric linear shear is not the gradient of a scalar convex quadratic, even though it can be invertible and can push one smooth distribution to another.
A weak map theorem is not automatic smooth regularity
If source and target have smooth positive densities and the optimal gradient map is sufficiently differentiable and locally invertible, change of variables gives target density evaluated at ∇ψ times det(D²ψ) equal to the source density. This is the Monge–Ampère equation. Those regularity and invertibility assumptions are additional: Brenier’s basic conclusion alone does not guarantee a classical Hessian or a smooth solution of that equation.
For practical calculations, first prove that a proposed convex-gradient map has the correct pushforward and finite cost. These facts certify optimality under the theorem’s setting without solving a nonlinear PDE numerically. Scaling a one-dimensional uniform distribution is a transparent example. Gaussian covariance examples similarly reduce to linear algebra when the relevant map is a symmetric positive matrix. Keep the one-half convention visible: the optimal coupling is unchanged if the cost is doubled, but the reported minimum value doubles.
Work it through
Uniform expansion as a convex gradient
Find the quadratic optimal map from the uniform law on [0, 1] to the uniform law on [0, 2], and compute both the half-squared transport cost and W₂ squared.
The map T(x) = 2x has the correct pushforward: a target interval of length ℓ inside [0, 2] has preimage length ℓ/2, matching the target density one-half.
T is the derivative of the convex potential ψ(x) = x². The source has a density and both distributions have finite second moment, so this convex-gradient pushforward is the unique quadratic optimal map almost everywhere.
The displacement is x, so the half-squared cost is (1/2)∫₀¹x² dx = 1/6. Without the factor one-half, the value W₂² is ∫₀¹x² dx = 1/3.
The optimal map is T(x) = 2x, the half-squared cost is 1/6, and W₂² is 1/3.
The living notebook
Explore the model
Interactive lab · matrix
A convex-gradient stretch versus a shear
For each run Pause and Reset, then set both controls. Choose diagonal scale a = 2 and shear b = 0; predict the symmetric positive matrix diag(2, 1), a convex-gradient stretch. In a fresh run choose a = 1 and b = 1; its nonsymmetric shear matrix is not a gradient map on an open region. The diagram displays one linear family, not an OT solver; final readouts remain tied to the selected matrix during interpolation.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forInvertibility alone does not supply the convex-gradient structure of quadratic optimal transport.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let μ be the centered two-dimensional Gaussian with identity covariance and ν the centered Gaussian with covariance diag(4, 9). Propose the optimal map, identify a convex potential, and compute W₂² and the half-squared cost.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A diagonal positive scaling transforms the independent coordinate variances, and the source coordinates each have second moment one.
Compare with the solution
Take T(x₁, x₂) = (2x₁, 3x₂), the gradient of ψ = x₁² + (3/2)x₂². It pushes the independent standard coordinates to variances four and nine, so the target is correct. The source is absolutely continuous and the potential is convex, giving quadratic optimality. The displacement is (x₁, 2x₂), so its mean squared norm is 1 + 4 = 5. Therefore W₂² = 5 and the half-squared transport cost is 5/2. The map is unique μ-almost everywhere.
Keep the idea
Take it with you
- An absolutely continuous source and finite second moments yield a unique quadratic optimal plan induced by a convex gradient.
- Uniqueness of the map almost everywhere does not make the potential globally unique.
- Classical Monge–Ampère regularity requires more than the basic transport theorem.
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