Brenier's theorem

Summarize Timeline Fact Check

Let ${\displaystyle {\mathcal {P}}_{2}(\mathbb {R} ^{n})}$ denote the set of Borel probability measures on ${\displaystyle \mathbb {R} ^{n}}$ with finite second moment. If ${\displaystyle \mu ,\nu \in {\mathcal {P}}_{2}(\mathbb {R} ^{n})}$, a measurable map ${\displaystyle T:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}$ is said to push ${\displaystyle \mu }$ forward to ${\displaystyle \nu }$, written

${\displaystyle T_{\#}\mu =\nu ,}$

if

${\displaystyle \nu (B)=\mu (T^{-1}(B))}$

for every Borel set ${\displaystyle B\subseteq \mathbb {R} ^{n}}$. The Monge problem for the quadratic cost is to minimize

${\displaystyle \int _{\mathbb {R} ^{n}}|x-T(x)|^{2}\,d\mu (x)}$

among all measurable maps ${\displaystyle T}$ such that ${\displaystyle T_{\#}\mu =\nu }$.

One form of Brenier's theorem is the following.

Theorem. Let ${\displaystyle \mu ,\nu \in {\mathcal {P}}_{2}(\mathbb {R} ^{n})}$, and suppose that ${\displaystyle \mu }$ is absolutely continuous with respect to the Lebesgue measure. Then there exists a convex function ${\displaystyle \varphi$ :\mathbb {R} ^{n}\to (-\infty ,+\infty ]} such that

${\displaystyle T=\nabla \varphi }$

pushes ${\displaystyle \mu }$ forward to ${\displaystyle \nu }$ and solves the quadratic Monge problem. The map ${\displaystyle T}$ is unique ${\displaystyle \mu }$-almost everywhere. Equivalently, the optimal transport plan for the corresponding Kantorovich problem is unique and is concentrated on the graph of ${\displaystyle \nabla \varphi }$.^[1][2][3]

The function ${\displaystyle \varphi }$ is called a Brenier potential, and the map ${\displaystyle \nabla \varphi }$ is called the Brenier map from ${\displaystyle \mu }$ to ${\displaystyle \nu }$.

If both ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are absolutely continuous, then the Brenier map from ${\displaystyle \mu }$ to ${\displaystyle \nu }$ has an inverse in the almost-everywhere sense. If

${\displaystyle T=\nabla \varphi }$

is the Brenier map from ${\displaystyle \mu }$ to ${\displaystyle \nu }$, then the Brenier map from ${\displaystyle \nu }$ back to ${\displaystyle \mu }$ is given by

${\displaystyle S=\nabla \varphi ^{*},}$

where ${\displaystyle \varphi ^{*}}$ is the convex conjugate of ${\displaystyle \varphi }$. The maps satisfy

${\displaystyle S(T(x))=x}$ for ${\displaystyle \mu }$-almost every ${\displaystyle x}$,

and

${\displaystyle T(S(y))=y}$ for ${\displaystyle \nu }$-almost every ${\displaystyle y}$.^[2][3]