# An optimal transport problem with backward martingale constraints motivated
by insider trading.

Dmitry Kramkov111Carnegie Mellon University, Department of Mathematical
Sciences, 5000 Forbes Avenue, Pittsburgh, PA, 15213-3890, USA. Email:
<EMAIL_ADDRESS>Yan Xu222Carnegie Mellon University, Department of
Mathematical Sciences, 5000 Forbes Avenue, Pittsburgh, PA, 15213-3890, USA.
Email<EMAIL_ADDRESS>
###### Abstract

We study a single-period optimal transport problem on $\mathbb{R}^{2}$ with a
covariance-type cost function $c(x,y)=(x_{1}-y_{1})(x_{2}-y_{2})$ and a
backward martingale constraint. We show that a transport plan $\gamma$ is
optimal if and only if there is a maximal monotone set $G$ that supports the
$x$-marginal of $\gamma$ and such that $c(x,y)=\min_{z\in G}c(z,y)$ for every
$(x,y)\in\operatorname{supp}\gamma$. We obtain sharp regularity conditions for
the uniqueness of an optimal plan and for its representation in terms of a
map. Our study is motivated by a variant of the classical Kyle model of
insider trading from Rochet and Vila (1994).

Keywords:

martingale optimal transport, Kyle equilibrium.

AMS Subject Classification (2010):

60G42, 91B24, 91B52.

##  1 Introduction

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and
$Y=(Y_{1},Y_{2})$ be a $2$-dimensional random variable with finite second
moment: $Y\in\mathcal{L}^{2}(\Omega,\mathcal{F},\mathbb{P})$. Our goal is to

$\text{minimize}\quad\mathbb{E}\left(c(X,Y)\right)\quad\text{over}\quad
X\in\mathcal{X}(Y),$ (1)

for the cost function $c(x,y)=(x_{1}-y_{1})(x_{2}-y_{2})$,
$x,y\in\mathbb{R}^{2}$, and the domain $\mathcal{X}(Y)$ that consists of
$Y$-measurable random variables $X=(X_{1},X_{2})$ such that $(X,Y)$ is a
martingale: $\mathbb{E}\left(\left.Y\right\lvert{X}\right)=X$. A relaxation of
the $Y$-measurability constraint on $X$ leads to the optimal transport
problem:

$\text{minimize}\int c(x,y)d\gamma\quad\text{over}\quad\gamma\in\Gamma(\nu),$
(2)

where $\nu=\operatorname{{Law}}(Y)$ and $\Gamma(\nu)$ is the family of
probability measures $\gamma=\gamma(dx,dy)$ on
$\mathbb{R}^{2}\times\mathbb{R}^{2}$ that have $\nu$ as their $y$-marginal:
$\gamma(\mathbb{R}^{2},dy)=\nu(dy)$, and make a martingale out of the
canonical process: $\gamma(y|x)=x$. In view of the martingale constraint,
problem (2) admits an equivalent formulation:

$\displaystyle\text{maximize}\quad\int
x_{1}x_{2}d\gamma\quad\text{over}\quad\gamma\in\Gamma(\nu),$

and thus, has a natural connection to the classical Fréchet-Hoeffding
inequality and the Wasserstein $2$-distance.

Problem (2) exhibits a _backward_ structure in the sense that the initial
marginal $\mu(dx)=\gamma(dx,\mathbb{R}^{2})$ is part of the solution. In this
regard, it differs from the “standard” single-period martingale transport
problem in Beiglböck and Juillet (2016), Beiglböck et al. (2017), Henry-
Labordère and Touzi (2016), and Ghoussoub et al. (2019), among others, where
both the initial and terminal marginals are fixed. We point out that for our
cost function $c(x,y)=(x_{1}-y_{1})(x_{2}-y_{2})$, the standard problem is
trivial, as every martingale measure $\gamma=\gamma(dx,dy)$ with given
marginals $\mu=\mu(dx)$ and $\nu=\nu(dy)$ produces the same average cost:

$\int c(x,y)d\gamma=\int y_{1}y_{2}d\nu-\int x_{1}x_{2}d\mu.$

Our work is motivated by the classical Kyle (1985) equilibrium with insider
from financial economics. More precisely, we consider the model from Rochet
and Vila (1994), where the insider observes _both_ the terminal value $V$ of
the risky asset and the order flow $U$ of the noise traders; see Section 6 for
details. Setting $Y=(U,V)$ we establish in Theorem 6.3 the equivalence between
the existence of equilibrium and that of an optimal _map_ $X$ for (1) such
that $\gamma=\operatorname{{Law}}{(X,Y)}$ is an optimal _plan_ for (2).
Moreover, the components of $X=(R,S)$ are naturally identified as
equilibrium’s total order $R$ and price $S$. To the best of our knowledge, the
connection between the Kyle equilibrium and a martingale optimal transport is
new.

The main results of the paper are Theorems 2.2 and 4.6. In Theorem 2.2 we
prove the existence of an optimal plan for (2) and characterize its support.
We show that $\gamma\in\Gamma(\nu)$ is optimal if and only if there is a
_maximal monotone set_ $G$ in $\mathbb{R}^{2}$ that supports the $x$-marginal
of $\gamma$ and such that

$c(x,y)=\phi_{G}(y)\triangleq\inf_{z\in
G}c(z,y),\quad(x,y)\in\operatorname{supp}\gamma.$ (3)

Geometrically, the support of ${\gamma}$ has the _hyperbolic tangent property_
: it connects $y\not\in G$ to those $x\in G$, that are touched by the
hyperbola

$H=\left\\{{z\in\mathbb{R}^{2}}:\;c(y,z)=\phi_{G}(y)\right\\}=\left\\{{z\in\mathbb{R}^{2}}:\;z_{2}=y_{2}+\frac{\phi_{G}(y)}{z_{1}-y_{1}}\right\\};$

see Figure 1. Surprisingly, as a consequence of (3), the optimal plan $\gamma$
possesses properties of solutions to classical unconstrained problems. By
Corollary 2.3, the $x$-marginal of $\gamma$ is a Fréchet-Hoeffding coupling
between its first and second coordinates, while, by Corollary 2.5, $\gamma$ is
a classical optimal coupling between its $x$\- and $y$-marginals.

In Theorem 3.2 we show that the set $G$ from (3) is a solution of the _dual_
problem:

$\text{maximize}\quad\int\phi_{G}d\nu\quad\text{over}\quad G\in\mathfrak{M},$
(4)

where $\mathfrak{M}$ is the family of all maximal monotone sets in
$\mathbb{R}^{2}$, and that primal and dual problems (2) and (4) have identical
values. The dual problem appears in (Rochet and Vila, 1994, Eq. (2.3)), where
$G$ stands for the graph of a pricing rule. When $\nu=\operatorname{{Law}}(Y)$
has a Gaussian or, more generally, elliptically contoured distribution, $G$
becomes a line with strictly positive slope; see Example 5.1.

In Theorem 4.1, we show that optimal map and plan problems (1) and (2) have
identical values, provided that $\nu$ is atomless. The result is similar to
that of Pratelli (2007) for the classical unconstrained case. The existence of
an optimal map $X$ for (1) that induces an optimal plan
$\gamma=\operatorname{{Law}}(X,Y)$ for (2) is obtained in Theorem 4.5 under
the condition that $\nu$ gives zero mass to the graphs of _strictly
decreasing_ Lipschitz functions. This assumption is weaker than the standard
regularity condition of the Brenier theorem, see (Ambrosio and Gigli, 2013,
Theorem 1.26), that requires $\nu$ to assign zero mass to rotations of the
graphs of Lipschitz functions. Our second main result, Theorem 4.6,
establishes the uniqueness of solutions to (1) and (2) if, in addition, the
(one-dimensional) distribution functions of $Y_{1}$ and $Y_{2}$ are
continuous. Examples 5.2 and 5.3 show that the conditions of Theorems 4.1 and
4.5 are sharp.

Being applied to the model of Rochet and Vila (1994), Theorems 4.5 and 4.6
yield sufficient conditions for the existence and uniqueness of equilibria,
which are stated in Theorem 6.7. These assumptions generalize those in Rochet
and Vila (1994), where $Y=(U,V)$ is required to have a continuous compactly
supported density in $\mathbb{R}^{2}$. Rochet and Vila (1994) work with dual
problem (4) and rely on the properties of the space of closed graph
correspondences endowed with the Hausdorff topology.

Finally, Appendix A contains a density result for the Wasserstein spaces, for
which we could not find a ready reference, while Appendix B collects the
properties of the function $\phi_{G}$ from (3).

##  2 A backward martingale optimal transport problem

We denote by $\mathcal{P}_{2}(\mathbb{R}^{d})$ the family of Borel probability
measures with finite second moments and by $\mathcal{B}(\mathbb{R}^{d})$ the
Borel $\sigma$-algebra on $\mathbb{R}^{d}$. For a Borel probability measure
$\mu$ on $\mathbb{R}^{d}$, a $\mu$-integrable $m$-dimensional Borel function
$f=(f_{1},\dots,f_{m})$, and an $n$-dimensional Borel function
$g=(g_{1},\dots,g_{n})$, the notation $\mu(f|g)$ stands for the
$m$-dimensional vector of conditional expectations of $f_{i}$ given $g$ under
$\mu$:

$\mu(f|g)=(\mu(f_{1}|g_{1},\dots,g_{n}),\dots,\mu(f_{m}|g_{1},\dots,g_{n})).$

Similarly, $\int fd\mu=(\int f_{1}d\mu,\dots,\int f_{m}d\mu)$. We write a
point in $\mathbb{R}^{4}=\mathbb{R}^{2}\times\mathbb{R}^{2}$ as $(x,y)$, where
$x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$ belong to $\mathbb{R}^{2}$, and think
about $x$ and $y$ as the initial and terminal values of the canonical two-
dimensional process.

Let $\nu=\nu(dy)\in\mathcal{P}_{2}(\mathbb{R}^{2})$. We denote by
$\Gamma(\nu)$ the family of probability measures
$\gamma=\gamma(dx,dy)\in\mathcal{P}_{2}(\mathbb{R}^{2}\times\mathbb{R}^{2})$
that have $\nu$ as their $y$-marginal and make a martingale out of the
canonical process:

$\Gamma(\nu)\triangleq\left\\{{\gamma\in\mathcal{P}_{2}(\mathbb{R}^{2}\times\mathbb{R}^{2})}:\;\gamma(\mathbb{R}^{2},dy)=\nu(dy)\text{
and }\gamma(y|x)=x\right\\}.$

Our goal is to

$\text{minimize}\quad\int
c(x,y)d\gamma\quad\text{over}\quad\gamma\in\Gamma(\nu)$ (5)

for the _covariance-type_ cost function

$c(x,y)=(x_{1}-y_{1})(x_{2}-y_{2}),\quad x,y\in\mathbb{R}^{2}.$

Problem (5) belongs to the class of optimal transport problems with _backward_
martingale constraints, in the sense that the initial $x$-marginal is part of
the solution. As we shall see in Section 6, such problem naturally arises in
the study of the Kyle-type equilibrium with insider.

######  Remark 2.1.

Problem (5) admits several equivalent formulations. For instance, it has same
solutions as the one, where we

$\text{maximize}\int
x_{1}x_{2}d\gamma\quad\text{over}\quad\gamma\in\Gamma(\nu).$ (6)

The justification comes from the identity

$\displaystyle\int c(x,y)d\gamma$
$\displaystyle=\int(x_{1}-y_{1})(x_{2}-y_{2})d\gamma=\int(y_{1}y_{2}-x_{1}x_{2})d\gamma$
$\displaystyle=\int y_{1}y_{2}d\nu-\int x_{1}x_{2}d\gamma,$

where the second equality holds by the martingale property of
$\gamma\in\Gamma(\nu)$.

For a Borel probability measure $\gamma$ on $\mathbb{R}^{d}$ we denote by
$\operatorname{supp}{\gamma}$ its _support_ , that is, the smallest closed set
with full measure. We recall that a set $G\subset\mathbb{R}^{2}$ is _monotone_
if

$c(r,s)=(r_{1}-s_{1})(r_{2}-s_{2})\geq 0,\quad r,s\in G.$

A monotone set $G$ is _maximal_ if it is not a proper (or strict) subset of a
monotone set. We denote by $\mathfrak{M}$ the family of maximal monotone sets
in $\mathbb{R}^{2}$. It is well-known that $G\in\mathfrak{M}$ if and only if
$G$ is the graph of the subdifferential of a proper closed convex function on
$\mathbb{R}$.

For $G\in\mathfrak{M}$ we define a function

$\phi_{G}(y)\triangleq\inf_{x\in G}c(x,y)=\inf_{x\in
G}(x_{1}-y_{1})(x_{2}-y_{2}),\quad y\in\mathbb{R}^{2}.$

Such functions $\phi_{G}$ will play a key role in our study. Their properties
are collected in Appendix B. In particular, Lemma B.1 states that $\phi_{G}$
takes values in $[-\infty,0]$ and
$G=\left\\{{x\in\mathbb{R}^{2}}:\;\phi_{G}(x)=0\right\\}$.

The main results of the paper are Theorems 2.2 and 4.6. Theorem 2.2
establishes the existence of an optimal plan $\gamma$ for (5) and shows the
structure of its support. Theorem 4.6 contains a uniqueness result.

######  Theorem 2.2.

Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{2})$. An optimal plan for (5) exists.
For a probability measure $\gamma\in\Gamma(\nu)$ the following conditions are
equivalent:

  1. (a)

$\gamma$ is an optimal plan for (5).

  2. (b)

If points $(x^{0},y^{0})$ and $(x^{1},y^{1})$ belong to
$\operatorname{supp}{\gamma}$, then

$(1-t)c(x^{0},y^{0})+tc(x^{1},y^{1})\leq t(1-t)c(y^{0},y^{1}),\quad
t\in[0,1].$ (7)

  3. (c)

There is $G\in\mathfrak{M}$ such that

$c(x,y)\leq\phi_{G}(y),\quad(x,y)\in\operatorname{supp}{\gamma}.$ (8)

Moreover, if $G$ is a maximal monotone set satisfying (8) and $\mu$ is the
$x$-marginal of $\gamma$, then $G$ contains $\operatorname{supp}{\mu}$ and

$c(x,y)=\phi_{G}(y)=\min_{z\in
G}c(z,y),\quad(x,y)\in\operatorname{supp}{\gamma}.$

$x^{1}$$y^{1}$$x^{0}$$y^{0}$$H^{1}$$H^{0}$$G$ Figure 1: Hyperbolic non-
crossing and tangent properties of the support of optimal plan.

Figure 1 illustrates the properties of the support of an optimal plan $\gamma$
stated in Theorem 2.2. Let $(x^{0},y^{0})$ and $(x^{1},y^{1})$ belong to
$\operatorname{supp}{\gamma}$ and be such that $x^{0}\not=x^{1}$ and the
points $y^{0}$ and $y^{1}$ lie, respectively, strictly above and strictly
below the maximal monotone set $G$ from item (c). As Lemma 2.11 shows, item
(b) means that the hyperbolas

$\displaystyle
H^{0}=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y^{0})=c(x^{0},y^{0}),\;z_{1}>y^{0}_{1}\right\\},$
$\displaystyle
H^{1}=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y^{1})=c(x^{1},y^{1}),\;z_{1}<y^{1}_{1}\right\\},$

do _not cross_. The geometric interpretation of item (c) is that these
hyperbolas are _tangent_ to $G$.

Before proceeding with the proof of Theorem 2.2, we establish rather
surprising connections between an optimal martingale plan for (5) and the
solutions of classical unconstrained optimal transport problems. If $\mu$ and
$\nu$ are Borel probability measures on $\mathbb{R}^{d}$, then $\Pi(\mu,\nu)$
denotes the family of all couplings of $\mu$ and $\nu$, that is, the family of
Borel probability measures $\pi$ on
$\mathbb{R}^{d}\times\mathbb{R}^{d}=\left\\{{(x,y)}:\;x,y\in\mathbb{R}^{d}\right\\}$
with $x$-marginal $\mu$ and $y$-marginal $\nu$. For
$\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$, the Wasserstein 2-metric is given
by

$W_{2}(\mu,\nu)\triangleq\inf_{\pi\in\Pi(\mu,\nu)}\sqrt{\int\left\lvert
x-y\right\rvert^{2}d\pi}.$ (9)

######  Corollary 2.3.

Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{2})$, $\mu$ be the $x$-marginal of an
optimal plan $\gamma$ for (5), and $\mu_{i}$ be the $x_{i}$-marginal of $\mu$,
$i=1,2$. Then $\mu$ is a solution of the optimal transport problem:

$\text{maximize}\quad\int
x_{1}x_{2}d\pi\quad\text{over}\quad\pi\in\Pi(\mu_{1},\mu_{2}),$ (10)

or, equivalently,

$W_{2}(\mu_{1},\mu_{2})=\sqrt{\int\left\lvert
x_{1}-x_{2}\right\rvert^{2}d\mu}.$

###### Proof.

It is well-known that problems (9) and (10) have same solutions and that an
element of $\Pi(\mu_{1},\mu_{2})$ is such a solution if and only if its
support belongs to a cyclically monotone set. By Theorem 2.2, there is a
monotone set $G$ that contains the support of $\mu$. Since every monotone set
in $\mathbb{R}^{2}$ is also cyclically monotone, the result follows. ∎

######  Remark 2.4.

Let $G$ be a maximal monotone set from Theorem 2.2 and $P_{1}$ be its
projection on $x_{1}$-coordinate. If $\mu_{1}$ is atomless, then the
increasing function

$f(x_{1})=\inf\left\\{{x_{2}\in\mathbb{R}}:\;(x_{1},x_{2})\in G\right\\},\quad
x_{1}\in P_{1},$

taking values in $\mathbb{R}\cup\left\\{{-\infty}\right\\}$, defines an
optimal map solution to (10):

$\mu(B)=\mu_{1}\left\\{{t\in\mathbb{R}}:\;(t,f(t))\in B\right\\},\quad
B\in\mathcal{B}(\mathbb{R}^{2}).$

The function $f$ is a _pricing rule_ in a version of the Kyle equilibrium with
insider studied in Section 6.

######  Corollary 2.5.

Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{2})$, $\gamma$ be an optimal plan for
(5), and $\mu$ be the $x$-marginal of $\gamma$. Then $\gamma$ is a solution of
the optimal transport problem:

$\text{minimize}\quad\int c(x,y)d\pi\quad\text{over}\quad\pi\in\Pi(\mu,\nu).$
(11)

###### Proof.

By Theorem 2.2, there is $G\in\mathfrak{M}$ such that

$\int\phi d\nu=\int c(x,y)d\gamma,$

where $\phi=\phi_{G}$. Lemma B.1 shows that the $c$-conjugate function

$\phi^{c}(x)\triangleq\inf_{y\in\mathbb{R}^{2}}(c(x,y)-\phi(y)),\quad
x\in\mathbb{R}^{2},$

takes values in $[-\infty,0]$ and
$G=\left\\{{x\in\mathbb{R}^{2}}:\;\phi^{c}(x)=0\right\\}$. By Theorem 2.2,
$\operatorname{supp}{\mu}\subset G$ and thus,

$\int\phi^{c}d\mu=0.$

Since $\phi^{c}(x)+\phi(y)\leq c(x,y)$, we deduce that

$\displaystyle\int c(x,y)d\gamma=\int\phi d\nu+\int\phi^{c}d\mu\leq\int
c(x,y)d\pi,\quad\pi\in\Pi(\mu,\nu),$

and the optimality of $\gamma$ for (11) follows. ∎

######  Remark 2.6.

We point out that the assertions of the corollaries are not sufficient for the
optimality of $\gamma\in\Gamma(\nu)$. Indeed, let $\gamma$ be the simplest
martingale measure, whose $x$-marginal $\mu$ is the Dirac measure concentrated
at the mean $\int yd\nu\in\mathbb{R}^{2}$. In this case, the families
$\Pi(\mu_{1},\mu_{2})$ and $\Pi(\mu,\nu)$ are singletons and hence, $\mu$ and
$\gamma$ are trivial solutions to (10) and (11). An elementary analysis of (8)
shows that such $\gamma$ is optimal for (5) if and only if the support of
$\nu$ belongs to a line with negative or infinite slope.

The rest of the section is devoted to the proof of Theorem 2.2, which we
divide into lemmas. We start with the existence part and recall some basic
facts on the Wasserstein distance $W_{2}$; see (Ambrosio and Gigli, 2013,
Theorem 2.7 and Proposition 2.4). If $(\mu_{n})$ and $\mu$ are in
$\mathcal{P}_{2}(\mathbb{R}^{d})$, then $W_{2}(\mu_{n},\mu)\to 0$ if and only
if $\int f(x)d\mu_{n}\to\int f(x)d\mu$ for every continuous function $f=f(x)$
on $\mathbb{R}^{d}$ with quadratic growth:

$\left\lvert f(x)\right\rvert\leq K(1+\left\lvert x\right\rvert^{2}),\quad
x\in\mathbb{R}^{d},$

where $K=K(f)>0$ is a constant. A set
$A\subset\mathcal{P}_{2}(\mathbb{R}^{d})$ is _pre-compact_ under $W_{2}$ if
and only if

$\sup_{\mu\in A}\int_{\left\lvert x\right\rvert\geq K}\left\lvert
x\right\rvert^{2}d\mu\to 0,\quad K\to\infty.$

######  Lemma 2.7.

The family $\Gamma(\nu)$ is a convex compact set in
$\mathcal{P}_{2}(\mathbb{R}^{2}\times\mathbb{R}^{2})$ under the Wasserstein
metric $W_{2}$.

###### Proof.

The martingale property $\gamma(y|x)=x$ of $\gamma\in\Gamma(\nu)$ is
equivalent to the identity

$\int f(x)(y-x)d\gamma=0$

for every bounded and continuous function $f=f(x)$ on $\mathbb{R}^{2}$. The
convexity and the closedness of $\Gamma(\nu)$ under $W_{2}$ readily follow. It
only remains to be shown that $\Gamma(\nu)$ is pre-compact under $W_{2}$ or,
equivalently, that

$\sup_{\gamma\in\Gamma(\nu)}\int(\left\lvert x\right\rvert^{2}+\left\lvert
y\right\rvert^{2})1_{\left\\{{\left\lvert x\right\rvert+\left\lvert
y\right\rvert\geq K}\right\\}}d\gamma\to 0,\quad K\to\infty.$

For $\gamma\in\Gamma(\nu)$ we have that

$\displaystyle\frac{1}{2}\int(\left\lvert x\right\rvert^{2}+\left\lvert
y\right\rvert^{2})1_{\left\\{{\left\lvert x\right\rvert+\left\lvert
y\right\rvert\geq 2K}\right\\}}d\gamma$
$\displaystyle\leq\int\left(\left\lvert
x\right\rvert^{2}1_{\left\\{{\left\lvert x\right\rvert\geq
K}\right\\}}+\left\lvert y\right\rvert^{2}1_{\left\\{{\left\lvert
y\right\rvert\geq K}\right\\}}\right)d\gamma$
$\displaystyle=\int\left\lvert\gamma(y|x)\right\rvert^{2}1_{\left\\{{\left\lvert
x\right\rvert\geq K}\right\\}}d\gamma+\int\left\lvert
y\right\rvert^{2}1_{\left\\{{\left\lvert y\right\rvert\geq K}\right\\}}d\nu$
$\displaystyle\leq\int\left\lvert y\right\rvert^{2}1_{\left\\{{\left\lvert
x\right\rvert\geq K}\right\\}}d\gamma+\int\left\lvert
y\right\rvert^{2}1_{\left\\{{\left\lvert y\right\rvert\geq K}\right\\}}d\nu$

and then that

$\displaystyle\int\left\lvert y\right\rvert^{2}1_{\left\\{{\left\lvert
x\right\rvert\geq K}\right\\}}d\gamma$ $\displaystyle\leq\int\left\lvert
y\right\rvert^{2}1_{\left\\{{\left\lvert x\right\rvert\geq K>\left\lvert
y\right\rvert^{2}}\right\\}}d\gamma+\int\left\lvert
y\right\rvert^{2}1_{\left\\{{\left\lvert y\right\rvert^{2}\geq
K}\right\\}}d\nu$ $\displaystyle\leq K\gamma(\left\lvert x\right\rvert\geq
K)+\int\left\lvert y\right\rvert^{2}1_{\left\\{{\left\lvert
y\right\rvert\geq\sqrt{K}}\right\\}}d\nu.$

Finally, we obtain that

$\displaystyle K\gamma(\left\lvert x\right\rvert\geq
K)\leq\frac{1}{K}\int\left\lvert
x\right\rvert^{2}d\gamma=\frac{1}{K}\int\left\lvert\gamma(y|x)\right\rvert^{2}d\gamma\leq\frac{1}{K}\int\left\lvert
y\right\rvert^{2}d\nu,$

and the result follows. ∎

######  Lemma 2.8.

An optimal plan $\gamma$ for (5) exists.

###### Proof.

Let $(\gamma_{n})$ be a sequence in $\Gamma(\nu)$ such that

$\lim_{n\to\infty}\int c(x,y)d\gamma_{n}=\inf_{\zeta\in\Gamma(\nu)}\int
c(x,y)d\zeta.$

By Lemma 2.7, $\Gamma(\nu)$ is compact under $W_{2}$. Hence, there is a
subsequence $(\gamma_{n_{k}})\subset(\gamma_{n})$ that converges to
$\gamma\in\Gamma(\nu)$ under $W_{2}$. Since the cost function $c=c(x,y)$ is
continuous and has quadratic growth, we deduce that

$\int c(x,y)d\gamma=\lim_{k\to\infty}\int
c(x,y)d\gamma_{n_{k}}=\inf_{\zeta\in\Gamma}\int c(x,y)d\zeta.$

Thus, $\gamma$ is an optimal plan. ∎

The implication (a)$\implies$(b) of Theorem 2.2 is proved in Lemma 2.10 and
relies on the following first-order optimality condition.

######  Lemma 2.9.

Let $\gamma$ be an optimal plan for (5). Then

$\int c(x,y)d\eta\leq\int y_{1}y_{2}d\eta-\int y_{1}d\eta\int y_{2}d\eta,$
(12)

for every $\eta\in\mathcal{P}_{2}(\mathbb{R}^{2}\times\mathbb{R}^{2})$ such
that $\operatorname{supp}\eta\subset\operatorname{supp}\gamma$.

###### Proof.

We first establish (12) for a Borel probability measure $\eta$ on
$\mathbb{R}^{2}\times\mathbb{R}^{2}$ that has a bounded density with respect
to $\gamma$:

$V(x,y)=\frac{d\eta}{d\gamma}\in\mathcal{L}^{\infty}(\mathbb{R}^{2}\times\mathbb{R}^{2}).$

We choose a non-atom $q\in\mathbb{R}^{2}$ of
$\mu(dx)=\gamma(dx,\mathbb{R}^{2})$ and define the probability measure

$\zeta(dx,dy)=\delta_{q}(dx)\eta(\mathbb{R}^{2},dy),$

where $\delta_{q}$ is the Dirac measure concentrated at $q$. For sufficiently
small $\varepsilon>0$ the probability measure

$\widetilde{\gamma}=\gamma+\varepsilon(\zeta-\eta)$

is well-defined and has the same $y$-marginal $\nu$ as $\gamma$. We define the
conditional expectation $\widetilde{X}(x)=\widetilde{\gamma}(y|x)$ and observe
that the law of $(\widetilde{X},y)$ under $\widetilde{\gamma}$ belongs to
$\Gamma(\nu)$. The optimality of $\gamma$ for (5) or, equivalently, for (6)
implies that

$\int\widetilde{X}_{1}\widetilde{X}_{2}d\widetilde{\gamma}\leq\int
x_{1}x_{2}d\gamma.$ (13)

Standard computations based on Bayes formula show that

$\displaystyle\widetilde{X}(x)=1_{\left\\{{x\not=q}\right\\}}\frac{x-\varepsilon
R(x)}{1-\varepsilon U(x)}+1_{\left\\{{x=q}\right\\}}m,$

where $U(x)=\gamma(V|x)$, $R(x)=\gamma(Vy|x)$, and $m=\int yd\eta$. Since
$\left\lvert V\right\rvert\leq K$ for some constant $K>0$, we deduce that
$\left\lvert U\right\rvert\leq K$ and $\left\lvert R\right\rvert\leq
K\gamma(\left\lvert y\right\rvert|x)$. It follows that

$\displaystyle\int\widetilde{X}_{1}\widetilde{X}_{2}d\widetilde{\gamma}$
$\displaystyle=\int\frac{(x_{1}-\varepsilon R_{1})(x_{2}-\varepsilon
R_{2})}{(1-\varepsilon U)^{2}}(1-\varepsilon V)d\gamma+\varepsilon m_{1}m_{2}$
$\displaystyle=\int\frac{(x_{1}-\varepsilon R_{1})(x_{2}-\varepsilon
R_{2})}{1-\varepsilon U}d\gamma+\varepsilon m_{1}m_{2}$ $\displaystyle=\int
x_{1}x_{2}d\gamma+\varepsilon\left(m_{1}m_{2}+\int(x_{1}x_{2}U-x_{1}R_{2}-x_{2}R_{1})d\gamma\right)+O(\varepsilon^{2}).$

In view of (13), the first-order term is negative. It can be written as

$\displaystyle 0$ $\displaystyle\geq
m_{1}m_{2}+\int(x_{1}x_{2}U-x_{1}R_{2}-x_{2}R_{1})d\gamma$ $\displaystyle=\int
y_{1}d\eta\int y_{2}d\eta+\int(x_{1}x_{2}-x_{1}y_{2}-x_{2}y_{1})Vd\gamma$
$\displaystyle=\int y_{1}d\eta\int
y_{2}d\eta+\int(x_{1}x_{2}-x_{1}y_{2}-x_{2}y_{1})d\eta$ $\displaystyle=\int
y_{1}d\eta\int y_{2}d\eta+\int(c(x,y)-y_{1}y_{2})d\eta$

and the result follows.

In the general case, where
$\eta\in\mathcal{P}_{2}(\mathbb{R}^{2}\times\mathbb{R}^{2})$ and
$\operatorname{supp}{\eta}\subset\operatorname{supp}{\gamma}$, we use the
approximation result from Appendix A. By Theorem A.1, there are Borel
probability measures $(\eta_{n})$ on $\mathbb{R}^{2}\times\mathbb{R}^{2}$ that
have bounded densities with respect to $\gamma$ and converge to $\eta$ under
$W_{2}$. By what we have already proved,

$\int c(x,y)d\eta_{n}\leq\int y_{1}y_{2}d\eta_{n}-\int y_{1}d\eta_{n}\int
y_{2}d\eta_{n},\quad n\geq 1.$

Since the integrands are continuous functions with quadratic growth, we can
pass to the limit as $n\to\infty$ and obtain (12). ∎

######  Lemma 2.10.

Let $\gamma$ be an optimal plan for (5). Then condition (b) of Theorem 2.2
holds.

###### Proof.

Lemma 2.9 yields inequality (12) for the probability measure

$\eta(dx,dy)=(1-t)\delta_{(x^{0},y^{0})}(dx,dy)+t\delta_{(x^{1},y^{1})}(dx,dy),$

where $t\in[0,1]$, $(x^{i},y^{i})\in\operatorname{supp}{\gamma}$, and
$\delta_{(x^{i},y^{i})}$ is the Dirac measure concentrated at $(x^{i},y^{i})$,
$i=0,1$. Elementary computations show that for such $\eta$ (12) becomes (7). ∎

The equivalence of assertions (b) and (c) of Theorem 2.2 is a special case of
Lemma 2.12, whose proof relies on the following geometric interpretation of
(7). Figure 1 visualizes the arguments.

######  Lemma 2.11.

Let $x^{i}$ and $y^{i}$, $i=0,1$, be points in $\mathbb{R}^{2}$ such that
$y^{0}_{1}<y^{1}_{1}$. Then (7) holds if and only if for all
$a_{i}<c(x^{i},y^{i})$, $i=0,1$, the graphs of the hyperbolas

$\displaystyle h^{0}(s)$
$\displaystyle=y^{0}_{2}+\frac{a_{0}}{s-y^{0}_{1}},\quad s>y^{0}_{1},$
$\displaystyle h^{1}(s)$
$\displaystyle=y^{1}_{2}+\frac{a_{1}}{s-y^{1}_{1}},\quad s<y^{1}_{1},$

do not intersect:

$h^{0}(s)<h^{1}(s),\quad s\in(y^{0}_{1},y^{1}_{1}).$

###### Proof.

The result follows from the identity:

$\displaystyle(h^{1}(s)-h^{0}(s))(s-y^{0}_{1})(y^{1}_{1}-s)$
$\displaystyle\qquad=(y^{1}_{2}-y^{0}_{2})(s-y^{0}_{1})(y^{1}_{1}-s)-a_{0}(y^{1}_{1}-s)-a_{1}(s-y^{0}_{1})$
$\displaystyle\qquad=(y^{1}_{1}-y^{0}_{1})\left(t(1-t)c(y^{0},y^{1})-(1-t)a_{0}-ta_{1}\right),$

where $s\in(y^{0}_{1},y^{1}_{1})$ and $t=(s-y^{0}_{1})/(y^{1}_{1}-y^{0}_{1})$.
∎

######  Lemma 2.12.

For a set $A\subset\mathbb{R}^{2}\times\mathbb{R}^{2}$ the following
conditions are equivalent:

  1. (i)

If points $(x^{0},y^{0})$ and $(x^{1},y^{1})$ belong to $A$, then (7) holds.

  2. (ii)

There is $G\in\mathfrak{M}$ such that

$c(x,y)\leq\phi_{G}(y),\quad(x,y)\in A.$

###### Proof.

We observe first that under either (i) or (ii),

$c(x,y)\leq 0,\quad(x,y)\in A.$

Indeed, under (i) this inequality follows from (7), while under (ii) it holds
because $\phi_{G}\leq 0$. We define the open sets

$\displaystyle B^{i}=\cup_{(x,y)\in A}B^{i}(x,y),\quad i=0,1,$

where, for $(x,y)\in A$,

$\displaystyle B^{0}(x,y)$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y)<c(x,y),\;z_{1}>y_{1}\right\\}$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;z_{2}<y_{2}+\frac{c(x,y)}{z_{1}-y_{1}},\;z_{1}>y_{1}\right\\},$
$\displaystyle B^{1}(x,y)$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y)<c(x,y),\;z_{1}<y_{1}\right\\}$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;z_{2}>y_{2}+\frac{c(x,y)}{z_{1}-y_{1}},\;z_{1}<y_{1}\right\\}.$

The boundaries of $B^{i}$, $i=0,1$, are, respectively, upper and lower
envelopes of the graphs of increasing hyperbolas and thus, are maximal
monotone sets.

By Lemma 2.11, item (i) holds if and only if the sets $B^{0}$ and $B^{1}$ are
disjoint:

$B^{0}\cap B^{1}=\emptyset.$ (14)

On the other hand, item (ii) holds if and only if the closed set

$C=\cap_{(x,y)\in A}\left\\{{z\in\mathbb{R}^{2}}:\;c(x,y)\leq c(z,y)\right\\}$

contains a maximal monotone set $G$. As

$C=\mathbb{R}^{2}\setminus\left(B^{0}\cup B^{1}\right),$

every such set $G$ separates $B^{0}$ and $B^{1}$. Hence, its existence yields
(14) and then (i). Conversely, if the sets $B^{0}$ and $B^{1}$ are disjoint,
then their boundaries belong to $C$. As the boundaries are maximal monotone
sets, we obtain (ii). ∎

The remaining assertions of the theorem follow from Lemma 2.14. A key role is
played by inequality (15).

######  Lemma 2.13.

Let $\gamma\in\Gamma(\nu)$ and $\mu$ be its $x$-marginal. For every
$G\in\mathfrak{M}$ we have that

$\gamma(\phi_{G}(y)-c(x,y)|x)\leq\phi_{G}(x)\leq 0,\quad\gamma\text{-a.s.},$

and then that

$\int\phi_{G}(y)d\nu-\int c(x,y)d\gamma\leq\int\phi_{G}(x)d\mu\leq 0.$ (15)

###### Proof.

We only need to prove the inequality with conditional expectations. From Lemma
B.1 we know that $\phi_{G}\leq 0$. As $x=\gamma(y|x)$, we deduce that for
every $r\in\mathbb{R}^{2}$:

$\gamma(c(y,r)-c(x,y)|x)=c(x,r),\quad\gamma\text{-a.s.},$

and taking $\inf$ over a dense countable set of $r\in G$ obtain the result. ∎

The following lemma completes the proof of the theorem.

######  Lemma 2.14.

Let $\gamma\in\Gamma(\nu)$, $\mu$ be its $x$-marginal, and $G\in\mathfrak{M}$
be such that

$\int c(x,y)d\gamma\leq\int\phi_{G}(y)d\nu.$ (16)

Then in (16) we actually have the equality, $\gamma$ is an optimal plan for
(5), the set $G$ contains $\operatorname{supp}{\mu}$, and

$c(x,y)=\phi_{G}(y)=\min_{r\in
G}c(r,y),\quad(x,y)\in\operatorname{supp}{\gamma}.$ (17)

###### Proof.

We shall write $\phi$ for $\phi_{G}$. From (15) and (16) we deduce that
$\gamma$ is a solution to (5), that in (16) we have an equality, and that

$\int\phi(x)d\mu=0.$

Lemma B.1 states that $\phi\leq 0$ and
$G=\left\\{{x\in\mathbb{R}^{2}}:\;\phi(x)=0\right\\}$. It follows that
$\mu(G)=1$. Being a closed set, $G$ contains $\operatorname{supp}{\mu}$. In
particular, if $(x,y)\in\operatorname{supp}{\gamma}$, then $x\in G$. It
follows that

$c(x,y)\geq\inf_{r\in
G}c(r,y)=\phi(y),\quad(x,y)\in\operatorname{supp}{\gamma}.$

Accounting for (16), we deduce that

$c(x,y)=\phi(y),\quad\gamma\text{-a.s.}.$

Hence, for every $(x,y)\in\operatorname{supp}{\gamma}$ we can find a sequence
$\left\\{{(x_{n},y_{n})}\right\\}\subset\operatorname{supp}{\gamma}$ that
converges to $(x,y)$ and such that $c(x_{n},y_{n})=\phi(y_{n})$, $n\geq 1$.
Being a pointwise infinum of continuous functions, the function $\phi$ is
upper semi-continuous. It follows that

$c(x,y)=\lim_{n\to\infty}c(x_{n},y_{n})=\lim_{n\to\infty}\phi(y_{n})\leq\phi(y),$

and we obtain (17). ∎

##  3 Dual problem

In view of Theorem 2.2 and Lemma 2.13, a natural dual problem to (5) is to

$\text{maximize}\quad\int\phi_{G}d\nu\quad\text{over}\quad G\in\mathfrak{M}.$
(18)

Such problem appears in Rochet and Vila (1994) in connection to their study of
Kyle-type equilibrium with insider; see Section 6. They use a direct method
based on the properties of the space of closed graph correspondences and
assume that $\nu$ has a compactly supported density.

We recall that $G=\left\\{{x\in\mathbb{R}^{2}}:\;\phi_{G}(x)=0\right\\}$,
$G\in\mathfrak{M}$, and thus the family $\mathfrak{M}$ of all maximal monotone
sets in $\mathbb{R}^{2}$ is in one-to-one correspondence with the family of
functions

$\Phi\triangleq\left\\{{\phi_{G}}:\;G\in\mathfrak{M}\right\\}.$

Hence, (18) is equivalent to the problem, where we

$\text{maximize}\quad\int\phi d\nu\quad\text{over}\quad\phi\in\Phi.$

A technical inconvenience of the set $\Phi$ is the absence of convexity. It
turns out that the set of functions dominated by the elements of $\Phi$ is not
only convex, but also admits a self-contained description related to item (b)
of Theorem 2.2.

######  Lemma 3.1.

Let $\phi:\;\mathbb{R}^{2}\rightarrow[-\infty,0]$ be a Borel function. Then
$\phi\leq\phi_{G}$ for some $G\in\mathfrak{M}$ if and only if

$(1-t)\phi(y^{0})+t\phi(y^{1})\leq t(1-t)c(y^{0},y^{1}),\quad
y^{0},y^{1}\in\mathbb{R}^{2},\;t\in[0,1].$ (19)

The set of such functions $\phi$ is convex.

###### Proof.

The result follows directly from Lemma 2.12, where we take

$A=\left\\{{(x,y)\in\mathbb{R}^{2}\times\mathbb{R}^{2}}:\;c(x,y)=\phi(y)\right\\}.$

Clearly, the family of functions $\phi$ satisfying (19) is convex. ∎

######  Theorem 3.2.

Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{2})$. We have that

$\min_{\gamma\in\Gamma(\nu)}\int
c(x,y)d\gamma=\max_{G\in\mathfrak{M}}\int\phi_{G}d\nu,$

where the lower and upper bounds are attained at respective solutions to (5)
and (18). A probability measure $\gamma\in\Gamma(\nu)$ and a maximal monotone
set $G$ are such solutions if and only if

$c(x,y)=\phi_{G}(y),\quad(x,y)\in\operatorname{supp}{\gamma}.$ (20)

In this case, $G$ contains the support of the $x$-marginal of $\gamma$.
Moreover, $\phi_{G}$ and $G$ are uniquely defined on
$\operatorname{supp}{\nu}$, that is,

$\displaystyle\phi_{G}(y)$ $\displaystyle=\phi_{\widetilde{G}}(y),\quad
y\in\operatorname{supp}{\nu},$ $\displaystyle G\cap\operatorname{supp}{\nu}$
$\displaystyle=\widetilde{G}\cap\operatorname{supp}{\nu},$

for any other solution $\widetilde{G}$ to (18). In particular, $\phi_{G}$ and
$G$ are unique if $\operatorname{supp}{\nu}=\mathbb{R}^{2}$.

###### Proof.

With an exception of the uniqueness part, all other assertions follow directly
from Theorem 2.2 and Lemmas 2.13 and 2.14.

Let $\gamma$ be a solution to (5), $G$ and $\widetilde{G}$ be solutions to
(18) and denote $\phi=\phi_{G}$ and $\widetilde{\phi}=\phi_{\widetilde{G}}$.
From (20) we deduce that the functions $\phi$ and $\widetilde{\phi}$ coincide
on $P_{y}$, the projection of $\operatorname{supp}{\gamma}$ on
$y$-coordinates. Since every $y\in\operatorname{supp}{\nu}$ is the limit of a
sequence $(y_{n})\subset P_{y}$, Lemma B.3 yields that

$\phi(y)=\lim_{n\to\infty}\phi(y_{n})=\lim_{n\to\infty}\widetilde{\phi}(y_{n})=\widetilde{\phi}(y).$

We have proved the uniqueness of $\phi_{G}$ on $\operatorname{supp}{\nu}$. The
uniqueness of $G$ on $\operatorname{supp}{\nu}$ holds as
$G=\left\\{{x\in\mathbb{R}^{2}}:\;\phi_{G}(x)=0\right\\}$. ∎

##  4 Optimal maps

For simplicity of notations, we slightly modify the setup. We start with a
2-dimensional random variable $Y=(Y_{1},Y_{2})$ having a finite second moment:
$Y\in\mathcal{L}^{2}=\mathcal{L}^{2}(\Omega,\mathcal{F},\mathbb{P})$. As
usual, we identify random variables that differ only on a set of measure zero.
Our goal is to

$\text{minimize}\quad\mathbb{E}\left(c(X,Y)\right)\quad\text{over}\quad
X\in\mathcal{X}(Y)$ (21)

for the same cost function $c(x,y)=(x_{1}-y_{1})(x_{2}-y_{2})$ and the domain

$\mathcal{X}(Y)\triangleq\left\\{{X=(X_{1},X_{2})\in\mathcal{L}^{2}}:\;X\text{
is $Y$-measurable and
}\mathbb{E}\left(\left.Y\right\lvert{X}\right)=X\right\\}.$

We denote $\nu=\operatorname{{Law}}(Y)$ and observe that
$\operatorname{{Law}}(X,Y)\in\Gamma(\nu)$ for every $X\in\mathcal{X}(Y)$.
Thus, optimal _plan_ problem (5) may be viewed as a Kantorovich-type
relaxation of optimal _map_ problem (21). In general,

$\min_{\gamma\in\Gamma(\nu)}\int
c(x,y)d\gamma\leq\inf_{X\in\mathcal{X}(Y)}\mathbb{E}\left(c(X,Y)\right),$ (22)

and the inequality may be strict and an optimal map may not exist as Examples
5.2 and 5.3 show.

The main results of this sections are Theorems 4.1, 4.5, and 4.6. Theorem 4.1
yields the equality in (22) provided that $\nu=\operatorname{{Law}}(Y)$ is
atomless. Theorem 4.5 shows the existence of an optimal map if $\nu$ is
$\mathcal{D}$-regular in the sense of Definition 4.4. Theorem 4.6 establishes
the uniqueness of optimal plan and map if, in addition, every component
$Y_{i}$ has a continuous distribution function. The last two theorems play a
key role in the study of equilibrium in Section 6.

We shall use the notations from Appendix B related to the function
$\phi=\phi_{G}$, where $G\in\mathfrak{M}$. In particular,
$D^{c}=(D^{c}_{1},D^{c}_{2})$ stands for the differential operator associated
with the cost function $c=c(x,y)$:

$\displaystyle D^{c}_{1}\phi(y)=y_{1}-\frac{\partial\phi}{\partial
y_{2}}(y),\quad D^{c}_{2}\phi(y)=y_{2}-\frac{\partial\phi}{\partial
y_{1}}(y),\quad y\in\operatorname{dom}{\nabla\phi},$

where $\operatorname{dom}{\nabla\phi}$ is the set of points where $\phi$ is
differentiable. We denote by $E^{G}=E_{1}^{G}\cup E_{2}^{G}$ the union of the
vertical and horizontal line segments of $G$:

$\begin{split}E_{i}^{G}(t)&=\left\\{{x=(x_{1},x_{2})\in
G}:\;x_{i}=t\right\\},\quad t\in\mathbb{R},\\\
\mathcal{T}_{i}^{G}&=\left\\{{t\in\mathbb{R}}:\;{E_{i}^{G}(t)}\;\text{has more
than one point}\right\\},\\\
E_{i}^{G}&=\cup_{t\in\mathcal{T}_{i}^{G}}E_{i}^{G}(t),\quad i=1,2.\end{split}$
(23)

Clearly, the sets $(\mathcal{T}^{G}_{i})$ are countable. Finally, we define

$\displaystyle\operatorname{Arg}_{G}(y)$
$\displaystyle=\operatorname*{arg\>min}_{x\in G}c(x,y)=\left\\{{x\in
G}:\;\phi_{G}(y)=c(x,y)\right\\},$
$\displaystyle\operatorname{dom}{\operatorname{Arg}_{G}}$
$\displaystyle=\left\\{{y\in\mathbb{R}^{2}}:\;\operatorname{Arg}_{G}(y)\not=\emptyset\right\\}.$

The following result is similar to that of Pratelli (2007) obtained for the
classical unconstrained optimal transport problem.

######  Theorem 4.1.

Let $Y=(Y_{1},Y_{2})\in\mathcal{L}^{2}$ and suppose that
$\nu=\operatorname{{Law}}(Y)$ is atomless. Then plan and map problems (5) and
(21) have identical values:

$\min_{\gamma\in\Gamma(\nu)}\int
c(x,y)d\gamma=\inf_{X\in\mathcal{X}(Y)}\mathbb{E}\left(c(X,Y)\right).$

The proof of the theorem relies on some lemmas.

######  Lemma 4.2.

Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{2})$, $\gamma\in\Gamma(\nu)$ be an
optimal plan for (5), and $G\in\mathfrak{M}$ be a maximizer for (18). If
$\nu(G)>0$, then the probability measure

$\eta(dx,dy)=\frac{1}{\nu(G)}1_{\left\\{{y\in G}\right\\}}\gamma(dx,dy)$

has the martingale property: $\eta(y|x)=x$.

###### Proof.

We write $\phi$ for $\phi_{G}$. We shall show that $\eta(y_{1}|x)=x_{1}$, that
is, that

$\int f(x)(y_{1}-x_{1})d\eta=\frac{1}{\nu(G)}\int
f(x)(y_{1}-x_{1})1_{\left\\{{y\in G}\right\\}}d\gamma=0,$ (24)

for every bounded Borel function $f$ on $\mathbb{R}^{2}$. The martingale
property for the second coordinate has a similar proof.

Let $E_{2}=E_{2}^{G}=\cup_{t\in\mathcal{T}_{2}}E_{2}(t)$ be the union of the
horizontal line segments of $G$. If $(x,y)\in\operatorname{supp}{\gamma}$,
then Theorem 3.2 yields that $x\in G$ and $c(x,y)=\phi(y)$. If, in addition,
$y\in G\setminus E_{2}$, then $c(x,y)=\phi(y)=0$ and, subsequently,
$x_{1}=y_{1}$. Hence, (24) holds if

$\int f(x)(y_{1}-x_{1})1_{\left\\{{y\in E_{2}(t)}\right\\}}d\gamma=0,\quad
t\in\mathcal{T}_{2}.$ (25)

Hereafter, we fix $t\in\mathcal{T}_{2}$. Let
$(x,y)\in\operatorname{supp}{\gamma}$. If $y\in E_{2}(t)$, then
$c(x,y)=\phi(y)=0$ and thus, $x\in E_{2}(t)$. Conversely, if
$x\in\operatorname*{ri}{E_{2}(t)}$, the relative interior of $E_{2}(t)$, then
Lemma B.4 yields that $y\in G$ and then, as $c(x,y)=\phi(y)=0$, that $y\in
E_{2}(t)$. Hence,

$\displaystyle 1_{\left\\{{y\in E_{2}(t)}\right\\}}$
$\displaystyle=1_{\left\\{{x,y\in E_{2}(t)}\right\\}}$
$\displaystyle=1_{\left\\{{x\in\operatorname*{ri}{E_{2}(t)}}\right\\}}+1_{\left\\{{x=a(t),y\in
E_{2}(t)}\right\\}}+1_{\left\\{{x=b(t),y\in E_{2}(t)}\right\\}},$

where $a(t)$ and $b(t)$ are the boundary points of $E_{2}(t)$ such that
$a_{1}(t)<b_{1}(t)$. Accounting for the martingale property of $\gamma$, we
obtain that

$\displaystyle\int
f(x)(y_{1}-x_{1})1_{\left\\{{x\in\operatorname*{ri}{E_{2}(t)}}\right\\}}d\gamma=0.$

Let $y\in\mathbb{R}^{2}$ be such that $\phi(y)=c(b(t),y)$. If $y\not\in G$,
then Lemma B.4 yields that $b_{1}(t)>y_{1}$. If $y\in G\setminus E_{2}(t)$,
then $c(b(t),y)=\phi(y)=0$ and thus, $b_{1}(t)=y_{1}$. Finally, if $y\in
E_{2}(t)$, then $b_{1}(t)\geq y_{1}$. It follows that

$\displaystyle\int\left\lvert x_{1}-y_{1}\right\rvert 1_{\left\\{{x=b(t),y\in
E_{2}(t)}\right\\}}d\gamma\leq\int(x_{1}-y_{1})1_{\left\\{{x=b(t)}\right\\}}d\gamma=0,$

where at the last step we used the martingale property of $\gamma$. The case
of the left boundary $a(t)$ is similar. We have proved (25). ∎

######  Lemma 4.3.

Let $G\in\mathfrak{M}$ and $X=(X_{1},X_{2})$ and $Y=(Y_{1},Y_{2})$ be random
variables such that $X$ takes values in $G$, $X1_{\left\\{{Y\in
G}\right\\}}=Y1_{\left\\{{Y\in G}\right\\}}$, and $c(X,Y)=\phi_{G}(Y)$. If the
law of $Y$ is atomless, then for every $\epsilon>0$ there is a random variable
$Z=Z(\epsilon)$ such that $\operatorname{{Law}}(Z)=\operatorname{{Law}}(Y)$,
$\left\lvert Z-Y\right\rvert\leq\epsilon$, and $X$ is $Z$-measurable.

###### Proof.

We fix $\epsilon>0$ and denote $\phi=\phi_{G}$,
$\operatorname{Arg}=\operatorname{Arg}_{G}$, and

$D=\operatorname{dom}{\operatorname{Arg}}\setminus(\operatorname{dom}{\nabla\phi}\cup
G).$

Theorems B.6 and B.12 show that $D=\cup_{n}D_{n}$, where $D_{n}$ is either a
point or the graph of a strictly decreasing function. Of course, we can choose
the sets $(D_{n})$ so that

$\operatorname{diam}{D_{n}}\triangleq\sup_{x,y\in D_{n}}\left\lvert
x-y\right\rvert\leq\epsilon,\quad n\geq 1.$

For every $n\geq 1$ we shall construct a two-dimensional random variable
$Z^{n}=(Z^{n}_{1},Z^{n}_{2})$ and a Borel function $f^{n}:\;D_{n}\rightarrow
G$ such that

$\begin{split}\operatorname{{Law}}(Z^{n}1_{\left\\{{Y\in
D_{n}}\right\\}})&=\operatorname{{Law}}(Y1_{\left\\{{Y\in
D_{n}}\right\\}}),\\\ X1_{\left\\{{Y\in
D_{n}}\right\\}}&=f^{n}(Z^{n})1_{\left\\{{Y\in D_{n}}\right\\}}.\end{split}$
(26)

Given the sequence of such pairs $(Z^{n},f^{n})$, $n\geq 1$, we define

$\displaystyle Z$
$\displaystyle=Y1_{\left\\{{Y\in\operatorname{dom}{\nabla\phi}\cup
G}\right\\}}+\sum_{n}Z^{n}1_{\left\\{{Y\in F_{n}}\right\\}},$ $\displaystyle
f(y)$ $\displaystyle=y1_{\left\\{{y\in
G}\right\\}}+D^{c}\phi(y)1_{\left\\{{y\in\operatorname{dom}{\nabla\phi}\setminus
G}\right\\}}+\sum_{n}f^{n}(y)1_{\left\\{{y\in F_{n}}\right\\}}.$

where $F_{n}=D_{n}\setminus\cup_{k<n}D_{k}$. We have that
$\operatorname{{Law}}(Z)=\operatorname{{Law}}(Y)=\nu$ and $\left\lvert
Z-Y\right\rvert\leq\epsilon$. Moreover, in view of Theorem B.6, $X=f(Z)$.
Hence, (26) is all we need to obtain.

Using the conditional probabilities with respect to events $\left\\{{Y\in
D_{n}}\right\\}$, we can reduce the general case to the situation where

$Y\in D=\left\\{{(t,h(t))}:\;t\in[t_{0},t_{1}]\right\\},$

for some strictly decreasing function $h=h(t)$. Since $\nu$ is atomless, every
component $Y_{i}$ has a continuous distribution function
$a_{i}(t)=\mathbb{\mathbb{P}}\left(Y_{i}\leq t\right)$, $i=1,2$. It follows
that $U=a_{1}(Y_{1})$ has the uniform distribution on $[0,1]$ and
$Y_{1}=a_{1}^{-1}(U)$, where $a_{1}^{-1}$ is the pseudo-inverse function to
$a_{1}$:

$a_{1}^{-1}(t)=\inf\left\\{{s\in[t_{0},t_{1}]}:\;a_{1}(s)\geq t\right\\},\quad
t\in[0,1].$

In particular, $Y=(Y_{1},h(Y_{1}))$ is $U$-measurable.

Lemma B.16 yields Borel functions $g_{i}:\;D\rightarrow G$, $i=1,2$, such that
either $X=g_{1}(Y)$ or $X=g_{2}(Y)$. As the functions

$\displaystyle b_{i}(t)=\mathbb{\mathbb{P}}\left(U\leq
t,X=g_{i}(Y)\right),\quad t\in[0,1],\;i=1,2,$

are continuous and increasing, the random variable

$V=b_{1}(U)1_{\left\\{{X=g_{1}(Y)}\right\\}}+(b_{1}(1)+b_{2}(U))1_{\left\\{{X=g_{2}(Y)}\right\\}}$

has the uniform distribution on $[0,1]$. Clearly, $U$ and the indicators
$(1_{\left\\{{X=g_{i}(Y)}\right\\}})$ are $V$-measurable. It follows that $Y$
and $X$ are also $V$-measurable. Setting

$Z_{1}=a_{1}^{-1}(V),\quad Z_{2}=h(Z_{1}),$

we obtain that $Z=(Z_{1},Z_{2})$ has the same law as $Y$, that
$V=a_{1}(Z_{1})$, and that $X$ is $Z$-measurable. ∎

###### Proof of Theorem 4.1.

Let $\gamma$ be an optimal plan for (5). By extending, if necessary, the
underlying probability space we can assume that
$\gamma=\operatorname{{Law}}(X,Y)$ for some random variable $X$. As
$\gamma(y|x)=x$, we have that
$X=\mathbb{E}\left(\left.Y\right\lvert{X}\right)$. Theorem 2.2 yields
$G\in\mathfrak{M}$ such that $X\in G$ and $c(X,Y)=\phi_{G}(Y)$.

We denote $\widetilde{X}=X1_{\left\\{{Y\not\in G}\right\\}}+Y1_{\left\\{{Y\in
G}\right\\}}$ and observe that
$\widetilde{\gamma}=\operatorname{{Law}}(\widetilde{X},Y)$ is another optimal
plan. Indeed, by Lemma 4.2,

$\mathbb{E}\left(\left.(Y-X)1_{\left\\{{Y\in
G}\right\\}}\right\lvert{X}\right)=0$

and therefore, for a bounded Borel function $g=g(x)$ on $\mathbb{R}^{2}$,

$\mathbb{E}\left((Y-\widetilde{X})g(\widetilde{X})\right)=\mathbb{E}\left((Y-X)g(X)1_{\left\\{{Y\not\in
G}\right\\}}\right)=\mathbb{E}\left((Y-X)g(X)\right)=0.$

It follows that
$\mathbb{E}\left(\left.Y\right\lvert{\widetilde{X}}\right)=\widetilde{X}$ and
thus, $\widetilde{\gamma}\in\Gamma(\nu)$. By the construction of
$\widetilde{X}$, we have that $c(X,Y)=c(\widetilde{X},Y)$ and the optimality
of $\widetilde{\gamma}$ follows. This fact allows us to assume from the start
that $X1_{\left\\{{Y\in G}\right\\}}=Y1_{\left\\{{Y\in G}\right\\}}$. Then,
$X$ and $Y$ satisfy the assumptions of Lemma 4.3.

Let $\epsilon>0$ and $Z=Z(\epsilon)$ be the random variable yielded by Lemma
4.3. As $X$ is $Z$-measurable, the conditional expectation
$V\triangleq\mathbb{E}\left(\left.Z\right\lvert{X}\right)$ is also
$Z$-measurable. Thus, there is a Borel function
$f:\;\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ such that $V=f(Z)$. Since $Y$
and $Z$ have identical laws, $U\triangleq
f(Y)=\mathbb{E}\left(\left.Y\right\lvert{U}\right)$. As

$\left\lvert
V_{i}-X_{i}\right\rvert=\left\lvert\mathbb{E}\left(\left.Z_{i}-Y_{i}\right\lvert{X}\right)\right\rvert\leq\mathbb{E}\left(\left.\left\lvert
Z-Y\right\rvert\right\lvert{X}\right)\leq\epsilon,\quad i=1,2,$

we deduce that

$\displaystyle\mathbb{E}\left(c(U,Y)\right)$
$\displaystyle=\mathbb{E}\left(c(V,Z)\right)=\mathbb{E}\left(Z_{1}Z_{2}\right)-\mathbb{E}\left(V_{1}V_{2}\right)$
$\displaystyle\leq\mathbb{E}\left(Y_{1}Y_{2}\right)-\mathbb{E}\left(X_{1}X_{2}\right)+\epsilon\mathbb{E}\left(\left\lvert
X_{1}\right\rvert+\left\lvert V_{2}\right\rvert\right)$
$\displaystyle\leq\mathbb{E}\left(c(X,Y)\right)+\epsilon\mathbb{E}\left(\left\lvert
Y_{1}\right\rvert+\left\lvert Y_{2}\right\rvert\right).$

The result follows, because $\epsilon$ is any positive number. ∎

Let $\mathcal{D}$ be the family of graphs of strictly decreasing functions
$f=f(t)$ defined on closed intervals of $\mathbb{R}$ such that both $f$ and
its inverse $f^{-1}$ are Lipschitz functions:

$\frac{1}{K}(t-s)\leq f(s)-f(t)\leq K(t-s),\quad s<t,$

for some constant $K=K(f)>0$. To make statements shorter we allow for a
degenerate case where the domain of $f$ is just a point. Thus,
$\mathbb{R}^{2}\subset\mathcal{D}$.

######  Definition 4.4.

A Borel probability measure $\mu$ on $\mathbb{R}^{2}$ is _$\mathcal{D}$
-regular_ if $\mu(D)=0$, $D\in\mathcal{D}$.

The following theorem establishes the existence of optimal maps that induce
optimal plans.

######  Theorem 4.5.

Let $Y=(Y_{1},Y_{2})\in\mathcal{L}^{2}$ and suppose that
$\nu=\operatorname{{Law}}(Y)$ is $\mathcal{D}$-regular. Let $G\in\mathfrak{M}$
be a maximizer for (18) and denote $\phi=\phi_{G}$ and $E=E^{G}$. Then

$X=Y1_{\left\\{{Y\in E}\right\\}}+D^{c}\phi(Y)1_{\left\\{{Y\not\in
E}\right\\}}$

is an optimal map for (21), $\gamma=\operatorname{{Law}}(X,Y)$ is an optimal
plan for (5), and the law of $X$ is $\mathcal{D}$-regular. Moreover, if
$\widetilde{X}$ is an optimal map and $\widetilde{\gamma}$ is an optimal plan,
then

$\displaystyle X1_{\left\\{{Y\not\in E}\right\\}}$
$\displaystyle=\widetilde{X}1_{\left\\{{Y\not\in E}\right\\}},$ (27)
$\displaystyle 1_{\left\\{{y\not\in E}\right\\}}\gamma(dx,dy)$
$\displaystyle=1_{\left\\{{y\not\in E}\right\\}}\widetilde{\gamma}(dx,dy).$
(28)

###### Proof.

Let $\widetilde{\gamma}\in\Gamma(\nu)$ be an optimal plan. From Theorem 3.2 we
deduce that if $(x,y)\in\operatorname{supp}{\widetilde{\gamma}}$, then
$y\in\operatorname{dom}{\operatorname{Arg}}=\operatorname{dom}{\operatorname{Arg}_{G}}$.
In particular, $\nu(\operatorname{dom}{\operatorname{Arg}})=1$. By Theorem
B.12, the exception set
$\operatorname{dom}{\operatorname{Arg}}\setminus\operatorname{dom}{\nabla\phi}$
belongs to the union of $E=E^{G}$ and of a countable family of sets from
$\mathcal{D}$. Since $\nu=\operatorname{{Law}}(Y)$ is $\mathcal{D}$-regular,
we have that

$\nu(\operatorname{dom}{\nabla\phi}\setminus
E)=\nu(\operatorname{dom}{\operatorname{Arg}}\setminus E)=1-\nu(E).$ (29)

It follows that the random variable $X=(X_{1},X_{2})$ is well-defined.

Theorem B.6 shows that if $y\in\operatorname{dom}{\nabla\phi}\setminus E$,
then $D^{c}\phi(y)$ is the only element of $\operatorname{Arg}(y)$. It follows
that $X\in G$ and $\phi(Y)=c(X,Y)$. In view of Theorem 3.2,
$\gamma\triangleq\operatorname{{Law}}(X,Y)$ is an optimal plan if it has the
martingale property: $\gamma(y|x)=x$.

If $(x,y)\in\operatorname{supp}{\widetilde{\gamma}}$ and
$y\in\operatorname{dom}{\nabla\phi}\setminus E$, then Theorems 3.2 and B.6
yield that $x=D^{c}\phi(y)$. Since $\gamma$ and $\widetilde{\gamma}$ have
common $y$-marginal $\nu$ satisfying (29), they coincide outside of
$\mathbb{R}^{2}\times E$, that is, (28) holds.

Let $f=f(x)$ be a bounded Borel function on $\mathbb{R}^{2}$. As
$(Y-X)1_{\left\\{{Y\in G}\right\\}}=0$, we deduce that

$\displaystyle\int(y-x)f(x)d\gamma$
$\displaystyle=\int(y-x)f(x)1_{\left\\{{y\not\in
G}\right\\}}d\gamma=\int(y-x)f(x)1_{\left\\{{y\not\in
G}\right\\}}d\widetilde{\gamma}$ $\displaystyle=-\int(y-x)f(x)1_{\left\\{{y\in
G}\right\\}}d\widetilde{\gamma}=0,$

where the last two equalities follow from the martingale property of
$\widetilde{\gamma}$ and Lemma 4.2, respectively. Thus, $\gamma(y|x)=x$. We
have proved that $\gamma$ is an optimal plan and, in particular, that $X$ is
an optimal map. The uniqueness property (27) for optimal maps follows directly
from the corresponding property (28) for optimal plans.

It only remains to be shown that $\mu\triangleq\operatorname{{Law}}(X)$ is
$\mathcal{D}$-regular. As $\operatorname{supp}{\mu}\subset G$ and the
intersection of $G$ with any set from $\mathcal{D}$ is a point, $\mu$ is
$\mathcal{D}$-regular if and only if it is atomless. Assume to the contrary,
that $\mu(\left\\{{r}\right\\})>0$ for some $r\in G$ and define the Borel
probability measure

$\eta(dy)=\frac{1}{\mu(\left\\{{r}\right\\})}\gamma(\left\\{{r}\right\\},dy).$

Being $\mathcal{D}$-regular, the measure $\nu$ is atomless. Hence,

$\displaystyle\mu(\left\\{{r}\right\\})\eta(G)$
$\displaystyle=\gamma(\left\\{{r}\right\\}\times
G)=\mathbb{\mathbb{P}}\left(X=r,Y\in G\right)$
$\displaystyle=\mathbb{\mathbb{P}}\left(Y=r\right)=\nu(\left\\{{r}\right\\})=0.$

From the optimality of $\gamma$ we deduce that

$\operatorname{supp}{\eta}\subset
D(r)\triangleq\left\\{{y\in\mathbb{R}^{2}}:\;\phi(y)=c(r,y)\right\\}$

and then that $\eta(D(r)\setminus G)=1>0$. The martingale property of $\gamma$
yields that $\int yd\eta=r$. The last two properties of $\eta$ and the fact
that $\phi<0$ outside of $G$ imply the existence of $y^{0},y^{1}\in
D(r)\setminus G$ such that

$y^{0}_{1}<r_{1}<y^{1}_{1},\quad y^{1}_{2}<r_{2}<y^{0}_{2}.$

By Lemma B.5, $D(r)$ belongs to the graph of a strictly decreasing linear
function and thus, belongs to $\mathcal{D}$. As $\nu$ is
$\mathcal{D}$-regular, we arrive to a contradiction:
$\mu(\left\\{{r}\right\\})=\gamma(\left\\{{r}\right\\}\times
D(r))\leq\nu(D(r))=0$. ∎

We now state the main uniqueness result of the paper, which can be viewed as
an adaptation of the classical Brenier theorem to our setting. We point out
that our regularity assumption on $\nu$ is weaker than the standard condition
of the Brenier theorem, which requires $\nu$ to assign zero mass to rotations
of the graphs of Lipschitz functions.

######  Theorem 4.6.

Let $Y=(Y_{1},Y_{2})\in\mathcal{L}^{2}$ and suppose that
$\nu=\operatorname{{Law}}(Y)$ is $\mathcal{D}$-regular and the (one-
dimensional) laws of $Y_{1}$ and $Y_{2}$ are atomless. Let $G\in\mathfrak{M}$
be a maximizer for (18) and denote $\phi=\phi_{G}$. Then $X=D^{c}\phi(Y)$ or,
in more detail,

$\displaystyle X_{1}$
$\displaystyle=D^{c}_{1}\phi(Y)=Y_{1}-\frac{\partial\phi}{\partial y_{2}}(Y),$
$\displaystyle X_{2}$
$\displaystyle=D^{c}_{2}\phi(Y)=Y_{2}-\frac{\partial\phi}{\partial y_{1}}(Y),$

is the unique optimal map for (21) and the law of $(X,Y)$ is the unique
optimal plan for (5). Moreover, the law of $X$ is $\mathcal{D}$-regular and
the laws of $X_{1}$ and $X_{2}$ are atomless.

###### Proof.

We omit $G$ from the notations (23) related to its vertical and horizontal
line segments. As the law of $Y_{i}$ is atomless and the set $\mathcal{T}_{i}$
is countable, we deduce that

$\nu(E)=\mathbb{\mathbb{P}}\left(Y\in
E\right)\leq\sum_{i=1}^{2}\mathbb{\mathbb{P}}\left(Y\in
E_{i}\right)=\sum_{i=1}^{2}\mathbb{\mathbb{P}}\left(Y_{i}\in\mathcal{T}_{i}\right)=0.$

Except the continuity of the distribution functions for $X_{1}$ and $X_{2}$,
all other assertions follow directly from Theorem 4.5.

We shall prove that the law of $X_{2}$ is atomless. If
$t\not\in\mathcal{T}_{2}$, then the set ${E}_{2}(t)$ is a singleton:
$E_{2}(t)=\left\\{{z}\right\\}$. By Theorem 4.5, the law of $X$ is
$\mathcal{D}$-regular and, in particular, atomless. It follows that
$\mathbb{\mathbb{P}}\left(X_{2}=t\right)=\mathbb{\mathbb{P}}\left(X=z\right)=0$.

Let $t\in\mathcal{T}_{2}$. Lemma B.4 shows that if
$x\in\operatorname*{ri}{E_{2}(t)}$ and $\phi(y)=c(x,y)$, then $c(x,y)=0$ and
subsequently, $y\in E_{2}(t)$. As $X\in G$, $\phi(Y)=c(X,Y)$, and the law of
$X$ is atomless, we obtain that

$\displaystyle\mathbb{\mathbb{P}}\left(X_{2}=t\right)$
$\displaystyle=\mathbb{\mathbb{P}}\left(X\in
E_{2}(t)\right)=\mathbb{\mathbb{P}}\left(X\in\operatorname*{ri}{E_{2}(t)}\right)$
$\displaystyle\leq\mathbb{\mathbb{P}}\left(Y\in
E_{2}(t)\right)=\mathbb{\mathbb{P}}\left(Y_{2}=t\right)=0,$

where the last step holds by the continuity of the law of $Y_{2}$. ∎

##  5 Examples

######  Example 5.1 (Linear optimal map).

Let $Y=(Y_{1},Y_{2})$ be a random variable in $\mathcal{L}^{2}$ such that
$\mathbb{E}\left(Y_{i}\right)=0$,
$\mathbb{E}\left(Y_{i}^{2}\right)=\sigma_{i}^{2}>0$, and

$\mathbb{E}\left(\left.Y_{i}\right\lvert{Z}\right)=\frac{\mathbb{E}\left(Y_{i}Z\right)}{\mathbb{E}\left(Z^{2}\right)}Z\quad\text{for
all}\quad Z=a_{1}Y_{1}+a_{2}Y_{2},\;a_{j}\in\mathbb{R}.$

The latter property holds if the distribution of $Y$ is Gaussian or, more
generally, elliptically contoured.

We denote $\lambda=\frac{\sigma_{2}}{\sigma_{1}}>0$ and define
$G=\left\\{{x\in\mathbb{R}^{2}}:\;x_{2}=\lambda x_{1}\right\\}$ and

$\displaystyle X_{1}=\frac{1}{2}Y_{1}+\frac{1}{2\lambda}Y_{2},\quad
X_{2}=\lambda X_{1}=\frac{\lambda}{2}Y_{1}+\frac{1}{2}Y_{2}.$

Elementary computations show that
$\mathbb{E}\left(\left.Y\right\lvert{X}\right)=X=(X_{1},X_{2})$ and

$\phi_{G}(Y)=\inf_{x\in
G}c(x,Y)=\inf_{x_{1}\in\mathbb{R}}(Y_{1}-x_{1})(Y_{2}-\lambda x_{1})=c(X,Y).$

Being the graph of an increasing linear function, $G\in\mathfrak{M}$. Setting
$\nu=\operatorname{{Law}}(Y)$, we deduce from Theorem 3.2 that $G$ and
$\gamma=\operatorname{{Law}}(X,Y)$ are respective solutions to (18) and (5).
Moreover, as $X$ is the only element of $G$ such that $\phi_{G}(Y)=c(X,Y)$,
the characteristic property (20) yields that $\gamma$ is the unique optimal
plan. In particular, $X$ is the unique optimal map for (21).

######  Example 5.2 (Optimal map may not yield optimal plan).

Let $Y$ be a random variable taking values in $y^{0}=(-1,1)$, $y^{1}=(0,-1)$,
and $y^{2}=(1,0)$ with probability $\frac{1}{3}$. Direct computations show
that the points

$z^{i}=\frac{1}{3}y^{0}+\frac{2}{3}y^{i}=(-1)^{i}\big{(}\frac{1}{3},\frac{1}{3}\big{)},\quad
i=1,2,$

belong to the set
$G=\left\\{{x\in\mathbb{R}^{2}}:\;c(x,y^{0})=-\frac{8}{9},\;x_{1}>y_{1}^{0}\right\\}$,
that

$\phi_{G}(y^{i})=\min_{x\in G}c(y^{i},x)=c(y^{i},z^{i}),\quad i=1,2,$

and that the probability measure

$\gamma=\sum_{i=1}^{2}\left(\frac{1}{3}\delta_{(z^{i},y^{i})}+\frac{1}{6}\delta_{(z^{i},y^{0})}\right)$

belongs to $\Gamma(\nu)$, where $\nu=\operatorname{{Law}}(Y)$. Being the graph
of an increasing hyperbola, $G\in\mathfrak{M}$. By Theorem 2.2, $\gamma$ is an
optimal plan for (5). The value of this problem is

$\int
c(x,y)d\gamma=\sum_{i=1}^{2}\left(\frac{1}{3}c(z^{i},y^{i})+\frac{1}{6}c(z^{i},y^{0})\right)=-\frac{4}{9}.$

On the other hand, let $X\in\mathcal{X}(Y)$, that is, $X$ is $Y$-measurable
and $X=\mathbb{E}\left(\left.Y\right\lvert{X}\right)$. We write
$x^{i}=X(y^{i})$, $i=1,2,3$. If all $(x^{i})$ are distinct, then $X=Y$ and
$c(X,Y)=0$. If they are the same point, then $X=\mathbb{E}\left(Y\right)=0$
and

$\mathbb{E}\left(c(X,Y)\right)=\frac{1}{3}\sum_{i=0}^{2}c(0,y^{i})=\frac{1}{3}\sum_{i=0}^{2}y^{i}_{1}y^{i}_{2}=-\frac{1}{3}.$

Finally, if precisely two of the elements of $(x^{i})$ coincide:
$x^{k}=x^{l}\not=x^{m}$, where $(k,l,m)$ is a permutation of $(0,1,2)$, then
$x^{k}=x^{l}=\frac{1}{2}(y^{k}+y^{l})$, $x^{m}=y^{m}$, and

$\displaystyle\mathbb{E}\left(c(X,Y)\right)=\frac{1}{3}\left(c(y^{k},\frac{1}{2}(y^{k}+y^{l}))+c(y^{l},\frac{1}{2}(y^{k}+y^{l}))\right)=\frac{1}{6}c(y^{k},y^{l}).$

As $c(y^{0},y^{1})=c(y^{0},y^{2})=-2$ and $c(y^{1},y^{2})=1$, the value
function of the optimal map problem (21) is given by $-\frac{1}{3}$, which is
strictly less than $-\frac{4}{9}$, the value of the optimal plan problem (5).

######  Example 5.3 (Optimal map may not exist).

Let $U$ and $V$ be independent symmetric random variables in $\mathcal{L}^{2}$
with $U$ having a continuous distribution function and $V$ taking values in
$\left\\{{-1,1}\right\\}$. We define a 2-dimensional random variable

$Y=\big{(}\frac{U}{3}(1-2V),U(1+2V)\big{)}1_{\left\\{{U<0}\right\\}}+\big{(}U(1+2V),\frac{U}{3}(1-2V)\big{)}1_{\left\\{{U\geq
0}\right\\}}.$

The components $Y_{1}$ and $Y_{2}$ have continuous distribution functions and,
in particular, $\nu=\operatorname{{Law}}(Y)$ is atomless. By Theorem 4.1, the
plan and map problems (5) and (21) have identical values. We shall prove that
there is a unique optimal plan, which is not induced by a ($Y$-measurable
martingale) map, and hence, shall show that an optimal map does not exist.

To this end, we define a 2-dimensional random variable

$\displaystyle
X=(\frac{1}{3}U,U)1_{\left\\{{U<0}\right\\}}+(U,\frac{1}{3}U)1_{\left\\{{U\geq
0}\right\\}}.$

We observe that $X$ takes values in the set

$G=\left\\{{x_{2}=3x_{1},\;x_{1}<0}\right\\}\cup\left\\{{x_{2}=\frac{1}{3}x_{1},\;x_{1}\geq
0}\right\\}$

consisting of two upward-slopping lines and thus, belonging to $\mathfrak{M}$.
Direct computations show that
$\mathbb{E}\left(\left.Y\right\lvert{X}\right)=X$ and

$c(X,Y)=\phi_{G}(Y)\triangleq\inf_{x\in G}c(x,Y).$

By Theorem 2.2, the law of $(X,Y)$ is an optimal plan and $G$ is a dual
maximizer. We shall proceed to show that this is the only optimal plan and
that it is not induced by a map from $\mathcal{X}(Y)$.

From the construction of $Y$ we deduce the equality of the sets:

$\left\\{{V=-1}\right\\}=\left\\{{Y_{1}=-Y_{2}}\right\\}=\left\\{{Y=(-\left\lvert
U\right\rvert,\left\lvert U\right\rvert)}\right\\}.$

It follows that

$\displaystyle\mathbb{E}\left(\left.X\right\lvert{Y}\right)1_{\left\\{{Y_{1}=-Y_{2}}\right\\}}$
$\displaystyle=\mathbb{E}\left(\left.X\right\lvert{\left\lvert
U\right\rvert}\right)1_{\left\\{{Y_{1}=-Y_{2}}\right\\}}=\frac{1}{3}(\left\lvert
U\right\rvert,-\left\lvert U\right\rvert)1_{\left\\{{Y_{1}=-Y_{2}}\right\\}}$
$\displaystyle=-\frac{1}{3}Y1_{\left\\{{Y_{1}=-Y_{2}}\right\\}}\not=X1_{\left\\{{Y_{1}=-Y_{2}}\right\\}}.$

Hence, $X$ is not $Y$-measurable.

Let $\gamma\in\Gamma(\nu)$ be an optimal plan and $\mu$ be its $x$-marginal.
By Theorem 3.2, $\operatorname{supp}{\mu}\subset G$ and

$c(x,y)=\phi(y),\quad(x,y)\in\operatorname{supp}{\gamma}.$

The random variable $Y$ takes values in $F=F_{1}\cup F_{2}$, where

$\displaystyle F_{1}$
$\displaystyle=\left\\{{y_{2}=-y_{1},\;y_{1}<0}\right\\},$ $\displaystyle
F_{2}$ $\displaystyle=\left\\{{y_{2}=-9y_{1}\text{ or
}y_{2}=-\frac{1}{9}y_{1},\;y_{1}\geq 0}\right\\}.$

Elementary computations show that for $x\in G$ the set of $y\in F$ such that
$c(x,y)=\phi_{G}(y)$ consists of two points $g(x)$ and $f(x)$ such that

$\displaystyle f(x)$
$\displaystyle=(-x_{1},9x_{1})1_{\left\\{{x_{1}<0}\right\\}}+(3x_{1},-\frac{1}{3}x_{1})1_{\left\\{{x_{1}\geq
0}\right\\}},$ $\displaystyle g(x)$
$\displaystyle=(3x_{1},-3x_{1})1_{\left\\{{x_{1}<0}\right\\}}+(-x_{1},x_{1})1_{\left\\{{x_{1}\geq
0}\right\\}}.$

For $x\not=0$, $x\in G$, the three points $\left\\{{g(x),x,f(x)}\right\\}$ are
distinct and

$x=\frac{1}{2}(g(x)+f(x)).$

On the other hand, by the martingale property of $\gamma$ and the fact that
$\nu(\left\\{{0}\right\\})=0$, we have that

$x=\gamma(y|x)=f(x)\gamma(y=f(x)|x)+g(x)\gamma(y=g(x)|x),\quad\gamma\text{-a.s.},$

and therefore, the conditional probabilities

$\gamma(y=f(x)|x)=\gamma(y=g(x)|x)=\frac{1}{2},\quad\gamma\text{-a.s.}.$

For a bounded Borel function $h=h(x,y)$ on
$\mathbb{R}^{2}\times\mathbb{R}^{2}$ we then obtain that

$\displaystyle\int h(x,y)d\gamma$
$\displaystyle=\int(h(x,g(x))1_{\left\\{{y=g(x)}\right\\}}+h(x,f(x))1_{\left\\{{y=f(x)}\right\\}})d\gamma$
$\displaystyle=\frac{1}{2}\int(h(x,g(x))+h(x,f(x)))d\mu.$

Hence, $\gamma$ is unique if and only if $\mu$ is unique. We observe now that
the map $f:\;G\rightarrow F_{2}$ is one-to-one. Thus, for a Borel set
$B\in\mathbb{R}^{2}$,

$\displaystyle\frac{1}{2}\mu(B)=\int 1_{\left\\{{x\in
B}\right\\}}1_{\left\\{{y=f(x)}\right\\}}d\gamma=\int 1_{\left\\{{y\in
f(B)}\right\\}}d\gamma=\nu(f(B)),$

and the uniqueness of $\mu$ follows.

##  6 Equilibrium with insider

We consider a single-period financial market. There are a bank account with
zero interest rate and a stock. The stock value at maturity $t=1$ is
represented by a random variable $V$. The stock price $S$ at initial time
$t=0$ is the result of the interaction between _noise traders_ , an _insider_
, and _market makers_ , where

  1. 1.

The noise traders place an order for $U$ stocks; $U$ is a random variable.

  2. 2.

The insider knows the value of both $U$ and $V$ and places an order for $Q$
stocks. The _trading strategy_ $Q$ is a $(U,V)$-measurable random variable.

  3. 3.

The market makers observe only the total order $R=Q+U$. They quote the price
$S=f(R)$ according to a _pricing rule_ $f=f(r)$, which is a Borel function
$f:\;\mathbb{R}\rightarrow\overline{\mathbb{R}}\triangleq\mathbb{R}\cup\left\\{{-\infty}\right\\}\cup\left\\{{\infty}\right\\}$.

######  Definition 6.1.

An _equilibrium_ $(Q,f)$ is defined by a trading strategy $Q$ and a pricing
rule $f=f(r)$ such that

  1. 1.

Given the total order $R=Q+U$, the price $S=f(R)$ is _efficient_ in the sense
that

$S=\mathbb{E}\left(\left.V\right\lvert{R}\right).$

  2. 2.

Given the pricing rule $f=f(r)$, the order $Q$ _maximizes_ insider’s profit:

$Q(V-f(Q+U))=\max_{q\in\mathbb{R}}q(V-f(q+U)),$

with the convention $0\times\infty=0$.

######  Remark 6.2.

Up to minor technical differences, our notion of equilibrium coincides with
the one in Rochet and Vila (1994). It differs from the classical equilibrium
from Kyle (1985) in the ability of the insider to observe noise traders’ order
flow $U$. In the model of Kyle (1985), the insider maximizes
$\mathbb{E}\left(\left.Q(V-f(Q+U)\right\lvert{V}\right)$ over all
$V$-measurable random variable $Q$.

The following result links the existence of equilibrium with the existence of
an optimal map for (21) that induces an optimal plan for (5).

######  Theorem 6.3.

Let $Y=(U,V)\in\mathcal{L}^{2}$ and denote $\nu=\operatorname{{Law}}(Y)$. An
equilibrium $(Q,f)$ exists if and only if there is an optimal map $X$ for (21)
such that the law of $(X,Y)$ is an optimal plan for (5). Insider’s profit is
unique and given by

$Q(V-f(Q+U))=-c(X,Y)=-\phi_{G}(U,V),$

where $G\in\mathfrak{M}$ is a maximizer for (18).

Moreover, there are equilibrium $(Q,f)$ and optimal map $X=(R,S)$ such that
the pricing rule $f:\;\mathbb{R}\rightarrow\overline{\mathbb{R}}$ is an
increasing function, the total order $Q+U=R$, and the price $f(Q+U)=S$.

We divide the proof of the theorem into lemmas.

######  Lemma 6.4.

Let $Y=(U,V)\in\mathcal{L}^{2}$, $\nu=\operatorname{{Law}}(Y)$, and $X=(R,S)$
be an optimal map for (21) such that $S$ is $R$-measurable and the law of
$(X,Y)$ is an optimal plan for (5). Then there is an increasing function
$f:\;\mathbb{R}\rightarrow\overline{\mathbb{R}}$ such that $S=f(R)$ and
$(Q,f)$ is an equilibrium with $Q=R-U$.

###### Proof.

By construction, $S=\mathbb{E}\left(\left.V\right\lvert{R}\right)$. Hence, we
only need to verify the profit maximization condition for the order $Q=R-U$.
Theorem 2.2 yields $G\in\mathfrak{M}$ such that $(R,S)\in G$ and

$(U-R)(V-S)=\min_{(r,s)\in G}(U-r)(V-s)=\phi_{G}(U,V).$

Let $P_{1}$ be the projection of $G$ on the first or $r$-coordinate. Clearly,
$P_{1}$ is an interval. As $S$ is $R$-measurable, there is an _increasing_
function $f=f(r)$ on $P_{1}$ such that $S=f(R)$ and $(r,f(r))\in G$ for $r\in
P_{1}$. By construction,

$(U-R)(V-S)=\min_{r\in P_{1}}(U-r)(V-f(r)).$

We now extend $f$ to an increasing function from $\mathbb{R}$ to
$\overline{\mathbb{R}}$ by setting its values to $-\infty$ on the left and to
$+\infty$ on the right of $P_{1}$. As $\phi_{G}(U,V)>-\infty$, Lemma B.2
yields that $U$ takes values in the closure of $P_{1}$. Under the standing
convention: $0\times\infty=0$, we obtain that

$\phi_{G}(U,V)=(U-R)(V-S)=\min_{r\in\mathbb{R}}(U-r)(V-f(r)).$

Hence, $(Q,f)$ is an equilibrium with $Q=R-U$. ∎

######  Lemma 6.5.

Let $f:\;\mathbb{R}\rightarrow\overline{\mathbb{R}}$ be a Borel function and

$\phi(y)=\inf_{r\in\mathbb{R}}(y_{1}-r)(y_{2}-f(r))\in[-\infty,0],\quad
y\in\mathbb{R}^{2},$

with the convention: $0\times\infty=0$. Then there is $G\in\mathfrak{M}$ such
that $\phi\leq\phi_{G}$.

###### Proof.

Given $y^{0},y^{1}\in\mathbb{R}^{2}$ and $t\in[0,1]$, we denote
$r=y^{0}_{1}+t(y^{1}_{1}-y^{0}_{1})$ and deduce that

$\displaystyle(1-t)\phi(y^{0})+t\phi(y^{1})$
$\displaystyle\leq(1-t)\min((y^{0}_{1}-r)(y^{0}_{2}-f(r)),0)$
$\displaystyle\quad+t\min((y^{1}_{1}-r)(y^{1}_{2}-f(r)),0)$ $\displaystyle\leq
t(1-t)(y^{1}_{1}-y^{0}_{1})(y^{1}_{2}-y^{0}_{2}),$

where in the middle we used the negative parts to account for the possibility
that $\left\lvert f(r)\right\rvert=\infty$. The result now follows from Lemma
3.1. ∎

######  Lemma 6.6.

Let $Y=(U,V)\in\mathcal{L}^{2}$, $\nu=\operatorname{{Law}}(Y)$, and $(Q,f)$ be
an equilibrium with the total order $R=Q+U$ and the price $S=f(R)$. Then
$X=(\widetilde{R},S)$ with
$\widetilde{R}=\mathbb{E}\left(\left.U\right\lvert{R}\right)$ is an optimal
map for (21), the law of $(X,Y)$ is an optimal plan for (5), and

$Q(f(Q+U)-V)=(R-U)(S-V)=(\widetilde{R}-U)(S-V).$ (30)

###### Proof.

From the definition of the equilibrium we obtain that

$\phi(U,V)=Q(f(Q+U)-V)=(R-U)(S-V),$

where $\phi(u,v)=\inf_{r\in\mathbb{R}}(u-r)(v-f(r))$,
$(u,v)\in\mathbb{R}^{2}$. We claim that

$\mathbb{E}\left(\phi(U,V)\right)=\mathbb{E}\left((U-\widetilde{R})(V-S)\right).$
(31)

As the integrability properties of $R$ are unknown, we use a localization
argument. For $n\geq 1$ from the martingale properties
$\mathbb{E}\left(\left.V\right\lvert{R}\right)=S$ and
$\mathbb{E}\left(\left.U\right\lvert{R}\right)=\widetilde{R}$ we deduce that

$\displaystyle\mathbb{E}\left(\phi(U,V)1_{\left\\{{\left\lvert
R\right\rvert\leq n}\right\\}}\right)$
$\displaystyle=\mathbb{E}\left((U-R)(V-S)1_{\left\\{{\left\lvert
R\right\rvert\leq n}\right\\}}\right)$
$\displaystyle=\mathbb{E}\left(U(V-S)1_{\left\\{{\left\lvert R\right\rvert\leq
n}\right\\}}\right)$
$\displaystyle=\mathbb{E}\left((U-\widetilde{R})(V-S)1_{\left\\{{\left\lvert
R\right\rvert\leq n}\right\\}}\right).$

Taking the limit as $n\to\infty$, we obtain (31) by the dominated convergence
theorem.

Lemma 6.5 yields $G\in\mathfrak{M}$ such that $\phi\leq\phi_{G}$. From Lemma
2.13 we deduce that

$\mathbb{E}\left(\phi(U,V)\right)\leq\mathbb{E}\left(\phi_{G}(U,V)\right)=\int\phi_{G}d\nu\leq\int
c(x,y)d\gamma,\quad\gamma\in\Gamma(\nu).$

In view of (31) and since
$\widetilde{\gamma}=\operatorname{{Law}}(\widetilde{R},S,U,V)$ belongs to
$\Gamma(\nu)$, we obtain that

$\mathbb{E}\left((U-\widetilde{R})(V-S)\right)=\mathbb{E}\left(\phi(U,V)\right)=\mathbb{E}\left(\phi_{G}(U,V)\right)=\int
c(x,y)d\widetilde{\gamma}.$

It follows that $\widetilde{\gamma}$ is an optimal plan, $(\widetilde{R},S)$
is an optimal map, and $\phi(U,V)=\phi_{G}(U,V)$. Finally, Theorem 3.2 yields
that

$\phi_{G}(U,V)=c(X,Y)=(\widetilde{R}-U)(S-V),$

and we obtain (30). ∎

###### Proof of Theorem 6.3.

If $X=(R,S)$ is an optimal map, then $\widetilde{X}=(R,\widetilde{S})$ with
$\widetilde{S}=\mathbb{E}\left(\left.V\right\lvert{R}\right)=\mathbb{E}\left(\left.S\right\lvert{R}\right)$
is an optimal map as well and $\widetilde{S}$ is $R$-measurable. By Theorem
3.2,

$c(X,Y)=c(\widetilde{X},Y)=\phi_{G}(Y)=\phi_{G}(U,V)$

for every maximizer $G\in\mathfrak{M}$ to (18). In particular, $c(X,Y)$ is the
same random variable for every optimal map $X$. After these observations, the
proof follows from Lemmas 6.4 and 6.6. ∎

We now state sufficient conditions for the existence and uniqueness of
equilibrium. Theorem 6.7 generalizes a result from Rochet and Vila (1994),
where the distribution of $(U,V)$ has a compact support and a continuous
density.

######  Theorem 6.7.

Let $Y=(U,V)\in\mathcal{L}^{2}$ and suppose that the law of ${Y}$ is
$\mathcal{D}$-regular. Then an equilibrium $(Q,f)$ exists.

If, in addition, the laws of $U$ and $V$ are atomless, then insider’s order
$Q$, the total order $R=Q+U$, and the price $S=f(R)$ are unique. Moreover,
$X=(R,S)$ is the unique optimal map for (21) and
$\gamma=\operatorname{{Law}}(X,Y)$ is the unique optimal plan for (5).

For the proof we need a lemma.

######  Lemma 6.8.

Let $f:\;\mathbb{R}\rightarrow\overline{\mathbb{R}}$ be a Borel function and

$\phi(u,v)=\inf_{r\in\mathbb{R}}(u-r)(v-f(r))\in[-\infty,0],\quad(u,v)\in\mathbb{R}^{2},$

with the convention: $0\times\infty=0$. There is a countable set
$A\subset\mathbb{R}$ such that if $u,v\notin A$ and $\phi(u,v)=0$, then

$f^{-1}(v)\triangleq\left\\{{r\in\mathbb{R}}:\;f(r)=v\right\\}=\left\\{{u}\right\\}.$

###### Proof.

If $\phi(u,v)=\inf_{r\in\mathbb{R}}(u-r)(v-f(r))=0$, then

$g(u)\triangleq\sup_{r<u}f(r)\leq v\leq\inf_{r>u}f(r)\triangleq h(u).$

Clearly,

$\inf_{r\geq u}f(r)\leq f(u)\leq\sup_{r\leq u}f(r).$

Thus, if the increasing functions $g$ and $h$ are continuous and strictly
increasing at $u$, then $f^{-1}(v)=\left\\{{u}\right\\}$. To conclude the
proof we just observe that the set of arguments, where an increasing function
is discontinuous, and the set of values, where it is not strictly increasing,
are countable. ∎

###### Proof of Theorem 6.7.

If $\nu=\operatorname{{Law}}(Y)$ is $\mathcal{D}$-regular, then Theorem 4.5
yields an optimal map $X$ such that the law of $(X,Y)$ is an optimal plan. By
Theorem 6.3, there is an equilibrium $(Q,f)$.

If the laws of $U=Y_{1}$ and $V=Y_{2}$ are atomless, then, by Theorem 4.6, the
optimal map $X=(X_{1},X_{2})$ is unique. Lemma 6.6 shows that $S=X_{2}$ is the
unique equilibrium price: $S=f(R)$, where $R=Q+U$.

Let $\phi$ be the function defined in Lemma 6.8 and $G\in\mathfrak{M}$ be a
maximizer for (18). From the definition of the equilibrium and Theorem 6.3 we
deduce that

$(U-R)(V-f(R))=(U-R)(V-S)=\phi(U,V)=\phi_{G}(U,V).$

If $\phi(U,V)<0$, then the total order $R$ is clearly unique. If
$\phi(U,V)=0$, then $R=U$ by Lemma 6.8 and the continuity of the distributions
of $U$ and $V$. Thus, the total order $R$ and insider’s order $Q=R-U$ are
unique. By Theorem 6.3, the uniqueness of $R$ and $S$ implies that $(R,S)$ is
an optimal map. Hence $X_{1}=R$. ∎

##  Appendix A Closure of probability measures with bounded densities in
$\mathcal{W}_{p}(\mathbb{R}^{d})$

Let $p\geq 1$. We denote by $\mathcal{W}_{p}(\mathbb{R}^{d})$ the space of
Borel probability measures on $\mathbb{R}^{d}$ with finite $p$-th moments
equipped with the Wasserstein metric:

$W_{p}(\mu,\nu)=\left\\{\inf_{\gamma\in\Pi(\mu,\nu)}\int|x-y|^{p}d\gamma\right\\}^{1/p},$

where $\Pi(\mu,\nu)$ is the family of Borel probability measures $\gamma$ on
$\mathbb{R}^{d}\times\mathbb{R}^{d}=\left\\{{(x,y)}:\;x,y\in\mathbb{R}^{d}\right\\}$
with $x$-marginal $\mu$ and $y$-marginal $\nu$. We recall that
$\mathcal{W}_{p}(\mathbb{R}^{d})$ is a complete separable metric space and
that $\mu_{n}\rightarrow\mu$ in $\mathcal{W}_{p}(\mathbb{R}^{d})$ if and only
if $\int f(x)d\mu_{n}\rightarrow\int f(x)d\mu$ for every continuous function
$f$ with polynomial $p$-th growth:

$|f(x)|\leq K(1+|x|^{p}),\quad x\in\mathbb{R}^{d}.$

Let $\nu\in\mathcal{W}_{p}(\mathbb{R}^{d})$ and $\mathcal{Q}_{\infty}(\nu)$ be
the family of Borel probability measures on $\mathbb{R}^{d}$ that have bounded
densities with respect to $\nu$:

$\mathcal{Q}_{\infty}(\nu)\triangleq\left\\{{\mu\ll\nu}:\;\frac{d\mu}{d\nu}\in\mathcal{L}^{\infty}(\mathbb{R}^{d})\right\\}.$

Clearly, $\mathcal{Q}_{\infty}(\nu)\subset\mathcal{W}_{p}(\mathbb{R}^{d})$.
The following result, used in the proof of our main Theorem 2.2, describes the
closure of $\mathcal{Q}_{\infty}(\nu)$ under $W_{p}$.

######  Theorem A.1.

Let $p\geq 1$ and $\nu\in\mathcal{W}_{p}(\mathbb{R}^{d})$. Then the closure of
$\mathcal{Q}_{\infty}(\nu)$ in $\mathcal{W}_{p}(\mathbb{R}^{d})$ has the form:

$\mathcal{S}_{p}(\nu)=\left\\{{\mu\in\mathcal{W}_{p}(\mathbb{R}^{d})}:\;\operatorname{supp}{\mu}\subset\operatorname{supp}{\nu}\right\\}.$

###### Proof.

If $\mu_{n}\to\mu$ in $\mathcal{W}_{p}(\mathbb{R}^{d})$, then $\mu_{n}\to\mu$
weakly and thus,

$\mu(C)\geq\limsup_{n\to\infty}\mu_{n}(C),$

for every closed set $C$. In particular, if
$(\mu_{n})\subset\mathcal{S}_{p}(\nu)$, then

$\mu(\operatorname{supp}{\nu})\geq\limsup_{n\to\infty}\mu_{n}(\operatorname{supp}{\nu})=1$

and hence, $\mu\in\mathcal{S}_{p}(\nu)$. It follows that
$\mathcal{S}_{p}(\nu)$ is closed in $\mathcal{W}_{p}(\mathbb{R}^{d})$.
Clearly, $\mathcal{S}_{p}(\nu)$ is convex.

If $\operatorname{supp}{\nu}$ is compact, then restricted to
$\mathcal{S}_{p}(\nu)$ the convergence under $W_{p}$ is equivalent to the weak
convergence and thus, the family of probability measures in
$\mathcal{S}_{p}(\nu)$ with _finite_ support is dense. Being a closed convex
set, $\mathcal{S}_{p}(\nu)$ is then the closure of $\mathcal{Q}_{\infty}(\nu)$
in $\mathcal{W}_{p}(\mathbb{R}^{d})$ if and only if every Dirac measure
$\delta_{y}=\delta_{y}(dx)$ concentrated at $y\in\operatorname{supp}{\nu}$ is
the weak limit of a sequence $(\mu_{n})\subset\mathcal{Q}_{\infty}(\nu)$. The
sequence $(\mu_{n})$ with

$\frac{d\mu_{n}}{d\nu}(x)=\frac{1}{\nu(B_{1/n}(y))}1_{\left\\{{x\in
B_{1/n}(y)}\right\\}},\quad n\geq 1,$

where $B_{r}(y)$ is the ball of radius $r>0$ centered at $y$, has the required
properties.

If $\operatorname{supp}{\nu}$ is not compact, we approximate
$\mu\in\mathcal{S}_{p}(\nu)$ by the sequence $(\mu_{n})$ given by

$\frac{d\mu_{n}}{d\mu}(x)=\frac{1}{\mu(B_{n}(y))}1_{\left\\{{x\in
B_{n}(y)}\right\\}},\quad n\geq 1,$

for some $y\in\operatorname{supp}{\mu}$. We have that
$(\mu_{n})\subset\mathcal{S}_{p}(\nu)$ and $\mu_{n}\to\mu$ under $W_{p}$. By
what we have already proved, each $\mu_{n}$ belongs to the closure of
$\mathcal{Q}_{\infty}(\nu_{n})$ in $\mathcal{W}_{p}(\mathbb{R}^{d})$, where

$\frac{d\nu_{n}}{d\nu}(x)=\frac{1}{\nu(B_{n}(y))}1_{\left\\{{x\in
B_{n}(y)}\right\\}},\quad n\geq 1.$

As $\mathcal{Q}_{\infty}(\nu_{n})\subset\mathcal{Q}_{\infty}(\nu)$, $n\geq 1$,
we deduce that the sequence $(\mu_{n})$ belongs to the closure of
$\mathcal{Q}_{\infty}(\nu)$ in $\mathcal{W}_{p}(\mathbb{R}^{d})$. Same
property holds for its $W_{p}$-limit $\mu$ and the result follows. ∎

##  Appendix B Properties of the function $\phi_{G}$

Let $G$ be a maximal monotone set: $G\in\mathfrak{M}$. In this appendix, we
collect the properties of the function

$\phi=\phi_{G}(y)\triangleq\inf_{x\in G}c(x,y)=\inf_{x\in
G}(x_{1}-y_{1})(x_{2}-y_{2}),\quad y\in\mathbb{R}^{2},$

used throughout the paper.

######  Lemma B.1.

The function $\phi=\phi_{G}$ and its $c$-conjugate

$\phi^{c}(x)=\inf_{y\in\mathbb{R}^{2}}(c(x,y)-\phi(y)),\quad
x\in\mathbb{R}^{2},$

take values in $[-\infty,0]$, $\phi^{c}\leq\phi$, and

$G=\left\\{{y\in\mathbb{R}^{2}}:\;\phi(y)=0\right\\}=\left\\{{x\in\mathbb{R}^{2}}:\;\phi^{c}(x)=0\right\\}.$

###### Proof.

Let $y\in G$. As $G\in\mathfrak{M}$, we have that $c(x,y)\geq 0$, $x\in G$,
and thus, $\phi(y)=0$. Conversely, if $y\not\in G$, then the maximal monotone
set $G$ crosses the interior of either the upper-left or the lower-right
quadrants relative to $y$. If $z\in G$ belongs to such intersection, then

$\phi(y)\leq c(z,y)=(z_{1}-y_{1})(z_{2}-y_{2})<0.$

We have shown that $\phi<0$ on $\mathbb{R}^{2}\setminus G$ and $\phi=0$ on
$G$. It follows that

$\phi^{c}(x)\leq\inf_{y\in G}(c(x,y)-\phi(y))=\inf_{y\in G}c(x,y)=\phi(x)\leq
0.$

If $\phi^{c}(x)=0$, then $\phi(x)=0$ and thus, $x\in G$. Conversely, if $x\in
G$, then $c(x,y)-\phi(y)\geq 0$, $y\in\mathbb{R}^{2}$, and therefore,
$\phi^{c}(x)=0$. ∎

We associate with $\phi$ the closed convex function

$\psi(y)=\psi_{G}(y)=y_{1}y_{2}-\phi(y)=\sup_{x\in
G}(x_{1}y_{2}+x_{2}y_{1}-x_{1}x_{2}),\quad y\in\mathbb{R}^{2}.$

Clearly, $\phi$ and $\psi$ have same domains:

$\operatorname{dom}{\phi}=\left\\{{y\in\mathbb{R}^{2}}:\;\phi(y)>-\infty\right\\}=\left\\{{y\in\mathbb{R}^{2}}:\;\psi(y)<\infty\right\\}=\operatorname{dom}{\psi}.$

For a convex set $A\subset\mathbb{R}^{d}$ we denote by
$\operatorname*{cl}{A}$, $\operatorname*{int}{A}$, $\operatorname*{ri}{A}$,
and
$\operatorname{\partial}{A}=\operatorname*{cl}{A}\setminus\operatorname*{ri}{A}$
its respective closure, interior, relative interior and relative boundary.

######  Lemma B.2.

The domain of $\phi$ is convex. If $G$ is either horizontal or vertical line,
then $\operatorname{dom}{\phi}=G$. Otherwise, $\operatorname{dom}{\phi}$ has a
non-empty interior:

$\operatorname*{int}{\operatorname{dom}{\phi}}=\operatorname*{int}{P_{1}}\times\operatorname*{int}{P_{2}},$
(32)

where $P_{i}$ is the projection of $G$ on $x_{i}$-coordinate, $i=1,2$. If
$y\in\operatorname{\partial}{\operatorname{dom}{\phi}}\cap\operatorname{dom}{\phi}$,
then the relative interiors of the horizontal and vertical parts of
$\operatorname{\partial}{\operatorname{dom}{\phi}}$ containing $y$ also belong
to $\operatorname{dom}{\phi}$.

###### Proof.

Being convex, the function $\psi=\psi_{G}$ has convex domain. As
$\operatorname{dom}{\phi}=\operatorname{dom}{\psi}$, the domain of $\phi$ is
also convex.

We observe that $P_{i}$ is either a point or an interval. If
$P_{1}=\left\\{{a_{1}}\right\\}$, then $G$ is a vertical line:
$G=\left\\{{x\in\mathbb{R}^{2}}:\;x_{1}=a_{1}\right\\}$. For $y\not\in G$ we
have that $\left\lvert y_{1}-a_{1}\right\rvert>0$ and

$\phi(y)=\inf_{x_{2}\in\mathbb{R}}(y_{1}-a_{1})(y_{2}-x_{2})=-\infty.$

Thus, $\operatorname{dom}\phi=G$. The case where $P_{2}$ is a point and thus,
$G$ is a horizontal line is identical.

We assume now that $\operatorname*{int}{P_{i}}=(a_{i},b_{i})$, where
$-\infty\leq a_{i}<b_{i}\leq\infty$. If
$y=(y_{1},y_{2})\in(a_{1},b_{1})\times(a_{2},b_{2})$, then the set
$C\triangleq\left\\{{x\in G}:\;c(x,y)\leq 0\right\\}$ is bounded and
therefore,

$\phi(y)=\inf_{x\in G}c(x,y)=\inf_{x\in C}c(x,y)>-\infty.$

Conversely, suppose that $y$ does not belong to the closure of $P_{1}\times
P_{2}$, say $y_{1}<a_{1}$; other cases are covered similarly. Then
$a_{2}=-\infty$ and hence,

$\phi(y)=\inf_{x\in G}c(x,y)\leq(a_{1}-y_{1})(a_{2}-y_{2})=-\infty.$

We have proved (32).

For the last assertion of the lemma, we assume that $a_{1}>-\infty$ and take
$y=(a_{1},y_{2})$ and $z=(a_{1},z_{2})$ with $z_{2}<b_{2}$. Given that
$\phi(y)>-\infty$, we have to show that $\phi(z)>-\infty$. Indeed, otherwise
there is a sequence $(x^{n})\subset G$ such that

$\lim_{n\rightarrow\infty}(a_{1}-x_{1}^{n})(z_{2}-x_{2}^{n})=-\infty.$

Since $z_{2}<b_{2}$, the sequence $(x_{1}^{n})$ is bounded and
$x_{2}^{n}\rightarrow a_{2}=-\infty$. It follows that

$\phi(y)\leq\limsup_{n\rightarrow\infty}(a_{1}-x_{1}^{n})(y_{2}-z_{2})+\lim_{n\rightarrow\infty}(a_{1}-x_{1}^{n})(z_{2}-x_{2}^{n})=-\infty,$

and we obtain a contradiction. ∎

The closed convex function $\psi=\psi_{G}$ is lower semi-continuous on
$\mathbb{R}^{2}$ and is continuous on the interior of its domain. The
following result shows that $\phi$ and $\psi$ are continuous relative to their
full domains.

######  Lemma B.3.

If $(y^{n})\subset\operatorname{dom}{\psi}=\operatorname{dom}{\phi}$ and
$y^{n}\to y$, then $\psi(y^{n})\to\psi(y)$ and $\phi(y^{n})\to\phi(y)$.

###### Proof.

It is sufficient to consider the case of the function $\psi$ and take
$y\in\operatorname{\partial}{\operatorname{dom}{\psi}}$. If $\psi(y)=\infty$,
then the result holds by the lower semi-continuity:

$\liminf_{n\to\infty}\psi(y^{n})\geq\psi(y)=\infty.$

Thus, we assume that $\psi(y)<\infty$ or, equivalently, that
$y\in\operatorname{\partial}{\operatorname{dom}{\psi}}\cap\operatorname{dom}{\psi}$.
By Lemma B.2, the relative interiors of the horizontal and vertical parts of
$\partial\operatorname{dom}{\psi}$ containing $y$ belong to
$\operatorname{dom}{\psi}$. Hence, there is a closed triangle in
$\operatorname{dom}{\psi}$ that contains $(y^{n})_{n\geq n_{0}}$, for
sufficiently large $n_{0}$. Being convex, the function $\psi$ is continuous on
this triangle and the result follows. ∎

We define a multi-valued function

$\displaystyle{\operatorname{Arg}}_{G}(y)\triangleq\operatorname*{arg\>min}_{x\in
G}c(x,y)=\left\\{{x\in G}:\;\phi(y)=c(x,y)\right\\},\quad y\in\mathbb{R}^{2},$

taking values in the closed (possibly empty) subsets of $G$, and denote

$\displaystyle\operatorname{dom}{{\operatorname{Arg}}_{G}}\triangleq\left\\{{y\in\mathbb{R}^{2}}:\;{\operatorname{Arg}}_{G}(y)\not=\emptyset\right\\}.$

Let $E^{G}_{i}=\cup_{t\in\mathcal{T}^{G}_{i}}E^{G}_{i}(t)$, $i=1,2$, be the
union of vertical and horizontal line segments of $G$; see (23). As the set
$G$ is fixed, we write simply

$\operatorname{Arg}={\operatorname{Arg}}_{G},\quad E_{i}=E^{G}_{i},\quad
E_{i}(t)=E^{G}_{i}(t),\quad\mathcal{T}_{i}=\mathcal{T}^{G}_{i}.$

The following lemma shows that for
$y\in\operatorname{dom}{\operatorname{Arg}}\setminus G$ the set
$\operatorname{Arg}(y)$ can intersect $E_{i}(t)$ only at
$\operatorname{\partial}{E_{i}(t)}$. We denote $\left\langle
x,y\right\rangle\triangleq\sum_{i=1}^{2}x_{i}y_{i}$, the scalar product of
$x,y\in\mathbb{R}^{2}$.

######  Lemma B.4.

Let $i\in\left\\{{1,2}\right\\}$ and $t\in\mathcal{T}_{i}$. If
$y\in\operatorname{dom}{\operatorname{Arg}}\setminus G$ and $x\in
E_{i}(t)\cap\operatorname{Arg}(y)$, then $x$ belongs to the boundary of
${E_{i}(t)}$ and

$\left\langle z-x,y-x\right\rangle>0,\quad z\in\operatorname*{ri}{E_{i}(t)}.$

###### Proof.

Without loss of generality we can assume that $i=2$ and that $y$ stays above
$G$. Then the increasing hyperbola

$H=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y)=\phi(y),\;z_{1}>y_{1}\right\\}$

contains $x$ and lays below $G$, which is only possible if $x$ is the right
boundary of the horizontal line segment $E_{2}(t)$. In this case,

$\left\langle z-x,y-x\right\rangle=(z_{1}-x_{1})(y_{1}-x_{1})>0,\quad
z\in\operatorname*{ri}{E_{2}(t)},$

and the result follows. ∎

For $x,y\in\mathbb{R}^{2}$ we denote by $L(x,y)$ the line segment connecting
$x$ and $y$:

$L(x,y)\triangleq\left\\{{tx+(1-t)y}:\;t\in[0,1]\right\\}.$

######  Lemma B.5.

Let $y^{0}$ and $y^{1}$ belong to
$\operatorname{dom}{\operatorname{Arg}}\setminus G$ and stay above and below
$G$, respectively. If
$x\in\operatorname{Arg}(y^{0})\cap\operatorname{Arg}(y^{1})$, then $x$ belongs
to the line segment $L(y^{0},y^{1})$ connecting $y^{0}$ and $y^{1}$.

###### Proof.

The conditions of the lemma imply that the increasing hyperbolas

$\displaystyle H^{0}$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;c(y^{0},z)=\phi(y^{0}),\;z_{1}>y^{0}_{1}\right\\},$
$\displaystyle H^{1}$
$\displaystyle=\left\\{{z\in\mathbb{R}^{2}}:\;c(y^{1},z)=\phi(y^{1}),\;z_{1}<y^{1}_{1}\right\\},$

contain $x$ and stay below and above $G$, respectively. Hence, they have
identical tangent lines at $x$. Elementary computations show that the slope of
the tangent line is given by

$\frac{x_{2}-y^{0}_{2}}{y^{0}_{1}-x_{1}}=\frac{x_{2}-y^{1}_{2}}{y^{1}_{1}-x_{1}}$

and the result follows. ∎

For $y\in\operatorname*{int}{\operatorname{dom}{\phi}}$ the derivative
$\nabla\phi(y)$ is defined in the classical sense. For
$y\in\operatorname{\partial}{\operatorname{dom}{\phi}}$ the derivative
$\nabla\phi(y)$ exists if it is the limit:
$\nabla\phi(y^{n})\to\nabla\phi(y)$, for every sequence
$(y^{n})\subset\operatorname*{int}{\operatorname{dom}{\phi}}\cap\operatorname{dom}{\nabla\phi}$
that converges to $y$. We write

$\displaystyle\operatorname{dom}{\nabla\phi}\triangleq\left\\{{y\in\operatorname{dom}{\phi}}:\;\nabla\phi(y)\;\text{exists}\right\\}.$

By $D^{c}\triangleq(D^{c}_{1},D^{c}_{2})$ we denote the differential operator
associated with the cost function $c=c(x,y)$:

$\displaystyle D^{c}_{1}\phi(y)\triangleq y_{1}-\frac{\partial\phi}{\partial
y_{2}}(y),\quad D^{c}_{2}\phi(y)\triangleq y_{2}-\frac{\partial\phi}{\partial
y_{1}}(y).$

Finally, let $E=E^{G}=E^{G}_{1}\cup E^{G}_{2}$ be the union of the vertical
and horizontal line segments of $G$ and denote

$\displaystyle\operatorname{\widehat{\operatorname{dom}}}{{\operatorname{Arg}}}\triangleq\left\\{{y\in\mathbb{R}^{2}}:\;{\operatorname{Arg}}(y)\text{
is a singleton}\right\\}.$

We observe that

$E=G\setminus\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}.$
(33)

######  Theorem B.6.

We have that

$\operatorname{dom}{\nabla\phi}\setminus\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}\subset\operatorname{\partial}{\operatorname{dom}{\phi}}\cap
G=\operatorname{\partial}{\operatorname{dom}{\phi}}\cap E.$ (34)

Conversely, the set difference
$\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}\setminus\operatorname{dom}{\nabla\phi}$
has at most two points and these points belong to different linear parts of
$\operatorname{\partial}{\operatorname{dom}{\phi}}$. If
$y\in\operatorname{dom}{\nabla\phi}\cap\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}$,
then $D^{c}\phi(y)$ is the only element of $\operatorname{Arg}(y)$ and

$\phi(y)=c(D^{c}\phi(y),y)=\frac{\partial\phi}{\partial
x_{1}}(y)\frac{\partial\phi}{\partial x_{2}}(y).$ (35)

We divide the proof of the theorem into lemmas. We write $x\leq y$ if
$x_{i}\leq y_{i}$, $i=1,2$. If $x,y\in G$ and $x\leq y$, then $G(x,y)$ denotes
the segment of $G$ bounded by $x$ and $y$:

$G(x,y)\triangleq\left\\{{z\in G}:\;x\leq z\leq y\right\\}.$

######  Lemma B.7.

Let $y\in\operatorname*{int}{\operatorname{dom}{\phi}}$. Then
$y\in\operatorname{dom}{\nabla\phi}$ if and only if
$y\in\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}$. In this
case, $D^{c}\phi(y)$ is the only element of $\operatorname{Arg}(y)$.

###### Proof.

From the structure of $\operatorname*{int}{\operatorname{dom}{\phi}}$ in Lemma
B.2 we deduce the existence of $x^{0},x^{1}\in G$ such that $x^{0}\leq x^{1}$,
$y\in\operatorname*{int}{R(x^{0},x^{1})}$, and
$R(x^{0},x^{1})\subset\operatorname*{int}{\operatorname{dom}{\phi}}$, where
$R(x^{0},x^{1})$ is the rectangle with the diagonal $L(x^{0},x^{1})$. Every
$x\in G$ such that $c(x,y)\leq 0$ belongs to
$G(x^{0},x^{1})=R(x^{0},x^{1})\cap G$. Hence,

$\operatorname{Arg}(y)=\operatorname*{arg\>min}_{z\in
G(x^{0},x^{1})}c(z,y)=\operatorname*{arg\>max}_{z\in
G(x^{0},x^{1})}(y_{1}z_{2}+y_{2}z_{1}-z_{1}z_{2}).$

As $G(x^{0},x^{1})$ is compact, $\operatorname{Arg}(y)$ is non-empty. If
$x\in\operatorname{Arg}(y)$, then

$\displaystyle\psi(u)-\psi(y)$ $\displaystyle=\sup_{z\in
G}(u_{1}z_{2}+u_{2}z_{1}-z_{1}z_{2})-(y_{1}x_{2}+y_{2}x_{1}-x_{1}x_{2})$
$\displaystyle\geq x_{2}(u_{1}-y_{1})+x_{1}(u_{2}-y_{2}),\quad
u\in\mathbb{R}^{2}.$

It follows that $(x_{2},x_{1})$ belongs to $\partial\psi(y)$, the
subdifferential of $\psi=\psi_{G}$ at $y$. Differentiability of $\phi$
(equivalently, of $\psi$) at $y$ then implies that $\operatorname{Arg}(y)$ is
a singleton and

$x_{1}=\frac{\partial\psi}{\partial
y_{2}}(y)=y_{1}-\frac{\partial\phi}{\partial y_{2}}(y)=D^{c}_{1}\phi(y),\quad
x_{2}=D^{c}_{2}\phi(y).$

Conversely, let $x$ be the only element of $\operatorname{Arg}(y)$ and
$\widetilde{x}=(\widetilde{x}_{1},\widetilde{x}_{2})\in\mathbb{R}^{2}$ be such
that $(\widetilde{x}_{2},\widetilde{x}_{1})\in\partial\psi(y)$. We have to
show that $x=\widetilde{x}$. We take a unit vector $e=(e_{1},e_{2})$ in
$\mathbb{R}^{2}$ and define a sequence $(y^{n})$ in $\mathbb{R}^{2}$ such that

$y^{n}_{1}=y_{1}+\frac{1}{n}e_{2},\;y^{n}_{2}=y_{2}+\frac{1}{n}e_{1},\quad
n\geq 1.$

Let $n_{0}$ be an index such that
$y^{n}\in\operatorname*{int}{R(x^{0},x^{1})}$, $n\geq n_{0}$. By the first
part of the proof, for $n\geq n_{0}$ the set $\operatorname{Arg}(y^{n})$ is
non-empty and belongs to the compact $G(x^{0},x^{1})$. Moreover, if
$x^{n}\in\operatorname{Arg}(y^{n})$ then
$(x^{n}_{2},x^{n}_{1})\in\partial\psi(y^{n})$. It follows that

$\displaystyle x^{n}_{2}(y^{n}_{1}-y_{1})+x^{n}_{1}(y^{n}_{2}-y_{2})$
$\displaystyle\geq\psi(y^{n})-\psi(y)$
$\displaystyle\geq\widetilde{x}_{2}(y^{n}_{1}-y_{1})+\widetilde{x}_{1}(y^{n}_{2}-y_{2}),$

and then that $\left\langle
x^{n},e\right\rangle\geq\left\langle\widetilde{x},e\right\rangle$. As $x$ is
the only element of $\operatorname{Arg}(y)$ and

$\psi(y^{n})=y^{n}_{1}x^{n}_{2}+y^{n}_{2}x^{n}_{1}-x^{n}_{1}x^{n}_{2}\to\psi(y)=y_{1}x_{2}+y_{2}x_{1}-x_{1}x_{2},$

every convergent subsequence of $(x^{n})$ goes to $x$ and then $x^{n}\to x$.
Hence, $\left\langle
x,e\right\rangle\geq\left\langle\widetilde{x},e\right\rangle$ and, as $e$ is
an arbitrary unit vector in $\mathbb{R}^{2}$, we obtain that
$x=\widetilde{x}$. ∎

######  Lemma B.8.

Let $y\in\operatorname{dom}{\operatorname{Arg}}\setminus G$ and
$x\in\operatorname{Arg}(y)$. Then

$\operatorname*{ri}{L(x,y)}=\left\\{{ty+(1-t)x}:\;t\in(0,1)\right\\}\subset\operatorname{dom}{\nabla\phi}$

and $D^{c}\phi(z)=x$, $z\in\operatorname*{ri}{L(x,y)}$. The slope of the line
segment $L(x,y)$ is negative and has the form:

$\frac{y_{2}-x_{2}}{y_{1}-x_{1}}=\frac{z_{2}-x_{2}}{z_{1}-x_{1}}=\frac{\left(\frac{\partial\phi}{\partial
y_{1}}(z)\right)^{2}}{\phi(z)}=\frac{\phi(z)}{\left(\frac{\partial\phi}{\partial
y_{2}}(z)\right)^{2}},\quad z\in\operatorname*{ri}{L(x,y)}.$

###### Proof.

We fix $t\in(0,1)$ and denote $y(t)=ty+(1-t)x$. From the description of
$\operatorname*{int}{\operatorname{dom}{\phi}}$ in Lemma B.2 we deduce that
$y(t)\in\operatorname*{int}{\operatorname{dom}{\phi}}$. Without loss in
generality we can assume that $y_{1}<x_{1}$. Then the hyperbola

$H=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y)=\phi(y),\;z_{1}>y_{1}\right\\}$

contains $x$ and stays below $G$, while the hyperbola

$H(t)=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y(t))=c(x,y(t)),\;z_{1}>y_{1}(t)\right\\}$

contains $x$ and stays below $H$. It follows that $x$ is the only element of
$\operatorname{Arg}(y(t))$. Lemma B.7 yields that $D^{c}\phi(y(t))=x$. The
last part of the lemma follows directly from the definition of $D^{c}\phi$ and
the fact that $\phi(y(t))=c(x,y(t))=c(D^{c}\phi(y(t)),y(t))$. ∎

The following corollary of Lemma B.8 will also be used in the proof of Theorem
B.12.

######  Lemma B.9.

Let $y^{0}$ and $y^{1}$ be distinct points in
$\operatorname{dom}{\operatorname{Arg}}\setminus G$ and
$x^{i}\in\operatorname{Arg}(y^{i})$, $i=1,2$. Then either $x^{0}=x^{1}$ or the
line segments $L(x^{0},y^{0})$ and $L(x^{1},y^{1})$ do not intersect.

###### Proof.

If $L(x^{0},y^{0})$ and $L(x^{1},y^{1})$ have common _interior_ point $z$,
then Lemma B.8 yields that $x^{0}=D^{c}\phi(z)=x^{1}$. ∎

######  Lemma B.10.

Let $y\in\operatorname{dom}{\nabla\phi}\setminus G$. Then $D^{c}\phi(y)$ is
the only element of $\operatorname{Arg}(y)$.

###### Proof.

In view of Lemma B.7 we can further assume that
$y\in\operatorname{\partial}{\operatorname{dom}{\phi}}$. Let $(y^{n})$ be a
sequence in
$\operatorname*{int}{\operatorname{dom}{\phi}}\cap\operatorname{dom}{\nabla\phi}$
that converges to $y$. By Lemma B.7, $D^{c}\phi(y^{n})$ is the only element of
$\operatorname{Arg}(y^{n})$. From the construction of $\nabla\phi$ on
$\operatorname{\partial}{\operatorname{dom}{\phi}}$ and Lemma B.3 we deduce
that

$\displaystyle D^{c}\phi(y)$
$\displaystyle=\lim_{n\to\infty}D^{c}\phi(y^{n})\in G,$ $\displaystyle\phi(y)$
$\displaystyle=\lim_{n\to\infty}\phi(y^{n})=\lim_{n\to\infty}c(D^{c}\phi(y^{n}),y^{n})=c(D^{c}\phi(y),y).$

Hence, $D^{c}\phi(y)\in\operatorname{Arg}(y)$. On the other hand, if
$x\in\operatorname{Arg}(y)$, then Lemma B.8 allows us to choose the sequence
$(y^{n})$ so that $D^{c}\phi(y^{n})=x$. Hence, $x=D^{c}\phi(y)$. ∎

######  Lemma B.11.

The set difference
$\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}\setminus\operatorname{dom}{\nabla\phi}$
has at most two points and these points belong to different linear parts of
$\operatorname{\partial}{\operatorname{dom}{\phi}}$.

###### Proof.

From Lemma B.2 we deduce that
$\operatorname*{int}{\operatorname{dom}{\phi}}=(a_{1},b_{1})\times(a_{2},b_{2})$,
where $-\infty\leq a_{i}<b_{i}\leq\infty$ and $(a_{i},b_{i})$ is the interior
of the projection of $G$ on the $x_{i}$-coordinate. Without loss of generality
we can assume that $a_{1}>-\infty$. Let $y^{0}$ and $y^{1}$ be such that
$y^{0}_{1}=y^{1}_{1}=a_{1}$, $y^{0}_{2}<y^{1}_{2}<b_{2}$ and
$y^{0}\in\operatorname{dom}{\operatorname{Arg}}$,
$y^{1}\in\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}$. We
are going to show that $y^{1}\in\operatorname{dom}{\nabla\phi}$. By doing so,
we shall prove that the interior of each linear part of
$\operatorname{\partial}{\operatorname{dom}{\phi}}$ has at most one element of
$\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}/\operatorname{dom}{\nabla\phi}$.

Let $(z^{n})$ be a sequence in
$\operatorname*{int}{\operatorname{dom}{\phi}}\cap\operatorname{dom}{\nabla\phi}$
that converges to $y^{1}$. Then $\sup_{n}z^{n}_{2}<b_{2}$ and there is $w\in
G$ such that $\sup_{n}z^{n}_{2}\leq w_{2}<b_{2}$. In view of Lemma B.7,
$u^{n}=D^{c}\phi(z^{n})$ is the only element of $\operatorname{Arg}(z^{n})$.
If $x^{0}\in\operatorname{Arg}(y^{0})$, then $y^{1}$ stays strictly above the
line segment $L(x^{0},y^{0})$ and, as $z^{n}\to y^{1}$, we can assume that
same property holds for $(z^{n})$. By Lemmas B.8 and B.10, the line segment
$L(z^{n},u^{n})$ has negative slope and can intersect $L(y^{0},x^{0})$ only at
$x^{0}$. It follows that $u^{n}$ belongs to the compact set $G(x^{0},w)$.
Continuity of $\phi=\phi_{G}$ from Lemma B.3 yields that any convergent
subsequence of $(u^{n})$ goes to the unique
$x^{1}\in\operatorname{dom}{\operatorname{Arg}(y^{1})}$. Hence,
$y^{1}\in\operatorname{dom}{\nabla\phi}$ and $x^{1}=D^{c}\phi(y^{1})$, by the
definition of $\nabla\phi$ on
$\operatorname{\partial}{\operatorname{dom}{\phi}}$.

Similar arguments show that if the “corner” point
$\widehat{y}=(a_{1},b_{2})\in\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}$
and there are $z^{0},z^{1}\in\operatorname{dom}{\operatorname{Arg}}$ that
belong to the interiors of different linear parts of
$\operatorname{\partial}{\operatorname{dom}{\phi}}$, then
$\widehat{y}\in\operatorname{dom}{\nabla\phi}$. ∎

###### Proof of Theorem B.6.

From Lemmas B.7 and B.10 we deduce that

$\operatorname{dom}{\nabla\phi}\setminus\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}\subset\operatorname{\partial}{\operatorname{dom}{\phi}}\cap
G.$

Lemma B.2 shows that the boundary of $\operatorname{dom}{\phi}$ is contained
in the union of two lines and that each of these lines is either vertical or
horizontal. It follows that

$\operatorname{\partial}{\operatorname{dom}{\phi}}\cap
G=\operatorname{\partial}{\operatorname{dom}{\phi}}\cap E$

and we obtain (34). Lemma B.11 states the structure of
$\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}/\operatorname{dom}{\nabla\phi}$.
Let
$y\in\operatorname{dom}{\nabla\phi}\cap\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}$.
Accounting for (33) we deduce that

$y\not\in\operatorname{\partial}{\operatorname{dom}{\phi}}\cap
E=\operatorname{\partial}{\operatorname{dom}{\phi}}\cap G.$

Lemmas B.7 and B.10 now yield that $D^{c}\phi(y)$ is the only element of
$\operatorname{Arg}(y)$. Finally, identity (35) holds by the definition of
$D^{c}\phi$. ∎

We recall that $\mathcal{D}$ denotes the family of graphs of strictly
decreasing functions $h=h(t)$ defined on closed intervals of $\mathbb{R}$ such
that $h$ and its inverse $h^{-1}$ are Lipschitz functions. We allow for a
degenerate case where the domain of $h$ is just a point. Thus,
$\mathbb{R}^{2}\subset\mathcal{D}$.

######  Theorem B.12.

The exception set

$\operatorname{dom}{\operatorname{Arg}}\setminus\left(\operatorname{dom}{\nabla\phi}\cap\operatorname{\widehat{\operatorname{dom}}}{\operatorname{Arg}}\right)=D\cup
E,$ (36)

where $D$ is a countable union of sets in $\mathcal{D}$ and $E=E^{G}$ is the
union of horizontal and vertical line segments of $G$.

We divide the proof into lemmas. For
$y\in\operatorname{dom}{\operatorname{Arg}}\setminus G$ and the points $r\leq
s$ in $\operatorname{Arg}(y)$, we denote by $\Delta(y,r,s)$ the closed curved
triangle bounded by the line segments $L(r,y)$, $L(y,s)$, and the segment
$G(r,s)$ of $G$; see Figure 2. If $r=s$, then $\Delta(y,r,s)=L(r,y)=L(s,y)$;
otherwise $\operatorname*{int}{\Delta(y,r,s)}\not=\emptyset$.

######  Lemma B.13.

Let $y^{0},y^{1}$ be distinct points in
$\operatorname{dom}{\operatorname{Arg}}\setminus G$, let $r^{i}\leq s^{i}$ be
in $\operatorname{Arg}(y^{i})$, and denote
$\Delta^{i}\triangleq\Delta(y^{i},r^{i},s^{i})$, $i=0,1$.

  1. (a)

If $y^{0}\in\Delta^{1}$, then $\Delta^{0}\subset\Delta^{1}$.

  2. (b)

If $y^{0}\not\in\Delta^{1}$ and $y^{1}\not\in\Delta^{0}$, then the
intersection of ${\Delta^{0}}$ and ${\Delta^{1}}$ is at most one point, which
is then either $r^{1}=s^{0}$ or $s^{1}=r^{0}$.

###### Proof.

If either (a) or (b) fails to hold, then there are line segments
$L^{i}\in\left\\{{L(r^{i},y^{i}),L(s^{i},y^{i})}\right\\}$, $i=0,1$, that
intersect only at an interior point. We obtain a contradiction with Lemma B.9.
∎

Lemma B.13 (a) yields a partial order relation on
$\operatorname{dom}{\operatorname{Arg}}\setminus G$: $y^{0}\prec y^{1}$ if
$y^{0}\in\Delta(y^{1},r^{1},s^{1})$ for some $r^{1}\leq s^{1}$ in
$\operatorname{Arg}(y^{1})$.

######  Lemma B.14.

If $y^{0},y^{1}$ belong to $\operatorname{dom}{\operatorname{Arg}}\setminus G$
and $y^{0}\prec y^{1}$, then

$D(y^{0},y^{1})\triangleq\left\\{{y\in\operatorname{dom}{\operatorname{Arg}}\setminus
G}:\;y^{0}\prec y\prec y^{1}\right\\}\in\mathcal{D},$

that is, $D(y^{0},y^{1})$ is the graph of a strictly decreasing function
$h=h(t)$ on $[y^{0}_{1},y^{1}_{1}]$ such that $h$ and its inverse $h^{-1}$ are
Lipschitz functions.

$r^{1}$$y^{1}$$s^{1}$$r^{0}$$y^{0}$$s^{0}$$G$$z^{1}$$z^{0}$$D$ Figure 2: The
curve $D=D(y^{0},y^{1})$ separates the parts of the $c$-gradient flow of
$\phi$ ending on the segments $G(r^{1},r^{0})$ and $G(s^{0},s^{1})$.

###### Proof.

We illustrate the proof on Figure 2. Without loss of generality we can assume
that $y^{0}$ and $y^{1}$ are distinct points that stay above $G$. Let
$r^{i}\leq s^{i}$ be in $\operatorname{Arg}(y^{i})$. We have that $r^{1}\leq
r^{0}\leq s^{0}\leq s^{1}$. If $y^{0}$ belongs to the line segment
$L(y^{1},s^{1})$, then Lemma B.8 yields that

$s^{1}=r^{0}=s^{0}=D^{c}\phi(y^{0})$

and then that $D=L(y^{0},y^{1})$. Same lemma shows that the line segment
$L(y^{0},y^{1})$ has a negative slope and thus, belongs to $\mathcal{D}$. The
case, where $y^{0}\in L(y^{1},r^{1})$ is identical.

Hereafter, we assume that $y^{0}\not\in L(y^{1},s^{1})\cup L(y^{1},r^{1})$ or,
equivalently, that $y^{0}\in\operatorname*{int}{\Delta(y^{1},r^{1},s^{1})}$.
Being a chord of the concave hyperbola

$H^{1}=\left\\{{z\in\mathbb{R}^{2}}:\;c(z,y^{1})=\phi(y^{1}),\quad
z_{1}>y_{1}\right\\},$

which touches $G$ from below, the line segment $L(r^{1},s^{1})$ stays below
$G$. It follows that $y^{0}$ belongs to the interior of the triangle with
vertices $\left\\{{r^{1},y^{1},s^{1}}\right\\}$. Hence, there are unique
$z^{1}\in\operatorname*{ri}{L(y^{1},s^{1})}$ and
$z^{0}\in\operatorname*{ri}{L(r^{1},y^{1})}$ such that the line segments
$L(r^{1},z^{1})$ and $L(s^{1},z^{0})$ intersect at $y^{0}$.

We observe that the convex polygon $P$ with the vertices
$\left\\{{z^{1},y^{1},z^{0},y^{0}}\right\\}$ contains every
$y\in\operatorname{dom}{\operatorname{Arg}}\setminus G$ such that $y^{0}\prec
y\prec y^{1}$ and thus, contains $D$. Being convex, $\psi$ is bounded on $P$.
Hence, $\phi$ is bounded on $P$ as well. Moreover, as $P$ stays away from
$G=\left\\{{x\in\operatorname{dom}{\phi}}:\;\phi(x)=0\right\\}$, same
boundedness property holds for $1/\phi$. If $y\in
P\cap\operatorname{dom}{\nabla\phi}$, then Lemmas B.9 and B.10 show that
$D^{c}\phi(y)$ belongs to the union of $G(r^{1},r^{0})$ and $G(s^{0},s^{1})$.
In particular, $D^{c}\phi$ and then also $\nabla\phi$ are bounded on
$P\cap\operatorname{dom}{\nabla\phi}$. From Lemma B.8 we deduce the existence
of negative constants $a$ and $b$ such that

$-\infty<a\leq\frac{y_{2}-x_{2}}{y_{1}-x_{1}}\leq b<0,\quad y\in
P,\;x\in\operatorname{Arg}(y).$

Let $y,z\in D$ be distinct. Lemma B.13 yields that either $y\prec z$ or
$z\prec y$. Assuming that $z\prec y$ we deduce the existence of
$r,s\in\operatorname{Arg}(y)$ such that $r\leq s$ and $z\in\Delta(y,r,s)$. The
slope of $L(y,z)$ is then bounded from below by the slope of $L(y,r)$ and from
above by the slope of $L(y,s)$, and thus is bounded in between by $a$ and $b$:

$-\infty<a\leq\frac{y_{2}-z_{2}}{y_{1}-z_{1}}\leq b<0.$

Hence, the set $D$ has the required Lipschitz properties.

It remains to be shown that the set $D$ is connected or, equivalently, that
for every pair of distinct points $w^{0}\prec w^{1}$ in $D$ there is $w\in D$,
which is different from $w^{0}$ and $w^{1}$ and such that $w^{0}\prec w\prec
w^{1}$. Without loss of generality we can take $w^{0}=y^{0}$ and
$w^{1}=y^{1}$. We shall find the required $w$ in $L(z^{0},z^{1})$.

Let $z(t)=(1-t)z^{0}+tz^{1}$, $t\in[0,1]$. From the non-intersection property
of Lemma B.9 and the continuity of $\phi$ on its domain, we deduce that

  1. 1.

If $t\in(0,1)$ and $\operatorname{Arg}(z(t))\cap
G(r^{1},r^{0})\not=\emptyset$, then $\operatorname{Arg}(z(s))\subset
G(r^{1},r^{0})$, $0\leq s<t$.

  2. 2.

If $(t_{n})\in[0,1]$ is such that $t_{n}\to t$ and
$\operatorname{Arg}(z(t_{n}))\cap G(r^{1},r^{0})\not=\emptyset$, $n\geq 1$,
then $\operatorname{Arg}(z(t))\cap G(r^{1},r^{0})\not=\emptyset$.

Similar properties (with obvious modifications in the first item) hold when
$G(r^{1},r^{0})$ is replaced with $G(s^{0},s^{1})$. These properties readily
yield the unique $t^{*}\in(0,1)$ such that $\operatorname{Arg}(z(t^{*}))$
intersects with _both_ $G(r^{1},r^{0})$ and $G(s^{0},s^{1})$. Clearly,
$w=z(t^{*})$ is different from both $y^{0}$ and $y^{1}$ and $y^{0}\prec w\prec
y^{1}$, thanks to Lemma B.13. ∎

######  Lemma B.15.

The set

$D\triangleq\left\\{{y\in\operatorname{dom}{\operatorname{Arg}}\setminus
G}:\;\operatorname{Arg}(y)\;\text{contains at least 2 points}\right\\},$ (37)

if not empty, is a countable union of sets in $\mathcal{D}$. More precisely,

$D=\cup_{n\geq 1}D(u^{n},v^{n})=\cup_{n\geq 1}\left\\{{y\in D}:\;u^{n}\prec
y\prec v_{n}\right\\},$

for some $u^{n}\leq v^{n}$ in $D$, $n\geq 1$.

###### Proof.

Clearly, $D=\cup_{n\geq 1}D(\frac{1}{n})$, where

$D(\epsilon)\triangleq\left\\{{y\in D}:\;\left\lvert
x^{0}-x^{1}\right\rvert\geq\epsilon\text{ for some
}x^{0},x^{1}\in\operatorname{Arg}(y)\right\\},\quad\epsilon>0.$

Let $\epsilon>0$. We denote by $\widehat{D}(\epsilon)$ the set of _minimal_
elements of $D(\epsilon)$ with respect to the order relation $\prec$. In other
words, $\widehat{y}\in\widehat{D}(\epsilon)$ if any $y\in D(\epsilon)$ such
that $y\prec\widehat{y}$ coincides with $\widehat{y}$. From Lemma B.13 we
deduce that $\widehat{D}(\epsilon)$ is countable. Let $y\in D(\epsilon)$. If
$y$ is not a minimal element, then there is $y^{\prime}\in D(\epsilon)$ such
that $y^{\prime}\prec y$, $y\not=y^{\prime}$. Being contained in
$\Delta(y,u,v)$ for some $u\leq v$ in $\operatorname{Arg}{(y)}$, the set
$\left\\{{z\in D(\epsilon)}:\;z\prec y^{\prime}\right\\}$ is bounded. By the
continuity of $\phi=\phi_{G}$, this set is closed and hence, contains some
$\widehat{y}\in\widehat{D}(\epsilon)$. It follows that

$D(\epsilon)=\cup_{\widehat{y}\in\widehat{D}(\epsilon)}\left\\{{y\in
D}:\;\widehat{y}\prec y\right\\}.$

Finally, for $y\in D$, Lemmas B.13 and B.14 show that $\left\\{{z\in
D}:\;y\prec z\right\\}$ is the graph of a strictly decreasing function $h$
such that $h$ and $h^{-1}$ are locally Lipschitz. The result readily follows.
∎

###### Proof of Theorem B.12.

By Theorem B.6 representation (36) holds if we add to the set $D$ given by
(37) at most 2 points. Lemma B.15 yields the result. ∎

######  Lemma B.16.

Let $D$ be given by (37) and

$\displaystyle
S=\left\\{{y\in\operatorname{dom}{\operatorname{Arg}}}:\;\operatorname{Arg}(y)\;\text{contains
at least 3 points}\right\\}.$

Then $S$ is countable and there are Borel functions $g_{i}:\;D\rightarrow G$,
$i=1,2$, such that

$\operatorname{Arg}(y)=\left\\{{g_{1}(y),g_{2}(y)}\right\\},\;g_{1}(y)\not=g_{2}(y),\quad
y\in D\setminus S.$

###### Proof.

In view of Lemma B.15, it is sufficient to prove the result for the sets
$D^{\prime}=D(u,v)$ and $S^{\prime}=S\cap D^{\prime}$, where $u\prec v$ in
$D$. Let $r\leq s$ be distinct elements of $\operatorname{Arg}(u)$. The
functions

$\displaystyle g_{1}(y)$
$\displaystyle=\max\left\\{{x\in\operatorname{Arg}(y)}:\;x\leq r\right\\},$
$\displaystyle g_{2}(y)$
$\displaystyle=\min\left\\{{x\in\operatorname{Arg}(y)}:\;x\geq s\right\\},$

map $D^{\prime}$ to $G$ and are monotone with respect to the order relations
$\prec$ on $D^{\prime}$ and $\leq$ on $G$. Thus, their respective sets
$(R_{i})$ of discontinuities are countable. From Lemma B.13 we deduce that
$S^{\prime}\subset R_{1}\cup R_{2}$ and from the continuity of $\phi$ that
$g_{i}(y)\in\operatorname{Arg}(y)$, $y\in D^{\prime}$. The proof readily
follows. ∎

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