a nonlocal monge-kantorovich problemmate.dm.uba.ar/~jrossi/monge.pdf · 2009. 11. 25. · a...

A NONLOCAL MONGE-KANTOROVICH PROBLEM

N. IGBIDA, J. M. MAZON, J. D. ROSSI AND J. TOLEDO

Abstract. This paper is concerned with a nonlocal version of the classical Monge-Kantorovichmass transport problem in which we replace the Euclidean distance with a discrete distance tomeasure the transport cost. We fix the length of a step and the distance that measures the cost ofthe transport depends of the number of steps that is needed to transport the involved mass fromits origin to its destination. For this problem we deal with the issues of existence of Monge maps,Kantorovich potentials and optimal transport plans. We find that, even in one dimension, andfor absolutely continuous measures respect to the Lebesgue measure, the existence of an optimaltransport map depends on the involved masses. We also prove that when the length of the steptends to zero these nonlocal problems give an approximation to the classical Monge-Kantorovichmass transport problem. Finally, we study how the approach developed by Evans and Gangbofor the classical case works in this context. We find an equation for the potentials, obtained as alimit of nonlocal p−Laplacian problems, and we use it to construct optimal transport plans. Wealso obtain the transport density for the classical Monge-Kantorovich problem by rescaling.

To the memory of Fuensanta Andreu, our friend and colleague.

Contents

1. Introduction and main results 21.1. Introduction 21.2. Preliminaries 41.3. Statement of the main results 72. A nonlocal Monge-Kantorovich problem 102.1. Kantorovich potentials 122.2. Equality between Monge’s infimun and Kantorovich’s minimum 183. The one-dimensional case 203.1. Kantorovich potentials 203.2. Nonexistence of optimal transport maps 213.3. Optimal transport plans 234. Approximation of the classical Monge-Kantorovich problem by the nonlocal problems 315. The nonlocal version of the Evans-Gangbo approach 395.1. Characterizing the Euler-Lagrange equation 395.2. Constructing optimal transport plans 435.3. Construction of the transport density for the classical MK-problem by rescaling 48References 52

Key words and phrases. Mass transport, nonlocal problems, Monge-Kantorovich problems.2000 Mathematics Subject Classification. 49J20,49J45,45G10.

1

2 N. IGBIDA, J. M. MAZON, J. D. ROSSI AND J. TOLEDO

1. Introduction and main results

1.1. Introduction. The Monge mass transport problem, as proposed by Monge in 1781, consistsin finding the optimal way of moving points from one mass distribution to another so that thework done is minimized. In general, the total work is proportional to some cost function. In theclassical Monge problem the cost function is the Euclidean distance, and this problem has beenintensively studied and generalized in different directions that correspond to different classes ofcost functions. We refer to the surveys and books [1], [3], [10], [17], [19], [20] for further discussionof Monge’s problem, its history, and applications.

However, even being the case of discontinuous cost functions very interesting for concretesituations and applications, it seems not to be well covered in the literature. For instance, assumethat you want to transport an amount of bricks to a hole, then you count the number of steps thatyou have to move each brick to its final destination in the hole and try to move the total amountof bricks making as less as possible steps. This amounts to the classical Monge-Kantorovichproblem for the discrete distance (that count the number of steps)

d1(x, y) =

0 if x = y,

[[|x− y|]] + 1 if x 6= y,

where |.| is the Euclidean norm and [[r]] is defined for r > 0 by [[r]] := n, if n < r ≤ n + 1, n =0, 1, 2, . . . . That is,

d1(x, y) =

0 if x = y,1 if 0 < |x− y| ≤ 1,2 if 1 < |x− y| ≤ 2,...

This transport problem with this discrete distance also appears naturally when one considers atransport problem between cities in which the cost is measured by the toll in the road (that in asimplified version is a discrete function of the number of kilometers). We want to mention anothermotivation for the study of this mass transport problem that comes from a nonlocal model forsandpiles studied in [5] (see also [14]), and from which we have decided to add the word nonlocalto this transport problem.

In [10] Evans gives an interpretation of the sandpile model introduced by Aronsson, Evans andWu in [6], in terms of Monge-Kantorovich mass transport theory for the particular case of theEuclidean distance d|·| in RN as cost. The model for the evolution of a sandpile with source f ≥ 0is the evolution equation

(1.1)

f − ut ∈ ∂IKd|·| (R

N )(u), t > 0,

u(0, x) = 0,

where u(t, x) is the height of the sandpile at time t in x and

Kd|·|(RN ) := L2(RN ) ∩ Lip1(RN , d|·|) =

u ∈ L2(RN ) ∩W 1,∞(RN ) : |∇u| ≤ 1 a.e.

.

The interpretation reads as follows: at each moment of time, the mass µ+ = f(t, ·) dx is instantlyand optimally transported downhill by the potential u(t, ·) into the mass µ− = ut(t, ·) dy. In otherwords, the height function u(t, ·) of the sandpile is deemed also to be the potential generating theMonge-Kantorovich reallocation of µ+ to µ−.

A NONLOCAL MONGE-KANTOROVICH PROBLEM 3

In [5] it is studied the following nonlocal version of the sandpile model given in [6]. Let

(1.2)

J : RN → R be a nonnegative continuous radial function with

supp(J) = B(0, 1), J(0) > 0 and∫

RN

J(x) dx = 1,

and setKJ

1 :=u ∈ L2(RN ) : |u(x)− u(y)| ≤ 1 for x− y ∈ supp(J) a.e.

.

The nonlocal problem is written as

(1.3)

f(t, ·)− ut(t, ·) ∈ ∂IKJ1(u(t, ·)) a.e. t ∈ (0, T ),

u(x, 0) = u0(x).

To give a mass transport interpretation of this model, we consider the previously mentioneddistance d1. Observe that KJ

1 = L2(RN ) ∩ Lip1(RN , d1). The following result is proved in [5]:the solution u(t, ·) of the limit problem (1.3) is a solution of the Kantorovich dual problem

maxu∈KJ

1

∫

RN

u(f+ − f−)

when the involved measures are the source term f+ = f(t, x) and the time derivative of thesolution f− = ut(t, x). Hence, as for the local model, at each moment of time, the mass µ+ =f(t, ·) dx is instantly and optimally transported downhill by the potential u(t, ·) into the massµ− = ut(t, ·) dy, but in this case the discrete function d1 is what gives the transport cost.

The aim of this paper is, on the one hand, the study of the Monge and Kantorovich problemsfor the discrete cost function d1(x, y). And, on the other hand, the study of rescaled problemswith costs dε in which the size of the step is ε (see (1.21) for a precise definition) in order toapproach the Monge and Kantorovich problems for the Euclidean distance as the parameter εgoes to 0. We perform a detailed study of the one dimensional case with several concrete exampleswhich illustrate the general obstructions to existence and the difficulties that may appear withthis particular cost. We also find an equation for the potentials, obtained as a limit of nonlocalp−Laplacian problems, and we use it to construct optimal transport plans. Moreover, we obtainthe transport density for the classical Monge-Kantorovich problem by rescaling.

It is clear that our problem falls into the scope of lower semi-continuous cost functions, so thatstandard results, like the existence of a solution for the relaxed problem, the so called Monge-Kantorovich problem, or the Kantorovich duality remain true for d1. Moreover, since d1 is ametric, it is well known that one of the interesting features of the theory is the existence ofa potential, the so called Kantorovich potential. However, since d1 is discrete, the connectionbetween Monge and Monge-Kantorovich problems, the characterization of the potential transport,the Evans-Gangbo approach, as well as the computation of the optimal plan and/or map transportare not covered in the literature for this concrete distance. In particular, the potential can notbe characterized in a standard way, i.e., by using standard differentiation. In this work, wecharacterize the Kantorovich potential by using a nonlocal equation and we develop a completetheory concerning the Monge problem with the discrete distance d1. Moreover, we give theconnection between the Monge-Kantorovich problem with the discrete distance and the classicalMonge-Kantorovich problem with the Euclidean distance, proving that when the length of the steptends to zero these nonlocal problems give an approximation to the classical Monge-Kantorovichmass transport problem.


1.2. Preliminaries. To fix notation and for convenience of the reader, let us start recalling withsome detail what are the problems in which we are interested and also state some known resultsand useful properties of their solutions. Although most of them hold true in a more generalframework we will state them for Ω a bounded convex domain in RN which will be the frameworkof our results. Whenever T is a map from a measure space (X, µ) to an arbitrary space Y , wedenote by T#µ the pushforward measure of µ by T . Explicitly, (T#µ)[B] = µ[T−1(B)]. Whenwe write T#f = g, where f and g are nonnegative functions, this means that the measure havingdensity f is pushed-forward to the measure having density g.

The Monge problem. Fix c : RN × RN → [0, +∞] a lower semi-continuous cost function,and two non-negative Borel function f+, f− ∈ L1(Ω) satisfying the mass balance condition

(1.4)∫

Ωf+(x) dx =

∫

Ωf−(y) dy.

Let A(f+, f−) be the set of Borel maps T : Ω → Ω which push the measure µ+ := f+(x) dxforward to µ− := f−(y) dy, that is

T#f+ = f− ⇐⇒∫

T−1(B)f+(x) dx =

∫

Bf−(y) dy, ∀B ⊂ Ω Borel.

The Monge problem is to find a map T ∗ ∈ A(f+, f−) which minimizes the cost functional

(1.5) Fc(T ) :=∫

Ωc(x, T (x))f+(x) dx

in the set A(f+, f−). A map T ∗ ∈ A(f+, f−) satisfying

Fc(T ∗) = minFc(T ) : T ∈ A(f+, f−),is called an optimal transport map pushing f+ to f−.

The original problem studied by Monge corresponds to the cost function

c(x, y) = d|·|(x, y) := |x− y|,the Euclidean distance. In general, the Monge problem is ill-posed. To overcome the difficultiesof the Monge problem, in 1942 L. V. Kantorovich, [15], proposed to study a relaxed version ofthe Monge problem and, what is more relevant here, introduced a dual variational principle.

We will use the usual convention of denoting by πi : RN × RN the projections, π1(x, y) := x,π2(x, y) := y. Given a Radon measure µ in Ω×Ω, its marginals are defined by projx(µ) := π1#µ,projy(µ) := π2#µ. The relaxed problem is the following.

The Monge-Kantorovich problem. Fix c, f+ and f−as above. Let

π(f+, f−) := Nonnegative Radon measures µ in Ω× Ω :

projx(µ) = f+(x) dx, projy(µ) = f−(y) dy

,

and consider the functional

(1.6) Kc(µ) :=∫

Ω×Ωc(x, y) dµ(x, y).


The Monge-Kantorovich problem is to find a measure µ∗ ∈ π(f+, f−) which minimizes the costfunctional Kc(µ), that is, to find a solution to the minimization problem

(1.7) Kc(µ∗) = minKc(µ) : µ ∈ π(f+, f−).

The elements µ ∈ π(f+, f−) are called transport plans between f+ and f−, and µ∗ satisfying(1.7) is called an optimal transport plan between f+ and f−. Remark that we say plans betweenf+ and f− since this problem is reversible, which is not true in general for the Monge problem.Given two non-negative Borel function f+, f− ∈ L1(Ω) satisfying the mass balance condition(1.4), there exists always the trivial transport plan, f+ £ f−, defined as

(f+ £ f−)(A×B) :=

∫

A×Bf+(x)f−(y) dx dy

∫

Ωf+(x) dx

=

∫

A×Bf+(x)f−(y) dx dy

∫

Ωf−(y) dy

.

The relation between transport maps and transport plans is given in the following result (see,for instance, [1, Propostion 2.1]).

Proposition 1.1. Any Borel transport map T ∈ A(f+, f−) induces a transport plan µT ∈π(f+, f−) defined by µT := (Id × T )#f+ = f+(x)δ[y=T (x)], that is,

∫Ω×Ω c(x, y) dµT (x, y) =∫

Ω c(x, T (x))f+(x) dx. Conversely, a transport plan µ ∈ π(f+, f−) is induced by a transport mapif µ is supported on a µ-measurable graph.

An immediate consequence of the above result is that

(1.8) infKc(µ) : µ ∈ π(f+, f−) ≤ infFc(T ) : T ∈ A(f+, f−).Under the above assumptions it is not difficult to show the existence of an optimal transport

plan (see [1, 16] and the references therein).

Proposition 1.2. Let c : RN × RN → [0, +∞] a lower semi-continuous cost function, and letf+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balance condition (1.4).Then, there exists an optimal transport plan µ∗ ∈ π(f+, f−) solving the Monge-Kantorovichproblem

Kc(µ∗) = minKc(µ) : µ ∈ π(f+, f−).Moreover, if c is continuous we have that (1.8) is, in fact, an equality:

(1.9) minKc(µ) : µ ∈ π(f+, f−) = infFc(T ) : T ∈ A(f+, f−).The dual formulation by Kantorovich of the Monge problem is the following.

The Kantorovich dual problem. Fix c : RN × RN → [0, +∞] a lower semi-continuouscost function, and two non-negative Borel function f+, f− ∈ L1(Ω) satisfying the mass balancecondition (1.4). The dual problem is to find (u∗, v∗) that maximize

(1.10) D(u, v) :=∫

Ωu(x)f+(x) dx +

∫

Ωv(y)f−(y) dy

in the set

Φc(f+, f−) :=

(u, v) ∈ L1(Ω, f+(x) dx)× L1(Ω, f−(y) dy) : u(x) + v(y) ≤ c(x, y)

.


We have the following result (see for instance [3, Theorem 3.1] or [19, Theorem 1.3]).

Theorem 1.3. (Kantorovich duality). Let c : RN × RN → [0, +∞] a lower semi-continuouscost function, and let f+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4). Then,

(1.11) minKc(µ) : µ ∈ π(f+, f−) = supD(u, v) : (u, v) ∈ Φc(f+, f−).Furthermore, it does not change the value of the supremum in the right-hand side of (1.11) if onerestricts the definition of Φc(f+, f−) to those functions (u, v) which are bounded and continuous.

When the cost function is a metric, c(x, y) = d(x, y) on RN , then there is more structure in theKantorovich duality principle. We denote by Kd(Ω) the set of functions in Ω which are Lipschitzcontinuous with Lipschitz constant no greater than one,

Kd(Ω) := Lip1(Ω, d) =u ∈ L1(Ω) : |u(x)− u(y)| ≤ d(x, y) a.e.

.

We have the following result (see for instance [19, Theorem 1.14]).

Theorem 1.4. (Kantorovich-Rubinstein Theorem). Let d be a lower semi-continuous met-ric on RN , and let f+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4). Then,

(1.12) minKd(µ) : µ ∈ π(f+, f−) = supPf+,f−(u) : u ∈ Kd(Ω)

,

wherePf+,f−(u) :=

∫

Ωu(x)(f+(x)− f−(x)) dx.

The maximizers u∗ of (1.12) are called Kantorovich (transport) potentials.

Now, given a convex set K ⊂ L2(Ω), if we denote by IK its indicator function, that is, thefunction defined as

IK(u) :=

0 if u ∈ K,+∞ if u 6∈ K,

we have that the Euler-Lagrange equation associated with the variational problem

max∫

Ωu(x)(f+(x)− f−(x)) dx : u ∈ Lip1(Ω, d)

is the equation

(1.13) f+ − f− ∈ ∂IKd(Ω)(u).

That is, the Kantorovich potentials are solutions of (1.13).In the particular case of the Euclidean distance d(x, y) = d|·|(x, y) = |x− y| and for adequate

masses f+ and f−, Evans and Gangbo in [11] find a solution of (1.13), that is, a Kantorovichpotential for the mass transport problem of f+ to f−, as a limit as p → ∞ of solutions to thelocal p−Laplace equation with Dirichlet boundary conditions in a sufficiently large ball B(0, R),

−∆pup = f+ − f− B(0, R),

up = 0, ∂B(0, R),

Moreover, they characterize the solutions to the limit equation (1.13) by means of a PDE.


Theorem 1.5. (Evans-Gangbo Theorem). Let d|·| be the Euclidean distance on RN , andlet f+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balance condition (1.4).Assume additionally that f+ and f− are Lipschitz continuous functions with compact supportsuch that supp(f+) ∩ supp(f−) = ∅. Then, there exists u∗ ∈ Kd|·| such that

∫

Ωu∗(x)(f+(x)− f−(x)) dx = max

∫

Ωu(x)(f+(x)− f−(x)) dx : u ∈ Kd|·|

;

and there exists 0 ≤ a ∈ L∞(Ω) such that

(1.14) f+ − f− = −div(a∇u∗) in D′(Ω).

Furthermore|∇u∗| = 1 a.e. on the set a > 0.

The function a that appear in the previous result is the Lagrange multiplier corresponding tothe constraint |∇u∗| ≤ 1, and it is called the transport density.

Moreover, what is very important from the point of view of mass transport, Evans and Gangbouse this PDE to find a proof of the existence of an optimal transport map for the classical Mongeproblem, different to the first one given by Sudakov in 1979 by means of probability methods([18], see also [1] and [3]).

1.3. Statement of the main results. Let us state the main results of this paper. We use thesame notation previously introduced.

Since d1 is lower semi-continuous, then, for any f+, f− ∈ L1(Ω) satisfying the mass balancecondition (1.4), we deduce that

• there exists an optimal plan transport µ∗ minimizing the Monge-Kantorovich problemKd1(µ

∗) = minKd1(µ) : µ ∈ π(f+, f−).• The Kantorocich duality remains true, i.e.,

minKd1(µ) : µ ∈ π(f+, f−) = supDd1(u, v) : (u, v) ∈ Φc(f+, f−).Moreover, since d1 is a metric, the interesting way to study the problem is by studing the Kan-torovich potential, indeed the Kantorovich-Rubinstein Theorem states that

minKd(µ) : µ ∈ π(f+, f−) = supPf+,f−(u) : u ∈ Kd1(Ω)

.

The metric d1 makes the problem independent from processes related to differentiation. Here,we characterize the potential u∗ by means of a nonlocal equation. Following the ideas from [11],we first construct Kantorovich potentials for the cost function d1(x, y) taking limit as p → ∞in some nonlocal p−Laplacian type problems. This limit procedure is nontrivial due to the lackof regularity that these nonlocal problems enjoy. Moreover this result is a remarkable fact ofthe parallelism between local and nonlocal p–Laplacian problems and their approach to Monge-Kantorovich problems.

Next, we prove existence of Kantorovich potentials with a finite number of jumps of size one.This result is of special interest for searching/constructing optimal transport maps and plans.

Theorem 1.6. Given f+, f− ∈ L∞(Ω) two non-negative Borel functions satisfying the massbalance condition (1.4) and such that

(1.15) |supp(f+) ∩ supp(f−)| = 0,


there exists a Kantorovich potential u∗ such that u∗(Ω) ⊂ Z and takes a finite number of values.More precisely, there exists a partition Ajk

j=0 of Ω such that

u∗ =k∑

j=0

j χAj .

In addition, we havef+ χA0 = f− χAk

= 0.

Next, we deal with the equality between Monge’s infimun and Kantorovich’s minimum. It iswell known (see [16]) that equality between Monge’s infimun and Kantorovich’s minimum is nottrue in general if the cost function is not continuous. The example given by Pratelli in [16] canbe adapted to get a counterexample also for the case of the cost function given by the metric d1,see Section 2. We prove that the cause of the above counterexample is not the discontinuity ofd1 being the cause the fact that the measures f+ and f− are not absolutely continuous respectthe Lebesgue measure. Indeed, we get the following result.

Theorem 1.7. Let f+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4). Then,

(1.16) minKd1(µ) : µ ∈ π(f+, f−) = infFd1(T ) : T ∈ A(f+, f−)

.

We study the existence and nonexistence for optimal transport maps, there are several examplesthat show that the existence of a transport map is a delicate issue that depends strongly on theconfiguration of the data f+ and f− and the distance d1. For example, we obtain the followingresult.

Theorem 1.8. In one space dimension, let f+ = Lχ[0,1] and f− = χ

[−L,0], then there exists anoptimal transport map T if and only if L 6∈ N or L = 1.

Another main task is to show how the Evans-Gangbo approach, studied in [11] for the Euclid-ean distance, works in the setting of the discrete distance d1. Recall that for the Euclideandistance, the Monge-Kantorovich equation (1.14) contains all the information concerning the op-timal transportation of f+ into f−. In our case, we use an analogous analysis by providing a newnonlocal equation and proving how this enables to construct the optimal plan transportation off+ into f−. Actually, we characterize the solutions of the Euler-Lagrange equation

(1.17) f+ − f− ∈ ∂IKd1(Ω)(u),

as the functions u∗ ∈ K1, for which there exists an antisymmetric bounded measure σ∗ in Ω, suchthat

(1.18)[σ∗]+ (x, y) ∈ Ω× Ω : u∗(x)− u∗(y) = 1, |x− y| ≤ 1,[σ∗]− (x, y) ∈ Ω× Ω : u∗(y)− u∗(x) = 1, |x− y| ≤ 1,

satisfying

(1.19)∫

Ωdσ∗(x, y) = f+(x)− f−(x), x ∈ Ω.


and

(1.20) |σ∗|(Ω× Ω) = 2∫

Ω(f+(x)− f−(x))u∗(x) dx.

We see that the measure σ∗ jointly with the potential u∗ contain all the information needed toconstruct an optimal transport plan and we show how to find it. Using Theorem 1.6 and (1.18),we have that the positive part of σ∗ may be written as

[σ∗]+ =k∑

j=0

[σ∗]+ (Aj ×Aj−1)

and, we use the sequence of measures [σ∗]+ (Aj × Aj−1), jointly with the Gluing Lemma 5.4,that follows from [19, Lemma 7.6], to give an exact description of an optimal transport of f+

into f− (see Section 5.2 for the details).Note that, as we will show, there is no optimal transport map in general, therefore, we can not

expect to obtain such a map from the limit equation, however, we find a general construction ofoptimal transport plans.

The last task is the connection between our optimal mass transport problem and the classicalMonge transport problem. Actually, we approximate the usual Monge-Kantorovich problem withthe Euclidean distance with nonlocal Monge-Kantorovich problems when we reduce the step size.To this end let us set the definition of the discrete distance for step of size ε,

(1.21) dε(x, y) =

0 if x = y,

ε([[ |x−y|

ε

]]+ 1

)if x 6= y.

Recall that for the Euclidean distance, we have

infFd|.|(T ) : T ∈ A(f+, f−)

= minKd|.|(µ) : µ ∈ π(f+, f−)

= supPf+,f−(u) : u ∈ Kd|.|

=: Wd|.| .

For the discrete distance dε, thanks to Theorem 1.16, we deduce that

infFε(T ) : T ∈ A(f+, f−)

= minKε(µ) : µ ∈ π(f+, f−)= sup

Pf+,f−(u) : u ∈ Kε

=: Wε.

And we prove the following result.

Theorem 1.9. Let f+, f− ∈ L2(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4).(1) The sequence (Wε)ε>0 is nonincreasing as ε → 0 and

0 ≤ Wε −Wd|.| ≤ ε

∫f+(x) dx for any ε > 0.

(2) The optimal transport plans µε converge as ε → 0 to an optimal transport plan for the Monge-Kantorovich problem with respect to the Euclidean distance: there exists µ∗ ∈ π(f+, f−), suchthat, after a subsequence,

µε µ∗ as measures


andKd|·|(µ

∗) = minKd|·|(µ) : µ ∈ π(f+, f−).(3) Convergence for the Kantorovich potentials: let u∗ε be a Kantorovich potential associated withthe metric dε, then, after a subsequence,

u∗ε u∗ in L2 as ε → 0,

where u∗ is a Kantorovich potential associated with the Euclidean distance.

Finally, we obtain the transport density for the classical Monge-Kantorovich problem by rescal-ing. We prove that there exists a subsequence εn → 0, such that

µn := (π1, Tεn)#[εnσ∗εn]+ σ+ weakly as measures,

whereTε(x, y) =

y − x

ε,

anda = π1#σ+

is the transport density for the classical Monge-Kantorovich problem given in Theorem 1.5.

2. A nonlocal Monge-Kantorovich problem

Given ε > 0, we consider the convex set

(2.1) Kε :=u ∈ L2(Ω) : |u(x)− u(y)| ≤ ε if |x− y| ≤ ε a.e.

.

For the distance dε defined in (1.21) it is easy to see that

Kε = Kdε(Ω) =u ∈ L2(Ω) : |u(x)− u(y)| ≤ dε(x, y) a.e.

,

that is,Kε = Lip1(Ω, dε).

Given f+, f− ∈ L1(Ω) two non-negative Borel functions satisfying the mass balance condition(1.4), we have that the Monge problem associated with the distance dε is to find an optimaltransport map T ∗ ∈ A(f+, f−) satisfying

(2.2) Fε(T ∗) = minFε(T ) : T ∈ A(f+, f−),where Fε(T ) := Fdε(T ). We will see in Example 3.2 that, in general and even for masses forwhich the classical Monge problem has solution, the Monge problem associated with the distancedε does not have a solution.

The Monge-Kantorovich problem associated with the distance dε is to find and optimal planµ∗ ∈ π(f+, f−) solving the minimization problem

(2.3) Kε(µ∗) = minKε(µ) : µ ∈ π(f+, f−),where Kε(µ) := Kdε(µ). By Proposition 1.2, we have that there is an optimal plan µ∗ ∈ π(f+, f−)solving the Monge-Kantorovich problem (2.3). Now, since dε is not continuous, we do not know,in principle, if

minKε(µ) : µ ∈ π(f+, f−) = infFε(T ) : T ∈ A(f+, f−).


On the other hand, from Theorems 1.3 and 1.4, it follows that, for f+, f− ∈ L2(Ω),

(2.4) minKε(µ) : µ ∈ π(f+, f−) = sup∫

Ωu(x)(f+(x)− f−(x)) dx : u ∈ Kε

.

The aim of this section is the study of (2.2), (2.3) and the Kantorovich potentials in (2.4). Letus first prove, working as in the proof of [9, Lemma 6], the following Dual Criteria for Optimality.

Lemma 2.1. Fix u∗ ∈ Kε and let T ∗ ∈ A(f+, f−). If

(2.5) u∗(x)− u∗(T ∗(x)) = dε(x, T ∗(x)) for almost all x ∈ supp(f+),

then

(i) u∗ is a Kantorovich potential for the metric dε.(ii) T ∗ is an optimal map for the Monge problem associated to the metric dε.(iii)


= sup

Pf+,f−(u) : u ∈ Kε

.

(iv) Every optimal map T for the Monge problem associated to the metric dε and Kantorovichpotential u for the metric dε satisfy (2.5).

Proof. We will write P = Pf+,f− for simplicity. For every T ∈ A(f+, f−) and u ∈ Kε, we have

Fε(T ) =∫

Ωdε(x, T (x))f+(x) dx ≥

∫

Ω(u(x)− u(T (x))) f+(x) dx

=∫

Ωu(x)f+(x) dx−

∫

Ωu(y) f−(y) dy = P(u).

Hence,

(2.6) infFε(T ) : T ∈ A(f+, f−)

≥ sup P(u) : u ∈ Kε .

On the other hand, the assumption (2.5) imply Fε(T ∗) = P(u∗). Therefore,

P(u∗) = Fε(T ∗) ≥ infFε(T ) : T ∈ A(f+, f−)

≥ sup P(u) : u ∈ Kε ≥ P(u∗),

and consequently (iii) holds. Moreover, we also get

P(u∗) = max P(u) : u ∈ Kε ,

from where it follows (i), and

Fε(T ∗) = minFε(T ) : T ∈ A(f+, f−)

,

from where (ii) follows.Finally, let T be an optimal map for the Monge problem associated to the metric dε and u a

Kantorovich potential for the metric dε. Then, Fε(T ) = Fε(T ∗) and P(u) = P(u∗). Thus, by(iii), we obtain that Fε(T ) = P(u), and consequently,∫

Ωdε(x, T (x))f+(x) dx =

∫

Ω(u(x)− u(T (x))) f+(x) dx.

From where it follows that

u(x)− u(T (x)) = dε(x, T (x)) for almost all x ∈ supp(f+),

since dε(x, T (x)) ≥ u(x)− u(T (x)) and f+ ≥ 0. ¤


Remark 2.2. Observe also that if u∗ is a Kantorovich potential for the metric d, from (1.12)and the inequality u∗(x)− u∗(y) ≤ d(x, y) it follows that, if µ∗ ∈ π(f+, f−),

(2.7) µ∗ is optimal ⇐⇒ u∗(x)− u∗(y) = d(x, y) µ∗ − a.e. in Ω× Ω.

2.1. Kantorovich potentials. We will show, like in [11], that it is possible to obtain Kantorovichpotentials associated with the metric d1 taking limit as p → ∞ in some related nonlocal p–Laplacian type problems. Let J be as in (1.2), f ∈ L∞(Ω), and consider the functional

(2.8) Fp(u) =12p

∫

Ω

∫

ΩJ(x− y)|u(y)− u(x)|p dy dx−

∫

Ωf(x)u(x) dx

in the space

(2.9) Sp =

u ∈ Lp(Ω) :∫

Ωu(x) dx = 0

.

We will use the following Poincare type inequality from [4].

Proposition 2.3. Given p ≥ 1, J and Ω, there exists βp−1 = βp−1(J,Ω, p) > 0 such that

(2.10) βp−1

∫

Ω

∣∣∣∣u−1|Ω|

∫

Ωu

∣∣∣∣p

≤ 12

∫

Ω

∫

ΩJ(x− y)|u(y)− u(x)|p dy dx ∀u ∈ Lp(Ω).

Using this proposition we can show that there is a unique minimum of Fp in Sp .

Theorem 2.4. There exists a unique up ∈ Sp such that

(2.11) Fp(up) = minv∈Sp

Fp(v).

Proof. Let un be a minimizing sequence. Hence, Fp(un) ≤ C, that is12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx−

∫

Ωf(x)un(x) dx ≤ C

Then,12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx ≤

∫

Ωf(x)un(x) dx + C

From the Poincare inequality (2.10) and Holder’s inequality, we get12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx ≤ ‖f‖Lp′ (Ω)‖un‖Lp + C

≤ C(f)(

12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx

)1/p

+ C.

Therefore, we have that12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx ≤ C.

Then, applying again Poincare’s inequality (2.10), we have un : n ∈ N is bounded in Lp(Ω).Hence, we can extract a subsequence that converges weakly in Lp(Ω) to some u (that clearly hasto verify

∫Ω u = 0) and we obtain

lim infn→+∞

12p

∫

Ω

∫

ΩJ(x− y)|un(y)− un(x)|p dy dx ≥ 1

2p

∫

Ω

∫

ΩJ(x− y)|u(y)− u(x)|p dy dx


andlim

n→+∞

∫

Ωf(x)un(x) dx =

∫

Ωf(x)u(x) dx

Therefore, u is a minimizer of Fp.Uniqueness is a direct consequence of the fact that Fp is strictly convex. ¤

Now we want to study the limit as p →∞ of the sequence up of minimizers of Fp in Sp.

Theorem 2.5. Let f+, f− ∈ L∞(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4). Let up the solution of (2.11) for f = f+ − f−. Then, there exists a subsequenceupnn∈N having a weak limit u which is a Kantorovich potential associated with the metric d1,that is, ∫

Ωu(x)(f+(x)− f−(x)) dx = max

v∈K1

∫

Ωv(x)(f+(x)− f−(x)) dx.

Proof. For 1 ≤ q, we set

|||u|||q :=(∫

Ω

∫

ΩJ(x− y)|u(y)− u(x)|q dxdy

) 1q

.

Let us write f := f+ − f−. Since Fp(up) ≤ Fp(0) = 0,

|||up|||pp ≤ 2p

∫

Ωf(x)up(x) dx ≤ 2p ‖f‖∞|||up|||1.

For 1 ≤ q < p, by Holder’s and Poincare’s inequalities, we have

|||up|||q ≤(∫

Ω

∫

ΩJ(x− y)|up(y)− up(x)|p dxdy

) 1p

(∫

Ω

∫

ΩJ(x− y) dxdy

) p−qpq

≤ (2p)1p ‖f‖

1p∞|||up|||

1p

1

(∫

Ω

∫

ΩJ(x− y) dxdy

) p−qpq

≤(

2p ‖f‖∞β0

) 1p

|||up|||1pq

(∫

Ω

∫

ΩJ(x− y) dxdy

) p−1pq

.

Consequently,

(2.12) |||up|||q ≤(

2p ‖f‖∞β0

) 1p−1

(∫

Ω

∫

ΩJ(x− y) dxdy

) 1q

.

Then, |||up|||q : p > q is bounded. Hence, by Poincare’s inequality (2.10), we have thatup : p > q is bounded in Lq(Ω). Therefore, we can assume that up u weakly in Lq(Ω). Bya diagonal process, we have there is a sequence pn → ∞, such that upn u weakly in Lm(Ω),as n → +∞, for all m ∈ N. Thus, u ∈ L∞(Ω). Since the functional v 7→|||v|||q is weakly lowersemi-continuous, having in mind (2.12), we have

|||u|||q ≤(∫

Ω

∫

ΩJ(x− y) dxdy

) 1q

.

Therefore,lim

q→+∞ |||u|||q ≤ 1,


from where it follows that|u(x)− u(y)| ≤ d1(x, y) a.e.

Let us see that u is a Kantorovich potential associated with the metric d1. Fix v ∈ K1. Then,

−∫

Ωfup ≤ 1

2p

∫

Ω

∫

ΩJ(x− y)|up(y)− up(x)|p dxdy −

∫

Ωf(x)up(x) dx

= Fp(up) ≤ Fp

(v − 1

|Ω|∫

Ωv

)

=12p

∫

Ω

∫

ΩJ(x− y)|v(y)− v(x)|pdxdy −

∫

Ωf(x)v(x) dx

≤ 12p

∫

Ω

∫

ΩJ(x− y)dxdy −

∫

Ωf(x)v(x) dx,

where we have used∫Ω f = 0 for the second equality and the fact that v ∈ K1 for the last

inequality.Hence, taking limit as p →∞, we obtain that∫

Ωu(x)(f+(x)− f−(x)) dx ≥

∫

Ωv(x)(f+(x)− f−(x)) dx.

Therefore, u is a Kantorovich potential associated with the metric d1. ¤

Our next step is to study the existence of some special Kantorovich potentials which will playan important role in the rest of the paper. We will see that, under quite general assumptions,there is a Kantorovich potential u∗ associated with d1 such that its image has a finite number ofvalues in Z, that is, u∗ : Ω 7→ Z with u∗(Ω) finite. To this end, let us first prove the followingresult.

Lemma 2.6. Assume that v ∈ K1 takes a finite number of values. Then there exists u ∈ K1 thatalso takes a finite number of values but with jumps of length 1, the number of points in its imageis less or equal than the number of points in the image of v and improves in the maximizationproblem, that is, ∫

Ωu(x)(f+(x)− f−(x)) dx ≥

∫

Ωv(x)(f+(x)− f−(x)) dx.

Proof. The proof runs by induction in the number of levels of v. Take f := f+− f− and supposethat v ∈ K1 is given by, without loss of generality,

v(x) = a0χA0 + a1χA1 + · · ·+ akχAk,

a0 = 0, |Ai| > 0, Ai ∩Aj = ∅ for any i 6= j.

Set s := Sign(∫

A0

f

), where

Sign(r) =

1 if r ≥ 0,

−1 if r < 0,

and consider

t0 = maxt ≥ 0 : ut := (a0 + st)χA0 + a1χA1 + · · ·+ akχAk∈ K1.


So, t0 is such that ∃ i 6= 0, dist(Ai, A0) ≤ 1 and |a0 + st0 − ai| = 1 and∫

Ωf(x)v(x) dx ≤

∫

Ωf(x)ut(x) dx.

Hence, replacing v by ut0 , we can assume that Ai are disjoint sets, dist(A0, A1) ≤ 1 and|u0 − u1| = 1.

Now, we set s := Sign(∫

A0∪A1

f

)and we consider

t0 = maxt ≥ 0 ; ut := (a0 + st)χA0 + (a1 + st)χA1 + a2χA2 + · · ·+ akχAk∈ K1.

So, t0 is such that ∃ i ∈ 0, 1 and ∃ ji 6∈ 0, 1 such that dist(Ai, Aji) ≤ 1, |ai + st0 − aji | = 1and ∫

Ωf(x)v(x) dx ≤

∫

Ωf(x)ut(x) dx.

Hence, replacing v by ut0 , we can assume that Ai are disjoint sets and |ui − uj | ∈ 0, 1, 2, forany i, j ∈ 0, 1, 2.

Now, by induction assume that we have u = a0χA0 + · · ·+ alχAl+ · · ·+ akχAk

, where Ai aredisjoint sets, and |ai− aj | ∈ N, for any i, j = 0, 1, . . . , l, and let us prove that we can assume thatAi are disjoint compact sets, and |ai − aj | ∈ N, for any i, j ∈ 0, 1, . . . , l + 1. We set

s := Sign(∫

A0∪...∪Al

f

),

and we consider

t0 = maxt ≥ 0 ; ut := (a0 + st)χA0 + · · ·+ (al + st)χAl+ al+1χAl+1

+ · · ·+ akχAk∈ K1.

So, t0 is such that ∃ i ∈ 0, 1, . . . , l and ∃ ji 6∈ 0, 1, . . . , l such that dist(Ai, Aji) ≤ 1, |ui +st0 − uji | = 1 and ∫

Ωf(x)u(x) dx ≤

∫

Ωf(x)ut(x) dx.

Hence, replacing u by ut0 , we can assume that Ai are disjoint sets, and |ai − aj | ∈ N, for anyi, j ∈ 0, 1, . . . , l + 1.

At last, by induction we deduce that we can assume that Ai are disjoint compact sets, and|ai − aj | ∈ N, for any i, j ∈ 0, 1, . . . , k. ¤

Theorem 2.7. Let f+, f− ∈ L∞(Ω) two non-negative Borel functions satisfying the mass balancecondition (1.4) and such that |supp(f+) ∩ supp(f−)| = 0. Then there exists is a Kantorovichpotential u∗, associated with the metric d1, such that u∗(Ω) ⊂ Z and takes a finite number ofvalues.

Proof. Take f := f+− f−. By density, we have that there exists a maximizing sequence vn ∈ K1

such that vn takes a finite number of values and∫

Ωvnf → max

w∈K1

∫

Ωwf.


Thanks to the previous lemma, there exists un ∈ K1 such that all the jumps of the levels ofun are of length one and it is a new maximizing sequence,

(2.13)∫

Ωunf → max

w∈K1

∫

Ωwf.

Notice that the number of levels of un ∈ K1 is bounded by a constant that only depends on Ω.Indeed, if u ∈ K1 is like

u(x) = jχC1 + (j + 1)χC2 + · · ·+ (j + k)χCk,

with |Ci| > 0, Ci ∩ Cj = ∅ for i 6= j, then, there exists a null set N ⊂ Ω× Ω such that

|x− y| > 1 ∀(x, y) ∈ (Ci−1 × Ci+1) \N, ∀i,otherwise u 6∈ K1. Therefore, since Ω has finite diameter, this provides a bound for the numberof possible sets k that is the same as the number of possible levels.

Let us take k an uniform bound for the number of levels of functions in K1 and let us supposealso that the lower level for the minimizing functions is 0. By Fatou’s Lemma and having in mind(2.13) we get

maxw∈K1

∫

Ωwf ≤

∫

Ωlim sup

n→∞(unf).

Now, since supp(f+) ∩ supp(f−) is null,

lim supn→∞

(fun) ≤ f+ lim supn→∞

un − f− lim infn→∞ un

= f+k∑

i=0

iχAi − f−k∑

i=0

iχBi = fk∑

i=0

iχCi ,

whereCi =

(Ai ∩ f+(x) > 0) ∪ (

Bi ∩ f−(x) > 0) , i > 0,

and

C0 = Ω \k⋃

i=0

Ci.

So, setting

u∗ =k∑

i=0

iχCi ,

we have

maxw∈K1

∫

Ωwf ≤

∫

Ωfu∗.

To finish the proof let us see that u∗ ∈ K1. Indeed, let Nn ⊂ Ω× Ω the null set where

|un(x)− un(y)| > d1(x, y),

and set N := ∪n∈NNn. Take (x, y) ∈ (Ω × Ω) \ N such h that |x − y| ≤ 1. Let us supposethat x ∈ Ai ∩ f+ > 0 and y ∈ Bj ∩ f− > 0 (the other cases being similar), then since|un(x)− un(y)| ≤ 1 for all n ∈ N, by the definition of lim sup and lim inf, we have

|u∗(x)− u∗(y)| = |i− j| ≤ 1.


If not, that is, if |i− j| > 1, assuming for instance that i < j, we have that there exists 0 < ε < 1such that

i < i + ε < j − ε < j

and there exist n ∈ N such that un(x) ∈ [i, i + ε], and un(y) ∈ [j − ε, j], that is, un(x) = i andun(y) = j, which contradicts that |un(x) − un(y)| ≤ d1(x, y) (we have taken (x, y) ∈ (Ω × Ω) \Nn). ¤

Remark 2.8. Note that with a similar proof, for the distance dε we obtain potentials that takefinite values in jε : j ∈ Z.Remark 2.9. By Lemma 2.1, for an optimal transport map T ∗ pushing f+ to f−, we have

(2.14) u∗(x)− u∗(T ∗(x)) = d1(x, T ∗(x)) for almost all x ∈ supp(f+),

being u∗ a Kantorovich potential. This relation says that a point x ∈ supp(f+) has to be carriedto a point T ∗(x) in such a way that the number of steps is given by the jumps of u∗, that is thenumber of steps is equal to u∗(x) − u∗(T ∗(x)). To search or to construct an optimal transportmap T ∗ we must have this observation in mind.

Let us make another observation. If we assume that u∗ takes only the values j, j + 1, j +2, ..., j + k, j ∈ Z, that is, u∗ = jχA0 + (j + 1)χA1 + (j + 2)χA2 + .... + (j + k)χAk

, then,

(2.15) |Ak ∩ supp(f−)| = 0 and |A0 ∩ supp(f+)| = 0.

In fact, if not, just redefine u∗ to be

u∗(x) =

j + k − 1 in Ak ∩ supp(f−),u∗(x) otherwise,

and we get that u∗ ∈ K1 with ∫

Ωu∗f <

∫

Ωu∗f,

a contradiction.We also observe that

(2.16)∫

Ak

f+ ≥∫

Ak−1

f−.

In fact, if not, we define

u∗(x) =

j + k − 1 in Ak,j + k − 2 in Ak−1 ∩ supp(f−),u∗(x) otherwise,

and we get that u∗ ∈ K1 with ∫

Ωu∗f <

∫

Ωu∗f,

a contradiction. The inequality (2.16) will be of special interest.


2.2. Equality between Monge’s infimun and Kantorovich’s minimum. It is well known(see, [16]) that equality (1.9) between Monge’s infimun and Kantorovich’s minimum is not truein general if the cost function is not continuous. The example given by Pratelli in [16], can beadapted to get a counterexample also for the case of the cost function given by the metric d1.

Example 2.10. Consider R, S and T are the parallel segments in R2, given by R := (−1, y) :y ∈ [−1, 1], S := (0, y) : y ∈ [−1, 1], Q := (1, y) : y ∈ [−1, 1]. Let f+ := 2H1 S andf− := H1 R +H1 Q. It is not difficult to see that

minKd1(µ) : µ ∈ π(f+, f−) = 2

and the the minimum is achieved by the transport plan splitting the central segment S in twoparts and translating them on the left and on the right. On the other hand, we claim that

(2.17) infF1(T ) : T ∈ A(f+, f−)

≥ 4.

To prove (2.17), fix T ∈ A(f+, f−) and consider I(T ) := x ∈ S : d1(x, T (x)) = 1. If we seethat

(2.18) f+(I(T )) = 0,

then

F1(T ) =∫

Sd1(x, T (x)) df+(x) = 2

∫

S\I(T )d1(x, T (x)) dH1(x) ≥ 4,

and (2.17) follows. Finally, let us see that (2.18) holds. If we define

I(T )R := x ∈ I(T ) : T (x) ∈ R and I(T )Q := x ∈ I(T ) : T (x) ∈ Q,we have I(T ) = I(T )R ∪ I(T )Q and I(T )R ∩ I(T )Q = ∅, and by the definition of I(T ), if E =T (I(T )), it is easy to see that

H1(E) = H1(E ∩R) +H1(E ∩Q) = H1(I(T )R) +H1(I(T )R) = H1(I(T )).

Therefore,f+(I(T )) = 2f−(E).

But since T ∈ A(f+, f−) one has

f−(E) = f+(T−1(E)) ≥ f+(I(T )) = 2f−(E),

that implies f+(I(T )) = 0 and (2.18) is proved.

Let us see that the cause of the above counterexample is not the discontinuity of d1 being thecause the fact that the measures f+ and f− are not absolutely continuous respect the Lebesguemeasure.

Theorem 2.11. Let f+, f− ∈ L1(Ω) two non-negative Borel function satisfying the mass balancecondition (1.4). Then,

(2.19) minKd1(µ) : µ ∈ π(f+, f−) = infF1(T ) : T ∈ A(f+, f−)

.

Proof. For n ∈ N, n ≥ 2, let ϕn be the function defined by

ϕn(r) :=

m if m ≤ r ≤ m + 1− 1

n ,

n(r − (m + 1)) + m + 1 if m + 1− 1n < r < m + 1,


with m = 0, 1, 2, . . .. Consider the continuous cost function

cn(x, y) :=

0 if x = y,ϕn(|x− y|) + 1 if x 6= y.

By construction, c1 ≥ c2 ≥ . . . cn ≥ d1 and limn→+∞ cn(x, y) = d1(x, y). Let µ∗ ∈ π(f+, f−),such that

Kd1(µ∗) = minKd1(µ) : µ ∈ π(f+, f−).

Since cn is continuous, by Proposition 1.2, we have there exists µn ∈ π(f+, f−), such that

Kcn(µn) = minKcn(µ) : µ ∈ π(f+, f−) = inf∫

Ωcn(x, T (x))f+(x) dx : T ∈ A(f+, f−)

.

Therefore, for every n ≥ 2, there exists Tn ∈ A(f+, f−) such that∫

Ω

∫

Ωcn(x, y) dµ∗(x, y) ≥

∫

Ω

∫

Ωcn(x, y) dµn(x, y) ≥

∫

Ωcn(x, Tn(x))f+(x) dx− 1

n.

Then, by the Monotone Convergence Theorem, we get∫

Ω

∫

Ωd1(x, y) dµ∗(x, y) = lim

n→+∞

∫

Ω

∫

Ωcn(x, y) dµ∗(x, y) ≥ lim sup

n→+∞

∫

Ωcn(x, Tn(x))f+(x) dx

and since cn ≥ d1, we have

lim supn→+∞

∫

Ωd1(x, Tn(x))f+(x) dx ≤ lim sup

n→+∞

∫

Ωcn(x, Tn(x))f+(x) dx

≤∫

Ω

∫

Ωd1(x, y) dµ∗(x, y) = Kd1(µ

∗).

On the other hand,

infF1(T ) : T ∈ A(f+, f−)

≤∫

Ωd1(x, Tn(x))f+(x) dx,

hence

Kd1(µ∗) ≤ inf

F1(T ) : T ∈ A(f+, f−) ≤ lim inf

n→+∞

∫

Ωd1(x, Tn(x))f+(x) dx.

Consequently,

Kd1(µ∗) = inf

F1(T ) : T ∈ A(f+, f−)

= limn→+∞

∫

Ωd1(x, Tn(x))f+(x) dx.

This ends the proof. ¤

Remark 2.12. Let us remark that the results we have obtained are also true if in the metric dε

we change the Euclidean norm by any norm ‖ ‖ of RN , that is if we consider the distance

dε(x, y) =

0 if x = y,

ε([[‖x−y‖

ε

]]+ 1

)if x 6= y.

Especially interesting is the case in which we consider the ‖ · ‖∞ norm since in this case it countsthe maximum of steps moving parallel to the coordinate axes. That is, in this case we measurethe distance cost as the number of blocks that the taxi has to cover going from x to y in a city.


3. The one-dimensional case

In this section we will see that in the one dimensional case we can be more precise and providea better description of the abstract situation together with some examples that illustrate thedifficulties of this problem.

3.1. Kantorovich potentials. We assume first that the functions f+ and f− are L∞-functionssatisfying

(3.1) f− = f−χ[a,0], f+ = f+χ

[c,d], c ≥ 0, supp(f±) ⊂ [−L,L], for some L ∈ N.

Set Ω any interval containing [−L,L].By Theorem 2.7, there exists a Kantorovich potential u∗ associated with the metric d1, such

that u∗(Ω) ⊂ Z and takes a finite number of values. It is easy to see that we can take

(3.2) u∗(x) = θα(x) :=

...−1 if α− 2 < x ≤ α− 1,0 if α− 1 < x ≤ α,1 if α < x ≤ α + 1,...

for some 0 < α ≤ 1. In order to find which α’s give the Kantorovich potential, we need tomaximize

∫

Ωu∗(x)(f+(x)− f−(x)) dx = −

∫ 0

−Lu∗(x)f−(x) dx +

∫ L

0u∗(x)f+(x) dx

= −−1∑

j=−L

∫ 1

0(θα(x) + j)f−(x + j) dx +

L−1∑

j=0

∫ 1

0(θα(x) + j)f+(x + j) dx

= −−1∑

j=−L

∫ 1

0θα(x)f−(x + j) dx +

L−1∑

j=0

∫ 1

0θα(x)f+(x + j) dx

−−1∑

j=−L

∫ 1

0jf−(x + j) dx +

L−1∑

j=0

∫ 1

0jf+(x + j) dx.

Since the last two integrals are independent of θα, we only need to maximize

−−1∑

j=−L

∫ 1

0(θα(x))f−(x + j) dx +

L−1∑

j=0

∫ 1

0(θα(x))f+(x + j) dx

=∫ 1

0θα(x)M(x) dx =

∫ 1

αM(x) dx,

for 0 < α ≤ 1, where

(3.3) M(x) = −−1∑

j=−L

f−(x + j) +L−1∑

j=0

f+(x + j), 0 < x ≤ 1.


Observe that ∫ 1

0M(x) dx =

∫(f+ − f−) = 0.

If M(x) is non decreasing, it is clear that, for 0 < x ≤ 1,

θα(x) =

0 if M(x) < 0,

1 if M(x) > 0,

is the best choice (unique for points where M(x) 6= 0). If M(x) is non increasing, it is also clearthat α = 1 is the best choice.

Remark 3.1. Let us suppose now that the supports of the masses are not ordered. For example,let us search for a Kantorovich potential associated with the metric d1 for f− = f1 + f2, f1 =f−1 χ

(a1,a2), f2 = f−2 χ(c1,c2), and f+ = f+χ

(b1,b2), with a1 < a2 < b1 < b2 < c1 < c2. Letb ∈ (b1, b2) be such that

∫f1 =

∫fχ

(b1,b) and∫

f2 =∫

fχ(b,b2). Let us call f+

1 := fχ(b1,b) and

f+2 := fχ

(b,b2). By the previous example we construct a non decreasing stair function, θ1, asKantorovich potential for f+

1 and f−1 with value at b equals to some λ fixed, and a non increasingstair function, θ2, as Kantorovich potential for f+

2 and f−2 with the same value λ at b . Then,θ = θ1χ(a1,b) + θ2χ(b,c2) gives a Kantorovich potential for f+ and f−. This construction can bedone for any configuration f+ =

∑mi=1

χ(b1,i,b2,i) and f− =

∑ni=1

χ(c1,i,c2,i).

3.2. Nonexistence of optimal transport maps. Let us see with a simple example that, ingeneral, an optimal transport map does not exist for d1 as cost function. Let us point out thatfor the Euclidean distance it is well known (see for instance [1] or [19]) the existence of an optimaltransport map in the case f± ∈ L1(a, b), even more, the existence of an unique optimal transportmap in the class of non-decreasing functions which is given by

(3.4) T0(x) := sup

y ∈ R :∫ y

af−(t) dt ≤

∫ x

af+(t) dt

if x ∈ (a, b).

Example 3.2. Let f+ = Lχ[0,1] and f− = χ

[−L,0] with L ∈ R. Set Ω an interval containing[−L,L]. Let us see that if L ∈ N, L ≥ 2, then there is no optimal transport map T with distanced1 pushing f+ to f−, nevertheless we will see later in Example 3.5 that if L /∈ N then there is anoptimal transport map pushing f+ to f−. With a similar proof it can be proved that there is notransport map T between f+ = Lχ

[0,1] and f− = χ[−L,0] with L ∈ N if one considers the distance

d1/k with k ∈ N.Arguing by contradiction assume there is an optimal transport map T pushing f+ to f−. Let

Ai = x ∈]0, 1[ : d1(x, T (x)) = i, i = 1, ..., L.

A Kantorovich potential for this configuration of masses f+ and f− is given by

u∗(x) =

0, x ∈ (0, 1)−1, x ∈ (−1, 0]

...−L, x ∈ (−L,−L + 1].


From Lemma 2.1 we have the equality

u∗(x)− u∗(T (x)) = d1(x, T (x)).

Therefore, if x ∈ Ai then T (x) ∈ (−i,−i + 1] and hence we can conclude that

|Ai| = 1/L,

in fact, Ai = T−1((−i,−i + 1]). Moreover,

T (x) ≥ x− i for all x ∈ Ai.

Now, we claim that

T (x) = x− i for all x ∈ Ai, for every i = 1, . . . , L.

Therefore |T (Ai)| = 1/L which gives a contradiction with the fact that |T ([0, 1])| = L.To prove that claim we argue as follows: assume, without lose of generality, that there is a set

of positive measure K ⊂ A1 such that T (x) > x−1 in K. Then, it is easy to see that there existsθ ∈ (0, 1) such that

|T−1((−1, θ − 1))| < |A1 ∩ (0, θ)|.Therefore, since

T−1((−i, θ − i)) ⊂ Ai ∩ (0, θ) for all i,

we have

θ =1L

∣∣∣∣∣L⋃

i=1

(−i, θ − i)

∣∣∣∣∣ =

∣∣∣∣∣T−1

(L⋃

i=1

(−i, θ − i)

)∣∣∣∣∣ =

∣∣∣∣∣L⋃

i=1

T−1 ((−i, θ − i))

∣∣∣∣∣ <L⋃

i=1

|Ai ∩ (0, θ)| = θ,

and we arrive to a contradiction.Nevertheless, since d1 is lower semi-continuous, by Proposition 1.2, there exists an optimal

plan µ∗ ∈ π(f+, f−) solving the Monge-Kantorovich problem

K1(µ∗) = minK1(µ) : µ ∈ π(f+, f−) = sup P(u) : u ∈ K1

=∫

Ωu∗(x)(f+(x)− f−(x)) dx = 1 + 2 + 3 + · · ·+ L =

L(L + 1)2

.

Now, if define the measure µ∗ in Ω× Ω by

µ∗(x, y) := Lχ[0,1](x)

(1L

δ[y=−1+x] +1L

δ[y=−2+x] + · · ·+ 1L

δ[y=−L+x]

),

then µ∗ ∈ π(f+, f−) and, moreover, since

K1(µ∗) =∫

Ω×Ωd1(x, y) dµ∗(x, y)

= L

∫ 1

0

(1L

d1(x,−1 + x) +1L

d1(x,−2 + x) + · · ·+ 1L

d1(x,−L + x))

dx =L(L + 1)

2,

we have that µ∗ is an optimal plan.Let us finish this example with the following construction. We consider L = 2 for simplicity.

By Theorem 2.11 we know that there exists tn ∈ A(f+, f−) such that

Fd1(tn) =∫

Ωd1(x, tn(x))f+(x) dx

n→0+→ 3.


Nevertheless we will find an explicit approximation tn from which we can infer a different approachfor the proof of Theorem 2.11. This tn can be constructed following the subsequent ideas. Pushf+χh

1− 12n+1 ,1

i to f−χ[−2,−2+ 12n ] with a plan induced by a map (see the picture bellow), paying 3

2n ,

and f+χh0,1− 1

2n+1

i to f−χ[−2+ 12n ,0] with a plan induced also by a map (see the picture bellow),

paying 3− 22n . Therefore,

Fd1(tn) =∫

Ωd1(x, tn(x))f+(x) dx = 3 +

12n

n→0+→ 3,

and the Monge and Kantorovich problems have the same cost.

Support of 2χ[0,1](x)δ[y=tn(x)] (n = 2)

x

y

78

− 74

− 14

34

12

− 32

14

− 34

180 1

0

-1

-2

Observe that all the segments have slope 2.

3.3. Optimal transport plans. Let us construct an optimal transport plan under the assump-tions (3.1). This will show, on account of Remark 3.1, how to work in a more general situation.


Let u∗ = θα the Kantorovich potential given by (3.2) and construct a new configuration ofequal masses as follows,

f+0 (x) =

L−1∑

j=0

f+(x + j)

χ

]0,1[(x), f−0 (x) =

L−1∑

j=0

f−(x− j)

χ

]−1,0[(x).

For these masses, the same u∗ is a Kantorovich potential. Moreover,

(3.5)

∫ L

−Lu∗(x)(f+(x)− f−(x)) dx =

∫ 1

−1u∗(x)(f+

0 (x)− f−0 (x)) dx

+L−1∑

j=0

∫ 1

0j f+(x + j) dx +

L−1∑

j=0

∫ 0

−1j f−(x− j) dx.

In fact, ∫ L

−Lu∗(y)f+(y) dy =

L−1∑

j=0

∫ j+1

ju∗(y)f+(y) dy

=L−1∑

j=0

∫ 1

0u∗(x + j)f+(x + j) dx =

L−1∑

j=0

∫ 1

0(u∗(x) + j) f+(x + j) dx

=L−1∑

j=0

∫ 1

0u∗(x)f+(x + j) dx +

L−1∑

j=0

∫ 1

0j f+(x + j) dx

=∫ 1

0u∗f+

0 +L−1∑

j=0

∫ 1

0j f+(x + j) dx.

Similarly, ∫ L

−Lu∗(y)f−(y) dy =

∫ 0

−1u∗f−0 −

L−1∑

j=0

∫ 0

−1j f−(x− j) dx.

And (3.5) follows. By (2.16) there exists β ∈ [α, 1] such that

(3.6)∫ β

αf+0 =

∫ 0

−1+αf−0 .

Consider the smallest of such β. Take also the smallest γ ∈ [−1,−1 + α] such that

(3.7)∫ 1

βf+0 =

∫ γ

−1f−0 .

For x ∈ (0, 1), we define T0 by

T0(x) =

sup

y ∈ R :∫ y

−1+αf−0 =

∫ x

αf+0

if x ∈ (α, β),

sup

y ∈ R :∫ y

−1f−0 =

∫ x

βf+0

if x ∈ (β, 1),

sup

y ∈ R :∫ y

γf−0 =

∫ x

0f+0

if x ∈ (0, α).


Observe that T0(0) ≥ T0(1) = γ, T0(α) = −1 + α and

−1 ≤ limx→β+

T0(x) ≤ limx→β−

T0(x) ≤ 0.

Moreover,d1(x, T0(x)) = 1 for all x ∈ [0, β],

andd1(x, T0(x)) = 2 for all x ∈ (β, 1].

In fact, as M(x) = f+0 (x)− f−0 (x− 1) (see (3.3)), we have, for any s ∈ (α, 1),

∫ 1

α(f+

0 (x)− f−0 (x− 1)) dx ≥∫ 1

s(f+

0 (x)− f−0 (x− 1)) dx;

therefore, ∫ s

αf+0 (x) dx ≥

∫ s

αf−0 (x− 1) dx =

∫ −1+s

−1+αf−0 (x) dx,

and consequently, ∫ T0(s)

−1+αf−0 (x) dx ≥

∫ −1+s

−1+αf−0 (x) dx,

which implies −1 + s ≤ T0(s) < 0. Similarly, one can obtain the other cases. Then,

µ00(x, y) = f+0 (x)δ[y=T0(x)]

is an optimal transport plan between f+0 and f−0 for the cost function d1.

x

y

γ

α β0 1

−1

0

The straight lines are only illustrative.

In fact, ∫ 1

−1u∗(f+

0 − f−0 ) =∫ γ

−11 · f−0 +

∫ −1+α

γ1 · f−0 +

∫ 0

−1+α0 · f−0

+∫ α

00 · f+

0 +∫ β

α1 · f+

0 +∫ 1

β1 · f+

0 .


Now, since ∫ γ

−1f−0 =

∫ 1

βf+0 and

∫ −1+α

γf−0 =

∫ α

0f+0 ,

we have ∫ 1

−1u∗(f+

0 − f−0 ) =∫ β

01 · f+

0 +∫ 1

β2 · f+

0 =∫ 1

0d1(x, T0(x))f+

0 (x) dx.

Also,

µ0(x, y) =L−1∑

j=0

f+(x)χ(j,j+1)(x)δ[y=T0(x−j)]

is an optimal transport plan between f+ and f−0 for the cost function d1. In fact,∫ L

−1u∗(f+ − f−0 ) =

∫ 1

−1u∗(f+

0 − f−0 ) +L−1∑

j=0

∫ 1

0j f+(x + j) dx

=∫ 1

0d1(x, T0(x))f+

0 (x)dx +L−1∑

j=0

∫ 1

0j f+(x + j) dx

=L−1∑

j=0

∫ 1

0d1(x, T0(x))f+(x + j) dx +

L−1∑

j=0

∫ 1

0j f+(x + j) dx

=L−1∑

j=0

∫ 1

0(d1(x, T0(x)) + j) f+(x + j) dx

=L−1∑

j=0

∫ j+1

j(d1(x− j, T0(x− j)) + j) f+(x) dx

=L−1∑

j=0

∫ j+1

jd1(x, T0(x− j))f+(x) dx =

∫

Ω×Ωd1(x, y)µ0(x, y).

A remarkable observation is that

these µ00 and µ0 are induced by optimal transport maps.

Now, by splitting the mass

(3.8) f+(x)χ(j,j+1)(x) =L−1∑

i=0

gi,j(x), j = 0, 1, ..., L− 1,

is such a way that, for i = 0, 1, ..., L− 1,

(3.9)L−1∑

j=0

∫ x+j

jgi,j =

∫ T0(x)−i

γ−if− if x ∈ (0, β),

and

(3.10)L−1∑

j=0

∫ x+j

β+jgi,j =

∫ T0(x)−i

−1−if− if x ∈ (β, 1),


we have that

µ(x, y) =L−1∑

i=0

L−1∑

j=0

gi,j(x)χ(j,j+1)(x)δ[y=−i+T0(x−j)]

is a transport plan between f+ and f− for the cost function d1. Let us see that µ is an optimaltransport plan. In fact, taking x = β in (3.9), and x = 1 in (3.10), respectively, we get

L−1∑

j=0

∫ β+j

jgi,j =

∫ γ−i

γ−if− and

L−1∑

j=0

∫ 1+j

β+jgi,j =

∫ γ−i

−1−if−.

Adding the last two equalities, we obtain

L−1∑

j=0

∫ 1+j

jgi,j(x) dx =

∫ −i

−1−if−(x) dx =

∫ 0

−1f−(x− i) dx.

Hence,

∫ L

−Lu∗(f+ − f−) =

∫ ∫d1(x, y)µ0(x, y) +

L−1∑

j=0

∫ 0

−1j f−(x− j) dx

=L−1∑

j=0

∫ j+1

jd1(x, T0(x− j))f+(x) +

L−1∑

j=0

i

∫ 0

−1f−(x− i) dx

=L−1∑

j=0

∫ j+1

jd1(x, T0(x− j))

(L−1∑

i=0

gi,j(x)

)dx +

L−1∑

i=0

iL−1∑

j=0

∫ j+1

jgi,j(x) dx

=L−1∑

i=0

L−1∑

j=0

∫ j+1

j(d1(x, T0(x− j))− i) gi,j(x) dx

=L−1∑

i=0

L−1∑

j=0

∫ j+1

jd1(x,−i + T0(x− j))gi,j(x) dx =

∫

Ω×Ωd1(x, y)µ(x, y).

In the following example, µ(x, y) = f+(x)δ[y=T ∗1 (x)] draws the above construction.

Example 3.3. Set f− = 14χ

]−1,0[ and f+ = χ] 74,2[. Then M = −1

4χ

]0, 34[ + 3

4χ

] 34,1[ and therefore

u∗(x) =

...

0 if − 54 < x < −1

4 ,

1 if − 14 < x < 3

4 ,

2 if 34 < x < 7

4 ,

3 if 74 < x < 11

4 ,

...


is (up to adding a constant) the unique Kantorovich potential associated with the metric d1 forf− and f+, and

∫u∗(f+ − f−) =

1116

.

Nevertheless, there exist infinitely many optimal transport maps. For example, the following twoare optimal transport maps,

(3.11) T1∗(x) =

4x− 294 if 28

16 < x < 2916 ,

4x− 334 if 29

16 < x < 2,x otherwise,

T2∗(x) =

4x− 294 if 28

16 < x < 5732 ,

−4x + 578 if 57

32 < x < 2916 ,

−4x + 7 if 2916 < x < 2,

x otherwise.

Observe that both push the mass f+χ] 74, 2916

[ toward f−χ]− 1

4,0[ paying, after 2 steps, 2 × 1

16 , and

push the rest from f+χ] 2916

,2[ toward f−χ]−1,− 1

4[ paying, after 3 steps, 3× 3

16 . Therefore the totalcost is, as known,

2× 116

+ 3× 316

=1116

.

f−T1∗x f+

−1 − 14− 1

4 0 ... 2816

2916

2916 2


nothing

Support of f+(x)δ[y=T∗1 (x)]

x

y

0 1 2

0

-1

y = 4x− 294

y = 4x− 334

− 14

2916

Remark 3.4. Observe that the unique non-decreasing optimal transport map, T0, for the Euclid-ean distance as cost function that pushes f+ forward to f−, given by (3.4), in this particular caseis T0(x) = 4x− 8. Now, T0 is not an optimal transport map for d1, the transport cost with thismap is, in fact, 12

16 . However, it is well known (see [3]) that if the cost function c(x, y) is equal toφ(|x−y|) with φ non-decreasing and convex then T0 is an optimal transport, but in our situationφ fails to be convex.

On the other hand, the following transport plan, not induced by a map, between f+ and f− isvery simple and it is optimal: µ = χ

( 74,2)(x)

(14δ[y=x−2] + 1

4δ[y=x− 94] + 1

4δ[y=x− 104

] + 14δ[y=x− 11

4]

).

Example 3.5. Let f+ = Lχ[0,1] and f− = χ

[−L,0] with L /∈ N. Let us see that there is optimaltransport map T with distance d1 pushing f+ to f− (compare this example with Example 3.2).

In order to simplify the exposition we take 2 < L < 3. This particular case shows clearly howto handle the general case with any other L.

Using the procedure introduced in Subsection 3.3 we have that (α = 1 = β and γ = −1)

T0(x) =

L2 x− 1 if 0 < x < 2(3−L)

L ,

L3 (x− 1) if 2(3−L)

L < x < 1,


is an optimal transport map pushing f+0 to f−0 .

x

y

2− L

2(3−L)L0 1

−1

0

Now, in order to obtain an optimal transport map T pushing f+ to f−, we do the splittingprocedure (3.8) (there are many different ways) in the following adequate way. For x < 2(3−L)

Lwe have to distribute the mass f+ in two equiweighted parts, so, set the rectangles with cornercoordinates,

upper-left, uli = (xi+1, yi),upper-right, uri = (xi, yi),lower-left, lli = (xi+1, yi+1),lower-right, lri = (xi, yi+1),

i = 1, 2, . . . , where

x1 =2(3− L)

L, y1 = 2− L,

yi+1 = xi − 1, xi+1 = xi − 2L

(yi − yi+1) =2L

(yi+1 + 1).

Observe that lri ∈ [y = x−1] and lli, uri ∈ [y = L2 x−1]. In each rectangle we can trace 2 parallel

segments of slope L defined by the lines

y = L(x− xi) + yi,

and

y = L(x− xi) + yi, xi = xi − xi − xi+1

2.

Then

Ti(x) = f+(x)χ]xi,xi[(x)δ[y=L(x−xi)+yi] + f+(x)χ]xi+1,xi[(x)δ[y=L(x−xi)+yi−1]

push in an optimal way f+χ]xi+1,xi[ to f−χ

]yi+1,yi[∪]yi+1−1,yi−1[, for i = 1, 2, . . .


For x > 2(3−L)L we have to distribute the mass f+ in three equiweighted parts, so, set the

rectangles with corner coordinates,

lower-left, lli = (xi, yi),lower-right, lri = (xi+1, yi),upper-left, uli = (xi, yi+1),upper-right, uri = (xi+1, yi+1),

i = 1, 2, . . . , where now

x1 =2(3− L)

L, y1 = 2− L,

xi+1 = yi + 1, yi+1 = yi +L

3(xi+1 − xi) =

L

3(xi+1 − 1).

Observe that lri ∈ [y = x − 1] and lli, uri ∈ [y = L3 (x − 1)]. In each rectangle we can trace

3 parallel segments of slope L defined by the lines

y = L(x− xi) + yi,

y = L(x− xi) + yi, xi = xi +xi+1 − xi

3,

andy = L(x− xi) + yi, xi = xi + 2

xi+1 − xi

3.

ThenTi(x) = f+(x)χ(xi,xi)(x)δ[y=L(x−xi)+yi] + f+(x)χ(xi,xi)(x)δ[y=L(x−xi)+yi−1]

+f+(x)χ(xi,xi+1)(x)δ[y=L(x−xi)+yi−2]

push in an optimal way f+χ(xi,xi+1) to f−χ

(yi,yi+1)∪(yi−1,yi+1−1)∪(yi−2,yi+1−2), for i = 1, 2, . . .

4. Approximation of the classical Monge-Kantorovich problem by the nonlocalproblems

Consider the classical Monge problem, with the Euclidean distance d|·| as cost, for two non-negative Borel function f+, f− ∈ L2(Ω) satisfying the mass balance condition∫

Ωf+(x) dx =

∫

Ωf−(y) dy.

To approach an optimal transport map T ∗ ∈ A(f+, f−) which minimizes the cost functional

Fd|·|(T ) =∫

Ω|x− T (x)| f+(x) dx.

by means of nonlocal optimal transport maps is going to be quite difficult since even in 1−dimen-sion we have proved that there is not optimal transport map for some concrete configurations ofthe data with respect to the metric dε, even if one takes ε small. However, as we will see below,the Monge infimum for the nonlocal problems converges to the Monge infimum for the Euclideandistance.

On the other hand, we will approach µ∗, an optimal plan, and u∗, a Kantorovich potential, forwhich

Fd|·|(T∗) = Kd|·|(µ

∗) = P(u∗),


by nonlocal optimal transport plans and nonlocal Kantorovich potentials, respectively. And,moreover, we will see in the one dimensional case that the local optimal transport plan inducedby the unique non decreasing local optimal transport map can be approximated by nonlocaloptimal transport plans.

First of all, in the following result we state the convergence to the Monge/Kantorovich prob-lems.

Proposition 4.1. We have, for the dual problems,

(4.1) limε→0+

supPf+,f−(u) : u ∈ Kε

= sup

Pf+,f−(u) : u ∈ Kd|·|

,

or equivalently, for the relaxed problems,

(4.2) limε→0+

minKε(µ) : µ ∈ π(f+, f−)

= min

Kd|·|(µ) : µ ∈ π(f+, f−)

,

and also, for the primal problems,

(4.3) limε→0+

infFε(µ) : µ ∈ π(f+, f−)

= inf

Fd|·|(T ) : T ∈ A(f+, f−)

.

Moreover,

(4.4) 0 ≤ supPf+,f−(u) : u ∈ Kε

− supPf+,f−(u) : u ∈ Kd|·|

≤ ε

∫

Ωf+(x) dx

and

(4.5) 0 ≤ minKε(µ) : µ ∈ π(f+, f−)

−minKd|·|(µ) : µ ∈ π(f+, f−)

≤ ε

∫

Ωf+(x) dx.

Proof. We have (see [5]) that Kd|·| ⊂ Kε1 ⊂ Kε2 , if ε1 < ε2. Therefore, for any ε > 0,


≥ supPf+,f−(u) : u ∈ Kd|·|

.

Hence

(4.6) lim infε→0


≥ supPf+,f−(u) : u ∈ Kd|·|

.

For a given µ ∈ π(f+, f−) we have

limε→0

∫

Ω×Ωdε(x, y) dµ(x, y) =

∫

Ω×Ωd|·|(x, y) dµ(x, y),

that islimε→0

Kε(µ) = Kd|·|(µ).

This holds since for any two points x, y in Ω we have

(4.7)∣∣dε(x, y)− |x− y|∣∣ ≤ ε, dε(x, y) ≤ 2|x− y| for |x− y| ≥ ε

and we have µ has compact support.Hence, if µ∗ is an optimal plan for Kd|·| , we get

lim supε→0


= lim sup

ε→0min

Kε(µ) : µ ∈ π(f+, f−)

≤ lim supε→0

Kε(µ∗) = Kd|·|(µ∗) = sup

Pf+,f−(u) : u ∈ Kd|·|

.


Then, by (4.6), we obtain (4.1) and (4.2). On the other hand, we also have

(4.8) limε→0+


= inf

Fd|·|(T ) : T ∈ A(f+, f−)

.

Take now T ′ a transport map. Thanks to (4.7),

lim supε→0


= lim supε→0

inf∫

Ωdε(x, T (x))f+(x) dx : T ∈ A(f+, f−)

≤ lim supε→0

∫

Ωdε(x, T ′(x))f+(x) dx =

∫

Ω|x− T ′(x)|f+(x) dx.

Therefore,

(4.9) lim supε→0


≤ infFd|·|(T ) : T ∈ A(f+, f−)

.

On the other hand, by (1.8),


≤ infFε(T ) : T ∈ A(f+, f−)

,

taking lim infε→0 and taking into account (4.8) and (4.9) we obtain (4.3).By Kantorovich-Rubinstein’s Theorem (Theorem 1.4), to finished the proof is enough to prove

(4.5). Sincedε(x, y)− ε ≤ d|·|(x, y) ≤ dε(x, y),

given µ ∈ π(f+, f−), we have∫

Ω×Ω(dε(x, y)− ε) dµ(x, y) ≤

∫

Ω×Ωd|·|(x, y) dµ(x, y) ≤

∫

Ω×Ωdε(x, y) dµ(x, y).

Then, taking the minimun over all µ ∈ π(f+, f−), and having in mind that∫

Ω×Ωdµ(x, y) =

∫

Ωf+(x) dx,

we obtain (4.5). ¤Remark 4.2. Observe that in the above proof we have not made use of Theorem 2.11. Also notethat since dε ≤ dε′ , the sequence of costs is nonincreasing as ε decreases to zero.

Let us now proceed to approach optimal transport plans. Since dε is lower semi-continuous, byProposition 1.2 we have there exists an optimal transport plans µε, that is, µε ∈ π(f+, f−) and

Kε(µε) = minKε(µ) : µ ∈ π(f+, f−).Let us see that µε converge as ε → 0 to an optimal transport plan for the Monge-Kantorovichproblem with respect to the euclidean distance.

Proposition 4.3. There exists µ∗ ∈ π(f+, f−), such that, after a subsequence,

µε µ∗ as measures

andKd|·|(µ

∗) = minKd|·|(µ) : µ ∈ π(f+, f−).


Proof. To prove this we just observe that

d|·|(x, y) = |x− y| ≤ dε(x, y) ≤ |x− y|+ ε.

Therefore

dε(x, y) → |x− y| uniformly as ε → 0.

Hence,∫

Ω×Ω|x− y| dµε(x, y) ≤

∫

Ω×Ωdε(x, y) dµε(x, y) ≤

∫

Ω×Ω(|x− y|+ ε) dµε(x, y).

On the other hand, by Prokhorov’s Theorem, we can assume that µε converges weakly∗ in thesense of measures to a limit µ∗. Therefore, we conclude that

∫

Ω×Ω|x− y| dµ∗(x, y) = lim

ε→0

∫

Ω×Ωdε(x, y) dµε(x, y).

Finally, by Proposition 4.1 we obtain that µ∗ is a minimizer for the usual euclidean distance. ¤

Example 4.4. Let f+, f− the masses considered in Example 3.2 for L = 2. Let us see that foreach n ∈ N, there is µn ∈ π(f+, f−), optimal transport plan for the distance d 1

2n, that is,

K 12n

(µn) = minK 12n

(µ) : µ ∈ π(f+, f−),

such that

µn f+ £ f− weakly∗.

For the above masses the trivial plan f+ £ f− is an optimal transport plan between f+ and f−for the euclidean distance.

In fact, for n ∈ N and m = 0, . . . , 3 · 2n − 1, we define the functions

sm(x) =(x− m

2n

)χ(0, m

2n )(x), m = 1, . . . , 2n,

sm(x) =(x− m

2n

)χ

(0,1)(x), m = 2n + 1, . . . , 2 · 2n,

sm(x) =(x− m

2n

)χ(m−2·2n

2n ,1)(x), m = 2 · 2n + 1, . . . , 3 · 2n − 1.

Using these functions we construct the measure µn ∈ π(f+, f−), as

µn(x, y) =12n

3·2n−1∑

m=0

δ[y=sm(x)].


Here is a picture where we schematize the support of µn,x

y

0 10

−2

Let us see that µn f+ £ f−, weakly∗. If ϕ ∈ Cc(Ω× Ω), we have

limn→∞

∫

Ω×Ωϕ(x, y) dµn(x, y) = lim

n→∞12n

3·2n−1∑

m=1

∫ 1

0ϕ(x, sm(x)) dx

= limn→∞

12n

2n∑

m=1

∫ m2n

0ϕ

(x, x− m

2n

)dx + lim

n→∞12n

2·2n∑

m=2n+1

∫ 1

0ϕ

(x, x− m

2n

)dx

+ limn→∞

12n

3·2n∑

m=2·2n+1

∫ 1

m−2·2n

2n

ϕ(x, x− m

2n

)dx.

Now, if

φ1(y) :=∫ y

0ϕ(x, x− y) dx,

using Fubini’s Theorem, we have

limn→∞

12n

2n∑

m=1

∫ m2n

0ϕ

(x, x− m

2n

)dx = lim

n→∞12n

2n∑

m=1

φ1

( m

2n

)=

∫ 1

0φ1(y) dy

=∫ 1

0

(∫ y

0ϕ(x, x− y) dx

)dy =

∫ 1

0

(∫ 1

xϕ(x, x− y) dy

)dx

=∫ 1

0

(∫ 0

x−1ϕ(x, z) dz

)dx.

Similarly, if

φ2(y) :=∫ 1

0ϕ(x, x− 1− y) dx,


we have

limn→∞

12n

2·2n∑

m=2n+1

∫ 1

0ϕ

(x, x− m

2n

)dx = lim

n→∞12n

2n∑

k=1

∫ 1

0ϕ

(x, x− 1− k

2n

)dx

= limn→∞

12n

2n∑

k=1

φ2

(k

2n

)=

∫ 1

0φ2(y) dy =

∫ 1

0

(∫ 1

0ϕ(x, x− 1− y) dx

)dy

=∫ 1

0

(∫ 1

0ϕ(x, x− 1− y) dy

)dx =

∫ 1

0

(∫ x−1

x−2ϕ(x, z) dz

)dx.

Finally, when

φ3(y) :=∫ 1

yϕ(x, x− 2− y) dx,

we obtain

limn→∞

12n

3·2n∑

m=2·2n+1

∫ 1

m−2·2n

2n

ϕ(x, x− m

2n

)dx = lim

n→∞12n

2n−1∑

k=1

∫ 1

k2n

ϕ

(x, x− 2− k

2n

)dx

= limn→∞

12n

2n−1∑

k=1

φ3

(k

2n

)=

∫ 1

0φ3(y) dy =

∫ 1

0

(∫ 1

yϕ(x, x− 2− y) dx

)dy

=∫ 1

0

(∫ x

0ϕ(x, x− 2− y) dy

)dx =

∫ 1

0

(∫ x−2

−2ϕ(x, z) dz

)dx.

Consequently, we get that

limn

∫

Ω×Ωϕ(x, y) dµn(x, y)

=∫ 1

0

(∫ 0

x−1ϕ(x, z) dz

)dx +

∫ 1

0

(∫ x−1

x−2ϕ(x, z) dz

)dx +

∫ 1

0

(∫ x−2

−2ϕ(x, z) dz

)dx

=∫ 1

0

(∫ 0

−2ϕ(x, z) dz

)dx =

∫

Ω×Ωϕ(x, y) dµ(x, y),

where µ = f+ £ f−.Let us see now that for each n ∈ N, there is µn ∈ π(f+, f−), optimal transport plan for the

distance d 12n

, such that µn µ = f+(x)δ[y=T (x)] weakly∗, where T (x) = 2x − 2 is the uniquenondecreasing optimal transport map for the euclidean distance between f+ and f−. Set, forn ∈ N,

µn(x, y) = χ[ 2

n−12n ,1](x)δ[y=x−1] +

2n−1∑

m=1

χ[ 2

n−m−12n , 2

n−m+12n ](x)δ[y=x−1− m

2n ] + χ[0, 1

2n ](x)δ[y=x−2].


A DrawSupport of µn, n = 2

x

y

0 10

−2

Then, for ϕ ∈ Cc(Ω× Ω), we have

limn→∞

∫

Ω×Ωϕ(x, y) dµn(x, y) = lim

n→∞

2n−1∑

m=1

∫ 2n−m+12n

2n−m−12n

ϕ(x, x− 1− m

2n

)dx

= limn→∞

2n−1∑

m=1

∫ 12n

− 12n

ϕ(x + 1− m

2n, x− 2

m

2n

)dx

= limn→∞

12n

2n−1∑

m=1

2n

2

∫ 12n

− 12n

2ϕ(x + 1− m

2n, x− 2

m

2n

)dx,

and, by the continuity of ϕ,

limn→∞

12n

2n−1∑

m=1

2n

2

∫ 12n

− 12n

2ϕ(x + 1− m

2n, x− 2

m

2n

)dx

=∫ 1

02ϕ(1− z,−2z)dz =

∫ 1

02ϕ(x, 2x− 2) dx =

∫

Ω×Ωϕ(x, y) dµ(x, y).

Remark 4.5. The above example illustrates how to design an splitting process (3.8) which allowsto obtain dε–optimal transport plans between f+ and f− that converge to the optimal transportplan for the euclidean distance induced by the unique optimal transport map pushing f+ to f−,T0, which is nondecreasing. This splitting process should charge the maximum of mass from theright to the left and from the top to the bottom. Let us see how it works for ε = 1.

Since T0 is nondecreasing we have that

∫ −j+T0(x)

−j+γf− and

∫ −j+T0(x)

−jf−

are functions of bounded variation with positive locally integrable derivatives, and we can chosein the splitting process (3.8) the following functions,


g0,L−1(x) =

min

f+(x),

d

dx

∫ T0(x)

γ

f−

for x ∈ (L− 1, L− 1 + β),

min

f+(x),

d

dx

∫ T0(x)

−1

f−

for x ∈ (L− 1 + β, L),

g1,L−1(x) =

min

f+(x)− g0,L−1(x),

d

dx

∫ −1+T0(x)

−1+γ

f−

for x ∈ (L− 1, L− 1 + β),

min

f+(x)− g0,L−1(x),

d

dx

∫ −1+T0(x)

−2

f−

for x ∈ (L− 1 + β, L),

...

g0,L−2(x) =

min

f+(x),

d

dx

∫ T0(x)

γ

f− − g0,L−1(x)

for x ∈ (L− 2, L− 2 + β),

min

f+(x),

d

dx

∫ T0(x)

−1

f− − g0,L−1(x)

for x ∈ (L− 2 + β, L− 1),

g1,L−2(x) =

min

f+(x)− g0,L−2(x),

d

dx

∫ T0(x)

γ

f− − g1,L−1(x)

for x ∈ (L− 2, L− 2 + β),

min

f+(x)− g0,L−2(x),

d

dx

∫ T0(x)

−1

f− − g1,L−1(x)

for x ∈ (L− 2 + β, L− 1),

...

Let us finish this section with a convergence result for Kantorovich potentials.

Theorem 4.6. Let u∗ε be a Kantorovich potential associated with the metric dε. Then, after asubsequence,

u∗ε u∗ in L2 as ε → 0,where u∗ is a Kantorovich potential associated with the metric d|·|.

Proof. It is an obvious fact that uε is L∞-bounded, then, there exists a sequence,

u∗εn v in L2.

Therefore,

limn→+∞

∫

Ωu∗εn

(x)(f+(x)− f−(x)) dx =∫

Ωv(x)(f+(x)− f−(x)) dx.

Now, since ∫

Ωu∗εn

(x)(f+(x)− f−(x)) dx = supPf+,f−(u) : u ∈ Kεn

,

by Proposition 4.1, we conclude that∫

Ωv(x)(f+(x)− f−(x)) dx = sup

Pf+,f−(u) : u ∈ Kd|·|

.


In order to have that the limit v is a maximizer u∗ we need to show that v ∈ Kd|.| , and thisfollows by the Mosco-convergence of IKε to IKd|.|

(see [5]). ¤

Remark 4.7. For 0 < ε ≤ 1, let u∗ε be a Kantorovich potential for f = f+ − f− respect to thedistance dε, that is,

Pf+,f−(u∗ε) = maxPf+,f−(u) : u ∈ Kε

.

Let us see what are the relation between u∗ε and u∗1 for a fixed 0 < ε < 1. Given u ∈ Kε, we have∫u(x)f

(x

ε

)dx = εN

∫u(εy)f(y) dy = εN+1

∫1εu(εy)f(y) dy

≤ εN+1

∫u∗1(y)f(y) dy =

∫εu∗1

(x

ε

)f

(x

ε

)dx

since the map y 7→ 1εu(εy) is an element of K1. Therefore, we have shown that if u∗1 is a

Kantorovich potential for f respect to the distance d1, then εu∗1(

xε

)is a Kantorovich potential

for f(

xε

)respect to the distance dε.

5. The nonlocal version of the Evans-Gangbo approach

Let us consider again the Euler-Lagrange equation

(5.1) f+ − f− ∈ ∂IKd(Ω)(u),

Kd(Ω) = Lip1(Ω, d), associated with the variational problem

(5.2) max∫

Ωu(x)(f+(x)− f−(x)) dx : u ∈ Kd(Ω)

.

Recall that we assume that Ω is convex. In the particular case of the Euclidean distanced(x, y) = d|·|(x, y) in Ω, and as we have mentioned in the Introduction, Evans and Gangbo,in [11], characterize solutions to this equation by means of a PDE and use this PDE to find aproof of the existence of an optimal transport map for the classical Monge problem. The aim ofthis section is to perform a similar approach with the step distance dε.

5.1. Characterizing the Euler-Lagrange equation. The Euler-Lagrange equation to be con-sidered is

(5.3) f+ − f− ∈ ∂IKε(Ω)(u),

where Kε(Ω) = Kdε(Ω). We characterize ∂IKε by introducing the following operator Bε in L2(Ω).Let

Mab (Ω× Ω) := bounded antisymmetric Radon measures in Ω× Ω .

We define the operator Bε in L2(Ω) as follows, (u, v) ∈ Bε if and only if u ∈ Kε, v ∈ L2(Ω), andthere exists σ ∈Ma

b (RN × Ω) such that

σ (x, y) ∈ Ω× Ω : |x− y| ≤ ε,∫

Ω

∫

Ωξ(x) dσ(x, y) =

∫

Ωξ(x)v(x) dx, ∀ ξ ∈ Cc(Ω),

and|σ|(Ω× Ω) ≤ 2

ε

∫

Ωv(x)u(x) dx.


Theorem 5.1. We have∂IKε = Bε.

Proof. Let us first see that Bε ⊂ ∂IKε . Let (u, v) ∈ Bε, to see that (u, v) ∈ ∂IKε we need to provethat

(5.4) 0 ≤∫

Ωv(x)(u(x)− ξ(x)) dx, ∀ ξ ∈ Kε.

Using an approximation procedure taking a convolution, we can assume that ξ ∈ Kε is continuous.Then, ∫

Ωv(x)(u(x)− ξ(x)) dx ≥ ε

2|σ|(Ω× Ω)−

∫

Ωv(x)ξ(x) dx

=ε

2|σ|(Ω× Ω)−

∫

Ω

∫

Ωξ(x) dσ(x, y)

=ε

2|σ|(Ω× Ω)− 1

2

∫

Ω

∫

Ω(ξ(x)− ξ(y)) dσ(x, y) ≥ 0,

where in the last equality we have used the antisymmetry of σ. Therefore, we have Bε ⊂ ∂IKε .Since ∂IKε is a maximal monotone operator, to see that the operators are equal we only need toshow that for every f ∈ L2(Ω) there exists u ∈ Kε such that

(5.5) u + Bε(u) = f.

Let J : RN → R as in (1.2) and

Jε(z) =1

εNJ

(z

ε

).

By results in [5], given p > N and f ∈ L2(Ω) there exists a unique solution up ∈ L∞(Ω) of thenonlocal p-Laplacian problem

(5.6) up(x) +∫

ΩJε(x− y)

∣∣∣∣up(y)− up(x)

ε

∣∣∣∣p−2

(up(y)− up(x)) dy = Tp(f)(x) ∀x ∈ Ω,

where Tk(r) := maxmink, r,−r. And we also know that there exists u ∈ Kε such that

(5.7) up → u in L2(Ω) as p → +∞,

withu + ∂IKε(u) 3 f,

from where it follows that∫

Ω(f(x)− u(x))(w(x)− u(x)) dx ≤ 0, ∀w ∈ Kε,

and consequently, u = PKε(f).Multiplying (5.6) by up and integrating, we get

(5.8)∫

Ω(Tp(f)(x)− up(x))up(x) dx =

12εp−2

∫

Ω×ΩJε(x− y) |up(y)− up(x)|p dxdy,

from where it follows that

(5.9)1

εp−2

∫

Ω×ΩJε(x− y) |up(y)− up(x)|p dxdy +

∫

Ω|up(x)|2 dx ≤ ‖f‖2

L2(Ω).


If we set

σp(x, y, ε) :=1

εp−2Jε(x− y) |up(y)− up(x)|p−2 (up(y)− up(x)),

by Holder’s inequality,∫

Ω×Ω|σp(x, y, ε)| dxdy =

1εp−2

∫

Ω×ΩJε(x− y) |up(y)− up(x)|p−1 dxdy

≤ 1εp−2

(∫

Ω×ΩJε(x− y) |up(y)− up(x)|p dxdy

) p−1p

(∫

Ω×ΩJε(x− y) dxdy

) 1p

=1

εp−2

(∫


) p−1p

.

Now, by (5.9), we have

∫

Ω×Ω|σp(x, y, ε)| dxdy ≤

(‖f‖2

L2(Ω)

) p−1p

ε2−p

p .

Hence, σp : p ≥ 2 is bounded in L1(Ω× Ω), and consequently we can assume that

(5.10) σp(., ., ε) πε =: σ weakly∗ in Mb(Ω× Ω).

Obviously, since each σp is antisymmetric, σ ∈ Mab (Ω × Ω). Moreover, since supp(Jε) = Bε(0),

we have

σ (x, y) ∈ Ω× Ω : |x− y| ≤ ε.On the other hand, given ξ ∈ Cc(Ω), by (5.6), (5.7) and (5.10), we get

∫

Ω

∫

Ωξ(x) dσ(x, y) = lim

p→+∞

∫

Ω

∫

Ωξ(x)σp(x, y, ε) dx dy

= limp→+∞

1εp−2

∫

Ω

∫

ΩJε(x− y) |up(y)− up(x)|p−2 (up(y)− up(x))ξ(x) dx dy

= limp→+∞

∫

Ω(Tp(f)(x)− up(x))ξ(x) dx =

∫

Ω(f(x)− u(x))ξ(x) dx.

Then, to prove (5.5), we only need to show that

|σ|(Ω× Ω) ≤ 2ε

∫

Ω(f(x)− u(x))u(x) dx.

In fact, by (5.10), we have

|σ|(Ω× Ω) ≤ lim infp→+∞

∫

Ω

∫

Ω|σp(x, y)| dx dy.


Now, by (5.8),

∫

Ω×Ω|σp(x, y)| dx dy ≤ 1

εp−2

(∫


) p−1p

=1

εp−2

(2εp−2

∫

Ω(Tp(f)(x)− up(x))up(x) dx

) p−1p

= 2p−1

p ε2−p

p

(∫

Ω(Tp(f)(x)− up(x))up(x) dx

) p−1p

.

Therefore,

|σ|(Ω× Ω) ≤ 2ε

∫

Ω(f(x)− u(x))u(x) dx.

This ends the proof. ¤

We can rewrite the operator Bε as follows.

Corollary 5.2. (u, v) ∈ Bε if and only if u ∈ Kε, v ∈ L2(Ω), and there exists σ ∈ Mab (Ω × Ω)

such thatσ+ (x, y) ∈ Ω× Ω : |x− y| ≤ ε, u(x)− u(y) = ε,σ− (x, y) ∈ Ω× Ω : |x− y| ≤ ε, u(y)− u(x) = ε,∫

Ω

∫

Ωξ(x)dσ(x, y) =

∫

Ωξ(x)v(x) dx, ∀ ξ ∈ Cc(Ω),

and

|σ|(Ω× Ω) =2ε

∫

Ωv(x)u(x) dx.

Proof. Let (u, v) ∈ Bε, then

(5.11)∫

Ω

∫

Ωξ(x)dσ(x, y) =

∫

Ωξ(x)v(x) dx, ∀ ξ ∈ Cc(Ω).

Hence, ∫

Ωdσ(x, y) = v(x).

Therefore, by approximation, we can take ξ ∈ L2(Ω) in (5.11) and∫Ω

∫Ω ξ(x)dσ(x, y) has this

sense. Taking ξ = u in (5.11) and using the antisymmetric of σ and the previous result we get

|σ|(Ω× Ω) ≥∫

Ω

∫

Ω

(u(x)− u(y))ε

dσ(x, y)

=2ε

∫

Ω

∫

Ωu(x)dσ(x, y) =

2ε

∫

Ωu(x)v(x) dx ≥ |σ|(Ω× Ω),

from where the result follows. ¤


As consequence of the above result, we have that the Kantorovich potentials u∗ε associated withthe metric dε, that is, the functions u∗ε ∈ Kε such that

(5.12)∫

Ωu∗ε(x)(f+(x)− f−(x)) dx = max

∫

Ωu(x)(f+(x)− f−(x)) dx : u ∈ Kε

are the solutions of the nonlocal equation

(5.13) f+ − f− ∈ Bε(u).

In other words, the functions u∗ε ∈ Kε, for which there exists σ∗ε ∈Mab (Ω× Ω), such that

(5.14)[σ∗ε ]+ (x, y) ∈ Ω× Ω : u∗ε(x)− u∗ε(y) = ε, |x− y| ≤ ε,[σ∗ε ]− (x, y) ∈ Ω× Ω : u∗ε(y)− u∗ε(x) = ε, |x− y| ≤ ε,

satisfying

(5.15)∫

Ωdσ∗ε(x, y) = f+(x)− f−(x), x ∈ Ω.

and

(5.16) |σ∗ε |(Ω× Ω) =2ε

∫

Ω(f+(x)− f−(x))u∗ε(x) dx,

that is,

|σ∗ε |(Ω× Ω) =2εP(u∗ε).

We want to highlight that (5.13) plays the role of (1.14). Moreover we will see in the nextsubsection that we can construct optimal transport plans from it, more precisely, we shall seethat the potential u∗ε and the measure σ∗ε encode all the information that we need to constructan optimal transport plan associated with the problem.

5.2. Constructing optimal transport plans. Let us begin by constructing σ∗ for some con-crete masses f+ and f− and trying to obtain from it and the potential, information to constructan optimal transport plan between them.

Example 5.3. In the Example 3.3, for the masses f+ = χ[7/4,2] and f− = 1

4χ

[−1,0], we have seenthat

infF1(T ) : T ∈ A(f+, f−) = minK1(µ) : µ ∈ π(f+, f−)

= supPf+,f−(u) : u ∈ K1

=

1116

,

is attained at

u∗(x) =

...

0 if − 54 < x < −1

4 ,

1 if − 14 < x < 3

4 ,

2 if 34 < x < 7

4 ,

...


For these masses, we have that the following antisymmetric measure satisfies the above condi-tions (5.14), (5.15) and (5.16),

σ∗(x, y) = χ(7/4,2)(x)δ[y=x−1] − χ

(3/4,1)(x)δ[y=x+1]

+14χ

( 34,1)(x)

(δ[y=x−1] + δ[y=x− 3

4] + δ[y=x− 2

4] + δ[y=x− 1

4]

)

−14χ

(− 14,0)(x)δ[y=x+1] −

14χ

(0, 14)(x)δ[y=x+ 3

4] −

14χ

( 14, 24)(x)δ[y=x+ 2

4] −

14χ

( 24, 34)(x)δ[y=x+ 1

4]

+14χ

(0, 34)(x)δ[y=x−1] −

14χ

(−1,− 14)(x)δ[y=x+1].

Let us point out that also the antisymmetric measure

ν∗(x, y) = χ(7/4,2)(x)δ[y=x−1] − χ

(3/4,1)(x)δ[y=x+1] −14χ

(−1,0)(x)δ[y=x+1] +14χ

(0,1)(x)δ[y=x−1]

+χ(3/4,1)(x)χ(0,3/4)(y)− χ

(0,3/4)(x)χ(3/4,1)(y)

satisfies the above conditions.We try now to recover the optimal transport plan µ∗ given by

µ∗(x, y) = χ(7/4,2)(x)

(14δ[y=x−2] +

14δ[y=x− 9

4] +

14δ[y=x− 10

4] +

14δ[y=x− 11

4]

)

from the measure σ∗(x, y). To do that we proceed as follows. Let µ3 be the restriction of [σ∗]+to the set where u∗(x) = 3, that is, µ3(x, y) = χ

(7/4,2)(x)δ[y=x−1]. Then, given ϕ,ψ ∈ Cc(R), wehave∫

R×R(ϕ(x) + ψ(y)) dµ3(x, y) =

∫ 2

7/4(ϕ(x) + ψ(x− 1)) dx =

∫ 2

7/4ϕ(x) dx +

∫ 1

3/4ψ(y) dy.

Thereforeµ3 ∈ π

(f+, χ(3/4,1)

).

Moreover, sinceχ

(7/4,2)(x)− χ(3/4,1)(y) = 1 = d1(x, y) µ3 − a.e.,

by (2.7), we have µ3 is an optimal transport plan.Consider now µ2 the restriction of [σ∗]+ to the set where u∗(x) = 2, that is,

µ2(x, y) = χ( 34,1)(x)

(14δ[y=x−1] +

14δ[y=x− 3

4] +

14δ[y=x− 2

4] +

14δ[y=x− 1

4]

).

Then, for ϕ,ψ ∈ Cc(R), we have∫

R×R(ϕ(x) + ψ(y)) dµ2(x, y) =

∫ 1

3/4ϕ(x) dx +

14

∫ 3/4

−1/4ψ(y) dy.

Therefore

µ2 ∈ π(

χ(3/4,1),

14χ

(−1/4,3/4)

),

and again by (2.7), we have µ2 is an optimal transport plan.


Now, we split the mass 14χ

(−1/4,3/4) as follows

14χ

(−1/4,3/4) =14χ

(−1/4,0) +14χ

(0,3/4) = f−χ(−1/4,0) +

14χ

(0,3/4).

Let µ1 be the restriction of [σ∗]+ to the set where u∗(x) = 1, that is, µ1 = 14χ

(0,3/4)(x)δ[y=x−1].It is easy to see that

µ1 ∈ π(

14χ

(0,3/4), f−χ

(−1,−1/4)

),

and that it is an optimal transport plan.

Let us see how to glue together these transport plans to find an optimal transport plan betweenf+ and f−. We will first perform the above construction in general and then we will glue in thegeneral situation.

A general construction of optimal transport plans.Given f+, f− ∈ L∞(Ω) two non-negative Borel functions satisfying the mass balance condi-

tion (1.4) and such that |supp(f+) ∩ supp(f−)| = 0, by Theorems 1.4 and 1.6, there exists aKantorovich potential u∗ such that u∗(Ω) ⊂ Z, taking a finite number of values, and

minKd1(µ) : µ ∈ π(f+, f−) = Kd1(µ∗) =

∫

Ωu∗(x)(f+(x)− f−(x)) dx.

Now, u∗ verifies the Euler-Lagrange equation

(5.17) f+ − f− ∈ ∂IK1(Ω)(u∗).

Then, by Theorem 5.1, there exists σ ∈Mab (Ω× Ω) such that

(5.18)

σ+ (x, y) ∈ Ω× Ω : |x− y| ≤ 1, u∗(x)− u∗(y) = 1,

σ− (x, y) ∈ Ω× Ω : |x− y| ≤ 1, u∗(y)− u∗(x) = 1,∫

Ω

∫

Ωξ(x)dσ(x, y) =

∫

Ωξ(x)(f+(x)− f−(x)) dx ∀ ξ ∈ Cc(Ω),

and

|σ|(Ω× Ω) = 2∫

Ωu∗(x)(f+(x)− f−(x)) dx = 2 minKd1(µ) : µ ∈ π(f+, f−).

We are going to give a method to obtain an optimal transport plan µ∗ from the measure σ. Todo that we shall use the Gluing Lemma (see Lemma 7.6 in [19]), which permits to glue togethertwo transport plans in an adequate way. As it is remarked in [19], it is possible to state thefollowing result, that we present for the distance d1.

Lemma 5.4. Let f1, f2, g be three positive measures in Ω. If µ1 ∈ π(f1, g) and µ2 ∈ π(g, f2),there exits a measure G(µ1, µ2) ∈ π(f1, f2) such that

(5.19) Kd1(G(µ1, µ2)) ≤ Kd1(µ1) +Kd1(µ2).


We divide the construction in two steps. We assume without loss of generality that

u∗ = 0χA0 + 1χA1 + · · ·+ kχAk,

with Ai = x ∈ Ω : u∗(x) = i.Step 1. How the measures σ+ (Aj ×Aj−1) work.Taking into account the antisymmetry of σ and (5.18), we have that

projx(σ+)− projy(σ+) = f+ − f−,

which impliesg := projx(σ+)− f+ = projy(σ+)− f−.

By (2.15), projx(σ+) Ak = f+ χAkand projx(σ+) A0 = f+ χA0 = 0, then

g Ak = g A0 = 0.

Moreover we have

projx(σ+ (Aj ×Aj−1)) = projx(σ+) Aj and projy(σ+ (Aj ×Aj−1)) = projx(σ+) Aj−1,

thenprojx(σ+ (Aj ×Aj−1)) = f+ χAj + g Aj

andprojy(σ+ (Aj ×Aj−1)) = f− χAj−1 + g Aj−1.

Let us callµj := σ+ (Aj ×Aj−1).

Let us briefly comment what these measures do. The first one, µk, transports f+χAkinto f−χAk−1

plus something else, that is g Ak−1. Afterwards, µj transports f+χAj + g Aj into f−χAj−1

again plus something else, that is g Aj−1. The last one, µ1, transports f+χA1 +g A1 to f−χA0 .Step 2. The Gluing.Now, we would like to glue this transportations, and, in order to apply the Gluing Lemma, weconsider the measures

µlk(x, y) := µk(x, y) + f+(x)χAk−1

(x)δ[y=x],

andµr

k−1(x, y) := µk−1(x, y) + f−(x)χAk−1(x)δ[y=x].

It is easy to see that

µlk ∈ π(f+χAk

+ f+χAk−1, f−χAk−1

+ projx(σ+) Ak−1)

andµr

k−1 ∈ π(f−χAk−1+ projx(σ+) Ak−1, f

−χAk−1+ f−χAk−2

+ g Ak−2).Therefore, by the Gluing Lemma,

G(µlk, µ

rk−1) ∈ π(f+χAk

+ f+χAk−1, f−χAk−1

+ f−χAk−2+ g Ak−2).

Let us now consider the measures

µlk−1(x, y) := G(µl

k, µrk−1)(x, y) + f+(x)χAk−2

(x)δ[y=x]

andµr

k−2(x, y) := µk−2(x, y) +(f−(x)χAk−1

(x) + f−(x)χAk−2(x)

)δ[y=x].


Then we have

µlk−1 ∈ π(f+χAk

+ f+χAk−1+ f+χAk−2

, f−χAk−2+ f−χAk−1

+ projx(σ+) Ak−2)

and

µrk−2 ∈ π(f−χAk−2

+ f−χAk−1+ projx(σ+) Ak−2, f

−χAk−1+ f−χAk−2

+ f−χAk−3+ g Ak−3).

Consequently,

G(µlk−1, µ

rk−2) ∈ π(f+χAk

+ f+χAk−1+ f+χAk−2

, f−χAk−1+ f−χAk−2

+ f−χAk−3+ g Ak−3).

Proceeding in this way we arrive to the construction of

µl2(x, y) = G(µl

3, µr2)(x, y) + f+(x)χA1(x)δ[y=x],

µr1(x, y) = µ1(x, y) +

k−1∑

i=1

f−(x)χAi(x)δ[y=x]

andµ∗ = G(µl

2, µr1) ∈ π(f+, f−),

which is, in fact, an optimal transport plan since, by (5.19),

Kd1(µ∗) = Kd1(G(µl

2, µr1)) ≤ Kd1(µ

l2) + Kd1(µ

r1) = Kd1(G(µl

3, µr2)) + Kd1(µ1)

≤ Kd1(µl3) + Kd1(µ

r2) + Kd1(µ1) = Kd1(G(µl

4, µr3)) + Kd1(µ2) + Kd1(µ1) ≤ . . .

≤ Kd1(µlk) +

k−1∑

j=1

Kd1(µj) =k∑

j=1

Kd1(µj).

Hence,

Kd1(µ∗) ≤

k∑

j=1

Kd1(µk) =k∑

j=1

∫ ∫

Ω×Ωdσ+ (Aj ×Aj−1) =

∫ ∫

Ω×Ωdσ+

=12|σ|(Ω× Ω) = minKd1(µ) : µ ∈ π(f+, f−).

We want to remark that the a similar construction works for any Kantorovich potential u∗,without assuming that u∗(Ω) ⊂ Z. We have presented the above one for simplicity.

With this general gluing construction it is easy to find an optimal measure in Example 5.3.Let us present another example.

Example 5.5. Let us see how the above method works in the particular case of Example 3.2, forwhich we know that, for the masses f+ = 2χ

[0,1] and f− = χ[−2,0], there is no optimal transport

map pushing f+ to f− for the cost function d1. A Kantorovich potential for this configurationof masses is given by

u∗(x) =

2, x ∈ (0, 1),

1, x ∈ (−1, 0),

0, x ∈ (−2,−1).


For these masses, we have that the following antisymmetric measure satisfies the above condi-tions (5.14), (5.15) and (5.16),

σ∗(x, y) = 2χ(0,1)(x)δ[y=x−1] − 2χ

(−1,0)(x)δ[y=x+1] − χ(−2,−1)(x)δ[y=x+1] + χ

(−1,0)(x)δ[y=x−1].

In this case, µ2, the restriction of [σ∗]+ to the set (0, 1)× (−1, 0), is

µ2(x, y) = 2χ(0,1)(x)δ[y=x−1] ∈ π

(2χ

(0,1), f−χ

(−1,0) + χ(−1,0)

),

and µ1, the restriction of [σ∗]+ to the set (−1, 0)× (−2,−1), is

µ1(x, y) = χ(−1,0)(x)δ[y=x−1] ∈ π

(χ

(−1,0), χ(−2,−1)

).

Therefore, for µl2 = µ2 and µr

1 = µ1 + f−χ(−1,0), we have that

µ = G(µl2, µ

r1)

is an optimal transport plan between f+ and f−.

5.3. Construction of the transport density for the classical MK-problem by rescaling.Under the hypothesis of Theorem 1.5 we know that there exists a transport density 0 ≤ a ∈ L∞(Ω)such that

(5.20)

−div(a∇u) = f+ − f− in D′(Ω)

|∇u| = 1 a.e. on the set a > 0,where u is a Kantorovich potential associated with the Euclidean distance d|·|.

By Theorem 4.6, if u∗ε is a Kantorovich potential associated with the metric dε, then, after asubsequence,

(5.21) u∗ε u∗ in L2(Ω) as ε → 0,

where u∗ is a Kantorovich potential associated with the metric d|·|.In this subsection we shall see that also it is possible to get the transport density a by rescaling,

to be more precise from the measures σ∗ε associated with the Euler-Lagrange equation (5.13).Let us fix

(5.22) Ω′ ⊂⊂ Ω′′ ⊂⊂ Ω

be such that |x− y| > r = diam(supp(f+ − f−)) for any x ∈ supp(f+ − f−) and any y ∈ Ω \Ω′.By Theorem 5.1 we have that there exists σ∗ε ∈Ma

b (Ω×Ω) satisfying (5.14), (5.15) and (5.16).Hence,

(5.23)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫

Ω

∫

Ωξ(x) dσ∗ε(x, y), ∀ ξ ∈ Cc(Ω).

If we take ξ ∈ C1c (Ω), from (5.23) and taking into account the antisymmetry of σ∗ε , we have

that

(5.24)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫

Ω

∫

Ωξ(x) dσ∗ε(x, y) =

∫ ∫ξ(x)− ξ(y)

εd

(ε

2σ∗ε(x, y)

),

and

(5.25)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫

Ω

∫

Ωξ(x) dσ∗ε(x, y) =

∫ ∫ξ(x)− ξ(y)

εd(ε[σ∗ε ]

+(x, y)).


Now observe that for ϕ ∈ Cc(Ω× Ω), if φ(x, z) = ϕ(x, x + εz) and Tε(x, y) = y−xε , then

∫ ∫ϕ(x, y) d[σ∗ε ]

+(x, y) =∫ ∫

φ((π1, Tε)(x, y)) d[σ∗ε ]+(x, y)

=∫ ∫

φ(x, z)d((π1, Tε)#[σ∗ε ]+)(x, z) =

∫ ∫ϕ(x, x + εz) d((π1, Tε)#[σ∗ε ]

+)(x, z).

Therefore, we can rewrite (5.25) as

(5.26)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫ ∫ξ(x)− ξ(x + εz)

εd

((π1, Tε)#[εσ∗ε ]

+)(x, z).

On the other hand, by (5.16),∣∣(π1, Tε)#[εσ∗ε ]

+∣∣ is bounded by a constant independent of ε.

Therefore there exists a subsequence εn → 0, such that

(5.27) µn := (π1, Tεn)#[εnσ∗εn]+ σ+ weakly as measures.

Then, taking limit in (5.26), for ε = εn, as n goes to infinity, we obtain

(5.28)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫

Ω

∫

RN

∇ξ(x) · (−z) dσ+(x, z).

Since[εnσ∗εn

]+ (x, y) ∈ Ω× Ω : u∗εn(x)− u∗εn

(y) = εn, |x− y| ≤ εn,and (π1, Tε) is one to one and continuous, we have

µn (π1, Tεn)((x, y) ∈ Ω× Ω : u∗εn

(x)− u∗εn(y) = εn, |x− y| ≤ εn

)

that is,µn (x, z) : x ∈ Ω, x + εnz ∈ Ω, |z| ≤ 1, u∗εn

(x)− u∗εn(x + εnz) = εn,

and consequentlyσ+ (x, z) : x ∈ Ω, |z| ≤ 1.

Now, by disintegration of the measure σ+ (see [2]),

σ+ = (σ+)x ⊗ µ,

withµ = π1#σ+,

that is a nonnegative measure. Moreover, if we define

ν(x) :=∫

RN

(−z) d(σ+)x(z), x ∈ Ω,

then, ν ∈ L1µ(Ω,RN ) and we can rewrite (5.28) as

(5.29)∫

Ωξ(x)(f+(x)− f−(x)) dx =

∫

Ω∇ξ(x) · ν(x) dµ(x), ∀ ξ ∈ C1

c (Ω).

Let us see that

(5.30) supp(µ) ⊂⊂ Ω.


The proof of (5.30) follows the argument of [1, Lemma 5.1] (we include this argument here forthe sake of completeness). In fact, let x0 ∈ supp(f+−f−) be a minimum point for the restrictionof u∗ to supp(f+ − f−) and define

w(x) := min(u∗(x)− u∗(x0))+,dist(x,Ω \ Ω′),where Ω′ verifies (5.22). Then, w(x) = u∗(x) − u∗(x0) on supp(f+ − f−) and w ≡ 0 on Ω \ Ω′.On the other hand,

(5.31)

µ(Ω) = σ+(Ω× RN ) ≤ lim infε→0

µε(Ω× RN ) ≤ lim infε→0

ε [σ∗ε ]+(Ω× RN )

= lim infε→0

∫

Ωu∗ε(x) (f+(x)− f−(x)) dx =

∫

Ωu∗(x) (f+(x)− f−(x)) dx.

and, for a regularizing sequence ρ 1n, on account of (5.29) and using that |ν(x)| ≤ 1, we have

∫

Ωu∗(x) (f+(x)− f−(x)) dx =

∫

Ω(u∗(x)− u∗(x0)) (f+(x)− f−(x)) dx

= limn

∫

Ω(w ∗ ρ 1

n)(x) (f+(x)− f−(x)) dx = lim

n

∫

Ω∇(w ∗ ρ 1

n)(x) · ν(x) dµ(x) ≤ µ(Ω′′),

where Ω′′ verifies (5.22). So, µ(Ω \ Ω′′) = 0, and (5.30) is satisfied.Let us now recall some tangential calculus for measures (see [7], [8]). We introduce the tangent

space Tµ to the measure µ which is defined µ-a.e. by setting Tµ(x) := N⊥µ (x) where:

Nµ(x) = ξ(x) : ξ ∈ Nµ being

Nµ = ξ ∈ L∞µ (Ω,RN ) : ∃un smooth, un → 0 uniformly, ∇un ξ weakly∗ in L∞µ .In [7], given u ∈ D(Ω), for µ–a.e. x ∈ Ω, the tangential derivative ∇µu(x) is defined as theprojection of ∇u(x) on Tµ(x). Now, by [8, Proposition 3.2], there is an extension of the linearoperator ∇µ to Lip1(Ω, d|·|) the set of Lipschitz continuous functions.

Let us see that

(5.32) ν(x) ∈ Tµ(x) µ− a.e. x ∈ Ω.

For that we need to show that

(5.33)∫

Ων(x) · ξ(x) dµ(x) = 0, ∀ ξ ∈ Nµ.

In fact, given ξ ∈ Nµ, there exists un smooth, un → 0 uniformly, ∇un ξ weakly∗ in L∞µ . Then,taking ξ = un in (5.29), which is possible on account of (5.30), we obtain∫

Ωun(x)(f+(x)− f−(x)) dx =

∫

Ω∇un(x) · ν(x) dµ(x),

from here, taking limit as n → +∞, we get∫

Ωv(x)ξ(x) · ν(x) dµ(x) = 0, ∀v ∈ D(Ω),

from where (5.33) follows.Now, if we set Φ := νµ, by (5.29) we have

(5.34) −div(Φ) = f+ − f− in D′(Ω).


Then, having in mind (5.32), by [8, Proposition 3.5], we get

(5.35)∫

Ωu∗(x)(f+(x)− f−(x)) dx =

∫

Ων(x)∇µu∗(x) dµ(x),

where ∇µu∗ is the tangential derivative. Then, since |ν(x)| ≤ 1 and |∇µu∗(x)| ≤ 1 for µ-a.ex ∈ Ω, from (5.35) and (5.31), we obtain that

ν(x) = ∇µu∗(x) and |∇µu∗(x)| = 1 µ− a.e x ∈ Ω.

Therefore, we have

(5.36)

−div(µ∇µu∗) = f+ − f− in D′(Ω),

|∇µu∗(x)| = 1 µ− a.e x ∈ Ω.

Now, by the regularity results given in [12] (see also [1] and [13]) , since f+, f− ∈ L∞(Ω), wehave that the transport density µ ∈ L∞(Ω).

Consequently we conclude that the density transport of Evans-Gangbo is represented by

a = π1#σ+,

for any σ+ obtained as in (5.27).

Example 5.6. By Example 3.2, we know that there is not optimal transport map betweenf+ = 2χ

[0,1] and f− = χ[−2,0] for the cost function dε, ε = 1/2n, n = 1, 2, ..., and that

infFε(T ) : T ∈ A(f+, f−) = minKε(µ) : µ ∈ π(f+, f−)

= supPf+,f−(u) : u ∈ Kε

= 3,

being the supremun attained at u∗ε(x) =∑

j εjχ(εj,εj+ε](x). Also we have that the followingantisymmetric measure satisfies the above conditions (5.14), (5.15) and (5.16),

σ∗12n

(x, y) =2n∑

i=1

2(2n − i + 1)χ( i−12n , i

2n )(x)δ[y=x− 12n ] −

2n∑

i=1

2(2n − i + 1)χ( i−22n , i−1

2n )(x)δ[y=x+ 12n ]

+2·2n−1∑

i=1

(2 · 2n − i)χ(−i2n ,−i+1

2n )(x)δ[y=x− 12n ] −

2·2n−1∑

i=1

(2 · 2n − i)χ(−i−12n ,−i

2n )(x)δ[y=x+ 12n ].


Then, for ξ ∈ C1c ((−2, 1)) we have∫ ∫

ξ(x)d(σ∗1

2n(x, y)

)

=2n∑

i=1

2(2n − i + 1)∫ ∫

(ξ(x)− ξ(y))χ( i−1

2n , i2n )(x)δ[y=x− 1

2n ]

+2·2n−1∑

i=1

(2 · 2n − i)∫ ∫

(ξ(x)− ξ(y))χ(−i2n ,−i+1

2n )(x)δ[y=x− 12n ]

=2n∑

i=1

2(2n − i + 1)∫ i

2n

i−12n

(ξ(x)− ξ

(x− 1

2n

))

+2·2n−1∑

i=1

(2 · 2n − i)∫ −i+1

2n

−i2n

(ξ(x)− ξ

(x− 1

2n

))

=12n

2n∑

i=1

22n − i + 1

2nξ′

(i

2n

)

+1

2 · 2n

2·2n−1∑

i=1

2 · 2n − i

2 · 2n4ξ′

(−2

i

2 · 2n

)+ o(n).

Hence

limn→∞

∫ ∫ξ(x)d

(σ∗1

2n(x, y)

)=

∫ 1

02(1− x)ξ′(x)dx−

∫ 1

04(1− x)ξ′(−2x)dx

=∫ 1

−2ξ′(x)a(x)dx =

∫ 1

−2ξ′(x)a(x)u∗′(x)dx,

where

a(x) =

2 + x if − 2 < x < 0,

2− 2x if 0 < x < 1,

andu∗(x) = x.

Observe that for ξ ∈ C1c ((−2, 1)),

∫ 1

−2ξ′(x)a(x)u∗′(x) dx = 2

∫ 1

0ξ(x) dx−

∫ 0

−2ξ(x) dx =

∫ 1

−2ξ(x)(f+(x)− f−(x)) dx.

Acknowledgments: Theorem 5.1 was obtained jointly with Prof. B. Andreianov and F. Andreuin the bosom of an interesting working week at Universitat of Valencia. N.I., J.M.M. and J.T. havebeen partially supported by the Spanish MEC and FEDER, project MTM2008-03176. J.D.R.has been partially supported by UBA X196 and by CONICET, Argentina.

References

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Noureddine IgbidaLAMFA, CNRS-UMR 6140, Universite de Picardie Jules VerneAmiens, France.

E-mail address: [email protected]

Jose M. Mazon and Julian ToledoDepartament d’Analisi Matematica, Universitat de ValenciaValencia, Spain.

E-mail address: [email protected], [email protected]

Julio D. RossiDepartamento de Matematica, FCEyN, Universidad de Buenos Aires,Buenos Aires, Argentina.

E-mail address: [email protected]

a nonlocal monge-kantorovich problemmate.dm.uba.ar/~jrossi/monge.pdf · 2009. 11. 25. · a...

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