On a solution to the Monge transport problem on the real line arising from the strictly concave case
Nicolas Juillet
Abstract
It is well-known that the optimal transport problem on the real line for the classical distance cost may not have a unique solution. In this paper we recover uniqueness by considering the transport problems where the costs are a power smaller than one of the distance, and letting this parameter tend to one. A complete construction of this solution that we call excursion coupling is given. This is reminiscent to the one in the convex case. It is also characterized as the solution of secondary transport problems. Moreover, a combinatoric/geometric characterization of the routes used for this transport plan is provided.
We first introduce the mass transport problem and quickly arrive to our Main Theorem. The reason why it can appear so fast is that it concerns the most basic setting – together with the finite discrete setting – where the optimal transport problem can be introduced, namely the real line for a power cost.
Definition 0.1. Let(X,d)be a metric space andpbe a positive real number.
We denote byPp(X)the set of Borel probability measuresµ∈ P(X)such that Rd(x0, x)pdµ(x)<+∞for some (and in fact any)x0∈ X and say thatµ has finite moment of order p. The set Marg(µ, ν)is the (convex) set of measures π∈ P(X × X)with first marginalµand second marginal ν, i.e (proj1)#π=µ and(proj2)#π=νwhereprojiis thei-th coordinate function. These definitions naturally extend to positive measuresπ, µ, ν of finite positive massµ(X).
Forµandν two measures withµ(X) =ν(X)we call “Lptransport problem”
(of Monge and Kantorovich), the problem to minimize Cp:π∈Marg(µ, ν)7→
Z Z
d(x, y)pdπ(x, y)∈R∪ {∞}.
We denote byMarg∗p(µ, ν)the (convex) set of solutions to this problem.
The spaceMarg(µ, ν)is equipped with the weak convergence topology and it is compact according to Prokhorov Theorem, see [Vil03, §1.1.7]. MoreoverCp
is continuous onMarg(µ, ν), so that Marg∗p(µ, ν)is not empty. One may have a look at page 23 where this basic continuity result and a more evolved one are stated and proved. Note that ifµand ν have finite moment of order pthe
1
value of the problem is finite, with minπ∈Marg(µ,ν)Cp(π) ≤Cp(µ⊗ν)< +∞.
Hence, we will assume thatµ and ν have finite moment of order1 permitting a finite value for all the Lp transport problems, for p ∈]0,1]. Note moreover that for everyp∈]0,1]sinceCp is continuous andMarg(µ, ν)compact, the set Marg∗p(µ, ν)is not empty.
Let us review some of the known facts concerning the solutions of the Lp transport problem on the real lineX =R. See also Remark 5.8 for the connexion of the L1 transport problem on Rwith the L1 transport problem on geodesic spaces.
• Forp >1, except from degeneracy occurring whenminπ∈Marg(µ,ν)Cp =∞ the set of solutions Marg∗p(µ, ν) is reduced to a single element, known as commonotonic coupling, quantile coupling, monotone rearrangement, Hoeffding-Fréchet transport or any combination of these vocables. The universality and usefulness of the notion is reflected by this number of names. If the atomic part of µ is zero, i.e µ(x) = 0 for every x ∈ R the monotone rearrangement π writes π = (id, T)#µ where T is a non increasing map namely Fν◦Fµ−1 : Fµ(R) ⊂ R → R in the notation of Section 1. Such aT is called anoptimal transport mapsince it minimizes S 7→ R
dp(x, S(x)) dµ(x) among all transport maps, i.e the maps S that satisfyS#µ=ν.
• The value p = 1 is the one in the original problem of Monge of 1781 [Mon81] (except that Monge considersX =R2orR3). For this parameter the quantile transport is still one of the solutions but it may not be the unique one. There is actually a broad class of pairs(µ, ν)such that the set of solutionsMarg∗1(µ, ν)is the whole set of transport plansMarg(µ, ν), see Remark 5.3. Note also that the solutions of the Monge problem in many geodesic space X can be disintegrated along transport rays, i.e parts of X isometric to intervals where the obtained measures are solution of the Monge problem on the real line. This disintegration is the fundament for many powerful recent applications, see Remark 5.8
• The problem for the values p∈]0,1[may be less known and understood asp≥1. However, Gangbo-McCann [GM96] and McCann [McC01] thor- oughly explored this range of the parameterpand discovered a paradoxal hierarchical behavior. They say: “[...] the concavity of the cost function favours a long trip and a shorter trip over two trips of average length; as a result, it can be efficient for two trucks carrying the same commodity to pass each other travelling opposite directions on the highway [...]. In optimal solutions, such ‘pathologies” may nest on many scales, leading to a natural hierarchy among the regions [...].” Unlike the monotone rear- rangement the orientation an optimal transport plan can reverse the order of two points, typically making it locally decreasing if it is a map. How- ever, the orientation does not have to be globally reversed. In particular the antimonotonic coupling is not but for very special pairs(µ, ν).
3
For a class of absolutely continuous measures µ and ν, McCann found an algorithm to restrict the search for the solution – it is unique under the absolute continuity assumption – to a finite number of classes, where regions ofRconcerningµare mapped onto other regions ofRconcerning ν, the frontiers between the regions having to be determined. Note that for general values of (µ, ν), the solution space Marg∗p(µ, ν) may not be reduced to a single value. Moreover it is not constant as a function of p∈]0,1[. See Example 2.2 about these two facts. See also Lemma 2.4: the maximal amount of mass that can stay must also stay on place.
A few recent more works also deal with the ‘concave costs’: a corollary of [PPS15] for instance yields that if the atomic part ofµ is reduced to zero and µ and ν are mutually singular there exists a unique optimal transport plan π and this one can be represented by a transport plan, i.e π = (id, T)#µ. This improves µ(Spt(ν)) = 0 that was the sufficient condition in Gangbo–McCann’s work. On the discrete side, an algorithm given in [DSS12] provides the optimal assignment in the case of finitely many atoms of same mass.
• Our paper being concerned with the valuep= 1 and its limitsp→ 1±, we interrupt here the description of the power costs. It is continued in Remarks 5.1 and 5.2 where we give an account on p = 0 and, beyond, p <0.
The novelty of this paper is that for the critical and historical parameter p= 1we distinguish a special element ofMarg∗p(µ, ν)that we call theexcursion coupling. This element may be seen as a solution for p→1−, the counterpart of the quantile coupling that would be the solution forp→1+.
Main Theorem. Let µ and ν be two probability mesures in P1(R), π ∈ Marg(µ, ν)andq∈]0,1[. The following assumptions are equivalent.
1. (Solution of the L1− limit transport problem) There exists a sequence (πn, pn)n ∈ N with pn < 1 and πn ∈ Marg∗pn(µ, ν) for every n, such that (πn, pn)→n (π,1).
2. (Solution of theL1,qsecondary transport problem)πis inMarg∗1(µ, ν)and in this space it minimizes γ∈Marg∗1(µ, ν)7→RR
|y−x|qdπ(x, y).
3. (Monotonicity)πis concentrated on a setS⊂R×Rwhose arches do not cross, do not connect and have the same orientation when they are nested (see Definition 0.3 and Figure 4).
4. (Excursion coupling) πis the excursion coupling of µandν as defined in Section 1.
In Section 4 we will prove the Main Theorem as well as the following corollary based on the uniqueness of a coupling satisfying1or 2in the Main Theorem.
Corollary 0.2. Letµandνbe two probability mesures inP1(R). The excursion couplingπ∈Marg(µ, ν)satisfies the two following statements.
1’. For(πn, pn)n∈N satisfying pn →1 (with pn <1) andπn ∈Marg∗pn(µ, ν) it holdsπn→π.
2’. For every q0 < 1, the map γ ∈ Marg∗1(µ, ν) 7→ RR
|y−x|q0dπ(x, y) is minimized byπ.
The paper is organized as follow: To give the Main Theorem a complete meaning we first briefly make 3. more precise in Definition 0.3 and define in Section 1 the excursion coupling attached to a pair(µ, ν). Note that this defi- nition relies on several facts concerning functions with bounded variations that will be recalled and established in §3.1. Then we prove step by step the follow- ing implications: We prove1⇒3and2⇒3in Section 2. We prove 3⇒4in Section 3. The fact that there exists at least one solution to theL1− problem and theL1,q secondary problem completes the proof (see Section 4). In Section 5 we provide some more comments.
Definition 0.3(Arches in the Main Theorem). LetSbe a subsetR×R, whose elements we calltransport routes. In what follows[a, b]means[min(a, b),max(a, b)]
and the routes are seen as arches (half-circles) over the real line.
• The routes of S are saidnon-crossing arches if for every (x, y) ∈S and (x0, y0)∈S at least one of the three happens.
– [x, y]∩[x0, y0] =∅
– [x, y]∩[x0, y0] ={z}for somez∈R
– One of the two,[x, y]or[x0, y0], is included in the other.
• The arches do not connect means that if (x, y) ∈ S, (x0, y0) ∈ S and min(|y−x|,|y0−x0|)>0 theny6=x0.
• The nested arches have the same orientation if for every(x, y) ∈S and (x0, y0)∈S such that[x0, y0]⊂]x, y[we have (y−x)(y0−x0)≥0.
The first property can be summed up saying that in the plane the two circles of diameter|y−x|and|y0−x0|linking xandy, andx0 andy0, respectively, do not cross. The second property states that a point can not be at the same time a starting and arriving point. In the last property is stated that transporting xtoy, andx0 toy0 must be done in the same direction when one of the arches is included in the other. See Figure 4 for a representation of the forbidden configuration and the authorized rerouting – where (x, y0) and (x0, y) are the new routes.
Aknowledgements. I warmly thank Guillaume Carlier, Augusto Gerolin, Martin Huesmann, Filippo Santambrogio, Dario Trevisan and the referee for discussions, bibliographic or editorial suggestions.
1 Definition of the excursion coupling 5
1 Definition of the excursion coupling
Given two probability measuresµ andν, we consider the signed measureσ = µ−ν and its cumulative distribution function
Fσ:x∈R7→σ]− ∞, x] =Fµ(x)−Fν(x). (1) Since it is the difference of Fµ and Fν, the function Fσ is càdlàg (right con- tinuous and with left limits at any point) and has limits0 in ±∞. The graph Graph(Fσ) = {(x, Fσ(x)) ∈ R2 : x ∈ R)} of Fσ can be completed at dis- continuity pointsx by vertical segments [(x, Fσ(x−)),(x, Fσ(x))] ⊂R2 where Fσ(x−) = lim
ε→0, ε>0Fσ(x−ε)denotes the left limit atx. We denote byFσ∗ the multivalued map defined byFσ∗(x) = [Fσ(x−), Fσ(x)] at discontinuity pointsx andFσ∗(x) ={Fσ(x)}at continuity points. Its graph is
Graph(Fσ∗) ={(x, h)∈R2: h∈Fσ∗(x)}
and we may also denote it byGraph∗(Fσ). If(x, h)∈Graph(Fσ∗)the realxis called a generalized solution ofFσ=h.
We further introduce the following subsets of Graph∗(Fσ): Graph∗,+(Fσ) is the set of increasing points (x, h), i.e such that in a neighborhood Ux of x any point (x0, h0) ∈ Graph∗(Fσ) with x0 ∈ Ux\ {x} satisfies (h0−h)(x0 − x) > 0. Graph∗,−(Fσ) is the set of decreasing points (x, y), i.e such that in a neighborhood Ux of x any point (x0, h0) ∈ Graph∗(Fσ) with x0 ∈ Ux\ {x}
satisfies(h0−h)(x0−x)<0.
Theorem 1.1(Excursion couplings can be defined). Letµandν be probability measures andσ=µ−ν,FσandGraph(Fσ∗)be defined as above. One can define a transport of Marg(µ, ν) as described in what follows, the results implicitly stated during this construction (see Remark 1.2 for a list) are correct and we call excursion couplingthe resulting coupling.
We distinguish two cases (the first one is a special case of the second).
1. Assume that µ and ν are singular measures (µ ⊥ ν). We define a cou- pling (X, Y) where X ∼ µ and Y ∼ ν. Let θ be the measure with den- sity h 7→ i±(h) = card[(R× {h})∩Graph∗,+(Fσ)] + card[(R× {h})∩ Graph∗,−(Fσ)]. Let H be a random variable with law θ/2 and note(R× {H})∩Graph∗(Fσ) = {x1, . . . , xN} × {H} where N =i±(H) and x1 <
· · ·< xN. Conditionally onHandH >0, the random vector(X, Y)is de- fined to be uniform on{(x1, x2),(x3, x4), . . . ,(xN−1, xN)}. Conditionally onH andH <0 it is uniform on{(x2, x1),(x4, x3), . . . ,(xN, xN−1)}.
2. If µ = η+µ0 and ν = η+ν0 with µ0 ⊥ ν0, with probability η(R) the random vector(X, Y)satisfiesX =Y andX ∼η. On the complementary event, with probability 1−η(R) =µ0(R)it is distributed as the coupling of the singular probability measuresµ0(R)−1µ0 andν0(R)−1ν0 defined in the first item.
Remark 1.2. To make Theorem 1.1 a rigorous definition of the excursion cou- pling we will have to prove that θ/2(dh) is a probability density, that N is almost surely both finite and even with(R× {h})∩Graph∗(Fσ) = (R× {h})∩ (Graph∗,−(Fσ)∪Graph∗,+(Fσ)). Moreover, we must prove that the laws ofX andY areµandν, respectively.
To illustrate the definition of the excursion coupling given in Theorem 1.1 we provide a few naive examples that we encourage the reader to test by drawing.
They should in particular permit her to understand the coupling in terms of transport maps. Transport maps are often more intuitive than plans. However, they are less symmetric and general. The excursion coupling can in particular unfortunately not always be represented by a transport map.
Example 1.3. Letµ be the uniform measure on[0,1]∪[2,3]andν uniform on [1,2]∪[2,4]. ThenFσ is continuous and defined piecewise by
Fσ(t) =
t/2 ift∈[0,1]
1−t/2 ift∈[1,2]
−1 +t/2 ift∈[2,3]
2−t/2 ift∈[3,4]
0 otherwise.
For(X, Y, H)the coupling in Theorem 1.1 we have H uniform on [0,1/2]and H = Fσ(X). Moreover, the excursion coupling is induced by a map T, i.e π= (id, T)#µ= Law(X, T(X))withLaw(X) =µ. The mapT is defined by
T(x) =
(2−x ifx∈[0,1]
6−x ifx∈[2,3].
This map is locally non increasing on each of the two intervals but not globally.
Finally, one can check thatY =T(X)has lawν =T#µ.
Example 1.4. More generally, letabe in[0,1)and assume thatµis uniform on [0,1 +a]∪[2,3−a]andν on[1 +a,2]∪[3−a,4]. We have
Fσ(t) =
t/2 ift∈[0,1 +a]
1 +a−t/2 ift∈[1 +a,2]
a−1 +t/2 ift∈[2,3−a]
2−t/2 ift∈[3−a,4]
0 otherwise.
Again the excursion coupling isπ= (id, T)#µ= Law(X, T(X)). The random variableH has density1on[0, a], and2on[a,(1 +a)/2]. The mapT is defined by
T(x) =
4−x ifx∈[0,2a) 2 + 2a−x ifx∈[2a,1 +a]
6−2a−x ift∈[2,3−a]
1 Definition of the excursion coupling 7
The three following examples illustrate that the excursion coupling may fail to be represented by a transport mapT from µtoν. This occurs in particular whenµpossesses a non trivial atomic part (see Examples 1.5 and 1.6) or when µandν are not mutually singular as in Example 1.7.
Example 1.5. Letµ=δ0be the Dirac mass in zero andµbe uniform on[−1,1].
ThereforeFσ is given by
Fσ(t) =
−t/2 ift∈[−1,0[
(1−t)/2 ift∈[0,1]
0 otherwise.
This is completed be the vertical segment between −1/2 and 1/2 on the y- axis. The excursion coupling is not given by a map as before. Instead of that, we uniformly choose hin [−1/2,1/2]on the y-axis and map the mass at 0 to Th(0) =−1−2hifh <0andTh(0) = 1−2hifh≥0. Therefore, we haveX = 0 andY =TH(X). The transport mapT of the previous examples is replaced by a random mapTH.
Example 1.6. For a more complex excursion coupling with atoms we consider µ= (1/3)(δ−1+1[−1,0]dx+δ0)andν uniform on[0,1]. In this case TH(X) = (1−X)/3ifX∈(−1,0)which correspond to a deterministic transport (notice H= (2+X)/3in this case). However, conditioned on{X =−1}, the heightHis uniform on[0,1/3]. It is uniform on[2/3,1]ifX = 0and we haveTH(X) = 1−H both forX= 0 andX= 1. This correspond to a randomized multivalued map (probabilists say a kernel).
Note that in this example the excursion coupling can be represented by a transport map fromν toµ–in place of fromµto ν.
Example 1.7. Let nowµbe uniform on[0,1]andν uniform on[0,3]. Then the common measureµ∧ν = 131[0,1]dx represents the mass that stays during the transport. The information on the moving mass is contained in the function
Fσ(t) =
2t/3 ift∈[0,1[
1−t/3 ift∈[1,3]
0 otherwise.
that permits one to consider the mapTmove :x∈[0,1]7→3−2x. Finally the excursion transport is multivalued. With probability 1/3 the mass is staying Tstay(x) = x, with probability 2/3 it is moving with Tmove. This writes π = (id, Tstay)#(µ∧ν) + (id, Tmove)#(µ−(µ∧ν)).
Remark 1.8. Note that ifµ⊥ν= 0 andµhas atomic part zero, almost surely Hsimply writesH =Fσ(X)and since the arches corresponding to the heightH are almost surely properly separated,Y is a function ofX. This behavior that was illustrated in Examples 1.3 and 1.4 corresponds to the one observed in the strictly concave case, as proved in [PPS15]. Ifµ⊥ν 6= 0andµ(orµ−µ∧ν) still has atomic part zero we are in the situation of Example 1.7 where the excursion coupling writesπ= (id, Tstay)#(µ∧ν) + (id, Tmove)#(µ−(µ∧ν)). This is again the shape obtained in [GM96, PPS15].
2 Monotonicity of the solutions of the L
1,qand L
1−transport problems
In this section we prove the implications1⇒3and2⇒3of the Main Theorem.
The title “(cyclical-)monotonicity” is a generic name in Optimal Transport that in this section is represented by property 3. (Cyclical-)monotonicity results are variations of the following simple swapping lemma that concerns the Lp transport problem for cycles of length two. For theL1− transport problem we will firstly interpret Lemma 2.1 forp <1 and secondly let pgo to1. For the L1,q transport problem will need a more deep result than Lemma 2.1: It will be Lemma 2.5 on page 13. Wheras the first lemma is classical and could be skipped, the second is not. We state and prove both results for the sake of completeness and because the proof of Lemma 2.1 is required in Lemma 2.5.
Lemma 2.1 (Swapping lemma). Let p be positive and π be in Marg∗p(µ, ν).
Assume moreover Cp(π) < +∞. Consider a set S ⊂R2 such that π(S) = 1 and for(x, y)∈S any neighborhood of (x, y)has positive measure (Notice that the support ofπsatisfies these conditions, so that we may choose S = Spt(π)).
For any(x, y)and(x0, y0) inS, the following holds:
|y−x|p+|y0−x0|p ≤ |y−x0|p+|y0−x|p. (2) Proof. Striking for a contradiction, suppose that the opposite identity holds.
Then for some(x, y),(x0, y0)∈Sthere existsε >0such that for every(a, b, a0, b0) withmax(|a−x|,|a0−x0|,|b−y|,|b0−y0|)≤εone has|b−a0|p+|b0−a|p<|b−a|p+
|b0−a0|p. The two balls (in the∞-norm) of radiusεcentered in(x, y)and(x0, y0) have positive measure. We chooseεsmall enough to make their intersection the empty set. Then it is easily possible to replace π by a competitor π0 that coincide with it outside the ballsB∞((a, b), ε),B∞((a0, b0), ε),B∞((a, b0), ε)and B∞((a0, b), ε), has marginals µ and ν and satisfies Cp(π0) < Cp(π) (see e.g [GM96, page 129] or the end of the proof of Lemma 2.5), a contradiction.
This simple principle allows for a complete characterization ofπin the case p >1. We provide it now for the sake of completeness and for further comparison with theL1− limit andL1,q secondary problems.
TheLp transport problem forp >1 In this case the identity|y−x|p+|y0− x0|p ≤ |y−x0|p+|y0−x|p obtained by applying Lemma 2.1 is equivalent to (y0−y)(x0−x)≥0. This may be graphically represented by the condition that segments inR2 connecting for every(x, y)∈S the point(x,1)to(y,0)are not allowed to cross each other, see Figure 1. These segments may be interpreted as transport routes betweenµconcentrated on a line parallel to thex-axis and ν concentrated on the x-axis. The striking fact is that, given µ and ν, the elementsπofMarg(µ, ν)being concentrated on such a setSare in fact reduced to a single transport plan, called thequantile transport plan. The latter is the law of (Gµ, Gν) on the probability space (]0,1[, λ) of quantile levels, where λ
2 Monotonicity of the solutions of theL1,qandL1− transport problems 9
Fig. 1: Left: the quantile transport is based on simultaneous simulation through the cumulative distribution functionsFµandFν. The dashed horizontal lines (below on the figure) are chosen uniformly in]0,1[. The transport routes are materialized by arrows (above on the figure) that do not cross.
Fig. 2: Right: the moving mass of the excursion coupling is simulated through the intersections with Fν −Fν ; the commun massµ∧ν stays on the on the same place. Given the horizontal line the excursion is chosen uniformly (among two or one pairs for the lines on the figure). The increasing intersection corresponds to µ and the decreasing one to ν.
The transport routes are materialized by arches that do not cross.
is the Lebesgue measure and for any a real probability measureη, the quantile functionGη is defined as a pseudo-inverse ofFη :x7→η(]− ∞, x]), namely
Gη(α) = inf{x∈R: Fη(x)≥α}
(this infimum is a minimum).
2.1 The L
ptransport problem for p < 1 and the L
1−limit transport problem
Important preliminary comparaison to the convex casep >1 In the previous section we recalled that forp >1, provided the Lp transport problem admits a
4 4 9
1
Fig. 3: For transporting the two crosses to the two circles, each symbol being of mass1/2, the left pattern is optimal forp≤1/2, the right is forp≥1/2.
solutionπp withCp(πp)<∞, this transport planπp is theunique solution and it is the quantile coupling. In particular it isindependent of the value ofp.
The two last assumptions areboth false in the case p < 1, which we prove in the following example.
Example 2.2. Setµ= 12(δ0+δ5)andν= 12(δ4+δ9). The set of transport plans is easily described as
Marg(µ, ν) ={πλ=λπ00+ (1−λ)π0: λ∈[0,1]}
whereπ0= 12(δ0,4+δ5,9)andπ00= 12(δ0,9+δ5,4). In the casep= 1/2every trans- port planπgives the same global costCp(πλ) =λ12(√
4 +√
4) + (1−λ)12(√
√ 9 +
1) = 2. This proves thenon-uniqueness; Marg∗1/2(µ, ν) = Marg(µ, ν). More- over, with the same marginals, depending whether p < 1/2 or p > 1/2 the measureπ0=π0orπ00=π1, respectively, is the unique optimal transport plan.
Hence the solutiondepends on the value ofp.
Example 2.3. With this example strongly inspired by Example 1.1 in [McC01]
we again illustrate that the solution (here it is unique) depends on p∈ (0,1).
Consider as before (see Example 1.3) µ the uniform measure on [0,1]∪[2,3]
andν uniform on[2,3]∪[3,4]. In this case the algorithm developed in [McC01]
shows that the optimal transport plan is induced by a map, namely the map
T(x) =
4−x ifx∈[0, ap) 2−x ifx∈[ap,1]
4−x ift∈[2,2 +ap) 6−x ift∈[2 +ap,3].
As we announced this map depends on pthrough the solution ap > 0 of the equation(4−2ap)p+ (2ap)p= 2(2−2ap)p. It is not a surprise thatap goes to zero as ptends to 1 since the formula forap = 0 corresponds to the excursion coupling. Note that on each interval the map is decreasing.
In the two next paragraphs we prove that in theLp transport problem (for p <1) arches do not connect and do not cross. In the third next paragraph this will be transmitted to theL1− (limit) transport problem where we also prove that nested arches have the same orientation, completing the proof of1⇒3in the Main Theorem.
2 Monotonicity of the solutions of theL1,qandL1− transport problems 11
Conclusion concerning coinciding points forp <1 LetSsatisfying (2) in the swapping lemma, Lemma 2.1 for p <1 and (x, y, x0, y0)be in S×S. Suppose moreover that y and x0 coincide meaning that one can take a route from x to y and a second one from y = x0 to y0. Then studying the variations of t∈[0,+∞[7→(t+h)p−tpand consideringh=|y−y0|we see that this can only occur in one of the degenerate case x=y or x0 =y0 – compare with the two most right subfigures in Figure 4. With respect to the terminology of Definition 0.3 we have proved that a solutionπ of the Lp transport problem (p < 1) is concentrated on a set whosearches do not connect. This well-known situation has been studied be Gangbo and McCann in [GM96, Proposition 2.9] more than 20 years ago. It yields thatπcan be decomposed as the sumπ=π∆+π0where π∆= (id×id)#(µ∧ν). Let us remind the argument for the sake of completeness.
Lemma 2.4. If π ∈ Marg(µ, ν) is concentrated on a set S whose arches do not connect, it can be decomposed as follows: π=π∆+π0 where π∆ = (id× id)#(µ∧ν)and, for ∆ ={(x, y)∈R2: y=x} denoting the diagonal, it holds π0(∆) = 0. Consequently,π0 is a transport plan with the two marginals singular with respect to each other.
Proof. Let us write π=π∆+π0 whereπ0 is concentrated onS\∆and π∆ is concentrated on ∆. Therefore π∆ writes (id×id)#η and the marginals of π0
areµ−ηandν−η. They are respectively concentrated on the two projections {x0∈R: ∃y06=x0,(x0, y0)∈S\∆}and{y∈R: ∃x6=y,(x, y)∈S\∆}of the setS\∆. The fact that these two sets do not intersect is the direct consequence of our assumption. It followsµ−η⊥ν−ηwhich is equivalent toη=µ∧ν. Interpretation of the swapping lemma for p <1 and non coinciding points Forp <1the use of the swapping lemma furnishes, compared to(y0−y)(x0−x)≥ 0, less direct information. Equation (2) may be seen, similarly as in Example 2.2, as a competition of two transport plans, each transporting two points in two other points, in one or the other way. For this reduced transport problem if |y−x|p+|y0−x0|p 6= |y0−x|p+|y−x0|p at most one of the two is true (x, y, x0, y0)∈ S×S or (x, y0, x0, y)∈ S×S. Unlike the situation studied for p > 1, determining which transport is better does not only depend on the relative positions ofxandy with respect tox0 andy0, respectively, but on the 4!relative positions of the four points. Moreover, even though this ranking of the four point is important and yields the conclusion in configurations xxyy andxyyxwe know since Example 2.2 and Figure 3 that it does not permit to conclude in configurationxyxy.
Notation. The notation xyxy denotes the configurations of two routes (x, y) and(x0, y0)where x < y < x0 < y0 or x < y0 < x0 < y or x0 < y < x < y0 or x0 < y0 < x < y, the different alternatives being signed asxyx0y0,xy0x0y,xy0xy0 andx0yxy0, respectively.
Since theLpproblem inMarg(µ, ν)is in bijection with the one inMarg(ν, µ), when we study configuration xyxy we in fact also study yxyx. In order to consider all configurations without coinciding points we finally only have to
look at xxyy, xyyx and xyxy. Here is the conclusion for these three cases.
They are also illustrated on Figure 4 (corresponding, in the same order, to the first three pairs of patterns, from the left).
• In the first casexx0y0y is allowed andxx0yy0 forbidden. To see that, one can study the variations of t ∈[0,+∞[7→(t+h)p−tp whereh > 0 and apply it toh=|y−y0|.
• In the second casexyy0x0is allowed andxy0yx0forbidden. This is a simple consequence of the fact thatd∈R+7→dp is increasing.
• In the last case, as observed in Example 2.2, it depends on p whether xyx0y0 orxy0x0y is forbidden or authorized.
With the two first points we have proved that the arches of S do not cross.
Let us give another, less descriptive argument. When four points bound three intervals of lengtha,bandc, in this order, the cost for a crossing corresponds to (a+b)p+(b+c)p. It is easy to check that it smaller thanap+cpor(a+b+c)p+bp, that is the cost of the swapped (and uncrossed) configuration.
From theLp to theL1− transport problem In this paragraph we prove that the solutions of the L1− transport problem have arches that do not connect, do not cross and have the same orientation when they are nested. This is implication1⇒3of Main Theorem.
Considerπsuch that there exists a sequence(πn)n∈N weakly converging to πwhereπn∈Marg∗pn(µ, ν)forpn →1−. It also holdsπn⊗πn→π⊗π, actually an equivalent fact. Hence, ifF ⊂R4 is a closed set and πn⊗πn(F) = 1, the equation goes to the limit. In particular this holds for the complementary setF1
of the open set{(x, y, x0, y0)∈R4: x < x0 < y < y0 orx0 < x < y0 < y} that encodes the condition on non-intersecting arches in configuration xxyy (first condition of Definition 0.3). As it is satisfied by πn it is also satisfied by π.
The same argument applies to the other configurations: yyxx,xyyxandyxxy.
Therefore,πis concentrated on a set whose archesdo not cross.
Suppose by contradiction that the setF3c⊂R4of pairs of nested arches that do not have the same orientation has positive measure forπ⊗πand without loss of generality we assume that it happens for the configurationxy0x0y. Due to the countable additivity ofπ⊗πthere exists rational numbers x < y0< x0< yand ε∈Q+ withmin(|y0−x|,|x0−y0|,|y−x0|)> ε such thatU =]x±ε/10[×]y± ε/10[×]x0±ε/10[×]y0±ε/10[has positive measure. Since
[(y0−x)+2ε/10]pn+[(y−x0)+2ε/10]pn ≤[(y−x)−2ε/10]pn+[(x0−y0)−2ε/10]pn forpn close enough to 1, the open setU has measure zero for the corresponding measuresπn⊗πn. Therefore, it has measure zero forπ⊗πas well, a contradic- tion. We conclude thatπ concentrated on a set whose nested arches have the same orientation.
The implication1⇒3finally amounts to prove thatπis concentrated on a set satisfying the second condition of Definition 0.3: the arches do not connect.
2 Monotonicity of the solutions of theL1,qandL1− transport problems 13
Recall from Lemma 11 thatπn= (id×id)#(µ∧ν) +π0nwhereπn0 has marginals µ0 = µ−(µ∧ν), ν0 =ν −(µ∧ν)and µ0 ⊥ ν0. Thus π is the limit of the sequence if and only if it can be written(id×id)#(µ∧ν) +π0 whereπ0is the limit of(πn0)n∈N. Therefore,π0 has marginals µ0 and ν0. Thus there existsA andB =Acsuch thatπis concentrated onE= [(A×R)∩(R×B)]∪∆. This exactly means that it is concentrated on a set whose arches do not connect.
2.2 The L
1,qsecondary transport problem
Swapping lemma for theL1,qsecondary transport problem and interpretation We furnish now a swapping lemma corresponding to theL1,qtransport problem.
Lemma 2.5(Swapping lemma for the secondary problem). Letπbe a transport plan such thatC1(π)<+∞andπ∈Marg∗∗1,q(µ, ν)for some fixed q <1. There exists a setS ⊂R2 with π(S) = 1 such that for any(x, y) and(x0, y0) in S it holds
|y−x|+|y0−x0| ≤ |y−x0|+|y0−x| (3) and if|y−x|+|y0−x0|=|y−x0|+|y−x0|, the following holds:
|y−x|q+|y0−x0|q ≤ |y−x0|q+|y0−x|q. (4) Proof. Let π be an optimal transport plan for the L1 primary and the L1,q secondary transport problem. Let Ny,− be {(x, y) ∈ Spt(π) : ∃ε > 0, π([x− ε, x+ε]×[y−ε, y]) = 0}. It is the set of routes in the support ofπsuch that at some sizeε >0 no mass is transported from the ball of centerxand radius εto the left of y at distance less thanε. We have π(Ny,−) = 0. The proof of this result is postponed to Lemma 2.6. The symmetrically defined setsNy,+, Nx,− andNx,+ have measure zero too (for instanceNx,+ here denotes the set {(x, y)∈Spt(π) : ∃ε >0, π([x, x+ε]×[y−ε, y+ε]) = 0}). We define now S asS0\(Ny,−∪Ny,+∪Nx,−∪Nx,+)whereS0is any set, as for instanceSpt(π), satisfying the condition in Lemma 2.1. Observe thatπ(S) = 1.
We are now ready for the proof. It is almost identical to that of Lemma 2.1 that we invite the reader to read again: the principle is thatπis compared to a competitorπ0 defined rerouting part of the mass around(x, y)and(x0, y0)to mass aroud (x, y0) and (x0, y). From Lemma 2.1 we note that (3) is satisfied for any(x, y, x0, y0)∈S×S. Aiming for a contradiction, suppose that for some routes(x, y)and(x0, y0)inSit holds at the same time|y−x|+|y0−x0|=|y−x0|+
|y0−x|and|y0−x|q+|y−x0|q <|y−x|q+|y0−x0|q. Until the end of the paragraph let us see that without loss of generality we can assume x < x0 ≤ y < y0: The first equation induces ‘max(x, x0)≤min(y, y0)or max(y, y0)≤min(x, x0)’.
Without loss of generality we can assume the first case. The second inequality impliesx6=x0andy6=y0and without loss of generality we assumex < x0. This finally impliesy < y0.
Apparently the swapping method used in the proof of Lemma 2.1 that con- sists in picking mass in equal quantity around the pointsx,x0,y andy0 can be
applied without problem providing a competitorπ0∈Marg(µ, ν)withCq(π0)<
Cq(π), a contradiction. However, this argument only works if x0 < y and not directly ifx0 =ybecauseπ0∈Marg∗1(µ, ν)can not be certified: during the swap the routes(x, y)and(x0, y0)withx < x0 =y < y0 have a priori in their neigh- borhood routes(˜x,y)˜ and(˜x0,y˜0)with˜x <y <˜ x˜0 <y˜0 that do no longer satisfy the preliminary condition for (4) (we mean|y−x|+|y0−x0|=|y−x0|+|y−x0|), so that C1(π0) > C1(π) is made possible. The relation Cq(π0) < Cq(π) is no longer a contradiction becauseπ0 ∈/Marg∗1(µ, ν), i.eC1(π0) =C1(π)is false. In this critical situation let us callzthe pointx0=y. As(x, y)and(x0, y0)are not inNy,−∪Ny,+∪Nx,−∪Nx,+we can swap selecting the mass on the proper side ofz. More precisely forπthere is some mass traveling from a neighborhood (as small as we want) of xto a small right neighborhood of z. There is the same mass on a small left neighborhood ofztransported to a small neighborhood of y0. We swap for defining π0: the mass aroundxis transported around y0 and the mass directly on the left ofz is transported directly to the right ofz. We obtainCq(π0)< Cq(π)and keep C1(π0) =C1(π), a contradiction. Let us refor- mulate this argument in a less descriptive manner. There existsε >0 as well as two positive measuresγ andγ0 with the same mass that satisfies a bunch of properties. Due to our assumptions we can in fact assume the following: γ≤π and γ is concentrated on [x−ε, x+ε]×[z, z+ε]; in a parallel way γ0 ≤ π and γ0 is concentrated on [z−ε, z]×[y0−ε, y0 +ε1]. Moreover, this can be done for everyε close enough to zero. Settingm =γ0(R2) = γ0(R2)we note that ζ0 = (1/m)(proj1#γ×proj2#γ0) + (1/m)(proj1#γ0×proj2#γ)has the same marginals as ζ=γ+γ0. Let us now denote π−ζ+ζ0 by π0. One can easily check that for ε small enough it holds at the same time C1(π0) = C1(π) and Cq(π0)> Cq(π).
Lemma 2.6. Let π be a solution to the L1,q secondary optimal transport and Ny,− be {(x, y) ∈ Spt(π) : ∃ε > 0, π([x−ε, x+ε]×[y−ε, y]) = 0}. Then π(Ny,−) = 0.
Proof. Fixa < btwo rational numbers. Consider the setAa,b of points(x, y)∈ Spt(π|]a,b[×R)such that there existsε >0withπ([x−ε, x+ε]×[y−ε, y]) = 0 and moreover x ∈]a, b[⊂ [x−ε, x+ε]. Then every point y ∈ proj2(Aa,b) is in the support of the measure νa,b defined by (proj2)#π|]a,b[×R, but it is also isolated on the left in the sense that νa,b([y −h, y]) = 0 for h small enough.
One can easily convince that there are countably many such pointsyand they are not atoms. Therefore, ν(proj2(Aa,b)) = 0 andπ(Aa,b) =π|]a,b[×R(Aa,b)≤ ν(projy(Aa,b)) = 0. Finally, sinceNy,−=S
a<b∈QAa,bwe findπ(Ny,−) = 0.
We can now give the geometric meaning for the routes of setsSas in Lemma 2.5. As for theL1− problem we compare the different configurations with their alternative – recall the list on page 12 – and again these comparaisons are illustrated on the three most left patterns of Figure 4.
3 Transport plans concentrated on monotone sets of arches are the excursion coupling 15
Fig. 4: Forbidden (first line) versus authorized (second line) configurations in theL1,q secondary transport problem, wherep <1.
• Patternxx0y0y is allowed and xx0yy0 forbidden. To see that, since |y0− x0|+|y−x|=|y0−x|+|y−x0|we have to look at the secondary problem.
• Patternxyy0x0 is allowed andxy0yx0 forbidden because as in theLpprob- lem, (3) is a strict inequality.
• Finally, pattern xyx0y0 is allowed and xy0x0y is forbidden because as in theLp problem, (3) is a strict inequality.
Some of the points x, y, x0, y0 may be equal. Swapping does not change the cost ifx=x0 ory=y0. We have only to look atx=y0 andx06=yand see that it is never better thanx=y and x0 6=y0 (see on Figure 4 the two patterns on the right).
• For the patternx(yx)y, meaningx < y=x < y, we have|y0−x0|+|y−x|=
|y0−x|+|y−x0|but the secondary problem tells us to choosexx0y0y in place ofxyx0y0.
• For(xy)xy, from the primary problem we choosex=y < x0 < y0 in place ofx=y0 < x0< y.
Finally as for theL1− limit transport problem, we have proved that if πis a solution of theL1,q secondary transport problem it is concentrated on a set S whose arches do not cross, do not connected, and have the same orientation when they are nested.
3 Transport plans concentrated on monotone sets of arches are the excursion coupling
3.1 Proof of Theorem 1.1 defining the excursion coupling
We need to explain why Theorem 1.1 can define the excursion coupling. We only need to investigate the construction in case1 whereµand ν are singular (µ⊥ν), which we thus assume in the present subsection.
With the following lemma we will be able to handle with Fσ as if it were a continuous function.
Lemma 3.1(Generalized intermediate value theorem). For any càdlàg function F, anyx0, x1∈Randh∈Rsuch thatx0< x1and(F(x0)−h)(F(x1)−h)<0, there existsx∈]x0, x1]such that (x, h)∈Graph∗(F).
Proof. Let F, x0, x1 be as in the statement. Without loss of generality we assumeh= 0, F(x0)<0 and F(x1) >0. Letxbe the infimum of A={x∈ [x0, x1] : F(x)≥0}. As A3x1 it is a not empty set. As moreoverF is right continuous we havex∈Aandx6=x0. Due to the definitions ofA andx, any x0 < xsatisfies F(x0)<0. IfF is left-continuous at xwe haveF(x) = 0. If it is not, asF(x−)<0 we also have(x,0)∈Graph(F∗).
A result by Bertoin and Yor establishes a relation between the occupation measure in a setB ⊂ R of a function F of finite variation and its variations when its values are inB, see Remark 5.6 for details. In particular we can apply their Theorem 1 in [BY14] (see also their §5) toF=Fσ andB=R, and relate the total variation (without its saltus part) with the number of solutions of the equationsFσ=h, wherehgoes overR: For any pointss≤tin R
T Vst(Fσ)− X
x∈]s,t]
|σ(x)|= Z
R
card{x∈[s, t] : (x, h)∈Graph(Fσ)}dh (5)
= Z
R
card[(R× {h})∩Graph(Fσ)] dh
= Z
R
i]s,t](h) dh
where
iA:h∈R7→iA(h) = card{x∈A: Fσ(x) =h}
= card{x∈A: (x, h)∈Graph(Fσ)}
is the so-called Banach indicatrix, after [Ban25]. Notice that in [BY14] the result is stated fors = 0and t ∈ [0,∞[. Our statement on a general interval ]s, t] is a trivial generalization. Another difference is that we only apply the formula to the occupation measure ofFσ in B whenB =R. It is hardly more than a simple exercise to rewrite (5) for generalized solutions ofFσ=h, which permits at the same time to forget about the saltus part. For this purpose we introduce the generalized Banach indicatrix. We set
i∗A:h∈R7→i∗A(h) = card{x∈A: h∈Fσ∗(x)}
= card{x∈A: (x, h)∈Graph(Fσ∗)} ∈N∪ {∞}.
Therefore, (5) yields
T Vst(Fσ) = Z
i∗]s,t](h) dh. (6)
3 Transport plans concentrated on monotone sets of arches are the excursion coupling 17
A consequence of Theorem 1 in [BY14] specifies that not all intersections with the graph need to be considered. Recall first thatGraph+(Fσ∗)andGraph−(Fσ∗) have been defined in Section 1. Bertoin and Yor proved that almost surely for h, the (generalized) Banach indicatrixi∗]s,t] equals i∗,±]s,t]:=i∗,+]s,t]+i∗,−]s,t] where
i∗,+]s,t] :h∈R7→i+(h) = card{x∈R: (x, h)∈Graph+(Fσ∗)} ∈N∪ {∞}, i∗,−]s,t] :h∈R7→i−(h) = card{x∈R: (x, h)∈Graph−(Fσ∗)} ∈N∪ {∞}
(with obvious notation Bertoin and Yor in fact proved the equalityi =i± for the non yet generalized indicatrixi=i++i−). This also comes from Theorem 1 in [BY14] (wherenx(t)dx=λx(t)dx):
T Vst(Fσ) = Z
i∗,±]s,t](h) dh≤ Z
i∗]s,t](h) dh=T Vst(Fσ). (7) With the next result we go further in the analysis.
Proposition 3.2. For almost every h ∈ R, the set (R× {h})∩Graph∗Fσ
has cardinality a finite and even integer. In fact for almost every h ∈ R it holds i∗(h) = i∗,+(h) +i∗,−(h) with i∗,+(h) = i∗,−(h). Moreover if h 6= 0 the generalized solutions of Fσ = h are alternatively crossing positively and negatively, starting with the most-left solution(x, h)∈ Graph∗,+(Fσ) if h >0 and(x, h)∈Graph∗,−(Fσ)if h <0.
Proof. Due to (6) the generalized Banach indicatrix is finite for almost every h ∈ R and with (7) for almost every h we have (R× {h})∩Graph∗Fσ = [(R× {h})∩Graph∗,+Fσ]∪[(R× {h})∩Graph∗,−Fσ]. With the generalized intermediate value theorem (Lemma 3.1) and aslim∞Fσ = lim−∞Fσ = 0 we conclude that i∗(h) is an even number for almost everyh. Let us clarify this.
Let h be as before and x¯0 < x¯1 satisfy (¯x0, h) ∈ Graph∗(Fσ) and (¯x1, h) ∈ Graph∗(Fσ). We moreover assume
(x, h)∈/ Graph∗(Fσ)for everyx∈(¯x0,x¯1). (8) Therefore Fσ > h or Fσ < h on all (¯x0,x¯1). If not the generalized inter- mediate value theorem applied to the proper interval [x, x0] ⊂ (¯x0,x¯1) fur- nishes a contradiction with (8). Since now Fσ > h on (¯x0,x¯1) – the oppo- site assumption Fσ > h yields the symmetric conclusion – and (R× {h})∩ Graph∗Fσ = [(R× {h})∩Graph∗,+Fσ]∪[(R× {h})∩Graph∗,−Fσ] we find (¯x0, h) ∈ Graph∗,+Fσ and (¯x1, h) ∈ Graph∗,−Fσ. A similar argument leads to the fact thath−Fσ has a unique sign in the neighborhood of−∞ and in the neighborhood of +∞. It is the sign of h (where we assume h 6= 0). Fi- nallyh−Fσcan only change an even number of times its sign. More precisely, due to the generalized intermediate value theorem applied toFσ and reminding that this function has limit zero in ∞ and +∞ the points x1 < x01 < x2 <
· · · < x0n−1 < xn < x0n of {x ∈ R : (x, y) ∈ Graph∗(Fσ)} are ordered with (x1, h), . . . ,(xn, h) ∈ Graph∗,+(Fσ) and (x01, h), . . . ,(x0n, h) ∈ Graph∗,−(Fσ) if h >0,(x1, h), . . . ,(xn, h)∈Graph∗,−(Fσ)and(x01, h), . . . ,(x0n, h)∈Graph∗,+(Fσ) ifh <0.
Coming back to (6), another direct generalization of Bertoin and Yor’s study is
Fσ(t)−Fσ(s) = Z
R
i∗,+]s,t]−i∗,−]s,t]dh. (9) It will be useful for recoveringFµ andFν fromFσ.
For any measurable setG⊂R2, letζGdenote the following positive measure ζG:E∈ T(R2)7→ζG(E) =
Z +∞
−∞
card(x∈R: (x, h)∈E∩G) dh.
We consider in particular G = Graph∗(Fσ) and G+ = Graph∗,+(Fσ) and G− = Graph∗,−(Fσ) and call ζ, ζ+ and ζ− the corresponding measures. As a consequence of Proposition 3.2 and of the concerned definitions we can al- ready stateζ=ζ++ζ− andproj2#ζ+= proj2#ζ−=i∗R(h)/2 dh. The next result indicates the other projections.
Proposition 3.3. The measuresζ+andζ−defined as in the previous paragraph satisfyproj1#ζ+=µ andproj1#ζ−=ν.
Proof. From (6) and (9), computing the half sum and the half difference we already have
(T V+)ts(Fσ) = Z
R
i∗,+]s,t](h) dh and (T V−)ts(Fσ) = Z
R
i∗,−]s,t](h) dh. (10) Therefore,
(T V+)ts(Fσ) = Z
R
card(x∈]s, t] : (x, h)∈Graph+(Fσ∗)) dh (11)
= Z +∞
−∞
card(x∈R: (x, h)∈(]s, t]×R)∩Graph+(Fσ∗)) dh (12)
= proj1#ζ+(]s, t]) (13)
and(T V−)ts(Fσ) = proj1#ζ−(]s, t]). Recall moreover the definitions
T Vst(F) = sup
m∈N, s=r0<...<rm+1=t m
X
j=0
|F(rk+1)−F(rk)|
(T V+)ts(F) = sup
m∈N, s=r0<...<rm+1=t m
X
j=0
[F(rk+1)−F(rk)]+
(T V−)ts(F) = sup
m∈N, s=r0<...<rm+1=t m
X
j=0
[F(rk+1)−F(rk)]−
(14)
where [u]+ = 12(|u|+u) and [u]− = 12(|u| −u) are the positive and negative parts ofu. Asµ⊥ν, there existsP withµ(P) =µ(R)andν(P) = 0. By outer
