Variational representations for N-cyclically monotone vector fields

0.15 MB·English

by  Alfred Galichon

N
Variational representations for -cyclically monotone vector fields
Alfred Galichon∗ Nassif Ghoussoub†
Economics Department Department of Mathematics
Ecole Polytechnique University of British Columbia
2 91128 Palaiseau cedex France Vancouver BC Canada V6T 1Z2
1 [email protected] [email protected]
0
2
July 11, 2012

l
u
J

0
1 Abstract

Givenadomain ΩinRd,weshowthataboundedmeasurable vectorfieldu:Ω→Rd isN-cyclically
]
C monotone for some N ≥ 2, if and only if there exists a Hamiltonian H, which is concave in the first
O variable, convex in the last (N −1) variables such that
h. (−u(x),0,...,0,u(x))∈∂H(x,x,...,x) for x∈Ω.
t
a N−1
m Moreover, H is N-subsymmetric, meaning that P H(σix) ≤ 0 for all x = (x1,...,xN) ∈ ΩN, σ being
i=0
[ the cyclic permutation on Rd defined by σ(x1,x2,...,xN) = (x2,x3,...,xN,x1). Furthermore, H is N-
symmetricin thefollowing sense
1
v
8 H(x1,x2,...,xN)+H2,...,N−1(x1,x2,...,xN)=0,
0 N−1
4 where H2,...,N is the concavification of the function K(x)= P H(σix) with respect to the last N −1
2 i=1
variables. This can be seen as an extension of a theorem of Krauss, which associates to any monotone
.
7 operator, a concave-convexanti-symmetric saddle function.
0
2
1 1 Introduction
:
v
i GivenadomainΩ inRd, recallthata single-valuedmapu fromΩ to Rd is saidto be N-cyclically monotone
X
if for every cycle x ,...,x ,x =x of points in Ω, one has
1 N N+1 1
r
a
N
hu(x ),x −x i≥0. (1)
i i i+1
i=1
X
A classical theorem of Rockafellar [10] states that a single-valued map u from Ω to Rd is N-cyclically
monotone for every N ≥2 if and only if
u(x)=∇φ(x) for all x∈Ω, (2)
where φ : Rd → R is a convex function. On the other hand, a result of E. Krauss [9] yields that u is a
monotone map, i.e., a 2-cyclically monotone map, if and only if
u(x)=∇ H(x,x) for all x∈Ω, (3)
2
∗GalichongratefullyacknowledgessupportfromChaireAxa“AssuranceetRisquesMajeurs”,ChaireEDF-Calyon“Finance
andD´eveloppement Durable”andFiME,LaboratoiredeFinancedesMarch´esdel’Energie(www.fime-lab.org).
†PartiallysupportedbyagrantfromtheNaturalSciences andEngineeringResearchCouncilofCanada.
1
where H is a concave-convexanti-symmetric Hamiltonian on Rd×Rd.
In this paper, we extend the result of Krauss to N-cyclically monotone vector fields, where N ≥ 3. To
formulate variational principles for such vector fields, we let σ to be the cyclic permutation on Rd, defined
by
σ(x1,x2,...xN−1,xN)=(x2,x3,...,xN,x1),
and consider the family of continuous N-cyclically symmetric Hamiltonians on ΩN, that is
N−1
H (Ω)={H ∈C(ΩN); H(σi(x ,...,x ))=0}
N 1 N
i=0
X
We shall say that H is N-cyclically subsymmetric if the equality above is replaced by ≤. We associate to
any function H on ΩN, the following functional on Ω×(Rd)N−1,
N−1
LH(x,p1,...,pN−1)=sup hpi,yii−H(x,y1,...,yN−1);yi ∈Ω . (4)
( )
i=1
X
Note that if Ω is convexandif H is convexinthe lastN−1 variables,then L is nothing but the Legendre
H
transform of H˜ with respect to the last N −1 variables, where H˜ = H on ΩN and = +∞ outside of ΩN.
Since H(x,...,x)=0 for any H ∈HN(Ω), then for any such H, we have for x∈Ω and p1,...,pN−1 ∈Rd,
N−1
LH(x,p1,...,pN−1)≥ hx,pii. (5)
i=1
X
We also consider the class of σ-invariant probability measures on ΩN, which are those π ∈ P(ΩN) such
that for all f ∈L1(ΩN,dπ), we have
f(x ,...,x )dπ = f(σ(x ,...,x ))dπ. (6)
1 N 1 N
ZΩN ZΩN
We then denote
P (ΩN)={π ∈P(ΩN); π σ-invariant probability on ΩN}. (7)
sym
For a given probability measure µ on Ω, we also consider the class
Pµ (ΩN)={π ∈P (ΩN); proj π =µ}, (8)
sym sym 1
i.e., the set of all π ∈P (ΩN) with a given first marginal µ, meaning that
sym
f(x )dπ(x ,...,x )= f(x )dµ(x ) for every f ∈L1(Ω,µ). (9)
ΩN 1 1 N Ω 1 1
Consider now the setRS(Ω) of µ-measure preserRving transformations on Ω, which can be identified with a
closedsubsetofthesphereofL2(Ω,Rd). WeshallalsoconsiderthesubsetofS(Ω)consistingofN-involutions,
that is
S (Ω)={S ∈S(Ω); SN =I a.e.}
N
The following lemma deals with those probabilities in Pµ (ΩN), that are carried by graphs of functions
sym
from Ω to ΩN.
Lemma 1 Let S :Ω→Ω be a µ-measurable map, then the following properties are equivalent:
1. The image of µ by the map x→(x,Sx,...,SN−1x) belongs to Pµ (ΩN).
sym
2. S is µ-measure preserving and SN(x)=x µ-a.e.
3. H(x,Sx,...,SN−1x)dµ(x)=0 for every H ∈L1(ΩN,⊗ dµ) that is N-cyclically symmetric on Ω.
Ω N
R
2
Proof. Itisclearthat1)implies3)since H(x)dπ(x)=0foranyN-cyclicallysymmetricHamiltonian
ΩN
H and any π ∈P (ΩN).
sym
2)implies1)isalsostraightforwardsinceRifπisthepush-forwardofµbythemapx→(x,Sx,...,SN−1x),
where S is µ-measure preserving and SN(x)=x µ-a.e., and for all f ∈L1(ΩN,dπ), we have
f(x ,...,x )dπ = f(x,Sx,...,SN−1x)dµ(x)= f(Sx,S2x,...,SN−1x,SNx)dµ(x)
1 N
ZΩN ZΩN ZΩN
= f(Sx,S2x,...,SN−1x,x)dµ(x) = f(σ(x ,...,x ))dπ.
1 N
ZΩN ZΩN
We now prove that 2) and 3) are equivalent. Assuming first that S is measure preserving such that SN =I
a.e, then for every N-symmetric H in L1(ΩN), we have
H(x,Sx,S2x,...SN−1x)dµ = H(Sx,S2x,...SN−1x,x)dµ
ZΩ ZΩ
= ...= H(SN−1x,x,Sx,...,SN−2x)dµ.
ZΩ
Since H is N-cyclically symmetric, then
H(x,Sx,...SN−1x)+H(Sx,S2x,...,SN−1x,x)+...H(SN−1x,x,Sx,..,SN−2x)=0.
It follows that N H(x,Sx,S2x,..,SN−1x)dµ=0.
Ω
For the reversRe implication, assume H(x,Sx,S2x,...,SN−1x)dµ = 0 for every N-cyclically antisym-
Ω
metric Hamiltonian H. By using that identity with the Hamiltonians
R
H(x ,x ,...,x )=f(x )−f(x )
1 2 N 1 i
where f is any continuous function on Ω, one gets that S is measure preserving. Now take the Hamiltonian
H(x ,x ,...,x )=|x −Sx |−|Sx −x |−|x −Sx |+|Sx −x |.
1 2 N 1 N 1 2 2 1 2 3
Note that H ∈ H (Ω) since it is of the form H(x ,...,x ) = f(x ,x ,x )−f(x ,x ,x ). Now apply the
N 1 N 1 2 N 2 3 1
identity for such an H to obtain
0= H(x,Sx,S2x,...,SN−1x)dµ= |x−SSN−1x|dµ.
ZΩ ZΩ
It follows that SN =I µ a.e., and we are done. (cid:3)
2 Characterizations of monotone vector fields
In order to simplify the exposition, we shall always assume in the sequel that dµ is Lebesgue measure dx
normalized to be a probability on Ω, and µ can and will then be dropped from all notation. We shall also
assume that Ω is convex and that its boundary has measure zero.
Theorem 2 Let u : Ω → Rd be a bounded measurable vector field. The following properties are then
equivalent:
1. u is N-cyclically monotone a.e.
2. The infimum of the following Monge-Kantorovich problem
inf{ hu(x ),x −x idπ(x); π ∈P (ΩN) and proj π =dx} (10)
ΩN 1 1 2 sym 1
is equal to zero, and isRtherefore attained by the push-forward of dx by the map x→(x,x,...,x).
3
3. The vector field u is in the polar of S (Ω), that is
N
inf{ hu(x),x−Sxidx;S ∈S (Ω)}=0. (11)
N
ZΩ
4. The projection of u on S (Ω) is the identity map, that is
N
inf{ |u(x)−Sx|2dx;S ∈S (Ω)}= |u(x)−x|2dx. (12)
N
ZΩ ZΩ
5. There exists a N-subsymmetric Hamiltonian H which is concave in the first variable, convex in the
last (N −1) variables such that
(−u(x),0,...,0,u(x))=∂H(x,x,...,x) for x∈Ω. (13)
Moreover, H is N-symmetric in the following sense
H(x1,x2,...,xN)+H2,...,N−1(x1,x2,...,xN)=0, (14)
N−1
whereH2,...,N−1 is theconcavification of thefunction K(x)= H(σix) withrespect tothelast N−1
i=1
variables. P
6. The following duality holds:
inf{ L (x,0,...,0,u(x))dx; H ∈H (Ω)}=sup{ hu(x),Sxidx;S ∈S (Ω)}
H N N
ZΩ ZΩ
and the latter is attained at the identity map.
Remark 3 NotethatinthecaseN =2,K(x)=H(x ,x )isconcavewithrespecttox ,henceH (x ,x )=
2 1 2 2 1 2
H(x ,x ), and (14) becomes
2 1
H(x ,x )+H(x ,x )=0,
1 2 2 1
thus H is skew-symmetric, recovering well-known results [9], [4], [7], [8].
Proof: To show that (1) implies (2), it suffices to notice that if π is a σ-invariant probability measure on
ΩN such that proj π =dx, then
1
N
1
hu(x ),x −x idπ(x ,...,x ) = hu(x ),x −x idπ(x ,...,x )
1 1 2 1 N N i i i+1 1 N
ZΩN i=1ZΩN
X
N
1
= hu(x ),x −x i dπ(x ,...,x ),
N i i i+1 1 N
ZΩN i=1 !
X
whichisnon-negativewheneveruisN-cyclicallymonotone. Ontheotherhand,ifπ istheinvariantmeasure
obtained by taking the image of µ:=dx by x→(x,...,x), then
hu(x ),x −x idπ(x ,...,x )=0.
1 1 2 1 N
ZΩN
To showthat(2)implies (3),let S be aµ-measurepreservingtransformationonΩ suchthat SN =Idµ a.e.
Then the image π of µ by the map
S
x→ x,Sx,S2x,...,SN−1x
is σ-invariant, hence (cid:0) (cid:1)
hu(x ),x −x idπ (x ,...,x )= hu(x),x−Sxidx≥0.
1 1 2 S 1 N
ZΩN ZΩ
4
Again, by taking S =I, we get that the infimum is necessarily zero.
That (3) and (4) are equivalent is a standard consequence of the Hahn-Banach theorem.
We now show that (3) implies (1). For that take N points x ,x ,...x in Ω, and let R>0 be such that
1 2 N
B(x ,R)⊂Ω. Consider the transformation
i
x−x +x for x∈B(x ,R)
1 2 1
x−x +x for x∈B(x ,R)
 2 3 2
S (x)= ...
R  x−xN +x1 for x∈B(xN,R)
x otherwise
It is easy to see that S is a measure preserving transformation and that SN =Id. We then have
R R
N
0≤ hu(x),x−S xidx≤ hu(x ),x −x idx.
R i i+1 i
ZΩ i=1ZB(xi,R)
X
Letting R→0, we get from Lebesgue’s density theorem, that
1
hu(x),x −x idx→hu(x ),x −x i,
|B(x ,R)| i i+1 i i i+1
i ZB(xi,R)
from which follows that T is N-cyclically monotone a.e.
In order to prove that (1) implies (5), we shall need the following lemma.
Lemma 4 Let f(x,y)=hu(x),x−yi and let f1(x,y) be its convexification in x for fixed y, that is
n n n
f1(x,y)=inf λ f(x ,y):λ ≥0, λ =1, λ x =x . (15)
k k k k k k
( )
k=1 k=1 k=1
X X X
Then, f1 satisfies the following properties:
1. f ≥f1 on Ω,
2. f1 is convex in the first variable and concave with respect to the second,
3. f1(x,x)=0 for each x∈Ω,
4. f1 is N-cyclically supersymmetric in the following sense: for any cyclic family x ,...,x ,x = x
1 N N+1 1
in Ω, we have
N
f1(x ,x )≥0. (16)
i i+1
i=1
X
Proof. Clearlyf ≥f1 andf1 isconvexwithrespecttox. f1 isalsoconcavewithrespecttoy,sincef itself
isconcave(actuallylinear)withrespecttoy. Notealsothatf(x,x)=0andsinceuisN-cyclicallymonotone,
the function f(x,y)=hu(x),x−yi satisfies N f(x ,x )≥0 for any cyclic family x ,...,x ,x =x
i=1 i i+1 1 N N+1 1
in Ω.
Fix now x ,x ,...,x ,x = x in ΩPand consider (xk)n in Ω, and (λ ) such λ ≥ 0 such that
1 2 N N+1 1 1 k=1 k k k
n λ =1 and n λ xk =x . For each k, we have
k=1 k k=1 k 1 1
P P N−1
f(xk,x )+ f(x ,x )+f(x ,xk)≥0,
1 2 i i+1 N 1
i=2
X
5
thus, summing over k, and using that f is linear in the second variable, we have
n N−1
λ f(xk,x )+ f(x ,x )+f(x ,x )≥0,
k 1 2 i i+1 N 1
k=1 i=2
X X
hence
N
f1(x ,x )+ f(x ,x )≥0.
1 2 i i+1
i=2
X
n
Let now λ ≥0, xk ∈Ω be such that n λ =1 and λ xk =x . We have
k N k=1 k k N N
k=1
P P
N−2
f1(x1,x2)+ f(xi,xi+1)+f(xN−1,xkN)+f(xkN,x1)≥0.
i=2
X
Multiplying by λ , summing over k and using again that f is linear in the second variable, we obtain
k
N−1 n
f1(x ,x )+ f(x ,x )+ λ f(xk ,x )≥0.
1 2 i i+1 k N 1
i=2 k=1
X X
Hence, by taking the infimum over all possible such choices, we get
N−1
f1(x ,x )+ f(x ,x )+f1(x ,x )≥0.
1 2 i i+1 N 1
i=2
X
Repeat this procedure with xN−1,xN−2,...,x3 until we get
N
f1(x ,x )+f(x ,x )+ f1(x ,x )≥0.
1 2 2 3 i i+1
i=3
X
Note that f1 is not necessarilylinear in the second variable,we cannot apply the same reasoningonx , but
2
we can argue in the following way. Since
N
f(x ,x )≥−f1(x ,x )− f1(x ,x ),
2 3 1 2 i i+1
i=3
X
andsincef1isconcaveinthesecondvariable,wehaveforfixedx ,x ,...,x ,thefunctionx →−f1(x ,x )−
1 3 N 2 1 2
N f1(x ,x ) is a convex minorant of x →f(x ,x ). It follows that
i=3 i i+1 2 2 3
P N
f(x ,x )≥f1(x ,x )≥−f1(x ,x )− f1(x ,x ),
2 3 2 3 1 2 i i+1
i=3
X
which finally implies that N f (x ,x )≥0.
i=1 1 i i+1
This clearly implies that f1(x,x)≥0 for any x∈Ω. On the other hand, since f1(x,x)≤f(x,x)=0,we
get that f1(x,x)=0 for alPl x∈Ω.
Proof of 1) implies 5) Let ψ(x,y)=−f1(y,x) and note that
(i) x→ψ(x,y) is convex,
(ii) y →ψ(x,y) is concave,
(iii) ψ(x,y)≥−f(y,x)=hu(y),y−xi, and
6
(iv) ψ is N-subsymmetric, meaning that ψ(x,x)=0 and
N
ψ(x ,x )≤0 for all cyclic families x ,...,x ,x =x in Ω. (17)
i i+1 1 N N+1 1
i=1
P
Consider now the family H of functions H :ΩN →R such that
1. H(x ,x ,...,x )≥ψ(x ,x ) for all x ,...,x in Ω.
1 2 N N 1 1 N
2. H is concave in the first variable,
3. H is jointly convex in the last N −1 variables.
4. H is N-subsymmetric in the following sense: H(x,...,x)=0 and
N
H(σi−1x)≤0 for all x=(x ,...,x )∈ΩN. (18)
1 N
i=1
P
Note that H =6 ∅ since H(x ,x ,...,x ) := ψ(x ,x ) belongs to H. Moreover, by N-subsymmetry, any
1 2 N N 1
H ∈H satisfies for all x=(x ,...,x )∈ΩN,
1 N
N−1 N−1
H(x)≤− H(σi(x)≤− ψ(x ,x ). (19)
i i+1
i=1 i=1
X X
This also yields that
N−1
hu(x ),x −x i≤H(x)≤− hu(x ),x −x i, (20)
1 N 1 i+1 i i+1
i=1
X
which means that H(x,x,...,x)=0 for every H ∈H and any x∈Ω.
On the other hand, it is easy to see that every directed family (Hi)i in H has a supremum H∞ ∈ H,
meaning that H is a Zorn family, and therefore it has a maximal element H.
Consider now the function
(N −1)H(x)− N−1H(σi(x)
H¯(x)= i=1 ,
N
P
and note that
(i) H¯ is N-symmetric, since
N−1
1
H¯(x)= [H(x)−H(σix)],
N
i=1
X
and each K (x):=H(x)−H(σix) is N-symmetric.
i
(ii) H¯ ≥H on ΩN, since
N−1
N[H¯(x)−H(x)]=− H(σi(x)≥0,
i=0
X
because H itself is N-subsymmetric.
The maximality of H would have implied that H = H¯ is N-symmetric if only H¯ was jointly convex in
the last N −1-variables, but since this is not necessarily the case, we consider for x = (x ,x ,...,x ), the
1 2 N
function
N−1
K(x ,x ,...,x )=K(x):=− H(σi(x)),
1 2 N
i=1
X
7
which is already concave in the first variable x . Its convexification in the last (N −1)-variables, that is
1
n n
K2,...N(x)=inf λ K(x ,xi,...,xi ); λ ≥0, λ (xi,...,xi ,1)=(x ,...,x ,1) ,
i 1 2 N i i 2 N 2 N
( )
i=1 i=1
X X
is still concave in the first variable, but is now convex in the last N −1 variables. Moreover,
N−1
H ≤K2,...N ≤K =− H ◦σi. (21)
i=1
X
Indeed, K2,...N ≤ K from the definition of K2,...N, while H ≤ K2,...N because H ≤ K and H is already
convex in the last (N −1)-variables.
It follows that
N−1
(N −1)H − H ◦σi
(N −1)H +K2,...N (N −1)H +K
H ≤ ≤ = i=1 =H¯.
N N N P
The function H′ = (N−1)H+K2,...N belongs to the family H and therefore H =H′ by the maximality of H.
N
This finally yields that H is N-subsymmetric, that H(x,x,x)=0 for all x∈Ω and that
H(x)+H (x)=0 for every x∈ΩN,
2,...,N
where H = −K2,...N, which is the concavification of x → N−1H(σix) with respect to the last
2,...,N i=1
N −1-variables.
P
Note now that since for any x ,...,x in Ω, H(x ,...x ) ≥ hu(x ),x −x i, and H(x ,x ,x ) = 0, we
1 N 1 N 1 N 1 1 1 1
have
H(x ,...x )−H(x ,...,x )≥hu(x ),x −x i.
1 N 1 1 1 N 1
Since H is convex in the last N −1 variables, this means that for all x∈Ω, we have
(0,...,0,u(x))∈∂ H(x,x,...,x). (22)
2,3
On the other hand, (20) gives that for any x ,...,x in Ω,
1 N
N−1
hu(x ),x −x i≤H(x)≤− hu(x ),x −x i, (23)
1 N 1 i+1 i i+1
i=1
X
which means that for any x,y in Ω, we have
H(y,x,x,...x)−H(x,x,x)≤−hu(x),x−yi.
Since H is concave in the first variable, we deduce that
−u(x)∈∇ H(x,x,x), (24)
1
and consequently
(−u(x),0,...,0,u(x))∈∂H(x,x,...,x). (25)
To prove that 5) implies 6) note that
N−1
LH(x,p1,...,pN−1)+H(x,y1,...,yN−1)≥ hpi,yii,
i=1
X
which yields that
[L (x,0,...,0,u(x))dx+H(x,SN−1x,...,Sx)]dx≥ hu(x),Sxi.
H
ZΩ ZΩ
8
If now H ∈H (Ω) and S ∈S (Ω), we then have H(x,SN−1x,...,Sx)dx=0, and therefore
N N Ω
R
L (x,0,...,0,u(x))dx≥ hu(x),Sxidx.
H
ZΩ ZΩ
IfnowH istheN-subsymmetricHamiltonianH obtainedby5),whichisconcaveinthefirstvariable,convex
in the last N −1 variables and such that
(−u(x),0,...,0,u(x))=∇H(x,x,...,x) for all x∈Ω,
then
L (x,0,...,0,u(x))+H(x,x,...x)=hu(x),xi for all x∈Ω,
H
and therefore L (x,0,...,0,u(x))dx= hu(x),xidx.
Ω H Ω
Consider now
R (RN −1)H(x)− N−1H(σi(x))
H¯(x)= i=1 .
N
P
As in the proof of 1) implies 5), we have that H¯ ∈ HN(Ω) and H¯ ≥ H. Since LH¯ ≤ LH, we have that
ΩLH¯(x,0,...,0,u(x))dx= Ωhu(x),xidx and (6) is proved.
R R
Finally, note that (6) readily implies (3), which means that u is then N-cyclically monotone.
Remark 5 Note that the sets of measure preserving N-involutions (S (Ω)) do not form a nested family,
N N
that is S (Ω) is not necessarily included in S (Ω), whenever N ≤ M, unless of course M is a multiple of
N M
N. On the other hand, the above theorem shows that their polar sets, that is
S (Ω)0 ={u∈L2(Ω,Rd); hu(x),x−Sxidx≥0 for all S ∈S (Ω)},
N Ω N
which coincide with the N-cyclically monotonRe maps, satisfy for every N ≥1,
S (Ω)0 ⊂S (Ω)0.
N+1 N
Thiscanalsobe seendirectly. Indeed,itis clearthata2-involutionisa4-involutionbutnotnecessarilya3-
involution. Ontheotherhand,assumethatuis3-cyclicallymonotoneoperator,thenforanytransformation
S :Ω→Ω, we have
hu(x),x−Sxidx+ hu(Sx),Sx−S2xidx+ hu(S2x),S2x−xidx≥0.
ZΩ ZΩ ZΩ
If now S is measure preserving, we have
hu(x),x−Sxidx+ hu(x),x−Sxidx+ hu(S2x),S2x−xidx≥0,
ZΩ ZΩ ZΩ
and if S2 = I, then hu(x),x−Sxidx ≥ 0, which means that u ∈ S (Ω)0. Similarly, one can show that
Ω 2
any (N +1)-cyclically monotone operator belongs to S (Ω)0. In other words, S (Ω)0 ⊂ S (Ω)0 for all
N N+1 N
N ≥2. Note that S (RΩ)0 =L2(Ω,Rd), while
1
S(Ω)0 =∩ S (Ω)0 ={u∈L2(Ω,Rd),u=∇φ where φ is a convex function in W1,2(Rd)},
N N
in view of classical results of Rockafellar [11] and Brenier [1].
References
[1] Y. Brenier, Polar factorization and monotone rearrangement of vector-valued functions, Comm. Pure
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9
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