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Sweeping Processes Perturbed by Rough Signals

Charles Castaing, Nicolas Marie, Paul Raynaud de Fitte

To cite this version:

Charles Castaing, Nicolas Marie, Paul Raynaud de Fitte. Sweeping Processes Perturbed by Rough Sig-

nals. Séminaire de Probabilités LI, Lecture Notes in Mathematics, Springer, inPress. �hal-01738241v3�

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Sweeping Processes Perturbed by Rough Signals

Charles CASTAING, Nicolas MARIE and Paul RAYNAUD DE FITTE

Abstract This paper deals with the existence, the uniqueness and an approximation scheme of the solution to sweeping processes perturbed by a continuous signal of finite p-variation with p ∈ [ 1 , 3 [ . It covers pathwise stochastic noises directed by a fractional Brownian motion of Hurst parameter greater than 1 / 3.

1 Introduction

Consider a multifunction C : [ 0 , T] ⇒ R e with e ∈ N . Roughly speaking, the Moreau sweeping process (see Moreau [20]) associated to C is the path X , living in C , such that when it hits the frontier of C , a minimal force is applied to X in order to keep it inside of C . Precisely, X is a solution to the following differential inclusion:

 

 

− dDX

d |DX | (t) ∈ N C(t) (X(t)) |DX | -a.e.

X ( 0 ) = a ∈ C( 0 )

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where DX is the differential measure associated with the continuous function of bounded variation X , |DX | is its variation measure, and N C(t) (X (t)) is the normal cone of C(t) at X (t) . This problem has been deeply studied by many authors. For instance, the reader can refer to Moreau [20], Valadier [27] or Monteiro Marques [19].

Charles CASTAING

Département de Mathématiques, Université Montpellier 2, Montpellier, France Nicolas MARIE

Laboratoire Modal’X, Université Paris Nanterre / ESME Sudria, Paris, France e-mail:

nmarie@parisnanterre.fr Paul RAYNAUD DE FITTE

Laboratoire Raphaël Salem, Université de Rouen Normandie, Rouen, France e-mail: prf@univ- rouen.fr

1

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Several authors studied some perturbed versions of Problem (1), in particular by a stochastic multiplicative noise in Itô’s calculus framework (see Revuz and Yor [22]). For instance, the reader can refer to Bernicot and Venel [3] or Castaing et al. [5]. On reflected diffusion processes, which are perturbed sweeping processes with constant constraint set, the reader can refer to Kang and Ramanan [13].

The general way to formulate the perturbed Problem (1) when the perturbation has unbounded variation is to split the unknown X in two parts, following ideas of Skorokhod [24, 25]: one part has bounded variation and represents the “pure”

sweeping process, and the other one may have unbounded variation and represents the perturbed part. In this line, we consider the perturbed sweeping process

 

 

 

 

 

X(t) = H(t) + Y (t) H(t) = ∫ t

0 f (X(s))dZ (s)

− dDY

d |DY | (t) ∈ N C

H

(t) (Y (t)) | DY| -a.e. with Y ( 0 ) = a

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where X, Y and H are unknown, C H (t) = C(t) − H(t) , t ∈ [ 0 ,T] (thus N C

H

( t ) (Y (t)) = N C ( t ) (X (t)) ), Z : [ 0 ,T ] → R d is a continuous signal of finite p-variation with d ∈ N and p ∈ [ 1 , ∞[ , f ∈ Lip γ ( R e , M e,d ( R )) with γ > p , and the integral against Z is taken in the sense of rough paths (see Section 2 for precise definitions)). On the rough integral, the reader can refer to Lyons [16], Friz and Victoir [11] or Friz and Hairer [9]. Throughout the paper, the multifunction C satisfies the following assumption.

Assumption 1 C is a convex compact valued multifunction, continuous for the Haus- dorff distance, and there exists a continuous selection γ : [ 0 ,T ] → R e satisfying

B e (γ(t),r) ⊂ int (C(t)) ; ∀t ∈ [ 0 ,T ],

where B e (γ(t),r) denotes the closed ball of radius r centered at γ(t).

This assumption is equivalent to saying that C(t) has nonempty interior for every t ∈ [ 0 ,T ] , see [5, Lemma 2.2].

In Falkowski and Słomiński [8], when p ∈ [ 1 , 2 [ and C(t) is a cuboid of R e for

every t ∈ [ 0 ,T ] , the authors proved the existence and uniqueness of the solution of

Problem (2). Furthermore, several authors studied the existence and uniqueness of

the solution for reflected rough differential equations. In [1], M. Besalú et al. proved

the existence and uniqueness of the solution for delayed rough differential equations

with non-negativity constraints. Recently, S. Aida gets the existence of solutions for

a large class of reflected rough differential equations in [2] and [1]. Finally, in [6],

A. Deya et al. proved the existence and uniqueness of the solution for 1-dimensional

reflected rough differential equations. An interesting remark related to these refer-

ences is that when C is not a cuboid, moving or not, it is a challenge to get the

uniqueness of the solution for reflected rough differential equations and sweeping

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processes.

For p ∈ [ 1 , 3 [ , the purpose of this paper is to prove the existence of solutions to Problem (2) when C satisfies Assumption 1, and a necessary and sufficient condition for uniqueness, close to the monotonicity of the normal cone which allows to prove the uniqueness in the case when p = 1 and there is an additive continuous signal of finite q -variation with q ∈ [ 1 , 3 [ . In that, the convergence of an approximation scheme is also proved.

Section 2 deals with some preliminaries on sweeping processes and the rough integral. Section 3 is devoted to the existence of solutions to Problem (2) when Z is a moderately irregular signal (i.e. p ∈ [ 1 , 2 [ ) and when Z is a rough signal (i.e. p ∈ [ 2 , 3 [ ). Section 4 deals with some uniqueness results. The convergence of an approximation scheme based on Moreau’s catching up algorithm is proved in Section 5 in the case when p = 1 and there is an additive continuous signal of finite q -variation with q ∈ [ 1 , 3 [ . Finally, Section 6 deals with sweeping processes perturbed by a pathwise stochastic noise directed by a fractional Brownian motion of Hurst parameter greater than 1 / 3.

The following notations, definitions and properties are used throughout the paper.

Notations and elementary properties:

1. C h (t) := C(t) − h(t) for every function h : [ 0 ,T ] → R e .

2. N C (x) is the normal cone of C at x , for any closed convex subset C of R e and any x ∈ R e (recall that N C (x) = ∅ if x < C ).

3. ∆ T := {(s,t) ∈ [ 0 ,T ] 2 : s < t } and ∆ s,t := {(u, v) ∈ [s, t] 2 : u < v} for every (s,t) ∈ ∆ T .

4. For every function x from [ 0 , T] into R d and (s,t) ∈ ∆ T , x(s,t) := x(t) − x(s) . 5. Consider (s, t) ∈ ∆ T . The vector space of continuous functions from [s, t] into R d

is denoted by C 0 ([s, t], R d ) and equipped with the uniform norm k.k ∞ ,s,t defined by

k xk ∞ ,s,t := sup

u ∈[ s,t ]

kx(u)k

for every x ∈ C 0 ([s, t], R d ) , or the semi-norm k.k 0,s,t defined by kxk 0,s,t := sup

u,v∈[s,t] kx(v) − x(u)k

for every x ∈ C 0 ([s, t], R d ) . Moreover, k.k ∞,T := k.k ∞,0,T , k.k 0,T := k.k 0,0,T and C 0 0 ([s,t], R d ) := { x ∈ C 0 ([s,t], R d ) : x( 0 ) = 0 }.

6. Consider (s, t) ∈ ∆ T . The set of all dissections of [s,t] is denoted by D [s,t] and

the set of all strictly increasing sequences (s n ) n ∈ N of [s,t] such that s 0 = s and

lim ∞ s n = t is a denoted by D ∞,[s,t] .

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7. Consider (s, t) ∈ ∆ T . A function x : [s, t] → R d has finite p -variation if and only if,

k xk p-var,s,t := sup

 

 

n − 1

Õ

k =1

kx(t k , t k+1 )k p

1 /p

; n ∈ N and (t k ) k ∈ n1,n o ∈ D [s,t]

 

 

< ∞.  Consider the vector space

C p -var ([s,t], R d ) := { x ∈ C 0 ([s,t], R d ) : k xk p-var,s,t < ∞}.

The map k.k p-var,s,t is a semi-norm on C p-var ([s, t], R d ) . Moreover, k.k p-var,T := k.k p-var,0,T .

Remarks:

a. For every q, r ∈ [ 1 ,∞[ such that q > r,

∀x ∈ C r-var ([s,t], R d ) , k xk q-var,s,t 6 k xk r-var,s,t .

In particular, any continuous function of bounded variation on [s, t] belongs to C q-var ([s, t], R d ) for every q ∈ [ 1 , ∞[ .

b. For every (s, t) ∈ ∆ T and x ∈ C 1-var ([s,t], R ) , k xk 1-var,s,t = ∫ t s

|Dx|,

where |Dx| is the variation measure of the differential measure Dx associated with x .

8. The vector space of Lipschitz continuous maps from R e into M e,d ( R ) is denoted by Lip ( R e , M e,d ( R )) and equipped with the Lipschitz semi-norm k.k Lip defined by

kϕk Lip := sup kϕ(y) − ϕ(x)k

ky − xk ; x, y ∈ R e and x , y

for every ϕ ∈ Lip ( R e , M e,d ( R )) . 9. For every λ ∈ R,

bλc := max {n ∈ Z : n < λ}

and {λ} := λ − bλc .

10. Consider γ ∈ [ 1 , ∞[ . A continuous map ϕ : R e → M d,e ( R ) is γ -Lipschitz in the sense of Stein if and only if,

k ϕk Lip

γ

:= k D bγc ϕk { γ } -Höl ∨ max {kD k ϕk ∞ ; k ∈ n0 , bγc o }

< ∞.

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Consider the vector space

Lip γ ( R e ,M e,d ( R )) := {ϕ ∈ C 0 ( R e ,M e,d ( R )) : kϕk Lip

γ

< ∞}.

The map k.k Lip

γ

is a norm on Lip γ ( R e , M e,d ( R )) . Remarks :

a. If ϕ ∈ Lip γ ( R e ,M e,d ( R )) , then ϕ ∈ Lip ( R e ,M e,d ( R )) .

b. If ϕ ∈ C bγc +1 ( R e ,M e,d ( R )) is bounded with bounded derivatives, then ϕ ∈ Lip γ ( R e , M e,d ( R )) .

2 Preliminaries

This section deals with some preliminaries on sweeping processes and the rough integral. The first subsection states some fundamental results on unperturbed sweep- ing processes coming from Moreau [20], Valadier [27] and Monteiro Marques [19].

A continuity result of Castaing et al. [5], which is the cornerstone of the proofs of Theorem 3 and Theorem 4, is also stated. The second subsection deals with the integration along rough paths. In this paper, definitions and propositions are stated as in Friz and Hairer [9], in accordance with M. Gubinelli’s approach (see Gubinelli [12]).

2.1 Sweeping processes

The following theorem, due to Monteiro Marques [17, 18, 19] using an estimation due to Valadier (see [4, 27]), states a sufficient condition of existence and uniqueness of the solution of the unperturbed sweeping process defined by Problem (1).

Proposition 1 Assume that C is a convex compact valued multifunction, continuous for the Hausdorff distance, and such that there exists (x,r) ∈ R e ×] 0 , ∞[ satisfying

B e (x,r) ⊂ C(t) ; ∀t ∈ [ 0 ,T ].

Then Problem (1) has a unique continuous solution of finite 1 -variation y : [ 0 ,T ] → R e such that

k yk 1-var,T 6 l(r, ka − xk), where l : R 2 + → R + is the map defined by

l(s, S) :=

 

 

 max

0 , S 2 − s 2 2 s

if e > 1

max { 0 , S − s} if e = 1 ; ∀s, S ∈ R + .

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This proposition is a consequence of the two following ones. These two propositions are also used in Section 5.

Proposition 2 Under Assumption 1, a map y : [ 0 ,T ] → R e is a solution of Problem (1) if it satisfies the two following conditions:

1. For every t ∈ [ 0 ,T ], y(t) ∈ C(t).

2. For every (s,t) ∈ ∆ T and z ∈ ∩ τ∈[s,t] C(τ), hz, y(t) − y(s)i > 1

2 (ky(t)k 2 − ky(s)k 2 ).

Let us denote by proj A (.) the orthogonal projection on a convex set A ⊂ R e . Proposition 3 Consider n ∈ N , (t 0 n , . . . , t n n ) the dissection of [ 0 ,T ] of constant mesh T/n and the step function Y n defined by

 

 

Y 0 n := a

Y k+1 n = proj C(t

kn+1

) (Y n

k ) ; k ∈ n0 , n − 1o Y n (t) := Y k n ; t ∈ [t n

k ,t k+1 n [, k ∈ n0 , n − 1o .

1. Under the conditions of Proposition 1 on C, kY n k 1-var,T 6 l(r, ka − xk).

2. Under Assumption 1, for every m ∈ n N and t ∈ [ 0 ,T ], there exist i ∈ n1 ,n o and j ∈ n1 , m o such that t ∈ [t i− n 1 ,t i n [, t ∈ [t m j− 1 ,t j m [ and

kY n (t ) − Y m (t)k 2 6 2 d H (C(t i n ), C(t m j ))(kY n k 1-var,t + kY m k 1-var,t ).

See Monteiro Marques [19], Chapter 2 for the proofs of the three previous proposi- tions.

Let h be a continuous function from [ 0 , T] into R e such that h( 0 ) = 0. If it ex- ists, a Skorokhod decomposition of (C, a, h) is a couple (v h , w h ) such that:

 

 

v h (t) = h(t) + w h (t)

− dDw h

d |Dw h | (t) ∈ N C

h

( t ) (w h (t)) |Dw h | -a.e. with w h ( 0 ) = a (3) where v h and w h are continuous, and w h has bounded variation. Since N C

h

(t) (x) = ∅ when x < C h (t) , the system (3) implies that, |Dw h | -a.e., w h (t ) ∈ C h (t) , that is, v h (t) ∈ C(t) . Under Assumption 1, by Proposition 1 together with Castaing et al. [5, Lemma 2.2], (C,a, h) has a unique Skorokhod decomposition (v h , w h ) .

Theorem 1 Under Assumption 1, if (h n ) n∈ N is a sequence of continuous functions from [ 0 ,T] into R e which converges uniformly to h ∈ C 0 ([ 0 ,T ], R e ), then

sup

n∈ N

kw h

n

k 1-var,T < ∞

and

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(v h

n

, w h

n

) −−−−−→ k . k

∞,T

n→∞ (v h ,w h ).

See Castaing et al. [5, Theorem 2.3].

Under Assumption 1, note that there exist R > 0, N ∈ N and a dissection (t 0 , . . . , t N ) of [ 0 ,T ] such that

B e (γ(t k ), R) ⊂ C(u) (4) for every k ∈ n0 , N − 1o and u ∈ [t k , t k +1 ] .

Proposition 4 Under Assumption 1:

1. The map (v . ,w . ) is continuous from

C 0 0 ([ 0 ,T ], R e ) to C 0 ([ 0 ,T ], R e ) × C 1-var ([ 0 ,T ], R e ).

2. Consider (s,t) ∈ ∆ T and ρ ∈] 0 , R/ 2 ] where R is defined in (4). For every h ∈ C 0 0 ([s, t], R e ) such that khk 0,s,t 6 ρ,

kw h k 1-var,s,t 6 M(ρ) with

M(ρ) := N 2 ρ sup

u∈[ 0,T ]

x,y∈C(u) sup k x − yk 2 .

Proof Refer to Castaing et al. [5, Lemma 5.3] for a proof of the first point.

Let us insert s and t in the dissection (t 0 , . . . , t N ) of [ 0 ,T ] and define k(s), k(t) ∈ n0 , N + 2o by

t k (s) := s and t k(t) := t . Consider k ∈ n k(s), k(t) − 1o and u ∈ [t k , t k+1 ] . On the one hand,

B e (γ(t k ) − h(t k ), ρ) ⊂ B e (γ(t k ) − h(u), R) ⊂ C h (u).

So,

B e (γ(t k ) − v h (t k ), ρ) ⊂ C(u) − h(t k ,u) − v h (t k ).

On the other hand,

v h (t k , u) = h(t k , u) + w h,t

k

(u) with

w h,t

k

(u) := w h (u) − w h (t k ).

Moreover,

− dDw h

d |Dw h | (u) ∈ N C(u)−h(u) (w h (u)) | Dw h | -a.e.

and then,

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 

 

− dDw h,t

k

d| Dw h,t

k

| (u) ∈ N C(u)−h(t

k

,u)−v

h

(t

k

) (w h,t

k

(u)) |Dw h,t

k

| -a.e.

w h,t

k

(t k ) = 0 . So, by Proposition 1:

kw h k 1-var,t

k

,t

k+1

= kw h,t

k

k 1-var,t

k

,t

k+1

6 l(ρ, kγ(t k ) − v h (t k )k).

Therefore,

kw h k 1-var,s,t =

k(t)− 1

Õ

k=k(s)

kw h k 1-var,t

k

,t

k+1

6 N sup

u ∈[s,t] l(ρ, kγ(u) − v h (u)k)

6 M(ρ).

2.2 Young’s integral, rough integral

The first part of the subsection deals with the definition and some basic properties of Young’s integral which allow to integrate a map y ∈ C r-var ([ 0 ,T],M e,d ( R )) with respect to z ∈ C q -var ([ 0 ,T ], R d ) when q, r ∈ [ 1 , ∞[ and 1 /q + 1 /r > 1. The second part of the subsection deals with the rough integral which extends Young’s integral when the condition 1 /q + 1 /r > 1 is not satisfied anymore. The signal z has to be enhanced as a rough path.

Definition 1 A map ω : ∆ T → R + is a control function if and only if, 1. ω is continuous.

2. ω(s, s) = 0 for every s ∈ [ 0 ,T] . 3. ω is super-additive:

ω(s, u) + ω(u, t) 6 ω(s, t) for every s, t, u ∈ [ 0 , T] such that s 6 u 6 t .

Example. Let p > 1. For every z ∈ C p -var ([ 0 ,T ], R d ) , the map ω p,z : (s,t) ∈ ∆ T 7−→ ω p,z (s, t) := kzk p-var,s,t p is a control function.

Proposition 5 Let p > 1 . Consider x ∈ C 0 ([ 0 ,T ], R d ) and a sequence (x n ) n∈ N of elements of C p-var ([ 0 ,T ], R d ) such that

n→∞ lim kx n − xk ∞ ,T = 0 and sup

n∈ N

kx n k p-var,T < ∞.

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Then x ∈ C p-var ([ 0 , T], R d ) and

n→∞ lim k x n − xk ( p+ε ) -var,T = 0 ; ∀ε > 0 . See Friz and Victoir [11, Lemma 5.12 and Lemma 5.27] for a proof.

Proposition 6 (Young’s integral) Consider q,r ∈ [ 1 ,∞[ such that 1 /q +1 /r > 1 , and two maps y ∈ C r-var ([ 0 ,T ], M e,d ( R )) and z ∈ C q-var ([ 0 ,T ], R d ). For every n ∈ N and (t n

k ) k ∈ n1,no ∈ D [ 0,T ] , the limit

n→∞ lim

n− 1

Õ

k=1

y(t k n )z(t k n , t k+1 n )

exists and does not depend on the dissection (t k n ) k ∈ n1,n o . That limit is denoted by

∫ T

0 y(s)dz(s)

and called Young’s integral of y with respect to z on [ 0 ,T]. Moreover, there exists a constant c(q,r) > 0, depending only on q and r, such that for every (s, t) ∈ ∆ T ,

∫ .

0 y(s)dz(s)

r-var,s,t 6 c(q,r)kzk q-var,s,t (k yk r-var,s,t + k yk ∞,s,t ).

See Lyons [16, Theorem 1.16], Lejay [14, Theorem 1] or Friz and Victoir [11, Theorem 6.8] for a proof.

Proposition 7 Consider q, r ∈ [ 1 , ∞[ such that 1 /q + 1 /r > 1 , two maps y ∈ C r-var ([ 0 ,T ], M e,d ( R )) and z ∈ C q-var ([ 0 ,T], R d ), and a sequence (y n ) n∈ N of elements of C r-var ([ 0 ,T],M e,d ( R )) such that:

n→∞ lim ky n − yk ∞,T = 0 and sup

n∈ N

ky n k r-var,T < ∞.

Then,

n→∞ lim

∫ .

0 y n (s)dz(s) −

∫ . 0

y(s)dz(s) ∞,T = 0 . See Friz and Victoir [11, Proposition 6.12] for a proof.

Consider p ∈ [ 2 , 3 [ and let us define the rough integral for continuous functions of finite p-variation.

Remark. In the sequel, the reader has to keep in mind that:

1. M e,d ( R ) R e ⊗ R d . 2. M d,1 ( R ) M 1,d ( R ) R d .

3. L( R d , M e,d ( R )) L( R d , L( R d , R e )) L( R d ⊗ R d , R e ) .

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Definition 2 Consider z ∈ C 1-var ([ 0 ,T ], R d ) . The step-2 signature of z is the map S 2 (z) : ∆ T → R d × M d ( R ) defined by

S 2 (z)(s,t) :=

z(s,t), ∫

s<u<v<t dz(v) ⊗ dz(u)

for every (s, t) ∈ ∆ T .

Notation. S T ( R d ) := {S 2 (z)( 0 , .) ; z ∈ C 1-var ([ 0 ,T ], R d )} .

Definition 3 The geometric p -rough paths metric space GΩ p,T ( R d ) is the closure of S T ( R d ) in C p -var ([ 0 ,T ], R d ) × C p/ 2-var ([ 0 ,T ], M d ( R )) .

Definition 4 For z ∈ C p-var ([ 0 ,T ], R d ) , a map y ∈ C p-var ([ 0 ,T ], M e,d ( R )) is con- trolled by z if and only if there exists y 0 ∈ C p-var ([ 0 ,T ], L( R d ,M e,d ( R ))) such that

y(s, t) = y 0 (s)z(s,t) + R y (s,t) ; ∀(s,t ) ∈ ∆ T

with kR y k p/ 2-var,T < ∞ . For fixed z , the pairs (y, y 0 ) as above define a vector space denoted by D z p/ 2 ([ 0 ,T ], M e,d ( R )) and equipped with the semi-norm k.k z,p/ 2,T such that k(y, y 0 )k z,p / 2,T := k y 0 k p -var,T + kR y k p / 2-var,T

for every (y, y 0 ) ∈ D z p/ 2 ([ 0 ,T ], M e,d ( R )) .

Theorem 2 (Rough integral) Consider z := (z, Z ) ∈ GΩ p,T ( R d ) and (y, y 0 ) ∈ D z p / 2 ([ 0 ,T],M e,d ( R )). For every n ∈ N and (t n

k ) k ∈ n1,no ∈ D [ 0,T ] , the limit

n→∞ lim

n− 1

Õ

k=1

(y(t k n )z(t k n ,t k+1 n ) + y 0 (t k n ) Z (t k n ,t k n +1 ))

exists and does not depend on the dissection (t k n ) k ∈ n1,n o . That limit is denoted by

∫ T

0 y(s)d z(s)

and called rough integral of y with respect to z on [ 0 ,T ]. Moreover,

1. There exists a constant c(p) > 0 , depending only on p, such that for every (s,t) ∈ ∆ T ,

∫ t s

y(u)dz(u) − y(s)z(s, t) − y 0 (s) Z (s, t)

6 c(p)(kzk p -var,s,t kR y k p / 2-var,s,t

+ k Z k p/ 2-var,s,t k y 0 k p-var,s,t ).

2. The map

(y, y 0 ) 7−→

∫ .

0 y(s)dz(s), y

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is continuous from D z p/ 2 ([ 0 ,T],M e,d ( R )) into D z p/ 2 ([ 0 ,T ], R e ).

See Friz and Shekhar [10, Theorem 34] for a proof with the p -variation topology, and see Gubinelli [12, Theorem 1] or Friz and Hairer [9, Theorem 4.10] for a proof with the 1 /p -Hölder topology.

Proposition 8 Consider z := (z, Z ) ∈ GΩ p,T ( R d ), a continuous map (y, y 0 ) : [ 0 ,T ] −→ M e,d ( R ) × L( R d , M e,d ( R )), and a sequence (y n , y n 0 ) n ∈ N of elements of D z p / 2 ([ 0 ,T ], M e,d ( R )) such that

(y n 0 , R y

n

) −−−−→ d

∞,T

n →∞ (y 0 , R y ) and sup

n ∈ N

k(y n , y n 0 )k z,p/ 2,T < ∞.

Then, (y, y 0 ) ∈ D z p/ 2 ([ 0 , T],M e,d ( R )) and

n→∞ lim

∫ .

0 y n (s)dz(s) −

∫ .

0 y(s)dz(s)

,T = 0 . Proof On the one hand, since

(y n 0 , R y

n

) −−−−→ d

∞,T

n→∞ (y 0 , R y ),

the function y is the uniform limit of the sequence (y n ) n∈ N . Moreover, since sup

n∈ N

k(y n , y n 0 )k z,p/ 2,T < ∞, by Proposition 5,

y 0 ∈ C p-var ([ 0 ,T ], L( R d , M e,d ( R ))) and R y ∈ C p / 2-var ([ 0 ,T ], M e,d ( R )).

So, (y, y 0 ) ∈ D z p/ 2 ([ 0 ,T ],M e,d ( R )) .

On the other hand, also by Proposition 5, for any ε > 0 such that p + ε ∈ [ 2 , 3 [ ,

n lim →∞ k(y n , y n 0 ) − (y, y 0 )k z,(p+ε)/ 2,T = lim

n →∞ k y n 0 − y 0 k (p+ε) -var,T + lim

n→∞ k R y

n

− R y k (p+ε)/ 2-var,T

= 0 .

So, by continuity of the rough integral (see Theorem 2),

n→∞ lim

∫ .

0 y n (s)d z(s), y n

∫ .

0 y(s)dz(s), y

z,(p+ε)/ 2,T = 0 .

Therefore, in particular:

(13)

n→∞ lim

∫ .

0 y n (s)dz(s) −

∫ .

0 y(s)dz(s)

∞,T = 0 .

Proposition 9 Consider z := (z, Z ) ∈ GΩ p,T ( R d ), (x, x 0 ) ∈ D z p/ 2 ([ 0 ,T ], R e ) and ϕ ∈ Lip γ 1 ( R e , R d ). The couple of maps (ϕ(x), ϕ(x) 0 ), defined by

ϕ(x)(t) := ϕ(x(t)) and ϕ(x) 0 (t) := Dϕ(x(t))x 0 (t) for every t ∈ [ 0 ,T ], belongs to D z p / 2 ([ 0 ,T ], M e,d ( R )).

Remark. By Theorem 2 and Proposition 9 together,

∫ .

0 ϕ(x(u))dz(u) is defined. For every (s,t) ∈ ∆ T , consider

I ϕ,z,x (s,t) :=

∫ t s

ϕ(x(u))dz(u) − ϕ(x(s))z(s, t) − Dϕ(x(s))x 0 (s) Z (s,t) .

For every (s,t) ∈ ∆ T , since

k ϕ(x)k p-var,s,t 6 kϕk Lip

γ−1

k xk p-var,s,t ,

kϕ(x) 0 k p-var,s,t 6 kϕk Lip

γ−1

(kx 0 k p-var,s,t + kx 0 k ∞,s,t k xk p-var,s,t ) and kR ϕ(x) k p/ 2-var,s,t 6 kϕk Lip

γ−1

(kxk 2 p-var,s,t + k R x k p/ 2-var,s,t ),

by Theorem 2,

I ϕ,z,x (s,t) 6 c(p)kϕk Lip

γ−1

(kx 0 k p-var,s,t + kx 0 k ∞,s,t kxk p-var,s,t + kxk 2 p-var,s,t + kR x k p/ 2-var,s,tp,z (s, t) 1 /p , where ω p,z : ∆ T → R + is the control function defined by

ω p,z (u, v) := 2 p 1 (kzk p-var,u,v p + k Z k p

p / 2-var,u,v ) ; ∀(u, v) ∈ ∆ T .

Proposition 10 Consider z := (z, Z ) ∈ GΩ p,T ( R d ), ϕ ∈ Lip γ 1 ( R e , R d ), a continu- ous map

(x, x 0 ) : [ 0 ,T ] −→ R e × L( R d , R e ), and a sequence (x n , x n 0 ) n ∈ N of elements of D z p/ 2 ([ 0 ,T ], R e ) such that

(x n 0 , R x

n

) −−−−→ d

∞,T

n→∞ (x 0 , R x ) and sup

n∈ N

k(x n , x n 0 )k z,p/ 2,T < ∞.

Then, (ϕ(x), ϕ(x) 0 ) ∈ D z p/ 2 ([ 0 ,T ], M e,d ( R )) and

(14)

n→∞ lim

∫ .

0 ϕ(x n (s))dz(s) −

∫ .

0 ϕ(x(s))dz(s)

∞,T = 0 . Proof Since

(x n 0 , R x

n

) −−−−→ d

∞,T

n→∞ (x 0 , R x ) and sup

n∈ N

k(x n , x n 0 )k z,p/ 2,T < ∞, by Friz and Hairer [9, Theorem 7.5] together with Proposition 5,

(ϕ(x n ) 0 , R ϕ(x

n

) ) −−−−→ d

∞,T

n→∞ (ϕ(x) 0 , R ϕ(x) ) and sup

n ∈ N

k(ϕ(x n ), ϕ(x n ) 0 )k z,p/ 2,T < ∞.

So, by Proposition 8,

n→∞ lim

∫ .

0 ϕ(x n (s))dz(s) −

∫ .

0 ϕ(x(s))dz(s)

∞,T = 0 .

3 Existence of solutions

The existence of a solution to Problem (2) is established in Theorem 3 when p ∈ [ 1 , 2 [ , and in Theorem 4 when p ∈ [ 2 , 3 [ .

Theorem 3 Under Assumption 1, if p ∈ [ 1 , 2 [ and f ∈ Lip γ ( R e , M e,d ( R )), then Problem (2) has at least one solution which belongs to C p-var ([ 0 ,T ], R e ).

Proof Consider the discrete scheme

 

 

 

 

 

X n (t) = H n (t) + Y n (t) H n (t) = ∫ t

0 f (X n− 1 (s))dZ (s)

− dDY n

d |DY n | (t) ∈ N C

Hn

(t) (Y n (t)) | DY n | -a.e. with Y n ( 0 ) = a

(5)

for Problem (2), initialized by

 

 

− dDX 0

d| DX 0 | (t ) ∈ N C(t) (X 0 (t)) | DX 0 | -a.e.

X 0 ( 0 ) = a

(6)

Since the map kZ k p-var,0,. is continuous from [ 0 ,T ] into R + , and since kZ k p-var,0,0 = 0, there exists τ 0 ∈ [ 0 ,T ] such that

k Z k p-var,τ

0

6 µ := m

c(p, p)k f k Lip

γ

(m + M + 1 ) ,

(15)

where m := R/ 2 and M := M(R/ 2 ) (see Proposition 4.(2)). Let us show that for every n ∈ N,

 

 

k X n k p-var,τ

0

6 m + M kH n k p-var,τ

0

6 m

kY n k 1-var,τ

0

6 M

(7) By (6) together with Proposition 4,

k X 0 k p-var,τ

0

6 M .

Assume that Condition (7) is satisfied for n ∈ N arbitrarily chosen. By Proposition 6, and since kZ k p-var,0,. is an increasing map,

kH n+1 k p-var,τ

0

6 c(p, p)kZ k p-var,τ

0

(kD f k kX n k p-var,τ

0

+ k f ◦ X n k ∞,τ

0

) 6 µc(p, p)k f k Lip

γ

(m + M + 1 )

6 m.

Since Y n +1 = w H

n+1

, by Proposition 4,

kY n+1 k 1-var,τ

0

6 M . Therefore,

k X n+1 k p-var,τ

0

6 k H n+1 k p-var,τ

0

+ kY n+1 k p-var,τ

0

6 m + M.

By induction, (7) is satisfied for every n ∈ N.

For every t ∈ [ 0 ,T ] , the map k Z k p-var,t,. is continuous from [t,T] into R + and kZ k p-var,t,t = 0. Moreover, the constant µ depends only on p , m , M and k f k Lip

γ

. So, since [ 0 , T] is compact, there exist N ∈ N and (τ k ) k n0, N o ∈ D

0

,T ] such that

k Z k p-var,τ

k

k+1

6 µ ; ∀k ∈ n0 , N − 1o . Since for every n ∈ N the maps

(s,t) ∈ ∆ T 7−→ k X n k p p-var,s,t , (s,t) ∈ ∆ T 7−→ k H n k p-var,s,t p and

(s,t) ∈ ∆ T 7−→ kY n k 1-var,s,t

are control functions, recursively, the sequence (H n , X n ,Y n ) n ∈ N

is bounded in C p,1

T := C p-var ([ 0 ,T ], R e ) × C p-var ([ 0 ,T ], R e ) × C 1-var ([ 0 ,T], R e ).

By Proposition 6, for every n ∈ N and (s, t) ∈ ∆ T , kH n (t) − H n (s)k 6 c(p, p)

k D f k ∞ sup

n∈ N

kX n k p-var,T + k f k ∞

kZ k p-var,s,t .

(16)

Since (s,t) ∈ ∆ T 7→ kZ k p-var,s,t is a continuous map such that kZ k p-var,t,t = 0 for every t ∈ [ 0 ,T ] , (H n ) n ∈ N

is equicontinuous. Therefore, by Arzelà-Ascoli’s theo- rem together with Proposition 5, there exists an extraction ϕ : N → N such that (H ϕ ( n ) ) n∈ N

converges uniformly to an element H of C p-var ([ 0 ,T], R e ) .

Since (H ϕ (n) ) n ∈ N

converges uniformly to H , by Theorem 1, (X ϕ (n) ,Y ϕ (n) ) n ∈ N

con- verges uniformly to (X,Y ) := (v H , w H ) . So, for every t ∈ [ 0 , T] ,

 

 

X(t) = H(t) + Y (t)

− dDY

d |DY | (t) ∈ N C

H

(t) (Y (t)) |DY | -a.e. with Y ( 0 ) = a, and by Proposition 5,

X ∈ C p-var ([ 0 ,T ], R e ) and Y ∈ C 1-var ([ 0 ,T ], R e ).

Moreover, since (X ϕ(n) ) n ∈ N

converges uniformly to X , by Proposition 7,

n→∞ lim

H ϕ ( n ) −

∫ .

0 f (X(s))dZ(s) ∞,T = 0 . Therefore, since (H ϕ(n) ) n∈ N

converges also to H in C 0 ([ 0 ,T ], R e ) ,

H(t) = ∫ t

0 f (X(s))dZ (s) ; ∀t ∈ [ 0 ,T ].

In the sequel, assume that there exists Z : [ 0 ,T ] → M d ( R ) such that Z := (Z, Z ) ∈ GΩ p,T ( R d ) .

Theorem 4 Under Assumption 1, if p ∈ [ 2 , 3 [ and f ∈ Lip γ ( R e , M e,d ( R )), then Problem (2) has at least one solution which belongs to C p-var ([ 0 ,T ], R e ).

Proof Consider the discrete scheme

 

 

 

 

 

X n (t) = H n (t) + Y n (t) H n (t) = ∫ t

0 f (X n − 1 (s))dZ(s)

− dDY n

d |DY n | (t) ∈ N C

Hn

(t) (Y n (t)) | DY n | -a.e. with Y n ( 0 ) = a

(8)

for Problem (2), initialized by

 

 

− dDX 0

d| DX 0 | (t ) ∈ N C ( t ) (X 0 (t)) | DX 0 | -a.e.

X 0 ( 0 ) = a

(9)

At step n ∈ N of the scheme, the integral involved in the definition of H n (t) ,

t ∈ [ 0 ,T ] , is the rough integral on [ 0 ,T ] of f (X n− 1 (·)) with respect to the geometric

(17)

rough path Z over Z (see Definition 2).

Since the map ω p,Z ( 0 , .) is continuous from [ 0 ,T ] into R + , and since ω p,Z ( 0 , 0 ) = 0, there exists τ 0 ∈ [ 0 ,T ] such that

ω p,Z ( 0 , τ 0 ) 6 m C

c(p, 1 ) p k f k Lip p

γ

(M C + 1 ) p ∧ m C

(c 2 ∨ c 6 ) p ( 1 + µ C + M R + µ 2 C ) p

∧ 1

1 + µ p C + M R p + µ 2p C , where m C := R/ 2, M C := M(R/ 2 ) , µ C := m C + M C ,

M R := (c 1 M C 4 1 /p ) ∨ (c 5C p + 1 ) 1 /p )

and the positive constants c 1 , c 2 , c 5 and c 6 , depending only on p and k f k Lip

γ

, are defined in the sequel.

First of all, let us control the solution of the discrete scheme for n ∈ { 0 , 1 } :

• ( n = 0) By (9) together with Proposition 4:

k X 0 k 1-var,τ

0

6 M C .

• ( n = 1) Since X 0 ∈ C 1-var ([ 0 ,T ], R e ) , by Proposition 6:

kH 1 k p -var,τ

0

6 c(p, 1 )ω p,Z ( 0 , τ 0 ) 1 /p k f k Lip

γ

(M C + 1 ) 6 m C .

Since Y 1 = w H

1

, by Proposition 4:

kY 1 k 1-var,τ

0

6 M C . Therefore,

kX 1 k p-var,τ

0

6 kH 1 k p-var,τ

0

+ kY 1 k p-var,τ

0

6 µ C . Let us show that for every n ∈ N \{ 0 , 1 } ,

(X n− 1 , f (X n− 2 )) ∈ D p/ Z 2 ([ 0 , τ 0 ], R e ) (10) and

 

 

 

 

k X n k p-var,τ

0

6 µ C kH n k p -var,τ

0

6 m C

kY n k 1-var,τ

0

6 M C

kR X

n

k p / 2-var,τ

0

6 M R

(11)

Set X 1 0 := f (X 0 ) . For every (s, t) ∈ ∆ τ

0

,

(18)

R X

1

(s,t ) = X 1 (s,t) − X 1 0 (s)Z(s, t)

= Y 1 (s, t) + ∫ t s

f (X 0 (u))dZ (u) − f (X 0 (s))Z (s,t).

By Young-Love estimate (see Friz and Victoir [11, Theorem 6.8], or [7, Section 3.6 and the interesting historical notes pages 212-213]), for every (s, t) ∈ ∆ τ

0

,

k R X

1

(s,t)k 6 kY 1 k p/ 2-var,s,t + 1

1 − 2 1 3 /p k f k Lip

γ

k Z k p-var,τ

0

kX 0 k p/ 2-var,s,t . By super-additivity of the control functions kY 1 k p/ 2

p/ 2-var,. and k X 0 k p/ 2

p/ 2-var,. , there exists a constant c 1 > 0, depending only on p and k f k Lip

γ

, such that

kR X

1

k p/ 2-var,τ

0

6 c 1 (kY 1 k p/ 2

p/ 2-var,τ

0

+ kZ k p-var,τ p/ 2

0

k X 0 k p/ 2

p/ 2-var,τ

0

) 2 /p 6 c 1 M C ( 1 + ω p,Z ( 0 , τ 0 ) 1 / 2 ) 2 / p .

Then, kR X

1

k p/ 2-var,τ

0

6 c 1 M C 4 1 /p 6 M R and

(X 1 , f (X 0 )) ∈ D Z p / 2 ([ 0 , τ 0 ], R e ).

So, the rough integral

H 2 := ∫ .

0 f (X 1 (s))dZ(s) is well defined. For every (s,t) ∈ ∆ T ,

kH 2 (s,t)k 6 k f k Lip

γ

k Z k p-var,s,t + k f k Lip 2

γ

k Z k p/ 2-var,s,t + I f ,Z,X

n

(s, t) 6 (k f k Lip

γ

∨ k f k 2 Lip

γ

p,Z (s,t) 1 /p

+ c(p)k f k Lip

γ

( 1 ∨ k f k Lip

γ

)(kX 0 k p-var,s,t + k X 1 k p-var,s,t + k X 1 k 2 p-var,s,t + kR X

1

k p/ 2-var,s,tp,Z (s, t) 1 / p

6 c 2 ( 1 + µ C + M R + µ 2 Cp,Z (s, t) 1 /p ,

where c 2 > 0 is a constant depending only on p and k f k Lip

γ

. By super-additivity of the control function ω p,Z :

kH 2 k p-var,τ

0

6 c 2 ( 1 + µ C + M R + µ 2 Cp,Z ( 0 , τ 0 ) 1 /p 6 m C .

So, by Proposition 4,

kY 2 k p-var,τ

0

6 M C and

kX 2 k p -var,τ

0

6 kH 2 k p -var,τ

0

+ kY 2 k p -var,τ

0

6 µ C .

(19)

Therefore, Conditions (10)-(11) hold true for n = 2.

Assume that Conditions (10)-(11) hold true until n ∈ N \{ 0 , 1 } arbitrarily chosen.

Set X n 0 := f (X n− 1 ) . For every (s, t) ∈ ∆ τ

0

, R X

n

(s,t) = X n (s,t) − X n 0 (s)Z(s, t)

= Y n (s, t) + ∫ t s

f (X n− 1 (u))dZ(u) − f (X n− 1 (s))Z (s, t).

So, for every (s, t) ∈ ∆ τ

0

,

kR X

n

(s, t)k 6 kY n (s,t)k + k D f (X n − 1 (s)) f (X n − 2 (s)) Z (s, t)k + I f ,Z,X

n−1

(s, t) 6 kY n k p/ 2-var,s,t + k f k Lip 2

γ

k Z k p/ 2-var,s,t

+ c(p)(kZ k p-var,s,t kR f (X

n−1

) k p/ 2-var,s,t + k Z k p / 2-var,s,t kD f (X n− 1 (.)) f (X n− 2 )k p -var,s,t ) 6 kY n k p/ 2-var,s,t + k f k Lip 2

γ

k Z k p/ 2-var,s,t

+ c(p)k f k Lip

γ

(kZ k p-var,τ

0

ω n (s, t) 2 /p + M(n, τ 0 )k Z k p/ 2-var,s,t ), where

M (n, τ 0 ) := k f (X n − 2 )k p-var,τ

0

+ k f (X n − 2 )k ∞ ,τ

0

k X n − 1 k p-var,τ

0

6 k f k Lip

γ

(k X n− 2 k p-var,τ

0

+ kX n− 1 k p-var,τ

0

) and ω n : ∆ τ

0

→ R + is the control function defined by

ω n (u, v) := 2 p/ 2 1 (k X n − 1 k p-var,u,v p + kR X

n−1

k p/ 2

p/ 2-var,u,v ) for every (u, v) ∈ ∆ τ

0

. By super-additivity of the control functions

kY n k p / 2

p/ 2-var,. , k Z k p / 2

p/ 2-var,. and ω n ,

there exist three constants c 3 , c 4 ,c 5 > 0, depending only on p and k f k Lip

γ

, such that k R X

n

k p/ 2-var,τ

0

6 c 3 (kY n k p/ 2

p / 2-var,τ

0

+ k Z k p/ 2

p / 2-var,τ

0

+ kZ k p-var,τ p/ 2

0

ω n ( 0 , τ 0 ) + M (n, τ 0 ) p/ 2 k Z k p/ 2

p/ 2-var,τ

0

) 2 /p 6 c 4 (kY n k p

p / 2-var,τ

0

+ ( 1 + ω n ( 0 , τ 0 ) + M(n, τ 0 ) p/ 2 ) 2 ω p,Z ( 0 , τ 0 )) 1 /p 6 c 5C p + ( 1 + µ C p + kR X

n−1

k p

p / 2-var,τ

0

+ µ 2p Cp,Z ( 0 , τ 0 )) 1 / p 6 c 5C p + ( 1 + µ C p + M R p + µ 2p Cp,Z ( 0 , τ 0 )) 1 / p .

Then, kR X

n

k p/ 2-var,τ

0

6 c 5C p + 1 ) 1 /p 6 M R and

(20)

(X n , f (X n− 1 )) ∈ D p/ 2

Z ([ 0 , τ 0 ], R e ).

So, the rough integral

H n+1 := ∫ .

0 f (X n (s))dZ(s) is well defined. For every (s,t) ∈ ∆ T ,

kH n+1 (s, t)k 6 k f k Lip

γ

k Z k p-var,s,t + k f k Lip 2

γ

k Z k p/ 2-var,s,t + I f ,Z,X

n

(s,t) 6 (k f k Lip

γ

∨ k f k Lip 2

γ

p,Z (s,t) 1 /p

+ c(p)k f k Lip

γ

( 1 ∨ k f k Lip

γ

)(kX n − 1 k p-var,s,t + k X n k p-var,s,t + k X n k 2 p-var,s,t + kR X

n

k p / 2-var,s,t )ω p,Z (s, t) 1 /p

6 c 6 ( 1 + µ C + M R + µ 2 Cp,Z (s,t) 1 /p ,

where c 6 > 0 is a constant depending only on p and k f k Lip

γ

. By super-additivity of the control function ω p,Z :

kH n+1 k p-var,τ

0

6 c 6 ( 1 + µ C + M R + µ 2 Cp,Z ( 0 , τ 0 ) 1 /p 6 m C .

So, by Proposition 4,

kY n+1 k p-var,τ

0

6 M C and

kX n+1 k p-var,τ

0

6 kH n+1 k p-var,τ

0

+ kY n+1 k p-var,τ

0

6 µ C .

By induction, Conditions (10)-(11) are satisfied for every n ∈ N \{ 0 , 1 } . As in the proof of Theorem 3, the sequence (H n , X n ,Y n ) n ∈ N \{ 0,1 } is bounded in C T p,1 . In addi- tion, the sequence (R X

n

) n∈ N \{ 0,1 } is bounded in C p / 2-var ([ 0 ,T ], R e ) .

For every n ∈ N \{ 0 , 1 } and (s,t) ∈ ∆ T , kH n (s, t)k 6 (k f k Lip

γ

∨ k f k Lip 2

γ

+c(p)k f k Lip

γ

( 1 ∨ k f k Lip

γ

)( sup

n ∈ N

kX n− 2 k p -var,T + sup

n ∈ N

kX n− 1 k p -var,T + sup

n ∈ N

k X n− 1 k p-var,T 2 + sup

n ∈ N

k R X

n−1

k p/ 2-var,T ))ω p,Z (s, t) 1 /p . Since ω p,Z is a control function, (H n ) n ∈ N \{ 0,1 } is equicontinuous. Therefore, by Arzelà-Ascoli’s theorem together with Proposition 5, there exists an extraction ϕ : N \{ 0 , 1 } → N \{ 0 , 1 } such that (H ϕ (n) ) n∈ N \{ 0,1 } converges uniformly to an element H of C p-var ([ 0 ,T ], R e ) .

Since (H ϕ(n) ) n∈ N \{ 0,1 } converges uniformly to H , by Theorem 1, the sequence

(X ϕ(n) ,Y ϕ(n) ) n∈ N \{ 0,1 } converges uniformly to (X,Y ) := (v H , w H ) . So, for every

t ∈ [ 0 ,T ] ,

(21)

 

 

X(t) = H(t) + Y (t)

− dDY

d |DY | (t) ∈ N C

H

( t ) (Y (t)) |DY | -a.e. with Y ( 0 ) = a, and by Proposition 5,

X ∈ C p-var ([ 0 ,T ], R e ) and Y ∈ C 1-var ([ 0 ,T ], R e ).

Denoting X 0 := f (X) , X 0 (resp. R X ) is the uniform limit of (X ϕ 0 (

n ) ) n ∈ N \{ 0,1 }

(resp. (R X

ϕ(n)

) n∈ N \{ 0,1 } ). So, by Proposition 10:

n→∞ lim

H ϕ ( n ) −

∫ .

0 f (X (s))d Z(s)

∞,T = 0 . Therefore, since (H ϕ(n) ) n∈ N

converges also to H in C 0 ([ 0 ,T ], R e ) ,

H(t) = ∫ t

0 f (X(s))dZ(s) ; ∀t ∈ [ 0 ,T ].

4 Some uniqueness results

When p = 1 and there is an additive continuous signal of finite q -variation with q ∈ [ 1 , 3 [ , the uniqueness of the solution to Problem (2) is established in Proposition 11 below. Proposition 12 and Proposition 13 provide necessary and sufficient conditions for uniqueness of the solution when p ∈ [ 1 , 2 [ and p ∈ [ 2 , 3 [ respectively. These conditions are close to the monotonicity of the normal cone which allows to prove the uniqueness when p = 1 (see Proposition 11). The criteria of Propositions 12 and 13 seem difficult to apply.

Proposition 11 Assume that p = 1 and f ∈ Lip 1 ( R e , M e,d ( R )). Consider the Sko- rokhod problem

 

 

 

 

 

X(t) = H(t) + Y (t) H(t) = ∫ t

0 f (X(s))dZ (s) + W (t)

− dDY

d |DY | (t) ∈ N C

H

(t) (Y (t)) | DY| -a.e. with Y ( 0 ) = a

(12)

where W ∈ C q -var ([ 0 ,T ], R e ) with q ∈ [ 1 , 3 [. Under Assumption 1, Problem (12) has a unique solution which belongs to C q-var ([ 0 ,T ], R e ).

Proof Consider two solutions (X,Y ) and (X ,Y ) of Problem (2) on [ 0 ,T] . Since

(s, t) ∈ ∆ T 7→ kZ k 1-var,s,t is a control function, there exists n ∈ N and (τ k ) k ∈ n0,no ∈

D [ 0,T ] such that

(22)

k Z k 1-var,τ

k

k+1

6 M := 1

4 k f k Lip

γ

; ∀k ∈ n0 , n − 1o (13) For every t ∈ [ 0 , τ 1 ] ,

kX(t) − X (t)k 2 = kH(t) − H (t)k 2 + 2 ∫ t 0

hY (s) − Y (s), d(Y − Y )(s)i +2 ∫ t

0 hH(t) − H (t), d(Y − Y )(s)i 6 m 11 ) 2 + 2 m 2 (t) + 2 m 3 (t), with m 11 ) := kH − H k ∞,τ

1

,

m 2 (t) := ∫ t

0 hX (s) − X (s), d(Y − Y )(s)i, and

m 3 (t) := ∫ t

0 hH(t) − H (t) − (H(s) − H (s)), d (Y − Y )(s)i . Consider t ∈ [ 0 , τ 1 ] . By Friz and Victoir [11], Proposition 2.2:

kH(t) − H (t)k =

∫ t

0 ( f (X(s)) − f (X (s)))dZ (s) 6 k f k Lip

γ

k X − X k ∞,τ

1

k Z k 1-var,τ

1

. So,

m 11 ) 6 1

4 k X − X k ∞,τ

1

(14)

Since the map x ∈ C(t) 7→ N C ( t ) (x) is monotone, m 2 (t) 6 0. By the integration by parts formula,

m 3 (t) = ∫ t

0 hY (s) − Y (s), d(H − H )(s)i

= ∫ t 0

hX(s) − X (s) − (H(s) − H (s)), ( f (X(s)) − f (X (s)))dZ (s)i . So, by Friz and Victoir [11], Proposition 2.2 and Inequality (14),

m 3 (t) 6 kD f k ∞ k X − X k ∞,τ

1

(kX − X k ∞,τ

1

+ kH − H k ∞,τ

1

)k Z k 1-var,τ

1

6 k f k Lip

γ

k Z k 1-var,τ

1

( 1 + k f k Lip

γ

k Z k 1-var,τ

1

)k X − X k ∞,τ 2

1

6 5 / 16 k X − X k ∞,τ 2

1

. Therefore,

kX − X k 2

1

6 11

16 k X − X k 2

1

.

(23)

Necessarily, (X,Y ) = (X ,Y ) on [ 0 , τ 1 ] .

For k ∈ n0 , n − 1o, assume that (X,Y ) = (X ,Y ) on [ 0 , τ k ] . By Equation (13) and exactly the same ideas as on [ 0 , τ 1 ] :

k X − X k ∞,τ 2

k

k+1

6 11

16 k X − X k ∞,τ 2

k

k+1

.

So, (X,Y ) = (X ,Y ) on [ 0 , τ k+1 ] . Recursively, Problem (2) has a unique solution on

[ 0 ,T ] .

Remark. The cornerstone of the proof of Proposition 11 is that

∫ t

0 hX(s) − X (s), d(Y − Y )(s)i 6 0 ; ∀t ∈ [ 0 ,T] (15) Thanks to the monotonicity of the map x ∈ C(t) 7→ N C(t) (x) ( t ∈ [ 0 ,T ] ), Inequality (15) is true. When p ∈] 1 , 3 [ , it is not possible to get inequalities involving only the uniform norm of X − X . In that case, the construction of the Young/rough integral suggests to use ideas similar to those of the proof of Proposition 11, but using the p-variation norm of X − X .

In a probabilistic setting, uniqueness up to equality almost everywhere can be ob- tained for Brownian motion, with p > 2, in the frame of Itô calculus, using the martingale property of stochastic integrals and Doob’s inequality, see [26, 15, 23]

for a fixed convex set C and [3, 5] for a moving set.

The two following propositions show that when p ∈] 1 , 3 [ , there exist some con- ditions close to Inequality (15), ensuring the uniqueness of the solution to Problem (2). These criteria seem quite difficult to apply, and we have no example where they do. However we think they are interesting for themselves because they shed a light on the open problem of uniqueness of the solution to Problem (2) when p > 1.

Proposition 12 Consider (s, t) ∈ ∆ T , p ∈ [ 1 , 2 [, f ∈ Lip γ ( R e , M e,d ( R )) and two solutions (X,Y ) and (X ,Y ) to Problem (2) under Assumption 1. On [s,t], (X,Y ) = (X ,Y ) if and only if X (s) = X (s) and

∫ v u

hX (u,r) − X (u,r), d(Y − Y )(r)i 6 0 ; ∀(u, v) ∈ ∆ s,t (16) Proof For the sake of simplicity, the proposition is proved on [ 0 ,T ] instead of [s, t] , with (s, t) ∈ ∆ T .

First of all, if (X,Y ) = (X ,Y ) on [s,t] , then

∫ v u

h X(u,r) − X (u, r), d(Y − Y )(r)i = 0 ; ∀(u, v) ∈ ∆ s,t .

(24)

Now, let us prove that if X(s) = X (s) and Inequality (16) is true, then (X,Y ) = (X ,Y ) .

For every (s,t) ∈ ∆ T ,

kX(s, t) − X (s, t) k 2 = kH (s, t) − H (s, t) k 2 +2 ∫ t

s

hY(s, u) − Y (s, u), d(Y − Y )(u)i +2 ∫ t

s

hH(s, t) − H (s, t), d(Y − Y )(u)i

= kH (s, t) − H (s, t) k 2 +2 ∫ t

s

hX(s, u) − X (s, u), d(Y − Y )(u)i +2 ∫ t

s

hH(s, t) − H(s, u) − (H (s, t) − H (s, u)), d(Y − Y )(u)i 6 kH − H k 2 p -var,s,t + 2 m(s, t)

with

m(s,t) := ∫ t s

hH(u, t) − H (u,t), d(Y − Y )(u)i.

Let (s,t) ∈ ∆ T be arbitrarily chosen.

On the one hand,

m(s, t) 6 2 e · c(p, p)kH − H k p-var,s,t kY − Y k p-var,s,t . So, there exists a constant c 1 > 0, not depending on s and t , such that

k X(s, t) − X (s, t)k p 6 (kH − H k 2 p-var,s,t + 2 m(s,t)) p/ 2 6 c 1 (kH − H k p-var,s,t p

+ (kH − H k p-var,s,t p ) 1 / 2 (kY − Y k p-var,s,t p ) 1 / 2 ).

Since 1 / 2 + 1 / 2 = 1, the right-hand side of the previous inequality defines a control function (see Friz and Victoir [11], Exercice 1.9), and then there exists a constant c 2 > 0, not depending on s and t , such that

k X − X k p p-var,s,t 6 c 1 (kH − H k p p-var,s,t + kH − H k p-var,s,t p/ 2 kY − Y k p-var,s,t p/ 2 ) 6 c 2 (kH − H k p p-var,s,t

+ kH − H k p-var,s,t p/ 2 kX − X k p-var,s,t p/ 2 ) (17) The right-hand side of the previous inequality defines a control function.

On the other hand, since X ( 0 ) = X ( 0 ) :

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