Brahim El Asri
email: b.elasri@uiz.ac.ma

Said HAMADENE † Khalid Oufdil
email: khalid.ofdil@gmail.com.

On the Stochastic Control-Stopping Problem

Keywords: Reected Backward stochastic dierential equations, Mixed stochastic control, control-stopping problem, stochastic Lipschitz condition, Hamilton-Jacobi-Bellman equation, Viscosity solution. MSC2010 Classication: 93E20, 49J40, 49L25

We study the stochastic control-stopping problem when the data are of polynomial growth. The approach is based on backward stochastic dierential equations (BSDEs for short). The problem turns into the study of a specic reected BSDE with a stochastic Lipschitz coecient for which we show existence and uniqueness of the solution. We then establish its relationship with the value function of the control-stopping problem. The optimal strategy is exhibited. Finally in the Markovian framework we prove that the value function is the unique viscosity solution of the associated Hamilton-Jacobi-Bellman equation.

Introduction

The main objective of this paper is to deal with the nite horizon control-stopping problem in its weak formulation (see e.g. [START_REF] Benes | Existence of optimal stochastic control laws[END_REF][START_REF] Davis | Optimal play in stochastic dierential game[END_REF][START_REF] Davis | Dynamic programming conditions for partially observable stochastic systems[END_REF][START_REF] Hamadène | Refected BSDEs and mixed game problems[END_REF] to quote a few, and the references therein) which we describe hereafter.

Let us consider a controlled system whose dynamics (x s ) s≤T is a weak solution of the following functional stochastic dierential equation: dx s = f (s, x, u s )ds + σ(s, x)dB u s , s ∈ [0, T ] and x 0 ∈ R d xed;

(1.1)

u := (u s ) s≤T is a stochastic process by which the controller intervenes on the system by choosing the law P u , equivalent to a reference probability P, under which B u := (B u s ) s≤T is a Brownian motion. On the other hand, at her/his convenience, the controller chooses also the time τ to stop controlling the system. As a result this incurs a payo, which is a reward, J(u, τ ) given by: J(u, τ ) := E u τ 0 Γ(s, x, u s )ds + L τ 1 {τ <T } + g(x)1 {τ =T } .

(1.2) Here: i) Γ is the instantaneous reward; ii) L τ is the reward if the controller decides to stop at τ before the terminal time T ; iii) g is the reward at the terminal T .

This problem is termed of control-stopping (or mixed) type because it combines control and stopping. It has been considered in several papers including [START_REF] Bayraktar | Optimal stopping for non-linear expectations Part II[END_REF][START_REF] Karoui | Les aspects probabilistes du contrôle stochastique[END_REF][START_REF] Hamadène | Refected BSDEs and mixed game problems[END_REF][START_REF] Karatzas | Martingale approach to stochastic control with discretionary stopping[END_REF]. There are at least two methods to tackle this problem. One is based on the martingale approach ( [START_REF] Karoui | Les aspects probabilistes du contrôle stochastique[END_REF][START_REF] Karatzas | Martingale approach to stochastic control with discretionary stopping[END_REF]) and the other one uses the notion of backward stochastic dierential equations ( [START_REF] Bayraktar | Optimal stopping for non-linear expectations Part II[END_REF][START_REF] Hamadène | Refected BSDEs and mixed game problems[END_REF]). However in all those works there are technical restrictions on the data (f, Γ, L, g, etc.) which dene this problem. Actually, in [START_REF] Karoui | Les aspects probabilistes du contrôle stochastique[END_REF][START_REF] Karatzas | Martingale approach to stochastic control with discretionary stopping[END_REF], f is supposed of linear growth w.r.t. x and Γ, L, g bounded while in [START_REF] Bayraktar | Optimal stopping for non-linear expectations Part II[END_REF] this latter assumption of boundedness is relaxed to linear growth. On the other hand, in [START_REF] Hamadène | Refected BSDEs and mixed game problems[END_REF], f is supposed bounded while the other functions can have an arbitrary polynomial growth w.r.t. x. Therefore the main objective of this work is to unify those frameworks, i.e., to consider the control-stopping problem when f , on one hand, and Γ, L, g, on the other hand, are of linear and polynomial growths respectively. The approach is based on reected BSDEs.

To deal with the control-stopping in our general framework we are led to study the following reected BSDE (1.3) and to prove that it has a solution:

     Y t = g(x) + T t H * (s, x, Z s )ds + K T -K t - T t Z s dB s , t ≤ T ;
L ≤ Y and T 0 (Y s -L s ) dK s = 0, (1.3) where H * (s, x, z) = sup u∈A H(s, x, z, u) with H(t, x, z, u) = zσ -1 (t, x)f (t, x, u) + Γ(t, x, u).

The function H is the Hamiltonian of the problem and A the set of values of the controls. First let us notice that if σ -1 f is bounded then obviously the function H * is Lipschitz w.r.t. z and consequently the existence-uniqueness of a solution for the above reected BSDE (1.3) is obtained from the classical result by El-Karoui [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF]. Now in the case when σ -1 f is not bounded and satises a linear growth condition only, H * verifes the following property which is called the stochastic Lipschitz condition:

∀ z, z ∈ R d , |H * (t, x, z) -H * (t, x, z)| ≤ C(1 + x t )|z -z | ( x t = sup s≤t |x s |).
( 1.4) which means that the reected BSDE (1.3) is not standard and of stochastic Lipschitz type. These latter RBSDEs are already considered in some papers including [START_REF] Marzougue | Double barrier reected BSDEs with stochastic Lipschitz coecient[END_REF][START_REF] Wen | Reected BSDE with stochastic Lipschitz coecient[END_REF] (for non reected ones, see [START_REF] Bender | BSDEs with stochastic Lipschitz condition[END_REF]). However the results of these works do not allow to deduce satisfactorily the existence of a solution to (1.3) since they have been stated in frameworks which do not t completely to ours. Indeed, had we applied those results, we would have been led to assume restrictive conditions on the horizon T of the problem which, among other conditions, should be small. Therefore the rst task is to show that the reected BSDE (1.3) has a solution for arbitrary nite horizon T . Under condition (1.4) on H * and the polynomial growth of Γ, g and L with respect to x, we show that the reected BSDE (1.3) has a unique solution. In this proof, one point plays a crucial role which is the existence of a constant p > 1 such that The paper is organized as follows. In Section 2, we formulate the problem and consider a specic reected BSDE with stochastic Lipschitz coecient whose solution provides appropriate estimates for the solution of the reected BSDE (1.3). In Section 3, we show that equation (1.3) has a unique solution (Y, Z, K). This solution is constructed as a limit (twice) of an approximating scheme obtained by truncating H * twice. Namely we show that

Y = lim m→∞ lim n→∞ Y n,m
where (Y n,m , Z n,m , K n,m ) is the solution of the standard reected BSDE associated with (H * n,m , g(x), L) with H * n,m (t, x, z) := max{H * (t, x, z), 0}1 { x t≤n} -max{-H * (t, x, z), 0}1 { x t≤m} which is Lipschitz w.r.t. z since H * veries (1.4). Another property which plays an important role in this construction is the comparison of solutions of reected BSDEs. We then show that Y 0 is nothing else but the optimal payo of the control-stopping problem. The optimal pair of control and stopping time is exhibited.

In Section 4, we consider the Markovian framework of the mixed control problem, i.e., roughly speaking, for (t, x) ∈ [0, T ] × R d , the dynamics of the controlled system is given by: dx t,x s = f (t, x t,x s , u s )ds + σ(s, x t,x s )dB u s , t ∈ [0, T ] and x t,x t = x ∈ R d xed.

(

The payo J t (u, τ ) on the time interval [t, T ], is dened by: J t (u, τ ) := E u τ t Γ(s, x t,x s , u s )ds + h(τ, x t,x τ )1 {τ <T } + g(x t,x T )1 {τ =T } .

(1.6)

We show that the deterministic function

u(t, x) := Y t,x t = sup (u,τ ) J t (u, τ ),
where (Y t,x , Z t,x , K t,x ) is the unique solution of the reected BSDE (1.3) in this Markovian framework, is the unique viscosity solution of the Hamilton-Jacobi-Bellman associated with the control-stopping problem, i.e.,

   min [u(t, x) -h(t, x), -∂ t u(t, x) -Lu(t, x) -H * (t, x, ∇ x u(t, x)σ(t, x))] = 0, (t, x) ∈ [0, T [×R d ; u(T, x) = g(x), x ∈ R d .
(1.7) Moreover u is continuous and of polynomial growth. The main diculty is to show continuity of ūm (t, x) = lim n→∞ Y t,x,n,m t . Due to the polynomial growths of Γ, h and g, this continuity cannot be obtained by the usual characterization by means of Snell envelope of processes. To overcome this diculty we have shown that the comparison principle holds for the partial dierential equation (PDE for short) associated with the RBSDE veried by Y t,x,m = lim n→∞ Y t,x,n,m (see (4.47)) and that ūm (t, x) is a solution. Consequently it is continuous and unique in the classe of functions with polynomial growth. Finally, similarly, we deduce that u(t, x) is continuous and is the unique solution of (1.7) in the classe of functions with polynomial growth. Let σ be a function from

[0, T ] × Ω into R d×d such that: A1) σ is F 0 t -progressively measurable. A2) There exists a constant C > 0 such that: a) For every t ∈ [0, T ] and ω, ω ∈ Ω, |σ(t, ω) -σ(t, ω )| ≤ C||ω -ω || t , b) σ is bounded, invertible and its inverse σ -1 is bounded.
Let P be a probability measure on Ω such that (Ω, P) carries a d-dimensional Brownian motion (B t ) 0≤t≤T and for any x 0 ∈ R d , the process (x t ) 0≤t≤T is the unique solution of the following stochastic dierential equation:

x t = x 0 + t 0 σ(s, x)dB s , t ∈ [0, T ].
(2.1) Such a pair (P, B) exists thanks to Proposition 4.6 in ( [START_REF] Karatzas | Brownian Motion and Stochastic Calculus[END_REF], pp.315) since σ satises (A2). Moreover, for any p ≥ 1,

E x p T ≤ C p (2.2)
where C p depends only on p, T , the initial value x 0 and the linear growth constant of σ (see [START_REF] Karatzas | Brownian Motion and Stochastic Calculus[END_REF], pp. 306). Again, since σ satises (A2), F 0 t is the same as σ{B s , s ≤ t} for any t ≤ T . We denote by F := (F t ) 0≤t≤T the completion of (F 0 t ) t≤T with the P-null sets of Ω. Finally let P be the F-progressively measurable σ-algebra on Ω × [0, T ].

Let us introduce some notations. In the following, for q, k ≥ 1, we denote by:

i) L q :={ξ is an F T -measurable random variable s.t E[|ξ| q ] < +∞}. ii) H q,k := {Z := (Z s ) s≤T , P-measurable, R k -valued process s.t. E T 0 |Z s | 2 q 2 < +∞}.
iii) S q :={Y := (Y t ) t≤T , P-measurable, R-valued and continuous process s.t. E sup 0≤t≤T |Y t | q < +∞ }. iv) By convention inf{∅} = +∞.

Let us now consider the following functions g, ϕ and obstacle (L t ) t≤T .

(i) g : C → R is a Borel measurable function of polynomial growth, i.e., there exist constants C and p such that:

|g(x)| ≤ C(1 + x p T ), ∀x ∈ C.
(ii) The process (L t ) t≤T is P-measurable, continuous, R-valued. Moreover there exist positive constants C and p such that ∀t ∈ [0, T ],

|L t | ≤ C(1 + x p t ).
(2.3)

We moreover assume that P-a.s. L T ≤ g(x).

(iii) The function ϕ is dened as follows: For some p ≥ 1, for any

(t, x, z) ∈ [0, T ] × C × R d , ϕ(t, x, z) = c(1 + x t )|z| + c(1 + x p t ).
We now introduce a property of exponential martingales which plays an important role in this work. Let ϑ be a function from [0, T ] × C into R d which is P-measurable. For t ≤ T we set:

ζ t = e t 0 ϑ(s,ω)σ -1 (s,ω)dBs-1 2 t 0 |ϑ(s,ω)σ -1 (s,ω)| 2 ds .
We then have:

Lemma 2.1. i) Assume that ϑ is bounded. Then (ζ t ) t≤T is a martingale such that for any ≥ 1, E[(ζ T ) ] ≤ C ,ϑ,T
where C ,ϑ,T is a constant which depends only on T , and the constant of boundedness of ϑσ -1 .

ii) Assume that ϑ is of linear growth, i.e., for any t ≤ T , |ϑ(t, ω)| ≤ c ϑ (1 + |ω| t ). Then there exist constants q > 1 and which depend only on T , σ, σ -1 and c ϑ such that

E[(ζ T ) q ] ≤ .
(2.4) Moreover (ζ t ) t≤T is a martingale. iii) In both cases, the measure Q such that dQ = ζ T .dP is a probability on (Ω, F).

Proof: Point i) is classical and based on the fact that when ϑ is bounded, the familly of ζ's are martingales. The proof of point ii) is given in ( [START_REF] Haussmann | On the existence of optimal controls for partially observed diusions[END_REF], Equation (2.8), pp. 3). Finally point iii) is obvious since ζ is a positive martingale.

Next let

Φ be a function from [0, T ] × C × R d into R which is P ⊗ B(R d )-measurable. First
we dene the notion of a solution of the reected BSDE associated with a triplet (g, Φ, L) which we consider throughout this paper. Denition 2.1. A triplet (Y, Z, K) := (Y t , Z t , K t ) t≤T of P-measurable and R 1+d+1 -valued processes is a solution of the reected BSDE associated with the lower reecting barrier L, the terminal value g(x) and generator Φ if the followings hold:

i) Y is continuous, K is continuous non-decreasing (K 0 = 0), P-a.s. (Z t (ω)) t≤T is dt-square integrable; ii) Y t = g(x) + T t Φ(s, x, Z s )ds + K T -K t - T t Z s dB s , 0 ≤ t ≤ T ; iii) Y t ≥ L t , 0 ≤ t ≤ T and T 0 (Y s -L s )dK s = 0.
We are going to show that the reected BSDE associated with (ϕ, L, g) has a solution which also satises other integrability properties. As it is mentionned previously, ϕ is not a standard generator which does not enter neither in the framework of [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF] nor in the one of [START_REF] Bender | BSDEs with stochastic Lipschitz condition[END_REF]. Proposition 2.1. There exist P-measurable processes (Y, Z, K) valued in R 1+d+1 and a stationary non-decreasing sequence of stopping times (τ k ) k≥1 such that: i) (Y, Z, K) is a solution for the BSDE associated with (ϕ, L, g). ii) For any constant γ ≥ 1 and τ a stopping time valued in [0, T ],

E[|Y τ | γ ] < +∞. (2.5) iii) For any k ≥ 1, E[sup s≤T |Y s∧τ k | γ + K γ τ k + τ k 0 |Z s | 2 ds] < +∞, (2.6)
where τ k depends only on g, L and x.

Proof: For n ≥ 0 let us set

ϕ n (t, x, z) = [c(1 + x t ) ∧ n]|z| + c(1 + x p t ).
Then ϕ n is Lipschitz w.r.t z. Therefore by El-Karoui et al.'s result in [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF], there exists a triplet of processes (Y n , Z n , K n ) ∈ S 2 × H 2,d × S 2 (K n is non-decreasing and K n 0 = 0) such that :

Y n t = g(x) + T t ϕ n (s, x, Z n s )ds + K n T -K n t - T t Z n s dB s , t ∈ [0, T ] ; L t ≤ Y n t , ∀t ∈ [0, T ] and T 0 (Y n s -L s ) dK n s = 0. (2.7)
Next by the comparison Theorem (see e.g. [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF]) we have, for any n ≥ 0, Y n ≤ Y n+1 and dK n+1 ≤ dK n since ϕ n ≤ ϕ n+1 and the barrier L is xed. We now show the following Lemmas (which are steps forward in the proof of Proposition 2.1) related to estimations of the processes Y n , n ≥ 0, and the convergence of the sequence

(Y n ) n≥0 .
Lemma 2.2. There exists a P-measurable RCLL (for right continuous with left limits) process Y := (Y t ) t≤T , R-valued such that P-a.s. for any t ≤ T , Y n t Y t . Moreover for any γ ≥ 1 and any stopping times τ ∈ [0, T ],

E [|Y τ | γ ] ≤ c
where c is a constant independent of τ .

Proof: Let P n be the probability, equivalent to P, dened as follows:

dP n = L n T dP
where for any t ≤ T ,

L n t := exp{ t 0 {c(1 + ||x|| s ) ∧ n} (Z n s )dB s -1 2 t 0 {c(1 + x s ) ∧ n} (Z n s ) 2 ds}
with is a bounded measurable function such that (z).z = |z|, ∀z = (z i ) i=1,...,d ∈ R d . Such a function exists and it is enough to take (z) = ( i (z)) i=1,...,d where for i = 1, . . . , d -1

i (z) = ( z 2 i + ... + z 2 d -z 2 i+1 + ... + z 2 d )z -1 i 1 {z i =0} and d (z) = |z d |z -1 d 1 {z d =0} .
Then by Girsanov's Theorem, the process (B n t := B t -t 0 {c(1 + x s ) ∧ n} (Z n s )ds) t≥0 is a Brownian motion under P n and the triplet (Y n , Z n , K n ) veries: For any t ∈ [0, T ],

Y n t = g(x) + T t c(1 + x p s )ds + K n T -K n t - T t Z n s dB n s ; L t ≤ Y n t and T 0 (Y n s -L s ) dK n s = 0.
Therefore for any stopping time τ ≤ T , P-a.s., we have:

Y n τ = esssup σ≥τ E P n σ τ c(1 + x p s )ds + L σ 1 {σ<T } + g(x T )1 {σ=T } |F τ
(one can see [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF] for this characterization). Next by using polynomial growth of g and L and the fact that P and P n are equivalent, we deduce that:

|Y n τ | ≤ E P n T τ c(1 + x p s )ds + 2c(1 + x p T )|F τ ≤ c(1 + E P n x p T |F τ ).
Next let γ ≥ 1. Then by conditional Jensen's inequality we have:

|Y n τ | γ ≤ c(1 + E P n x γp T |F τ ) (2.8) = c(1 + E L n τ,T x γp T |F τ )
where for any t ≤ T ,

L n t,T := L n T ÷ L n t = exp{ T t {c(1 + ||x|| s ) ∧ n} (Z n s )dB s - 1 2 T t {c(1 + x s ) ∧ n} (Z n s ) 2 ds} = exp{ T 0 1 {t≤s≤T } {c(1 + ||x|| s ) ∧ n} (Z n s )dB s - 1 2 {c(1 + x s ) ∧ n} (Z n s ) 2 ds }. (2.9)
Now by Lemma 2.1, there exist constants c and q > 1, which do not depend on n and τ such that E[(L n τ,T ) q ] ≤ c. Next by Young's inequality we get from (2.8):

|Y n τ | γ ≤ c(1 + E q -1 (L n τ,T ) q + q-1 x γpq T |F τ ),
where q -1 + q-1 = 1. Taking now into account of (2.2) we deduce that: It remains to show that Y is RCLL. But this a direct consequence of the fact that Y n is a continuous supermartingale which converges increasingly and pointwise to Y and then the limiting process Y is also RCLL (see [START_REF] Dellacherie | Probabilités et Potentiel, chap. V-VIII[END_REF], pp.86).

∀n ≥ 0, E[|Y n τ | γ ] ≤ c (2.
Next let θ = |x 0 |+|L 0 |+|Y 0 0 |+|Y 0 | ( θ is a constant) and for k ≥ 1 let us dene the sequence of stopping times (τ k ) k≥1 by:

τ k := inf{t ≥ 0, |Y t | + x t + |L t | + |Y 0 t | ≥ θ + k} ∧ T.
The sequence of stopping times (τ k ) k≥1 is non-decreasing, of stationary type (i.e. constant after some rank k 0 (ω)) converging to T since the process Y is RCLL and Y 0 , x and L are continuous. Moreover for any k ≥ 1,

max{sup t≤τ k |L t |, sup t≤τ k |Y t |, sup t≤τ k |Y n t |} ≤ θ + k := θk .
Next we have the following result: Lemma 2.3.

i) The process Y is continuous.

ii) There exist P-measurable processes (K, Z) valued in R 1+d such that (Y, Z, K) is a solution of the reected BSDE associated with (ϕ, g, L) and veries (2.6).

Proof: For any k ≥ 1 and n ≥ 0 we have:

∀t ∈ [0, T ]    Y n t∧τ k = Y n τ k + τ k t∧τ k ϕ n (s, x, Z n s )ds + K n τ k -K n t∧τ k - τ k t∧τ k Z n s dB s ; L t∧τ k ≤ Y n t∧τ k and τ k 0 (Y n s -L s ) dK n s = 0.
By using the Itô formula with (Y n t∧τ k ) 2 and taking into account of (2.2), we classicaly deduce the existence of a constant C k , which depends on k, such that uniformly on n, we have

E τ k 0 |Z n s | 2 ds ≤ C k (2.
∀t ∈ [0, T ], (Y n t∧τ k -Y m t∧τ k ) 2 = (Y n τ k -Y m τ k ) 2 + 2 τ k t∧τ k (Y n s -Y m s )(ϕ n (s, x, Z n s ) -ϕ m (s, x, Z m s ))ds + 2 τ k t∧τ k (Y n s -Y m s )d(K n s -K m s ) - τ k t∧τ k |Z n s -Z m s | 2 ds -2 τ k t∧τ k (Y n s -Y m s )(Z n s -Z m s )dB s .
The denition of ϕ n and the fact that

(Y n s -Y m s )d(K n s -K m s ) ≤ 0 imply that: ∀t ≤ T , (Y n t∧τ k -Y m t∧τ k ) 2 + τ k t∧τ k |Z n s -Z m s | 2 ds ≤ (Y n τ k -Y m τ k ) 2
(2.12)

+ 2 τ k t∧τ k |Y n s -Y m s |C(1 + x s )(|Z n s | + |Z m s |)ds -2 τ k t∧τ k (Y n s -Y m s )(Z n s -Z m s )dB s .
Take expectation on both hand-sides to deduce that

E[ τ k t∧τ k |Z n s -Z m s | 2 ds] ≤ E[(Y n τ k -Y m τ k ) 2 ] + 2E[ τ k t∧τ k |Y n s -Y m s |C(1 + x s )(|Z n s | + |Z m s |)ds].
Now the denition of τ k , estimate (2.11) and the Cauchy-Schwarz inequality yield:

E[ τ k t∧τ k |Z n s -Z m s | 2 ds] → 0 as n, m → +∞. (2.13)
Consequently the sequence ((Z n s 1 {0≤s≤τ k } ) s≤T ) n≥0 is of Cauchy type in H 2,d and then there exists a process Z k which belongs to H 2,d such that

((Z n s 1 {0≤s≤τ k } ) s≤T ) n≥0 → n Z k (s) in H 2,d .
Next going back to (2.12), take the supremum over t, make use of Burkholder-Davis-Gundy (see e.g. [START_REF] Karatzas | Brownian Motion and Stochastic Calculus[END_REF][START_REF] Revuz | Continuous Martingales and Brownian Motion[END_REF] and BDG for short) inequality and nally take the expectation to deduce that

E[sup s≤T |Y n s∧τ k -Y m s∧τ k | 2 ] → 0 as n, m → +∞. (2.14) As (Y n ) n≥0 → Y then the process (Y s∧τ k ) s≥0 is continuous for any k ≥ 1 and E[sup s≤T |Y n s∧τ k -Y s∧τ k | 2 ] → 0 as n → +∞. (2.15)
But the sequence (τ k ) k≥1 is of stationary type, therefore the process Y is continuous. Now by (2.11) and (2.13), for any k ≥ 1, the sequence of processes

((ϕ n (s, x s , Z n s )1 {s≤τ k } ) s≤T ) n≥0 converges in H 2,d to (ϕ(s, x, Z k (s))1 {s≤τ k } ) s≤T ). Therefore if we set, for k ≥ 1 and t ≤ T , K k (t) = Y 0 -Y t∧τ k - t∧τ k 0 ϕ(s, x, Z k (s))ds + t∧τ k 0 Z k (s)dB s we obtain that E[sup s≤T |K n s∧τ k -K k (s)| 2 ] → 0 as n → +∞. (2.16)
Finally the uniform convergences (2.16) and (2.15) imply that, in view of Helly's Convergence Theorem (see [START_REF] Kolmogorov | Introductory Real Analysis[END_REF], pp. 370),

τ k 0 (Y s -L s )dK k (s) = 0.
It means that for any k ≥ 1 we have:

∀t ≤ T ,    Y t∧τ k = Y τ k + τ k t∧τ k ϕ(s, x, Z k (s))ds + K k (τ k ) -K k (t ∧ τ k ) - τ k t∧τ k Z k (s)dB s ; L t∧τ k ≤ Y t∧τ k and τ k 0 (Y s -L s ) dK k (s) = 0.
(2.17)

Take now the reected BSDE satised by (Y, Z k+1 , K k+1 ) on [0, τ k ] (since τ k ≤ τ k+1 ) yields: for any t ≤ T ,    Y t∧τ k = Y τ k + τ k t∧τ k ϕ(s, x, Z k+1 (s))ds + K k+1 (τ k ) -K k+1 (t ∧ τ k ) - τ k t∧τ k Z k+1 (s)dB s ; L t∧τ k ≤ Y t∧τ k and τ k 0 (Y s -L s ) dK k+1 (s) = 0.
(2.18) By uniqueness (which holds since on [0, τ k ], ϕ is Lipschitz w.r.t z), we have for any k ≥ 1 :

Z k (s)1 {s≤τ k } = Z k+1 (s)1 {s≤τ k } ds ⊗ dP -a.e and K k+1 (s ∧ τ k ) = K k (s ∧ τ k ), ∀s ≤ T.
Then let us dene (by concatenation) the processes Z and K as follows: ∀t ≤ T ,

Z t = Z 1 (t)1 {t≤τ 1 } + k≥2 Z k (t)1 {τ k-1 <t≤τ k } and K t =    K 1 (t) if t ≤ τ 1 ; (K k+1 (t) -K k+1 (τ k )) + K(τ k ) for τ k < t ≤ τ k+1 , k ≥ 1.
Note that the processes Z and K are well-dened since the sequence (τ k ) k≥1 is of stationary type. On the other hand, K is continuous non-decreasing and P-a.s., K T (ω) < +∞ and (Z s (ω)) s≤T is ds-square integrable. Finally for any k ≥ 1, it holds:

   Y t∧τ k = Y τ k + τ k t∧τ k ϕ(s, x, Z s )ds + K τ k -K t∧τ k - τ k t∧τ k Z s dB s ; L t∧τ k ≤ Y t∧τ k and τ k 0 (Y s -L s ) dK s = 0.
(2.19) Take now k great enough and since once more (τ k ) k≥1 is of stationary type to obtain: ∀t ≤ T ,

     Y t = g(x) + T t ϕ(s, x, Z(s))ds + K T -K t - T t Z s dB s ; L t ≤ Y t and T 0 (Y s -L s ) dK s = 0 (2.20)
which completes the proof.

Study of the stochastic mixed control problem

Let A be a compact metric space and let U be the space of P-measurable processes u := (u t ) t≤T with values in A. We rst introduce the objects and assumptions which we need in this section.

(1

) f : [0, T ] × Ω × A → R d is a function such that: i) For each a ∈ A, the function (t, ω) → f (t, ω, a) is P-measurable. ii) For each (t, ω), the mapping a → f (t, ω, a) is continuous.
iii) f is of linear growth, i.e., there exists a real constant C > 0 such that:

|f (t, ω, a)| ≤ C(1 + ||ω|| t ), ∀ 0 ≤ t ≤ T, ω ∈ Ω, a ∈ A. (3.1) (2) Γ : [0, T ] × Ω × A → R is a P ⊗ B(A)-measurable function such that for each (t, ω) the mapping a → Γ(t, ω, a)
is continuous. In addition there exist positive constants C > 0 and p such that:

|Γ(t, ω, a)| ≤ C(1 + ω p t ), ∀ t ≤ T, w ∈ Ω, a ∈ A. (3.2)
Next let u := (u t ) t≤T be an adapted A-valued stochastic process and P u the probability on C (which is actually one) equivalent to P such that:

dP u = M T dP
where for any t ≤ T ,

M t := e t 0 σ -1 (s,x)f (s,x,us)dBs-1 2 t 0 σ -1 (s,x)f (s,x,us) 2 ds .
Under P u the process (x t ) t≤T veries:

x t = x 0 + t 0 f (s, x, u s )ds + t 0 σ(s, x)dB u s , t ≤ T, (3.3)
where

B u t := B t - t 0 σ -1 (s, x)f (s, x, u s )ds, t ≤ T, (3.4)
is a Brownian motion under P u . It means that (x t ) t≤T is a weak solution of the standard functional SDE (3.3).

Next let τ be a stopping time. We associate with the pair (u, τ ) a payo J(u, τ ) given by :

J(u, τ ) = E u τ 0 Γ(s, x, u s )ds + L τ 1 {τ <T } + g(x)1 {τ =T } . (3.5)
The problem is to nd a pair (u * , τ * ) which maximizes J(u, τ ), i.e., J(u * , τ * ) ≥ J(u, τ ) for any (u, τ ). (3.6) This is the mixed control problem associated with x, f , Γ and g. It combines control and stopping. One can think of an agent who controls a system by choosing the probability P u which xes its dynamics, weak solution of (3.3), up to the time when she/he makes the decision to stop control at time τ . Therefore its payo is given by J(u, τ ) and the problem is to nd an optimal strategy (u * , τ * ), i.e., which satises (3.6). Now let us introduce the Hamiltonian function H associated with the mixed control problem which is given by:

H(t, x, z, u) := zσ -1 (t, x)f (t, x, u) + Γ(t, x, u), ∀(t, x, z, u) ∈ [0, T ] × Ω × R d × A. (3.7)
Note that for any z ∈ R d and u ∈ U, the process (H(t, x, z, u t )) t≤T is P-measurable. On the other hand thanks to Benes's selection Theorem ([4], Lemma 5, pp. 460), there exists a

P⊗B(R d )/B(A)-measurable function u * (t, x, z) such that for any given (t, x, z) ∈ [0, T ]×Ω×R d , H * (t, x, z) := sup u∈A H(t, x, z, u) = H(t, x, z, u * (t, x, z)). (3.8)
First we are going to show that the reected BSDE associated with (H * , g, L) has a unique solution which moreover veries some integrability properties. The following result is a step forward the proof of this result. Proposition 3.1. For any m ≥ 1, there exist P-measurable processes (Y * m , Z * m , K * m ) in R 1+d+1 such that: i) For any γ ≥ 1 and any stopping time τ ∈ [0.T ],

E [|Y * m τ | γ ] ≤ c, ∀m ≥ 1, (3.9)
where c is a constant independent of m and τ .

ii) The triple (Y * m , Z * m , K * m ) is a solution of the reected BSDE associated with (H * m , g, L) (in the sense of Def. 2.1) where

H * m (s, x, z) = H * + (s, x, z) -H * -(s, x, z)1 {1+||x||s≤m} ,
with for any α ∈ R, α + = α ∨ 0 and α -= (-α) ∨ 0.

Proof: For any m, n ≥ 1 let us set,

H * n,m (t, x, z) = H * + (s, x, z)1 {1+||x||s≤n} -H * -(s, x, z)1 {1+||x||s≤m} .
(3.10)

Then H * n,m is Lipschitz w.r.t z, therefore by El-Karoui et al's result [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF], there exists a triplet of processes (Y * n,m , Z * n,m , K * n,m ) ∈ S 2 × H 2 × S 2 that satises:

Y * n,m t = g(x) + T t H * n,m (s, x, Z * n,m s )ds + K * n,m T -K * n,m t - T t Z * n,m s dB s , L t ≤ Y * n,m t , ∀t ∈ [0, T ] and T 0 (Y * n,m s -L s )dK * n,m s = 0. (3.11)
Thus from the denition of H * n,m , we can clearly see that it is a non-decreasing (resp. nonincreasing) sequence of functions w.r.t n (resp. m). Then once again by comparison (see [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF]) we deduce that

Y * n,m ≤ Y * n+1,m and Y * n,m+1 ≤ Y * n,m . (3.12)
On the other hand we have

H * n,m (s, x, Z * n,m s ) ≤ C[(1 + ||x|| s ) ∧ n]|Z * n,m s | + C[(1 + ||x|| p s ).
Then again by comparison we have:

∀t ∈ [0, T ] L t ≤ Y * n,m t ≤ Y n t ≤ Y t , ∀n, m ≥ 1, (3.13)
where Y n (resp. Y ) is the process of (2.7) (resp. of Proposition 2.1).

Next we will divide the proof into two steps. In the rst (resp. second) one we will prove i) (resp. ii)).

Proof of i): From the growth condition on L (2.3), there exist positive constant C and p such that ∀t ∈ [0, T ] and x ∈ Ω

|L t | ≤ C(1 + ||x|| p t ).
From Lemma 2.2 we have that for any γ ≥ 1 and any stopping times τ ∈ [0, T ],

E [|Y τ | γ ] ≤ c,
where c is a constant independent of τ . Then from (3.13), we deduce that for any γ ≥ 1 and any stopping time τ ∈ [0.T ],

E [|Y * n,m τ | γ ] ≤ c, ∀n, m ≥ 1 (3.14)
where c is a constant that does not depend neither on n nor m. Next for m ≥ 1, let us set

Y * m = lim inf n→+∞ Y * n,m .
Then by Fatou's Lemma and (3.14) we have:

∀m ≥ 1, E [|Y * m τ | γ ] ≤ c and then |Y * m | < ∞, P -a.s.. (3.15)
On the other hand from (3.12) it holds that

L ≤ Y * ,m+1 ≤ Y * ,m ≤ Y, ∀m ≥ 1.
(3.16)

Proof of ii): Let (τ * k ) k≥1 be the sequence of stopping times dened as follow:

∀k ≥ 1, τ * k := inf{t ≥ 0, |Y t | + x t + |L t | ≥ θ * k := θ * + k} ∧ T,
where

θ * = |x 0 | + |L 0 | + |Y 0 |.
The sequence of stopping times (τ * k ) k≥1 is non-decreasing, of stationary type converging to T since the processes Y , x and L are continuous. Moreover for any k ≥ 1,

max{ sup t≤τ * k |L t |, sup t≤τ * k |Y t |, sup t≤τ * k |Y * n,m t |} ≤ θ * k
Let us now consider the following reected BSDE. For any n, m ≥ 1 and any k ≥ 1 we have:

∀t ∈ [0, T ],      Y * n,m t∧τ * k = Y * n,m τ * k + τ * k t∧τ * k H * n,m (s, x, Z * n,m s )ds + K * n,m τ * k -K * n,m t∧τ * k - τ * k t∧τ * k Z * n,m s dB s ; L t∧τ * k ≤ Y * n,m t∧τ * k and τ * k 0 Y * n,m s -L s dK * n,m s = 0.
Now we take into account (2.2) and the fact that

∀t ∈ [0, T ], |Y * n,m t∧τ * k | ≤ θ * k and Y * 0,m ≤ Y * n,m ≤ Y * m , then using Itô's formula with Y * n,m t∧τ * k 2 to conclude that there exists a constant C k such that : E τ * k 0 |Z * n,m s | 2 ds ≤ C k .
(3.17)

Next by Itô's formula we have:

∀t ∈ [0, T ] (Y * n 1 ,m t∧τ * k -Y * n 2 ,m t∧τ * k ) 2 = (Y * n 1 ,m τ * k -Y * n 2 ,m τ * k ) 2 -2 τ * k t∧τ * k (Y * n 1 ,m s -Y * n 2 ,m s )(Z * n 1 ,m s -Z * n 2 ,m s )dB s + 2 τ * k t∧τ * k (Y * n 1 ,m s -Y * n 2 ,m s )(H * n 1 ,m (s, x, Z * n 1 ,m s ) -H * n 2 ,m (s, x, Z * n 2 ,m s ))ds + 2 τ * k t∧τ * k (Y * n 1 ,m s -Y * n 2 ,m s )d(K * n 1 ,m s -K * n 2 ,m s ) - τ * k t∧τ * k |Z * n 1 ,m s -Z * n 2 ,m s | 2 ds.
Obviously for any t ≤ T ,

τ * k t∧τ * k (Y * n 1 ,m s -Y * n 2 ,m s )d(K * n 1 ,m s -K * n 2 ,m s ) ≤ 0.
On the other hand,

|H * n,m (s, x, Z * n,m s )| ≤ |H * (s, x, Z * n,m s )| ≤ C(1 + ||x|| s )|Z * n,m s | + C(1 + ||x|| p s ).
Therefore,

(Y * n 1 ,m t∧τ * k -Y * n 2 ,m t∧τ * k ) 2 + τ * m k t∧τ * k |Z * n 1 ,m s -Z * n 2 ,m s | 2 ds ≤ (3.18) (Y * n 1 ,m τ * k -Y * n 2 ,m τ * k ) 2 -2 τ * k t∧τ * k (Y * n 1 ,m s -Y * n 2 ,m s )(Z * n 1 ,m s -Z * n 2 ,m s )dB s +2 τ * k t∧τ * k |Y * n 1 ,m s -Y * n 2 ,m s | [C(1 + ||x|| s ) (|Z * n 1 ,m s | + |Z * n 2 ,m s |) + 2C(1 + x p s )] ds.
Take now expectation in (3.18) to deduce that:

E τ * k t∧τ * k |Z * n 1 ,m s -Z * n 2 ,m s | 2 ds ≤ E (Y * n 1 ,m τ * k -Y * n 2 ,m τ * k ) 2 +2E τ * k t∧τ * k |Y * n 1 ,m s -Y * n 2 ,m s | [C(1 + ||x|| s ) (|Z * n 1 ,m s | + |Z * n 2 ,m s |) + 2C(1 + x p s )] ds .
Then we conclude that

E τ * k t∧τ * k |Z * n 1 ,m s -Z * n 2 ,m s | 2 ds → 0 as n 1 , n 2 → +∞. (3.19)
This is due to estimate (3.17), the denition of τ * k and Cauchy-Schwarz inequality. Thus there exists a process

Z * m k ∈ H 2,d such that (Z * n,m t 1 {0≤t≤τ * k } ) t≤T n≥1 → (Z * m k (t)) t≤T in H 2,d as n → +∞, ∀ m ≥ 1.
Next by (3.18) and the use of BDG inequality we obtain

E sup s≤T |Y * n 1 ,m s∧τ * k -Y * n 2 ,m s∧τ * k | 2 → 0 as n 1 , n 2 → +∞. Since (Y * n,m ) n≥0 → Y * m , the process (Y * m t∧τ * k
) t≥0 is continuous for every k and m, also

E sup s≤T |Y * n,m s∧τ * k -Y * m s∧τ * k | 2 → 0 as n → +∞.
(3.20)

Next due to the fact that (τ * k ) is of stationary type, then the process Y * m is continuous. Now for m ≥ 1, let us set

K * m k (t) = Y * m 0 -Y * m t∧τ * k - t∧τ * k 0 H * m (s, x, Z * m k (s))ds + t∧τ * k 0 Z * m k (s)dB s .
In taking into account (3.17) and (3.19) we deduce that for any k ≥ 1, m ≥ 1,

{(H * n,m (t, x, Z * n,m t )1 {t≤τ * k } ) t≤T } n≥1 converges in H 2,d to (H * m (t, x, Z * m t )1 {t≤τ * k } ) t≤T ,
Z * m k (s)1 {s≤τ * k } = Z * m k+1 (s)1 {s≤τ * k } dt ⊗ dP -a.e and K * m k+1 (s ∧ τ * k ) = K * m k (s ∧ τ * k ), ∀s ≤ T.
So let us dene Z * m and K * m by concatenation as follows: ∀t ≤ T and m ≥ 1

Z * m t = Z * m 1 (t)1 {t≤τ * 1 } + k≥2 Z * m k (t)1 {τ * k-1 <t≤τ * k } (3.22)
and

K * m t =    K * m 1 (t) if t ≤ τ * 1 ; (K * m k+1 (t) -K * m k+1 (τ * k )) + K * m (τ * k ) for τ * k < t ≤ τ * k+1 , k ≥ 1.
(3.23)

Then for any k ≥ 1, and any m ≥ 1: ∀t ≤ T ,

     Y * m t∧τ * k = Y * m τ * k + τ * k t∧τ * k H * m (s, x, Z * m s )ds + K * m τ * k -K * m t∧τ * k - τ * k t∧τ * k Z * m s dB s ; L t∧τ k ≤ Y * m t∧τ k and τ k 0 (Y * m s -L s ) dK * m s = 0. (3.24)
It is worth noticing that the processes Z * m and K * m are well dened since the sequence

(τ * k ) k≥1
is of stationary type. Moreover K * m is continuous, non-decreasing and P-a.s. K * m T (ω) < +∞ and P-a.s. (Z * m (t)) t≤T is dt-square integrable. Finally going back to equation (3.24), take the limit when k → +∞ to obtain: ∀m ≥ 1

     Y * m t = g(x) + T t H * m (s, x, Z * m s )ds + K * m T -K * m t - T t Z * m s dB s ; L ≤ Y * m and T 0 (Y * m s -L s ) dK * m s = 0.
(3.25)

The proof is now complete.

In the following result we show that the reected BSDE (1.3) has a unique solution. Then after, we address the question of the link between the component Y * of the solution and the value function of the control-stopping problem.

Theorem 3.1. There exist P-measurable processes

(Y * , Z * , K * ) in R 1+d+1 such that: i) Y * is continuous, K * is continuous non-decreasing (K * 0 = 0) and P -a.s., (Z * m (t)) t≤T is dt-square integrable. ii) For any t ∈ [0, T ], Y * t = g(x) + T t H * (s, x, Z * s )ds + K * T -K * t - T t Z * s dB s , L t ≤ Y * t and T 0 (Y * s -L s )dK * s = 0. (3.26)
iii) For any γ ≥ 1 and any stopping time τ ∈ [0, T ],

E [|Y * τ | γ ] ≤ c, (3.27)
where c which does not depend on τ . iv) If ( Ȳ , Z, K) is another triple which satises i), ii) and iii), then ( Ȳ , Z, K) = (Y * , Z * , K * ),

i.e., the solution of the reected BSDE associated with (g, H * , L) is unique to satisfy i)-iii).

Proof: From Proposition 3.1, we have that for any m ≥ 1, there exists a triplet (Y * m , Z * m , K * m ) that satises the following reected BSDE: ∀t ≤ T ,

Y * m t = g(x) + T t H * m (t, x, Z * m s )ds + K * m T -K * m t - T t Z * m s dB s , L t ≤ Y * m t , and T 0 (Y * m s -L s )dK * m s = 0.
where

H * m (s, x, z) = H * + (s, x, z) -H * -(s, x, z)1 {1+||x||s≤m} . From (3.16) we know that for any m ≥ 1, L ≤ Y * ,m+1 ≤ Y * ,m ≤ Y . So, let us set for t ≤ T , Y * t = lim m Y * ,m t .
Therefore L ≤ Y * ≤ Y and then by (2.2), (2.3) and (2.5), Y * veries (3.27).

Next let (τ * k ) k≥1 be the sequence of stopping times dened as follows:

∀k ≥ 1, τ * k := inf{t ≥ 0, |Y t | + x t + |L t | ≥ θ * + k} ∧ T,
where

θ * = |x 0 | + |L 0 | + |Y 0 |.
Thanks to the continuity of Y , L and x, the sequence of stopping times (τ * k ) k≥1 is increasing of stationary type and converges to T . In the same way as in the proof of Proposition 3.1, there exists a constant C k (depending on k) such that:

E[ τ * k 0 |Z * m s | 2 ds] ≤ C k .
This inequality follows in a classic way after using Itô's formula with Y * m t∧τ * k 2 and taking into account (2.2). Next we apply again Itô's formula and we have:

(Y * m t∧τ * k -Y * n t∧τ * k ) 2 = (Y * m τ * k -Y * n τ * k ) 2 -2 τ * k t∧τ * k (Y * m s -Y * n s )(Z * m s -Z * n s )dB s + 2 τ * k t∧τ * k (Y * m s -Y * n s )(H * m (s, x, Z * m s ) -H * n (s, x, Z * n s ))ds + 2 τ * k t∧τ * k (Y * m s -Y * n s )d(K * m s -K * n s ) - τ * k t∧τ * k |Z * m s -Z * n s | 2 ds.
It follows that (in the same way as in the proof of Proposition 3.1): for any k ≥ 1, a) There exists a process

Z * k ∈ H 2,d , such that ((Z * m s 1 {0≤s≤τ * k } ) s≤T ) m≥1 → m Z * k (s) in H 2,d , which is a consequence of the fact that ((Z * m 1 {0≤s≤τ * k } ) s≤T ) m≥1 is a Cauchy sequence in H 2,d . b) lim m→+∞ E sup s≤T |Y * m s∧τ * k -Y * s∧τ * k | 2 = 0 and the process Y * is continuous. Next we set K * k (t) = Y * 0 -Y * t∧τ * k - t∧τ * k 0 H * (s, x, Z * k (s))ds + t∧τ * k 0 Z * k (s)dB s , t ≤ T,
and then we have,

lim m→+∞ E sup s≤T |K * m s∧τ * k -K * k (s)| 2 = 0.
(3.28)

It follows, from the uniform convergence of (Y * m ) m≥1 and (K * m ) m≥1 (see [START_REF] Kolmogorov | Introductory Real Analysis[END_REF], pp. 370), that:

τ * k 0 (Y * s -L s )dK s = 0
Now considering the reected BSDE satised by (Y * , Z * k , K * k ) and the one satised by

(Y * , Z * k+1 , K * k+1 ) on [0, τ * k ], yields. For any t ≤ T ,      Y * t∧τ * k = Y τ * k + τ * k t∧τ * k H * (s, x, Z * k (s))ds + K * k (τ * k ) -K * k (t ∧ τ * k ) - τ * k t∧τ * k Z * k (s)dB s ; L t∧τ * k ≤ Y * t∧τ * k and τ k 0 (Y * s -L s ) dK * k (s) = 0 (3.29) and      Y * t∧τ * k = Y * τ * k + τ * k t∧τ * k H * (s, x, Z * k+1 (s))ds + K * k+1 (τ * k ) -K * k+1 (t ∧ τ * k ) - τ * k t∧τ * k Z * k+1 (s)dB s ; L t∧τ * k ≤ Y * t∧τ * k and τ * k 0 (Y * s -L s ) dK * k+1 (s) = 0.
(3.30) Therefore for any k ≥ 1, by uniqueness, we have:

Z * k (s)1 {s≤τ * k } = Z * k+1 (s)1 {s≤τ * k } dt ⊗ dP -a.e and K * k+1 (s ∧ τ * k ) = K * k (s ∧ τ * k ), ∀s ≤ T.
Finally by concatenation let us dene the processes Z * and K * as follows: ∀t ≤ T ,

Z * t = Z * 1 (t)1 {t≤τ * 1 } + k≥2 Z * k (t)1 {τ * k-1 <t≤τ * k }
and

K * t =    K * 1 (t) if t ≤ τ * 1 ; (K * k+1 (t) -K * k+1 (τ * k )) + K * (τ * k ) for τ * k < t ≤ τ * k+1 , k ≥ 1.
The processes Z * and K * are well-dened. This is due to the fact that the sequence (τ * k ) k≥1 is of stationary type. Moreover K * is continuous non-decreasing and P-a.s. K * T (ω) < +∞ and

(Z * s (ω))
s≤T is ds-square integrable. Finally for any k ≥ 1 it holds:

     Y * t∧τ * k = Y τ * k + τ * k t∧τ * k H * (s, x, Z * s )ds + K * τ * k -K * t∧τ * k - τ * k t∧τ * k Z * s dB s ; L t∧τ * k ≤ Y * t∧τ * k and τ * k 0 (Y * s -L s ) dK * s = 0.
(3.31)

We now take the limit when k → +∞ and since once more (τ * k ) k≥1 is of stationary type, we have:

∀t ≤ T ,      Y * t = g(x) + T t H * (s, x, Z * s )ds + K * T -K * t - T t Z * s dB s ; L t ≤ Y * t and T 0 (Y * s -L s ) dK * s = 0, (3.32)
which completes the proof of ii).

We will now prove iv). Let ( Ȳ , Z, K) be another triple which satises i), ii) and iii). Then, using Itô's formula, we obtain:

(Y * t -Ȳt ) 2 = -2 T t (Y * s -Ȳs )(Z * s -Zs )dB s +2 T t (Y * s -Ȳs )(H * (s, x, Z * s ) -H * (s, x, Zs ))ds +2 T t (Y * s -Ȳs )d(K * s -Ks ) - T t |Z * s -Zs | 2 ds.
(3.33)

Next let P * be the probability, equivalent to P, dened as follows:

dP * = L T dP with L T := e T 0 H * (s,x,Z * s )-H * (s,x, Zs) Z * s -Zs 1 {Z * s -Zs =0} dBs-1 2 T 0 H * (s,x,Z * s )-H * (s,x, Zs) Z * s -Zs 1 {Z * s -Zs =0} 2 ds
and where

∆H * (s) := H * (s, x, Z * s ) -H * (s, x, Zs ) Z * s -Zs 1 {Z * s -Zs =0} is a P-measurable, R d -valued stochastic process such that ∀s ≤ T, H * (s, x, Z * s ) -H * (s, x, Zs ) = ∆H * (s) × (Z * s -Zs ).
As

|H * (s, x, z) -H * (s, x, z)| ≤ C(1 + x s )|z -z| then ∀s ≤ T, |∆H * (s)| ≤ C(1 + x s ).
It means that P * is actually a probability equivalent to P (by Lemma 2.1). Next for k ≥ 1, let τ k be the following stopping time:

τ k := inf{t ≥ 0, |Y * t | + | Ȳt | + t 0 |Z * s |ds + t 0 | Zs |ds ≥ k + |Y * 0 | + | Ȳ0 |} ∧ T.
As (Y * s -Ȳs )d(K * s -Ks ) ≤ 0 then going back to (3.33) to obtain:

(Y * t∧τ k -Ȳt∧τ k ) 2 ≤ (Y * τ k -Ȳτ k ) 2 + 2 τ k t∧τ k ( Ȳs -Y * s )(Z * s -Zs )d B s
where ( B t := B t -t 0 ∆H * (s)ds) t≤T is a Brownian motion under P * . Thus for any t ≤ T and

k ≥ 1, E P * [(Y * t∧τ k -Ȳt∧τ k ) 2 ] ≤ E P * [(Y * τ k -Ȳτ k ) 2 ]. ( 3
E P * [(Y * τ k -Ȳτ k ) 2 ] → k E P * [(Y * T -ȲT ) 2 ] = 0.
Going back now to (3.34), using Fatou's Lemma to obtain E P * [(Y * t -Ȳt ) 2 ] = 0 for any t ≤ T . It implies that Y * = Ȳ , P * and P-a.s. since the probabilities are equivalent. Thus we have also Z = Z and K = K, i.e. uniqueness.

We then have the following result: Theorem 3.2. Let (Y * , Z * , K * ) be the solution of the reected BSDE associated with (H * , g(x), L),

u * := (u * (t, x, Z * t )) t≤T and τ * = inf{t ∈ [0, T ], Y * t ≤ L t } ∧ T . Then, Y * 0 = J(u * , τ * ) = sup u∈U ,τ ≥0 J(u, τ ) (3.35)
i.e., (u * , τ * ) is an optimal strategy of the mixed control problem.

Proof: Recall that (Y * , Z * , K * ) veries: ∀t ≤ T ,

Y * t = g(x) + T t H * (s, x, Z * s )ds + K * T -K * t - T t Z * s dB s , L ≤ Y * and T 0 (Y * s -L s )dK * s = 0.
(3.36)

Next for k ≥ 1, let us set

γ k = inf{s ≥ 0, |L s | + |Y * s | + s 0 |Z * r | 2 dr ≥ k + |L 0 | + |Y * 0 |} ∧ T.
Now since Y * 0 is a deterministic constant we have:

Y * 0 = E u * Y * τ * ∧γ k + τ * ∧γ k 0 H * (s, x, Z * s )ds + K * τ * ∧γ k - τ * ∧γ k 0 Z * s dB s , = E u * Y * τ * ∧γ k + τ * ∧γ k 0 Γ(s, x, u * (s, x, Z * s ))ds + K * τ * ∧γ k - τ * ∧γ k 0 Z * s dB u * s .
From the denition of τ * and the properties of reected BSDEs we know that the process K * τ *

does not increase between 0 and τ * then K * τ * ∧γ k = 0. On the other hand, using the Burkholder-Davis-Gundy inequality and the assumptions on f we deduce that

{ t 0 Z * s dB u * s , t ∈ [0, γ k ]} is an P u * -martingale. Then, Y * 0 = E u * τ * ∧γ k 0 Γ(s, x, u * (s, x, Z * s ))ds + L τ * ∧γ k 1 {τ * ∧γ k <T } + g(x)1 {τ * ∧τ k =T } .
But the sequence of stopping times (γ k ) k≥1 is increasing, of stationary type and converges to T, then by taking the limit when k → +∞ we deduce that,

Y * 0 = J(u * , τ * ).
(3.37)

Now, let u be an admissible control and τ be an arbitrary stopping time. Since P and P u are equivalent probabilities on (Ω, F) we have:

Y * 0 = E u [Y * 0 ] = E u Y * τ ∧γ k + τ ∧γ k 0 H * (s, x, Z * s )ds + K * τ ∧γ k - τ ∧γ k 0 Z * s dB s , = E u Y * τ ∧γ k + τ ∧γ k 0 Γ(s, x, u s )ds + K * τ ∧γ k - τ ∧γ k 0 Z * s dB u s +E u τ ∧γ k 0 (H * (s, x, Z * s ) -H(s, x, Z * s , u s )) . But K * τ ∧γ k ≥ 0, H * (s, x, Z * s )-H(s, x, Z * s , u s ) ≥ 0 and { t 0 Z * s dB u s , t ∈ [0, γ k ]} is a P u -martingale, then, Y * 0 ≥ E u τ ∧γ k 0 Γ(s, x, u s )ds + L τ ∧γ k 1 {τ ∧τ k <T } + g(x)1 {τ ∧γ k =T } .
The sequence of stopping times (γ k ) k≥1 is increasing, of stationary type and converges to T, then by taking the limit as k → +∞ we get,

Y * 0 = J(u * , τ * ) ≥ E u τ 0 Γ(s, x, u s )ds + L τ 1 {τ <T } + g(x)1 {τ =T } = J(u, τ )
which is the claim.

Remark 3.1. The process Y * is the value function of the mixed optimal control problem, i.e., for any t ∈ [0, T ],

Y * t = esssup τ ≥t,u∈U E u τ t Γ(s, x, u s )ds + L τ 1 {τ <T } + g(x)1 {τ =T } |F t . (3.38)
This is another way to show uniqueness of the solution of (3.26). [START_REF] Benes | Existence of optimal stochastic control laws[END_REF] The Markovian framework: The HJB equation associated with the mixed control problem

In this section we are going to restrict the previous framework to the Markovian one and study the properties of the value function and the Hamilton-Jacobi-Bellman equation associated with the mixed control problem. We will show that the rst one provides the unique solution in viscosity sense of the second.

To begin with let us introduce the following spaces: i) For any γ ≥ 0, Π γ pg be the following space of functions Ψ dened on [0, T ] × R d , R-valued and such that

∀(t, x) ∈ [0, T ] × R d , |Ψ(t, x)| ≤ C(1 + |x| γ ) for some constants C ≥ 0.
ii) Π pg = ∪ γ≥0 Π γ pl . Next let us specify the functions σ, f , g, h and Γ, with their properties, in the following assumptions which we assume satised hereafter:

(HM): the functions σ, f , g, h and Γ verify:

a) σ : [0, T ] × R d → R d×d
Γ : [0, T ] × R d × A → R is continuous. In addition sup a∈A Γ belongs to Π pg . d) f : [0, T ] × R d × A → R d is continuous and the mapping x ∈ R d → f (t, x, a) is Lipschitz uniformly w.r.t (t, a).
Moreover it is of linear growth, i.e., sup a∈A f belongs to Π 1 pg and then for any

(t, x, a), |f (t, x, a)| ≤ C(1 + |x|) for some constant C ≥ 0. (4.1) Now for (t, x) ∈ [0, T ] × R d , let X t,x := (X t,x s
) s≤T be the solution of the following standard dierential equation:

dX t,x s = σ(s, X t,x s )dB s , s ∈ [t, T ]; X t,x s = x, s ≤ t.
(4.2)

In the following result we collect some properties of X t,x .

Proposition 4.1. (see e.g. [START_REF] Revuz | Continuous Martingales and Brownian Motion[END_REF]) The process X t,x satises the following estimates:

(i) For any q ≥ 2, there exists a constant C such that

E[ sup 0≤s≤T |X t,x s | q ] ≤ C(1 + |x| q ). ( 4

.3)

(ii) There exists a constant C such that for any t, t ∈ [0, T ] and x, x ∈ R d ,

E[ sup 0≤s≤T |X t,x s -X t ,x s | 2 ] ≤ C(1 + |x| 2 )(|x -x | 2 + |t -t |).2 (4.4)
Next for (t, x, z) ∈ [0, T ] × R d+d , let us introduce the function H * which is the same as the one given in (3.8) in this Markov setting:

H * (t, x, z) := sup a∈A H(t, x, z, a) (4.5)
where (with a ∈ A), H(t, x, z, a) := zσ -1 (t, x)f (t, x, a) + Γ(t, x, a).

(4.6)

The function H * is continuous in all its arguments (see Lemma 4.4 below), property which is needed later to deal with the HJB equation associated with the mixed control problem. Now to proceed, for (t, x, z) ∈ [0, T ] × R d+d and n, m ≥ 0, let us dene H * n,m by

H * n,m (t, x, z) := H * + (t, x, z)ρ n (x) -H * -(t, x, z)ρ m (x).
where for an x ∈ R d ,

ρ m (x) = 1 {|x|≤m} + (x + 1 + m)1 {-m-1≤x≤-m} + (-x + 1 + m)1 {m≤x≤m+1} .
The function H * n,m is also a truncation of H * which is moreover continuous w.r.t (t, x, z). It is uniformly Lipschitz in z and have the same monotonicity properties as H * n,m dened in (3.10), i.e., H * n,m is also increasing (resp. decreasing) w.r.t. n (resp. m) and lim m→∞ lim n→∞ H * n,m = lim n→∞ lim m→∞ H * n,m = H * since ρ m (x) 1. Finally the following estimate holds true:

| H * n,m (t, x, z)| ≤ φ(t, x, z) := C(1 + |x|)|z| + C(1 + |x| p ), ∀(t, x, z) ∈ [0, T ] × R d+d . (4.7)
Next for n, m ≥ 0, let ( Ȳ t,x,n,m , Zt,x,n,m , Kt,x,n,m ) be the unique solution of the following BSDE associated with ( H * n,m , g, h): ∀s ≤ T ,

         Ȳ t,x,
n,m , Kt,x,n,m ∈ S 2 and Zt,x,n,m ∈ H 2,d ( Kt,x,n,m increasing and Kt,x,n,m

0 = 0); Ȳ t,x,n,m s = g(X t,x T ) + T s H * n,m (u, X t,x u , Zt,x,n,m u )du + Kt,x,n,m T -Kt,x,n,m s - T s Zt,x,n,m u dB u ; h(s, X t,x s ) ≤ Ȳ t,x,n,m s and T 0 ( Ȳ t,x,n,m u -h(u, X t,x u ))d Kt,x,n,m u = 0.
(4.8) By comparison we have: ∀n, m ≥ 0, Ȳ t,x,n,m ≤ Ȳ t,x,n+1,m ≤ Ȳ t,x,n+1,m-1 .

(4.9)

On the other hand by Theorem 8.5 in [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF], since the framework is Markovian, then: i) ūn,m (t, x) = Ȳ t,x,n,m t , (t, x) ∈ [0, T ] × R d , is a deterministic continuous function which is moreover solution in viscosity sense of the following parabolic PDE: (4.13) By Proposition 3.1, there exist a P-measurable process Zt,x,m and an increasing continuous process Kt,x,m such that ( Ȳ t,x,m , Zt,x,m , Kt,x,m ) is a solution of the reected BSDE associated with ( H * m (s, X t,x s , z), g(X t,x T ), h(s, X t,x s )), i.e., ∀s ≤ T ,

       min ūn,m (t, x) -h(t, x), -∂ t ūn,m (t, x) -Lū n,m (t, x) -H * n,m (t, x, σ(t, x)∇ x ūn,m (t, x)) = 0, (t, x) ∈ [0, T ) × R d ; ūn,m (T, x) = g(x), x ∈ R d . (4.10) ii) For any s ∈ [t, T ], Ȳ t,x,n,m s = ūn,m (s, X t,x s ). (4.11) Next for m ≥ 0 let us set H * m (t, x, z) := lim n→∞ H * n,m (t, x, z) = H * + (t, x, z) -H * -(t, x, z)ρ m (x).
     Ȳ t,x,m s = g(X t,x T ) + T s H * m (r, X t,x r , Zt,x,m r )dr + Kt,x,m T -Kt,x,m s - T s Zt,x,m r dB r ; h(s, X t,x s ) ≤ Y t,x,m s and T 0 (Y t,x,m r -h(r, X t,x r ))dK t,x,m r = 0.
Note that Ȳ t,x,m veries the estimate (3.9), and P -a.s, Zt,x,m is ds-square integrable and Kt,x,m T < ∞. The inequalities (4.9) imply that for any

(t, x) ∈ [0, T ] × R d , ūn,m (t, x) ≤ ūn+1,m (t, x) ≤ ūn+1,m-1 (t, x).

Now for any

(t, x) ∈ [0, T ] × R d and m ≥ 0, let us set ūm (t, x) = lim n→∞ ūn,m (t, x).
Then by (4.11) and (4.13), it holds that for any

(t, x) ∈ [0, T ] × R d Ȳ m,t,x s = ūm (s, X t,x s ), ∀s ∈ [t, T ]. (4.14)
Next once more as a consequence of (4.9), the sequence ( Ȳ m,t,x ) m≥0 is decreasing and then so is the sequence of deterministic functions (ū m ) m≥0 . So let us dene the process Y t,x and the deterministic function u by:

Y t,x = lim m→∞ Ȳ t,x,m and u(t, x) = lim m→∞ ūm (t, x).
By (4.14), those latter objects are connected by the following relation:

Ȳ t,x s = u(s, X t,x s ), ∀s ∈ [t, T ]. (4.15)
As in Theorem 3.1, there exists a pair of processes

(Z t,x , K t,x ) valued in R d+1 such that (Y t,x , Z t,x , K t,x )
is the unique solution of the following reected BSDE: ∀s ∈ [0, T ],

     Y t,x s = g(X t,x T ) + T s H * (r, X t,x r , Z t,x r )dr + K t,x T -K t,x s - T s Z t,x r dB r ; h(s, X t,x s ) ≤ Y t,x s and T 0 (h(r, X t,x r ) -Y t,x r )dK t,x r = 0. (4.16)
Lemma 4.1. There exist two positive constants C and p such that for any

(t, x) ∈ [0, T ] × R d and m ≥ 0: i) |ū m (t, x)| + |u(t, x)| ≤ C(1 + |x| p ). (4.17)
ii)

E[ T 0 {| Zt,x,m r | 2 + |Z t,x r | 2 }dr] ≤ C(1 + |x| p ).
Proof: i) Recall the function φ introduced above in (4.7):

φ(t, x, z) := C(1 + |x|)|z| + C(1 + |x| p ).
By Proposition 2.1, there exists a triplet of processes (y t,x , z t,x , k t,x ) such that:

a) (y t,x , k t,x ) is R 1+1 -valued, k t,
x is non-decreasing and k t,x 0 = 0, (z t,x s (ω)) s≤T is R d -valued and ds-square integrable P -a.s. b) For any constant γ ≥ 1 and τ a stopping time valued in [0, T ],

E[|y t,x τ | γ ] < +∞. (4.18) c) For any s ≤ T ,      y t,x s = g(X t,x T ) + T s {C|z t,x r |(1 + |X t,x r |) + C(1 + |X t,x r | p )}dr - T s z t,x r dB r + k t,x T -k t,x s ; y t,x s ≥ l t,x s := h(s, X t,x
s ) and T 0 (y t,x r -l t,x r )dk t,x r = 0.

(4.19) Now by the standard comparison result of solutions of reected BSDEs and thanks to (4.7), one has:

l t,x s ≤ Ȳ t,x,n,m s ≤ y t,x
s , ∀s ≤ T.

(4.20) Next let Pt,x be the probability, equivalent to P, dened as follows:

d Pt,x = M t,x
T dP where

M t,x T := exp{ T 0 Cσ -1 (r, X t,x r ) (z t,x r )C(1 + |X t,x r |)dB r -1 2 T 0 Cσ -1 (r, X t,x r ) (z t,x r )(1 + |X t,x r |) 2 dr}
where is bounded measurable function such that (z).z = |z|, ∀z = (z i ) i=1,...,d ∈ R d (see the proof of Lemma 2.2 for its dention). Under Pt,x , X t,x is a weak solution of the following SDE:

dX t,x s = C (z t,x s )C(1 + |X t,x s |)ds + σ(s, X t,x s )d Bs , s ∈ [t, T ]; X t,x s = x, s ≤ t. (4.21)
were B is a Brownian motion under Pt,x . Then it veries the following estimate:

∀q ≥ 2, Ēt,x [ sup 0≤s≤T |X t,x s | q ] ≤ C(1 + |x| q ) (4.22)
where Ēt,x is the expectation under Pt,x ; the constant C does not depend on t, x. Next writing the reected BSDE(4.19) under the probability Pt,x reads: For any s ≤ T ,

     y t,x s = g(X t,x T ) + T s C(1 + |X t,x r | p )dr - T s z t,x r d Br + k t,x T -k t,x s ;
y t,x s ≥ l t,x s := h(s, X t,x s ) and T 0 (y t,x r -l t,x r )dk t,x r = 0.

(4.23)

Taking now into account of (2.5), we obtain as previously (see the proof of Theorem 3.2 or Remark 3.1) the following representation for y t,x : ∀s ≤ T ,

y t,x s = ess sup τ ≥s Ēt,x [ τ s C(1 + |X t,x r | p )dr + l t,x τ 1 {τ <T } + g(X t,x T )1 {τ =T } |F s ].
Next by the polynomial growth of h and g we deduce that: ∀s ≤ T ,

|y t,x s | ≤ C(1 + Ēt,x [sup s≤T |X t,x s | q |F s ])
for some constants C and q. Then by (4.22), we deduce that

Ēt,x [|y t,x t |] ≤ C(1 + |x| p )
for some xed C and p. Going back now to (4.20), take s = t, expectation w.r.t Pt,x (which is equivalent to P) and since Y t,x,n,m t is deterministic one deduces that

h(t, x) ≤ ūn,m (t, x) ≤ Ēt,x [y t,x t ] ≤ C(1 + |x| p ).
The proof is now completed in taking the limit w.r.t n then m since h is polynomial growth .

ii) It is obtained by the use of Itô's formula with |Y t,x s | 2 and | Ȳ m,t,x s | 2 and in taking into account the representations (4.14) and (4.15), the polynomial growths (4.17) of u and ūm , estimate (4.3) on X t,x and nally the fact that Z t,x s = Zm,t,x

s = 0 for s ∈ [0, t].
Remark 4.1. By the polynomial growth of u and estimate (4.3), we have also

E[sup s≤T |Y t,x s | p ] < ∞, ∀p ≥ 1, (4.24)
Next let Φ(t, x, z) be a function from [0, T ] × R d+d into R which we assume continuous in all its arguments. Let us now consider the following PDE with obstacle:

   min v(t, x) -h(t, x), -∂v ∂t (t, x) -Lv(t, x) -Φ(t, x, ∇ x v(t, x)σ(t, x)) = 0, (t, x) ∈ [0, T [×R d ; v(T, x) = g(x), x ∈ R d , (4.25
) where ∇ x is the derivative w.r.t. x and L is second order partial dierential operator associated with X t,x .

i.e

L = 1 2 i,j=1,d (σσ ) ij (t, x)∂ 2 x i x j .
In the case when we take Φ = H * , we obtain the HJB equation associated with the mixed control problem in the Markov framework.

Next let us dene the notion of a solution of (4.25). Let v be a funtion dened on [0, T ]×R d which is moreover locally bounded on each (t, x). We dene the upper (resp. lower) semicontinuous enveloppe of v by: ∀ The denitions of the limiting parabolic superjet (resp. subjet) J2,+ v (resp. J2,-v) of an upper (resp. a lower) semi-continuous function v dened on [0, T ] × R d are given e.g. [START_REF] Crandall | User's guide to viscosity solutions of second order partial dierential equations[END_REF], pp.47, 11.

(t, x) ∈ [0, T ] × R d , v * (t
(i) v * (T, x) ≥ g(x) (resp. v * (T, x) ≤ g(x)), ∀x ∈ R d ; (ii) For any (t, x) ∈ [0, T ) × R d and (p, q, X) ∈ J2,-v * (t, x) (resp. J2,+ v * (t, x)), min v * (t, x) -h(x), -p - 1 2 T r[σ Xσ(t, x)] -Φ(t, x, qσ(t, x)) ≥ 0 (resp. min v * (t, x) -h(x), -p - 1 2 T r[σ Xσ(t, x)] -Φ(t, x, qσ(t, x)) ≤ 0). (4.26)
b) It is called a viscosity solution if it is both a viscosity subsolution and supersolution.

Remark 4.2. There is another denition of the notion of viscosity solution which uses test functions which we will use sometimes later on (one can see e.g. [START_REF] Crandall | User's guide to viscosity solutions of second order partial dierential equations[END_REF] in pp. [START_REF] Karoui | Reected solutions of backward SDE'S, A related obstacle problems for PDE[END_REF][START_REF] Karoui | Les aspects probabilistes du contrôle stochastique[END_REF] for more details.

To begin with we are going to deal with the issue of comparison principle, and consequently uniqueness of the solution, for the PDE with obstacle (4.25). For that let us introduce the following assumptions on the function Φ:

(HΦ): ii) For any κ > 0, there exists a function Ψ κ from [-2κ, 2κ] into R + , continuous, Ψ κ (0) = 0 and such that for any (t, z, z

) ∈ [0, T ] × R 2d , |x| ≤ κ, |y| ≤ κ, Φ(t, x, zσ(t, x)) -Φ(t, y, z σ(t, y)) ≤ C κ (|z| + |z |)|x -y| + C κ |z -z | + Ψ κ (|x -y|) (4.28)
where C κ is a positive constant which may depend on κ.

As a preliminary result we have: Lemma 4.2. Assume that the function Φ veries (HΦ)-i). If v is a supersolution of (4.25) which belongs to Π pg , i.e.,

∀(t, x) ∈ [0, T ] × R d , |v(t, x)| ≤ C(1 + |x| 2γ )
for some constants C and γ non negative. Then there exists λ 0 > 0 such that for any λ ≥ λ 0 and θ > 0, (v(t, x) + θe -λt (1+ | x | 2γ+2 ) is a supersolution for (4.25) . P roof. We assume w.l.o.g. that the function v(t, x) is lsc. First note that the condition at T holds true since θe -λT (1+ | x | 2γ+2 ) ≥ 0. Next let t < T and ϕ ∈ C 1,2 be such that the function ϕ -(v + θe -λt (1+ | x | 2γ+2 ) has a local maximum in (t, x) which is equal to 0. Since v(t, x) is a supersolution for (4.25), then we have:

min v(t, x) -h(t, x), -∂ t ϕ(t, x) -θe -λt (1+ | x | 2γ+2 ) - 1 2 T r σ.σ (t, x)D 2 xx ϕ(t, x) -θe -λt | x | 2γ+2 -Φ(t, x, ∇ x (ϕ(t, x) -θe -λt | x | 2γ+2 )σ(t, x)) ≥ 0.
Hence

(v(t, x) + θe -λt (1+ | x | 2γ+2 )) -h(t, x) ≥ v(t, x) -h(t, x) ≥ 0.
(4.29)

On the other hand:

-∂ t ϕ(t, x) -θe -λt (1+ | x | 2γ+2 ) -1 2 T r σ.σ (t, x)D 2 xx ϕ(t, x) -θe -λt | x | 2γ+2 -Φ(t, x, ∇ x (ϕ(t, x) -θe -λt | x | 2γ+2 )σ(t, x)) ≥ 0. Therefore -∂ t ϕ(t, x) - 1 2 T r σ.σ (t, x)D 2 xx ϕ(t, x) -Φ(t, x, ∇ x (ϕ(t, x))σ(t, x)) ≥ θλe -λt (1+ | x | 2γ+2 ) - 1 2 θe -λt T r σ.σ (t, x)D 2 xx | x | 2γ+2 + Φ(t, x, ∇ x (ϕ(t, x) -θe -λt | x | 2γ+2 )σ(t, x)) -Φ(t, x, ∇ x (ϕ(t, x))σ(t, x)) ≥ θλe -λt (1+ | x | 2γ+2 ) - 1 2 θe -λt T r σ.σ (t, x)D 2 xx | x | 2γ+2 -C(1 + |x|)|∇ x (-θe -λt | x | 2γ+2 )σ(t, x))| ≥ θλe -λt (1+ | x | 2γ+2 ) - 1 2 θe -λt (C σ ) 2 (2γ + 2)(2γ + 1)|x| 2γ -C.C σ (1 + |x|)(2γ + 2)θe -λt | x | 2γ+1 (4.30)
where C σ is the constant of boundedness of σ. But there exists a constant λ 0 > 0 such that for any λ ≥ λ 0 , the right-hand side is positive for any θ > 0. Consequently for any λ ≥ λ 0 and θ > 0, (v + θe -λt

(1+ | x | 2γ+2
) is a supersolution of (4.25).

In the follwowing lemma, for which we omit the proof since it is classical, we transform the PDE (4.25) into another one which is more adapted to show uniqueness of the solution of (4.25). Lemma 4.3. Let v(t, x) be an R-valued locally bounded function dened on [0, T ] × R d . The function v is a viscosity subsolution (resp. supersolution) of (4.25) if and only if v(t, x) = e t v(t, x), (t, x) ∈ [0, T ] × R d , is a viscosity subsolution (resp. supersolution) of the following PDE with obstacle:

       min v(t, x) -e t h(t, x), -∂ t v(t, x) + v(t, x) -Lv(t, x) -e t Φ(t, x, e -t ∇ x v(t, x)σ(t, x)) = 0, (t, x) ∈ [0, T ) × R d ; v(T, x) = e T g(x).
We now address the question of comparison of subsolutions and supersolutions of the PDE (4.25). Proposition 4.2. Assume that Φ veries (HΦ)-i), ii). Let u (resp. v) be a subsolution (resp. supersolution) of (4.25). If u, v belong to Π pg , then u ≤ v.

Proof. First let γ be a positive constant such that

|u(t, x)| + |v(t, x)| ≤ C(1 + |x| 2γ ).
We now that there exists λ 0 such that for any λ ≥ λ 0 and θ > 0 such that v(t, x) + θe -λt (1 + |x| 2γ+2 ) still a supersolution of (4.25). Therefore it is enough to show that u(t, x) ≤ v(t, x) + θe -λt (1 + |x| 2γ+2 ) and then to take the limit as θ → 0 to obtain the desired result. Next the growth condition on u and v implies the existence of a positive constant R such that for any t ∈ [0, T ], |x| ≥ R, u(t, x) -(v(t, x) + θe -λt (1 + |x| 2γ+2 )) < 0. Finally by Lemma 4.3, e t u (resp. e t (v + θe -λt (1 + |x| 2γ+2 ))) is a viscosity subsolution (resp. supersolution) of (4.26) such that

|e t u(t, x)| ≤ C(1 + |x| 2γ ) and |e t (v(t, x) + θe -λt (1 + |x| 2γ+2 ))| ≤ C(1 + |x| 2γ+2 ).
Therefore to obtain the proof it is enough to show that if u (resp. w) is a subsolution (resp. supersolution) of (4.26) such that: i) There exits R > 0 such that u(t, x) -w(t, x) < 0 for any |x| ≥ R and t ∈ [0, T ]; ii)

|u(t, x)| ≤ C(1 + |x| 2γ ) and |w(t, x)| ≤ C(1 + |x| 2γ+2 ). Then u(t, x) ≤ w(t, x), for any (t, x) ∈ [0, T ] × R d .
We will proceed by contradiction and suppose that there exists (t 0 , x 0 ) ∈ [0, T ) × R d such that u(t 0 , x 0 ) -w(t 0 , x 0 ) > 0 ; w.l.o.g we assume u usc and w lsc. So let ( t, x) be such that:

max (t,x)∈[0,T ]×R d (u(t, x) -w(t, x)) = max (t,x)∈[0,T [×B R (u(t, x) -w(t, x)) = (u( t, x) -w( t, x)) = η > 0, (4.31)
where

B R := {x ∈ R d ; |x| < R} and ( t, x) ∈ [0, T [×B R .
Now let us take θ and β ∈ (0, 1], and w.l.o.g we assume γ ≥ 2. Then, for a small > 0, let us dene:

Φ (t, x, y) = u(t, x) -w(t, y) - 1 2 |x -y| 2γ -θ(|x -x| 2γ+2 + |y -x| 2γ+2 ) -β(t -t) 2 . (4.32)
Since u is usc and w is lsc, then there exists a (t , x , y ) ∈ [0, T ] × BR × BR such that:

Φ (t , x , y ) = max (t,x,y)∈[0,T ]× BR × BR Φ (t, x, y)
where BR is the closure of B R . Therefore from the inequality 2Φ (t , x , y ) ≥ Φ (t , x , x ) + Φ (t , y , y ), we deduce (4.33)

Consequently 1 |x -y | 2γ is bounded (thanks to the growth conditions on u and w), and as → 0, |x -y | → 0. By the boundedness of the sequences, one can substract subsequences which we still index by such that (x ) (resp. (y ) , resp. (t ) ) converges to x (resp. x, resp. t) when → 0. Next u( t, x) -w( t, x) ≤ Φ (t , x , y ) ≤ u(t , x ) -w(t , y ).

(4.34)

As u is usc and w is lsc, then we have:

u( t, x) -w( t, x) ≤ lim inf →0 Φ (t , x , y ) ≤ lim sup →0 Φ (t , x , y ) ≤ lim sup →0 (u(t , x ) -w(t , y )) ≤ u(t, x) -w(t, x) ≤ u( t, x) -w( t, x). ( 4

.35)

It follows that: i) the sequence (Φ (t , x , y )) is convergent to u( t, x) -w( t, x); ii) lim sup →0 (u(t , x ) -w(t , y )) = u( t, x) -w( t, x) and then taking into account of (4.34), we deduce that lim →0 (u(t , x ) -w(t , y )) = u( t, x) -w( t, x);

iii) From (4.32), we deduce that (t , x , y ) → ( t, x, x).

(4.36)

Next lim →0 (u(t , x ) -w(t , y )) = u( t, x) -w( t, x) = η > 0.
Therefore there exists a subsequence of ( ) such that

(u(t , x ) -w(t , y )) ≥ η 2 .
But w is a supersolution then w(t , y ) ≥ e t h(t , y ) and by continuity of h one can nd a subsequence such that |e t h(t , y ) -e t h(t , x )| < η 4 . Therefore for this last subsequence it holds u(t , x ) ≥ e t h(t , x ) + η 4 .

(4.37)

To proceed let us consider this latter subsequence and let us denote by

ϕ (t, x, y) = 1 2 |x -y| 2γ + θ(|x -x| 2γ+2 + |y -x| 2γ+2 ) + β(t -t) 2 .
(4.38)

Then we have: at the point (t , x , y ) (we choose small enough in such a way that t < T , and |x | < R and |y | < R), for any υ > 0 we can nd c, c 1 ∈ R, q 1 , q 2 ∈ R d and X, Y ∈ S d , such that:

                           D t ϕ (t, x, y) = 2β(t -t), D x ϕ (t, x, y) = γ (x -y)|x -y| 2γ-2 + θ(2γ + 2)(x -x)|x -x| 2γ , D y ϕ (t, x, y) = -γ (x -y)|x -y| 2γ-2 + θ(2γ + 2)(y -x)|y -x| 2γ , B(t, x, y) := D 2 x,y ϕ (t, x, y) = 1 a 1 (x, y) -a 1 (x, y) -a 1 (x, y) a 1 (x, y) + a 2 (x) 0 0 a 2 (y) with a 1 (x, y) = γ|x -y| 2γ-2 I + γ(2γ -2)(x -y)(x -y) * |x -y| 2γ-4 and a 2 (x) = θ(2γ + 2)|x -x| 2γ I + 2 θγ(2γ + 2)(x -x)(x -x) * |x -x| 2γ-2 . ( 4
               (c, q 1 , X) ∈ J2,+ u(t , x ) and (-c 1 , q 2 , Y ) ∈ J2,-w(t , y ), q 1 = D x ϕ (t , x , y ) = γ (x -y )|x -y | 2γ-2 + θ(2γ + 2)(x -x)|x -x| 2γ , q 2 = -D y ϕ (t , x , y ) = γ (x -y )|x -y | 2γ-2 -θ(2γ + 2)(y -x)|y -x| 2γ , c + c 1 = D t ϕ (t , x , y ) = 2β(t -t) and nally -( 1 υ + ||B(t , x , y )||)I ≤ X 0 0 -Y ≤ B(t , x , y ) + υB(t , x , y ) 2 .
(4.40)

Taking now into account (4.37), and the denition of viscosity solution, we get:

-c -1 2 T r[σ (t , x )Xσ(t , x )] + u(t , x ) -e t Φ(t , x , e -t q 1 σ(t , x )) ≤ 0 and c 1 -1 2 T r[σ (t , y )Y σ(t , y )] + w(t , y
) -e t Φ(t , y , e -t q 2 σ(t , y )) ≥ 0.

Then

u(t , x ) -w(t , y ) -c -c 1 ≤ 1 2 T r[σ (t , x )Xσ(t , x ) -σ (t , y )Y σ(t , y )] +e t Φ(t ,
x , e -t q 1 σ(t , x )) -e t Φ(t , y , e -t q 2 σ(t , y )). It follows that:

B + υB 2 ≤ C( 1 |x -y | 2γ-2 + 1 2 |x -y | 4γ-4 ) I -I -I I + C 1 θI. ( 4

.42)

Choosing now υ = , yields:

B + B 2 ≤ C (|x -y | 2γ-2 + |x -y | 4γ-4 ) I -I -I I + C 1 θI. (4.43)
From the Lipschitz continuity of σ, (4.40) and (4.43) we have:

1 2 T r[σ (t , x )Xσ(t , x ) -σ (t , y )Y σ(t , y )] ≤ C (|x -y | 2γ + |x -y | 4γ-2 ) + C 1 θ. (4.44)
Next taking into account of (HΦ) we have:

e t Φ(t , x , e -t q 1 σ(t , x )) -e t Φ(t , y , e -t q 2 σ(t , y ))

≤ e t {C R (|e -t q 1 | + |e -t q 2 |)|x -y | + C R |e -t q 1 -e -t q 2 | + Ψ R (|x -y |)} But lim →0 (|q 1 | + |q 2 |)|x -y | = 0 and lim →0 |q 1 -q 2 | = 0. It follows that lim sup →0 e t Φ(t ,
x , e -t q 1 σ(t , x )) -e t Φ(t , y , e -t q 2 σ(t , y )) ≤ 0.

Next go back to (4.41) take the superior limit w.r.t on each hand-side and take into account of (4.44) to obtain:

u( t, x) -w( t, x) ≤ C 1 θ.
Send now θ → 0 to obtain that u( t, x) -w( t, x) = 0 which is contradictory. The proof is now complete.

As a by-product we have Corollary 4.1. Under (HΦ), if the PDE (4.25) has a solution in Π pg , then it is unique and continuous.

Let us now go back to the HJB equation associated with the mixed control problem, i.e., equation (4.25) when Φ is replaced with H * whihc reads Proof: i) Let (t, x, z) and (t , x , z ) be xed. Without loss of generality we assume that Next as f , σ -1 and Γ are continuous and A is compact then the right-hand side of the previous inequality goes to 0 when (t , x , z ) → (t, x, z). Thus H * is continuous. where Ψ Γ κ is the modulus of continuity of Γ on [0, T ] × B(0, κ) × A ( B(0, κ) is the closure in R d of the open ball centered in 0 and of radius κ). In the last inequality we have used the fact that f is Lipschitz w.r.t x. The proof now follows since |x| ≤ κ.

   min v(t, x) -h(t, x), -∂v ∂t (t, x) -Lv(t, x) -H * (t, x, ∇ x v(t, x)σ(t, x)) = 0, (t, x) ∈ [0, T [×R d ; v(T, x) = g(x), x ∈ R d . ( 4
|t -t | + |x -x | + |z -z | ≤ 1. An easy computation shows that |H * (t, x, z) -H * (t , x , z )| = |H * (t, x, z) -H * (t , x , z) + H * (t , x , z) -H * (t , x , z )| ≤ |z| sup a∈A |σ -1 (t, x)f (t, x, a) -σ -1 (t , x )f (t ,
We now have the following result related to ūm . Proposition 4.3. For any m ≥ 0, ūm is continuous and is the unique viscosity solution in Π g of the following PDE with obstacle: Proof: First recall tha by (4.17), ūm belongs to Π pg . Next let us show that ūm is a supersolution of (4.47). The function ūm is lsc and ūm (T, x) = g(x). Thus the terminal condition is veried. On the other hand, ūm * = ūm . So let t < T and (p, q, X) ∈ J2,-ūm (t, x). As ūm = lim n ūn,m then thanks to Lemma 6.1 in [START_REF] Crandall | User's guide to viscosity solutions of second order partial dierential equations[END_REF], there exist sequences:

          
n j → +∞, (t j , x j ) → (t, x), (p j , q j , X j ) ∈ J2,ūn j ,m (t j , x j ), such that (p j , q j , X j ) → (p, q, X).

But for any j, -p j -1 2 T r[σ (t j , x j )X j σ(t j , x j )] ≥ H * n j ,m (t j , x j , q j σ(t j , x j ))

since ūn,m is a viscosity solution of (4.10). But by Dini's Theorem, ( H * n,m ) n converges uniformly to H * m on compact subsets. Therefore take the limit w.r.t. j in each hand-side of the previous inequality to obtain -p - Finally as ūm (t, x) = Ȳ m,t,x t ≥ h(t, x), then ūm is a viscosity supersolution of (4.47). Let us now show that ūm * is a subsolution of (4.47). Let (t, x) ∈ [0, T ) × R d . We obvioulsy have ūm * (t, x) ≥ h(t, x) since h is continuous. So assume that ūm * (t, x) > h(t, x) and let (p, q, X) ∈ J 2,-ūm * (t, x). As ūm = lim n ūn,m and ūn,m is continuous then (see [START_REF] Barles | Solutions de Viscosité des équations de Hamilton-Jacobi[END_REF], pp.91). Then once more by Lemma 6.1 in [START_REF] Crandall | User's guide to viscosity solutions of second order partial dierential equations[END_REF], there exist sequences such that n j → +∞, (t j , x j , ūn j ,m (t j , x j )) → (t, x, ūm * (t, x)), (p j , q j , X j ) ∈ J 2,-ūn j ,m (t j , x j ), and (p j , q j , X j ) → (p, q, X).

But there exists a subsequence which we still index by j such that for any j, ūn j ,m (t j , x j ) > h(t j , x j ) since ūm * (t, x) > h(t, x). The subsolution property of ūn j ,m implies that:

-p j -1 2 T r[σ (t j , x j )X j σ(t j , x j )] ≤ H * n j ,m (t j , x j , q j σ(t j , x j )).

Hence, as for the supersolution property, taking the limit as j → +∞, we conclude that: Proof: The proof is obtained in the same way as we did for ūm in the previous proposition since: i) by (4.17) we now that u belongs to Π pg ; ii) ūm veries the PDE (4.53), is continuous and the sequence (ū m ) m is decreasing and converges to u; iii) H * veries (4.27) and (4.28). The details are left to the care of the reader.

Remark 4.3. The characterization of u (see Remark 3.1) as the value function of the mixed control problem allows to show directly that u is continuous even if this proof is rather tedious. However there is no way to show that ūm is continuous without using the PDEs as we did previously in Proposition 4.3.

-

  L s ) dK * m k (s) = 0.Now if we consider the reected BSDE satised by(Y * m , Z * m k , K * m k ) on [0, τ * k ] and the reected BSDE satised by (Y * m , Z * m k+1 , K * m k+1 ) on [0, τ * k ],we get by uniqueness: ∀k ≥ 1, ∀m ≥ 1

  is a bounded continuous function, Lipschitz w.r.t x invertible and its inverse σ -1 (t, x) is bounded and continuous. b) h : [0, T ] × R d → R and g : R d → R. They are continuous and belong to Π pg . c)

  x,m = lim n→∞ Ȳ t,x,n,m .

Denition 4 . 1 .

 41 Let v be function dened on [0, T ] × R d , R-valued and locally bounded: a) It is said a viscosity supersolution (resp. subsolution) of (4.25) if:

  i) For any (t, x, z, z ) ∈ [0, T ] × R d+d+d , Φ(t, x, z) -Φ(t, x, z ) ≥ -C(1 + |x|)|z -z |.

( 4 .

 4 41)But from (4.39) there exist two constants C and C 1 (which may change from line to line) such that:||a 1 (x , y )|| ≤ C|x -y | 2γ-2 and (||a 2 (x )|| ∨ ||a 2 (y )||) ≤ C 1 θ. On the other hand we have (B := B(t , x , y )) B ≤ C |x -y | 2γ-2 I -I -I I + C 1 θI.

  x , a)| + sup a∈A |Γ(t, x, a) -Γ(t , x , a)| + C(1 + |x |)|z -z |.

  ii) |H * (t, x, z) -H * (t, x, z )| = | sup a∈A {zσ -1 (t, x)f (t, x, a) + Γ(t, x, a)} -sup a∈A {z σ -1 (t, x)f (t, x, a) + Γ(t, x, a)}| ≤ sup a∈A |z -z ||σ -1 (t, x)f (t, x, a)| ≤ C(1 + |x|)|z -z | since σ -1 isbounded and f of linear growth. Thus H * veries (4.27). Finally let us show that H * veries (4.28). Let κ be xed and (t, z, z ) ∈ [0, T ] × R 2d , x, y ∈ R d such that |x| ≤ κ and |y| ≤ κ. H * (t, x, zσ(t, x)) -H * (t, y, z σ(t, y))= sup a∈A {zf (t, x, a) + Γ(t, x, a)} -sup a∈A {z f (t, y, a) + Γ(t, y, a)} ≤ sup a∈A {zf (t, x, a) -z f (t, y, a)} + sup a∈A {Γ(t, x, a) -Γ(t, y, a)} ≤ C(1 + |x|)|z -z | + sup a∈A {z (f (t, x, a) -f (t, y, a))} + sup a∈A {Γ(t, x, a) -Γ(t, y, a)} ≤ C(1 + |x|)|z -z | + C|z ||x -y| + Ψ Γ κ (|x -y|).

  min [ū m (t, x) -h(t, x), -∂ t ūm (t, x) -Lū m (t, x) -H * m (t, x, σ(t, x)∇ x ūm (t, x)) = 0, (t, x) ∈ [0, T ) × R d ; ū(T, x) = g(x).

  ūm * = lim n→∞ sup * ūn,m where lim n→∞ sup * ūn,m (t, x) = lim sup n→∞,(t ,x )→(t,x),t <T ūn,m (t , x )

  σ (t, x)Xσ(t, x)] -H * m (t, x, qσ(t, x)) ≤ 0, i.e. ūm veries the subsolution property in (t, x) ∈ [0, T ) × R d . It remains to show the terminal condition is satised i.e. ūm * (T, x) = g(x). It is classical, however we give it in its main steps for completeness. First note that for any x ∈ R d , ūm * (T, x) ≥ ūn 0 ,m (T, x) = g(x).

( 4 .

 4 48)Assume now that for somex 0 ∈ R d , ūm * (T, x 0 ) -g(x 0 ) = 2 > 0 (4.49)Finally to show continuity of ūm and uniqueness of the solution it is enough to show that

Theorem 4 . 1 .

 41 The function u is continuous and is the unique viscosity solution in Π pg of the following PDE with obstacle:(t, x) -h(t, x), -∂ t u(t, x) -Lu(t, x) -H * (t, x, σ(t, x)∇ x u(t, x))] = 0, (t, x) ∈ [0, T ) × R d ;u(T, x) = g(x).

2

  Formulation of the problem. Study of a specic reected BSDE Let Ω = C [0, T ]; R d be the space of R d -valued continuous function on [0, T ] endowed with the metric of uniform convergence on [0, T ]. Denote by F the Borel σ-eld over Ω. Next for ω ∈ Ω and t ≤ T , let us set ||ω|| t := sup 0≤s≤t |ω s |. Let x := (x s ) s≤T be the coordinate process on Ω, i.e., x s (ω) = ω(s) and denote by (F 0

t := σ(x s , s ≤ t)) t∈[0,T ] , the ltration on Ω generated by x.

  10) where c is a constant which does not depend on n. As for any n ≥ 0, Y n ≤ Y n+1 then P-a.s. for any t ≤ T , Y n t → n Y t = lim inf n→+∞ Y n t . Therefore P-a.s. Y n τ → n Y τ and by Fatou's Lemma and (2.10) we have E[|Y τ | γ ] ≤ c.

  11) since |Y n t∧τ k | ≤ θk , for any t ≤ T , and Y 0 ≤ Y n ≤ Y . Next once more by Itô's formula we obtain:

  .34) But for any γ ≥ 1 and τ stopping time, E[|Y τ | γ + |Y * τ | γ ] ≤ C and by Lemma 2.1 there exists a constant p > 1 such that E[L p T ] < ∞. Then there exists a constant C such that for any stopping time τ , E P * [(Y * τ -Ȳτ ) 2 ] ≤ C. Consequently, the process (Y * -Ȳ ) 2 is of class [D] under the probability P * . Therefore (one can see e.g. [8], Theorem 21, pp. 36)

  .45) To begin with we will focus on the properties of the function H * . Lemma 4.4. i) The function H * is continuous in (t, x, z). ii) H * veries (4.27) and (4.28).

  H * m satises the assumptions (HΦ). But for any (t, x, z, z )| H * m (t, x, z) -H * m (t, x, z )| ≤ 2|H * (t, x, z) -H * (t, x, z )| ≤ 2C(1 + |x|)|z -z | since f is of linear growth, σ -1 is bounded and |ρ m | ≤ 1. Thus (4.27) is satised. On the other hand, H * m (t, x, zσ(t, x)) -H * m (t, y, z σ(t, y)) = H * + (t, x, zσ(t, x)) -H * + (t, y, z σ(t, y)) * -(t, x, zσ(t, x)) -H * -(t, y, z σ(t, y))) m (x) -H * -(t, y, z σ(t, y)) ρ m (x) -ρ m (y)). Now to conclude it is enough to remark that: i) H * veries (4.28) and to use the inequality u + -v + ≤ (u -v) + with (I); ii) H * veries (4.28) and to use the inequality u --v -≤ (v -u) + with (II); iii) H * -veries (4.27) and ρ m is continuous Lipschitz. The proof is now complete.We are now ready to state the main result of this section.

	+
	(I)
	{-(H (II)	(III)

ρ (

and let us construct a contradiction. Let (t k , x k ) k≥1 be a sequence in [0, T ) × R d such that:

(t k , x k ) → (T, x 0 ) and ūm (t k , x k ) → ūm * (T, x 0 ) as k → ∞.

Since ūm * is usc and of polynomial growth and taking into account of (4.48) and the inequalilty g(x 0 ) ≥ h(T, x 0 ), we can nd a sequence

n → ūm * and, on some neighbourhood B n of (T, x 0 ) we have:

(4.50)

After possibly passing to a subsequence of (t k , x k ) k≥1 we can then assume that it holds on

small enough in such a way that B n k ⊂ B. Now since ūm * is locally bounded then there exists ζ > 0 such that |ū m * | ≤ ζ on B n . We can then assume that n ≥ -2ζ on B n . Next let us dene:

)} ∧ T. Using Itô's formula and taking into account (4.50), (4.51) and (4.52) to obtain:

)ds] = 0. Therefore taking the limit in the previous inequalities yields: But this is contradictory since n → ūm * pointwise as n → ∞. Thus for any x ∈ R d we have ūm * (T, x) = g(x) and this completes the proof.