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A multidimensional singular stochastic control problem on a finite time horizon

A multidimensional singular stochastic control problem on a finite time horizon A singular stochastic control problem in n dimensions with timedependent coefficients on a finite time horizon is considered. We show thahe value function for this problem is a generalized solution of the corresponding HJB equation with locally bounded second derivatives with respeco the space variables and the first derivative with respeco time. Moreover, we prove that an optimal control exists and is unique. Singular stochastic control is a class of problems in which one is allowed to change the drift of a Markov process (usually a diffusion) at a price proportional to the variation of the control used. Admissible controls do not have to be absolutely continuous with respeco the Lebesgue measure and they may have jumps. This setup is natural for many problems of practical interest, including portfolio selection in finance, control of queueing networks and spacecraft control, to mention just a few examples. The reader is referred to Chapter VIII of [5] for more ormation and basic references. One-dimensional singular stochastic control problems are well understood by now, see, e.g., [2] and the references given there. In this case, if the running cost is convex, the optimal control makes the underlying process a reflected diffusion ahe boundary of the so-called nonaction region C. In the case of a diffusion with time-independent coefficients and discounted cost on the inite time horizon, C is just an interval and the value function enjoys 2010 Mathematics Subject Classification. Primary: 93E20; Secondary: 35Q93. Key wor and phrases. Singular stochastic control, generalized derivative, HJB equation, optimal control. M. Boryc and L. Kruk C 2 -regularity (smooth fit). Both C 2 -regularity of the value function and the characterization of the optimally controlled process have been extended to the case of singular control for the two-dimensional Brownian motion [14]. In n 3 dimensions, except for "close to one-dimensional" cases of a single push direction [15, 16] and the radially symmetric running cost [9], only partial results are known. For example, for optimal control of the Brownian motion on the inite time horizon, regularity of the boundary of C away from some "coer points" was shown in [17] and a characterization of the optimal control as a solution of the corresponding modified Skorokhod problem was given in [8]. In this paper we consider a n-dimensional singular stochastic control problem on a finite time horizon in which state is goveed by a linear stochastic differential equation with time-dependent coefficients, the running cost is convex and controls may act in any direction. We provide estimates for the corresponding value function. These estimates imply thahe value function has locally bounded generalized derivatives of the second order with respeco the space variable and of the first order with respeco the time variable. These properties are needed to consider the value function as a solution of the corresponding parabolic Hamilton­Jacobi­Bellman (HJB) equation in some generalized sense and to show existence and uniqueness of an optimal control. Similar results have been shown in Theorem 2.1 and Theorem 3.4 from [2] in the one-dimensional case with a single push direction. The corresponding results for a multidimensional singular stochastic control problem on the inite time horizon with time-independent drift, covariance, cost (i.e., for the elliptic case) can be found in [11]. Our article contains a generalization (or adjustment) of the approach of [2, 11] to an n-dimensional parabolic problem. Ius ouhat while the main ideas from those papers may be applied in our case, a mathematically rigorous analysis of our problem is somewhat delicate and nee rather careful arguments. Our motivation for pursuing this project is the hope thahe results given here will allow for a characterization of the optimal policy in the parabolic case as a solution to the corresponding Skorokhod problem for a domain with time-dependent (moving) boundary, which would be an analog of the main theorem from [8]. Indeed, the analysis of [8] used the results from [11] as the starting point, so it is plausible thaheir analogs will be useful in proving the corresponding result on a finite time horizon. Such a characterization would address a long-standing open problem on the structure of the optimal control in the case under consideration. We hope to address this issue in a subsequent paper. Existence results for multidimensional singular control problems closely related to our work may be found in [1, 3, 6]. Apparently, in spite of their considerable generality, none of them contains our existence result A multidimensional singular stochastic control problem... as a special case. Indeed, in these papers optimal weak solutions to the corresponding SDEs are constructed, while we are conceed about finding an optimal strong solution, i.e., for the given (as opposed to some) filtration and underlying Brownian motion. Moreover, the problem considered in [1] is elliptic and the allowable control directions lie in a cone, the opening of which cannot be too wide. In [3, 6] the time horizon is finite, buhe problem considered in [3] has the final cost instead of the running cost, while in [6] the drift of the controlled diffusion is bounded, which excludes its linear dependence on the state. The structure of this paper is as follows. In Section 1 we pose the singular stochastic control problem, give definitions and prove lemmas needed in further considerations. In Section 2 we prove estimates for the value function. In Section 3 we consider the Bellman's dynamic programming principle (DPP) and the HJB equation related to this problem. Section 4 contains proofs of existence and uniqueness of an optimal control. 1. Notation, assumptions and lemmas. Let Mn×n denote the set of matrices of dimension n × n with the operator norm, i.e. ||A|| = {|Ax| : x , |x| = 1}. Le > 0 be a fixed number representing our time horizon. For a function u = u(x, t) : × [0, T ] R we denote the gradient and the Hessian of u with respeco the space variables (i.e., xi ) by Du and D2 u, respectively. Let (Wt , t 0) be a standard n-dimensional Brownian motion defined on a complete probability space (, F, P ). Let (Ft , t 0) be the augmentation of the filtration generated by W (see [7], p. 89). Denote by V the set of controls v which are left-continuous, adapted to the filtration (Ft , t 0) random processes acting from [0, T ] into , with P -a.s. bounded variation and s.t. v(0) = 0 P -a.s. We note thahese processes are also progressively measurable (see [7], Th. 1.1.13). As it is customary in singular stochastic t control theory (see, e.g., [8]), we write v(t) = 0 (s), where |(t)| = 1 for every t [0, T ] and is nondecreasing and left-continuous. In other wor, (t) is the total variation of v on the time interval [0, t] and (t) is the Radon­Nikodym derivative of the vector-valued measure induced by v on [0, T ] with respeco its total variation . Consider the state process described by the stochastic integral equation (1) yxt (s) = x + a(r)yxt (r) + b(r) dr + (r)dWr-t + v(s - t), s [t, T ], where t [0, T ] is an initial time, x is an initial position, b : [0, T ] and a, : [0, T ] Mn×n stand for the drift and the covariance terms. Note that (yxt (s))s[t,T ] is a random process adapted to (Fs-t )s[t,T ] . M. Boryc and L. Kruk To each control v V, we associate a cost given by the payoff functional (2) Jxt (v) = E s t (r)dr + s t (r)dr d(s - t) , where f, and c are respectively the running cost, the discount factor and the instantaneous cost per unit of "fuel". Our purpose is to characterize the optimal cost, the so-called value function (3) u(x, t) = {Jxt (v) : v V}. It is often convenieno consider the following penalized problem associated with (3): (4) u (x, t) = {Jxt (v) : v V }, where > 0 and V is the set of all controls v V which are Lipschitz continuous and | dv (t)| 1 for almost every t [0, T ] almosurely. dt Definition 1.1. We say thahe finite time horizon stochastic control problem has the dynamic programming property in the weak sense if for every 0 x , t, t [0, T ] s.t. t < t and yxt (s) given by (1) with v 0 we have (5) u(x, t) E 0 s t (r)dr 0 + u(yxt (t ), t )e- (r)dr Let us assume the following: · , c are Lipschitz continuous from [0, T ] into [0, ) with constant L > 0, · b is Lipschitz continuous from [0, T ] into with the same constant L > 0, · a, are Lipschitz continuous from [0, T ] into Mn×n with the same constant L > 0, · there exists c0 > 0 such that c(t) c0 for all t [0, T ], ~ · f : × [0, T ] [0, ) and there exist constants p > 1, C0 , C0 > 0 n and (0, 1) we have such that for all t, t [0, T ], x, x R (6) (7) (8) (9) ~ C0 |x|p - C0 f (x, t) C0 (1 + |x|p ) , |f (x, t) - f (x + x , t)| C0 1 + f (x, t) + f (x + x , t) |f (x, t) - f (x, t )| C0 (1 + |x| )|t - t | , p |x | , 0 < f (x + x , t) - 2f (x, t) + f (x - x , t) C0 2 (1 + f (x, t))q , q = (1 - 2/p)+ . The last assumption implies strict convexity of the function f with respeco x. A multidimensional singular stochastic control problem... Let us denote by cmax and max the maximum of the function c, , respectively. Moreover, by amax , max , max and bmax we denote the maximum over t [0, T ] of the norms of the matrices a(t), (t), (t) and the vector b(t) respectively, where (t) = (t) T (t). Now we give lemmas needed for the proofs of the Theorems 2.1 and 2.2. The first one is well known. Lemma 1.2. For all x, y 0 we have xp + y p (x + y)p 2p-1 (xp + y p ), if p 1, 2p-1 (xp + y p ) (x + y)p xp + y p , if p (0, 1). Lemma 1.3 (See [10], Corollary 2.5.12). Consider an n-dimensional process described by a stochastic integral equation x(t) = x0 + t 0 g(x(s), s) + t 0 h(x(s), s)dWs , t 0, where x0 , g : × [0, ) and h : × [0, ) Mn×n . We assume thahere exists a constant C such that for all x and t 0 (10) ||h(x, t)|| + |g(x, t)| C(1 + |x|). Then for every q > 0 there exists a constant C11 > 0 depending only on q, C such that for all t 0 (11) E |x(s)|q C11 eC11 t (1 + |x0 |)q . 0st Remark 1.4. For the process yxt defined by (1) with v 0 the assumption (10) hol Indeed, is Lipschitz continuous, independent of x and defined on a finite time interval [0, T ], so it is bounded. We conclude the same about a, b, so |g(x, t)| = |a(t)·x+b(t)| C(1+|x|), where C = max{||a(t)||, |b(t)| : t [0, T ]}. Lemma 1.5. Let x, x , t [0, T ] and g(s) = yxt (s) - yx t (s) for s [t, T ]. Then dg s [t, T ], (s) = a(s)g(s), |g(s)| C12 |x - x |, (12) where C12 = 1 + amax T eamax T . Proof. In view of (1) we have g(s) = x - x + a(r)(yxt (r) - yx t (r))dr = x - x + a(r)g(r)dr. Taking the derivative d/ of both sides, we gehe differential equation dg (s) = a(s)g(s) with initial data g(t) = x - x . The solution of this problem satisfies |g(s)| |x - x | + |a(r)g(r)|dr |x - x | + amax s t |g(r)|dr. M. Boryc and L. Kruk Using the Gronwall's inequality (see [4], p. 625), we gehe second part of (12). Lemma 1.6. pose that for some x , t [0, T ], v V we have E s t (r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then (13) E f (yxt (s), s) C13 (1 + |x|p ), where C13 = C · e T 0 (r)dr . , we Proof. Indeed, multiplying both sides of our assumption by e get E (r)dr f (yxt (s), s)e (r)dr Ce (r)dr (1 + |x|p ) C13 (1 + |x|p ). Of course, the left-hand side is nomaller than E f (yxt (s), s) Lemma 1.7 (Compare a statement in [17], p. 181). The function Jxt (v) is convex with respeco (x, v), more precisely, for all x1 , x2 , t [0, T ], v1 , v2 V and [0, 1], Jx1 +(1-)x2 ,t (v1 + (1 - )v2 ) Jx1 ,t (v1 ) + (1 - )Jx2 ,t (v2 ) . v Proof. First, we note thahe set V is obviously convex. Let yxt (s) be the solution of (1) corresponding to a control v. Denote v0 = v1 + (1 - )v2 and x0 = x1 + (1 - )x2 . In view of the definition of Jxt (v), iuffices to prove two following inequalities (14) (15) v v v f yx0 ,t (s), s f (yx1 ,t (s), s) + (1 - )f (yx2 ,t (s), s), s [t, T ] , 0 1 2 d0 (s - t) d1 (s - t) + (1 - ) d2 (s - t) , where 0 , 1 , 2 are the total variations of v0 , v1 , v2 respectively. The latter inequality is a consequence of the fachahe variation of the sum of functions is not greater than the sum of their variations. So 0 1 + (1 - )2 . Because 0 , 1 , 2 are nondecreasing and 0 (0) = 1 (0) = 2 (0) = 0 P -a.s., we conclude that (15) is true. To prove (14) we show firshat (16) v v v yx0 ,t (s) = yx1 ,t (s) + (1 - )yx2 ,t (s) . 0 1 2 Indeed, using (1) we get v yxi ,t (s) = xi + i s t v a(r)yxi ,t (r) + b(r) dr + i s t (r)dWr-t + vi (s - t) , A multidimensional singular stochastic control problem... v v v i = 0, 1, 2. Let g(s) = yx0 ,t (s) - yx1 ,t (s) - (1 - )yx2 ,t (s). Then 0 1 2 g(s) = v v v a(r) yx0 ,t (r) - yx1 ,t (r) - (1 - )yx2 ,t (r) dr = 0 1 2 a(r)g(r)dr . Taking the derivative d/ of both sides, we gehe differential equation dg (s) = a(s)g(s) with initial data g(t) = x0 - x1 - (1 - )x2 = 0. The solution of this problem is g(s) 0, so (16) hol Using (16) and convexity of f we have (14). Lemma 1.8. pose that for some t [0, T ], x , v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t . Then there exists a constant C17 > 0 independent of x, uch that (17) E(T - t ) C17 (1 + |x|p ). Proof. Indeed, multiplying both sides of our assumption by e and using the lower bound of c, we get c0 E(T - t ) = c0 E Ce T 0 (t +r)dr E (1 + |x|p ). c(t + s)e T - (t +r)dr (r)dr Lemma 1.9. pose that for some x , t [0, T ], v V we have E s 0 C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then there exists a constant C18 > 0 independent of x, uch that (18) E (1 + |yxt (t + s)|p ) C18 (1 + |x|p ). Proof. From Lemma 1.6 we know that E Using (6), we get E Hence ~ C0 E 0 f (yxt (t + s), t + s) C13 (1 + |x|p ). ~ C0 |yxt (t + s)|p - C0 C13 (1 + |x|p ). T -t |yxt (t + s)|p (C13 + C0 T )(1 + |x|p ) M. Boryc and L. Kruk and finally ~ C0 E ~ (1 + |yxt (t + s)|p ) (C13 + C0 T + C0 T )(1 + |x|p ) . Lemma 1.10. Let 0 t t T and pose that for some x , v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t, t . Then there exists a constant C19 > 0 independent of x, t, uch that (19) E f (yxt (t + s), t + s) C19 (1 + |x|p ). Proof. We observe that using (8) we have f (yxt (t + s), t + s) |f (yxt (t + s), t + s) - f (yxt (t + s), t + s)| + f (yxt (t + s), t + s) C0 |t - t |(1 + |yxt (t + s)|p ) + f (yxt (t + s), t + s). Hence, in view of Lemma 1.6 and Lemma 1.9, we get E f (yxt (t + s), t + s) C0 |t - t |C18 (1 + |x|p ) + C13 (1 + |x|p ) C19 (1 + |x|p ), where C19 = C0 T C18 + C13 . Lemma 1.11. Let 0 t t T , x , v V. Assume that E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t , t. Then there exists a constant C20 > 0 independent of x, t , uch that for all s [0, T - t] we have (20) E|yxt (t + s) - yxt (t + s)|p C20 |t - t |p (1 + |x|p ) . s 0 s 0 s 0 s 0 Proof. For s [0, T - t], we have yxt (t + s) = x + yxt (t +s) = x+ so (21) a(t + r)yxt (t + r) + b(t + r) dr + a(t +r)yxt (t +r)+b(t +r) dr+ (t + r)dWr + v(s), (t +r)dWr +v(s), yxt (t + s) - yxt (t + s) = As + Bs + Ms , A multidimensional singular stochastic control problem... where As = A1 = s A2 = s Bs = Ms = s 0 0 0 0 0 a(t + r)yxt (t + r) - a(t + r)yxt (t + r) dr = A1 + A2 , s s a(t + r) yxt (t + r) - yxt (t + r) dr, a(t + r) - a(t + r) yxt (t + r)dr, b(t + r) - b(t + r) dr, (t + r) - (t + r) dWr . Recall that a, b, are Lipschitz continuous with the constant L. The process Ms is a martingale with quadratic variation [M ]s = s 0 (t + r) - (t + r) dr L2 |t - t |2 s. This, together with the Burkholder­Davis­Gundy inequalities (see [7], Theorem 3.3.28), implies the existence of a constant Cp , depending only on p, such that (22) Clearly, (23) |Ms |p Cp Lp T 2 |t - t |p . |Bs | LT |t - t |. By the H¨lder's inequality, for q = p/(p - 1) we have o (24) (25) |A1 |p (aq s) q s max s 0 |yxt (t + r) - yxt (t + r)|p dr, p q |A2 |p (L|t - t |)q s s s 0 |yxt (t + r)|p dr. By Lemma 1.9, the inequality (18) hol for every t [0, T ]. Lemma 1.2 and the relations (18), (21)­(25) imply thahe random variable |yxt (t + s) - yxt (t + s)|p is integrable and hence, by the Lebesgue dominated convergence theorem, the function F (s) = E|yxt (t + s) - yxt (t + s)|p is continuous on [0, T - t]. From Lemma 1.2 and (18), (21)­(25) we also have, for each s [0, T - t], F (s) c1 |t - t |p (1 + |x|p ) + c2 s 0 F (r)dr, M. Boryc and L. Kruk p p p where c1 = 22p-2 (Lp T p + Cp Lp T 2 + (LT ) q C18 ), c2 = 22p-2 ap T q . This, max together with the Gronwall's inequality (see, e.g., [7], Problem 5.2.7), implies that for all s [0, T - t], F (s) c1 |t - t |p (1 + |x|p ) 1 + c2 We have obtained (20) with C20 = c1 (1 + c2 T 0 s 0 ec2 (s-r) dr) . ec2 (T -r) dr). Lemma 1.12. pose that for some x , t [0, T ], v V we have E s t (r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then there exists a constant C26 > 0 independent of x, x , uch that for every x , (26) E f (yx+x ,t (s), s) C26 (1 + |x|p + |x + x |p ). Proof. From (6) and Lemma 1.2 we have E f (yx+x ,t (s), s) E t C0 (1 + |yx+x ,t (s)|p ) T C0 + C0 2p-1 E |yx+x ,t (s) - yx,t (s)|p + C0 2p-1 E |yx,t (s)|p . Now using Lemma 1.5, Lemma 1.9 and Lemma 1.2 again, we get E p f (yx+x ,t (s), s) T C0 + C0 2p-1 T · C12 |x |p + C0 2p-1 C18 (1 + |x|p ) p T C0 + C0 22p-2 T · C12 |x + x|p + |x|p + C0 2p-1 C18 (1 + |x|p ) C26 (1 + |x|p + |x + x |p ), p where C26 = C0 T + 22p-2 T · C12 + 2p-1 C18 . Lemma 1.13. pose that for some x , t [0, T ], v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t . Then there exists a constant C27 > 0 independent of x, t , uch that for every t [t , T ], (27) E f (yxt (t + s), t + s) C27 (1 + |x|p ). A multidimensional singular stochastic control problem... Proof. Using (6) and Lemma 1.2, we have E f (yxt (t + s), t + s) E C0 (1 + |yxt (t + s)|p ) C0 T + 2p-1 C0 E + 2p-1 C0 E |yxt (t + s)|p |yxt (t + s) - yxt (t + s)|p . In view of Lemma 1.9, the Fubini's theorem and Lemma 1.11, we get E f (yxt (t + s), t + s) C0 T + 2p-1 C0 C18 (1 + |x|p ) + 2p-1 C0 T C20 |t - t |p (1 + |x|p ) C27 (1 + |x|p ), where C27 = C0 T + 2p-1 C18 + 2p-1 T p+1 C20 . The nexwo definitions and lemma refer to mollification of a given function (see [4], p. 629­630). Definition 1.14. Define C ( ) by (28) (x) = C28 exp 0 1 |x|2 -1 if |x| < 1 if |x| 1, where the constant C28 is selected so that (x)dx = 1. For each m N set m (x) = mn · (mx). We call the standard mollifier. The functions m belong to the class C ( ) and satisfy m (x)dx = 1. Definition 1.15. Fix t [0, T ]. For each m N we define mollification of the function u(·, t ) by um (x) = m (y)u(x - y, t )dy, x , where B(0, r) = {x : |x| < r}. Lemma 1.16. For each m N we have um C ( ). Moreover, if u(·, t ) is continuous, then um (x) u(x, t ) uniformly on compacubsets of as m . 2. Estimates for the value function. Lehe assumptions from Section 1 appearing immediately after Definition 1.1 hold. Theorem 2.1. Let u be the value function defined by (3). Then for some positive constants C29 , C30 , C31 , the same p > 1 as in the assumptions (6)­ (9) and every t [0, T ], x, x and (0, 1), the following estimates M. Boryc and L. Kruk hold: (29) (30) (31) 0 u(x, t) C29 (1 + |x|p ) , |u(x, t) - u(x + x , t)| C30 (1 + |x|p-1 + |x + x |p-1 )|x | , 0 u(x + x , t) - 2u(x, t) + u(x - x , t) C31 2 (1 + |x|)(p-2) . Proof: Proof of (29). Nonnegativity of u is the consequence of nonnegativity of f and c. Next, taking the control v 0 and using (6), the Fubini's theorem, Lemma 1.3 and Lemma 1.2, we get u(x, t) Jxt (0) = E 0 s t (r)dr 0 C0 (1 + |yxt (s)|p ) = C0 T 0 E(1 + |yxt (s)|p ) C0 C0 0 E 1 + C11 eC11 (s-t) (1 + |x|)p (1 + C11 eC11 T 2p-1 )(1 + |x|p ) = C0 T (1 + C11 eC11 T 2p-1 )(1 + |x|p ) = C29 (1 + |x|p ), where C29 depen only on C0 , T, C11 , p, so (29) is proved. Proof of (30). Now we note that u(x + x , t) - u(x, t) = Jx+x ,t (v ) - Jx,t (v) v V Jx+x ,t (v) - Jx,t (v) . Hence u(x + x , t) - u(x, t) |Jxt (v) - Jx+x ,t (v)| E t |f (yxt (s), s) - f (yx+x ,t (s), s)|e- s t (r)dr Applying (7), we can estimate the last expression from above by E C0 1 + f (yxt (s), s) + f (yx+x ,t (s), s) · |yxt (s) - yx+x ,t (s)| Using Lemma 1.5, we have u(x + x , t) - u(x, t) C0 C12 |x | · E p-1 p 1 + f (yxt (s), s) + f (yx+x ,t (s), s) A multidimensional singular stochastic control problem... We use the H¨lder's inequality with exponent o pression above by (32) C0 C12 |x | · E p p-1 to estimate the last exp-1 p 1 + f (yxt (s), s) + f (yx+x ,t (s), s) T p. s t (r)dr for some arbitrary that > 0. From (32), Lemma 1.6 and Lemma 1.12 we see u(x + x , t) - u(x, t) C0 C12 |x | T + C13 (1 + |x|p ) + C26 (1 + |x|p + |x + x |p ) C30 |x | 1 + |x|p + |x + x |p p-1 p p-1 p Tp , where C30 = T 1/p · C0 C12 (T + C13 + C26 ) . Finally using Lemma 1.2, we get u(x + x , t) - u(x, t) C30 (1 + |x|p-1 + |x + x |p-1 )|x |. In an analogous manner we gehe same estimate for u(x, t)-u(x+x , t). Proof of (31). We observe that u(x + x , t) + u(x - x , t) - 2u(x, t) Jx+x ,t (v) + Jx-x ,t (v) - 2Jxt (v) = E t f (yx+x ,t (s), s) + f (yx-x ,t (s), s) - 2f (yxt (s), s) e- s t (r)dr In view of (12) we can apply (9) to get u(x + x , t) + u(x - x , t) - 2u(x, t) E (1-2/p)+ C0 2 1 + f (yxt (s), s) . If p 2 we have u(x + x , t) + u(x - x , t) - 2u(x, t) C0 T 2 . If p > 2 p we use the H¨lder inequality with exponent p-2 to get o u(x + x , t) + u(x - x , t) - 2u(x, t) C0 2 E 1-2/p (1 + f (yxt (s), s) T 2/p . M. Boryc and L. Kruk s t (r)dr for some arbitrary > 0. From Lemma 1.6 and Lemma 1.2 we see that u(x + x , t) + u(x - x , t) - 2u(x, t) C0 2 T + C13 (1 + |x|p ) C31 2 (1 + |x|p )1-2/p C31 2 (1 + |x|)p-2 , where C31 = T 2/p C0 (T + C13 )1-2/p . We have proved the upper bound of (31). To prove the lower bound of (31), it clearly suffices to prove convexity of u(x, t) with respeco the first variable. In view of the definition of u we know that for every > 0, x1 , x2 , t [0, T ], [0, 1] there exist v1 , v2 V such that Jxi ,t (vi ) u(xi , t) + , i = 1, 2. Using Lemma 1.7, we get u(x1 + (1 - )x2 , t) Jx1 +(1-)x2 ,t (v1 + (1 - )v2 ) Jx1 ,t (v1 ) + (1 - )Jx2 ,t (v2 ) u(x1 , t) + (1 - )u(x2 , t) + . Because > 0 is arbitrary, we get convexity of u(x, t) with respeco the first variable. Theorem 2.2. Lehe assumptions of Theorem 2.1 be satisfied. Assume thahe dynamic programming property in the weak sense hol (Definition 1.1). Then for some constant C33 > 0 and every t, t [0, T ], x , we have (33) |u(x, t) - u(x, t )| C33 (1 + |x|p )|t - t | . 1-2/p T 2/p Proof. We note that u(x, t) - u(x, t ) = Jxt (v) - Jxt (v ) v V v V Jxt (v ) - Jxt (v ) . A multidimensional singular stochastic control problem... For t he difference Jxt (v) - Jxt (v) is equal to E s 0 - + - - T - - - -t s 0 (t +r)dr s 0 s 0 - s 0 (t +r)dr s 0 (t +r)dr (t +r)dr Let us denote the expectations of the firswo integrals in the last expression by A and B, respectively. Because the laswo integrals are nonnegative we get (34) Jxt (v) - Jxt (v) A + B . s 0 s 0 We can estimate B as follows: BE - (t +r)dr . Adding and subtracting 0 (t +r)dr under the absolute value sign and using the triangle inequality and positivity of , we get BE cmax |e- s 0 - e- s 0 (t +r)dr | + |c(t + s) - c(t + s)| . Because |ex - ey | |x - y| for x, y 0 and c, are Lipschitz continuous, we have BE s cmax |(t + r) - (t + r)|dr + |c(t + s) - c(t + s)| (cmax T + 1)L|t - t |E = (cmax T + 1)L|t - t |E(T - t). s 0 (t +r)dr (C29 + )(1 + |x|p ) for some arbitrary > 0. Using Lemma 1.8, we get E(T -t) E(T -t ) C17 (1 + |x|p ) and (35) B C35 |t - t |(1 + |x|p ) , where C35 = (cmax T + 1)LC17 . M. Boryc and L. Kruk Now we estimate A: AE s 0 s 0 - +E (t +r)dr s 0 s 0 (t +r)dr - = A 1 + A2 . (t +r)dr Using the inequality |ex - ey | |x - y| for x, y 0 again, we get A1 E (36) f (yxt (t + s), t + s) s 0 |(t + r) - (t + r)|dr T L|t - t | E f (yxt (t + s), t + s) . By virtue of (29) we can consider only those controls v for which (37) E s 0 (t +r)dr for some arbitrary (38) to E > 0. Using (36) and Lemma 1.13, we get A1 C38 |t - t |(1 + |x|p ) , where C38 = T LC27 . To estimate A2 we use (7)­(8) and we have that A2 is less than or equal f (yxt (t + s), t + s) - f (yxt (t + s), t + s) + f (yxt (t + s), t + s) - f (yxt (t + s), t + s) E C0 1 + f (yxt (t + s), t + s) + f (yxt (t + s), t + s) × |yxt (t + s) - yxt (t + s)| +E C0 (1 + |yxt (t + s)|p )|t - t | = A3 + A4 . Using the H¨lder's inequality and the Fubini's theorem, we get o A3 C0 E × 1 + f (yxt (t + s), t + s) + f (yxt (t + s), t + s) 1/p p E|yxt (t + s) - yxt (t + s)| A multidimensional singular stochastic control problem... From this together with (37), Lemma 1.13, Lemma 1.10 and Lemma 1.11 we have A3 C0 {(T + C27 + C19 )(1 + |x|p )} · T C20 |t - t |p (1 + |x|p ) Because 1 + |x|p (1 + |x|)p , we get A3 C0 (T + C27 + C19 ) (1 + |x|)p-1 · (T C20 )1/p |t - t |(1 + |x|). Hence, from Lemma 1.2, (39) A3 C39 |t - t |(1 + |x|p ), 1/p where C39 = C0 (T + C27 + C19 ) (T C20 )1/p 2p-1 . Furthermore, from Lemma 1.9 we get (40) A4 C40 |t - t |(1 + |x|p ), where C40 = C0 C18 . u(x, t) - u(x, t ) C41 |t - t |(1 + |x|p ), In view of (34)­(35) and (38)­(40) we get for t t, (41) where C41 = C35 + C38 + C39 + C40 . To obtain a similar inequality for t < t we proceed as follows. Let 0 yxt (s) s[t,T ] be a solution of (1) with v 0. We can write the i-th coors t n 0 dinate of yxt (s) as follows 0 (42) yxt (s)i = xi + 0 aij (r)yxt (r)j + bi (r) dr + n j=1 t j ij (r)dWr-t , j=1 i = 1, . . . , n, where subscripts denote the corresponding coordinates. Let {um (·)}nN be a sequence of mollifications of the function u(·, t ) (see Def. 1.15). Applying the It^'s formula ([7], Th. 3.3.6), we get o 0 Eum (yxt (t )) n = um (x) + E i=1 n t 0 um (yxt (s)) n j=1 0 aij (s)yxt (s)j + bi (s) xi (43) +E i=1 t n 0 um (yxt (s)) xi t j ij (s)dWs-t j=1 0 0 d[yxt (s)i , yxt (s)j ] 1 E 2 2u 0 m (yxt (s)) i,j=1 t xi xj = um (x) + A + B + C . We need the following lemma. M. Boryc and L. Kruk Lemma 2.3. We assume (29)­(31). Le [0, T ] be fixed. Then there exist constants C45 , C46 > 0 such that for all x , m N and i, j {1, . . . , n}, (44) (45) lim um (x) = u(x, t ), um (x) C45 (1 + |x|)p-1 , xi 0 2 um (x) C46 (1 + |x|p ). xi xj n 0 aij (s)yxt (s)j + bi (s) j=1 0 n||a(s)|| · |yxt (s)| + |b(s)| (46) We estimate A as follows AE i=1 n t 0 um (yxt (s)) · xi 0 um (yxt (s)) xi i=1 t Using Lemma 2.3 and Lemma 1.2, we see that A is not greater than E (47) i=1 t 0 0 C45 (1 + |yxt (s)|)p-1 n(amax + bmax )(1 + |yxt (s)|) 0 (1 + |yxt (s)|p ), C47 E n2 C p-1 . where C47 = 45 (amax + bmax )2 Now we show that B = 0. Indeed, n B=E i,j=1 Zij (t ), where Zij (s) = 0 um (yxt (r)) j ij (r)dWr-t for s [t, t ]. xi From properties of the It^'s integrals (see [7], Section 3.2) the process o Zij (s) s[t,t ] is a martingale provided that E 0 um (yxt (s)) ij (s) xi 2 2 . Using Lemma 2.3 and Lemma 1.2, we have E 0 um (yxt (s)) ij (s) xi 0 (1 + |yxt (s)|2p ) E 2 0 2 C45 (1 + |yxt (s)|)2p-2 max 2 2 C45 max E 0 (1 + |yxt (s)|)2p 2 2 C45 max 22p-1 E A multidimensional singular stochastic control problem... Using the Fubini's theorem and Lemma 1.3, we get E 0 um (yxt (s)) ij (s) xi 2 2 2 C45 max 22p-1 1 + C11 eC11 T (1 + |x|)2p < . Hence Zij is a martingale and EZij (t ) = EZij (t) = 0. So (48) B=E i,j=1 Zij (t ) = 0. Now, using the conventional "multiplication rules" (see [7], p. 154), we know that = 0, i dWs = 0, i i dWs dWs = , i j dWs dWs = 0 for i = j. So in view of (42) we can write n 0 0 d[yxt (s)i , yxt (s)j ] n k ik (s)dWs-t k=1 n l jl (s)dWs-t l=1 k=1 ik (s)jk (s) From Lemma 2.3 we have C= (49) 1 2 1 2 i,j=1 n t 0 2 um (yxt (s)) xi xj ik (s)jk (s) k=1 i,j=1 t 0 2 C46 (1 + |yxt (s)|p )nmax = C49 E 1 0 2 (1 + |yxt (s)|p ), where C49 = C46 max n3 . 2 In summary, in view of (43) and (47)­(49) 0 Eum (yxt (t )) um (x) + (C47 + C49 )E 0 (1 + |yxt (s)|p ) Taking the limit as n and using the Fatou's lemma, we get (50) 0 Eu(yxt (t ), t ) u(x, t ) + C50 E 0 (1 + |yxt (s)|p ), C50 = C47 + C49 . Furthermore, from Lemma 1.2 and Lemma 1.3 we have for each s [t, T ] (51) 0 E 1 + |yxt (s)|p C51 (1 + |x|p ), C51 = C11 eC11 T 2p-1 + 1 . M. Boryc and L. Kruk Next, from (5), (6), (50), (51) and the Fubini's theorem we conclude u(x, t) E 0 s t (r)dr 0 + u(yxt (t ), t )e- (r)dr C0 0 0 E 1 + |yxt (s)|p + Eu(yxt (t ), t ) C0 C51 + C50 C51 |t - t |(1 + |x|p ) + u(x, t ). Hence, for t < t (52) u(x, t) - u(x, t ) C52 |t - t |(1 + |x|p ), C52 = C51 (C0 + C50 ) . It is clear that (41) and (52) imply (33). Now we give the proof of Lemma 2.3. Proof: Proof of (44). The continuity of u(·, t ) is a consequence of (30). So in view of Lemma 1.16 we conclude that limm um (x) = u(x, t ). Proof of (45). Let x , 0 |x | < 1. From Definitions 1.14, 1.15 and (30) we get |um (x) - um (x + x )| = m (y) u(x - y, t ) - u(x + x - y, t ) dy mn · (my)|u(x - y, t ) - u(x + x - y, t )|dy 1 + |x - y|p-1 + |x + x - y|p-1 dy. C28 C30 |x |mn Because |x | < 1 and |y| 1 m 1, we have (2 + 2p-1 )(1 + |x|)p-1 . 1 + |x - y|p-1 + |x + x - y|p-1 1 + (1 + |x|)p-1 + (2 + |x|)p-1 Furthermore (see [4], p. 615), dy = n/2 1 · n , where (t) = n ( 2 + 1) m st-1 e-s , for t > 0. In summary, |um (x) - um (x + x )| n/2 C28 C30 n (2 + 2p-1 )(1 + |x|)p-1 . |x | ( 2 + 1) Taking the limit as |x | 0 on both sides, we conclude (45). A multidimensional singular stochastic control problem... Proof of (46). Let x and (0, 1). We have um (x + x ) - 2um (x) + um (x - x ) = m (y) u(x + x , t ) - 2u(x, t ) + u(x - x , t ) dy. u From (31) and nonnegativity of m we conclude that xim (x) 0. On the xj other hand, using (31) and mimicking the proof of (45), we see that um (x + x ) - 2um (x) + um (x - x ) mn · (my)C31 2 (1 + |x|)(p-2) dy n/2 + (1 + |x|)(p-2) . n ( 2 + 1) 2 C28 C31 For p (1, 2], (1 + |x|)(p-2) = 1 (1 + |x|p ) 2p-1 (1 + |x|p ). For p > 2, in + view of Lemma 1.2, (1+|x|)(p-2) = (1+|x|)p-2 (1+|x|)p 2p-1 (1+|x|p ). Thus, for all p > 1 we have um (x + x ) - 2um (x) + um (x - x ) n/2 C28 C31 2p-1 (1 + |x|p ). 2 ( n + 1) 2 Taking the limit as 0 , we can conclude (46). Remark 2.4. Theorems 2.1 and 2.2 are true for functions u (see (4)) instead of u. Indeed, in view of the proofs we see thahe constants C29 , C30 , C31 , C33 do not depend on . Remark 2.5. It follows from (44)­(46) that for every t [0, T ] Dum (· ; t ) converges to Du(·, t ) the distributional gradient of u with respeco x almost uniformly as m (see the proof of Theorem 3.5, to follow, for a similar argument, with um replaced by u m ). This implies differentiability of u with respeco x in the classical sense (see, e.g., Theorem 7.17 in [13]), so Du is the classical gradient of u with respeco x at any point (x, t ) × [0, T ]. Moreover, by (46) Dum are locally Lipschitz in x uniformly in m, so Du is also locally Lipschitz in x. Thus Theorems 2.1, 2.2 and their proofs imply thahe value function u(x, t) has generalized derivatives of the first order with respeco t and of the second order with respeco x. These generalized derivatives belongs to the space L ( × [0, T ]) of all loc functions essentially bounded on every open bounded subset of the domain. Proposition 2.6. For all x and t [0, T ] we have u(x, t) (cmax + C29 )(1 + |x|). M. Boryc and L. Kruk Proof. Let x be arbitrary. Consider controls for which lims0+ vs = x. In view of (2) and (3) we have u(x , t) = {Jx t (v) : v V} c(t)|x| + {Jx+x ,t (v) : v V} = c(t)|x| + u(x + x , t). So u(x , t) - u(x + x , t) c(t)|x|. Similarly u(x + x , t) - u(x , t) c(t)|x|, so (53) |u(x + x , t) - u(x , t)| c(t)|x|. Taking x = 0, we get |u(x, t) - u(0, t)| c(t)|x|. From (29) we see that u(0, t) C29 so u(x, t) c(t)|x| + u(0, t) cmax |x| + C29 (cmax + C29 )(1 + |x|). Remark 2.7. The proof of Proposition 2.6 is not valid for u instead of u, because if a control v V , then it is continuous, so the condition lims0+ vs = x is invalid for x = 0. Remark 2.8. The value function u(x, t) satisfies |Du(x, t)| c(t) for all (x, t) × [0, T ]. Indeed, the gradient exists for all (x, t) × [0, T ] in view of Remark 2.5. From (53) we see thahe first derivative of u(x, t) with respeco x in any direction is bounded by c(t). Hence, the norm of the gradient Du(x, t) is bounded by c(t), too. 3. Dynamic Programming Principle and HJB equation. To consider the DPP and the HJB equation for our problem we will first prove the pointwise convergence of u to u if 0+ . For this purpose we need an integral form of the Gronwall's inequality with locally finite measures. Lemma 3.1 (see [18]). Let be a locally finite measure on the Borel algebra of [t, T ], where 0 t T . We consider a measurable function T defined on [t, T ] such tha |(r)|(dr) < . We assume thahere exists a Borel function 0 on [t, T ] such that for all s [t, T ], (s) (s) + Then for all s [t, T ], (s) (s) + Theorem 3.2. For all (x, t) (r)e([r,s)) (dr). we have lim 0+ [t,s) n ×[0, T ] R (r)(dr). [t,s) u (x, t) = u(x, t). Proof. Fix x and t [0, T ]. Consider an arbitrary v V such that Jxt (v) < . Step 1. We show firshat v Lp ( × [0, T - t], P Leb ), where Leb denotes the Lebesgue's measure. Since Jxt (v) < , we have E f (yxt (s), s) < A multidimensional singular stochastic control problem... and from (6) we get (54) E |yxt (s)|p < . From (1) we can write for s [t, T ] (55) v(s - t) = yxt (s) - x - b(r)dr - (r)dWr-t - a(r)yxt (r)dr. Using (54) and properties of the normal distribution, we know that each term from the line above, maybe except for the last one, belongs to the space Lp ( × [0, T - t]). Buhe laserm belongs to this space, too. Indeed, E p a(r)yxt (r)dr ap max |yxt (r)|dr Using the H¨lder's inequality and (54), we can estimate the last expression o above by ap E max |yxt (r)|p dr · |T - t|p/q |yxt (r)|p dr < , ap T 1+p/q E max 1 p 1 q where + = 1. Hence, from (55) we see that v Lp ( × [0, T - t]). Step 2. Now we define a sequence of bounded controls {vR , R > 0} such that vR is convergeno v in the space Lp ( × [0, T - t]) and the total variation of vR is pointwise convergeno the total variation of v from below. Let v(s), |v(s)| R vR (s) = v(s) |v(s)| · R, |v(s)| > R . We see that for all s [0, T - t] limR vR (s) = v(s) and |vR (s)| |v(s)|. Hence, from Lemma 1.2 and Step 1, E |v(s) - vR (s)|p 2p E |v(s)|p < and using the Lebesgue's dominated convergence theorem, we get |v(s) - vR (s)|p = E lim |v(s) - vR (s)|p = 0. The convergence in Lp is proved. Moreover, if (s), R (s) denote the total variations on the interval [0, s] of the functions v, vR respectively, then for all s [0, T - t], (56) R (s) (s) and lim R (s) = (s). M. Boryc and L. Kruk vR v Step 3. Let yxt , yxt denote the state processes (see (1)) corresponding to vR the controls v, vR respectively. We wano show that {yxt } is convergent v in the space Lp ( × [t, T ]). First we observe that for s [t, T ], to yxt vR v yxt (s) - yxt (s) = t vR v Denoting zR (s) = yxt (s) - yxt (s) and uR (s) = v(s - t) - vR (s - t), we s can rewrite the last equality in the form zR (s) = t a(r)zR (r)dr + uR (s). s Hence |zR (s)| t |zR (r)|amax dr + |uR (s)|. Using Lemma 3.1 with = |zR |, = |uR | and = amax · Leb , we ge vR v a(r) yxt (r) - yxt (r) dr + v(s - t) - vR (s - t). |zR (s)| |uR (s)| + (57) |uR (r)|eamax (s-r) dr s t |uR (s)| + C57 |uR (r)|dr, where C57 = eamax T . So from Lemma 1.2 and the H¨lder's inequality o |zR (s)| 2 p p-1 |uR (s)| + p C57 s t |uR (r)|dr s t (58) p 2p-1 |uR (s)|p + C57 (s - t)p/q |uR (r)|p dr C58 |uR (s)|p + 1 where p + we have 1 q |uR (r)|p dr , p = 1 and C58 = 2p-1 (1 + C57 T p/q ). Finally, in view of Step 2 |zR (s)|p lim C58 E R |uR (s)|p + |uR (r)|p dr = 0. lim C58 E |uR (s)|p + C58 T E |uR (r)|p dr Step 4. The nextep is to show that Jxt (vR ) Jxt (v) if R . Indeed, |Jxt (v) - Jxt (vR )| E vR v f (yxt (s), s) - f (yxt (s), s) e- (r)dr + E (r)dr d( - R )(s - t) = AR + BR . A multidimensional singular stochastic control problem... In view of (56), BR cmax E d( - R )(s - t) = cmax E (T - t) - R (T - t) . Using (56) again and the assumption that Jxt (v) < , we see that E (T -t) -R (T - t) E(T - t) < . Hence, from the Lebesgue's dominated convergence theorem we get lim BR lim cmax E (T - t) - R (T - t) = cmax E lim (T - t) - R (T - t) = 0. Using (7) and the H¨lder's inequality, we have o AR E E t vR v f (yxt (s), s) - f (yxt (s), s) vR v 1 + f (yxt (s), s) + f (yxt (s), s) vR v |yxt (s) - yxt (s)| 1/p vR v 1+f (yxt (s), s)+f (yxt (s), s) vR v |yxt (s)-yxt (s)|p In view of Step 3, the second factor in the last expression goes to 0 if R . We mushow thahe first factor is bounded. Indeed, from (6) and Lemma 1.2 we can write E vR v 1 + f (yxt (s), s) + f (yxt (s), s) vR v v v 2 + |yxt (s)|p + 2p-1 |yxt (s)|p + 2p-1 |yxt (s) - yxt (s)|p vR v 2 + |yxt (s)|p + |yxt (s)|p (1 + C0 )E (1 + C0 )E Using (54) and Step 3 again, we conclude thahe last expression is bounded uniformly in R. Hence limR AR = 0. Summarizing Steps 1-4, we know that Jxt (vR ) goes to Jxt (v) if R , so we can consider only bounded controls. Step 5. Consider v V such that ||v|| < R for some R > 0. We will construct a sequence of controls {vn , n N} convergeno v in Lp ( × [0, T - t]) and such that vn V1/(2nR) for all n. Besides we shall prove thahe variation of vn is pointwise convergeno the variation of v from below. s Let vn (s) = n (s-1/n)0 v(r)dr, s [0, T - t]. We observe that vn is a progressively measurable continuous random process such that ||vn || R, so vn Lp ( × [0, T - t]). From left-continuity of v we know that (59) s[0,T -t] lim vn (s) = v(s). M. Boryc and L. Kruk Using the Lebesgue's dominated convergence theorem, we get |v(s) - vn (s)|p = E lim |v(s) - vn (s)|p = 0, so Lp -convergence is proved. Now we wano check that vn V1/(2nR) . Indeed, d d n vn (s) = v(r)dr (s-1/n)0 = n v(s) - v (s - 1/n) 0 2nR. Let n (s), (s) denote the variations on the interval [0, s] of the functions vn , v respectively. For convenience, we define v(r) 0 for r < 0. Then s vn (s) = n s-1/n v(r)dr, s [0, T - t]. Fix , s (0, T - t]. Let = {s0 , s1 , . . . , sk } be a partition of the interval [0, s], where 0 = s0 < s1 < · · · < sk = s. Then k k |vn (si ) - vn (si-1 )| = n i=1 k 1/n 0 i=1 si si -1/n v(r)dr - si-1 v(r)dr si-1 -1/n =n i=1 v(si + r - 1/n) - v(si-1 + r - 1/n) dr v(si + r - 1/n) - v(si-1 + r - 1/n) dr n n 1/n k 0 1/n 0 i=1 (s)dr = (s). Letting |||| 0, we get (60) n (s) (s). k k On the other hand, from (59) we see that |v(si ) - v(si-1 )| = i=1 i=1 n k lim vn (si ) - lim vn (si-1 ) = lim |vn (si ) - vn (si-1 )| lim n (s). i=1 n Letting |||| 0 and using (60), we have (s) lim n (s) lim n (s) (s) n n lim n (s) = (s). Step 6. In view of Step 5 we can mimic Steps 3 and 4 to conclude that Jxt (vn ) Jxt (v) if n , where ||v|| < R for some R > 0. From this A multidimensional singular stochastic control problem... and Step 4, remembering that vn V1/(2nR) , we can write (61) and lim 0+ Jxt (v) = >0 Jxt (v) u (x, t) = u(x, t). Theorem 3.3 (Bellman's dynamic programming principle). Let x , v t [0, T ] and let yxt denote the state process corresponding to a control v V. Let [0, T - t] be a Markov time with respeco {Ft }. Then u(x, t) = E + v (r)dr t+ (r)dr v d(s - t) + u(yxt (t + ), t + ) . Proof. For convenience let us denote Jxt (v, ) = E + v s t t+ (r)dr + s t (r)dr d(s-t) . It is known that DPP hol for regular stochastic control problems (see, e.g., [10], Th. 3.1.6). Hence we have for each > 0, (62) u (x, t) = v Jxt (v, ) + Eu (yxt (t + ), t + ) . Considering any v V , we have ~ v ~ v u (x, t) Jxt (~, ) + Eu (yxt (t + ), t + ). If 0+ , from Theorem 3.2 and the Lebesgue's dominated convergence theorem we get v ~ v u(x, t) Jxt (~, ) + Eu(yxt (t + ), t + ). Because (63) > 0 and v V are arbitrary we can conclude that ~ u(x, t) >0 V On the other hand, from (62) u (x, t) Letting (64) >0 0+ , we get u(x, t) >0 V The inequalities (63), (64) and an argumenimilar to the proof of Theorem 3.2 (see (61)) imply that u(x, t) = M. Boryc and L. Kruk Corollary 3.4. The dynamic programming property in the weak sense hol (see Definition 1.1) and hence the value function satisfies (33). Denote Au(x, t) = -u(x, t) 1 - (t) D2 u(x, t) - a(t)x + b(t) Du(x, t) t 2 + (t)u(x, t), where denotes the scalar product of vectors and matrices respectively. Theorem 3.5 (The HJB equation). The value function u satisfies almost everywhere (a.e.) the following second-order differential equation: (65) max Au(x, t) - f (x, t) , |Du(x, t)| - c(t) = 0. Proof. An application of the DPP for regular stochastic control problems yiel for > 0 the following equation (see [5], Chapter IV.3): (66) Au (x, t) + 1 |Du (x, t)| - c(t) = f (x, t) a.e. In view of Theorems 2.1, 2.2, Remark 2.4, Theorem 3.2, Corollary 3.4 and the Arzela­Ascoli's theorem ([7], Th. 2.4.9) we see that u u uniformly on every compacet if 0+ . Fix t [0, T ]. From (31) and Remark 2.4 we see that D2 u (·, t) are locally uniformly bounded for all > 0 in their domains, so using the Arzela­Ascoli's theorem from every sequence { m }mN convergeno 0, we can choose a subsequence {~m }mN such that Du~m (·, t) v = (v1 , . . . , vn ) almost uniformly if m . But v must be equal to Du(·, t) in the distribution sense. Indeed, for any function Cc ( ) and any k = 1, . . . , n we have (x) u~m (x, t)dx = - xk (x) u~m (x, t) dx. xk Letting m , we get (x) u(x, t)dx = - xk (x)vk (x)dx, u so vk (·) = u(·,t) almost everywhere. Since xk and vk are Lipschitz continxk uous, the equality hol for all x . Thus, v does not depend on the choice of the subsequence {~m }, so (67) t[0,T ] Du (·, t) Du (·, t) almost uniformly if 0+ . A multidimensional singular stochastic control problem... Let = (1 + |x|)-2p-n-1 . From (29)­(33) we conclude that |Au (x, t)| is not greater than 1 + (C33 + C29 max )(1 + |x|p ) + max n2 C31 (1 + |x|)(p-2) 2 + amax |x| + bmax nC30 (1 + 2|x|p-1 ) for almost every (x, t) ×[0, T ]. Using Lemma 1.2, we have the estimate (68) |Au (x, t)| C68 (1 + |x|)p a.e. for some constant C68 > 0 depending only on C29 , C30 , C31 , C33 , n, p, amax , bmax , max , max . Hence |Au (x, t)|2 (x) 2 2 C68 (1 + |x|)2p C68 = (1 + |x|)2p+n+1 (1 + |x|)n+1 a.e. The same estimate hol for u instead u . So |Au |2 , |Au|2 L1 ( × [0, T ]). Moreover, Au , Au are uniformly bounded in the space L2 , where L2 = v : v 2 L1 ( × [0, T ]) = L2 Leb ( × [0, T ]). · From the Banach­Alaoglu theorem ([12], Th. 3.15) we know that balls in the space L2 are weakly compact. So for each sequence { m }mN convergent v in L2 if to 0, there exists a subsequence {~m }mN such that Au~m m . We will show that v = Au in the distribution sense. Indeed, for any function belonging to the class Cc ( × [0, T ]), we have (Au~m )dxdt = 1 u~m dxdt - t 2 T 0 (t) D2 u~m dxdr a(t) u~m dxdt + + 0 T 0 a(t)x + b(t) D u~m dxdt + (t)u~m dxdt. 1 udxdt - t 2 Letting m , we get vdxdt = + + T 0 T 0 (t) D2 udxdt a(t)x + b(t) D udxdt + (t)udxdt = tr a(t) udxdt (Au)dxdt. Hence Au~m conclude (69) Au in L2 if m . From uniqueness of the limit we Au Au in L2 if 0+ . M. Boryc and L. Kruk In view of (66) we have Au f a.e. From this and (69) we see that Au(x, t) f (x, t) a.e. This, together with Remark 2.8 ensure us that (70) max Au(x, t) - f (x, t) , |Du(x, t)| - c(t) 0 a.e. Take a sequence n 0 and let D be the set of (x, t) × [0, T ] such that (66) hol at (x, t) for all n . Then Leb ( × [0, T ]) \ D = 0. Choose t [0, T ] such that for almost every x we have (x, t) D. Since |Du n (x, t)| |Du(x, t)| as n (see (67)), I{|Du n (x,t)|<c(t)} = I{|Du(x,t)|<c(t)} for n large enough (depending on (x, t)). We have from this and (66) that (71) I{|Du(x,t)|<c(t)} Au n (x, t) I{|Du(x,t)|<c(t)} f (x, t) I{|Du(x,t)|<c(t)} Au n (x, t) I{|Du(x,t)|<c(t)} Au(x, t) a.e. in L2 , On the other hand, (69) yiel (72) so the sequence I{|Du(x,t)|<c(t)} Au n (x, t) is bounded in L2 and thus it is uniformly integrable in L1 . This, together with (71), implies that for every Cc ( × [0, T ]) L2 , I{|Du(x,t)|<c(t)} (Au n )(x, t)dxdt (73) I{|Du(x,t)|<c(t)} (f )(x, t)dxdt, which, together with (72) implies that I{|Du(x,t)|<c(t)} Au(x, t) = I{|Du(x,t)|<c(t)} f (x, t) a.e. 4. Existence and uniqueness of the optimal control. The results of this section are analogous to Theorems 7 and 8 from [11]. Fix (t, x) [0, T ) × (for t = he only admissible control is v(0) = 0 a.s.). Let mt be the measure on [t, T ] × , B([t, T ]) F equal to the product of the Lebesgue's measure and P . Remark 4.1. If a process X is a modification of a process Y and both processes have left-continuous sample paths a.s., then the processes X, Y are indistinguishable (compare Problem 1.1.5, [7]). Theorem 4.2. The optimal control v V, if it exists, is unique up to the indistinguishability. Proof. pose there are v1 , v2 V for which u(x, t) = Jxt (v1 ) = Jxt (v2 ). Put v0 = (v1 + v2 )/2. Of course v0 V. From Lemma 1.7 we have (74) u(x, t) - Jxt (v0 ) = 1 Jxt (v1 ) + Jxt (v2 ) - Jxt (v) 0. 2 A multidimensional singular stochastic control problem... 0 1 2 Let yxt , yxt , yxt be the solutions of (1) corresponding to v = v0 , v1 , v2 respectively. In view of the proof of Lemma 1.7 and strict convexity of the running cost function f , we have 1 1 0 1 2 (75) f yxt (s), s < f yxt (s), s + f yxt (s), s 2 2 1 2 provided that yxt (s) = yxt (s). Assume that v1 , v2 are not indistinguishable. Then there exists s (t, T ] such that P (A) > 0, where A = {v1 (s ) = v2 (s )} (see Remark 4.1). Because v1 , v2 have left-continuous sample paths a.s., there exists s () (t, s ) such 1 2 that v1 (s) = v2 (s) for all s [s (), s ], A. Thus, yxt (s) = yxt (s) on some mt -nonzero set. This facogether with (75) and the definition of Jxt imply thahe inequality (74) is strict, so we get a contradiction. We conclude that v1 , v2 must be indistinguishable. Lemma 4.3. Let {zn }nN be a sequence in Lp (mt ). If zn 0 in Lp (mt ), then T zn 0 in Lp (mt ), where T zn (s, ) = zn (s, ) - a(r)zn (r, )dr. s t a(r)zn (r, )dr Proof. By the H¨lder's inequality, the function g(s, ) = o satisfies ||g||p p ap E max L |zn (r, )|p · T p/q dr ap · T 1+p/q · ||zn ||p p , max L so T is a bounded operator from Lp (mt ) into Lp (mt ). Theorem 4.4. There exists an optimal control v V. Proof. Let {vk }kN be a sequence of admissible controls such that Jxt (vk ) k u(x, t) as k and let yxt be the solution of (1) corresponding to v = vk . Then Jxt (vk ) are uniformly bounded in k. By Lemma 1.9 the sequence k {yxt }kN is bounded in Lp (mt ) and hence, by the Banach­Alaoglu theorem k ([12], Th. 3.15), there exists a subsequence (still denoted by {yxt }) and a k yxt in Lp (mt ). process yxuch that yxt i Fix k N. Since the sequence {yxt }ik is also convergeno yxt , by the Mazur theorem there exists n(k) k zxt n(k) k,i · i yxt , k,i 0, k,i = 1, k n(k) < k k such that ||zxt - yxt ||Lp 1/k. In particular zxt yxt in Lp (mt ). Let n(k) k k = k,i · vi be the control corresponding to zxt in (1). Then k V. Moreover by Lemma 1.7, n(k) u(x, t) Jxt (k ) k,i · Jxt (vi ) ,...,n(k) max Jxt (vi ) - u(x, t). M. Boryc and L. Kruk For s [t, T ] we have k m zxt (s) - zxt (s) - t k k m Because {zxt } is convergent in Lp (mt ), zxt - zxt goes to 0 in Lp (mt ) as k, m . Using Lemma 4.3 we conclude that {k (· - t, ·)}kN is a Cauchy sequence in Lp (mt ) so it is convergeno a process v Lp (mt ). Without loss of generality we may assume that v(0) 0. Now we choose a subsequence (still denoted by k) such that k (s, ) v(s, ) as k for (s, ) A, where (Leb × P )(A) = T - t. For and s [0, T - t], we define s k m a(r) zxt (r) - zxt (r) dr = k (s - t) - m (s - t). A = s [0, T - t] : (s, ) A , As = : (s, ) A . Note that P (A0 ) = 1 because k (0) = v(0) = 0 P -a.s. Furthermore, let ~ = : Leb (A ) = T - t , S = s [0, T - t] : P (As ) = 1 . ~ Then P () = 1 and Leb (S) = T - t. Let N be a countable subset of S, dense in [0, T -t], including 0 and let AN = sN As . We have P (AN ) = 1. Let k (s) denote the total variation of k on the interval [0, s]. Because Jxt (k ) are uniformly bounded in k, there exists a constant C > 0 such that Ek (T - t) C for all k N. In view of the Fatou's lemma, E lim k k (T - t) lim k Ek (T - t) C, so lim k k (T - t) is finite a.s. Fix and let A , = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t. Let kn = kn () be a sequence of natural numbers such that limkn kn (T - t) = lim k k (T - t). Then v(ti+1 ) - v(ti ) = lim kn kn kn (ti+1 ) - kn (ti ) k lim kn (T - t) = lim k (T - t). Thus, v|A has bounded variation and hence it has left-hand and right-hand ~ ~ limits at each point. Let v = 0 on the P -zero set \ (AN ). On AN let 0 = v(0), s=0 v (s) = lim v(u) = lim v(u), s (0, T - t] . A us N us Then v is progressively measurable, left-continuous and v (0) = 0. More~ over, for AN and for each partition = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t, we can choose {tk }kN A such thak ti as i i A multidimensional singular stochastic control problem... k , i = 1, 2, . . . , m. Therefore, v (ti+1 ) - v (ti ) = lim v(tk ) - v(tk ) i+1 i k V ar(v, [0, T - t] A ) lim k (T - t), has bounded variation. This ensures us that V. ~ For AN , the set A {s [0, T - t] : v(s, ) = v (s, )} is countable, so its Lebesgue's measure is equal to 0. Therefore v = v mt -a.e. In particular, k v in Lp (mt ). Proceeding as in Steps 3­4 in the proof k v of Theorem 3.2 we can show that yxt yxt in Lp (mt ) and hence so (76) k k s t (r)dr = E v s t (r)dr To finish the proof we need to check that (77) E t T s t (r)dr d (s - t) lim E k t s t (r)dr dk (s - t), ~ is the total variation of v . Fix AN and let 0 s1 where s2 T - t. Let = {t0 , t1 , . . . , tm }, s1 = t0 < t1 < · · · < tm = s2 and let {tk }kN A be such thak ti as k , i = 0, 1, . . . , m. Then for i i every k0 N v (ti+1 ) - v (ti ) = lim v(tk ) - v(tk ) i+1 i V ar(v, [tk0 , s2 ] A ). 0 Letting k0 , we get and hence (78) m-1 v (ti+1 ) - v (ti ) V ar(v, [s1 , s2 ] A ) V ar(v , [s1 , s2 ]) V ar(v, [s1 , s2 ] A ). Let (s) = V ar(v, [0, s]A ), s [0, T -t]. Restricting = {t0 , t1 , . . . , tm }, s1 = t0 < t1 < · · · < tm = s2 so that A (in particular assuming s1 , s2 A ), we get v(ti+1 ) - v(ti ) = lim k (ti+1 ) - k (ti ) lim k (s2 ) - k (s1 ) . As |||| 0 we get (79) (s2 ) - (s1 ) lim k (s2 ) - k (s1 ) . Now take = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t, N . In particular, A for all AN . For every interval [ti , ti+1 ], M. Boryc and L. Kruk s t i = 0, 1, . . . , m - 1, let li = min ~ AN , by (78)­(79), we have (r)dr : s [ti + t, ti+1 + t] . For li · (ti+1 ) - (ti ) li · (ti+1 ) - (ti ) m-1 lim k li · k (ti+1 ) - k (ti ) . ~ This together with the Fatou's lemma and the fachat P (AN ) = 1 yiel li · (ti+1 ) - (ti ) E lim k m-1 li · k (ti+1 ) - k (ti ) li · k (ti+1 ) - k (ti ) lim E lim E s t (r)dr dk (s - t). Letting |||| 0, tm T - o that each partition in the sequence is contained in the next one, by the monotone convergence theorem, we get (77). From (76) and (77) Jxt (v ) lim Jxt (k ) = u(x, t). On the other hand, Jxt u(x, t) because v V and hence Jxt (v ) = is an optimal control. u(x, t) so v http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annales UMCS, Mathematica de Gruyter

A multidimensional singular stochastic control problem on a finite time horizon

Boryc, Marcin; Kruk, Łukasz
Annales UMCS, Mathematica , Volume 69 (1) – Jun 1, 2015
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Abstract

A singular stochastic control problem in n dimensions with timedependent coefficients on a finite time horizon is considered. We show thahe value function for this problem is a generalized solution of the corresponding HJB equation with locally bounded second derivatives with respeco the space variables and the first derivative with respeco time. Moreover, we prove that an optimal control exists and is unique. Singular stochastic control is a class of problems in which one is allowed to change the drift of a Markov process (usually a diffusion) at a price proportional to the variation of the control used. Admissible controls do not have to be absolutely continuous with respeco the Lebesgue measure and they may have jumps. This setup is natural for many problems of practical interest, including portfolio selection in finance, control of queueing networks and spacecraft control, to mention just a few examples. The reader is referred to Chapter VIII of [5] for more ormation and basic references. One-dimensional singular stochastic control problems are well understood by now, see, e.g., [2] and the references given there. In this case, if the running cost is convex, the optimal control makes the underlying process a reflected diffusion ahe boundary of the so-called nonaction region C. In the case of a diffusion with time-independent coefficients and discounted cost on the inite time horizon, C is just an interval and the value function enjoys 2010 Mathematics Subject Classification. Primary: 93E20; Secondary: 35Q93. Key wor and phrases. Singular stochastic control, generalized derivative, HJB equation, optimal control. M. Boryc and L. Kruk C 2 -regularity (smooth fit). Both C 2 -regularity of the value function and the characterization of the optimally controlled process have been extended to the case of singular control for the two-dimensional Brownian motion [14]. In n 3 dimensions, except for "close to one-dimensional" cases of a single push direction [15, 16] and the radially symmetric running cost [9], only partial results are known. For example, for optimal control of the Brownian motion on the inite time horizon, regularity of the boundary of C away from some "coer points" was shown in [17] and a characterization of the optimal control as a solution of the corresponding modified Skorokhod problem was given in [8]. In this paper we consider a n-dimensional singular stochastic control problem on a finite time horizon in which state is goveed by a linear stochastic differential equation with time-dependent coefficients, the running cost is convex and controls may act in any direction. We provide estimates for the corresponding value function. These estimates imply thahe value function has locally bounded generalized derivatives of the second order with respeco the space variable and of the first order with respeco the time variable. These properties are needed to consider the value function as a solution of the corresponding parabolic Hamilton­Jacobi­Bellman (HJB) equation in some generalized sense and to show existence and uniqueness of an optimal control. Similar results have been shown in Theorem 2.1 and Theorem 3.4 from [2] in the one-dimensional case with a single push direction. The corresponding results for a multidimensional singular stochastic control problem on the inite time horizon with time-independent drift, covariance, cost (i.e., for the elliptic case) can be found in [11]. Our article contains a generalization (or adjustment) of the approach of [2, 11] to an n-dimensional parabolic problem. Ius ouhat while the main ideas from those papers may be applied in our case, a mathematically rigorous analysis of our problem is somewhat delicate and nee rather careful arguments. Our motivation for pursuing this project is the hope thahe results given here will allow for a characterization of the optimal policy in the parabolic case as a solution to the corresponding Skorokhod problem for a domain with time-dependent (moving) boundary, which would be an analog of the main theorem from [8]. Indeed, the analysis of [8] used the results from [11] as the starting point, so it is plausible thaheir analogs will be useful in proving the corresponding result on a finite time horizon. Such a characterization would address a long-standing open problem on the structure of the optimal control in the case under consideration. We hope to address this issue in a subsequent paper. Existence results for multidimensional singular control problems closely related to our work may be found in [1, 3, 6]. Apparently, in spite of their considerable generality, none of them contains our existence result A multidimensional singular stochastic control problem... as a special case. Indeed, in these papers optimal weak solutions to the corresponding SDEs are constructed, while we are conceed about finding an optimal strong solution, i.e., for the given (as opposed to some) filtration and underlying Brownian motion. Moreover, the problem considered in [1] is elliptic and the allowable control directions lie in a cone, the opening of which cannot be too wide. In [3, 6] the time horizon is finite, buhe problem considered in [3] has the final cost instead of the running cost, while in [6] the drift of the controlled diffusion is bounded, which excludes its linear dependence on the state. The structure of this paper is as follows. In Section 1 we pose the singular stochastic control problem, give definitions and prove lemmas needed in further considerations. In Section 2 we prove estimates for the value function. In Section 3 we consider the Bellman's dynamic programming principle (DPP) and the HJB equation related to this problem. Section 4 contains proofs of existence and uniqueness of an optimal control. 1. Notation, assumptions and lemmas. Let Mn×n denote the set of matrices of dimension n × n with the operator norm, i.e. ||A|| = {|Ax| : x , |x| = 1}. Le > 0 be a fixed number representing our time horizon. For a function u = u(x, t) : × [0, T ] R we denote the gradient and the Hessian of u with respeco the space variables (i.e., xi ) by Du and D2 u, respectively. Let (Wt , t 0) be a standard n-dimensional Brownian motion defined on a complete probability space (, F, P ). Let (Ft , t 0) be the augmentation of the filtration generated by W (see [7], p. 89). Denote by V the set of controls v which are left-continuous, adapted to the filtration (Ft , t 0) random processes acting from [0, T ] into , with P -a.s. bounded variation and s.t. v(0) = 0 P -a.s. We note thahese processes are also progressively measurable (see [7], Th. 1.1.13). As it is customary in singular stochastic t control theory (see, e.g., [8]), we write v(t) = 0 (s), where |(t)| = 1 for every t [0, T ] and is nondecreasing and left-continuous. In other wor, (t) is the total variation of v on the time interval [0, t] and (t) is the Radon­Nikodym derivative of the vector-valued measure induced by v on [0, T ] with respeco its total variation . Consider the state process described by the stochastic integral equation (1) yxt (s) = x + a(r)yxt (r) + b(r) dr + (r)dWr-t + v(s - t), s [t, T ], where t [0, T ] is an initial time, x is an initial position, b : [0, T ] and a, : [0, T ] Mn×n stand for the drift and the covariance terms. Note that (yxt (s))s[t,T ] is a random process adapted to (Fs-t )s[t,T ] . M. Boryc and L. Kruk To each control v V, we associate a cost given by the payoff functional (2) Jxt (v) = E s t (r)dr + s t (r)dr d(s - t) , where f, and c are respectively the running cost, the discount factor and the instantaneous cost per unit of "fuel". Our purpose is to characterize the optimal cost, the so-called value function (3) u(x, t) = {Jxt (v) : v V}. It is often convenieno consider the following penalized problem associated with (3): (4) u (x, t) = {Jxt (v) : v V }, where > 0 and V is the set of all controls v V which are Lipschitz continuous and | dv (t)| 1 for almost every t [0, T ] almosurely. dt Definition 1.1. We say thahe finite time horizon stochastic control problem has the dynamic programming property in the weak sense if for every 0 x , t, t [0, T ] s.t. t < t and yxt (s) given by (1) with v 0 we have (5) u(x, t) E 0 s t (r)dr 0 + u(yxt (t ), t )e- (r)dr Let us assume the following: · , c are Lipschitz continuous from [0, T ] into [0, ) with constant L > 0, · b is Lipschitz continuous from [0, T ] into with the same constant L > 0, · a, are Lipschitz continuous from [0, T ] into Mn×n with the same constant L > 0, · there exists c0 > 0 such that c(t) c0 for all t [0, T ], ~ · f : × [0, T ] [0, ) and there exist constants p > 1, C0 , C0 > 0 n and (0, 1) we have such that for all t, t [0, T ], x, x R (6) (7) (8) (9) ~ C0 |x|p - C0 f (x, t) C0 (1 + |x|p ) , |f (x, t) - f (x + x , t)| C0 1 + f (x, t) + f (x + x , t) |f (x, t) - f (x, t )| C0 (1 + |x| )|t - t | , p |x | , 0 < f (x + x , t) - 2f (x, t) + f (x - x , t) C0 2 (1 + f (x, t))q , q = (1 - 2/p)+ . The last assumption implies strict convexity of the function f with respeco x. A multidimensional singular stochastic control problem... Let us denote by cmax and max the maximum of the function c, , respectively. Moreover, by amax , max , max and bmax we denote the maximum over t [0, T ] of the norms of the matrices a(t), (t), (t) and the vector b(t) respectively, where (t) = (t) T (t). Now we give lemmas needed for the proofs of the Theorems 2.1 and 2.2. The first one is well known. Lemma 1.2. For all x, y 0 we have xp + y p (x + y)p 2p-1 (xp + y p ), if p 1, 2p-1 (xp + y p ) (x + y)p xp + y p , if p (0, 1). Lemma 1.3 (See [10], Corollary 2.5.12). Consider an n-dimensional process described by a stochastic integral equation x(t) = x0 + t 0 g(x(s), s) + t 0 h(x(s), s)dWs , t 0, where x0 , g : × [0, ) and h : × [0, ) Mn×n . We assume thahere exists a constant C such that for all x and t 0 (10) ||h(x, t)|| + |g(x, t)| C(1 + |x|). Then for every q > 0 there exists a constant C11 > 0 depending only on q, C such that for all t 0 (11) E |x(s)|q C11 eC11 t (1 + |x0 |)q . 0st Remark 1.4. For the process yxt defined by (1) with v 0 the assumption (10) hol Indeed, is Lipschitz continuous, independent of x and defined on a finite time interval [0, T ], so it is bounded. We conclude the same about a, b, so |g(x, t)| = |a(t)·x+b(t)| C(1+|x|), where C = max{||a(t)||, |b(t)| : t [0, T ]}. Lemma 1.5. Let x, x , t [0, T ] and g(s) = yxt (s) - yx t (s) for s [t, T ]. Then dg s [t, T ], (s) = a(s)g(s), |g(s)| C12 |x - x |, (12) where C12 = 1 + amax T eamax T . Proof. In view of (1) we have g(s) = x - x + a(r)(yxt (r) - yx t (r))dr = x - x + a(r)g(r)dr. Taking the derivative d/ of both sides, we gehe differential equation dg (s) = a(s)g(s) with initial data g(t) = x - x . The solution of this problem satisfies |g(s)| |x - x | + |a(r)g(r)|dr |x - x | + amax s t |g(r)|dr. M. Boryc and L. Kruk Using the Gronwall's inequality (see [4], p. 625), we gehe second part of (12). Lemma 1.6. pose that for some x , t [0, T ], v V we have E s t (r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then (13) E f (yxt (s), s) C13 (1 + |x|p ), where C13 = C · e T 0 (r)dr . , we Proof. Indeed, multiplying both sides of our assumption by e get E (r)dr f (yxt (s), s)e (r)dr Ce (r)dr (1 + |x|p ) C13 (1 + |x|p ). Of course, the left-hand side is nomaller than E f (yxt (s), s) Lemma 1.7 (Compare a statement in [17], p. 181). The function Jxt (v) is convex with respeco (x, v), more precisely, for all x1 , x2 , t [0, T ], v1 , v2 V and [0, 1], Jx1 +(1-)x2 ,t (v1 + (1 - )v2 ) Jx1 ,t (v1 ) + (1 - )Jx2 ,t (v2 ) . v Proof. First, we note thahe set V is obviously convex. Let yxt (s) be the solution of (1) corresponding to a control v. Denote v0 = v1 + (1 - )v2 and x0 = x1 + (1 - )x2 . In view of the definition of Jxt (v), iuffices to prove two following inequalities (14) (15) v v v f yx0 ,t (s), s f (yx1 ,t (s), s) + (1 - )f (yx2 ,t (s), s), s [t, T ] , 0 1 2 d0 (s - t) d1 (s - t) + (1 - ) d2 (s - t) , where 0 , 1 , 2 are the total variations of v0 , v1 , v2 respectively. The latter inequality is a consequence of the fachahe variation of the sum of functions is not greater than the sum of their variations. So 0 1 + (1 - )2 . Because 0 , 1 , 2 are nondecreasing and 0 (0) = 1 (0) = 2 (0) = 0 P -a.s., we conclude that (15) is true. To prove (14) we show firshat (16) v v v yx0 ,t (s) = yx1 ,t (s) + (1 - )yx2 ,t (s) . 0 1 2 Indeed, using (1) we get v yxi ,t (s) = xi + i s t v a(r)yxi ,t (r) + b(r) dr + i s t (r)dWr-t + vi (s - t) , A multidimensional singular stochastic control problem... v v v i = 0, 1, 2. Let g(s) = yx0 ,t (s) - yx1 ,t (s) - (1 - )yx2 ,t (s). Then 0 1 2 g(s) = v v v a(r) yx0 ,t (r) - yx1 ,t (r) - (1 - )yx2 ,t (r) dr = 0 1 2 a(r)g(r)dr . Taking the derivative d/ of both sides, we gehe differential equation dg (s) = a(s)g(s) with initial data g(t) = x0 - x1 - (1 - )x2 = 0. The solution of this problem is g(s) 0, so (16) hol Using (16) and convexity of f we have (14). Lemma 1.8. pose that for some t [0, T ], x , v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t . Then there exists a constant C17 > 0 independent of x, uch that (17) E(T - t ) C17 (1 + |x|p ). Proof. Indeed, multiplying both sides of our assumption by e and using the lower bound of c, we get c0 E(T - t ) = c0 E Ce T 0 (t +r)dr E (1 + |x|p ). c(t + s)e T - (t +r)dr (r)dr Lemma 1.9. pose that for some x , t [0, T ], v V we have E s 0 C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then there exists a constant C18 > 0 independent of x, uch that (18) E (1 + |yxt (t + s)|p ) C18 (1 + |x|p ). Proof. From Lemma 1.6 we know that E Using (6), we get E Hence ~ C0 E 0 f (yxt (t + s), t + s) C13 (1 + |x|p ). ~ C0 |yxt (t + s)|p - C0 C13 (1 + |x|p ). T -t |yxt (t + s)|p (C13 + C0 T )(1 + |x|p ) M. Boryc and L. Kruk and finally ~ C0 E ~ (1 + |yxt (t + s)|p ) (C13 + C0 T + C0 T )(1 + |x|p ) . Lemma 1.10. Let 0 t t T and pose that for some x , v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t, t . Then there exists a constant C19 > 0 independent of x, t, uch that (19) E f (yxt (t + s), t + s) C19 (1 + |x|p ). Proof. We observe that using (8) we have f (yxt (t + s), t + s) |f (yxt (t + s), t + s) - f (yxt (t + s), t + s)| + f (yxt (t + s), t + s) C0 |t - t |(1 + |yxt (t + s)|p ) + f (yxt (t + s), t + s). Hence, in view of Lemma 1.6 and Lemma 1.9, we get E f (yxt (t + s), t + s) C0 |t - t |C18 (1 + |x|p ) + C13 (1 + |x|p ) C19 (1 + |x|p ), where C19 = C0 T C18 + C13 . Lemma 1.11. Let 0 t t T , x , v V. Assume that E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t , t. Then there exists a constant C20 > 0 independent of x, t , uch that for all s [0, T - t] we have (20) E|yxt (t + s) - yxt (t + s)|p C20 |t - t |p (1 + |x|p ) . s 0 s 0 s 0 s 0 Proof. For s [0, T - t], we have yxt (t + s) = x + yxt (t +s) = x+ so (21) a(t + r)yxt (t + r) + b(t + r) dr + a(t +r)yxt (t +r)+b(t +r) dr+ (t + r)dWr + v(s), (t +r)dWr +v(s), yxt (t + s) - yxt (t + s) = As + Bs + Ms , A multidimensional singular stochastic control problem... where As = A1 = s A2 = s Bs = Ms = s 0 0 0 0 0 a(t + r)yxt (t + r) - a(t + r)yxt (t + r) dr = A1 + A2 , s s a(t + r) yxt (t + r) - yxt (t + r) dr, a(t + r) - a(t + r) yxt (t + r)dr, b(t + r) - b(t + r) dr, (t + r) - (t + r) dWr . Recall that a, b, are Lipschitz continuous with the constant L. The process Ms is a martingale with quadratic variation [M ]s = s 0 (t + r) - (t + r) dr L2 |t - t |2 s. This, together with the Burkholder­Davis­Gundy inequalities (see [7], Theorem 3.3.28), implies the existence of a constant Cp , depending only on p, such that (22) Clearly, (23) |Ms |p Cp Lp T 2 |t - t |p . |Bs | LT |t - t |. By the H¨lder's inequality, for q = p/(p - 1) we have o (24) (25) |A1 |p (aq s) q s max s 0 |yxt (t + r) - yxt (t + r)|p dr, p q |A2 |p (L|t - t |)q s s s 0 |yxt (t + r)|p dr. By Lemma 1.9, the inequality (18) hol for every t [0, T ]. Lemma 1.2 and the relations (18), (21)­(25) imply thahe random variable |yxt (t + s) - yxt (t + s)|p is integrable and hence, by the Lebesgue dominated convergence theorem, the function F (s) = E|yxt (t + s) - yxt (t + s)|p is continuous on [0, T - t]. From Lemma 1.2 and (18), (21)­(25) we also have, for each s [0, T - t], F (s) c1 |t - t |p (1 + |x|p ) + c2 s 0 F (r)dr, M. Boryc and L. Kruk p p p where c1 = 22p-2 (Lp T p + Cp Lp T 2 + (LT ) q C18 ), c2 = 22p-2 ap T q . This, max together with the Gronwall's inequality (see, e.g., [7], Problem 5.2.7), implies that for all s [0, T - t], F (s) c1 |t - t |p (1 + |x|p ) 1 + c2 We have obtained (20) with C20 = c1 (1 + c2 T 0 s 0 ec2 (s-r) dr) . ec2 (T -r) dr). Lemma 1.12. pose that for some x , t [0, T ], v V we have E s t (r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t. Then there exists a constant C26 > 0 independent of x, x , uch that for every x , (26) E f (yx+x ,t (s), s) C26 (1 + |x|p + |x + x |p ). Proof. From (6) and Lemma 1.2 we have E f (yx+x ,t (s), s) E t C0 (1 + |yx+x ,t (s)|p ) T C0 + C0 2p-1 E |yx+x ,t (s) - yx,t (s)|p + C0 2p-1 E |yx,t (s)|p . Now using Lemma 1.5, Lemma 1.9 and Lemma 1.2 again, we get E p f (yx+x ,t (s), s) T C0 + C0 2p-1 T · C12 |x |p + C0 2p-1 C18 (1 + |x|p ) p T C0 + C0 22p-2 T · C12 |x + x|p + |x|p + C0 2p-1 C18 (1 + |x|p ) C26 (1 + |x|p + |x + x |p ), p where C26 = C0 T + 22p-2 T · C12 + 2p-1 C18 . Lemma 1.13. pose that for some x , t [0, T ], v V we have E s 0 (t +r)dr C(1 + |x|p ) for a suitable constant C > 0 independent of x, t . Then there exists a constant C27 > 0 independent of x, t , uch that for every t [t , T ], (27) E f (yxt (t + s), t + s) C27 (1 + |x|p ). A multidimensional singular stochastic control problem... Proof. Using (6) and Lemma 1.2, we have E f (yxt (t + s), t + s) E C0 (1 + |yxt (t + s)|p ) C0 T + 2p-1 C0 E + 2p-1 C0 E |yxt (t + s)|p |yxt (t + s) - yxt (t + s)|p . In view of Lemma 1.9, the Fubini's theorem and Lemma 1.11, we get E f (yxt (t + s), t + s) C0 T + 2p-1 C0 C18 (1 + |x|p ) + 2p-1 C0 T C20 |t - t |p (1 + |x|p ) C27 (1 + |x|p ), where C27 = C0 T + 2p-1 C18 + 2p-1 T p+1 C20 . The nexwo definitions and lemma refer to mollification of a given function (see [4], p. 629­630). Definition 1.14. Define C ( ) by (28) (x) = C28 exp 0 1 |x|2 -1 if |x| < 1 if |x| 1, where the constant C28 is selected so that (x)dx = 1. For each m N set m (x) = mn · (mx). We call the standard mollifier. The functions m belong to the class C ( ) and satisfy m (x)dx = 1. Definition 1.15. Fix t [0, T ]. For each m N we define mollification of the function u(·, t ) by um (x) = m (y)u(x - y, t )dy, x , where B(0, r) = {x : |x| < r}. Lemma 1.16. For each m N we have um C ( ). Moreover, if u(·, t ) is continuous, then um (x) u(x, t ) uniformly on compacubsets of as m . 2. Estimates for the value function. Lehe assumptions from Section 1 appearing immediately after Definition 1.1 hold. Theorem 2.1. Let u be the value function defined by (3). Then for some positive constants C29 , C30 , C31 , the same p > 1 as in the assumptions (6)­ (9) and every t [0, T ], x, x and (0, 1), the following estimates M. Boryc and L. Kruk hold: (29) (30) (31) 0 u(x, t) C29 (1 + |x|p ) , |u(x, t) - u(x + x , t)| C30 (1 + |x|p-1 + |x + x |p-1 )|x | , 0 u(x + x , t) - 2u(x, t) + u(x - x , t) C31 2 (1 + |x|)(p-2) . Proof: Proof of (29). Nonnegativity of u is the consequence of nonnegativity of f and c. Next, taking the control v 0 and using (6), the Fubini's theorem, Lemma 1.3 and Lemma 1.2, we get u(x, t) Jxt (0) = E 0 s t (r)dr 0 C0 (1 + |yxt (s)|p ) = C0 T 0 E(1 + |yxt (s)|p ) C0 C0 0 E 1 + C11 eC11 (s-t) (1 + |x|)p (1 + C11 eC11 T 2p-1 )(1 + |x|p ) = C0 T (1 + C11 eC11 T 2p-1 )(1 + |x|p ) = C29 (1 + |x|p ), where C29 depen only on C0 , T, C11 , p, so (29) is proved. Proof of (30). Now we note that u(x + x , t) - u(x, t) = Jx+x ,t (v ) - Jx,t (v) v V Jx+x ,t (v) - Jx,t (v) . Hence u(x + x , t) - u(x, t) |Jxt (v) - Jx+x ,t (v)| E t |f (yxt (s), s) - f (yx+x ,t (s), s)|e- s t (r)dr Applying (7), we can estimate the last expression from above by E C0 1 + f (yxt (s), s) + f (yx+x ,t (s), s) · |yxt (s) - yx+x ,t (s)| Using Lemma 1.5, we have u(x + x , t) - u(x, t) C0 C12 |x | · E p-1 p 1 + f (yxt (s), s) + f (yx+x ,t (s), s) A multidimensional singular stochastic control problem... We use the H¨lder's inequality with exponent o pression above by (32) C0 C12 |x | · E p p-1 to estimate the last exp-1 p 1 + f (yxt (s), s) + f (yx+x ,t (s), s) T p. s t (r)dr for some arbitrary that > 0. From (32), Lemma 1.6 and Lemma 1.12 we see u(x + x , t) - u(x, t) C0 C12 |x | T + C13 (1 + |x|p ) + C26 (1 + |x|p + |x + x |p ) C30 |x | 1 + |x|p + |x + x |p p-1 p p-1 p Tp , where C30 = T 1/p · C0 C12 (T + C13 + C26 ) . Finally using Lemma 1.2, we get u(x + x , t) - u(x, t) C30 (1 + |x|p-1 + |x + x |p-1 )|x |. In an analogous manner we gehe same estimate for u(x, t)-u(x+x , t). Proof of (31). We observe that u(x + x , t) + u(x - x , t) - 2u(x, t) Jx+x ,t (v) + Jx-x ,t (v) - 2Jxt (v) = E t f (yx+x ,t (s), s) + f (yx-x ,t (s), s) - 2f (yxt (s), s) e- s t (r)dr In view of (12) we can apply (9) to get u(x + x , t) + u(x - x , t) - 2u(x, t) E (1-2/p)+ C0 2 1 + f (yxt (s), s) . If p 2 we have u(x + x , t) + u(x - x , t) - 2u(x, t) C0 T 2 . If p > 2 p we use the H¨lder inequality with exponent p-2 to get o u(x + x , t) + u(x - x , t) - 2u(x, t) C0 2 E 1-2/p (1 + f (yxt (s), s) T 2/p . M. Boryc and L. Kruk s t (r)dr for some arbitrary > 0. From Lemma 1.6 and Lemma 1.2 we see that u(x + x , t) + u(x - x , t) - 2u(x, t) C0 2 T + C13 (1 + |x|p ) C31 2 (1 + |x|p )1-2/p C31 2 (1 + |x|)p-2 , where C31 = T 2/p C0 (T + C13 )1-2/p . We have proved the upper bound of (31). To prove the lower bound of (31), it clearly suffices to prove convexity of u(x, t) with respeco the first variable. In view of the definition of u we know that for every > 0, x1 , x2 , t [0, T ], [0, 1] there exist v1 , v2 V such that Jxi ,t (vi ) u(xi , t) + , i = 1, 2. Using Lemma 1.7, we get u(x1 + (1 - )x2 , t) Jx1 +(1-)x2 ,t (v1 + (1 - )v2 ) Jx1 ,t (v1 ) + (1 - )Jx2 ,t (v2 ) u(x1 , t) + (1 - )u(x2 , t) + . Because > 0 is arbitrary, we get convexity of u(x, t) with respeco the first variable. Theorem 2.2. Lehe assumptions of Theorem 2.1 be satisfied. Assume thahe dynamic programming property in the weak sense hol (Definition 1.1). Then for some constant C33 > 0 and every t, t [0, T ], x , we have (33) |u(x, t) - u(x, t )| C33 (1 + |x|p )|t - t | . 1-2/p T 2/p Proof. We note that u(x, t) - u(x, t ) = Jxt (v) - Jxt (v ) v V v V Jxt (v ) - Jxt (v ) . A multidimensional singular stochastic control problem... For t he difference Jxt (v) - Jxt (v) is equal to E s 0 - + - - T - - - -t s 0 (t +r)dr s 0 s 0 - s 0 (t +r)dr s 0 (t +r)dr (t +r)dr Let us denote the expectations of the firswo integrals in the last expression by A and B, respectively. Because the laswo integrals are nonnegative we get (34) Jxt (v) - Jxt (v) A + B . s 0 s 0 We can estimate B as follows: BE - (t +r)dr . Adding and subtracting 0 (t +r)dr under the absolute value sign and using the triangle inequality and positivity of , we get BE cmax |e- s 0 - e- s 0 (t +r)dr | + |c(t + s) - c(t + s)| . Because |ex - ey | |x - y| for x, y 0 and c, are Lipschitz continuous, we have BE s cmax |(t + r) - (t + r)|dr + |c(t + s) - c(t + s)| (cmax T + 1)L|t - t |E = (cmax T + 1)L|t - t |E(T - t). s 0 (t +r)dr (C29 + )(1 + |x|p ) for some arbitrary > 0. Using Lemma 1.8, we get E(T -t) E(T -t ) C17 (1 + |x|p ) and (35) B C35 |t - t |(1 + |x|p ) , where C35 = (cmax T + 1)LC17 . M. Boryc and L. Kruk Now we estimate A: AE s 0 s 0 - +E (t +r)dr s 0 s 0 (t +r)dr - = A 1 + A2 . (t +r)dr Using the inequality |ex - ey | |x - y| for x, y 0 again, we get A1 E (36) f (yxt (t + s), t + s) s 0 |(t + r) - (t + r)|dr T L|t - t | E f (yxt (t + s), t + s) . By virtue of (29) we can consider only those controls v for which (37) E s 0 (t +r)dr for some arbitrary (38) to E > 0. Using (36) and Lemma 1.13, we get A1 C38 |t - t |(1 + |x|p ) , where C38 = T LC27 . To estimate A2 we use (7)­(8) and we have that A2 is less than or equal f (yxt (t + s), t + s) - f (yxt (t + s), t + s) + f (yxt (t + s), t + s) - f (yxt (t + s), t + s) E C0 1 + f (yxt (t + s), t + s) + f (yxt (t + s), t + s) × |yxt (t + s) - yxt (t + s)| +E C0 (1 + |yxt (t + s)|p )|t - t | = A3 + A4 . Using the H¨lder's inequality and the Fubini's theorem, we get o A3 C0 E × 1 + f (yxt (t + s), t + s) + f (yxt (t + s), t + s) 1/p p E|yxt (t + s) - yxt (t + s)| A multidimensional singular stochastic control problem... From this together with (37), Lemma 1.13, Lemma 1.10 and Lemma 1.11 we have A3 C0 {(T + C27 + C19 )(1 + |x|p )} · T C20 |t - t |p (1 + |x|p ) Because 1 + |x|p (1 + |x|)p , we get A3 C0 (T + C27 + C19 ) (1 + |x|)p-1 · (T C20 )1/p |t - t |(1 + |x|). Hence, from Lemma 1.2, (39) A3 C39 |t - t |(1 + |x|p ), 1/p where C39 = C0 (T + C27 + C19 ) (T C20 )1/p 2p-1 . Furthermore, from Lemma 1.9 we get (40) A4 C40 |t - t |(1 + |x|p ), where C40 = C0 C18 . u(x, t) - u(x, t ) C41 |t - t |(1 + |x|p ), In view of (34)­(35) and (38)­(40) we get for t t, (41) where C41 = C35 + C38 + C39 + C40 . To obtain a similar inequality for t < t we proceed as follows. Let 0 yxt (s) s[t,T ] be a solution of (1) with v 0. We can write the i-th coors t n 0 dinate of yxt (s) as follows 0 (42) yxt (s)i = xi + 0 aij (r)yxt (r)j + bi (r) dr + n j=1 t j ij (r)dWr-t , j=1 i = 1, . . . , n, where subscripts denote the corresponding coordinates. Let {um (·)}nN be a sequence of mollifications of the function u(·, t ) (see Def. 1.15). Applying the It^'s formula ([7], Th. 3.3.6), we get o 0 Eum (yxt (t )) n = um (x) + E i=1 n t 0 um (yxt (s)) n j=1 0 aij (s)yxt (s)j + bi (s) xi (43) +E i=1 t n 0 um (yxt (s)) xi t j ij (s)dWs-t j=1 0 0 d[yxt (s)i , yxt (s)j ] 1 E 2 2u 0 m (yxt (s)) i,j=1 t xi xj = um (x) + A + B + C . We need the following lemma. M. Boryc and L. Kruk Lemma 2.3. We assume (29)­(31). Le [0, T ] be fixed. Then there exist constants C45 , C46 > 0 such that for all x , m N and i, j {1, . . . , n}, (44) (45) lim um (x) = u(x, t ), um (x) C45 (1 + |x|)p-1 , xi 0 2 um (x) C46 (1 + |x|p ). xi xj n 0 aij (s)yxt (s)j + bi (s) j=1 0 n||a(s)|| · |yxt (s)| + |b(s)| (46) We estimate A as follows AE i=1 n t 0 um (yxt (s)) · xi 0 um (yxt (s)) xi i=1 t Using Lemma 2.3 and Lemma 1.2, we see that A is not greater than E (47) i=1 t 0 0 C45 (1 + |yxt (s)|)p-1 n(amax + bmax )(1 + |yxt (s)|) 0 (1 + |yxt (s)|p ), C47 E n2 C p-1 . where C47 = 45 (amax + bmax )2 Now we show that B = 0. Indeed, n B=E i,j=1 Zij (t ), where Zij (s) = 0 um (yxt (r)) j ij (r)dWr-t for s [t, t ]. xi From properties of the It^'s integrals (see [7], Section 3.2) the process o Zij (s) s[t,t ] is a martingale provided that E 0 um (yxt (s)) ij (s) xi 2 2 . Using Lemma 2.3 and Lemma 1.2, we have E 0 um (yxt (s)) ij (s) xi 0 (1 + |yxt (s)|2p ) E 2 0 2 C45 (1 + |yxt (s)|)2p-2 max 2 2 C45 max E 0 (1 + |yxt (s)|)2p 2 2 C45 max 22p-1 E A multidimensional singular stochastic control problem... Using the Fubini's theorem and Lemma 1.3, we get E 0 um (yxt (s)) ij (s) xi 2 2 2 C45 max 22p-1 1 + C11 eC11 T (1 + |x|)2p < . Hence Zij is a martingale and EZij (t ) = EZij (t) = 0. So (48) B=E i,j=1 Zij (t ) = 0. Now, using the conventional "multiplication rules" (see [7], p. 154), we know that = 0, i dWs = 0, i i dWs dWs = , i j dWs dWs = 0 for i = j. So in view of (42) we can write n 0 0 d[yxt (s)i , yxt (s)j ] n k ik (s)dWs-t k=1 n l jl (s)dWs-t l=1 k=1 ik (s)jk (s) From Lemma 2.3 we have C= (49) 1 2 1 2 i,j=1 n t 0 2 um (yxt (s)) xi xj ik (s)jk (s) k=1 i,j=1 t 0 2 C46 (1 + |yxt (s)|p )nmax = C49 E 1 0 2 (1 + |yxt (s)|p ), where C49 = C46 max n3 . 2 In summary, in view of (43) and (47)­(49) 0 Eum (yxt (t )) um (x) + (C47 + C49 )E 0 (1 + |yxt (s)|p ) Taking the limit as n and using the Fatou's lemma, we get (50) 0 Eu(yxt (t ), t ) u(x, t ) + C50 E 0 (1 + |yxt (s)|p ), C50 = C47 + C49 . Furthermore, from Lemma 1.2 and Lemma 1.3 we have for each s [t, T ] (51) 0 E 1 + |yxt (s)|p C51 (1 + |x|p ), C51 = C11 eC11 T 2p-1 + 1 . M. Boryc and L. Kruk Next, from (5), (6), (50), (51) and the Fubini's theorem we conclude u(x, t) E 0 s t (r)dr 0 + u(yxt (t ), t )e- (r)dr C0 0 0 E 1 + |yxt (s)|p + Eu(yxt (t ), t ) C0 C51 + C50 C51 |t - t |(1 + |x|p ) + u(x, t ). Hence, for t < t (52) u(x, t) - u(x, t ) C52 |t - t |(1 + |x|p ), C52 = C51 (C0 + C50 ) . It is clear that (41) and (52) imply (33). Now we give the proof of Lemma 2.3. Proof: Proof of (44). The continuity of u(·, t ) is a consequence of (30). So in view of Lemma 1.16 we conclude that limm um (x) = u(x, t ). Proof of (45). Let x , 0 |x | < 1. From Definitions 1.14, 1.15 and (30) we get |um (x) - um (x + x )| = m (y) u(x - y, t ) - u(x + x - y, t ) dy mn · (my)|u(x - y, t ) - u(x + x - y, t )|dy 1 + |x - y|p-1 + |x + x - y|p-1 dy. C28 C30 |x |mn Because |x | < 1 and |y| 1 m 1, we have (2 + 2p-1 )(1 + |x|)p-1 . 1 + |x - y|p-1 + |x + x - y|p-1 1 + (1 + |x|)p-1 + (2 + |x|)p-1 Furthermore (see [4], p. 615), dy = n/2 1 · n , where (t) = n ( 2 + 1) m st-1 e-s , for t > 0. In summary, |um (x) - um (x + x )| n/2 C28 C30 n (2 + 2p-1 )(1 + |x|)p-1 . |x | ( 2 + 1) Taking the limit as |x | 0 on both sides, we conclude (45). A multidimensional singular stochastic control problem... Proof of (46). Let x and (0, 1). We have um (x + x ) - 2um (x) + um (x - x ) = m (y) u(x + x , t ) - 2u(x, t ) + u(x - x , t ) dy. u From (31) and nonnegativity of m we conclude that xim (x) 0. On the xj other hand, using (31) and mimicking the proof of (45), we see that um (x + x ) - 2um (x) + um (x - x ) mn · (my)C31 2 (1 + |x|)(p-2) dy n/2 + (1 + |x|)(p-2) . n ( 2 + 1) 2 C28 C31 For p (1, 2], (1 + |x|)(p-2) = 1 (1 + |x|p ) 2p-1 (1 + |x|p ). For p > 2, in + view of Lemma 1.2, (1+|x|)(p-2) = (1+|x|)p-2 (1+|x|)p 2p-1 (1+|x|p ). Thus, for all p > 1 we have um (x + x ) - 2um (x) + um (x - x ) n/2 C28 C31 2p-1 (1 + |x|p ). 2 ( n + 1) 2 Taking the limit as 0 , we can conclude (46). Remark 2.4. Theorems 2.1 and 2.2 are true for functions u (see (4)) instead of u. Indeed, in view of the proofs we see thahe constants C29 , C30 , C31 , C33 do not depend on . Remark 2.5. It follows from (44)­(46) that for every t [0, T ] Dum (· ; t ) converges to Du(·, t ) the distributional gradient of u with respeco x almost uniformly as m (see the proof of Theorem 3.5, to follow, for a similar argument, with um replaced by u m ). This implies differentiability of u with respeco x in the classical sense (see, e.g., Theorem 7.17 in [13]), so Du is the classical gradient of u with respeco x at any point (x, t ) × [0, T ]. Moreover, by (46) Dum are locally Lipschitz in x uniformly in m, so Du is also locally Lipschitz in x. Thus Theorems 2.1, 2.2 and their proofs imply thahe value function u(x, t) has generalized derivatives of the first order with respeco t and of the second order with respeco x. These generalized derivatives belongs to the space L ( × [0, T ]) of all loc functions essentially bounded on every open bounded subset of the domain. Proposition 2.6. For all x and t [0, T ] we have u(x, t) (cmax + C29 )(1 + |x|). M. Boryc and L. Kruk Proof. Let x be arbitrary. Consider controls for which lims0+ vs = x. In view of (2) and (3) we have u(x , t) = {Jx t (v) : v V} c(t)|x| + {Jx+x ,t (v) : v V} = c(t)|x| + u(x + x , t). So u(x , t) - u(x + x , t) c(t)|x|. Similarly u(x + x , t) - u(x , t) c(t)|x|, so (53) |u(x + x , t) - u(x , t)| c(t)|x|. Taking x = 0, we get |u(x, t) - u(0, t)| c(t)|x|. From (29) we see that u(0, t) C29 so u(x, t) c(t)|x| + u(0, t) cmax |x| + C29 (cmax + C29 )(1 + |x|). Remark 2.7. The proof of Proposition 2.6 is not valid for u instead of u, because if a control v V , then it is continuous, so the condition lims0+ vs = x is invalid for x = 0. Remark 2.8. The value function u(x, t) satisfies |Du(x, t)| c(t) for all (x, t) × [0, T ]. Indeed, the gradient exists for all (x, t) × [0, T ] in view of Remark 2.5. From (53) we see thahe first derivative of u(x, t) with respeco x in any direction is bounded by c(t). Hence, the norm of the gradient Du(x, t) is bounded by c(t), too. 3. Dynamic Programming Principle and HJB equation. To consider the DPP and the HJB equation for our problem we will first prove the pointwise convergence of u to u if 0+ . For this purpose we need an integral form of the Gronwall's inequality with locally finite measures. Lemma 3.1 (see [18]). Let be a locally finite measure on the Borel algebra of [t, T ], where 0 t T . We consider a measurable function T defined on [t, T ] such tha |(r)|(dr) < . We assume thahere exists a Borel function 0 on [t, T ] such that for all s [t, T ], (s) (s) + Then for all s [t, T ], (s) (s) + Theorem 3.2. For all (x, t) (r)e([r,s)) (dr). we have lim 0+ [t,s) n ×[0, T ] R (r)(dr). [t,s) u (x, t) = u(x, t). Proof. Fix x and t [0, T ]. Consider an arbitrary v V such that Jxt (v) < . Step 1. We show firshat v Lp ( × [0, T - t], P Leb ), where Leb denotes the Lebesgue's measure. Since Jxt (v) < , we have E f (yxt (s), s) < A multidimensional singular stochastic control problem... and from (6) we get (54) E |yxt (s)|p < . From (1) we can write for s [t, T ] (55) v(s - t) = yxt (s) - x - b(r)dr - (r)dWr-t - a(r)yxt (r)dr. Using (54) and properties of the normal distribution, we know that each term from the line above, maybe except for the last one, belongs to the space Lp ( × [0, T - t]). Buhe laserm belongs to this space, too. Indeed, E p a(r)yxt (r)dr ap max |yxt (r)|dr Using the H¨lder's inequality and (54), we can estimate the last expression o above by ap E max |yxt (r)|p dr · |T - t|p/q |yxt (r)|p dr < , ap T 1+p/q E max 1 p 1 q where + = 1. Hence, from (55) we see that v Lp ( × [0, T - t]). Step 2. Now we define a sequence of bounded controls {vR , R > 0} such that vR is convergeno v in the space Lp ( × [0, T - t]) and the total variation of vR is pointwise convergeno the total variation of v from below. Let v(s), |v(s)| R vR (s) = v(s) |v(s)| · R, |v(s)| > R . We see that for all s [0, T - t] limR vR (s) = v(s) and |vR (s)| |v(s)|. Hence, from Lemma 1.2 and Step 1, E |v(s) - vR (s)|p 2p E |v(s)|p < and using the Lebesgue's dominated convergence theorem, we get |v(s) - vR (s)|p = E lim |v(s) - vR (s)|p = 0. The convergence in Lp is proved. Moreover, if (s), R (s) denote the total variations on the interval [0, s] of the functions v, vR respectively, then for all s [0, T - t], (56) R (s) (s) and lim R (s) = (s). M. Boryc and L. Kruk vR v Step 3. Let yxt , yxt denote the state processes (see (1)) corresponding to vR the controls v, vR respectively. We wano show that {yxt } is convergent v in the space Lp ( × [t, T ]). First we observe that for s [t, T ], to yxt vR v yxt (s) - yxt (s) = t vR v Denoting zR (s) = yxt (s) - yxt (s) and uR (s) = v(s - t) - vR (s - t), we s can rewrite the last equality in the form zR (s) = t a(r)zR (r)dr + uR (s). s Hence |zR (s)| t |zR (r)|amax dr + |uR (s)|. Using Lemma 3.1 with = |zR |, = |uR | and = amax · Leb , we ge vR v a(r) yxt (r) - yxt (r) dr + v(s - t) - vR (s - t). |zR (s)| |uR (s)| + (57) |uR (r)|eamax (s-r) dr s t |uR (s)| + C57 |uR (r)|dr, where C57 = eamax T . So from Lemma 1.2 and the H¨lder's inequality o |zR (s)| 2 p p-1 |uR (s)| + p C57 s t |uR (r)|dr s t (58) p 2p-1 |uR (s)|p + C57 (s - t)p/q |uR (r)|p dr C58 |uR (s)|p + 1 where p + we have 1 q |uR (r)|p dr , p = 1 and C58 = 2p-1 (1 + C57 T p/q ). Finally, in view of Step 2 |zR (s)|p lim C58 E R |uR (s)|p + |uR (r)|p dr = 0. lim C58 E |uR (s)|p + C58 T E |uR (r)|p dr Step 4. The nextep is to show that Jxt (vR ) Jxt (v) if R . Indeed, |Jxt (v) - Jxt (vR )| E vR v f (yxt (s), s) - f (yxt (s), s) e- (r)dr + E (r)dr d( - R )(s - t) = AR + BR . A multidimensional singular stochastic control problem... In view of (56), BR cmax E d( - R )(s - t) = cmax E (T - t) - R (T - t) . Using (56) again and the assumption that Jxt (v) < , we see that E (T -t) -R (T - t) E(T - t) < . Hence, from the Lebesgue's dominated convergence theorem we get lim BR lim cmax E (T - t) - R (T - t) = cmax E lim (T - t) - R (T - t) = 0. Using (7) and the H¨lder's inequality, we have o AR E E t vR v f (yxt (s), s) - f (yxt (s), s) vR v 1 + f (yxt (s), s) + f (yxt (s), s) vR v |yxt (s) - yxt (s)| 1/p vR v 1+f (yxt (s), s)+f (yxt (s), s) vR v |yxt (s)-yxt (s)|p In view of Step 3, the second factor in the last expression goes to 0 if R . We mushow thahe first factor is bounded. Indeed, from (6) and Lemma 1.2 we can write E vR v 1 + f (yxt (s), s) + f (yxt (s), s) vR v v v 2 + |yxt (s)|p + 2p-1 |yxt (s)|p + 2p-1 |yxt (s) - yxt (s)|p vR v 2 + |yxt (s)|p + |yxt (s)|p (1 + C0 )E (1 + C0 )E Using (54) and Step 3 again, we conclude thahe last expression is bounded uniformly in R. Hence limR AR = 0. Summarizing Steps 1-4, we know that Jxt (vR ) goes to Jxt (v) if R , so we can consider only bounded controls. Step 5. Consider v V such that ||v|| < R for some R > 0. We will construct a sequence of controls {vn , n N} convergeno v in Lp ( × [0, T - t]) and such that vn V1/(2nR) for all n. Besides we shall prove thahe variation of vn is pointwise convergeno the variation of v from below. s Let vn (s) = n (s-1/n)0 v(r)dr, s [0, T - t]. We observe that vn is a progressively measurable continuous random process such that ||vn || R, so vn Lp ( × [0, T - t]). From left-continuity of v we know that (59) s[0,T -t] lim vn (s) = v(s). M. Boryc and L. Kruk Using the Lebesgue's dominated convergence theorem, we get |v(s) - vn (s)|p = E lim |v(s) - vn (s)|p = 0, so Lp -convergence is proved. Now we wano check that vn V1/(2nR) . Indeed, d d n vn (s) = v(r)dr (s-1/n)0 = n v(s) - v (s - 1/n) 0 2nR. Let n (s), (s) denote the variations on the interval [0, s] of the functions vn , v respectively. For convenience, we define v(r) 0 for r < 0. Then s vn (s) = n s-1/n v(r)dr, s [0, T - t]. Fix , s (0, T - t]. Let = {s0 , s1 , . . . , sk } be a partition of the interval [0, s], where 0 = s0 < s1 < · · · < sk = s. Then k k |vn (si ) - vn (si-1 )| = n i=1 k 1/n 0 i=1 si si -1/n v(r)dr - si-1 v(r)dr si-1 -1/n =n i=1 v(si + r - 1/n) - v(si-1 + r - 1/n) dr v(si + r - 1/n) - v(si-1 + r - 1/n) dr n n 1/n k 0 1/n 0 i=1 (s)dr = (s). Letting |||| 0, we get (60) n (s) (s). k k On the other hand, from (59) we see that |v(si ) - v(si-1 )| = i=1 i=1 n k lim vn (si ) - lim vn (si-1 ) = lim |vn (si ) - vn (si-1 )| lim n (s). i=1 n Letting |||| 0 and using (60), we have (s) lim n (s) lim n (s) (s) n n lim n (s) = (s). Step 6. In view of Step 5 we can mimic Steps 3 and 4 to conclude that Jxt (vn ) Jxt (v) if n , where ||v|| < R for some R > 0. From this A multidimensional singular stochastic control problem... and Step 4, remembering that vn V1/(2nR) , we can write (61) and lim 0+ Jxt (v) = >0 Jxt (v) u (x, t) = u(x, t). Theorem 3.3 (Bellman's dynamic programming principle). Let x , v t [0, T ] and let yxt denote the state process corresponding to a control v V. Let [0, T - t] be a Markov time with respeco {Ft }. Then u(x, t) = E + v (r)dr t+ (r)dr v d(s - t) + u(yxt (t + ), t + ) . Proof. For convenience let us denote Jxt (v, ) = E + v s t t+ (r)dr + s t (r)dr d(s-t) . It is known that DPP hol for regular stochastic control problems (see, e.g., [10], Th. 3.1.6). Hence we have for each > 0, (62) u (x, t) = v Jxt (v, ) + Eu (yxt (t + ), t + ) . Considering any v V , we have ~ v ~ v u (x, t) Jxt (~, ) + Eu (yxt (t + ), t + ). If 0+ , from Theorem 3.2 and the Lebesgue's dominated convergence theorem we get v ~ v u(x, t) Jxt (~, ) + Eu(yxt (t + ), t + ). Because (63) > 0 and v V are arbitrary we can conclude that ~ u(x, t) >0 V On the other hand, from (62) u (x, t) Letting (64) >0 0+ , we get u(x, t) >0 V The inequalities (63), (64) and an argumenimilar to the proof of Theorem 3.2 (see (61)) imply that u(x, t) = M. Boryc and L. Kruk Corollary 3.4. The dynamic programming property in the weak sense hol (see Definition 1.1) and hence the value function satisfies (33). Denote Au(x, t) = -u(x, t) 1 - (t) D2 u(x, t) - a(t)x + b(t) Du(x, t) t 2 + (t)u(x, t), where denotes the scalar product of vectors and matrices respectively. Theorem 3.5 (The HJB equation). The value function u satisfies almost everywhere (a.e.) the following second-order differential equation: (65) max Au(x, t) - f (x, t) , |Du(x, t)| - c(t) = 0. Proof. An application of the DPP for regular stochastic control problems yiel for > 0 the following equation (see [5], Chapter IV.3): (66) Au (x, t) + 1 |Du (x, t)| - c(t) = f (x, t) a.e. In view of Theorems 2.1, 2.2, Remark 2.4, Theorem 3.2, Corollary 3.4 and the Arzela­Ascoli's theorem ([7], Th. 2.4.9) we see that u u uniformly on every compacet if 0+ . Fix t [0, T ]. From (31) and Remark 2.4 we see that D2 u (·, t) are locally uniformly bounded for all > 0 in their domains, so using the Arzela­Ascoli's theorem from every sequence { m }mN convergeno 0, we can choose a subsequence {~m }mN such that Du~m (·, t) v = (v1 , . . . , vn ) almost uniformly if m . But v must be equal to Du(·, t) in the distribution sense. Indeed, for any function Cc ( ) and any k = 1, . . . , n we have (x) u~m (x, t)dx = - xk (x) u~m (x, t) dx. xk Letting m , we get (x) u(x, t)dx = - xk (x)vk (x)dx, u so vk (·) = u(·,t) almost everywhere. Since xk and vk are Lipschitz continxk uous, the equality hol for all x . Thus, v does not depend on the choice of the subsequence {~m }, so (67) t[0,T ] Du (·, t) Du (·, t) almost uniformly if 0+ . A multidimensional singular stochastic control problem... Let = (1 + |x|)-2p-n-1 . From (29)­(33) we conclude that |Au (x, t)| is not greater than 1 + (C33 + C29 max )(1 + |x|p ) + max n2 C31 (1 + |x|)(p-2) 2 + amax |x| + bmax nC30 (1 + 2|x|p-1 ) for almost every (x, t) ×[0, T ]. Using Lemma 1.2, we have the estimate (68) |Au (x, t)| C68 (1 + |x|)p a.e. for some constant C68 > 0 depending only on C29 , C30 , C31 , C33 , n, p, amax , bmax , max , max . Hence |Au (x, t)|2 (x) 2 2 C68 (1 + |x|)2p C68 = (1 + |x|)2p+n+1 (1 + |x|)n+1 a.e. The same estimate hol for u instead u . So |Au |2 , |Au|2 L1 ( × [0, T ]). Moreover, Au , Au are uniformly bounded in the space L2 , where L2 = v : v 2 L1 ( × [0, T ]) = L2 Leb ( × [0, T ]). · From the Banach­Alaoglu theorem ([12], Th. 3.15) we know that balls in the space L2 are weakly compact. So for each sequence { m }mN convergent v in L2 if to 0, there exists a subsequence {~m }mN such that Au~m m . We will show that v = Au in the distribution sense. Indeed, for any function belonging to the class Cc ( × [0, T ]), we have (Au~m )dxdt = 1 u~m dxdt - t 2 T 0 (t) D2 u~m dxdr a(t) u~m dxdt + + 0 T 0 a(t)x + b(t) D u~m dxdt + (t)u~m dxdt. 1 udxdt - t 2 Letting m , we get vdxdt = + + T 0 T 0 (t) D2 udxdt a(t)x + b(t) D udxdt + (t)udxdt = tr a(t) udxdt (Au)dxdt. Hence Au~m conclude (69) Au in L2 if m . From uniqueness of the limit we Au Au in L2 if 0+ . M. Boryc and L. Kruk In view of (66) we have Au f a.e. From this and (69) we see that Au(x, t) f (x, t) a.e. This, together with Remark 2.8 ensure us that (70) max Au(x, t) - f (x, t) , |Du(x, t)| - c(t) 0 a.e. Take a sequence n 0 and let D be the set of (x, t) × [0, T ] such that (66) hol at (x, t) for all n . Then Leb ( × [0, T ]) \ D = 0. Choose t [0, T ] such that for almost every x we have (x, t) D. Since |Du n (x, t)| |Du(x, t)| as n (see (67)), I{|Du n (x,t)|<c(t)} = I{|Du(x,t)|<c(t)} for n large enough (depending on (x, t)). We have from this and (66) that (71) I{|Du(x,t)|<c(t)} Au n (x, t) I{|Du(x,t)|<c(t)} f (x, t) I{|Du(x,t)|<c(t)} Au n (x, t) I{|Du(x,t)|<c(t)} Au(x, t) a.e. in L2 , On the other hand, (69) yiel (72) so the sequence I{|Du(x,t)|<c(t)} Au n (x, t) is bounded in L2 and thus it is uniformly integrable in L1 . This, together with (71), implies that for every Cc ( × [0, T ]) L2 , I{|Du(x,t)|<c(t)} (Au n )(x, t)dxdt (73) I{|Du(x,t)|<c(t)} (f )(x, t)dxdt, which, together with (72) implies that I{|Du(x,t)|<c(t)} Au(x, t) = I{|Du(x,t)|<c(t)} f (x, t) a.e. 4. Existence and uniqueness of the optimal control. The results of this section are analogous to Theorems 7 and 8 from [11]. Fix (t, x) [0, T ) × (for t = he only admissible control is v(0) = 0 a.s.). Let mt be the measure on [t, T ] × , B([t, T ]) F equal to the product of the Lebesgue's measure and P . Remark 4.1. If a process X is a modification of a process Y and both processes have left-continuous sample paths a.s., then the processes X, Y are indistinguishable (compare Problem 1.1.5, [7]). Theorem 4.2. The optimal control v V, if it exists, is unique up to the indistinguishability. Proof. pose there are v1 , v2 V for which u(x, t) = Jxt (v1 ) = Jxt (v2 ). Put v0 = (v1 + v2 )/2. Of course v0 V. From Lemma 1.7 we have (74) u(x, t) - Jxt (v0 ) = 1 Jxt (v1 ) + Jxt (v2 ) - Jxt (v) 0. 2 A multidimensional singular stochastic control problem... 0 1 2 Let yxt , yxt , yxt be the solutions of (1) corresponding to v = v0 , v1 , v2 respectively. In view of the proof of Lemma 1.7 and strict convexity of the running cost function f , we have 1 1 0 1 2 (75) f yxt (s), s < f yxt (s), s + f yxt (s), s 2 2 1 2 provided that yxt (s) = yxt (s). Assume that v1 , v2 are not indistinguishable. Then there exists s (t, T ] such that P (A) > 0, where A = {v1 (s ) = v2 (s )} (see Remark 4.1). Because v1 , v2 have left-continuous sample paths a.s., there exists s () (t, s ) such 1 2 that v1 (s) = v2 (s) for all s [s (), s ], A. Thus, yxt (s) = yxt (s) on some mt -nonzero set. This facogether with (75) and the definition of Jxt imply thahe inequality (74) is strict, so we get a contradiction. We conclude that v1 , v2 must be indistinguishable. Lemma 4.3. Let {zn }nN be a sequence in Lp (mt ). If zn 0 in Lp (mt ), then T zn 0 in Lp (mt ), where T zn (s, ) = zn (s, ) - a(r)zn (r, )dr. s t a(r)zn (r, )dr Proof. By the H¨lder's inequality, the function g(s, ) = o satisfies ||g||p p ap E max L |zn (r, )|p · T p/q dr ap · T 1+p/q · ||zn ||p p , max L so T is a bounded operator from Lp (mt ) into Lp (mt ). Theorem 4.4. There exists an optimal control v V. Proof. Let {vk }kN be a sequence of admissible controls such that Jxt (vk ) k u(x, t) as k and let yxt be the solution of (1) corresponding to v = vk . Then Jxt (vk ) are uniformly bounded in k. By Lemma 1.9 the sequence k {yxt }kN is bounded in Lp (mt ) and hence, by the Banach­Alaoglu theorem k ([12], Th. 3.15), there exists a subsequence (still denoted by {yxt }) and a k yxt in Lp (mt ). process yxuch that yxt i Fix k N. Since the sequence {yxt }ik is also convergeno yxt , by the Mazur theorem there exists n(k) k zxt n(k) k,i · i yxt , k,i 0, k,i = 1, k n(k) < k k such that ||zxt - yxt ||Lp 1/k. In particular zxt yxt in Lp (mt ). Let n(k) k k = k,i · vi be the control corresponding to zxt in (1). Then k V. Moreover by Lemma 1.7, n(k) u(x, t) Jxt (k ) k,i · Jxt (vi ) ,...,n(k) max Jxt (vi ) - u(x, t). M. Boryc and L. Kruk For s [t, T ] we have k m zxt (s) - zxt (s) - t k k m Because {zxt } is convergent in Lp (mt ), zxt - zxt goes to 0 in Lp (mt ) as k, m . Using Lemma 4.3 we conclude that {k (· - t, ·)}kN is a Cauchy sequence in Lp (mt ) so it is convergeno a process v Lp (mt ). Without loss of generality we may assume that v(0) 0. Now we choose a subsequence (still denoted by k) such that k (s, ) v(s, ) as k for (s, ) A, where (Leb × P )(A) = T - t. For and s [0, T - t], we define s k m a(r) zxt (r) - zxt (r) dr = k (s - t) - m (s - t). A = s [0, T - t] : (s, ) A , As = : (s, ) A . Note that P (A0 ) = 1 because k (0) = v(0) = 0 P -a.s. Furthermore, let ~ = : Leb (A ) = T - t , S = s [0, T - t] : P (As ) = 1 . ~ Then P () = 1 and Leb (S) = T - t. Let N be a countable subset of S, dense in [0, T -t], including 0 and let AN = sN As . We have P (AN ) = 1. Let k (s) denote the total variation of k on the interval [0, s]. Because Jxt (k ) are uniformly bounded in k, there exists a constant C > 0 such that Ek (T - t) C for all k N. In view of the Fatou's lemma, E lim k k (T - t) lim k Ek (T - t) C, so lim k k (T - t) is finite a.s. Fix and let A , = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t. Let kn = kn () be a sequence of natural numbers such that limkn kn (T - t) = lim k k (T - t). Then v(ti+1 ) - v(ti ) = lim kn kn kn (ti+1 ) - kn (ti ) k lim kn (T - t) = lim k (T - t). Thus, v|A has bounded variation and hence it has left-hand and right-hand ~ ~ limits at each point. Let v = 0 on the P -zero set \ (AN ). On AN let 0 = v(0), s=0 v (s) = lim v(u) = lim v(u), s (0, T - t] . A us N us Then v is progressively measurable, left-continuous and v (0) = 0. More~ over, for AN and for each partition = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t, we can choose {tk }kN A such thak ti as i i A multidimensional singular stochastic control problem... k , i = 1, 2, . . . , m. Therefore, v (ti+1 ) - v (ti ) = lim v(tk ) - v(tk ) i+1 i k V ar(v, [0, T - t] A ) lim k (T - t), has bounded variation. This ensures us that V. ~ For AN , the set A {s [0, T - t] : v(s, ) = v (s, )} is countable, so its Lebesgue's measure is equal to 0. Therefore v = v mt -a.e. In particular, k v in Lp (mt ). Proceeding as in Steps 3­4 in the proof k v of Theorem 3.2 we can show that yxt yxt in Lp (mt ) and hence so (76) k k s t (r)dr = E v s t (r)dr To finish the proof we need to check that (77) E t T s t (r)dr d (s - t) lim E k t s t (r)dr dk (s - t), ~ is the total variation of v . Fix AN and let 0 s1 where s2 T - t. Let = {t0 , t1 , . . . , tm }, s1 = t0 < t1 < · · · < tm = s2 and let {tk }kN A be such thak ti as k , i = 0, 1, . . . , m. Then for i i every k0 N v (ti+1 ) - v (ti ) = lim v(tk ) - v(tk ) i+1 i V ar(v, [tk0 , s2 ] A ). 0 Letting k0 , we get and hence (78) m-1 v (ti+1 ) - v (ti ) V ar(v, [s1 , s2 ] A ) V ar(v , [s1 , s2 ]) V ar(v, [s1 , s2 ] A ). Let (s) = V ar(v, [0, s]A ), s [0, T -t]. Restricting = {t0 , t1 , . . . , tm }, s1 = t0 < t1 < · · · < tm = s2 so that A (in particular assuming s1 , s2 A ), we get v(ti+1 ) - v(ti ) = lim k (ti+1 ) - k (ti ) lim k (s2 ) - k (s1 ) . As |||| 0 we get (79) (s2 ) - (s1 ) lim k (s2 ) - k (s1 ) . Now take = {t0 , t1 , . . . , tm }, 0 = t0 < t1 < · · · < tm T - t, N . In particular, A for all AN . For every interval [ti , ti+1 ], M. Boryc and L. Kruk s t i = 0, 1, . . . , m - 1, let li = min ~ AN , by (78)­(79), we have (r)dr : s [ti + t, ti+1 + t] . For li · (ti+1 ) - (ti ) li · (ti+1 ) - (ti ) m-1 lim k li · k (ti+1 ) - k (ti ) . ~ This together with the Fatou's lemma and the fachat P (AN ) = 1 yiel li · (ti+1 ) - (ti ) E lim k m-1 li · k (ti+1 ) - k (ti ) li · k (ti+1 ) - k (ti ) lim E lim E s t (r)dr dk (s - t). Letting |||| 0, tm T - o that each partition in the sequence is contained in the next one, by the monotone convergence theorem, we get (77). From (76) and (77) Jxt (v ) lim Jxt (k ) = u(x, t). On the other hand, Jxt u(x, t) because v V and hence Jxt (v ) = is an optimal control. u(x, t) so v

Journal

Annales UMCS, Mathematica – de Gruyter

Published: Jun 1, 2015

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