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A Direct Solution Method for Stochastic Impulse Control Problems of One-dimensional Diffusions

We consider stochastic impulse control problems where the process is driven by a general one-dimensional diffusion. We shall show a new mathematical characterization of the value function as a linear function in a certain transformed space. Our approach can relieve us from the burden of guessing and proving the optimal strategy, (2) present a simple method to find the value function and the corresponding control policies, and (3) handle systematically a broader class of reward and cost functions since the existence of the value function can be shown in much a simpler way.

The mathematical framework to these types of problems is in Bensoussan and Lions (1984). Impulse control has been studied widely in inventory control (Harrison et al. (1983)), exchange rate problem (Jeanblanc-Picqué (1993), Mundaca and Øksendal (1998), Cadenillas and Zapatero (2000)), dividend payout problems (Jeanblac-Picqué and Shiryaev (1995)), and portfolio optimization with transaction costs (Korn (1998), Morton and Pliska (1995)). Korn (1999) surveys the applications in mathematical finance. Also see Chancelier et al.(2002) for a combination of optimal stopping and impulse control problems. In many economic and financial applications where the controlled process is described as an Itô diffusion, the solution to the problem demands a through study of a related Hamilton-Jacobi-Bellman equation and quasi-variational inequalities. The method of quasi-variational inequalities split by a guess the state space into intervention and no intervention (continuation) regions. One guesses the form of (a) continuation region, (b) associated optimal policy, and (c) the value function. Then optimality of the candidate policy must be verified. Both steps are often very difficult and the success depends heavily on the form of the controlled process, reward and cost functions.
Alternatively, an impulse control problem can be viewed as a sequence of optimal stopping problems. The connection between impulse control and optimal stopping has been investigated by Davis (1992) and  among others. In this setting, the value functions of a sequence of optimal stopping problems converge to the value function of the impulse control problem under suitable conditions.
In this paper, we utilize this connection together with a novel method of Dayanik and Karatzas (2003) for optimal stopping problems. We use it to identify a new and useful characterization of the solution of the original impulse control problem. At the end we get rid of the sequence of optimal stopping problems altogether: the new characterization allow us to propose a new direct solution method for impulse control problems.
In the next section, we briefly go over the solution method for optimal stopping problems of one-dimensional diffusions. We describe the impulse control problem and its solution in Section 3. Examples are presented in Section 4. Finally, extensions and concluding remarks are in Section 5.

Summary of the Key Results of Optimal Stopping
Let (Ω, F, P) be a complete probability space with a standard Brownian motion W = {W t ; t ≥ 0} and consider the diffusion process X 0 with state pace I ⊆ R and dynamics for some Borel functions µ : I → R and σ : I → (0, ∞). We emphasize here that X 0 is an uncontrolled process. We assume that I is an interval with endpoints −∞ ≤ a < b ≤ +∞, and that X 0 is regular in (a, b); in other words, X 0 reaches y with positive probability starting at x for every x and y in (a, b). We shall denote by F = {F t } the natural filtration generated by X 0 . Let α ≥ 0 be a real constant and h(·) a Borel function such that E x [e −ατ h(X 0 τ )] is well-defined for every F-stopping time τ and x ∈ I. Let τ y be the first hitting time of y ∈ I by X 0 , and let c ∈ I be a fixed point of the state space. We set: x ≤ c, Then F (·) is continuous and strictly increasing. It should be noted that ψ(·) and ϕ(·) consist of an increasing and a decreasing solution of the second-order differential equation (A − α)u = 0 in I where A is the infinitesimal generator of X 0 . They are linearly independent positive solutions and uniquely determined up to multiplication. For the complete characterization of ψ(·) and ϕ(·) corresponding to various types of boundary behavior, refer to Itô and McKean (1974). Let F : [c, d] → R be a strictly increasing function. A real valued function u is called F -concave on [c, d] if, for every a ≤ l < r ≤ b and x ∈ [l, r], We denote by the value function of the optimal stopping problem with the reward function h(·) where the supremum is taken over the class S of all F-stopping times. Then we have the following results, the proofs of which we refer to Dayanik and Karatzas (2003).

Impulse control problems and its solution
Suppose that at any time t ∈ R + and any state x ∈ R + , we can intervene and give the system an impulse ξ ∈ R. Once the system gets intervened, the point moves from x to y ∈ R + with associated rewards earned. An impulse control for the system is a double sequence, where 0 ≤ T 1 < T 2 < .... are an increasing sequence of F-stopping times and ξ 1 , ξ 2 ... are F T imeasurable random variables representing impulses exercised at the corresponding intervention times T i with ξ i ∈ Z for all i where Z ∈ R is a given set of admissible impulse values. The controlled process is, in general, described as follows: with some mapping Γ : R × R → R.
In this section, we consider the absorbing boundary problem. Let 0 be the absorbing state, without loss of generality, and τ 0 inf{t : X t = 0} the ruin time. With the absorbing state at 0, it is natural to consider a set of problems where Z ∈ R + (i.e., ξ i = x i − y i > 0 for all i) and We shall comment on cases where interventions are allowed in both positive and negative directions in section 5.) With each pair (T i , ξ i ), we associate the interventions where K(x, y) : R × R → R is a given continuous function in the first and second argument that represents benefit/cost at interventions. Our result below does not depend on the specification of K(·). We assume that, for any point x ∈ R, due to the fixed cost incurred. We consider the following performance measure with ν ∈ V, a collection of admissible strategies, where P ∈ R − is a constant penalty at the ruin time and f : R → R is a continuous function, satisfying : Our goal is to find the optimal strategy ν * (T i , ξ i ) i≥0 and the corresponding value function, Let us briefly go over our plan. In section 3.1 we shall characterize optimal intervention times T i as exit times of the process X from an interval by implementing recursive optimal stopping scheme that eventually solves the original impulse control problem. Using the results, in section 3.2, we consider a special case where the mapping x → K ϕ (x) : R + → R + is F -concave. We show, under this assumption, that the optimal intervention times T i are characterized as exit times from an interval, say (0, b * ) for every i. Then we characterize the value function for impulse control problems and present a solution method based on the characterization of the intervention times and value function. In section 3.4, we consider the general case where the F -concavity assumption above does not hold.

A sequence of optimal stopping problems
In this subsection, we consider a recursive optimal stopping with a view to characterizing intervention times for the impulse control problems. Here we assume that no absorbing boundary exists. As we will see in the next subsection, the existence of an absorbing state is easily incorporated. Hence by using the same v(x), we consider the problem, and define the set S n and the objective function v n as follows: In other words, we are allowed to make at most n interventions. For this recursive approach, see, for example, Davis (1992) and . We use the following simple notation: Let H denote the space of all Borel functions. Define the two operators M : H → H and L : H → H as follows: and for u ∈ H. From the definition of the two operators, a 1 (x) ≤ a 2 (x) for x ∈ R, a 1 (·), a 2 (·) ∈ H implies Ma 1 (x) ≤ Ma 2 (x) and La 1 (x) ≤ La 2 (x) for all x ∈ R. Consider the following recursive formula: which is equivalent to (3.14) by applying the strong Markov property with (3.6) to the integral term. In fact, this derivation is explained in detail in subsection 3.2. By defining φ w − g, and adding and subtracting g(X τ ) on the right hand side of (3.14), it becomes It should be noted that, for each n, this is an optimal stopping problem over τ and can be written, by using the operator defined in (3.12), φ n+1 (x) = Lφ n (x).
(3.15) Let us start this recursive scheme with w 0 (x) g(x) (i.e., no interventions are allowed, equivalently φ 0 (x) = 0) and define recursively φ n (x) On the other hand, The last equation is due to the fact that only one intervention is allowed. Hence we have w 1 (x) = v 1 (x). By the definition of the recursive scheme, w n is an increasing sequence (i.e, w 1 (x) ≤ w 2 (x) ≤ .... for all x ∈ R). In fact, we shall prove that w n = v n for all n in Lemma 3.2. Before that, we need the following lemma to relate this recursive scheme with the method described in Section 2.
This lemma guarantees that we can use Proposition 2.1 to 2.3 to identify the value function and an optimal stopping rule for each of the recursive optimal stopping problems (3.13). Let us define, for notational convenience, K(x, y) K(x, y) − (g(x) − g(y)). (3.16) Further, we prove the following properties of the recursive optimization scheme.
Lemma 3.2. If we define w n by (3.13) (with w 0 = g) and v n by (3.9), then Moreover, w is the smallest solution majorizing g of the functional equation w − g = L(w − g).
Proof. The proof is given in Appendix.
Hence if we solve the optimal stopping problem recursively for each n, then we obtain φ( Summarizing the above argument, we have the following proposition: Proposition 3.1. The value function v(x) for (3.8) is given by the smallest solution majorizing g of the functional equation v − g = L(v − g), and v−g ϕ (= φ ϕ ) is always F -concave.
Proof. The first statement comes from Lemma 3.2. By the recursive method that we described above, we are solving a series of optimal stopping problems for each φ n . Hence Lemma 3.1 and Proposition 2.1 give the second statement.

Characterization of the Intervention Times and the Value Function: F -Concave Reward Case
Based on the results in the previous subsection, we first consider a special case where the mapping The argument in the previous subsection is modified to incorporate the existence of the ruin state. Instead of (3.9) and (3.13), we define, respectively, Then by defining the operator L : H → H instead of (3.12), we have the same recursion formula as in (3.15). We can obtain the same results as in Lemma 3.1 and Lemma 3.2. Proposition 3.1 also holds with one change that the value function is given by the smallest solution majorizing g 0 of the functional equation where L : H → H is given by (3.18). Now we consider the characterization of the intervention times.
Proposition 3.2. If the mapping x →K ϕ (x) : R + → R + is F -concave and 0 is an absorbing state, then the optimal intervention times T * i are given, for some b * ∈ R + , by Proof. Our proof is constructive, describing the procedure of recursive optimization steps. For any in the transformed space) always exists. Indeed, since we consider the case of ξ i > 0, i.e, x > y for K(x, y) and we should check whether the concave majorant exists, namely, holds when y ↓ 0. Note that lim y↓0 g 0 (y) = P and g(x) → g(0) as x → 0 due to the continuity of f . Hence (3.19) holds in the neighborhood of y = 0 because of (3.4). In the subsequent iterations, we consider We should check if the expression inside the supremum operator becomes less than P − g(0) as x ↓ 0 and y ↓ 0. Since lim y↓0 φ 1 (y) = φ 1 (0) = P − g(0) by the concavity (hence continuity) of φ 1 and since lim x↓0 g(x) = lim y↓0 g(y), we have in the neighborhood of y = 0, holds. Hence the concave majorant always exist also in the subsequent iterations. Now the F -concavity of φ n is obviously maintained for all n. The limit function, φ(x) lim n→∞ φ n (x) exists and is also F -concave. Accordingly,K(x, y)/ϕ(x) + φ(y) is F -concave. Hence φ andK + φ meet once and only once. Recall that the value function satisfies φ = Lφ. This implies that the continuous region is in the form of (0, b * ) for some b * ∈ R + , which completes the proof. By using the above characterization of intervention times, we next want to characterize the value function and reduce the impulse control problem (3.7) to some optimal stopping problem. Moreover, we shall present a method that does not have to go through the iteration scheme. Let us first simplify J ν : (3.20) This is just a reproduction of (3.5). Let us split the right hand side of (3.20) into pieces and use the strong Markov property (together with the shift operator θ(·)) to each of them. The first term becomes The second and third terms become where S i T 1 +T i •θ(T 1 ) and the index i runs from 1 for the sum in the second equality. Combining the three terms and rearranging, we have For any F stopping time τ , the strong Markov property with our assumption (3.6) gives us where g(·) is defined as in (3.10). We apply this result to (3.21) by reading τ = T 1 and τ = τ 0 to derive Noting that g(X 0 T 1 ) = g(X T 1 − , adding and subtracting g(X T 1 ) and further defining and we consider the maximization of this u(·) function and add back g( . Note that this simplification leading to (3.23) does not depend on the F -concavity assumption.
Since we have confirmed that optimal intervention times are exit times of the process from an interval, let us use a simpler notation that X T i − = b i and X T i = a i for all i. We can denote we have The second equation of (3.25) is obtained from (3.23) by noticing that, on x ∈ [b, ∞), P x (T 1 < τ 0 ) = 1. Indeed, in this case, we immediately jump to a, so that X T 1 − = x and X T 1 = a. Since a ∈ (0, b), u(a) = u 0 (a). Now let us note that we have the following representations in (3.23) where τ l inf{t > 0; X t = l} and τ r inf{t > 0; X t = r} and ϕ(·) and ψ(·) defined in the previous section. Finally, with F (·) being defined as in (2.2), we have a characterization of u(x), This represents a linear function that passes a fixed point, A (F (0), W (F (0)).
To discuss how to find the optimal pair (a * , b * ), we write u(x) as u a,b (x) to emphasize the dependence on a, b, then on x ∈ [0, b], ( 3.28) This can be considered as a two-stage optimization problem. First, let a be fixed. For each a, the inner maximization of (3.28) becomes and, among a ′ s, choose an optimal a in the sense,ṽ(x) sup a V a (x) for any x. It should be pointed out that V a (x) may take negative values if P − g(0) does. Now, we discuss a solution method of the first stage optimization (3.29). For this purpose, we need a lemma: for some Borel function h : R → R and with condition (3.6), then, for γ 1 > γ 2 ≥ 0, for any x.
Proof. The left hand side of (3.30) is well-defined due to (3.6). It is clear that Consider the bound of G(x, γ) for x fixed: The first term on the right hand side is constant in γ and the second term is linear in γ and the E x [e −ατ ] ≤ 1 for any τ ∈ S. Due to the convexity of G(x, γ) in γ, for the above inequality to hold, D + γ G(x, γ) ≤ 1 for all γ ∈ R. On account of (3.31), we have (3.30).
Coming back to (3.29), we need some care because the value function V a (x) contains its value at a, V a (a) in the definitive equation. Let us consider a family of optimal stopping problem parameterized by γ ∈ R.
Obviously, this parameterized problem can be solved by using Proposition 2.1 to 2.3. Now we link this parameterized optimal stopping problem to (3.29).
Lemma 3.4. For a > 0 given, if there exists a solution to (3.32), then there always exists unique γ such that γ = V γ a (a) holds, provided that (3.4) holds.
Proof. Without loss of generality, we need only to consider the case where for some a > 0. Indeed, suppose that there is no such a and let us consider a sequence of optimal stopping scheme. In each iteration, the value function for the optimal stopping problem takes negative values, so that φ n (·) < 0 for all n. Then in the next iteration,K(x, y) function will be shifted downwards, leading to φ n+1 (·) < 0. Hence the "no interventions" strategy is trivially optimal. In (3.32), since γ is some constant parameter, we benefit from Proposition 2.1 and claim that V γ a (x) is characterized as the smallest F -concave majorant of r γ (·, a) that passes F (0), P −g (0) ϕ (0) . In terms of the notation of Proposition 2.3, if we define W γ a (·) such that then W γ a (·) passes through the fixed point A = (F (0), W γ a (F (0)) and is the smallest concave . Now fix a. Our approach here is by starting with γ = 0, we move γ and evaluate V γ a (a) and try to find γ such that γ = V γ a (a). Due to (3.33), we have W 0 a (F (a)) > 0. By the monotonicity of F , it is equivalent to saying that V 0 a (a) > 0 = γ. As γ increases, W γ a (F (a)) increases monotonically by the right hand side of (3.32). Lemma 3.3 implies that for γ 1 > γ 2 ≥ 0, for any x ∈ R + . Note that W γ a (F (a)) ≥ H γ (F (a), a). However, since V has less than the linear growth in γ as demonstrated by (3.34), there is a certain γ ′ large enough such that W γ a (F (a)) = where the inequality is due to the assumption (3.4). For this γ ′ , we have V γ ′ a (a) < γ ′ .
The monotonicity and continuity of W γ (F (a)) (due to the convexity of V γ a (·)) with respect to γ, together with (3.34), implies that, for any a, there exists one and only one γ such that V γ a (a) = γ.
3.3 Methodology to find v(x) and (a * , b * ) Using (3.26), namely the characterization of u a,b , we describe an optimization procedure based on Proposition 2.2 and 2.3.
(3.36) 2. First stage optimization: For each slope β(a), we can calculate the value of W a (F (a)), but we have to find the W a (·) function such that, at some point F (b(a)), we have where we write b(a) ≡ b for notational simplicity. This requirement is equivalent to finding γ in (3.32) in Lemma 3.4 such that γ ϕ(a) = W γ a (F (a)).
By Proposition 3.2, C a (0, b(a)) is the continuation region. If R is a differentiable function with respect to the first argument, we can find the optimal point b(a) analytically. In effect, it is to find a point b(a) such that the linear majorant and the shifted function R(F (x), a) + W a (F (a)) ϕ(a) ϕ(x) have a tangency point. This is equivalent to calculating the maximum slope that majorizes R(F (x), a) after it is shifted. Explicitly, we solve (3.38) For the absorbing boundary case, these equations can be easily modified. Let us denote D W a (F (0)) = (P − g (0))/ϕ(0). Then (3.37) and (3.38) become respectively.
3. Second stage optimization: Now, let a vary and choose, among β(a), find β * , if exists, and also the corresponding b(a * ) and a * . Due to the characterization of the value function, these a * and b * b(a * ) must be the solution to (3.7). Suppose thatK(x, a) is a decreasing function of a. As a becomes closer to 0, the quantityK(x, a) becomes larger, while W a (F (a)) smaller. Hence we can expect the existence of a * that maximizes the slope β.
Remark 3.1. With respect to the third point of the proposed method above, we should check if there exists a concave majorant as a ↓ 0. Namely, we consider whether holds in the neighborhood of a = 0. Since lim x↓0 g(x) = lim a↓0 g(a) and lim a→0 u(a) = u(0) = P − g(0) by the continuity of u, the last inequality holds due to (3.4). ♦

Characterization of the Intervention Times and the Value Function: General Case
Let us move on to a general case where the mapping x →K ϕ (x) : R + → R + is not necessarily F -concave. First, we extend Proposition 3.2 to characterize optimal intervention times.
Proposition 3.3. The value function v(x) for (3.7) is given by the smallest solution majorizing g of v − g = L(v − g) and optimal intervention times T * i are given by exit times from an interval if and only if, for all y ∈ R + , x →K(x, y) is continuous and q lim sup Proof. For any given a ∈ R + , if we can find the smallest linear majorant ofK (F −1 (·),a)+γ for an arbitrary γ ∈ R + , we can find γ = ϕ(a)W a (F (a)) by Lemma 3.4. Due to the constancy of γ, it suffices to show that condition (3.41) is necessary and sufficient for the existence of concave majorant ofK ϕ • F −1 on F (I). The sufficiency is immediate. For the necessity, we assume that q = +∞. We can take a sequence of points {x k } ⊂ R such that x k → ∞ and D − K ϕ • F −1 (x k ) → ∞ as k → ∞.
If necessary, by taking a subsequence, we can make this sequence {x k } monotone. Consider the smallest concave majorant ofK ϕ There is no optimal intervention policy.
Suppose that the F -concavity of the reward function is violated, so that the intervention point may be multiple. Let us consider a strategy that we have two intervention points, b 1 and b 2 being arbitrarily chosen such that 0 < b 1 < b 2 . We want to characterize function J ν (x) as in (3.5) again. Recall that there are no controls in a way that the process is pulled up to avoid ruin. In other words, P x [τ 0 < ∞] = 1. Assume, for the moment, that we always apply control at these boundaries b 1 and b 2 and then, once applied, the process moves to a 1 < b 1 and a 2 < b 2 , respectively.
If we start with a point x ∈ [0, b 1 ], the problem is equivalent to the case we considered already, since the process cannot go beyond the level b 1 . Hence following (3.25), we have for , there are two strategies available: (A) Let X t move along. (It either hits b 1 or b 2 first.) (B) Apply the control immediately (t = 0) by moving the process from x to a 1 (the post-control point that corresponds to b 1 ) and let the process start at a 1 . (Recall that we do not let X enter into (b 1 , ∞) after moving to a 1 .)

Consider strategy (A) first. Let us define
Using the strong Markov property, we can reduce J ν 2 to a simpler form. For any (a 1 , b 1 ) and (a 2 , b 2 ), we have We shall use u 1 (x) = J ν 1 (x) − g(x) in the first term. Now let us define u 2 (x) J ν 2 (x) − g(x). Then the last equation becomes 2 ) + u 2 (a 2 ))] (3.42) on x ∈ [b 1 , b 2 ]. By identifyingK(b 2 , a 2 ) + u 2 (a 2 ) = u 2 (b 2 ) and u 2 (b 1 ) =K(b 1 , a 1 ) + u 1 (a 1 ) = u 1 (b 1 ) that shows u 1 (x) and u 2 (x) are connected at x = b 1 . Thus, (3.43) To summarize this result, if we define W i (·) u i ϕ • F −1 (·) for i = 1, 2 on F (I), this is again a linear function for each i. Hence by defining we have a piecewise linear function on F (I). Moreover, since we can treat b 1 as an absorbing boundary, we have a 1 < b 1 < a 2 < b 2 . Next consider strategy (B), whose value function is (3.44)

Lemma 3.5. (A) is better than (B) only if
Proof. Since the value function of strategy (B) is (3.44), choosing (A) over (B) is equivalent to If W 1 (F (x)) majorizes W 1 (F (x)) on x ∈ [0, ∞), then this problem reduces to F -concavity case discussed in the previous subsection. Hence we consider the case where there exists some There are two cases to consider: (1) If W 2 (F (x)) majorizes W 1 (F (x)) on x ∈ [b 1 , ∞), then we adopt the point b 2 as an intervention point. In this case, β 2 > β 1 holds. However, this implies that if we connect A (F (0), W 1 (F (0)) and C (F (b 2 ), W 2 (F (b 2 )), then this line segment AC is above the line segment connecting, piece by piece, points A, B (F (b 1 ), W 1 (F (b 1 )) and C. We can show that there exists a point b ′ ≥ b 2 such that its corresponding linear majorant W ′ (F (x)) satisfies W ′ (F (x)) > W 1 (F (x)) on x ∈ [0, b 1 ] and W ′ (F (x)) > W 2 (F (x)) on [b 1 , b 2 ]. The proof of the existence of a post-intervention point a ′ corresponding to this point b ′ follows in a similar manner to Lemma 3.4.
(2) If W 2 (F (x)) does not majorize W 1 (F (x)), we can find another pointb, instead of b 1 , such that the linear (not piecewise linear) function W (F (x)) corresponds tob majorizes R(F (x),ā) + W (F (ā)) ϕ(ā) ϕ(x) on x ∈ R + by Proposition 3.3. In either case, the value function in the transformed space should be a linear function that attains the largest slope among all the possible linear majorant. This argument holds true for any b 1 and b 2 with b 1 < b 2 . We can continue this argument inductively to the case of n intervention points, (b 1 , ...b n ). We here summarize our argument up to this point as a main proposition: Proposition 3.4. Suppose that (3.41) holds and the optimal continuation region is connected. The value function corresponding to (3.5) of the impulse problem described in (3.3)∼ (3.7) is written as (3.45) where W * (·) is the line segment connects (F (0), W * (F (0))) and (F (b * ), W * (F (b * ))) and satisfy the following: 1. W * (F (·)) is the smallest linear majorant of W * (F (a * )) ϕ(a * ) ϕ(·) + R(F (·), a * ) and meets with 2. The slope of W * (·), denoted as β * , is the largest slope among β(a)'s of all the possible linear majorants W a (·).
Moreover, if the mapping x →K ϕ (x) : R + → R + is F -concave, then the optimal continuation region (0, b * ) is uniquely determined.
Remark 3.2. If the F -concavity ofK is violated, there are two possible cases (and combination of them) of multiple continuation regions.
1. For some a * i with i = 1, 2, ..., we have the common β * . This is the case which we shall show in the next example. In this case, the continuation region is C Each time the process hits one of the points {b * i }, the control pulls the process back to the corresponding a * i .
2. Another case is that, for the unique optimal a * , there exists non-unique b * 1 and b * 2 . In this case, the continuation region is C = {(0, b * 1 ), (b * 1 , b * 2 )}, and the stopping region is Γ = {{b * 1 }, [b * 2 , ∞)}. If the process hits b * 1 or b * 2 , then the control pulls the process back to a * in either situation. It makes sense to continue in the region (b * 1 , b * 2 ) because there is a positive probability that one can extractK(b * 2 , a * )(>K(b * 1 , a * )) within a finite time.

No absorbing boundary case
Next, we extend our argument to a problem without the absorbing boundary. Hence the process can move along in the state space in an infinite amount of time. The problem becomes We can characterize intervention times as exit times from certain boundary and simplify the performance measure (3.46) The second equation is easily obtained in the same way as in the previous section by noting P x (T 1 < ∞) = 1. The last term does not depend on controls, so we define u(x) J ν (x) − g(x): Again, we consider the F -concave case with the notation T i − = τ b for all i and we have By defining W = (u/ϕ) • F −1 , we have We should note that F (c) F (c+) = ψ(c+)/ϕ(c+) = 0 and for any a ∈ (c, d]. For more detailed mathematical meaning of this value l c , we refer the reader to Dayanik and Karatzas(2003). We can effectively consider (F (c), l c ) as the absorbing boundary.

Examples
In this section, we work out some examples from financial engineering problems. For this purpose, we recall some useful observations. If h(·) is twice-differentiable at x ∈ I and y F (x), then with strict inequality if H ′′ (y) = 0. These identities are of practical use in identifying the concavities of H(·) when it is hard to calculate its derivatives explicitly.
Example 4.1. Øksendal (1999) considers the following problem: where X 0 t = B t is a standard Brownian motion and c > 0 and λ ≥ 0 are constants. The Brownian motion represents the exchange rate of some currency and each impulse represents an interventions taken by the central bank in order to keep the exchange rate in a given target zone. Here we are only allowed to give the system impulses ζ with values in (0, +∞). By reducing a level from b to a (i.e., b > a) through interventions, one can save continuously incurred cost (which is high if the process is at a high level). The problem is to minimize the expected total discounted cost v o (x) = inf ν J ν o (x). We want to solve its sup version and change the sign The continuous cost rate f (x) = −x 2 and the intervention cost is K(x, y) = −c − λ(x − y) in our terminology. By solving the equation 2α . Following our characterization of the value function, we obtain where g(x) can be calculated by Fubini's theorem: Note that when b > a, g(a) − g(b) > 0 is the source of cost savings. Let us fix a > 0 and consider h(x) −c − λ(x − a) + x 2 −a 2 α and H(y) (h/ϕ)(F −1 (y)), y > 0. By the first equation in (4.1), the sign of h ϕ ′ (x) will lead us to conclude that H(F (x)) is increasing from a certain point, say x = p on (p, ∞), so is H(F (x)). Also, by direct calculation, H ′ (+∞) = 0, from which we can assert that the value function is finite by Proposition 3.3. If we set p(x) −x 2 + a 2 + λα(x − a) + αc + 1/α, then (A − α)h(x) = p(x) for every x > 0. This quadratic function p(x) possibly has one or two positive roots. Let k be the largest one. Since lim x→∞ p(x) = −∞, by the second inequality in (4.1), H(·) is concave on (F (k), +∞). Hence H(y, a) is increasing and concave on y ∈ (F (k), ∞). Since the cost function in the transformed space is increasing and concave from a certain point on, there is a linear majorant that touches the cost function once and only once. We can conclude that for any a > 0 and the parameter set, we have a connected continuation region in the form of (0, b * ).
For this fixed a, let us define W a (·) such that V a (x) = ϕ(x)W a (F (x)) and r(x, a) = −c if x < a and −c − λ(x − a) + x 2 α − a 2 α if x ≥ a. Then we have for any a > 0, Recall that the left boundary −∞ is natural for a Brownian motion. Hence W a (y) that passes the origin of the transformed space is the straight-line majorant of R(·, a) + W a (F (a))/ϕ(F −1 (·)) where R(·, a) is defined in (3.35): We can represent W a as W (y) = βy. Since R(x, a) is differentiable with respect to x on x ≥ a, we can use (3.37) to find b(a) and corresponding β(a). Then varying a, one can find the optimal (a * , b * , β * ). Going back to the original space, on .
Finally, flip the sign and obtain the optimal cost function which coincides with the solution given by Øksendal (1999). Figure 1 displays the solution with parameters (c, λ, α) = (150, 50, 0.2).
Example 4.2. This example is a dividend payout problem where the underlying process follows an Ornstein-Uhlenbeck process. This problem was originally studied by Cadenillas et al.(2003) in an ingenious way, but the existence of the finite value function and the connectedness of the continuation region were left open. Suppose that X 0 has the dynamics where δ > 0, σ > 0 and m ∈ R. Only positive impulse is allowed in this problem. We consider the impulse control problem, with some positive constant K, k and the risk-aversion parameter γ ∈ (0, 1]. Since ξ ∈ R + , we havē The functions ψ(·) and ϕ(·) are positive, increasing and decreasing solutions of the differential We denote, byψ(·) andφ(·), the functions of the fundamental solutions for the auxiliary process Z t (X t − m)/σ, t ≥ 0, which satisfies dZ t = −δZ t dt + dW t . For every x ∈ R, and ψ(x) =ψ((x − m)/σ) and ϕ(x) =φ((x − m)/σ), where D ν (·) is the parabolic cylinder function; (see Borodin and Salminen (2002, Appendices 1.24 and 2.9) and Carmona and Dayanik (2003, Section 6.3)). By using the relation in terms of the Hermite function H ν of degree ν and its integral representation (see for example, Lebedev(1972, pp284, 290)). Let us consider the function Since the function h(·) is increasing, the function H(y) = (h/ϕ) • F −1 (y), y ∈ (0, ∞) is also increasing. Let us define the function which satisfies (A − α)h(x) = p(x). By using (4.1), H ′′ (y) and p(F −1 (y)) have the same sign at every y where h is twice-differentiable. Hence we study the (positive) roots of p(x) = 0. We have to divide two cases: (1) γ = 1 and (2) γ < 1. In either case, it can be shown that H ′ (+∞) = 0 by using (4.4) and (4.5) and the identity H ′ ν (z) = 2νH ν−1 (z), z ∈ R. Therefore, by Proposition 3.3, the finiteness of the value function is proved.
Let us move on to finding an optimal continuation region. By fixing a > 0, let us define r(x, a) = 0 if When we solve (3.37) for this a, it is not easy (at least analytically) to solve F −1 (y) explicitly. We can bypass this difficulty by using the first identity of (4.1) so that (3.39) with D = 0 becomes As in the previous examples, W a (·) is a straight line passing (F (0), 0) in the form of W a (y) = β(y − F (0)). The value function Therefore, the solution to the problem is This solves the problem. See Figure 2-(b) for the value function in case of parameters δ = 0.1, m = 0.9, σ = 0.35, α = 0.105 for the diffusions. As for the reward/cost function parameters, k = 0.7, K = 0.1 and γ = 0.75. The solution is (a * , b * , β) = (0.2192, 0.6220, 0.5749).
Example 4.3. We show a simple example where we have multiple continuation regions, the first case of Remark 3.2. Let the uncontrolled process is a standard Brownian motion B t and let α = 0, f = 0 and K(x, y) = −c(sin x − sin y) − δ with c ∈ R + and δ ∈ R + being some constant parameters. We want to solve In this case F (x) = x and let us define By solving (3.37) with some parameter (c, δ) = (10, 0.35), we find that a * k = 2.75 + 4kπ and b * k = 3.52 + 4kπ with k = 0, 1, 2.... For all these pairs, β * has a common value of 9.30. Hence all these pairs are optimal. This implies that if the initial state x ∈ (b * k , b * k+1 ), then we let the process move until it reaches b * k or b * k+1 . If it reaches b * k first, then an intervention is made to a * k . Now we are in the interval (b * k−1 , b * k ). We continue until the process is absorbed at x = 0.

Conclusions
Before we conclude this article, we shall mention an immediate extension to two boundary impulse control problems: for all x ∈ R, where ν = (T 1 , T 2 , ....; ζ 1 , ζ 2 , ....; S 1 , S 2 , ....; η 1 , η 2 , .....) with ζ i > 0 corresponds to interventions at the upper boundary at intervention time T i and η j < 0 at the lower boundary at intervention time S j . Examples of this type include the storage model analyzed by Harrison et al. (1983) and foreign exchange rate model studied by Jeanblanc-Picqué (1993). The former problem, for example, is that a controller continuously monitors the inventory so that the inventory level will not fall below the zero level. He is allowed to make interventions by increasing and decreasing the inventory by paying costs associated with interventions. In this case, the process remains within some band(s). In other words, the optimal intervention times are characterized as exit times from an interval in the form of (p * , b * ) for 0 ≤ p * < b * . See Korn (1999) for survey.
Under suitable assumptions, we can develop a similar argument to the previous chapters. Among others, the intervention times can be characterized as exit times from an interval (p * , b * ). We can also simplify the performance measure, where g(x) = E x ∞ 0 e −αs f (X 0 s )ds as usual. Again, the last term does not depend on controls, we define u(x) as u( where T 1 = τ b and S 1 = τ p and it follows that Hence if we define W u ϕ • F −1 , we have linear characterization again in the transformed space; (5.5) and the solution to the problem is described as for i = 1 and 2.
We have studied impulse control problems. The intervention times are characterized as exit times of the process from a finite union of disjoint intervals on the real line. A sufficient condition is given for the connectedness of the continuation region. The value function is shown to be linear in certain transformed space and a direct calculation method is described for it. This method can handle impulse control problems with non-smooth reward and cost functions. The finiteness of the value function is shown to be equivalent to the existence of a concave majorant of a suitable transformation. The latter is easier to check by using geometric arguments.
The new characterization of the value function and optimal strategies can be extended to other optimization problems, such as optimal switching, singular stochastic control and combined problems of optimal stopping and impulse control. If an optimal strategy exists in the class of exit times, then the problem can be reduced to a sequence of optimal stopping problems and an effective characterization of the value function is possible.
6 Appendix 6.1 Proof of Lemma 3.2 To make the proof more intuitive, we will work with (3.13) rather than with (3.14) where the integration part is converted to g functions. For this purpose, it is convenient to define the following two operators Hence we can proceed with the arguments developed in Davis (1992). In terms of the two operators just defined, (3.13) becomes (1) w n = v n for all n: Let us now prove v n (x) = w n (x) for all n ∈ N. We show already v 1 (x) = w 1 (x). We assume, to make an induction argument, that v n (x) = w n (x) and prove v n+1 (x) = w n+1 (x). We should note that, for each n, the optimization problem in (6.2) is an optimal stopping problem. Hence by Proposition 2.3, we can confine the set of strategy S n in (3.9) into a smaller set, i.e. barrier strategies;S n {ν ∈ S n : T i , i ∈ N is an exit time from some interval.}. for all x ∈ R. The second equality is due to the strong Markov property justified by (6.5). The third equality is by the induction hypothesis. This proves the first statement of the lemma.
(2) v(x) = lim n→∞ w n (x): Since w n is monotone increasing, the limit w(x) = lim n→ w n (x) exists. Since S n ⊂ S, w n (x) ≤ v(x). Hence w(x) ≤ v(x). To show the reverse inequality, we define S * be a set of interventions such that S * = {ν ∈ S : J ν (x) < ∞ for all x ∈ R}.
Let us assume that v(x) < +∞ and consider strategy ν * ∈ S * and another strategy ν n that coincides with ν * up to and including time T n and then takes no further interventions.
As n → +∞, the right hand side goes to 0 by the dominated convergence theorem. Hence it is shown v(x) = sup so that v(x) ≤ w(x). Next, consider the case of v(x) = +∞. By Proposition 3.3, we have q = +∞ in this case. Then by the recursive method described in Section 3.1, we see that v 1 (x) = w 1 (x) = ∞. By the first statement of this lemma, we can conclude v n (x) = w n (x) = ∞ for all n ∈ N, obtaining v(x) = lim n→∞ w n (x). This completes the proof of the second statement.
(3) w = L o w : Since w n ↑ w, we have the following chain of equalities: In view of this, if we take the limit on the both sides of (6.3) as n → ∞, by the monotone convergence theorem, e −αs f (X s )ds + e −ατ M o w(X τ ) .
This shows that w = L o w. Suppose w ′ (x) satisfies w ′ = L o w ′ and majorizes g(x) = v 0 (x). Then w ′ = L o w ′ ≥ L o v 0 = w 1 . If we assume that w ′ ≥ v n , then By the induction argument, we have w ′ ≥ w n for all n, leading to w ′ ≥ lim n→∞ w n = w. Thus it shows that w is the smallest solution majorizing g of the functional equation, w − g = L(w − g). This completes the third statement of the lemma.