The recent COVID-19 pandemic reaffirms the importance of dynamic decision-making under uncertainty. Due to the sudden increase in patients, a collapse in healthcare has occurred in several cities worldwide. With the limited number of available beds, ventilators, and other medical resources, hospitals must control the number of inpatients to secure sufficient care for each and also prevent in-hospital infections through costly measures, such as setting a cap on the number of patients to accept, sending patients having mild symptoms to other hospitals/accommodations, and applying risky medications that may have potential side effects.
These problems can be modeled in the framework of inventory control (see [3, 12, 13, 22, 24] ), in which the objective is to solve the tradeoff between minimizing the cost of inventory and the cost of controlling. As a continuous-time model, (uncontrolled) inventory is typically modeled by a Brownian motion, a compound Poisson process with negative jumps, or their combination. The assumption that the jumps of the process are only one-sided greatly simplifies the analysis and makes the problem more manageable. The optimal strategy, which minimizes the total expected running and controlling costs, is often shown to be a barrier or (s, S) type depending on whether fixed costs are considered.
Regarding the decision-making faced by hospitals, however, the number of patients is expected to experience both positive and negative jumps. It jumps upward, for example, when cluster infections of a disease occur. On the other hand, it jumps downward when the epidemic is stabilized and when new medications or vaccines become available. As evidenced by the case of COVID-19, these jumps in the number of patients have significant effects on decision-making for which one must secure sufficient resources for each patient and minimize the possibility of in-hospital infections. However, a sudden decrease in patients can potentially damage the financial condition of the hospital.
Decision-making in finance is also modeled similarly. For example, in the control of the exchange rate such as [15, 19] , the rate is modified by (costly) means to stabilize a currency in response to escalating or depreciating values against other currencies. Similar problems have also been studied in cash/international reserve management (see [2, 8, 9, 11] ). For these problems, most existing studies focus on the Brownian motion case. Although many analytical tools have recently become available to extend the Brownian motion model to a Lévy model with one-sided jumps (spectrally one-sided Lévy processes), the assumption of the one-sided jumps unfortunately does not suit financial applications well. Different from the inventory process in the classical setting, financial asset values are empirically known to contain both positive and negative jumps (see, e.g., [10] ). For realistic applications in finance, the use of stochastic processes having jumps in both directions is indeed desirable.
Motivated by these applications, this paper revisits the singular control problem without fixed costs and aims to show, under a mild condition, the optimality of a barrier strategy when driven by a general Lévy process with both positive and negative jumps.
Given a Lévy process X = (X t ) t≥0 modeling the (uncontrolled) inventory level of a single item, the decision-maker chooses a strategy π = (R π t ) t≥0 representing the aggregate replenishment of an item until time t. The objective is to minimize the total expected values of the inventory cost ∞ 0 e −qt f (U π t )dt and the controlling cost [0,∞) e −qt dR π t where U π := X + R π is the controlled inventory when a strategy π is applied. In the healthcare management and exchange rate control examples described previously, "inventory" (or its negative value) corresponds to the number of patients and the currency rate. These problems can be modeled essentially in the same way (by flipping the processes if necessary). This is a classical formulation, and complete solutions have been obtained for several stochastic processes, even for cases with fixed costs. In particular, Bensoussan et al. [4, 5] solved the problem when X is a combination of Brownian motion and a compound Poisson process with negative jumps; these results were generalized to a general spectrally negative Lévy case by [25] . There are several variations, such as that of [14, 21] , which deal with cases under restricted sets of admissible strategies. As a related study, the optimal dividend problem, in which the time horizon is given by a ruin time and the running cost is not considered, has been studied actively in the insurance mathematics community (see, in particular, [1, 18] ). These results focus on spectrally one-sided Lévy processes, and, to the best of our knowledge, existing results for a general Lévy process are significantly limited. Under the convexity (or a slightly more relaxed condition) of the running cost function f and some additional technical assumptions, a barrier strategy that reflects the inventory process at a certain boundary has been shown to be optimal for a subset of Lévy processes. It is a natural conjecture that the same conclusion can be drawn for a wider class. In this paper, we verify the conjecture that a barrier strategy is optimal for a general Lévy process.
In particular, we show that the optimal barrier, which we call b * in this paper, can be characterized concisely as the solution to
is the reflected Lévy process with lower barrier b of the Lévy process X started at b, and C ∈ R is the unit cost/reward of controlling. Thanks to the convexity assumption of f , the left-hand side of (1.1) is monotone in b, and hence the root b * can be obtained easily by bisection.
To show the optimality of the proposed strategy, the key observations are that the derivatives of the expected costs of inventory E
with respect to the starting value x and barrier b, can be written concisely in terms of the first downcrossing time of X:
Similar observations have been used in the optimal dividend problem with capital injections in [20] . However, to the best of our knowledge, this is the first result applied to the singular control problem with running costs. The conjectured optimal strategy is rigorously verified by showing that the candidate value function satisfies certain smoothness conditions and solves the variational inequalities. To this end, we first show that the strategy of reflecting the process at b * is optimal among barrier strategies. Using this, its optimality over all admissible strategies is shown via contradiction arguments by adapting the techniques of [1] .
The main contribution of the current paper is that we show the optimality for a general Lévy process without focusing on a particular type. This includes cases when X has paths of bounded and unbounded variations and, thus, can be applied in various settings in inventory, healthcare management, and finance models. Thanks to the explicit and concise expression of the optimal barrier b * as a solution to (1.1), it can be computed generally via a standard Monte Carlo simulation. In this paper, we also demonstrate that this can be done efficiently through a sequence of numerical examples.
The rest of the paper is organized as follows. Section 2 formulates the problem. Section 3 defines the barrier strategy and shows the optimality of that with barrier b * over the set of barrier strategies. Section 4 shows its optimality over all admissible strategies. Section 5 gives numerical results. Several long proofs are deferred to the appendix.
2.1. Inventory models. In this paper, we consider an inventory model where the cumulative demand D = {D π t : t ≥ 0} follows a general Lévy process. With its initial value x ∈ R, the inventory (in the absence of control) follows a (one-dimensional) Lévy process
We let (Ω, F, P) be a probability space hosting the Lévy process X = {X t : t ≥ 0}. For x ∈ R, we denote by P x the law of X when its initial value is x. In particular, we denote P = P 0 . Throughout the paper, let Ψ be the characteristic exponent of X that satisfies
which is known to admit the form
where γ ∈ R, σ ≥ 0, and Π is a Lévy measure on R\{0} satisfying the integrability condition:
Recall that the process X has bounded variation paths if and only if σ = 0 and |z|<1 |z| Π(dz) < ∞.
When this holds, we can write
where δ := γ − (−1,1)\{0} zΠ(dz). We consider the inventory control problem with proportional replenishment costs (without fixed costs). Let F := {F t : t ≥ 0} be the natural filtration generated by X. A strategy, representing the cumulative amount of replenishment, π = {R π t : t ≥ 0} is a nondecreasing, right-continuous, and F-adapted process starting at R π 0− = 0. The corresponding controlled inventory process U π becomes
We fix a discount factor q > 0 and a unit cost/reward of controlling C ∈ R (cost if it is positive and reward if negative). Associated with each strategy π ∈ A, the cost of inventory is modeled by ∞ 0 e −qt f (U π t )dt for a measurable inventory cost function f : R → R and that of controlling is given by
The problem is to minimize their expected sum
over the set of all admissible strategies A that satisfy all the constraints described above and the integrability condition:
The problem is to compute the value function
and to obtain the optimal strategy π * that attains it, if such a strategy exists.
Remark 2.1. In the healthcare management and finance examples described in Section 1, the process X can be understood as the number of patients and the currency rate that change over time. In the healthcare management example where the control variable is the number of the reduction (not the increment), the controlled process becomes the difference (not the sum) of X and the control process R π . This can be handled in the same way by flipping all the processes, and the resulting analysis will be the same.
Standing assumptions. Throughout the paper, we assume the following on the function f and the unit cost/reward C. Note that this is commonly assumed in the literature (see, e.g., [4, 5, 14, 25] ). Assumption 2.1 (Assumption on f and C). We assume the following throughout the paper:
(i) The function f is convex. This guarantees that the right-and left-hand derivative f + (x) and f − (x), respectively, exist for all x ∈ R. (ii) The function f has at most polynomial growth in the tail. That is to say, there exist k 1 , k 2 and N ∈ N such that |f (x)| ≤ k 1 + k 2 |x| N for all x ∈ R.
f + (∞) := lim x→∞ f + (x) ∈ (−∞, ∞], which exist by (i).
Assumption 2.2 (Assumption on X). We assume the following regarding the Lévy process X.
(i) The process X is not a driftless compound Poisson process.
(ii) There exists θ > 0 such that
This and [17, Theorem 3.6] guarantee that E[e θ|X 1 | ] < ∞ and, since e x ≥ x for x ≥ 0, we also have E[|X 1 |] < ∞.
For b ∈ R, we write the first down-crossing time:
As addressed in the introduction, our key tools are the expressions of the derivatives of the expected costs under the barrier strategy in terms of this random variable.
By Assumption 2.2(i), we have the following lemma (see Section A.1 for its proof).
Lemma 2.1. We have the following three facts.
(i) 0 is regular for R\{0} (i.e. P(inf{t > 0 : X t = 0} = 0) = 1). Assumption 2.2(ii) is also needed in the verification step when taking limits (see Theorem 3.1).
As is commonly pursued in classical singular control problems, our objective is to show the optimality of a barrier strategy π b for some b ∈ R with
It is well known that the resulting inventory process
becomes the reflected Lévy process with a lower barrier b. For the rest of the paper, let
whose finiteness is verified in Lemma 3.1. Note by the strong Markov property that
Lemma 3.1. For b ∈ R, the strategy π b satisfies the integrability condition (2.1) and is hence admissible.
The function v b (x) grows linearly as x ↓ −∞ as in (3.1). In addition, we have the following.
Proof. See Appendix A.4.
As discussed in the introduction, the objective of this current paper is to show that a barrier strategy with the barrier
is optimal. To see this is well-defined and finite, by the convexity of f , the mapping ρ(b) is nondecreasing. In addition, by Lemma 3.3, monotone convergence gives
3.1. Optimality over barrier strategies. We shall first show the optimality of the barrier strategy π b * over all barrier strategies
The rest of this subsection is devoted to the proof of Theorem 3.1. The following lemma gives key expressions of the right-hand derivative of v b with respect to the barrier b, written in terms of the first downcrossing time (2.2).
In addition, by replacing b and b+ε with b−ε and b, respectively, we obtain, for
Putting these together, for any ε ∈ R, we have for all t ≥ 0 (3.9) , the convexity of f and Lemma 3.3, for b ∈ R and ε ∈ (−1, 1) the triangle inequality and the mean value theorem give
Here, since f is convex and by (3.6), we have
. This limit together with Lemma 3.3 and the dominated convergence theorem shows, for
and stays at ε afterwards, and therefore
Now let ε > 0 and define
.
, as desired. Finally, (iii) is immediate by (i) and (ii).
We are now ready to complete the proof of Theorem 3.1.
Proof of Theorem 3.1. By (3.3) and the strong Markov property and the fact that
This contradicts with the fact that (3.11) when setting b = a 3 < b * is less than 0.
(ii) From the same argument we can prove that
3.2. Slopes and convexity of v b .
is continuous on R and has the right-and left-hand derivatives given by
Proof. (i) For y ∈ R and t ≥ 0, we write X (y) t := X t + y and τ (y) a
be the corresponding barrier strategy with barrier b and inventory process corresponding to X (y) . Then we have, for
Here, similar to the arguments in the proof of Lemma 3.4, we focus on the pathwise behaviors of
that imply
Here, since f is convex and by (3.18), we have
where the last limit holds because τ
. Therefore, by Lemma 3.3 and the dominated convergence theorem, taking a limit in (3.14) gives
On the other hand, by (3.16), (3.17), and (3.19) , the difference (3.15) is bounded with
By the dominated convergence theorem and using again τ
This shows for the right-hand derivative. Similar arguments show for the left-hand derivative for x = b.
The above arguments also show the continuity of x → v b (x) for x = b and hence it remains to show the continuity for x = b. In view of (3.16), (3.17) , (3.18) , (3.19) and (3.20) , by replacing b and b + ε with b − ε and b, respectively, we have, for ε ∈ R and t ≥ 0,
Thus we get the continuity in the same way as the proof of Lemma 3.4.
(ii) To show the differentiability (i.e. the right-and left-hand derivatives obtained in (i) coincide)
which is zero by [6, Proposition I.15 ] (note that this holds even for
The following lemma states that with the selection b * as in (3.2), the function v b * is continuously differentiable on R.
Lemma 3.6. For x ∈ R, the function v b * is convex and belongs to C 1 (R) with its derivative given by
Proof. See Appendix A.5.
In Theorem 3.1, we showed the optimality of π b * over the set of barrier strategies A bar . In this section, we strengthen the result and show the optimality over all admissible strategies A.
For n = 1, 2, let C n poly be a subset of n-times differentiable functions g in C n (R) satisfying, for some b 1 , b 2 > 0 and M ∈ N,
From Lemmas 3.2 and 3.6, we already know v b * ∈ C 1 poly . We need further assumption on the smoothness for the case X has paths of unbounded variation.
poly ) when X is of unbounded variation.
Below, we give a sufficient condition for Assumption 4.1. The proof is given in Appendix A.6.
Lemma 4.1. Suppose the running cost function f ∈ C 2 (R) and f has polynomial growth in the tail. Then, function v b * belongs to C 2 (and hence C 2 poly ).
We note that Assumption 4.1 is expected to hold more generally. In Section 5, we consider examples where the condition in Lemma 4.1 fails but Assumption 4.1 is shown to hold numerically.
Our main result of this paper is as follows.
Theorem 4.1. If Assumption 4.1 is satisfied, then the barrier strategy π b * is optimal over A.
The rest of this section is devoted to the proof of Theorem 4.1.
Let L be the infinitesimal generator associated with the process X applied to g ∈ C 1 poly (resp., C 2 poly ) for the case in which X is of bounded (resp., unbounded) variation with
Also, we write (L − q)g(x) := Lg(x) − qg(x), and for any càdlàg process Y , let ∆Y t := Y t − Y t− , t ≥ 0.
Suppose that X has bounded (resp., unbounded) variation paths, and let w be a function on R belonging to C 1 poly (resp., C 2 poly ) and satisfying, for x ∈ R,
Proof. Let π ∈ A be any admissible strategy that satisfies (2.1). By an application of the Itô formula (see, e.g., [23, Theorem II.31 or II.32]), we have
With R π,c t the continuous part of R π t such that U π t = X t + R π,c t + 0≤s≤t ∆R π s , t ≥ 0, we have
where {M t : t ≥ 0} is a local martingale satisfying and together with (4.1), we have
Because M is a local martingale, we can take a sequence of stopping times {T n } n∈N that is a localizing sequence for M with T n ↑ ∞ almost surely. Then, taking an expectation, we have
From the proof of [25, Theorem 2], we have lim t↑∞,n↑∞ E x e −q(t∧Tn) w(U π t∧Tn ) = 0. By taking the limit as t ↑ ∞ and n ↑ ∞ and the dominated convergence theorem thanks to (2.1), the proof is complete.
We now prove that the candidate value function v b * satisfies the conditions (4.1) and (4.2) in Proposition 4.1 (the opposite inequality v(x) ≤ v b * (x) holds because π b * is admissible as in Lemma 3.1). The latter condition is immediate as follows. In view of Proposition 4.1, we are now left to show that v b * satisfies (4.1).
By the same argument, {M [2] t : t ≥ 0} and {M [3] t : t ≥ 0} where, for t ≥ 0,
are martingales under P x for x ∈ (b * , ∞). By these and (4.4), the process {M [4] 
is a martingale under P x . By the same reasoning as that of the proof of [7, (12) ], we have (4.3) for x ∈ (b * , ∞). This also holds for x = b * by the continuity of (4.3), thanks to the assumed smoothness of v b * .
Proof. We follow the proof of [1, Lemma 5] and modify it for our results.
We write g(
From the convexity of v b * (as in Lemma 3.6) and f , the function g is convex on (−∞, b * ).
Next we fix b < b * and show the next identity:
Because U b is a semimartingale and v b * is sufficiently smooth, we can follow similar steps as the proof of Proposition 4.1 via the Itô formula. With the continuous part of
Because v b * is of polynomial growth by Lemma 3.
the proof of [25, Theorem 2] . In addition, when R b,c s increases U b s = b, and when ∆R b s > 0 (i.e.
where the last equality holds because, using the fact that
we have, using Lemma 4.2, [25, Theorem 2] . Since π b is admissible (satisfying (2.1)) and using the dominated convergence theorem, we have v t,n b (x) t,n↑∞ − −− →v b (x). In addition, the function g is continuous on R and hence finite on a finite interval, and thus g(U b s )1 (−∞,b * ) (U b s ) is bounded for s ≥ 0. Therefore, using the dominated convergence theorem, we have (4.5).
We are now ready to complete the proof. Because g is convex and g(b * ) = 0, if there exists a < b * such that g(a) < 0, then necessarily g(x) < 0 for all x ∈ (a, b * ). Setting b = a in (4.5) and by Theorem 3.1,
On the other hand, because U a t ≥ a a.s., ∞ 0 e −qs g(U a s )1 {−∞<U a s <b * } ds is nonpositive and also strictly negative with a positive probability when the starting point x is less than b * . This contradicts with (4.6) and the proof is complete.
In this section, we confirm the results obtained in the previous sections through numerical experiments via Monte Carlo simulation. While there exist many simulation techniques for Lévy processes, it is not our focus to evaluate and compare the performance of these techniques. Here, in order to confirm that our results are easily implementable, we use a classical Euler scheme focusing on the case X is the sum of drifted Brownian motion and a compound Poisson process with two-sided jumps (i.e. Π is a finite measure). An obvious shortcoming of the Euler scheme is that it fails to be accurate when f is not continuous (for the computation of b * ), and hence here we focus on the cases f is continuous. This issue may potentially be resolved by using other methods, such as the Wiener-Hopf simulation [16] which is implementable if the Wiener-Hopf factorization is explicitly known.
In order to confirm that the results hold for a wide class of Lévy processes, we purposefully choose a non-standard Lévy process. We consider X of the form
where (B t ; t ≥ 0) is a standard Brownian motion and (N + t ; t ≥ 0) and (N − t ; t ≥ 0) are Poisson processes with arrival rates 0.4 and 0.6, respectively. The upward and downward jumps (Z + n ; n = 1, 2, . . .) and (Z − n ; n = 1, 2, . . .) are i.i.d. sequences of (folded) normal random variables with mean zero and variance 1 and Weibull random variables with shape parameter 2 and scale parameter 1, respectively. These processes are assumed mutually independent. For other parameters, we set q = 0.05 and C = 1. For each realization, we truncate the time horizon to T = 100 and discretize [0, T ] using N = 10, 000 equally-spaced points with distance ∆ t := T /N .
For the inventory cost function f , we consider the following three cases:
for x ∈ R, which are convex and continuously differentiable on R. Here, we consider f 2 and f 3 as examples where they fail to be C 2 at 0 and 1, respectively (i.e. the condition in Lemma 4.1 does not hold), and hence the C 2 property of v b * as in Assumption 4.1 needs to be verified numerically.
For the approximation of the expectation, we first obtain a set of M := 10, 000 sample paths of X started at zero, say X := ( X (1) , . . . , X (M ) ) with X (m) = ( X . As shown in Section 3, ρ(b) is monotone and hence b * can be obtained by classical bisection. While ρ M (b) for each b is an approximated value, because we are using the same sample paths X, the monotonicity of b → ρ M (b) is still preserved, causing no problem in using bisection methods. With the approximated optimal barrier b * , we shall now confirm the optimality by comparing the expected total costs v b * with v b under suboptimal choices of b. In order to compute these, we continue using the set of paths X. The reflected path with lower barrier b under P x becomes U b,(m) Figure 2 shows the results. It can be confirmed that the selection b * indeed minimizes the total expected cost for all starting points. Finally, we confirm the smoothness of the value function v b * . We compute the derivative v b * as in (3.21) via simulation using the same sample paths X. In order to evaluate the second derivative, we compute (v b * (x + δ) − v b * (x))/δ where δ = 0.01. These results are summarized in Figure 3 . While To show that (A.1) implies that 0 is regular for R\{0}, suppose to derive a contradiction that 0 is not regular for R\{0}. Then, Blumenthal's zero-one law gives P(T R\{0} > 0)= 1, where T R\{0} := inf{t > 0 : X t = 0}.
Since 1 = P(T R\{0} > 0) = lim n↑∞ P(T R\{0} > 1 n ) by the dominated convergence theorem, there exists > 0 such that P(T R\{0} > ) > 0. On {T R\{0} > }, we have X t = 0 for t ∈ [0, ] which implies that sup s∈[0,t] X s = inf s∈[0,t] X s = 0 for t ∈ [0, ], contradicting (A.1).
(ii) We now show the continuity of x → τ − x . For x > 0, we have τ − x = 0 P-a.s., and so the continuity is obvious. Thus we assume x ≤ 0 for the rest of the proof.
From the definition of τ − x as in (2.2) , it is immediate that P-a.s. as ε ↓ 0,
where T − x := inf{t ≥ 0 : X t ≤ x}. This shows the left-continuity for x ≤ 0 (including x = 0). It now remains to show for x < 0 that the left-and right-limits coincide. Here, we want to prove τ − x = T − x , P-a.s. for x ∈ (−∞, 0). By the strong Markov property, we have
When 0 is regular for (−∞, 0), then, because
x < ∞} and thus the right hand side of (A.2) is equal to E e −T − x which, together with the fact that τ − x ≥ T − x a.s., implies that τ − x = T − x , P-a.s. When 0 is irregular for (−∞, 0), then X has bounded variation paths and has a non-negative drift by [17, Theorem 6.5] . If the process jumps downward onto x at T − x , then by the irregularity it immediately goes up and τ −
Hence, the following lemma completes the proof for the case 0 is irregular for (−∞, 0).
Proof. We recall some properties of the ladder height processes (see, e.g., [17, Section 6] ). Let H = {H t : t ≥ 0} and H = { H t : t ≥ 0}, respectively, be ascending and descending ladder height processes of X. Then, the processes H and H are subordinators, possibly killed and sent to the cemetery state +∞ at some independent exponential random variable. Below, let P H and P H be the laws of H and H when they start at zero. Since 0 is regular for (0, ∞) for X, it is easy to check that 0 is regular for (0, ∞) for H as well. Thus by [17, Theorem 5.4 ] the potential measure
has no atoms on (0, ∞).
Let Π H be the Lévy measure of H. Since 0 is irregular for (−∞, 0) for X and by the definition of the descending ladder height processes, Π H is a finite measure and H has no drift (see [17, Theorem 6.6 and Section 6.2]). In addition, the measure Π H has no atoms. Indeed, by [17, Theorem 7.8] and since Π (denoting the Lévy measure of the dual process −X) has atoms at most countable points and U H has no atoms, for y > 0, we have by the dominated convergence theorem, for some k > 0,
Since 0 is irregular for (−∞, 0) for X and by the definition of the ladder height process, we have, with
By the compensation formula of the Poisson point processes and since H has no drift, we have, with N H the Poisson random measure on ([0, ∞) × R, B[0, ∞) × (0, ∞), ds × Π H (dx)) associated with the jumps of H,
Since Π H is a finite measure and has no atoms, and by [17, Theorem 5.4] , the value (A.3) is equal to 0, as desired.
(iii) The proof of (iii) comes from the identity E x e −qτ − 0 = E e −qτ − −x , (ii) and the dominated convergence theorem.
A.2. The proof of Lemma 3.1. (i) We first prove
Without loss of generality, we assume b = 0. Because the strong Markov property gives Using the strong Markov property, we have
Here, we have
On the other hand, for n ≥ 1, U 0 T (n−1) − = 0 and hence (T (n)
.
Combining these,
Thus, it suffices to prove that E e −qT (1) [1, 2] h(z) < ∞. By the compensation formula of the Poisson point processes, with U 0 t the running supremum of U 0 , and N the Poisson random measure associated with the jumps of X as in the proof of Proposition 4.1,
which is finite since h is of polynomial growth from Remark 2.2 and by Assumption 2.2. Hence (A.4) holds.
(ii) Fix any arbitrary constant u > 0. We have
From the same argument as [20, Lemma 3.2] , this is also finite.
By (i) and (ii), the proof is complete.
A.3. The proof of Lemma 3.2. Since f is convex, either f monotonically decreases to a nonnegative value or otherwise there exists α > b such that |f | is nondecreasing on (α, ∞). Thus, for (3.6) ) and R b ≥ R b−x and hence by integration by parts
From Assumption 2.1(ii), we have for some k 1 , k 2 and N ∈ N,
which is of polynomial growth because E ∞ 0 e −qt U b t N −l dt is finite by Lemma 3.1.
A.4. The proof of Lemma 3.3. Fix ε > 0. Since f is a convex function, we have
This is finite by (A.4) , as desired.
A.5. The proof of Lemma 3.6. Since f + is right-continuous and by (3.6) , the map b → E b ∞ 0 e −qt f + (U b t )dt is right-continuous. In addition, we have f − (x) = lim y↑x f + (x) and thus
In view of (3.2) and because f + is nondecreasing, for δ > 0,
Hence, taking δ ↓ 0, 
where the second equality holds because E x
e −qt f ε * (X t )dt as in the proof of Lemma 3.5(ii) and we have U b * t = X t for t < τ − b * . Since f ε * is nondecreasing and v b * is continuous, v b * is convex.
To complete the proof, we now confirm the continuity of v b * ,+ . Since v b * ,+ is right derivative and v b * is convex, v b * ,+ is right-continuous and thus it suffices to prove lim δ↓0 (v b * ,+ (x) − v b * ,+ (x − δ)) = 0 for x ∈ R. Here we use the same notations as the proof in Lemma 3.5 for b = b * . For x ∈ R and δ > 0, we have
By (3.16) , (3.18) and (3.20) with b changed to b * , x changed to x − δ and x + ε changed to x and since f ε * is nondecreasing, by (A.6),
By the convexity of f , we have lim y ↑y f ε * (y ) = f − (y) for y ∈ R. Thus, by the monotone convergence theorem, with R (q) (x, ·) the measure as defined in the proof of Lemma 3.5(ii),
which is zero since f ε * and f − differ only at countable points by the convexity of f . Therefore, by the convexity of v b * , v b * ,+ (x) = v b * (x) and v b * belongs to C 1 (R).
A.6. Proof of Lemma 4.1. Fix x ∈ R. By Lemma 3.6 and because f is differentiable by assumption,
where U (x),b * is as defined in the proof of Lemma 3.5 for b = b * . Here, applying (3.20) and then (3.16) ,
We first prove