Violation of the Hamilton-Jacobi-Bellman Equation in Economic Dynamics
aa r X i v : . [ ec on . T H ] F e b Violation of the Hamilton-Jacobi-BellmanEquation in Economic Dynamics
Yuhki Hosoya ∗ Faculty of Economics, Chuo University † March 1, 2021
Abstract
We consider an extension of the classical capital accumulationmodel, and present an example in which the Hamilton-Jacobi-Bellman(HJB) equation is neither necessary nor sufficient for a function to bethe value function. Next, we present assumptions under which theHJB equation becomes a necessary and sufficient condition for a func-tion to be the value function, and using this result, we propose a newmethod for solving the original problem using the solution of the HJBequation. Our assumptions are so mild that many macroeconomicgrowth models satisfy them. Therefore, our results ensure that thesolution of the HJB equation is rigorously the value function in manymacroeconomic models, and present a new solving method for thesemodels.
JEL codes : C02, C61, E13.
Keywords : Capital Accumulation Model, Hamilton-Jacobi-Bellmanequation, Classical Solution, Linear Technology, Nonlinear Technol-ogy, Subdifferential Calculus.
The Hamilton-Jacobi-Bellman (HJB) equation has many applications in eco-nomics and is used in a variety of fields, both micro and macro. The HJB ∗ E-mail: hosoya(at)tamacc.chuo-u.ac.jp † We first show that the HJB equation is neither a necessary nor a sufficientcondition for a function to be the value function in this model. (Section 2.)Specifically, for a given time preference rate, the HJB equation has infinitelymany solutions. On the other hand, at another time preference rate, thevalue function is not a solution to the HJB equation.Let us add a few words about the latter result. First, the HJB equationis a differential equation for a real-valued function, so naturally, if the valuefunction is infinite, it does not satisfy the HJB equation. However, in theabove model, although the value function is finite, it is not a solution to theHJB equation. Second, the HJB equation is a differential equation of degen- This model was originated by Ramsey (1928), and modified by Cass (1965) and Koop-mans (1965). Because this model is widely known, many textbooks treat it. See, for ex-ample, Acemoglu (2009), Barro and Sala-i-Martin (2003), Blanchard and Fischer (1989),and Romer (2011). In the above model, however, the value function is not even aviscosity solution of the HJB equation.Therefore, the first problem mentioned above, “the absence of mathe-matical foundations,” is serious. Perhaps we are analyzing the HJB equationand the value function assuming that there is a relationship between them,even though there is no relationship at all. To dispel this concern, we showthat, under economically reasonable assumptions, nothing like the above ex-ample can occur. Specifically, we first show that under some assumptionsin the optimal capital accumulation models, the value function solves theHJB equation. (Proposition 3, Theorem 1) Second, we show that under anadditional assumption, any solution of the HJB equation that is increasing,concave, and continuously differentiable, and satisfies the growth conditionmust be the same as the value function. (Theorem 2) To combine these re-sults, we conclude that under several assumptions, solving the HJB equationis a necessary and sufficient condition for a function to be the value function.Our assumptions are sufficiently weak to allow us to deal with a muchwider variety of problems than the usual RCK model. First, we do not as-sume the Inada condition, nor do we assume its negation. Thus, we can dealwith models that satisfy the Inada condition, or conversely, we can deal withproblems that do not satisfy the Inada condition, such as the AK model.We also do not assume that the speed of capital accumulation is linear withrespect to consumption. Furthermore, we allow the instantaneous utilityfunction to depend not only on consumption but also on capital accumula-tion. Thus, the results of this paper can be applied to a considerable numberof macroeconomic dynamic models.Additionally, in the process of deriving our results, we derive an approachto compute the solution of the original problem using the solution of the HJBequation. (Proposition 4, Corollary 1) This result is useful even in the RCKmodel in the following sense. In the usual RCK model, the analysis usesphase diagrams for the simultaneous differential equation of the capital ac-cumulation equations and the Euler equations. This system is semi-stable.By setting the initial value of consumption on a stable manifold, we can ob-tain a solution of this problem. However, because the system is semi-stable,if we apply usual methods for approximating the solution of differential equa-tions such as the Runge-Kutta method, the approximation error will increase This notion was introduced by Lions (1982), and used by Crandall and Lions (1983)for analyzing the HJB equation. c ( k, p ), whichcorresponds to the policy function. Therefore, Proposition 4 and Corollary 1correspond to the result that one can derive the solution of the original prob-lem using the policy function. These results are all in Chapter 4 of Stokeyand Lucas (1989), and our results can be seen as a continuous-time versionof those.Finally, we consider the logarithmic AK model as an example of applyingour results. In this model, we can solve the HJB equation quite easily anduse it to obtain a solution to the original problem. Although this solution hasalready been obtained by Barro and Sara-i-Martin (2003), it is interesting inthe sense that it is derived via a completely different path.In Section 2, we present an example in which the HJB equation doesnot characterize the value function. In Section 3, we elaborate on the setupof our model and define a number of terms and assumptions. In Section4, we discuss the main results. In Section 5, we explain the computationalexamples, and discuss several topics. In Section 6, we present concludingremarks. Proofs of all results are given in Section 7.4 A Fatal Counterexample
We consider the traditional RCK model with linear technology and linearinstantaneous utility. That is, we consider the following model:max Z ∞ e − ρt u ( c ( t )) dt subject to. c ( t ) is locally integrable ,k ( t ) ≥ , c ( t ) ≥ t ≥ , (1)˙ k ( t ) = f ( k ( t )) − c ( t ) a.e. ,k (0) = ¯ k, where ρ >
0, ¯ k > u ( c ) = c , and f ( k ) = k . Let V ρ denote the value functionof the above model. Note that, the HJB equation is the following differentialequation. sup c ≥ { ( f ( k ) − c ) V ′ ( k ) + u ( c ) } − ρV ( k ) = 0 . We first show the following facts. Fact 1 . If ρ = 1, then there exist infinitely many solutions of the HJBequation. Proof of Fact 1 . Let V ( k ) = ak for some a ≥
1. Then,sup c ≥ { ( f ( k ) − c ) V ′ ( k ) + u ( c ) } = ak = ρV ( k ) . Fact 2 . The function V is finite, concave, and neither the classical norviscosity solution of the HJB equation. Proof of Fact 2 . We first show that V (¯ k ) = ¯ k . To show this, consider thefollowing pair of functions: k ( t ) ≡ ¯ k, c ( t ) ≡ ¯ k. Then, this pair ( k ( t ) , c ( t )) satisfies all constraints of the problem (1), andsatisfies the Euler equation ddt ( u ′ ◦ c )( t ) = ( ρ − f ′ ( k ( t )))( u ′ ◦ c )( t ) , We present the formal definitions of several technical terms, including the value func-tion and classical and viscosity solutions of the HJB equation, in the next section. Becausethis counterexample is most important in this paper, we present this before introducingthe formal definitions. t →∞ e − ρt u ′ ( c ( t )) f ( k ( t )) = 0 , if ρ = 1. Therefore, ( k ( t ) , c ( t )) is a solution of (1) if ρ = 1, and thus, V (¯ k ) = Z ∞ e − t ¯ kdt = ¯ k. Second, we show that 0 ≤ V (¯ k ) ≤ V (¯ k ) = ¯ k . Because u ( c ) ≥ c ≥
0, we have taht V (¯ k ) ≥
0. Choose any pair ( k ( t ) , c ( t )) that satisfies allconstraints of problem (1). Then, because u ( c ( t )) ≥ t ≥ Z ∞ e − t u ( c ( t )) dt ≤ Z ∞ e − t u ( c ( t )) dt ≤ V (¯ k ) = ¯ k, and thus, we have that V (¯ k ) ≤ ¯ k as desired.Therefore, V ( k ) is finite for all k >
0. Third, we show that the function V is concave. Choose k , k > s ∈ [0 , k = (1 − s ) k + sk .Suppose that a pair ( k i ( t ) , c i ( t )) satisfies all constraints of the problem (1)with ¯ k = k i for i ∈ { , } . Define k ( t ) = (1 − s ) k ( t ) + sk ( t ) , c ( t ) =(1 − s ) c ( t ) + sc ( t ). Then, ( k ( t ) , c ( t )) satisfies all constraints of the problem(1) with ¯ k = k . Therefore, V ( k ) ≥ Z ∞ e − t u ( c ( t )) dt = (1 − s ) Z ∞ e − t u ( c ( t )) dt + s Z ∞ e − t u ( c ( t )) dt. Because ( k i ( t ) , c i ( t )) is arbitrary, V ( k ) ≥ (1 − s ) V ( k ) + sV ( k ) , which implies that V is concave. Note that, because V is concave, thisfunction is locally Lipschitz.Fourth, we show that the function V is not a classical solution of theHJB equation. Suppose that V is a classical solution of the HJB equation.Then, V is continuously differentiable, and the following equation is satisfiedfor all k >
0: sup c ≥ { ( k − c ) V ′ ( k ) + c } − V ( k ) = 0 . (2)Because the left-hand side becomes infinite if V ′ ( k ) <
1, we have that V ′ ( k ) ≥ k >
0. Then, the supremum of the left-hand side is attained at c = 0.Therefore, (2) can be modified to the following equation: kV ′ ( k ) = 2 V ( k ) . For the proof of this fact, see Theorem 4 of Hosoya (2019a). V ( k ) = V (1) k . Therefore, V ′ ( k ) < k > V is not a classical solution of the HJB equation.Last, we show that the function V is not a viscosity solution of the HJBequation. Suppose that V ( k ) is a viscosity solution of the HJB equation.Note that, because V is locally Lipschitz, it is absolutely continuous oneach compact interval in R ++ , and thus, it is differentiable a.e.. If V isdifferentiable at k , then (2) holds at k . Therefore, by the same reason as inthe last paragraph, we have that V ′ ( k ) ≥ kV ′ ( k ) = 2 V ( k ) . Hence, again V ( k ) is a solution of the above linear differential equation, andthus, by Carath´eodory-Picard-Lindel¨of’s uniqueness theorem of the solution,we have V ( k ) = V (1) k . Thus, V is differentiable everywhere, and V ′ ( k ) < k > (cid:4) What is wrong in the above counterexample?
In related studies, the reason why the value function V (¯ k ) solves the HJBequation is told as follows. First, for every t >
0, we can obtain the followingequation: V (¯ k ) = sup k ( s ) ,c ( s ) (cid:26)Z t e − ρs u ( c ( s )) ds + e − ρt V ( k ( t )) (cid:27) . This equation can be transformed intosup k ( s ) ,c ( s ) (cid:26)Z t e − ρs u ( c ( s )) ds + e − ρt V ( k ( t )) − e − ρ V ( k (0)) (cid:27) = 0 . If we can assume that V is differentiable and c ( s ) is continuous, then the See our arguments on the viscosity solution of the HJB equation in Section 3. V ◦ k is differentiable at s = 0, and thus0 = sup k ( s ) ,c ( s ) (cid:26)Z t e − ρs u ( c ( s )) ds + e − ρt V ( k ( t )) − e − ρ V ( k (0)) (cid:27) = sup k ( s ) ,c ( s ) (cid:26) t × ddT (cid:20)Z T e − ρs u ( c ( s )) ds + e − ρT V ( k ( T )) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) T =0 + o ( t ) (cid:27) (3)= sup k ( s ) ,c ( s ) { t [ u ( c (0)) − ρV ( k (0)) + V ′ ( k (0)) ˙ k (0)] + o ( t ) } . Because k (0) = ¯ k and ˙ k (0) = f (¯ k ) − c (0), the last line depends only on c (0),and thus we have that0 = sup c { t [( f (¯ k ) − c ) V ′ (¯ k ) + u ( c )] + o ( t ) } − ρV (¯ k ) t. Dividing the right-hand side by t and taking the limit t →
0, we have theHJB equation.However, we have already obtained a counterexample V . Therefore, theabove “proof” is somewhat incorrect. Upon closer inspection, we find thatthere are at least two gaps in the above “proof”.First, we said that “the last line of (3) depends only on c (0)”. However,in the “last line”, there is a Landau’s little-o symbol o ( t ). To be precise, thisline just summarizes all the parts that depend on something other than c (0)in the form of o ( t ), and o ( t ) still depends on the form of the functions k ( s )and c ( s ). Additionally, the ‘ o ( t )’ part is o ( t ) if we fix a pair of functions k ( s )and c ( s ), but when we take the supremum over such functions, we are notreally sure if it is actually o ( t ).Second, for the derivation of (3), we made some assumptions that assuresthe differentiability of V ◦ k at s = 0. Specifically, we assumed the differen-tiability of V and the continuity of c ( s ). However, in our model, c ( s ) may bediscontinuous, and there is no guarantee that V is differentiable. Therefore,it is uncertain whether we can derive the equation (3) in the first place.We cannot determine which of these two gaps is more critical. However,we know that both gaps are almost never a problem when there is a solution ofthe original problem. The first gap seems to be a problem that can be solvedusing a solution ( k ∗ ( s ) , c ∗ ( s )) to evaluate o ( t ). Moreover, if a solution exists,then it will satisfy the Euler equation, and thus the optimal consumption path c ( s ) will be continuous in many cases. Furthermore, the “inner solution”condition of Benveniste and Scheinkman (1979) will be satisfied in mostmodels, and therefore their Theorem 2 can be applied and the value functionwill be differentiable. Hence, we think that if the original problem has asolution, both gaps will be filled in most cases. In other words, we can8magine that the problem in our example (1) probably arises because thereis no solution in the original problem.So, does this solve our problem? We think not. This is because in manystudies that apply the HJB equation, the solution of the original problem isnot guaranteed to exist at the stage of using the HJB equation. Moreover,another problem that there may be more than one solution to the HJB equa-tion (Fact 1) has not been solved. In many cases, we want to obtain a valuefunction by solving the HJB equation, and what is needed for this is notthat the value function solves the HJB equation, but rather that a solutionof the HJB equation is actually the value function. Fact 1 would be a morepowerful obstacle than Fact 2 when trying to achieve such a result.Thus, many studies that use the HJB equation are in a very precariousposition. A way to solve this problem is first to consider the solution of theHJB equation as a “candidate” of the value function, rather than immediatelyconsidering it as the value function, and then to prove that this “candidate”is indeed the value function. If we follow this procedure, we can use the HJBequation appropriately. However, it is uncertain whether we can follow theabove procedure at any time. Perhaps proving this “candidate” is really thevalue function may be tremendously difficult.Another way to solve this problem is to prove in a general framework thata solution of the HJB equation is the value function under several assump-tions. In the remainder of the paper, we will address this issue, and showthat the value function is a solution of the HJB equation, and that there isno other solution provided we do not deviate from the general assumptionsused in macroeconomic theory. This result will give many studies a crediblefoundation and greatly increase their persuasiveness.9 The Model and Definitions
The model that we argue in this paper is the following. max Z ∞ e − ρt u ( c ( t ) , k ( t )) dt subject to. c ( t ) is locally integrable ,k ( t ) ≥ , c ( t ) ≥ , (4) Z ∞ e − ρt u ( c ( t ) , k ( t )) dt can be defined , ˙ k ( t ) = F ( k ( t ) , c ( t )) a.e. ,k (0) = ¯ k. Note that, if the instantaneous utility function u ( c, k ) is independent of k andthe technology function F ( k, c ) = f ( k ) − dk − c , then the problem (4) is thetraditional RCK model. In this paper, we do not need the Inada condition,and thus this model admits the divergence of optimal growth paths. Notealso that, under the above constraints, k ( t ) must be absolutely continuous inevery compact interval in R + , because the solution of the differential equationmust be absolutely continuous in every compact interval.We make several assumptions of the above model. Assumption 1 . ρ > Assumption 2 . The instantaneous utility function u : R → R ∪ {−∞} is a continuous and concave function on R . Moreover, u ( c, k ) is nonde-creasing on R , and increasing in c and continuously differentiable on R . Furthermore, there exists c > u ( c, > −∞ . Assumption 3 . The technology function F : R → R is a continuous andconcave function that satisfies F (0 ,
0) = 0. Moreover, F is decreasing in c and there exist d , d ≥ F ( k, c ) > − d k − d c for every ( k, c )such that k > c ≥
0, and for every c ≥
0, there exists k > F ( k, c ) > F (0 , c ). The statement “ R ∞ e − ρt u ( c ( t ) , k ( t )) dt can be defined” admits that the value of thisintegral may be ±∞ . Because u is increasing in c on R , if ( c, k ) ∈ R , then u ( c, k ) > u (2 − c, k ) ≥ −∞ .Therefore, u ( c, k ) ∈ R for every ( c, k ) ∈ R . The last assumption means that for every ¯ c >
0, there exists k ∗ > ssumption 4 . The function ∂u∂c ( c, k ) is decreasing in c on R . Moreover,lim c → ∂u∂c ( c, k ) = + ∞ and lim c →∞ ∂u∂c ( c, k ) = 0 for every k >
0. Furthermore,for every k >
M >
0, the function c ∂u∂k ( c, k ) is bounded on ]0 , M ]. Assumption 5 . The technology function F is continuously differentiable in c on R . There are two types of optimal capital accumulation models: centralizedand decentralized. In this paper, we deal with a centralized model, whichis connected to the corresponding decentralized model by the fundamentaltheorem of welfare economics. In the centralized macroeconomic dynamicmodel, k ( t ) represents the capital stock and c ( t ) represents the consumption,and both are assumed to be non-negative. Some models assume that k ( t ) iscontinuously differentiable and c ( t ) is continuous, while others admit “jump”on c ( t ). In this paper, we take the latter position, and thus only assume that c ( t ) is integrable on any compact interval. As a result, we cannot assumecontinuous differentiability for the capital stock either, and only assume that k ( t ) is absolutely continuous on any compact interval.Of course, the model (4) includes the RCK model, which is the mostcommon in the literature on optimal capital accumulation models. We shallcheck this fact. In the RCK model, an instantaneous utility function u ( c )is given, which does not depend on k . Because u ( c ) may be a logarithmicfunction log c , boundedness of u is not usually assumed, and u (0) = −∞ is allowed. On the other hand, u ′ > , u ′′ <
0, lim c → u ′ ( c ) = + ∞ , andlim c → u ′ ( c ) = 0 are usually assumed. With a little checking, readers willnotice that this model satisfies our Assumptions 2 and 4.Next, in the RCK model, the relationship between the capital stock andconsumption is determined as follows:˙ k ( t ) = f ( k ( t )) − dk ( t ) − c ( t ) , where d ≥ f is called the productionfunction, and it indicates that if the capital stock is k , then the output is f ( k ). This output is separated into consumption and investment. If we let every c ∈ [0 , ¯ c ], k F ( k, c ) is increasing on [0 , k ∗ ]. Actually, we can choose k ∗ = min c ∈ [0 , ¯ c ] min k arg max { F ( k, c ) | ≤ k ≤ } . denote the consumption and i denote the investment, we obtain the firstrelationship: f ( k ( t )) = c ( t ) + i ( t ) . On the other hand, the speed of increase of the capital stock ˙ k ( t ) is equal to i ( t ) minus capital depletion dk ( t ). Thus, we obtain the second relationship:˙ k ( t ) = i ( t ) − dk ( t ) . By connecting these two relationships, we have that˙ k ( t ) = f ( k ( t )) − dk ( t ) − c ( t ) , as desired.Usually, it is assumed that f is concave, increasing, and continuous, f (0) = 0, and there exists k > f ( k ) > dk . By setting F ( k, c ) = f ( k ) − dk − c , d = d , and d = 1, readers can check that this model satisfiesour Assumptions 3 and 5. Thus, our model (4) includes the RCK model.Moreover, unlike many other studies, our assumptions do not assume eitherdifferentiability of the production function f or the Inada condition. There-fore, the scope of our model is sufficiently wide at this point. Of course, we can also deal with many models that are not RCK models.First, we can deal with a model in which the capital stock increases theinstantaneous utility. One typical example of such a utility is u ( c, k ) =( c ρ + k ρ ) Aρ , where 0 < A < ρ < , ρ = 0. This is a variant of the CESutility function. We should explain the assumption of the existence of c > u ( c, > −∞ . Economically, this assumption means that a lack of capi-tal stock is not fatal to people, and only lack of consumption can be fa-tal. Mathematically, this assumption is needed for step 4 of the proof ofProposition 3. Note that, this requirement prohibits some functions, such as u ( c, k ) = log c + log k .Second, we can deal with different technology structures from that inthe RCK model. For example, we can handle a model in which too highconsumption has a strong negative impact on the growth rate of k , and amodel in which choosing too high investment cannot be digested at once andused for production. The former model is given by the function F ( k, c ) = Because f ′ ( k ) < d can occur in the RCK model, we do not assume that F ( k, c ) isincreasing in k in Assumption 3. If ρ <
0, then we define u ( c,
0) = u (0 , k ) = 0. Note that, if A = 1, thenlim c →∞ ∂u∂c ( c, k ) = 1 when ρ >
0, and lim c → ∂u∂c ( c, k ) = 1 when ρ <
0, and thus As-sumption 4 is violated in any cases. ( k ) − dk − h ( c ), where h is an increasing, continuously differentiable, convexfunction that satisfies h (0) = 0 and has the Lipschitz constant d . The lattermodel is given by the function F ( k, c ) = h ( f ( k ) − c ) − dk , where h is anincreasing, continuously differentiable, and concave function that satisfies h (0) = 0 and has the Lipschitz constant d . In both cases, it is easy to provethat our Assumptions 3 and 5 are satisfied.Finally, we note that the problem (1) satisfies Assumptions 1-3 and 5.Only Assumption 4 is violated. We say that a pair of real-valued functions ( k ( t ) , c ( t )) defined on R + admis-sible if k ( t ) is absolutely continuous on every compact interval, c ( t ) is locallyintegrable, k ( t ) ≥ , c ( t ) ≥ R ∞ e − ρt u ( c ( t ) , k ( t )) dt can be defined, and˙ k ( t ) = F ( k ( t ) , c ( t )) a.e. . (5)Note that, if k ( t ) is absolutely continuous on every compact interval, it isdifferentiable almost everywhere and R ba ˙ k ( t ) dt = k ( b ) − k ( a ) for all a, b with0 ≤ a < b .Let A ¯ k denote the set of all admissible pairs such that k (0) = ¯ k . Usingthe notation of A ¯ k , we can simplify the model (4) as follows:max Z ∞ e − ρt u ( c ( t ) , k ( t )) dt subject to. ( k ( t ) , c ( t )) ∈ A ¯ k . Let ¯ V (¯ k ) = sup (cid:26) Z ∞ e − ρt u ( c ( t ) , k ( t )) dt (cid:12)(cid:12)(cid:12)(cid:12) ( k ( t ) , c ( t )) ∈ A ¯ k (cid:27) . We call this function ¯ V the value function of the problem (4). We call a pair ( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k a solution if and only if the followingtwo requirements hold. First, Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt ∈ R . Second, for every pair ( k ( t ) , c ( t )) ∈ A ¯ k , Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt ≥ Z ∞ e − ρt u ( c ( t ) , k ( t )) dt. Note that ¯ V is defined on R + . Later, we will show that, under mild assumptions, A ¯ k is nonempty for all ¯ k ≥ Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt = ¯ V (¯ k ) ∈ R . Note that, for all ( k ( t ) , c ( t )) ∈ A ¯ k such that R ∞ e − ρt u ( c ( t ) , k ( t )) dt ∈ R and T >
0, we have Z T e − ρt u ( c ( t ) , k ( t )) dt = Z ∞ e − ρt u ( c ( t ) , k ( t )) dt − Z ∞ T e − ρt u ( c ( t ) , k ( t )) dt ≥ Z ∞ e − ρt u ( c ( t ) , k ( t )) dt − e − ρT ¯ V ( k ( T )) . Particularly, if R ∞ e − ρt u ( c ( t ) , k ( t )) dt > M , then for all T > Z T e − ρt u ( c ( t ) , k ( t )) dt > M − e − ρT ¯ V ( k ( T )) . Therefore, if ¯ V (¯ k ) is finite, then for every ε >
0, there exists a pair ( k ( t ) , c ( t )) ∈ A ¯ k such that Z T e − ρt u ( c ( t ) , k ( t )) dt > ¯ V (¯ k ) − e − ρT ¯ V ( k ( T )) − ε for every T > The HJB equation is given by the following equation.sup c ≥ { F ( k, c ) V ′ ( k ) + u ( c, k ) } − ρV ( k ) = 0 . (6)A continuous function V : R + → R ∪ {−∞} is called a classical solution of the HJB equation if and only if V is continuously differentiable on R ++ and equation (6) holds for every k > V : R + → R ∪ {−∞} is called a viscosity subsolution of (6) if and only if it is real-valued and upper semi-continuous on R ++ , and for every k > ϕ defined on a neighborhood of k such that ϕ ( k ) = V ( k ) and ϕ ( k ′ ) ≥ V ( k ′ ) whenever both sides are defined,sup c ≥ { F ( k, c ) ϕ ′ ( k ) + u ( c, k ) } − ρV ( k ) ≤ . V : R + → R ∪{−∞} is called a viscosity supersolution of (6) if and only if it is real-valued and lower semi-continuous on R ++ , andfor every k > ϕ defined on aneighborhood of k such that ϕ ( k ) = V ( k ) and ϕ ( k ′ ) ≤ V ( k ′ ) whenever bothsides are defined, sup c ≥ { F ( k, c ) ϕ ′ ( k ) + u ( c, k ) } − ρV ( k ) ≥ . If a continuous function V : R + → R ∪ {−∞} is both viscosity sub- andsupersolution of (6), then V is called a viscosity solution of (6).Suppose that V is a viscosity solution of the HJB equation and is differ-entiable at k >
0. Then, it is known thatsup c ≥ { F ( k, c ) V ′ ( k ) + u ( c ) } − ρV ( k ) = 0 . See Proposition 1.9 of ch.2 of Bardi and Dolcetta (2008).
Consider the following differential equation.˙ k ( t ) = F ( k ( t ) , , k (0) = ¯ k. (7)Let k + ( t, ¯ k ) denote the solution of the above equation defined on R + . Thisfunction k + ( t, ¯ k ) is called the pure accumulation path . Later we will showthe following lemma. Lemma 1 . Under Assumption 3, the pure accumulation path k + is definedon the set R + × R ++ , and inf t ≥ k + ( t, ¯ k ) > k > V : R ++ → R . The following requirement of V is called the growthcondition . lim T →∞ e − ρT V ( k + ( T, ¯ k )) = 0 for all ¯ k > . (8)Let V be the space of all functions V : R + → R that is continuous, increasing,and concave, and satisfies the growth condition. Note that, by concavity,every V ∈ V is locally Lipschitz on R ++ . In this paper, we heavily use subdifferential calculus. We introduce thenotion of subdifferential and several results. For the proofs of these results,see textbooks on convex analysis, such as Rockafeller (1996).15uppose that a function G : U → R is concave, U ⊂ R n is convex, andthe interior V ⊂ U is nonempty. Choose any x ∈ V . We define ∂G ( x ) = { p ∈ R n | G ( y ) − G ( x ) ≤ p · ( y − x ) for all y ∈ U } . Then, we can show that ∂G ( x ) is nonempty. The set-valued mapping ∂G iscalled the subdifferential of G . It is known that G is differentiable at x if and only if ∂G ( x ) is a singleton, and if so, ∂G ( x ) = { DG ( x ) } .If n = 1, then define the left- and right-derivatives D − G ( x ) , D + G ( x )such as D − G ( x ) = lim y ↑ x G ( y ) − G ( x ) y − x , D + G ( x ) = lim y ↓ x G ( y ) − G ( x ) y − x . Note that, if G is concave, then G ( y ) − G ( x ) y − x is nonincreasing in y , and thus D − G ( x ) = inf t> G ( x − t ) − G ( x ) − t , D + G ( x ) = sup t> G ( x + t ) − G ( x ) t , which implies that both D − G ( x ) , D + G ( x ) are defined and real numbers. Itis known that ∂G ( x ) = [ D + G ( x ) , D − G ( x )].Recall that, under Assumption 3, our F is concave. In this case, thefunctions c F ( k, c ) and k F ( k, c ) are also concave, and thus the ‘partial’subdifferential can be considered. Let ∂ k F ( k, c ) = { p | F ( k ′ , c ) − F ( k, c ) ≤ p ( k ′ − k ) for all k ′ ≥ } ,∂ c F ( k, c ) = { p | F ( k, c ′ ) − F ( k, c ) ≤ p ( c ′ − c ) for all c ′ ≥ } . The partial left- and right-derivatives can be defined in the same manner.For example, D k, + F ( k, c ) = sup t> F ( k + t, c ) − F ( k, c ) t ,D k, − F ( k, c ) = inf t> F ( k − t, c ) − F ( k, c ) − t . We note one more fact. Suppose that p ≥ r ≥ q, p ∈ ∂ k F ( k , c ) , q ∈ ∂ k F ( k , c ) and k < k . We show that there exists k ∈ [ k , k ] such that r ∈ ∂ k F ( k, c ). If r ≥ D k, + F ( k , c ), then r ∈ ∂ k F ( k , c ). If r ≤ D k, − F ( k , c ), then r ∈ ∂ k F ( k , c ). Therefore, we assume that D k, − F ( k , c ) < r < D k, + F ( k , c ).Define g ( k ) = F ( k, c ) − rk . Then, D + g ( k ) > D − g ( k ) <
0, and thus,there exists k ∈ ] k , k [ such that g ( k ) = max k ′ ∈ [ k ,k ] g ( k ′ ). By the definitionof subdifferential, we have that 0 ∈ ∂g ( k ), and thus r ∈ ∂ k F ( k, c ), as desired. Formally, the subdifferential is defined for not concave but convex functions, and thusthe inequality in the definition is reversed. In this view, the name ‘subdifferential’ may benot appropriate, and ‘superdifferential’ may be more suitable. However, in the literatureof economics, these two notions are not distinguished, and thus our ∂G is traditionallycalled ‘subdifferential’. .7 Propositions The following propositions can be proved using only Assumptions 1-4.
Proposition 1 . Suppose that Assumptions 1-4 hold. Then, there exists apositive continuous function c ∗ ( p, k ) defined on R such that for all ( p, k ) ∈ R , F ( k, c ∗ ( p, k )) p + u ( c ∗ ( p, k ) , k ) = sup c ≥ { F ( k, c ) p + u ( c, k ) } . Note that, if F ( k, c ) = f ( k ) − dk − c and u ( c, k ) = u ( c ), then by thefirst-order condition, we have that c ∗ ( p, k ) = ( u ′ ) − ( p ), and thus c ∗ ( p, k ) isindependent of k . This result can be extended. If F ( k, c ) = f ( k ) − dk − h ( c ),then by the first-order condition, we have that c ∗ ( p, k ) is the unique c suchthat h ′ ( c ) p = u ′ ( c ), and thus c ∗ ( p, k ) is also independent of k . Proposition 2 . Suppose that Assumptions 1-3 hold. Then, ¯ V (¯ k ) > −∞ forevery ¯ k > V is nondecreasing and concave.By Proposition 2, we have ¯ V ( k ) ∈ R for some k > V ( k ) ∈ R for every k >
0. We say that ¯ V is finite if ¯ V ( k ) ∈ R for every k > Readers may think that Proposition 2 is obvious. However, the proof ofthis proposition is quite difficult. See the proof section.
In this section, we analyze the HJB equation for a characterization of thevalue function. We prohibit ourselves from making assumptions on the so-lution explicitly. Assumptions must be made for properties of primitives ρ, u, F in our model (4), and, for example, assumption for the existence ofthe solution in (4) is not appropriate. The reason why we restrict ourselvesis simple: in many situation, to ensure the existence of a solution of (4)is tremendously difficult. Hence, we prohibit ourselves from assuming theexistence of a solution, even though under the existence assumption of thesolution, the proofs of results become quite easy.
First, we show the following result. Note that, ¯ V (0) may be −∞ even if ¯ V is finite. roposition 3 . Suppose that Assumptions 1-4 hold. If the value function ¯ V is finite, then it is increasing on R + , and it is a viscosity solution of the HJBequation. If, in addition, Assumption 5 holds, then ¯ V is a classical solutionof the HJB equation.Note that our counterexample (1) satisfies Assumptions 1-3 and 5, andin this example, the value function V is finite. Therefore, the violation ofAssumption 4 is crucial for (1).Because Proposition 3 requires the finiteness of the value function, wewant an additional assumption that ensures the finiteness of the value func-tion. First, define the CRRA function as follows: u θ ( x ) = ( x − θ − − θ if θ = 1 , log x if θ = 1 , where θ > Assumption 6 . There exist k ∗ > , k + > , c ∗ ≥ , γ > , δ > , θ > , a > , b ≥ , C ∈ R such that F ( k ∗ , c ∗ ) > , (9)( γ, − δ ) ∈ ∂F ( k ∗ , c ∗ ) , < D k, + F ( k + , ≤ γ, (10) ρ − (1 − θ ) γ > , (11) u ( c, k ) ≤ au θ ( c ) + bu θ ( k ) + C for all c > , k > . (12)We stress that Assumption 6 is not strong. For example, in the RCKmodel, F ( k, c ) = f ( k ) − dk − c and u ( c, k ) = u ( c ). Suppose that f ( k ) isdifferentiable. Then, the condition ( γ, − δ ) ∈ ∂F ( k ∗ , c ∗ ) means that γ = f ′ ( k ∗ ) − d and δ = 1, and because D k, + F ( k + ,
0) = f ′ ( k + ) − d , (10) is satisfiedif k ∗ = k + and f ′ ( k ∗ ) > d . Because there is no restriction on c ∗ , requirements(9) and (10) are satisfied if k ∗ is sufficiently small and c ∗ = 0. Therefore, theactual restrictions are only (11) and (12). Moreover, if f satisfies the Inada This function is the solution of the following differential equation: − x u ′′ ( x ) u ′ ( x ) = θ, u (1) = 0 , u ′ (1) = 1 , where the left-hand side is sometimes called the relative risk aversion. Therefore, thisfunction is called the constant relative risk aversion function, and abbreviatedly, theCRRA function. we can choose γ so small that ρ − (1 − θ ) γ >
0, and thus theactual restriction is the existence of θ > u ( c ) ≤ au θ ( c ) + C forall c >
0. This is just a mild restriction on u . Additionally, if we can choose θ ≥
1, then (11) is automatically satisfied and vanishes even when f ( k ) doesnot satisfy the Inada condition.Note that, if u ( c, k ) = u θ ( c ) and F ( k, c ) = γk − c , then (11) provides anecessary and sufficient condition for the finiteness of the value function. Sufficiency is proved in ch.4 of Barro and Sara-i-Martin (2003), and we caneasily show necessity. Therefore, we guess that (11) is a crucial requirementfor the finiteness of the value function.On the other hand, if sup c,k u ( c, k ) < + ∞ , the requirement (12) is auto-matically satisfied for every θ ∈ ]0 , u θ (0) > −∞ when 0 < θ < u ( c, k ) = ( c ρ + k ρ ) Aρ and ρ < , ρ = 0 , < A <
1. If ρ >
0, then u ( c, k ) ≤ ( c ρ + k ρ ) A − ρ ρ − ( c + k − , and we can easily show that the right-hand side also satisfies (12) for θ =1 − A . If ρ <
0, then because x x Aρ is decreasing, by the formula forarithmetic-geometric means, we have that u ( c, k ) ≤ Aρ ( ck ) A ≤ Aρ − ( c A + k A ) , and the right-hand side satisfies (12) for θ = 1 − A . Therefore, there aresufficiently many models that satisfy Assumption 6.The following is our first main result. Theorem 1 . Suppose that Assumptions 1-6 hold. Then ¯ V ∈ V , and ¯ V is aclassical solution of the HJB equation.We make some remarks on Theorem 1. First, in many related studies onthe HJB equation in control problems, u is assumed to be bounded. Underthis condition, the value function is automatically bounded and finite, andthus there are many techniques that can be used in the proof. See, forexample, ch.3 of Bardi and Dolcetta (2008). However, we want to considerthe case in which u ( c ) = log c , and the logarithmic function is not bounded.Hence, we cannot assume the boundedness of u in this paper. Note that The function f satisfies the Inada condition if and only if f ′ ( R ++ ) = R ++ . Be-cause f is concave, it is equivalent to the following two requirements: lim k → f ′ ( k ) =+ ∞ , lim k →∞ f ′ ( k ) = 0. The technology F ( k, c ) = γk − c is called the AK technology . ( c ) = log c is not excluded for our assumptions: set θ = 1 and check (11)and (12) of Assumption 6.In Theorem 1, the value function automatically satisfies the growth con-dition (8). If (4) violates Assumptions 1-6, then (8) may be not satisfiedeven when the value function is a classical solution of the HJB equation. Forexample, let ρ = 1 , u ( c, k ) = c and F ( k, c ) = k − c . Then, we have alreadyobtained the value function ¯ V (¯ k ) = ¯ k in section 2. Because k + ( t, ¯ k ) = ¯ ke t ,we have that (8) is violated. However, we think that this is a pathologicalexample.Meanwhile, if ρ = 2 , u ( c, k ) = c and F ( k, c ) = k − c , then we have alreadyshown that 0 ≤ ¯ V (¯ k ) ≤ ¯ k and k + ( t, ¯ k ) = ¯ ke t , which implies that ¯ V satisfies(8). However, in this case ¯ V does not solve the HJB equation. Therefore,we have also obtained a model in which ¯ V ∈ V and ¯ V violates the HJBequation. This is another pathological example. In this subsection, we examine the sufficiency of the HJB equation to be thevalue function. We need an additional assumption.
Assumption 7 . Both ∂F∂c and ∂u∂c are continuously differentiable in k on R .Moreover, there exists k > c ≥ D k, + F ( k, c ) > ρ .Note that, in the RCK model, u ( c, k ) = u ( c ) and F ( k, c ) = f ( k ) − dk − c .Thus, the first assertion of Assumption 7 is automatically satisfied even if f ( k ) is not differentiable. If, in addition, there exists k such that f ′ ( k ) > ρ ,then the second assertion of Assumption 7 is also satisfied. Therefore, thisassumption is also not strong.The next proposition is crucial for our next main result. Proposition 4 . Suppose that Assumptions 1-5 and 7 hold. Suppose alsothat V ∈ V is a classical solution of the HJB equation. Choose any ¯ k > k ( t ) = F ( k ( t ) , c ∗ ( V ′ ( k ( t )) , k ( t ))) , k (0) = ¯ k. (13)Then, there exists a solution k ∗ ( t ) of the above equation defined on R + , andfor every such solution, inf t ≥ k ∗ ( t ) >
0. If we define c ∗ ( t ) = c ∗ ( V ′ ( k ∗ ( t )) , k ∗ ( t )),then c ∗ ( t ) is continuous and ( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k . Moreover, V (¯ k ) = Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt, k ( t ) , c ( t )) ∈ A ¯ k , if inf t ≥ k ( t ) >
0, then Z ∞ e − ρt u ( c ( t ) , k ( t )) dt ≤ V (¯ k ) . To combine this result and the preceding results, we obtain the followingtheorem.
Theorem 2 . Suppose that Assumptions 1-7 hold. Then, ¯ V is the uniquesolution of the HJB equation in V .Therefore, under Assumptions 1-7, the HJB equation is the perfect char-acterization for the value function in the function space V .Using Theorem 2 and Proposition 4, we obtain the following corollary. Corollary 1 . Suppose that Assumptions 1-7 hold. Consider the followingdifferential equation:˙ k ( t ) = F ( k ( t ) , c ∗ ( ¯ V ′ ( k ( t )) , k ( t ))) , k (0) = ¯ k. (14)Then, there exists a solution k ∗ ( t ) of (14) defined on R + . Moreover, if wedefine c ∗ ( t ) = c ∗ ( ¯ V ′ ( k ∗ ( t )) , k ∗ ( t )), then ( k ∗ ( t ) , c ∗ ( t )) is a solution of the model(4).We believe that the equation (14) is useful for analyzing the model, evenif the model is the RCK model with the Inada condition. In the RCK model,the solution is characterized by the simultaneous differential equation of thetechnology constraint (5) and the following Euler equation: ddt ( u ′ ◦ c )( t ) = [ ρ − f ′ ( k ( t ))]( u ′ ◦ c )( t ) . In this simultaneous system, the initial value of c ( t ) is not specified, and thischaracterization is incomplete. Therefore, to solve the model, we need anadditional requirement, and usually the transversality condition is used asthis requirement.However, there is a problem. In the RCK model, there uniquely existsa steady state, and this is not stable but only semistable . In a semistablesystem, any simple numerical computation method for the solution such asthe Runge-Kutta method does not work well, because the error divergesas t increases. Meanwhile, the equation (14) is a simple one-dimensionaldifferential equation, and the steady state is stable. Therefore, the error ofnumerical calculation converges to zero as t increases. Hence, using (14), wecan easily obtain an approximate solution by a usual numerical computationmethod. 21 Discussions
In this subsection, we provide an example that demonstrates that our resultswork well.
Example (logarithmic AK model). Let u ( c, k ) = log c and F ( k, c ) = γk − c ,where γ > ρ . By Theorem 2, ¯ V is the unique solution of the HJB equationin V . By the first-order condition, c ∗ ( p, k ) = 1 p . Therefore, the HJB equation is γk ¯ V ′ ( k ) − − log ¯ V ′ ( k ) = ρ ¯ V ( k ) . We guess that ¯ V ( k ) is twice continuously differentiable. Thus, to differentiateboth sides, we have γ ¯ V ′ ( k ) + γk ¯ V ′′ ( k ) − ¯ V ′′ ( k )¯ V ′ ( k ) = ρ ¯ V ′ ( k ) , and thus, ¯ V ′ ( k ) solves the following differential equation:( γk − ( x ( k )) − ) x ′ ( k ) = − ( γ − ρ ) x ( k ) . We guess that ( x ( k )) − = ρk . Then, x ( k ) = ρ − k − and x ′ ( k ) = − ρ − k − .We can easily verify that this function is actually the solution of the aboveequation. Hence, we obtain ¯ V ′ ( k ) = ρ − k − , and thus, ¯ V ( k ) = ρ − log k + C for some constant C . Using the HJB equation itself, we have that γρ − − ρ + log k = log k + ρC, and thus, C = ρ − [log ρ + γρ − − . In this case, the differential equation (14) is˙ k ( t ) = γk ( t ) − ρk ( t ) = ( γ − ρ ) k ( t ) , k (0) = ¯ k, and thus, by Corollary 1, k ∗ ( t ) = ¯ ke ( γ − ρ ) t , c ∗ ( t ) = ρ ¯ ke ( γ − ρ ) t is the solution of the problem (4). This completes the calculation.22 .2 The Magic of Capital Suppose that ¯ V (0) = −∞ . If ( k ( t ) , c ( t )) ∈ A ¯ k and k ( T ) = 0, then Z ∞ T e − ρt u ( c ( t ) , k ( t )) dt ≤ e − ρT ¯ V (0) = −∞ , and thus, if R ∞ e − ρt u ( c ( t ) , k ( t )) dt > −∞ , then k ( T ) > T > V (0) = −∞ , it is very useful to control thebehavior of k ( t ) when we analyze the problem. In this connection, readersmay have a conjecture that if u (0 ,
0) = −∞ , then ¯ V (0) = −∞ . However,Hosoya (2019b) found that this conjecture is incorrect.Suppose that ρ = 1 , F ( k, c ) = √ k − c and u ( c, k ) = − √ c . Then, As-sumptions 1-7 hold. Consider a pair ( k ( t ) , c ( t )) defined as follows: k ( t ) = t , c ( t ) = t . We can easily check that ( k ( t ) , c ( t )) ∈ A . Moreover,0 ≥ Z ∞ e − ρt u ( c ( t ) , k ( t )) dt ≥ − √ (cid:20)Z √ t dt + Z ∞ e − t dt (cid:21) = − √ − √ e − > −∞ , which implies that ¯ V (0) > −∞ .In the above example, the technology is a sort of RCK technology with f ( k ) = √ k , which corresponds to a sort of Cobb=Douglas technology func-tion. In this technology, f (0) = 0, and thus it is thought that if the initialcapital stock is absent, then people cannot produce anything. However, theabove example shows that actually people can produce something from noth-ing. We call this phenomenon the magic of capital . Because the abovetechnology is Cobb=Douglas and the instantaneous utility function is a pos-itive affine transform of a CRRA function with θ = , we have that thismagic is a commonplace event.The magic of capital makes our proofs in this paper difficult. For example,suppose that ( k ( t ) , c ( t )) ∈ A ¯ k and R ∞ e − ρt u ( c ( t ) , k ( t )) dt ≥ M for some M .If ¯ V (0) = −∞ , then k ( t ) > t ≥
0. However, ¯ V (0) may be greaterthan −∞ even when u (0 , k ) = −∞ . Therefore, we cannot prohibit k ( t ) = 0even in this case. 23 .3 Comparison with Related Literatures Usually, the HJB equation is written as a second-order degenerate elliptic dif-ferential equation. In fact, in stochastic economic dynamic models, the HJBequation becomes a second-order differential equation. Malliaris and Brock(1988), for example, deal with this kind of equation. This is probably be-cause of the use of Ito’s formula in the middle of the derivation. On the otherhand, since there is no stochastic variation in the dynamic model treated inthis paper, the HJB equation is just a first-order differential equation.The case in which the HJB equation does not have a classical solution hasbeen pointed out in many studies. Therefore, since Lions (1982), it has beencommon to use the viscosity solution as a solution concept in the study ofdegenerate elliptic differential equations. In this paper, however, we showedthat the HJB equation has a classical solution under assumptions that arenot so strong in macroeconomic dynamics. It is unclear whether this is alsotrue in other fields, such as models of search theory.As discussed in the introduction, the HJB equation has mostly been stud-ied in variational problems with finite time-interval. However, in economics,finite time-interval models are usually undesirable. Bardi and Dolcetta (2008)are one of the few studies that rigorously treat the HJB equation in varia-tional problems with infinite time-interval, where the instantaneous utilityfunction is assumed to be bounded. This assumption is by far the easiestto use, and it dramatically simplifies the proof of our Proposition 3, for ex-ample. However, the most widely used instantaneous utility functions inmacroeconomic dynamic models are the class of CRRA functions that in-clude the logarithmic function. Any assumption that excludes this class, nomatter how useful it may be, cannot be placed. This is the main reasonwhy the proofs in this paper is by far the most difficult. To the best of ourknowledge, the results in this paper are the first theoretical foundations ofthe HJB equation in macroeconomic dynamics that can be applied to a classof instantaneous utility functions, including the logarithmic function.
In this paper, we presented an example of economic dynamics in which theHJB equation is neither necessary nor sufficient for a function to be thevalue function. Because this problem is serious, we present a mathematicalfoundation of the HJB equation in economic dynamics, and showed thatunder Assumptions 1-7, the HJB equation gives a perfect characterization ofthe value function in some class of functions. Moreover, we presented a new24ethod for obtaining the solution of the model using the value function.There are several future tasks. First, in this paper, we assumed theexistence of the global Lipschitz constant d > F with respect to c . However, we may be able to omit this requirement usingthe additional result for the existence of the solution of differential equations.Second, we want to extend our result to some multidimensional modelsand stochastic models.Third, we want to obtain a simple method for gaining an approximatesolution of the value function. In discrete-time models, there is a famousapproximation method that uses Blackwell’s inequality and the contractionmapping theorem. We want to obtain a counterpart to this result in acontinuous-time model.Fourth, we need Assumptions 4 and 6 for solving the problem. However,these restrictions are a little strong and any sort of CES function cannot betreated. Hence, we want to relax these constraints. In this section, we frequently use knowledge on ODEs. Hence, we note basicknowledge on ODE for readers.First, consider the following differential equation:˙ x ( t ) = h ( t, x ( t )) , (15)where ˙ x denotes dxdt . We assume that h : U → R n , U ⊂ R + × R n , and therelative interior of U in R + × R n is nonempty (denoted by V ). We call a set I ⊂ R an interval if and only if I is a convex set of R that includes at leasttwo points. We say that a function x : I → R n is a solution of (15) if andonly if, 1) I is an interval, 2) x ( t ) is absolutely continuous on every compactsubinterval of I , and 3) ˙ x ( t ) = h ( t, x ( t )) for almost all t ∈ I . Suppose that( t ∗ , x ∗ ) ∈ U . If a solution x ( t ) of (15) satisfies 1) t ∗ ∈ I and 2) x ( t ∗ ) = x ∗ ,then x ( t ) is called a solution with initial value condition x ( t ∗ ) = x ∗ , orsimply, a solution of the following differential equation:˙ x ( t ) = h ( t, x ( t )) , x ( t ∗ ) = x ∗ . (16)Now, suppose that h : U → R n satisfies the following requirements: 1) forevery t ∈ R , x h ( t, x ) is continuous, and 2) for every x ∈ R n , t h ( t, x )25s measurable. Then, we say that h satisfies Carath´eodory’s condition .If, in addition, for a set C ⊂ U , there exists L > k h ( t, x ) − h ( t, x ) k ≤ L k x − x k for every ( t, x , x ) with ( t, x ) , ( t, x ) ∈ C , then h is called Lipschitz in x on C .The following facts are well known.1) Suppose that h satisfies Carath´eodory’s condition. Moreover, supposethat ( t ∗ , x ∗ ) ∈ V and there exist ε > r : [ t ∗ − ε, t ∗ + ε ] → R + such that k h ( t, x ) k ≤ r ( t ) for all ( t, x ) ∈ U with k ( t, x ) − ( t ∗ , x ∗ ) k ≤ ε . Then, there exists a solution x : I → R n of (16),where I is relatively open in R + .
2) Suppose that h satisfies Carath´eodory’s condition, and for every compactset C ⊂ V , h is Lipschitz in x on C . Moreover, suppose that ( t ∗ , x ∗ ) ∈ V and there exists a convex neighborhood of t ∗ such that t h ( t, x ∗ ) isintegrable on this neighborhood. Then, there exists a solution x : I → R n of (16), where I is relatively open in R + .3) Suppose that h satisfies Carath´eodory’s condition, and for every compactset C ⊂ V , h is Lipschitz in x on C . Moreover, suppose that for every( t + , x + ) ∈ V , there exists a convex neighborhood of t + such that t h ( t, x + ) is integrable on this neighborhood. Choose any ( t ∗ , x ∗ ) ∈ V .Suppose that x ( t ) , x ( t ) are two solutions of (16) such that ( t, x i ( t )) ∈ V for every t ∈ I i , where I i is the domain of x i ( t ). Then, x ( t ) = x ( t ) forevery t ∈ I ∩ I .
4) Suppose that h is continuous. Then, any solution x ( t ) of (15) is continu-ously differentiable. Next, suppose that h satisfies all requirements in 3) and ( t ∗ , x ∗ ) ∈ V .Choose a solution x : I → R n of (16). A solution y : J → R n is called extension of x if and only if 1) I ⊂ J , and 2) y ( t ) = x ( t ) for all t ∈ I . Then, x ( t ) is called nonextendable if and only if there is no extension except x ( t )itself.Then, the following facts are well known. This result is called Carath´eodory-Peano’s existence theorem. For the proof, see ch.2of Coddington and Levinson (1984). Results 2) and 3) are known as Carath´eodory-Picard-Lindel¨of’s existence theorem.For the proof, see Section 0.4 of Ioffe and Tikhomirov (1979). See ch.2 of Hartman (1997). Fact 5) can be proved easily. The proof of 6) is in ch.2 of Coddington and Levinson(1984).
26) In addition to the requirements of 3), suppose that V = U . Then, thereuniquely exists a nonextendable solution x ( t ) of (16). Moreover, the do-main I of x ( t ) is relatively open in R + .6) Suppose that all requirements of 5) hold, and let x : I → R n be thenonextendable solution of (16). Choose any compact set C ⊂ V . Then,there exists t + such that if t + ≤ t ∈ I , then ( t, x ( t )) / ∈ C .Finally, suppose that h ( t, x ) = a ( t ) x + b ( t ), where a ( t ) , b ( t ) are locally inte-grable functions defined on R + . Then, the solution of (15) is determined bythe following formula: x ( t ) = e R t a ( τ ) dτ (cid:20) x (0) + Z t e − R s a ( τ ) dτ b ( s ) ds (cid:21) . (17)This is called the formula of the solution of linear ODEs. Let g ( c, k, p ) = F ( k, c ) p + u ( c, k ) . Then, g is strictly concave with respect to c . Fix p, k >
0. Choose ε > p − ε > k − ε >
0. Define ε ∗ = min k ′ ∈ [ k − ε,k + ε ] [ F ( k, − F ( k, . Then, for every k ′ ∈ [ k − ε, k + ε ] , q ∈ [ p − ε, p + ε ] and c ≥ D c, − g ( c, k ′ , q ) ≤ max k ′′ ∈ [ k − ε,k + ε ] ∂u∂c ( c, k ′′ ) − ( p − ε ) ε ∗ . (18)Because ∂u∂c ( c, k ′′ ) → c → ∞ , we have that there exists ¯ c > c ≥ ¯ c . Then, for such k ′ and q , we have that g ( c, k ′ , q ) < g (¯ c, k ′ , q ) for all c > ¯ c , and thus,sup c ≥ g ( c, k ′ , q ) = max ≤ c ≤ ¯ c g ( c, k ′ , q ) . The rest of the proof on the continuity of the function c ∗ at ( p, k ) is a simpleapplication of Berge’s theorem. The positivity of c ∗ ( p, k ) follows from theassumption lim c → ∂u∂c ( c, k ) = + ∞ . (cid:4) Use Dini’s theorem. .3 Lemmas and Their Proofs We should prove Lemma 1. However, to derive Lemma 1, we must introducetwo more results on differential equations. Because these are also useful forlater arguments, we provide these results as lemmas.
Lemma 2 . Consider the following two ODEs:˙ k ( t ) = h i ( t, k ) , k (0) = ¯ k i , (19)where i ∈ { , } . Suppose that each h i is a real valued function defined onsome convex neighborhood U ⊂ R + × R of (0 , ¯ k i ) and h i satisfies Carath´eodory’scondition. Then, the following results hold.i) If there exists a locally integrable function r ( t ) such thatsup k :( t,k ) ∈ U | h i ( t, k ) | ≤ r ( t ) , then there exists T > k i :[0 , T ] → R . Moreover, if h i ( t, k ) is continuous, then k i ( t ) is continuouslydifferentiable.ii) Suppose that ¯ k ≤ ¯ k , h ( t, k ) ≤ h ( t, k ) for every ( t, k ) ∈ U , and forsome i ∗ ∈ { , } , there exists L > t, k ) , ( t, k ) ∈ U , then | h i ∗ ( t, k ) − h i ∗ ( t, k ) | ≤ L | k − k | . Suppose also that h i ∗ ( t, ¯ k ) is locally integrable, and there exist solutions k i : [0 , T ] → R of the above equations for i ∈ { , } . Then, k ( t ) ≤ k ( t )for all t ∈ [0 , T ]. Proof . Assertion i) is just a corollary of Carath´eodory-Peano’s existencetheorem, and thus we omit the proof.Suppose that assertion ii) does not hold. Then, there exists t ∗ > k ( t ∗ ) > k ( t ∗ ). Let t + = inf { t ∈ [0 , t ∗ ] | k ( s ) > k ( s ) for all s ∈ [ t, t ∗ ] } .Because k (0) = ¯ k ≤ ¯ k = k (0), we have k ( t + ) = k ( t + ). We treat onlythe case i ∗ = 2, because the case i ∗ = 1 can be treated symmetrically. Define k ( t ) = k ( t + ) + Z tt + h ( s, k ( s )) ds. If such an
L > k = ¯ k = 0 , h ( t, k ) = p | k | − t , h ( t, k ) = p | k | , k ( t ) = t , and k ( t ) ≡ k = ¯ k and h = h , then this claim immediately implies the uniqueness ofthe solution. For this view, this lemma is an extension of Carath´eodory-Picard-Lindel¨of’suniqueness result in theory of ODEs. h satisfies Carath´eodory’s condition, the mapping t h ( t, k ( t ))is measurable, and | h ( t, k ( t )) | ≤ | h ( t, k ( t )) − h ( t, ¯ k ) | + | h ( t, ¯ k ) | ≤ L | k ( t ) − ¯ k | + | h ( t, ¯ k ) | , and thus h ( t, k ( t )) is locally integrable, which implies that k ( t ) is definedon [ t + , t ∗ ]. Moreover, k ( t ) ≥ k ( t + ) + Z tt + h ( s, k ( s )) ds = k ( t ) ≥ k ( t )for every t ∈ [ t + , t ∗ ]. Meanwhile, k ( t ) − k ( t ) = Z tt + [ h ( s, k ( s )) − h ( s, k ( s ))] ds ≤ Z tt + L [ k ( s ) − k ( s )] ds ≤ L ( t − t + ) max s ∈ [ t + ,t ] ( k ( s ) − k ( s )) . Fix some t ∈ [ t + , t ∗ ] with 0 < t − t + < L − , and let s ∗ ∈ arg max { k ( s ) − k ( s ) | s ∈ [ t + , t ] } . Because k ( t + ) = k ( t + ) and k ( s ) > k ( s ) for s ∈ ] t + , t ∗ ],we have s ∗ > t + . Then, k ( s ∗ ) − k ( s ∗ ) ≤ L ( s ∗ − t + )( k ( s ∗ ) − k ( s ∗ )) < k ( s ∗ ) − k ( s ∗ ) , and thus k ( s ∗ ) < k ( s ∗ ), which is a contradiction. (cid:4) Lemma 3 . Consider the following ODE:˙ k ( t ) = h ( t, k ( t )) , k (0) = ¯ k > , (20)where h : R + × R ++ → R satisfies Carath´eodory’s condition, and for every( t + , k + ) ∈ R + × R ++ , there exists ε > r ( t ) defined on a neighborhood of t + such that if | k − k + | ≤ ε , then | h ( t, k ) | < r ( t ) for every t such that r ( t ) is defined. For either X = [0 , ˜ T ]or X = R + , suppose that there exist two positive continuous functions ˆ k ( t )and ¯ k ( t ) defined on X such that ¯ k ( t ) ≤ k ( t ) ≤ ˆ k ( t ) for every solution k ( t )defined on [0 , T ] and t ∈ [0 , T ] ∩ X . Then, this equation (20) has a solutiondefined on X itself. Proof . We treat only the case X = R + . The proof in the case X = [0 , ˜ T ] isalmost the same. See Section 8.1 of Ioffe and Tikhomirov (1979).
29y Lemma 2, we have that there exists
T > k ( t ) defined on [0 , T ]. By assumption, ¯ k ( t ) ≤ k ( t ) ≤ ˆ k ( t ) forall t ∈ [0 , T ].Next, let Y be the set of all solutions of equation (20) defined on either R + or [0 , T ′ ] for some T ′ >
0, and let k ( · ) (cid:23) k ( · ) if the domain of k ( · )includes that of k ( · ) and k ( t ) = k ( t ) when both are defined at t . Clearly, (cid:23) is a partial order on Y . For k ( · ) ∈ Y , let I k ( · ) be the domain of k ( · ). Chooseany chain C ⊂ Y of (cid:23) . If sup ∪ k ( · ) ∈ C I k ( · ) = + ∞ , then we can define k + ( t ) = k ( t ) if t ∈ I k ( · ) , and k + ( · ) is an upper bound of C . Otherwise, let T ∗ = sup ∪ k ( · ) ∈ C I k ( · ) .Define k + ( t ) = k ( t ) if t ∈ I k ( · ) . Then, k + ( t ) is a solution of (20) defined on [0 , T ∗ [. By the continuity ofˆ k ( t ) , ¯ k ( t ), there exist ε > δ > k ( T ∗ ) > ε and if 0
0. Suppose thatthe solution k ( t ) of the above equation satisfies k ( T ) < min { ¯ k, k } for some T >
0, and define t ∗ = inf { t ∈ [0 , T ] |∀ t ′ ∈ [ t, T ] , k ( t ′ ) < min { ¯ k, k }} . Because k (0) = ¯ k ≥ min { ¯ k, k } , we have k ( t ∗ ) = min { ¯ k, k } . Because F is concave and F (0 ,
0) = 0, we have F ( k ( t ∗ ) , >
0. Thus, there exists t ∈ ] t ∗ , T [ such that k ( t ) > min { ¯ k, k } , which is a contradiction. Therefore, k ( t ) ≥ min { ¯ k, k } for every solution k ( t ) of the above equation. Meanwhile, let p ∈ ∂ k F ( k, F ( k ′ , ≤ p ( k ′ − k ) + F ( k, k ( t ) ≤ ( e pt (cid:16) ¯ k + ( e − pt − pk − F ( k, p (cid:17) if p = 0 , ¯ k + tF ( k,
0) if p = 0 , (21)for every solution k ( t ) of the above equation. Therefore, we can apply Lemma3, and thus k + ( t, ¯ k ) is well-defined on R + × R ++ , and inf t ≥ k + ( t, ¯ k ) ≥ min { ¯ k, k } > k >
0. This completes the proof of Lemma 1. (cid:4)
Note that, if h ( t, k ) = F ( k, g ( t, k )) for some non-negative function g ( t, k ),then we can use k + ( t, ¯ k ) for ˆ k ( t ) in Lemma 3. Note also that, if g ( t, k ) = c ( t )for some locally integrable function c : R + → R + , then for all k ∈ [0 , ¯ k + 1], | h ( t, k ) | ≤ d (¯ k + 1) + d c ( t ) + max { F ( k ′ , | ≤ k ′ ≤ ¯ k + 1 } ≡ r ( t ) , (22)and because r ( t ) is locally integrable, the requirement of Lemma 3 is auto-matically satisfied. We will use these facts frequently.Later, we often need the boundedness of c ( t ) in the proof. Therefore, wepresent two more lemmas. One might think that if Lemma 5 is correct, then we can relax the existence assumptionof d ≥ d ( c ) such that d (0) = 0 and F ( k, c ) > − d k − d ( c ). If this conjecture is correct, we can treat technologyfunctions such as F ( k, c ) = √ k − c . However, we should use the above r ( t ) in the proof ofLemma 5. Because d ( c ( t )) may be not locally integrable even if c ( t ) is locally integrableand d ( c ) is continuous, our proof is broken, and thus this conjecture is not true. emma 4 . Suppose that Assumption 3 holds, and choose any nonnegativeand locally integrable function c ( t ) such that sup t ≥ c ( t ) < + ∞ . Define h ( t, k ) = F ( k, c ( t )) . Then, h satisfies Carath´eodory’s condition on R and for every compact set C ⊂ R + × R ++ , it is Lipschitz in k on C . Proof . The fact that h satisfies Carath´eodory’s condition can easily beshown. Choose any compact set C ⊂ R + × R ++ . Then, there exist T > k, ˆ k > C ⊂ [0 , T ] × [˜ k, ˆ k ]. Define c ∗ = sup t ≥ c ( t ) and˜ F ( k, c ) = ( F ( k, c ) if c ≥ ,F ( k,
0) if c < . Then, we have that ˜ F is defined on R ++ × R . We can easily check that ˜ F isconcave. By Theorem 24.7 of Rockafeller (1996), we have that the set X = { ( p, q ) | ( p, q ) ∈ ∂ ˜ F ( k, c ) for ( k, c ) ∈ [˜ k, ˆ k ] × [0 , c ∗ ] } is compact. Thus, for t ≥ k , k ∈ [˜ k, ˆ k ], | h ( t, k ) − h ( t, k ) | = | ˜ F ( k , c ( t )) − ˜ F ( k , c ( t )) |≤ max {| p || ( p, q ) ∈ X } × | k − k | , as desired. This completes the proof. (cid:4) Lemma 5 . Suppose that Assumptions 1-3 hold. Choose any ¯ k > k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k such that R ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt > −∞ , and define c n ( t ) = min { c ∗ ( t ) , n } for n ∈ N . Then, there exists a function k n ( t ) de-fined on R + such that k n ( t ) ≥ k ∗ ( t ) for all t , and if u ( n, > −∞ , then( k n ( t ) , c n ( t )) ∈ A ¯ k and R ∞ e − ρt u ( c n ( t ) , k n ( t )) dt > −∞ . Proof . First, define h ( t, k ) = F ( k, c ∗ ( t )) and h n ( t, k ) = F ( k, c n ( t )). Then, h n ( t, k ) ≥ h ( t, k ) for all t ≥
0. Consider the following differential equation:˙ k ( t ) = h n ( t, k ( t )) , k (0) = ¯ k. We show that there exists a solution k n ( t ) of the above equation defined on R + such that k n ( t ) ≥ k ∗ ( t ) for every t ≥ Note that [˜ k, ˆ k ] × [0 , c ] is in the relative interior of the domain of ˜ F , and thus Theorem24.7 of Rockafeller (1996) is applicable for ˜ F . Y be the set of all solutions k ( · ) of the above equation such that thedomain I k ( · ) is either R + or [0 , T ] for some T >
0, and k ( t ) ≥ k ∗ ( t ) for all t ∈ I k ( · ) . By Lemma 4, we have that h n satisfies all requirements in Lemma2, and thus Y is nonempty.Define a partial order (cid:23) on Y such that k ( · ) (cid:23) k ( · ) if and only if I k ( · ) ⊂ I k ( · ) and k ( t ) = k ( t ) for every t ∈ I k ( · ) . Let C be any chain of (cid:23) . Ifsup ∪ k ( · ) ∈ C I k ( · ) = + ∞ , we can define k + ( t ) = k ( t ) for t ∈ I k ( · ) , and k + ( · ) is anupper bound of C . If sup ∪ k ( · ) ∈ C I k ( · ) = T < + ∞ , we can define k + ( t ) = k ( t )for t ∈ I k ( · ) , and k + ( t ) is defined on [0 , T [. Because k + ( t ) ≤ k + ( t, ¯ k ), wehave that k + ( T ) = lim sup t → T k + ( t ) ∈ [0 , k + ( T, ¯ k )] can be defined. Let r ( t )be defined as in (22), where we use max t ∈ [0 ,T ] k + ( t, ¯ k ) instead of ¯ k and c ∗ ( t )instead of c ( t ). Then, we have that | k + ( t ) − k + ( T ) | ≤ R Tt r ( s ) ds , and thus k + ( · ) is continuous. Because k + ( t ) = ¯ k + Z t h n ( s, k + ( s )) dt for every t ∈ [0 , T ], we have that k + ( · ) is a solution of the above equation.Because both k + ( · ) , k ∗ ( · ) are continuous, we have that k + ( t ) ≥ k ∗ ( t ) forevery t ∈ [0 , T ], and thus k + ( · ) is an upper bound of C . Therefore, byZorn’s lemma, there exists a maximal element ˜ k ( · ). Suppose that I ˜ k ( · ) =[0 , T ] = R + . If ˜ k ( T ) >
0, then by using Lemma 2, we can easily showthat ˜ k ( · ) is not maximal, which is a contradiction. Therefore, we have that˜ k ( T ) = k ∗ ( T ) = 0. Choose any k ∗ > k F ( k, c ) is increasingon [0 , k ∗ ] for all c ∈ [0 , n ]. Define r ( t ) as in (22), where we use k ∗ instead of¯ k and c ∗ ( t ) instead of c ( t ). Choose ε > k ∗ ( t ) ≤ k ∗ whenever t ∈ [ T, T + ε ], and Z T + εT r ( t ) dt < k ∗ . Let I = [ T, T + ε ] and X be the space of all continuous functions x : I → [0 , k ∗ ]such that x ( T ) = 0, x ( t ) ≥ k ∗ ( t ) for all t ∈ I , and for every t , t ∈ I with t < t , | x ( t ) − x ( t ) | ≤ R t t r ( t ) dt . Because k ∗ ( · ) satisfies all requirements, wehave that X is nonempty. Clearly, X is convex. By Ascoli-Arzela’s theorem, X is compact with respect to sup norm. Define P ( x ( · ))( t ) = Z tT h n ( s, x ( s )) ds. That is, r ( t ) = d ( max t ∈ [0 ,T ] k + ( t, ¯ k ) + 1) + d c ∗ ( t ) + max { F ( k, | ≤ k ≤ max t ∈ [0 ,T ] k + ( t, ¯ k ) + 1 } . Such a construction of r ( t ) will be frequently used later. x ( · ) ∈ X and y ( · ) = P ( x ( · )), then y ( T ) = 0 ,y ( t ) ≥ Z tT h n ( s, k ∗ ( s )) ds ≥ Z tT h ( s, k ∗ ( s )) ds = k ∗ ( t ) , and if t , t ∈ I and t < t , then | y ( t ) − y ( t ) | ≤ Z t t | h n ( t, x ( t )) | dt ≤ Z t t r ( t ) dt. Therefore, we have that y ( t ) ≤ Z T + εT r ( t ) dt < k ∗ , and thus y ( · ) ∈ X , which implies that P is a mapping from X into X . Let( x m ( · )) be the convergent sequence of X on the sup norm and x ( · ) is thelimit. Define y m ( · ) = P ( x m ( · )) and y ( · ) = P ( x ( · )). Choose any ε ′ > A m = (cid:26) t ∈ I (cid:12)(cid:12)(cid:12)(cid:12) | h n ( t, x m ( t )) − h n ( t, x ( t )) | ≥ ε ′ ε (cid:27) ,B m = ∪ ℓ ≥ m A ℓ . Because F is continuous, we have that B m ↓ ∅ , and thuslim m →∞ Z B m r ( t ) dt = 0 . Hence, there exists m ∈ N such that if m ≥ m , then Z B m r ( t ) dt < ε ′ . Then, for every t ∈ I , if m ≥ m , | y m ( t ) − y ( t ) | ≤ Z T + εT | h n ( τ, x m ( τ )) − h n ( τ, x ( τ )) | dτ = Z I \ B m | h n ( τ, x m ( τ )) − h n ( τ, x ( τ )) | dτ + Z B m | h n ( τ, x m ( τ )) − h n ( τ, x ( τ )) | dτ ≤ Z I \ B m ε ′ ε dτ + Z B m r ( τ ) dτ< ε ′ ε ′ ε ′ , P is continuous with respect to the sup norm. Therefore,by Schauder’s fixed point theorem, there exists a fixed point x ∗ ( · ) ∈ X of P .Define ˆ k ( t ) = ( ˜ k ( t ) if 0 ≤ t ≤ T,x ∗ ( t ) if t ∈ I. Then, ˆ k ( · ) ∈ Y and ˆ k ( · ) ≻ ˜ k ( · ), which contradicts the maximality of ˜ k ( · ).Therefore, we have that the domain of ˜ k ( · ) is R + , and thus we can define k n ( t ) = ˜ k ( t ).Now, suppose that u ( n, > −∞ . Because R ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt > −∞ , we have Z { t | c ∗ ( t ) ≤ n } e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt > −∞ , and thus, −∞ < Z { t | c ∗ ( t ) ≤ n } e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt + Z { t | c ∗ ( t ) >n } e − ρt u ( n, k ∗ ( t )) dt = Z ∞ e − ρt u ( c n ( t ) , k ∗ ( t )) dt ≤ Z ∞ e − ρt u ( c n ( t ) , k n ( t )) dt, and thus R ∞ e − ρt u ( c n ( t ) , k n ( t )) dt is defined and is not −∞ , which impliesthat ( k n ( t ) , c n ( t )) ∈ A ¯ k . This completes the proof. (cid:4) Note that, if ( k ∗ ( t ) , c ∗ ( t )) , ( k n ( t ) , c n ( t )) ∈ A ¯ k are as in Lemma 5, thenwe have that R ∞ e − ρt u ( c n ( t ) , k ∗ ( t )) dt > −∞ . By the monotone convergencetheorem, we have thatlim inf n →∞ Z ∞ e − ρt u ( c n ( t ) , k n ( t )) dt ≥ Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt. In particular, if there exists ( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k such that R ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt >M , then we can assume that c ∗ ( t ) is bounded. We separate the proof into three steps.
Step 1 . ¯ V (¯ k ) > −∞ for every ¯ k > Proof of step 1 . We have assumed that F (0 ,
0) = 0 and there exists k ∗ > F ( k ∗ , >
0. If ¯ k > F (¯ k, >
0, then there exists ¯ c > F (¯ k, ¯ c ) = 0 because F is continuous, concave, and decreasing in c . Hence, ( k ( t ) , c ( t )) ≡ (¯ k, ¯ c ) ∈ A ¯ k , and thus ¯ V (¯ k ) > −∞ . If ¯ k > F (¯ k, ≤
0, then by the concavity of F , we have ¯ k > k ∗ . Choose any c ∗ > F ( k ∗ , c ∗ ) >
0, and k ( t ) is a solution of the differential equation˙ k ( t ) = F ( k ( t ) , c ∗ ) , k (0) = ¯ k, defined on [0 , T ]. Suppose that k ( t + ) < k ∗ for some t + >
0. Let t ∗ =inf { t ∈ [0 , t + ] |∀ s ∈ [ t, t + ] , k ( s ) < k ∗ } . Because k (0) = ¯ k > k ∗ , we have that k ( t ∗ ) = k ∗ , and thus F ( k ( t ∗ ) , c ∗ ) >
0. Therefore, there exists t ∈ ] t ∗ , t + ] suchthat k ( t ) > k ∗ , which is a contradiction. Hence, we have that k ( t ) ≥ k ∗ forall t ∈ [0 , T ], and thus by Lemma 3, there exists a solution k ∗ ( t ) of the aboveequation defined on R + . Let c ∗ ( t ) ≡ c ∗ . Then, ( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k , and thus¯ V (¯ k ) > −∞ . This completes the proof. (cid:4) Step 2 . ¯ V is nondecreasing. Proof of Step 2 . Choose any k , k ≥ k < k . If k >
0, then wehave proved in the proof of step 1 that there exists a pair ( k ( t ) , c ( t )) ∈ A k .If k = 0, then ( k ( t ) , c ( t )) ≡ (0 , ∈ A k . Therefore, A k is nonempty forevery k ≥
0. If ¯ V ( k ) = −∞ , then clearly ¯ V ( k ) ≥ ¯ V ( k ). Suppose that¯ V ( k ) > −∞ . Let ( k ( t ) , c ( t )) ∈ A k and Z ∞ e − ρt u ( c ( t ) , k ( t )) dt > min { ¯ V ( k ) − n − , n } . Because of Lemma 5, we can assume that c ( t ) is bounded. Define h ( t, k ) = F ( k, c ( t ))for all ( t, k ) ∈ R + × R ++ , and consider the following two ODEs: ˙ κ ( t ) = h ( t, κ ( t )) , κ (0) = k , (23)˙ κ ( t ) = F ( κ ( t ) , c ( t )) , κ (0) = k . (24)Suppose that there does not exist a solution κ : R + → R + of (24) such that κ ( t ) ≥ k ( t ) for every t ∈ R + . Because all assumptions in 5) of Subsection7.1 hold for equation (23), there exists the unique nonextendable solution By this assumption, we have that lim c →∞ F (¯ k, c ) = −∞ . Note that, equation (24) is different from equation (23), because the right-hand sideof (23) is defined only on R + × R ++ , while that of (24) is defined on R . The reason whywe need (23) is as follows: the equation (24) does not satisfy the requirement U = V in5) of subsection 7.1, and thus 5) cannot be directly applied. : I → R of (23). By our assumption, we have that sup { t ∈ I | κ ( t ) ≥ k ( t ) } = T < + ∞ . If [0 , T ] ⊂ I , then we have that κ ( T ) = k ( T ), and thus ifwe define ˜ k ( t ) = ( κ ( t ) if 0 ≤ t < T,k ( t ) otherwise , then ˜ k ( t ) is a solution of (24) defined on R + such that ˜ k ( t ) ≥ k ( t ) for all t ∈ R + , which is a contradiction. Therefore, we must have that I = [0 , T [.Because κ is bounded from above by the argument in subsection 7.3, by 6) insubsection 7.1, for every n ∈ N , there exists t n > t n < t < T ,then κ ( t ) < n − . Therefore, we have that lim sup t → T κ ( t ) = 0, and thuswe can define κ ( T ) = 0. Because κ ( t ) ≥ k ( t ) for all t ∈ I , we have that k ( T ) = 0, and thus we can define ˜ k ( t ) as above, which leads a contradiction.Therefore, there exists a solution κ ( t ) of (24) defined on R + such that κ ( t ) ≥ k ( t ) for every t ≥
0, which implies that ( κ ( t ) , c ( t )) ∈ A k . Hence,¯ V ( k ) ≥ Z ∞ e − ρt u ( c ( t ) , k ( t )) dt, and thus, ¯ V ( k ) ≥ ¯ V ( k ) , as desired. This completes the proof of step 2. (cid:4) Step 3 . ¯ V is concave. Proof of step 3 . Because ¯ V is nondecreasing, it suffices to show that¯ V ((1 − s ) k + sk ) ≥ (1 − s ) ¯ V ( k ) + s ¯ V ( k ) for every k , k > s ∈ ]0 , k , k > s ∈ ]0 , k ni ( t ) , c ni ( t )) ∈ A k i for i ∈ { , } suchthat Z ∞ e − ρt u ( c ni ( t ) , k ni ( t )) dt > min { ¯ V ( k i ) − n − , n } . By Lemma 5, we can assume that c ni ( t ) is bounded for i ∈ { , } . Define c ns ( t ) = (1 − s ) c n ( t ) + sc n ( t ) , and consider the following differential equation:˙ k ( t ) = F ( k ( t ) , c ns ( t )) , k (0) = (1 − s ) k + sk . We will show that there exists a solution k ns ( t ) of this equation defined on R + such that k ns ( t ) ≥ (1 − s ) k ( t ) + sk ( t ) for every t ≥ k ( t ) of the above equation defined on I suchthat k ( t ) > t ∈ I . Because F ( k, c ns ( t )) ≥ (1 − s ) F ( k, c n ( t )) + sF ( k, c n ( t )) , by Lemma 2, we have that k ( t ) ≥ (1 − s ) k n ( t ) + sk n ( t ) for all t ∈ I . Let Y be the set of all solutions k ( · ) of the above equation defined on I k ( · ) such that k ( t ) ≥ (1 − s ) k n ( t ) + sk n ( t ) for all t ∈ I k ( · ) , where either I k ( · ) = [0 , T ] for some T > I k ( · ) = R + . By the above argument, we have that Y is nonempty.Define a partial order (cid:23) on Y such that ˆ k ( · ) (cid:23) ˜ k ( · ) if and only if I ˜ k ( · ) ⊂ I ˆ k ( · ) and ˆ k ( t ) = ˜ k ( t ) for all t ∈ I ˜ k ( · ) . Let C be a chain in Y . If ∪ k ( · ) ∈ C I k ( · ) = R + ,then we can define k ∗ ( t ) = k ( t ) for t ∈ I k ( · ) , and k ∗ ( · ) is an upper bound of C .If sup ∪ k ( · ) ∈ C I k ( · ) = T < + ∞ , define k ∗ ( t ) = k ( t ) for t ∈ I k ( · ) . Then, k ∗ ( · ) isa solution of the above equation defined on [0 , T [. Because k ∗ ( t ) ≤ k + ( t, ¯ k ),we can define k ∗ ( T ) = lim sup t → T k ∗ ( t ) ∈ [0 , k + ( T, ¯ k )]. Define r ( t ) as in (22),where we use max t ∈ [0 ,T ] k + ( t, ¯ k ) instead of ¯ k and c ns ( t ) instead of c ( t ). Then, | k ∗ ( t ) − k ∗ ( T ) | ≤ R Tt r ( τ ) dτ , and thus k ∗ ( · ) is continuous, and because k ∗ ( t ) = ¯ k + Z t F ( k ∗ ( τ ) , c ns ( τ )) dτ, and (1 − s ) k n ( t ) + sk n ( t ) is continuous, we have that k ∗ ( · ) ∈ Y , and thus k ∗ ( · ) is an upper bound of C . Therefore, by Zorn’s lemma, we have thatthere is a maximal element k ∗ ( · ) ∈ Y . Suppose that I k ∗ ( · ) = [0 , T ] = R + . If k ∗ ( T ) >
0, then by Lemma 2, there exists ˜ k ( · ) ∈ Y such that ˜ k ( · ) ≻ k ∗ ( · ),which contradicts the maximality of k ∗ ( · ). Hence, we have that k ∗ ( T ) = k n ( T ) = k n ( T ) = 0. Let c ∗ = max { sup t ≥ c n ( t ) , sup t ≥ c n ( t ) } , and chooseany k ∗ > k F ( k, c ) is increasing on [0 , k ∗ ] for all c ∈ [0 , c ∗ ].Define r ( t ) as in (22), where we use k ∗ instead of ¯ k and c ns ( t ) instead of c ( t ).Choose a sufficiently small ε > i ∈{ , } ,t ∈ [ T,T + ε ] k ni ( t ) ≤ k ∗ , Z T + εT r ( t ) dt ≤ k ∗ , and let I = [ T, T + ε ] and X be the set of all continuous functions x : I → [0 , k ∗ ] such that x ( T ) = 0, x ( t ) ≥ (1 − s ) k n ( t ) + sk n ( t ), and for all t , t ∈ I with t < t , | x ( t ) − x ( t ) | ≤ R t t r ( t ) dt . Because (1 − s ) k n ( · ) + sk n ( · ) ∈ X ,we have that X is nonempty. The convexity of X is clear. By Ascoli-Arzela’stheorem, it is compact with respect to the sup norm. Define P ( x ( · ))( t ) = Z tT F ( x ( τ ) , c ns ( τ )) dτ t ∈ I . If x ( · ) ∈ X and y ( · ) = P ( x ( · )), then y ( T ) = 0 and y ( t ) ≥ Z tT F ((1 − s ) k n ( τ ) + sk n ( τ ) , c ns ( τ )) dτ ≥ Z tT [(1 − s ) F ( k n ( τ ) , c n ( τ )) + sF ( k n ( τ ) , c n ( τ ))] dτ = (1 − s ) k n ( t ) + sk n ( t )and y ( t ) ≤ Z tT r ( s ) ds ≤ k ∗ for all t ∈ I . If T ≤ t < t ≤ T + ε , then | y ( t ) − y ( t ) | ≤ Z t t | F ( x ( t ) , c ns ( t )) | dt ≤ Z t t r ( t ) dt. Therefore, we have that y ( · ) ∈ X , and thus P is a mapping from X into X . Choose any convergent sequence ( x m ( · )) on X and let x ( · ) be the limit.Define y m ( · ) = P ( x m ( · )) and y ( · ) = P ( x ( · )). Fix any ε ′ >
0, and define A m = (cid:26) t ∈ I (cid:12)(cid:12)(cid:12)(cid:12) | F ( x m ( t ) , c ns ( t )) − F ( x ( t ) , c ns ( t )) | ≥ ε ′ ε (cid:27) ,B m = ∪ ℓ ≥ m A ℓ . Because F is continuous, we have that B m ↓ ∅ . Therefore,lim m →∞ Z B m r ( t ) dt = 0 . Hence, there exists m ∈ N such that if m ≥ m , then Z B m r ( t ) dt < ε ′ . t ∈ I , if m ≥ m , then | y m ( t ) − y ( t ) | ≤ Z T + εT | F ( x m ( τ ) , c ns ( τ )) − F ( x ( τ ) , c ns ( τ )) | dτ ≤ Z I \ B m | F ( x m ( τ ) , c ns ( τ )) − F ( x ( τ ) , c ns ( τ )) | dτ + Z B m | F ( x m ( τ ) , c ns ( τ )) − F ( x ( τ ) , c ns ( τ )) | dτ ≤ Z I \ B m ε ′ ε dt + Z B m r ( τ ) dτ< ε ′ ε ′ ε ′ . Hence, P is continuous with respect to the sup norm. By Schauder’s fixedpoint theorem, we have that P has a fixed point x ∗ : I → R + . Define˜ k ( t ) = ( k ∗ ( t ) if 0 ≤ t ≤ T,x ∗ ( t ) if T ≤ t ≤ T + ε. Then, ˜ k ( · ) ∈ Y is an extension of k ∗ ( · ), which contradicts the maximalityof k ∗ ( · ). Hence, the domain of k ∗ ( · ) is R + , and thus we can define k ns ( t ) as k ∗ ( t ).Thus, for every M > V ((1 − s ) k + sk ) ≥ Z ∞ e − ρt u ( c ns ( t ) , k ns ( t )) dt ≥ (1 − s ) Z ∞ e − ρt u ( c n ( t ) , k n ( t )) dt + s Z ∞ e − ρt u ( c n ( t ) , k n ( t )) dt ≥ min { (1 − s ) ¯ V ( k ) + s ¯ V ( k ) − n − , M } , for all sufficiently large n ∈ N , which implies that¯ V ((1 − s ) k + sk ) ≥ (1 − s ) ¯ V ( k ) + s ¯ V ( k ) , as desired. This completes the proof. (cid:4) We separate the proof into eight steps.40 tep 1 . ¯ V is increasing. Proof of step 1 . Suppose not. Then, there exist k , k > k < k and ¯ V ( k ) = ¯ V ( k ). Because ¯ V is concave and nondecreasing, wehave ¯ V ( k ) = ¯ V ( k ) for every k > k .Choose any ( k ( t ) , c ( t )) ∈ A k . Because u is concave, for every ( k ∗ , c ∗ ) ∈ R , Z e − ρt [ u ( c ∗ , k ∗ ) − u ( c ( t ) , k ( t ))] dt ≥ Z e − ρt (cid:20) ∂u∂c ( c ∗ , k ∗ )( c ∗ − c ( t )) + ∂u∂k ( c ∗ , k ∗ )( k ∗ − k ( t )) (cid:21) dt = ∂u∂c ( c ∗ , k ∗ ) (cid:20) c ∗ ρ [1 − e − ρ ] − Z e − ρt c ( t ) dt (cid:21) + ∂u∂k ( c ∗ , k ∗ ) (cid:20) k ∗ ρ [1 − e − ρ ] − Z e − ρt k ( t ) dt (cid:21) ≥ ∂u∂c ( c ∗ , k ∗ ) (cid:20) c ∗ ρ [1 − e − ρ ] − Z c ( t ) dt (cid:21) + ∂u∂k ( c ∗ , k ∗ ) (cid:20) k ∗ ρ [1 − e − ρ ] − Z k ( t ) dt (cid:21) . Define N = max { F ( k, | ≤ k ≤ max t ∈ [0 , k + ( t ; k ) } ,ε = min { F ( k, − F ( k, | ≤ k ≤ max t ∈ [0 , k + ( t ; k ) } ,ε = max { F ( k, − F ( k, | ≤ k ≤ max t ∈ [0 , k + ( t ; k ) } . Because F is concave, we have F ( k, − F ( k, c ) ≥ ε ( c −
1) for every k ∈ [0 , max t ∈ [0 , k + ( t ; k )] and c >
1. Since 0 ≤ k ( t ) ≤ k + ( t ; k ) for every t ∈ [0 , F ( k ( t ) , c ( t )) = F ( k ( t ) ,
0) + ( F ( k ( t ) , c ( t )) − F ( k ( t ) , F ( k ( t ) , − F ( k ( t ) , ≤ N − ε + F ( k ( t ) , c ( t )) − F ( k ( t ) , ≤ N − ε + ε − ε ( c ( t ) − N + ε − ε c ( t ) , ≤ k (1)= k + Z ˙ k ( t ) dt = k + Z F ( k ( t ) , c ( t )) dt ≤ k + Z ( N + ε − ε c ( t )) dt = k + N + ε − ε Z c ( t ) dt. Therefore, Z c ( t ) dt ≤ k + N + ε ε . Moreover, Z k ( t ) dt ≤ Z k + ( t, k ) dt, and thus, for fixed sufficiently large c ∗ > k ∗ > k , there exists ε > Z e − ρt [ u ( c ∗ , k ∗ ) − u ( c ( t ) , k ( t ))] dt ≥ ε for every ( k ( t ) , c ( t )) ∈ A k . Choose any ( k ( t ) , c ( t )) ∈ A k such that c ( t ) isbounded and Z ∞ e − ρt u ( c ( t ) , k ( t )) dt > ¯ V ( k ) − ε , and define ˜ c ( t ) = ( c ∗ if 0 ≤ t ≤ ,c ( t ) otherwise . By almost the same arguments as in step 2 of the proof of Proposition 2, wecan show that there exists ˆ k > k ( t ) defined on R + such that ˜ k ( t ) ≥ k ∗ for all t ∈ [0 , k ( t ) ≥ k ( t ) for all t ≥
0, and (˜ k ( t ) , ˜ c ( t )) ∈ A ˆ k . Then,¯ V (ˆ k ) ≥ Z ∞ e − ρt u (˜ c ( t ) , ˜ k ( t )) dt > ¯ V ( k ) , which is a contradiction. This completes the proof of step 1. (cid:4) Step 2 . If ¯ V is differentiable at ¯ k >
0, thensup c ≥ { F (¯ k, c ) ¯ V ′ (¯ k ) + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) ≤ . roof of step 2 . Fix any c >
0. By Lemma 2, there exists ε > k ( t ) = F ( k ( t ) , c ) , k (0) = ¯ k has a positive and continuously differentiable solution k ( t ) on [0 , ε ]. For all t ∈ ]0 , ε [, Z t e − ρs u ( c, k ( s )) ds ≤ ¯ V (¯ k ) − e − ρt ¯ V ( k ( t )) . Thus, ¯ V (¯ k ) − ¯ V ( k ( t )) ≥ Z t e − ρs u ( c, k ( s )) ds + ( e − ρt −
1) ¯ V ( k ( t )) . Dividing both sides by t and taking the limit t →
0, we have − ¯ V ′ (¯ k ) F (¯ k, c ) ≥ u ( c, ¯ k ) − ρ ¯ V (¯ k ) , which implies that F (¯ k, c ) ¯ V ′ (¯ k ) + u ( c, ¯ k ) − ρ ¯ V (¯ k ) ≤ . Because the left-hand side is continuous in c , the above inequality holds forevery c ≥
0. Therefore,sup c ≥ { F (¯ k, c ) ¯ V ′ (¯ k ) + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) ≤ , as desired. This completes the proof of step 2. (cid:4) Consider the following modified problem of (4):max Z ∞ e − ρt u ( c ( t ) , k ( t )) dt subject to. ( k ( t ) , c ( t )) ∈ A ¯ k , (25) c ( t ) ≤ n for all t ≥ , where n >
0. Define A n ¯ k = { ( k ( t ) , c ( t )) ∈ A ¯ k | c ( t ) ≤ n for all t ≥ } , and let V n (¯ k ) = sup (cid:26) Z ∞ e − ρt u ( c ( t ) , k ( t )) dt (cid:12)(cid:12)(cid:12)(cid:12) ( k ( t ) , c ( t )) ∈ A n ¯ k (cid:27) . We call the function V n the value function of the problem (25). By almostthe same arguments as in the proof of Proposition 2, we can show that V n
43s concave and nondecreasing, and V n (¯ k ) > −∞ for all ¯ k >
0. Because V n (¯ k ) ≤ ¯ V (¯ k ) < + ∞ , we have that V n is finite. Step 3 . For every ¯ k > p ∈ ∂V n (¯ k ), sup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } − ρV n (¯ k ) ≥ . (26) Proof of step 3 . First, choose any
T > k > d nd ( e d T − . By Lemma 3, for all measurable functions c : [0 , T ] → [0 , n ], we have thatthere exists a solution k ( t ) of the following differential equation˙ k ( t ) = F ( k ( t ) , c ( t ))defined on [0 , T ] such that for every t ∈ [0 , T ], k ( t ) ≥ e − d T (cid:20) ¯ k − d nd ( e d T − (cid:21) > . (27)Choose any ( k ( t ) , c ( t )) ∈ A n ¯ k such that there exists h > Z T e − ρt u ( c ( t ) , k ( t ) − h ) dt ∈ R . Define c m ( t ) = max { c ( t ) , m − } . As we argued above, there exists a solution k m ( t ) of the following ODE:˙ k ( t ) = F ( k ( t ) , c m ( t )) , k (0) = ¯ k defined on [0 , T ]. Moreover, because F is Lipschitz on every compact set, wehave that there exists L > t ∈ [0 , T ], | k ( t ) − k m ( t ) | ≤ Z t | F ( k ( s ) , c ( s )) − F ( k m ( s ) , c m ( s )) | ds ≤ Z t L [ | k ( s ) − k m ( s ) | + | c ( s ) − c m ( s ) | ] ds. Recall that p ∈ ∂V n (¯ k ) if and only if V n ( k ) − V n (¯ k ) ≤ p ( k − ¯ k ) for all k ≥ We use the boundedness of the function c ∂u∂k ( c, k ) on ]0 , M ] for only proving thisstep. we have that lim m →∞ max t ∈ [0 ,T ] | k m ( t ) − k ( t ) | = 0. We show that for every t ∈ ]0 , T ],lim m →∞ Z t e − ρs u ( c m ( s ) , k m ( s )) ds = Z t e − ρs u ( c ( s ) , k ( s )) ds. Let h > m →∞ Z t e − ρs u ( c m ( s ) , k ( s ) − h ) ds = Z t e − ρs u ( c ( s ) , k ( s ) − h ) ds. Therefore, we have thatlim inf m →∞ Z t e − ρs u ( c m ( s ) , k m ( s )) ds ≥ lim h ↓ Z t e − ρs u ( c ( s ) , k ( s ) − h ) ds = Z t e − ρs u ( c ( s ) , k ( s )) ds. Meanwhile, by Lemma 2, we have that k m ( t ) ≤ k ( t ) for all t ∈ [0 , T ]. There-fore, lim sup m →∞ Z t e − ρs u ( c m ( s ) , k m ( s )) ds ≤ Z t e − ρs u ( c ( s ) , k ( s )) ds, which implies that our claim is correct.Fix sufficiently small t ∈ ]0 , T ] and ε >
0. Because V n is concave, it iscontinuous, and thus, there exists δ > k ( s ) , c ( s )) ∈ A n ¯ k − δ such that Z t e − ρs u ( c ( s ) , ˜ k ( s )) ds > V n (¯ k ) − e − ρt V n (˜ k ( t )) − tε. By (27), we can assume that inf s ∈ [0 ,t ] ˜ k ( s ) >
0. Consider the following ODE:˙ k ( s ) = F ( k ( s ) , c ( s )) , k (0) = ¯ k. By Lemma 3, there exists a solution k ( s ) of the above ODE defined on [0 , t ]such that k ( s ) ≥ ˜ k ( s ) for every s ∈ [0 , t ]. Because of Carath´eodory-Picard-Lindel¨of’s theorem, we must have that k ( s ) > ˜ k ( s ) for every s ∈ [0 , t ].Therefore, ( k ( t ) , c ( t )) ∈ A n ¯ k and there exists h > Z t e − ρs u ( c ( s ) , k ( s ) − h ) ds > V n (¯ k ) − e − ρt V n ( k ( t )) − tε. See the first part of Lemma 3 of Hosoya (2019a).
45y our previous argument, we can assume that inf s ∈ [0 ,t ] c ( s ) > By thedefinition of p , Z t − F ( k ( s ) , c ( s )) pds = p (¯ k − k ( t )) ≤ V n (¯ k ) − V n ( k ( t )) ≤ Z t e − ρs u ( c ( s ) , k ( s )) ds + ( e − ρt − V n ( k ( t )) + tε. Thus, Z t [ F (¯ k, c ( s )) p + u ( c ( s ) , ¯ k )] ds + Z t [ u ( c ( s ) , k ( s )) − u ( c ( s ) , ¯ k )] ds + Z t ( e − ρs − u ( c ( s ) , k ( s )) ds + Z t [ F ( k ( s ) , c ( s )) − F (¯ k, c ( s ))] pds + ( e − ρt − V n ( k ( t )) ≥ − tε. Because Z t [ F (¯ k, c ( s )) p + u ( c ( s ) , ¯ k )] ds ≤ sup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } t, we havesup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } t + ( e − ρt − V n ( k ( t )) ≥ − tε + Z t (1 − e − ρs ) u ( c ( s ) , k ( s )) ds + Z t [ F (¯ k, c ( s )) − F ( k ( s ) , c ( s ))] pds + Z t [ u ( c ( s ) , ¯ k ) − u ( c ( s ) , k ( s ))] ds ≡ − tε + J ( t ) + J ( t ) + J ( t ) . This requirement is needed for ensuring the integrability of the functions s u ( c ( s ) , ¯ k )and s u ( c ( s ) , k ( s )). ≥ Z t [ u ( c ( s ) , k ( s )) − u ( n, k + ( s, ¯ k ))] ds ≥ e ρt Z t e − ρs [ u ( c ( s ) , k ( s )) − u ( n, k + ( s, ¯ k ))] ds ≥ e ρt V n (¯ k ) − V n ( k ( t )) − e ρt tε − e ρt Z t e − ρs u ( n, k + ( s, ¯ k )) ds (28) ≥ e ρt V n (¯ k ) − V n ( k + ( t, ¯ k )) − e ρt tε − e ρt Z t e − ρs u ( n, k + ( s, ¯ k )) ds. Moreover, | J ( t ) | ≤ Z t (1 − e − ρs ) | u ( n, k + ( s, ¯ k )) | ds + (1 − e − ρt ) (cid:12)(cid:12)(cid:12)(cid:12)Z t [ u ( c ( s ) , k ( s )) − u ( n, k + ( s, ¯ k ))] ds (cid:12)(cid:12)(cid:12)(cid:12) , and thus, by (28), we have thatlim t → J ( t ) t = 0 . Let k ∗ ( s ) be the solution of the following differential equation:˙ k ( s ) = F ( k ( s ) , n ) , k (0) = ¯ k. Then, | J ( t ) | ≤ Z t p max {| F ( k, c ) − F (¯ k, c ) || k ∗ ( s ) ≤ k ≤ k + ( s, ¯ k ) , ≤ c ≤ n } ds, and thus lim t → J ( t ) t = 0 . Finally, because u is concave, u ( c ( s ) , ¯ k ) − u ( c ( s ) , k ( s )) ≥ ∂u∂k ( c ( s ) , ¯ k )(¯ k − k + ( s, ¯ k )) , and by Assumption 4, we have that c ∂u∂k ( c, ¯ k ) is bounded on ]0 , n ]. There-fore, lim sup t → J ( t ) t ≥ . Note that, the pair of functions ( k ( s ) , c ( s )) depends on t >
0. This problem makesthe following evaluation difficult.
47n conclusion, we have thatsup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } − ρV n (¯ k ) ≥ − ε, and because ε > (cid:4) Step 4 . For every ¯ k > V n (¯ k ) ↑ ¯ V (¯ k ) as n → ∞ . Proof of step 4 . It is clear by Lemma 5. (cid:4)
Step 5 . If ¯ V is differentiable at ¯ k >
0, thensup c ≥ { F (¯ k, c ) ¯ V ′ (¯ k ) + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) ≥ . Proof of step 5 . For any sufficiently large n , choose any p n ∈ ∂V n (¯ k ). Fixany δ >
0, and choose k , k > k < ¯ k < k and¯ V ′ (¯ k ) − δ < ¯ V ( k ) − ¯ V (¯ k ) k − ¯ k ≤ ¯ V ′ (¯ k ) ≤ ¯ V ( k ) − ¯ V (¯ k ) k − ¯ k < ¯ V ′ (¯ k ) + δ. Because V n ( k i ) ↑ ¯ V ( k i ) as n → ∞ for i = 1 , V n (¯ k ) ↑ ¯ V (¯ k ), we havemax i =1 , (cid:12)(cid:12)(cid:12)(cid:12) V n ( k i ) − V n (¯ k ) k i − ¯ k − ¯ V ( k i ) − ¯ V (¯ k ) k i − ¯ k (cid:12)(cid:12)(cid:12)(cid:12) < δ for any sufficiently large n . Therefore, | p n − ¯ V ′ (¯ k ) | ≤ δ for sufficiently large n , and thus lim n →∞ p n = ¯ V ′ (¯ k ).Hence,0 ≤ sup c ≥ { F (¯ k, c ) p n + u ( c, ¯ k ) } − ρV n (¯ k )= F (¯ k, c ∗ ( p n , ¯ k )) p n + u ( c ∗ ( p n , ¯ k ) , ¯ k ) − ρV n (¯ k ) → F (¯ k, c ∗ ( ¯ V ′ (¯ k ) , ¯ k )) ¯ V ′ (¯ k ) + u ( c ∗ ( ¯ V ′ (¯ k ) , ¯ k ) , ¯ k ) − ρ ¯ V (¯ k )= sup c ≥ { F (¯ k, c ) ¯ V ′ (¯ k ) + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) , as desired. This completes the proof of step 5. (cid:4) tep 6 . For all ¯ k > p ∈ ∂ ¯ V (¯ k ),sup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) = 0 . Proof of step 6 . First, let p = D − ¯ V ( k ) , k m ↑ k as m → ∞ , and ¯ V ′ ( k m )can be defined. Because ¯ V is concave, ¯ V ′ ( k m ) is decreasing in m , and thusconverges to some number p ∗ ≥ p . If p ∗ > p , then there exists k ′ < k suchthat ¯ V ( k ′ ) − ¯ V ( k ) k ′ − k < p ∗ . Because k m ↑ k as m → ∞ , there exists m such that k m > k ′ , ¯ V ( k ′ ) − ¯ V ( k m ) k ′ − k m < p ∗ . This implies that ¯ V ′ ( k m ) < p ∗ , which is a contradiction. Therefore, ¯ V ′ ( k m )converges to p . Similarly, if p = D + ¯ V ( k ) , k m ↓ k and ¯ V ′ ( k m ) can be defined,then ¯ V ′ ( k m ) converges to p . In particular, if ¯ V is differentiable around ¯ k ,then ¯ V ′ is continuous at ¯ k .Choose any ¯ k >
0. If ¯ V is differentiable at ¯ k , then ∂ ¯ V (¯ k ) = { ¯ V ′ (¯ k ) } , andthe claim of step 6 follows from steps 2 and 5. Therefore, suppose that ¯ V isnot differentiable at ¯ k , and p = D − ¯ V (¯ k ) , p = D + ¯ V (¯ k ). By assumption, wehave p > p , and ∂ ¯ V (¯ k ) = [ p , p ]. Choose sequences ( k m ) , ( k m ) such that k m ↑ ¯ k, k m ↓ ¯ k as m → ∞ , and ¯ V is differentiable at k mi for all m . By theabove arguments, we have that ¯ V ′ ( k m ) ↓ p and ¯ V ′ ( k m ) ↑ p as m → ∞ . Bysteps 2 and 5, we havesup c ≥ { F (¯ k, c ) p i + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) = 0 . Next, choose any p with p < p < p , and define ε = min { p − p, p − p } . Asin the proof of step 5, we can show that for any m and i ∈ { , } , q ∈ ∂V n ( k mi ) ⇒ | q − ¯ V ′ ( k mi ) | < ε for sufficiently large n . Choose n m as such an n . We can assume that n m isincreasing in m . Because | ¯ V ′ ( k mi ) − p | ≥ ε , we have there exists k m ∈ [ k m , k m ]such that p ∈ ∂V n m ( k m ). Then, by step 3,sup c ≥ { F ( k m , c ) p + u ( c, k m ) } − ρV n m ( k m ) ≥ . V n m ( k ) ↑ ¯ V ( k ) as m → ∞ pointwise, by Dini’s theorem, V n m → ¯ V uniformly as m → ∞ on any compact set in R ++ . Therefore,sup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) ≥ . Now, define g ( p ) = sup c ≥ { F (¯ k, c ) p + u ( c, ¯ k ) } . Suppose that p, q ∈ [ p , p ] and r = (1 − t ) p + tq for t ∈ [0 , g ( r ) = F (¯ k, c ∗ ( r, ¯ k )) r + u ( c ∗ ( r, ¯ k ) , ¯ k )= (1 − t )[ F (¯ k, c ∗ ( r, ¯ k )) p + u ( c ∗ ( r, ¯ k ) , ¯ k )] + t [ F (¯ k, c ∗ ( r, ¯ k )) q + u ( c ∗ ( r, ¯ k ) , ¯ k )] ≤ (1 − t ) g ( p ) + tg ( q ) . Therefore, g is convex. Suppose that p ∈ [ p , p ]. Because g ( p ) = g ( p ) = ρ ¯ V (¯ k ) and g ( p ) ≥ ρ ¯ V (¯ k ), we have that g ( p ) = ρ ¯ V (¯ k ). This completes theproof of step 6. (cid:4) Step 7 . ¯ V is a viscosity solution of the HJB equation. Proof of step 7 . Choose any ¯ k >
0. Suppose that ϕ ( k ) is a continuouslydifferentiable function defined on a neighborhood of ¯ k such that ϕ (¯ k ) = ¯ V (¯ k )and either ϕ ( k ) ≥ ¯ V ( k ) for all k or ϕ ( k ) ≤ ¯ V ( k ) for all k . Then, ¯ k is anextremal point of ϕ ( k ) − ¯ V ( k ), and thus ϕ ′ (¯ k ) ∈ ∂ ¯ V (¯ k ). This implies thatsup c ≥ { F (¯ k, c ) ϕ ′ (¯ k ) + u ( c, ¯ k ) } − ρ ¯ V (¯ k ) = 0 . Therefore, ¯ V is a viscosity solution of the HJB equation. This completes theproof of step 7. (cid:4) Step 8 . If Assumption 5 holds, then ¯ V is continuously differentiable on R ++ . Proof of step 8 . As we have argued in the proof of step 6, it suffices to showthat ¯ V is differentiable at any ¯ k >
0. Suppose that ¯ V is not differentiableat ¯ k >
0, and let p = D − ¯ V (¯ k ) and p = D + ¯ V (¯ k ). Then, we have p < p .Define a function g as in the proof of step 6, and choose p ∈ ] p , p [. Then,for every q around p , g ( q ) ≥ F (¯ k, c ∗ ( p, ¯ k )) q + u ( c ∗ ( p, ¯ k ) , ¯ k ) , g ( p ) = F (¯ k, c ∗ ( p, ¯ k )) p + u ( c ∗ ( p, ¯ k ) , ¯ k ) . Because g is constant around p , we must have F (¯ k, c ∗ ( p, ¯ k )) = 0 for such p .Therefore, there exists c ∗ > c ∗ ( p, ¯ k ) = c ∗ for every p ∈ ] p , p [.50owever, this implies that ∂F∂c (¯ k, c ∗ ) p + ∂u∂c ( c ∗ , ¯ k ) = 0 for every p ∈ ] p , p [,which is impossible. This completes the proof of step 8. (cid:4) Steps 1-8 show that all the claims in Proposition 3 are correct. Thiscompletes the proof. (cid:4)
First, consider the following problem:max Z ∞ e − ρt u ( c ( t ) , k ( t )) dt subject to. c ( t ) is locally integrable ,k ( t ) ≥ , c ( t ) ≥ , Z ∞ e − ρt u ( c ( t ) , k ( t )) dt can be defined , ˙ k ( t ) = γ ( k ( t ) − k ∗ ) − δ ( c ( t ) − c ∗ ) + F ( k ∗ , c ∗ ) a.e. ,k (0) = ¯ k. Define A L ¯ k as the set of all pairs ( k ( t ) , c ( t )) such that k ( t ) is nonnegative andabsolutely continuous on every compact set, c ( t ) is nonnegative and locallyintegrable, R ∞ e − ρt u ( c ( t ) , k ( t )) dt can be defined, k (0) = ¯ k and˙ k ( t ) = γ ( k ( t ) − k ∗ ) − δ ( c ( t ) − c ∗ ) + F ( k ∗ , c ∗ )for almost all t ≥
0. Let V (¯ k ) = sup (cid:26) Z ∞ e − ρt u ( c ( t ) , k ( t )) dt (cid:12)(cid:12)(cid:12)(cid:12) ( k ( t ) , c ( t )) ∈ A L ¯ k (cid:27) . Step 1 . V (¯ k ) ≥ ¯ V (¯ k ). Proof of step 1 . By Proposition 2, ¯ V (¯ k ) > −∞ . For every ε > N >
0, there exists ( k ( t ) , c ( t )) ∈ A ¯ k such that c ( t ) is bounded and Z ∞ e − ρt u ( c ( t ) , k ( t )) dt ≥ min { ¯ V (¯ k ) − ε, N } . Consider the following differential equation:˙ k ( t ) = γ ( k ( t ) − k ∗ ) − δ ( c ( t ) − c ∗ ) + F ( k ∗ , c ∗ ) , k (0) = ¯ k. k ( t ) = e γt (cid:20) ¯ k − Z t e − γs ( γk ∗ + δ ( c ( s ) − c ∗ ) − F ( k ∗ , c ∗ )) ds (cid:21) . Because ( γ, − δ ) ∈ ∂F ( k ∗ , c ∗ ), we have F ( k, c ) ≤ γ ( k − k ∗ ) − δ ( c − c ∗ ) + F ( k ∗ , c ∗ ) for all ( k, c ), and thus, by Lemma 2, ˆ k ( t ) ≥ k ( t ) for every t ≥ k ( t ) , c ( t )) ∈ A L ¯ k , and thus V (¯ k ) ≥ min { ¯ V (¯ k ) − ε, N } . Because ε, N are arbitrary, we have that V (¯ k ) ≥ ¯ V (¯ k ). This completes theproof of step 1. (cid:4) Second, consider the following problem:max Z ∞ e − ρt [ au θ ( c ( t )) + bu θ ( k ( t ))] dt subject to. c ( t ) is locally integrable, k ( t ) ≥ , c ( t ) ≥ , Z ∞ e − ρt [ au θ ( c ( t )) + bu θ ( k ( t ))] dt can be defined , ˙ k ( t ) = γ ( k ( t ) − k ∗ ) − δ ( c ( t ) − c ∗ ) + F ( k ∗ , c ∗ ) a.e. ,k (0) = ¯ k. Define A L k as the set of all pairs ( k ( t ) , c ( t )) such that k ( t ) is absolutelycontinuous in every compact set, c ( t ) is locally integrable, R ∞ e − ρt [ au θ ( c ( t ))+ bu θ ( k ( t ))] dt can be defined, k (0) = ¯ k and˙ k ( t ) = γ ( k ( t ) − k ∗ ) − δ ( c ( t ) − c ∗ ) + F ( k ∗ , c ∗ )for almost all t ≥
0. Let V (¯ k ) = sup (cid:26) Z ∞ e − ρt [ au θ ( c ( t )) + bu θ ( k ( t ))] dt (cid:12)(cid:12)(cid:12)(cid:12) ( k ( t ) , c ( t )) ∈ A L k (cid:27) . Step 2 . V (¯ k ) + Cρ ≥ V (¯ k ). Proof of step 2 . By Proposition 2 and step 1, V (¯ k ) > −∞ . For every ε > N >
0, there exists ( k ( t ) , c ( t )) ∈ A L ¯ k such that Z ∞ e − ρt u ( c ( t ) , k ( t )) dt ≥ min { V (¯ k ) − ε, N } . Z ∞ e − ρt min { au θ ( c ( t ))+ bu θ ( k ( t ))+ C, } dt ≥ Z ∞ e − ρt min { u ( c ( t ) , k ( t )) , } dt > −∞ , we have that Z ∞ e − ρt [ au θ ( c ( t )) + bu θ ( k ( t ))] dt can be defined. Hence, ( k ( t ) , c ( t )) ∈ A L k , and thus, V (¯ k ) + Cρ ≥ min { V (¯ k ) − ε, N } . Because ε, N are arbitrary, we have that V (¯ k ) + Cρ ≥ V (¯ k ). This completesthe proof of step 2. (cid:4) Define C ∗ = ρ − (1 − θ ) γθδ (cid:20) ¯ k − k ∗ + δc ∗ + F ( k ∗ , c ∗ ) γ (cid:21) . Note that, − k ∗ + δc ∗ + F ( k ∗ , c ∗ ) γ = 1 γ [ γ (0 − k ∗ ) − δ (0 − c ∗ ) + F ( k ∗ , c ∗ )] ≥ , and thus C ∗ > k >
0. Define V (¯ k ) = ( ( C ∗ ) − θ θ (1 − θ )( ρ − (1 − θ ) γ ) − ρ (1 − θ ) , if θ = 1 , log C ∗ ρ + γ − ρρ if θ = 1and V (¯ k ) = (cid:16) ¯ k − k ∗ + δc ∗ + F ( k ∗ ,c ∗ ) γ (cid:17) − θ (1 − θ )( ρ − (1 − θ ) γ ) − ρ (1 − θ ) if θ = 1 , log (cid:16) ¯ k − k ∗ + δc ∗ + F ( k ∗ ,c ∗ ) γ (cid:17) ρ + γρ , if θ = 1 . Step 3 . aV (¯ k ) + bV (¯ k ) ≥ V (¯ k ). Proof of Step 3 . Let c ∗ ( t ) = C ∗ e γ − ρθ t ,k ∗ ( t ) = e γt (cid:20) ¯ k − Z t e − γs [ γk ∗ + δ ( c ∗ ( s ) − c ∗ ) − F ( k ∗ , c ∗ )] ds (cid:21) . Note that, ˙ k ∗ ( t ) = γ ( k ∗ ( t ) − k ∗ ) − δ ( c ∗ ( t ) − c ∗ ) + F ( k ∗ , c ∗ ) , dt ( u ′ θ ( c ∗ ( t ))) = ( ρ − γ ) u ′ θ ( c ∗ ( t )) . Then, for every ( k ( t ) , c ( t )) ∈ A L k , Z T e − ρt ( u θ ( c ( t )) − u θ ( c ∗ ( t ))) dt ≤ Z T e − ρt u ′ θ ( c ∗ ( t ))( c ( t ) − c ∗ ( t )) dt = δ − Z T e − ρt u ′ θ ( c ∗ ( t ))[ γ ( k ( t ) − k ∗ ( t )) − ( ˙ k ( t ) − ˙ k ∗ ( t ))] dt = δ − e − ρT u ′ θ ( c ∗ ( T ))( k ∗ ( T ) − k ( T )) ≤ δ − e − ρT u ′ θ ( c ∗ ( T )) k ∗ ( T )= δ − ( C ∗ ) − θ (cid:20) ¯ k − (cid:18) k ∗ − δc ∗ + F ( k ∗ , c ∗ ) γ (cid:19) (1 − e − γT ) − θδC ∗ ρ − (1 − θ ) γ (1 − e (1 − θ ) γ − ρθ T ) (cid:21) → T → ∞ ) . Moreover, k ( t ) ≤ (cid:18) ¯ k − k ∗ + δc ∗ + F ( k ∗ , c ∗ ) γ (cid:19) e γt ≡ ˆ ke γt , and thus, we have Z ∞ e − ρt [ au θ ( c ( t )) + bu θ ( k ( t ))] dt ≤ a Z ∞ e − ρt u θ ( c ∗ ( t )) dt + b Z ∞ e − ρt u θ (ˆ ke γt ) dt = aV (¯ k ) + bV (¯ k ) , which completes the proof. (cid:4) By steps 1-3, we have ¯ V (¯ k ) < + ∞ for every ¯ k > Step 4 . For every ¯ k > e − ρt ¯ V ( k + ( t, ¯ k )) → t → ∞ . Proof . We have already shown that inf t ≥ k + ( t, ¯ k ) > k >
0, andthus it suffices to show that for i ∈ { , } ,lim sup t →∞ e − ρt V i ( k + ( t, ¯ k )) ≤ , Define ˆ k ( t ) = e γt (cid:20) ¯ k − (cid:18) k ∗ − δc ∗ + F ( k ∗ , c ∗ ) γ (cid:19)(cid:21) .
54y Lemma 2, we have that ˆ k ( t ) ≥ k + ( t, ¯ k ) for all t ≥
0, and it suffices toshow that for i ∈ { , } , lim t →∞ e − ρt V i (ˆ k ( t )) = 0 . If θ = 1, then V i (ˆ k ( t )) = A log( e γt + B ) + C for some A > , B ≥
0, and C ∈ R , and e − ρt V i (ˆ k ( t )) = Ae − ρt log( e γt + B ) + e − ρt C → t → ∞ . If θ = 1, then V i (ˆ k ( t )) = A ( e γt + B ) − θ + C, for some A, B, C ∈ R such that A (1 − θ ) > B ≥
0. If θ <
1, then e − ρt V i (ˆ k ( t )) = A ( e (1 − θ ) γ − ρ − θ t + e − ρ − θ t B ) − θ + e − ρt C → t → ∞ . If θ >
1, then e − ρt V i (ˆ k ( t )) = e − ρt [ A ( e γt + B ) − θ + C ] → t → ∞ . Thus, in any case, our claim is correct. This completes the proof. (cid:4) We separate the proof into three steps.
Step 1 . Suppose that V ∈ V is a classical solution of the HJB equation.Then, there exists a solution k ∗ ( t ) of the differential equation (13) definedon R + . Moreover, for any such solution, inf t k ∗ ( t ) > Proof of step 1 . Let ε > c ≥ D k, + F ( ε, c ) > ρ, ε < ¯ k. We show that for any solution k ( t ) of (13) defined on [0 , T ], k ( t ) ≥ ε .Suppose that this claim is incorrect. Then, there exists t ∗ > < k ( t ∗ ) < ε . By the mean value theorem, we can assume that t ∗ < T k ( t ∗ ) <
0. By Alexandrov’s theorem, V is twice differentiable almosteverywhere, and thus we can assume that V ′′ ( k ( t ∗ )) can be defined. Define c ( t ) = c ∗ ( V ′ ( k ( t )) , k ( t ))on [0 , T ]. Then, c ( t ) is continuous and positive. Because V is a solution ofthe HJB equation, F ( k ( t ) , c ( t )) V ′ ( k ( t )) + u ( c ( t ) , k ( t )) = ρV ( k ( t )) , for every t around t ∗ . Because ˙ k ( t ∗ ) <
0, for any sufficiently small h > k , k ∈ [ k ( t ∗ + h ) , k ( t ∗ )] , θ , θ ∈ [0 , p ∈ ∂ k F ( k , c ( t ∗ + h ))and q = ∂F∂c ( k ( t ∗ ) , c ( t ∗ + θ h )) such that ρV ′ ( k )( k ( t ∗ + h ) − k ( t ∗ ))= ρ ( V ( k ( t ∗ + h )) − V ( k ( t ∗ )))= F ( k ( t ∗ + h ) , c ( t ∗ + h )) V ′ ( k ( t ∗ + h )) + u ( c ( t ∗ + h ) , k ( t ∗ + h )) − F ( k ( t ∗ ) , c ( t ∗ )) V ′ ( k ( t ∗ )) − u ( c ( t ∗ ) , k ( t ∗ ))= ( F ( k ( t ∗ + h ) , c ( t ∗ + h )) − F ( k ( t ∗ ) , c ( t ∗ + h ))) V ′ ( k ( t ∗ + h ))+ ( F ( k ( t ∗ ) , c ( t ∗ + h )) − F ( k ( t ∗ ) , c ( t ∗ ))) V ′ ( k ( t ∗ + h ))+ F ( k ( t ∗ ) , c ( t ∗ ))( V ′ ( k ( t ∗ + h )) − V ′ ( k ( t ∗ )))+ u ( c ( t ∗ + h ) , k ( t ∗ + h )) − u ( c ( t ∗ + h ) , k ( t ∗ ))+ u ( c ( t ∗ + h ) , k ( t ∗ )) − u ( c ( t ∗ ) , k ( t ∗ ))= pV ′ ( k ( t ∗ + h ))( k ( t ∗ + h ) − k ( t ∗ )) + ˙ k ( t ∗ )( V ′ ( k ( t ∗ + h )) − V ′ ( t ∗ ))+ ∂u∂k ( c ( t ∗ + h ) , k ( t ∗ + θ h ))( k ( t ∗ + h ) − k ( t ∗ ))+ (cid:18) ∂u∂c ( c ( t ∗ + θ h ) , k ( t ∗ + θ h )) + qV ′ ( k ( t ∗ + h )) (cid:19) ( c ( t ∗ + h ) − c ( t ∗ ))+ (cid:18) ∂u∂c ( c ( t ∗ + θ h ) , k ( t ∗ )) − ∂u∂c ( c ( t ∗ + θ h ) , k ( t ∗ + θ h )) (cid:19) ( c ( t ∗ + h ) − c ( t ∗ )) See Alexandrov (1939) or Howard (1998).
56y mean value theorem. To modify this equation,( ρV ′ ( k ) − pV ′ ( k ( t ∗ + h )))( k ( t ∗ + h ) − k ( t ∗ )) h = ˙ k ( t ∗ )( V ′ ( k ( t ∗ + h )) − V ′ ( k ( t ∗ ))) h + ∂u∂k ( c ( t ∗ + h ) , k ( t ∗ + θ h ))( k ( t ∗ + h ) − k ( t ∗ )) h (29)+ (cid:0) ∂u∂c ( c ( t ∗ + θ h ) , k ∗ ( t ∗ + θ h )) + qV ′ ( k ( t ∗ + h )) (cid:1) ( c ( t ∗ + h ) − c ( t )) h + (cid:0) ∂u∂c ( c ( t ∗ + θ h ) , k ∗ ( t ∗ )) − ∂u∂c ( c ( t ∗ + θ h ) , k ∗ ( t ∗ + θ h )) (cid:1) ( c ( t ∗ + h ) − c ( t ∗ )) h . Because k ( t ∗ + h ) ≤ k ( t ∗ ) < ε , we have p > inf c ≥ D k, + F ( k ( t ∗ ) , c ) > ρ , andthus lim inf h ↓ ( ρV ′ ( k ) − pV ′ ( k ( t ∗ + h )))( k ( t ∗ + h ) − k ( t ∗ )) h ≥ ( ρ − inf c ≥ D k, + F ( k ( t ∗ ) , c )) V ′ ( k ( t ∗ )) ˙ k ( t ∗ ) > . Meanwhile, the first and second terms of the right-hand side of (29) are alwaysnonpositive. Because of the definition of c ( t ) and the first-order condition,we have ∂u∂c ( c ( t ∗ + θ h ) , k ( t ∗ + θ h )) = − rV ′ ( k ( t ∗ + θ h )) , where r = ∂F∂c ( k ( t ∗ + θ h ) , c ( t ∗ + θ h )), and thus the absolute value of thethird term of the right-hand side of (29) is bounded from − q | V ′ ( k ( t ∗ )) − V ′ ( k ( t ∗ + h )) | × | c ( t ∗ + h ) − c ( t ∗ ) | h + | r − q | V ′ ( k ( t ∗ + θ h )) × | c ( t ∗ + h ) − c ( t ∗ ) | h , where, V ′ ( k ( t ∗ )) − V ′ ( k ( t ∗ + h )) h → V ′′ ( k ( t ∗ )) ˙ k ( t ∗ ) as h ↓ ,r − qh = ∂ F∂k∂c ( k ( t ∗ + θ ′ h ) , c ( t ∗ + θ h )) ˙ k ( t ∗ + θ ′ h ) for some θ ′ ∈ [0 , θ ] . Therefore, the third term of the right-hand side converges to zero as h ↓ ∂u∂c is differentiable in k on R , we have that the fourthterm of the right-hand side of (29) is ∂ u∂k∂c ( c ( t ∗ + θ h ) , k ( t ∗ + θ ′′ h )) ˙ k ( t ∗ + θ ′′ h )( c ( t ∗ + h ) − c ( t ∗ ))57or some θ ′′ ∈ [0 , θ ]. Therefore, the fourth term also converges to 0, andthus the limsup of the right-hand side of (29) is nonpositive, which is acontradiction.Hence, k ( t ) ≥ ε for every t ∈ [0 , T ]. By Lemma 3, we have that thereexists a solution k ∗ ( t ) of (13) defined on R + . Clearly, we have that k ∗ ( t ) ≥ ε for every t ≥
0, and thus inf t ≥ k ∗ ( t ) >
0. This completes the proof of step1. (cid:4)
Step 2 . Under the assumptions in step 1, choose a solution k ∗ ( t ) of equa-tion (13) defined on R + , and define c ∗ ( t ) = c ∗ ( V ′ ( k ∗ ( t )) , k ∗ ( t )). Then,( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k and V (¯ k ) = Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt. Proof of step 2 . By step 1, we have there exists ε > ε ≤ k ∗ ( t ) ≤ k + ( t, ¯ k )for all t ≥
0. Because V is a classical solution of the HJB equation, F ( k ∗ ( t ) , c ∗ ( t )) V ′ ( k ∗ ( t )) + u ( c ∗ ( t ) , k ∗ ( t )) = ρV ( k ∗ ( t )) , for every t ≥
0. Therefore, Z T e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt = Z T e − ρt [ F ( k ∗ ( t ) , c ∗ ( t )) V ′ ( k ∗ ( t )) + u ( c ∗ ( t ) , k ∗ ( t ))] dt − Z T e − ρt V ′ ( k ∗ ( t )) ˙ k ∗ ( t ) dt = − Z T [ − ρe − ρt V ( k ∗ ( t )) + e − ρt V ′ ( k ∗ ( t )) ˙ k ∗ ( t )] dt = − Z T ddt [ e − ρt V ( k ∗ ( t ))] dt = V (¯ k ) − e − ρT V ( k ∗ ( T )) , and thus, V (¯ k ) − e − ρT V ( ε ) ≥ V (¯ k ) − Z T e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt ≥ V (¯ k ) − e − ρT V ( k + ( T, ¯ k )) . Hence, Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt = V (¯ k ) , R ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt can be defined. Therefore, ( k ∗ ( t ) , c ∗ ( t )) ∈ A ¯ k , which completes the proof of step 2. (cid:4) Step 3 . If ( k ( t ) , c ( t )) ∈ A ¯ k and inf t ≥ k ( t ) >
0, then Z ∞ e − ρt u ( c ( t ) , k ( t )) dt ≤ V (¯ k ) . Proof of step 3 . Suppose that ( k ( t ) , c ( t )) ∈ A ¯ k and inf t ≥ k ( t ) >
0. Because V is a classical solution of the HJB equation, Z T e − ρt u ( c ( t ) , k ( t )) dt = Z T e − ρt [ F ( k ( t ) , c ( t )) V ′ ( k ( t )) + u ( c ( t ) , k ( t ))] dt − Z T e − ρt V ′ ( k ( t )) ˙ k ( t ) dt ≤ − Z T [ − ρe − ρt V ( k ( t )) + e − ρt V ′ ( k ( t )) ˙ k ( t )] dt = Z T ddt [ − e − ρt V ( k ( t ))] dt = V (¯ k ) − e − ρT V ( k ( T )) . Because inf t ≥ k ( t ) > V satisfies (8), the right-hand side converges to V (¯ k ) as T → ∞ . Therefore, Z ∞ e − ρt u ( c ( t )) dt ≤ V (¯ k ) , as desired. This completes the proof. (cid:4) Because Assumptions 1-7 hold, we have that ¯ V ∈ V , and it is a classical so-lution of the HJB equation. By Proposition 4, there exists a solution k ∗ ( t ) of(14) defined on R + , where inf t ≥ k ∗ ( t ) >
0. Define c ∗ ( t ) = c ∗ ( ¯ V ′ ( k ∗ ( t )) , k ∗ ( t )).Then, by Proposition 4, we have that¯ V (¯ k ) = Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt, which implies that ( k ∗ ( t ) , c ∗ ( t )) is a solution of (4).59eanwhile, suppose that V ∈ V is also a classical solution of the HJBequation. Because inf t ≥ k ∗ ( t ) >
0, by Proposition 4, we have that V (¯ k ) ≥ Z ∞ e − ρt u ( c ∗ ( t ) , k ∗ ( t )) dt = ¯ V (¯ k ) . Meanwhile, by Proposition 4, there exists a solution k + ( t ) of (13) defined on R + such that inf t ≥ k + ( t ) >
0. Define c + ( t ) = c ∗ ( V ′ ( k + ( t )) , k + ( t )). Then, byProposition 4, we have that ( k + ( t ) , c + ( t )) ∈ A ¯ k , and thus V (¯ k ) = Z ∞ e − ρt u ( c + ( t ) , k + ( t )) dt ≤ ¯ V (¯ k ) . Hence, we conclude that V = ¯ V . This completes the proof. (cid:4) Acknowledgements
The research agenda pursued in this paper was proposed by Hiroyuki Ozaki.We are grateful to him for pointing out an important issue in economicdynamics. This research started from an acute idea proposed by SusumuKuwata. We received comments from Alexander Zaslavsky and MonikaMotta at the 12th AIMS Conference on Dynamical Systems, DifferentialEquations and Applications. We also received very good advice from KenjiMiyazaki at the 2018 Fall Meeting of the Japanese Economic Association.Keiichi Morimoto also gave us sharp comments at a workshop at Meisei Uni-versity. Makoto Hanazono made a useful suggestion at the 2020 meeting ofthe Japanese Society for Mathematical Economics. Additionally, we receivedmany comments and suggestions from Nobusumi Sagara, Toru Maruyama,and Chaowen Yu through private discussions. We would like to express ourgratitude to all of them. Of course, the author is responsible for all remainingerrors. Finally, this work was supported by JSPS KAKENHI Grant NumberJP18K12749.