# Violation of the Hamilton-Jacobi-Bellman Equation in Economic Dynamics

Yuhki Hosoya
Faculty of Economics, Chuo University E-mail:
hosoya(at)tamacc.chuo-u.ac.jp742-1 Higashinakano, Hachioji-shi, Tokyo
192-0393, Japan.

###### Abstract

We consider an extension of the classical capital accumulation model, and
present an example in which the Hamilton-Jacobi-Bellman (HJB) equation is
neither necessary nor sufficient for a function to be the value function.
Next, we present assumptions under which the HJB equation becomes a necessary
and sufficient condition for a function to be the value function, and using
this result, we propose a new method for solving the original problem using
the solution of the HJB equation. Our assumptions are so mild that many
macroeconomic growth models satisfy them. Therefore, our results ensure that
the solution of the HJB equation is rigorously the value function in many
macroeconomic models, and present a new solving method for these models.

JEL codes: C02, C61, E13.

Keywords: Capital Accumulation Model, Hamilton-Jacobi-Bellman equation,
Classical Solution, Linear Technology, Nonlinear Technology, Subdifferential
Calculus.

##  1 Introduction

The Hamilton-Jacobi-Bellman (HJB) equation has many applications in economics
and is used in a variety of fields, both micro and macro. The HJB equation is
paired with a parameterized variational problem, and in many applied studies,
it is unconditionally believed that the solution of the HJB equation coincides
with the value function of the problem. Therefore, these studies first obtain
the value function by solving the HJB equation, and then use the value
function to analyze the original problem.

However, after a little survey, one can find that there seems to be no
mathematical basis for the above results that could be used in economic
dynamics. There are many reasons for this, but the most important one is the
following. The HJB equation is a mathematical tool that has evolved together
with calculus of variations, and its main application is in the natural
sciences, including physics and engineering. Most dynamic models in these
fields have finite time-interval. Meanwhile, most dynamic models in economics
deal with infinite time-interval. Consequently, there are few mathematical
foundations for the HJB equation that can be used in economics.

In recent years, mathematicians who have addressed this problem have made some
progress in the study of the HJB equation for variational problems with
infinite time-interval. For example, Bardi and Dolcetta (2008) deeply studied
such a problem. However, in these studies, strong restrictions, such as
boundedness, are imposed on the instantaneous utility function. In
macroeconomic dynamics, a typical example of an instantaneous utility function
is the CRRA function, which cannot be assumed to be bounded. Currently, to the
best of our knowledge, there is no rigorous mathematical basis for the HJB
equation that can be applied to macroeconomic models.

In this paper, we first present an example of an optimal growth model to show
the seriousness of the absence of a basis for the HJB equation. This model
deviates only slightly from the common assumptions of the Ramsey-Cass-Koopmans
capital accumulation model (hereafter, the RCK model).111This model was
originated by Ramsey (1928), and modified by Cass (1965) and Koopmans (1965).
Because this model is widely known, many textbooks treat it. See, for example,
Acemoglu (2009), Barro and Sala-i-Martin (2003), Blanchard and Fischer (1989),
and Romer (2011). We first show that the HJB equation is neither a necessary
nor a sufficient condition for a function to be the value function in this
model. (Section 2.) Specifically, for a given time preference rate, the HJB
equation has infinitely many solutions. On the other hand, at another time
preference rate, the value function is not a solution to the HJB equation.

Let us add a few words about the latter result. First, the HJB equation is a
differential equation for a real-valued function, so naturally, if the value
function is infinite, it does not satisfy the HJB equation. However, in the
above model, although the value function is finite, it is not a solution to
the HJB equation. Second, the HJB equation is a differential equation of
degenerate elliptic type, and typically, such an equation does not have a
“classical” solution. Therefore, mathematicians create the notion of the
“viscosity solution,” which are slightly weaker than usual, and use it to
analyze this kind of equation.222This notion was introduced by Lions (1982),
and used by Crandall and Lions (1983) for analyzing the HJB equation. In the
above model, however, the value function is not even a viscosity solution of
the HJB equation.

Therefore, the first problem mentioned above, “the absence of mathematical
foundations,” is serious. Perhaps we are analyzing the HJB equation and the
value function assuming that there is a relationship between them, even though
there is no relationship at all. To dispel this concern, we show that, under
economically reasonable assumptions, nothing like the above example can occur.
Specifically, we first show that under some assumptions in the optimal capital
accumulation models, the value function solves the HJB equation. (Proposition
3, Theorem 1) Second, we show that under an additional assumption, any
solution of the HJB equation that is increasing, concave, and continuously
differentiable, and satisfies the growth condition must be the same as the
value function. (Theorem 2) To combine these results, we conclude that under
several assumptions, solving the HJB equation is 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 much wider
variety of problems than the usual RCK model. First, we do not assume the
Inada condition, nor do we assume its negation. Thus, we can deal with models
that satisfy the Inada condition, or conversely, we can deal with problems
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 with respect to
consumption. Furthermore, we allow the instantaneous utility function to
depend not only on consumption but also on capital accumulation. Thus, the
results of this paper can be applied to a considerable number of macroeconomic
dynamic models.

Additionally, in the process of deriving our results, we derive an approach to
compute the solution of the original problem using the solution of the HJB
equation. (Proposition 4, Corollary 1) This result is useful even in the RCK
model in the following sense. In the usual RCK model, the analysis uses phase
diagrams for the simultaneous differential equation of the capital
accumulation equations and the Euler equations. This system is semi-stable. By
setting the initial value of consumption on a stable manifold, we can obtain a
solution of this problem. However, because the system is semi-stable, if we
apply usual methods for approximating the solution of differential equations
such as the Runge-Kutta method, the approximation error will increase due to
the unstable manifold, and the calculation will eventually fail. On the other
hand, what we have derived is a one-dimensional ordinary differential equation
(ODE) for calculating the optimal capital accumulation, and in the case of the
RCK model, its steady state is stable. Therefore, the above problem does not
occur and a usual approximate method can be used to obtain an approximate
solution.

The model we deal with in this paper does not have stochastic shocks. This is
because the first example has no stochastic shocks. In other words, even in
the optimal capital accumulation model without stochastic shocks, the HJB
equation has been discussed without any rigorous foundation to date. Of
course, in a capital accumulation model with stochastic shocks, the problem
would be much more difficult.

Although our model is a continuous-time model, our results are very similar to
those of dynamic programming in discrete-time models. For example, Proposition
3 and Theorem 1 correspond to the result that the value function solves the
Bellman equation. Theorem 2, conversely, corresponds to the result that the
solution to the Bellman equation is the value function. In the process of
producing these results, we define a function $c(k,p)$, which corresponds to
the policy function. Therefore, Proposition 4 and Corollary 1 correspond to
the result that one can derive the solution of the original problem using the
policy function. These results are all in Chapter 4 of Stokey and Lucas
(1989), and our results can be seen as a continuous-time version of those.

Finally, we consider the logarithmic AK model as an example of applying our
results. In this model, we can solve the HJB equation quite easily and use it
to obtain a solution to the original problem. Although this solution has
already been obtained by Barro and Sara-i-Martin (2003), it is interesting in
the sense that it is derived via a completely different path.

In Section 2, we present an example in which the HJB equation does not
characterize the value function. In Section 3, we elaborate on the setup of
our model and define a number of terms and assumptions. In Section 4, we
discuss the main results. In Section 5, we explain the computational examples,
and discuss several topics. In Section 6, we present concluding remarks.
Proofs of all results are given in Section 7.

##  2 A Fatal Counterexample

We consider the traditional RCK model with linear technology and linear
instantaneous utility. That is, we consider the following model:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t))dt$ subject to. $\displaystyle~{}c(t)\mbox{ is locally integrable},$
$\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0\mbox{ for all }t\geq 0,$ (1)
$\displaystyle~{}\dot{k}(t)=f(k(t))-c(t)\mbox{ a.e.},$
$\displaystyle~{}k(0)=\bar{k},$

where $\rho>0$, $\bar{k}>0$, $u(c)=c$, and $f(k)=k$. Let $V_{\rho}$ denote the
value function of the above model. Note that, the HJB equation is the
following differential equation.

$\sup_{c\geq 0}\\{(f(k)-c)V^{\prime}(k)+u(c)\\}-\rho V(k)=0.$

We first show the following facts.333We present the formal definitions of
several technical terms, including the value function and classical and
viscosity solutions of the HJB equation, in the next section. Because this
counterexample is most important in this paper, we present this before
introducing the formal definitions.

Fact 1. If $\rho=1$, then there exist infinitely many solutions of the HJB
equation.

Proof of Fact 1. Let $V(k)=ak$ for some $a\geq 1$. Then,

$\sup_{c\geq 0}\\{(f(k)-c)V^{\prime}(k)+u(c)\\}=ak=\rho V(k).$

Fact 2. The function $V_{2}$ is finite, concave, and neither the classical nor
viscosity solution of the HJB equation.

Proof of Fact 2. We first show that $V_{1}(\bar{k})=\bar{k}$. To show this,
consider the following pair of functions:

$k(t)\equiv\bar{k},\ c(t)\equiv\bar{k}.$

Then, this pair $(k(t),c(t))$ satisfies all constraints of the problem (1),
and satisfies the Euler equation

$\frac{d}{dt}(u^{\prime}\circ c)(t)=(\rho-f^{\prime}(k(t)))(u^{\prime}\circ
c)(t),$

and the transversality condition

$\lim_{t\to\infty}e^{-\rho t}u^{\prime}(c(t))f(k(t))=0,$

if $\rho=1$. Therefore, $(k(t),c(t))$ is a solution of (1) if $\rho=1$,444For
the proof of this fact, see Theorem 4 of Hosoya (2019a). and thus,

$V_{1}(\bar{k})=\int_{0}^{\infty}e^{-t}\bar{k}dt=\bar{k}.$

Second, we show that $0\leq V_{2}(\bar{k})\leq V_{1}(\bar{k})=\bar{k}$.
Because $u(c)\geq 0$ for all $c\geq 0$, we have taht $V_{2}(\bar{k})\geq 0$.
Choose any pair $(k(t),c(t))$ that satisfies all constraints of problem (1).
Then, because $u(c(t))\geq 0$ for all $t\geq 0$,

$\int_{0}^{\infty}e^{-2t}u(c(t))dt\leq\int_{0}^{\infty}e^{-t}u(c(t))dt\leq
V_{1}(\bar{k})=\bar{k},$

and thus, we have that $V_{2}(\bar{k})\leq\bar{k}$ as desired.

Therefore, $V_{2}(k)$ is finite for all $k>0$. Third, we show that the
function $V_{2}$ is concave. Choose $k_{1},k_{2}>0$ and $s\in[0,1]$, and let
$k=(1-s)k_{1}+sk_{2}$. Suppose that a pair $(k_{i}(t),c_{i}(t))$ satisfies all
constraints of the problem (1) with $\bar{k}=k_{i}$ for $i\in\\{1,2\\}$.
Define $k(t)=(1-s)k_{1}(t)+sk_{2}(t),\ c(t)=(1-s)c_{1}(t)+sc_{2}(t)$. Then,
$(k(t),c(t))$ satisfies all constraints of the problem (1) with $\bar{k}=k$.
Therefore,

$V_{2}(k)\geq\int_{0}^{\infty}e^{-2t}u(c(t))dt=(1-s)\int_{0}^{\infty}e^{-2t}u(c_{1}(t))dt+s\int_{0}^{\infty}e^{-2t}u(c_{2}(t))dt.$

Because $(k_{i}(t),c_{i}(t))$ is arbitrary,

$V_{2}(k)\geq(1-s)V_{2}(k_{1})+sV_{2}(k_{2}),$

which implies that $V_{2}$ is concave. Note that, because $V_{2}$ is concave,
this function is locally Lipschitz.

Fourth, we show that the function $V_{2}$ is not a classical solution of the
HJB equation. Suppose that $V_{2}$ is a classical solution of the HJB
equation. Then, $V_{2}$ is continuously differentiable, and the following
equation is satisfied for all $k>0$:

$\sup_{c\geq 0}\\{(k-c)V_{2}^{\prime}(k)+c\\}-2V_{2}(k)=0.$ (2)

Because the left-hand side becomes infinite if $V_{2}^{\prime}(k)<1$, we have
that $V_{2}^{\prime}(k)\geq 1$ for all $k>0$. Then, the supremum of the left-
hand side is attained at $c=0$. Therefore, (2) can be modified to the
following equation:

$kV_{2}^{\prime}(k)=2V_{2}(k).$

Because this differential equation is linear, we can use the formula of the
solution of linear ODEs, and we have that

$V_{2}(k)=V_{2}(1)k^{2}.$

Therefore, $V_{2}^{\prime}(k)<1$ if $k>0$ is sufficiently small, which is a
contradiction. Thus, $V_{2}$ is not a classical solution of the HJB equation.

Last, we show that the function $V_{2}$ is not a viscosity solution of the HJB
equation. Suppose that $V_{2}(k)$ is a viscosity solution of the HJB equation.
Note that, because $V_{2}$ is locally Lipschitz, it is absolutely continuous
on each compact interval in $\mathbb{R}_{++}$, and thus, it is differentiable
a.e.. If $V_{2}$ is differentiable at $k$, then (2) holds at $k$.555See our
arguments on the viscosity solution of the HJB equation in Section 3.
Therefore, by the same reason as in the last paragraph, we have that
$V_{2}^{\prime}(k)\geq 1$ and

$kV_{2}^{\prime}(k)=2V_{2}(k).$

Hence, again $V_{2}(k)$ is a solution of the above linear differential
equation, and thus, by Carathéodory-Picard-Lindelöf’s uniqueness theorem of
the solution, we have

$V_{2}(k)=V_{2}(1)k^{2}.$

Thus, $V_{2}$ is differentiable everywhere, and $V_{2}^{\prime}(k)<1$ if $k>0$
is sufficiently small, which is a contradiction. This completes the proof.
$\blacksquare$

What is wrong in the above counterexample?

In related studies, the reason why the value function $V(\bar{k})$ solves the
HJB equation is told as follows. First, for every $t>0$, we can obtain the
following equation:

$V(\bar{k})=\sup_{k(s),c(s)}\left\\{\int_{0}^{t}e^{-\rho s}u(c(s))ds+e^{-\rho
t}V(k(t))\right\\}.$

This equation can be transformed into

$\sup_{k(s),c(s)}\left\\{\int_{0}^{t}e^{-\rho s}u(c(s))ds+e^{-\rho
t}V(k(t))-e^{-\rho 0}V(k(0))\right\\}=0.$

If we can assume that $V$ is differentiable and $c(s)$ is continuous, then the
composition $V\circ k$ is differentiable at $s=0$, and thus

$\displaystyle 0=$
$\displaystyle~{}\sup_{k(s),c(s)}\left\\{\int_{0}^{t}e^{-\rho
s}u(c(s))ds+e^{-\rho t}V(k(t))-e^{-\rho 0}V(k(0))\right\\}$ $\displaystyle=$
$\displaystyle~{}\sup_{k(s),c(s)}\left\\{t\times\frac{d}{dT}\left.\left[\int_{0}^{T}e^{-\rho
s}u(c(s))ds+e^{-\rho T}V(k(T))\right]\right|_{T=0}+o(t)\right\\}$ (3)
$\displaystyle=$ $\displaystyle~{}\sup_{k(s),c(s)}\\{t[u(c(0))-\rho
V(k(0))+V^{\prime}(k(0))\dot{k}(0)]+o(t)\\}.$

Because $k(0)=\bar{k}$ and $\dot{k}(0)=f(\bar{k})-c(0)$, the last line depends
only on $c(0)$, and thus we have that

$0=\sup_{c}\\{t[(f(\bar{k})-c)V^{\prime}(\bar{k})+u(c)]+o(t)\\}-\rho
V(\bar{k})t.$

Dividing the right-hand side by $t$ and taking the limit $t\to 0$, we have the
HJB equation.

However, we have already obtained a counterexample $V_{2}$. Therefore, the
above “proof” is somewhat incorrect. Upon closer inspection, we find that
there 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,
this line 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 not really sure if it is actually $o(t)$.

Second, for the derivation of (3), we made some assumptions that assures the
differentiability of $V\circ k$ at $s=0$. Specifically, we assumed the
differentiability of $V$ and the continuity of $c(s)$. However, in our model,
$c(s)$ may be discontinuous, 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 of the
original problem. The first gap seems to be a problem that can be solved using
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 most models, and therefore their Theorem 2 can be applied and the
value function will be differentiable. Hence, we think that if the original
problem has a solution, both gaps will be filled in most cases. In other
words, we can imagine that the problem in our example (1) probably arises
because there is no solution in the original problem.

So, does this solve our problem? We think not. This is because in many studies
that apply the HJB equation, the solution of the original problem is not
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 equation (Fact 1)
has not been solved. In many cases, we want to obtain a value function by
solving the HJB equation, and what is needed for this is not that the value
function solves the HJB equation, but rather that a solution of the HJB
equation is actually the value function. Fact 1 would be a more powerful
obstacle than Fact 2 when trying to achieve such a result.

Thus, many studies that use the HJB equation are in a very precarious
position. A way to solve this problem is first to consider the solution of the
HJB equation as a “candidate” of the value function, rather than immediately
considering 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 HJB
equation appropriately. However, it is uncertain whether we can follow the
above procedure at any time. Perhaps proving this “candidate” is really the
value function may be tremendously difficult.

Another way to solve this problem is to prove in a general framework that a
solution of the HJB equation is the value function under several assumptions.
In the remainder of the paper, we will address this issue, and show that the
value function is a solution of the HJB equation, and that there is no other
solution provided we do not deviate from the general assumptions used in
macroeconomic theory. This result will give many studies a credible foundation
and greatly increase their persuasiveness.

##  3 The Model and Definitions

###  3.1 The Model

The model that we argue in this paper is the following.666The statement
“$\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt$ can be defined” admits that the
value of this integral may be $\pm\infty$.

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ subject to. $\displaystyle~{}c(t)\mbox{ is locally
integrable},$ $\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$ (4)
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\mbox{ can be
defined},$ $\displaystyle~{}\dot{k}(t)=F(k(t),c(t))\mbox{ a.e.},$
$\displaystyle~{}k(0)=\bar{k}.$

Note that, if the instantaneous utility function $u(c,k)$ is independent of
$k$ and the technology function $F(k,c)=f(k)-dk-c$, then the problem (4) is
the traditional RCK model. In this paper, we do not need the Inada condition,
and thus this model admits the divergence of optimal growth paths. Note also
that, under the above constraints, $k(t)$ must be absolutely continuous in
every compact interval in $\mathbb{R}_{+}$, because the solution of the
differential equation must be absolutely continuous in every compact interval.

We make several assumptions of the above model.

Assumption 1. $\rho>0$.

Assumption 2. The instantaneous utility function
$u:\mathbb{R}^{2}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is a continuous and
concave function on $\mathbb{R}^{2}_{+}$. Moreover, $u(c,k)$ is nondecreasing
on $\mathbb{R}^{2}_{+}$, and increasing in $c$ and continuously differentiable
on $\mathbb{R}^{2}_{++}$.777Because $u$ is increasing in $c$ on
$\mathbb{R}^{2}_{++}$, if $(c,k)\in\mathbb{R}^{2}_{++}$, then
$u(c,k)>u(2^{-1}c,k)\geq-\infty$. Therefore, $u(c,k)\in\mathbb{R}$ for every
$(c,k)\in\mathbb{R}^{2}_{++}$. Furthermore, there exists $c>0$ such that
$u(c,0)>-\infty$.

Assumption 3. The technology function $F:\mathbb{R}^{2}_{+}\to\mathbb{R}$ is a
continuous and concave function that satisfies $F(0,0)=0$. Moreover, $F$ is
decreasing in $c$ and there exist $d_{1},d_{2}\geq 0$ such that
$F(k,c)>-d_{1}k-d_{2}c$ for every $(k,c)$ such that $k>0$ and $c\geq 0$, and
for every $c\geq 0$, there exists $k>0$ such that $F(k,c)>F(0,c)$.888The last
assumption means that for every $\bar{c}>0$, there exists $k^{*}>0$ such that
for every $c\in[0,\bar{c}]$, $k\mapsto F(k,c)$ is increasing on $[0,k^{*}]$.
Actually, we can choose
$k^{*}=\min_{c\in[0,\bar{c}]}\min_{k}\arg\max\\{F(k,c)|0\leq k\leq 1\\}.$

Assumption 4. The function $\frac{\partial u}{\partial c}(c,k)$ is decreasing
in $c$ on $\mathbb{R}^{2}_{++}$. Moreover, $\lim_{c\to 0}\frac{\partial
u}{\partial c}(c,k)=+\infty$ and $\lim_{c\to\infty}\frac{\partial u}{\partial
c}(c,k)=0$ for every $k>0$. Furthermore, for every $k>0$ and $M>0$, the
function $c\mapsto\frac{\partial u}{\partial k}(c,k)$ is bounded on $]0,M]$.

Assumption 5. The technology function $F$ is continuously differentiable in
$c$ on $\mathbb{R}^{2}_{++}$.

###  3.2 Interpretation of Our Model

There are two types of optimal capital accumulation models: centralized and
decentralized. In this paper, we deal with a centralized model, which is
connected to the corresponding decentralized model by the fundamental theorem
of welfare economics. In the centralized macroeconomic dynamic model, $k(t)$
represents the capital stock and $c(t)$ represents the consumption, and both
are assumed to be non-negative. Some models assume that $k(t)$ is continuously
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 assume
continuous 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 most common in
the literature on optimal capital accumulation models. We shall check 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 logarithmic function
$\log c$, boundedness of $u$ is not usually assumed, and $u(0)=-\infty$ is
allowed. On the other hand, $u^{\prime}>0,u^{\prime\prime}<0$, $\lim_{c\to
0}u^{\prime}(c)=+\infty$, and $\lim_{c\to 0}u^{\prime}(c)=0$ are usually
assumed. With a little checking, readers will notice that this model satisfies
our Assumptions 2 and 4.

Next, in the RCK model, the relationship between the capital stock and
consumption is determined as follows:

$\dot{k}(t)=f(k(t))-dk(t)-c(t),$

where $d\geq 0$ denotes the capital wastage rate. There are two economic
relationships behind this equation. First, the function $f$ is called the
production function, 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 $c$ denote the consumption and $i$ denote the
investment, we obtain the first relationship:

$f(k(t))=c(t)+i(t).$

On the other hand, the speed of increase of the capital stock $\dot{k}(t)$ is
equal to $i(t)$ minus capital depletion $dk(t)$. Thus, we obtain the second
relationship:

$\dot{k}(t)=i(t)-dk(t).$

By connecting these two relationships, we have that

$\dot{k}(t)=f(k(t))-dk(t)-c(t),$

as desired.

Usually, it is assumed that $f$ is concave, increasing, and continuous,
$f(0)=0$, and there exists $k>0$ such that $f(k)>dk$. By setting
$F(k,c)=f(k)-dk-c$, $d_{1}=d$, and $d_{2}=1$, readers can check that this
model satisfies our Assumptions 3 and 5. Thus, our model (4) includes the RCK
model. Moreover, unlike many other studies, our assumptions do not assume
either differentiability of the production function $f$ or the Inada
condition. Therefore, the scope of our model is sufficiently wide at this
point.999Because $f^{\prime}(k)<d$ can occur in the RCK model, we do not
assume that $F(k,c)$ is increasing in $k$ in Assumption 3.

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 the
instantaneous utility. One typical example of such a utility is
$u(c,k)=(c^{\rho}+k^{\rho})^{\frac{A}{\rho}}$, where $0<A<1$ and
$\rho<1,\rho\neq 0$. This is a variant of the CES utility function.101010If
$\rho<0$, then we define $u(c,0)=u(0,k)=0$. Note that, if $A=1$, then
$\lim_{c\to\infty}\frac{\partial u}{\partial c}(c,k)=1$ when $\rho>0$, and
$\lim_{c\to 0}\frac{\partial u}{\partial c}(c,k)=1$ when $\rho<0$, and thus
Assumption 4 is violated in any cases.

We should explain the assumption of the existence of $c>0$ such that
$u(c,0)>-\infty$. Economically, this assumption means that a lack of capital
stock is not fatal to people, and only lack of consumption can be fatal.
Mathematically, this assumption is needed for step 4 of the proof of
Proposition 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 in the RCK
model. For example, we can handle a model in which too high consumption has a
strong negative impact on the growth rate of $k$, and a model in which
choosing too high investment cannot be digested at once and used for
production. The former model is given by the function $F(k,c)=f(k)-dk-h(c)$,
where $h$ is an increasing, continuously differentiable, convex function that
satisfies $h(0)=0$ and has the Lipschitz constant $d_{2}$. The latter model is
given by the function $F(k,c)=h(f(k)-c)-dk$, where $h$ is an increasing,
continuously differentiable, and concave function that satisfies $h(0)=0$ and
has the Lipschitz constant $d_{2}$. In both cases, it is easy to prove that
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.

###  3.3 Admissibility and the Value Function

We say that a pair of real-valued functions $(k(t),c(t))$ defined on
$\mathbb{R}_{+}$ admissible if $k(t)$ is absolutely continuous on every
compact interval, $c(t)$ is locally integrable, $k(t)\geq 0,c(t)\geq 0$,
$\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt$ can be defined, and

$\dot{k}(t)=F(k(t),c(t))\ \mbox{a.e.}.$ (5)

Note that, if $k(t)$ is absolutely continuous on every compact interval, it is
differentiable almost everywhere and $\int_{a}^{b}\dot{k}(t)dt=k(b)-k(a)$ for
all $a,b$ with $0\leq a<b$.

Let $A_{\bar{k}}$ denote the set of all admissible pairs such that
$k(0)=\bar{k}$. Using the notation of $A_{\bar{k}}$, we can simplify the model
(4) as follows:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ subject to. $\displaystyle~{}(k(t),c(t))\in A_{\bar{k}}.$

Let

$\bar{V}(\bar{k})=\sup\left\\{\left.\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\right|(k(t),c(t))\in A_{\bar{k}}\right\\}.$

We call this function $\bar{V}$ the value function of the problem
(4).111111Note that $\bar{V}$ is defined on $\mathbb{R}_{+}$. Later, we will
show that, under mild assumptions, $A_{\bar{k}}$ is nonempty for all
$\bar{k}\geq 0$.

We call a pair $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$ a solution if and only if
the following two requirements hold. First,

$\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt\in\mathbb{R}.$

Second, for every pair $(k(t),c(t))\in A_{\bar{k}}$,

$\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt\geq\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt.$

These two requirements can be summarized by the following formula:

$\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt=\bar{V}(\bar{k})\in\mathbb{R}.$

Note that, for all $(k(t),c(t))\in A_{\bar{k}}$ such that
$\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\in\mathbb{R}$ and $T>0$, we have

$\displaystyle\int_{0}^{T}e^{-\rho t}u(c(t),k(t))dt=$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt-\int_{T}^{\infty}e^{-\rho t}u(c(t),k(t))dt$
$\displaystyle\geq$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt-e^{-\rho T}\bar{V}(k(T)).$

Particularly, if $\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt>M$, then for all
$T>0$,

$\int_{0}^{T}e^{-\rho t}u(c(t),k(t))dt>M-e^{-\rho T}\bar{V}(k(T)).$

Therefore, if $\bar{V}(\bar{k})$ is finite, then for every $\varepsilon>0$,
there exists a pair $(k(t),c(t))\in A_{\bar{k}}$ such that

$\int_{0}^{T}e^{-\rho t}u(c(t),k(t))dt>\bar{V}(\bar{k})-e^{-\rho
T}\bar{V}(k(T))-\varepsilon$

for every $T>0$.

###  3.4 The HJB equation

The HJB equation is given by the following equation.

$\sup_{c\geq 0}\\{F(k,c)V^{\prime}(k)+u(c,k)\\}-\rho V(k)=0.$ (6)

A continuous function $V:\mathbb{R}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is
called a classical solution of the HJB equation if and only if $V$ is
continuously differentiable on $\mathbb{R}_{++}$ and equation (6) holds for
every $k>0$.

It is known that, in many models of the dynamic control problem, there exists
no classical solution of the HJB equation. Hence, we should extend the notion
of the solution. First, a function
$V:\mathbb{R}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is called a viscosity
subsolution of (6) if and only if it is real-valued and upper semi-continuous
on $\mathbb{R}_{++}$, and for every $k>0$ and every continuously
differentiable function $\varphi$ defined on a neighborhood of $k$ such that
$\varphi(k)=V(k)$ and $\varphi(k^{\prime})\geq V(k^{\prime})$ whenever both
sides are defined,

$\sup_{c\geq 0}\\{F(k,c)\varphi^{\prime}(k)+u(c,k)\\}-\rho V(k)\leq 0.$

Second, a function $V:\mathbb{R}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is called
a viscosity supersolution of (6) if and only if it is real-valued and lower
semi-continuous on $\mathbb{R}_{++}$, and for every $k>0$ and every
continuously differentiable function $\varphi$ defined on a neighborhood of
$k$ such that $\varphi(k)=V(k)$ and $\varphi(k^{\prime})\leq V(k^{\prime})$
whenever both sides are defined,

$\sup_{c\geq 0}\\{F(k,c)\varphi^{\prime}(k)+u(c,k)\\}-\rho V(k)\geq 0.$

If a continuous function $V:\mathbb{R}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is
both viscosity sub- and supersolution of (6), then $V$ is called a viscosity
solution of (6).

Suppose that $V$ is a viscosity solution of the HJB equation and is
differentiable at $k>0$. Then, it is known that

$\sup_{c\geq 0}\\{F(k,c)V^{\prime}(k)+u(c)\\}-\rho V(k)=0.$

See Proposition 1.9 of ch.2 of Bardi and Dolcetta (2008).

###  3.5 Pure Accumulation Path

Consider the following differential equation.

$\dot{k}(t)=F(k(t),0),\ k(0)=\bar{k}.$ (7)

Let $k^{+}(t,\bar{k})$ denote the solution of the above equation defined on
$\mathbb{R}_{+}$. This function $k^{+}(t,\bar{k})$ is called the pure
accumulation path. Later we will show the following lemma.

Lemma 1. Under Assumption 3, the pure accumulation path $k^{+}$ is defined on
the set $\mathbb{R}_{+}\times\mathbb{R}_{++}$, and $\inf_{t\geq
0}k^{+}(t,\bar{k})>0$ for every $\bar{k}>0$.

Let $V:\mathbb{R}_{++}\to\mathbb{R}$. The following requirement of $V$ is
called the growth condition.

$\lim_{T\to\infty}e^{-\rho T}V(k^{+}(T,\bar{k}))=0\mbox{ for all }\bar{k}>0.$
(8)

Let $\mathscr{V}$ be the space of all functions
$V:\mathbb{R}_{+}\to\mathbb{R}$ that is continuous, increasing, and concave,
and satisfies the growth condition. Note that, by concavity, every
$V\in\mathscr{V}$ is locally Lipschitz on $\mathbb{R}_{++}$.

###  3.6 Subdifferentials and Left- and Right-Derivatives

In this paper, we heavily use subdifferential calculus. We introduce the
notion of subdifferential and several results. For the proofs of these
results, see textbooks on convex analysis, such as Rockafeller (1996).

Suppose that a function $G:U\to\mathbb{R}$ is concave,
$U\subset\mathbb{R}^{n}$ is convex, and the interior $V\subset U$ is nonempty.
Choose any $x\in V$. We define

$\partial G(x)=\\{p\in\mathbb{R}^{n}|G(y)-G(x)\leq p\cdot(y-x)\mbox{ for all
}y\in U\\}.$

Then, we can show that $\partial G(x)$ is nonempty. The set-valued mapping
$\partial G$ is called the subdifferential of $G$.121212Formally, the
subdifferential is defined for not concave but convex functions, and thus the
inequality in the definition is reversed. In this view, the name
‘subdifferential’ may be not appropriate, and ‘superdifferential’ may be more
suitable. However, in the literature of economics, these two notions are not
distinguished, and thus our $\partial G$ is traditionally called
‘subdifferential’. It is known that $G$ is differentiable at $x$ if and only
if $\partial G(x)$ is a singleton, and if so, $\partial G(x)=\\{DG(x)\\}$.

If $n=1$, then define the left- and right-derivatives $D_{-}G(x),\ D_{+}G(x)$
such as

$D_{-}G(x)=\lim_{y\uparrow x}\frac{G(y)-G(x)}{y-x},\
D_{+}G(x)=\lim_{y\downarrow x}\frac{G(y)-G(x)}{y-x}.$

Note that, if $G$ is concave, then $\frac{G(y)-G(x)}{y-x}$ is nonincreasing in
$y$, and thus

$D_{-}G(x)=\inf_{t>0}\frac{G(x-t)-G(x)}{-t},\
D_{+}G(x)=\sup_{t>0}\frac{G(x+t)-G(x)}{t},$

which implies that both $D_{-}G(x),D_{+}G(x)$ are defined and real numbers. It
is known that $\partial G(x)=[D_{+}G(x),D_{-}G(x)]$.

Recall that, under Assumption 3, our $F$ is concave. In this case, the
functions $c\mapsto F(k,c)$ and $k\mapsto F(k,c)$ are also concave, and thus
the ‘partial’ subdifferential can be considered. Let

$\partial_{k}F(k,c)=\\{p|F(k^{\prime},c)-F(k,c)\leq p(k^{\prime}-k)\mbox{ for
all }k^{\prime}\geq 0\\},$ $\partial_{c}F(k,c)=\\{p|F(k,c^{\prime})-F(k,c)\leq
p(c^{\prime}-c)\mbox{ for all }c^{\prime}\geq 0\\}.$

The partial left- and right-derivatives can be defined in the same manner. For
example,

$D_{k,+}F(k,c)=\sup_{t>0}\frac{F(k+t,c)-F(k,c)}{t},$
$D_{k,-}F(k,c)=\inf_{t>0}\frac{F(k-t,c)-F(k,c)}{-t}.$

We note one more fact. Suppose that $p\geq r\geq
q,p\in\partial_{k}F(k_{1},c),q\in\partial_{k}F(k_{2},c)$ and $k_{1}<k_{2}$. We
show that there exists $k\in[k_{1},k_{2}]$ such that $r\in\partial_{k}F(k,c)$.
If $r\geq D_{k,+}F(k_{1},c)$, then $r\in\partial_{k}F(k_{1},c)$. If $r\leq
D_{k,-}F(k_{2},c)$, then $r\in\partial_{k}F(k_{2},c)$. Therefore, we assume
that $D_{k,-}F(k_{2},c)<r<D_{k,+}F(k_{1},c)$. Define $g(k)=F(k,c)-rk$. Then,
$D_{+}g(k_{1})>0$ and $D_{-}g(k_{2})<0$, and thus, there exists
$k\in]k_{1},k_{2}[$ such that
$g(k)=\max_{k^{\prime}\in[k_{1},k_{2}]}g(k^{\prime})$. By the definition of
subdifferential, we have that $0\in\partial g(k)$, and thus
$r\in\partial_{k}F(k,c)$, as desired.

###  3.7 Propositions

The following propositions can be proved using only Assumptions 1-4.

Proposition 1. Suppose that Assumptions 1-4 hold. Then, there exists a
positive continuous function $c^{*}(p,k)$ defined on $\mathbb{R}^{2}_{++}$
such that for all $(p,k)\in\mathbb{R}^{2}_{++}$,

$F(k,c^{*}(p,k))p+u(c^{*}(p,k),k)=\sup_{c\geq 0}\\{F(k,c)p+u(c,k)\\}.$

Note that, if $F(k,c)=f(k)-dk-c$ and $u(c,k)=u(c)$, then by the first-order
condition, we have that $c^{*}(p,k)=(u^{\prime})^{-1}(p)$, and thus
$c^{*}(p,k)$ is independent 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 $c$ such that $h^{\prime}(c)p=u^{\prime}(c)$, and
thus $c^{*}(p,k)$ is also independent of $k$.

Proposition 2. Suppose that Assumptions 1-3 hold. Then,
$\bar{V}(\bar{k})>-\infty$ for every $\bar{k}>0$ and the function $\bar{V}$ is
nondecreasing and concave.

By Proposition 2, we have $\bar{V}(k)\in\mathbb{R}$ for some $k>0$ if and only
if $\bar{V}(k)\in\mathbb{R}$ for every $k>0$. We say that $\bar{V}$ is finite
if $\bar{V}(k)\in\mathbb{R}$ for every $k>0$.131313Note that, $\bar{V}(0)$ may
be $-\infty$ even if $\bar{V}$ is finite.

Readers may think that Proposition 2 is obvious. However, the proof of this
proposition is quite difficult. See the proof section.

##  4 Results

In this section, we analyze the HJB equation for a characterization of the
value function. We prohibit ourselves from making assumptions on the solution
explicitly. Assumptions must be made for properties of primitives $\rho,u,F$
in our model (4), and, for example, assumption for the existence of the
solution in (4) is not appropriate. The reason why we restrict ourselves is
simple: in many situation, to ensure the existence of a solution of (4) is
tremendously difficult. Hence, we prohibit ourselves from assuming the
existence of a solution, even though under the existence assumption of the
solution, the proofs of results become quite easy.

###  4.1 The Necessity of the HJB Equation

First, we show the following result.

Proposition 3. Suppose that Assumptions 1-4 hold. If the value function
$\bar{V}$ is finite, then it is increasing on $\mathbb{R}_{+}$, and it is a
viscosity solution of the HJB equation. If, in addition, Assumption 5 holds,
then $\bar{V}$ is a classical solution of the HJB equation.

Note that our counterexample (1) satisfies Assumptions 1-3 and 5, and in this
example, the value function $V_{2}$ is finite. Therefore, the violation of
Assumption 4 is crucial for (1).

Because Proposition 3 requires the finiteness of the value function, we want
an additional assumption that ensures the finiteness of the value function.
First, define the CRRA function as follows:

$u_{\theta}(x)=\begin{cases}\frac{x^{1-\theta}-1}{1-\theta}&\mbox{if
}\theta\neq 1,\\\ \log x&\mbox{if }\theta=1,\end{cases}$

where $\theta>0$.141414This function is the solution of the following
differential equation:  $-x\frac{u^{\prime\prime}(x)}{u^{\prime}(x)}=\theta,\
u(1)=0,\ u^{\prime}(1)=1,$ where the left-hand side is sometimes called the
relative risk aversion. Therefore, this function is called the constant
relative risk aversion function, and abbreviatedly, the CRRA function.

Assumption 6. There exist $k^{*}>0,k^{+}>0,c^{*}\geq
0,\gamma>0,\delta>0,\theta>0,a>0,b\geq 0,C\in\mathbb{R}$ such that

$\displaystyle~{}F(k^{*},c^{*})>0,$ (9)
$\displaystyle~{}(\gamma,-\delta)\in\partial F(k^{*},c^{*}),\
0<D_{k,+}F(k^{+},0)\leq\gamma,$ (10)
$\displaystyle~{}\rho-(1-\theta)\gamma>0,$ (11) $\displaystyle~{}u(c,k)\leq
au_{\theta}(c)+bu_{\theta}(k)+C\mbox{ for all }c>0,k>0.$ (12)

We stress that Assumption 6 is not strong. For example, in the RCK model,
$F(k,c)=f(k)-dk-c$ and $u(c,k)=u(c)$. Suppose that $f(k)$ is differentiable.
Then, the condition $(\gamma,-\delta)\in\partial F(k^{*},c^{*})$ means that
$\gamma=f^{\prime}(k^{*})-d$ and $\delta=1$, and because
$D_{k,+}F(k^{+},0)=f^{\prime}(k^{+})-d$, (10) is satisfied if $k^{*}=k^{+}$
and $f^{\prime}(k^{*})>d$. Because there is no restriction on $c^{*}$,
requirements (9) and (10) are satisfied if $k^{*}$ is sufficiently small and
$c^{*}=0$. Therefore, the actual restrictions are only (11) and (12).
Moreover, if $f$ satisfies the Inada condition,151515The function $f$
satisfies the Inada condition if and only if
$f^{\prime}(\mathbb{R}_{++})=\mathbb{R}_{++}$. Because $f$ is concave, it is
equivalent to the following two requirements: $\lim_{k\to
0}f^{\prime}(k)=+\infty,\ \lim_{k\to\infty}f^{\prime}(k)=0$. we can choose
$\gamma$ so small that $\rho-(1-\theta)\gamma>0$, and thus the actual
restriction is the existence of $\theta>0$ such that $u(c)\leq
au_{\theta}(c)+C$ for all $c>0$. This is just a mild restriction on $u$.
Additionally, if we can choose $\theta\geq 1$, then (11) is automatically
satisfied and vanishes even when $f(k)$ does not satisfy the Inada condition.

Note that, if $u(c,k)=u_{\theta}(c)$ and $F(k,c)=\gamma k-c$, then (11)
provides a necessary and sufficient condition for the finiteness of the value
function.161616The technology $F(k,c)=\gamma k-c$ is called the AK technology.
Sufficiency is proved in ch.4 of Barro and Sara-i-Martin (2003), and we can
easily show necessity. Therefore, we guess that (11) is a crucial requirement
for the finiteness of the value function.

On the other hand, if $\sup_{c,k}u(c,k)<+\infty$, the requirement (12) is
automatically satisfied for every $\theta\in]0,1[$, because
$u_{\theta}(0)>-\infty$ when $0<\theta<1$. Therefore, again our requirements
are automatically satisfied. Next, suppose that
$u(c,k)=(c^{\rho}+k^{\rho})^{\frac{A}{\rho}}$ and $\rho<1,\ \rho\neq 0,\
0<A<1$. If $\rho>0$, then

$u(c,k)\leq(c^{\rho}+k^{\rho})^{\frac{A-1}{\rho}}2^{\frac{1}{\rho}-1}(c+k-2),$

and we can easily show that the right-hand side also satisfies (12) for
$\theta=1-A$. If $\rho<0$, then because $x\mapsto x^{\frac{A}{\rho}}$ is
decreasing, by the formula for arithmetic-geometric means, we have that

$u(c,k)\leq 2^{\frac{A}{\rho}}(ck)^{\frac{A}{2}}\leq
2^{\frac{A}{\rho}-1}(c^{A}+k^{A}),$

and the right-hand side satisfies (12) for $\theta=1-A$. Therefore, there are
sufficiently many models that satisfy Assumption 6.

The following is our first main result.

Theorem 1. Suppose that Assumptions 1-6 hold. Then $\bar{V}\in\mathscr{V}$,
and $\bar{V}$ is a classical solution of the HJB equation.

We make some remarks on Theorem 1. First, in many related studies on the HJB
equation in control problems, $u$ is assumed to be bounded. Under this
condition, the value function is automatically bounded and finite, and thus
there are many techniques that can be used in the proof. See, for example,
ch.3 of Bardi and Dolcetta (2008). However, we want to consider the 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 $u(c)=\log c$ is
not excluded for our assumptions: set $\theta=1$ and check (11) and (12) of
Assumption 6.

In Theorem 1, the value function automatically satisfies the growth condition
(8). If (4) violates Assumptions 1-6, then (8) may be not satisfied even when
the value function is a classical solution of the HJB equation. For example,
let $\rho=1,u(c,k)=c$ and $F(k,c)=k-c$. Then, we have already obtained the
value function $\bar{V}(\bar{k})=\bar{k}$ in section 2. Because
$k^{+}(t,\bar{k})=\bar{k}e^{t}$, we have that (8) is violated. However, we
think that this is a pathological example.

Meanwhile, if $\rho=2,u(c,k)=c$ and $F(k,c)=k-c$, then we have already shown
that $0\leq\bar{V}(\bar{k})\leq\bar{k}$ and $k^{+}(t,\bar{k})=\bar{k}e^{t}$,
which implies that $\bar{V}$ satisfies (8). However, in this case $\bar{V}$
does not solve the HJB equation. Therefore, we have also obtained a model in
which $\bar{V}\in\mathscr{V}$ and $\bar{V}$ violates the HJB equation. This is
another pathological example.

###  4.2 Sufficiency of the HJB Equation and Construction of the Solution

In this subsection, we examine the sufficiency of the HJB equation to be the
value function. We need an additional assumption.

Assumption 7. Both $\frac{\partial F}{\partial c}$ and $\frac{\partial
u}{\partial c}$ are continuously differentiable in $k$ on
$\mathbb{R}^{2}_{++}$. Moreover, there exists $k>0$ such that $\inf_{c\geq
0}D_{k,+}F(k,c)>\rho$.

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
$f^{\prime}(k)>\rho$, then the second assertion of Assumption 7 is also
satisfied. Therefore, this assumption 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 also that
$V\in\mathscr{V}$ is a classical solution of the HJB equation. Choose any
$\bar{k}>0$, and consider the following differential equation:

$\dot{k}(t)=F(k(t),c^{*}(V^{\prime}(k(t)),k(t))),\ k(0)=\bar{k}.$ (13)

Then, there exists a solution $k^{*}(t)$ of the above equation defined on
$\mathbb{R}_{+}$, and for every such solution, $\inf_{t\geq 0}k^{*}(t)>0$. If
we define $c^{*}(t)=c^{*}(V^{\prime}(k^{*}(t)),k^{*}(t))$, then $c^{*}(t)$ is
continuous and $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$. Moreover,

$V(\bar{k})=\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt,$

and for every $(k(t),c(t))\in A_{\bar{k}}$, if $\inf_{t\geq 0}k(t)>0$, then

$\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\leq V(\bar{k}).$

To combine this result and the preceding results, we obtain the following
theorem.

Theorem 2. Suppose that Assumptions 1-7 hold. Then, $\bar{V}$ is the unique
solution of the HJB equation in $\mathscr{V}$.

Therefore, under Assumptions 1-7, the HJB equation is the perfect
characterization for the value function in the function space $\mathscr{V}$.

Using Theorem 2 and Proposition 4, we obtain the following corollary.

Corollary 1. Suppose that Assumptions 1-7 hold. Consider the following
differential equation:

$\dot{k}(t)=F(k(t),c^{*}(\bar{V}^{\prime}(k(t)),k(t))),\ k(0)=\bar{k}.$ (14)

Then, there exists a solution $k^{*}(t)$ of (14) defined on $\mathbb{R}_{+}$.
Moreover, if we define $c^{*}(t)=c^{*}(\bar{V}^{\prime}(k^{*}(t)),k^{*}(t))$,
then $(k^{*}(t),c^{*}(t))$ is a solution of the model (4).

We believe that the equation (14) is useful for analyzing the model, even if
the model is the RCK model with the Inada condition. In the RCK model, the
solution is characterized by the simultaneous differential equation of the
technology constraint (5) and the following Euler equation:

$\frac{d}{dt}(u^{\prime}\circ c)(t)=[\rho-f^{\prime}(k(t))](u^{\prime}\circ
c)(t).$

In this simultaneous system, the initial value of $c(t)$ is not specified, and
this characterization is incomplete. Therefore, to solve the model, we need an
additional requirement, and usually the transversality condition is used as
this requirement.

However, there is a problem. In the RCK model, there uniquely exists a steady
state, and this is not stable but only semistable. In a semistable system, any
simple numerical computation method for the solution such as the Runge-Kutta
method does not work well, because the error diverges as $t$ increases.
Meanwhile, the equation (14) is a simple one-dimensional differential
equation, and the steady state is stable. Therefore, the error of numerical
calculation converges to zero as $t$ increases. Hence, using (14), we can
easily obtain an approximate solution by a usual numerical computation method.

##  5 Discussions

###  5.1 Example of the Actual Computation

In this subsection, we provide an example that demonstrates that our results
work well.

Example (logarithmic AK model). Let $u(c,k)=\log c$ and $F(k,c)=\gamma k-c$,
where $\gamma>\rho$. By Theorem 2, $\bar{V}$ is the unique solution of the HJB
equation in $\mathscr{V}$. By the first-order condition,

$c^{*}(p,k)=\frac{1}{p}.$

Therefore, the HJB equation is

$\gamma k\bar{V}^{\prime}(k)-1-\log\bar{V}^{\prime}(k)=\rho\bar{V}(k).$

We guess that $\bar{V}(k)$ is twice continuously differentiable. Thus, to
differentiate both sides, we have

$\gamma\bar{V}^{\prime}(k)+\gamma
k\bar{V}^{\prime\prime}(k)-\frac{\bar{V}^{\prime\prime}(k)}{\bar{V}^{\prime}(k)}=\rho\bar{V}^{\prime}(k),$

and thus, $\bar{V}^{\prime}(k)$ solves the following differential equation:

$(\gamma k-(x(k))^{-1})x^{\prime}(k)=-(\gamma-\rho)x(k).$

We guess that $(x(k))^{-1}=\rho k$. Then, $x(k)=\rho^{-1}k^{-1}$ and
$x^{\prime}(k)=-\rho^{-1}k^{-2}$. We can easily verify that this function is
actually the solution of the above equation. Hence, we obtain

$\bar{V}^{\prime}(k)=\rho^{-1}k^{-1},$

and thus,

$\bar{V}(k)=\rho^{-1}\log k+C$

for some constant $C$. Using the HJB equation itself, we have that

$\gamma\rho^{-1}-1+\log\rho+\log k=\log k+\rho C,$

and thus,

$C=\rho^{-1}[\log\rho+\gamma\rho^{-1}-1].$

In this case, the differential equation (14) is

$\dot{k}(t)=\gamma k(t)-\rho k(t)=(\gamma-\rho)k(t),\ k(0)=\bar{k},$

and thus, by Corollary 1,

$k^{*}(t)=\bar{k}e^{(\gamma-\rho)t},\ c^{*}(t)=\rho\bar{k}e^{(\gamma-\rho)t}$

is the solution of the problem (4). This completes the calculation.

###  5.2 The Magic of Capital

Suppose that $\bar{V}(0)=-\infty$. If $(k(t),c(t))\in A_{\bar{k}}$ and
$k(T)=0$, then

$\int_{T}^{\infty}e^{-\rho t}u(c(t),k(t))dt\leq e^{-\rho
T}\bar{V}(0)=-\infty,$

and thus, if $\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt>-\infty$, then
$k(T)>0$ for all $T>0$. Therefore, if we know that $\bar{V}(0)=-\infty$, it is
very useful to control the behavior of $k(t)$ when we analyze the problem. In
this connection, readers may have a conjecture that if $u(0,0)=-\infty$, then
$\bar{V}(0)=-\infty$. However, Hosoya (2019b) found that this conjecture is
incorrect.

Suppose that $\rho=1,\ F(k,c)=\sqrt{k}-c$ and $u(c,k)=-\frac{1}{\sqrt{c}}$.
Then, Assumptions 1-7 hold. Consider a pair $(k(t),c(t))$ defined as follows:

$k(t)=\frac{t^{2}}{16},\ c(t)=\frac{t}{8}.$

We can easily check that $(k(t),c(t))\in A_{0}$. Moreover,

$\displaystyle 0\geq$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ $\displaystyle\geq$
$\displaystyle~{}-\sqrt{8}\left[\int_{0}^{1}\frac{1}{\sqrt{t}}dt+\int_{1}^{\infty}e^{-t}dt\right]$
$\displaystyle=$ $\displaystyle~{}-\sqrt{32}-\sqrt{8}e^{-1}>-\infty,$

which implies that $\bar{V}(0)>-\infty$.

In the above example, the technology is a sort of RCK technology with
$f(k)=\sqrt{k}$, which corresponds to a sort of Cobb=Douglas technology
function. In this technology, $f(0)=0$, and thus it is thought that if the
initial capital stock is absent, then people cannot produce anything. However,
the above example shows that actually people can produce something from
nothing. We call this phenomenon the magic of capital. Because the above
technology is Cobb=Douglas and the instantaneous utility function is a
positive affine transform of a CRRA function with $\theta=\frac{3}{2}$, we
have that this magic is a commonplace event.

The magic of capital makes our proofs in this paper difficult. For example,
suppose that $(k(t),c(t))\in A_{\bar{k}}$ and $\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\geq M$ for some $M$. If $\bar{V}(0)=-\infty$, then $k(t)>0$
for all $t\geq 0$. However, $\bar{V}(0)$ may be greater than $-\infty$ even
when $u(0,k)=-\infty$. Therefore, we cannot prohibit $k(t)=0$ even in this
case.

###  5.3 Comparison with Related Literatures

Usually, the HJB equation is written as a second-order degenerate elliptic
differential equation. In fact, in stochastic economic dynamic models, the HJB
equation becomes a second-order differential equation. Malliaris and Brock
(1988), for example, deal with this kind of equation. This is probably because
of the use of Ito’s formula in the middle of the derivation. On the other
hand, since there is no stochastic variation in the dynamic model treated in
this paper, the HJB equation is just a first-order differential equation.

The case in which the HJB equation does not have a classical solution has been
pointed out in many studies. Therefore, since Lions (1982), it has been common
to use the viscosity solution as a solution concept in the study of degenerate
elliptic differential equations. In this paper, however, we showed that the
HJB equation has a classical solution under assumptions that are not so strong
in macroeconomic dynamics. It is unclear whether this is also true in other
fields, such as models of search theory.

As discussed in the introduction, the HJB equation has mostly been studied 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 variational
problems with infinite time-interval, where the instantaneous utility function
is assumed to be bounded. This assumption is by far the easiest to use, and it
dramatically simplifies the proof of our Proposition 3, for example. However,
the most widely used instantaneous utility functions in macroeconomic dynamic
models are the class of CRRA functions that include the logarithmic function.
Any assumption that excludes this class, no matter how useful it may be,
cannot be placed. This is the main reason why the proofs in this paper is by
far the most difficult. To the best of our knowledge, the results in this
paper are the first theoretical foundations of the HJB equation in
macroeconomic dynamics that can be applied to a class of instantaneous utility
functions, including the logarithmic function.

##  6 Concluding Remarks

In this paper, we presented an example of economic dynamics in which the HJB
equation is neither necessary nor sufficient for a function to be the value
function. Because this problem is serious, we present a mathematical
foundation of the HJB equation in economic dynamics, and showed that under
Assumptions 1-7, the HJB equation gives a perfect characterization of the
value function in some class of functions. Moreover, we presented a new method
for obtaining the solution of the model using the value function.

There are several future tasks. First, in this paper, we assumed the existence
of the global Lipschitz constant $d_{2}>0$ of the technology function $F$ with
respect to $c$. However, we may be able to omit this requirement using the
additional result for the existence of the solution of differential equations.

Second, we want to extend our result to some multidimensional models and
stochastic models.

Third, we want to obtain a simple method for gaining an approximate solution
of the value function. In discrete-time models, there is a famous
approximation method that uses Blackwell’s inequality and the contraction
mapping theorem. We want to obtain a counterpart to this result in a
continuous-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 be
treated. Hence, we want to relax these constraints.

##  7 Proofs

###  7.1 Knowledge on Ordinary Differential Equations

In this section, we frequently use knowledge on ODEs. Hence, we note basic
knowledge on ODE for readers.

First, consider the following differential equation:

$\dot{x}(t)=h(t,x(t)),$ (15)

where $\dot{x}$ denotes $\frac{dx}{dt}$. We assume that
$h:U\to\mathbb{R}^{n}$, $U\subset\mathbb{R}_{+}\times\mathbb{R}^{n}$, and the
relative interior of $U$ in $\mathbb{R}_{+}\times\mathbb{R}^{n}$ is nonempty
(denoted by $V$). We call a set $I\subset\mathbb{R}$ an interval if and only
if $I$ is a convex set of $\mathbb{R}$ that includes at least two points. We
say that a function $x:I\to\mathbb{R}^{n}$ is a solution of (15) if and only
if, 1) $I$ is an interval, 2) $x(t)$ is absolutely continuous on every compact
subinterval of $I$, and 3) $\dot{x}(t)=h(t,x(t))$ for almost all $t\in I$.
Suppose that $(t^{*},x^{*})\in U$. If a solution $x(t)$ of (15) satisfies 1)
$t^{*}\in I$ and 2) $x(t^{*})=x^{*}$, then $x(t)$ is called a solution with
initial value condition $x(t^{*})=x^{*}$, or simply, a solution of the
following differential equation:

$\dot{x}(t)=h(t,x(t)),\ x(t^{*})=x^{*}.$ (16)

Now, suppose that $h:U\to\mathbb{R}^{n}$ satisfies the following requirements:
1) for every $t\in\mathbb{R}$, $x\mapsto h(t,x)$ is continuous, and 2) for
every $x\in\mathbb{R}^{n}$, $t\mapsto h(t,x)$ is measurable. Then, we say that
$h$ satisfies Carathéodory’s condition. If, in addition, for a set $C\subset
U$, there exists $L>0$ such that

$\|h(t,x_{1})-h(t,x_{2})\|\leq L\|x_{1}-x_{2}\|$

for every $(t,x_{1},x_{2})$ with $(t,x_{1}),(t,x_{2})\in C$, then $h$ is
called Lipschitz in $x$ on $C$.

The following facts are well known.

  1. 1)

Suppose that $h$ satisfies Carathéodory’s condition. Moreover, suppose that
$(t^{*},x^{*})\in V$ and there exist $\varepsilon>0$ and a positive integrable
function $r:[t^{*}-\varepsilon,t^{*}+\varepsilon]\to\mathbb{R}_{+}$ such that
$\|h(t,x)\|\leq r(t)$ for all $(t,x)\in U$ with
$\|(t,x)-(t^{*},x^{*})\|\leq\varepsilon$. Then, there exists a solution
$x:I\to\mathbb{R}^{n}$ of (16), where $I$ is relatively open in
$\mathbb{R}_{+}$.171717This result is called Carathéodory-Peano’s existence
theorem. For the proof, see ch.2 of Coddington and Levinson (1984).

  2. 2)

Suppose that $h$ satisfies Carathéodory’s condition, and for every compact set
$C\subset V$, $h$ is Lipschitz in $x$ on $C$. Moreover, suppose that
$(t^{*},x^{*})\in V$ and there exists a convex neighborhood of $t^{*}$ such
that $t\mapsto h(t,x^{*})$ is integrable on this neighborhood. Then, there
exists a solution $x:I\to\mathbb{R}^{n}$ of (16), where $I$ is relatively open
in $\mathbb{R}_{+}$.

  3. 3)

Suppose that $h$ satisfies Carathéodory’s condition, and for every compact set
$C\subset V$, $h$ is Lipschitz in $x$ on $C$. Moreover, suppose that for every
$(t^{+},x^{+})\in V$, there exists a convex neighborhood of $t^{+}$ such that
$t\mapsto h(t,x^{+})$ is integrable on this neighborhood. Choose any
$(t^{*},x^{*})\in V$. Suppose that $x_{1}(t),x_{2}(t)$ are two solutions of
(16) such that $(t,x_{i}(t))\in V$ for every $t\in I_{i}$, where $I_{i}$ is
the domain of $x_{i}(t)$. Then, $x_{1}(t)=x_{2}(t)$ for every $t\in I_{1}\cap
I_{2}$.181818Results 2) and 3) are known as Carathéodory-Picard-Lindelöf’s
existence theorem. For the proof, see Section 0.4 of Ioffe and Tikhomirov
(1979).

  4. 4)

Suppose that $h$ is continuous. Then, any solution $x(t)$ of (15) is
continuously differentiable.191919See ch.2 of Hartman (1997).

Next, suppose that $h$ satisfies all requirements in 3) and $(t^{*},x^{*})\in
V$. Choose a solution $x:I\to\mathbb{R}^{n}$ of (16). A solution
$y:J\to\mathbb{R}^{n}$ is called extension of $x$ if and only if 1) $I\subset
J$, and 2) $y(t)=x(t)$ for all $t\in I$. 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.202020Fact 5) can be proved easily.
The proof of 6) is in ch.2 of Coddington and Levinson (1984).

  1. 5)

In addition to the requirements of 3), suppose that $V=U$. Then, there
uniquely exists a nonextendable solution $x(t)$ of (16). Moreover, the domain
$I$ of $x(t)$ is relatively open in $\mathbb{R}_{+}$.

  2. 6)

Suppose that all requirements of 5) hold, and let $x:I\to\mathbb{R}^{n}$ be
the nonextendable solution of (16). Choose any compact set $C\subset V$. Then,
there exists $t^{+}$ such that if $t^{+}\leq t\in I$, then $(t,x(t))\notin C$.

Finally, suppose that $h(t,x)=a(t)x+b(t)$, where $a(t),b(t)$ are locally
integrable functions defined on $\mathbb{R}_{+}$. Then, the solution of (15)
is determined by the following formula:

$x(t)=e^{\int_{0}^{t}a(\tau)d\tau}\left[x(0)+\int_{0}^{t}e^{-\int_{0}^{s}a(\tau)d\tau}b(s)ds\right].$
(17)

This is called the formula of the solution of linear ODEs.

###  7.2 Proof of Proposition 1

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
$\varepsilon>0$ so small that $p-\varepsilon>0$ and $k-\varepsilon>0$. Define

$\varepsilon^{*}=\min_{k^{\prime}\in[k-\varepsilon,k+\varepsilon]}[F(k,0)-F(k,1)].$

Then, for every $k^{\prime}\in[k-\varepsilon,k+\varepsilon],\
q\in[p-\varepsilon,p+\varepsilon]$ and $c\geq 1$,

$D_{c,-}g(c,k^{\prime},q)\leq\max_{k^{\prime\prime}\in[k-\varepsilon,k+\varepsilon]}\frac{\partial
u}{\partial c}(c,k^{\prime\prime})-(p-\varepsilon)\varepsilon^{*}.$ (18)

Because $\frac{\partial u}{\partial c}(c,k^{\prime\prime})\to 0$ as
$c\to\infty$, we have that there exists $\bar{c}>1$ such that the right-hand
side of (18) is negative for all $c\geq\bar{c}$.212121Use Dini’s theorem.
Then, for such $k^{\prime}$ and $q$, we have that
$g(c,k^{\prime},q)<g(\bar{c},k^{\prime},q)$ for all $c>\bar{c}$, and thus,

$\sup_{c\geq 0}g(c,k^{\prime},q)=\max_{0\leq c\leq\bar{c}}g(c,k^{\prime},q).$

The rest of the proof on the continuity of the function $c^{*}$ at $(p,k)$ is
a simple application of Berge’s theorem. The positivity of $c^{*}(p,k)$
follows from the assumption $\lim_{c\to 0}\frac{\partial u}{\partial
c}(c,k)=+\infty$. $\blacksquare$

###  7.3 Lemmas and Their Proofs

We should prove Lemma 1. However, to derive Lemma 1, we must introduce two
more results on differential equations. Because these are also useful for
later arguments, we provide these results as lemmas.

Lemma 2. Consider the following two ODEs:

$\dot{k}(t)=h_{i}(t,k),\ k(0)=\bar{k}_{i},$ (19)

where $i\in\\{1,2\\}$. Suppose that each $h_{i}$ is a real valued function
defined on some convex neighborhood $U\subset\mathbb{R}_{+}\times\mathbb{R}$
of $(0,\bar{k}_{i})$ and $h_{i}$ satisfies Carathéodory’s condition. Then, the
following results hold.

  1. i)

If there exists a locally integrable function $r(t)$ such that

$\sup_{k:(t,k)\in U}|h_{i}(t,k)|\leq r(t),$

then there exists $T>0$ such that these equations (19) have solutions
$k_{i}:[0,T]\to\mathbb{R}$. Moreover, if $h_{i}(t,k)$ is continuous, then
$k_{i}(t)$ is continuously differentiable.

  2. ii)

Suppose that $\bar{k}_{1}\leq\bar{k}_{2}$, $h_{1}(t,k)\leq h_{2}(t,k)$ for
every $(t,k)\in U$, and for some $i^{*}\in\\{1,2\\}$, there exists $L>0$ such
that if $(t,k_{1}),(t,k_{2})\in U$, then

$|h_{i^{*}}(t,k_{1})-h_{i^{*}}(t,k_{2})|\leq L|k_{1}-k_{2}|.$

Suppose also that $h_{i^{*}}(t,\bar{k})$ is locally integrable, and there
exist solutions $k_{i}:[0,T]\to\mathbb{R}$ of the above equations for
$i\in\\{1,2\\}$. Then, $k_{1}(t)\leq k_{2}(t)$ for all $t\in[0,T]$.222222If
such an $L>0$ is absent, then this lemma does not hold. For example, consider
$\bar{k}_{1}=\bar{k}_{2}=0,\ h_{1}(t,k)=\sqrt{|k|}-\frac{t}{8},\
h_{2}(t,k)=\sqrt{|k|},\ k_{1}(t)=\frac{t^{2}}{16},$ and $k_{2}(t)\equiv 0$.
Note that if $\bar{k}_{1}=\bar{k}_{2}$ and $h_{1}=h_{2}$, then this claim
immediately implies the uniqueness of the solution. For this view, this lemma
is an extension of Carathéodory-Picard-Lindelöf’s uniqueness result in theory
of ODEs.

Proof. Assertion i) is just a corollary of Carathéodory-Peano’s existence
theorem, and thus we omit the proof.

Suppose that assertion ii) does not hold. Then, there exists $t^{*}>0$ such
that $k_{1}(t^{*})>k_{2}(t^{*})$. Let
$t^{+}=\inf\\{t\in[0,t^{*}]|k_{1}(s)>k_{2}(s)\mbox{ for all
}s\in[t,t^{*}]\\}$. Because $k_{1}(0)=\bar{k}_{1}\leq\bar{k}_{2}=k_{2}(0)$, we
have $k_{1}(t^{+})=k_{2}(t^{+})$. We treat only the case $i^{*}=2$, because
the case $i^{*}=1$ can be treated symmetrically. Define

$k_{3}(t)=k_{1}(t^{+})+\int_{t^{+}}^{t}h_{2}(s,k_{1}(s))ds.$

Because $h_{2}$ satisfies Carathéodory’s condition, the mapping $t\mapsto
h_{2}(t,k_{1}(t))$ is measurable,232323See Section 8.1 of Ioffe and Tikhomirov
(1979). and

$|h_{2}(t,k_{1}(t))|\leq|h_{2}(t,k_{1}(t))-h_{2}(t,\bar{k})|+|h_{2}(t,\bar{k})|\leq
L|k_{1}(t)-\bar{k}|+|h_{2}(t,\bar{k})|,$

and thus $h_{2}(t,k_{1}(t))$ is locally integrable, which implies that
$k_{3}(t)$ is defined on $[t^{+},t^{*}]$. Moreover,

$k_{3}(t)\geq k_{1}(t^{+})+\int_{t^{+}}^{t}h_{1}(s,k_{1}(s))ds=k_{1}(t)\geq
k_{2}(t)$

for every $t\in[t^{+},t^{*}]$. Meanwhile,

$\displaystyle k_{3}(t)-k_{2}(t)=$
$\displaystyle~{}\int_{t^{+}}^{t}[h_{2}(s,k_{1}(s))-h_{2}(s,k_{2}(s))]ds$
$\displaystyle\leq$ $\displaystyle~{}\int_{t^{+}}^{t}L[k_{1}(s)-k_{2}(s)]ds$
$\displaystyle\leq$
$\displaystyle~{}L(t-t^{+})\max_{s\in[t^{+},t]}(k_{1}(s)-k_{2}(s)).$

Fix some $t\in[t^{+},t^{*}]$ with $0<t-t^{+}<L^{-1}$, and let
$s^{*}\in\arg\max\\{k_{1}(s)-k_{2}(s)|s\in[t^{+},t]\\}$. Because
$k_{1}(t^{+})=k_{2}(t^{+})$ and $k_{1}(s)>k_{2}(s)$ for $s\in]t^{+},t^{*}]$,
we have $s^{*}>t^{+}$. Then,

$k_{3}(s^{*})-k_{2}(s^{*})\leq
L(s^{*}-t^{+})(k_{1}(s^{*})-k_{2}(s^{*}))<k_{1}(s^{*})-k_{2}(s^{*}),$

and thus $k_{3}(s^{*})<k_{1}(s^{*})$, which is a contradiction. $\blacksquare$

Lemma 3. Consider the following ODE:

$\dot{k}(t)=h(t,k(t)),\ k(0)=\bar{k}>0,$ (20)

where $h:\mathbb{R}_{+}\times\mathbb{R}_{++}\to\mathbb{R}$ satisfies
Carathéodory’s condition, and for every
$(t^{+},k^{+})\in\mathbb{R}_{+}\times\mathbb{R}_{++}$, there exists
$\varepsilon>0$ and a positive and locally integrable function $r(t)$ defined
on a neighborhood of $t^{+}$ such that if $|k-k^{+}|\leq\varepsilon$, then
$|h(t,k)|<r(t)$ for every $t$ such that $r(t)$ is defined. For either
$X=[0,\tilde{T}]$ or $X=\mathbb{R}_{+}$, suppose that there exist two positive
continuous functions $\hat{k}(t)$ and $\bar{k}(t)$ defined on $X$ such that
$\bar{k}(t)\leq k(t)\leq\hat{k}(t)$ for every solution $k(t)$ defined on
$[0,T]$ and $t\in[0,T]\cap X$. Then, this equation (20) has a solution defined
on $X$ itself.

Proof. We treat only the case $X=\mathbb{R}_{+}$. The proof in the case
$X=[0,\tilde{T}]$ is almost the same.

By Lemma 2, we have that there exists $T>0$ such that equation (20) has a
solution $k(t)$ defined on $[0,T]$. By assumption, $\bar{k}(t)\leq
k(t)\leq\hat{k}(t)$ for all $t\in[0,T]$.

Next, let $Y$ be the set of all solutions of equation (20) defined on either
$\mathbb{R}_{+}$ or $[0,T^{\prime}]$ for some $T^{\prime}>0$, and let
$k_{1}(\cdot)\succeq k_{2}(\cdot)$ if the domain of $k_{1}(\cdot)$ includes
that of $k_{2}(\cdot)$ and $k_{1}(t)=k_{2}(t)$ when both are defined at $t$.
Clearly, $\succeq$ is a partial order on $Y$. For $k(\cdot)\in Y$, let
$I_{k(\cdot)}$ be the domain of $k(\cdot)$. Choose any chain $C\subset Y$ of
$\succeq$. If $\sup\cup_{k(\cdot)\in C}I_{k(\cdot)}=+\infty$, then we can
define

$k^{+}(t)=k(t)\mbox{ if }t\in I_{k(\cdot)},$

and $k^{+}(\cdot)$ is an upper bound of $C$. Otherwise, let
$T^{*}=\sup\cup_{k(\cdot)\in C}I_{k(\cdot)}$. Define

$k^{+}(t)=k(t)\mbox{ if }t\in I_{k(\cdot)}.$

Then, $k^{+}(t)$ is a solution of (20) defined on $[0,T^{*}[$. By the
continuity of $\hat{k}(t),\bar{k}(t)$, there exist $\varepsilon>0$ and
$\delta>0$ such that $\bar{k}(T^{*})>\varepsilon$ and if $0<T^{*}-t<\delta$,
then $k^{+}(t)\in[\bar{k}(T^{*})-\varepsilon,\hat{k}(T^{*})+\varepsilon]$.
Hence, we can define

$k^{+}(T^{*})=\limsup_{t\to
T^{*}}k^{+}(t)\in[\bar{k}(T^{*})-\varepsilon,\hat{k}(T^{*})+\varepsilon].$

Because $[\bar{k}(T^{*})-\varepsilon,\hat{k}(T^{*})+\varepsilon]$ is compact,
we have that if $\delta>0$ is sufficiently small, then there exists a positive
and locally integrable function $r(t)$ such that $|h(t,k)|<r(t)$ for every
$(t,k)$ such that $0<T^{*}-t<\delta$ and
$k\in[\bar{k}(T^{*})-\varepsilon,\hat{k}(T^{*})+\varepsilon]$. Thus, we have
that $|k^{+}(t)-k^{+}(T^{*})|\leq\int_{t}^{T^{*}}r(s)ds$, and hence

$\lim_{t\to T^{*}}k^{+}(t)=k^{+}(T^{*}),$

which implies that $k^{+}(t)$ is continuous on $[0,T^{*}]$. Because

$k^{+}(t)=\bar{k}+\int_{0}^{t}h(s,k^{+}(s))ds$

for all $t\in[0,T^{*}]$, we conclude that $k^{+}(t)$ is a solution of equation
(20) defined on $[0,T^{*}]$, and thus $k^{+}(\cdot)\in Y$ and it is an upper
bound of $C$. Therefore, by Zorn’s lemma, there exists a maximal element
$k^{*}(\cdot)\in Y$ of $\succeq$. If the domain $I_{k^{*}(\cdot)}=[0,T]$, then
$k^{*}(T)\in[\bar{k}(T),\hat{k}(T)]$, and thus by Lemma 2, there exists a
solution $\tilde{k}(t)$ of the following equation:

$\dot{k}(t)=h(t+T,k(t)),\ k(0)=k^{*}(T),$

defined on $[0,T^{\prime}]$. Define

$\kappa(t)=\begin{cases}k^{*}(t)&\mbox{if }t\in[0,T],\\\
\tilde{k}(t-T)&\mbox{if }t\in[T,T+T^{\prime}].\end{cases}$

Then $\kappa(\cdot)\in Y$ and $\kappa(\cdot)\succ k^{*}(\cdot)$, which is a
contradiction. Therefore, the domain of $k^{*}(t)$ must be $\mathbb{R}_{+}$.
This completes the proof of Lemma 3. $\blacksquare$

Proof of Lemma 1. Suppose that Assumption 3 holds, and consider the following
ODE:

$\dot{k}(t)=F(k(t),0),\ k(0)=\bar{k}.$

By Assumption 3, there exists $k>0$ such that $F(k,0)>0$. Suppose that the
solution $k(t)$ of the above equation satisfies $k(T)<\min\\{\bar{k},k\\}$ for
some $T>0$, and define $t^{*}=\inf\\{t\in[0,T]|\forall
t^{\prime}\in[t,T],k(t^{\prime})<\min\\{\bar{k},k\\}\\}$. Because
$k(0)=\bar{k}\geq\min\\{\bar{k},k\\}$, we have $k(t^{*})=\min\\{\bar{k},k\\}$.
Because $F$ is concave and $F(0,0)=0$, we have $F(k(t^{*}),0)>0$. Thus, there
exists $t\in]t^{*},T[$ such that $k(t)>\min\\{\bar{k},k\\}$, which is a
contradiction. Therefore,

$k(t)\geq\min\\{\bar{k},k\\}$

for every solution $k(t)$ of the above equation. Meanwhile, let
$p\in\partial_{k}F(k,0)$. Then, $F(k^{\prime},0)\leq p(k^{\prime}-k)+F(k,0)$,
and thus, by Lemma 2 and the formula (17),

$k(t)\leq\begin{cases}e^{pt}\left(\bar{k}+(e^{-pt}-1)\frac{pk-F(k,0)}{p}\right)&\mbox{if
}p\neq 0,\\\ \bar{k}+tF(k,0)&\mbox{if }p=0,\end{cases}$ (21)

for every solution $k(t)$ of the above equation. Therefore, we can apply Lemma
3, and thus $k^{+}(t,\bar{k})$ is well-defined on
$\mathbb{R}_{+}\times\mathbb{R}_{++}$, and $\inf_{t\geq
0}k^{+}(t,\bar{k})\geq\min\\{\bar{k},k\\}>0$ for every $\bar{k}>0$. This
completes the proof of Lemma 1. $\blacksquare$

Note that, if $h(t,k)=F(k,g(t,k))$ for some non-negative function $g(t,k)$,
then we can use $k^{+}(t,\bar{k})$ for $\hat{k}(t)$ in Lemma 3. Note also
that, if $g(t,k)=c(t)$ for some locally integrable function
$c:\mathbb{R}_{+}\to\mathbb{R}_{+}$, then for all $k\in[0,\bar{k}+1]$,

$|h(t,k)|\leq d_{1}(\bar{k}+1)+d_{2}c(t)+\max\\{F(k^{\prime},0)|0\leq
k^{\prime}\leq\bar{k}+1\\}\equiv r(t),$ (22)

and because $r(t)$ is locally integrable, the requirement of Lemma 3 is
automatically satisfied. We will use these facts frequently.

Later, we often need the boundedness of $c(t)$ in the proof. Therefore, we
present two more lemmas.242424One might think that if Lemma 5 is correct, then
we can relax the existence assumption of $d_{2}\geq 0$ as follows: there
exists an increasing and continuous function $d_{2}(c)$ such that $d_{2}(0)=0$
and $F(k,c)>-d_{1}k-d_{2}(c)$. If this conjecture is correct, we can treat
technology functions such as $F(k,c)=\sqrt{k}-c^{2}$. However, we should use
the above $r(t)$ in the proof of Lemma 5. Because $d_{2}(c(t))$ may be not
locally integrable even if $c(t)$ is locally integrable and $d_{2}(c)$ is
continuous, our proof is broken, and thus this conjecture is not true.

Lemma 4. Suppose that Assumption 3 holds, and choose any nonnegative and
locally integrable function $c(t)$ such that $\sup_{t\geq 0}c(t)<+\infty$.
Define

$h(t,k)=F(k,c(t)).$

Then, $h$ satisfies Carathéodory’s condition on $\mathbb{R}_{+}^{2}$ and for
every compact set $C\subset\mathbb{R}_{+}\times\mathbb{R}_{++}$, it is
Lipschitz in $k$ on $C$.

Proof. The fact that $h$ satisfies Carathéodory’s condition can easily be
shown. Choose any compact set $C\subset\mathbb{R}_{+}\times\mathbb{R}_{++}$.
Then, there exist $T>0$ and $\tilde{k},\hat{k}>0$ such that
$C\subset[0,T]\times[\tilde{k},\hat{k}]$. Define $c^{*}=\sup_{t\geq 0}c(t)$
and

$\tilde{F}(k,c)=\begin{cases}F(k,c)&\mbox{if }c\geq 0,\\\ F(k,0)&\mbox{if
}c<0.\end{cases}$

Then, we have that $\tilde{F}$ is defined on
$\mathbb{R}_{++}\times\mathbb{R}$. We can easily check that $\tilde{F}$ is
concave. By Theorem 24.7 of Rockafeller (1996), we have that the set

$X=\\{(p,q)|(p,q)\in\partial\tilde{F}(k,c)\mbox{ for
}(k,c)\in[\tilde{k},\hat{k}]\times[0,c^{*}]\\}$

is compact.252525Note that $[\tilde{k},\hat{k}]\times[0,c]$ is in the relative
interior of the domain of $\tilde{F}$, and thus Theorem 24.7 of Rockafeller
(1996) is applicable for $\tilde{F}$. Thus, for $t\geq 0$ and
$k_{1},k_{2}\in[\tilde{k},\hat{k}]$,

$\displaystyle|h(t,k_{1})-h(t,k_{2})|=$
$\displaystyle~{}|\tilde{F}(k_{1},c(t))-\tilde{F}(k_{2},c(t))|$
$\displaystyle\leq$ $\displaystyle~{}\max\\{|p||(p,q)\in
X\\}\times|k_{1}-k_{2}|,$

as desired. This completes the proof. $\blacksquare$

Lemma 5. Suppose that Assumptions 1-3 hold. Choose any $\bar{k}>0$ and
$(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$ such that $\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt>-\infty$, and define $c_{n}(t)=\min\\{c^{*}(t),n\\}$
for $n\in\mathbb{N}$. Then, there exists a function $k_{n}(t)$ defined on
$\mathbb{R}_{+}$ such that $k_{n}(t)\geq k^{*}(t)$ for all $t$, and if
$u(n,0)>-\infty$, then $(k_{n}(t),c_{n}(t))\in A_{\bar{k}}$ and
$\int_{0}^{\infty}e^{-\rho t}u(c_{n}(t),k_{n}(t))dt>-\infty$.

Proof. First, define $h(t,k)=F(k,c^{*}(t))$ and $h_{n}(t,k)=F(k,c_{n}(t))$.
Then, $h_{n}(t,k)\geq h(t,k)$ for all $t\geq 0$. Consider the following
differential equation:

$\dot{k}(t)=h_{n}(t,k(t)),\ k(0)=\bar{k}.$

We show that there exists a solution $k_{n}(t)$ of the above equation defined
on $\mathbb{R}_{+}$ such that $k_{n}(t)\geq k^{*}(t)$ for every $t\geq 0$.

Let $Y$ be the set of all solutions $k(\cdot)$ of the above equation such that
the domain $I_{k(\cdot)}$ is either $\mathbb{R}_{+}$ or $[0,T]$ for some
$T>0$, and $k(t)\geq k^{*}(t)$ for all $t\in I_{k(\cdot)}$. By Lemma 4, we
have that $h_{n}$ satisfies all requirements in Lemma 2, and thus $Y$ is
nonempty.

Define a partial order $\succeq$ on $Y$ such that $k_{1}(\cdot)\succeq
k_{2}(\cdot)$ if and only if $I_{k_{2}(\cdot)}\subset I_{k_{1}(\cdot)}$ and
$k_{1}(t)=k_{2}(t)$ for every $t\in I_{k_{2}(\cdot)}$. Let $C$ be any chain of
$\succeq$. If $\sup\cup_{k(\cdot)\in C}I_{k(\cdot)}=+\infty$, we can define
$k^{+}(t)=k(t)$ for $t\in I_{k(\cdot)}$, and $k^{+}(\cdot)$ is an upper bound
of $C$. If $\sup\cup_{k(\cdot)\in C}I_{k(\cdot)}=T<+\infty$, we can define
$k^{+}(t)=k(t)$ for $t\in I_{k(\cdot)}$, and $k^{+}(t)$ is defined on $[0,T[$.
Because $k^{+}(t)\leq k^{+}(t,\bar{k})$, we have that $k^{+}(T)=\limsup_{t\to
T}k^{+}(t)\in[0,k^{+}(T,\bar{k})]$ can be defined. Let $r(t)$ be defined as in
(22), where we use $\max_{t\in[0,T]}k^{+}(t,\bar{k})$ instead of $\bar{k}$ and
$c^{*}(t)$ instead of $c(t)$.262626That is,
$r(t)=d_{1}(\max_{t\in[0,T]}k^{+}(t,\bar{k})+1)+d_{2}c^{*}(t)+\max\\{F(k,0)|0\leq
k\leq\max_{t\in[0,T]}k^{+}(t,\bar{k})+1\\}.$ Such a construction of $r(t)$
will be frequently used later. Then, we have that
$|k^{+}(t)-k^{+}(T)|\leq\int_{t}^{T}r(s)ds$, and thus $k^{+}(\cdot)$ is
continuous. Because

$k^{+}(t)=\bar{k}+\int_{0}^{t}h_{n}(s,k^{+}(s))dt$

for every $t\in[0,T]$, we have that $k^{+}(\cdot)$ is a solution of the above
equation. Because both $k^{+}(\cdot),k^{*}(\cdot)$ are continuous, we have
that $k^{+}(t)\geq k^{*}(t)$ for every $t\in[0,T]$, and thus $k^{+}(\cdot)$ is
an upper bound of $C$. Therefore, by Zorn’s lemma, there exists a maximal
element $\tilde{k}(\cdot)$. Suppose that
$I_{\tilde{k}(\cdot)}=[0,T]\neq\mathbb{R}_{+}$. If $\tilde{k}(T)>0$, then by
using Lemma 2, we can easily show that $\tilde{k}(\cdot)$ is not maximal,
which is a contradiction. Therefore, we have that $\tilde{k}(T)=k^{*}(T)=0$.
Choose any $k^{*}>0$ such that $k\mapsto F(k,c)$ is increasing on $[0,k^{*}]$
for all $c\in[0,n]$. Define $r(t)$ as in (22), where we use $k^{*}$ instead of
$\bar{k}$ and $c^{*}(t)$ instead of $c(t)$. Choose $\varepsilon>0$ so small
that $k^{*}(t)\leq k^{*}$ whenever $t\in[T,T+\varepsilon]$, and

$\int_{T}^{T+\varepsilon}r(t)dt<k^{*}.$

Let $I=[T,T+\varepsilon]$ and $X$ be the space of all continuous functions
$x:I\to[0,k^{*}]$ such that $x(T)=0$, $x(t)\geq k^{*}(t)$ for all $t\in I$,
and for every $t_{1},t_{2}\in I$ with $t_{1}<t_{2}$,
$|x(t_{1})-x(t_{2})|\leq\int_{t_{1}}^{t_{2}}r(t)dt$. Because $k^{*}(\cdot)$
satisfies all requirements, we have that $X$ is nonempty. Clearly, $X$ is
convex. By Ascoli-Arzela’s theorem, $X$ is compact with respect to sup norm.
Define

$P(x(\cdot))(t)=\int_{T}^{t}h_{n}(s,x(s))ds.$

If $x(\cdot)\in X$ and $y(\cdot)=P(x(\cdot))$, then

$y(T)=0,$
$y(t)\geq\int_{T}^{t}h_{n}(s,k^{*}(s))ds\geq\int_{T}^{t}h(s,k^{*}(s))ds=k^{*}(t),$

and if $t_{1},t_{2}\in I$ and $t_{1}<t_{2}$, then

$|y(t_{1})-y(t_{2})|\leq\int_{t_{1}}^{t_{2}}|h_{n}(t,x(t))|dt\leq\int_{t_{1}}^{t_{2}}r(t)dt.$

Therefore, we have that

$y(t)\leq\int_{T}^{T+\varepsilon}r(t)dt<k^{*},$

and thus $y(\cdot)\in X$, which implies that $P$ is a mapping from $X$ into
$X$. Let $(x_{m}(\cdot))$ be the convergent sequence of $X$ on the sup norm
and $x(\cdot)$ is the limit. Define $y_{m}(\cdot)=P(x_{m}(\cdot))$ and
$y(\cdot)=P(x(\cdot))$. Choose any $\varepsilon^{\prime}>0$ and define

$A_{m}=\left\\{t\in
I\left||h_{n}(t,x_{m}(t))-h_{n}(t,x(t))|\geq\frac{\varepsilon^{\prime}}{2\varepsilon}\right.\right\\},$
$B_{m}=\cup_{\ell\geq m}A_{\ell}.$

Because $F$ is continuous, we have that $B_{m}\downarrow\emptyset$, and thus

$\lim_{m\to\infty}\int_{B_{m}}r(t)dt=0.$

Hence, there exists $m_{0}\in\mathbb{N}$ such that if $m\geq m_{0}$, then

$\int_{B_{m}}2r(t)dt<\frac{\varepsilon^{\prime}}{2}.$

Then, for every $t\in I$, if $m\geq m_{0}$,

$\displaystyle|y_{m}(t)-y(t)|\leq$
$\displaystyle~{}\int_{T}^{T+\varepsilon}|h_{n}(\tau,x_{m}(\tau))-h_{n}(\tau,x(\tau))|d\tau$
$\displaystyle=$ $\displaystyle~{}\int_{I\setminus
B_{m}}|h_{n}(\tau,x_{m}(\tau))-h_{n}(\tau,x(\tau))|d\tau$
$\displaystyle~{}+\int_{B_{m}}|h_{n}(\tau,x_{m}(\tau))-h_{n}(\tau,x(\tau))|d\tau$
$\displaystyle\leq$ $\displaystyle~{}\int_{I\setminus
B_{m}}\frac{\varepsilon^{\prime}}{2\varepsilon}d\tau+\int_{B_{m}}2r(\tau)d\tau$
$\displaystyle<$
$\displaystyle~{}\frac{\varepsilon^{\prime}}{2}+\frac{\varepsilon^{\prime}}{2}=\varepsilon^{\prime},$

which implies that $P$ is continuous with respect to the sup norm. Therefore,
by Schauder’s fixed point theorem, there exists a fixed point $x^{*}(\cdot)\in
X$ of $P$. Define

$\hat{k}(t)=\begin{cases}\tilde{k}(t)&\mbox{if }0\leq t\leq T,\\\
x^{*}(t)&\mbox{if }t\in I.\end{cases}$

Then, $\hat{k}(\cdot)\in Y$ and $\hat{k}(\cdot)\succ\tilde{k}(\cdot)$, which
contradicts the maximality of $\tilde{k}(\cdot)$. Therefore, we have that the
domain of $\tilde{k}(\cdot)$ is $\mathbb{R}_{+}$, and thus we can define
$k_{n}(t)=\tilde{k}(t)$.

Now, suppose that $u(n,0)>-\infty$. Because $\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt>-\infty$, we have

$\int_{\\{t|c^{*}(t)\leq n\\}}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt>-\infty,$

and thus,

$\displaystyle-\infty<$ $\displaystyle~{}\int_{\\{t|c^{*}(t)\leq n\\}}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt+\int_{\\{t|c^{*}(t)>n\\}}e^{-\rho t}u(n,k^{*}(t))dt$
$\displaystyle=$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c_{n}(t),k^{*}(t))dt$ $\displaystyle\leq$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c_{n}(t),k_{n}(t))dt,$

and thus $\int_{0}^{\infty}e^{-\rho t}u(c_{n}(t),k_{n}(t))dt$ is defined and
is not $-\infty$, which implies that $(k_{n}(t),c_{n}(t))\in A_{\bar{k}}$.
This completes the proof. $\blacksquare$

Note that, if $(k^{*}(t),c^{*}(t)),(k_{n}(t),c_{n}(t))\in A_{\bar{k}}$ are as
in Lemma 5, then we have that $\int_{0}^{\infty}e^{-\rho
t}u(c_{n}(t),k^{*}(t))dt>-\infty$. By the monotone convergence theorem, we
have that

$\liminf_{n\to\infty}\int_{0}^{\infty}e^{-\rho
t}u(c_{n}(t),k_{n}(t))dt\geq\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt.$

In particular, if there exists $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$ such that
$\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt>M$, then we can assume
that $c^{*}(t)$ is bounded.

###  7.4 Proof of Proposition 2

We separate the proof into three steps.

Step 1. $\bar{V}(\bar{k})>-\infty$ for every $\bar{k}>0$.

Proof of step 1. We have assumed that $F(0,0)=0$ and there exists $k^{*}>0$
such that $F(k^{*},0)>0$. If $\bar{k}>0$ and $F(\bar{k},0)>0$, then there
exists $\bar{c}>0$ such that $F(\bar{k},\bar{c})=0$ because $F$ is continuous,
concave, and decreasing in $c$.272727By this assumption, we have that
$\lim_{c\to\infty}F(\bar{k},c)=-\infty$. Hence,
$(k(t),c(t))\equiv(\bar{k},\bar{c})\in A_{\bar{k}}$, and thus
$\bar{V}(\bar{k})>-\infty$. If $\bar{k}>0$ and $F(\bar{k},0)\leq 0$, then by
the concavity of $F$, we have $\bar{k}>k^{*}$. Choose any $c^{*}>0$ such that
$F(k^{*},c^{*})>0$, and $k(t)$ is a solution of the differential equation

$\dot{k}(t)=F(k(t),c^{*}),\ k(0)=\bar{k},$

defined on $[0,T]$. Suppose that $k(t^{+})<k^{*}$ for some $t^{+}>0$. Let
$t^{*}=\inf\\{t\in[0,t^{+}]|\forall s\in[t,t^{+}],k(s)<k^{*}\\}$. Because
$k(0)=\bar{k}>k^{*}$, we have that $k(t^{*})=k^{*}$, and thus
$F(k(t^{*}),c^{*})>0$. Therefore, there exists $t\in]t^{*},t^{+}]$ such that
$k(t)>k^{*}$, which is a contradiction. Hence, we have that $k(t)\geq k^{*}$
for all $t\in[0,T]$, and thus by Lemma 3, there exists a solution $k^{*}(t)$
of the above equation defined on $\mathbb{R}_{+}$. Let $c^{*}(t)\equiv c^{*}$.
Then, $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, and thus
$\bar{V}(\bar{k})>-\infty$. This completes the proof. $\blacksquare$

Step 2. $\bar{V}$ is nondecreasing.

Proof of Step 2. Choose any $k_{1},k_{2}\geq 0$ with $k_{1}<k_{2}$. If
$k_{1}>0$, then we have proved in the proof of step 1 that there exists a pair
$(k(t),c(t))\in A_{k_{1}}$. If $k_{1}=0$, then $(k(t),c(t))\equiv(0,0)\in
A_{k_{1}}$. Therefore, $A_{k_{1}}$ is nonempty for every $k_{1}\geq 0$. If
$\bar{V}(k_{1})=-\infty$, then clearly $\bar{V}(k_{2})\geq\bar{V}(k_{1})$.
Suppose that $\bar{V}(k_{1})>-\infty$. Let $(k(t),c(t))\in A_{k_{1}}$ and

$\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt>\min\\{\bar{V}(k_{1})-n^{-1},n\\}.$

Because of Lemma 5, we can assume that $c(t)$ is bounded. Define

$h(t,k)=F(k,c(t))$

for all $(t,k)\in\mathbb{R}_{+}\times\mathbb{R}_{++}$, and consider the
following two ODEs:282828Note that, equation (24) is different from equation
(23), because the right-hand side of (23) is defined only on
$\mathbb{R}_{+}\times\mathbb{R}_{++}$, while that of (24) is defined on
$\mathbb{R}^{2}_{+}$. The reason why we need (23) is as follows: the equation
(24) does not satisfy the requirement $U=V$ in 5) of subsection 7.1, and thus
5) cannot be directly applied.

$\displaystyle\dot{\kappa}(t)=$ $\displaystyle~{}h(t,\kappa(t)),\
\kappa(0)=k_{2},$ (23) $\displaystyle\dot{\kappa}(t)=$
$\displaystyle~{}F(\kappa(t),c(t)),\ \kappa(0)=k_{2}.$ (24)

Suppose that there does not exist a solution
$\kappa:\mathbb{R}_{+}\to\mathbb{R}_{+}$ of (24) such that $\kappa(t)\geq
k(t)$ for every $t\in\mathbb{R}_{+}$. Because all assumptions in 5) of
Subsection 7.1 hold for equation (23), there exists the unique nonextendable
solution $\kappa:I\to\mathbb{R}$ of (23). By our assumption, we have that
$\sup\\{t\in I|\kappa(t)\geq k(t)\\}=T<+\infty$. If $[0,T]\subset I$, then we
have that $\kappa(T)=k(T)$, and thus if we define

$\tilde{k}(t)=\begin{cases}\kappa(t)&\mbox{if }0\leq t<T,\\\
k(t)&\mbox{otherwise},\end{cases}$

then $\tilde{k}(t)$ is a solution of (24) defined on $\mathbb{R}_{+}$ such
that $\tilde{k}(t)\geq k(t)$ for all $t\in\mathbb{R}_{+}$, which is a
contradiction. Therefore, we must have that $I=[0,T[$. Because $\kappa$ is
bounded from above by the argument in subsection 7.3, by 6) in subsection 7.1,
for every $n\in\mathbb{N}$, there exists $t_{n}>0$ such that if $t_{n}<t<T$,
then $\kappa(t)<n^{-1}$. Therefore, we have that $\limsup_{t\to
T}\kappa(t)=0$, and thus we can define $\kappa(T)=0$. Because $\kappa(t)\geq
k(t)$ for all $t\in I$, we have that $k(T)=0$, and thus we can define
$\tilde{k}(t)$ as above, which leads a contradiction.

Therefore, there exists a solution $\kappa(t)$ of (24) defined on
$\mathbb{R}_{+}$ such that $\kappa(t)\geq k(t)$ for every $t\geq 0$, which
implies that $(\kappa(t),c(t))\in A_{k_{2}}$. Hence,

$\bar{V}(k_{2})\geq\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt,$

and thus,

$\bar{V}(k_{2})\geq\bar{V}(k_{1}),$

as desired. This completes the proof of step 2. $\blacksquare$

Step 3. $\bar{V}$ is concave.

Proof of step 3. Because $\bar{V}$ is nondecreasing, it suffices to show that
$\bar{V}((1-s)k_{0}+sk_{1})\geq(1-s)\bar{V}(k_{0})+s\bar{V}(k_{1})$ for every
$k_{0},k_{1}>0$ and $s\in]0,1[$. Fix such $k_{0},k_{1}>0$ and $s\in]0,1[$. Let
$(k_{i}^{n}(t),c_{i}^{n}(t))\in A_{k_{i}}$ for $i\in\\{0,1\\}$ such that

$\int_{0}^{\infty}e^{-\rho
t}u(c_{i}^{n}(t),k_{i}^{n}(t))dt>\min\\{\bar{V}(k_{i})-n^{-1},n\\}.$

By Lemma 5, we can assume that $c_{i}^{n}(t)$ is bounded for $i\in\\{0,1\\}$.
Define

$c_{s}^{n}(t)=(1-s)c_{0}^{n}(t)+sc_{1}^{n}(t),$

and consider the following differential equation:

$\dot{k}(t)=F(k(t),c_{s}^{n}(t)),\ k(0)=(1-s)k_{0}+sk_{1}.$

We will show that there exists a solution $k_{s}^{n}(t)$ of this equation
defined on $\mathbb{R}_{+}$ such that
$k_{s}^{n}(t)\geq(1-s)k_{0}(t)+sk_{1}(t)$ for every $t\geq 0$.

First, choose any solution $k(t)$ of the above equation defined on $I$ such
that $k(t)>0$ for all $t\in I$. Because

$F(k,c_{s}^{n}(t))\geq(1-s)F(k,c_{0}^{n}(t))+sF(k,c_{1}^{n}(t)),$

by Lemma 2, we have that $k(t)\geq(1-s)k_{0}^{n}(t)+sk_{1}^{n}(t)$ for all
$t\in I$. Let $Y$ be the set of all solutions $k(\cdot)$ of the above equation
defined on $I_{k(\cdot)}$ such that $k(t)\geq(1-s)k_{0}^{n}(t)+sk_{1}^{n}(t)$
for all $t\in I_{k(\cdot)}$, where either $I_{k(\cdot)}=[0,T]$ for some $T>0$
or $I_{k(\cdot)}=\mathbb{R}_{+}$. By the above argument, we have that $Y$ is
nonempty. Define a partial order $\succeq$ on $Y$ such that
$\hat{k}(\cdot)\succeq\tilde{k}(\cdot)$ if and only if
$I_{\tilde{k}(\cdot)}\subset I_{\hat{k}(\cdot)}$ and $\hat{k}(t)=\tilde{k}(t)$
for all $t\in I_{\tilde{k}(\cdot)}$. Let $C$ be a chain in $Y$. If
$\cup_{k(\cdot)\in C}I_{k(\cdot)}=\mathbb{R}_{+}$, then we can define
$k^{*}(t)=k(t)$ for $t\in I_{k(\cdot)}$, and $k^{*}(\cdot)$ is an upper bound
of $C$. If $\sup\cup_{k(\cdot)\in C}I_{k(\cdot)}=T<+\infty$, define
$k^{*}(t)=k(t)$ for $t\in I_{k(\cdot)}$. Then, $k^{*}(\cdot)$ is a solution of
the above equation defined on $[0,T[$. Because $k^{*}(t)\leq
k^{+}(t,\bar{k})$, we can define $k^{*}(T)=\limsup_{t\to
T}k^{*}(t)\in[0,k^{+}(T,\bar{k})]$. Define $r(t)$ as in (22), where we use
$\max_{t\in[0,T]}k^{+}(t,\bar{k})$ instead of $\bar{k}$ and $c_{s}^{n}(t)$
instead of $c(t)$. Then, $|k^{*}(t)-k^{*}(T)|\leq\int_{t}^{T}r(\tau)d\tau$,
and thus $k^{*}(\cdot)$ is continuous, and because

$k^{*}(t)=\bar{k}+\int_{0}^{t}F(k^{*}(\tau),c_{s}^{n}(\tau))d\tau,$

and $(1-s)k_{0}^{n}(t)+sk_{1}^{n}(t)$ is continuous, we have that
$k^{*}(\cdot)\in Y$, and thus $k^{*}(\cdot)$ is an upper bound of $C$.
Therefore, by Zorn’s lemma, we have that there is a maximal element
$k^{*}(\cdot)\in Y$. Suppose that $I_{k^{*}(\cdot)}=[0,T]\neq\mathbb{R}_{+}$.
If $k^{*}(T)>0$, then by Lemma 2, there exists $\tilde{k}(\cdot)\in Y$ such
that $\tilde{k}(\cdot)\succ k^{*}(\cdot)$, which contradicts the maximality of
$k^{*}(\cdot)$. Hence, we have that $k^{*}(T)=k_{0}^{n}(T)=k_{1}^{n}(T)=0$.
Let $c^{*}=\max\\{\sup_{t\geq 0}c_{0}^{n}(t),\sup_{t\geq 0}c_{1}^{n}(t)\\}$,
and choose any $k^{*}>0$ such that $k\mapsto F(k,c)$ is increasing on
$[0,k^{*}]$ for all $c\in[0,c^{*}]$. Define $r(t)$ as in (22), where we use
$k^{*}$ instead of $\bar{k}$ and $c_{s}^{n}(t)$ instead of $c(t)$. Choose a
sufficiently small $\varepsilon>0$ such that

$\max_{i\in\\{0,1\\},t\in[T,T+\varepsilon]}k_{i}^{n}(t)\leq k^{*},\
\int_{T}^{T+\varepsilon}r(t)dt\leq k^{*},$

and let $I=[T,T+\varepsilon]$ and $X$ be the set of all continuous functions
$x:I\to[0,k^{*}]$ such that $x(T)=0$,
$x(t)\geq(1-s)k_{0}^{n}(t)+sk_{1}^{n}(t)$, and for all $t_{1},t_{2}\in I$ with
$t_{1}<t_{2}$, $|x(t_{2})-x(t_{1})|\leq\int_{t_{1}}^{t_{2}}r(t)dt$. Because
$(1-s)k_{0}^{n}(\cdot)+sk_{1}^{n}(\cdot)\in X$, we have that $X$ is nonempty.
The convexity of $X$ is clear. By Ascoli-Arzela’s theorem, it is compact with
respect to the sup norm. Define

$P(x(\cdot))(t)=\int_{T}^{t}F(x(\tau),c_{s}^{n}(\tau))d\tau$

for all $t\in I$. If $x(\cdot)\in X$ and $y(\cdot)=P(x(\cdot))$, then $y(T)=0$
and

$\displaystyle y(t)\geq$
$\displaystyle~{}\int_{T}^{t}F((1-s)k_{0}^{n}(\tau)+sk_{1}^{n}(\tau),c_{s}^{n}(\tau))d\tau$
$\displaystyle\geq$
$\displaystyle~{}\int_{T}^{t}[(1-s)F(k_{0}^{n}(\tau),c_{0}^{n}(\tau))+sF(k_{1}^{n}(\tau),c_{1}^{n}(\tau))]d\tau$
$\displaystyle=$ $\displaystyle~{}(1-s)k_{0}^{n}(t)+sk_{1}^{n}(t)$

and

$y(t)\leq\int_{T}^{t}r(s)ds\leq k^{*}$

for all $t\in I$. If $T\leq t_{1}<t_{2}\leq T+\varepsilon$, then

$\displaystyle|y(t_{2})-y(t_{1})|\leq$
$\displaystyle~{}\int_{t_{1}}^{t_{2}}|F(x(t),c_{s}^{n}(t))|dt$
$\displaystyle\leq$ $\displaystyle~{}\int_{t_{1}}^{t_{2}}r(t)dt.$

Therefore, we have that $y(\cdot)\in X$, and thus $P$ is a mapping from $X$
into $X$. Choose any convergent sequence $(x_{m}(\cdot))$ on $X$ and let
$x(\cdot)$ be the limit. Define $y_{m}(\cdot)=P(x_{m}(\cdot))$ and
$y(\cdot)=P(x(\cdot))$. Fix any $\varepsilon^{\prime}>0$, and define

$A_{m}=\left\\{t\in
I\left||F(x_{m}(t),c_{s}^{n}(t))-F(x(t),c_{s}^{n}(t))|\geq\frac{\varepsilon^{\prime}}{2\varepsilon}\right.\right\\},$
$B_{m}=\cup_{\ell\geq m}A_{\ell}.$

Because $F$ is continuous, we have that $B_{m}\downarrow\emptyset$. Therefore,

$\lim_{m\to\infty}\int_{B_{m}}r(t)dt=0.$

Hence, there exists $m_{0}\in\mathbb{N}$ such that if $m\geq m_{0}$, then

$\int_{B_{m}}2r(t)dt<\frac{\varepsilon^{\prime}}{2}.$

For every $t\in I$, if $m\geq m_{0}$, then

$\displaystyle|y_{m}(t)-y(t)|\leq$
$\displaystyle~{}\int_{T}^{T+\varepsilon}|F(x_{m}(\tau),c_{s}^{n}(\tau))-F(x(\tau),c_{s}^{n}(\tau))|d\tau$
$\displaystyle\leq$ $\displaystyle~{}\int_{I\setminus
B_{m}}|F(x_{m}(\tau),c_{s}^{n}(\tau))-F(x(\tau),c_{s}^{n}(\tau))|d\tau$
$\displaystyle~{}+\int_{B_{m}}|F(x_{m}(\tau),c_{s}^{n}(\tau))-F(x(\tau),c_{s}^{n}(\tau))|d\tau$
$\displaystyle\leq$ $\displaystyle~{}\int_{I\setminus
B_{m}}\frac{\varepsilon^{\prime}}{2\varepsilon}dt+\int_{B_{m}}2r(\tau)d\tau$
$\displaystyle<$
$\displaystyle~{}\frac{\varepsilon^{\prime}}{2}+\frac{\varepsilon^{\prime}}{2}=\varepsilon^{\prime}.$

Hence, $P$ is continuous with respect to the sup norm. By Schauder’s fixed
point theorem, we have that $P$ has a fixed point $x^{*}:I\to\mathbb{R}_{+}$.
Define

$\tilde{k}(t)=\begin{cases}k^{*}(t)&\mbox{if }0\leq t\leq T,\\\
x^{*}(t)&\mbox{if }T\leq t\leq T+\varepsilon.\end{cases}$

Then, $\tilde{k}(\cdot)\in Y$ is an extension of $k^{*}(\cdot)$, which
contradicts the maximality of $k^{*}(\cdot)$. Hence, the domain of
$k^{*}(\cdot)$ is $\mathbb{R}_{+}$, and thus we can define $k_{s}^{n}(t)$ as
$k^{*}(t)$.

Thus, for every $M>0$,

$\displaystyle~{}\bar{V}((1-s)k_{0}+sk_{1})$ $\displaystyle\geq$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c_{s}^{n}(t),k_{s}^{n}(t))dt$
$\displaystyle\geq$ $\displaystyle~{}(1-s)\int_{0}^{\infty}e^{-\rho
t}u(c_{0}^{n}(t),k_{0}^{n}(t))dt+s\int_{0}^{\infty}e^{-\rho
t}u(c_{1}^{n}(t),k_{1}^{n}(t))dt$ $\displaystyle\geq$
$\displaystyle\min\\{(1-s)\bar{V}(k_{0})+s\bar{V}(k_{1})-n^{-1},M\\},$

for all sufficiently large $n\in\mathbb{N}$, which implies that

$\bar{V}((1-s)k_{0}+sk_{1})\geq(1-s)\bar{V}(k_{0})+s\bar{V}(k_{1}),$

as desired. This completes the proof. $\blacksquare$

###  7.5 Proof of Proposition 3

We separate the proof into eight steps.

Step 1. $\bar{V}$ is increasing.

Proof of step 1. Suppose not. Then, there exist $k_{1},k_{2}>0$ such that
$k_{1}<k_{2}$ and $\bar{V}(k_{1})=\bar{V}(k_{2})$. Because $\bar{V}$ is
concave and nondecreasing, we have $\bar{V}(k)=\bar{V}(k_{1})$ for every
$k>k_{1}$.

Choose any $(k(t),c(t))\in A_{k_{1}}$. Because $u$ is concave, for every
$(k^{*},c^{*})\in\mathbb{R}^{2}_{++}$,

$\displaystyle~{}\int_{0}^{1}e^{-\rho t}[u(c^{*},k^{*})-u(c(t),k(t))]dt$
$\displaystyle\geq$ $\displaystyle~{}\int_{0}^{1}e^{-\rho
t}\left[\frac{\partial u}{\partial c}(c^{*},k^{*})(c^{*}-c(t))+\frac{\partial
u}{\partial k}(c^{*},k^{*})(k^{*}-k(t))\right]dt$ $\displaystyle=$
$\displaystyle~{}\frac{\partial u}{\partial
c}(c^{*},k^{*})\left[\frac{c^{*}}{\rho}[1-e^{-\rho}]-\int_{0}^{1}e^{-\rho
t}c(t)dt\right]$ $\displaystyle~{}+\frac{\partial u}{\partial
k}(c^{*},k^{*})\left[\frac{k^{*}}{\rho}[1-e^{-\rho}]-\int_{0}^{1}e^{-\rho
t}k(t)dt\right]$ $\displaystyle\geq$ $\displaystyle~{}\frac{\partial
u}{\partial
c}(c^{*},k^{*})\left[\frac{c^{*}}{\rho}[1-e^{-\rho}]-\int_{0}^{1}c(t)dt\right]$
$\displaystyle~{}+\frac{\partial u}{\partial
k}(c^{*},k^{*})\left[\frac{k^{*}}{\rho}[1-e^{-\rho}]-\int_{0}^{1}k(t)dt\right].$

Define

$N=\max\\{F(k,0)|0\leq k\leq\max_{t\in[0,1]}k^{+}(t;k_{1})\\},$
$\varepsilon_{1}=\min\\{F(k,0)-F(k,1)|0\leq
k\leq\max_{t\in[0,1]}k^{+}(t;k_{1})\\},$
$\varepsilon_{2}=\max\\{F(k,0)-F(k,1)|0\leq
k\leq\max_{t\in[0,1]}k^{+}(t;k_{1})\\}.$

Because $F$ is concave, we have $F(k,1)-F(k,c)\geq\varepsilon_{1}(c-1)$ for
every $k\in[0,\max_{t\in[0,1]}k^{+}(t;k_{1})]$ and $c>1$. Since $0\leq
k(t)\leq k^{+}(t;k_{1})$ for every $t\in[0,1]$, we have

$\displaystyle F(k(t),c(t))=$
$\displaystyle~{}F(k(t),0)+(F(k(t),c(t))-F(k(t),1))+(F(k(t),1)-F(k(t),0))$
$\displaystyle\leq$ $\displaystyle~{}N-\varepsilon_{1}+F(k(t),c(t))-F(k(t),1)$
$\displaystyle\leq$
$\displaystyle~{}N-\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{1}(c(t)-1)$
$\displaystyle=$ $\displaystyle~{}N+\varepsilon_{2}-\varepsilon_{1}c(t),$

and thus,

$\displaystyle 0\leq$ $\displaystyle~{}k(1)$ $\displaystyle=$
$\displaystyle~{}k_{1}+\int_{0}^{1}\dot{k}(t)dt$ $\displaystyle=$
$\displaystyle~{}k_{1}+\int_{0}^{1}F(k(t),c(t))dt$ $\displaystyle\leq$
$\displaystyle~{}k_{1}+\int_{0}^{1}(N+\varepsilon_{2}-\varepsilon_{1}c(t))dt=k_{1}+N+\varepsilon_{2}-\varepsilon_{1}\int_{0}^{1}c(t)dt.$

Therefore,

$\int_{0}^{1}c(t)dt\leq\frac{k_{1}+N+\varepsilon_{2}}{\varepsilon_{1}}.$

Moreover,

$\int_{0}^{1}k(t)dt\leq\int_{0}^{1}k^{+}(t,k_{1})dt,$

and thus, for fixed sufficiently large $c^{*}>0$ and $k^{*}>k_{1}$, there
exists $\varepsilon>0$ such that

$\int_{0}^{1}e^{-\rho t}[u(c^{*},k^{*})-u(c(t),k(t))]dt\geq\varepsilon$

for every $(k(t),c(t))\in A_{k_{1}}$. Choose any $(k(t),c(t))\in A_{k_{1}}$
such that $c(t)$ is bounded and

$\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt>\bar{V}(k_{1})-\frac{\varepsilon}{2},$

and define

$\tilde{c}(t)=\begin{cases}c^{*}&\mbox{if }0\leq t\leq 1,\\\
c(t)&\mbox{otherwise}.\end{cases}$

By almost the same arguments as in step 2 of the proof of Proposition 2, we
can show that there exists $\hat{k}>0$ and $\tilde{k}(t)$ defined on
$\mathbb{R}_{+}$ such that $\tilde{k}(t)\geq k^{*}$ for all $t\in[0,1]$,
$\tilde{k}(t)\geq k(t)$ for all $t\geq 0$, and $(\tilde{k}(t),\tilde{c}(t))\in
A_{\hat{k}}$. Then,

$\bar{V}(\hat{k})\geq\int_{0}^{\infty}e^{-\rho
t}u(\tilde{c}(t),\tilde{k}(t))dt>\bar{V}(k_{1}),$

which is a contradiction. This completes the proof of step 1. $\blacksquare$

Step 2. If $\bar{V}$ is differentiable at $\bar{k}>0$, then

$\sup_{c\geq
0}\\{F(\bar{k},c)\bar{V}^{\prime}(\bar{k})+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})\leq
0.$

Proof of step 2. Fix any $c>0$. By Lemma 2, there exists $\varepsilon>0$ such
that the differential equation

$\dot{k}(t)=F(k(t),c),\ k(0)=\bar{k}$

has a positive and continuously differentiable solution $k(t)$ on
$[0,\varepsilon]$. For all $t\in]0,\varepsilon[$,

$\int_{0}^{t}e^{-\rho s}u(c,k(s))ds\leq\bar{V}(\bar{k})-e^{-\rho
t}\bar{V}(k(t)).$

Thus,

$\bar{V}(\bar{k})-\bar{V}(k(t))\geq\int_{0}^{t}e^{-\rho
s}u(c,k(s))ds+(e^{-\rho t}-1)\bar{V}(k(t)).$

Dividing both sides by $t$ and taking the limit $t\to 0$, we have

$-\bar{V}^{\prime}(\bar{k})F(\bar{k},c)\geq
u(c,\bar{k})-\rho\bar{V}(\bar{k}),$

which implies that

$F(\bar{k},c)\bar{V}^{\prime}(\bar{k})+u(c,\bar{k})-\rho\bar{V}(\bar{k})\leq
0.$

Because the left-hand side is continuous in $c$, the above inequality holds
for every $c\geq 0$. Therefore,

$\sup_{c\geq
0}\\{F(\bar{k},c)\bar{V}^{\prime}(\bar{k})+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})\leq
0,$

as desired. This completes the proof of step 2. $\blacksquare$

Consider the following modified problem of (4):

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ subject to. $\displaystyle~{}(k(t),c(t))\in A_{\bar{k}},$
(25) $\displaystyle~{}c(t)\leq n\mbox{ for all }t\geq 0,$

where $n>0$. Define $A^{n}_{\bar{k}}=\\{(k(t),c(t))\in A_{\bar{k}}|c(t)\leq
n\mbox{ for all }t\geq 0\\}$, and let

$V_{n}(\bar{k})=\sup\left\\{\left.\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\right|(k(t),c(t))\in A^{n}_{\bar{k}}\right\\}.$

We call the function $V_{n}$ the value function of the problem (25). By almost
the same arguments as in the proof of Proposition 2, we can show that $V_{n}$
is concave and nondecreasing, and $V_{n}(\bar{k})>-\infty$ for all
$\bar{k}>0$. Because $V_{n}(\bar{k})\leq\bar{V}(\bar{k})<+\infty$, we have
that $V_{n}$ is finite.

Step 3. For every $\bar{k}>0$ and $p\in\partial V_{n}(\bar{k})$,292929Recall
that $p\in\partial V_{n}(\bar{k})$ if and only if $V_{n}(k)-V_{n}(\bar{k})\leq
p(k-\bar{k})$ for all $k\geq 0$.

$\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}-\rho V_{n}(\bar{k})\geq 0.$
(26)

Proof of step 3.303030We use the boundedness of the function
$c\mapsto\frac{\partial u}{\partial k}(c,k)$ on $]0,M]$ for only proving this
step. First, choose any $T>0$ such that

$\bar{k}>\frac{d_{2}n}{d_{1}}(e^{d_{1}T}-1).$

By Lemma 3, for all measurable functions $c:[0,T]\to[0,n]$, we have that there
exists a solution $k(t)$ of the following differential equation

$\dot{k}(t)=F(k(t),c(t))$

defined on $[0,T]$ such that for every $t\in[0,T]$,

$k(t)\geq
e^{-d_{1}T}\left[\bar{k}-\frac{d_{2}n}{d_{1}}(e^{d_{1}T}-1)\right]>0.$ (27)

Choose any $(k(t),c(t))\in A^{n}_{\bar{k}}$ such that there exists $h>0$ such
that

$\int_{0}^{T}e^{-\rho t}u(c(t),k(t)-h)dt\in\mathbb{R}.$

Define $c_{m}(t)=\max\\{c(t),m^{-1}\\}$. As we argued above, there exists a
solution $k_{m}(t)$ of the following ODE:

$\dot{k}(t)=F(k(t),c_{m}(t)),\ k(0)=\bar{k}$

defined on $[0,T]$. Moreover, because $F$ is Lipschitz on every compact set,
we have that there exists $L>0$ such that for every $t\in[0,T]$,

$\displaystyle|k(t)-k_{m}(t)|\leq$
$\displaystyle~{}\int_{0}^{t}|F(k(s),c(s))-F(k_{m}(s),c_{m}(s))|ds$
$\displaystyle\leq$
$\displaystyle~{}\int_{0}^{t}L[|k(s)-k_{m}(s)|+|c(s)-c_{m}(s)|]ds.$

Therefore, by Gronwall’s inequality,313131See the first part of Lemma 3 of
Hosoya (2019a). we have that
$\lim_{m\to\infty}\max_{t\in[0,T]}|k_{m}(t)-k(t)|=0$. We show that for every
$t\in]0,T]$,

$\lim_{m\to\infty}\int_{0}^{t}e^{-\rho
s}u(c_{m}(s),k_{m}(s))ds=\int_{0}^{t}e^{-\rho s}u(c(s),k(s))ds.$

Let $h>0$ be sufficiently small. Then, by the monotone comvergence theorem,

$\lim_{m\to\infty}\int_{0}^{t}e^{-\rho
s}u(c_{m}(s),k(s)-h)ds=\int_{0}^{t}e^{-\rho s}u(c(s),k(s)-h)ds.$

Therefore, we have that

$\displaystyle~{}\liminf_{m\to\infty}\int_{0}^{t}e^{-\rho
s}u(c_{m}(s),k_{m}(s))ds$ $\displaystyle\geq$
$\displaystyle~{}\lim_{h\downarrow 0}\int_{0}^{t}e^{-\rho
s}u(c(s),k(s)-h)ds=\int_{0}^{t}e^{-\rho s}u(c(s),k(s))ds.$

Meanwhile, by Lemma 2, we have that $k_{m}(t)\leq k(t)$ for all $t\in[0,T]$.
Therefore,

$\limsup_{m\to\infty}\int_{0}^{t}e^{-\rho
s}u(c_{m}(s),k_{m}(s))ds\leq\int_{0}^{t}e^{-\rho s}u(c(s),k(s))ds,$

which implies that our claim is correct.

Fix sufficiently small $t\in]0,T]$ and $\varepsilon>0$. Because $V_{n}$ is
concave, it is continuous, and thus, there exists $\delta>0$ and
$(\tilde{k}(s),c(s))\in A^{n}_{\bar{k}-\delta}$ such that

$\int_{0}^{t}e^{-\rho s}u(c(s),\tilde{k}(s))ds>V_{n}(\bar{k})-e^{-\rho
t}V_{n}(\tilde{k}(t))-t\varepsilon.$

By (27), we can assume that $\inf_{s\in[0,t]}\tilde{k}(s)>0$. Consider the
following ODE:

$\dot{k}(s)=F(k(s),c(s)),\ k(0)=\bar{k}.$

By Lemma 3, there exists a solution $k(s)$ of the above ODE defined on $[0,t]$
such that $k(s)\geq\tilde{k}(s)$ for every $s\in[0,t]$. Because of
Carathéodory-Picard-Lindelöf’s theorem, we must have that $k(s)>\tilde{k}(s)$
for every $s\in[0,t]$. Therefore, $(k(t),c(t))\in A^{n}_{\bar{k}}$ and there
exists $h>0$ such that

$\int_{0}^{t}e^{-\rho s}u(c(s),k(s)-h)ds>V_{n}(\bar{k})-e^{-\rho
t}V_{n}(k(t))-t\varepsilon.$

By our previous argument, we can assume that
$\inf_{s\in[0,t]}c(s)>0$.323232This requirement is needed for ensuring the
integrability of the functions $s\mapsto u(c(s),\bar{k})$ and $s\mapsto
u(c(s),k(s))$. By the definition of $p$,

$\displaystyle\int_{0}^{t}-F(k(s),c(s))pds=$ $\displaystyle~{}p(\bar{k}-k(t))$
$\displaystyle\leq$ $\displaystyle~{}V_{n}(\bar{k})-V_{n}(k(t))$
$\displaystyle\leq$ $\displaystyle~{}\int_{0}^{t}e^{-\rho
s}u(c(s),k(s))ds+(e^{-\rho t}-1)V_{n}(k(t))+t\varepsilon.$

Thus,

$\displaystyle~{}\int_{0}^{t}[F(\bar{k},c(s))p+u(c(s),\bar{k})]ds+\int_{0}^{t}[u(c(s),k(s))-u(c(s),\bar{k})]ds$
$\displaystyle~{}+\int_{0}^{t}(e^{-\rho
s}-1)u(c(s),k(s))ds+\int_{0}^{t}[F(k(s),c(s))-F(\bar{k},c(s))]pds$
$\displaystyle~{}+(e^{-\rho t}-1)V_{n}(k(t))$ $\displaystyle\geq$
$\displaystyle~{}-t\varepsilon.$

Because

$\int_{0}^{t}[F(\bar{k},c(s))p+u(c(s),\bar{k})]ds\leq\sup_{c\geq
0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}t,$

we have

$\displaystyle~{}\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}t+(e^{-\rho
t}-1)V_{n}(k(t))$ $\displaystyle\geq$
$\displaystyle~{}-t\varepsilon+\int_{0}^{t}(1-e^{-\rho s})u(c(s),k(s))ds$
$\displaystyle~{}+\int_{0}^{t}[F(\bar{k},c(s))-F(k(s),c(s))]pds+\int_{0}^{t}[u(c(s),\bar{k})-u(c(s),k(s))]ds$
$\displaystyle\equiv$
$\displaystyle~{}-t\varepsilon+J_{1}(t)+J_{2}(t)+J_{3}(t).$

Now, we have that333333Note that, the pair of functions $(k(s),c(s))$ depends
on $t>0$. This problem makes the following evaluation difficult.

$\displaystyle 0\geq$
$\displaystyle~{}\int_{0}^{t}[u(c(s),k(s))-u(n,k^{+}(s,\bar{k}))]ds$
$\displaystyle\geq$ $\displaystyle~{}e^{\rho t}\int_{0}^{t}e^{-\rho
s}[u(c(s),k(s))-u(n,k^{+}(s,\bar{k}))]ds$ $\displaystyle\geq$
$\displaystyle~{}e^{\rho t}V_{n}(\bar{k})-V_{n}(k(t))-e^{\rho
t}t\varepsilon-e^{\rho t}\int_{0}^{t}e^{-\rho s}u(n,k^{+}(s,\bar{k}))ds$ (28)
$\displaystyle\geq$ $\displaystyle~{}e^{\rho
t}V_{n}(\bar{k})-V_{n}(k^{+}(t,\bar{k}))-e^{\rho t}t\varepsilon-e^{\rho
t}\int_{0}^{t}e^{-\rho s}u(n,k^{+}(s,\bar{k}))ds.$

Moreover,

$\displaystyle|J_{1}(t)|\leq$ $\displaystyle~{}\int_{0}^{t}(1-e^{-\rho
s})|u(n,k^{+}(s,\bar{k}))|ds$ $\displaystyle~{}+(1-e^{-\rho
t})\left|\int_{0}^{t}[u(c(s),k(s))-u(n,k^{+}(s,\bar{k}))]ds\right|,$

and thus, by (28), we have that

$\lim_{t\to 0}\frac{J_{1}(t)}{t}=0.$

Let $k^{*}(s)$ be the solution of the following differential equation:

$\dot{k}(s)=F(k(s),n),\ k(0)=\bar{k}.$

Then,

$|J_{2}(t)|\leq\int_{0}^{t}p\max\\{|F(k,c)-F(\bar{k},c)||k^{*}(s)\leq k\leq
k^{+}(s,\bar{k}),0\leq c\leq n\\}ds,$

and thus

$\lim_{t\to 0}\frac{J_{2}(t)}{t}=0.$

Finally, because $u$ is concave,

$u(c(s),\bar{k})-u(c(s),k(s))\geq\frac{\partial u}{\partial
k}(c(s),\bar{k})(\bar{k}-k^{+}(s,\bar{k})),$

and by Assumption 4, we have that $c\mapsto\frac{\partial u}{\partial
k}(c,\bar{k})$ is bounded on $]0,n]$. Therefore,

$\limsup_{t\to 0}\frac{J_{3}(t)}{t}\geq 0.$

In conclusion, we have that

$\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}-\rho
V_{n}(\bar{k})\geq-\varepsilon,$

and because $\varepsilon>0$ is arbitrary, equation (26) holds. This completes
the proof of step 3. $\blacksquare$

Step 4. For every $\bar{k}>0$, $V_{n}(\bar{k})\uparrow\bar{V}(\bar{k})$ as
$n\to\infty$.

Proof of step 4. It is clear by Lemma 5. $\blacksquare$

Step 5. If $\bar{V}$ is differentiable at $\bar{k}>0$, then

$\sup_{c\geq
0}\\{F(\bar{k},c)\bar{V}^{\prime}(\bar{k})+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})\geq
0.$

Proof of step 5. For any sufficiently large $n$, choose any $p_{n}\in\partial
V_{n}(\bar{k})$. Fix any $\delta>0$, and choose $k_{1},k_{2}>0$ with
$k_{1}<\bar{k}<k_{2}$ and

$\bar{V}^{\prime}(\bar{k})-\delta<\frac{\bar{V}(k_{2})-\bar{V}(\bar{k})}{k_{2}-\bar{k}}\leq\bar{V}^{\prime}(\bar{k})\leq\frac{\bar{V}(k_{1})-\bar{V}(\bar{k})}{k_{1}-\bar{k}}<\bar{V}^{\prime}(\bar{k})+\delta.$

Because $V_{n}(k_{i})\uparrow\bar{V}(k_{i})$ as $n\to\infty$ for $i=1,2$ and
$V_{n}(\bar{k})\uparrow\bar{V}(\bar{k})$, we have

$\max_{i=1,2}\left|\frac{V_{n}(k_{i})-V_{n}(\bar{k})}{k_{i}-\bar{k}}-\frac{\bar{V}(k_{i})-\bar{V}(\bar{k})}{k_{i}-\bar{k}}\right|<\delta$

for any sufficiently large $n$. Therefore,

$|p_{n}-\bar{V}^{\prime}(\bar{k})|\leq 2\delta$

for sufficiently large $n$, and thus
$\lim_{n\to\infty}p_{n}=\bar{V}^{\prime}(\bar{k})$.

Hence,

$\displaystyle 0\leq$ $\displaystyle~{}\sup_{c\geq
0}\\{F(\bar{k},c)p_{n}+u(c,\bar{k})\\}-\rho V_{n}(\bar{k})$ $\displaystyle=$
$\displaystyle~{}F(\bar{k},c^{*}(p_{n},\bar{k}))p_{n}+u(c^{*}(p_{n},\bar{k}),\bar{k})-\rho
V_{n}(\bar{k})$ $\displaystyle\to$
$\displaystyle~{}F(\bar{k},c^{*}(\bar{V}^{\prime}(\bar{k}),\bar{k}))\bar{V}^{\prime}(\bar{k})+u(c^{*}(\bar{V}^{\prime}(\bar{k}),\bar{k}),\bar{k})-\rho\bar{V}(\bar{k})$
$\displaystyle=$ $\displaystyle~{}\sup_{c\geq
0}\\{F(\bar{k},c)\bar{V}^{\prime}(\bar{k})+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k}),$

as desired. This completes the proof of step 5. $\blacksquare$

Step 6. For all $\bar{k}>0$ and $p\in\partial\bar{V}(\bar{k})$,

$\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})=0.$

Proof of step 6. First, let $p_{1}=D_{-}\bar{V}(k),k_{m}\uparrow k$ as
$m\to\infty$, and $\bar{V}^{\prime}(k_{m})$ can be defined. Because $\bar{V}$
is concave, $\bar{V}^{\prime}(k_{m})$ is decreasing in $m$, and thus converges
to some number $p^{*}\geq p_{1}$. If $p^{*}>p_{1}$, then there exists
$k^{\prime}<k$ such that

$\frac{\bar{V}(k^{\prime})-\bar{V}(k)}{k^{\prime}-k}<p^{*}.$

Because $k_{m}\uparrow k$ as $m\to\infty$, there exists $m$ such that

$k_{m}>k^{\prime},\
\frac{\bar{V}(k^{\prime})-\bar{V}(k_{m})}{k^{\prime}-k_{m}}<p^{*}.$

This implies that $\bar{V}^{\prime}(k_{m})<p^{*}$, which is a contradiction.
Therefore, $\bar{V}^{\prime}(k_{m})$ converges to $p_{1}$. Similarly, if
$p_{2}=D_{+}\bar{V}(k),k_{m}\downarrow k$ and $\bar{V}^{\prime}(k_{m})$ can be
defined, then $\bar{V}^{\prime}(k_{m})$ converges to $p_{2}$. In particular,
if $\bar{V}$ is differentiable around $\bar{k}$, then $\bar{V}^{\prime}$ is
continuous at $\bar{k}$.

Choose any $\bar{k}>0$. If $\bar{V}$ is differentiable at $\bar{k}$, then
$\partial\bar{V}(\bar{k})=\\{\bar{V}^{\prime}(\bar{k})\\}$, and the claim of
step 6 follows from steps 2 and 5. Therefore, suppose that $\bar{V}$ is not
differentiable at $\bar{k}$, and
$p_{1}=D_{-}\bar{V}(\bar{k}),p_{2}=D_{+}\bar{V}(\bar{k})$. By assumption, we
have $p_{1}>p_{2}$, and $\partial\bar{V}(\bar{k})=[p_{2},p_{1}]$. Choose
sequences $(k_{1}^{m}),(k_{2}^{m})$ such that
$k_{1}^{m}\uparrow\bar{k},k_{2}^{m}\downarrow\bar{k}$ as $m\to\infty$, and
$\bar{V}$ is differentiable at $k_{i}^{m}$ for all $m$. By the above
arguments, we have that $\bar{V}^{\prime}(k_{1}^{m})\downarrow p_{1}$ and
$\bar{V}^{\prime}(k_{2}^{m})\uparrow p_{2}$ as $m\to\infty$. By steps 2 and 5,
we have

$\sup_{c\geq 0}\\{F(\bar{k},c)p_{i}+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})=0.$

Next, choose any $p$ with $p_{2}<p<p_{1}$, and define
$\varepsilon=\min\\{p_{1}-p,p-p_{2}\\}$. As in the proof of step 5, we can
show that for any $m$ and $i\in\\{1,2\\}$,

$q\in\partial
V_{n}(k_{i}^{m})\Rightarrow|q-\bar{V}^{\prime}(k_{i}^{m})|<\varepsilon$

for sufficiently large $n$. Choose $n_{m}$ as such an $n$. We can assume that
$n_{m}$ is increasing in $m$. Because
$|\bar{V}^{\prime}(k_{i}^{m})-p|\geq\varepsilon$, we have there exists
$k^{m}\in[k_{1}^{m},k_{2}^{m}]$ such that $p\in\partial V_{n_{m}}(k^{m})$.
Then, by step 3,

$\sup_{c\geq 0}\\{F(k^{m},c)p+u(c,k^{m})\\}-\rho V_{n_{m}}(k^{m})\geq 0.$

Because $V_{n_{m}}(k)\uparrow\bar{V}(k)$ as $m\to\infty$ pointwise, by Dini’s
theorem, $V_{n_{m}}\to\bar{V}$ uniformly as $m\to\infty$ on any compact set in
$\mathbb{R}_{++}$. Therefore,

$\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})\geq 0.$

Now, define

$g(p)=\sup_{c\geq 0}\\{F(\bar{k},c)p+u(c,\bar{k})\\}.$

Suppose that $p,q\in[p_{2},p_{1}]$ and $r=(1-t)p+tq$ for $t\in[0,1]$. Then,

$\displaystyle g(r)=$
$\displaystyle~{}F(\bar{k},c^{*}(r,\bar{k}))r+u(c^{*}(r,\bar{k}),\bar{k})$
$\displaystyle=$
$\displaystyle~{}(1-t)[F(\bar{k},c^{*}(r,\bar{k}))p+u(c^{*}(r,\bar{k}),\bar{k})]+t[F(\bar{k},c^{*}(r,\bar{k}))q+u(c^{*}(r,\bar{k}),\bar{k})]$
$\displaystyle\leq$ $\displaystyle~{}(1-t)g(p)+tg(q).$

Therefore, $g$ is convex. Suppose that $p\in[p_{2},p_{1}]$. Because
$g(p_{1})=g(p_{2})=\rho\bar{V}(\bar{k})$ and $g(p)\geq\rho\bar{V}(\bar{k})$,
we have that $g(p)=\rho\bar{V}(\bar{k})$. This completes the proof of step 6.
$\blacksquare$

Step 7. $\bar{V}$ is a viscosity solution of the HJB equation.

Proof of step 7. Choose any $\bar{k}>0$. Suppose that $\varphi(k)$ is a
continuously differentiable function defined on a neighborhood of $\bar{k}$
such that $\varphi(\bar{k})=\bar{V}(\bar{k})$ and either
$\varphi(k)\geq\bar{V}(k)$ for all $k$ or $\varphi(k)\leq\bar{V}(k)$ for all
$k$. Then, $\bar{k}$ is an extremal point of $\varphi(k)-\bar{V}(k)$, and thus
$\varphi^{\prime}(\bar{k})\in\partial\bar{V}(\bar{k})$. This implies that

$\sup_{c\geq
0}\\{F(\bar{k},c)\varphi^{\prime}(\bar{k})+u(c,\bar{k})\\}-\rho\bar{V}(\bar{k})=0.$

Therefore, $\bar{V}$ is a viscosity solution of the HJB equation. This
completes the proof of step 7. $\blacksquare$

Step 8. If Assumption 5 holds, then $\bar{V}$ is continuously differentiable
on $\mathbb{R}_{++}$.

Proof of step 8. As we have argued in the proof of step 6, it suffices to show
that $\bar{V}$ is differentiable at any $\bar{k}>0$. Suppose that $\bar{V}$ is
not differentiable at $\bar{k}>0$, and let $p_{1}=D_{-}\bar{V}(\bar{k})$ and
$p_{2}=D_{+}\bar{V}(\bar{k})$. Then, we have $p_{2}<p_{1}$. Define a function
$g$ as in the proof of step 6, and choose $p\in]p_{2},p_{1}[$. Then, for every
$q$ around $p$,

$g(q)\geq F(\bar{k},c^{*}(p,\bar{k}))q+u(c^{*}(p,\bar{k}),\bar{k}),\
g(p)=F(\bar{k},c^{*}(p,\bar{k}))p+u(c^{*}(p,\bar{k}),\bar{k}).$

Because $g$ is constant around $p$, we must have
$F(\bar{k},c^{*}(p,\bar{k}))=0$ for such $p$. Therefore, there exists
$c^{*}>0$ such that $c^{*}(p,\bar{k})=c^{*}$ for every $p\in]p_{2},p_{1}[$.
However, this implies that $\frac{\partial F}{\partial
c}(\bar{k},c^{*})p+\frac{\partial u}{\partial c}(c^{*},\bar{k})=0$ for every
$p\in]p_{2},p_{1}[$, which is impossible. This completes the proof of step 8.
$\blacksquare$

Steps 1-8 show that all the claims in Proposition 3 are correct. This
completes the proof. $\blacksquare$

###  7.6 Proof of Theorem 1

First, consider the following problem:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ subject to. $\displaystyle~{}c(t)\mbox{ is locally
integrable},$ $\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\mbox{ can be
defined},$
$\displaystyle~{}\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})\mbox{
a.e.},$ $\displaystyle~{}k(0)=\bar{k}.$

Define $A^{L}_{\bar{k}}$ as the set of all pairs $(k(t),c(t))$ such that
$k(t)$ is nonnegative and absolutely continuous on every compact set, $c(t)$
is nonnegative and locally integrable, $\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ can be defined, $k(0)=\bar{k}$ and

$\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})$

for almost all $t\geq 0$. Let

$V_{1}(\bar{k})=\sup\left\\{\left.\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\right|(k(t),c(t))\in A^{L}_{\bar{k}}\right\\}.$

Step 1. $V_{1}(\bar{k})\geq\bar{V}(\bar{k})$.

Proof of step 1. By Proposition 2, $\bar{V}(\bar{k})>-\infty$. For every
$\varepsilon>0$ and $N>0$, there exists $(k(t),c(t))\in A_{\bar{k}}$ such that
$c(t)$ is bounded and

$\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\geq\min\\{\bar{V}(\bar{k})-\varepsilon,N\\}.$

Consider the following differential equation:

$\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*}),\
k(0)=\bar{k}.$

The solution of the above equation is:

$\hat{k}(t)=e^{\gamma t}\left[\bar{k}-\int_{0}^{t}e^{-\gamma s}(\gamma
k^{*}+\delta(c(s)-c^{*})-F(k^{*},c^{*}))ds\right].$

Because $(\gamma,-\delta)\in\partial F(k^{*},c^{*})$, we have
$F(k,c)\leq\gamma(k-k^{*})-\delta(c-c^{*})+F(k^{*},c^{*})$ for all $(k,c)$,
and thus, by Lemma 2, $\hat{k}(t)\geq k(t)$ for every $t\geq 0$. Therefore,
$(\hat{k}(t),c(t))\in A^{L}_{\bar{k}}$, and thus

$V_{1}(\bar{k})\geq\min\\{\bar{V}(\bar{k})-\varepsilon,N\\}.$

Because $\varepsilon,N$ are arbitrary, we have that
$V_{1}(\bar{k})\geq\bar{V}(\bar{k})$. This completes the proof of step 1.
$\blacksquare$

Second, consider the following problem:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt$ subject to.
$\displaystyle~{}c(t)\mbox{ is locally integrable,}$ $\displaystyle~{}k(t)\geq
0,\ c(t)\geq 0,$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt\mbox{ can be defined},$
$\displaystyle~{}\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})\mbox{
a.e.},$ $\displaystyle~{}k(0)=\bar{k}.$

Define $A^{L2}_{\bar{k}}$ as the set of all pairs $(k(t),c(t))$ such that
$k(t)$ is absolutely continuous in every compact set, $c(t)$ is locally
integrable, $\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt$ can be defined, $k(0)=\bar{k}$ and

$\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})$

for almost all $t\geq 0$. Let

$V_{2}(\bar{k})=\sup\left\\{\left.\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt\right|(k(t),c(t))\in
A^{L2}_{\bar{k}}\right\\}.$

Step 2. $V_{2}(\bar{k})+\frac{C}{\rho}\geq V_{1}(\bar{k})$.

Proof of step 2. By Proposition 2 and step 1, $V_{1}(\bar{k})>-\infty$. For
every $\varepsilon>0$ and $N>0$, there exists $(k(t),c(t))\in A^{L}_{\bar{k}}$
such that

$\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt\geq\min\\{V_{1}(\bar{k})-\varepsilon,N\\}.$

Because

$\int_{0}^{\infty}e^{-\rho
t}\min\\{au_{\theta}(c(t))+bu_{\theta}(k(t))+C,0\\}dt\geq\int_{0}^{\infty}e^{-\rho
t}\min\\{u(c(t),k(t)),0\\}dt>-\infty,$

we have that

$\int_{0}^{\infty}e^{-\rho t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt$

can be defined. Hence, $(k(t),c(t))\in A^{L2}_{\bar{k}}$, and thus,

$V_{2}(\bar{k})+\frac{C}{\rho}\geq\min\\{V_{1}(\bar{k})-\varepsilon,N\\}.$

Because $\varepsilon,N$ are arbitrary, we have that
$V_{2}(\bar{k})+\frac{C}{\rho}\geq V_{1}(\bar{k})$. This completes the proof
of step 2. $\blacksquare$

Define

$C^{*}=\frac{\rho-(1-\theta)\gamma}{\theta\delta}\left[\bar{k}-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right].$

Note that,

$-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}=\frac{1}{\gamma}[\gamma(0-k^{*})-\delta(0-c^{*})+F(k^{*},c^{*})]\geq
0,$

and thus $C^{*}>0$ for all $\bar{k}>0$. Define

$V_{3}(\bar{k})=\begin{cases}\frac{(C^{*})^{1-\theta}\theta}{(1-\theta)(\rho-(1-\theta)\gamma)}-\frac{1}{\rho(1-\theta)},&\mbox{if
}\theta\neq 1,\\\ \frac{\log
C^{*}}{\rho}+\frac{\gamma-\rho}{\rho^{2}}&\mbox{if }\theta=1\end{cases}$

and

$V_{4}(\bar{k})=\begin{cases}\frac{\left(\bar{k}-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right)^{1-\theta}}{(1-\theta)(\rho-(1-\theta)\gamma)}-\frac{1}{\rho(1-\theta)}&\mbox{if
}\theta\neq 1,\\\ \frac{\log\left(\bar{k}-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right)}{\rho}+\frac{\gamma}{\rho^{2}},&\mbox{if
}\theta=1.\end{cases}$

Step 3. $aV_{3}(\bar{k})+bV_{4}(\bar{k})\geq V_{2}(\bar{k})$.

Proof of Step 3. Let

$c^{*}(t)=C^{*}e^{\frac{\gamma-\rho}{\theta}t},$ $k^{*}(t)=e^{\gamma
t}\left[\bar{k}-\int_{0}^{t}e^{-\gamma s}[\gamma
k^{*}+\delta(c^{*}(s)-c^{*})-F(k^{*},c^{*})]ds\right].$

Note that,

$\dot{k}^{*}(t)=\gamma(k^{*}(t)-k^{*})-\delta(c^{*}(t)-c^{*})+F(k^{*},c^{*}),$
$\frac{d}{dt}(u_{\theta}^{\prime}(c^{*}(t)))=(\rho-\gamma)u_{\theta}^{\prime}(c^{*}(t)).$

Then, for every $(k(t),c(t))\in A^{L2}_{\bar{k}}$,

$\displaystyle~{}\int_{0}^{T}e^{-\rho
t}(u_{\theta}(c(t))-u_{\theta}(c^{*}(t)))dt$ $\displaystyle\leq$
$\displaystyle~{}\int_{0}^{T}e^{-\rho
t}u_{\theta}^{\prime}(c^{*}(t))(c(t)-c^{*}(t))dt$ $\displaystyle=$
$\displaystyle~{}\delta^{-1}\int_{0}^{T}e^{-\rho
t}u_{\theta}^{\prime}(c^{*}(t))[\gamma(k(t)-k^{*}(t))-(\dot{k}(t)-\dot{k}^{*}(t))]dt$
$\displaystyle=$ $\displaystyle~{}\delta^{-1}e^{-\rho
T}u_{\theta}^{\prime}(c^{*}(T))(k^{*}(T)-k(T))$ $\displaystyle\leq$
$\displaystyle~{}\delta^{-1}e^{-\rho T}u_{\theta}^{\prime}(c^{*}(T))k^{*}(T)$
$\displaystyle=$
$\displaystyle~{}\delta^{-1}(C^{*})^{-\theta}\left[\bar{k}-\left(k^{*}-\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right)(1-e^{-\gamma T})-\frac{\theta\delta
C^{*}}{\rho-(1-\theta)\gamma}(1-e^{\frac{(1-\theta)\gamma-\rho}{\theta}T})\right]$
$\displaystyle\to$ $\displaystyle~{}0\mbox{ (as $T\to\infty$)}.$

Moreover,

$k(t)\leq\left(\bar{k}-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right)e^{\gamma t}\equiv\hat{k}e^{\gamma t},$

and thus, we have

$\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt$ $\displaystyle\leq$
$\displaystyle~{}a\int_{0}^{\infty}e^{-\rho
t}u_{\theta}(c^{*}(t))dt+b\int_{0}^{\infty}e^{-\rho
t}u_{\theta}(\hat{k}e^{\gamma t})dt$ $\displaystyle=$
$\displaystyle~{}aV_{3}(\bar{k})+bV_{4}(\bar{k}),$

which completes the proof. $\blacksquare$

By steps 1-3, we have $\bar{V}(\bar{k})<+\infty$ for every $\bar{k}>0$.

Step 4. For every $\bar{k}>0$, $e^{-\rho t}\bar{V}(k^{+}(t,\bar{k}))\to 0$ as
$t\to\infty$.

Proof. We have already shown that $\inf_{t\geq 0}k^{+}(t,\bar{k})>0$ for all
$\bar{k}>0$, and thus it suffices to show that for $i\in\\{3,4\\}$,

$\limsup_{t\to\infty}e^{-\rho t}V_{i}(k^{+}(t,\bar{k}))\leq 0,$

Define

$\hat{k}(t)=e^{\gamma t}\left[\bar{k}-\left(k^{*}-\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}\right)\right].$

By Lemma 2, we have that $\hat{k}(t)\geq k^{+}(t,\bar{k})$ for all $t\geq 0$,
and it suffices to show that for $i\in\\{3,4\\}$,

$\lim_{t\to\infty}e^{-\rho t}V_{i}(\hat{k}(t))=0.$

If $\theta=1$, then

$V_{i}(\hat{k}(t))=A\log(e^{\gamma t}+B)+C$

for some $A>0,\ B\geq 0$, and $C\in\mathbb{R}$, and

$e^{-\rho t}V_{i}(\hat{k}(t))=Ae^{-\rho t}\log(e^{\gamma t}+B)+e^{-\rho t}C\to
0$

as $t\to\infty$. If $\theta\neq 1$, then

$V_{i}(\hat{k}(t))=A(e^{\gamma t}+B)^{1-\theta}+C,$

for some $A,B,C\in\mathbb{R}$ such that $A(1-\theta)>0$ and $B\geq 0$. If
$\theta<1$, then

$e^{-\rho
t}V_{i}(\hat{k}(t))=A(e^{\frac{(1-\theta)\gamma-\rho}{1-\theta}t}+e^{-\frac{\rho}{1-\theta}t}B)^{1-\theta}+e^{-\rho
t}C\to 0$

as $t\to\infty$. If $\theta>1$, then

$e^{-\rho t}V_{i}(\hat{k}(t))=e^{-\rho t}[A(e^{\gamma t}+B)^{1-\theta}+C]\to
0$

as $t\to\infty$. Thus, in any case, our claim is correct. This completes the
proof. $\blacksquare$

###  7.7 Proof of Proposition 4

We separate the proof into three steps.

Step 1. Suppose that $V\in\mathscr{V}$ is a classical solution of the HJB
equation. Then, there exists a solution $k^{*}(t)$ of the differential
equation (13) defined on $\mathbb{R}_{+}$. Moreover, for any such solution,
$\inf_{t}k^{*}(t)>0$.

Proof of step 1. Let $\varepsilon>0$ satisfy that

$\inf_{c\geq 0}D_{k,+}F(\varepsilon,c)>\rho,\ \varepsilon<\bar{k}.$

We show that for any solution $k(t)$ of (13) defined on $[0,T]$,
$k(t)\geq\varepsilon$.

Suppose that this claim is incorrect. Then, there exists $t^{*}>0$ such that
$0<k(t^{*})<\varepsilon$. By the mean value theorem, we can assume that
$t^{*}<T$ and $\dot{k}(t^{*})<0$. By Alexandrov’s theorem, $V$ is twice
differentiable almost everywhere, and thus we can assume that
$V^{\prime\prime}(k(t^{*}))$ can be defined.343434See Alexandrov (1939) or
Howard (1998). Define

$c(t)=c^{*}(V^{\prime}(k(t)),k(t))$

on $[0,T]$. Then, $c(t)$ is continuous and positive. Because $V$ is a solution
of the HJB equation,

$F(k(t),c(t))V^{\prime}(k(t))+u(c(t),k(t))=\rho V(k(t)),$

for every $t$ around $t^{*}$. Because $\dot{k}(t^{*})<0$, for any sufficiently
small $h>0$, there exists $k_{1},k_{2}\in[k(t^{*}+h),k(t^{*})],\
\theta_{1},\theta_{2}\in[0,1]$, $p\in\partial_{k}F(k_{2},c(t^{*}+h))$ and
$q=\frac{\partial F}{\partial c}(k(t^{*}),c(t^{*}+\theta_{2}h))$ such that

$\displaystyle~{}\rho V^{\prime}(k_{1})(k(t^{*}+h)-k(t^{*}))$ $\displaystyle=$
$\displaystyle~{}\rho(V(k(t^{*}+h))-V(k(t^{*})))$ $\displaystyle=$
$\displaystyle~{}F(k(t^{*}+h),c(t^{*}+h))V^{\prime}(k(t^{*}+h))+u(c(t^{*}+h),k(t^{*}+h))$
$\displaystyle~{}-F(k(t^{*}),c(t^{*}))V^{\prime}(k(t^{*}))-u(c(t^{*}),k(t^{*}))$
$\displaystyle=$
$\displaystyle~{}(F(k(t^{*}+h),c(t^{*}+h))-F(k(t^{*}),c(t^{*}+h)))V^{\prime}(k(t^{*}+h))$
$\displaystyle~{}+(F(k(t^{*}),c(t^{*}+h))-F(k(t^{*}),c(t^{*})))V^{\prime}(k(t^{*}+h))$
$\displaystyle~{}+F(k(t^{*}),c(t^{*}))(V^{\prime}(k(t^{*}+h))-V^{\prime}(k(t^{*})))$
$\displaystyle~{}+u(c(t^{*}+h),k(t^{*}+h))-u(c(t^{*}+h),k(t^{*}))$
$\displaystyle~{}+u(c(t^{*}+h),k(t^{*}))-u(c(t^{*}),k(t^{*}))$
$\displaystyle=$
$\displaystyle~{}pV^{\prime}(k(t^{*}+h))(k(t^{*}+h)-k(t^{*}))+\dot{k}(t^{*})(V^{\prime}(k(t^{*}+h))-V^{\prime}(t^{*}))$
$\displaystyle~{}+\frac{\partial u}{\partial
k}(c(t^{*}+h),k(t^{*}+\theta_{1}h))(k(t^{*}+h)-k(t^{*}))$
$\displaystyle~{}+\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}+\theta_{2}h))+qV^{\prime}(k(t^{*}+h))\right)(c(t^{*}+h)-c(t^{*}))$
$\displaystyle~{}+\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}))-\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}+\theta_{2}h))\right)(c(t^{*}+h)-c(t^{*}))$

by mean value theorem. To modify this equation,

$\displaystyle~{}\frac{(\rho
V^{\prime}(k_{1})-pV^{\prime}(k(t^{*}+h)))(k(t^{*}+h)-k(t^{*}))}{h}$
$\displaystyle=$
$\displaystyle~{}\frac{\dot{k}(t^{*})(V^{\prime}(k(t^{*}+h))-V^{\prime}(k(t^{*})))}{h}$
$\displaystyle~{}+\frac{\frac{\partial u}{\partial
k}(c(t^{*}+h),k(t^{*}+\theta_{1}h))(k(t^{*}+h)-k(t^{*}))}{h}$ (29)
$\displaystyle~{}+\frac{\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}+\theta_{2}h))+qV^{\prime}(k(t^{*}+h))\right)(c(t^{*}+h)-c(t))}{h}$
$\displaystyle~{}+\frac{\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}))-\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}+\theta_{2}h))\right)(c(t^{*}+h)-c(t^{*}))}{h}.$

Because $k(t^{*}+h)\leq k(t^{*})<\varepsilon$, we have $p>\inf_{c\geq
0}D_{k,+}F(k(t^{*}),c)>\rho$, and thus

$\displaystyle~{}\liminf_{h\downarrow 0}\frac{(\rho
V^{\prime}(k_{1})-pV^{\prime}(k(t^{*}+h)))(k(t^{*}+h)-k(t^{*}))}{h}$
$\displaystyle\geq$ $\displaystyle~{}(\rho-\inf_{c\geq
0}D_{k,+}F(k(t^{*}),c))V^{\prime}(k(t^{*}))\dot{k}(t^{*})>0.$

Meanwhile, the first and second terms of the right-hand side of (29) are
always nonpositive. Because of the definition of $c(t)$ and the first-order
condition, we have

$\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}+\theta_{2}h))=-rV^{\prime}(k(t^{*}+\theta_{2}h)),$

where $r=\frac{\partial F}{\partial
c}(k(t^{*}+\theta_{2}h),c(t^{*}+\theta_{2}h))$, and thus the absolute value of
the third term of the right-hand side of (29) is bounded from

$\displaystyle~{}\frac{-q|V^{\prime}(k(t^{*}))-V^{\prime}(k(t^{*}+h))|\times|c(t^{*}+h)-c(t^{*})|}{h}$
$\displaystyle~{}+\frac{|r-q|V^{\prime}(k(t^{*}+\theta_{2}h))\times|c(t^{*}+h)-c(t^{*})|}{h},$

where,

$\frac{V^{\prime}(k(t^{*}))-V^{\prime}(k(t^{*}+h))}{h}\to
V^{\prime\prime}(k(t^{*}))\dot{k}(t^{*})\mbox{ as }h\downarrow 0,$
$\frac{r-q}{h}=\frac{\partial^{2}F}{\partial k\partial
c}(k(t^{*}+\theta^{\prime}h),c(t^{*}+\theta_{2}h))\dot{k}(t^{*}+\theta^{\prime}h)\mbox{
for some }\theta^{\prime}\in[0,\theta_{2}].$

Therefore, the third term of the right-hand side converges to zero as
$h\downarrow 0$. Finally, because $\frac{\partial u}{\partial c}$ is
differentiable in $k$ on $\mathbb{R}^{2}_{++}$, we have that the fourth term
of the right-hand side of (29) is

$\frac{\partial^{2}u}{\partial k\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}+\theta^{\prime\prime}h))\dot{k}(t^{*}+\theta^{\prime\prime}h)(c(t^{*}+h)-c(t^{*}))$

for some $\theta^{\prime\prime}\in[0,\theta_{2}]$. Therefore, the fourth term
also converges to $0$, and thus the limsup of the right-hand side of (29) is
nonpositive, which is a contradiction.

Hence, $k(t)\geq\varepsilon$ for every $t\in[0,T]$. By Lemma 3, we have that
there exists a solution $k^{*}(t)$ of (13) defined on $\mathbb{R}_{+}$.
Clearly, we have that $k^{*}(t)\geq\varepsilon$ for every $t\geq 0$, and thus
$\inf_{t\geq 0}k^{*}(t)>0$. This completes the proof of step 1. $\blacksquare$

Step 2. Under the assumptions in step 1, choose a solution $k^{*}(t)$ of
equation (13) defined on $\mathbb{R}_{+}$, and define
$c^{*}(t)=c^{*}(V^{\prime}(k^{*}(t)),k^{*}(t))$. Then, $(k^{*}(t),c^{*}(t))\in
A_{\bar{k}}$ and

$V(\bar{k})=\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt.$

Proof of step 2. By step 1, we have there exists $\varepsilon>0$ such that

$\varepsilon\leq k^{*}(t)\leq k^{+}(t,\bar{k})$

for all $t\geq 0$. Because $V$ is a classical solution of the HJB equation,

$F(k^{*}(t),c^{*}(t))V^{\prime}(k^{*}(t))+u(c^{*}(t),k^{*}(t))=\rho
V(k^{*}(t)),$

for every $t\geq 0$. Therefore,

$\displaystyle\int_{0}^{T}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt=$
$\displaystyle~{}\int_{0}^{T}e^{-\rho
t}[F(k^{*}(t),c^{*}(t))V^{\prime}(k^{*}(t))+u(c^{*}(t),k^{*}(t))]dt$
$\displaystyle~{}-\int_{0}^{T}e^{-\rho t}V^{\prime}(k^{*}(t))\dot{k}^{*}(t)dt$
$\displaystyle=$ $\displaystyle~{}-\int_{0}^{T}[-\rho e^{-\rho
t}V(k^{*}(t))+e^{-\rho t}V^{\prime}(k^{*}(t))\dot{k}^{*}(t)]dt$
$\displaystyle=$ $\displaystyle~{}-\int_{0}^{T}\frac{d}{dt}[e^{-\rho
t}V(k^{*}(t))]dt$ $\displaystyle=$ $\displaystyle~{}V(\bar{k})-e^{-\rho
T}V(k^{*}(T)),$

and thus,

$V(\bar{k})-e^{-\rho T}V(\varepsilon)\geq V(\bar{k})-\int_{0}^{T}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt\geq V(\bar{k})-e^{-\rho T}V(k^{+}(T,\bar{k})).$

Hence,

$\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt=V(\bar{k}),$

which implies that $\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt$ can be
defined. Therefore, $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, which completes the
proof of step 2. $\blacksquare$

Step 3. If $(k(t),c(t))\in A_{\bar{k}}$ and $\inf_{t\geq 0}k(t)>0$, then

$\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\leq V(\bar{k}).$

Proof of step 3. Suppose that $(k(t),c(t))\in A_{\bar{k}}$ and $\inf_{t\geq
0}k(t)>0$. Because $V$ is a classical solution of the HJB equation,

$\displaystyle\int_{0}^{T}e^{-\rho t}u(c(t),k(t))dt=$
$\displaystyle~{}\int_{0}^{T}e^{-\rho
t}[F(k(t),c(t))V^{\prime}(k(t))+u(c(t),k(t))]dt$
$\displaystyle~{}-\int_{0}^{T}e^{-\rho t}V^{\prime}(k(t))\dot{k}(t)dt$
$\displaystyle\leq$ $\displaystyle~{}-\int_{0}^{T}[-\rho e^{-\rho
t}V(k(t))+e^{-\rho t}V^{\prime}(k(t))\dot{k}(t)]dt$ $\displaystyle=$
$\displaystyle~{}\int_{0}^{T}\frac{d}{dt}[-e^{-\rho t}V(k(t))]dt$
$\displaystyle=$ $\displaystyle~{}V(\bar{k})-e^{-\rho T}V(k(T)).$

Because $\inf_{t\geq 0}k(t)>0$ and $V$ satisfies (8), the right-hand side
converges to $V(\bar{k})$ as $T\to\infty$. Therefore,

$\int_{0}^{\infty}e^{-\rho t}u(c(t))dt\leq V(\bar{k}),$

as desired. This completes the proof. $\blacksquare$

###  7.8 Proof of Theorem 2

Because Assumptions 1-7 hold, we have that $\bar{V}\in\mathscr{V}$, and it is
a classical solution of the HJB equation. By Proposition 4, there exists a
solution $k^{*}(t)$ of (14) defined on $\mathbb{R}_{+}$, where $\inf_{t\geq
0}k^{*}(t)>0$. Define $c^{*}(t)=c^{*}(\bar{V}^{\prime}(k^{*}(t)),k^{*}(t))$.
Then, by Proposition 4, we have that

$\bar{V}(\bar{k})=\int_{0}^{\infty}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt,$

which implies that $(k^{*}(t),c^{*}(t))$ is a solution of (4).

Meanwhile, suppose that $V\in\mathscr{V}$ is also a classical solution of the
HJB equation. Because $\inf_{t\geq 0}k^{*}(t)>0$, by Proposition 4, we have
that

$V(\bar{k})\geq\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt=\bar{V}(\bar{k}).$

Meanwhile, by Proposition 4, there exists a solution $k^{+}(t)$ of (13)
defined on $\mathbb{R}_{+}$ such that $\inf_{t\geq 0}k^{+}(t)>0$. Define
$c^{+}(t)=c^{*}(V^{\prime}(k^{+}(t)),k^{+}(t))$. Then, by Proposition 4, we
have that $(k^{+}(t),c^{+}(t))\in A_{\bar{k}}$, and thus

$V(\bar{k})=\int_{0}^{\infty}e^{-\rho
t}u(c^{+}(t),k^{+}(t))dt\leq\bar{V}(\bar{k}).$

Hence, we conclude that $V=\bar{V}$. This completes the proof. $\blacksquare$

## Acknowledgements