# The Hamilton-Jacobi-Bellman Equation in Economic Dynamics with a Non-Smooth
Fiscal Policy

Yuhki Hosoya
Faculty of Economics, Chuo University E-mail: hosoya(at)tamacc.chuo-u.ac.jp,
ORCID ID: 0000-0002-8581-4518742-1 Higashinakano, Hachioji-shi, Tokyo
192-0393, Japan.

###### Abstract

We consider a class of economic growth models that includes the classical
Ramsey–Cass–Koopmans capital accumulation model and verify that, under several
assumptions, the value function of the model is the unique viscosity solution
to the Hamilton–Jacobi–Bellman equation. Moreover, we discuss a solution
method for these models using differential inclusion, where the
subdifferential of the value function plays an important role. Next, we
present an assumption under which the value function is a classical solution
to the Hamilton–Jacobi–Bellman equation, and show that many economic models
satisfy this assumption. In particular, our result still holds in an economic
growth model in which the government takes a non-smooth Keynesian policy rule.

MSC2020 codes: 35F21, 49L25, 91B62.

Keywords: Capital Accumulation Model, Hamilton–Jacobi–Bellman Equation,
Viscosity Solution, Keynesian Fiscal Policy, Subdifferential Calculus.

##  1 Introduction

Since the 1970s, dynamic theory in macroeconomics has commonly been modelled
using the calculus of variations or optimal control theory. The most important
reason for the emergence of such a culture was the observation in the
developed economies of the 1970s of phenomena that could not be explained by
the traditional theory of fiscal and monetary policy based on the work of
Philips [23]. Lucas [20] explained that this stemmed from a flaw in the
previous theory, and this was considered the most persuasive view in
economics. The best method for removing the flaw that Lucas criticized was
seen to be the use of optimal control theory in the model, so this style
became popular. In particular, the so-called Ramsey–Cass–Koopmans (RCK) model,
constructed by Ramsey [24] and modified by Cass [9] and Koopmans [18] for ease
of use, is the most fundamental of these models, and is frequently used in
textbooks of advanced macroeconomics: see, for example, Acemoglu [1], Barro
and Sara-i-Martin [6], Blanchard and Fischer [8], or Romer [26].

Unfortunately, the models treated in modern theories of macroeconomic dynamics
are far more complex than the RCK model and, in most cases, cannot be solved
directly. Therefore, analysis using the Hamilton–Jacobi–Bellman (HJB) equation
is commonly applied. Achdou et al. [2] is a typical example of such analysis.
In fact, optimal control problems in economics have a different form than that
classically treated by Lions [19] and Crandall and Lions [11]. For such a
class of problems, Barles [5] pointed out that the HJB equation may behave in
an unusual way. In economics, however, a ‘proof’ of the relationship between
the value function of the problem and the HJB equation was provided early on,
as dealt with in Marrialis and Brock [22]. Therefore, it was believed that the
problem noted by Barles did not occur in economics.

Hosoya [15] identified a gap of this ‘proof’ of the relationship between the
value function and the HJB equation in economics. This gap is so serious that
we can construct an economic model in which the HJB equation has no
relationship with the value function. Therefore, the problem specified by
Barles is important even in economics, making it necessary to find conditions
under which the problem vanishes. Hosoya [14] addressed this problem and
showed that, under certain conditions, the value function is the unique
classical solution to the HJB equation in some functional space.

However, Hosoya [14] assumed that the functions being treated in the model are
smooth. In economic models, it may be possible to assume that the utility
function (which expresses the consumer’s preference, and the technology
function, which describes the production process) are smooth, but we cannot be
sure that the fiscal policy rule owned by the government is smooth. Indeed, in
this paper, we construct an example of an economic model in which the
government uses a Keynesian fiscal policy rule that is non-smooth (see
Subsection 4.2). Therefore, the assumption of the smoothness is undesirable
for functions in the control problem of economic models that take fiscal
policy into account.

This is exactly the problem we focus on in this paper. In other words, our
purpose is to derive a relationship between the HJB equation and the value
function without the differentiability assumptions of the technology function
considered by Hosoya [14]. In this case, because the model is not
differentiable, it is natural that the solution of the HJB equation is also
not differentiable. Therefore, unlike Hosoya [14], we mainly use the notion of
viscosity solutions. In Hosoya [14], it was proved that the value function is
a viscosity solution to the HJB equation under several assumptions. In this
paper, we show a similar result under weaker conditions (Theorem 1).

Furthermore, this paper attempts to determine a condition under which there is
no other viscosity solution to the HJB equation than the value function. This
problem is quite non-trivial. As treated in Bardi and Capuzzo-Dolcetta [4],
the viscosity solutions to the HJB equation are usually discussed in the class
of bounded, uniformly continuous functions. In this case, for any pair of a
viscosity subsolution and a viscosity supersolution, the former is less than
the latter, from wihch the uniqueness of the viscosity solution can easily be
derived. However, this argument is possible because boundedness can be
naturally introduced into the objective function. The most typical objective
function in economic control problems is $\int_{0}^{\infty}e^{-\rho t}\log
c(t)dt$, where $c(t)$ is only assumed to be non-negative and locally
integrable. In this case, the value function is usually unbounded in the first
place. Thus, the above class of functions cannot be used as the set of
candidates for solutions. Fortunately, in most economic models, the value
function is concave. However, when considering the class of concave functions,
there are often multiple solutions to the HJB equation (see the beginning of
Subsection 3.5).

Therefore, as the set of candidates for solutions, we consider the space of
increasing and concave functions that satisfy a sort of growth condition. All
functions included in this class have positive subderivatives. Therefore, we
can evaluate these functions using differential inclusion defined in terms of
subderivatives. For problems satisfying the assumptions treated in Theorem 1
and one additional assumption, we show that the value function is the unique
viscosity solution to the HJB equation in this class (Theorem 2).

The differential inclusion that we consider is interesting in its own right.
Using this differential inclusion and the value function, we can obtain the
solution to the original optimal control problem (Corollary 1). Furthermore,
under an additional assumption, the value function becomes differentiable,
whereupon the differential inclusion becomes a usual differential equation.
Hence, we can prove that a solution to the optimal control problem can be
taken continuously (Corollary 2). These are the main results in this paper.

Finally, we confirm that our results are correctly applicable to economic
models. We do not assume differentiability in our model, but it is, of course,
possible to apply our results to models that do assume differentiability.
Therefore, we first consider the usual RCK model and check that our
assumptions are very weak. Next, we introduce a non-smooth Keynesian fiscal
policy for such models. Even in such a case, our assumptions hold naturally,
and thus the results mentioned above are all applicable. Surprisingly, even
though our constructed fiscal policy function is not differentiable, if
capital depreciation is absent, then the value function is differentiable.
However, this is not always the case in models with positive capital
depreciation.

The remaining of this paper is organized as follows. First, Section 2
discusses the model and its underlying assumptions, and presents rigorous
definitions of terms such as the value function. Section 3 derives the main
results. Section 4 confirms the applicability of our results to economic
models. Section 5 discusses the relationship with related studies. Section 6
presents the conclusion.

##  2 Model and Definitions

###  2.1 Model

In this paper, we use the following notation. First, we define
$\mathbb{R}^{n}_{+}=\\{x\in\mathbb{R}^{n}|x_{i}\geq 0\mbox{ for all }i\\}$ and
$\mathbb{R}^{n}_{++}=\\{x\in\mathbb{R}^{n}|x_{i}>0\mbox{ for all }i\\}$. If
$n=1$, we simply write these sets as $\mathbb{R}_{+}$ and $\mathbb{R}_{++}$,
respectively.

The model that we discuss in this paper is the following.111The 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. $\displaystyle~{}c(\cdot)\in W,$
$\displaystyle~{}k(t)\mbox{ is locally absolutely continuous},$
$\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$ (2.1)
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\mbox{ can be
defined},$ $\displaystyle~{}\dot{k}(t)=F(k(t),c(t))\mbox{ a.e.},$
$\displaystyle~{}k(0)=\bar{k},$

where $W$ denotes some set of functions. The symbol $c(t)$ means the amount of
consumption, and $k(t)$ denotes the amount of capital. If the instantaneous
utility function $u(c,k)$ is independent of $k$ and the technology function
$F(k,c)$ is $f(k)-dk-c$, then problem (2.1) coincides with the traditional RCK
model.222The constant $d>0$ denotes the capital depreciation rate. In the
context of the RCK model, the requirement that
$\mathbb{R}_{++}=f^{\prime}(\mathbb{R}_{++})$ (called the Inada condition) is
sometimes used. Later, we treat such a case (see Subsection 4.1).

We implicitly consider the following model: there exists a fiscal policy rule
$G(c,k)$ that is exogenously determined, and the government expenditure $g(t)$
is determined by the equation $g(t)=G(c(t),k(t))$. Moreover, $F(k,c)=f(k)-dk-
c-G(c,k)$, and $u(c,k)=v(c,G(c,k))$, where $v$ is some function. In economic
models, $f$ and $v$ are usually assumed to be differentiable, but $G$ may not
be differentiable. Hence, $F$ and $u$ may not be differentiable. However, the
non-differentiability of $u$ significantly increases the difficulty of the
analysis, and thus, we later make a compromise and assume that $u$ is
differentiable.333This can be naturally justified if $v$ is independent of
$g$. Later, we discuss the meaning of such an assumption.

Let $W_{1}$ be the set of all functions $c:\mathbb{R}_{+}\to\mathbb{R}_{+}$
that are locally integrable, and $W_{2}$ be the set of all functions
$c:\mathbb{R}_{+}\to\mathbb{R}_{+}$ that are measurable and locally bounded.
Note that $W_{2}\subset W_{1}$.

Assumption 1. $\rho>0$, and $W$ is either $W_{1}$ or $W_{2}$.

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 $\mathbb{R}^{2}_{++}$.444Because $u$ 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 $d_{1}>0$ and an increasing function
$\delta_{2}:\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that $\delta_{2}(0)=0$ and
$F(k,c)>-d_{1}k-\delta_{2}(c)$ for every $(k,c)$ such that $k>0$ and $c\geq
0$. If $W=W_{1}$, then there exists $d_{2}\geq 0$ such that
$\delta_{2}(c)=d_{2}c$ for all $c\geq 0$. Furthermore, for every $c\geq 0$,
there exists $k>0$ such that $F(k,c)>F(0,c)$.555The 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 $[0,k^{*}]$. In fact, we
can choose  $k^{*}=\min_{c\in[0,\bar{c}]}\min\arg\max\\{F(k,c)|0\leq k\leq
1\\}.$

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 $k>0$,
$\limsup_{c\downarrow 0}\frac{\partial u}{\partial k}(c,k)<+\infty$.

###  2.2 Admissibility and the Value Function

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

$\dot{k}(t)=F(k(t),c(t))\ \mbox{a.e.}.$ (2.2)

Note that, if $k(t)$ is absolutely continuous on every compact interval, then
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 $A_{\bar{k}}$, we can simplify model (2.1)
as follows:

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

For each $\bar{k}>0$, 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 problem (2.1).666We can
easily show that $A_{\bar{k}}$ is nonempty for all $\bar{k}\geq 0$. Therefore,
we can define $\bar{V}$ on $\mathbb{R}_{+}$. However, in this paper, we
consider the domain of $\bar{V}$ to be $\mathbb{R}_{++}$ for some technical
reasons; thus, $\bar{V}(0)$ is not treated throughout this paper, although it
can be defined.

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:

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

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 $T>0$, if $k(T)>0$, then

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

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

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

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 either $k(T)=0$ or

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

for every $T>0$.

###  2.3 HJB Equation

The HJB equation is given as follows.

$\sup_{c\geq 0}\\{F(k,c)V^{\prime}(k)+u(c,k)\\}-\rho V(k)=0.$ (2.3)

A function $V:\mathbb{R}_{++}\to\mathbb{R}\cup\\{-\infty\\}$ is called a
classical solution to the HJB equation if and only if $V$ is continuously
differentiable and equation (2.3) holds for every $k>0$.

In many models of the dynamic control problem, there exists no classical
solution to the HJB equation. Hence, we should extend the notion of the
solution. First, a function $V:\mathbb{R}_{++}\to\mathbb{R}$ is called a
viscosity subsolution to (2.3) if and only if it is upper semi-continuous, and
for every $k>0$ and every continuously differentiable function $\varphi$
defined on a neighbourhood of $k$ such that $\varphi(k)=V(k)$ and
$\varphi(k^{\prime})\leq V(k^{\prime})$ whenever $k^{\prime}$ is in the domain
of $\varphi$,777Usually, the condition for $\varphi$ is that
$\varphi(k^{\prime})\geq V(k^{\prime})$, because this equation is considered
for some minimization problem. In this paper, however, we treat a maximization
problem (2.1), and thus the inequality is reversed.

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

Second, a function $V:\mathbb{R}_{++}\to\mathbb{R}$ is called a viscosity
supersolution to (2.3) if and only if it is lower semi-continuous, and for
every $k>0$ and every continuously differentiable function $\varphi$ defined
on a neighbourhood of $k$ such that $\varphi(k)=V(k)$ and
$\varphi(k^{\prime})\geq V(k^{\prime})$ whenever $k^{\prime}$ is in the domain
of $\varphi$,

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

If a continuous function $V:\mathbb{R}_{++}\to\mathbb{R}$ is both a viscosity
sub- and supersolution to (2.3), then $V$ is called a viscosity solution to
(2.3).

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

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

See Proposition 1.9 of Ch.2 of Bardi and Capuzzo-Dolcetta [4].

Note: In many studies, it is proved that for any viscosity subsolution $v_{1}$
and viscosity supersolution $v_{2}$, $v_{1}\leq v_{2}$. If so, it is trivial
that if there exists a viscosity solution, then such a solution is unique.
However, in (2.1), such a result cannot be proved. For a simple counter-
example, see the first paragraph of Subsection 3.5.

###  2.4 Subdifferentials and Left- and Right-Derivatives

In this paper, we heavily use subdifferential calculus. We introduce the
notion of the subdifferential and several results. For the proofs of these
results, see textbooks on convex analysis, such as Rockafeller [25].

Suppose that a function $G:U\to\mathbb{R}$ is concave,
$U\subset\mathbb{R}^{n}$ is convex, and the interior $V$ of $U$ 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 $G$.888Formally, 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 not be 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 the
‘subdifferential’. It is known that $G$ is differentiable at $x$ if and only
if $\partial G(x)$ is a singleton, and if so, $\partial G(x)=\\{\nabla
G(x)\\}$.

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

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

Now, suppose that $f$ is concave and continuous, and $k_{1}<k_{2}$. Define
$r=\frac{f(k_{2})-f(k_{1})}{k_{2}-k_{1}}$. Let $p=D_{+}f(k_{1})$ and
$q=D_{-}f(k_{2})$. If $p=r$ or $q=r$, then $f^{\prime}(k)\equiv r$ for all
$k\in]k_{1},k_{2}[$. Suppose that $p>r>q$. Define $g(k)=f(k)-rk$. 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
the subdifferential, $0\in\partial g(k)$, and thus $r\in\partial f(k)$. In
conclusion, we obtain the following mean value theorem: if $f$ is concave and
$k_{1}<k_{2}$, then there exist $k\in]k_{1},k_{2}[$ and $r\in\partial f(k)$
such that $f(k_{2})-f(k_{1})=r(k_{2}-k_{1})$.

Suppose that $G:I\to\mathbb{R}$ is a concave function, where $I$ is some open
interval in $\mathbb{R}$. We prove the upper hemi-continuity of $\partial
G(x)$. Suppose not. Then, there exist $\varepsilon>0$ and a sequence $(x_{m})$
such that $x_{m}\to x$ as $m\to\infty$ and either $D_{+}G(x_{m})\leq
D_{+}G(x)-2\varepsilon$ or $D_{-}G(x_{m})\geq D_{-}G(x)+2\varepsilon$ for each
$m$. We only treat the case in which the former holds for all $m$, because the
other cases can be treated symmetrically. By the definition of the right-
derivative, there exists $h>0$ such that

$G(x+h)-G(x)\geq h[D_{+}G(x)-\varepsilon].$

Therefore, for any sufficiently large $m$,

$G(x_{m}+h)-G(x_{m})>h[D_{+}G(x)-2\varepsilon]\geq hD_{+}G(x_{m}),$

which is a contradiction.

We use these facts in the proof of Proposition 4.

###  2.5 Pure Accumulation Path

Consider the following differential equation:

$\dot{k}(t)=F(k(t),0),\ k(0)=\bar{k}.$ (2.4)

Let $k^{+}(t,\bar{k})$ denote the solution to the above equation defined on
$\mathbb{R}_{+}$. This function $k^{+}(t,\bar{k})$ is called the pure
accumulation path. The following lemma has been proved in Hosoya [14].999Note
that, under Assumption 3, there exists $k>0$ such that $F(k,0)>0$.

######  Lemma 1.

Under Assumption 3, the pure accumulation path $k^{+}$ is uniquely defined on
the set $\mathbb{R}_{+}\times\mathbb{R}_{++}$. Moreover, if $F(k,0)>0$ and
$\gamma\in\partial_{k}F(k,0)$, then $k^{+}(t,\bar{k})\geq\min\\{k,\bar{k}\\}$
and

$k(t)\leq\begin{cases}e^{\gamma t}\left(\bar{k}+(e^{-\gamma t}-1)\frac{\gamma
k-F(k,0)}{\gamma}\right)&\mbox{if }\gamma\neq 0,\\\ \bar{k}+tF(k,0)&\mbox{if
}\gamma=0,\end{cases}$ (2.5)

for all $t\geq 0$.

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

$\lim_{T\to\infty}e^{-\rho T}V(k^{+}(T,\bar{k}))=0\mbox{ for all }\bar{k}>0.$
(2.6)

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

The following lemma shows that (2.6) is not strong.

######  Lemma 2.

Suppose that there exist $k>0$ and $\gamma\in\mathbb{R}$ such that
$\gamma\in\partial_{k}F(k,0)$ and $\gamma<\rho$. Then, $\mathscr{V}$ coincides
with the set of all increasing and concave real-valued functions on
$\mathbb{R}_{++}$.

###### Proof.

Without loss of generality, we can assume that $F(k,0)>0$. Therefore, by
(2.5), there exist $A,B,C\in\mathbb{R}$ such that $k^{+}(t,\bar{k})\leq
Ae^{\gamma t}+Bt+C$. Choose any increasing and concave function
$V:\mathbb{R}_{++}\to\mathbb{R}$ and $p\in\partial V(\bar{k})$. Then,

$\displaystyle-\infty<$ $\displaystyle~{}V(\inf_{t\geq 0}k^{+}(t,\bar{k}))\leq
V(k^{+}(T,\bar{k}))$ $\displaystyle\leq$
$\displaystyle~{}V(\bar{k})+p(k^{+}(T,\bar{k})-\bar{k}),$ $\displaystyle\leq$
$\displaystyle~{}V(\bar{k})+p(Ae^{\gamma T}+BT+C-\bar{k}),$

which implies that $V$ automatically satisfies (2.6). Thus, the requirement
(2.6) vanishes and $\mathscr{V}$ coincides with the set of all increasing and
concave functions on $\mathbb{R}_{++}$. This completes the proof. ∎

##  3 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 (2.1), and, for example, the assumption for the existence of the
solution to (2.1) is not appropriate. The reason why we restrict ourselves is
simple: in many cases, ensuring the existence of a solution to (2.1) is
tremendously difficult. Hence, we prohibit ourselves from assuming the
existence of a solution, although under the existence assumption of the
solution, the proofs of results become quite easy.101010See, for example,
Hosoya [15] for detailed arguments.

###  3.1 Knowledge on Ordinary Differential Equations

In this section, we frequently use knowledge on ordinary differential
equations (ODEs). Hence, we note basic knowledge on ODEs for readers.

First, consider the following differential equation:

$\dot{x}(t)=h(t,x(t)),$ (3.1)

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 to (3.1) if and only
if, 1) $I$ is an interval, 2) $x(t)$ is absolutely continuous on every compact
subinterval of $I$, 3) the graph of $x(t)$ is included in $U$, and 4)
$\dot{x}(t)=h(t,x(t))$ for almost all $t\in I$. Suppose that $(t^{*},x^{*})\in
U$. If a solution $x(t)$ to (3.1) satisfies 1) $t^{*}\in I$ and 2)
$x(t^{*})=x^{*}$, then $x(t)$ is called a solution with the initial value
condition $x(t^{*})=x^{*}$, or simply, a solution to the following
differential equation:

$\dot{x}(t)=h(t,x(t)),\ x(t^{*})=x^{*}.$ (3.2)

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, additionally, 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 $(t,x_{1},x_{2})$ such that $(t,x_{1}),(t,x_{2})\in C$, then $h$ is
said to be 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 an 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
$x:I\to\mathbb{R}^{n}$ to (3.2), where $I$ is relatively open in
$\mathbb{R}_{+}$.111111This result is called the Carathéodory–Peano existence
theorem. For a proof, see Ch.2 of Coddington and Levinson [10].

  2. 2)

Suppose that $h$ satisfies Carathéodory’s condition, and that 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 neighbourhood of $t^{*}$
such that $t\mapsto h(t,x^{*})$ is integrable on this neighbourhood. Then,
there exists a solution $x:I\to\mathbb{R}^{n}$ to (3.2), where $I$ is
relatively open in $\mathbb{R}_{+}$.

  3. 3)

Suppose that $h$ satisfies Carathéodory’s condition, and that 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 neighbourhood of
$t^{+}$ such that $t\mapsto h(t,x^{+})$ is integrable on this neighbourhood.
Choose any $(t^{*},x^{*})\in V$. Suppose that $x_{1}(t),x_{2}(t)$ are two
solutions to (3.2) 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 $t\in
I_{1}\cap I_{2}$.121212Results 2) and 3) are known as the
Carathéodory–Picard–Lindelöf existence theorem. For a proof, see Section 0.4
of Ioffe and Tikhomirov [17].

  4. 4)

Suppose that $h$ is continuous. Then, any solution $x(t)$ to (3.1) is
continuously differentiable.131313See Ch.2 of Hartman [12].

Next, suppose that $h$ satisfies all requirements in 3) and $(t^{*},x^{*})\in
V$. Choose a solution $x:I\to\mathbb{R}^{n}$ to (3.2). A solution
$y:J\to\mathbb{R}^{n}$ is called an 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 said to be
nonextendable if and only if there is no extension except $x(t)$ itself.

The following facts are well known.141414Fact 5) can be proved easily. The
proof of 6) is in Ch.2 of Coddington and Levinson [10].

  1. 5)

In addition to the requirements of 3), suppose that $V=U$. Then, there
uniquely exists a nonextendable solution $x(t)$ to (3.2). 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 to (3.2). Choose any compact set $C\subset V$.
Then, there exists $t^{+}\in I$ 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 $\mathbb{R}_{+}$ and $a(t)$ is bounded. Then,
the solution to (3.1) is determined by the following formula:

$x(t)=e^{\int_{0}^{t}a(\tau)d\tau}\left[x(0)+\int_{0}^{t}e^{-\int_{0}^{s}a(\tau)d\tau}b(s)ds\right].$
(3.3)

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

###  3.2 Basic Lemma and Propositions

In this subsection, we introduce a lemma and several propositions that have
been proved in Hosoya [14]. Because all results are proved in Hosoya [14], we
omit the proofs.

######  Lemma 3.

Consider the following two ODEs:

$\dot{k}(t)=h_{i}(t,k),\ k(0)=\bar{k}_{i},$ (3.4)

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

  1. i)

For some $i\in\\{1,2\\}$, 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 this equation $(\ref{eq:eq13})$ has a
solution $k_{i}:[0,T]\to\mathbb{R}$. 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 $i^{*}\in\\{1,2\\}$, $h_{i^{*}}$ is Lipschitz
in $k$ on $U$. Suppose also that $h_{i^{*}}(t,\bar{k})$ is locally integrable,
and there exist solutions $k_{i}:[0,T]\to\mathbb{R}$ to the above equations
for $i\in\\{1,2\\}$. Then, $k_{1}(t)\leq k_{2}(t)$ for all
$t\in[0,T]$.151515If 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. From this perspective, this lemma is
an extension of the Carathéodory–Picard–Lindelöf uniqueness result in the
theory of ODEs.

######  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
$(p,k)\in\mathbb{R}^{2}_{++}$,

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

######  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, $\bar{V}(k)\in\mathbb{R}$ 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 $k>0$.

######  Proposition 3.

Suppose that Assumptions 1-4 hold. If the value function $\bar{V}$ is finite,
then it is increasing, and it is a viscosity solution to the HJB equation.

###  3.3 Necessity of the HJB Equation

Because Proposition 3 requires the finiteness of the value function, we want
an additional assumption that ensures the finiteness of the value function.
However, we cannot assume that $u(c,k)$ is bounded, because a typical example
of $u(c,k)$ is $\log c$. Therefore, we need an alternative condition for $u$.

First, define the constant relative risk aversion (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 $\theta>0$. This function is the unique solution to 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 of this equation is sometimes called the relative
risk aversion of $u$.

Using this function, we present an additional assumption.

Assumption 5. There exist $k^{*}>0,c^{*}\geq
0,\gamma>0,\delta>0,\theta>0,a>0,b\geq 0,C\in\mathbb{R}$ such that

$\displaystyle~{}(\gamma,-\delta)\in\partial F(k^{*},c^{*}),$ (3.5)
$\displaystyle~{}\rho-(1-\theta)\gamma>0,$ (3.6) $\displaystyle~{}u(c,k)\leq
au_{\theta}(c)+bu_{\theta}(k)+C\mbox{ for all }c>0,k>0.$ (3.7)

Then, we obtain the following result.

######  Lemma 4.

Suppose that Assumptions 1-5 hold. Define161616Because  $-k^{*}+\frac{\delta
c^{*}+F(k^{*},c^{*})}{\gamma}=\frac{1}{\gamma}[\gamma(0-k^{*})-\delta(0-c^{*})+F(k^{*},c^{*})]\geq
0,$ we have that $\hat{k},C^{*}>0$ for all $\bar{k}>0$.

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

and

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

Then,

$aV_{3}(\bar{k})+bV_{4}(\bar{k})+\frac{C}{\rho}\geq\bar{V}(\bar{k})$

for all $\bar{k}>0$.

###### Proof.

First, consider the following problem:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t),k(t))dt$ subject to. $\displaystyle~{}c(t)\in W_{1},$
$\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt\mbox{ can be
defined},$
$\displaystyle~{}\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})\mbox{
a.e.},$ $\displaystyle~{}k(0)=\bar{k}.$

Define $A^{L}_{\bar{k}}$ as the set of all pairs $(k(t),c(t))$ of nonnegative
functions such that $k(t)$ is absolutely continuous on every compact set,
$c(t)\in W_{1}$, $\int_{0}^{\infty}e^{-\rho t}u(c(t),k(t))dt$ 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.

$V_{1}(\bar{k})\geq\bar{V}(\bar{k})$ for all $\bar{k}>0$.

###### 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 to 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 $(\gamma,-\delta)\in\partial F(k^{*},c^{*})$,
$F(k,c)\leq\gamma(k-k^{*})-\delta(c-c^{*})+F(k^{*},c^{*})$ for all $(k,c)$,
and thus, by Lemma 3, $\hat{k}(t)\geq k(t)$ 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. ∎

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.
$\displaystyle~{}c(t)\in W_{1},$ $\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt\mbox{ can be defined},$
$\displaystyle~{}\dot{k}(t)=\gamma(k(t)-k^{*})-\delta(c(t)-c^{*})+F(k^{*},c^{*})\mbox{
a.e.},$ $\displaystyle~{}k(0)=\bar{k}.$

Define $A^{L2}_{\bar{k}}$ as the set of all pairs $(k(t),c(t))$ of nonnegative
functions such that $k(t)$ is absolutely continuous in every compact set,
$c(t)\in W_{1}$, $\int_{0}^{\infty}e^{-\rho
t}[au_{\theta}(c(t))+bu_{\theta}(k(t))]dt$ 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.

$V_{2}(\bar{k})+\frac{C}{\rho}\geq V_{1}(\bar{k})$ for all $\bar{k}>0$.

###### 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, $V_{2}(\bar{k})+\frac{C}{\rho}\geq
V_{1}(\bar{k})$. This completes the proof of Step 2. ∎

######  Step 3.

$aV_{3}(\bar{k})+bV_{4}(\bar{k})\geq V_{2}(\bar{k})$ for all $\bar{k}>0$.

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

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

Moreover, because $\gamma k^{*}-\delta c^{*}-F(k^{*},c^{*})\leq 0$,

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

and thus,

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

which completes the proof. ∎

Steps 1-3 show that our claim is correct. This completes the proof.

∎

Combining Proposition 3 and Lemma 4, we obtain the following result.

######  Theorem 1.

Suppose that Assumptions 1-5 hold. Then $\bar{V}\in\mathscr{V}$, and $\bar{V}$
is a viscosity solution to the HJB equation.

###### Proof.

By Lemma 4, we have that $\bar{V}$ is finite, and by Propositions 2 and 3, we
have that $\bar{V}$ is an increasing and concave viscosity solution to the HJB
equation. It suffices to show that $\bar{V}$ satisfies the growth condition
(2.6).

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

$\hat{k}(t)=\hat{k}e^{\gamma t}.$

By Lemma 3 and the calculation in Step 3 of the proof of Lemma 4,
$\hat{k}(t)\geq k^{+}(t,\bar{k})$ 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. ∎

###  3.4 Differential Inclusions

In the previous subsection, we showed that under Assumptions 1-5, the value
function is a viscosity solution to the HJB equation. We require the converse:
that is, we need to show that, under some additional assumption, $\bar{V}$ is
the unique viscosity solution to the HJB equation in $\mathscr{V}$. However,
to derive this result, we need the aid of differential inclusions.

Hence, in this subsection, we introduce several properties of differential
inclusions. First, consider the following autonomous differential inclusion:

$\dot{k}(t)\in\Gamma(k(t)),\ k(0)=\bar{k}>0,$ (3.8)

where $\Gamma:\mathbb{R}_{++}\twoheadrightarrow\mathbb{R}$ is a nonempty-
valued set function. A function $k:I\to\mathbb{R}_{++}$ is called a solution
to (3.8) if and only if, 1) $I$ is an interval that includes $0$, 2) $k(t)$ is
absolutely continuous on any compact subinterval of $I$, 3) $k(0)=\bar{k}$,
and 4) $\dot{k}(t)\in\Gamma(k(t))$ for almost every $t\in I$. It is known that
if $\Gamma$ is a compact- and convex-valued upper hemi-continuous mapping,
then there exists at least one solution $k(t)$ to (3.8) defined on $[0,T]$ for
some $T>0$.171717If the domain of $\Gamma$ is $\mathbb{R}$, then we can apply
Theorem 1 of Maruyama [21] directly. It is easy to extend this result to our
case. Define $\Phi(k)=\Gamma(k)$ if $x>\bar{k}/2$ and
$\Phi(k)=\Gamma(\bar{k}/2)$ otherwise. Then,
$\Phi:\mathbb{R}\twoheadrightarrow\mathbb{R}$ is a nonempty-, compact-, and
convex-valued upper hemi-continuous mapping, and thus, the inclusion
$\dot{k}(t)\in\Phi(k(t)),\ k(0)=\bar{k}>0,$ has a solution
$k:[0,T]\to\mathbb{R}$ such that $k(t)\geq\bar{k}/2$ for any $t\in[0,T]$. This
$k(t)$ is also a solution to (3.8).

We need the following lemmas.

######  Lemma 5.

Consider the differential inclusion $(\ref{DI})$, and the differential
equation

$\dot{k}(t)=h(k(t)),\ k(0)=\bar{k}>0,$ (3.9)

where $h:\mathbb{R}_{++}\to\mathbb{R}$ is a locally Lipschitz function such
that, for any $k\in\mathbb{R}_{++}$ and $y\in\Gamma(k)$, $y\leq h(k)$. Let
$k_{1}(t)$ be a solution to $(\ref{DEI})$ and $k_{2}(t)$ be a solution to
$(\ref{DI})$ defined on $[0,T]$. Then, $k_{1}(t)\geq k_{2}(t)$ for all
$t\in[0,T]$.

###### Proof.

Suppose not. Then, there exists $t^{+}>0$ such that
$k_{1}(t^{+})<k_{2}(t^{+})$. Let $t^{*}=\inf\\{t\in[0,T]|\forall
s\in[t,t^{+}],\ k_{1}(s)<k_{2}(s)\\}$. Because $k_{1}(0)=k_{2}(0)=\bar{k}$, we
have that $k_{1}(t^{*})=k_{2}(t^{*})$. Let

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

Because $\dot{k}_{2}(t)\in\Gamma(k_{2}(t))$, we have that
$h(k_{2}(t))\geq\dot{k}_{2}(t)$ almost everywhere, and thus,

$k_{3}(t)\geq k_{2}(t)>k_{1}(t)$

for all $t\in]t^{*},t^{+}]$. Let $L>0$ be the Lipschitz constant on a compact
set $C$ that includes $\\{k_{i}(t)|i\in\\{1,2\\},\ t\in[t^{*},t^{+}]\\}$.
Then,

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

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

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

and thus, $k_{3}(s^{*})<k_{2}(s^{*})$, which is a contradiction. This
completes the proof. ∎

######  Lemma 6.

Consider the differential inclusion $(\ref{DI})$, where
$\Gamma:\mathbb{R}_{++}\twoheadrightarrow\mathbb{R}$ is a nonempty-, compact-,
and convex-valued upper hemi-continuous mapping. Suppose that there exist two
positive continuous functions $\hat{k}(t)$ and $\bar{k}(t)$ defined on
$\mathbb{R}_{+}$ such that $\bar{k}(t)\leq k(t)\leq\hat{k}(t)$ for any
solution $k(t)$ to $(\ref{DI})$ defined on $[0,T]$ for some $T>0$ and
$t\in[0,T]$. Then, this inclusion $(\ref{DI})$ has a solution defined on
$\mathbb{R}_{+}$ itself.

###### Proof.

Let $Y$ be the set of all solutions to (3.8) defined on either
$\mathbb{R}_{+}$ or $[0,T^{\prime}]$ for some $T^{\prime}$. As mentioned
above, $Y$ is nonempty. For $k_{1}(\cdot),k_{2}(\cdot)\in Y$, 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 to (3.8) 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
and $\Gamma$ is upper hemi-continuous, there exists $r>0$ such that $|y|<r$
for every $k\in[\bar{k}(T^{*})-\varepsilon,\hat{k}(T^{*})+\varepsilon]$ and
$y\in\Gamma(k)$. Thus, $|k^{+}(t)-k^{+}(T^{*})|\leq r(T^{*}-t)$, and hence

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

which implies that $k^{+}(t)$ is continuous on $[0,T^{*}]$. Then, $k^{+}(t)$
is differentiable almost everywhere on $[0,T^{*}]$ and

$k^{+}(t)=\bar{k}+\int_{0}^{t}\dot{k}^{+}(t)ds$

for all $t\in[0,T^{*}]$, and thus, we conclude that $k^{+}(t)$ is a solution
to equation (3.8) defined on $[0,T^{*}]$. Hence, $k^{+}(\cdot)\in Y$ 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
as mentioned above, there exists a solution $k_{1}(t)$ to the following
inclusion:

$\dot{k}(t)\in\Gamma(k(t)),\ k(0)=k^{*}(T),$

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

$\kappa(t)=\begin{cases}k^{*}(t)&\mbox{if }t\in[0,T],\\\ k_{1}(t-T)&\mbox{if
}t\in[T,T+T^{\prime}].\end{cases}$

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. ∎

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

In this subsection, we examine the sufficiency of the HJB equation to
determine the value function. It can be easily shown that if $u(c)=-1/c$ and
$F(k,c)=\sqrt{k}-c$, then $V(k)\equiv 0$ is automatically a classical solution
to the HJB equation. However, it is not increasing. By Theorem 1, the value
function $\bar{V}$ is an increasing solution to the HJB equation, and thus
$\bar{V}\neq V$. This indicates that the usual uniqueness argument cannot be
applied to our model, and, at least, the increasing requirement is crucial for
the uniqueness result.

We need two additional lemmas.

######  Lemma 7.

Suppose that Assumptions 1-4 hold, and that $V:\mathbb{R}_{++}\to\mathbb{R}$
is a concave function. Then, $V$ is a viscosity solution to the HJB equation
if and only if, for any $k>0$ and $p\in\partial V(k)$,

$\sup_{c\geq 0}\\{F(k,c)p+u(c,k)\\}=\rho V(k).$ (3.10)

###### Proof.

First, suppose that $V$ is a viscosity solution to the HJB equation. If $V$ is
differentiable at $k$, then $p=V^{\prime}(k)$, and thus (3.10) holds. Suppose
that $\partial V(k)=[D_{+}V(k),D_{-}V(k)]$ and $D_{+}V(k)<D_{-}V(k)$. Because
$\partial V$ is upper hemi-continuous and $V$ is locally Lipschitz, there
exist sequences $(k_{m}^{1}),(k_{m}^{2})$ such that $V$ is differentiable at
$k_{m}^{i}$, and

$\lim_{m\to\infty}V^{\prime}(k_{m}^{1})=D_{+}V(k),\
\lim_{m\to\infty}V^{\prime}(k_{m}^{2})=D_{-}V(k).$

Then,

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

and letting $m\to\infty$, we have that (3.10) holds for $p=D_{+}V(k)$ or
$p=D_{-}V(k)$. Now, define

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

Then, we can easily check that $g$ is convex. Because
$g(D_{+}V(k))=g(D_{-}V(k))=\rho V(k)$, we have that $g(p)\leq\rho V(k)$ for
all $p\in\partial V(k)$. Now, choose any $p\in\partial V(k)$, and define
$\varphi(k^{\prime})=V(k)+p(k^{\prime}-k)$. Then, $\varphi(k)=V(k)$ and
$\varphi^{\prime}(k^{\prime})\geq V(k^{\prime})$ for all $k^{\prime}$. Because
$V$ is a viscosity supersolution, we have that
$g(p)=g(\varphi^{\prime}(k))\geq\rho V(k)$, and thus $g(p)\equiv\rho V(k)$ on
$\partial V(k)$, as desired.

Conversely, suppose that (3.10) holds for any $p\in\partial V(k)$. Choose any
continuously differentiable function $\varphi$ defined on some neighbourhood
of $k$ such that $k$ is either a minimum point or a maximum point of
$\varphi-V$. Then, $\varphi^{\prime}(k)\in\partial V(k)$, and thus (3.10)
holds for $p=\varphi^{\prime}(k)$, which implies that $V$ is a viscosity
solution to the HJB equation. This completes the proof. ∎

######  Lemma 8.

Suppose that $U\subset\mathbb{R}$ is an open interval, $H:U\to\mathbb{R}$ is
concave, $A<B$, and $x:[A,B]\to U$ is absolutely continuous. Let
$\psi(t)=H(x(t))$. Then, $\psi(t)$ is also absolutely continuous. Moreover, if
both $\dot{\psi}(t)$ and $\dot{x}(t)$ are defined, then for any $p\in\partial
H(x(t))$, $\dot{\psi}(t)=p\dot{x}(t)$.

###### Proof.

First, let $C=x([A,B])$. Then, $C$ is a compact and convex subset of $U$, and
thus $H$ has a Lipschitz constant $L>0$ on $C$. Choose any $\varepsilon>0$.
Then, there exists $\delta>0$ such that if $s_{1}<t_{1}\leq
s_{2}<t_{2}\leq...\leq s_{n}<t_{n}$ and $\sum_{i=1}^{n}|t_{i}-s_{i}|<\delta$,
then $\sum_{i=1}^{n}|x(t_{i})-x(s_{i})|<L^{-1}\varepsilon$, which implies that

$\sum_{i=1}^{n}|\psi(t_{i})-\psi(s_{i})|\leq
L\sum_{i=1}^{n}|x(t_{i})-x(s_{i})|<\varepsilon,$

as desired. Therefore, $\psi$ is absolutely continuous.

Second, suppose that $\dot{\psi}(t)$ and $\dot{x}(t)$ are defined, and
$p\in\partial H(x(t))$. Then,

$\displaystyle\dot{\psi}(t)-p\dot{x}(t)=$ $\displaystyle~{}\lim_{h\downarrow
0}\frac{\psi(t+h)-\psi(t)}{h}-p\dot{x}(t)$ $\displaystyle\leq$
$\displaystyle~{}\lim_{h\downarrow 0}p\frac{x(t+h)-x(t)}{h}-p\dot{x}(t)=0,$
$\displaystyle\dot{\psi}(t)-p\dot{x}(t)=$ $\displaystyle~{}\lim_{h\uparrow
0}\frac{\psi(t+h)-\psi(t)}{h}-p\dot{x}(t)$ $\displaystyle\geq$
$\displaystyle~{}\lim_{h\uparrow 0}p\frac{x(t+h)-x(t)}{h}-p\dot{x}(t)=0,$

which implies that $\dot{\psi}(t)=p\dot{x}(t)$. This completes the proof. ∎

We now introduce an additional assumption.

Assumption 6. There exists $\varepsilon_{0}>0$ such that $\frac{\partial
u}{\partial c}$ is continuously differentiable in $k$ on
$\mathbb{R}_{++}\times]0,\varepsilon_{0}[$, and there exists a continuously
differentiable function
$H:]0,\varepsilon_{0}[\times\mathbb{R}_{++}\to\mathbb{R}$ such that
$\frac{\partial H}{\partial k}(k,c)>0,\ \frac{\partial H}{\partial c}(k,c)<0$
for all $(k,c)\in]0,\varepsilon_{0}[\times\mathbb{R}_{++}$, and if
$k<\varepsilon_{0}$ and $H(k,c)\neq 0$, then $F$ is continuously
differentiable in $c$ and $\frac{\partial F}{\partial c}$ is continuously
differentiable in $k$ around $(k,c)$. Moreover, there exists $k>0$ such that
$\inf_{c\geq 0}D_{k,+}F(k,c)>\rho$.

We define some additional notation. Let $B_{\bar{k}}$ denote the set of all
pairs of nonnegative functions $(k(t),c(t))$ defined on $\mathbb{R}_{+}$ such
that $k(t)$ is absolutely continuous on any compact interval, $c(t)\in W$,
$\lim_{T\to\infty}\int_{0}^{T}e^{-\rho t}u(c(t),k(t))dt$ exists,
$k(0)=\bar{k}$, and

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

for almost all $t\geq 0$. Clearly, $A_{\bar{k}}\subset B_{\bar{k}}$, but it is
unknown whether $A_{\bar{k}}=B_{\bar{k}}$.

Now, suppose that $V:\mathbb{R}_{++}\to\mathbb{R}$ is an increasing and
concave function. As we discussed in Subsection 2.4, $\partial V$ is a
nonempty-, compact-, and convex-valued upper hemi-continuous mapping.
Therefore, the mapping $F(k,c^{*}(\partial V(k),k))$ is also a nonempty-,
compact-, and convex-valued upper hemi-continuous mapping. Note also that, the
mapping $c^{*}(\partial V(k),k)$ is also compact-valued and upper hemi-
continuous, and thus, this mapping is measurable, and for any continuous
function $k:I\to\mathbb{R}_{++}$ defined on an interval $I$, there is a
measurable selection $c(t)$ of $c^{*}(\partial V(k(t)),k(t))$.181818See
Section 8.1 of Ioffe and Tikhomirov [17].

The next proposition is crucial for our next main result.

######  Proposition 4.

Suppose that Assumptions 1-4 and 6 hold. Suppose also that $V\in\mathscr{V}$
is a viscosity solution to the HJB equation. Choose any $\bar{k}>0$, and
consider the following differential inclusion:

$\dot{k}(t)\in F(k(t),c^{*}(\partial V(k(t)),k(t))),\ k(0)=\bar{k}.$ (3.11)

Then, there exists a solution $k^{*}(t)$ to the above equation defined on
$\mathbb{R}_{+}$, and for any such solution, $\inf_{t\geq 0}k^{*}(t)>0$.
Moreover, if we define

$c^{*}(t)=\arg\min\\{|\limsup_{n\to\infty}n(k^{*}(t+n^{-1})-k^{*}(t))-c||c\in
c^{*}(\partial V(k^{*}(t)),k^{*}(t))\\},$ (3.12)

then, $c^{*}(t)$ is locally bounded, and $(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$.
Furthermore, for any such pair $(k^{*}(t),c^{*}(t))$,

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

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

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

###### Proof.

We separate the proof into three steps.

######  Step 1.

Suppose that $V\in\mathscr{V}$ is a viscosity solution to the HJB equation.
Then, there exists a solution $k^{*}(t)$ to the differential inclusion
$(\ref{SOL})$ defined on $\mathbb{R}_{+}$. Moreover, for any such solution,
$\inf_{t\geq 0}k^{*}(t)>0$.

###### Proof of Step 1.

First, as we have mentioned, the differential inclusion (3.11) has at least
one solution $k(t)$ defined on $[0,T]$.

Let $\varepsilon>0$ satisfy that

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

We show that, for any solution $k(t)$ to (3.11) defined on $[0,T]$,
$k(t)\geq\varepsilon$.

Suppose that this claim is incorrect. Then, there exists $t^{+}>0$ such that
$0<k(t^{+})<\varepsilon$. Let $I\subset[0,T[$ be the set of all $t$ such that
1) $k(t)<\varepsilon$, 2) $\dot{k}(t)<0$ and (3.11) holds at $t$, 3) $V$ is
differentiable at $k(t)$, and 4) there exists an Alexandrov Hessian
$L\in\mathbb{R}$ of $V$ at $k(t)$.191919For a concave function $f$, the number
$L$ is called an Alexandrov Hessian of $f$ at $x$ if and only if, for all
$\varepsilon^{\prime}>0$, there exists $\delta^{\prime}>0$ such that if
$|x^{\prime}-x|<\delta^{\prime}$, then
$|y^{\prime}-y-L(x^{\prime}-x)|\leq\varepsilon|x^{\prime}-x|$ for all
$y\in\partial f(x),\ y^{\prime}\in\partial f(x^{\prime})$. For a detailed
argument, see Alexandrov [3] or Howard [16]. We show that $I$ is nonempty.
First, we show that if $F(k,c^{*}(\partial V(k),k))\neq\\{0\\}$, then $V$ is
differentiable at $k$. Suppose not. Then, $\partial
V(k)=[D_{+}V(k),D_{-}V(k)]$, where $D_{+}V(k)<D_{-}V(k)$. Choose
$p\in]D_{+}V(k),D_{-}V(k)[$. By Lemma 7,

$\displaystyle F(k,c^{*}(p,k))p+u(c^{*}(p,k),k)=$ $\displaystyle~{}\rho V(k),$
$\displaystyle F(k,c^{*}(p,k))q+u(c^{*}(p,k),k)\leq$ $\displaystyle~{}\rho
V(k)$

for any $q\in\partial V(k)$. Therefore, $F(k,c^{*}(p,k))=0$. By the continuity
of $F$ and $c^{*}$, we have that $F(k,c^{*}(\partial V(k),k))=\\{0\\}$, which
is a contradiction. Therefore, $V$ is differentiable at $k$. Now, it is easy
to show that there exists $t\in[0,T[$ such that $k(t)<\varepsilon$,
$\dot{k}(t^{*})<0$, and $\dot{k}(t^{*})\in F(k(t^{*}),c^{*}(\partial
V(k(t^{*})),k(t^{*})))$. This implies that $F(k(t^{*}),c^{*}(\partial
V(k(t^{*})),k(t^{*})))\neq\\{0\\}$, and thus $V$ is differentiable at
$k(t^{*})$. Because $\partial V$ is upper hemi-continuous, there exists a
neighbourhood $U$ of $k(t^{*})$ such that if $k\in U$, then
$F(k,c^{*}(\partial V(k),k))\neq\\{0\\}$, which implies that $V$ is
differentiable around $k(t^{*})$. Again, because $\partial V$ is upper hemi-
continuous, we have that $V$ is continuously differentiable around $k(t^{*})$.
Therefore, there exists an open neighbourhood of $t^{*}$ such that $k(t)$ is a
solution to the following differential equation:

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

which implies that $k(t)$ is continuously differentiable around $t^{*}$.
Because of Rademacher’s theorem and Alexandrov’s theorem, we have that for
almost all $t$ near to $t^{*}$, $\dot{k}(t)<0$, $V$ is differentiable at
$k(t)$, and there exists an Alexandrov Hessian $L\in\mathbb{R}$ of $V$ at
$k(t)$, as desired. Moreover, we have shown that if $t\in I$, then there
exists an open neighbourhood $U$ of $t$ and an open neighbourhood $W$ of
$k(t)$ such that $V$ is continuously differentiable on $W$, $k(\cdot)$ is
continuously differentiable on $U$, and $t^{\prime}\in I$ for almost all
$t^{\prime}\in U$.

Choose $c(T)\in c^{*}(\partial V(k(T)),k(T))$, and if $0\leq t<T$, define

$c(t)=\arg\min\\{|\limsup_{n\to\infty}n(k(t+n^{-1})-k(t))-c||c\in
c^{*}(\partial V(k(t)),k(t))\\}.$

Then, $c(t)$ is a measurable, positive, and locally bounded function defined
on $[0,T]$, and $c(t)\in c^{*}(\partial V(k(t)),k(t))$ for all $t\in[0,T]$.
Note that, if $t\in I$, then $\partial V(k(t))=\\{V^{\prime}(k(t))\\}$, and
thus $c(t)$ is continuous at $t$.

Choose $t^{*}\in I$. Because $\dot{k}(t^{*})<0$, for any sufficiently small
$h>0$, $k(t^{*}+h)<k(t^{*})$ and $V$ is differentiable on
$[k(t^{*}+h),k(t^{*})]$. Thus, for any $t\in[t^{*},t^{*}+h]$,

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

and there exist $k_{1},k_{2}\in[k(t^{*}+h),k(t^{*})],\
\theta_{1},\theta_{2}\in]0,1[$, $p\in\partial_{k}F(k_{2},c(t^{*}+h))$, and
$q\in\partial_{c}F(k(t^{*}),c(t^{*}+\theta_{2}h))$ such that,202020See the
mean value theorem in Subsection 2.4.

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

To modify this equation,

$\displaystyle~{}\frac{(\rho
V^{\prime}(k_{1})-pV^{\prime}(k(t^{*}+h)))(k(t^{*}+h)-k(t^{*}))}{h}$
$\displaystyle=$
$\displaystyle~{}\frac{\dot{k}(t^{*})(V^{\prime}(k(t^{*}+h))-V^{\prime}(k(t^{*})))}{h}$
$\displaystyle~{}+\frac{\frac{\partial u}{\partial
k}(c(t^{*}+h),k(t^{*}+\theta_{1}h))(k(t^{*}+h)-k(t^{*}))}{h}$ (3.13)
$\displaystyle~{}+\frac{\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}+\theta_{2}h))+qV^{\prime}(k(t^{*}+h))\right)(c(t^{*}+h)-c(t^{*}))}{h}$
$\displaystyle~{}+\frac{\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}))-\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k^{*}(t^{*}+\theta_{2}h))\right)(c(t^{*}+h)-c(t^{*}))}{h}.$

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

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

On the other hand, the first and second terms of the right-hand side of (3.13)
are always nonpositive. Because $\frac{\partial u}{\partial c}$ is
differentiable in $k$ on $\mathbb{R}_{++}\times]0,\varepsilon_{0}[$, the
fourth term of the right-hand side of (3.13) is

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

for some $\theta^{\prime\prime}\in[0,\theta_{2}]$. Therefore, the fourth term
converges to $0$ as $h\to 0$.

We claim that there exists $t^{*}\in I$ such that the limsup of the third term
is not greater than $0$: that is, we show that there exists $t^{*}\in I$ such
that

$\limsup_{h\downarrow 0}\frac{\left(\frac{\partial u}{\partial
c}(c(t^{*}+\theta_{2}h),k(t^{*}+\theta_{2}h))+qV^{\prime}(k(t^{*}+h))\right)(c(t^{*}+h)-c(t^{*}))}{h}\leq
0.$ (3.14)

Because of the definition of $c(t)$ and the first-order condition,

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

where $r\in\partial_{c}F(k(t^{*}+\theta_{2}h),c(t^{*}+\theta_{2}h))$.

We separate the proof into two cases. First, suppose that there exists
$t^{*}\in I$ such that $H(k(t^{*}),c(t^{*}))\neq 0$. Then, $F$ is
differentiable in $c$ and $\frac{\partial F}{\partial c}$ is differentiable in
$k$ around $(k(t^{*}),c(t^{*}))$. Hence, if $h>0$ is sufficiently small, then
$q=\frac{\partial F}{\partial c}(k(t^{*}),c(t^{*}+\theta_{2}h))$ and
$r=\frac{\partial F}{\partial c}(k(t^{*}+\theta_{2}h),c(t^{*}+\theta_{2}h))$.
In this case, the absolute value of the third term of the right-hand side of
(3.13) is bounded from

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

where

$\frac{V^{\prime}(k(t^{*}+h))-V^{\prime}(k(t^{*}))}{h}\to
L\dot{k}(t^{*})\mbox{ as }h\downarrow 0$

and

$\frac{q-r}{h}=-\frac{\partial^{2}F}{\partial k\partial
c}(k(t^{*}+\theta^{\prime}h),c(t^{*}+\theta_{2}h))\frac{k(t^{*}+\theta_{2}h)-k(t^{*})}{h}$

for some $\theta^{\prime}\in[0,\theta_{2}]$. Therefore, (3.14) holds. Next,
suppose that there is no $t^{*}\in I$ such that $H(k(t^{*}),c(t^{*}))\neq 0$.
Then, there exists $h>0$ such that $H(k(t),c(t))=0$ for all
$t\in[t^{*},t^{*}+h]$. By the implicit function theorem, there exists a
continuously differentiable function $c^{+}(k)$ such that $c(t)=c^{+}(k(t))$
for all $t\in[t^{*},t^{*}+h]$. This implies that $c(t)$ is continuously
differentiable around $t^{*}$. Because $(c^{+})^{\prime}(k)>0$, we have that
$c(t)$ is decreasing around $t^{*}$. In this case, the value of the third term
of the right-hand side of (3.13) is bounded from

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

The first term of the above formula converges to $0$ because of the same
arguments as above. Moreover, $F$ is continuously differentiable at
$(k(t^{*}),c(t^{*}+\theta_{2}h))$, and

$q=\frac{\partial F}{\partial c}(k(t^{*}),c(t^{*}+\theta_{2}h))\to
D_{c,-}F(k(t^{*}),c(t^{*}))\mbox{ as }h\downarrow 0.$

Moreover, because $\theta_{2}<1$,

$r\leq D_{c,+}F(k(t^{*}+\theta_{2}h),c(t^{*}+h)).$

We show that

$\limsup_{h\downarrow 0}D_{c,+}F(k(t^{*}+\theta_{2}h),c(t^{*}+h))\leq
D_{c,-}F(k(t^{*}),c(t^{*})).$

Suppose not. Then, there exist $\delta>0$ and a sequence $(k_{m},c_{m})$ such
that $(k_{m},c_{m})\to(k(t^{*}),c(t^{*}))$ as $m\to\infty$, and
$D_{c,+}F(k_{m},c_{m})>D_{c,-}F(k(t^{*}),c(t^{*}))+\delta$ for all $m$. Then,
there exists $a>0$ such that

$\frac{F(k(t^{*}),c(t^{*})-a)-F(k(t^{*}),c(t^{*}))}{-a}<D_{c,-}F(k(t^{*}),c(t^{*}))+\delta.$

Therefore, for sufficiently large $m$,

$D_{c,+}F(k_{m},c_{m})\leq\frac{F(k_{m},c_{m}-a)-F(k_{m},c_{m})}{-a}<D_{c,-}F(k(t^{*}),c(t^{*}))+\delta,$

which is a contradiction. Combining the above inequalities,

$\limsup_{h\downarrow
0}\frac{(q-r)V^{\prime}(k(t^{*}+\theta_{2}h))(c(t^{*}+h)-c(t^{*}))}{h}\leq 0,$

and thus, (3.14) holds. This implies that the limsup of the right-hand side of
(3.13) is nonpositive, which is a contradiction.

Hence, $k(t)\geq\varepsilon$ for every $t\in[0,T]$. By Lemma 5, $k(t)\leq
k^{+}(t,\bar{k})$ for every $t\in[0,T]$, and by Lemma 6, there exists a
solution $k^{*}(t)$ to (3.11) defined on $\mathbb{R}_{+}$. Clearly,
$k^{*}(t)\geq\varepsilon$ for every $t\geq 0$, and thus $\inf_{t\geq
0}k^{*}(t)>0$. This completes the proof of Step 1. ∎

######  Step 2.

Under the assumptions in Step 1, choose a solution $k^{*}(t)$ to equation
$(\ref{SOL})$ defined on $\mathbb{R}_{+}$, and define $c^{*}(t)$ by (3.12).
Then, $(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$ and

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

###### Proof of Step 2.

By Step 1, there exists $\varepsilon>0$ such that

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

for all $t\geq 0$. Moreover, by the definition of $c^{*}(t)$,

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

for almost every $t\in\mathbb{R}_{+}$. Let

$p^{*}(t)=\min\arg\min\\{|\rho
V(k^{*}(t))-u(c^{*}(t),k^{*}(t))-F(k^{*}(t),c^{*}(t))p||p\in\partial
V(k^{*}(t))\\}.$

Then, $p^{*}(t)$ is measurable and positive, and

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

for all $t\geq 0$. Choose any $T>0$. Let $W^{*}(t)=V(k^{*}(t))$. By Lemma 8,
$W^{*}(t)$ is absolutely continuous on $[0,T]$ and
$\dot{W}^{*}(t)=p^{*}(t)\dot{k}^{*}(t)$ almost everywhere, and thus,

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

and hence,

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

Therefore, by (2.6),

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

which implies that $(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$. This completes the
proof of Step 2. ∎

######  Step 3.

If $(k(t),c(t))\in B_{\bar{k}}$ and $\inf_{t\geq 0}k(t)>0$, then

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

###### Proof of Step 3.

Suppose that $(k(t),c(t))\in B_{\bar{k}}$ and $\inf_{t\geq 0}k(t)>0$. Choose
$T>0$. Define $W(t)=V(k(t))$. By Lemma 8, $W(t)$ is absolutely continuous on
$[0,T]$ and $\dot{W}(t)=D_{-}V(k(t))\dot{k}(t)$ almost everywhere. Moreover,
by Lemma 7,

$\displaystyle~{}F(k(t),c(t))D_{-}V(k(t))+u(c(t),k(t))$ $\displaystyle\leq$
$\displaystyle~{}\sup_{c\geq 0}\\{F(k(t),c)D_{-}V(k(t))+u(c,k(t))\\}=\rho
V(k(t)).$

Therefore,

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

Because $\inf_{t\geq 0}k(t)>0$ and $V$ satisfies (2.6), the right-hand side
converges to $V(\bar{k})$ as $T\to\infty$. Therefore,

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

as desired. This completes the proof of Step 3. ∎

Steps 1-3 state that all of our claims in Proposition 4 are correct. This
completes the proof. ∎

We now consider the following differential inclusion:

$\dot{k}(t)\in F(k(t),c^{*}(\partial\bar{V}(k(t)),k(t))),\ k(0)=\bar{k},$
(3.15)

and the corresponding definition of the function:

$c^{*}(t)=\arg\min\\{|\limsup_{n\to\infty}n(k^{*}(t+n^{-1})-k^{*}(t))-c||c\in
c^{*}(\partial\bar{V}(k^{*}(t)),k^{*}(t))\\}.$ (3.16)

######  Theorem 2.

Suppose that Assumptions 1-6 hold. Then, $\bar{V}$ is the unique viscosity
solution to the HJB equation in $\mathscr{V}$.

###### Proof.

Because Assumptions 1-6 hold, $\bar{V}\in\mathscr{V}$, and it is a viscosity
solution to the HJB equation. By Proposition 4, there exists a solution
$k^{*}(t)$ to (3.15) defined on $\mathbb{R}_{+}$ such that $\inf_{t\geq
0}k^{*}(t)>0$, and if we define $c^{*}(t)$ by (3.16), then
$(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$ and

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

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

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

By Proposition 4, there exists a solution $k^{+}(t)$ to (3.11) defined on
$\mathbb{R}_{+}$ such that $\inf_{t\geq 0}k^{+}(t)>0$. Define $c^{+}(t)$ as
(3.12), where $k^{*}(t)$ is replaced with $k^{+}(t)$. Then,
$(k^{+}(t),c^{+}(t))\in B_{\bar{k}}$, and

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

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

Therefore, under Assumptions 1-6, the HJB equation is the perfect
characterization for the value function in the functional space $\mathscr{V}$.

In the proof of Theorem 2, if $\int_{0}^{\infty}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt$ is defined in the sense of the Lebesgue integral,
then $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, and thus it is a solution to (2.1).
Because $\inf_{t\geq 0}k^{*}(t)>0$, if $\inf_{t\geq 0}c^{*}(t)>0$, then
$(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$. However, this is not so easily verified.
The following corollary presents three appropriate sufficient conditions for
$(k^{*}(t),c^{*}(t))$ to be a solution to (2.1).

######  Corollary 1.

Suppose that Assumptions 1-6 hold, $k^{*}(t)$ is a solution to $(\ref{SOL2})$
defined on $\mathbb{R}_{+}$, and $c^{*}(t)$ is defined by $(\ref{CONSUM2})$.
Suppose that one of the following three conditions holds.

  1. 1)

$u(c,k)$ is bounded from above or below.

  2. 2)

$k^{*}(t)$ is bounded.

  3. 3)

$\liminf_{k\to\infty}c^{*}(p,k)>0$ for all $p>0$.

Then, $(k^{*}(t),c^{*}(t))$ is a solution to $(\ref{MODEL})$.

###### Proof.

By Proposition 4 and Theorem 1, $\bar{V}$ is a solution to the HJB equation,
$(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$, and

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

Therefore, it suffices to show that $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$.

For 1), if $u(c,k)$ is either bounded from above or below, then
$A_{\bar{k}}=B_{\bar{k}}$, and thus our claim holds.

For 2), suppose that $k^{*}(t)$ is bounded. As we proved in Proposition 4,
$\inf_{t\geq 0}k^{*}(t)>0$. Therefore, the trajectory of $k^{*}(t)$ is
included in some compact set $C\subset\mathbb{R}_{++}$. This implies that
$\inf_{t\geq 0}c^{*}(t)>0$, and thus $e^{-\rho t}u(c^{*}(t),k^{*}(t))$ is
bounded from below. Hence, $(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, as desired.

For 3), let $0<p_{1}<p_{2}$, and $c_{i}=c^{*}(p_{i},k)$. Then,

$F(k,c_{i})p_{i}+u(c_{i},k)=\max_{c\geq 0}\\{F(k,c)p_{i}+u(c,k)\\}.$

By the first-order condition, there exists $r_{i}\in\partial_{c}F(k,c_{i})$
such that

$r_{i}p_{i}=-\frac{\partial u}{\partial c}(c_{i},k).$

If $c_{1}<c_{2}$, then

$r_{1}\geq D_{c,+}F(k,c_{1})\geq D_{c,-}F(k,c_{2})\geq r_{2},$

and thus,212121Note that $F$ is decreasing in $c$, and thus $r_{i}<0$.

$\frac{\partial u}{\partial c}(c_{2},k)=-r_{2}p_{2}>-r_{2}p_{1}\geq-
r_{1}p_{1}=\frac{\partial u}{\partial c}(c_{1},k),$

which contradicts the concavity of $u$. Therefore, $c^{*}(p,k)$ is
nonincreasing in $p$. Define $\varepsilon=\inf_{t\geq 0}k^{*}(t)$. Then,

$c^{*}(t)\geq
c^{*}(D_{-}\bar{V}(k^{*}(t)),k^{*}(t))\geq\inf_{k\geq\varepsilon}c^{*}(D_{-}\bar{V}(\varepsilon),k)>0,$

and thus $u(c^{*}(t),k^{*}(t))$ is bounded from below. Hence, again
$(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, as desired. This completes the proof. ∎

Note that, the requirement in Corollary 1 is not strong. For example, suppose
that $u(c,k)=au_{\theta}(c)+bu_{\theta}(k)$ for $a,b\geq 0$ for a CRRA
function $u_{\theta}$ with $\theta\neq 1$, then $u(c,k)$ is either bounded
from above or below, and 1) holds. Next, suppose that $F(k,c)=f(k)-dk-c$,
$d>0$, $f$ is continuously differentiable, and
$\lim_{k\to\infty}f^{\prime}(k)<d$. Then, $k^{+}(t,\bar{k})$ is bounded, and
thus $k^{*}(t)$ is also bounded, and 2) holds. Third, if $F(k,c)=f(k)-h(c)$
for a convex function $h$ and $u(c,k)=u(c)$, then $c^{*}(p,k)$ is independent
of $k$. In this case, $\liminf_{k\to\infty}c^{*}(p,k)>0$ is trivially
satisfied, and 3) holds.

The converse of Corollary 1 holds unconditionally. We state this result as a
proposition.

######  Proposition 5.

Suppose that Assumptions 1-5 hold, and that there exists a solution
$(k^{*}(t),c^{*}(t))$ to $(\ref{MODEL})$ such that $\inf_{t\geq 0}k^{*}(t)>0$.
Then, $k^{*}(t)$ is a solution to $(\ref{SOL2})$, and $c^{*}(t)\in
c^{*}(\partial\bar{V}(k^{*}(t)),k^{*}(t))$ for almost all $t\geq 0$.

###### Proof.

Suppose that $k^{*}(t)$ violates (3.15). Then, there exists $T>0$ such that
the set $I=\\{t\in[0,T]|c^{*}(t)\notin
c^{*}(\partial\bar{V}(k^{*}(t)),k^{*}(t))\\}$ is not a null set. By Theorem 1,
$\bar{V}$ is a viscosity solution to the HJB equation. Define
$W(t)=\bar{V}(k^{*}(t))$. By Lemma 8, $W(t)$ is absolutely continuous on
$[0,T]$ and $\dot{W}(t)=D_{-}\bar{V}(k^{*}(t))\dot{k}^{*}(t)$ almost
everywhere. Therefore,

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

By the same arguments, we have that for every $T^{\prime}\geq T$,

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

By (3.17), there exists $\varepsilon>0$ such that

$\int_{0}^{T}e^{-\rho t}u(c^{*}(t),k^{*}(t))dt<\bar{V}(\bar{k})-e^{-\rho
T}\bar{V}(k^{*}(T))-2\varepsilon.$

Choose $\delta>0$ such that $\inf_{t\geq 0}k^{*}(t)\geq\delta$. Then, for
every $T^{\prime}\geq T$ such that222222Note that, because of Theorem 1,
$\bar{V}$ satisfies (2.6).

$\max\\{|e^{-\rho T^{\prime}}\bar{V}(k^{+}(T^{\prime},\bar{k}))|,|e^{-\rho
T^{\prime}}\bar{V}(\delta)|\\}<\varepsilon,$

we have that

$\int_{0}^{T^{\prime}}e^{-\rho
t}u(c^{*}(t),k^{*}(t))dt<\bar{V}(\bar{k})-e^{-\rho
T^{\prime}}\bar{V}(k^{*}(T^{\prime}))-2\varepsilon<\bar{V}(\bar{k})-\varepsilon.$

Because $T^{\prime}$ is arbitrary,

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

which is a contradiction. Therefore, $k^{*}(t)$ is a solution to (3.15).
Define $c^{+}(t)$ by (3.16). Then,

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

for almost all $t\geq 0$, and

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

for almost all $t\geq 0$, which implies that $c^{*}(t)=c^{+}(t)$ almost
everywhere, and thus $c^{*}(t)\in c^{*}(\partial\bar{V}(k^{*}(t)),k^{*}(t))$
almost everywhere. This completes the proof. ∎

Finally, we discuss the differentiability of $\bar{V}(k)$.

######  Proposition 6.

Suppose that Assumptions 1-4 hold and $\bar{V}$ is finite. Suppose also that
if $F(k,c)=0$, then $F$ is differentiable in $c$ at $(k,c)$. Then, $\bar{V}$
is continuously differentiable, and thus it is a classical solution to the HJB
equation.

###### Proof.

Because of Proposition 3, $\bar{V}$ is an increasing and concave viscosity
solution to the HJB equation. Moreover, because $\partial\bar{V}$ is upper
hemi-continuous, if it is single-valued, then it is continuous. Hence, it
suffices to show that $\bar{V}$ is differentiable everywhere. Suppose that
$\bar{V}$ is not differentiable at $k$. Then,
$\partial\bar{V}(k)=[D_{+}\bar{V}(k),D_{-}\bar{V}(k)]$, where
$D_{+}\bar{V}(k)<D_{-}\bar{V}(k)$. Choose
$p\in]D_{+}\bar{V}(k),D_{-}\bar{V}(k)[$. By Lemma 7,

$F(k,c^{*}(p,k))p+u(c^{*}(p,k),k)=\rho\bar{V}(k),$
$F(k,c^{*}(p,k))q+u(c^{*}(p,k),k)\leq\rho\bar{V}(k),$

for all $q\in\partial\bar{V}(k)$. Therefore, $F(k,c^{*}(p,k))=0$. By the
continuity of $F$ and $c^{*}$, we have that $F(k,c^{*}(p,k))=0$ for all
$p\in\partial\bar{V}(k)$, and thus there exists $\bar{c}>0$ such that,
$c^{*}(p,k)=\bar{c}$ for all $p\in\partial\bar{V}(k)$. By assumption, $F$ is
differentiable in $c$ at $(k,\bar{c})$, and by the first-order condition,

$\frac{\partial F}{\partial c}(k,\bar{c})p+\frac{\partial u}{\partial
c}(\bar{c},k)=0$

for all $p\in\partial\bar{V}(k)$, which is impossible because $\frac{\partial
F}{\partial c}(k,\bar{c})<0$. This completes the proof. ∎

As a corollary, we obtain the following result.

######  Corollary 2.

Suppose that Assumptions 1-6 hold, and that if $F(k,c)=0$, then $F$ is
continuously differentiable in $c$ at $(k,c)$. Then, $\bar{V}$ is a classical
solution to the HJB equation in $\mathscr{V}$, and there is no other viscosity
solution to the HJB equation in $\mathscr{V}$. Moreover, the following ODE

$\dot{k}(t)=F(k(t),c^{*}(\bar{V}^{\prime}(k(t)),k(t))),\ k(0)=\bar{k}$ (3.18)

has a solution $k^{*}(t)$ defined on $\mathbb{R}_{+}$ such that $\inf_{t\geq
0}k^{*}(t)>0$, and if we define
$c^{*}(t)=c^{*}(\bar{V}^{\prime}(k^{*}(t)),k^{*}(t))$, then
$(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$ and

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

Furthermore, if one of the three conditions in Corollary 1 holds, then
$(k^{*}(t),c^{*}(t))$ is a solution to $(\ref{MODEL})$ such that $k^{*}(t)$ is
continuously differentiable and $c^{*}(t)$ is continuous.

###### Proof.

By Proposition 6 and Theorem 2, $\bar{V}$ is the unique classical solution to
the HJB equation in $\mathscr{V}$. Therefore,
$\partial\bar{V}(k)=\\{\bar{V}^{\prime}(k)\\}$, and thus (3.18) is equivalent
to (3.15). Hence, this has a solution $k^{*}(t)$ defined on $\mathbb{R}_{+}$
such that $\inf_{t\geq 0}k^{*}(t)>0$. Because the right-hand side of (3.18) is
continuous in $k$, we have that $k^{*}(t)$ is continuously differentiable.
Now, $c^{*}(t)$ is the same as that defined by (3.16). Therefore,
$(k^{*}(t),c^{*}(t))\in B_{\bar{k}}$ and

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

If one of the three conditions in Corollary 1 holds, then
$(k^{*}(t),c^{*}(t))\in A_{\bar{k}}$, and thus it is a solution to (2.1). By
definition, $c^{*}(t)$ is continuous. This completes the proof. ∎

##  4 Examples

###  4.1 Case of the RCK model

Our model (2.1) is much more complex than the most commonly used class of
models in economics. This makes it difficult to understand what our
assumptions imply. In this section, we examine the assumptions of the
traditional RCK model, and clarify the implications of Assumptions 1-6.

The model we discuss in this section is as follows.

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t))dt$ subject to. $\displaystyle~{}c(\cdot)\in W_{1},$
$\displaystyle~{}k(t)\mbox{ is absolutely continuous on any compact
interval},$ $\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$ (4.1)
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t))dt\mbox{ can be defined},$
$\displaystyle~{}\dot{k}(t)=f(k(t))-dt-c(t)\mbox{ a.e.},$
$\displaystyle~{}k(0)=\bar{k}.$

We make the following assumptions.

Assumption 1’. $\rho>0$.

Assumption 2’. The instantaneous utility function
$u:\mathbb{R}_{+}\to\mathbb{R}\cup\\{-\infty\\}$ is a continuous, concave, and
increasing function. Moreover, $u$ is continuously differentiable on
$\mathbb{R}_{++}$, $u^{\prime}(c)$ is decreasing, and $\lim_{c\to
0}u^{\prime}(c)=+\infty$, $\lim_{c\to\infty}u^{\prime}(c)=0$.

Assumption 3’. The production function $f:\mathbb{R}_{+}\to\mathbb{R}_{+}$ is
a continuous, concave, and increasing function that satisfies $f(0)=0$.
Moreover, $d\geq 0$ and there exists $k>0$ such that $f(k)>dk$.

######  Proposition 7.

Suppose that Assumptions 1’-3’ holds. Define $u(c,k)=u(c)$ and
$F(k,c)=f(k)-dk-c$. Then, Assumptions 1-4 holds, and $F$ is continuously
differentiable in $c$. In particular, the value function $\bar{V}$ is a
nondecreasing and concave function such that $\bar{V}(k)>-\infty$ for all
$k>0$, and if $\bar{V}$ is finite, then it is an increasing function that is a
classical solution to the HJB equation.

###### Proof.

First, because $W=W_{1}$, Assumption 1 holds. Second, Assumption 2 requires
that $u$ is a continuous and concave function on $\mathbb{R}_{+}$ that is
increasing and continuously differentiable on $\mathbb{R}_{++}$. Assumption 4
requires that $u^{\prime}$ is decreasing in $c$, $\lim_{c\to
0}u^{\prime}(c)=+\infty$, and $\lim_{c\to\infty}u^{\prime}(c)=0$. All
requirements are satisfied under Assumption 2’. Third, Assumption 3 requires
that $f$ is a continuous and concave function that satisfies $f(0)=0$, $d\geq
0$, and there exists $k>0$ such that $f(k)>dk$. All requirements are satisfied
under Assumption 3’. Moreover, for our $F(k,c)$, $\frac{\partial F}{\partial
c}(k,c)\equiv-1$. By Proposition 2, $\bar{V}$ is a nondecreasing and concave
function such that $\bar{V}(k)>-\infty$ for all $k>0$. Moreover, by
Propositions 3 and 6, if $\bar{V}$ is finite, then it is an increasing
function that is a classical solution to the HJB equation, as desired. This
completes the proof. ∎

We say that $f$ satisfies the Inada condition if $f$ is continuously
differentiable, strictly concave, and
$f^{\prime}(\mathbb{R}_{++})=\mathbb{R}_{++}$. Then, the following proposition
holds.

######  Proposition 8.

Suppose that Assumptions 1’-3’ hold, and that $f$ satisfies the Inada
condition. Define $u(c,k)=u(c)$ and $F(k,c)=f(k)-dk-c$. Then, Assumption 6
holds. Moreover, if there exist $a>0,C\in\mathbb{R}$, and $\theta>0$ such that

$u(c)\leq au_{\theta}(c)+C$ (4.2)

for all $c\geq 0$, then Assumption 5 holds, and thus, $\bar{V}$ is an
increasing and concave classical solution to the HJB equation, and there is no
other increasing and concave viscosity solution to the HJB equation.
Furthermore, a solution $(k^{*}(t),c^{*}(t))$ to $(\ref{MODEL2})$ can be
obtained from $(\ref{SOL2})$ and $(\ref{CONSUM2})$, and $k^{*}(t)$ is
continuously differentiable and $c^{*}(t)$ is continuous.

###### Proof.

First, $\frac{\partial^{2}u}{\partial k\partial c}(c,k)\equiv 0$. By the Inada
condition, there exists $k>0$ such that $f^{\prime}(k)>\rho$. Because
$\frac{\partial^{2}F}{\partial k\partial c}(k,c)\equiv 0$ for all $(k,c)$, we
can choose $\varepsilon_{0}=1,\ H(k,c)\equiv k-c$. Therefore, Assumption 6
holds.232323Note that, we do not prohibit the differentiability of $F$ at
$(k,c)$ with $H(k,c)=0$. For Assumption 6, it is only necessary that if
$k<\varepsilon_{0}$ and $H(k,c)\neq 0$, then $F$ satisfies some
differentiability requirements, and the converse is not needed. Second,
suppose that there exist $a>0,C\in\mathbb{R}$, and $\theta>0$ such that (4.2)
holds. By the Inada condition, there exists $k>0$ such that $f^{\prime}(k)>d$
and $\rho-(1-\theta)(f^{\prime}(k)-d)>0$. Clearly,
$(f^{\prime}(k),-1)\in\partial F(k,1)$. If we choose $b=0$, then all the
requirements in Assumption 5 holds. Therefore, Assumptions 1-6 holds, and by
Theorem 2, $\bar{V}$ is the unique viscosity solution to the HJB equation in
$\mathscr{V}$. Note that, by Lemma 2, $\mathscr{V}$ is the set of all
increasing and concave functions, and thus there is no increasing and concave
viscosity solution to the HJB equation other than $\bar{V}$. By Proposition 7,
$\bar{V}$ is a classical solution to the HJB equation. Finally, because
$c^{*}(p,k)=(u^{\prime})^{-1}(p)$, statement 3) of Corollary 1 holds, and thus
Corollary 2 can be applied. Hence, a solution $(k^{*}(t),c^{*}(t))$ to (4.1)
is obtained by (3.18) and the equality $c^{*}(t)=f(k^{*}(t))-\dot{k}^{*}(t)$,
and $k^{*}(t)$ is continuously differentiable and $c^{*}(t)$ is continuous.
This completes the proof. ∎

Hence, in the usual RCK model with the Inada condition, (4.2) is the unique
requirement for ensuring that the value function is the unique classical
solution to the HJB equation.

###  4.2 The Non-Smooth Fiscal Policy

In this section, we treat an economic example of the model in which
Assumptions 1-6 are satisfied, but $F(k,c)$ is not differentiable in $c$. In
this model, $\bar{V}$ may not be differentiable, and thus it may not be a
classical solution to the HJB equation, despite being a viscosity solution to
the HJB equation.

The model is as follows:

$\displaystyle\max~{}~{}~{}~{}~{}$ $\displaystyle~{}\int_{0}^{\infty}e^{-\rho
t}u(c(t))dt$ subject to. $\displaystyle~{}c(\cdot)\in W_{1},$
$\displaystyle~{}k(t)\mbox{ is absolutely continuous on any compact
interval},$ $\displaystyle~{}k(t)\geq 0,\ c(t)\geq 0,$ (4.3)
$\displaystyle~{}\int_{0}^{\infty}e^{-\rho t}u(c(t))dt\mbox{ can be defined},$
$\displaystyle~{}\dot{k}(t)=f(k(t))-dk(t)-c(t)-g(k(t),c(t))\mbox{ a.e.},$
$\displaystyle~{}k(0)=\bar{k},$

where $g(k,c)=\max\\{[Af(k)-c]B,0\\}$ and $0<A,B<1$.

Let us explain the background story of this model. As in the RCK model, $f$
denotes the production function and $d\geq 0$ is the capital depreciation
rate. The difference from the RCK model is the presence of the fiscal policy
rule $g(k,c)$. We assume that the government conducts a Keynesian policy to
attempt to buoy the economy when consumption is too small relative to
production. In this model, the boundary value of consumption at which the
government decides to spend is $100A$% of the output, and government
expenditure is determined as $100B$% of the shortfall by which consumption is
below that boundary value. The function $g(k,c)$ reflects this policy rule.
Through this fiscal policy, the government produces some ‘public’ commodity
that increases the consumer’s utility. However, in the decentralized economy
behind this model, the action of the government is independently determined,
and the consumer does not consider changing government spending through
his/her own action. Thus, the consumer’s behavioural decision does not reflect
the utility derived from the ‘public’ commodity produced by the government,
and as a result, it is excluded from the model’s objective function.
Therefore, the instantaneous utility function is assumed to be a function of
$c$ only.

Let $u(c,k)=u(c)$ and $F(k,c)=f(k)-c-g(k,c)$. Suppose that Assumptions 1’-3’
hold, $f$ satisfies the Inada condition, and there exist $a>0$,
$C\in\mathbb{R}$, and $\theta>0$ such that (4.2) holds. We check that model
(4.3) satisfies Assumptions 1-6. First, by the same arguments as in the proof
of Propositions 7 and 8, Assumptions 1, 2, 4, and 5 are clearly satisfied. For
Assumption 3, consider

$G(y,c)=y-c-\max\\{(Ay-c)B,0\\}.$

It is easy to show that $G$ is concave and increasing in $y$. Then,

$\displaystyle~{}F((1-t)(k_{1},c_{1})+t(k_{2},c_{2}))$ $\displaystyle=$
$\displaystyle~{}G(f((1-t)k_{1}+tk_{2}),(1-t)c_{1}+tc_{2})-d[(1-t)k_{1}+tk_{2}]$
$\displaystyle\geq$
$\displaystyle~{}G((1-t)(f(k_{1}),c_{1})+t(f(k_{2}),c_{2}))-d[(1-t)k_{1}+tk_{2}]$
$\displaystyle\geq$
$\displaystyle~{}(1-t)[G(f(k_{1}),c_{1})-dk_{1}]+t[G(f(k_{2}),c_{2})-dk_{2}]$
$\displaystyle=$ $\displaystyle~{}(1-t)F(k_{1},c_{1})+tF(k_{2},c_{2}),$

and thus $F$ is concave. The rest of the claim in Assumption 3 clearly holds
with $d_{1}=d,\delta_{2}(c)=c$. Finally, let $H(k,c)=Af(k)-c$. Then, we can
easily check that Assumption 6 holds. Hence, we have that the value function
$\bar{V}$ in (4.3) is the unique viscosity solution to the HJB equation in
$\mathscr{V}$. Because $F(k,0)=(1-AB)f(k)-dk$, Lemma 2 implies that
$\mathscr{V}$ is the set of all increasing and concave real-valued functions
on $\mathbb{R}_{++}$.

We can easily show that

$c^{*}(p,k)=\begin{cases}(u^{\prime})^{-1}(p)&\mbox{if }p\leq
u^{\prime}(Af(k)),\\\ Af(k)&\mbox{if }p\geq u^{\prime}(Af(k))\geq(1-B)p,\\\
(u^{\prime})^{-1}((1-B)p)&\mbox{if }(1-B)p\geq u^{\prime}(Af(k)),\end{cases}$

and thus, statement 3) of Corollary 1 holds. Therefore, a solution
$(k^{*}(t),c^{*}(t))$ to (4.3) can be obtained from (3.18) and the equality
$c^{*}(t)=c^{*}(\bar{V}^{\prime}(k^{*}(t)),k^{*}(t))$.

Surprisingly, if $d=0$, then $\bar{V}$ is continuously differentiable. Indeed,
we can easily check that if $F(k,c^{*}(p,k))=0$, then $c^{*}(p,k)=f(k)>Af(k)$,
and thus $F$ is differentiable at $(k,c^{*}(p,k))$. Hence, Proposition 6
implies that $\bar{V}$ is continuously differentiable.

However, if $d>0$, then there exists $k^{*}$ such that
$f(k^{*})-dk^{*}=Af(k^{*})$. In this case, it is unknown whether $\bar{V}$ is
differentiable at $k^{*}$.

##  5 Discussion

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. For example, Malliaris
and Brock [22] dealt with this type of equation. Probably, the use of Ito’s
formula in the middle of the derivation makes the HJB equation second-order.
On the other hand, since there is no stochastic variation in the dynamic model
considered in this paper, the HJB equation is only a first-order differential
equation.

The case in which the HJB equation does not have a classical solution has been
highlighted in many studies. Therefore, following Lions [19], the viscosity
solution is typically taken as the solution concept in the study of degenerate
elliptic differential equations. However, as in Proposition 6, the condition
for the value function to be differentiable is very mild, and thus, in many
economic models, the value function is a classical solution to the HJB
equation. For such a model, the net contribution of this paper is to show the
uniqueness of the viscosity, not classical, solution to the HJB equation
rigorously.

As noted in the introduction, the properties of the HJB equation presented in
many previous studies are not applicable to the model considered in this
paper. There are several reasons for this. First, in the model considered
herein, the time interval is infinite. To the best of our knowledge, this type
of problem was first addressed by Soner [27, 28], who made the assumption that
moving the control variable of the cost function does not have a significant
effect on the cost. Applying this assumption to economic models, we must
assume that $u(c,k)$ is bounded in $c$. However, the most typical example of
$u(c,k)$ in economics is $\log c$, which violates this assumption. Barles [5]
pointed out that when this boundedness requirement does not hold, a
pathological problem arises. Hosoya [14, 15] showed that this pathological
problem appears even in RCK models, and thus this problem is serious for
economic models.

This boundedness assumption is often applied in the modern theory of the HJB
equation for control problems with infinite time interval. Bardi and Capuzzo-
Dolcetta [4] is a typical example. In contrast, several recent studies have
not assumed the boundedness of $u$. For example, Hermosilla et al. [13]
treated such a case for a problem with a finite time interval. However, even
in that case, $u$ is assumed to be Lipschitz. This is a condition that rarely
applies to economic models, because any CRRA function is not Lipschitz around
$0$. To the best of our knowledge, no model other than that of Hosoya [14, 15]
has dealt with the case of $u(c,k)=\log c$. Hosoya [14] allows the
differentiability of $F$ to derive the results, but this is not assumed in the
present paper. Hosoya [15] assumes the existence of a continuous solution to
the original problem, but again, the present paper does not make such an
assumption. Hence, the results presented in this paper are independent of
those derived in previous studies.

On the other hand, in economics, if $c$ can be written as a function of $k$
and $\dot{k}$, the value function is differentiable when the original problem
has a solution such that $\dot{k}(t)$ is continuous. This is because of
Theorem 2 of Benveniste and Scheinkman [7]. Thus, the value function $\bar{V}$
becomes non-differentiable only if the optimal capital accumulation path
$k^{*}(t)$ is not continuously differentiable. In other words, the value
function can be non-differentiable only when (3.15) is truly a differential
inclusion. Unfortunately, (3.15) contains information on $\bar{V}$ itself, and
thus deriving the non-differentiability of $\bar{V}$ from the assumptions of
$u$ and $F$ involves considerable difficulty. However, there exist cases in
which it is not possible to judge whether the value function is
differentiable. In such cases, our Theorems 1-2 serve as powerful analytical
tools.

##  6 Concluding Remarks

In this paper, we considered a class of economic control problems, and
presented the conditions under which the value function is the unique
viscosity solution to the HJB equation. For such cases, we provided a method
for deriving a solution to the original problem using the HJB equation.
Furthermore, we presented a condition under which the value function is
differentiable. Our conditions are sufficiently weak that our results can be
applied to many economic applications.

There are several future tasks. First, we want to prove that Corollary 1 holds
unconditionally.

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

Third, we want to obtain a simple method for gaining an approximate solution
to the value function. In discrete-time models, there is a famous
approximation method that uses Blackwell’s inequality and the contraction
mapping theorem.242424See Stokey and Lucas [29] for more detailed arguments.
We want to obtain a counterpart to this result for a continuous-time model.

Fourth, we want to extend our result to decentralized models in economic
theory. Usually, macroeconomic dynamic models can be classified into two
categories: centralized and decentralized. We have only discussed centralized
models in this paper. If the government is absent, then by the fundamental
theorem of welfare economics, the results of the two models coincide. However,
if fiscal policy is introduced, then these two models can derive different
results. Hence, we want to extend our result to decentralized models.

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