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An elementary proof of the Euler equation in growth theory
Hosoya, Yuhki
Keio Economic Society, Keio University2023
Keio economic studies Vol.56, (2023. ) ,p.23- 38
In this paper, we derive the Euler equation in a continuous-time macroeconomic
dynamic model using only elementary mathematical knowledge. We also provide a
proof that the Euler equation and transversality condition are sufficient conditions
for the optimal solution. To understand the proofs in this paper, no knowledge ofLebesgue integrals is required. Readers require only some basic mathematicalknowledge.
Journal Article
https://koara.lib.keio.ac.jp/xoonips/modules/xoonips/detail.php?koara_id=AA00260
492-20230000-0023

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KEIO ECONOMIC STUDIES 56, 23–38 (2023)

AN ELEMENTARY PROOF OF THE EULER EQUATION IN
GROWTH THEORY
Yuhki H OSOYA
Faculty of Economics, Chuo University, Tokyo, JapanFirst version received November 2021; final version accepted December 2021

Abstract: In this paper, we derive the Euler equation in a continuous-time macroeconomic dynamic model using only elementary mathematical knowledge. We also provide
a proof that the Euler equation and transversality condition are sufficient conditions forthe optimal solution. To understand the proofs in this paper, no knowledge of Lebesgue
integrals is required. Readers require only some basic mathematical knowledge.
Key words:

Continuous-time growth model, Euler equation, transversality condition, du Bois-Reymond’s

lemma.
JEL Classification Number:C61, A22, E13.

1.

INTRODUCTION

In classical continuous-time macroeconomic dynamic models, the Euler equation
plays a very important role. However, to the best of our knowledge, the derivationof the Euler equation is not adequately explained in almost all textbooks. For example,
Romer (2012) derived the Euler equation by “pretending” that Lagrange’s multiplier
rule in finite-dimensional space could be applied. Blanchard and Fischer (1989) derived the Euler equation by introducing and applying Pontryagin’s maximum principle
in an appendix. Similar explanations has been done by Acemoglu (2009) and Barro andSala-i-Martin (2003). However, deriving Lagrange’s multiplier rule in function spaces
requires deep knowledge of Banach spaces. Moreover, Pontryagin’s maximum principle is nightmarishly difficult to prove. As a result, most undergraduate and graduate
students are unable to gain effective knowledge of the techniques for deriving the Euler
equation from textbooks.1
This paper explains how to derive the Euler equation for a Ramsey-Cass-Koopmans
E-mail: ukki(at)gs.econ.keio.ac.jp
1 Chapter 1 of Ioffe and Tikhomirov (1979) gives a detailed argument of Lagrange’s multiplier rule for
abstract problems, but one requires a lot of preparation to read and understand this textbook. The simplest
proof of Pontryagin’s maximum principle that we know of is given by Luenburger (1997), which also requires
some basic theorems on Banach spaces.

Copyright©2023, by the Keio Economic Society23

24

KEIO ECONOMIC STUDIES

optimal capital accumulation model.2 As the primary readers for this paper, we envision
undergraduate students in the third year and above, and graduate students in the mastercourse. The only knowledge required is the usual knowledge of differential and integral
calculus, a very elementary knowledge of linear algebra, a basic knowledge of the ε-δmethod, and a simple knowledge of the properties of concave functions. In particular,
as for integrals, only Riemann integrals are treated in this paper. Even readers withno knowledge of Lebesgue integral theory will be able to understand the proofs in thispaper.
There are two types of optimal capital accumulation models: decentralized and centralized. Both of which will be treated in this paper. In both models, the Euler equation
can be derived using exactly the same method, in which a very simple lemma of linear
algebra is crucial. In the centralized model, it is known that the Euler equation andtransversality condition are sufficient conditions for the solution, which is also treatedin this paper. In the appendix, we have added a discussion of how to derive the Euler
equation for more difficult problems.
The assumptions regarding the utility function and the production technology usedin this paper are far fewer than in many macroeconomics textbooks. For example, we
do not require twice differentiability of the utility function. Although this assumption
is popular, it is not necessary at all if we just want to derive the Euler equation. On theother hand, there are some unfamiliar assumptions in the model, such as the definabilityof the objective function or the real-valuedness of the solution. However, these are essentially necessary for a rigorous discussion in this context, and in this sense, textbooks
do not describe the problem correctly.
The remainder of this paper is organized as follows. In Section 2, we describe thecentralized capital accumulation model, and explain terms such as solutions and theEuler equation. In Section 3, we present proofs of the Leibniz integral rule and du BoisReymond’s lemma, and use them to show the main results. Section 4 deals with thedecentralized model and again derives the Euler equation. Section 5 is the conclusion,
where we discuss in detail what to do for more in-depth study. The appendix contains
an example of deriving the Euler equation for a more difficult problem.
2.

PRELIMINARIES

2.1. The Model
The first model we treat in this paper is as follows.
 ∞e−ρt u(c(t))dt,
max0

subject to. c(t) ≥ 0, k(t) ≥ 0,c(t) is continuous,k(t) is continuously differentiable,(1)

2 This model was formulated by Ramsey (1928), and modified by Cass (1965) and Koopmans (1965).

HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

 ∞

25

e−ρt u(c(t))dt is defined,

0k̇(t) = f (k(t)) − c(t),k(0) = k̄,∞where ρ > 0 and k̄ > 0. Note that, “ 0 e−ρt u(c(t))dt is defined” means that thisintegral is defined in the sense of an improper Riemann integral, in other words, that
T
limT →∞ 0 e−ρt u(c(t))dt exists, though it may be either +∞ or −∞. Note also that,the symbol k̇(t) denotes the derivative of k(t) at time t: in other words, k̇(t) indicatesthe speed of increase for the function k(t).
We should explain the equationk̇(t) = f (k(t)) − c(t).

(2)

Actually, there are three relationships behind this equation. The first relationship is thesimplified IS relationship: that is,y(t) = c(t) + i(t),where c(t) is consumption, i(t) is investment, and y(t) is the amount of production attime t. The second relationship is on the production technology:y(t) = g(k(t)),
where k(t) is the total amount of capital at time t and g(k) is the production function.
The third relationship determines the speed of capital accumulation:k̇(t) = i(t) − dk(t),where d ≥ 0 represents the capital depreciation rate. Using these three formulas, we
obtaink̇(t) = g(k(t)) − dk(t) − c(t),and if we define f (k) = g(k) − dk, then we obtain equation (2).
We assume the following.3
Assumption U. u : R+ → R ∪ {−∞} is a continuous and increasing function that iscontinuously differentiable on R++ . Moreover, u (c) > 0 for all c > 0.
Assumption F. g : R+ → R is a continuous and increasing function such that g(0) = 0
and g is continuously differentiable on R++ .
Assumption C. The functions u, g are concave.
Under Assumption F, we have that f (0) = 0, f is continuous, and continuously
differentiable on R++ . If, in addition, Assumption C holds, then f is also concave. Notethat f is not necessarily increasing, because the possibility of g  (k) < d is not excluded.
3 As usual, we define R as the set of nonnegative numbers and R+++ as the set of positive numbers,
respectively.

26

KEIO ECONOMIC STUDIES

Note also that we do not assume the Inada condition.4 Although this condition is very
popular in this context, we do not require this assumption for our purpose.
2.2. An Admissible Pair and a Solution
Let (k(t), c(t)) be a pair of functions from R+ into R+ . We call these functions
an admissible pair when all restrictions of (1) are met. In other words, (k(t), c(t)) isadmissible if 1) k(t) is continuously differentiable, 2) c(t) is continuous,3) k(0) = k̄, ∞ −ρt4) the capital accumulation equation (2) holds for all t ≥ 0, and 5) 0 e u(c(t))dt
can be defined.
Let (k ∗ (t), c∗ (t)) be an admissible pair. We call this pair a solution to problem (1) if
the following two requirements hold.
1. The value ∞e−ρt u(c∗ (t))dt0

is a real number.
2. For every admissible pair (k(t), c(t)),
 ∞
 ∞
−ρt
∗
e u(c (t))dt ≥e−ρt u(c(t))dt.
00Note that, by the first requirement, if there exists an admissible pair (k(t), c(t)) such∞that 0 e−ρt u(c(t))dt = +∞, then there is no solution to (1).
Finally, we define the notion of inner solutions. A solution (k(t), c(t)) to (1) is said
to be an inner solution if k(t) > 0, c(t) > 0 for all t ≥ 0.
2.3. The Euler Equation and Transversality Condition
Usually, the Euler equation of problem (1) is written as follows.
u (c(t))ċ(t) = (ρ − f  (k(t))) .
(3)u (c(t))However, this expression has at least two problems. First, we do not assume the differentiability of c(t). Second, we do not assume the twice differentiability of u. Therefore,
we mainly use the following alternative formula:d (u ◦ c)(t) = (ρ − f  (k(t)))(u ◦ c)(t).
(4)
dt
We say that an admissible pair (k(t), c(t)) is a solution to the Euler equation if k(t) > 0,c(t) > 0, u (c(t)) is continuously differentiable, and (4) holds for all t ≥ 0.
If u is twice continuously differentiable and u (c) = 0 on R++ , then the inverse
function of u is also continuously differentiable by the inverse function theorem, andthus, u (c(t)) is continuously differentiable if and only if c(t) is continuously differentiable. In this case, (4) is equivalent to (3). However, at least we do not require the twice
differentiability of u in the derivation of the Euler equation.
Next, we mention transversality condition. An admissible pair (k(t), c(t)) is said tosatisfy transversality condition if c(t) > 0 for all t ≥ 0 and
4 The Inada condition means that lim
k→0 g (k) = +∞ and limk→∞ g (k) = 0.

HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

27

lim e−ρT u (c(T ))k(T ) = 0.

(5)

T →∞

3.

RESULTS3.1. On the Leibniz Integral Rule
The Leibniz integral rule is a famous result in differential calculus, and plays a fundamental role in our proof of main results. However, this rule is not so elementary.
Therefore, we provide this rule and its proof rigorously. Readers who are familiar withthis result can skip this subsection.
The Leibniz Integral Rule. Suppose that a < b and f : [a, b] ×U → R is continuous,
where U is an open interval in R. Then, the followings hold.
1) The functionbg(x) =f (t, x)dta

is continuous on U .
2) If the function ∂f∂x (t, x) is also continuous, then b∂f
(t, x)dt.
g (x) =a ∂x
Proof. Choose any x ∈ U and ε > 0. Because f is continuous, for each t ∈ [a, b],there exists δ(t) > 0 such that if (s, y) ∈ [a, b] × U and max{|s − t|, |y − x|} < δ(t),then|f (s, y) − f (t, x)| < 2−1 (b − a)−1 ε.
Let V (t) = (t − δ(t), t + δ(t)). Because [a, b] is compact, there is a finite set {t1 , ..., tn }such that [a, b] ⊂ ∪ni=1 V (ti ). Defineδ = min{δ(t1 ), ..., δ(tn )}.
Suppose that t ∈ [a, b], y ∈ U and |y − x| < δ. Then, there exists i such that |t − ti | <δ(ti ). Hence,|f (t, y) − f (t, x)| ≤ |f (t, y) − f (ti , x)| + |f (ti , x) − f (t, x)| < (b − a)−1 ε.
Therefore, if y ∈ U and |y − x| < δ, then b|g(y) − g(x)| ≤|f (t, y) − f (t, x)|dt < ε,awhich implies that g is continuous.5 This completes the proof of 1).
For 2), choose any x ∈ U and ε > 0. By almost the same arguments as in the above
paragraph, we can show that there exists δ > 0 such that
5 The inequality

 


 b
b


f (x)dx  ≤|f (x)|dx


 a
a

can easily be shown, and thus we omit its proof.

28

KEIO ECONOMIC STUDIES



 ∂f

 (t, y) − ∂f (t, x) < (b − a)−1 ε
 ∂x

∂x
for any t ∈ [a, b] and y ∈ U such that |y − x| < δ. By the mean value theorem,f (t, x + h) − f (t, x) − ∂f (t, x)h < (b − a)−1 ε|h|∂xwhenever |h| < δ. Therefore, if |h| < δ, then


 b


∂f
g(x + h) − g(x) − h(t, x)dt 

a ∂x

 b


f (t, x + h) − f (t, x) − h ∂f (t, x) dt
≤


∂x
a
< ε|h|,
which implies thatg  (x) = baThis completes the proof. ∂f(t, x)dt.
∂x3.2. On du Bois-Reymond’s Lemma
Classically, in the derivation of the Euler equation, the result named du BoisReymond’s lemma is frequently used.6 This lemma states the following:
du Bois-Reymond’s Lemma. Suppose that b(t) is continuous, and Tb(t)x(t)dt = 00

for every continuous function x(t) such that x(0) = x(T ) = 0 and Tx(t)dt = 0.
0Then, b(t) is a constant function.
In many variational problems, this lemma plays a key role in deriving the Euler equation. We show this result using two lemmas.
Lemma 1. Suppose that V is a vector space, and let f0 and f1 be linear functionals
on V ; that is, each fi is a function from V into R such that for every v1 , v2 ∈ V and
a1 , a2 ∈ R, fi (a1 v1 + a2 v2 ) = a1 fi (v1 ) + a2 fi (v2 ). Define Ker fi = {v ∈ V |fi (v) =0}. Then, the following two statements are equivalent.
1) f0 = a1 f1 for some a1 ∈ R.
2) Ker f1 ⊂ Ker f0 .
6 Regarding this surname, Emil Heinrich du Bois-Reymond is quite famous. However, this lemma wasfound by his brother, Paul David Gustav du Bois-Reymond.

HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

29

Proof. Clearly, 1) implies 2). Therefore, it suffices to show that 2) implies 1).
If Ker f0 = V , then this claim is obvious. Hence, we assume that there exists v ∗ ∈ Vsuch that f0 (v ∗ ) = 0. Because Ker f1 ⊂ Ker f0 , we have that f1 (v ∗ ) = 0. Without lossof generality, we assume that f1 (v ∗ ) = 1. Define f0 (v ∗ ) = a1 . Choose any v ∈ V . If
f1 (v) = 0, then f0 (v) = 0, and thus f0 (v) = a1 f1 (v). If f1 (v) = 0, define b1 = f1 (v).
Then,
f1 (v ∗ − b1−1 v) = f1 (v ∗ ) − b1−1 f1 (v) = 0.
Therefore, v ∗ − b1−1 v ∈ Ker f1 ⊂ Ker f0 . Hence,

0 = f0 (v ∗ − b1−1 v) = f0 (v ∗ ) − b1−1 f0 (v),and thus,

a1 = f0 (v ∗ ) = b1−1 f0 (v),which implies that f0 (v) = a1 b1 = a1 f1 (v). In conclusion, we have that for eachv ∈ V , f0 (v) = a1 f1 (v), as desired. This completes the proof. 
Lemma 2. Let V be the set of all continuous functions x(t) from [0, T ] into R suchthat x(0) = x(T ) = 0. Suppose that b(t) is a continuous function such that Tb(t)x(t)dt = 00

for all x(t) ∈ V . Then, b(t) ≡ 0.
Proof. Suppose that b(t) = 0 at t ∈ [0, T ]. Without loss of generality, we assume that0 = t = T and b(t) > 0. Choose ε > 0 such that 0 < t − ε < t + ε < T and for everys ∈ [t − ε, t + ε], b(s) > b(t)/2. Definex(s) = max{0, 1 − ε−1 |t − s|}.
Then, x(0) = x(T ) = 0 and Tb(t)x(t)dt > 0,

which is a contradiction. 0

Proof of du Bois-Reymond’s Lemma. Let V be the same set as in Lemma 2. Notethat V is a linear space. Define
 Tb(t)x(t)dt,
0 (x(t)) =0

1 (x(t)) = Tx(t)dt.
0

By assumption of du Bois-Reymond’s lemma, Ker 1 ⊂ Ker 0 holds true, and thusby 1) of Lemma 1, we have that there exists a1 ∈ R such that0 = a1 1 ,which implies that for all x(t) ∈ V ,30

KEIO ECONOMIC STUDIES T(b(t) − a1 )x(t)dt = 0.

0

By Lemma 2, we have that b(t) − a1 ≡ 0, and thus b(t) ≡ a1 . This completes the proof.

3.3. Main Result I: Necessity of the Euler Equation
Our first main result is as follows.
Theorem 1. Suppose that Assumptions U and F hold, and let (k ∗ (t), c∗ (t)) be an inner
solution to (1). Then, this pair is a solution to the Euler equation.
Proof. First, we delete c(t) from (1) by using (2). Because of (2),c(t) = f (k(t)) − k̇(t).
Therefore, (1) is equivalent to the following problem: ∞max
e−ρt u(f (k(t)) − k̇(t))dt,0subject to. k(t) is continuously differentiable,k(t) ≥ 0, f (k(t)) − k̇(t) ≥ 0,
 ∞e−ρt u(f (k(t)) − k̇(t))dt is defined,(6)0k(0) = k̄.
By definition, k ∗ (t) is a solution to this model and c∗ (t) = f (k ∗ (t)) − k̇ ∗ (t) for allt ≥ 0.
Next, choose any T > 0, and define V as the set of all continuous functions x :T[0, T ] → R such that x(0) = x(T ) = 0. Choose x(t) ∈ V such that 0 x(t)dt = 0,
and define k(t) and ks (t) as follows: tk(t) =x(τ )dτ,0⎧⎨k ∗ (t) + sk(t) if 0 ≤ t ≤ T ,ks (t) =⎩k ∗ (t)if t ≥ T .
Then, ks (t) is continuously differentiable. Because (k ∗ (t), c∗ (t)) is an inner solution,
there exists δ > 0 such that if |s| < δ, then ks (t) ≥ 0 and f (ks (t)) − k̇s (t) ≥ 0 for allt ≥ 0. DefineTϕ(s) =
0e−ρt u(f (ks (t)) − k̇s (t))dt.

Because k ∗ (t) is a solution to (6), ϕ(0) = maxs∈(−δ,δ) ϕ(s), and thus ϕ  (0) = 0. By the
Leibniz integral rule, T[e−ρt u (c∗ (t))f  (k ∗ (t))k(t) − e−ρt u (c∗ (t))x(t)]dt.
(7)0 = ϕ  (0) =0HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

31

Now, recall the formula of integration by parts: b bf  (t)g(t)dt = [f (t)g(t)]ba −f (t)g  (t)dt.
a

T

a

e−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ into f (t) and k(t) into g(t), we obtain

Substituting − t Te−ρt u (c∗ (t))f  (k ∗ (t))k(t)dt

0


 T
T
−ρτ  ∗ ∗eu (c (τ ))f (k (τ ))dτ k(t)
= −
t

+

 T  T0= T  T00e−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ x(t)dtte−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ x(t)dt.

t

Therefore, by using the relation (7), we have that
 T  Te−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ − e−ρt u (c∗ (t)) x(t)dt = 0.
0t

By du Bois-Reymond’s lemma, we have that there exists a ∈ R such that Te−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ = e−ρt u (c∗ (t)) + a.

(8)t

Therefore,∗u (c (t)) = e

ρt

 T

e−ρτ u (c∗ (τ ))f  (k ∗ (τ ))dτ − a ,twhere the right-hand side is continuously differentiable. Therefore, u (c∗ (t)) is continuously differentiable. Differentiating both side of (8), we have that for all t ∈ [0, T ],d−ρe−ρt u (c∗ (t)) + e−ρt (u ◦ c∗ )(t) = −e−ρt u (c∗ (t))f  (k ∗ (t)).
dtwhich implies thatd  ∗(u ◦ c )(t) = (ρ − f  (k ∗ (t)))(u ◦ c∗ )(t).
dtThis equation is the same as (4). Because T > 0 is arbitrary, we have that (k ∗ (t), c∗ (t))
is a solution to the Euler equation. This completes the proof. 
3.4. Main Result II: Sufficiency of the Euler Equation and Transversality Condiition
Before arguing our second main result, we make a preparation. Suppose that L :U → R is a continuously differentiable and concave function, where U ⊂ R2 is convex.
Choose any (x ∗ , y ∗ ) ∈ U . For (x, y) ∈ U , define d(t) = L((1 − t)(x ∗ , y ∗ ) + t (x, y)).
Then, we can easily check that d(t) is concave, and thus d(1) − d(0) ≤ d  (0). By thechain rule, we have that
∂L ∗ ∗∂L ∗ ∗L(x, y) − L(x ∗ , y ∗ ) ≤(x , y )(x − x ∗ ) +(x , y )(y − y ∗ ).
(9)
∂x
∂y

32

KEIO ECONOMIC STUDIES

We use this formula in the proof of the following result.
Theorem 2. Suppose that Assumptions U, F, and C hold. Let (k ∗ (t), c∗ (t)) be an∞admissible pair such that k ∗ (t) > 0, c∗ (t) > 0 for all t ≥ 0, and 0 e−ρt u(c∗ (t))dt is∗∗a real number. If (k (t), c (t)) satisfies the Euler equation and transversality condition,then it is a solution to (1).
Proof. Again, we use the transformc(t) = f (k(t)) − k̇(t)throughout this proof. We defineL(x, y) = u(f (x) − y).
Then, the function L is continuously differentiable on the set
U = {(x, y) ∈ R2 |x > 0, f (x) − y > 0}.
Let t ∈ [0, 1] and (x1 , y1 ), (x2 , y2 ) ∈ U , and define (x, y) = (1 − t)(x1, y1 ) + t (x2, y2 ).
Then, x > 0 and
f (x) − y ≥ (1 − t)f (x1 ) + tf (x2 ) − (1 − t)y1 − ty2= (1 − t)[f (x1 ) − y1 ] + t[f (x2 ) − y2 ] > 0,
which indicates that (x, y) ∈ U . Moreover, because u is increasing and concave,
L(x, y) = u(f (x) − y)≥ u((1 − t)[f (x1 ) − y1 ] + t[f (x2 ) − y2 ])≥ (1 − t)u(f (x1 ) − y1 ) + tu(f (x2 ) − y2 )= (1 − t)L(x1 , y1 ) + tL(x2 , y2 ).
Therefore, we have that U is convex and L is concave, and the formula (9) can beapplied to L.
Choose any admissible pair (k(t), c(t)). Then, for T > 0,
 Te−ρt [u(c(t)) − u(c∗ (t))]dt0 Te−ρt [L(k(t), k̇(t)) − L(k ∗ (t), k̇ ∗ (t))]dt
0

 T
∂L ∗
∂L ∗
(k (t), k̇ ∗ (t))(k(t) − k ∗ (t)) +(k (t), k̇ ∗ (t))(k̇(t) − k̇ ∗ (t)) dt
≤
e−ρt
∂x
∂y
0
 T
e−ρt [u (c∗ (t))f  (k ∗ (t))(k(t) − k ∗ (t)) − u (c∗ (t))(k̇(t) − k̇ ∗ (t))]dt.
==

0Using the Euler equation, we have thate−ρt u (c∗ (t))f  (k ∗ (t))(k(t) − k ∗ (t)) − e−ρt u (c∗ (t))(k̇(t) − k̇ ∗ (t))=

d −ρt  ∗[e u (c (t))(k ∗ (t) − k(t))].
dt

HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

33

Because k ∗ (0) = k(0) = k̄ and k(T ) ≥ 0, T Td −ρt  ∗[e u (c (t))(k ∗ (t) − k(t))]dt
e−ρt [u(c(t)) − u(c∗ (t))]dt ≤dt00= e−ρT u (c∗ (T ))(k ∗ (T ) − k(T ))≤ e−ρT u (c∗ (T ))k ∗ (T ) → 0 (as T → ∞),where the last line follows from transversality condition. This implies that ∞
 ∞e−ρt u(c∗ (t))dt ≥e−ρt u(c(t))dt,00

as desired. This completes the proof. 
4.

DECENTRALIZED MODEL

Problem (1) is a sort of centralized model, in which interest rate and wage are absent.
However, in many macroeconomic models, the decentralized version of (1) is used andanalyzed. In these models, it is thought that Lagrange’s multiplier rule for infinite dimensional space is needed to derive the Euler equation. However, this is incorrect. In
this section, we try to derive this rule using only our lemmas.
The problem considered in this section is as follows: ∞max
e−ρt u(c(t))dt,0subject to. c(t) ≥ 0,c(t) is continuous, ∞e−ρt u(c(t))dt is defined, ∞0 ∞e−R(t )c(t)dt = k̄ +e−R(t )w(t)dt,
0

(10)

0

where w : R+ → R++ and r : R+ → R++ are given continuous functions, andtR(t) = 0 r(τ )dτ . The value w(t) represents wage and r(t) represents interest rate attime t. The last equality represents the consumer’s budget constraint.
As in section 2, we define the notion of admissible paths.
 ∞ A continuous function c(t)is said to be admissible if 1) c(t) ≥ 0 for all t ≥ 0, 2) 0 e−ρt u(c(t))dt is defined, and∞∞3) 0 e−R(t )c(t)dt = k̄ + 0 e−R(t )w(t)dt. An admissible function c∗ (t) is a solution
to (10) if∞e−ρt u(c∗ (t))dt0

is a real number, and for any admissible function c(t),
 ∞
 ∞
−ρt
∗
e u(c (t))dt ≥e−ρt u(c(t))dt.
00

If c∗ (t) is a solution to (10) and c∗ (t) > 0 for all t ≥ 0, then it is said to be an inner
solution.

34

KEIO ECONOMIC STUDIESIn this section, the Euler equation is the following:d (u ◦ c)(t) = (ρ − r(t))(u ◦ c)(t).
(11)
dt
A continuous function c(t) is said to be a solution to the Euler equation if u (c(t)) is
continuously differentiable and (11) holds for all t ≥ 0.
Theorem 3. Suppose that Assumption U holds. If c∗ (t) is an inner solution to (10),then it is a solution to the Euler equation.
Proof. Fix any T > 0. Let V be the set of all continuous functions x : [0, T ] → Rsuch that x(0) = x(T ) = 0. Choose any x(t) ∈ V such that
 T
e−R(t )x(t)dt = 0,0

and define cs (t) as follows:cs (t) =⎧⎨c∗ (t) + sx(t)if 0 ≤ t ≤ T ,⎩c∗ (t)

if t ≥ T .

Because c∗ (t) is an inner solution, there exists δ > 0 such that if |s| < δ, then cs (t) ≥ 0
for all t ≥ 0. DefineTϕ(s) =
0e−ρt u(cs (t))dt.

By the same derivation as (7), we have that T
0 = ϕ (0) =e−ρt u (c∗ (t))x(t)dt.
0Define T

e−ρt u (c∗ (t))x(t)dt,
 T
e−R(t )x(t)dt.
1 (x(t)) =0 (x(t)) =0

0

The above arguments say that Ker 1 ⊂ Ker 0 . By Lemma 1, there exists a ∈ R suchthat 0 = a1 . By Lemma 2,e−ρt u (c∗ (t)) = ae−R(t ).
Because the left-hand side is positive, we have that a > 0, andu (c∗ (t)) = aeρt −R(t ).
Since the right-hand side is continuously differentiable, we have that u (c∗ (t)) is also
continuously differentiable. Moreover,log u (c∗ (t)) = log a + ρt − R(t),and differentiating both sides, for every t ∈ [0, T ],

HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY

35

d  ∗
(u ◦ c )(t) = (ρ − r(t))(u ◦ c∗ )(t),dt
where this equation is the same as (11). Because T is arbitrary, we have that c∗ (t) is asolution to the Euler equation. This completes the proof. 
5.

CONCLUSION

This paper has mainly considered the derivation of the Euler equation in a classicalcontinuous-time optimal growth model. Additionally, we have provided a proof of thefact that the Euler equation and transversality condition are sufficient conditions for the
optimal solution in concave problems.
In this section, we discuss some matters that are necessary for further study. First, theEuler equation and transversality condition are often treated as “necessary and sufficientconditions” for the solution to a problem. In this paper, however, we have dealt withthe fact that “the Euler equation is a necessary condition for the solution” and that “theEuler equation and transversality condition are sufficient conditions for the solution,”
but we have not discussed whether “transversality condition is a necessary condition forthe solution” at all. This was discussed in detail in Kamihigashi (2001), and it seems
that transversality condition is a necessary condition for the solution in most cases.
However, we could not put this result into an elementary form. In particular, it is very
difficult to discuss the above papers while maintaining the continuity requirement ofc(t). For this reason, we have decided not to mention it at all in this paper.
Next, there is a problem related to the relaxation of the continuity of c(t) discussed
above. In fact, macroeconomic models often allow for the possibility of “jumps” in c(t).
In that case, the continuity requirement of c(t) is replaced by that of piecewise continuity or local integrability. Then, the solution to the capital accumulation equation (2) is
not necessarily continuously differentiable, and thus the requirement of k(t) changes topiecewise continuous differentiability or absolute continuity. In this case, we inevitably
need the dominated convergence theorem to derive the Leibniz integral rule. Additionally, the continuity of b(t) in du Bois-Reymond’s lemma is no longer available, andhence other arguments are needed. In any case, without a basic knowledge of Lebesgue
integrals, it will be difficult to go further.
∞
Third, there is a question of whether the quirky requirement that “ 0 e−ρt u(c(t))dt
is defined” can be removed. Actually, this requirement can be removed by considering
a slightly different kind of optimality problem called overtaking optimality, and our
results can be reproduced by almost the same argument. See Carlson et al. (2011) for
detailed arguments. As a matter of fact, Kamihigashi (2001) discusses the transversality
condition in this context.
Fourth, there is a problem of corner solutions. In this paper, we have dealt with the
Euler equation as a condition for “inner solutions.” This paper has nothing to say about
corner solutions, i.e., solutions for which c(t) = 0 is possible. In this regard, Hosoya(2019) dealt with the requirements of u for excluding corner solutions. However, evenin this manuscript, this problem was only partially solved, and it is very difficult to36

KEIO ECONOMIC STUDIES

eliminate corner solutions in general. Note that, since the derivation of equation (7)is not possible for corner solutions, the Euler equation is not a necessary condition forsuch solutions.
Finally, this paper has only dealt with classical optimal growth models, and somereaders may have concerns that the same approach may not work for slightly different
models that are not classical. For this, we assure you that similar techniques can be used
for a number of problems. As an example, the appendix presents a problem in which theobjective function u has two variables and there are multiple constraints of the integral
equations, and derives the Euler equations by almost the same method as in the proofof Theorem 3. Although this problem is somewhat artificial, it serves as an example of
how to apply our techniques to non-classical problems.
A.

APPENDIX : MULTIPLE RESTRICTIONS

For proving du Bois-Reymond’s lemma, Lemma 1 is sufficient. However, in macroeconomic model, there are problems in which Lemma 1 is insufficient for deriving theEuler equation. In this appendix, we provide a method for solving such problems.
Lemma 3. Suppose that V is a vector space, and let f0 , f1 , ..., fn be the family of
linear functionals on V . Define Ker fi = {v ∈ V |fi (v) = 0}. Then, the following two
statements are equivalent.
1) f0 = a1 f1 + ... + an fn for some a1 , ..., an ∈ R.
2) ∩ni=1 Ker fi ⊂ Ker f0 .
Proof. Clearly, 1) implies 2). Therefore, it suffices to show that 2) implies 1). We use
mathematical induction on n. The case in which n = 1 is just Lemma 1.
Next, choose any n ≥ 2, and suppose that for n − 1, this lemma is correct. If
n−1∩i=1 Ker fi ⊂ Ker f0 , then by assumption,
f0 = a1 f1 + ... + an−1 fn−1 + 0fn ,and 1) holds. Hence, we assume that ∩n−1
i=1 Ker fi ⊂ Ker f0 . Thus, there exists∗ ) = 1. Because ∩n Ker f ⊂ Ker f , we have
Ker
f
such
that
f
(v
v ∗ ∈ ∩n−1
i
0
i
0
i=1
i=1
that fn (v ∗ ) = 0. Define b1 = fn (v ∗ ) and an = b1−1 . By induction hypothesis, itsuffices to show that∩n−1
i=1 Ker fi ⊂ Ker (f0 − an fn ).
Choose any v ∈ ∩n−1
i=1 Ker fi . It suffices to show that f0 (v) = an fn (v). If fn (v) = 0,then f0 (v) = 0 by 2), and thus f0 (v) = an fn (v). If fn (v) = b2 = 0, then fn (v ∗ −b1 b2−1 v) = 0, and thus by 2), f0 (v ∗ − b1 b2−1 v) = 0. Therefore,
1 = f0 (v ∗ ) = b1 b2−1 f0 (v),and thus,
f0 (v) = an b2 = an fn (v),as desired. This completes the proof. HOSOYA: AN ELEMENTARY PROOF OF THE EULER EQUATION IN GROWTH THEORY37

We now consider an example in which Lemma 3 can be effectively applied. Consider
the following problem.
 ∞max
e−ρt u(c(t), a(t))dt,0subject to. c(t) ≥ 0, a(t) ≥ 0,c(t), a(t) are continuous, ∞e−ρt u(c(t), a(t))dt is defined,0 ∞e−R1 (t ) c(t)dt = M,0 ∞e−R2 (t ) a(t)dt = N.

(12)

0

We assume that u(c, a) is continuous on R2+ and continuously differentiable on R2++ ,tand that Ri (t) = 0 ri (τ )dτ , where ri (t) is continuous and positive for all t ≥ 0.
Suppose that (c∗ (t), a ∗ (t)) is a solution to (12) such that c∗ (t) > 0 and a ∗ (t) > 0 for
all t ≥ 0. Now, fix any T > 0, and let V be the set of all pairs of continuous functions(x, y) : [0, T ] → R2 such that x(0) = x(T ) = y(0) = y(T ) = 0. Choose any
TT(x(t), y(t)) ∈ V such that 0 e−R1 (t ) x(t)dt = 0 and 0 e−R2 (t ) y(t)dt = 0. Let Te−ρt u(c∗ (t) + sx(t), a ∗ (t) + sy(t))dt,g(s) =0∂u ∗∂u ∗(c (t), a ∗ (t)) = uc (t),(c (t), a ∗ (t)) = ua (t).
∂c∂aThen, by the same arguments as in the proof of Theorem 3, we have that T
0 = g (0) =e−ρt [uc (t)x(t) + ua (t)y(t)]dt.
0Define T

e−ρt [uc (t)x(t) + ua (t)y(t)]dt,
 T
1 (x(t), y(t)) =e−R1 (t )x(t)dt,0
 Te−R2 (t ) y(t)dt.
2 (x(t), y(t)) =0 (x(t), y(t)) =00Then, Lemma 3 implies that 0 = a1 1 + a2 2 . If y(t) ≡ 0, then T
 Te−ρt uc (t)x(t)dt =a1 e−R1 (t ) x(t)dt00

for all continuous function x : [0, T ] → R such that x(0) = x(T ) = 0. By Lemma
2, we have that e−ρt uc (t) = a1 e−R1 (t ) for all t. By the same reason, we obtain thate−ρt ua (t) = a2 e−R2 (t ) . To differentiate these equations, we obtain thatuc (t) = (ρ − r1 (t))uc (t), ua (t) = (ρ − r2 (t))ua (t),which is the Euler equation in this model.

38

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