Economic growth and economic development 287

Introduction to Modern Economic Growth
¯ (x) exists and is finite, we have
Proof. Since under Assumption 6.1 U
∞
X
¯
β t U (x (t) , x (t + 1))
U (x) =
t=0

= U (x (0) , x (1)) + β

∞
X

β s U (x (s + 1) , x (s + 2))

s=0

¯ 0)
= U (x (0) , x (1)) + U(x
Ô

as defined in the lemma.

We start with the proof of Theorem 6.1. Before providing this proof, it is useful
to be more explicit about what it means for V and V ∗ to be solutions to Problems
A1 and A2. Let us start with Problem A1. Using the notation introduced in this
section, we can write that for any x (0) ∈ X,
V ∗ (x (0)) =

sup

¯
U(x).

x∈Φ(x(0))

In view of Assumption 6.1, which ensures that all values are bounded, this immediately implies
¯
V ∗ (x (0)) ≥ U(x)
for all x ∈ Φ(x (0)),

(6.9)

since no other feasible sequence of choices can give higher value than the supremum, V ∗ (x (0)). However, if some v satisfies condition (6.9), so will αv for α > 1.
Therefore, this condition is not sufficient. In addition, we also require that
(6.10)

¯ 0 ) + ε,
for any ε > 0, there exists x0 ∈ Φ(x (0)) s.t. V ∗ (x (0)) ≤ U(x

The conditions for V (·) to be a solution to Problem A2 are similar. For any x (0) ∈

X,

(6.11)

V (x (0)) ≥ U(x (0) , y) + βV (y),

all y ∈ G(x (0)),

and
(6.12)
for any ε > 0, there exists y 0 ∈ G (x (0)) s.t. V (x (0)) ≤ U (x (0) , y 0 ) + βV (y) + ε.
We now have:
Proof of Theorem 6.1. If β = 0, Problems A1 and A2 are identical, thus the
result follows immediately. Suppose that β > 0 and take an arbitrary x (0) ∈ X and
273