Economic growth and economic development 289

Introduction to Modern Economic Growth
Next, let ε > 0 be a positive scalar. From (6.12), we have that for any ε0 =
ε (1 − β) > 0, there exists xε (1) ∈G (x (0)) such that
V (x (0)) ≤ U (x (0) , xε (1)) + βV (xε (1)) + ε0 .
Let xε (t) ∈ G (x (t − 1)), with xε (0) = x (0), and define xε ≡ (x (0) , xε (1) , xε (2) , ...).
Again substituting recursively for V (xε (1)), V (xε (2)),..., we obtain
V (x (0)) ≤

n
X

U (xε (t) , xε (t + 1)) + β n+1 V (x (n + 1)) + ε0 + ε0 β + ... + ε0 β n

t=0

¯ (xε ) + ε,
≤ U

P
t
where the last step follows using the definition of ε (in particular that ε = ε0 ∞
t=0 β )
Pn
¯ (xε ). This establishes that
U (xε (t) , xε (t + 1)) → U
and because as n ,
t=0

V (0) satisfies (6.10), and completes the proof.

Ô

In economic problems, we are often interested not in the maximal value of the
program but in the optimal plans that achieve this maximal value. Recall that
the question of whether the optimal path resulting from Problems A1 and A2 are
equivalent was addressed by Theorem 6.2. We now provide a proof of this theorem.
Proof of Theorem 6.2. By hypothesis x∗ ≡ (x (0) , x∗ (1) , x∗ (2) , ...) is a

solution to Problem A1, i.e., it attains the supremum, V ∗ (x (0)) starting from x (0).
Let x∗t ≡ (x∗ (t) , x∗ (t + 1) , ...).

We first show that for any t ≥ 0, x∗t attains the supremum starting from x∗ (t),

so that
(6.13)

¯ ∗t ) = V ∗ (x (t)) .
U(x

The proof is by induction. The base step of induction, for t = 0, is straightforward,
since, by definition, x∗0 = x∗ attains V ∗ (x (0)).
Next suppose that the statement is true for t, so that (6.13) is true for t, and we
will establish it for t + 1. Equation (6.13) implies that
(6.14)

¯ ∗t )
V ∗ (x∗ (t)) = U(x
¯ ∗t+1 ).
= U(x∗ (t) , x∗ (t + 1)) + β U(x
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