Economic growth and economic development 288

Introduction to Modern Economic Growth
some x (1) ∈ G (x (0)). In view of Assumption 6.1, V ∗ (x (0)) is finite. Moreover,

Assumptions 6.1 and 6.2 also enable us to apply Weierstrass theorem to Problem
A1, thus there exists x ∈Φ (x (0)) attaining V ∗ (x (0)) (see Mathematical Appendix).

A similar reasoning implies that there exists x0 ∈Φ (x (1)) attaining V ∗ (x (1)). Next,

since (x (0) , x0 ) ∈ Φ (x (0)) and V ∗ (x (0)) is the supremum in Problem A1 starting
with x (0), Lemma 6.1 implies

V ∗ (x (0)) ≥ U (x (0) , x (1)) + βV ∗ (x (1)) ,

= U (x (0) , x0 (1)) + βV ∗ (x0 (1)) ,

thus verifying (6.11).
Next, take an arbitrary ε > 0. By (6.10), there exists x0ε = (x (0) , x0ε (1) , x0ε (2) , ...) ∈Φ (x (0))

such that

¯ (x0ε ) ≥ V ∗ (x (0)) − ε.
U
Now since x00ε = (x0ε (1) , x0ε (2) , ...) ∈ Φ (x0ε (1)) and V ∗ (x0ε (1)) is the supremum in

Problem A1 starting with x0ε (1), Lemma 6.1 implies

¯ (x00ε ) ≥ V ∗ (x (0)) − ε
U (x (0) , x0ε (1)) + β U

U (x (0) , x0ε (1)) + βV ∗ (x0ε (1)) ≥ V ∗ (x (0)) − ε,
The last inequality verifies (6.12) since x0ε (1) ∈ G (x (0)) for any ε > 0. This proves

that any solution to Problem A1 satisfies (6.11) and (6.12), and is thus a solution
to Problem A2.
To establish the reverse, note that (6.11) implies that for any x (1) ∈ G (x (0)),
V (x (0)) ≥ U (x (0) , x (1)) + βV (x (1)) .
Now substituting recursively for V (x (1)), V (x (2)), etc., and defining x = (x (0) , x (1) , ...),
we have
V (x (0)) ≥

As n → ∞,

Pn

t=0

n
X

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

t=0

¯ (x) and since V (x) is finite for any x ∈ X,
U (x (t) , x (t + 1)) → U

β n+1 V (x (n + 1)) → 0, we obtain

¯ (x) ,
V (x (0)) ≥ U
for any x ∈Φ (x (0)), thus verifying (6.9).

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