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Introduction to Modern Economic Growth
for this to show how this conclusion can be reached either by looking at Problem
A1 or at Problem A2, and then exploiting their equivalence. The first proof is more
abstract and works directly on the sequence problem, Problem A1.
Proof of Theorem 6.3. (Version 1) Consider Problem A1. The choice set of
this problem Φ (0) is a subset of X ∞ (infinite product of X). From Assumption 6.1,
X is compact. By Tychonof’s Theorem (see Mathematical Appendix), the infinite
product of a sequence of compact sets is compact in the product topology. Since
again by Assumption 6.1, G (x) is compact-valued, the set Φ (x (0)) is bounded. A
bounded subset of a compact set, here X ∞ , is compact. From Assumption 6.2 and
the fact that β < 1, the objective function is continuous in the product topology.
Then from Weierstrass’ Theorem, an optimal path x (0) exists.
Ô
Proof of Theorem 6.3. (Version 2) Consider Problem A2. In view of
Assumptions 6.1 and 6.2, there exists some M < ∞, such that |U(x, y)| < M for
all (x, y) ∈ XG . This immediately implies that |V ∗ (x)| ≤ M/(1 − β), all x ∈ X.
Consequently, V ∗ ∈ C (X), where C (X) denotes the set of continuous functions
defined on X, endowed with the sup norm, kf k = supx∈X |f (x)|. Moreover, all
functions in C (X) are bounded since they are continuous and X is compact.
Over this set, define the operator T
(6.15)
T V (x) = max U(x, y) + βV (y).
y∈G(x)