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 > 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 > 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) 274