Economic growth and economic development 283

Introduction to Modern Economic Growth (Uniqueness) Suppose, to obtain a contradiction, that there exist zˆ, z ∈ S, such that T z = z and T zˆ = zˆ with zˆ 6= z This implies < d (ˆ z , z) = d(T zˆ, T z) ≤ βd(ˆ z , z), which delivers a contradiction in view of the fact that < and establishes uniqueÔ ness The Contraction Mapping Theorem can be used to prove many well-known results The next example and Exercise 6.4 show how it can be used to prove existence of unique solutions to differential equations Exercise 6.5 shows how it can be used to prove the Implicit Function Theorem (recall the Mathematical Appendix) Example 6.3 Consider the following one-dimensional differential equation (6.7) x˙ (t) = f (x (t)) , with a boundary condition x (0) = c ∈ R Suppose that f : R → R is Lipschitz continuous in the sense that it is continuous and also for some M < ∞, it satisfies the following boundedness condition, |f (x00 ) − f (x0 )| ≤ M |x00 − x0 | for all x0 , x00 ∈ R The Contraction Mapping Theorem, Theorem 6.7, can be used to prove the existence of a continuous function x∗ (t) that is the unique solution to this differential equation on any compact interval, in particular on [0, s] for some s ∈ R+ To this, consider the space of continuous functions on [0, s], C [0, s], and define the following operator, T such that for any g ∈ C [0, s], T g (z) = c + Z z f (g (x)) dx Notice that T is a mapping from the space of continuous functions on [0, s] into itself, i.e., T : C [0, s] → C [0, s] Moreover, it can be verified T is a contraction for some s This follows because for any z ∈ [0, s], we have ¯ Z z ¯Z z Z z ¯ ¯ ¯≤ ¯ f (g (x)) dx − f (˜ g (x)) dx M |g (x) − g˜ (x)| dx (6.8) ¯ ¯ 0 by the Lipschitz continuity of f (·) This implies that kT g (z) − T g˜ (z)k ≤ M × s × kg − g˜k , where recall that k·k denotes the sup norm, now defined over the space of functions Choosing s < 1/M establishes that for s sufficiently small, T is indeed a contraction 269