Economic growth and economic development 284

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Economic growth and economic development 284

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Introduction to Modern Economic Growth Then applying Theorem 6.7, we can conclude that there exists a unique fixed point of T over C [0, s] This fixed point is the unique solution to the differential equation and it is also continuous Exercise 6.4 will ask you to verify some of these steps and also suggest how the result can be extended so that it applies to C [0, s] for any s ∈ R+ The main use of the Contraction Mapping Theorem for us is that it can be applied to any metric space, so in particular to the space of functions Applying it to equation (6.1) will establish the existence of a unique value function V in Problem A2, greatly facilitating the analysis of such dynamic models Naturally, for this we have to prove that the recursion in (6.1) defines a contraction mapping We will see below that this is often straightforward Before doing this, let us consider another useful result Recall that if (S, d) is a complete metric space and S is a closed subset of S, then (S , d) is also a complete metric space Theorem 6.8 (Applications of Contraction Mappings) Let (S, d) be a complete metric space, T : S → S be a contraction mapping with T zˆ = zˆ (1) If S is a closed subset of S, and T (S ) ⊂ S , then zˆ ∈ S (2) Moreover, if T (S ) ⊂ S 00 ⊂ S , then zˆ ∈ S 00 Proof Take z0 ∈ S , and construct the sequence {T n z0 }∞ n=0 Each element of this sequence is in S by the fact that T (S ) ⊂ S Theorem 6.7 implies that T n z0 → zˆ Since S is closed, zˆ ∈ S , proving part in the theorem We know that zˆ ∈ S Then the fact that T (S ) ⊂ S 00 ⊂ S implies that zˆ = T zˆ ∈ T (S ) S 00 , establishing part Ô The second part of this theorem is very important to prove results such as strict concavity or that a function is strictly increasing This is because the set of strictly concave functions or the set of the strictly increasing functions are not closed (and complete) Therefore, we cannot apply the Contraction Mapping Theorem to these spaces of functions The second part of this theorem enables us to circumvent this problem 270

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