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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 0 is a closed subset of S, then (S 0 , 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 0 is a closed subset of S, and T (S 0 ) ⊂ S 0 , then zˆ ∈ S 0 .
(2) Moreover, if T (S 0 ) ⊂ S 00 ⊂ S 0 , then zˆ ∈ S 00 .
Proof. Take z0 ∈ S 0 , and construct the sequence {T n z0 }∞
n=0 . Each element
of this sequence is in S 0 by the fact that T (S 0 ) ⊂ S 0 . Theorem 6.7 implies that
T n z0 → zˆ. Since S 0 is closed, zˆ ∈ S 0 , proving part 1 in the theorem.
We know that zˆ ∈ S 0 . Then the fact that T (S 0 ) ⊂ S 00 ⊂ S 0 implies that
zˆ = T zˆ ∈ T (S 0 ) ⊂ S 00 , establishing part 2.