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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
0 < d (ˆ
z , z) = d(T zˆ, T z) ≤ βd(ˆ
z , z),
which delivers a contradiction in view of the fact that β < 1 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 do this, consider
the space of continuous functions on [0, s], C [0, s], and define the following operator,