homework6 2010 solutions - 2 Let X d be a complete metric...

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ECN/APEC 7240 Spring 2010 Homework 6 Solutions 1) Let f : R R be a continuously differentiable function with a fixed point x *. Suppose that | f ( x *)| < β < 1. a) There exists δ > 0 such that if x R and 0 < | x - x *| < δ then . *) ( ' *) ( ' * *) ( ) ( x f x f x x x f x f - < - - - β Thus by the reverse Triangle Inequality, , *) ( ' *) ( ' * *) ( ) ( *) ( ' * *) ( ) ( x f x f x x x f x f x f x x x f x f - < - - - - - - β so . * *) ( ) ( * *) ( ) ( x x x f x f x x x f x f - < - < - - β β Note also that 0 = | f ( x *) - f ( x *)| < β | x * - x *| = 0. Thus for all x V = ( x * - δ , x * + δ ), | f ( x ) - f ( x *)| < β | x - x *|. Furthermore, since f ( x *) = x *, , * * ) ( βδ β < - < - x x x x f so f ( x ) V . Now suppose f ( x *) 0. Then since f is continuous and | f ( x *)| < β there exists ε (0, δ ] such that | f ( x )| < β for all x U = ( x * - ε , x * + ε ). Let x , y U such that x < y . By the Mean-Value Theorem, there exists z ( x , y ) such that ). ( ' ) ( ) ( z f x y x f y f = - - Thus . ) ( ' ) ( ) ( x y x y z f x f y f - < - = - β Therefore f is a contraction with modulus β .
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ECN/APEC 7240 Spring 2010
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Unformatted text preview: 2) Let ( X , d ) be a complete metric space and T be a contraction mapping on X with modulus β ∈ (0, 1). Let x * be the unique fixed point of T . Let x ∈ X , and let n ∈ ϖ . Then by the Triangle Inequality, *). , ( ) , ( *) , ( 1 1 x x T d x T x T d x x T d n n n n + + + ≤ Since T is a contraction, we also have *). , ( *) , ( *) , ( 1 1 x x T d Tx x T d x x T d n n n ≤ = + + Thus, ). , ( 1 1 *) , ( ) , ( *) , ( ) 1 ( *). , ( ) , ( *) , ( 1 1 1 x T x T d x x T d x T x T d x x T d x x T d x T x T d x x T d n n n n n n n n n n + + +-≤ ≤-+ ≤...
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