Introduction to Mathematical Analysis I - Second Edition

154 Solutions and Hints for Selected Exercises (b) Apply the Extreme Value Theorem for the function g ( x ) = f ( x ) x on the interval [ a , b ] . Exercise 3.4.8 . First consider the case where f is monotone decreasing on [ 0 , 1 ] . By Exercise 3.4.5 , f has a fixed point in [ 0 , 1 ] , which means that there exists x 0 ∈ [ 0 , 1 ] such that f ( x 0 ) = x 0 . Since f is monotone decreasing, f has a unique fixed point. Indeed, suppose that there exists x 1 ∈ [ 0 , 1 ] such that f ( x 1 ) = x 1 . If x 1 < x 0 , then x 1 = f ( x 1 ) ≥ f ( x 0 ) = x 0 , which yields a contradiction. It is similar for the case where x 1 > x 0 . Therefore, x 0 is the unique point in [ 0 , 1 ] such that f ( x 0 ) = x 0 . Since f ( g ( x )) = g ( f ( x )) for all x ∈ [ 0 , 1 ] , we have f ( g ( x 0 )) = g ( f ( x 0 )) = g ( x 0 ) . Thus, g ( x 0 ) is also a fixed point of f and, hence, g ( x 0 ) = x 0 = f ( x 0 ) . The proof is complete in this case. Consider the case where f is monotone increasing. In this case, f could have several fixed points on [ 0 , 1 ] , so the previous argument does not work. However, by Exercise 3.4.5 , there exists c ∈ [ 0 , 1 ] such that g ( c ) = c . Define the sequence { x n } as follows: x 1 = c , x n + 1 = f ( x n ) for all n ≥ 1 . Since f is monotone increasing, { x n } is a monotone sequence. In fact, if x 1 ≤ x 2 , then { x n } is monotone increasing; if x 1 ≥ x 2 , then { x n } is monotone decreasing. Since f is bounded, by the monotone convergence theorem (Theorem 2.3.1 ) , there exists x 0 ∈ [ 0 , 1 ] such that lim n → ∞ x n = x 0 . Since f is continuous and x n + 1 = f ( x n ) for all n ∈ N , taking limits we have f ( x 0 ) = x 0 . We can prove by induction that g ( x n ) = x n for all n ∈ N . Then g ( x 0 ) = lim n → ∞ g ( x n ) = lim x n = x 0 . Therefore, f ( x 0 ) = g ( x 0 ) = x 0 . SECTION 3.5 Exercise 3.5.2 . (a) Let f : D → R . From Theorem 3.5.3 we see that if there exist two sequences { x n } and { y n } in D such that | x n − y n | → 0 as n → ∞ , but {| f ( x n ) − f ( y n ) |} does not converge to 0, then f is not uniformly continuous on D . Roughly speaking, in order for f to be uniformly continuous on D , if x and y are close to each other, then f ( x ) and f ( y ) must be close to each other. The behavior of the graph of the squaring function suggests the argument below to show that f ( x ) = x 2 is not uniformly continuous on R . Define two sequences { x n } and { y n } as follows: x n = n and y n = n + 1 n for n ∈ N . Then | x n − y n | = 1 n → 0 as n → ∞ . However, | f ( x n ) − f ( y n ) | = n + 1 n 2 − n 2 = 2 + 1 n 2 ≥ 2 for all n ∈ N .