164 Solutions and Hints for Selected Exercises Since f (g(x)) = g( f (x)) for all x ∈ [0,1], we have f (g(x0)) = g( f (x0)) = g(x0). Thus, g(x0) is also a fxed point of f and, hence, g(x0)= x0 = f (x0). The proof is complete in this case. Consider the case where f is monotone increasing. In this case, f could have several fxed 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. Defne the sequence {xn} as follows: x1 = c, xn+1 = f (xn) for all n ≥ 1. Since f is monotone increasing, {xn} is a monotone sequence. In fact, if x1 ≤ x2, then {xn} is monotone increasing; if x1 ≥ x2, then {xn} is monotone decreasing. Since f is bounded, by the monotone convergence theorem (Theorem 2.3.1), there exists x0 ∈ [0,1] such that lim xn = x0. n→∞ Since f is continuous and xn+1 = f (xn) for all n ∈ N, taking limits we have f (x0)= x0. We can prove by induction that g(xn)= xn for all n ∈ N. Then g(x0)= lim g(xn)= limxn = x0. n→∞ Therefore, f (x0)= g(x0)= x0. 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 {xn} and {yn} in D such that |xn − yn|→ 0 as n → ∞, but {| f (xn) − f (yn)|} 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)= x2 is not uniformly continuous on R. 1 Defne two sequences {xn} and {yn} as follows: xn = n and yn = n + for n ∈ N. Then n 1 |xn − yn| = → 0 as n → ∞. However, n 2 1 1 2 | f (x n) − f (yn)| = n + − n = 2 + n2 ≥ 2 for all n ∈ N. n Therefore, {| f (xn) − f (yn)|} does not converge to 0 and, hence, f is not uniformly continuous on R. r 1 √ In this solution, we can use xn = n+ and yn = n for n ∈ N instead. n (b) Use xn = 1 and yn = 1 π ,n ∈ N. π/2+ 2nπ 2n (c) Use xn = 1/n and yn = 1/(2n).
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