Introduction to Mathematical Analysis I - 3rd Edition

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 fixed 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 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 {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 n→∞ xn =x0. Since f is continuous andxn+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 n→∞ g(xn) =limxn =x0. 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 Dsuch 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 andy 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 onR. Define two sequences {xn} and {yn} as follows: xn =n and yn =n+ 1 n for n ∈N. Then |xn −yn| = 1 n →0 as n→∞. However, | f (xn)−f (yn)| = n+ 1 n 2 −n2 =2+ 1 n2 ≥ 2 for all n∈N. Therefore, {| f (xn)−f (yn)|}does not converge to 0 and, hence, f is not uniformly continuous on R. In this solution, we can use xn =rn+ 1 n andyn =√nfor n∈Ninstead. (b) Use xn = 1 π/2+2nπ andyn = 1 2nπ ,n∈N. (c) Use xn =1/nandyn =1/(2n).

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