18 100B Fall 2002 Homework 10 Due by Noon Thursday December 5 Rudin 1 Chapter 7 Problem 14 Solution There is a function f as described just set 0 t 31 1 f t 3 t 3 13 t 32 2 1 3 t 1 and for instance f 2 t f t for 1 t 2 and then f 2k t f t for all k N k 0 t 0 2 This gives a continuous function Consider 1 x t 2 n f 32n 1 t y t n 1 2 n f 32n t n 1 The nth term in the series for 1 is bounded 2 n f 32n 1 t 2 n By Theorem 7 10 the series converges uniformly Thus x t is continuous by Theorem 7 12 The same argument applies to y t so t x t y t is continuous by Theorem 4 10 Now to see that is surjective follow the hint Each point in 0 1 has a dyadic decomposition x 2 n bn bn 0 or 1 n 1 Indeed one can compute the successive bn n 1 k to arrange that k 2 n bn 2 k Then choose bk 1 0 or 1 depending on whether 0 x n 1 k 0 x the 2 n bn 2 k 1 or 2 k 1 x n 1 a2n 1 be k 2 n bn 2 k Now we let n 1 these numbers for x0 and a2n those for y0 It follows that t0 3 i 1 2ai converges to a number in 0 1 and i 1 that 3k t0 2N 32 ak tk where N is an integer ak 0 or 1 and tk 31 The properties of f show that f 3k t0 f 32 ak tk ak since as ak 0 or 1 then 32 ak tk 0 31 of 23 1 From the de nition of x t and y t it follows that t0 x0 y0 and hence is surjective These points t0 are the ones appearing in the Cantor set as de ned in section 2 44 2 Chapter 7 Problem 18 Solution If fn is a uniformly bounded sequence of functions on a b there exists a constant M such that fn x M for all x a b and all n Since the fn are Riemann integrable the functions x Fn x fn t dt a 1 2 are continuous We show that the sequence Fn is equicontinuous In fact if x x a b then x Fn x Fn x fn t dt M x x x Thus if x x M then Fn x Fn x showing the equicontinuity We also have a uniform bound Fn x b a M As a sequence of uniformly bounded and equicontinuous functions on a compact metric space we see from Theorem 7 25 that Fn has a uniformly convergent subsequence 3 Chapter 7 Problem 24 Solution We de ne a map X C X into the space of bounded continuous functions on X Fixing a point a X let f X p fp fp x d x p d x a By the triangle inequality d x p d x a d a p and d x a d x p d a p shows that fp x d a p Thus fp is a bounded function It is continuous again by the triangle inequality fp x fp y d x p d y p d x a d y a 2d x y Thus fp C X Now consider fp x fq x sup d x p d x a d x q d x a x X sup d x p d x q d p q x X The last inequality follows from the triangle inequality and the fact that fp q fq q d p q It follows that f is a continuous map It is an isometry namely the distance in C X is f g In particular this shows that f is 1 1 Let Y be the closure of f X C X Then we may regard X Y using f As a closed subset of C X Y is complete as a metric space Thus X is isometric to a subset of the complete space Y in which it is dense 4 Chapter 5 Problem 26 Solution We assume that f is di erentiable on a b f a 0 and there is a real number A such that f x A f x for all x a b Following the hint for a chosen x0 a b set M0 sup f x and M1 sup f x with the suprema over a x0 where the second exists because of the assumption It follows that M1 AM0 By the mean value theorem f x f a f y x a for some y a x0 so f x f a f x f y x a M1 xa AM0 x0 a Taking the sup over x a x0 we see that M0 AM0 x0 a Taking x0 a so small that A x0 a 1 we see M0 0 and hence M0 0 Thus f x 0 on a x0 Set z sup x0 f x 0 on a x0 If z b we are nished since f is continuous so f z 0 If z b then we may apply the argument again to nd a contradiction 5 Chapter 5 Problem 27 3 Solution I did this in class in a slightly di erent way so I am just asking you to write down the proof Namely if fi x are two solutions for i 1 2 then we may integrate the equation to see that x fi x c t fi t dt a It follows that the di erence f x f2 x f1 x sati es f a 0 f x x f1 x x f2 x 2 We see that f x A f x where A is the constant in the Lipschitz condition Applying the previous problem we conclude that f 0 1 Di erentiating y 14 x2 y 12 x y 2 shows that it is a solution in 1 2 0 1 Thus both y 0 and y 4 x are both solutions These are not the only solutions since if x0 0 and we de ne 0 x x0 y 1 2 x x x x0 0 4 we get a continuously di erentiable solution by the same argument Con1 versely every solution y 0 of y y 2 is of this form Indeed if y t 0 for some t 0 1 then being continuous it is positive nearby Thus we 1 1 d 2y 2 1 on any interval where can divide by y 2 and conclude that dx 1 y 0 This implies that 2y 2 x x0 for some constant x0 and hence that 2 holds in any interval where y 0 In principle there could be different values of x0 on di erent intervals However y in 2 is increasing so the set on which it …