18 100B Fall 2002 Homework 9 Due by Noon Tuesday November 26 Rudin 1 Chapter 6 Problem 12 Proof Suppose that f R let C 0 be such that f x C for all x a b Given 0 there exists a partition P of a b such that 1 U f P L f P n xi xi 1 Mi mi 2 2C i 1 2 where Mi and mi are the supremum and in mum of f over xi 1 xi Consider the function given in the hint xi t t xi 1 g t f xi 1 f xi t xi 1 xi xi xi 1 xi xi 1 Note that the value at t xi is independent of choice even if there are two intervals of which it is an end point On xi 1 xi g is continuous since it is linear there and it is continuous at each xi hence is continuous everywhere On xi 1 xi g takes values in mi Mi since its maximum and minimum occur at the ends it is linear and these are values of f Since f takes values in the same interval it follows that f g takes values in mi Mi Mi mi Thus f x g x 2 Mi mi 2 2C Mi mi on xi 1 xi Estimating the integral on each segment of the partition we see that f x g 2 d 2C xi xi 1 Mi mi 2 i I which implies that f g 2 2 Chapter 6 Problem 15 Solution By assumption f is real and continuously di erentiable on a b hence so is F x xf 2 x This has derivative f 2 x 2xf x f x so by the fundamental theorem of calculus b f 2 x 2xf x f x dx F b F a 0 a since f a f b 0 Thus b 1 b 2 1 xf x f x dx f x dx 2 a 2 a By Schwarz inequality 2 b b b 1 xf x f x dx f x 2 dx x2 f 2 x dx 4 a a a 3 Chapter 7 Problem 2 1 2 Proof If fn and gn converge uniformly on a set E then they are uniformly Cauchy Hence given 0 there exist N and N such that n m N fn x fm x 2 n m N gn x gm x 2 x E Taking N max N N we see that n m N fn x gn x fm x gm x x E so fn gn is uniformly Cauchy and hence uniformly convergent If both fn and gm are uniformly bounded with fn x gn x M for all x E and all n then fn x gn x fm x gm x fn x gn x fn x gm x fn x gm x fm x gm x M if n m N showing that fn gn is uniformly Cauchy and hence uniformly convergent 4 Chapter 7 Problem 6 Proof We may write the series as the sum of 2 1 n nx2 and 1 n1 n n The second series converges uniformly as a series of functions in x since it converges and does not depend on x The rst series converges uniformly 1 on any bounded interval using Theorem 7 10 and the convergence of n2 n It follows that the sum of the series converges uniformly using the triangle inequality m 1 n n p m m 2 1 x2 n nx 1 1 n 2 n2 n n n p n p 5 Chapter 7 Problem 8 Proof If cn converges then for any m n n m cj I x xj j n m cj x a b j n By Theorem 7 10 it follows that the series converges uniformly on a b Given 0 there exists N such that cj I y xj 3 y a b j N If x xn for any n then it follows that j N cj I y xj is continuous at x so there exists 0 such that x y cj I y xj 3 cj I x xj j N j N 3 Then we see that if x y f x f y cj I y xj cj I x xj j N j N j N Thus f is continuous at x cj I x xj cj I y xj j N