18 100B Fall 2002 Homework 8 solutions Was due by Noon Tuesday November 19 Rudin 1 Chapter 6 Problem 5 Solution 1 No it is not true that a bounded function f on a b with f 2 R is necessarily in R itself We need a counterexample to see this Take the function f 1 at rational points and f 1 at irrational points This is not integrable by the preceeding question the di erence between upper and lower sums is always 2 b a On the other hand f 2 1 2 If f is real valued and bounded and f 3 R then f R as follows from Theorem 6 11 with t t1 3 the unique real cube root 2 Chapter 6 Problem 7 Solution a If f R on 0 1 then 1 1 f x dx f x dx c 0 c f x dx 0 1 1 c and if f M then 0 f x dx 2M c so c f x dx 0 f x dx as 1 c 0 It is enough to say that c f x dx depends continuously on c by Theorem 6 20 b Consider g x x 3 2 x 0 and g 0 0 This is de nitely not integrable since it is not bounded Moverover the integral over c 1 does not converge since 1 1 x 3 2 dx 2 1 c 2 c as c 0 Now consider the function 1 1 x 3 2 2k x 2k 1 1 k f x 1 1 3 2 x 2k 1 x 2k For any c 0 this function is integrable on c 1 since it is bounded and has only a nite number of points of discontinuity The integral over any 1 1 1 1 1 of the intervals 2k 2k 1 is 2 2k 1 2 2k 2 and over 2k 1 2k is 1 1 1 2 2k 2 2k 1 2 Both of these are bounded in absolute value by Ck 2 1 Combining the two integrals shows that the integral over 2k 1 2k 1 is 1 1 1 2 2 2k 2 2k 1 2 2k 1 2 Ck 3 2 by Taylor s theorem applied 1 1 to x 0 for 2 1 x 2 1 x 2 with x 1 2k Thus if N is the largest integer such that 2N c then 1 1 1 f dx f dx CN 2 0 as N 1 2N 1 c and 1 1 2N 1 f dx N k 1 1 1 2k 1 1 2N 1 f dx 2 converges by comparison to k 1 k 3 2 This shows that 1 c f dx con verges as c 0 Note that if f is bounded and integrable on c 1 for every c 0 then it is integrable on 0 1 so you cannot do this with a bounded function 3 Chapter 6 Problem 10 a b and c 1 Proof a If u 0 or v 0 this is obvious so we can assume that both are q positive Since p and q are both positive and p q 1 both of them must lie in the interval 1 p Now divide through the inequality we want by v q and set a up v q It follows that uv 1 q a1 p since q p q 1 Thus we only need to show that 1 1 a1 p a 0 a p q The continuous function p1 a 1q a1 p is positive at 0 and tends to as a Thus if it has an interior minimum in 0 it will have to be at a point where the derivative vanishes namely p1 p1 a1 p 1 which is to say a 1 Since it takes the value 0 there it is in fact non negative meaning 1 holds This proves the inequality uv up vq p q with equality only where a 1 which is up v q including the case where both are zero b If 0 f R and 0 g R then f p and g q R by Theorem 6 11 It also follows that f g R and using b 1 b p 1 b q f gd f d g d 1 p a q a a c If f and g are complex valued in R then f and g are nonnegative elements of R and f g R Moverover b b f gd f g d b a b a If I a f p 0 and J a g q 0 then apply the conclusion of the previous part to f c and g d where cp I and dq J This gives the desired result 1 p 1 q b b b p q g d f gd cd f d a a a On the other hand if one of these intgrals vanishes say the rst since we can always reverse the roles of p and q then b 1 b q f c g d cq g d q a a b for any c 0 and sending c 0 shows that a f g d 0 so the inequality still holds 3 4 Chapter 6 Problem 11 Setting p q 2 in the previous problem we see that b b b 2 2 u d v 2 d uv d a a a Now multiply out b b b u v 2 d u 2 d u v uv d a a a a b v d 2 a b 21 2 u d a b 12 2 v d 2 This means u v 2 u 2 v 2 Now setting u f g and v h g gives the general case