Math630-9 - Real Analysis - Math 630 Homework Set #9 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Real Analysis - Math 630 Homework Set #9 - Chapters 5 and 6 by Bobby Rohde 11-10-00 Problem 12 Let f be absolutely continuous (AC) in the interval , 1 , > 0. Does the continuity of f at 0 imply that f is AC on 0, 1 ? What if f is also a bounded variation (BV) on 0, 1 ? Proof Does the continuity of f at 0 imply that f is AC on 0, 1 ? 1 No. Consider for example the function f x2 sin x2 . Over the interval , 1 , we know that f is x 1 2 1 2 x sin x2 expressable as f x f t t f , where f x cos x2 , which is welldefined over , 1 as is f . Thus by Theorem 14, f is AC over , 1 . However we know from the result of Problem 10a, that f is not BV on 0, 1 and hence by the contrapositive of Lemma 11, we know that f is not AC on 0, 1 . What if f is also a bounded variation (BV) on 0, 1 ? Yes, this suffices. Consider an f that is BV on 0, 1 , AC on , 1 > 0 and continuous at 0. We n essentially need to show that i 1 f xi f xi 0 over disjoint intervals xi , xi 0, as 0, as this would ensure that any additional intervals added to to an existing AC description could be bounded by choosing sufficiently smaller . Given that f is BV we then have that over any partition n 1 f xi f xi 1 . Also since it's i continuous at x 0, we know that limx 0 f x f 0 0. Since by Problem 8 we know that 0 b c b , with c Ta Ta Tc a, b , this says that, T1 T1 T 0 . And the above statement gives that n T 0 0, as 0. Hence i 1 f xi f xi 1 , over inteverals 0, goes to 0. Hence giving n that i 1 f xi f xi 0 as desired. So f is AC. QED. MATH630-9.nb 2 Problem 20 A function f is said to satisfy a Lipschitz condition (LC) on an interval if a constant M such that f x fy M x y , x and y in the interval. a) Show that a function satisfying a LC is AC. Proof Let f be an LC function with constant M. If M = 0 then f is constant so f is trivially AC. Thus we may assume M > 0. Then finite collections of non-overlapping intervals xi , xi we know that f xi f xi n n xi , by the LC on each term of the sum. Given > 0. If we take = i 1 xi i1 M f xi f xi n n n xi f xi . QED. i 1 f xi M , then > i 1 xi M= > i1 M b) Show that an AC function f satisfies a LC iff f Proof Suppose f x exists. Then we know that f x f Hence f limy x f yy x x . If f is LC then f x if f is LC then f If f fy M. M x y D fx limh 0 fxh fx h is bounded. = limy fy fx yx x . f limy x fy fx yx limy x Myx yx = M. f f M , then M f x = limy x f yy x x = limy x f yy x x , since fy fx 0, 0, such that 0 < y x < yx fy fx fx yx y x . Applyng the triangle inequality sided and rearranging terms we have that f y f x yx fy fx yx M . f f to f x exists. x , the left hand x . This expression is true for all y , such that 0 yx , however our choice of x is arbitrary so that there must exist a associated with any x which makes the above true. Further since yx y a a x, a x, y we know that f y f x yx M must hold for all x, y. Finally we may take the limit as 0. This gives that f y fx y x M . QED MATH630-9.nb 3 Problem 1 Show that Proof By definition inf N : m t : g t f g f N f = ess 0 , and g sup | f t | = inf M : m t : f t M 0, fg = inf L : m t : f t g t L 0. g = This last gives inf L : m t : f t g t L0 inf L : m t : f t gt L0 inf M N : m t : f t M t: g t N 0 , since this last union clearly includes all points with f t gt M N , as well as many others as well. M t: g t N 0 However, inf M N : m t : f t inf M N : m t : f t M m t: g t N 0 , by subadditivity. Further though we may require that each set have measure zero since measure is non-negative. Finally this allows us to M 0+ break it into two independant infimums yielding. inf M : m t : f t inf N : m t : g t N 0 which gives us that f g f g . QED. Problem 2 Let f be a bounded measurable function on 0, 1 . Then fp f lim p Proof f p 1 0 p 1 p = 1 p f . Clearly if we replace f f M 1 p with M = ess sup f we have f , 1 0 1 0 f p 1 p 1 p 0M =M= 0, 1 : f . Hence and D f p p. f p 1 p Take = E > 0. f p D Define f p 1 p E x 0, 1 1 p E. p p mE M M mE . We know that mE 0 since M is the ess sup of f. If Df 1 mE = 0 then there would be a lower essential sup M . However 0, p 1 as p . 1 So in the limit as p ,M mE p M . However since can be taken arbitratily small. We thus have that lim p f p M= f . Thus since each term in the limit is less than M. It must be the case that lim p fp f QED. MATH630-9.nb 4 Problem 3 Prove that Proof f g 1 1= 0 f f g g 1 f f 1 g 1 0 1 1 0 g= f 1 0 g= f 1 g 1. QED. Problem 4 If f Proof fg L1 and g f g L , then f f g f 1 g f= f g . QED. ess sup g = ess sup g 1 Extra Credit - Problem 5.21 Let g be a monotone increasing AC function on a, b with ga c, g b d. a) Show that for any open set O Proof Since O is an open set we know that it is expressible as the union of a finite collection of open intervals ci , di . Furthermore since g is AC that implies it is continuous, and since it is monotone increasing we know from elementary analysis that the pre-image of an open interval must be an Thus open interval (possiblity degenerate or empty), with endpoints g 1 ci , g 1 di . g 1 c, d , mO = g 1 O gx x O gx n gx x= i1 x t ga. ag t g 1 g 1 di g 1 ci gx x. However since g is AC, we further know that n Hence i 1 g 1 ci g x x = this is equal to n 1 di ci i di n 1d g i i 1 gg n i 1 l ci , di = mO. g 1 ci . By definition of inverse we have that QED. MATH630-9.nb 5 b) Let H x:g x 0 . If E is a subset of c, d with mE = 0, then 1 gE H has measure zero. Proof BWOC Suppose g 1 E H has measure > 0. Then g 1 E H g > 0. Since the integration is preformed over a set of measure > 0 where g is also > 0 everywhere. Since g is monotone increasing we may neglect the absolute value since the derivate must be 0. Further we know that g 1 E H g = g 1 E g , since when integrating over the complement of H we have that g = 0. So since g 1 E H has measure > 0, g 1 E has measure > 0, and thus some open interval , g 1E. gxg g , but g Which contradicts the fact that QED. , g 1 E gg 1E E that g , H g > 0. Hence we have that g 1 E g g 0. H has measure zero. c) If E is a measurable subset of c, d , then F b and mE = F g = a E g x gx x Proof F g 1 E H is measurable g = g 1 E H g . Clearly we may neglect the H, since if we are not in H then g 0 and contributes nothing to the integral. So F g = g 1 E g . We know that an open set O with O E and mE = mO. By part b) we may neglect the points in E ~ O since they are a set of measure 0. So considering g 1 O g and invoking part a) we get F g = mO = mE. But also g 1 g 1 O O g= gx . x= b ag O x E b ag x g 1 O g 1 x O x. x However clearly x x= b ag g 1 O iff g x O x x O gx x. Finally since adding a set of measure 0 does not change the value of the integral we may replace O with E. Leaving b ag x gx F g = mE. QED d) If f is a nonnegative measurable function on c, d , then f ° g g is b b measurable on a, b and a f y y g x x. af gx Proof I'm tired. Good night. ...
View Full Document

Ask a homework question - tutors are online