Math630-9

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

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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. ...
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## This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.

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