Math630-8 - Real Analysis Math 630 Homework Set#8 Chapter 5...

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Unformatted text preview: Real Analysis - Math 630 Homework Set #8 - Chapter 5 by Bobby Rohde 11-03-00 Problem 7 a) Let f be of bounded variation (BV) on a, b . Show that for each c a, b the limit of f x exists as x c and also as x c . Prove that a monotone function (and hence a function of BV) can have only a countable number of discountinuities. Proof Lemma: For a monotone BV function g on a, b , the number of discontinuies such that 1 0, g x gx ,n , is finite. n BWOC Suppose this number is infinite. Since it is a BV, we know that g is strictly real valued. However we can start at x a, and proceed towards b such that each discontinuity we choose succifiently small that infintely many disconinuties between the current location and b. Thus we experience arbitrarily many increases of at least 1 , which since g is monotonic implies that n gb , which is a contradiction. Hence the number is finite. The number of discontinuities of g as above is at most countable. 1 Let An x g x gx > 0}. It is clear that any discontinuity of a monotone n, function must be in some An for n sufficiently large. So the set of all discontinuities is n 1 An , which must be countable since it is a countable union of finite sets. Thus the number of discontinuities is countable. MATH630-8.nb 2 The limit of f x exists as x c and x c Consider a fixed c. Suppose > 0 such that f over c , c , contains no discontinuties, then f is continuous in that region and limit of f x as x c exists. Thus we may assume that > 0 a discontuinity in f over the region in question. We need to show that given > 0, > 0, such that x with 0 c x fx fc . However we know that f is BV and thus the sum of two monotone functions g, h. Thus by the first lemma there exist only finitely many points with g x g c n1 . So we may take sufficiently small such that all points of differance > 2 are more than away. Similarly for h. Since they contribute with opposite sign, f x f c , were is the minimum of the two 's needed. Hence limit x c exists. The proof however clearly proceeds likewise for x c . QED b) Construct a monotone function on a, b which is discontinuous at each rational point. Construction Let the sequence fr x Let f x r, x 0, x r . r x 1 2n ri i 1, be an enumeration of the rationals in the interval 0, 1 . Define i 1 fri . Claim that f is such a function as required. 1 1 1. Thus since the lefthand side is a monotonically increasing Clearly i 1 fri x 2n i 1 2n bounded series it must converge to some number less than infinity. Furthermore it should be clear that x, y 0, 1 with x y, we have that some q x, y , since rationals are dense. So 1 fy fx fq y 2N , for some N since all terms for which fri x 0 yields fri x fri y , and fq x 0 fq y . Thus f x is monotonicly increasing. Furthermore p , and h > 0. f p h f p h f p p h 21N , using the above property. However since p implies that f p p h fp p p. Thus h > 0, p p fp h fp h 0. Which implies that f p h f p h 0, 2N 2N , so limit h p limx p f x limx p f x 2N . Hence f x is discontinuous at p since the limits can not be equal. Thus f satisfies all the conditions desired. QED. MATH630-8.nb 3 Problem 8 a) Show that if a Proof b Ta c Ta b Tc c b b, then Ta c Ta b c Tc and that hence Ta b Ta . b By definition Ta sup t, where sup is taken over all possible partitions of a, b . k t f xi 1 . Then either c xi , for some i, or c xi , xi 1 , for some i. i 1 f xi However If c xi , then clearly we may split the partition over a, b into two partitions, one over a, c and b c b the other over c, b , such that ta is equal to ta + tc , for those partitions. xi , xi 1 then consider the partition a x0 x1 ... xi c xi 1 ... xk b. In this If c partition t2 is to the original t t1 , since f xi 1 f xi f xi 1 f c f c f xi f xi 1 fc f c f xi , by triangle inequality. If we break the new partition as above we have two partitions about a, c and c, b , respectively with t taken over these partitions summed = c b b t1 . Hence Ta Tc Ta . t2 b Ta c Ta b Tc Every partition p1 , p2 over a, c and c, b respectively can be used to form a partition over a, b b by joining the two at c. However Ta = sup t over all possible partitions, so this includes the b c partition formed by the union of p1 and p2 and hence Ta Ta Tcb . b Thus Ta c Ta b c Tc . Showing that Ta b Ta . b b) Show that Ta f g b Ta f b b Ta g , and Ta c f c b Ta f . Proof b Ta f g b Ta f b Ta g b Ta f g sup t f g , where k g xi f xi 1 g xi 1 i 1 f xi inequality, however this last term is just t f g sup t f k i tf g 1 f xi tg. g f xi k i1 1 f g xi f g xi g xi 1 g xi 1 = by the triangle Thus sup t f b t g , hence Ta f b Ta f b Ta g . QED MATH630-8.nb 4 b Ta c f c k i1 b Ta f b Ta c f =c sup t c f , where t c f f xi f xi 1 = c t f . sup c tf c k i1 c f xi c f xi 1 = k i1 c f xi f xi 1 Thus sup t c f b sup t f , hence Ta c f c b Ta f . QED Problem 10 a) Let f be defined by f 0 on 1, 1 ? Answer 0 and f x x2 sin 1 x2 , for x 0. Is f of BV 1 1 1 No. Over any interval a, b , the diffence from x2 sin x2 , will be = a2 sin a2 b2 sin b2 . 1 Suppose a is a minimum, thus a2 n 2 , for some integer n. Thus an adjacent maximum at 1 1 1 1 n So a2 sin a2 b2 sin b2 = n1 1 1= b such that b2 2. n 2 2n 1 2 2n 1 2 2 n = 4 n28 1 . However as the number of partitions gets arbitrarily large we may add arbitrarily many terms of this form with n increasing. So 8n 1 0f T1 , since the last series is a constant times the n 1 4 n2 1 n 12 n harmonic series. Therefore f is not a BV. MATH630-8.nb 5 b) Let g be defined by g 0 on 1, 1 ? Answer 0 and g x x2 sin 1 x , for x 0. Is g of BV Yes. Since g x is symmetric in x we know that it is suffiecent to show that g x is BV on 1, 0 . Consider a single transition for a minimum to an adjacent maximum. It is clear that it will proceed monotonically in this region. On inspection it is clear that t over a monotonic region will be independant of the choice of partition and have the value equal to the differance of the end points. (Either p or n is 0). b From problem 8a) and an obvious induction it is clear that Ta may be broken into finitely many pieces. We know that minima and maxima occur at sin T 1< 2 2 1 x 1 x= n 1 2 . It is obvious that b , purely by inspection and our knowledge of basic calculus. Proceed by breaking Ta into m pieces such that the first piece is T 1 , and the proceeding m-2 pieces follow for decreasing values of n, and the last piece represents the tail going to 0. Now we need to evaluate T over one these pieces that have been set up to run from a local max to a local min or vice versa. So the points in question have the form n 1 , n 1 1 . So n n 1 2 2 sin n 2 n1 2 2 2 2 2 2n 2 2 2 < n1 2 n2 1 4 n 12 n 2 1 2 2 2 sin n 1 2 n2 n2 1 4 4 2 = n 1 2 2 2 n1 1 2 2 2 = < , note that the inequalities depend on the fact that n < 0. Now we want to consider the total T 01 covered in m steps. Taking limit m the n 1 2 n2 n2 1 4 4 T 1+ 2 n m2 1 , the tail contribution must go to 0. So we only need that n1 2 n2 1 4 2 n4 2 n2 n2 1 4 4 0 0 Ttail, where Ttail is the part not < . However this may be rewritten as . However this is 3 bounded above by n2 for n large, which is a convergent sequence. Thus our sum must converge. Hence T 01 . So T 11 x2 sin 1 is BV. QED. x MATH630-8.nb 6 Problem 14 a) Show that the sum and differance of two AC functions are also AC. Proof Let f and g be AC. Let > 0. Then n g xi 2 finite collection i 1 g xi n xi min 1 , 2 . i 1 xi n However i 1 f xi inequlaity. Further n i1 1, 2 of such that intervals n i1 f xi xi, xi , f xi such 2 and that g xi f xi min f xi f xi 1, 2 g xi g xi f xi g xi < 2 n i1 f xi 2= . g xi g xi , by triangle Hence we may choose in order to satisfy the condition and show that f = f xi f xi g is AC. Since f xi f xi = f xi f xi that -f is AC. So f g must also be AC. , we know that f is AC implies b) Show that the product of two AC functions is AC Proof Lemma: f is AC implies that f 2 is AC. Let f be AC with respect to a, b . Then we know that f is BV, and hence that since a, b is a closed interval, that f is bounded. Let M > 0 be a bound on f . Let > 0. Since f is AC, >0 n such that f xi 2 M , finite collection of intervals xi , xi , such that i 1 f xi n xi . i 1 xi Consider n i 1 f xi Thus f 2 n i1 f xi f xi 2 f xi f xi f xi 2 n i1 = f xi n i1 f xi f xi f xi 2M < 2M f xi f xi 2M = . is AC. Concluding the Lemma. MATH630-8.nb 7 If f , g are AC over a, b then f g is AC. fg f g fg , but that both sums, differances and squares of AC functions are AC, so we 2 only need that constant multiples of AC functions are AC. But for c f , where f is AC it follows immediately that for any > 0 we may choose f constrained by c , and the c will carry through to make c f , AC. (Note c = 0 is obviously true.) 2 2 2 Hence f g can be gotten as a sequence of operations that perserve AC, so it is also AC. QED. c) If f is AC on a, b and if f is never zero there, then the function g also AC on a, b . Proof Since f is never 0 over a closed interval, we know that m > 0 such that f n > 0. Since f is AC, > 0, such that i 1 f xi f xi m2 . Consider Thus g 1 f n i1 1 f xi 1 f xi 1 f is m everywhere. Let = n i1 f xi f xi f xi f xi n i1 f xi f xi m2 < m2 m2 =. is AC. QED. ...
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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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