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Unformatted text preview: Functions of Bounded Variation and Rectifiable Curves Functions of bounded variation 6.1 Determine which of the follwoing functions are of bounded variation on 0,1 . (a) f x x 2 sin 1/ x if x 0, f 0. (b) f x x sin 1/ x if x 0, f 0. Proof : (a) Since f x 2 x sin 1/ x cos 1/ x for x 0,1 and f 0, we know that f x is bounded on 0,1 , in fact,  f x  3 on 0,1 . Hence, f is of bounded variation on 0,1 . (b) First, we choose n 1 be an even integer so that 1 2 n 1 1, and thus consider a partition P x , x 1 1 2 , x 2 1 2 2 ,..., x n 1 n 2 , x n 1 1 n 1 2 , x n 2 1 , then we have k 1 n 2  f k  2 2 k 1 n 1/ k . Since 1/ k diverges to , we know that f is not of bounded variation on 0,1 . 6.2 A function f , defined on a , b , is said to satisfy a uniform Lipschitz condition of order 0 on a , b if there exists a constant M 0 such that  f x f y  M  x y  for all x and y in a , b . (Compare with Exercise 5.1.) (a) If f is such a function, show that 1 implies f is constant on a , b , whereas 1 implies f is of bounded variation a , b . Proof : As 1, we consider, for x y , where x , y a , b ,  f x f y   x y  M  x y  1 . Hence, f x exists on a , b , and we have f x 0 on a , b . So, we know that f is constant. As 1, consider any partition P a x , x 1 ,..., x n b , we have k 1 n  f k  M k 1 n  x k 1 x k  M b a . That is, f is of bounded variation on a , b . (b) Give an example of a function f satisfying a uniform Lipschitz condition of order 1 on a , b such that f is not of bounded variation on a , b . Proof : First, note that x satisfies uniform Lipschitz condition of order , where 1. Choosing 1 such that 1 and let M k 1 1 k since the series converges. So, we have 1 1 M k 1 1 k . Define a function f as follows. We partition 0,1 into infinitely many subsintervals. Consider x 0, x 1 x 1 M 1 1 , x 2 x 1 1 M 1 2 ,..., x n x n 1 1 M 1 n ,.... And in every subinterval x i , x i 1 , where i 0,1,...., we define f x x x i x i 1 2 , then f is a continuous function and is not bounded variation on 0,1 since k 1 1 2 M 1 k diverges....
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 Fall '09
 Frade

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