ma108-2 - MA108 Homework 2 Seung Woo Shin January 25th 2009...

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Unformatted text preview: MA108 Homework 2 Seung Woo Shin January 25th 2009 1 First, note that by Theorem 13.9, v ( x ) = V x a f is right continuous on [ a,b ]. Fix > 0 and let P = a = x < x 1 < x 2 < ... < x n = b . Then, obviously there is > 0 such that V y x < / 2 n whenever | x- y | < . Now, we construct another partition Q using P as follows. Q = { y ,...,y 2 n } where y 2 k = x k and y 2 k +1 ( y 2 k ,y 2 k + ). Now, construct a continuous f ( x ) such that f ( x ) = sgn( ( y 2 k )- ( y 2 k- 1 )) when x [ y 2 k- 1 ,y 2 k ] and || f || 1 (it is obviously possible). Then, Z b a fd- V ( ,Q ) = 2 n X i =1 ( Z y i y i- 1 fd- V ( ,Q )) = n X k =1 sgn( ( y 2 k )- ( y 2 k- 1 ))( ( y 2 k )- ( y 2 k- 1 )) + n- 1 X k =0 Z y 2 k y 2 k +1 fd- V ( ,Q ) = n X k =1 | ( y 2 k )- ( y 2 k- 1 ) | + n- 1 X k =0 Z y 2 k y 2 k +1 fd- V ( ,Q ) =- n- 1 X k =0 | ( y 2 k +1 )- ( y 2 k ) | + n- 1 X k =0 Z y 2 k y 2 k +1 fd >- n / 2 n- n / 2 n =- (Note that | ( y 2 k +1 )- ( y 2 k ) | < / 2 n since y 2 k +1 ( y 2 k ,y 2 k + ) and that by Theorem 14.16, | R y 2 k y 2 k +1 fd | || f || V y 2 k y 2 k +1 V y 2 k y 2 k +1 < / 2 n ) Thus, we have Z b a fd V ( ,Q )- 1 Since Q P , V ( ,Q ) V ( ,P ). So, finally, Z b a fd V ( ,P )- Now, note that by Corollary 14.13, C [ a,b ] R [ a,b ]. Suppose P = { x ,...,x n } and T = { t 1 ,...,t n } is an appropriate selection of points for P . Then, for any f such that || f || 1, S ( f,P,T ) n X i =1 | ( x i )- ( x i- 1 ) | V b a ( ) So, sup { R b a fd : || f || 1 } V b a ( ) = sup V ( ,P ) sup { R b a fd : || f || 1 } . Hence, V b a ( ) = sup { Z b a fd : || f || 1 } 2 Let = . Ill first show that BV [ c,d ]. Let P be any partition of [ c,d ]. Then, if we let Q = ( P ), V ( ,P ) = V ( ,Q ) since is strictly increasing. So, BV [ c,d ].We want to show that g R [ c,d ]. That is, I R such that > 0, there is a partition P * of [ c,d ] such that | S ( g,P,T )- I | < for all refinements P P * and all selections of points T ....
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ma108-2 - MA108 Homework 2 Seung Woo Shin January 25th 2009...

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