chap4 - 48 BRUCE K. DRIVER 4. The Riemann Integral In this...

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48 BRUCE K. DRIVER 4. The Riemann Integral In this short chapter, the Riemann integral for Banach space valued functions is de f ned and developed. Our exposition will be brief, since the Lebesgue integral and the Bochner Lebesgue integral will subsume the content of this chapter. The following simple “Bounded Linear Transformation” theorem will often be used here andintheseque ltode f ne linear transformations. Theorem 4.1 (B .L .T .Th eo r em ) . Suppose that Z is a normed space, X is a Banach space, and S Z is a dense linear subspace of Z. If T : S X is a bounded linear transformation (i.e. there exists C< such that k Tz k C k z k for all z S ) , then T hasaun iqueex tens iontoane lemen t ¯ T L ( Z, X ) and this extension still satis f es ° ° ¯ Tz ° ° C k z k for all z ¯ S . Exercise 4.1. Prove Theorem 4.1. For the remainder of the chapter, let [ a, b ] be a f xed compact interval and X be a Banach space. The collection S = S ([ a, b ] ,X ) of step functions, f :[ a, b ] X, consists of those functions f which may be written in the form (4.1) f ( t )= x 0 1 [ a,t 1 ] ( t )+ n 1 X i =1 x i 1 ( t i ,t i +1 ] ( t ) , where π { a = t 0 <t 1 < ··· <t n = b } is a partition of [ a, b ] and x i X. For f as in Eq. (4.1), let (4.2) I ( f ) n 1 X i =0 ( t i +1 t i ) x i X. Exercise 4.2. Show that I ( f ) is well de f ned, independent of how f is represented as a step function. ( Hint: show that adding a point to a partition π of [ a, b ] does not change the right side of Eq. (4.2).) Also verify that I : S X is a linear operator. Proposition 4.2 (Riemann Integral) . The linear function I : S X extends uniquely to a continuous linear operator ¯ I from ¯ S (the closure of the step functions inside of c ([ a, b ] ,X )) to X and this operator satis f es, (4.3) k ¯ I ( f ) k ( b a ) k f k for all f ¯ S . Furthermore, C ([ a, b ] ,X ) ¯ S c ([ a, b ] ,X ) and for f , ¯ I ( f ) may be computed as (4.4) ¯ I ( f )= l im | π | 0 n 1 X i =0 f ( ± π i )( t i +1 t i ) where π { a = t 0 <t 1 < ··· <t n = b } denotes a partition of [ a, b ] , | π | =max {| t i +1 t i | : i =0 ,...,n 1 } is the mesh size of π and ± π i may be chosen arbitrarily inside [ t i ,t i +1 ] . Proof. Taking the norm of Eq. (4.2) and using the triangle inequality shows, (4.5) k I ( f ) k n 1 X i =0 ( t i +1 t i ) k x i k n 1 X i =0 ( t i +1 t i ) k
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ANALYSIS TOOLS WITH APPLICATIONS 49 The existence of ¯ I satisfying Eq. (4.3) is a consequence of Theorem 4.1. For
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This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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chap4 - 48 BRUCE K. DRIVER 4. The Riemann Integral In this...

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