104_Fall09_review_Final_solutions

104_Fall09_review_Final_solutions - Solutions to Review...

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Unformatted text preview: Solutions to Review Problems for Final 1. For f : [0 , 1] R , define k f k 2 = Z 1 f ( x ) 2 dx 1 2 . For f,g C [0 , 1], define d 2 ( f,g ) = k f- g k 2 . Prove that d 2 gives a metric on C [0 , 1] (you may use the Cauchy-Schwarz inequality, which is proven exactly the same as for Euclidean space - see p. 22 of Pugh): Z 1 f ( x ) g ( x ) dx k f k 2 k g k 2 Solution: We first prove that kk 2 is a norm. If f C [0 , 1], and k f k 2 = 0, then R 1 f ( x ) 2 dx = 0 implies f ( x ) 2 = 0, and therefore f ( x ) = 0 by Corollary 3.29, p. 169 (and conversely). We have k cf k 2 = R 1 ( cf ( x )) 2 dx 1 2 = | c | R 1 f ( x ) 2 dx 1 2 = | c |k f k 2 by Theorem 3.16. Finally, ( k f + g k 2 ) 2 = Z 1 ( f ( x ) + g ( x )) 2 dx = Z 1 f ( x ) 2 dx + 2 Z 1 f ( x ) g ( x ) dx + Z 1 g ( x ) 2 dx k f k 2 2 + 2 k f k 2 k g k 2 + k g k 2 2 = ( k f k 2 + k g k 2 ) 2 . Taking square roots, we see that k f + g k 2 k f k 2 + k g k 2 , so k k 2 is a norm (see p. 28). Then we see that d 2 ( f,g ) = k f- g k 2 is a metric on C [0 , 1], since for f,g,h C [0 , 1], d 2 ( f,g ) = k f- g k 2 = k f- h + h- g k 2 k f- h k 2 + k h- g k 2 = d 2 ( f,h ) + d 2 ( h,g ) , so the triangle inequality holds. Clearly d 2 is symmetric and non-degenerate as well by the first two properties of the norm. 2. Let ( M,d ) be a metric space. Given a set S M , define the characteristic function S : M { , 1 } of S as S ( x ) = ( 1 if x S if x / S....
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104_Fall09_review_Final_solutions - Solutions to Review...

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