hw4 - MA322 Spring 2008 Homework 4 Solutions Section 8.1 #9...

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Unformatted text preview: MA322 Spring 2008 Homework 4 Solutions Section 8.1 #9 (page 234): Prove that the 1 and norms also satisfy Theorem 8.6. That is, show that the norms (a) x 1 := n k =1 | x k | (b) x := max 1 k n | x k | satisfy the properties (i) x 0 with equality if and only if x = (ii) x = | | x (iii) x + y x + y for all x , y R n and all scalars R . Note: The reverse triangle inequality holds for any norm, so we need not prove it here. Proof: For the 1 norm: (i) Clearly x 1 = n k =1 | x k | 0, with equality if and only if x k = 0 for all k = 1 , . . . , n , i.e., if and only if x = . (ii) x 1 = n k =1 | x k | = n k =1 | | | x k | = | | n k =1 | x k | = | | x 1 . (iii) Using the triangle inequality for real numbers: x + y 1 = n k =1 | x k + y k | n k =1 ( | x k | + | y k | ) = n k =1 | x k | + n k =1 | y k | = x 1 + y 1 . Likewise, for the norm: (i) Clearly x = max 1 k n | x k | 0, with equality if and only if x k = 0 for all k = 1 , . . . , n , i.e., if and only if x = . (ii) x = max 1 k n | x k | = max 1 k n | | | x k | = | | max 1 k n | x k | = | | x . (iii) Using the triangle inequality for real numbers, for each k = 1 , . . . , n we have | x k + y k | | x k | + | y k | x + y . Since this hold for each k , it also holds when we take the maximum over k , i.e., x + y x + y ....
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This note was uploaded on 09/13/2009 for the course MA 322 taught by Professor Koch during the Spring '09 term at Nevada.

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hw4 - MA322 Spring 2008 Homework 4 Solutions Section 8.1 #9...

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