math403 hw2 - Solutions for HWK 2 Partially based on the...

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Solutions for HWK # 2 Partially based on the HWKs of my former student Russell Yanofsky September 18, 2010 Problem A Prove k x k 2 := q d i =1 x 2 i is a norm in R d . We have to prove that k x k 2 satis es the de nition of a norm. 1) k x k 2 0 x R d because x 2 i 0 x i R , and summing non-negative reals gives non-negative reals. 2) If x = 0 then k x k 2 = 0 because q d i =1 0 2 = 0 . If k x k 2 = 0 then x = 0 . To show this, assume x i s.t. x i 6 = 0 . Then, q d i =1 x 2 i 6 = 0 , which contradicts k x k 2 = 0 . Therefore, if k x k 2 = 0 then x i = 0 x i , and x = 0 . 3) k α x k 2 = | α |k x k 2 because k α x k 2 = q d i =1 ( αx i ) 2 = q α 2 d i =1 ( x i ) 2 = α 2 q d i =1 x 2 i = | α |k x k 2 . 4) k x + y k 2 ≤ k x k 2 + k y k 2 x , y R d because, by the Cauchy inequality d X i =1 x i y i v u u t d X i =1 x 2 i v u u t d X i =1 y 2 i , we have || x + y || 2 = v u u t d X i =1 ( x i + y i ) 2 = v u u t d X i =1 x 2 i + d X i =1 y 2 i + 2 d X i =1 x i y i v u u u t d X i =1 x 2 i + d X i =1 y 2 i + 2 v
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math403 hw2 - Solutions for HWK 2 Partially based on the...

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