EuclideanNorm - k a k 2-2 k a k 2 + k a k 4 k b k 2 ( a b )...

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Economics 511 Professor: John Nachbar October 19, 2009 Basic Properties of the Euclidean Norm. Given x R N , define k x k = ( x · x ) 1 / 2 . If N = 1 then k x k = | x | , the absolute value of x . k x k is called the Euclidean norm of x . It measures the distance, in the standard everyday use of the term, from x to the origin. For example, suppose that x = ( x 1 ,x 2 ) R 2 . Then the distance from the origin to x is the length of the hypotenuse of the right triangle with short sides given by ( x 1 , 0) and (0 ,x 2 ). By the Pythagorean theorem, the length of this hypotenuse is q x 2 1 + x 2 2 = k x k . It is almost immediate from the definition that the following properties hold. 1. k x k ≥ 0 for all x R N . 2. k x k = 0 iff x = 0. 3. For any λ R , λ > 0, k λx k = | λ |k x k . A deeper result is the following fact, called the Cauchy-Schwartz inequality . Theorem 1 (Cauchy-Schwartz) . For any a,b R N , | a · b | ≤ k a kk b k Proof of Theorem 1. If a · b = 0 then the result is immediate. Suppose a · b 6 = 0. For any λ 0, 0 ≤ k a - λb k 2 = k a k 2 - 2 λ ( a · b ) + λ 2 k b k 2 Since a · b 6 = 0, I can define λ = k a k 2 a · b . Substituting into the above yields
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Unformatted text preview: k a k 2-2 k a k 2 + k a k 4 k b k 2 ( a b ) 2 = k a k 2-1 + k a k 2 k b k 2 ( a b ) 2 ! . 1 Since a b 6 = 0, a 6 = 0, hence k a k > 0. Therefore, the above can be rearranged to yield ( a b ) 2 k a k 2 k b k 2 . Taking the square root yields the result. From the proof, it is clear that the Cauch-Schwartz inequality is, in fact, strict unless for some , a = b , meaning that a and b are collinear. Cauchy-Schwartz, in turn, implies the following. Theorem 2. For any a,b R N , k a + b k k a k + k b k . Proof k a + b k 2 = ( a + b ) ( a + b ) = k a k 2 + 2 a b k b k 2 k a k 2 + 2 | a b | + k b k 2 k a k 2 + 2 k a kk b k + k b k 2 = ( k a k + k b k ) 2 . The second inequality follows form Theorem 1. Taking square roots yields the re-sult. 2...
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EuclideanNorm - k a k 2-2 k a k 2 + k a k 4 k b k 2 ( a b )...

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