16-17-hilbert

16-17-hilbert - Lectures 16-17 6. Hilbert spaces In this...

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Unformatted text preview: Lectures 16-17 6. Hilbert spaces In this section we examine a special type of Banach spaces. Definition. Let K be one of the fields R or C , and let X be a vector space over K . An inner product on X is a map X X 3 ( , ) 7- ( ) K , with the following properties: ( ) 0, X ; if X satisfies ( ) = 0, then = 0; for any X , the map X 3 7- ( ) K is K-linear; ( ) = ( ) , xi, X . Comments. Combining the last two properties, one gets ( 1 + 2 ) = ( 1 ) + ( 2 ) , , 1 , 2 X , K ; ( 1 + 2 ) = ( 1 ) + ( 2 ) , 1 , 2 , X , K . In particular, one has (1) ( ) = ( ) = | | 2 ( ) , X , K . Proposition 6.1 (Cauchy-Bunyakowski-Schwartz Inequality) . Let ( ) be an inner product on the K-vector space X . Then (2) ( ) 2 ( ) ( ) , , X . Moreover, if equality holds then and are proportional, in the sense that either = 0 , or = 0 , or = . Proof. Fix , X . Assume 6 = 0. ( In the case when = 0, both statements are trivial ) . Choose a number K , with | | = 1, such that ( ) = ( ) = ( ) . Define the map F : K K by F ( z ) = ( z + z + ) , z K . A simple computation gives F ( z ) = z z ( ) + z ( ) + z ( ) + ( ) = = | z | 2 | | 2 ( ) + z ( ) + z ( ) + ( ) = = | z | 2 ( ) + z ( ) + z ( ) + ( ) , z R . 117 118 LECTURES 16-17 In particular, when we restrict F to R , it becomes a quadratic function: F ( t ) = at 2 + bt + c, t R , where a = ( ) > 0, b = 2 ( ) , c = ( ) . Notice that we have F ( t ) , t R . This forces b 2- 4 ac 0. This last inequality gives 4 ( ) 2- 4 ( ) ( ) , so we get ( ) 2 ( ) ( ) , and the inequality (2) is proven. Let us examine now when we have equality. The equality in (2) gives b 2- 4 ac = 0, which in terms of quadratic equations says that the equation F ( t ) = at 2 + bt + c = 0 has a ( unique ) solution t . This will give ( t + t + ) = F ( t ) = 0 , which forces t + = 0, i.e. = (- t ) . Corollary 6.1. Let ( ) be an inner product on the K-vector space X . Then the map X 3 7- q ( ) [0 , ) is a norm on X . Proof. Denote q ( ) simply by k k . The fact that k k is non-negative is clear. The implication k k = 0 = 0 is also clear. Using (1) we have k k = q ( ) = q | | 2 ( ) = | | q ( ) = | | k k , X , K ....
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16-17-hilbert - Lectures 16-17 6. Hilbert spaces In this...

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