This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lectures 1617 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 Klinear; ( ) = ( ) , 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 (CauchyBunyakowskiSchwartz Inequality) . Let ( ) be an inner product on the Kvector 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 1617 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 Kvector 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 nonnegative 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 ....
View Full
Document
 Three '10
 Smith
 Vector Space

Click to edit the document details