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 ( ξ, η ) 7-→ ( ξ η ) ∈ K , with the following properties: • ( ξ ξ ) ≥ 0, ∀ ξ ∈ X ; • if ξ ∈ X satisfies ( ξ ξ ) = 0, then ξ = 0; • for any ξ ∈ X , the map X 3 η 7-→ ( ξ η ) ∈ K is K-linear; • ( η ξ ) = ( ξ η ) , ∀ xi, η ∈ X . Comments. Combining the last two properties, one gets ( ξ λη 1 + η 2 ) = λ ( ξ η 1 ) + ( ξ η 2 ) , ∀ ξ, η 1 , η 2 ∈ X , λ ∈ K ; ( λξ 1 + ξ 2 η ) = λ ( ξ 1 η ) + ( ξ 2 η ) , ∀ ξ 1 , ξ 2 , η ∈ X , λ ∈ K . In particular, one has (1) ( λξ λξ ) = λλ ( ξ ξ ) = | λ | 2 · ( ξ ξ ) , ∀ ξ ∈ X , λ ∈ K . Proposition 6.1 (Cauchy-Bunyakowski-Schwartz Inequality) . Let ( · · ) be an inner product on the K-vector space X . Then (2) ( ξ η ) 2 ≤ ( ξ ξ ) · ( η η ) , ∀ ξ, η ∈ X . Moreover, if equality holds then ξ and η are proportional, in the sense that either ξ = 0 , or η = 0 , or ξ = λη . Proof. Fix ξ, η ∈ X . Assume η 6 = 0. ( In the case when η = 0, both statements are trivial ) . Choose a number λ ∈ K , with | λ | = 1, such that ( ξ η ) = λ ( ξ η ) = ( ξ λη ) . Define the map F : K → K by F ( z ) = ( zλη + ξ zλη + ξ ) , ∀ z ∈ K . A simple computation gives F ( z ) = zλ z λ ( η η ) + zλ ( ξ η ) + z λ ( η ξ ) + ( ξ ξ ) = = | z | 2 | λ | 2 ( η η ) + zλ ( ξ η ) + z λ ( ξ η ) + ( ξ ξ ) = = | z | 2 ( η η ) + z ( ξ η ) + z ( ξ η ) + ( ξ ξ ) , ∀ 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 = ( η η ) > 0, b = 2 ( ξ η ) , c = ( ξ ξ ) . Notice that we have F ( t ) ≥ , ∀ t ∈ R . This forces b 2- 4 ac ≤ 0. This last inequality gives 4 ( ξ η ) 2- 4 ( ξ ξ ) · ( η η ) ≤ , so we get ( ξ η ) 2 ≤ ( ξ ξ ) · ( η η ) , 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 λη + ξ t λη + ξ ) = F ( t ) = 0 , which forces t λη + ξ = 0, i.e. ξ = (- t λ ) η . Corollary 6.1. Let ( · · ) be an inner product on the K-vector space X . Then the map X 3 ξ 7-→ q ( ξ ξ ) ∈ [0 , ∞ ) is a norm on X . Proof. Denote q ( ξ ξ ) simply by k ξ k . The fact that k ξ k is non-negative is clear. The implication k ξ k = 0 ⇒ ξ = 0 is also clear. Using (1) we have k λξ k = q ( λξ λξ ) = q | λ | 2 ( ξ ξ ) = | λ | · q ( ξ ξ ) = | λ | · k ξ k , ∀ ξ ∈ X , λ ∈ K ....
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16-17-hilbert - Lectures 16-17 6 Hilbert spaces In this...

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