Completion Notes

Completion Notes - Completion Theorem 1(Completion If V V...

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Completion Theorem 1 (Completion) If ( V , , ·A V ) is any inner product space, then there exists a Hilbert space ( H , , , H ) and a map U : V → H such that (i) U is 1–1 (ii) U is linear (iii) a U x ,U y A H = a x , y A V for all x , y ∈ V (iv) U ( V ) = b U x v v x ∈ V B is dense in H . If V is complete, then U ( V ) = H . Remark 2 (a) H is called the completion of V . (b) U gives a one-to–one correspondance between elements of V and elements of U ( V ). So we can think of U as giving the new name U x to each x ∈ V and we can think of U ( V ) as being the same as V but with the names of the elements changed. Thus, we can think of V as being U ( V ) ⊂ H . Using this point of view, the above theorem says that any inner product space can be extended to a complete inner product space. I.e. can have its holes Flled in. Motivation. The hard part of the proof is to make a guess as to what H should be. That’s what we’ll do now. A good strategy is to work backwards. Suppose that, somehow, we have found a suitable H with V ⊂ H . If we can Fnd a way to describe each element of H purely in terms of elements of V , then we can turn around and take that as the deFnition of H . Because V is dense in H , each element of H may be written as the limit of a sequence in V , and each such sequence is Cauchy. Thus specifying an element of H is equivalent to specifying a Cauchy sequence in V . But there is not a one–to–one correspondance between elements of H and Cauchy sequences in V , because many di±erent Cauchy sequences in V converge to the same element of H . ²or example, if b x n B n IN ⊂ V converges to x ∈ H , then b x n + 1 2 n x 1 B n ⊂ V and b e 1 /n x n + ( 1) n n x 2 B n ⊂ V also converge to x . To get a one–to–one correspondance, we can identify each x ∈ H with the set of all Cauchy sequences in V that converge to x . Outline of Proof. ²irst deFne V = ± { x n } n v v v { x n } n is a Cauchy sequence in V ² That is, V is the set of all Cauchy sequences in V . Next we deFne two Cauchy sequences { x n } n and { y n } n in V to be “equivalent”, written { x n } n ∼ { y n } n , if and only if lim n →∞ b x n y n b V = 0 This deFnition is rigged so that any two convergent sequences have the same limit if and only if they are equivalent. Next, if { x n } n ∈ V , we deFne the “equivalence class of { x n } n ” to be the set ³ { x n } n ´ = ± { y n } n ∈ V v v v { y n } n ∼ { x n } n ² of all Cauchy sequences that are equivalent to { x n } n . We shall shortly prove September 7, 2011 Completion 1
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Lemma 3 is an equivalence relation (1) . In particular, if { x n } n IN , { y n } n ∈ V then either b { x n } n B = b { y n } n B or b { x n } n B b { y n } n B = If you think of a Cauchy sequence as one person and an equivalence class of Cauchy sequences as a “family” of related people, then the above Lemma says, that the whole world is divided into a collection of nonoverlapping families. Next, we deFne H = ± b { x n } n B v v v { x n } n ∈ V ² as the set of all “families” and prove
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