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Because v is dense in h each element of h may be

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Unformatted text preview: n around and take that as the definition 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 different Cauchy sequences in V converge to the same element of H. For example, if xn n∈IN ⊂ V converges to x ∈ H, then xn + 21 x1 n∈IN ⊂ V n n 1) and e1/n xn + (−n x2 n∈IN ⊂ 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. First define V= {xn }n∈IN {xn }n∈IN is a Cauchy sequence in V That is, V is the set of all Cauchy sequences in V . Next we define two Cauchy sequences {xn }n∈IN and {yn }n∈IN in V to be “equivalent”, written {xn }n∈IN ∼ {yn }n∈IN , if and only if lim n→∞ xn − yn...
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This note was uploaded on 02/13/2014 for the course MATH 511 taught by Professor Joelfeldman during the Spring '13 term at UBC.

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