Replace in 1 xm and ym by xn and yn respectively this

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Unformatted text preview: oth xn − xm V and yn − ym V converge to zero as m, n → ∞. So the same is true for xn , yn V − xm , ym V . So { xn , yn V }n∈IN is a Cauchy sequence of complex numbers. Since C is complete, this Cauchy sequence converges. Lemma 5 Define, for each {xn }n∈IN ∈ H, {yn }n∈IN ∈ H and α ∈ C, {xn }n∈IN + {yn }n∈IN = {xn + yn }n∈IN α {xn }n∈IN = {αxn }n∈IN {xn }n∈IN , {yn }n∈IN H = lim xn , yn n→∞ V Each of these operations is well–defined. September 7, 2011 Completion 3 Proof: Let {xn }n∈IN ∼ {xn }n∈IN and {yn }n∈IN ∼ {yn }n∈IN . Replace, in (1), xm and ym by xn and yn respectively. This gives xn , yn V − xn , yn V ≤ xn − xn Since {xn }n∈IN ∼ {xn }n∈IN , we have that lim n→∞ we have that lim n→∞ lim n→∞ xn , yn V yn − yn − xn , yn V V = 0. Since yn V xn − xn V V yn + xn V yn − yn V = 0. Since {yn }n∈IN ∼ {yn }n∈IN , and V n∈IN xn are bounded, V n∈IN = 0. So the inner product is well–defined. Similarly (xn + yn ) − (xn + yn ) αxn − αxn V V ≤ xn − xn + yn − yn V n→∞ V −→ 0 n→∞ ≤ |α| xn − xn V...
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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 The University of British Columbia.

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