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³x("x zl€x *0] } + ₂[€* *0} } − z16x *)) } – 2z|14|| = ; x 13-13-14-50-4

³x("x zl€x *0] } + ₂[€* *0} } − z16x *)) } – 2z|14|| = ; x 13-13-14-50-4

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.4: Ordered Integral Domains
Problem 1E: Complete the proof of Theorem 5.30 by providing the following statements, where and are arbitrary...
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THEOREM 10.1. Let S be a dense subset of the inner product space X, and let A = {x}, where a runs through the index set A, be an orthonormal set in X. If, for any y € S |ly||² = { 1(y,x,)1² (that is, Parseval's identity holds for every y € S), then A is complete. Proof. By part (1) of Theorem 9.11, we know that, if we can show Parseval's identity to hold for every vector in the space, we can conclude that A is complete, and this is what will be shown in this proof. To this end, let x be any vector in X. Since S is a dense subset, for any preascribed & > 0, there is some ye S such that ||y - x | <&. Since y € S, we can apply the hypothesis to assert that ||y||² = Σ 1(x, x91². Now let us note that, for any finite subset of A, J, |y – Ţ (», xDx.||² = 11×11² – Ę 1609, ×D1² − Ę 10, ×D1² + Ę 10, ×D1² !, \, || = ||y||² – Į |(y, ׂ)|² ≤||ly||² - 16, xD1²³1. (10.1) Now consider Request explain this step Since the series (10.1) is summable, the last term above can be made arbitrarily small, and by taking a suitable finite subset K of A, we can say that, for any finite subset J > K, |y- & α₁ x x < ² 20-0-0-xxx = { [(y – x, x,)|² ≤lly - x ||² < ε ², (10.2)
Transcribed Image Text:THEOREM 10.1. Let S be a dense subset of the inner product space X, and let A = {x}, where a runs through the index set A, be an orthonormal set in X. If, for any y € S |ly||² = { 1(y,x,)1² (that is, Parseval's identity holds for every y € S), then A is complete. Proof. By part (1) of Theorem 9.11, we know that, if we can show Parseval's identity to hold for every vector in the space, we can conclude that A is complete, and this is what will be shown in this proof. To this end, let x be any vector in X. Since S is a dense subset, for any preascribed & > 0, there is some ye S such that ||y - x | <&. Since y € S, we can apply the hypothesis to assert that ||y||² = Σ 1(x, x91². Now let us note that, for any finite subset of A, J, |y – Ţ (», xDx.||² = 11×11² – Ę 1609, ×D1² − Ę 10, ×D1² + Ę 10, ×D1² !, \, || = ||y||² – Į |(y, ׂ)|² ≤||ly||² - 16, xD1²³1. (10.1) Now consider Request explain this step Since the series (10.1) is summable, the last term above can be made arbitrarily small, and by taking a suitable finite subset K of A, we can say that, for any finite subset J > K, |y- & α₁ x x < ² 20-0-0-xxx = { [(y – x, x,)|² ≤lly - x ||² < ε ², (10.2)
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