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is a bounded sequence in H and since A is a compact operator, there is a subsequence (Yn₁) such that (A(Yn;)) converges in H. For j, k 1, 2, ..., we have = ||A*(xn;) — A*(xnx) || ² = = = (A*(xn, - Ink), A* (In, — Ink)) - (AA*(Znj – Ink), In, nh) (A(yn,) - A(ynk), In, - Ink) ≤ 2a||A(yn,) - A(Ynk)||. Request exple this step

is a bounded sequence in H and since A is a compact operator, there is a subsequence (Yn₁) such that (A(Yn;)) converges in H. For j, k 1, 2, ..., we have = ||A*(xn;) — A*(xnx) || ² = = = (A*(xn, - Ink), A* (In, — Ink)) - (AA*(Znj – Ink), In, nh) (A(yn,) - A(ynk), In, - Ink) ≤ 2a||A(yn,) - A(Ynk)||. Request exple this step

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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28.1 Theorem Let A € BL(H). (a) If R(A) is finite dimensional, then A is compact. (b) If each An is compact and || An – A|| → 0, then A is compact. (c) If A is a compact, then so is A*. (c) Let A be compact. To show that A* is compact, consider a sequence (n) in H such that ||xn|| ≤ a for all n and some a > 0. Let Yn A*(xn), n = 1,2,.... Since A* is a bounded operator, (yn) is a bounded sequence in H and since A is a compact operator, there is a subsequence (yn;) such that (A(Yn;)) converges in H. For j, k 1, 2, ..., we have ||A*(xn;) — A*(xnx)||² = - (A*(xn; — Xnk), A*(Xn; — Xnx)) = (AA* (In, Ink), In, — Xnx) - (A(Yn;) — A(Ynk), xn; — xnx) ≤ 2a||A(yn;) — A(Ynk)||. explain Request This shows that (A*(xn;)) is a Cauchy sequence in H. Since H is complete, it converges in H. Thus A* is compact. this stulp
Transcribed Image Text:28.1 Theorem Let A € BL(H). (a) If R(A) is finite dimensional, then A is compact. (b) If each An is compact and || An – A|| → 0, then A is compact. (c) If A is a compact, then so is A*. (c) Let A be compact. To show that A* is compact, consider a sequence (n) in H such that ||xn|| ≤ a for all n and some a > 0. Let Yn A*(xn), n = 1,2,.... Since A* is a bounded operator, (yn) is a bounded sequence in H and since A is a compact operator, there is a subsequence (yn;) such that (A(Yn;)) converges in H. For j, k 1, 2, ..., we have ||A*(xn;) — A*(xnx)||² = - (A*(xn; — Xnk), A*(Xn; — Xnx)) = (AA* (In, Ink), In, — Xnx) - (A(Yn;) — A(Ynk), xn; — xnx) ≤ 2a||A(yn;) — A(Ynk)||. explain Request This shows that (A*(xn;)) is a Cauchy sequence in H. Since H is complete, it converges in H. Thus A* is compact. this stulp
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