Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question
100%
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
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS