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26.2 Theorem Let H be a Hilbert space and A € BL(H). (a) Let A be self-adjoint. Then ||A|| = sup{|(A(x), x)| : x € H, ||x|| ≤ 1}. In particular, A = 0 if and only if (A(x), x) = 0 for all z €. H. Proof: (a) We have seen in Section 25 that for every A & BL(H), ||A|| = sup{|(A(x), y)| : x, y ≤ H, ||x|| ≤ 1, ||y|| ≤ 1}. Let a = sup{|(A(x), x) : x € H, ||x|| ≤ 1}. Clearly, a ≤ ||A||. To prove || A|| ≤ a, we note that for x, y € H, (A(x+y), x+y) — (A(x − y), x − y) - - = 2(A(x), y) +2(A(y), x) 4Re (A(x), y), since A is self-adjoint. Hence 4Re (A(x), y) ≤ a(||x + y||² + ||x − y||²) = 2x(||x||² + ||y||²) by the parallelogram law (21.2(b)). Let ||x|| ≤ 1 and ||y|| ≤ 1. Then it follows that Re (A(x), y) ≤ a. If (A(x), y) = reie for r≥ 0 and 0 € R, then let zo = e-ier, so that ||xo|| = ||*|| ≤ 1 and |(A(x), y)| = r = (A(xo), y) = Re (A(xo), y) ≤ a. Taking supremum over all x, y € H with ||*|| ≤ 1, ||y|| ≤ 1, we obtain ||A|| ≤ a, as desired. In particular, if (A(x), x) = 0 for all x € H, then a = 0, so that || A|| = 0, that is, A = 0. Conversely, if A = 0, then it is clear that (A(x), x) = 0 for all z € H. Requent clarity underlined portions

26.2 Theorem Let H be a Hilbert space and A € BL(H). (a) Let A be self-adjoint. Then ||A|| = sup{|(A(x), x)| : x € H, ||x|| ≤ 1}. In particular, A = 0 if and only if (A(x), x) = 0 for all z €. H. Proof: (a) We have seen in Section 25 that for every A & BL(H), ||A|| = sup{|(A(x), y)| : x, y ≤ H, ||x|| ≤ 1, ||y|| ≤ 1}. Let a = sup{|(A(x), x) : x € H, ||x|| ≤ 1}. Clearly, a ≤ ||A||. To prove || A|| ≤ a, we note that for x, y € H, (A(x+y), x+y) — (A(x − y), x − y) - - = 2(A(x), y) +2(A(y), x) 4Re (A(x), y), since A is self-adjoint. Hence 4Re (A(x), y) ≤ a(||x + y||² + ||x − y||²) = 2x(||x||² + ||y||²) by the parallelogram law (21.2(b)). Let ||x|| ≤ 1 and ||y|| ≤ 1. Then it follows that Re (A(x), y) ≤ a. If (A(x), y) = reie for r≥ 0 and 0 € R, then let zo = e-ier, so that ||xo|| = ||*|| ≤ 1 and |(A(x), y)| = r = (A(xo), y) = Re (A(xo), y) ≤ a. Taking supremum over all x, y € H with ||*|| ≤ 1, ||y|| ≤ 1, we obtain ||A|| ≤ a, as desired. In particular, if (A(x), x) = 0 for all x € H, then a = 0, so that || A|| = 0, that is, A = 0. Conversely, if A = 0, then it is clear that (A(x), x) = 0 for all z € H. Requent clarity underlined portions

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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26.2 Theorem Let H be a Hilbert space and A € BL(H). (a) Let A be self-adjoint. Then ||A|| = sup{|(A(x), x)| : x € H, ||x|| ≤ 1}. In particular, A = 0 if and only if (A(x), x) = 0 for all z €. H. Proof: (a) We have seen in Section 25 that for every A & BL(H), ||A|| = sup{|(A(x), y)| : x, y ≤ H, ||x|| ≤ 1, ||y|| ≤ 1}. Let a = sup{|(A(x), x) : x € H, ||x|| ≤ 1}. Clearly, a ≤ ||A||. To prove || A|| ≤ a, we note that for x, y € H, (A(x+y), x+y) — (A(x − y), x − y) - - = 2(A(x), y) +2(A(y), x) 4Re (A(x), y), since A is self-adjoint. Hence 4Re (A(x), y) ≤ a(||x + y||² + ||x − y||²) = 2x(||x||² + ||y||²) by the parallelogram law (21.2(b)). Let ||x|| ≤ 1 and ||y|| ≤ 1. Then it follows that Re (A(x), y) ≤ a. If (A(x), y) = reie for r≥ 0 and 0 € R, then let zo = e-ier, so that ||xo|| = ||*|| ≤ 1 and |(A(x), y)| = r = (A(xo), y) = Re (A(xo), y) ≤ a. Taking supremum over all x, y € H with ||*|| ≤ 1, ||y|| ≤ 1, we obtain ||A|| ≤ a, as desired. In particular, if (A(x), x) = 0 for all x € H, then a = 0, so that || A|| = 0, that is, A = 0. Conversely, if A = 0, then it is clear that (A(x), x) = 0 for all z € H. Requent clarity underlined portions
Transcribed Image Text:26.2 Theorem Let H be a Hilbert space and A € BL(H). (a) Let A be self-adjoint. Then ||A|| = sup{|(A(x), x)| : x € H, ||x|| ≤ 1}. In particular, A = 0 if and only if (A(x), x) = 0 for all z €. H. Proof: (a) We have seen in Section 25 that for every A & BL(H), ||A|| = sup{|(A(x), y)| : x, y ≤ H, ||x|| ≤ 1, ||y|| ≤ 1}. Let a = sup{|(A(x), x) : x € H, ||x|| ≤ 1}. Clearly, a ≤ ||A||. To prove || A|| ≤ a, we note that for x, y € H, (A(x+y), x+y) — (A(x − y), x − y) - - = 2(A(x), y) +2(A(y), x) 4Re (A(x), y), since A is self-adjoint. Hence 4Re (A(x), y) ≤ a(||x + y||² + ||x − y||²) = 2x(||x||² + ||y||²) by the parallelogram law (21.2(b)). Let ||x|| ≤ 1 and ||y|| ≤ 1. Then it follows that Re (A(x), y) ≤ a. If (A(x), y) = reie for r≥ 0 and 0 € R, then let zo = e-ier, so that ||xo|| = ||*|| ≤ 1 and |(A(x), y)| = r = (A(xo), y) = Re (A(xo), y) ≤ a. Taking supremum over all x, y € H with ||*|| ≤ 1, ||y|| ≤ 1, we obtain ||A|| ≤ a, as desired. In particular, if (A(x), x) = 0 for all x € H, then a = 0, so that || A|| = 0, that is, A = 0. Conversely, if A = 0, then it is clear that (A(x), x) = 0 for all z € H. Requent clarity underlined portions
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