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en A € BL(X). In fact, ||A|| ≤ (2j-11/3²) = x/√6. For =1,2,..., let en (0,...,0,1,0,0,...), where 1 occurs only in the h entry. If B BL(X) and (A(z), y) = (x, B(y)) for all x, y = H, en B(e₁)(n)= (en, B(e₁)) = (A(en), e₁) = 1/27 all n = 1,2,.... But this is impossible since B(e₁) € Coo- #0 n Request explain marked portion

en A € BL(X). In fact, ||A|| ≤ (2j-11/3²) = x/√6. For =1,2,..., let en (0,...,0,1,0,0,...), where 1 occurs only in the h entry. If B BL(X) and (A(z), y) = (x, B(y)) for all x, y = H, en B(e₁)(n)= (en, B(e₁)) = (A(en), e₁) = 1/27 all n = 1,2,.... But this is impossible since B(e₁) € Coo- #0 n Request explain marked portion

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
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We remark that if X is an inner product space which is not com- plete and A € BL(X), then there may not exist B € BL(X) such that (A(r), y) = (x, B(y)) for all x, y € H. For example, let X = coo the linear space of all scalar sequences having only a finite number of nonzero entries with the inner product defined in 21.3(b). For x € X, let A(T) = (£³² 0,0,0,...). Then A € BL(X). In fact, || A|| ≤ (₁¹/₁²) ¹/² = π/√6. For n=1,2,..., let en = (0,...,0, 1, 0, 0,...), where 1 occurs only in the nth entry. If B € BL(X) and (A(z), y) = (x, B(y)) for all x, y = H, then B(e₁)(n) = (en, B(er)) = (A(en), e₁) = 1 #0 for all n = 1,2,.... But this is impossible since B(e₁) € coo. n Request explain marked portion
Transcribed Image Text:We remark that if X is an inner product space which is not com- plete and A € BL(X), then there may not exist B € BL(X) such that (A(r), y) = (x, B(y)) for all x, y € H. For example, let X = coo the linear space of all scalar sequences having only a finite number of nonzero entries with the inner product defined in 21.3(b). For x € X, let A(T) = (£³² 0,0,0,...). Then A € BL(X). In fact, || A|| ≤ (₁¹/₁²) ¹/² = π/√6. For n=1,2,..., let en = (0,...,0, 1, 0, 0,...), where 1 occurs only in the nth entry. If B € BL(X) and (A(z), y) = (x, B(y)) for all x, y = H, then B(e₁)(n) = (en, B(er)) = (A(en), e₁) = 1 #0 for all n = 1,2,.... But this is impossible since B(e₁) € coo. n Request explain marked portion
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