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The Hilbert Bound of a Certain Doubly‐Infinite Matrix
Author(s) -
Hill C. K.
Publication year - 1960
Publication title -
journal of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.441
H-Index - 62
eISSN - 1469-7750
pISSN - 0024-6107
DOI - 10.1112/jlms/s1-35.1.128-s
Subject(s) - mathematics , reductio ad absurdum , combinatorics , binomial theorem , matrix (chemical analysis) , discrete mathematics , chemistry , computer science , interpretation (philosophy) , programming language , chromatography
In the statement of Theorem 5(iii) the inequality− 5 2 − 2 p ⩽ λ ⩽ − 1 2 − 2 pshould read − 5 2 − 2 p ⩽ λ ⩽ − 1 2 − 2 p . The equation a n +1 = η n a 0 , where 1 < η n < ∞ ( n = 1, 2, …), (1) in the proof of Theorem 5(ii) is false for −½⩽λ<, since l n < 0. Indeed, using the binomial theorem it is easily deduced from the equations b n = a n +1 − a n − a 0 ( l n +1 + l n ) = 0 ( n = 0, 1, 2, …) (2) that a n → 0 as n →∞ in contradiction to (1), and in fact the root vector a = k (λ) has precisely this property. The proof by reductio ad absurdum may be completed as follows. Since a is a Hilbert vector, a n → 0 as n → ∞. Hence, by (2),a k = ∑ n = k ∞( a n − a n + 1) = − a 0 ∑ n = k ∞( l n + 1 + l n )= − {a 0 / Γ ( λ )} ∑ n = k ∞{Γ ( n + 1 + λ ) / Γ ( n + 2 ) + Γ ( n + λ ) / Γ ( n + 1 )} ,and as k → + ∞ this is asymptotic to− {2 a 0 / Γ ( λ )} ∑ n = k ∞n λ − 1∼ − {2 a 0 / Γ ( λ )} ∫ k ∞n λ − 1d n = 2 a 0 k λ / Γ ( λ + 1 )which clearly contradicts the hypothesis that a is a non‐null Hilbert vector when λ⩾−½. Hence the theorem is correct. The results of Theorems 4 and 5 are, however, contained in Theorem 5 of Rosenblum's note “ On the Hilbert Matrix (II) ”, Proc. American Math. Soc ., 9 (1958), 581–585.