The Hilbert Bound of a Certain Doubly-Infinite Matrix
1960; Wiley; Volume: s1-35; Issue: 1 Linguagem: Inglês
10.1112/jlms/s1-35.1.128-s
ISSN1469-7750
Autores Tópico(s)Mathematics and Applications
ResumoIn the statement of Theorem 5(iii) the inequality − 5 2 − 2 p ⩽ λ ⩽ − 1 2 − 2 p should read − 5 2 − 2 p ⩽ λ ⩽ − 1 2 − 2 p . The equation an+1 = ηna0, where 1 < ηn < ∞ (n = 1, 2, …), (1) in the proof of Theorem 5(ii) is false for −½⩽λ<, since ln < 0. Indeed, using the binomial theorem it is easily deduced from the equations bn = an+1−an−a0(ln+1+ln) = 0 (n = 0, 1, 2, …) (2) that an → 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, an → 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 − 1 d 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.
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