tailieunhanh - New inequalities similar to Hardy-Hilbert’s inequality

In this paper, we establish a new inequality similar to Hardy-Hilbert’s inequality. As applications, some particular results and the equivalent form are derived. The integral analogues of the main results are also given. | Turk J Math 34 (2010) , 153 – 165. ¨ ITAK ˙ c TUB doi: New inequalities similar to Hardy-Hilbert’s inequality Namita Das and Srinibas Sahoo Abstract In this paper, we establish a new inequality similar to Hardy-Hilbert’s inequality. As applications, some particular results and the equivalent form are derived. The integral analogues of the main results are also given. Key Words: Hardy-Hilbert’s inequality; H¨ older’s inequality; β -function. 1. Introduction If p > 1, 1p + 1 q = 1, an , bn ≥ 0 satisfy 0 1, 1p + 1q = 1, f, g ≥ 0 satisfy 0 1, an ≥ 0 and An = a1 + a2 + . + an , then p ∞ An n n=1 1, f ≥ 0 and F (x) = ∞ x 0 F (x) x 0 f(t)dt, then p dx 0 for k = 1, 2, 3, 4, and lim f (k) (x) = x→∞ 0, for k = 0, 1, 2, 3, 4, then the following inequality holds: 1 − f(1) 2 , then ∞ mα−1 m=1 (m + n)λ 2 , β > 0 such that α + β 1 >− Again, by (), ∞ 1 ρ1 (x)fn (x)dx ∞ − n−λ . g1 (1) = − λ 12 12 (n + 1) 12 1 1 λ−α+1 λ − α + 1 −λ B(α, λ − α) ∞ nα+β−λ−1 − n=1 1 λ + α 12 ∞ nβ−λ−1 , n=1 ∞ nβ−λ−1 . n=1 2 Thus () is valid. This proves the lemma. 3. Main results In this section we prove our main result and derive some particular cases. Theorem Let p > 1, p1 + 1q = 1, 0 2, r + s = n n ∞ ∞ λ, an , bn ≥ 0, An = k=1 ak , Bn = k=1 bk . If 0 0 , take a ˜ n = n− ∞ 1+ε p 1+ε , ˜bn = n− q for n ≥ 1 . Then p1 ˜pn a n=1 ∞ 1q ˜bq n n=1 1 1, A˜m = m k=1 a ˜k > m−1 k+1 k=1 k x− 1+ε p dx = 1 m x− 1+ε p dx = 1 ε q mq −p − 1 . 1 − ε(q − .

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