Journal of Inequalities in Pure and Applied Mathematics A RELATION TO HARDY-HILBERT’S INTEGRAL INEQUALITY AND MULHOLLAND’S INEQUALITY volume 6, issue 4, article 112, 2005. BICHENG YANG Department of Mathematics Guangdong Institute of Education Guangzhou, Guangdong 510303 P. R. China. Received 14 November, 2004; accepted 25 August, 2005. Communicated by: L. Pick EMail: bcyang@pub.guangzhou.gd.cn Abstract Contents JJ J II I Home Page Go Back Close c 2000 Victoria University ISSN (electronic): 1443-5756 223-04 Quit Abstract This paper deals with a relation between Hardy-Hilbert’s integral inequality and Mulholland’s integral inequality with a best constant factor, by using the Beta function and introducing a parameter λ. As applications, the reverse, the equivalent form and some particular results are considered. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality 2000 Mathematics Subject Classification: 26D15. Key words: Hardy-Hilbert’s integral inequality; Mulholland’s integral inequality; β function; Hölder’s inequality. Bicheng Yang Contents 1 2 3 4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some Particular Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 The first reversible form . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The second reversible form . . . . . . . . . . . . . . . . . . . . . . 4.3 The form which does not have a reverse . . . . . . . . . . . References 3 6 9 16 16 19 23 Title Page Contents JJ J II I Go Back Close Quit Page 2 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 1. Introduction R∞ If p > 1, p1 + 1q = 1, f, g ≥ 0 satisfy 0 < 0 f p (x)dx < ∞ and 0 < R∞ q g (x)dx < ∞, then one has two equivalent inequalities as (see [1]): 0 Z ∞Z ∞ f (x)g(y) (1.1) dxdy x+y 0 0 Z ∞ p1 Z ∞ 1q π p q < f (x)dx g (x)dx ; 0 0 sin πp Z ∞ Z (1.2) 0 0 ∞ f (x) dx x+y p p dy < h sin π π p ip Z A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang ∞ p f (x)dx, 0 π π where the constant factors sin(π/p) and sin(π/p) are all the best possible. Inequality (1.1) is called Hardy- Hilbert’s integral inequality, which is important in analysisRand its applications (cf. Mitrinovic R ∞ 1 etq al. [2]). ∞ 1 p If 0 < 1 x F (x)dx < ∞ and 0 < 1 y G (y)dy < ∞, then the Mulholland’s integral inequality is as follows (see [1, 3]): Z ∞Z ∞ F (x)F (y) (1.3) dxdy xy ln xy 1 1 Z ∞ p p1 Z ∞ q 1q π F (x) G (y) < dx dy , x y 1 1 sin πp Title Page Contents JJ J II I Go Back Close Quit Page 3 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au π where the constant factor sin(π/p) is the best possible. Setting f (x) = F (x)/x, and g(y) = G(y)/y in (1.3), by simplification, one has (see [12]) Z ∞Z ∞ f (x)g(y) (1.4) dxdy ln xy 1 1 Z ∞ p1 Z ∞ 1q π < xp−1 f p (x)dx xq−1 g q (x)dx . π 1 1 sin p We still call (1.4) Mulholland’s integral inequality. In 1998, Yang [11] first introduced an independent parameter λ and the β function for given an extension of (1.1) (for p = q = 2). Recently, by introducing a parameter λ, Yang [8] and Yang et al. [10] gave some R ∞ extensions of (1.1) and (1.2)Ras: If λ > 2 − min{p, q}, f, g ≥ 0 satisfy 0 < 0 x1−λ f p (x)dx < ∞ ∞ and 0 < 0 x1−λ g q (x)dx < ∞, then one has two equivalent inequalities as: Z ∞Z ∞ f (x)g(y) (1.5) dxdy (x + y)λ 0 0 p1 Z ∞ 1q Z ∞ 1−λ p 1−λ q < kλ (p) x f (x)dx x g (x)dx 0 0 ∞ y (1.6) 0 (p−1)(λ−1) Z 0 ∞ f (x) dx (x + y)λ p Bicheng Yang Title Page Contents JJ J II I Go Back and Z A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality p Z dy < [kλ (p)] Close ∞ x 1−λ p f (x)dx, Quit 0 p p+λ−2 q+λ−2 , q p where the constant factors kλ (p) and [kλ (p)] (kλ (p) = B , B(u, v) is the β function) are all the best possible. By introducing a parameter Page 4 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au α, Kuang [5] gave an extension of (1.1), and [9] gave an improvement R ∞Yang (p−1)(1−α) p of [5] If α > 0, f, g ≥ 0 satisfy 0 < 0 x f (x)dx < ∞ and R ∞as: (q−1)(1−α) q 0< 0 x g (x)dx < ∞, then Z ∞ Z ∞ f (x)g(y) dxdy xα + y α 0 0 Z ∞ p1 Z ∞ 1q π < x(p−1)(1−α) f p (x)dx x(q−1)(1−α) g q (x)dx , π 0 0 α sin p (1.7) π where the constant α sin(π/p) is the best possible. Recently, Sulaiman [6] gave some new forms of (1.1) and Hong [14] gave an extension of Hardy-Hilbert’s inequality by introducing two parameters λ and α. Yang et al. [13] provided an extensive account of the above results. The main objective of this paper is to build a relation to (1.1) and (1.4) with a best constant factor, Rby Rintroducing the β function and a parameter λ, related b b f (x)g(y) to the double integral a a (u(x)+u(y)) λ dxdy (λ > 0). As applications, the reversion, the equivalent form and some particular results are considered. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 5 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 2. Some Lemmas First, we need the formula of the β function as (cf. Wang et al. [7]): Z ∞ 1 (2.1) B(u, v) := tu−1 dt = B(v, u) (u, v > 0). u+v (1 + t) 0 Lemma 2.1 (cf. [4]). If p > 1, p1 + 1q = 1, ω(σ) > 0, f, g ≥ 0, f ∈ Lpω (E) and g ∈ Lqω (E), then one has the Hölder’s inequality with weight as: Z Z ω(σ)f (σ)g(σ)dσ ≤ (2.2) E ω(σ)f p (σ)dσ p1 Z E ω(σ)g q (σ)dσ 1q ; E if p < 1 (p 6= 0), with the above assumption, one has the reverse of (2.2), where the equality (in the above two cases) holds if and only if there exists nonnegative real numbers c1 and c2 , such that they are not all zero and c1 f p (σ) = c2 g q (σ), a. e. in E. Lemma 2.2. If p 6= 0, 1, p1 + 1q = 1, φr = φr (λ) > 0 (r = p, q), φp +φq = λ, and u(t) is a differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞) such that u(a+) = 0 and u(b−) = ∞, for r = p, q, define ωr (x) as Z b 0 (u(y))φr −1 u (y) λ−φr (2.3) ωr (x) := (u(x)) dy (x ∈ (a, b)). λ a (u(x) + u(y)) Then for x ∈ (a, b), each ωr (x) is constant, that is (2.4) ωr (x) = B(φp , φq ) (r = p, q). A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 6 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au Proof. For fixed x, setting v = u(y) u(x) in (2.3), one has 0 b (u(y))φr −1 u (y) ωr (x) = (u(x)) dy λ λ a (u(x)) (1 + u(y)/u(x)) Z ∞ (vu(x))φr −1 λ−φr = (u(x)) u(x)dv (u(x))λ (1 + v)λ 0 Z ∞ φr −1 v = dv (r = p, q). (1 + v)λ 0 λ−φr Z By (2.1), one has (2.4). The lemma is proved. Lemma 2.3. If p > 1, p1 + 1q = 1, φr > 0 (r = p, q), satisfy φp + φq = λ, and u(t) is a differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞) satisfying u(a+) = 0 and u(b−) = ∞, then for c = u−1 (1) and 0 < ε < qφp , (2.5) ε 0 (u(x))φq − p −1 u (x) φp − qε −1 0 (u(y)) u (y)dxdy I := (u(x) + u(y))λ c c 1 ε ε > B φp − , φq + − O(1) ; ε q q Z bZ b if 0 < p < 1 (or p < 0), with the above assumption and 0 < ε < −qφq (or 0 < ε < qφp ), then ε ε 1 (2.6) I < B φp − , φq + . ε q q A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 7 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au Proof. For fixed x, setting v = b Z I := φq − pε −1 (u(x)) u(y) u(x) in I, one has "Z 0 u (x) c c Z b c Z ∞ # ε (u(y))φp − q −1 0 u (y)dy dx (u(x) + u(y))λ ε 1 v φp − q −1 dvdx λ (1 + v) 1 u(x) φp − qε −1 Z ∞ 0 u (x)dx v dv = 1+ε (1 + v)λ 0 c (u(x)) Z 1 φp − qε −1 Z b 0 u(x) v u (x) dvdx − 1+ε (1 + v)λ 0 c (u(x)) "Z 1 # ε Z Z b 0 u(x) ε 1 ∞ v φp − q −1 u (x) v φp − q −1 dv dx > dv − λ ε 0 (1 + v) 0 c (u(x)) −2 Z ∞ φp − qε −1 v ε 1 = dv − φp − . λ ε 0 (1 + v) q Z (2.7) 0 (u(x))−1−ε u (x) = b b By (2.1), inequality (2.5) is valid. If 0 < p < 1 (or p < 0), by (2.7), one has Z b Z ∞ 0 ε u (x) 1 I< dx v φp − q −1 dv, 1+ε λ (1 + v) c (u(x)) 0 and then by (2.1), inequality (2.6) follows. The lemma is proved. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 8 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 3. Main Results Theorem 3.1. If p > 1, p1 + 1q = 1, φr > 0 (r = p, q), φp + φq = λ, u(t) is a differentiable strict increasing function in (a, b) (−∞ ≤ a < b ≤ ∞), such that Rb p(1−φq )−1 u(a+) = 0 and u(b−) = ∞, and f, g ≥ 0 satisfy 0 < a (u(x)) f p (x)dx < 0 (u (x))p−1 Rb q(1−φp )−1 ∞ and 0 < a (u(x)) g q (x)dx < ∞, then 0 (u (x))q−1 Z bZ (3.1) a a b f (x)g(y) dxdy (u(x) + u(y))λ Z b p1 (u(x))p(1−φq )−1 p < B(φp , φq ) f (x)dx (u0 (x))p−1 a Z b 1q (u(x))q(1−φp )−1 q × g (x)dx , (u0 (x))q−1 a where the constant factor B(φp , φq ) is the best possible. If p < 1 (p 6= 0), {λ; φr > 0 (r = p, q), φp + φq = λ} 6= φ, with the above assumption, one has the reverse of (3.1), and the constant is still the best possible. Proof. By (2.2), one has Z bZ b f (x)g(y) dxdy J := λ a a (u(x) + u(y)) A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 9 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 0 (u(x))(1−φq )/q (u (y))1/p 1 = f (x) λ (u(y))(1−φp )/p (u0 (x))1/q a a (u(x) + u(y)) 0 (u(y))(1−φp )/p (u (x))1/q × g(y) dxdy (u(x))(1−φq )/q (u0 (y))1/p Z b Z b p1 0 (u(y))φp −1 u (y) (u(x))(p−1)(1−φq ) p ≤ dy f (x)dx λ (u0 (x))p−1 a a (u(x) + u(y)) 1q Z b Z b 0 (u(y))(q−1)(1−φp ) q (u(x))φq −1 u (x) × dx g (y)dy . λ (u0 (y))q−1 a a (u(x) + u(y)) Z bZ (3.2) b If (3.2) takes the form of equality, then by (2.2), there exist non-negative numbers c1 and c2 , such that they are not all zero and 0 0 u (x)(u(y))(q−1)(1−φp ) q u (y)(u(x))(p−1)(1−φq ) p f (x) = c g (y), c1 2 (u(y))1−φp (u0 (x))p−1 (u(x))1−φq (u0 (y))q−1 a.e. in (a, b) × (a, b). It follows that c1 (u(y))q(1−φp ) q (u(x))p(1−φq ) p f (x) = c g (y) = c3 , a.e. in (a, b) × (a, b), 2 0 (u (x))p (u0 (y))q where c3 is a constant. Without loss of generality, suppose c1 6= 0. One has 0 (u(x))p(1−φq )−1 p c3 u (x) f (x) = , a.e. in (a, b), 0 (u (x))p−1 c1 u(x) A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 10 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au which contradicts 0 < b Z (3.3) J < a Rb a (u(x))p(1−φq )−1 p f (x)dx (u0 (x))p−1 < ∞. Then by (2.3), one has p1 (u(x))p(1−φq )−1 p ωp (x) f (x)dx (u0 (x))p−1 1q Z ∞ (u(x))q(1−φp )−1 q × ωq (x) g (x)dx , (u0 (x))q−1 0 and in view of (2.4), it follows that (3.1) is valid. For 0 < ε < qφp , setting fε (x) = gε (x) = 0, x ∈ (a, c) (c = u−1 (1)); ε 0 fε (x) = (u(x))φq − p −1 u (x), ε 0 gε (x) = (u(x))φp − q −1 u (x), x ∈ [c, b), we find Z (3.4) a b p1 (u(x))p(1−φq )−1 p fε (x)dx (u0 (x))p−1 Z b 1q (u(x))q(1−φp )−1 q 1 × gε (x)dx = . 0 q−1 (u (x)) ε a If the constant factor B(φp , φq ) in (3.1) is not the best possible, then, there exists a positive constant k < B(φp , φq ), such that (3.1) is still valid if one replaces A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 11 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au B(φp , φq ) by k. In particular, by (2.6) and (3.4), one has ε ε B φp − , φq + − εO(1) q q Z bZ b fε (x)gε (y) dxdy <ε λ a a (u(x) + u(y)) Z b p1 Z b 1q (u(x))p(1−φq )−1 p (u(x))q(1−φp )−1 q < εk fε (x)dx gε (x)dx = k, (u0 (x))p−1 (u0 (x))q−1 a a and then B(φp , φq ) ≤ k (ε → 0+ ). This contradicts the fact that k < B(φp , φq ). Hence the constant factor B(φp , φq ) in (3.1) is the best possible. For 0 < p < 1 (or p < 0), by the reverse of (2.2) and using the same procedures, one can obtain the reverse of (3.1). For 0 < ε < −qφq (or 0 < ε < qφp ), setting fε (x) and gε (x) as the above, we still have (3.4). If the constant factor B(φp , φq ) in the reverse of (3.1) is not the best possible, then, there exists a positive constant K > B(φp , φq ), such that the reverse of (3.1) is still valid if one replaces B(φp , φq ) by K. In particular, by (2.7) and (3.4), one has ε ε B φp − , φq + q q Z bZ b fε (x)gε (y) >ε dxdy λ a a (u(x) + u(y)) Z b p1 Z b 1q (u(x))p(1−φq )−1 p (u(x))q(1−φp )−1 q > εK fε (x)dx gε (x)dx = K, (u0 (x))p−1 (u0 (x))q−1 a a A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 12 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au and then B(φp , φq ) ≥ K (ε → 0+ ). This contradiction concludes that the constant in the reverse of (3.1) is the best possible. The theorem is proved. Theorem 3.2. Let the assumptions of Theorem 3.1 hold. (i) If p > 1, p1 + follows b Z (3.5) a 1 q = 1, one obtains the equivalent inequality of (3.1) as 0 u (y) (u(y))1−pφp p f (x) dx dy λ a (u(x) + u(y)) Z b (u(x))p(1−φq )−1 p p < [B(φp , φq )] f (x)dx; (u0 (x))p−1 a Z b (ii) If 0 < p < 1, one obtains the reverse of (3.5) equivalent to the reverse of (3.1); A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents (iii) If p < 0, one obtains inequality (3.5) equivalent to the reverse of (3.1), where the constants in the above inequalities are all the best possible. Proof. Set JJ J II I Go Back 0 u (y) g(y) := (u(y))1−pφp Z a b f (x) dx (u(x) + u(y))λ p−1 Close , Quit Page 13 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au and use (3.1) to obtain Z b (u(y))q(1−φp )−1 q 0< g (y)dy (u0 (y))q−1 a p Z b Z b 0 u (y) f (x) = dx dy 1−pφp λ a (u(y)) a (u(x) + u(y)) Z bZ b f (x)g(y) = dxdy ≤ B(φp , φq ) λ a a (u(x) + u(y)) p1 Z b (u(x))p(1−φq )−1 p × f (x)dx (u0 (x))p−1 a Z b 1q (u(y))q(1−φp )−1 q (3.6) × g (y)dy ; (u0 (y))q−1 a A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents Z b q(1−φp )−1 1− 1q (u(y)) g q (y)dy 0 q−1 (u (y)) a (Z Z b p ) p1 0 b u (y) f (x) = dx dy 1−pφp λ a (u(y)) a (u(x) + u(y)) Z b p1 (u(x))p(1−φq )−1 p f (x)dx ≤ B(φp , φq ) < ∞. (u0 (x))p−1 a 0< (3.7) JJ J II I Go Back Close Quit Page 14 of 28 It follows that (3.6) takes the form of strict inequality by using (3.1); so does J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au (3.7). Hence one can get (3.5). On the other hand, if (3.5) is valid, by (2.2), Z bZ b f (x)g(y) dxdy λ a a (u(x) + u(y)) #" # 1 1 Z b Z b" 0 (u (y)) p (u(y)) p −φp f (x) = dx g(y) dy 1 1 λ a (u(y)) p −φp a (u(x) + u(y)) (u0 (y)) p (Z Z b p ) p1 0 b u (y) f (x) ≤ dx dy 1−pφ λ p a (u(x) + u(y)) a (u(y)) Z b 1q (u(y))q(1−φp )−1 q (3.8) × g (y)dy . (u0 (y))q−1 a Hence by (3.5), (3.1) yields. It follows that (3.1) and (3.5) are equivalent. If the constant factor in (3.5) is not the best possible, one can get a contradiction that the constant factor in (3.1) is not the best possible by using (3.8). Hence the constant factor in (3.5) is still the best possible. If 0 < p < 1 (or p < 0), one can get the reverses of (3.6), (3.7) and (3.8), and thus concludes the equivalence. By (3.6), for 0 < p < 1, one can obtain the reverse of (3.5); for p < 0, one can get (3.5). If the constant factor in the reverse of (3.5) (or simply (3.5)) is not the best possible, then one can get a contradiction that the constant factor in the reverse of (3.1) is not the best possible by using the reverse of (3.8). Thus the theorem is proved. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 15 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 4. Some Particular Results We point out that the constant factors in the following particular results of Theorems 3.1 – 3.2 are all the best possible. 4.1. The first reversible form Corollary 4.1. Let the assumptions of Theorems 3.1 – 3.2 hold. For 1 φr = 1 − (λ − 2) + 1 (r = p, q), r b Z 0< a and (u(x))1−λ p f (x)dx < ∞ (u0 (x))p−1 b (u(x))1−λ q g (x)dx < ∞, 0 q−1 a (u (x)) q+λ−2 setting kλ (p) = B p+λ−2 , , p q Z 0< (i) If p > 1, p1 + 1q = 1, λ > 2 − min{p, q} , then we have the following two equivalent inequalities: Z bZ (4.1) a a b A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit f (x)g(y) dxdy (u(x) + u(y))λ Page 16 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au Z < kλ (p) a b (u(x))1−λ p f (x)dx (u0 (x))p−1 p1 Z a b (u(x))1−λ q g (x)dx (u0 (x))q−1 1q and Z (4.2) a b 0 u (y) (u(y))(p−1)(1−λ) Z a b p f (x) dx dy (u(x) + u(y))λ Z b (u(x))1−λ p p f (x)dx. < [kλ (p)] 0 p−1 a (u (x)) A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality (ii) If 0 < p < 1 and 2 − p < λ < 2 − q, one obtains two equivalent reverses of (4.1) and (4.2), Bicheng Yang (iii) If p < 0 and 2 − q < λ < 2 − p, we have the reverse of (4.1) and the inequality (4.2), which are equivalent. In particular, by (4.1), Title Page α (a) setting u(x) = x (α > 0, x ∈ (0, ∞)), one has Z (4.3) 0 ∞ Z ∞ f (x)g(y) dxdy (xα + y α )λ 0 Z ∞ p1 1 p−1+α(2−λ−p) p < kλ (p) x f (x)dx α 0 Z ∞ 1q q−1+α(2−λ−q) q × x g (x)dx ; Contents JJ J II I Go Back Close Quit Page 17 of 28 0 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au (b) setting u(x) = ln x, x ∈ (1, ∞), one has Z ∞ ∞ Z f (x)g(y) dxdy (ln xy)λ 1 Z ∞ p1 p−1 1−λ p < kλ (p) x (ln x) f (x)dx (4.4) 1 1 Z × ∞ x q−1 1−λ q (ln x) 1q g (x)dx ; 1 (c) setting u(x) = ex , x ∈ (−∞, ∞), one has A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Z ∞ Z ∞ f (x)g(y) dxdy x y λ −∞ (e + e ) Z ∞ p1 (2−p−λ)x p < kλ (p) e f (x)dx (4.5) −∞ Title Page Contents −∞ Z ∞ × e(2−q−λ)x g q (x)dx 1q ; JJ J II I −∞ Go Back (d) setting u(x) = tan x, x ∈ (0, π2 ), one has Z π 2 Z (4.6) 0 0 π 2 f (x)g(y) dxdy (tan x + tan y)λ Close Quit Page 18 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au (Z π 2 < kλ (p) 0 tan1−λ x p f (x)dx sec2(p−1) x ) p1 (Z π 2 0 tan1−λ x q g (x)dx sec2(q−1) x ) 1q ; (e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has Z (4.7) 0 4.2. π 2 Z π 2 f (x)g(y) dxdy λ 0 (sec x + sec y − 2) ) p1 (Z π 2 (sec x − 1)1−λ f p (x)dx < kλ (p) p−1 0 (sec x tan x) ) 1q (Z π 2 (sec x − 1)1−λ . g q (x)dx × q−1 0 (sec x tan x) Corollary 4.2. Let the assumptions of Theorems 3.1 – 3.2 hold. For λ−1 1 + 2 r b Z 0< a Bicheng Yang Title Page Contents The second reversible form φr = A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality (r = p, q), 1−λ (u(x))p 2 p f (x)dx < ∞ (u0 (x))p−1 JJ J II I Go Back Close Quit Page 19 of 28 and Z 0< a b q 1−λ 2 (u(x)) g q (x)dx < ∞, 0 q−1 (u (x)) J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au setting keλ (p) = B pλ−p+2 qλ−q+2 , 2q 2p , (i) If p > 1, p1 + 1q = 1, λ > 1 − 2 min{ p1 , 1q } , then one can get two equivalent inequalities as follows: Z bZ (4.8) a a b f (x)g(y) dxdy (u(x) + u(y))λ ) p1 (Z b p 1−λ 2 (u(x)) < keλ (p) f p (x)dx 0 p−1 (u (x)) a ) 1q (Z 1−λ b (u(x))q 2 q ; g (x)dx × 0 q−1 a (u (x)) A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Z (4.9) a b 0 u (y) p 1−λ 2 (u(y)) Z a b f (x) dx (u(x) + u(y))λ Contents p dy ip Z e < kλ (p) h a b 1−λ (u(x))p 2 p f (x)dx, (u0 (x))p−1 JJ J II I Go Back Close (ii) If 0 < p < 1, 1 − p2 < λ < 1 − 2q , one can get two equivalent reversions of (4.8) and (4.9), (iii) If p < 0, 1 − 2q < λ < 1 − p2 , one can get the reversion of (4.8) and inequality (4.9), which are equivalent. In particular, by (4.8), Quit Page 20 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au (a) setting u(x) = xα (α > 0, x ∈ (0, ∞)), one has Z ∞ (4.10) 0 ∞ Z f (x)g(y) dxdy (xα + y α )λ 0 Z ∞ p1 1e p−1+α(1−p 1+λ ) p 2 < kλ (p) x f (x)dx α 0 1q Z ∞ q−1+α(1−q 1+λ ) q 2 × x g (x)dx ; 0 (b) setting u(x) = ln x, x ∈ (1, ∞), one has A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Z ∞ (4.11) 1 ∞ Z f (x)g(y) dxdy (ln xy)λ 1 p1 Z ∞ p p−1 p 1−λ e x (ln x) 2 f (x)dx < kλ (p) Title Page Contents 1 Z × ∞ q 1−λ 2 xq−1 (ln x) g q (x)dx 1q ; JJ J II I 1 Go Back (c) setting u(x) = ex , x ∈ (−∞, ∞), one has Z ∞ Z ∞ (4.12) −∞ −∞ f (x)g(y) dxdy (ex + ey )λ Close Quit Page 21 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au Z ∞ < keλ (p) (1−p 1+λ )x p 2 e p1 Z ∞ f (x)dx (1−q 1+λ )x q 2 e −∞ 1q g (x)dx ; −∞ (d) setting u(x) = tan x, x ∈ (0, π2 ), one has Z π 2 Z (4.13) 0 0 π 2 f (x)g(y) dxdy (tan x + tan y)λ ) p1 (Z π 1−λ 2 tanp 2 x f p (x)dx < keλ (p) 2(p−1) x 0 sec ) 1q (Z π 1−λ 2 tanq 2 x ; g q (x)dx × 2(q−1) x 0 sec A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page (e) setting u(x) = sec x − 1, x ∈ (0, Z (4.14) 0 π 2 Z π 2 π ), 2 one has f (x)g(y) dxdy λ 0 (sec x + sec y − 2) (Z π ) p1 1−λ 2 (sec x − 1)p 2 < keλ (p) f p (x)dx p−1 (sec x tan x) 0 ) 1q (Z π 1−λ 2 (sec x − 1)q 2 g q (x)dx × . q−1 (sec x tan x) 0 Contents JJ J II I Go Back Close Quit Page 22 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au 4.3. The form which does not have a reverse Corollary 4.3. Let the assumptions of Theorems 3.1 – 3.2 hold. For φr = λ 1 1 (r = p, q), if p > 1, + = 1, λ > 0, r p q Z b (u(x))(p−1)(1−λ) p 0< f (x)dx < ∞ (u0 (x))p−1 a and b (u(x))(q−1)(1−λ) q 0< g (x)dx < ∞, (u0 (x))q−1 a then one can get two equivalent inequalities as: Z bZ b f (x)g(y) (4.15) dxdy λ a a (u(x) + u(y)) Z b p1 λ λ (u(x))(p−1)(1−λ) p <B , f (x)dx p q (u0 (x))p−1 a Z b 1q (u(x))(q−1)(1−λ) q × g (x)dx ; (u0 (x))q−1 a Z A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Z (4.16) a b 0 u (y) (u(y))1−λ Z a b p f (x) dx dy (u(x) + u(y))λ p Z b λ λ (u(x))(p−1)(1−λ) p < B , f (x)dx. p q (u0 (x))p−1 a Quit Page 23 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au In particular, by (4.15), (a) setting u(x) = xα (α > 0; x ∈ (0, ∞)), one has Z ∞ (4.17) 0 ∞ Z f (x)g(y) dxdy (xα + y α )λ 0 Z ∞ p1 λ λ 1 (p−1)(1−αλ) p < B , x f (x)dx α p q 0 Z ∞ 1q (q−1)(1−αλ) q × x g (x)dx ; A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality 0 Bicheng Yang (b) setting u(x) = ln x, x ∈ (1, ∞), one has Z (4.18) ∞ ∞ Z f (x)g(y) dxdy (ln xy)λ 1 1 Z ∞ p1 λ λ p−1 (p−1)(1−λ) p <B , x (ln x) f (x)dx p q 1 Z ∞ 1q q−1 (q−1)(1−λ) q × x (ln x) g (x)dx ; 1 (c) setting u(x) = ex , x ∈ (−∞, ∞), one has Z ∞ Z ∞ (4.19) −∞ −∞ Title Page Contents JJ J II I Go Back Close Quit Page 24 of 28 f (x)g(y) dxdy (ex + ey )λ J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au <B λ λ , p q Z ∞ (1−p)λx p e p1 Z ∞ f (x)dx −∞ (1−q)λx q e g (x)dx 1q ; −∞ (d) setting u(x) = tan x, x ∈ (0, π2 ), one has Z π 2 (4.20) 0 Z π 2 f (x)g(y) dxdy λ 0 (tan x + tan y) ) p1 (Z π 2 tan(p−1)(1−λ) x λ λ f p (x)dx <B , 2(p−1) x sec p q 0 ) 1q (Z π 2 tan(q−1)(1−λ) x ; g q (x)dx × 2(q−1) x sec 0 (e) setting u(x) = sec x − 1, x ∈ (0, π2 ), one has Z (4.21) π 2 Z π 2 f (x)g(y) dxdy λ 0 0 (sec x + sec y − 2) ) p1 (Z π 2 (sec x − 1)(p−1)(1−λ) λ λ <B , f p (x)dx p−1 p q (sec x tan x) 0 (Z π ) 1q 2 (sec x − 1)(q−1)(1−λ) × g q (x)dx . q−1 (sec x tan x) 0 A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 25 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au Remark 1. For α = 1, (4.3) reduces to (1.5). For λ = 1, inequalities (4.3), (4.10) and (4.17) reduce to (1.7), and inequalities (4.4), (4.11) and (4.18) reduce to (1.4). It follows that inequality (3.5) is a relation between (1.4) and (1.7)(ro (1.1)) with a parameter λ. Still for λ = 1, (4.5), (4.12) and (4.19) reduce to Z ∞Z ∞ f (x)g(y) (4.22) dxdy x y −∞ −∞ e + e Z ∞ p1 Z ∞ 1q π (1−q)x q (1−p)x p e g (x)dx , < e f (x)dx −∞ −∞ sin πp (4.6), (4.13) and (4.20) reduce to Z πZ π 2 2 f (x)g(y) dxdy (4.23) 0 0 tan x + tan y (Z π ) p1 (Z π ) 1q 2 2 π < cos2(p−1) xf p (x)dx cos2(q−1) xg q (x)dx , π 0 0 sin p and (4.7), (4.14) and (4.21) reduce to Z πZ π 2 2 f (x)g(y) dxdy (4.24) 0 0 sec x + sec y − 2 (Z π ) p1 (Z π ) 1q p−1 q−1 2 2 2 cos2 x π cos x < f p (x)dx g q (x)dx . π sin x sin x 0 0 sin p A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 26 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au References [1] G. H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, Cambridge, 1952. [2] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving Functions and Their Integrals and Derivatives, Kluwer Academic Publishers, Boston, 1991. [3] H.P. MULHOLLAND, Some theorems on Dirichlet series with positive coefficients and related integrals, Proc. London Math. Soc., 29(2) (1929), 281–292. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality [4] JICHANG KUANG, Applied Inequalities, Shangdong Science and Technology Press, Jinan, 2004. Bicheng Yang [5] JICHANG KUANG, On a new extension of Hilbert’s integral inequality, J. Math. Anal. Appl., 235 (1999), 608–614. Title Page [6] W.T. SULAIMAN, On Hardy-Hilbert’s integral inequality, J. Inequal. in Pure and Appl. Math., 5(2) (2004), Art. 25. [ONLINE: http://jipam. vu.edu.au/article.php?sid=385] Contents JJ J II I [7] ZHUXI WANG AND DUNRIN GUO, An Introduction to Special Functions, Science Press, Beijing, 1979. Go Back [8] BICHENG YANG, On a general Hardy-Hilbert’s inequality with a best value, Chinese Annals of Math., 21A(4) (2000), 401–408. Quit [9] BICHENG YANG, On an extension of Hardy-Hilbert’s inequality, Chinese Annals of Math., 23A(2) (2002), 247–254. Close Page 27 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au [10] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J. Math. Anal. Appl., 261 (2001), 295–306. [11] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl., 220 (1998), 778–785. [12] BICHENG YANG, On new inequality similar to Hardy-Hilbert’s inequality, Math. Ineq. and Appl., 6(1) (2003), 37–44. [13] BICHENG YANG AND Th.M. RASSIAS, On the way of weight coefficient and research for the Hilbert-type inequalities, Math. Ineq. Appl., 6(4) (2003), 625–658. [14] HONG YONG, A extension and improvement of Hardy-Hilbert’s double series inequality, Math. Practice and Theory, 32(5) (2002), 850–854. A Relation to Hardy-Hilbert’s Integral Inequality and Mulholland’s Inequality Bicheng Yang Title Page Contents JJ J II I Go Back Close Quit Page 28 of 28 J. Ineq. Pure and Appl. Math. 6(4) Art. 112, 2005 http://jipam.vu.edu.au