ON THREE INEQUALITIES SIMILAR TO HARDY-HILBERT’S INTEGRAL INEQUALITY W. T. SULAIMAN Abstract. The following three inequalities, which are similar to Hardy-Hilbert integral inequality are proved. Z∞Z∞ β/q α/p x y F (x)G(y) λ p1−1/p q 1−1/q dxdy ≤ max{xλ , y λ } (α + 1)1/p (β + 1)1/q (p − 1)(q − 1) 0 0 0∞ 11/p 0 ∞ 11/q Z Z @ g q (t)dtA · @ f p (t)dtA . 0 0 where Zy Zx f (t)dt, F (x) = G(y) = 0 Z∞Z∞ 0 g(t)dt. 0 x1/p y 1/q F (x)G(y) pq dxdy < B 1/p (p, p)B 1/q (q, q) (x + y)4 (p − 1)(q − 1) 0 0∞ 11/p 0 ∞ 11/q Z Z p q @ A @ · f (t)dt g (t)dtA . 0 0 Received June 13, 2006. 2000 Mathematics Subject Classification. Primary 26D15. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Abstract. Z∞Z∞Z∞ 1/p 1/q 1/r p x y z xyzF (x)G(y)H(z) dxdydz (x + y + z)8 0 0 0 11/r 11/q 0 ∞ 11/p 0 ∞ 0∞ Z Z Z r/2 q/2 p/2 @ h (t)dtA @ g (t)dtA . (t)dtA < Kp Kq Kr @ f 0 0 0 where Zz s h(t)dt, H(z) = Kp = “p p” p/2 B 1/p , B 1/p (p, p). p/2 − 1 2 2 0 1. Introduction Let f, g ≥ 0 satisfy Z∞ 2 f (t)dt < ∞ 0< Z∞ and 0 0< g 2 (t)dt < ∞, 0 then Z∞ Z∞ (1) 0 0 ∞ 1/2 Z Z∞ f (x)g(y) dx dy < π f 2 (t)dt g 2 (t)dt , x+y 0 0 where the constant factor π is the best possible (cf. Hardy et al. [3]). Inequality (1) is well known as Hilbert’s integral inequality. This inequality had been extended by Hardy [1] as follows •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit If p > 1, 1 p + 1 q = 1, f, g ≥ 0 satisfy Z∞ 0< Z∞ f p (t)dt < ∞ and 0< 0 g q (t)dt < ∞, 0 then Z∞ Z∞ (2) 0 π f (x)g(y) dx dy < x+y sin(π/p) ∞ 1/p ∞ 1/q Z Z f p (t)dt g q (t)dt , 0 0 0 π sin(π/p) is the best possible. Inequality (2) is called Hardy-Hilbert’s integral inequality where the constant factor and is important in analysis and application (cf. Mitrinovic et al. [4]). B. Yang gave the following extension of (2) as follows: Theorem ([5]). If λ > 2 − min{p, q}, f, g ≥ 0, satisfy Z∞ 0< t Z∞ 1−λ p f (t)dt < ∞ and 0 0< t1−λ g q (t)dt < ∞, 0 then Z∞ Z∞ (3) 0 ∞ 1/p ∞ 1/q Z Z f (x)g(y) dx dy < kλ (p) t1−λ f p (t)dt t1−λ g q (t)dt , (x + y)λ 0 where the constant factor kλ (p) = B 0 p+λ−2 q+λ−2 , q p 0 is the best possible, B is the beta function. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit The object of this paper is that to give some new inequalities similar to that of Hardy-Hilbert’s inequality. We need the following result for our aim Theorem A ([2]). Let f be a nonnegative integrable function. Define Zx F (x) = f (t)dt. a Then Z∞ F (x) x p dx < p p−1 p Z∞ f p (x)dx, p > 1. 0 0 2. New results We state and prove the following: Ry Rx Theorem 1. Let f, g ≥ 0, F (x) = 0 f (t)dt, G(y) = 0 g(t)dt, p > 1, p1 + 1q = 1, p = λ − α − 1 > 1, q = λ − β − 1 > 1, α, β > −1 then Z∞ Z∞ β/q α/p x y F (x)G(y) λ p1−1/p q 1−1/q dx dy ≤ max{xλ , y λ } (α + 1)1/p (β + 1)1/q (p − 1)(q − 1) 0 0 ∞ 1/p ∞ 1/q Z Z · f p (t)dt g q (t)dt . 0 0 •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Proof. We have Z∞ Z∞ β/q α/p x y F (x)G(y) dx dy max{xλ , y λ } 0 0 Z∞ Z∞ Z∞ Z∞ y α/q F (x) · xβ/q G(y) dx dy 1/q (max{xλ , y λ }) 0 0 0 0 1/p ∞ ∞ 1/q ∞∞ Z Z Z Z β q α p x G (y) y F (x) dx dy dx dy ≤ max{xλ , y λ } max{xλ , y λ } 1/p (max{xλ , y λ }) 0 0 1/p =M N 0 1/q 0 . Observe that Z∞ M = 0 = F p (x)dx Z∞ yα dx dy = max{xλ , y λ } 0 Z∞ p F (x) λ dx = xλ−α−1 (α + 1)p 0 < x Z∞ α Z α y y F p (x)dx dy + dy xλ yλ 0 λ (α + 1)(λ − α − 1) pp−1 λ (α + 1) (p − 1)p Z∞ Z∞ x 0 Z∞ F (x) x p dx 0 f p (x)dx, 0 in view of Theorem A. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Similarly, It can be shown that λ q q−1 N< (β + 1) (q − 1)q Z∞ g q (x)dx. 0 Combining these inequalities to obtain Z∞ Z∞ 0 λ p1−1/p q 1−1/q xβ/q y α/p F (x)G(y) dx dy ≤ λ λ 1/p 1/q max{x , y } (α + 1) (β + 1) (p − 1)(q − 1) 0 1/q 1/p ∞ ∞ Z Z · f p (t)dt g q (t)dt . 0 0 Theorem 2. Let f, g ≥ 0, F (x) = Z∞ Z∞ 0 Rx 0 f (t)dt, G(y) = Ry 0 g(t)dt, p > 1, 1 p + 1 q = 1, then, we have x1/p y 1/q F (x)G(y) pq dxdy < B 1/p (p, p)B 1/q (q, q) (x + y)4 (p − 1)(q − 1) 0 Z∞ · 0 1/p ∞ 1/q Z f p (t)dt g q (t)dt . 0 •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Proof. Z∞ Z∞ 0 x1/p y 1/q F (x)G(y) dxdy (x + y)4 0 Z∞ Z∞ y 1/q F (x) x1/p G(y) · dx dy (x + y)2 (x + y)2 = 0 0 ∞∞ 1/p ∞ ∞ 1/q Z Z p/q p Z Z q/p q y F (x) x G (y) ≤ dx dy dx dy (x + y)2p (x + y)2q = P 0 0 1/p 1/q Q 0 0 . Now, we consider Z∞ P = F (x) x Z∞ p dx 0 0 < B(p, p) p p−1 y p−1 1 x x 2p dy 1 + xy p Z∞ Z∞ = 0 F (x) x Z∞ p dx up−1 du (1 + u)2p 0 f p (x)dx, 0 by Theorem A. Similarly, Q < B(q, q) q q−1 q R∞ g q (y)dy. 0 •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Therefore, we have Z∞ Z∞ 0 x1/p y 1/q F (x)G(y) pq dxdy < B 1/p (p, p)B 1/q (q, q) 4 (x + y) (p − 1)(q − 1) 0 ∞ 1/p ∞ 1/q Z Z · f p (t)dt g q (t)dt . 0 0 This completes the proof of the theorem. Theorem 3. Let f, g, h ≥ 0, p, q, r > 2 Z x F (x) = f (t)dt, 1 p + 1 q + 1 r = 1, Z y G(y) = g(t)dt, 0 Z H(z) = 1 z h(t)dt. 0 Then Z∞ Z∞ Z∞ 1/p 1/q 1/r p x y z xyzF (x)G(y)H(z) dxdydz (x + y + z)8 0 0 0 Z∞ < K p Kq Kr where 1/p f p/2 (x)dx 0 1/q ∞ 1/r Z g q/2 (y)dy hr/2 (z)dz . 0 s Kp = Z∞ 0 p p p/2 B 1/p , B 1/p (p, p). p/2 − 1 2 2 •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Proof. Z∞ Z∞ Z∞ 0 0 p x1/p y 1/q z 1/r xyzF (x)G(y)H(z) dxdydz (x + y + z)8 0 Z∞ Z∞ Z∞ = 0 0 0 p p F (x) z (1/2−1/q) x(1/p+1/r) G(y) · (x + y + z)2 p x(1/2−1/r) y (1/p+1/q) H(z) · dx dy dz (x + y + z)2 y (1/2−1/p) z (1/q+1/r) (x + y + z)2 ∞∞∞ 1/p Z Z Z (p/2−1) p−1 p/2 y z F (x) ≤ dx dy dz (x + y + z)2p 0 0 0 ∞∞∞ 1/q Z Z Z (q/2−1) q−1 q/2 z x G (y) · dx dy dz (x + y + z)2q 0 0 0 ∞∞∞ 1/r Z Z Z (r/2−1) r−1 r/2 x y H (z) · dx dy dz (x + y + z)2r 0 0 0 = A1/p B 1/q C 1/r . •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit Observe that Z∞ A= F p/2 (x) xp/2 Z∞ dx 0 0 Z∞ = y p/2−1 1 x px 1 + xy F (x) x Z∞ p/2 dx 0 Z∞ dy z x+y 0 up/2−1 du (1 + u)p 0 Z∞ p−1 1+ z x+y 1 x+y 2p dz v p−1 dv (1 + v)2p 0 p/2 Z∞ Z∞ p p F (x) p =B , B(p, p) dx < Kp f p/2 (x)dx, 2 2 x 0 0 in view of Theorem A. Similarly, B< Kqq Z∞ g 0 The proof is complete. p/2 (y)dy, C< Krr Z∞ hr/2 (z)dz. 0 1. Hardy G. H., Note on a theorem of Hilbert concerning series of positive terms, Proc. Math. Soc. 23(2) (1925), Records of Proc. XLV-XLVI. 2. , Note on a theorem of Hilbert, Math. Z., 6 (1920) 314–317. 3. Hardy G. H., Littlewood J. E. and Polya G., Inequalities, Cambridge University Press, Cambridge, 1952. 4. Mitrinovic D. S., Pecaric J. E. and Fink A. M., Inequalities involving functions and their integrals and derivatives, Kluwer Academic Publishers, Boston, 1991. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit 5. Yang B., On Hardy-Hilbert’s integral inequality, J. Math. Anal. Appl. 261 (2001) 295–306. W. T. Sulaiman, Department of Mathematics, College of Computer Sciences & Mathematics, University of Mosul, Mosul, Iraq, e-mail: waadsulaiman@hotmail.com •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit