Volume 10 (2009), Issue 1, Article 25, 8 pp. ON A DECOMPOSITION OF HILBERT’S INEQUALITY BICHENG YANG D EPARTMENT OF M ATHEMATICS G UANGDONG E DUCATION I NSTITUTE G UANGZHOU , G UANGDONG 510303, P.R. CHINA bcyang@pub.guangzhou.gd.cn Received 07 October, 2008; accepted 20 March, 2009 Communicated by B. Opic A BSTRACT. By using the Euler-Maclaurin’s summation formula and the weight coefficient, a pair of new inequalities is given, which is a decomposition of Hilbert’s inequality. The equivalent forms and the extended inequalities with a pair of conjugate exponents are built. Key words and phrases: Hilbert’s inequality; Weight coefficient; Equivalent form; Hilbert-type inequality. 2000 Mathematics Subject Classification. 26D15. 1. I NTRODUCTION InP 1908, H. Weyl publishedPthe following Hilbert inequality: If {an }, {bn } are real sequences, ∞ 2 2 0< ∞ n=1 an < ∞ and 0 < n=1 bn < ∞, then [1] (1.1) ∞ ∞ ∞ X ∞ X X X am b n <π a2n b2n m + n n=1 m=1 n=1 n=1 ! 12 , where the constant factor π is the best possible. In 1925, G. H. Hardy gave an extension of (1.1) by introducing one pair of conjugate exponents (p, q) ( p1 + 1q = 1) as [2]: If p > 1, an , bn ≥ 0, P P∞ q p 0< ∞ n=1 an < ∞ and 0 < n=1 bn < ∞, then (1.2) ∞ X ∞ X am b n π < m+n sin( πp ) n=1 m=1 ∞ X n=1 ! p1 apn ∞ X ! 1q bqn , n=1 where the constant factor π/ sin( πp ) is the best possible. We refer to (1.2) as the Hardy-Hilbert inequality. In 1934, Hardy et al. [3] gave some applications of (1.1) and (1.2). By introducing a pair of non-conjugate exponents (p, q) in (1.1), Hardy et al. [3] gave: If p, q > 1, p1 + 1q ≥ The author expresses his sincerest thanks to the referee for his thoughtful suggestions in improving this work. 277-08 2 B ICHENG YANG 1, 0 < λ = 2 − ( p1 + 1q ) ≤ 1, then (1.3) ∞ X ∞ X ∞ X am b n ≤ K(p, q) apn λ (m + n) n=1 m=1 n=1 ! p1 ∞ X ! 1q bqn , n=1 where the constant factor K(p, q) is the best value only for λ = 1. In 1951, Bonsall [4] considered (1.3) in the case of a general kernel. In 1991, Mitrinović et al. [5] summarized the above method and results. In 1997-1998, by using weight coefficients, Yang and Gao [6], [7] gave a strengthened version of (1.2) as: (∞ " # ) p1 ( ∞ " # ) 1q ∞ X ∞ X X X 1−γ p 1−γ q am b n π π − 1/p an − 1/q bn , < (1.4) m+n sin( πp ) n sin( πp ) n n=1 m=1 n=1 n=1 where, 1 − γ = 0.42278433+ (γ is the Euler constant). In 2001, Yang [8] gave an extension of (1.1) by introducing an independent parameter 0 < λ ≤ 4 as ! 12 X ∞ X ∞ ∞ ∞ X X am b n λ λ (1.5) <B , n1−λ a2n n1−λ b2n , λ (m + n) 2 2 n=1 m=1 n=1 n=1 where the constant factor B( λ2 , λ2 ) is the best possible (B(u, v) is the Beta function). In 2004, Yang [9] published the dual form of (1.2) as follows ! 1q ! p1 ∞ ∞ X ∞ ∞ X X X am b n π nq−2 bqn . (1.6) < np−2 apn π m + n sin( ) p n=1 n=1 m=1 n=1 For p = q = 2, both (1.6) and (1.2) reduce to (1.1). It means that there are two different best extensions of (1.1). To generalize (1.2) and (1.6), in 2005, Yang [10] gave an extension of (1.2) and (1.6) with two pairs of conjugate exponents (p, q), (r, s)(p, r > 1) and parameters α, λ > P P −1 p q (1− αλ −1 q p(1− αλ r ) s ) 0 (αλ ≤ min{r, s}) as: If 0 < ∞ an < ∞ and 0 < ∞ bn < ∞, n=1 n n=1 n then ) p1 ( ∞ ) 1q (∞ ∞ X ∞ X X X αλ αλ am bn (1.7) < kαλ (r) np(1− r )−1 apn nq(1− s )−1 bqn , α + nα )λ (m n=1 n=1 m=1 n=1 where the constant factor kαλ (r) = α1 B( λr , λs ) is the best possible. T. K. Pogány [11] also 1 considered a best extension of (1.2) with the non-homogeneous kernel as (λm +ρ µ (µ, λm , ρn > n) 0). We have a non-negative decomposition of kernel in (1.1): 1 max{m, n} min{m, n} = + (m, n ∈ N) m+n (m + n)2 (m + n)2 (N is the set of positive integer numbers). In this paper, by using the Euler-Maclaurin summation formula and the weight coefficient as in [8], we give a pair of new Hilbert-type inequalities as ! 12 ∞ X ∞ ∞ ∞ π X X X max{m, n} (1.8) am b n < +1 a2n b2n ; 2 (m + n) 2 n=1 m=1 n=1 n=1 ! 12 ∞ X ∞ ∞ ∞ π X X X min{m, n} (1.9) am b n < −1 a2n b2n , 2 (m + n) 2 n=1 m=1 n=1 n=1 J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ H ILBERT ’ S I NEQUALITY 3 where the sum of two best constant factors is π. The equivalent forms and extended inequalities with a pair of conjugate exponents are considered. 2. S OME L EMMAS Lemma 2.1 (Euler-Maclaurin’s summation formula, cf. [8, 12, Lemma 1]). If f (x) ∈ C 1 [1, ∞), then we have Z ∞ Z ∞ ∞ X 1 (2.1) f (k) = f (x)dx + f (1) + P1 (x)f 0 (x)dx, 2 1 1 k=1 where P1 (x) = x − [x] − 12 is the Bernoulli function of the first order; if g ∈ C 3 [1, ∞), (−1)i g (i) (x) > 0, g (i) (∞) = 0, (i = 0, 1, 2, 3), then Z n 1 (2.2) [g(n) − g(1)] < P1 (x)g(x)dx < 0, 12 1 Z ∞ 1 − g(n) < P1 (x)g(x)dx < 0. 12 n Lemma 2.2. If 1 2 ≤ α < 1, setting the weight coefficient ω(α, m) as ∞ X max{m, n}mα ω(α, m) := (2.3) (m ∈ N), (m + n)2 nα n=1 then we have (2.4) k(α) = Aα (m) + ω(α, m); ω 1 ,m 2 1 π <k = + 1, 2 2 where 1 k(α) := α Z 0 ∞ max u1/α , 1 1 −2 u α du (u1/α + 1)2 and Aα (m) = O(mα−1 ), (m → ∞). Proof. For fixed follows that (2.5) 1 2 ≤ α < 1, m ∈ N, setting f (x) := ω(α, m) = m α ∞ X max{m,x} , (m+x)2 xα x ∈ (0, ∞), then by (2.1), it f (n) n=1 =m =m α α Z ∞ 1 ∞ Z 1 f (x)dx + f (1) + 2 Z ∞ 0 P1 (x)f (x)dx 1 f (x)dx − mα ρ(α, m), P1 (x)f 0 (x)dx. 0 Z (2.6) ρ(α, m) := 0 1 1 f (x)dx − f (1) − 2 Z ∞ P1 (x)f 0 (x)dx. 1 We find 1 −m −1 1 − f (1) = = + , 2 2 2(m + 1) 2(m + 1) 2(m + 1)2 J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ 4 B ICHENG YANG and 1 1 Z 1 m m 1 f (x)dx = dx ≥ dx = ; 2 α 2 m+1 0 0 (m + x) x 0 (m + x) Z 1 Z 1 1 m f (x)dx ≤ dx = . 2 α (1 − α)m 0 0 m x Z Z For x ∈ (0, m), f (x) = f (x) = x1−α , (m+x)2 m , (m+x)2 xα it follows f 0 (x) = −2m (m+x)3 xα − αm ; (m+x)2 xα+1 for x ∈ (m, ∞), we find 1−α −2x1−α + (m + x)3 (m + x)2 xα −2(x + m − m) 1−α = + (m + x)3 xα (m + x)2 xα −2 2m 1−α = + + . (m + x)2 xα (m + x)3 xα (m + x)2 xα f 0 (x) = 1 1 In the following, it is obvious that g1 (x) = (m+x) 3 xα , g2 (x) = (m+x)2 xα+1 and g3 (x) = are suited to apply in (2.2). Then by (2.2), we obtain Z m (2.7) − P1 (x)f 0 (x)dx 1 Z m Z m 2mP1 (x)dx αmP1 (x)dx = + 3 α (m + x) x (m + x)2 xα+1 1 1 2m 1 1 αm 1 1 > − + − 12 8m3+α (m + 1)3 12 4m3+α (m + 1)2 α 2−α 1 α+1 = − − + ; 2+α 2 48m 12(m + 1) 12(m + 1) 6(m + 1)3 Z (2.8) 1 (m+x)2 xα ∞ − m P1 (x)f 0 (x)dx Z ∞ Z ∞ Z ∞ 2P1 (x)dx 2mP1 (x)dx P1 (x)dx = − − (1 − α) 2 α 3 α 2 α m (m + x) x m (m + x) x m (m + x) x Z ∞ −1 > − P1 (x)f 0 (x)dx 24m2+α 1 Z m Z ∞ 0 =− P1 (x)f (x)dx − P1 (x)f 0 (x)dx 1 m α−1 α 2−α 1 > − − + . 2+α 2 48m 12(m + 1) 12(m + 1) 6(m + 1)3 Hence by (2.6), for α = 12 , it follows that 1 −1 1 1 (2.9) ρ ,m > + + 2 2 2(m + 1) 2(m + 1) m+1 1 1 −1 2 − 12 1 2 + 2 2+1/2 − − + 2 48m 12(m + 1) 12(m + 1) 6(m + 1)3 J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ H ILBERT ’ S I NEQUALITY 5 11 9 −1 1 + + + 2 2+1/2 24(m + 1) 24(m + 1) 96m 6(m + 1)3 9 −1 1 11 ≥ + + + > 0. 2 2 24(m + 1) 96m 96m 6(m + 1)3 = By (2.7) and (2.8), we obtain Z ∞ Z 0 − P1 (x)f (x)dx = − 1 m Z 0 ∞ P1 (x)f (x)dx − 1 P1 (x)f 0 (x)dx m 1 1−α < + 2+α 48m 48m2+α 2−α = . 48m2+α Then by (2.6), it follows 0 < m1−α [mα ρ(α, m)] −m m 1 2−α < + + + 2 2(m + 1) 2(m + 1) 1 − α 48m1+α 1 1 → − (m → ∞). 1−α 2 (2.10) Setting u = (x/m)α , we find Z ∞ Z ∞ max{m, x} α α (2.11) dx m f (x)dx = m (m + x)2 xα 0 0 Z 1 ∞ max{u1/α , 1} 1 −2 = u α du = k(α), α 0 (u1/α + 1)2 Z ∞ Z 1 1 max{u2 , 1} du (2.12) =2 du = 4 k 2 2 2 (u2 + 1)2 0 0 (u + 1) Z π 4 π =4 cos2 θdθ = + 1. 2 0 Hence by (2.5), (2.9), (2.10) and (2.11), (2.4) is valid and the lemma is proved. Similar to Lemma 2.2, we still have Lemma 2.3. If 1 2 ≤ α < 1, setting the weight coefficient $(α, m) as $(α, m) := (2.13) ∞ X min{m, n}mα (m ∈ N), (m + n)2 nα n=1 then we have (2.14) e k(α) = Bα (m) + $(α, m); $ 1 ,m 2 1 π e <k = − 1, 2 2 where 1 e k(α) = α and Bα (m) = O(m α−2 Z 0 ∞ min{u1/α , 1} 1 −2 u α du (u1/α + 1)2 ), (m → ∞). J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ 6 B ICHENG YANG 3. M AIN R ESULTS AND THEIR E QUIVALENT F ORMS P P p q −1 p −1 q 2 2 Theorem 3.1. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 < ∞ an < ∞ and 0 < ∞ bn < n=1 n n=1 n ∞, then we have the following equivalent inequalities (3.1) (3.2) ! 1q ! p1 ∞ ∞ X ∞ ∞ π X X q X p max{m, n} ; n 2 −1 bqn I := am b n < +1 n 2 −1 apn 2 (m + n) 2 n=1 m=1 n=1 n=1 " ∞ #p ∞ ∞ π p X X X max{m, n} q p −1 J := am < +1 n 2 −1 apn , n2 2 (m + n) 2 m=1 n=1 n=1 where the constant factors π 2 + 1 and π 2 p + 1 are the best possible. Proof. By Hölder’s inequality and (2.3) – (2.4), we find " ∞ #p ( ∞ )p X max{m, n} X max{m, n} m1/(2q) n1/(2p) am = am (3.3) (m + n)2 (m + n)2 n1/(2p) m1/(2q) m=1 m=1 " ∞ # X max{m, n} mp/(2q) p ≤ a (m + n)2 n1/2 m m=1 " ∞ #p−1 X max{m, n} nq/(2p) × (m + n)2 m1/2 m=1 ∞ X 1 max{m, n} mp/(2q) p p−1 1− p2 ,n n a =ω 2 (m + n)2 n1/2 m m=1 ∞ π p−1 X max{m, n} mp/(2q) p 1− p2 ≤ +1 n a ; 2 (m + n)2 n1/2 m m=1 J≤ = = π 2 π 2 π 2 ∞ X ∞ p−1 X max{m, n} mp/(2q) p +1 am 2 1/2 (m + n) n n=1 m=1 "∞ # ∞ p−1 X X max{m, n} mp/(2q) +1 apm 2 1/2 (m + n) n m=1 n=1 ∞ ∞ π p−1 X p X p p 1 −1 p +1 ω , m m 2 am < +1 m 2 −1 apm . 2 2 m=1 m=1 Therefore (3.2) is valid. By Hölder’s inequality, we find that (3.4) I= ∞ X n=1 " 1 1 n2−p # ∞ h −1 1 i X 1 max{m, n} am n 2 + p b n ≤ J p 2 (m + n) m=1 ∞ X ! 1q q n 2 −1 bqn . n=1 Then by (3.2), we have (3.1). On the other hand, suppose that (3.1) is valid. Setting " ∞ #p−1 X max{m, n} p bn := n 2 −1 am , n ∈ N, (m + n)2 m=1 J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ H ILBERT ’ S I NEQUALITY 7 P q −1 q 2 then it follows ∞ bn = J. By (3.3), we confirm that J < ∞. If J = 0, then (3.2) is n=1 n naturally valid; if 0 < J < ∞, then by (3.1), we find ! p1 ∞ ! 1q ∞ ∞ π X X q X q p +1 n 2 −1 apn n 2 −1 bqn ; n 2 −1 bqn = J = I < 2 n=1 n=1 n=1 ! p1 ! p1 ∞ ∞ π X p X q 1 n 2 −1 bqn = Jp < , +1 n 2 −1 apn 2 n=1 n=1 and inequality (3.2) is valid, which is equivalent to (3.1). −1 −1 − pε − qε 2 2 e eb = {ebn }∞ as e For 0 < ε < 2q , setting e , b , for n ∈ N, if there exists a = {e an } ∞ , a n n n=1 n=1 π a constant 0 < k ≤ 2 + 1, such that (3.1) is still valid when we replace π2 + 1 by k, then we find ! p1 ∞ ! 1q ∞ X ∞ ∞ ∞ X X X X e p q max{m, n}e am bn 1 −1 p −1eq e 2 2 I := <k n e an n bn =k ; 2 1+ε (m + n) n n=1 m=1 n=1 n=1 n=1 # " ∞ ∞ X X max{m, n} −1 − ε −1 ε m 2 p n 2 −q Ie = 2 (m + n) n=1 m=1 1 ε ∞ ∞ X 1 X max{m, n}m 2 + q 1 1 ε = = ω + ,m . 1 ε 1+ε 1+ε 2n 2 + q m m 2 q (m + n) m=1 m=1 n=1 And then by (2.4) and the above results, it follows that X ∞ ∞ X 1 1 ε 1 1 A 1 + ε (m) (3.5) k >k + 1− 1+ε 1+ε 2 q 1 ε n 2 q m k 2+q n=1 m=1 X ∞ ∞ X 1 1 ε 1 1 =k + − A 1 ε (m) 1+ε 1+ε 2 + q 1 ε 2 q m m k + m=1 2 q m=1 P ∞ X 1 ∞ 1 ε (m)] A m=1 m1+ε 2 + q 1 ε 1 ; P =k + 1 − ∞ 1 ε 1 2 q m=1 m1+ε k 2+q m=1 m1+ε 1 ε P∞ 1 1 2−q O m=1 m1+ε m 1 ε . P k>k + 1− ∞ 1 2 q k 1+ε ∞ X 2 q m=1 m1+ε + qε , by Fatou’s Lemma, it follows that Z 1 ε 1 ∞ max{u1/α , 1} 1 −2 lim k + = lim+ u α du ε→0+ 2 q (u1/α + 1)2 α→ 12 α 0 Z ∞ 1 max{u1/α , 1} 1 −2 π ≥2 lim+ u α du = k = + 1. 1/α 2 (u + 1) 2 2 α→ 21 0 Setting α = 1 2 Then by (3.5), we have k ≥ π2 + 1 (ε → 0+ ). Hence k = π2 + 1 is the best value of (3.1). We confirm that the constant factor in (3.2) is the best, otherwise we would obtain a contradiction by (3.4) that the constant factor in (3.1) is not the best possible. The theorem is proved. In the same manner, by Lemma 2.3, we have: J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 25, 8 pp. http://jipam.vu.edu.au/ 8 B ICHENG YANG P P p q −1 p −1 q 2 2 Theorem 3.2. If p > 1, p1 + 1q = 1, an , bn ≥ 0, 0 < ∞ an < ∞ and 0 < ∞ bn < n=1 n n=1 n ∞, then we have the following equivalent inequalities ! p1 ∞ ! 1q ∞ X ∞ ∞ π X X X q p min{m, n} (3.6) am b n < −1 n 2 −1 apn ; n 2 −1 bqn 2 (m + n) 2 n=1 m=1 n=1 n=1 " ∞ #p ∞ ∞ X p X min{m, n} pX q π n 2 −1 a < − 1 n 2 −1 apn , m 2 (m + n) 2 n=1 m=1 n=1 where the constant factors π 2 − 1 and ( π2 − 1)p are the best possible. Remark 1. For p = q = 2, (3.1) reduces to (1.8) and (3.6) reduces to (1.9). R EFERENCES [1] H. WEYL, Singuläre Integralgleichungen mit besonderer Berücksichtigung des Fourierschen Integraltheorems, Göttingen : Inaugural–Dissertation, 1908. [2] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive term, Proceedings of the London Math. Society, 23 (1925),45–46. [3] G.H. HARDY, J.E. LITTLEWOOD Cambridge,1934. AND G. PÓLYA, Inequalities, Cambridge University Press, [4] F.F. BONSALL, Inequalities with non-conjugate parameter, J. Math. Oxford Ser., 2(2) (1951), 135–150. [5] D. 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