SOME NEW HILBERT-PACHPATTE’S INEQUALITIES Hilbert-Pachpatte’s Inequalities WENGUI YANG School of Mathematics and Computational Science Xiangtan University, Xiangtan, 411105 Hunan, P.R. China Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page EMail: yangwg8088@163.com Contents Received: 05 September, 2008 Accepted: 2 January, 2009 Communicated by: W.S. Cheung 2000 AMS Sub. Class.: 26D15. Key words: Hilbert-Pachpatte’s inequality; Hölder’s inequality; Jensen’s inequality; Nonnegative sequences. Abstract: Some new Hilbert-Pachpatte discrete inequalities and their integral analogues are established in this paper. Other inequalities are also given in remarks. JJ II J I Page 1 of 28 Go Back Full Screen Close Contents 1 Introduction 3 2 Main Results 6 3 Integral Analogues 17 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 2 of 28 Go Back Full Screen Close 1. Introduction Let p ≥ 1, q ≥ 1 and {am } and {bn } be two nonnegative sequences of real numbers defined for m = 1, 1, 2, . . . , r, where k and r are natural numbers P2, . . . , k and n = P n a and B = and define Am = m n s=1 s t=1 bt . Then ( k ) 12 k X r X X Apm Bnq 2 (1.1) ≤ C(p, q, k, r) (k − m + 1)(Ap−1 m am ) m + n m=1 n=1 m=1 ( r ) 12 X × , (r − n + 1)(Bnq−1 bn )2 n=1 1 pq 2 √ unless {am } or {bn } is null, where C(p, q, k, r) = kr. An integral analogue of (1.1) is given in the following result. Let p ≥ 1, q ≥ 1 and f (σ) ≥ 0, g(τ ) ≥ 0Rfor σ ∈ (0, x), τ ∈ (0, y), R t where x, y s are positive real numbers and define F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ, for s ∈ (0, x), t ∈ (0, y). Then Z x 21 Z xZ y p F (s)Gq (t)dsdt p−1 2 (1.2) ≤ D(p, q, x, y) (x − s)(F (s)f (s)) ds s+t 0 0 0 Z y 12 q−1 2 × (y − t)(G (t)g(t)) dt , 0 √ 1 pq xy. 2 unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y) = Inequalities (1.1) and (1.2) are the well known Hilbert-Pachpatte inequalities [1], which gave new estimates on Hilbert type inequalities [2]. It is well known that the Hilbert-Pachpatte inequalities play a dominant role in analysis, so the literature on such inequalities and their applications is vast [3] – [8]. Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 3 of 28 Go Back Full Screen Close Young-Ho Kim [9] gave new inequalities similar to the Hilbert-Pachpatte inequalities as follows. Let p ≥ 1, q ≥ 1, α > 0, and {am } and {bn } be two nonnegative sequences of real numbers defined for m = 1, 2, P.m. . , k and n = 1,P2,n . . . , r, where k and r are natural numbers and define Am = s=1 as and Bn = t=1 bt . Then (1.3) k X r X m=1 n=1 ( Apm Bnq (mα + 1 nα ) α ≤ C(p, q, k, r; α) k X ) 12 2 (k − m + 1)(Ap−1 m am ) Hilbert-Pachpatte’s Inequalities Wengui Yang m=1 × ( r X vol. 10, iss. 1, art. 26, 2009 ) 12 (r − n + 1)(Bnq−1 bn )2 , n=1 Title Page 1 √ unless {am } or {bn } is null, where C(p, q, k, r; α) = 12 α pq kr. An integral analogue of (1.3) is given in the following result. Let p ≥ 1, q ≥ 1, α > 0 and f (σ) ≥ 0, g(τ ) ≥ R0 for σ ∈ (0, x), τ ∈ (0,Ry), where s t x, y are positive real numbers and define F (s) = 0 f (σ)dσ and G(t) = 0 g(τ )dτ, for s ∈ (0, x), t ∈ (0, y). Then Z xZ y p F (s)Gq (t)dsdt (1.4) 1 (sα + tα ) α 0 0 Z x 12 p−1 2 ≤ D(p, q, x, y; α) (x − s)(F (s)f (s)) ds 0 Z × y (y − t)(G q−1 12 (t)g(t)) dt , 0 unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α) = 1 2 α1 √ pq xy. 2 Contents JJ II J I Page 4 of 28 Go Back Full Screen Close The purpose of the present paper is to derive some new generalized inequalities (1.1) and (1.2) that are similar to (1.3) and (1.4). By applying an elementary inequality, we also obtain some new inequalities similar to some results in [1, 9]. Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 5 of 28 Go Back Full Screen Close 2. Main Results Now we give our results as follows in this paper. Theorem 2.1. Let p ≥ 1, q ≥ 1, α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of real numbers defined for m = 1, P 2, . . . , k and n = 1, 2, P.n. . , r, m where k and r are natural numbers and define Am = s=1 as and Bn = t=1 bt . Then k X r X Apm Bnq (2.1) ≤ C(p, q, k, r; α, γ) (γ−1)(α+γ) (α−1)(α+γ) αγ αγ + αn m=1 n=1 γm ( k ) α1 ( r ) γ1 X X α , × (k − m + 1)(Ap−1 (r − n + 1)(Bnq−1 bn )γ m am ) m=1 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page n=1 unless {am } or {bn } is null, where C(p, q, k, r; α, γ) = Contents α−1 pq k α r α+γ γ−1 γ . Proof. The idea for the proof Theorem 2.1 comes from Theorem 1 of [1] and Theorem 2.1 of [9]. From the hypotheses of Theorem 2.1 and using the following inequality (see [10, 11]), ( n )β ( m )β−1 n X X X (2.2) zm ≤β zm zk , m=1 m=1 k=1 where β ≥ 1 is a constant and zm ≥ 0, (m = 1, 2, . . . , n), it is easy to observe that m X p (2.3) Am ≤ p Ap−1 m = 1, 2, . . . , k, s as , Bnq ≤ q s=1 n X t=1 Btq−1 bt , n = 1, 2, . . . , r. JJ II J I Page 6 of 28 Go Back Full Screen Close From (2.3) and Hölder’s inequality, we have ( m ) α1 m X X α−1 α α (2.4) Ap−1 , (Ap−1 s as ≤ m s as ) s=1 m = 1, 2, . . . , k, s=1 and (2.5) n X Btq−1 bt ≤n γ−1 γ ( n X t=1 ) γ1 Hilbert-Pachpatte’s Inequalities (Btq−1 bt )γ , n = 1, 2, . . . , r. t=1 Using the inequality of means [12] ) r1 ( n ( ) Ω1 n n X Y 1 ≤ ωi sri (2.6) sωi i Ωn i=1 i=1 P for r > 0, ωi > 0, ni=1 ωi = Ωn , we observe that (sω1 1 sω2 2 )r/(ω1 +ω2 ) ≤ (2.7) α−1 Let s1 = m , s2 = n and (2.7), we have (2.8) Apm Bnq ≤ pqm γ−1 α−1 α , ω1 = n γ−1 γ 1 , α ( m X ≤ pqαγ α+γ m ω2 = 1 γ Contents ) α1 ( n X t=1 + n (γ−1)(α+γ) αγ γ ) II J I Go Back ) γ1 (Btq−1 bt )γ JJ Page 7 of 28 and r = ω1 + ω2 , from (2.3) – (2.5) α (Ap−1 s as ) (α−1)(α+γ) αγ α Title Page 1 (ω1 sr1 + ω2 sr2 ) . ω1 + ω2 s=1 ( Wengui Yang vol. 10, iss. 1, art. 26, 2009 Full Screen Close × ( m X α (Ap−1 s as ) ) α1 ( n X s=1 ) γ1 (Btq−1 bt )γ , t=1 for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.8), we observe that Apm Bnq (2.9) γm (α−1)(α+γ) αγ + αn (γ−1)(α+γ) αγ pq ≤ α+γ Hilbert-Pachpatte’s Inequalities ( m X α (Ap−1 s as ) s=1 ) α1 ( n X Wengui Yang ) γ1 (Btq−1 bt )γ , t=1 for m = 1, 2, . . . , k, n = 1, 2, . . . , r. Taking the sum on both sides of (2.9) first over n from 1 to r and then over m from 1 to k of the resulting inequality and using Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order of summations, we observe that k X r X Apm Bnq (α−1)(α+γ) vol. 10, iss. 1, art. 26, 2009 (γ−1)(α+γ) γm αγ + αn αγ ( m ) α1 r ( n ) γ1 k X X X X pq α (Ap−1 ≤ (Btq−1 bt )γ s as ) α + γ m=1 s=1 n=1 t=1 Title Page Contents JJ II J I Page 8 of 28 m=1 n=1 ≤ pq α−1 k α α+γ ( k X m X m=1 s=1 ) α1 α (Ap−1 s as ) r γ−1 γ ( r n XX n=1 t=1 ) γ1 (Btq−1 bt )γ Go Back Full Screen Close = pq k α+γ α−1 α × r γ−1 γ ( k X ) α1 α (k − m + 1)(Ap−1 m am ) m=1 ( r X ) γ1 (r − n + 1)(Bnq−1 bn )γ . n=1 Hilbert-Pachpatte’s Inequalities Remark 1. In Theorem 2.1, setting α = γ = 2, we have (1.1). In Theorem 2.1, setting α1 + γ1 = 1, we have k X r X m=1 n=1 Apm Bnq ≤ C(p, q, k, r; α, γ) γmα−1 + αnγ−1 ( k ) α1 ( r ) γ1 X X α , × (k − m + 1)(Ap−1 (r − n + 1)(Bnq−1 bn )γ m am ) m=1 n=1 unless {am } or {bn } is null, where C(p, q, k, r; α, γ) = Remark 2. In Theorem 2.1, setting p = q = 1, we have (2.10) k X r X α−1 pq k α r α+γ γ−1 γ . (α−1)(α+γ) αγ Contents JJ II J I Page 9 of 28 Full Screen Am Bn γm Title Page Go Back (γ−1)(α+γ) + αn αγ ( k ) α1 ( r ) γ1 X X ≤ C(1, 1, k, r; α, γ) (k − m + 1)aαm (r − n + 1)bγn , m=1 n=1 Wengui Yang vol. 10, iss. 1, art. 26, 2009 m=1 unless {am } or {bn } is null, where C(1, 1, k, r; α, γ) = n=1 α−1 1 k α r α+γ γ−1 γ . Close In the following theorem we give a further generalization of the inequality (2.10) obtained in Remark 2. Before we give our result, we point out that {pm } and {qn } should be two positive sequences for m = 1, 2, . . . , k and n = 1, 2, . . . , r in Theorem 2.3 of [9]. Theorem 2.2. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of real numbers and {pm } and {qn } be positive sequences defined for m = 1,P 2, . . . , k and n P = 1, 2, . . . , r, where k and r are natural numbers and define Am = m s=1 as , P P n Bn = nt=1 bt , Pm = m p and Q = q . Let Φ and Ψ be two real-valued, s n t s=1 t=1 nonnegative, convex, and submultiplicative functions defined on R+ = [0, ∞). Then (2.11) k X r X Φ(Am )Ψ(Bn ) γm Title Page where M (k, r; α, γ) = Wengui Yang vol. 10, iss. 1, art. 26, 2009 (γ−1)(α+γ) αγ (α−1)(α+γ) αγ + αn ( k α ) α1 X am ≤ M (k, r; α, γ) (k − m + 1) pm Φ pm m=1 ( r γ ) γ1 X bn × (r − n + 1) qn Ψ , q n n=1 m=1 n=1 Hilbert-Pachpatte’s Inequalities Contents JJ II J I Page 10 of 28 Go Back Full Screen 1 α+γ ( γ−1 ( r γ ) α ) α−1 α k X Ψ(Qn ) γ−1 γ X Φ(Pm ) α−1 m=1 Pm n=1 Qn Close . Proof. From the hypotheses of Φ and Ψ and by using Jensen’s inequality and Hölder’s inequality, it is easy to observe that P Pm m ps as /ps s=1 Pm (2.12) Φ(Am ) = Φ s=1 ps Pm m ps as /ps Φ(Pm ) X as s=1 Pm ≤ Φ(Pm )Φ ≤ ps Φ Pm s=1 ps s=1 ps ( m α ) α1 as Φ(Pm ) α−1 X m α ps Φ , ≤ Pm p s s=1 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 and similarly, Title Page (2.13) Ψ(Bn ) ≤ Ψ(Qn ) γ−1 n γ Qn ( n γ ) γ1 X bt . qt Ψ q t t=1 Let s1 = mα−1 , s2 = nγ−1 , ω1 = α1 , ω2 = γ1 and r = ω1 + ω2 , from (2.7), (2.12) and (2.13), we have ( m α ) α1 X γ−1 α−1 Φ(Pm ) as Φ(Am )Ψ(Bn ) ≤ m α n γ ps Φ (2.14) Pm p s s=1 ( n γ ) γ1 X Ψ(Qn ) bt × qt Ψ Qn q t t=1 ( (α−1)(α+γ) (γ−1)(α+γ) ) αγ m αγ n αγ + ≤ α+γ α γ Contents JJ II J I Page 11 of 28 Go Back Full Screen Close ) α1 ( m ( n ) γ1 α γ X Φ(Pm ) X as bt Ψ(Qn ) × ps Φ qt Ψ Pm p Q q s n t s=1 t=1 for m = 1, 2, . . . , k, n = 1, 2, . . . , r. From (2.14), we observe that Φ(Am )Ψ(Bn ) (2.15) γm (α−1)(α+γ) αγ + αn (γ−1)(α+γ) αγ Hilbert-Pachpatte’s Inequalities ( m α ) α1 X as 1 Φ(Pm ) ≤ ps Φ α+γ Pm p s s=1 ( n γ ) γ1 X Ψ(Qn ) bt × qt Ψ Qn q t t=1 for m = 1, 2, . . . , k, n = 1, 2, . . . , r. Taking the sum on both sides of (2.15) first over n from 1 to r and then over m from 1 to k of the resulting inequality and using Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order of summations, we observe that k X r X (γ−1)(α+γ) γm αγ + αn αγ ( m ( n α ) α1 γ ) γ1 k r X X X X 1 Φ(Pm ) as Ψ(Qn ) bt ≤ ps Φ qt Ψ α + γ m=1 Pm p Q q s n t s=1 n=1 t=1 m=1 n=1 ≤ 1 α+γ ( α k X Φ(Pm ) α−1 m=1 Pm Title Page Contents JJ II J I Page 12 of 28 Go Back Φ(Am )Ψ(Bn ) (α−1)(α+γ) Wengui Yang vol. 10, iss. 1, art. 26, 2009 ) α−1 ( α α k X m X as ps Φ ps m=1 s=1 ) α1 Full Screen Close γ−1 ( ( r γ ) γ ) γ1 r X n X Ψ(Qn ) γ−1 γ X bt × qt Ψ Qn qt n=1 n=1 t=1 α−1 γ−1 ( k ) α ( r γ ) α α−1 γ−1 γ X X 1 Φ(Pm ) Ψ(Qn ) = α + γ m=1 Pm Qn n=1 1 ( k γ ) γ1 α ) α (X r X bt as × (r − n + 1) qt Ψ . (k − m + 1) ps Φ ps qt n=1 m=1 Remark 3. From the inequality (2.7), we obtain 1 (2.16) sω1 1 sω2 2 ≤ ω1 sω1 1 +ω2 + ω2 sω2 1 +ω2 ω1 + ω2 for ω1 > 0, ω2 > 0. If we apply the elementary inequality (2.16) on the right-hand sides of (2.1) in Theorem 2.1 and (2.11) in Theorem 2.2, then we get the following inequalities k X r X m=1 n=1 Apm Bnq γm (α−1)(α+γ) αγ + αn Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 13 of 28 Go Back (γ−1)(α+γ) αγ Full Screen ( ) α+γ αγ k X αγC(p, q, k, r; α, γ) 1 p−1 α ≤ (k − m + 1)(Am am ) α+γ α m=1 ( r ) α+γ αγ X 1 , + (r − n + 1)(Bnq−1 bn )γ γ n=1 Close where C(p, q, k, r; α, γ) = k X r X α−1 pq k α r α+γ γ−1 γ . Also, Φ(Am )Ψ(Bn ) (γ−1)(α+γ) + αn αγ ( α ) α+γ αγ k X am αγM (k, r; α, γ) 1 ≤ (k − m + 1) pm Φ α+γ α m=1 pm m=1 n=1 γm (α−1)(α+γ) αγ Hilbert-Pachpatte’s Inequalities Wengui Yang ( r γ ) α+γ αγ X bn 1 , + (r − n + 1) qn Ψ γ n=1 qn vol. 10, iss. 1, art. 26, 2009 Title Page where Contents 1 M (k, r; α, γ) = α+γ ( γ−1 ) α−1 ( r γ ) α α−1 γ−1 α k γ X X Φ(Pm ) Ψ(Qn ) m=1 Pm n=1 Qn . The following theorems deal with slight variants of the inequality (2.11) given in Theorem 2.2. Theorem 2.3. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of real numbers defined for m = 1, 2,P . . . , k and n = 1, 2, .P . . , r, where k and r are n 1 natural numbers and define Am = m1 m a and B = s n s=1 t=1 bt . Let Φ and Ψ be n two real-valued, nonnegative, convex functions defined on R+ = [0, ∞). Then k X r X m=1 n=1 mnΦ(Am )Ψ(Bn ) γm (α−1)(α+γ) αγ + αn (γ−1)(α+γ) αγ ≤ C(1, 1, k, r; α, γ) JJ II J I Page 14 of 28 Go Back Full Screen Close ( × k X ) α1 ( r ) γ1 X , (k − m + 1)Φα (am ) (r − n + 1)Ψγ (bn ) m=1 where C(1, 1, k, r; α, γ) = n=1 α−1 1 k α r α+γ γ−1 γ . Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy to observe that ! ( m ) α−1 m m α X X X 1 1 α−1 1 α α , Φ(Am ) = Φ as ≤ Φ(as ) ≤ m Φ (as ) m s=1 m s=1 m s=1 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page n Ψ(Bn ) = Ψ 1X bt n t=1 ! n 1X 1 γ−1 ≤ Ψ(bt ) ≤ n γ n t=1 n ( n X ) γ−1 γ Ψγ (bt ) Contents . t=1 The rest of the proof can be completed by following the same steps as in the proofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details. Theorem 2.4. Let α > 1, γ > 1 and {am } and {bn } be two nonnegative sequences of real numbers and {pm } and {qn } be positive sequences defined for m = 1,P 2, . . . , k and n P = 1, 2, . . . , r, whereP k and r are natural numbers and define Pm = m s=1 ps , P n 1 Qn = nt=1 qt , Am = P1m m p a and B = q b . Let Φ and Ψ be two n s=1 m s t=1 n t Qn real-valued, nonnegative, convex functions defined on R+ = [0, ∞). Then k X r X m=1 n=1 Pm Qn Φ(Am )Ψ(Bn ) γm (α−1)(α+γ) αγ + αn (γ−1)(α+γ) αγ JJ II J I Page 15 of 28 Go Back Full Screen Close ( ≤ C(1, 1, k, r; α, γ) k X ) α1 (k − m + 1) [pm Φ (am )]α m=1 × ( r X ) γ1 (r − n + 1) [qn Ψ (bn )]γ , n=1 where C(1, 1, k, r; α, γ) = α−1 1 k α r α+γ γ−1 γ Hilbert-Pachpatte’s Inequalities . Wengui Yang Proof. From the hypotheses and by using Jensen’s inequality and Hölder’s inequality, it is easy to observe that ! m 1 X p s as Φ(Am ) = Φ Pm s=1 ( m ) α−1 m α 1 X 1 α−1 X , ≤ ps Φ(as ) ≤ m α [ps Φ(as )]α Pm s=1 Pm s=1 vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 16 of 28 Ψ(Bn ) = Ψ n 1 X qt b t Qn t=1 ! n 1 X 1 γ−1 ≤ qt Ψ(bt ) ≤ n γ Qn t=1 Qn Go Back ( n X Full Screen ) γ−1 γ [qt Ψ(bt )]γ . t=1 The rest of the proof can be completed by following the same steps as in the proofs of Theorems 2.1 and 2.2 with suitable changes and hence we omit the details. Close 3. Integral Analogues Now we give the integral analogues of the inequalities in Theorems 2.1 – 2.4. An integral analogue of Theorem 2.1 is given in the following theorem. Theorem 3.1. Let p ≥ 0, q ≥ 0, α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥R 0 for σ ∈ s (0, x), τ ∈ (0, y), where x, y are positive real numbers, define F (s) = 0 f (σ)dσ, Rt G(t) = 0 g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Then Z xZ y F p (s)Gq (t) (3.1) dsdt (γ−1)(α+γ) (α−1)(α+γ) αγ 0 0 γs + αt αγ Z x α1 p−1 α ≤ D(p, q, x, y; α, γ) (x − s)(F (s)f (s)) ds Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page 0 Contents Z × y γ1 q−1 γ , (y − t)(G (t)g(t)) dt 0 unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α, γ) = α−1 pq x α y α+γ γ−1 γ . Proof. From the hypotheses of F (s) and G(t), it is easy to observe that Z s p (3.2) F (s) = p F p−1 (σ)f (σ)dσ, s ∈ (0, x), 0 Z t q G (t) = q Gq−1 (τ )g(τ )dτ, t ∈ (0, y). 0 From (3.2) and Hölder’s inequality, we have Z s α1 Z x α−1 (3.3) F p−1 (σ)f (σ)dσ ≤ s α (F p−1 (σ)f (σ))α dσ , 0 0 s ∈ (0, s), JJ II J I Page 17 of 28 Go Back Full Screen Close and Z (3.4) y Gq−1 (t)g(t)dt ≤ t γ−1 γ Z 0 t (Gq−1 (τ )g(τ ))γ dτ γ1 , t ∈ (0, t). 0 Let s1 = sα−1 , s2 = tγ−1 , ω1 = α1 , ω1 = γ1 , r = ω1 + ω2 , from (3.2) – (3.4) and (2.7), we observe that Hilbert-Pachpatte’s Inequalities (3.5) F p (s)Gq (t) α−1 α γ−1 Wengui Yang Z α1 Z s γ1 t (Gq−1 (τ )g(τ ))γ dτ t γ (F p−1 (σ)f (σ))α dσ 0 0 ( (α−1)(α+γ) (γ−1)(α+γ) ) pqαγ m αγ n αγ ≤ + α+γ α γ Z s α1 Z t γ1 p−1 α q−1 γ × (F (σ)f (σ)) dσ (G (τ )g(τ )) dτ ≤ pqs 0 Title Page Contents 0 for s ∈ (0, x), t ∈ (0, y). From (3.5), we observe that γs (α−1)(α+γ) αγ + αt II J I Go Back (γ−1)(α+γ) αγ pq ≤ α+γ JJ Page 18 of 28 F p (s)Gq (t) (3.6) vol. 10, iss. 1, art. 26, 2009 Z s (F 0 Full Screen p−1 α α1 Z (σ)f (σ)) dσ t (G q−1 γ γ1 (τ )g(τ )) dτ 0 for s ∈ (0, x), t ∈ (0, y). Taking the integral on both sides of (3.6) first over t from 0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order Close of integrals, we observe that Z xZ y F p (s)Gq (t) 0 ≤ ≤ (α−1)(α+γ) αγ γs "Z 0 pq α+γ x α−1 pq x α α+γ (γ−1)(α+γ) αγ dsdt s Z 0 + αt (F p−1 (σ)f (σ))α dσ α1 # "Z 0 x s Z (F p−1 (σ)f (σ))α dσds 0 0 γ−1 α−1 pq = x α y γ α+γ Z t Z (Gq−1 (τ )g(τ ))γ dτ ds 0 Z y α1 y γ−1 γ Z y 0 (x − s)(F p−1 α # dt 0 α1 Z x γ1 Z t γ1 (Gq−1 (τ )g(τ ))γ dτ dt 0 t (y − t)(G (s)f (s)) ds 0 q−1 γ1 . (t)g(t)) dt Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 γ 0 Title Page Contents Remark 4. In Theorem 3.1, setting α = γ = 2, we have (1.2). In Theorem 3.1, setting α1 + γ1 = 1, we have Z 0 x Z 0 y JJ II J I Page 19 of 28 F p (s)Gq (t) dsdt γsα−1 + αtγ−1 Go Back Z ≤ D(p, q, x, y; α, γ) x (x − s)(F p−1 (s)f (s))α ds α1 Full Screen 0 Z × y γ1 q−1 γ (y − t)(G (t)g(t)) dt , 0 unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(p, q, x, y; α, γ) = α−1 pq x α y α+γ γ−1 γ . Close Remark 5. In Theorem 3.1, setting p = q = 1, we have Z xZ y F (s)G(t) (3.7) dsdt (α−1)(α+γ) (γ−1)(α+γ) αγ 0 0 γs + αt αγ Z x α1 Z y γ1 α γ ≤ D(1, 1, x, y; α, γ) (x − s)f (s)ds (y − t)g (t)dt , 0 0 α−1 1 x α y α+γ γ−1 γ . unless f (σ) ≡ 0 or g(τ ) ≡ 0, where D(1, 1, x, y; α, γ) = In the following theorem we give a further generalization of the inequality (3.7) obtained in Remark 5. Theorem 3.2. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0, p(σ) > 0 and q(τ ) > 0 for real numbers. Define R s σ ∈ (0, x), τ ∈ (0, y), R t where x, y are positive Rs R t F (s) = f (σ)dσ and G(t) = 0 g(τ )dτ , P (s) = 0 p(σ)dσ and Q(t) = 0 q(τ )dτ for 0 s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, nonnegative, convex, and submultiplicative functions defined on R+ = [0, ∞). Then Z xZ y Φ(F (s))Ψ(G(t)) (3.8) dsdt (α−1)(α+γ) (γ−1)(α+γ) αγ 0 0 γs + αt αγ Z x α α1 f (s) ≤ L(x, y; α, γ) (x − s) p(s)Φ ds p(s) 0 Z y γ γ1 g(t) × (y − t) q(t)Ψ dt , q(t) 0 where (Z ) α−1 (Z ) γ−1 γ α α−1 γ−1 α γ x y 1 Φ(P (s)) Ψ(Q(t)) L(x, y; α, γ) = ds dt . α+γ P (s) Q(t) 0 0 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 20 of 28 Go Back Full Screen Close Proof. From the hypotheses of Φ and Ψ and by using Jensen’s inequality and Hölder’s inequality, it is easy to see that Rs (σ) dσ P (s) 0 p(σ) fp(σ) Rs (3.9) Φ(F (s))) = Φ p(σ)dσ 0 R s f (σ) p(σ) dσ p(σ) 0 Rs ≤ Φ(P (s))Φ p(σ)dσ 0 Z Φ(P (s)) s f (σ) ≤ p(σ)Φ dσ P (s) p(σ) 0 Z s α α1 Φ(P (s)) α−1 f (σ) dσ , ≤ s α p(σ)Φ P (s) p(σ) 0 and similarly, (3.10) Ψ(G(t)) ≤ α−1 Ψ(Q(t)) γ−1 t γ Q(t) γ−1 Let s1 = s , s2 = t (2.7), we observe that , ω1 = 1 , α Z t γ γ1 g(τ ) dτ . q(τ )Ψ q(τ ) 0 ω1 = 1 , γ r = ω1 + ω2 , from (3.9), (3.10) and Z s α α1 # f (σ) Φ(P (s)) p(σ)Φ dσ (3.11) Φ(F (s))Ψ(G(t)) ≤ s t P (s) p(σ) 0 " Z t γ γ1 # Ψ(Q(t)) g(τ ) × q(τ )Ψ dτ Q(t) q(τ ) 0 α−1 α γ−1 γ " Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 21 of 28 Go Back Full Screen Close αγ ≤ α+γ " ( m (α−1)(α+γ) αγ α + n (γ−1)(α+γ) αγ ) γ Z s α α1 # Φ(P (s)) f (σ) × p(σ)Φ dσ P (s) p(σ) 0 " Z t γ γ1 # g(τ ) Ψ(Q(t)) × q(τ )Ψ dτ Q(t) q(τ ) 0 Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 for s ∈ (0, x), t ∈ (0, y). From (3.11), we observe that Φ(F (s))Ψ(G(t)) (3.12) γs (α−1)(α+γ) αγ Title Page (γ−1)(α+γ) αγ + αt " Z s α α1 # Φ(P (s)) f (σ) 1 dσ ≤ p(σ)Φ α+γ P (s) p(σ) 0 " Z t γ γ1 # g(τ ) Ψ(Q(t)) × q(τ )Ψ dτ Q(t) q(τ ) 0 for s ∈ (0, x), t ∈ (0, y). Taking the integral on both sides of (3.12) first over t from 0 to y and then over s from 0 to x of the resulting inequality and using Hölder’s inequality with indices α, α/(α − 1) and γ, γ/(γ − 1) and interchanging the order of integrals, we observe that Z xZ y Φ(F (s))Ψ(G(t)) dsdt (α−1)(α+γ) (γ−1)(α+γ) αγ αγ 0 0 γs + αt Contents JJ II J I Page 22 of 28 Go Back Full Screen Close Z s α α1 # 1 Φ(P (s)) f (σ) ≤ p(σ)Φ dσ ds α+γ 0 P (s) p(σ) 0 "Z Z t γ γ1 # y g(τ ) Ψ(Q(t)) × q(τ )Ψ dτ dt Q(t) q(τ ) 0 0 (Z ) α−1 α α−1 α1 α α Z x Z s x Φ(P (s)) 1 f (σ) ≤ ds p(σ)Φ dσds α+γ P (s) p(σ) 0 0 0 (Z ) γ−1 γ γ γ1 γ−1 γ Z y Z t y Ψ(Q(t)) g(τ ) dτ dt × dt q(τ )Ψ q(τ ) Q(t) 0 0 0 ) γ−1 ) α−1 (Z (Z γ α γ α y x Ψ(Q(t)) γ−1 1 Φ(P (s)) α−1 dt ds = Q(t) α+γ P (s) 0 0 Z x α α1 Z y γ γ1 g(t) f (s) × ds (y − t) q(t)Ψ dt . (x − s) p(s)Φ p(s) q(t) 0 0 "Z x Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 23 of 28 Go Back Remark 6. From the inequality (2.7), we obtain (3.13) sω1 1 sω2 2 1 ω1 sω1 1 +ω2 + ω2 sω2 1 +ω2 ≤ ω1 + ω2 for ω1 > 0, ω2 > 0. If we apply the elementary inequality (3.13) on the right-hand sides of (3.1) in Theorem 3.1 and (3.8) in Theorem 3.2, then we get the following Full Screen Close inequalities Z xZ (3.14) y F p (s)Gq (t) dsdt (γ−1)(α+γ) + αt αγ " Z α+γ x αγ αγD(p, q, x, y; α, γ) 1 p−1 α ≤ (x − s)(F (s)f (s)) ds α+γ α 0 # Z y α+γ αγ 1 , + (y − t)(Gq−1 (t)g(t))γ dt γ 0 0 0 γs (α−1)(α+γ) αγ pq where D(p, q, x, y; α, γ) = α+γ x Z xZ y Φ(F (s))Ψ(G(t)) 0 0 γs (α−1)(α+γ) αγ + αt α−1 α y (γ−1)(α+γ) αγ γ−1 γ Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 . Also, Title Page Contents dsdt " Z α α+γ x αγ αγL(x, y; α, γ) 1 f (s) ≤ (x − s) p(s)Φ ds α+γ α p(s) 0 # Z y γ α+γ αγ 1 g(t) + (y − t) q(t)Ψ dt , γ q(t) 0 II J I Page 24 of 28 Go Back Full Screen where L(x, y; α, γ) = JJ x α Z Φ(P (s)) α−1 1 α+γ P (s) 0 ds α−1 ( Z α 0 y Ψ(Q(t)) Q(t) γ γ−1 Close ) γ−1 γ dt . The following theorems deal with slight variants of (3.8) Rgiven in Theorem 3.2. s Before we state our next theorem, we point out that “F (s) = 0 f (σ)dσ and G(t) = Rt g(τ )dτ ” are replaced by “F (s) = orem 3.4 in [9]. 0 1 s Rs 0 f (σ)dσ and G(t) = 1 t Rt 0 g(τ )dτ ” in The- Theorem 3.3. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0 for R s σ ∈ (0, x), τ ∈ (0, y), where x, y are positive real numbers. Define F (s) = 1s 0 f (σ)dσ, G(t) = R 1 t g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, nonnegative, t 0 convex functions defined on R+ = [0, ∞). Then Z xZ y stΦ(F (s))Ψ(G(t)) dsdt (γ−1)(α+γ) (α−1)(α+γ) αγ αγ 0 0 γs + αt γ1 Z x α1 Z y γ α (y − t)Ψ (g(t))dt , ≤ D(1, 1, x, y; α, γ) (x − s)Φ (f (s))ds 0 0 where D(1, 1, x, y; α, γ) = α−1 1 x α y α+γ γ−1 γ Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents . Theorem 3.4. Let α > 1, γ > 1 and f (σ) ≥ 0, g(τ ) ≥ 0, p(σ) > 0 and q(τ ) > 0 forR σ ∈ (0, x), τ ∈ R(0, y), where x, y are positive real numbers. DeRs s t fine P (s) = 0 p(σ)dσ, Q(t) = 0 q(τ )dτ , F (s) = P 1(s) 0 p(σ)f (σ)dσ and G(t) = Rt 1 q(τ )g(τ )dτ for s ∈ (0, x), t ∈ (0, y). Let Φ and Ψ be two real-valued, Q(t) 0 nonnegative, convex functions defined on R+ = [0, ∞). Then Z xZ y P (s)Q(t)Φ(F (s))Ψ(G(t)) dsdt (α−1)(α+γ) (γ−1)(α+γ) αγ 0 0 γs + αt αγ Z x α1 α ≤ D(1, 1, x, y; α, γ) (x − s) [p(s)Φ (f (s))] ds 0 Z × 0 y γ1 (y − t) [q(t)Ψ (g(t))] dt , γ JJ II J I Page 25 of 28 Go Back Full Screen Close where D(1, 1, k, r; α, γ) = α−1 1 x α y α+γ γ−1 γ . The proofs of Theorems 3.3 and 3.4 are similar to the proof of Theorem 3.2 and similar to the proofs of Theorems 2.3 and 2.4. Hence, we leave out the details. Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 26 of 28 Go Back Full Screen Close References [1] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. and Appl., 226 (1998), 116–179. [2] G.H. HARDY, J.E. LITTLEWOOD Cambridge University Press, 1952. AND G. POLYA, Inequalities, London: [3] G.D. HANDLEY, J.J. KOLIHA AND J. PEČARIĆ, Hilbert-Pachpatte type multidimensional integral inequalities, J. Ineq. Pure and Appl. Math., 5(2) (2004), Art. 34. [ONLINE http://jipam.vu.edu.au/article.php?sid= 389]. Hilbert-Pachpatte’s Inequalities [4] C.J. ZHAO AND W.S. CHEUNG, Inverses of new Hilbert-Pachatte type inequalities, J. Inequ. and Appl., 2006 (2006), 1–11. Title Page [5] B.G. PACHPATTE, Inequalities similar to certain extensions of Hilbert’s inequality, J. Math. Anal. and Appl., 243 (2000), 217–227. [6] Z.X. LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities, J. Ineq. Pure and Appl. Math., 4(2) (2003), Art. 33. [ONLINE http: //jipam.vu.edu.au/article.php?sid=271]. [7] J. PEČARIĆ, I. PERIĆ AND P. 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FINK, Classical and New Inequalities in Analysis, Kluwer Academic, Dordrecht, 1993. [13] YOUNG-HO KIM AND BYUNG-IL KIM, An analogue of Hilbert’s inequality and its extensions, Bull. Korean Math. Soc., 39(3) (2002), 377–388. Hilbert-Pachpatte’s Inequalities Wengui Yang vol. 10, iss. 1, art. 26, 2009 Title Page Contents JJ II J I Page 28 of 28 Go Back Full Screen Close