In this paper, by introducing some parameters and by employing

a sharpening of Hölder’s inequality, a new generalization of Hardy-Hilbert

integral inequality involving the Beta function is established. At the same

time, an extension of Widder’s theorem is given

Dragomir, Sever S

Leping, He

Qiaoshun, Yang

English

Victoria University Eprints Repository

A FURTHER GENERALIZATION OF HARDY-HILBERT’SINTEGRAL INEQUALITY WITH PARAMETER ANDAPPLICATIONSHE LEPING, SEVER S. DRAGOMIR, AND YANG QIAOSHUNAbstract. In this paper, by introducing some parameters and by employinga sharpening of Ho¨lder’s inequality, a new generalization of Hardy-Hilbertintegral inequality involving the Beta function is established. At the sametime, an extension of Widder’s theorem is given.1. IntroductionSuppose that p > 1, 1p +1q = 1, and f, g : (0,∞)→ (0,∞) are so that0 <∫ ∞0fp(t)dt <∞, 0 <∫ ∞0gq(t)dt <∞.Then we may state the following integral inequality(1.1)∫ ∞0∫ ∞0f(x)g(y)x+ ydxdy <pisin(pi/p)(∫ ∞0fp(t)dt) 1p(∫ ∞0gq(t)dt) 1q,in which the constant factor pisin(pi/p) is the best possible.The inequality (1.1) is well known in the litarature as the Hardy-Hilbert’s integralinequality.Recently, some improvements and generalizations of Hardy-Hilbert’s integral in-equality have been given. For instance, we refer the reader to the papers [2]–[7] andthe bibliography therein.The main purpose of this paper is to establish a new extended Hardy-Hilbert’stype inequality, which includes improvements and generalisations of the correspond-ing results from [2]–[3].2. Lemmas and their ProofsFor convenience, we firstly introduce some notations:(fr, gs) =∫ ∞αfr(x)gs(x)dx, ‖f‖p =(∫ ∞αfp(x)dx) 1p,‖f‖2 = ‖f‖, Sr (H,x) =(Hr/2, x)‖H‖−r/2r ,where x is a parametric variable unit vector. Clearly, Sr (H,x) = 0 when the vectorx selected is orthogonal to Hp2 .2000 Mathematics Subject Classification. 26D15.Key words and phrases. Hardy-Hilbert’s integral inequality, Ho¨lder’s inequality, Weight coef-ficient, Beta function, Widder’s theorem.12 HE LEPING, SEVER S. DRAGOMIR, AND YANG QIAOSHUNThroughout this paper, m is taken to be m = min{1p ,1q}.In order to state our results, we need to point out the following lemmas.Lemma 1. Let f(x), g(x) > 0, x ∈ (0,∞) , 1p + 1q = 1 and p > 1. If 0 < ||f ||p <∞, 0 < ||g||q <∞, then(2.1) (f, g) < ||f ||p||g||q (1−R)m ,where R = (Sp (f, h)− Sq (g, h))2 , ||h|| = 1, fp/2(x), gq/2(x) and h(x) are linearlyindependent.The lemma is proved in [4], and we omit the details.In the following, we definekλ = B(p+ λ− 2p,q + λ− 2q),θλ(r) =∫ 101(1 + u)λ(1u) (2−λ)rdu (r = p, q),whereB(u, v) =∫ 10t(−1+u)(1 + t)u+vdt (u, v > 0)is the Beta function.The following lemma also holds.Lemma 2. Let b < 1, λ > 0. Define the functionϕ(b, y) = y−1+b∫ y01(1 + u)λ(1u)bdu, y ∈ (0, 1].Then we have(2.2) ϕ(b, y) > ϕ(b, 1), (0 < y < 1).A proof of Lemma 2 is given in paper [5], and we omit it here.Another technical result that will be required in the following is:Lemma 3. Let p > 1, 1p +1q = 1, λ > 2 −min{p, q}, α ≥ −β. Define the weightfunction ωλ by(2.3) ωλ(α, β, r, x) =∫ ∞α1(x+ y + 2β)λ(x+ βy + β) (2−λ)rdy x ∈ (α,∞).(i) For α = −β,(2.4) ωλ(−β, β, r, x) = kλ(x+ β)1−λ x ∈ (−β,∞).(ii) For α > −β,(2.5) ωλ(α, β, r, x) <kλ − θλ(r)(α+ βx+ β)1+ (λ−2)r  (x+ β)1−λ x ∈ (α,∞).Proof. Setting u = (y + β)/(x+ β), we haveωλ(α, β, r, x) = (x+ β)1−λ∫ ∞α+βx+β1(1 + u)λ(1u) (2−λ)rdu.HARDY-HILBERT’S INTEGRAL INEQUALITY 3(i) For α = −β, (2.4) is valid.(ii) For α > −β, we haveωλ(α, β, r, x)= (x+ β)1−λ∫ ∞01(1 + u)λ(1u) (2−λ)rdu(2.6)−∫ α+βx+β01(1 + u)λ(1u) (2−λ)rdu= (x+ β)1−λkλ −(α+ βx+ β)1+ (λ−2)rϕ(2− λr,α+ βx+ β) .(2.7)Putting b = 2−λr , and since λ > 2−min{p, q}, b < 1 is valid, then by Lemma2 we get(2.8) ϕ(2− λr,α+ βx+ β)> ϕ(2− λr, 1)= θλ(r) (x ∈ (0,∞)).Substituting (2.8) into (2.7), we obtain (2.5). The proof is completed.Finally, the following result is needed as well.Lemma 4. Let an(n = 0, 1, 2, 3, . . . ) be complex numbers. IfA(z) :=∞∑n=0anznis analytic on unit disk |z| ≤ 1, andA∗(z) :=∞∑n=0anznn!is analytic on |z| <∞, then(2.9)∫ 10|A(x)|2 dx =∫ 10∣∣∣∣∫ ∞0e−s/xA∗(s)ds∣∣∣∣2 1x2 dx,where s ∈ (0,∞), x ∈ (0, 1].Proof. Since A∗(z) is analytic on the complex plane, the series∞∑n=0e−tan(xt)nn!4 HE LEPING, SEVER S. DRAGOMIR, AND YANG QIAOSHUNis uniformly convergent in (0,∞), and we obtain∫ ∞0e−tA∗(tx)dt =∫ ∞0e−t∞∑n=0an(xt)nn!dt=∞∑n=0anxnn!∫ ∞0tne−tdt=∞∑n=0anxn = A(x).Setting tx = s, thenA(x) =1x∫ ∞0e−s/xA∗(s)dswhence ∫ 10|A(x)|2 dx =∫ 10∣∣∣∣∫ ∞0e−s/xA∗(s)ds∣∣∣∣2 1x2 dx.The lemma is thus proved.3. Main ResultsFor the sake of convenience, we need the following notations:F (x, y) =f(x)(x+ y + 2β)λ/p(x+ βy + β) (2−λ)pq,G(x, y) =g(y)(x+ y + 2β)λ/q(y + βx+ β) (2−λ)pq,(3.1) φ(r, x) =∫ α+βx+β01(1 + u)λ(1u)(2−λ)/rdu,Sp(F, h) ={∫ ∞α∫ ∞αF p/2hdxdy}{∫ ∞α[kλ − φ(q, x)] (x+ β)1−λfp(x)dx}− 12,andSq(G, h) ={∫ ∞α∫ ∞αGq/2hdxdy}{∫ ∞α[kλ − φ(p, x)] (x+ β)1−λgq(x)dx}− 12,where h = h(x, y) is a unit vector satisfying the property||h|| ={∫ ∞α∫ ∞αh2(x, y)dxdy} 12= 1and F p/2, Gq/2, h are linearly independent.The first main result is incorporated in the following theorem.Theorem 1. Let p > 1, 1p +1q = 1, λ > 2 − min {p, q} , α ≥ −β, and f, g > 0.Assume also that0 <∫ ∞α(t+ β)1−λfp(t)dt <∞,and0 <∫ ∞α(t+ β)1−λgq(t)dt <∞.HARDY-HILBERT’S INTEGRAL INEQUALITY 5(i) If α > −β, then we have(3.2)∫ ∞α∫ ∞αf(x)g(y)(x+ y + 2β)λdxdy<{∫ ∞α(kλ − θλ(q)(α+ βt+ β)1+(λ−2)/q)(t+ β)1−λfp(t)dt} 1p×{∫ ∞α(kλ − θλ(p)(α+ βt+ β)1+(λ−2)/p)(t+ β)1−λgq(t)dt} 1q(1−Rλ)m.(ii) If α = −β, then we have(3.3)∫ ∞α∫ ∞αf(x)g(y)(x+ y + 2β)λdxdy< kλ(∫ ∞−β(t+ β)1−λfp(t)dt) 1p(∫ ∞−β(t+ β)1−λgq(t)dt) 1q(1−Rλ)m,whereRλ = (Sp(F, h)− Sq(G, h))2 ,while the function h is defined by(3.4) h(x, y) =(2pi) 12 eα−x(x+ y − 2α) 12(x− αy − α) 14.Proof. By Lemma 1 and the equality (2.3), we have∫ ∞α∫ ∞αf(x)g(y)(x+ y + 2β)λdxdy(3.5)=∫ ∞α∫ ∞αFGdxdy≤{∫ ∞α∫ ∞αF pdxdy} 1p{∫ ∞α∫ ∞αGqdxdy} 1q(1−Rλ)m=(∫ ∞αωλ(α, β, q, t)fp(t)dt) 1p(∫ ∞αωλ(α, β, p, t)gq(t)dt) 1q(1−Rλ)m.Substituting (2.5) and (2.4) into the inequality (3.5) respectively, the inequalities(3.2) and (3.3) follow.Next, let us discuss the expression Rλ.We can choose the function h indicated by (3.4). Setting s = x−α and t = y−α,we get‖h‖2 =∫ ∞α∫ ∞αh2(x, y)dxdy =2pi∫ ∞0e−2sds∫ ∞01s+ t(st) 12dt = 1.Hence, ‖h‖ = 1.6 HE LEPING, SEVER S. DRAGOMIR, AND YANG QIAOSHUNBy Lemma 1 and the given h, we have(3.6) Rλ ={(∫ ∞α∫ ∞αF p/2h dxdy)(∫ ∞α∫ ∞αF p dxdy)− 12−(∫ ∞α∫ ∞αGq/2h dxdy)(∫ ∞α∫ ∞αGq dxdy)− 12}2.Substituting (2.3), (2.6) and (3.1) into (3.6), we getRλ = (Sp(F, h)− Sq(G, h))2.It is obvious that F p/2, Gq/2 and h are linearly independent, so it is impossiblefor equality to hold in (3.5).The proof is thus completed.Owing to p, q > 1, when λ = 1, 2; the condition λ > 2 − min{p, q} is satisfied.We haveθ1(r) =∫ 1011 + u(1u) 1rdu >∫ 1011 + udu = ln 2, k1 = B(1p,1q)=pisin(pi/p),θ2(r) =∫ 101(1 + u)2du =12, k2 = B(p+ 2− 2p,q + 2− 2q)= B(1, 1) = 1.The following results are natural consequences of Theorem 1.Corollary 1. If p > 1, 1p +1q = 1, α ≥ β, f, g > 0,0 <∫ ∞αfp(t)dt <∞,and0 <∫ ∞αgq(t)dt <∞,then(3.7)∫ ∞α∫ ∞αf(x)g(y)x+ y + 2βdxdy<{∫ ∞α(pisin(pi/p)−(α+ βt+ β) 1p· ln2)fp(t)dt} 1p×{∫ ∞α(pisin(pi/p)−(α+ βt+ β) 1q· ln2)gq(t)dt} 1q(1−R1)m,(3.8)∫ ∞−β∫ ∞−βf(x)g(y)x+ y + 2βdxdy<pisin(pi/p)(∫ ∞−βfp(t)dt) 1p(∫ ∞−βgq(t)dt) 1q(1−R1)m,HARDY-HILBERT’S INTEGRAL INEQUALITY 7and(3.9)∫ ∞0∫ ∞0f(x)g(y)x+ ydxdy<pisin(pi/p)(∫ ∞0fp(t)dt) 1p(∫ ∞0gq(t)dt) 1q(1− r1)m.Remark 1. When p = q = 2, the inequality (3.9) reduces, after some simplecomputation, to an inequality obtained in [2].Corollary 2. If p > 1, 1p +1q = 1, α ≥ β, f, g > 0,0 <∫ ∞α(t+ β)−1fp(t)dt <∞and0 <∫ ∞α(t+ β)−1gq(t)dt <∞,then(3.10)∫ ∞α∫ ∞αf(x)g(y)(x+ y + 2β)2dxdy<{∫ ∞α(1− α+ β2(t+ β))1t+ βfp(t)dt} 1p×{∫ ∞α(1− α+ β2(t+ β))1t+ βgq(t)dt} 1q(1−R2)mand(3.11)∫ ∞−β∫ ∞−βf(x)g(y)(x+ y + 2β)2dxdy<(∫ ∞−β1t+ βfp(t)dt) 1p(∫ ∞−β1t+ βgq(t)dt) 1q(1−R2)m.Remark 2. The inequalities (3.2), (3.3) and (3.7) – (3.9) are generalizations of(1.1).Remark 3. We can also define h(x, y) ash(x, y) ={1 (x, y) ∈ [0, 1]× [0, 1]0 (x, y) ∈ (0,∞)× (0,∞)[0, 1]× [0, 1].In this case, the expression of Rλ will be much simpler. The details are omitted.4. ApplicationsWe start with the following result:Theorem 2. Suppose that an(n = 0, 1, 2, 3, . . . ) are complex numbers. Also, defineA(x) =∑∞n=0 anxn, A∗(x) =∑∞n=0anxnn! , and the function f as:(4.1) f(x) = e−xA∗(x), x ∈ (0,∞).8 HE LEPING, SEVER S. DRAGOMIR, AND YANG QIAOSHUNIf p > 1, 1p +1q = 1, then(4.2)∫ 10|A(x)|2 dx < pisin(pi/p)(∫ ∞0|f(x)|p dx) 1p(∫ ∞0|f(x)|q dx) 1q(1−R)m,whereR :=(Sp(F , h)− Sq (G, h))2 > 0,with ‖h‖ = 1, andF :=|f(s)|(s+ t)1p(st) 1pq,G :=|f(t)|(s+ t)1q(ts) 1pq; ψ(t) :=∫ ∞0e−s(s+ t)(ts) (q−p)2pqds,Sp(F , h) :=√2{∫ ∞0e−s |f(s)|p/2 ds}·{∫ ∞0|f(s)|p ds}− 12,Sq(G, h) :=√2 sin (pi/p)pi{∫ ∞0ψ(t) |f(t)|q/2 dt}·{∫ ∞0|f(s)|q ds}− 12and(F )p/2, (G)q/2, hare linearly independent.Proof. Setting y = 1x on the right-hand side of the equality (2.9), we have(4.3)∫ 10|A(x)|2 dx =∫ ∞1∣∣∣∣∫ ∞0e−syA∗(s)ds∣∣∣∣2 dy.Next, put u = y − 1. According to the equalities (4.1) and (4.3), we get∫ 10|A(x)|2 dx =∫ ∞0du∣∣∣∣∫ ∞0e−suf(s)ds∣∣∣∣2 .Using Hardy’s technique, we may state that∫ 10|A(x)|2 dx =∫ ∞0du∣∣∣∣∫ ∞0e−suf(s)ds∣∣∣∣2=∫ ∞0du∫ ∞0e−suf(s)ds∫ ∞0e−tuf(t)dt≤∫ ∞0∫ ∞0(∫ ∞0e−(s+t)udu)|f(s)| |f(t)| dsdt=∫ ∞0∫ ∞0|f(s)| |f(t)|s+ tdsdt<pisin(pi/p)(∫ ∞0|f(x)|p) 1p(∫ ∞0|f(x)|q) 1q(1−R)m.Let us choose the function h(s, t) to be defined byh(s, t) :=(2pi) 12 e−s(s+ t)12(st) 14,HARDY-HILBERT’S INTEGRAL INEQUALITY 9then‖h‖ ={∫ ∞0∫ ∞0h2(s, t)dsdt} 12= 1.Notice that k1(p) = B( 1q ,1p ) = pi/ sin(pip ), and, in a similar way to the one inTheorem 1, the expression of R is easily given. We omit the details.Remark 4. In particular, when p = q = 2, it follows from (4.2) that(4.4)∫ 10A2(x)dx = pi(1− r) 12∫ ∞0f2(x)dx.If r in (4.4) is replaced by zero, then Widder’s theorem (see [8]) can be recaptured.Remark 5. After simple computation, the inequality (4.4) is equivalent to theinequality (3.4) in [2]. Consequently, inequality (4.2) is an extension of (3.4) in[2].References[1] G. H. Hardy, J. E. Littlewood, and G. Polya, Inequalities, Cambridge Univ. Press, Cambridge,UK, 1952.[2] Gao Mingzhe, Tan Li and Lokenath Debnath, Some improvements on Hilbert’s integral in-equality, J. Math. Anal. Appl., Vol. 229(1999), 682-689.[3] Gao Mingzhe, On the Hilbert inequality, Zeitschrift f u¨r Analysis und ihre Anwendungen, Vol.18, No. 4 (1999), 1117-1122.[4] Leping He, Mingzhe Gao and Weijian Jia, On the improvement of the Hardy-Hilbert’s integralinequality with parameters, Journal of Inequalities in Pure and Applied Mathematics, Vol.4(2003), No. 5, Art. 94, [ONLINE: http://jipam.vu.edu.au/v4n5/108 03.html].[5] Yang Bicheng, On Hardy-Hilbert’s integral inequality, J. Math. Anal.Appl., Vol. 261(2001),295–306.[6] He Leping, Gao Mingzhe and Wei Shangrong, A note on Hilbert’s inequality, MathematicalInequalities & Applications, Vol. 6(2003), No. 2, 283-288.[7] Gao Mingzhe, Wei Shangrong and He Leping, On the Hilbert inequality with weights,Zeitschrift f u¨r Analysis und ihre Anwendungen, Vol. 21(2002), No. 1, 257-263.[8] D.V. Widder, An inequality related to one of Hilbert’s, J. London Math. Soc. Vol. 4(1924),194-198.(He Leping) Department of Mathematics and Computer Science, Jishou University,Jishou Hunan 416000, People’s Republic of ChinaE-mail address: lepinghe65729@163.com(S.S. Dragomir) School of Computer Science and Mathematics, Victoria Universityof Technology, PO Box 14428, Melbourne City M.C., Victoria 8001, Australia.E-mail address: sever.dragomir@vu.edu.auURL: http://rgmia.vu.edu.au/SSDragomirWeb.html(Yang Qiaoshun) Department of Mathematics and Computer Science, Normal College,Jishou University, Jishou, Hunan, 416000, People’s Republic of ChinaE-mail address: yqs244@163.com

An inequality related to one of Hilbert’s,

Gao Mingzhe and Wei Shangrong, A note on Hilbert’s inequality,

Inequalities,

On Hardy-Hilbert’s integral inequality,

On the Hilbert inequality, Zeitschrift f¨ ur Analysis und ihre

On the improvement of the Hardy-Hilbert’s integral inequality with parameters,

Tan Li and Lokenath Debnath, Some improvements on Hilbert’s integral inequality,

Wei Shangrong and He Leping, On the Hilbert inequality with weights,

A Further Generalization of Hardy-Hilbert's Integral Inequality with Parameter and Applications

A Further Generalization of Hardy-Hilbert's Integral Inequality with Parameter and Applications

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