JFSA Journal of Function Spaces and Applications 1758-4965 0972-6802 Hindawi Publishing Corporation 686404 10.1155/2013/686404 686404 Research Article A New Refinement of Generalized Hölder’s Inequality and Its Application http://orcid.org/0000-0002-0631-038X Tian Jingfeng Bui Huy Qui College of Science and Technology North China Electric Power University Baoding, Hebei 071051 China ncepu.edu.cn 2013 5 10 2013 2013 24 05 2013 02 09 2013 2013 Copyright © 2013 Jingfeng Tian. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We present a new refinement of generalized Hölder’s inequality due to Vasić and Pečarić. Moreover, the obtained result is used to improve Beckenbach-type inequality due to Wang.

1. Introduction

If ak0, bk0  (k=1,2,,  n), p>1, and (1/p)+(1/q)=1, then (1)k=1nakbk(k=1nakp)1/p(k=1nbkq)1/q. The sign of inequality is reversed for p<1, p0 (for p<0; we assume that ak,bk>0). Inequality (1) and its reversed version are called Hölder’s inequalities and are important in the study of inequalities and in the field of applied mathematics. The important inequalities have attracted interest of many mathematicians and have been improved as well as generalized in several different directions. For example, Barza et al.  presented matriceal versions of Hölder’s inequality. Nikolova and Varošanec  obtained some new refinements of the classical Hölder’s inequality by using a convex function. Tian and Hu  established a new reversed version of a generalized sharp Hölder’s inequality. For more detailed expositions, the interested reader may consult  and the references therein. Among various generalizations of (1), Vasić and Pečarić in  presented the following interesting theorem.

Theorem A.

Let Aij0  (i=1,2,,n,j=1,2,,m).

If βj are positive numbers, such that j=1m(1/βj)1, then (2)i=1nj=1mAijj=1m(i=1nAijβj)1/βj.

If β1>0, βj<0  (j=2,3,,m) and if j=1m(1/βj)1, then (3)i=1nj=1mAijj=1m(i=1nAijβj)1/βj.

If βj<0  (j=1,2,,m), then (4)i=1nj=1mAijj=1m(i=1nAijβj)1/βj.

The main objective of this paper is to build some new refinements of inequalities (2), (3), and (4). Moreover, the obtained results will be applied to improve Beckenbach-type inequality which is due to Wang .

2. A New Refinement of Generalized Hölder’s Inequality

In this section, we first prove the following lemma, which plays a crucial role in proving our main results.

Lemma 1.

Let Xij>0 and let 1-j=1mXijβj>0  (i=1,2,,n,j=1,2,,m).

If βj>0  (j=1,2,,m) and if   j=1m(1/βj)1, then (5)j=1m(1-i=1nXijβj)1/βj+i=1nj=1mXij[1-(i=1nXi2β2-i=1nXi1β1)2]1/β*,

where β*=max{β1,β2}.

If β1>0, βj>0  (j=2,3,,m) and if j=1m(1/βj)1, then (6)j=1m(1-i=1nXijβj)1/βj+i=1nj=1mXij[1-(i=1nXi2β2-i=1nXi1β1)2]1/β2.

If βj<0  (j=1,2,,m), then (7)j=1m(1-i=1nXijβj)1/βj+i=1nj=1mXij[1-(i=1nXi2β2-i=1nXi1β1)2]1/β2.

Proof.

(a) Without loss of generality, we assume that β1β2.

Case 1 (when 0<β1<β2). It implies that 1/β2>0 and (1/β1)-(1/β2)>0. According to (1/β2)+(1/β2)+((1/β1)-(1/β2))+(1/β3)++(1/βm)1, by using inequality (2), we have (8)[1-(i=1nXi2β2-i=1nXi1β1)2]1/β2=[(1-i=1nXi1β1)+i=1nXi2β2]1/β2×[(1-i=1nXi2β2)+i=1nXi1β1]1/β2×[(1-i=1nXi1β1)+i=1nXi1β1](1/β1)-(1/β2)×j=3m[(1-i=1nXijβj)+i=1nXijβj]1/βj(1-i=1nXi1β1)1/β2(1-i=1nXi2β2)1/β2×(1-i=1nXi1β1)(1/β1)-(1/β2)×j=3m(1-i=1nXi1β1)1/βj+i=1n[(Xi1β1)(1/β1)-(1/β2)j=3m(Xijβj)1/βj(Xi2β2)1/β2(Xi1β1)1/β2×(Xi1β1)(1/β1)-(1/β2)j=3m(Xijβj)1/βj]=j=3m(1-i=1nXijβj)1/βj+i=1nj=1mXij, which means that the desired inequality (5) holds for 0<β1<β2.

Case 2 (when β1=β2>0). By applying inequality (2), we obtain (9)[1-(i=1nXi2β2-i=1nXi1β1)2]1/β2=[(1-i=1nXi1β1)+i=1nXi2β2]1/β2×[(1-i=1nXi2β2)+i=1nXi1β1]1/β2×j=3m[(1-i=1nXijβj)+i=1nXijβj]1/βj(1-i=1nXi1β1)1/β2(1-i=1nXi2β2)1/β2×j=3m(1-i=1nXijβj)1/βj+i=1n[(Xi2β2)1/β2(Xi1β1)1/β2×j=3m(Xijβj)1/βj]=j=1m(1-i=1nXijβj)1/βj+i=1nj=1mXij. That is, inequality (5) is true for β1=β2>0.

(b) If β1>0, βj<0  (j=2,3,,m), then (1/β1)-(1/β2)>0, (1/βj)<0  (j=2,3,,m). By the same method as in Case 1, we obtain the desired inequality (6).

(c) The proof of inequality (7) is similar to the one of inequality (5), and we omit it.

The proof of Lemma 1 is completed.

Next, we present new refinements of inequalities (2), (3), and (4).

Theorem 2.

Let Aij0  (i=1,2,,n,j=1,2,,m), and let s be any given natural number (1sn).

If βj>0  (j=1,2,,m) and if j=1m(1/βj)1, then (10)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(As2β2i=1nAi2β2-As1β1i=1nAi1β1)2]1/β*,

where β*=max{β1,β2}.

If β1>0, βj<0  (j=2,3,,m) and if j=1m(1/βj)1, then (11)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(As2β2i=1nAi2β2-As1β1i=1nAi1β1)2]1/β2.

If βj<0  (j=1,2,,m), then (12)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(As2β2i=1nAi2β2-As1β1i=1nAi1β1)2]1/β2.

Proof.

Consider the following substitution: (13)Xij=Aij(k=1nAkjβj)1/βj  (i=1,2,,n,j=1,2,,m).

It is easy to see that, for any given natural number s  (1sn), the following inequalities hold: (14)Xij>0,1-1in,isXijβj>0.

Consequently, by using the substitution (13) and inequality (5), we have (15)j=1m[1-1in,is(Aijβjk=1nAkjβj)]1/βj+1in,is(j=1mAij(k=1nAkjβj)1/βj){1in,is(Ai1β1k=1nAk2β2)]21-[1in,is(Ai2β2k=1nAk2β2)-1in,is(Ai1β1k=1nAk2β2)]2}1/β* for βj>0  (j=1,2,,m), j=1m(1/βj)1, and thus we have (16)j=1mAsjj=1m(k=1nAkjβj)1/βj+1in,isj=1mAijj=1m(k=1nAkjβj)1/βj[1-(As2β2k=1nAk2β2-As1β1k=1nAk1β1)2]1/β*, that is, (17)i=1n(j=1mAij)j=1m(k=1nAkjβj)1/βj[1-(As2β2k=1nAk2β2-As1β1k=1nAk1β1)2]1/β*. So, we have the desired inequality (10). The proof of inequalities (11) and (12) is similar to the one of inequality (10), and we omit it. The proof of Theorem 2 is completed.

Putting s=1 in (10), (11), and (12), respectively, we obtain the following corollary.

Corollary 3.

Let Aij0  (i=1,2,,n,j=1,2,,m).

If βj>0  (j=1,2,,m) and if j=1m(1/βj)1, then (18)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(A12β2i=1nAi2β2-A11β1i=1nAi1β1)2]1/β*,  where  β*=max{β1,β2}.

If β1>0,  βj<0  (j=2,3,,m) and if j=1m(1/βj)1, then (19)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(A12β2i=1nAi2β2-A11β1i=1nAi1β1)2]1/β2.

If βj<0  (j=1,2,,m), then (20)i=1nj=1mAij[j=1m(i=1nAijβj)1/βj]×[1-(A12β2i=1nAi2β2-A11β1i=1nAi1β1)2]1/β2.

3. Application

In this section, we present a refinement of Beckenbach-type inequality by using Corollary 3. The classical Beckenbach inequality was proved by Beckenbach in . Since Beckenbach discovered this inequality, it has been discussed by many researchers, who either improved it using various techniques or generalized it in many different ways. The interested reader may refer to [7, 16] and references therein. In 1983, Wang  established the following Beckenbach-type inequality.

Theorem B.

Let f(x) and g(x) be positive integrable functions defined on [0,T], and let (1/p)+(1/q)=1. If qp>1, then, for any positive numbers a, b, and c, the inequality (21)(a+c0Thp(x)dx)1/pb+c0Th(x)g(x)dx(a+c0Tfp(x)dx)1/pb+c0Tf(x)g(x)dx holds, where h(x)=((ag(x))/b)q/p. The sign of the inequality in (21) is reversed if 0<p<1.

Theorem 4.

Let f(x) and g(x) be positive integrable functions defined on [0,T], and let (1/p)+(1/q)=1. If qp>1, then, for any positive numbers a, b, and c, the inequality (22)(a+c0Thp(x)dx)1/pb+c0Th(x)g(x)dx(a+c0Tfp(x)dx)1/pb+c0Tf(x)g(x)dx×[1-(a-(q/p)bqa-(q/p)bq+c0Tgq(x)dx-aa+c0Tfp(x)dx)2]1/q holds, where h(x)=((ag(x))/b)q/p. The sign of the inequality in (22) is reversed if 0<p<1.

Proof.

After some simple calculations, we have (23)(a+c0Thp(x)dx)1/pb+c0Th(x)g(x)dx=(a-(q/p)bq+c0Tgq(x)dx)-1/q. On the other hand, putting β1=p, β2=q, m=2 in (18), from the integral form of Hölder’s inequality (1) and Corollary 3, we obtain (24)b+c0Tf(x)g(x)dxb+c(0Tfp(x)dx)1/p(0Tgq(x)dx)1/q=a1/p(ba-1/p)+(c0Tfp(x)dx)1/p×(c0Tgq(x)dx)1/q(a+c0Tfp(x)dx)1/p×(a-(q/p)bq+c0Tgq(x)dx)1/q×[1-(a-(q/p)bqa-(q/p)bq+c0Tgq(x)dx-aa+c0Tfp(x)dx)2]1/q, that is, (25)(a-(q/p)bq+c0Tgq(x)dx)-1/q(a+c0Tfp(x)dx)1/pb+c0Tf(x)g(x)dx×[1-(a-(q/p)bqa-(q/p)bq+c0Tgq(x)dx-aa+c0Tfp(x)dx)2]1/q. Combining inequalities (23) and (25) yields inequality (22). In a similar way, we can prove that the reversed version of inequality (22) is true. Thus, the proof of Theorem 4 is complete.

Acknowledgments

The author would like to sincerely declare his special thanks to both the anonymous referees for their helpful comments and suggestions. This work was supported by the NNSF of China (Grant no. 61073121) and the Fundamental Research Funds for the Central Universities (Grant no. 13ZD19).

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