Refinements of Hardy-Type Inequalities

Abstract

Using Hu Ke's inequality, which is a sharped Hölder's inequality, we present some new refinements of Hardy-type inequalities proposed by Imoru.

1. Introduction

Let 𝑓(𝑥)≥0, 𝑓(𝑥)∈𝐿^𝑝(0,∞), 𝑝>1. Then the famous Hardy's inequality [1, Theorem 319] reads as ^∞₀^∞₀𝐾𝑥,𝑦𝑓(𝑥)𝑑𝑥^𝑝≤𝑑𝑦_∞0𝐾(𝑥,1)𝑥^−1/𝑝𝑑𝑥^𝑝∞₀𝑓^𝑝(𝑥)𝑑𝑥, (1) where 𝐾(𝑥,𝑦) is nonnegative and homogeneous of degree −1. The sign of the inequality in (1) is reversed if 0<𝑝<1. The special cases of inequality (1) are the subject of the following theorem, which is also due to Hardy et al. [1, Theorem 330].

Theorem A. Let 𝑝>1,𝑓(𝑥) be nonnegative and Lebesgue integrable on [0,𝑎] or [𝑎,∞) for every 𝑎>0, according to 𝑟>1 or 𝑟<1. Then ^∞₀𝑦^−𝑟𝐹^𝑝𝑦𝑝𝑑𝑦≤𝑟−1^𝑝∞₀𝑥^−𝑟𝑥𝑓(𝑥)^𝑝𝑑𝑥(𝑟>1_∞0),𝑦^−𝑟𝐹^𝑝𝑦𝑝𝑑𝑦≤1−𝑟^𝑝∞₀𝑥^−𝑟𝑥𝑓(𝑥)^𝑝𝑑𝑥(𝑟<1), (2)

where 𝐹𝑦=∫^𝑦₀𝑓(𝑥)()∫𝑑𝑥𝑟>1_∞𝑦𝑓(𝑥)𝑑𝑥(𝑟<1). (3) The signs of the inequalities are reversed if 0<𝑝<1.

As is well known, inequalities (2) play a very important role in both theory and applications. Ever since Hardy discovered inequalities (2), they have been studied by many authors, who either reproved them using various techniques or improved, generalized, and applied them in many different ways (see e.g. [2–22] and references therein). For further remarks concerning the improvements and properties of inequalities (2) and their generalizations, see for example, [23] or [24].

In the year 1977, Imoru [6] obtained the following integral inequalities which are related to Hardy's (see Theorem A).

Theorem B. Let 𝑔 be continuous and nondecreasing on [0,∞] with 𝑔(0)=0, 𝑔(𝑥)>0 for 𝑥>0 and 𝑔(∞)=∞. Let 𝑓(𝑥) be nonnegative and Lebesgue integrable with respect to 𝑔(𝑥) on [0,𝑏] or on [𝑎,∞) according to 𝑟>1 or 𝑟<1, where 𝑎, 𝑏>0. Suppose 𝐹(𝑥)=∫^𝑥₀𝑓(𝑡)(𝑡)∫𝑑𝑔)(𝑟>1_∞𝑥𝑓(𝑡)𝑑𝑔(𝑡)(𝑟<1). (4) If 𝑝≥1, then ^𝑏₀𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥)(𝑥)+𝑝𝑑𝑔𝑔𝑟−1^1−𝑟(𝑏)𝐹^𝑝(𝑏)≤𝑝𝑟−1^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟>1_∞𝑎),𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥)(𝑥)+𝑝𝑑𝑔𝑔1−𝑟^1−𝑟(𝑎)𝐹^𝑝(𝑎)≤𝑝1−𝑟^𝑝∞_𝑎𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟<1), (5) with both signs of inequalities reversed if  0<𝑝≤1.

Later, in 1981, Chan in [2] derived several exponential generalizations of the Imoru's inequalities (5). In 1985, Imoru in [7] presented further extensions of (5). Moreover, in 1988, Yang et al. [22] gave some new generalizations of (5). Recently, Oguntuase and Imoru in [10] obtained other generalizations of the Yang et al.'s results.

The main purpose of this work is to give some improvements of inequalities (5) by using Hu Ke's inequality which is a sharp Hölder's inequality.

2. A Set of Lemmas

In this section, we will prove lemmas, which play crucial roles in proving our main results.

Lemma 1 (see [25] Hu Ke's inequality). Let 𝑓(𝑥),𝑔(𝑥), and 𝑒(𝑥) be integrable functions defined on [0,+∞) and 𝑓(𝑥),𝑔(𝑥)≥0,1−𝑒(𝑥)+𝑒(𝑦)≥0  for all 𝑥,𝑦∈[0,+∞), and let 𝑝>1,1/𝑝+1/𝑞=1. Then ^∞₀𝑓(𝑥)𝑔(𝑥)≤𝑑𝑥_∞0𝑓^𝑝(𝑥)𝑑𝑥^1/𝑝∞₀𝑔^𝑞(𝑥)𝑑𝑥^1/𝑞×∫1−^∞0𝑓^𝑝(𝑥)𝑒(𝑥)𝑑𝑥∫^∞₀𝑓^𝑝(𝑥)−∫𝑑𝑥_∞0𝑔^𝑞(𝑥)𝑒(𝑥)𝑑𝑥∫^∞₀𝑔^𝑞(𝑥)𝑑𝑥^2𝜏/2, (6) where 𝜏=min{1/𝑝,1/𝑞}.

Lemma 2 (see [26]). Let 𝑓(𝑥),𝑔(𝑥), and 𝑒(𝑥) be integrable functions defined on [0,+∞) and 𝑓(𝑥)≥0,𝑔(𝑥)>0,1−𝑒(𝑥)+𝑒(𝑦)≥0 for all 𝑥,𝑦∈[0,+∞), let 𝑞<0,1/𝑝+1/𝑞=1, and let 𝑓(𝑥),𝑔(𝑥)∈𝐿^𝑞(0,∞). Then ^∞₀𝑓(𝑥)𝑔(𝑥)≥𝑑𝑥_∞0𝑓^𝑝(𝑥)𝑑𝑥^1/𝑝∞₀𝑔^𝑞(𝑥)𝑑𝑥^1/𝑞×∫1−^∞0𝑓(𝑥)𝑔(𝑥)𝑒(𝑥)𝑑𝑥∫^∞₀𝑓(𝑥)𝑔(𝑥)−∫𝑑𝑥_∞0𝑔^𝑞(𝑥)𝑒(𝑥)𝑑𝑥∫^∞₀𝑔^𝑞(𝑥)𝑑𝑥^2𝜌/2, (7) where 𝜌=max{−1,1/𝑞}.

Lemma 3. Let 𝑔 be continuous and nondecreasing on [𝑎,𝑏]. Let 𝜑(𝑥,𝑡) and 𝑒(𝑥) be integrable functions and 𝜑(𝑥,𝑡)≥0, 1−𝑒(𝑥)+𝑒(𝑦)≥0 for all 𝑥,𝑦,𝑡∈[0,+∞), and let 𝜙 be nondecreasing. If 𝑝≥1, then ^𝑏₀𝑔⁻¹(𝑥)^𝑥₀𝜑()(𝑡(𝑥)≥𝑥,𝑡𝑑𝜙)𝑑𝑔_𝑏0𝑔⁻¹(𝑥)^𝑥₀𝜑^1/𝑝(𝑥,𝑡)𝑑𝜙(𝑡)^𝑝×^𝑥₀𝑑𝜙(𝑡)^1−𝑝×∫1−^𝑥0𝜑(𝑥,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑥₀𝜑()(𝑡)−∫𝑥,𝑡𝑑𝜙_𝑥0𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑥₀𝑑𝜙(𝑡)^2𝛽/2𝑑𝑔_∞𝑎(𝑥),𝑔⁻¹(𝑥)^∞_𝑥𝜑()(𝑡(𝑥)≥𝑥,𝑡𝑑𝜙)𝑑𝑔_∞𝑎𝑔⁻¹(𝑥)^∞_𝑥𝜑^1/𝑝(𝑥,𝑡)𝑑𝜙(𝑡)^𝑝×^∞_𝑥𝑑𝜙(𝑡)^1−𝑝×∫1−^∞𝑥𝜑(𝑥,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝜑()(𝑡)−∫𝑥,𝑡𝑑𝜙_∞𝑥𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝑑𝜙(𝑡)^2𝛽/2𝑑𝑔(𝑥), (8)(9) where 𝛽=max{−1,1−𝑝}. If 0<𝑝<1, then ^∞_𝑎𝑔⁻¹(𝑥)^∞_𝑥𝜑(𝑥,𝑡)𝑑𝜙(𝑡)𝑑𝑔(𝑥)≤^∞_𝑎𝑔⁻¹(𝑥)^∞_𝑥𝜑^1/𝑝(𝑥,𝑡)𝑑𝜙(𝑡)^𝑝×^∞_𝑥𝑑𝜙(𝑡)^1−𝑝×∫1−^∞𝑥𝜑^1/𝑝(𝑥,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝜑^1/𝑝()(𝑡)−∫𝑥,𝑡𝑑𝜙_∞𝑥𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝑑𝜙(𝑡)^2𝛾/2𝑑𝑔_∞𝑎(𝑥),𝑔⁻¹(𝑥)^∞_𝑥𝜑()(𝑡(𝑥)≤𝑥,𝑡𝑑𝜙)𝑑𝑔_∞𝑎𝑔⁻¹(𝑥)^∞_𝑥𝜑^1/𝑝(𝑥,𝑡)𝑑𝜙(𝑡)^𝑝×^∞_𝑥𝑑𝜙(𝑡)^1−𝑝×∫1−^∞𝑥𝜑^1/𝑝(𝑥,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝜑^1/𝑝()(𝑡)−∫𝑥,𝑡𝑑𝜙_∞𝑥𝑒(𝑡)𝑑𝜙(𝑡)∫^∞_𝑥𝑑𝜙(𝑡)^2𝛾/2𝑑𝑔(𝑥), (10) where 𝛾=min{𝑝,1−𝑝}.

Proof. From Lemmas 1 and 2, the conclusion is easy to obtain.

Lemma 4. Let 𝑔 be continuous and nondecreasing on [0,∞] with 𝑔(0)=0, 𝑔(𝑥)>0 for 𝑥>0 and 𝑔(∞)=∞. Let 𝛿=(1−𝑟)/𝑝, 𝑟≠1, and 𝑓(𝑥), 𝑒(𝑥) be nonnegative and Lebesgue integrable with respect to 𝑔(𝑥) on [0,𝑏] or on [𝑎,∞) according to 𝑟>1 or 𝑟<1, where 𝑎, 𝑏>0, and let 1−𝑒(𝑥)+𝑒(𝑦)≥0 for all 𝑥, 𝑦∈[0,+∞). Suppose 𝜆(𝑥)=∫^𝑥₀𝑔(𝑡)^{(𝑝−1)(1+𝛿)}𝑓^𝑝(𝑡)(𝑡)∫𝑑𝑔)(𝑟>1_∞𝑥𝑔(𝑡)^{(𝑝−1)(1+𝛿)}𝑓^𝑝(𝑡)𝑑𝑔(𝑡)(𝑟<1). (11) If 𝑝≥1, then

^𝑏₀𝑔^𝛿−1(𝑥)𝜆(𝑥)(𝑥)≥𝑑𝑔−𝛿^−11−𝑝×^𝑏₀𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑥)(),𝑑𝑔𝑟>1 (12)^∞_𝑎𝑔^𝛿−1(𝑥)𝜆(𝑥)𝑑𝑔(𝑥)≥𝛿^𝑝−1×^∞_𝑎𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×−∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^𝛽/2−∫^∞_𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑥)(),𝑑𝑔𝑟<1 (13)

where 𝐹(𝑥) is as in Theorem B, 𝛽=max{−1,1−𝑝}. If 0<𝑝<1, then^𝑏₀𝑔^𝛿−1(𝑥)𝜆(𝑥)(𝑥)≤𝑑𝑔−𝛿^−11−𝑝×^𝑏₀𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^𝑥0𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑓(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛾/2(𝑥)(),𝑑𝑔𝑟>1_∞𝑎𝑔^𝛿−1(𝑥)𝜆(𝑥)𝑑𝑔(𝑥)≤𝛿^𝑝−1×^∞_𝑎𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^∞𝑥𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑓(𝑡)(𝑡)−∫𝑑𝑔_∞𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛾/2(𝑥)(),𝑑𝑔𝑟<1 (14)

where 𝐹(𝑥) is as in Theorem B, 𝛾=min{𝑝,1−𝑝}.

Proof. We only prove inequality (12); the proofs of (13) and (14) are similar. Let 𝜑(𝑥,𝑡)=𝑔^𝛿(𝑥)[𝑔(𝑡)]^𝑝(1+𝛿)𝑓^𝑝(𝑡), 𝑑𝜙(𝑡)=[𝑔(𝑡)]^−(1+𝛿)𝑑𝑔(𝑡) in inequality (8). Then, if 𝑟, 𝑝>1, we have^𝑏₀𝑔^𝛿−1(𝑥)𝜆(𝑥)(𝑥)=𝑑𝑔_𝑏0𝑔⁻¹(𝑥)^𝑥₀𝜑()(𝑡(𝑥)≥𝑥,𝑡𝑑𝜙)𝑑𝑔_𝑏0𝑔⁻¹(𝑥)^𝑥₀𝜑^1/𝑝(𝑥,𝑡)𝑑𝜙(𝑡)^𝑝×^𝑥₀𝑑𝜙(𝑡)^1−𝑝×∫1−^𝑥0𝜑(𝑥,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑥₀𝜑()(𝑡)−∫𝑥,𝑡𝑑𝜙_𝑥0𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑥₀𝑑𝜙(𝑡)^2𝛽/2(𝑥)=𝑑𝑔_𝑏0𝑔^−1+𝛿(𝑥)^𝑥₀𝑓(𝑡)𝑑𝑔(𝑡)^𝑝𝑥₀𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^1−𝑝×∫1−^𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑥)=𝑑𝑔−𝛿^−11−𝑝×^𝑏₀𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛽/2𝑑𝑔(𝑥). (15)
This proves inequality (12). Lemma 4 is proved.

Lemma 5. With notation as in Lemma 4, one has the results as follows. If 𝑝>1, then 𝑔^𝛿(𝑏)𝜆(𝑏)≥−𝛿^−11−𝑝𝑔^𝛿𝑝(𝑏)𝐹^𝑝(𝑏)×∫1−^𝑏0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_𝑏0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(),𝑔𝑟>1_𝛿(𝑎)𝜆(𝑎)≥𝛿^𝑝−1𝑔^𝛿𝑝(𝑎)𝐹^𝑝(𝑎)×∫1−^∞𝑎𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑎𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_∞𝑎𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑎𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑟<1). (16)(17) If 0<𝑝<1, then 𝑔^𝛿(𝑏)𝜆(𝑏)≤−𝛿^−11−𝑝𝑔^𝛿𝑝(𝑏)𝐹^𝑝(𝑏)×∫1−^𝑏0𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑓(𝑡)(𝑡)−∫𝑑𝑔_𝑏0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛾/2(),𝑔𝑟>1_𝛿(𝑎)𝜆(𝑎)≤𝛿^𝑝−1𝑔^𝛿𝑝(𝑎)𝐹^𝑝(𝑎)×∫1−^∞𝑎𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑎𝑓(𝑡)(𝑡)−∫𝑑𝑔_∞𝑎𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑎𝑔^−(1+𝛿)(𝑡)𝑑𝑔(𝑡)^2𝛾/2(𝑟<1). (18)

Proof. We only prove inequality (16); the proofs of (17) and (18) are similar. If 𝑟, 𝑝>1, by using inequality (8), we have 𝑔^𝛿(𝑏)𝜆(𝑏)=𝑔^𝛿(𝑏)^𝑏₀𝑔^{(𝑝−1)(1+𝛿)}(𝑡)𝑓^𝑝(𝑡)(𝑡)=𝑑𝑔_𝑏0𝜑()(𝑡)≥𝑏,𝑡𝑑𝜙_𝑏0𝜑^1/𝑝(𝑏,𝑡)𝑑𝜙(𝑡)^𝑝×^𝑏₀𝑑𝜙(𝑡)^1−𝑝×∫1−^𝑏0𝜑(𝑏,𝑡)𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑏₀𝜑()(𝑡)−∫𝑏,𝑡𝑑𝜙^𝑏0𝑒(𝑡)𝑑𝜙(𝑡)∫^𝑏₀𝑑𝜙(𝑡)^2𝛽/2=−𝛿^−11−𝑝𝑔^𝛿𝑝(𝑏)𝐹^𝑝(𝑏)×∫1−^𝑏0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_𝑏0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑏₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2, (19) where 𝜑(𝑏,𝑡)=𝑔^𝛿(𝑏)(𝑔(𝑡))^𝑝(1+𝛿)𝑓^𝑝(𝑡), 𝑑𝜙(𝑡)=(𝑔(𝑡))^−(1+𝛿)𝑑𝑔(𝑡). This proves inequality (16). Lemma 5 is proved.

Lemma 6 (see [23]). If 𝑥>−1, 𝛼>1, or 𝛼<0, then (1+𝑥)^𝛼≥1+𝛼𝑥. (20) The inequality is reversed for 0<𝛼<1.

3. Refinements of Hardy-Type Inequalities

Theorem 7. Let 𝑔 be continuous and nondecreasing on [0,∞] with 𝑔(0)=0, 𝑔(𝑥)>0 for 𝑥>0 and 𝑔(∞)=∞. Let 𝑓(𝑥), and 𝑒(𝑥) be nonnegative and Lebesgue integrable with respect to 𝑔(𝑥) on [0,𝑏] or on [𝑎,∞) according to 𝑟>1 or 𝑟<1, where 𝑎, 𝑏>0, and let 1−𝑒(𝑥)+𝑒(𝑦)≥0 for all 𝑥, 𝑦∈[0,+∞). Suppose 𝐹(𝑥) is as in Theorem B. If 𝑝≥1, then ^𝑏₀𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥𝛽)1−2𝜔²(𝑥)+𝑝𝑓,𝑔,𝑒;𝑥𝑑𝑔𝑔𝑟−1^1−𝑟(𝑏)𝐹^𝑝(𝑏𝛽)1−2𝜔²≤𝑝𝑓,𝑔,𝑒;𝑏𝑟−1^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟>1_∞𝑎),𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥𝛽)1−2𝜛²(𝑥)+𝑝𝑓,𝑔,𝑒;𝑥𝑑𝑔𝑔1−𝑟^1−𝑟(𝑏)𝐹^𝑝(𝑏𝛽)1−2𝜛²≤𝑝𝑓,𝑔,𝑒;𝑎1−𝑟^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟<1), (21)(22) where 𝛽=max{−1,1−𝑝}, 𝛿=(1−𝑟)/𝑝, 𝜔=∫𝑓,𝑔,𝑒;𝑥^𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)(𝑡),𝜛=∫𝑑𝑔𝑓,𝑔,𝑒;𝑥^∞𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_∞𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)(𝑡).𝑑𝑔 (23) If 0<𝑝<1, then ^𝑏₀𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥𝛾)1−2𝜇²(𝑥)+𝑝𝑓,𝑔,𝑒;𝑥𝑑𝑔𝑔𝑟−1^1−𝑟(𝑏)𝐹^𝑝(𝑏𝛾)1−2𝜇²≥𝑝𝑓,𝑔,𝑒;𝑏𝑟−1^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟>1_∞𝑎),𝑔^−𝑟(𝑥)𝐹^𝑝(𝑥𝛾)1−2𝜈²(𝑥)+𝑝𝑓,𝑔,𝑒;𝑥𝑑𝑔𝑔1−𝑟^1−𝑟(𝑏)𝐹^𝑝(𝑏𝛾)1−2𝜈²≥𝑝𝑓,𝑔,𝑒;𝑎1−𝑟^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑓(𝑥)𝑔(𝑥)^𝑝𝑑𝑔(𝑥)(𝑟<1), (24) where 𝛾=min{𝑝,1−𝑝}, 𝛿=(1−𝑟)/𝑝, 𝜇=∫𝑓,𝑔,𝑒;𝑥_𝑥0𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑓(𝑡)(𝑡)−∫𝑑𝑔^𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)(𝑡),𝜈=∫𝑑𝑔𝑓,𝑔,𝑒;𝑥_∞𝑥𝑓(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑓(𝑡)(𝑡)−∫𝑑𝑔^∞𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)(𝑡).𝑑𝑔 (25)

Proof. We only prove the case 𝑝≥1; the proof of case 0<𝑝<1 is similar.(i) When 𝑟>1, by using the nondecreasing property of 𝑔, we have 0<𝜆(𝑥)=^𝑥₀𝑔^{−(1−𝑝)(1+𝛿)}(𝑡)𝑓^𝑝(𝑡)(𝑡)=𝑑𝑔_𝑥0𝑔^{(𝑟−1)/𝑝}(𝑡)𝑔^𝑝−𝑟(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)≤𝑔^{(𝑟−1)/𝑝}(𝑥)^𝑥₀𝑔^𝑝−𝑟(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡), (26)and hence lim_𝑥→0+𝑔^𝛿(𝑥)𝜆(𝑥)=0, (27) from which and from inequality (12) we have, on using integration by parts,

^𝑏₀𝑔^𝛿−1(𝑥)𝜆(𝑥)𝑑𝑔(𝑥)=𝛿⁻¹𝑔^𝛿(𝑏)𝜆(𝑏)−𝛿^−1𝑏₀𝑔^𝛿𝑝−1(𝑥)𝑔(𝑥)𝑓(𝑥)^𝑝(𝑥)≥𝑑𝑔−𝛿^−11−𝑝×^𝑏₀𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×−∫^𝑥₀𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)(𝑡)−𝑑𝑔_𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)(𝑡)×𝑑𝑔_𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^−12𝛽/2𝑑𝑔(𝑥). (28)

That is,^𝑏₀𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^𝑥0𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_𝑥0𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^𝑥₀𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑥)𝑝𝑑𝑔≤−𝑟−1^𝑝𝑔^𝛿(𝑏)𝜆(𝑏)+𝑝𝑟−1^𝑝𝑏₀𝑔^−𝑟(𝑥)𝑔(𝑥)𝑓(𝑥)^𝑝𝑑𝑔(𝑥). (29)

Combining inequalities (16), (20), and (29) yields inequality (21).(ii) When 𝑟<1, by the same method as in case (i), we obtain 0<𝜆(𝑥)=^∞_𝑥𝑔^{−(1−𝑝)(1+𝛿)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)≤𝑔^{(𝑟−1)/𝑝}(𝑥)^∞_𝑥𝑔^𝑝−𝑟(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡), (30)and hence lim_𝑥→∞𝑔^𝛿(𝑥)𝜆(𝑥)=0, (31) from which and from inequality (13) we have, on using integration by parts,^∞_𝑎𝑔^𝛿−1(𝑥)𝜆(𝑥)(𝑥)=()𝑑𝑔−𝛿⁻¹𝑔^𝛿(𝑎)𝜆(𝑎)+𝛿^−1∞_𝑎𝑔^𝛿𝑝−1(𝑥)𝑔(𝑥)𝑓(𝑥)^𝑝𝑑𝑔(𝑥)≥𝛿^𝑝−1×^∞_𝑎𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^∞𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_∞𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^{[1+𝛿](𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2𝑑𝑔(𝑥). (32)

That is,^∞_𝑎𝑔(𝑥)^𝛿𝑝−1𝐹^𝑝(𝑥)×∫1−^∞𝑥𝑔^−(1+𝛿)(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^−(1+𝛿)(𝑡)(𝑡)−∫𝑑𝑔_∞𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑒(𝑡)𝑑𝑔(𝑡)∫^∞_𝑥𝑔^{(1+𝛿)(𝑝−1)}(𝑡)𝑓^𝑝(𝑡)𝑑𝑔(𝑡)^2𝛽/2(𝑥)𝑝𝑑𝑔≤−1−𝑟^𝑝𝑔^𝛿(𝑏)𝜆(𝑏)+𝑝1−𝑟^𝑝𝑏₀𝑔^−𝑟(𝑥)[𝑔(𝑥)𝑓(𝑥)]^𝑝𝑑𝑔(𝑥). (33)

Combining inequalities (17), (20), and (33) yields inequality (22). The proof of Theorem 7 is complete.

Acknowledgments

The authors would like to express hearty thanks to the anonymous referees for their great efforts to improve this paper. This work was supported by the NNSF of China (Grant no. 61073121) and the Fundamental Research Funds for the Central Universities (Grant no. 13ZD19).