An Opial-Type Inequality on Time Scales

Abstract

We establish some new Opial-type inequalities involving higher order delta derivatives on time scales. These extend some known results in the continuous case in the literature and provide new estimates in the setting of time scales.

1. Introduction

Opial's inequality appeared for the first time in 1960 in [1] and has been receiving continual attention throughout the years (cf., e.g., [2–7]). The inequality together with its numerous generalizations, extensions, and discretizations has been playing a fundamental role in the study of the existence and uniqueness properties of solutions of initial and boundary value problems for differential equations as well as difference equations [8, 9]. Two excellent surveys on these inequalities can be found in [10, 11].

In 1960, Opial established the following integral inequality.

Theorem A (see  [1]). If 𝑓∈𝐶¹[0,ℎ] satisfies 𝑓(0)=𝑓(ℎ)=0 and 𝑓(𝑥)>0 for all 𝑥∈(0,ℎ), then ^ℎ₀𝑓(𝑥)𝑓^′(𝑥ℎ)𝑑𝑥≤4^ℎ₀𝑓^′(𝑥)²𝑑𝑥. (1)

Shortly after the publication of Opial’s paper, Olech provided a modified version of Theorem A. His result is stated in the following.

Theorem B (see  [12]). If 𝑓 is absolutely continuous on [0,ℎ] with 𝑓(0)=0, then ^ℎ₀𝑓(𝑥)𝑓^′(𝑥ℎ)𝑑𝑥≤2^ℎ₀𝑓^′(𝑥)²𝑑𝑥. (2)
The equality in (2) holds if and only if 𝑓(𝑥)=𝑐𝑥, where 𝑐 is a constant.

The first natural extension of Opial’s inequality (1) involving higher order derivatives 𝑥^(𝑛)(𝑠)(𝑛≥1) is embodied in the following.

Theorem C (see  [10]). Let 𝑥(𝑡)∈𝐶^(𝑛)[0,𝑎] be such that 𝑥^(𝑖)(0)=0,0≤𝑖≤𝑛−1(𝑛≥1). Then the following inequality holds: ^𝑎₀𝑥(𝑡)𝑥^(𝑛)(𝑡1)𝑑𝑡≤2𝑎^𝑛𝑎₀𝑥^(𝑛)(𝑡)²𝑑𝑡. (3)

In 1997, Alzer [13] considered Opial-type inequalities which involve higher-order derivatives of two functions. These generalize earlier results of Agarwal and Pang [14].

In this paper, we consider the Opial-type inequality which involves higher-order delta derivatives of two functions on time scales. Our results in special cases yield some of the recent results on Opial’s inequality and provide some new estimates on such types of inequalities in this general setting.

2. Main Results

Let 𝕋 be a time scale; that is, 𝕋 is an arbitrary nonempty closed subset of real numbers. Let 𝑎, 𝑏∈𝕋. We suppose that the reader is familiar with the basic features of calculus on time scales for dynamic equations. Otherwise one can consult Bohner and Peterson’s book [15] for most of the materials needed.

We first quote the following elementary lemma and the delta time scales Taylor formula.

Lemma 1 (see  [16]). Let 𝑎≥0,𝑝≥1 be real constants. Then 𝑎^1/𝑝≤1𝑝𝑘^{(1−𝑝)/𝑝}𝑎+𝑝−1𝑝𝑘^1/𝑝 (4) for any 𝑘>0.

Lemma 2 (see [17]). Let 𝑓∈𝐶^𝑚_𝑟𝑑(𝕋)= the set of functions that are 𝑚 times differentiable with rd-continuous derivatives on 𝕋,𝑚∈ℕ. Then for any 𝑎,𝑏∈𝕋 and 𝑡∈[𝑎,𝑏]∩𝕋,𝑓(𝑡)=^𝑚−1_𝑘=0ℎ_𝑘(𝑡,𝑎)𝑓^Δ𝑘(𝑎)+^𝑡_𝑎ℎ_𝑚−1(𝑡,𝜎(𝜏))𝑓^Δ𝑚(𝜏)Δ𝜏, (5) where ℎ₀(𝑡,𝑠)∶=1,ℎ_𝑛+1(𝑡,𝑠)∶=∫^𝑡_𝑠ℎ_𝑛(𝜏,𝑠)Δ𝜏,𝑛∈ℕ.

Our main results are given in the following theorems.

Theorem 3. Let 0≤𝑟≤𝑠<𝑡, 𝑠>0, 𝑡>1 be real numbers, and let 𝑚, 𝑘 be integers with 0≤𝑘≤𝑚−1. Let 𝑝>0 and 𝑞≥0 be measurable functions on Υ∶=[𝑎,𝑏]∩𝕋. Further, let 𝑓,𝑔∈𝐶^𝑚−1_𝑟𝑑(Υ) with 𝑓^Δ𝑖(𝑎)=𝑔^Δ𝑖(𝑎)=0,𝑖=0,1,…,𝑚−1, and let𝑓^Δ𝑚−1,𝑔^Δ𝑚−1 be absolutely continuous on Υ such that the integrals ∫^𝑏_𝑎𝑝(𝑥)|𝑓^Δ𝑚(𝑥)|^𝑡Δ𝑥and ∫^𝑏_𝑎𝑝(𝑥)|𝑔^Δ𝑚(𝑥)|^𝑡Δ𝑥exist. Then one has ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠+𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠Δ𝑥≤2^1−𝛼𝑏_𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅𝛽𝑘^𝛽−1𝐹(𝑏)𝐺(𝑏)+1−𝛽𝑘^𝛽(𝐹(𝑏)+𝐺(𝑏))^𝛼, (6) where 𝛼∶=𝑠/𝑡, 𝛽∶=𝑟/𝑠, 𝑃(𝑥)∶=^𝑥_𝑎ℎ^{𝑡/(𝑡−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^1/(1−𝑡)ℎ(𝑥)(𝑥)𝑝(𝑥)Δ𝜏,∶=𝑞^−𝛼𝑃(𝑥)^{𝑟(𝑡−1)/𝑡},𝐹(𝑥)∶=^𝑥_𝑎𝑝(𝜏)𝑓^Δ𝑚(𝜏)^𝑡𝐺(𝑥)Δ𝜏,∶=^𝑥_𝑎𝑝(𝜏)𝑔^Δ𝑚(𝜏)^𝑡Δ𝜏. (7)

Proof. Since 𝑓^Δ𝑖(𝑎)=0,𝑖=0,1,…,𝑚−1, we obtain from Taylor's theorem that for all 𝑥∈Υ, 𝑓(𝑥)=^𝑥_𝑎ℎ_𝑚−1(𝑥,𝜎(𝜏))𝑓^Δ𝑚(𝜏)Δ𝜏, (8) and hence 𝑓^Δ𝑘(𝑥)=^𝑥_𝑎ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑓^Δ𝑚(𝜏)Δ𝜏. (9)
From (9) and Hölder's inequality we get 𝑓^Δ𝑘(𝑥)≤^𝑥_𝑎ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑓^Δ𝑚(𝜏=)Δ𝜏_𝑥𝑎ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^−1/𝑡𝑝(𝜏)^1/𝑡𝑓^Δ𝑚(𝜏≤)Δ𝜏_𝑥𝑎ℎ^{𝑡/(𝑡−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^1/(1−𝑡)Δ𝜏^1−1/𝑡⋅^𝑥_𝑎𝑝(𝜏)𝑓^Δ𝑚(𝜏)^𝑡Δ𝜏^1/𝑡=𝑃(𝑥)^1−1/𝑡𝐹(𝑥)^1/𝑡, (10) where 𝑃(𝑥)∶=∫^𝑥_𝑎ℎ^{𝑡/(𝑡−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^1/(1−𝑡)Δ𝜏,  𝐹(𝑥)∶=∫^𝑥_𝑎𝑝(𝜏)|𝑓^Δ𝑚(𝜏)|^𝑡Δ𝜏.
Let 𝐺(𝑥)∶=^𝑥_𝑎𝑝(𝜏)𝑔^Δ𝑚(𝜏)^𝑡Δ𝜏. (11)
Then we have 𝑔^Δ𝑚(𝑥)^𝑠=𝐺^Δ(𝑥)^𝑠/𝑡𝑝(𝑥)^−𝑠/𝑡. (12)
So (10) together with (12) implies 𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠≤ℎ(𝑥)𝐹(𝑥)^𝑟/𝑡𝐺^Δ(𝑥)^𝑠/𝑡, (13) where ℎ(𝑥)∶=𝑞(𝑥)𝑝(𝑥)^−𝑠/𝑡𝑃(𝑥)^{𝑟(𝑡−1)/𝑡}. Integrating both sides of (13) over Υ and making use of Hölder's inequality, we obtain^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠≤Δ𝑥_𝑏𝑎ℎ(𝑥)𝐹(𝑥)^𝑟/𝑡𝐺^Δ(𝑥)^𝑠/𝑡≤Δ𝑥_𝑏𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^{1−𝑠/𝑡𝑏}_𝑎𝐹(𝑥)^𝑟/𝑠𝐺^Δ(𝑥)Δ𝑥^𝑠/𝑡. (14)
Similarly, we get ^𝑏_𝑎𝑞(𝑥)𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠≤Δ𝑥_𝑏𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^{1−𝑠/𝑡}⋅^𝑏_𝑎𝐺(𝑥)^𝑟/𝑠𝐹^Δ(𝑥)Δ𝑥^𝑠/𝑡. (15)
Recall the elementary inequalities 𝑐_𝛼(𝐴+𝐵)^𝛼≤𝐴^𝛼+𝐵^𝛼≤𝑑_𝛼(𝐴+𝐵)^𝛼,(𝐴,𝐵≥0), (16) where 𝑐_𝛼2∶=1,0≤𝛼≤1,_1−𝛼𝑑,𝛼≥1,_𝛼2∶=^1−𝛼,0≤𝛼≤1,1,𝛼≥1. (17)
Let 𝛽=𝑟/𝑠. Since 𝛼=𝑠/𝑡∈(0,1) and 𝐹 is nondecreasing, from (14)–(16), we have ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠+𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠≤Δ𝑥_𝑏𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅^𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽(𝑥)Δ𝑥^𝛼+^𝑏_𝑎𝐺^Δ(𝑥)𝐹^𝛽(𝑥)Δ𝑥^𝛼≤^𝑏_𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅2^1−𝛼𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽(𝑥)+𝐺^Δ(𝑥)𝐹^𝛽(𝑥)Δ𝑥^𝛼≤^𝑏_𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅2^1−𝛼𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽(𝑥)+𝐺^Δ(𝑥)𝐹^𝛽(𝜎(𝑥))Δ𝑥^𝛼. (18)
By Lemma 1, we get ^𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽(𝑥)+𝐺^Δ(𝑥)𝐹^𝛽(𝜎(𝑥≤))Δ𝑥_𝑏𝑎𝛽𝑘^𝛽−1𝐺(𝑥)𝐹^Δ(𝑥)+1−𝛽𝑘^𝛽𝐹^Δ(𝑥)+𝛽𝑘^𝛽−1𝐹(𝜎(𝑥))𝐺^Δ(𝑥)+1−𝛽𝑘^𝛽𝐺^Δ(𝑥)Δ𝑥=𝛽𝑘^𝛽−1𝑏_𝑎𝐺(𝑥)𝐹^Δ(𝑥)+𝐹(𝜎(𝑥))𝐺^Δ(𝑥+𝑘)Δ𝑥1−𝛽^𝛽𝑏_𝑎𝐹^Δ(𝑥)Δ𝑥+^𝑏_𝑎𝐺^Δ(𝑥)Δ𝑥=𝛽𝑘^𝛽−1𝐹(𝑏)𝐺(𝑏)+1−𝛽𝑘^𝛽[𝐹(𝑏)+𝐺(𝑏)]. (19)
From (18) and (19), we conclude ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠+𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠Δ𝑥≤2^1−𝛼𝑏_𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅𝛽𝑘^𝛽−1𝐹(𝑏)𝐺(𝑏)+1−𝛽𝑘^𝛽(𝐹(𝑏)+𝐺(𝑏))^𝛼. (20)
The proof is complete.

Theorem 4. Let 𝑟≥0, 𝑠>0, 𝑠<𝑡, 𝑡>1 be real numbers, and let 𝑚, 𝑘 be integers with 0≤𝑘≤𝑚−1. Let 𝑝>0, and 𝑞≥0 be measurable functions on Υ∶=[𝑎,𝑏]∩𝕋. Further, let 𝑓, 𝑔∈𝐶^𝑚−1_𝑟𝑑(Υ) with let 𝑓^Δ𝑖(𝑎)=𝑔^Δ𝑖(𝑎)=0,𝑖=0,1,…,𝑚−1, and 𝑓^Δ𝑚−1,𝑔^Δ𝑚−1 be absolutely continuous on Υ such that the integrals ∫^𝑏_𝑎𝑝(𝑥)|𝑓^Δ𝑚(𝑥)|^𝑡Δ𝑥and ∫^𝑏_𝑎𝑝(𝑥)|𝑔^Δ𝑚(𝑥)|^𝑡Δ𝑥exist. Then one has ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠+𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠Δ𝑥≤2^1−𝛼𝑏_𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅𝑑_𝛽Γ(𝐺+𝐹)−Γ(𝐺)−Γ(𝐹)^𝛼, (21) where Γ(𝐻)∶=∫^𝑏_𝑎𝐻^𝛽Δ𝐻, 𝛽∶=𝑟/𝑠, 𝛼∶=𝑠/𝑡, ℎ(𝑥)∶=𝑞(𝑥)𝑝(𝑥)^−𝛼𝑃(𝑥)^{𝑟(𝑡−1)/𝑡}, 𝑃(𝑥)∶=∫^𝑥_𝑎ℎ^{𝑡/(𝑡−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^1/(1−𝑡)Δ𝜏, and 𝑑_𝛽2∶=^1−𝛽,0≤𝛽≤1,1,𝛽≥1. (22)

Proof. Following the proof of Theorem 3, we obtain ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑘(𝑥)^𝑟𝑔^Δ𝑚(𝑥)^𝑠+𝑔^Δ𝑘(𝑥)^𝑟𝑓^Δ𝑚(𝑥)^𝑠≤Δ𝑥_𝑏𝑎ℎ(𝑥)^{𝑡/(𝑡−𝑠)}Δ𝑥^1−𝛼⋅2^1−𝛼𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽𝑥+𝐺^Δ(𝑥)𝐹^𝛽(𝑥)Δ𝑥^𝛼. (23)
Using (16), ^𝑏_𝑎𝐹^Δ(𝑥)𝐺^𝛽(𝑥)+𝐺^Δ(𝑥)𝐹^𝛽(𝑥=)Δ𝑥_𝑏𝑎𝐺^𝛽(𝑥)+𝐹^𝛽(𝑥)𝐺^Δ(𝑥)+𝐹^Δ(𝑥)Δ𝑥−^𝑏_𝑎𝐺^𝛽(𝑥)𝐺^Δ(𝑥)+𝐹^𝛽(𝑥)𝐹^Δ(𝑥)Δ𝑥≤𝑑_𝛽^𝑏_𝑎(𝐺(𝑥)+𝐹(𝑥))^𝛽Δ(𝐺(𝑥)(𝑥−+𝐹))_𝑏𝑎𝐺^𝛽(𝑥)Δ𝐺(𝑥)−^𝑏_𝑎𝐹^𝛽(𝑥)Δ𝐹(𝑥)=𝑑_𝛽Γ(𝐺+𝐹)−Γ(𝐺)−Γ(𝐹). (24)
The proof is complete.

Remark 5. In the special case where 𝕋=ℝ, Theorem 4 reduces to Theorem 1 of [13].

Theorem 6. Let 𝑓∈𝐶^𝑚−1_𝑟𝑑(Υ), Υ∶=[𝑎,𝑏]∩𝕋 be such that 𝑓^Δ𝑖(𝑎)=0, 0≤𝑘≤𝑚−1, let 𝑓^Δ𝑚−1(𝑥) be absolutely continuous on Υ, and let ∫^𝑏_𝑎|𝑓^Δ𝑚(𝑥)|²Δ𝑥<∞. Then ^𝑏_𝑎𝑓^Δ𝑘(𝑥)𝑓^Δ𝑚(𝑥≤)Δ𝑥_{𝑏𝑎𝑥𝑎}ℎ²_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))Δ𝜏Δ𝑥^1/2⋅^𝑏_𝑎𝑓^Δ𝑚(𝜏)²Δ𝜏^1/2. (25)

Proof. From the hypotheses, we have 𝑓^Δ𝑘(𝑥)≤^𝑥_𝑎ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑓^Δ𝑚(𝜏)Δ𝜏. (26)
Multiplying (26) by |𝑓^Δ𝑚(𝑥)| and using Cauchy-Schwarz inequality, we obtain 𝑓^Δ𝑘(𝑥)𝑓^Δ𝑚(𝑥≤𝑓_Δ𝑚)(𝑥)^𝑥_𝑎ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑓^Δ𝑚(𝜏≤𝑓)Δ𝜏_Δ𝑚(𝑥)^𝑥_𝑎ℎ²_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))Δ𝜏^1/2⋅^𝑥_𝑎𝑓^Δ𝑚(𝜏)²Δ𝜏^1/2. (27)
Integrating both sides over 𝑥 from 𝑎 to 𝑏 and using Cauchy-Schwarz inequality, we observe ^𝑏_𝑎𝑓^Δ𝑘(𝑥)𝑓^Δ𝑚(𝑥≤)Δ𝑥_{𝑏𝑎𝑥𝑎}ℎ²_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))Δ𝜏Δ𝑥^1/2⋅^𝑏_𝑎𝑓^Δ𝑚(𝑥)^2𝑥_𝑎𝑓^Δ𝑚(𝜏)²Δ𝜏Δ𝑥^1/2=^𝑏_𝑎^𝑥_𝑎ℎ²_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))Δ𝜏Δ𝑥^1/2⋅^𝑏_𝑎𝑓^Δ𝑚(𝜏)²Δ𝜏^1/2. (28)
The proof is complete.

Theorem 7. Let 𝑝(𝑥)>0, 𝑞(𝑥) be nonnegative and measurable on Υ=[𝑎,𝑏]∩𝕋, and let 𝑓∈𝐶^𝑚−1_𝑟𝑑(Υ) be such that 𝑓^Δ𝑘(𝑎)=0, 0≤𝑘≤𝑚−1. If 𝑓^Δ𝑚−1(𝑥) is absolutely continuous on Υ, then for 𝑟>1, 𝑟_𝑘>0, and any 0≤𝑟_𝑚<𝑟, ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑚(𝑥)^𝑟𝑚𝑓^Δ𝑘(𝑥)^𝑟𝑘≤Δ𝑥_𝑏𝑎𝑞(𝑥)^{𝑟/(𝑟−𝑟𝑚)}𝑝(𝑥)^{−𝑟𝑚/(𝑟−𝑟𝑚)}×𝑄(𝑥)^{(𝑟−1)𝑟𝑚/(𝑟−𝑟𝑚)}Δ𝑥^{(𝑟−𝑟𝑚)/𝑟}Φ𝑦^{𝑟𝑚/𝑟}, (29) where 𝑄(𝑥)∶=∫^𝑥_𝑎ℎ^{𝑟/𝑟−1}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^{−(𝑟/𝑟−1)}Δ𝜏, Φ(𝑦)∶=∫^𝑏_𝑎𝑦^{𝑟𝑘/𝑟𝑚}Δ𝑦(𝑥),𝑦(𝑥)∶=∫^𝑥_𝑎𝑝(𝜏)|𝑓^Δ𝑚(𝜏)|^𝑟Δ𝜏.

Proof. Following the hypotheses, it is easy to see that (26) holds. By using Hölder's inequality with indices 𝑟 and 𝑟/(𝑟−1), we obtain 𝑓^Δ𝑘(𝑥≤_𝑥𝑎)ℎ_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^−1/𝑟𝑝(𝜏)^1/𝑟𝑓^Δ𝑚(𝜏≤)Δ𝜏_𝑥𝑎ℎ^{𝑟/(𝑟−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^{−𝑟/(𝑟−1)}Δ𝜏^{(𝑟−1)/𝑟}⋅^𝑥_𝑎𝑝(𝜏)𝑓^Δ𝑚(𝜏)^𝑟Δ𝜏^1/𝑟=𝑄(𝑥)^{(𝑟−1)/𝑟}𝑦(𝑥)^1/𝑟, (30) where 𝑄(𝑥)∶=∫^𝑥_𝑎ℎ^{𝑟/(𝑟−1)}_{𝑚−𝑘−1}(𝑥,𝜎(𝜏))𝑝(𝜏)^{−𝑟/(𝑟−1)}Δ𝜏, 𝑦(𝑥)∶=∫^𝑥_𝑎𝑝(𝜏)|𝑓^Δ𝑚(𝜏)|^𝑟Δ𝜏. So we get 𝑦^Δ(𝑥)=𝑝(𝑥)𝑓^Δ𝑚(𝑥)^𝑟, (31) and hence for any 𝑟_𝑚, 𝑓^Δ𝑚(𝑥)^𝑟𝑚=𝑝(𝑥)^{−𝑟𝑚/𝑟}𝑦^Δ(𝑥)^{𝑟𝑚/𝑟}. (32)
Thus for 𝑟_𝑘>0, 𝑞(𝑥)𝑓^Δ𝑚(𝑥)^𝑟𝑚𝑓^Δ𝑘(𝑥)^𝑟𝑘≤𝑞(𝑥)𝑝(𝑥)^{−𝑟𝑚/𝑟}𝑦^Δ(𝑥)^{𝑟𝑚/𝑟}×𝑄(𝑥)^{(𝑟−1)𝑟𝑘/𝑟}𝑦(𝑥)^{𝑟𝑘/𝑟}. (33)
Integrating both sides of (33) from 𝑎 to 𝑏 and applying Hölder's inequality with indices 𝑟/𝑟_𝑚 and 𝑟/(𝑟−𝑟_𝑚), we obtain ^𝑏_𝑎𝑞(𝑥)𝑓^Δ𝑚(𝑥)^𝑟𝑚𝑓^Δ𝑘(𝑥)^𝑟𝑘≤Δ𝑥_𝑏𝑎𝑞(𝑥)𝑝(𝑥)^{−𝑟𝑚/𝑟}𝑦^Δ(𝑥)^{𝑟𝑚/𝑟}𝑄(𝑥)^{(𝑟−1)𝑟𝑘/𝑟}𝑦(𝑥)^{𝑟𝑘/𝑟}≤Δ𝑥_𝑏𝑎𝑞(𝑥)^{𝑟/(𝑟−𝑟𝑚)}𝑝(𝑥)^{−𝑟𝑚/(𝑟−𝑟𝑚)}𝑄(𝑥)^{(𝑟−1)𝑟𝑘/(𝑟−𝑟𝑚)}Δ𝑥^{(𝑟−𝑟𝑚)/𝑟}⋅^𝑏_𝑎𝑦^Δ(𝑥)𝑦(𝑥)^{𝑟𝑘/𝑟𝑚}Δ𝑥^{𝑟𝑚/𝑟}=^𝑏_𝑎𝑞(𝑥)^{𝑟/(𝑟−𝑟𝑚)}𝑝(𝑥)^{−𝑟𝑚/(𝑟−𝑟𝑚)}×𝑄(𝑥)^{(𝑟−1)𝑟𝑘/(𝑟−𝑟𝑚)}Δ𝑥^{(𝑟−𝑟𝑚)/𝑟}Φ𝑦^{𝑟𝑚/𝑟}, (34) where Φ(𝑦)∶=∫^𝑏_𝑎𝑦(𝑥)^{𝑟𝑘/𝑟𝑚}Δ𝑦(𝑥). The proof is complete.

Remark 8. In the special case where 𝕋=ℝ, Theorems 6 and 7 reduce to Theorems 1 and 2 of [18].

Acknowledgments

The first author's research was supported by NSF of China (11071054), Natural Science Foundation of Hebei Province (A2011205012). The second author's research was partially supported by an HKU URG grant.