Hardy-Type Inequalities on Time Scale via Convexity in Several

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Jul 1, 2013 - We extend some Hardy-type inequalities with general kernels to arbitrary time scales using multivariable convex functions. Some classical and ...
Hindawi Publishing Corporation ISRN Mathematical Analysis Volume 2013, Article ID 903196, 9 pages http://dx.doi.org/10.1155/2013/903196

Research Article Hardy-Type Inequalities on Time Scale via Convexity in Several Variables Tzanko Donchev,1,2 Ammara Nosheen,2 and Josip PeIariT2,3 1

Department of Mathematics, University Al. I. Cuza, 700506 Ias¸i, Romania Abdus Salam School of Mathematical Sciences, Government College University, Lahore 54600, Pakistan 3 University of Zagreb, Faculty of Textile Technology, 10000 Zagreb, Croatia 2

Correspondence should be addressed to Josip Peˇcari´c; [email protected] Received 3 June 2013; Accepted 1 July 2013 Academic Editors: M. Lindstrom and S. Liu Copyright © 2013 Tzanko Donchev et al. 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 extend some Hardy-type inequalities with general kernels to arbitrary time scales using multivariable convex functions. Some classical and new inequalities are deduced seeking applications.

1. Introduction The significant Hardy inequality is published in [1] (1952) (both in the continuous and discrete settings). More general Hardy integral inequalities have been studied in continuous cases. We notice only [2–6] and the references therein. Recently, this inequality is studied in discrete case, and some variants of it are proved in case of time scales [7–9]. In [10], the authors study Hardy-type inequalities using convex functions of one variable with general kernels to arbitrary time scales. The aim of this paper is to provide Hardy-type inequalities using multivariable convex function with general kernels to arbitrary time scales. Notice that time scales include continuous and discrete time cases under unified approach. Firstly, we recall necessary preliminary facts needed afterward. The main results are given in Section 3. Section 4 is devoted to some inequalities with certain kernels. In the last section, we discuss some particular cases of Hardy-type inequalities.

2. Preliminaries First, we recall the basic concepts used in the paper and refer the interested reader to [11] for the theory of time scales.

A time scale is any nonempty closed subset of the real line T ⊂ R. On nondensity points we define forward, respectively, backward jump operators 𝜎, 𝜌 : T → T as 𝜎 (𝑡) = inf {𝑠 ∈ T : 𝑠 > 𝑡} ,

(1)

𝜌 (𝑡) = sup {𝑠 ∈ T : 𝑠 < 𝑡} .

The point 𝑡 is said to be right-scattered if 𝜎(𝑡) > 𝑡 and leftscattered if 𝜌(𝑡) < 𝑡, respectively. Clearly, 𝑡 is right-dense if 𝜎(𝑡) = 𝑡 and left dense if 𝜌(𝑡) = 𝑡, respectively. Let 𝑛 ∈ N; we define 𝑛-dimensional time scale by the Cartesian product of given time scales T𝑖 , 𝑖 ∈ {1, . . . , 𝑛}, as Ω𝑛 = {𝑎⃗ = (𝑎1 , 𝑎2 , . . . , 𝑎𝑛 ) : 𝑎𝑖 ∈ T𝑖 , 𝑖 ∈ {1, . . . , 𝑛}} .

(2)

Evidently, Ω𝑛 is a complete metric space with distance as follows: 1/2

𝑛

⃗ = (∑󵄨󵄨󵄨𝑏 − 𝑎 󵄨󵄨󵄨2 ) 𝑑 (𝑎,⃗ 𝑏) 𝑖󵄨 󵄨𝑖

,

𝑎, 𝑏 ∈ Ω𝑛 .

(3)

𝑖=1

Now we are going to describe the construction of Lebesgue measure in Ω𝑛 . We refer to [12–14] for the theory of measure spaces and measurable functions on time scales. Let F be the family of all 𝑛-dimensional time scale intervals in Ω𝑛 ; that is, ⃗ = [𝑎 , 𝑏 ) × [𝑎 , 𝑏 ) × ⋅ ⋅ ⋅ × [𝑎 , 𝑏 ) , (4) 𝑉 = [𝑎,⃗ 𝑏) 1

1

2

2

𝑛

𝑛

2

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with 𝑎⃗ = (𝑎1 , 𝑎2 , . . . , 𝑎𝑛 ), 𝑏⃗ = (𝑏1 , 𝑏2 , . . . , 𝑏𝑛 ) ∈ Ω𝑛 , and 𝑎𝑖 ≤ 𝑏𝑖 for all 𝑖 ∈ {1, . . . , 𝑛}. Let 𝑚 : F → [0, ∞) be the set function that assigns to each 𝑛-dimensional time scale interval 𝑉 = [𝑎, 𝑏) its volume as follows: 𝑛

𝑚 (𝑉) = ∏ (𝑏𝑖 − 𝑎𝑖 ) .

theory (see, e.g., [15]). The Lebesgue integral associated with the measure 𝜇Δ on Ω𝑛 is called the Lebesgue Δ-integral. For a Δ-measurable set 𝐸 ⊂ Ω𝑛 and a Δ-measurable function 𝑓 : 𝐸 → R, the corresponding Δ-integral of 𝑓 over 𝐸 will be denoted by

(5)

∫ 𝑓 (𝑡1 , . . . , 𝑡𝑛 ) Δ𝑡1 ⋅ ⋅ ⋅ Δ𝑡𝑛

𝑖=1

𝑛

Let 0 ≠ 𝐸 ⊂ Ω . If there exists finite or countable system of pairwise disjoint 𝑛-dimensional time scale intervals 𝑉𝑘 = [𝑎𝑘⃗ , 𝑏𝑘⃗ ) with 𝐸 ⊂ ⋃𝑘≥1 𝑉𝑘 , then the outer measure 𝑚∗ of 𝐸 is defined by 𝑚∗ (𝐸) = inf {∑ 𝑚 (𝑉𝑘 ) , 𝐸 ⊂ ⋃ 𝑉𝑘 } . 𝑘≥1

𝐸

𝑏

(6)

∫ 𝑓 (𝑡) Δ𝑡 = ∑ (𝜎 (𝑡) − 𝑡) 𝑓 (𝑡) . 𝑎



𝑚∗ (𝐸) = 𝑚∗ (𝐸 ∩ 𝐴) + 𝑚∗ (𝐸 ∩ 𝐴𝑐 ) 𝑛

𝑐

(7)

𝑛

𝜇Δ ({𝑡}) = ∏ (𝜎𝑖 (𝑡𝑖 ) − 𝑡𝑖 ) .

(8)

𝑖=1

Obviously, for all 𝑖 ∈ {1, . . . , 𝑛}, the set Ω𝑛0 = Ω𝑛 \ Ω𝑛∞ can be represented as a finite or countable union of intervals of the family F; hence it is Δ-measurable. Furthermore, the set Ω𝑛∞ = Ω𝑛 \ Ω𝑛0 is Δ-measurable being the difference of two Δ-measurable sets Ω𝑛 and Ω𝑛0 , but Ω𝑛∞ does not have a finite or countable covering intervals of F, that is, for any Δ-measurable subset 𝐴 of Ω𝑛 such that 𝐴 ∩ Ω∞ ≠ 0 has Δmeasure ∞. In particular, if 𝑎 ∈ T, where T is an arbitrary time scale, then the set [𝑎, ∞) = {𝑡 ∈ T : 𝑎 ≤ 𝑡} is Δmeasurable. The function 𝑓 : Ω𝑛 → R := [−∞, ∞] is said to be Δmeasurable if for every 𝛼 ∈ R, the set 𝑓−1 ([−∞, 𝛼)) = {𝑡 = (𝑡1 , . . . , 𝑡𝑛 ) ∈ Ω𝑛 : 𝑓(𝑡) < 𝛼} is Δ-measurable. It is easy to see that 𝑓 is Δ-measurable if and only if for each open set 𝐺 ⊂ R, the set 𝑓−1 (𝐺) = {𝑡 ∈ Ω𝑛 : 𝑓(𝑡) ∈ 𝐺} is Δ-measurable. Moreover, if 𝑓 : Ω𝑛 → R is Δ-measurable and Φ : 𝐼 ⊂ R → R is a continuous function, then Φ ∘ 𝑓 : Ω𝑛 → R is Δ-measurable. Having the 𝜎-additive measure 𝜇Δ on Ω𝑛 , we possess the corresponding integration theory for functions 𝑓 : 𝐸 ⊂ Ω𝑛 → R, according to the general Lebesgue integration

(10)

𝑡∈[𝑎,𝑏)

Let (Ω, M, 𝜇Δ ) and (Λ, L, 𝜆 Δ ) be two finite dimensional time scale measure spaces. We define the product measure space (Ω × Λ, M × L, 𝜇Δ × 𝜆 Δ ), where M × L is the product 𝜎-algebra generated by {𝐸 × 𝐹 : 𝐸 ∈ M, 𝐹 ∈ L} and

𝑛

holds for all 𝐸 ⊂ Ω , where 𝐴 = Ω \ 𝐴. The family M of all 𝑚∗ -measurable subsets of Ω𝑛 is a 𝜎-algebra generated by F. The restriction of 𝑚∗ to M, which we denote by 𝜇Δ , is a 𝜎-additive measure on M. Clearly, F ⊂ M and 𝜇Δ (𝑉) = 𝑚(𝑉) for each 𝑉 ∈ F. The measure 𝜇Δ (called the Lebesgue Δ-measure on Ω𝑛 ) is Carath´eodory extension of the original measure 𝑚 defined on F. We call (Ω𝑛 , M, 𝜇Δ ) an 𝑛dimensional time scale measure space. Denote 𝑎̃𝑖 = inf 𝑡∈T𝑖 𝑡. Similarly, ̃𝑏𝑖 = sup𝑡∈T𝑖 𝑡. Let Ω𝑛∞ be the set of all points 𝑏⃗ = (𝑏1 , 𝑏2 , . . . , 𝑏𝑛 ) ∈ Ω𝑛 for which there exists at least one 𝑏𝑖 such that 𝑏𝑖 = ̃𝑏𝑖 . From [11, Theorem 3.1], we know that if 𝑡 ⃗ = (𝑡1 , 𝑡2 , . . . , 𝑡𝑛 ) ∈ Ω𝑛 \ Ω𝑛∞ , then the singlepoint set {𝑡} is Δ-measurable and

(9)

𝐸

Notice that all theorems of the general Lebesgue integration theory, including the Lebesgue dominated convergence theorem, hold also for Lebesgue Δ-integrals on Ω𝑛 . If T is a time scale and the interval [𝑎, 𝑏) ⊂ T contains only isolated points, then

𝑘≥1

If there is no such covering of 𝐸, then 𝑚 (𝐸) = ∞. A subset 𝐴 of Ω𝑛 is said to be measurable (or Δmeasurable) if

∫ 𝑓 (𝑡) Δ𝑡.

or

(𝜇Δ × 𝜆 Δ ) (𝐸 × 𝐹) = 𝜇Δ (𝐸) 𝜆 Δ (𝐹) .

(11)

Fubini’s Theorem in Time Scales see ([16, Theorem 1.1]). If 𝑓 : Ω × Λ → R is a 𝜇Δ × 𝜆 Δ -integrable function and if we define the function 𝜑(𝑦) = ∫Ω 𝑓(𝑥, 𝑦)Δ𝑥 for a.e. 𝑦 ∈ Λ and 𝜓(𝑥) = ∫Λ 𝑓(𝑥, 𝑦)Δ𝑦 for a.e. 𝑥 ∈ Ω, then 𝜑 is 𝜆 Δ -integrable on Λ, 𝜓 is 𝜇Δ -integrable on Ω, and ∫ Δ𝑥 ∫ 𝑓 (𝑥, 𝑦) Δ𝑦 = ∫ Δ𝑦 ∫ 𝑓 (𝑥, 𝑦) Δ𝑥. Ω

Λ

Λ

(12)

Ω

3. Inequalities with General Kernels Let f(𝑦) = (𝑓1 (𝑦), . . . , 𝑓𝑚 (𝑦)) be 𝑚-tuple of functions such that 𝑓𝑗 (𝑦) are 𝜆 Δ -integrable for all 𝑗 ∈ {1, . . . , 𝑚}. Then ∫Λ f(𝑦)Δ(𝑦) denotes the 𝑚-tuple (∫Λ 𝑓1 (𝑦)Δ(𝑦), . . ., ∫Λ 𝑓𝑚 (𝑦)Δ(𝑦)); that is, Δ-integral acts on each component of f(𝑦). The following Jensen’s inequality on time scales is given in [17, Theorem 2.1]. Theorem 1. Let (Ω, M, 𝜇Δ ) and (Λ, L, 𝜆 Δ ) be two time scale measure spaces. Suppose 𝑈 ⊂ R𝑚 is a closed convex set, Φ ∈ 𝐶(𝑈, R) is convex, and f(Λ) ⊂ 𝑈. Moreover, let 𝑘 : Ω×Λ → R be nonnegative such that 𝑘(𝑥, ⋅) is 𝜆 Δ -integrable. Then one has Φ(

∫Λ 𝑘 (𝑥, 𝑦) f (𝑦) Δ𝑦 ∫Λ 𝑘 (𝑥, 𝑦) Δ𝑦

)≤

∫Λ 𝑘 (𝑥, 𝑦) Φ (f (𝑦)) Δ𝑦 ∫Λ 𝑘 (𝑥, 𝑦) Δ𝑦

. (13)

Throughout this section, we assume that the following hypotheses hold. H1 (Ω, M, 𝜇Δ ) and (Λ, L, 𝜆 Δ ) are two time scale measure spaces. H2 𝑘 : Ω × Λ → R+ is such that 𝐾(𝑥) := ∫Λ 𝑘(𝑥, 𝑦)Δ𝑦 < ∞, 𝑥 ∈ Ω. H3 𝜉 : Ω → R+ is such that 𝑤(𝑦) := ∫Ω (𝑘(𝑥, 𝑦)𝜉(𝑥)/ 𝐾(𝑥))Δ𝑥, 𝑦 ∈ Λ.

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3

Theorem 2. If 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous, then ∫ 𝜉 (𝑥) Φ ( Ω

1 ∫ 𝑘 (𝑥, 𝑦) f (𝑦) Δ𝑦) Δ𝑥 𝐾 (𝑥) Λ

(14)

≤ ∫ 𝑤 (𝑦) Φ (f (𝑦)) Δ𝑦 Λ

holds for all 𝜆 Δ -integrable functions f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈. Proof. Using Jensen’s inequality (13) for several variables and the Fubini theorem on time scales, we find that 1 ∫ 𝑘 (𝑥, 𝑦) f (𝑦) Δ𝑦) Δ𝑥 ∫ 𝜉 (𝑥) Φ ( 𝐾 (𝑥) Λ Ω = ∫ 𝜉 (𝑥) Φ ( Ω

Remark 5. In case that T = N and 𝑚 = 2, Corollary 4 is as [18, Corollary 1.2]. Remark 6. In case that 𝑚 = 2, we can use the results of Beck [19] (see also [20], page 194) as applications of Corollary 4, which corresponds to the generalizations of H¨older’s and Minkoski’s inequalities. In classical case, many authors have studied these types of generalizations; see, for example, [21– 23]. Furthermore, in the paper, we use 𝑝 > 1 with (1/𝑝) + (1/𝑝)́ = 1. Corollary 7. If f(𝑥) = (𝑓1 (𝑥), 𝑓2 (𝑥)), then ∫ 𝜉 (𝑥) ( Ω

1 1 ∫ 𝑘 (𝑥, 𝑦) 𝑓1 (𝑦) Δ𝑦, . . . , 𝐾 (𝑥) Λ 𝐾 (𝑥)

́

×(

× ∫ 𝑘 (𝑥, 𝑦) 𝑓𝑚 (𝑦) Δ𝑦) Δ𝑥 Λ

holds for all 𝜆 Δ -integrable 𝑓𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}. Proof. Use 𝑚 = 2, Ψ(𝑠1 , 𝑠2 ) = 𝑠1 𝑠2 , 𝐿 1 (𝑡1 ) = (𝑡1 )𝑝 , 𝐿 2 (𝑡2 ) = 1/𝑝 1/𝑝́ ́ (𝑡2 )𝑝 in Corollary 4; then Φ(𝑠1 , 𝑠2 ) = 𝑠1 𝑠2 is concave in Theorem 2.

= ∫ 𝑤 (𝑦) Φ (f (𝑦)) Δ𝑦. Λ

(15)

Corollary 8. If f(𝑥) = (𝑓1 (𝑥), 𝑓2 (𝑥)), then

The proof is therefore complete. Remark 3. If Φ is concave, then (14) holds in reverse direction.

∫ 𝜉 (𝑥) (( Ω

for all 𝑗 ∈ {1, . . . , 𝑚}. If Φ(𝑠1 , . . . , 𝑠𝑚 ) = 𝐿−1 𝑚 (𝑠𝑚 )) is convex, then

)

1/𝑝 1 ∫ 𝑘(𝑥, 𝑦)(𝑓1 (𝑦))𝑝 Δ𝑦) 𝐾 (𝑥) Λ 𝑝

1/𝑝 1 +( ∫ 𝑘(𝑥, 𝑦)(𝑓2 (𝑦))𝑝 Δ𝑦) ) Δ𝑥 (19) 𝐾 (𝑥) Λ

Corollary 4. Let Ψ : [𝑙1 , 𝑙1́ ) × ⋅ ⋅ ⋅ × [𝑙𝑚 , 𝑙𝑚́ ) → R+ be continuous function and define

𝐾 (𝑥)

(18)

Λ

𝑘 (𝑥, 𝑦) 𝜉 (𝑥) Δ𝑥) Δ𝑦 = ∫ Φ (f (𝑦)) (∫ 𝐾 (𝑥) Λ Ω

̃ 𝑗 (𝑓𝑗 , Λ) = 𝐿−1 ( 𝐿 𝑗

1/𝑝 1 ́ ∫ 𝑘(𝑥, 𝑦)(𝑓2 (𝑦))𝑝 Δ𝑦) Δ𝑥 𝐾 (𝑥) Λ

≥ ∫ 𝑤 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) Δ𝑦

𝜉 (𝑥) (∫ 𝑘 (𝑥, 𝑦) Φ (f (𝑦)) Δ𝑦) Δ𝑥 ≤∫ Ω 𝐾 (𝑥) Λ

∫Λ 𝑘 (𝑥, 𝑦) 𝐿 𝑗 (𝑓𝑗 (𝑦)) Δ𝑦

1/𝑝 1 ∫ 𝑘(𝑥, 𝑦)(𝑓1 (𝑦))𝑝 Δ𝑦) 𝐾 (𝑥) Λ

≥ ∫ 𝑤 (𝑦) (𝑓1 (𝑦) + 𝑓2 (𝑦))𝑝 Δ𝑦 (16)

Λ

holds for all 𝜆 Δ -integrable 𝑓𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}. Ψ(𝐿−1 1 (𝑠1 ), . . . ,

̃ 1 (𝑓1 , Λ) , . . . , 𝐿 ̃ 𝑚 (𝑓𝑚 , Λ)) Δ𝑥 ∫ 𝜉 (𝑥) Ψ (𝐿 Ω

(17) ≤ ∫ 𝑤 (𝑦) Ψ (𝑓1 (𝑦) , . . . , 𝑓𝑚 (𝑦)) Δ𝑦

Proof. Use 𝑚 = 2, Ψ(𝑠1 , 𝑠2 ) = (𝑠1 + 𝑠2 )𝑝 , 𝐿 1 (𝑡1 ) = (𝑡1 )𝑝 , 1/𝑝 1/𝑝 𝐿 2 (𝑡2 ) = (𝑡2 )𝑝 in Corollary 4; then Φ(𝑠1 , 𝑠2 ) = (𝑠1 + 𝑠2 )𝑝 is concave in Theorem 2. Remark 9. If 𝑝 < 1, then reverse inequalities hold in Corollaries 4, 7, and 8.

Λ

holds for all 𝑓𝑗 (Λ) ⊂ [𝑙𝑗 , 𝑙𝑗́ ) and continuous monotone functions 𝐿 𝑗 : [𝑙𝑗 , 𝑙𝑗́ ) → R such that 𝐿 𝑗 ∘ 𝑓𝑗 are 𝜆 Δ -integrable for all 𝑗 ∈ {1, . . . , 𝑚}. Proof. Replace in Theorem 2 𝑓𝑗 (𝑦) with 𝐿 𝑗 (𝑓𝑗 (𝑦)) for all −1 𝑗 ∈ {1, . . . , 𝑚} and Φ(𝑠1 , . . . , 𝑠𝑚 ) with Ψ(𝐿−1 1 (𝑠1 ), . . . , 𝐿 𝑚 (𝑠𝑚 )).

4. Inequalities with Special Kernels Throughout this section and in the next section, we assume that the following hypothesis holds. H4 Ω = Λ = [𝑎1 , 𝑏1 )T × [𝑎2 , 𝑏2 )T × ⋅ ⋅ ⋅ × [𝑎𝑛 , 𝑏𝑛 )T , 0 ≤ 𝑎𝑖 < 𝑏𝑖 ≤ ∞ for all 𝑖 ∈ {1, . . . , 𝑛}, where T is an arbitrary time scale.

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Corollary 10. Assume that

Proof. Statement follows from Theorem 2 using

𝜉 : Ω → R+ is such that 𝑏1

𝑘 (𝑥1 , . . . , 𝑥𝑛 , 𝑦1 , . . . , 𝑦𝑛 ) = 0,

𝑏𝑛

𝑘 (𝑥1 , . . . , 𝑥𝑛 , 𝑦) 𝜉 (𝑥1 , . . . , 𝑥𝑛 ) 𝑤 (𝑦) = ∫ ⋅ ⋅ ⋅ ∫ 𝐾 (𝑥1 , . . . , 𝑥𝑛 ) 𝑦1 𝑦𝑛 × Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 ,

for all 𝑖 ∈ {1, . . . , 𝑛} ,

(20)

𝐾 (𝑥) = ∫

𝑏1

𝜎(𝑥1 )

⋅⋅⋅∫

𝑏𝑛

𝜎(𝑥𝑛 )

𝑘 (𝑥, 𝑦1 , . . . , 𝑦𝑛 ) Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 .

𝑏1

𝑏𝑛

Theorem 12. Assume that

𝑎1

𝑎𝑛

𝜉 : Ω → R+ is such that

∫ ⋅ ⋅ ⋅ ∫ 𝜉 (𝑥1 , . . . , 𝑥𝑛 ) Φ ((𝐴 𝑘 f) (𝑥1 , . . . , 𝑥𝑛 )) Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 𝑏𝑛

𝑎1

𝑎𝑛

≤ ∫ ⋅ ⋅ ⋅ ∫ 𝑤 (𝑦1 , . . . , 𝑦𝑛 ) Φ (f (𝑦1 , . . . , 𝑦𝑛 ))

𝑏1

𝑏𝑛

𝑦1

𝑦𝑛

̃ (𝑦) = ∫ ⋅ ⋅ ⋅ ∫ 𝑤

𝜉 (𝑥1 , . . . , 𝑥𝑛 ) Δ𝑥 ⋅ ⋅ ⋅ Δ𝑥𝑛 , ∏𝑛𝑖=1 (𝜎 (𝑥𝑖 ) − 𝑎𝑖 ) 1

× Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 (21) holds for all 𝜆 Δ -integrable f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈, where (𝐴 𝑘 f) (𝑥) =

𝜎(𝑥1 ) 𝜎(𝑥𝑛 ) 1 ⋅⋅⋅∫ 𝑘 (𝑥, 𝑦1 , . . . , 𝑦𝑛 ) ∫ 𝐾 (𝑥) 𝑎1 𝑎𝑛

× f (𝑦1 , . . . , 𝑦𝑛 ) Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . (22)

if 𝑎𝑖 ≤ 𝑦𝑖 ≤ 𝜎 (𝑥𝑖 ) for all 𝑖 ∈ {1, . . . , 𝑛} ,

𝜎(𝑥1 )

𝑎1

⋅⋅⋅∫

𝜎(𝑥𝑛 )

𝑎𝑛

(23)

𝑏1

𝑏𝑛

𝑎1

𝑎𝑛

𝑏1

𝑏𝑛

𝑎1

𝑎𝑛

̃ (𝑦1 , . . . , 𝑦𝑛 ) Φ (f (𝑦1 , . . . , 𝑦𝑛 )) ≤ ∫ ⋅⋅⋅∫ 𝑤 × Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 holds for all 𝜆 Δ -integrable f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈, where ̃𝑘 f) (𝑥1 , . . . , 𝑥𝑛 ) = (𝐴

∏𝑛𝑖=1

1 (𝜎 (𝑥𝑖 ) − 𝑎𝑖 )

𝜎(𝑥1 )

×∫

𝑘 (𝑥, 𝑦1 , . . . , 𝑦𝑛 ) Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . (24)

𝑎1

⋅⋅⋅∫

𝜎(𝑥𝑛 )

𝑎𝑛

𝑎1

𝑎𝑛

𝑤 (𝑦) = ∫ ⋅ ⋅ ⋅ ∫

̃f (𝑦 , . . . , 𝑦 ) (31) 1 𝑛

× Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . Proof. Statement follows from Theorem 2 using

𝜉 : Ω 󳨀→ R+ is such that 𝑦𝑛

(29)

̃𝑘 f) (𝑥1 , . . . , 𝑥𝑛 )) Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 ∫ ⋅ ⋅ ⋅ ∫ 𝜉 (𝑥1 , . . . , 𝑥𝑛 ) Φ ((𝐴

Corollary 11. Assume that

𝑦1

𝑦 ∈ Λ.

(30)

since in this case 𝐾 (𝑥) = ∫

(28)

If 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous, then

Proof. Statement follows from Theorem 2 using 𝑘 (𝑥1 , . . . , 𝑥𝑛 , 𝑦1 , . . . , 𝑦𝑛 ) = 0,

(27)

since in this case

𝑦 ∈ Λ.

If 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous; then

𝑏1

if 𝑎𝑖 ≤ 𝜎 (𝑥𝑖 ) ≤ 𝑦𝑖

𝑘 (𝑥1 , . . . , 𝑥𝑛 , 𝑦) 𝜉 (𝑥1 , . . . , 𝑥𝑛 ) 𝐾 (𝑥1 , . . . , 𝑥𝑛 )

× Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 ,

𝑘 (𝑥1 , . . . , 𝑥𝑛 , 𝑦1 , . . . , 𝑦𝑛 ) (25)

𝑦 ∈ Λ.

If 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous, then (21) holds for all 𝜆 Δ -integrable f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈, and

1 if 𝑎𝑖 ≤ 𝑦𝑖 < 𝜎 (𝑥𝑖 ) ≤ 𝑏𝑖 , 𝑖 ∈ {1, . . . , 𝑛} ={ }, 0 otherwise, since in this case 𝐾 (𝑥1 , . . . , 𝑥𝑛 ) = ∫

𝜎(𝑥1 )

𝑎1

𝑛

𝑏1 𝑏𝑛 1 ⋅⋅⋅∫ 𝑘 (𝑥, 𝑦1 , . . . , 𝑦𝑛 ) (𝐴 𝑘 f) (𝑥) = ∫ 𝐾 (𝑥) 𝜎(𝑥1 ) 𝜎(𝑥𝑛 )

× f (𝑦1 , . . . , 𝑦𝑛 ) Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . (26)

(32)

⋅⋅⋅∫

𝜎(𝑥𝑛 )

𝑎𝑛

Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛

= ∏ (𝜎 (𝑥𝑖 ) − 𝑎𝑖 ) . 𝑖=1

̃𝑘 , 𝑤 = 𝑤 ̃. Thus 𝐴 𝑘 = 𝐴

(33)

ISRN Mathematical Analysis

5

Corollary 13. If 𝑎𝑖 = 0 for all 𝑖 ∈ {1, . . . , 𝑛} in H4 and 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous, then 𝑏1

𝑏𝑛

0

0

∫ ⋅ ⋅ ⋅ ∫ Φ ((𝐴 𝑘 f) (𝑥1 , . . . , 𝑥𝑛 ))

then ∞



0

0

∫ (∫

Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 𝑥1 ⋅ ⋅ ⋅ 𝑥𝑛

𝑏𝑛 𝑛

𝑏1

1 1 ≤ ∫ ⋅ ⋅ ⋅ ∫ ∏ ( − ) Φ (f (𝑦1 , . . . , 𝑦𝑛 )) Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 𝑦 𝑏𝑖 0 0 𝑖=1 𝑖 (34)

× (∫

0

1

𝜎(𝑥1 )

0

1/𝑝󸀠

󸀠

(𝑦/𝑥)1−1/𝑞 (𝑔2 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

Δ𝑥

(39)

≥ ∫ 𝐾2 (𝑦) 𝑔1 (𝑦) 𝑔2 (𝑦) Δ𝑦 0

holds for all 𝜆 Δ -integrable 𝑔𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}.

∏𝑛𝑖=1 𝜎 (𝑥𝑖 ) ×∫

1/𝑝

𝑝

(𝑔1 (𝑦)) Δ𝑦) 𝑥+𝑦



holds for all 𝜆 Δ -integrable f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈, where (𝐴 𝑘 f) (𝑥1 , . . . , 𝑥𝑛 ) :=



1−1/𝑞

(𝑦/𝑥)

⋅⋅⋅∫

𝜎(𝑥𝑛 )

0

f (𝑦1 , . . . , 𝑦𝑛 )

(35)

Proof. Use 𝜉(𝑥) = 𝐾1 (𝑥)/𝑥 and 𝑘(𝑥, 𝑦) (𝑦/𝑥)−1/𝑞 /(𝑥+𝑦), 𝑥 ≠ 0,𝑦 ≠ 0, 𝑥+𝑦 ≠ 0, { } in Corollary 7 to obtain 0, otherwise

=

× Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . Proof. The statement follows from Theorem 12 using 𝜉(𝑥1 , . . . , 𝑥𝑛 ) = 1/𝑥1 ⋅ ⋅ ⋅ 𝑥𝑛 , since in this case 𝑏1

𝑏𝑛

𝑦1

𝑦𝑛

𝑤 (𝑦1 , . . . , 𝑦𝑛 ) = ∫ ⋅ ⋅ ⋅ ∫ 𝑛

= ∏( 𝑖=1

1

Δ𝑥 ∏𝑛𝑖=1 𝑥𝑖 𝜎 (𝑥𝑖 ) 1





0

⋅ ⋅ ⋅ Δ𝑥𝑛

1 1 − ). 𝑦𝑖 𝑏𝑖

1/𝑝

∞ (𝑦/𝑥)−1/𝑞 (𝑓 (𝑦))𝑝 𝐾1 (𝑥) 1 1 ( Δ𝑦) ∫ 𝑥 𝐾1 (𝑥) 0 𝑥+𝑦 1/𝑝󸀠

󸀠

∞ (𝑦/𝑥)−1/𝑞 (𝑓 (𝑦))𝑝 1 2 ×( Δ𝑦) ∫ 𝐾1 (𝑥) 0 𝑥+𝑦

(36)

Δ𝑥

(40)



≥ ∫ 𝑤 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) Δ𝑦, 0

Example 14. If 𝑏𝑖 = ∞ for all 𝑖 ∈ {1, . . . , 𝑛} in addition to the assumptions of Corollary 13, then (34) takes the form ∞



0

0

∫ ⋅ ⋅ ⋅ ∫ Φ ((𝐴 𝑘 f) (𝑥1 , . . . , 𝑥𝑛 )) ∞



0

0

≤ ∫ ⋅ ⋅ ⋅ ∫ Φ (f (𝑦1 , . . . , 𝑦𝑛 ))

Δ𝑥1 ⋅ ⋅ ⋅ Δ𝑥𝑛 𝑥1 ⋅ ⋅ ⋅ 𝑥𝑛 Δ𝑦1 ⋅ ⋅ ⋅ Δ𝑦𝑛 . 𝑦1 ⋅ ⋅ ⋅ 𝑦𝑛

(37)

where 𝑤 (𝑦) = ∫



1 ∫ 𝑦 0

=





0

0

1/𝑝

(𝑦/𝑥)−1/𝑞 (𝑓1 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

In this section, firstly, we give Hilbert-type inequality on time scales. Theorem 17. If 𝑛 = 1, 𝑎1 = 0, 𝑎𝑛𝑑 𝑏1 = ∞ in H4 and for 𝑞 > 1 one defines 𝐾1 (𝑥) = ∫

(𝑦/𝑥)−1/𝑞 Δ𝑦, 𝑥+𝑦



0



𝐾2 (𝑦) = ∫

0

(𝑦/𝑥)1−1/𝑞 Δ𝑥, 𝑥+𝑦

(41)

𝐾 (𝑦) (𝑦/𝑥)1−1/𝑞 Δ𝑥 = 2 . 𝑥+𝑦 𝑦

Using this value in (40), we obtain

∫ (∫

5. Some Particular Cases

∞ 𝑘 (𝑥, 𝑦) 𝑘 (𝑥, 𝑦) 𝜉 (𝑥) Δ𝑥 = ∫ Δ𝑥 𝐾1 (𝑥) 𝑥 0

0

Remark 15. Clearly, if the left-hand side is ∞ in (34), then right-hand side is also ∞. Remark 16. For T = R and 𝑚 = 1, inequality (37) is proved in [4, 24].



× (∫



0



≥ ∫ 𝐾2 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) 0

󸀠

1/𝑝󸀠

(𝑦/𝑥)−1/𝑞 (𝑓2 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

Δ𝑥 𝑥

(42)

Δ𝑦 . 𝑦

(38) Now, if we replace 𝑓1 (𝑦) with 𝑔1 (𝑦)𝑦1/𝑝 and 𝑓2 (𝑦) with 󸀠 𝑔2 (𝑦)y1/𝑝 , we obtain (39).

6

ISRN Mathematical Analysis

Another Proof of (42). Consider L.H.S of (40) and applying H¨older inequality on time scale [25, Theorem 6.2] and Fubini theorem on time scale [16, Theorem 1.1], we have ∫



0

1/𝑝

∞ (𝑦/𝑥)−1/𝑞 (𝑓 (𝑦))𝑝 𝐾1 (𝑥) 1 1 ( Δ𝑦) ∫ 𝑥 𝐾1 (𝑥) 0 𝑥+𝑦

×(

∞ (𝑦/𝑥)−1/𝑞 (𝑓 (𝑦))𝑝 1 2 Δ𝑦) ∫ 𝐾1 (𝑥) 0 𝑥+𝑦 ∞



0

0

≥ ∫ (∫ ∞



0

0

=∫



0

(43)

0

𝜋 , sin (𝜋/𝑞)

𝑑𝑥 =

≥ ∫ 𝐾2 (𝑦) (𝑓1 (𝑦) + 𝑓2 (𝑦))𝑝 0

0

0

∫ (∫

(𝑦/𝑥)1−1/𝑞 (𝑓1 (𝑦))𝑝 𝑑𝑦) 𝑥+𝑦



(44)



∫ 𝜉 (𝑥) Φ (

× (∫

0

󸀠

1/𝑝

(𝑦/𝑥)1−1/𝑞 (𝑓2 (𝑦))𝑝 𝑑𝑦) 𝑥+𝑦

0

0

∫ ((∫

1/𝑝󸀠

𝑑𝑥

(45)

0

Corollary 22. Assume (29). Then 1/𝑝



∫ 𝜉 (𝑥) ( 𝑎

1/𝑝

+(∫

holds for all 𝜆 Δ -integrable f : Λ → R𝑚 such that f(Λ) ⊂ 𝑈. Proof. The statement follows from Theorem 12 using 𝑛 = 1.

(𝑦/𝑥)1−1/𝑞 (𝑓1 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦 ∞

𝜎(𝑥) 1 (𝑓1 (𝑦))𝑝 Δ𝑦) ∫ 𝜎 (𝑥) − 𝑎 𝑎 1/𝑝󸀠

1/𝑝

(𝑦/𝑥)1−1/𝑞 (𝑓2 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

(48)

𝑎

Theorem 19. If (38) is satisfied, then ∞

𝜎(𝑥) 1 f (𝑦) Δ𝑦) Δ𝑥 ∫ 𝜎 (𝑥) − 𝑎 𝑎

≤ ∫ 𝑤 (𝑦) Φ (f (𝑦)) Δ𝑦

In the rest of paper, we take 𝑛 = 1, 𝑎 ≥ 0, 𝑏 = ∞ in H4.



(47)

Δ𝑦 . 𝑦



∞ 𝜋 ∫ 𝑓1 (𝑦) 𝑓2 (𝑦) 𝑑𝑦. sin (𝜋/𝑞) 0



Δ𝑥 𝑥

Corollary 21. If (29) holds, and furthermore, if 𝑈 ⊂ R𝑚 is a closed convex set such that Φ : 𝑈 → R is convex and continuous, then

𝑎



)

Remark 20. (a) We can give another proof of (47) using Minkowski’s inequality on time scale [25, Theorem 7.2]. (b) If 𝑝 < 1, then we have reverse inequalities. Now we consider some generalizations of the P´olyaKnopp type inequalities.

for all 𝑥, 𝑦 ∈ R+ = (0, ∞) with 𝑞 > 1. If T = R; then from (39), we obtain ∞

𝑝

1/𝑝

(𝑦/𝑥)−1/𝑞 (𝑓2 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

Now, if we replace 𝑓1 (𝑦) with 𝑓1 (𝑦)𝑦1/𝑝 and 𝑓2 (𝑦) with 𝑓2 (𝑦)𝑦1/𝑝 , we obtain (46).

Example 18. It is known that (𝑦/𝑥) (𝑦/𝑥) 𝑑𝑦 = ∫ 𝑥+𝑦 𝑥+𝑦 0





Δ𝑦 . = ∫ 𝐾2 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) 𝑦 0



1/𝑝

0

(𝑦/𝑥)−1/𝑞 Δ𝑥 ) Δ𝑦 𝑥+𝑦 𝑥

1−1/𝑞

0

=

(𝑦/𝑥)−1/𝑞 (𝑓1 (𝑦))𝑝 Δ𝑦) 𝑥+𝑦

+(∫





0

∫ ((∫

∞ (𝑦/𝑥)1−1/𝑞 𝑓1 (𝑦) 𝑓2 (𝑦) (∫ Δ𝑥) Δ𝑦 𝑦 𝑥+𝑦 0

−1/𝑞



Δ𝑥

(𝑦/𝑥)−1/𝑞 𝑓1 (𝑦) 𝑓2 (𝑦) Δ𝑥 Δ𝑦) 𝑥+𝑦 𝑥

= ∫ 𝑓1 (𝑦) 𝑓2 (𝑦) (∫



1/𝑝󸀠

󸀠



Proof. Use 𝜉(𝑥) = 𝐾1 (𝑥)/𝑥 and 𝑘(𝑥, 𝑦) −1/𝑞 ≠ 0, 𝑥+𝑦 ≠ 0, { (𝑦/𝑥) 0,/(𝑥+𝑦), 𝑥 ≠ 0,𝑦 } in Corollary 7 to obtain otherwise

𝜎(𝑥) 󸀠 1 ×( (𝑓1 (𝑦))𝑝 Δ𝑦) ∫ 𝜎 (𝑥) − 𝑎 𝑎

𝑝

) Δ𝑥

(46)

Δ𝑥

(49)



≥ ∫ 𝑤 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) Δ𝑦 𝑎



≥ ∫ 𝐾2 (𝑦) (𝑓1 (𝑦) + 𝑓2 (𝑦))𝑝 Δ𝑦 0

holds for all 𝜆 Δ -integrable 𝑓𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}.

holds for all 𝜆 Δ -integrable 𝑓𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}. Proof. Statement follows from Corollary 7 using 𝑚 = 2.

ISRN Mathematical Analysis

7

Corollary 23. Assume (29). Then

Example 26. For T = ℎN = {ℎ𝑛 : 𝑛 ∈ N} with ℎ > 0, 𝑎 = 1, and 𝜉(𝑥) = 1/𝜎(𝑥), (51) takes the form ∞

𝑛 1 ( ∑ (𝑓1 (𝑘ℎ))𝑝 ) ∑ 𝑛=1 𝑛 (𝑛 + 1) 𝑘=1

1/𝑝



𝜎(𝑥) 1 (𝑓1 (𝑦))𝑝 Δ𝑦) ∫ 𝜉 (𝑥) (( ∫ 𝜎 (𝑥) − 𝑎 𝑎 𝑎

1/𝑝

𝜎(𝑥) 1 +( (𝑓2 (𝑦))𝑝 Δ𝑦) ∫ 𝜎 (𝑥) − 𝑎 𝑎

𝑝

𝑛

) Δ𝑥 (50)

𝑝󸀠

1/𝑝

1/𝑝󸀠

(53)

× ( ∑ (𝑓2 (𝑘ℎ)) ) 𝑘=1

∞ 1 ≥ ∑ 𝑓1 (𝑛ℎ) 𝑓2 (𝑛ℎ) . 𝑛 𝑛=1



≥ ∫ 𝑤 (𝑦) (𝑓1 (𝑦) + 𝑓2 (𝑦))𝑝 Δ𝑦 𝑎

Inequality (52) takes the form holds for all 𝜆 Δ -integrable 𝑓𝑗 : Λ → R+ , where 𝑗 ∈ {1, 2}.

1/𝑝



𝑛 1 (( ∑ (𝑓1 (𝑘ℎ))𝑝 ) ∑ 𝑛=1 𝑛 (𝑛 + 1) 𝑘=1

Proof. Statement follows from Corollary 8 using 𝑚 = 2. Example 24. When T consists of isolated points, then from Corollary 22, we have

𝑝

1/𝑝

𝑛

𝑝

+( ∑ (𝑓2 (𝑘ℎ)) )

)

(54)

𝑘=1

∞ 1 𝑝 ≥ ∑ (𝑓1 (𝑛ℎ) + 𝑓2 (𝑛ℎ)) . 𝑛 𝑛=1

𝜉 (𝑥) (𝜎 (𝑥) − 𝑥) 𝜎 (𝑥) − 𝑎 𝑥∈[𝑎,∞) ∑

Example 27. For T = N2 = {𝑛2 : 𝑛 ∈ N}, 𝑎 = 1, and

1/𝑝

×(

𝑝

(𝑓1 (𝑦)) (𝜎 (𝑦) − 𝑦))



𝑦∈[𝑎,𝜎(𝑥))

1/𝑝

×(

(51)

󸀠

(𝑓2 (𝑦)) (𝜎 (𝑦) − 𝑦))



2 𝑛=1 (2𝑛 + 1) (2𝑛 + 3)

𝑦∈[𝑎,𝜎(𝑥))



(55)

inequality (51) takes the form

𝑝󸀠



2 (𝜎 (𝑥) − 1) , (𝜎(𝑥) − 𝑥)2 (2√𝑥 + 3)

𝜉 (𝑥) =



∑ 𝑤 (𝑦) 𝑓1 (𝑦) 𝑓2 (𝑦) (𝜎 (𝑦) − 𝑦) ,

𝑦∈[𝑎,∞)

𝑛

1/𝑝

𝑝

2

× ( ∑ (2𝑘 + 1)(𝑓1 (𝑘 )) ) 𝑘=1

where 𝑤(𝑦) = ∑𝑥∈[𝑦,∞) (𝜉(𝑥)(𝜎(𝑥) − 𝑥)/(𝜎(𝑥) − 𝑎)).

𝑛

𝑝

(56)

1/𝑝󸀠

󸀠

× ( ∑ (2𝑘 + 1)(𝑓2 (𝑘2 )) )

Example 25. When T consists of isolated points, then from Corollary 23, we have

𝑘=1



≥ ∑ 𝑓1 (𝑘2 ) 𝑓2 (𝑘2 ) .

𝜉 (𝑥) (𝜎 (𝑥) − 𝑥) 𝜎 (𝑥) − 𝑎 𝑥∈[𝑎,∞)

𝑘=1



Inequality (52) takes the form 1/𝑝

× ((





2 + 1) (2𝑛 + 3) 𝑛=1 (2𝑛 ∑

𝑝

(𝑓1 (𝑦)) (𝜎 (𝑦) − 𝑦))

𝑦∈[𝑎,𝜎(𝑥))

𝑝

1/𝑝

+(



(𝑓2 (𝑦))𝑝 (𝜎 (𝑦) − 𝑦))

)

(52)

𝑛

2

𝑝

1/𝑝

× (( ∑ (2𝑘 + 1)(𝑓1 (𝑘 )) ) 𝑘=1

𝑦∈[𝑎,𝜎(𝑥))



𝑝

∑ 𝑤 (𝑦) (𝑓1 (𝑦) + 𝑓2 (𝑦)) (𝜎 (𝑦) − 𝑦) ,

𝑦∈[𝑎,∞)

𝑛

𝑘=1



where 𝑤(𝑦) is the same as in Example 24.

𝑝

+( ∑ (2𝑘 + 1)(𝑓2 (𝑘2 )) ) 𝑝

≥ ∑ (𝑓1 (𝑘2 ) + 𝑓2 (𝑘2 )) . 𝑘=1

𝑝

1/𝑝

)

(57)

8

ISRN Mathematical Analysis Example 28. For T = 𝑞N = {𝑞𝑛 : 𝑛 ∈ {N}, 𝑞 > 1}, 𝑎 = 𝑞 and

If we take 𝜎 (𝑥) − 1 , 𝑥𝜎 (𝑥)

𝜉 (𝑥) =



(𝑞 − 1) 𝑛 𝑘−1 𝑝 ( ∑ 𝑞 (𝑓1 (𝑞𝑘 )) ) ∑ 𝑛 𝑞 𝑛=1 𝑘=1

1/𝑝



𝑛 𝑝 (2𝑛 + 1) ( ∑ (2𝑘 + 1)(𝑓1 (𝑘2 )) ) ∑ 2 2 𝑛=1 𝑛 (𝑛 + 1) 𝑘=1 𝑛

𝑝

1/𝑝󸀠

󸀠

𝑛

× ( ∑ (2𝑘 + 1)(𝑓2 (𝑘2 )) )

(59)

𝑝󸀠



× ( ∑ 𝑞𝑘−1 (𝑓2 (𝑞𝑘 )) )

𝑛=1

(𝑞 − 1) 𝑞𝑛 𝑛=1 𝑛

(2𝑛 + 1) ∑ 2 2 𝑛 𝑛=1 (𝑛 + 1)

𝑝

1/𝑝

× (( ∑ 𝑞𝑘−1 (𝑓1 (𝑞𝑘 )) ) 𝑘=1

𝑛

𝑝

1/𝑝

𝑛

2

+( ∑ (2𝑘 + 1)(𝑓2 (𝑘 )) )

(60)

𝑝

1/𝑝

𝑝



)



𝑝 (2𝑘 + 1) (𝑓1 (𝑘2 ) + 𝑓2 (𝑘2 )) . 2 𝑘 𝑘=1 1/𝑝

𝑓1 (𝑘2 ), 𝑓2 (𝑘2 ) =

𝑓2 (𝑘2 ) in (59) and in (60), we have

𝑝 (2𝑛 + 1) ( ∑ 𝑘2 (𝑓1 (𝑘2 )) ) 2 (𝑛 + 1)2 𝑛 𝑛=1 𝑘=1 2

2

1/𝑝

𝑝󸀠

𝑝

𝑛=1

≥∑

𝑛

)

≥ ∑ (𝑓1 (𝑞𝑛 ) + 𝑓2 (𝑞𝑛 )) .

𝑘=1

By replacing 𝑓1 (𝑘2 ) = (𝑘2 /(2𝑘 + 1))

(64)

𝑘=1

𝑘=1

𝑛

𝑝

1/𝑝

𝑝

+( ∑ 𝑞𝑘−1 (𝑓2 (𝑞𝑘 )) )

× (( ∑ (2𝑘 + 1)(𝑓1 (𝑘2 )) )

Remark 29. (a) In classical case for ℎ = 1, inequalities (53) and (54) are the same as (1.7) and (1.9), also (63) and (64) are the same as (1.6) and (1.8) in [18, Corollary 1.3], respectively, while according to authors knowledge, (56), (57), (59), (60), and (61) are not existing in the literature. (b) For 𝑝 < 1, we get the reverse inequalities. Remark 30. The results given in Section 5 can analogously be proved for Ω = Λ = [𝑎1 , ∞)T × ⋅ ⋅ ⋅ × [𝑎𝑛 , ∞)T .

1/𝑝

Remark 31. The results given in Corollaries 7 and 8 and in their given applications can also be obtained for finite value of 𝑚 > 2 analogously.

󸀠

× ( ∑ 𝑘 (𝑓2 (𝑘 )) ) 𝑘=1



≥ ∑ 𝑓1 (𝑞𝑛 ) 𝑓2 (𝑞𝑛 ) .





𝑛

(63) ∞



Inequality (52) takes the form



1/𝑝

Inequality (52) takes the form

(2𝑘 + 1) 𝑓1 (𝑘2 ) 𝑓2 (𝑘2 ) . 2 𝑘 𝑘=1

≥∑



1/𝑝󸀠

𝑘=1

𝑘=1

(𝑘2 /(2𝑘 + 1))

(62)

The inequality (51) takes the form

Inequality (51) takes the form

1/𝑝́

𝜎 (𝑥) − 𝑎 , 𝜎 (𝑥) (𝜎 (𝑥) − 𝑥)

𝜉 (𝑥) =

(58)

Acknowledgments 2

2

≥ ∑ 𝑓1 (𝑘 ) 𝑓2 (𝑘 ) , 𝑘=1

(61)

1/𝑝



𝑛 𝑝 (2𝑛 + 1) (( 𝑘2 (𝑓1 (𝑘2 )) ) ∑ ∑ 2 2 𝑛=1 𝑛 (𝑛 + 1) 𝑘=1 𝑛

2

2

𝑝

1/𝑝

+( ∑ 𝑘 (𝑓2 (𝑘 )) )

𝑝

)

The research of the first author is supported by a grant of the Romanian National Authority for Scientific Research, CNCS UEFISCDI, project no. PN-II-ID-PCE-2011-3-0154. The research of the second author is partially supported by the Higher Education Commission, Pakistan, and the research of the third author is supported by the Croatian Ministry of Science, Education and Sports under the research Grant 1171170889-0888.

𝑘=1



𝑝

≥ ∑ (𝑓1 (𝑘2 ) + 𝑓2 (𝑘2 )) , 𝑘=1

respectively.

References [1] G. H. Hardy, J. E. Littlewood, and G. P´olya, Inequalities, Cambridge Mathematical Library, Cambridge University Press, Cambridge, UK, 1988, Reprint of the 1952 edition.

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9 [23] M. Krni´c, N. Lovriˇcevi´c, and J. E. Peˇcari´c, “On the properties of McShane’s functional and their applications,” Periodica Mathematica Hungarica, vol. 66, no. 2, pp. 159–180, 2013. [24] R. P. Boas, Jr. and C. O. Imoru, “Elementary convolution inequalities,” SIAM Journal on Mathematical Analysis, vol. 6, pp. 457–471, 1975. [25] M. Anwar, R. Bibi, M. Bohner, and J. Peˇcari´c, “Integral inequalities on time scales via the theory of isotonic linear functionals,” Abstract and Applied Analysis, vol. 2011, Article ID 483595, 16 pages, 2011.

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