Sharp Estimation Type Inequalities for Lipschitzian Mappings in Euclidean Sense on a Disk

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This is the Published version of the following publication

Rostamian Delavar, M, Dragomir, Sever S and De La Sen, M (2021) Sharp Estimation Type Inequalities for Lipschitzian Mappings in Euclidean Sense on a Disk. Journal of Function Spaces, 2021. ISSN 2314-8896

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Research Article

Sharp Estimation Type Inequalities for Lipschitzian Mappings in Euclidean Sense on a Disk

M. Rostamian Delavar ,¹S. S. Dragomir ,²and M. De La Sen ³

1Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, P. O. Box 1339, Bojnord 94531, Iran

2Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia

3Institute of Research and Development of Processes, University of Basque Country, Campus of Leioa (Bizkaia)-Aptdo. 644-Bilbao, Bilbao 48080, Spain

Correspondence should be addressed to M. Rostamian Delavar; [email protected] Received 14 December 2020; Accepted 1 February 2021; Published 8 February 2021

Academic Editor: Ismat Beg

Copyright © 2021 M. Rostamian Delavar 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.

Some sharp trapezoid and midpoint type inequalities for Lipschitzian bifunctions deﬁned on a closed disk in Euclidean sense are obtained by the use of polar coordinates. Also, bifunctions whose partial derivative is Lipschitzian are considered. A new presentation of Hermite-Hadamard inequality for convex function deﬁned on a closed disk and its reverse are given.

Furthermore, two mappingsHðtÞandhðtÞare considered to give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian in Euclidean sense on a disk.

1. Introduction and Preliminaries

Consider thatDðC,RÞis a closed disk in the plane centered at the pointC=ða,bÞhaving the radiusR> 0. In [1] (see also [2]), the Hermite-Hadamard inequality for a convex function deﬁned onDðC,RÞhas been obtained as follows:

Theorem 1. If the mappingF :DðC,RÞ→R is convex on DðC,RÞ, then one has the inequality

F Cð Þ≤ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy≤ 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þ,ψ ð1Þ where∂ðC,RÞis the circle centered at the pointC=ða,bÞwith radiusR. The above inequalities are sharp.

First of all, we give the following result which is including a new presentation of (1) and its reverse as well:

Theorem 2.For a continuous functionFdeﬁned on a convex subsetA⊂ℝ²,

(1) ifF is convex onA, then for anyDðC,RÞ⊂A, we have

ð ð

D C,Rð ÞFðx,yÞdA≤ 1 R ð

∂ðC,RÞFðx,yÞðy−bÞ²d∂, ð2Þ where∂ðC,RÞis the boundary ofDðC,RÞ

(2) if (2) holds for allDðC,RÞ⊂A, thenF is convex

Proof.

(1) Consider the change of coordinatesM:½a−R,a+ R×½0, 1→DðC,RÞdeﬁned as

Mðx,sÞ= x−a, 2sð −1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²−ðx−aÞ²

q +b

: ð3Þ

Hindawi

Journal of Function Spaces

Volume 2021, Article ID 6615626, 10 pages https://doi.org/10.1155/2021/6615626

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It follows that

ð ð ð

D C,Rð ÞFðx,yÞdA

= 2ð_a+R

a−R

ð₁

0F s x, ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²−ðx−aÞ²

q +b

+ 1ð −sÞ x,− ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²−ðx−aÞ²

q +b

× ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²−ðx−aÞ² q

dsdx≤2 ð_a+R

a−R

ð₁

0sF x, ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ²

q +b

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ² q

dsdx+ 2ð_a+R

a−R

ð₁

0ð1−sÞF x,− ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ² q

+b

R²−ðx−aÞ² q

dsdx= ð_a+R

a−RF x, ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ² q

+b

R²−ðx−aÞ² q

dx+ ð_a+R

a−RF x,− ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ² q

+b

R²−ðx−aÞ² q

dx:

ð4Þ Now consider y= ± ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R²−ðx−aÞ²

q +b in above inte-

grals with

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 +ð∂y/∂xÞ²

q =R/ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R²−ðx−aÞ²

q Þ=R/ðy−

bÞto obtain the desired result

(2) Suppose that there exist X₁,X₂∈A and s∈ð0, 1Þ such that

FðsX₁+ 1ð −sÞx₂Þ>sFð ÞX₁ + 1ð −sÞFð ÞX₂ ð5Þ SinceFis continuous onA, then there existsR> 0and a pointC₀=ða₀,b₀Þin convex combination ofX₁andX₂such that (5) holds on whole ofDðC₀,RÞ⊂A. Now, if we follow the proof of part (1) forFby the use of (5) onDðC₀,RÞand

∂ðC₀,RÞ, then we have ð ð

D Cð 0,RÞFðx,yÞdA> 1 R

ð

∂ðC0,RÞFðx,yÞðy−bÞ²d∂: ð6Þ This contradiction proves the convexity ofF onA. We remind that the classic form of Hermite-Hadamard inequality (see [3–5]) for a real valued convex function f deﬁned on½a,bis the following:

f a+b 2

b−a ð Þ≤ð_b

a

f xð Þdx≤ðb−aÞf að Þ+f bð Þ

2 : ð7Þ

Generally, in the literature associated to any Hermite- Hadamard type inequality, there exist two inequalities which we call them trapezoid and mid-point type inequalities. The names“trapezoid”and “midpoint”comes from two classic inequalities (due to their geometric interpretation) related to the Hermite-Hadamard inequality obtained in [6, 7], respectively:

∣ð_b

a

f xð Þdx−ðb−aÞf að Þ+f bð Þ 2 ∣ ≤1

8ðb−aÞ²∣f′ð Þ∣+∣fa ′ð Þb ∣ , ð8Þ

∣ð_b

a

f xð Þdx−ðb−aÞf a+b 2

∣ ≤1

8ðb−aÞ²∣f′ð Þa∣+∣f′ð Þb ∣ , ð9Þ where f :I^∘⊆ℝ→ℝ is a diﬀerentiable mapping on I^∘,a,b

∈I^∘witha<band∣f′∣ is convex on½a,b. For more results about convex functions, related inequalities, and generaliza- tions of (7)–(9), see [8–23] and references therein.

Recently, in [17], the authors obtained the trapezoid and midpoint type inequality related to (1) as follows, respectively:

Theorem 3. Consider a setI⊂ℝ² with DðC,RÞ⊂I^∘. Sup- pose that the mapping F :DðC,RÞ→ℝ has continuous partial derivatives in the diskDðC,RÞwith respect to the var- iablesρandφin polar coordinates. If for any constantφ∈½ 0,2π, the function∣∂F/∂ρ∣ is convex with respect to the var- iableρon½0,Rthen

∣ 1 2πR

ð

∂ðC,RÞFð Þdlγ ð Þγ − 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy∣

≤ 1 6π

ð

∂ðC,RÞ∣∂F

∂r ∣ð Þdlγ ð Þ,γ

ð10Þ

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy−f Cð Þ∣ ≤ 2 3π

ð

∂ðC,RÞ∣∂F

∂r ∣ð Þdlγ ð Þ:γ ð11Þ Note that inequality (10) is sharp.

As we can see in (8) and (9), the classic trapezoid and midpoint type inequalities have been obtained for the functions whose the ﬁrst derivative absolute values are convex.

In [22, 24], the authors considered Lipschitzian mappings instead of those whose the ﬁrst derivative absolute values are convex to obtain some midpoint and trapezoid type inequalities:

Theorem 4 [24].Let f :I^∘⊆ℝ→ℝbe an M -Lipschitzian mapping on I and a,b∈I with a<b. Then, we have the inequalities

∣ð_b

a

f xð Þdx−ðb−aÞf a+b 2

∣ ≤M 4 ðb−aÞ,

∣ð_b

a

f xð Þdx−ðb−aÞf að Þ+f bð Þ 2 ∣ ≤M

3ðb−aÞ:

ð12Þ

Motivated by above works and results, we obtain some trapezoid and midpoint type inequalities related to (1) for Lipschitzian mappings (in Euclidean sense) deﬁned on the disk DðC,RÞin a plane. Also we investigate trapezoid and mid-point type inequalities in the case that in polar coordinates ðρ,φÞ, the derivative of considered function with respect to the variable ρ is Lipschitzian. Furthermore, two mappings HðtÞ and hðtÞ are considered to give some

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generalized Hermite-Hadamard type inequalities in the case that the functions are Lipschitzian on a diskDðC,RÞ.

Here, we should mention that in [25], we canﬁnd some inequalities for the integral mean of Hölder continuous functions deﬁned on disks in a plane which in a special case leads to trapezoid and midpoint type inequalities for a kind of Lipschitzian mappings as the following:

Theorem 5.Iff :DðC,RÞ→ℝsatisﬁes the condition

∣f a,ð bÞ−f x,ð yÞ∣ ≤M₁∣x−a∣+M2jy−bj,ðx,yÞ∈D Cð ,RÞ, ð13Þ

whereM₁,M₂>0, then we have the inequalities

∣ 1 2πR

ð

∂ðC,RÞfð Þdlγ ð Þγ − 1 πR²

ð ð

D C,Rð Þf x,ð yÞdxdy∣ ≤2R

3πðM₁+M₂Þ, ð14Þ

∣ 1 πR²

ð ð

D C,Rð Þf x,ð yÞdxdy−f Cð Þ∣ ≤4R

3πðM₁+M₂Þ:

ð15Þ The main point is that the Euclidean Lipschitz condition used in this paper is a stronger condition than (13) in the case thatM₁=M₂ and so our results obtained in (20) and (29) will provide more accurate estimation compared to (14) and (15). Furthermore, we obtain new trapezoid and midpoint type inequalities of our function is Lipschitzian.

2. Main Results

In this section,ﬁrst, we obtain some trapezoid and midpoint type inequalities related to (1) for the case that our considered function is Lipschitzian (in Euclidean sense). Second, we obtain some trapezoid and mid-point type inequalities related to (1) for the case that the partial derivative of our function with respect to the variableρin polar coordinates ðρ,ϕÞis Lipschitzian (in Euclidean sense).

Deﬁnition 6[26]. A functionF :I⊂ℝ²→ℝis said to satisfy a Lipschitz condition (brieﬂy K-Lipschitzian) on I with respect to a normk∙k, if there exists a constantK> 0such that

∣Fð ÞX₁ −Fð ÞX₂ ∣ ≤KkX₁−X₂k, ð16Þ for anyX₁,X₂∈I.

If F:DðC,RÞ→ℝ is Lipschitzian with respect to a constant K> 0and the Euclidean norm k∙k, then for any X₁=ða+ρ₁cosφ₁,b+ρ₁sinφ₁Þand X₂=ða+ρ₂cosφ₂,b

+ρ₂sinφ₂Þ, we have

∣Fð ÞX₁ −Fð ÞX₂ ∣=∣Fða+ρ₁cosφ₁,b+ρ₁sinφ₁Þ

−Fða+ρ₂cosφ₂,b+ρ₂sinφ₂Þ∣

≤Kkðρ₁cosφ₁−ρ₂cosφ₂,ρ₁sinφ₁−ρ₂sinφ₂Þk

=K ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²1+ρ²2−2ρ₁ρ₂cosðφ₁−φ₂Þ

q ,

ð17Þ for anyρ₁,ρ₂∈½0,Randφ₁,φ₂∈½0, 2π. Also, it is obvious that ifF :I⊆ℝ²→ℝis Lipschitzian with respect to a con- stantK> 0onI, then, it is continuous and so integrable on I.

2.1. F Is Lipschitzian. Theﬁrst result of this section is the trapezoid type inequality related to (1) for the case that our considered function is Lipschitzian. We start with a lemma.

Lemma 7.Deﬁne a functionF :DðC,RÞ→ℝas

Fð ÞX =Fða+ρcosφ,b+ρsinφÞ=K Rð −ρÞ, ð18Þ forﬁxedK>0and all0≤ρ≤R,0≤φ≤2π. Then, the func- tionF isK-Lipschitzian.

Proof.ConsiderX₁=ða+ρ₁cosφ₁,b+ρ₁sinφ₁ÞandX₂= ða+ρ₂cosφ₂,b+ρ₂sinφ₂Þ, for ρ₁,ρ₂∈½0,R and φ₁,φ₂

∈½0, 2π. So

∣Fð ÞX₁ −Fð Þ∣X₂ =∣Fða+ρ₁cosφ₁,b+ρ₁sinφ₁Þ

−Fða+ρ₂cosφ₂,b+ρ₂sinφ₂Þ∣

=K∣ρ₂−ρ₁∣=K ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²₁+ρ²₂−2ρ₁ρ₂ q

≤K ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²₁+ρ²₂−2ρ₁ρ₂cosðφ₁−φ₂Þ q

=Kkða+ρ₁cosφ₁,b+ρ₁sinφ₁Þ

−ða+ρ₂cosφ₂,b+ρ₂sinφ₂Þk

=KkX₁−X₂k,

ð19Þ

for allX₁,X₂∈DðC,RÞ.

Theorem 8.Suppose that the mappingF :DðC,RÞ→ℝis Lipschitzian with respect to a constantK>0and the Euclid- ean normk∙k. Then,

∣ 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þψ − 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy∣ ≤KR 3 , ð20Þ where ∂ðC,RÞis the boundary ofDðC,RÞandψ:½0,2π

→ℝ²is its corresponding curve. Also, inequality (20) is sharp.

3 Journal of Function Spaces

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Proof.SinceF is Lipschitzian with respect toK> 0and the Euclidean norm onDðC,RÞ, then, we have

∣ð_2π

0

ð_R

0Fða+ρcosφ,b+ρsinφÞρdρdφ

−ð_2π

0

ð_R

0 Fða+Rcosφ,b+RsinφÞρdρdφ∣

≤ð_2π

0

ð_R

0 ∣Fða+ρcosφ,b+ρsinφÞ

−Fða+Rcosφ,b+RsinφÞ∣ρdρdφ

≤Kð_2π

0

ð_R

0kððρ−RÞcosφ,ðρ−RÞsinφÞkρdρdφ

=Kð_2π

0

ð_R

0ρ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²+R²−2ρR

q dρdφ

=Kð_2π

0

ð_R

0ρðR−ρÞdρdφ=KπR³

3 :

ð21Þ Now, consider the constantRand the curve ψ:½0, 2π

→ℝ²deﬁned by

ψ φð Þ: xð Þφ =a+Rcosφ, yð Þφ =b+Rsinφ, (

φ∈½0, 2π: ð22Þ

It is clear thatψð½0, 2πÞ=∂ðC,RÞ, and then by integrat- ing, we obtain that

ð

∂ðC,RÞFð Þdlψ ð Þψ =ð_2π

0 Fðxð Þ,φ yð Þφ Þ ∂xð Þφ

∂φ

2

+ ∂yð Þφ

∂φ 2!_1/2

dφ

=Rð_2π

0 Fða+Rcosφ,b+RsinφÞdφ,

ð23Þ where ½ð∂xðφÞÞ/∂φ²+½ð∂yðφÞÞ/∂φ²=ðR²sin²φ+R² cos²φÞ^1/2=R, and ð∂xðφÞÞ/∂φ,ð∂yðφÞÞ/∂φ are derivatives of“xðφÞ”and “yðφÞ”with respect toφ, respectively. So the fact that

ð_2π

0

ð_R

0 Fða+Rcosφ,b+RsinφÞρdρdφ

=R² 2

ð_2π

0 Fða+Rcosφ,b+RsinφÞdφ,

ð24Þ

implies that ð2π

0

ðR

0Fða+Rcosφ,b+RsinφÞρdρdφ=R 2

ð

∂ðC,RÞFð Þdlψ ð Þ:ψ ð25Þ

Also, by the use of polar coordinates, we get to ð ð

D C,Rð ÞFðx,yÞdxdy=ð_2π

0

ð_R

0Fða+ρcosφ,b+ρsinφÞρdρdφ:

ð26Þ

Finally, by replacing (25) and (26) in (21) and then divid- ing the result with“πR²,”we deduce the desired result. To prove sharpness of (20), consider the function F :DðC,R Þ→ℝdeﬁned by

Fða+ρcosφ,b+ρsinφÞ=K Rð −ρÞ, ð27Þ

forﬁxedK> 0and all0≤ρ≤R,0≤φ≤2π. The functionf isK-Lipschitzian by Lemma 7. It is not hard to see thatFð a+ρcosφ,b+ρsinφÞ≥0for all 0≤ρ≤Rand also for the case that ρ=R, we have Fða+Rcosφ,b+RsinφÞ= 0. Now applying these results in (21) implies that

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy− 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þ∣ψ

= 1

πR² ð_2π

0

ð_R

0 Fða+ρcosφ,b+ρsinφÞρdρdφ

= 1

πR² ð_2π

0

ð_R

0 K Rð −ρÞρdρdφ=KR 3 :

ð28Þ

The following result is the midpoint type inequality related to (1) for Lipschitzian functions deﬁned on a closed disk.

Theorem 9.Suppose that the mappingF :DðC,RÞ→ℝis Lipschitzian with respect to a constantK>0and the Euclid- ean normk∙k. Then,

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy−F Cð Þ∣ ≤2KR

3 : ð29Þ

Furthermore, inequality (29) is sharp.

Proof. Since the mappingF satisﬁes a Lipschitz condition with respect to a constant K> 0 and the Euclidean norm onDðC,RÞ, we have

∣Fða+ρcosφ,b+ρsinφÞ−Fða,bÞ∣ ≤Kkðρcosφ,ρsinφÞk=Kρ,

ð30Þ

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for allρ∈½0,Randφ∈½0, 2π. It follows that

∣ð_2π

0

ð_R

0Fða+ρcosφ,b+ρsinφÞρdρdφ−ð_2π

0

ð_R

0Fða,bÞρdρdφ∣

≤ð_2π

0

ð_R

0∣Fða+ρcosφ,b+ρsinφÞ−Fða,bÞ∣ρdρdφ

≤Kð_2π

0

ð_R

0ρ²dρdφ= 2KπR³

3 :

ð31Þ By the use of identity (26) in inequality (31), we obtain that

∣ð ð

D C,Rð ÞFðx,yÞdxdy−πR²fð Þ∣ ≤C 2KπR³

3 : ð32Þ Finally, it is enough to divide (32) with“πR^2”to get the result. For the sharpness of (29), consider the function F :DðC,RÞ→ℝdeﬁned by

Fða+ρcosφ,b+ρsinφÞ=Kρ, ð33Þ for K> 0, 0≤ρ≤R and 0≤φ≤2π. By a similar method used in the proof of Lemma 7, the function F is K -Lipschitzian. Also, it is obvious that Fða+ρcosφ,b+ρ

sinφÞ≥0andFða,bÞ= 0. So, we have

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy−fð Þ∣C

= 1

πR² ð_2π

0

ð_R

0 f að +ρcosφ,b+ρsinφÞρdρdφ

= 1

πR² ð_2π

0

ð_R

0 Kρ²dρdφ=2KR

3 ,

ð34Þ

showing that inequality (29) is sharp.

Corollary 10.Suppose thatU⊂ℝ²is an open set withDðC ,RÞ⊂U. IfF is a convex function deﬁned onU, then Theo- rem D of Section 41 in [26] implies thatFsatisﬁes a Lipschitz condition onDðC,RÞwith respect to a constantK>0and so from inequalities (20) and (29) along with inequality (1), we have the following results:

0≤ 1 2πR

ð

ð ð

D C,Rð ÞFðx,yÞdxdy≤KR 3 , 0≤ 1

πR² ð ð

D C,Rð ÞFðx,yÞdxdy−F Cð Þ≤2KR 3 :

ð35Þ In the following example, for a given function, it is illus- trated how we can obtain a Lipschitz constantK for a real valued bifunction deﬁned on a disk.

Example 11. ConsiderFðx,yÞ=ðx−aÞⁿ+ðy−bÞⁿ,ðx,yÞ∈ DðC,RÞ, n∈ℕ. We ﬁnd a Lipschitz constant for F as follows:

ForX₁,X₂∈DðC,RÞ, consider the pathη:½0, 1→Dð C,RÞfromX₂toX₁inDðC,RÞas

ηð Þs =sX₁+ 1ð −sÞX₂, ð36Þ for s∈½0, 1. The fundamental theorem of calculus implies that

∣Fð ÞX₁ −Fð Þ∣X₂ =∣Fðηð Þ1 Þ−Fðηð Þ0 Þ∣= ð₁

0

dFðηð Þs Þ ds ds

: ð37Þ

Also, the chain rule for diﬀerentiation implies that dFðηð Þs Þ

ds =∇Fðηð Þs Þ:η′ð Þs =∇Fðηð Þs ÞðX₁−X₂Þ, ð38Þ where∇f is the gradient vector ofF. So,

∣ð₁

0

dFðηð Þs Þ

ds ds∣=∣ð₁

0∇Fðηð ÞsÞðX₁−X₂Þds∣

≤kX₁−X₂kð₁

0k∇Fðηð Þs Þkds

≤kX₁−X₂k sup

w∈D C,Rð Þk∇Fð Þwk,

ð39Þ

which implies that

∣Fð ÞX₁ −Fð Þ∣ ≤X₂ kX₁−X₂k sup

w∈D C,Rð Þk∇Fð Þw k: ð40Þ Now, we conclude that K= sup_w∈DðC_,RÞk∇FðwÞk (if exists) is a Lipschitz constant forF. Therefore, for anyðx,y Þ∈DðC,RÞ, we have

∇Fðx,yÞ=n xð −aÞⁿ⁻¹,n yð −bÞⁿ⁻¹

, ð41Þ and then by the use of polar transformation, we get

∇Fð Þw

k k= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n xð −aÞⁿ⁻¹

2+n yð −bÞⁿ⁻¹2

q

=n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²cos²φ

ð Þⁿ⁻¹+ðρ²sin²φÞⁿ⁻¹ q

≤n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ²cos²φ+ρ²sin²φ

ð Þⁿ⁻¹

q =nρⁿ⁻¹≤nRⁿ⁻¹:

ð42Þ So, we can chooseK= sup_w∈DðC,RÞk∇FðwÞk=nRⁿ⁻¹as a Lipschitz constant forF onDðC,RÞ.

Remark 12. According to the above example, if we have a function F :DðC,RÞ→ℝ such that K= sup_w∈DðC,RÞk∇ FðwÞk<∞ with respect to the Euclidean norm k∙k, then 5 Journal of Function Spaces

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we can considerK as a Lipschitz constant and then obtain inequalities (20) and (29).

2.2. ∂F/∂ρ Is Lipschitzian. In this part, we investigate the trapezoid and midpoint type inequalities in the case that in polar coordinatesðρ,φÞ, the partial derivative of considered function with respect to the variableρis Lipschitzian in the Euclidean normk∙k.

Theorem 13.Consider a setI⊂ℝ²withDðC,RÞ⊂I^∘and a mappingF :DðC,RÞ→ℝsuch that∂F/∂ρ(partial derivative ofFwith respect to the variableρin polar coordinates) is Lipschitzian with respect to a constantK>0and the Euclid- ean normk∙k. Then,

∣ 1 2πR

ð

ð ð

D C,Rð ÞFðx,yÞdxdy∣ ≤KR² 4 : ð43Þ

Proof.For anyﬁxedφ∈½0, 2π, if we set xð Þρ =a+ρcosφ, yð Þρ =b+ρsinφ, (

ð44Þ

then we obtain that ð½ð∂xðρÞÞ/∂ρ²+½ð∂xðρÞÞ/∂ρ²Þ^1/2= ðsin²ðφÞ+ cos²ðφÞÞ^1/2= 1, where ð∂xðρÞÞ/∂ρ, ð∂yðρÞÞ/∂ρ are the derivatives of xðρÞ,yðρÞ, respectively, with respect to the variableρin½0,R. By the above facts, using integra- tion by parts and identities (25) and (26) obtained in Theo- rem 8, we get

ð_2π

0

ð_R

0

∂F

∂ρða+ρcosφ,b+ρsinφÞρ²dρdφ

=R²ð_2π

0 Fða+Rcosφ,b+RsinφÞdφ

−2ð_2π

0

ð_R

0 Fða+ρcosφ,b+ρsinφÞρdρdφ

=Rð

∂ðC,RÞFð Þdlψ ð Þψ −2ð ð

D C,Rð ÞFðx,yÞdxdy:

ð45Þ

On the other hand, we have ð_2π

0

ð_R

0

∂F

=ð_π

0

ð_R

0

∂F

∂ρða+ρcosφ,b+ρsinφÞρ²dρdφ +ð_2π

π

ð_R

0

∂F

=ð_π

0

ð_R

0

∂F

−ð_π

0

ð_R

0

∂F

∂ρða−ρcosφ,b−ρsinφÞρ²dρdφ:

ð46Þ

So from (45) and (46), we obtain that

∣Rð

∂ðC,RÞFð Þdlψ ð Þψ −2ð ð

D C,Rð ÞFðx,yÞdxdy∣

≤ð_π

0

ð_R

0 ∣∂F

∂ρða+ρcosφ,b+ρsinφÞ

−∂F

∂ρða−ρcosφ,b−ρsinφÞ∣ρ²dρdφ

≤Kð_π

0

ð_R

0 kð2ρcosφ, 2ρsinφÞkρ²dρdφ=πKR⁴

2 :

ð47Þ Finally, it is enough to divide (47) with“2πR²”to get the desired result.

The following is a trapezoid type inequality for the case that the partial derivative of considered function with respect to the variable“ρ”is Lipschitzian with respect to the Euclid- ean normk∙k.

Theorem 14.Consider a setI⊂ℝ²withDðC,RÞ⊂I^∘and a mappingF :DðC,RÞ→ℝsuch that∂F/∂ρ(partial derivative off with respect to the variableρin polar coordinates) is Lipschitzian with respect to a constantK>0and the Euclid- ean normk∙k. Then,

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy−F Cð Þ∣ ≤3KR²

4 : ð48Þ Proof. Using the description provided in the beginning of Theorem 13, it is not hard to see that

ð_2π

0

ð_R

0

∂F

∂ρða+ρcosφ,b+ρsinφÞdρdφ

=ð_2π

0 Fða+Rcosφ,b+RsinφÞdφ−2πF Cð Þ:

ð49Þ

Also by the use of (23) in Theorem 8, we have ð_2π

0 Fða+Rcosφ,b+RsinφÞdφ= 1 R

ð

∂ðC,RÞFð Þdlψ ð Þ:ψ ð50Þ

On the other hand, we have ð_2π

0

ð_R

0

∂F

=ð_π

0

ð_R

0

∂F

−ð_π

0

ð_R

0

∂F

∂ρða−ρcosφ,b−ρsinφÞdρdφ:

ð51Þ

(8)

Then, 1 R

ð

∂ðC,RÞFð Þdlψ ð Þψ −2πF Cð Þ

=ð_π

0

ð_R

0

∂F

∂ρða+ρcosφ,b+ρsinφÞ

−∂F

∂ρða−ρcosφ,b−ρsinφÞ

dρdφ:

ð52Þ

Since∂F/∂ρisK-Lipschitzian, we have that

∣ 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þψ −F Cð Þ∣ ≤KR²

2 : ð53Þ

Now triangle inequality and inequality (43) imply that

∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy−F Cð Þ∣

≤ ∣ 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy− 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þ∣ψ +∣ 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þψ −Fð Þ∣C

≤ KR²

4 +KR²

2 =3KR² 4 :

ð54Þ Here, we provide two examples in connection with results obtained in this subsection.

Example 15.Deﬁne a functionF:DðC,RÞ→ℝby Fða+ρcosφ,b+ρsinφÞ=Mρ², ð55Þ for M> 0,0≤ρ≤R, and0≤φ≤2π. Since∂F/∂ρ= 2Mρ, then according to Remark 12, we can consider K= sup_w∈DðC,RÞk∇FðwÞk= 2M<∞ as a Lipschitz constant with respect to the Euclidean normk∙k. On the other hand,

Fð ÞC =F a,ð bÞ= 0, ð ð

D C,Rð ÞFðx,yÞdxdy=MπR⁴

2 ,

ð

∂ðC,RÞFð Þdlψ ð Þψ = 2MπR³:

ð56Þ

So, 1 2πR

ð

∂ðC,RÞFð Þdlψ ð Þψ − 1 πR²

ð ð

D C,Rð ÞFðx,yÞdxdy=MR² 2 ,

ð57Þ

which shows that (43) is sharp. Also, 1

2πR ð

∂ðC,RÞFð Þdlψ ð Þψ −F Cð Þ=MR², ð58Þ which implies that (53) is sharp.

Example 16.Considera,b> 0,0 <R≤minfa,bgand0≤ρ

≤R. For n∈ℕ and polar function Fðρ,φÞ=ða−ρÞⁿ+ ðb−ρÞⁿwhich is deﬁned onDðða,bÞ,RÞ, by some calcula- tions we can conclude that

∇ ∂F

∂ρ

ρ,φ

ð Þ=n nð −1Þða−ρÞⁿ⁻²+ðb−ρÞⁿ⁻², 0 , ð59Þ

and then K= sup

0≤ρ≤R,0≤φ≤2π ∇ ∂F

∂ρ

ρ,φ

ð Þ

=n nð −1Þaⁿ⁻²+bⁿ⁻² , ð60Þ is a Lipschitz constant forF. Then, from inequality (53), we have that

∣A aðð −RÞⁿ,ðb−RÞⁿÞ−A að ⁿ,bⁿÞ∣ ≤n nð −1ÞA a ⁿ⁻²,bⁿ⁻² R²

2 ,

ð61Þ whereAða,bÞ=ða+bÞ/2is arithmetic mean ofaandb. Also for the functionF, by the use of (43) and (48), we can obtain other arithmetic mean type inequalities.

3. MappingsH andh

In this section, by the use of two mappingsHðtÞ: ½0, 1→ℝ andhðtÞ:½0, 1→ℝdeﬁned in [1], we give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian with respect to Euclidean normk∙kon a diskDðC,RÞ:

H tð Þ= 1 πR²

ð ð

D C,Rð ÞFðtC+ 1ð −tÞðx,yÞÞdxdy, h tð Þ=

1 2πtR

ð

∂ðC,tRÞ, Fð Þdlγ ðγð Þt Þ,t∈ð0, 1, F Cð Þ, t= 0:

8>

<

>:

ð62Þ

By the use of some properties for the mappingshandH, we give some reﬁnements for trapezoid and midpoint type inequalities obtained in previous sections for K-Lipschit- zian mappingsF :DðC,RÞ→ℝ.

Theorem 17.Suppose that the mappingF :DðC,RÞ→ℝis Lipschitzian with respect to a constantK>0and the Euclid- ean norm k∙k. Then, the mapping H is Lipschitzian with respect to“2KR/3”and the mappinghis Lipschitzian with respect to“KR.”The following inequalities also for allt∈ð 7 Journal of Function Spaces

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0,1Þhold.

∣ 1 πR²

ð ð

D C,Rð ÞFðtC+ð1−tÞðx,yÞÞdxdy−tF Cð Þ

− 1−t πR²

ð ð

D C,Rð ÞFðx,yÞdxdy∣ ≤4KRt 1ð −tÞ

3 ,

ð63Þ

∣ 1 2πtR

ð

∂ðC,tRÞFð Þdlγ ðγð ÞtÞ− 1 πR²

ð ð

D C,Rð ÞFðtC+ð1−tÞðx,yÞÞdxdy∣

≤2KRt 3 ,

ð64Þ

∣ 1 2πtR

ð

∂ðC,tRÞFð Þγdlðγð Þt Þ− t 2πR

ð

∂ðC,RÞFð Þγdlð Þγ −ð1−tÞF Cð Þ∣

≤2KRt 1ð −tÞ:

ð65Þ Proof. Consider the following relations for t₁,t₂∈½0, 1, which prove theﬁrst part of this theorem:

∣H tð Þ1 −H tð Þ₂ ∣ ≤ 1 πR²

ð ð

D C,Rð Þ∣Fðt₁C+ 1ð −t₁Þðx,yÞÞ

−Fðt₂C+ 1ð −t₂Þðx,yÞÞ∣dxdy

≤ K∣t₁−t₂∣ πR²

ð ð

D C,Rð Þkða−x,b−yÞkdxdy

= K∣t₁−t₂∣ πR²

ð_2π

0

ð_R

0 kðρcosφ,ρsinφÞkρdρdφ

= K∣t₁−t₂∣ πR²

ð_2π

0

ð_R

0 ρ²dρdφ= 2KR 3 jt₁−t₂j:

ð66Þ

Also,

∣h tð Þ₁ −h tð Þ∣₂ =∣ 1 2πt₁R

ð

∂ðC,t₁RÞFð Þdlγ ðγð Þt₁ Þ− 1 2πt₂R ð

∂ðC,t₂RÞFð Þdlγ ðγð Þt₂ Þ∣

= 1

πR²∣ ð_2π

0

ð_R

0Fða+t₁Rcosφ,b+t₁RsinφÞρdρdφ

−ð_2π

0

ð_R

0Fða+t₂Rcosφ,b+t₂RsinφÞρdρdφ∣

≤K∣t₁−t₂∣ πR²

ð_2π

0

ð_R

0kðRcosφ,RsinφÞkρdρdφ

=KRjt₁−t₂j:

ð67Þ For inequality (63), we use the deﬁnition of H and the fact thatF isK-Lipschitzian:

∣H tð Þ−tHð Þ1 −ð1−tÞHð Þ∣ ≤0 t∣H tð Þ−Hð Þ∣1 + 1ð −tÞ∣H tð Þ−Hð Þ∣0

≤ t πR²

ð ð

D C,Rð Þ∣FðtC+ 1ð −tÞðx,yÞÞ

−F Cð Þ∣dxdy+ 1−t πR²

ð ð

D C,Rð Þ

∣FðtC+ 1ð −tÞðx,yÞÞ−Fðx,yÞ∣dxdy

≤2Ktð1−tÞ πR²

ð ð

D C,Rð Þkðx−a,y−bÞkdxdy

=2Ktð1−tÞ πR²

ð2π 0

ðR

0ρ²dρdφ=4KRtð1−tÞ

3 :

ð68Þ To prove (64), ift∈ð0, 1, we consider the following identity presented in [1],

H tð Þ= 1 πR²t²

ð ð

D C,tRð ÞFðx,yÞdxdy: ð69Þ

Now, consider transformation

xð Þρ =a+tρcosφ; ρ∈½0,R,φ∈½0, 2π,t∈ð0, 1, yð Þρ =b+tρsinφ:

(

ð70Þ This implies that

H tð Þ= 1 πR²

ð_2π

0

ð_R

0 Fða+tρcosφ,b+tρsinφÞρdρdφ:

ð71Þ Also, we have

h tð Þ= 1 2πtR

ð

∂ðC,tRÞFð Þdlγ ðγð Þt Þ

= 1 2π

ð_2π

0 Fða+tRcosφ,b+tRsinφÞdφ

= 1

πR² ð_2π

0

ð_R

0 Fða+tRcosφ,b+tRsinφÞρdρdφ:

ð72Þ

So, we conclude that

∣h tð Þ−H tð Þ∣ ≤ 1 πR²

ð_2π

0

ð_R

0∣Fða+tRcosφ,b+tRsinφÞ

−Fða+tρcosφ,b+tρsinφÞ∣ρdρdφ

≤ K

πR² ð_2π

0

ð_R

0kðtðR−ρÞcosφ,tðR−ρÞsinφÞkρdρdφ

= Kt πR²

ð_2π

0

ð_R

0ðR−ρÞρdρdφ=KtR 3 :

ð73Þ