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

Authors: M. Rostamian Delavar; S. S. Dragomir; M. De La Sen
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

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 mappingsHt and ht are considered to give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian in Euclidean sense on a disk.


---

## Body

## 1. Introduction and Preliminaries

Consider thatDC,R is a closed disk in the plane centered at the point C=a,b having the radius R>0. In [1] (see also [2]), the Hermite-Hadamard inequality for a convex function defined on DC,R has been obtained as follows:Theorem 1.
If the mappingF:DC,R→R is convex on DC,R, then one has the inequality(1)FC≤1πR2∫∫DC,RFx,ydxdy≤12πR∫∂C,RFψdlψ,where ∂C,R is the circle centered at the point C=a,b with radius R. 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 functionF defined on a convex subset A⊂ℝ2,(1)
ifF is convex on A, then for any DC,R⊂A, we have(2)∫∫DC,RFx,ydA≤1R∫∂C,RFx,yy−b2d∂,where ∂C,R is the boundary of DC,R(2)
if (2) holds for all DC,R⊂A, then F is convexProof.
(1)
Consider the change of coordinatesM:a−R,a+R×0,1→DC,R defined as(3)Mx,s=x−a,2s−1R2−x−a2+b.
It follows that(4)∫∫∫DC,RFx,ydA=2∫a−Ra+R∫01Fsx,R2−x−a2+b+1−sx,−R2−x−a2+b×R2−x−a2dsdx≤2∫a−Ra+R∫01sFx,R2−x−a2+bR2−x−a2dsdx+2∫a−Ra+R∫011−sFx,−R2−x−a2+bR2−x−a2dsdx=∫a−Ra+RFx,R2−x−a2+bR2−x−a2dx+∫a−Ra+RFx,−R2−x−a2+bR2−x−a2dx.
Now considery=±R2−x−a2+b in above integrals with 1+∂y/∂x2=R/R2−x−a2=R/y−b to obtain the desired result(2)
Suppose that there existX1,X2∈A and s∈0,1 such that(5)FsX1+1−sx2>sFX1+1−sFX2
SinceF is continuous on A, then there exists R>0 and a point C0=a0,b0 in convex combination of X1 and X2 such that (5) holds on whole of DC0,R⊂A. Now, if we follow the proof of part (1) for F by the use of (5) on DC0,R and ∂C0,R, then we have(6)∫∫DC0,RFx,ydA>1R∫∂C0,RFx,yy−b2d∂.
This contradiction proves the convexity ofF on A.We remind that the classic form of Hermite-Hadamard inequality (see [3–5]) for a real valued convex function f defined on a,b is the following:(7)fa+b2b−a≤∫abfxdx≤b−afa+fb2.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:(8)∣∫abfxdx−b−afa+fb2∣≤18b−a2∣f′a∣+∣f′b∣,(9)∣∫abfxdx−b−afa+b2∣≤18b−a2∣f′a∣+∣f′b∣,where f:I∘⊆ℝ→ℝ is a differentiable mapping on I∘,a,b∈I∘ with a<b and ∣f′∣ is convex on a,b. For more results about convex functions, related inequalities, and generalizations 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.
Considera set I⊂ℝ2 with DC,R⊂I∘. Suppose that the mapping F:DC,R→ℝ has continuous partial derivatives in the disk DC,R with respect to the variables ρ and φ in polar coordinates. If for any constant φ∈0,2π, the function ∣∂F/∂ρ∣ is convex with respect to the variable ρ on 0,R then(10)∣12πR∫∂C,RFγdlγ−1πR2∫∫DC,RFx,ydxdy∣≤16π∫∂C,R∣∂F∂r∣γdlγ,(11)∣1πR2∫∫DC,RFx,ydxdy−fC∣≤23π∫∂C,R∣∂F∂r∣γdlγ.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].
Letf:I∘⊆ℝ→ℝ be an M-Lipschitzian mapping on I and a,b∈I with a<b. Then, we have the inequalities(12)∣∫abfxdx−b−afa+b2∣≤M4b−a,∣∫abfxdx−b−afa+fb2∣≤M3b−a.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 DC,R in a plane. Also we investigate trapezoid and mid-point type inequalities in the case that in polar coordinates ρ,φ, the derivative of considered function with respect to the variable ρ is Lipschitzian. Furthermore, two mappings Ht and ht are considered to give some generalized Hermite-Hadamard type inequalities in the case that the functions are Lipschitzian on a disk DC,R.Here, we should mention that in [25], we can find 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:DC,R→ℝ satisfies the condition(13)∣fa,b−fx,y∣≤M1∣x−a∣+M2y−b,x,y∈DC,R,where M1,M2>0, then we have the inequalities(14)∣12πR∫∂C,Rfγdlγ−1πR2∫∫DC,Rfx,ydxdy∣≤2R3πM1+M2,(15)∣1πR2∫∫DC,Rfx,ydxdy−fC∣≤4R3πM1+M2.The main point is that the Euclidean Lipschitz condition used in this paper is a stronger condition than (13) in the case that M1=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 variable ρ in polar coordinates ρ,ϕ is Lipschitzian (in Euclidean sense).Definition 6 [26].
A functionF:I⊂ℝ2→ℝ is said to satisfy a Lipschitz condition (briefly K-Lipschitzian) on I with respect to a norm ∙, if there exists a constant K>0 such that(16)∣FX1−FX2∣≤KX1−X2,for any X1,X2∈I.IfF:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙, then for any X1=a+ρ1cosφ1,b+ρ1sinφ1 and X2=a+ρ2cosφ2,b+ρ2sinφ2, we have(17)∣FX1−FX2∣=∣Fa+ρ1cosφ1,b+ρ1sinφ1−Fa+ρ2cosφ2,b+ρ2sinφ2∣≤Kρ1cosφ1−ρ2cosφ2,ρ1sinφ1−ρ2sinφ2=Kρ12+ρ22−2ρ1ρ2cosφ1−φ2,for any ρ1,ρ2∈0,R and φ1,φ2∈0,2π. Also, it is obvious that if F:I⊆ℝ2→ℝ is Lipschitzian with respect to a constant K>0 on I, then, it is continuous and so integrable on I.


### 2.1.F Is Lipschitzian

The first 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:DC,R→ℝ as(18)FX=Fa+ρcosφ,b+ρsinφ=KR−ρ,for fixed K>0 and all 0≤ρ≤R, 0≤φ≤2π. Then, the function F is K-Lipschitzian.Proof.
ConsiderX1=a+ρ1cosφ1,b+ρ1sinφ1 and X2=a+ρ2cosφ2,b+ρ2sinφ2, for ρ1,ρ2∈0,R and φ1,φ2∈0,2π. So(19)∣FX1−FX2∣=∣Fa+ρ1cosφ1,b+ρ1sinφ1−Fa+ρ2cosφ2,b+ρ2sinφ2∣=K∣ρ2−ρ1∣=Kρ12+ρ22−2ρ1ρ2≤Kρ12+ρ22−2ρ1ρ2cosφ1−φ2=Ka+ρ1cosφ1,b+ρ1sinφ1−a+ρ2cosφ2,b+ρ2sinφ2=KX1−X2,for all X1,X2∈DC,R.Theorem 8.
Suppose that the mappingF:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(20)∣12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy∣≤KR3,where ∂C,R is the boundary of DC,R and ψ:0,2π→ℝ2 is its corresponding curve. Also, inequality (20) is sharp.Proof.
SinceF is Lipschitzian with respect to K>0 and the Euclidean norm on DC,R, then, we have(21)∣∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ−∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ∣≤∫02π∫0R∣Fa+ρcosφ,b+ρsinφ−Fa+Rcosφ,b+Rsinφ∣ρdρdφ≤K∫02π∫0Rρ−Rcosφ,ρ−Rsinφρdρdφ=K∫02π∫0Rρρ2+R2−2ρRdρdφ=K∫02π∫0RρR−ρdρdφ=KπR33.
Now, consider the constantR and the curve ψ:0,2π→ℝ2 defined by(22)ψφ:xφ=a+Rcosφ,yφ=b+Rsinφ,φ∈0,2π.
It is clear thatψ0,2π=∂C,R, and then by integrating, we obtain that(23)∫∂C,RFψdlψ=∫02πFxφ,yφ∂xφ∂φ2+∂yφ∂φ21/2dφ=R∫02πFa+Rcosφ,b+Rsinφdφ,where ∂xφ/∂φ2+∂yφ/∂φ2=R2sin2φ+R2cos2φ1/2=R, and ∂xφ/∂φ,∂yφ/∂φ are derivatives of “xφ” and “yφ” with respect to φ, respectively. So the fact that(24)∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ=R22∫02πFa+Rcosφ,b+Rsinφdφ,implies that(25)∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ=R2∫∂C,RFψdlψ.
Also, by the use of polar coordinates, we get to(26)∫∫DC,RFx,ydxdy=∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ.
Finally, by replacing (25) and (26) in (21) and then dividing the result with “πR2,” we deduce the desired result. To prove sharpness of (20), consider the function F:DC,R→ℝ defined by(27)Fa+ρcosφ,b+ρsinφ=KR−ρ,for fixed K>0 and all 0≤ρ≤R, 0≤φ≤2π. The function f is K-Lipschitzian by Lemma 7. It is not hard to see that Fa+ρcosφ,b+ρsinφ≥0 for all 0≤ρ≤R and also for the case that ρ=R, we have Fa+Rcosφ,b+Rsinφ=0. Now applying these results in (21) implies that(28)∣1πR2∫∫DC,RFx,ydxdy−12πR∫∂C,RFψdlψ∣=1πR2∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ=1πR2∫02π∫0RKR−ρρdρdφ=KR3.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:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(29)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤2KR3.Furthermore, inequality (29) is sharp.Proof.
Since the mappingF satisfies a Lipschitz condition with respect to a constant K>0 and the Euclidean norm on DC,R, we have(30)∣Fa+ρcosφ,b+ρsinφ−Fa,b∣≤Kρcosφ,ρsinφ=Kρ,for all ρ∈0,R and φ∈0,2π. It follows that(31)∣∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ−∫02π∫0RFa,bρdρdφ∣≤∫02π∫0R∣Fa+ρcosφ,b+ρsinφ−Fa,b∣ρdρdφ≤K∫02π∫0Rρ2dρdφ=2KπR33.
By the use of identity (26) in inequality (31), we obtain that(32)∣∫∫DC,RFx,ydxdy−πR2fC∣≤2KπR33.
Finally, it is enough to divide (32) with “πR2” to get the result. For the sharpness of (29), consider the function F:DC,R→ℝ defined by(33)Fa+ρcosφ,b+ρsinφ=Kρ,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 Fa+ρcosφ,b+ρsinφ≥0 and Fa,b=0. So, we have(34)∣1πR2∫∫DC,RFx,ydxdy−fC∣=1πR2∫02π∫0Rfa+ρcosφ,b+ρsinφρdρdφ=1πR2∫02π∫0RKρ2dρdφ=2KR3,showing that inequality (29) is sharp.Corollary 10.
Suppose thatU⊂ℝ2 is an open set with DC,R⊂U. If F is a convex function defined on U, then Theorem D of Section 41 in [26] implies that F satisfies a Lipschitz condition on DC,R with respect to a constant K>0 and so from inequalities (20) and (29) along with inequality (1), we have the following results:(35)0≤12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy≤KR3,0≤1πR2∫∫DC,RFx,ydxdy−FC≤2KR3.In the following example, for a given function, it is illustrated how we can obtain a Lipschitz constantK for a real valued bifunction defined on a disk.Example 11.
ConsiderFx,y=x−an+y−bn, x,y∈DC,R, n∈ℕ. We find a Lipschitz constant for F as follows:
ForX1,X2∈DC,R, consider the path η:0,1→DC,R from X2 to X1 in DC,R as(36)ηs=sX1+1−sX2,for s∈0,1. The fundamental theorem of calculus implies that(37)∣FX1−FX2∣=∣Fη1−Fη0∣=∫01dFηsdsds.
Also, the chain rule for differentiation implies that(38)dFηsds=∇Fηs.η′s=∇FηsX1−X2,where ∇f is the gradient vector of F. So,(39)∣∫01dFηsdsds∣=∣∫01∇FηsX1−X2ds∣≤X1−X2∫01∇Fηsds≤X1−X2supw∈DC,R∇Fw,which implies that(40)∣FX1−FX2∣≤X1−X2supw∈DC,R∇Fw.Now, we conclude thatK=supw∈DC,R∇Fw (if exists) is a Lipschitz constant for F. Therefore, for any x,y∈DC,R, we have(41)∇Fx,y=nx−an−1,ny−bn−1,and then by the use of polar transformation, we get(42)∇Fw=nx−an−12+ny−bn−12=nρ2cos2φn−1+ρ2sin2φn−1≤nρ2cos2φ+ρ2sin2φn−1=nρn−1≤nRn−1.So, we can chooseK=supw∈DC,R∇Fw=nRn−1 as a Lipschitz constant for F on DC,R.Remark 12.
According to the above example, if we have a functionF:DC,R→ℝ such that K=supw∈DC,R∇Fw<∞ with respect to the Euclidean norm ∙, then we can consider K 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 coordinatesρ,φ, the partial derivative of considered function with respect to the variable ρ is Lipschitzian in the Euclidean norm ∙.Theorem 13.
Consider a setI⊂ℝ2 with DC,R⊂I∘ and a mapping F:DC,R→ℝ such that ∂F/∂ρ (partial derivative of F with respect to the variable ρ in polar coordinates) is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(43)∣12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy∣≤KR24.Proof.
For any fixedφ∈0,2π, if we set(44)xρ=a+ρcosφ,yρ=b+ρsinφ,then we obtain that ∂xρ/∂ρ2+∂xρ/∂ρ21/2=sin2φ+cos2φ1/2=1, where ∂xρ/∂ρ, ∂yρ/∂ρ are the derivatives of xρ,yρ, respectively, with respect to the variable ρ in 0,R. By the above facts, using integration by parts and identities (25) and (26) obtained in Theorem 8, we get(45)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=R2∫02πFa+Rcosφ,b+Rsinφdφ−2∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ=R∫∂C,RFψdlψ−2∫∫DC,RFx,ydxdy.
On the other hand, we have(46)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ+∫π2π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ−∫0π∫0R∂F∂ρa−ρcosφ,b−ρsinφρ2dρdφ.
So from (45) and (46), we obtain that(47)∣R∫∂C,RFψdlψ−2∫∫DC,RFx,ydxdy∣≤∫0π∫0R∣∂F∂ρa+ρcosφ,b+ρsinφ−∂F∂ρa−ρcosφ,b−ρsinφ∣ρ2dρdφ≤K∫0π∫0R2ρcosφ,2ρsinφρ2dρdφ=πKR42.
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 Euclidean norm ∙.Theorem 14.
Consider a setI⊂ℝ2 with DC,R⊂I∘ and a mapping F:DC,R→ℝ such that ∂F/∂ρ (partial derivative of f with respect to the variable ρ in polar coordinates) is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(48)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤3KR24.Proof.
Using the description provided in the beginning of Theorem13, it is not hard to see that(49)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ=∫02πFa+Rcosφ,b+Rsinφdφ−2πFC.
Also by the use of (23) in Theorem 8, we have(50)∫02πFa+Rcosφ,b+Rsinφdφ=1R∫∂C,RFψdlψ.
On the other hand, we have(51)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ−∫0π∫0R∂F∂ρa−ρcosφ,b−ρsinφdρdφ.
Then,(52)1R∫∂C,RFψdlψ−2πFC=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφ−∂F∂ρa−ρcosφ,b−ρsinφdρdφ.
Since∂F/∂ρ is K-Lipschitzian, we have that(53)∣12πR∫∂C,RFψdlψ−FC∣≤KR22.
Now triangle inequality and inequality (43) imply that(54)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤∣1πR2∫∫DC,RFx,ydxdy−12πR∫∂C,RFψdlψ∣+∣12πR∫∂C,RFψdlψ−FC∣≤KR24+KR22=3KR24.Here, we provide two examples in connection with results obtained in this subsection.Example 15.
Define a functionF:DC,R→ℝby(55)Fa+ρcosφ,b+ρsinφ=Mρ2,for M>0, 0≤ρ≤R, and 0≤φ≤2π. Since ∂F/∂ρ=2Mρ, then according to Remark 12, we can consider K=supw∈DC,R∇Fw=2M<∞ as a Lipschitz constant with respect to the Euclidean norm ∙. On the other hand,(56)FC=Fa,b=0,∫∫DC,RFx,ydxdy=MπR42,∫∂C,RFψdlψ=2MπR3.
So,(57)12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy=MR22,which shows that (43) is sharp. Also,(58)12πR∫∂C,RFψdlψ−FC=MR2,which implies that (53) is sharp.Example 16.
Considera,b>0, 0<R≤mina,b and 0≤ρ≤R. For n∈ℕ and polar function Fρ,φ=a−ρn+b−ρn which is defined on Da,b,R, by some calculations we can conclude that(59)∇∂F∂ρρ,φ=nn−1a−ρn−2+b−ρn−2,0,and then(60)K=sup0≤ρ≤R,0≤φ≤2π∇∂F∂ρρ,φ=nn−1an−2+bn−2,is a Lipschitz constant for F. Then, from inequality (53), we have that(61)∣Aa−Rn,b−Rn−Aan,bn∣≤nn−1Aan−2,bn−2R22,where Aa,b=a+b/2 is arithmetic mean of a and b. Also for the function F, by the use of (43) and (48), we can obtain other arithmetic mean type inequalities.


## 2.1.F Is Lipschitzian

The first 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:DC,R→ℝ as(18)FX=Fa+ρcosφ,b+ρsinφ=KR−ρ,for fixed K>0 and all 0≤ρ≤R, 0≤φ≤2π. Then, the function F is K-Lipschitzian.Proof.
ConsiderX1=a+ρ1cosφ1,b+ρ1sinφ1 and X2=a+ρ2cosφ2,b+ρ2sinφ2, for ρ1,ρ2∈0,R and φ1,φ2∈0,2π. So(19)∣FX1−FX2∣=∣Fa+ρ1cosφ1,b+ρ1sinφ1−Fa+ρ2cosφ2,b+ρ2sinφ2∣=K∣ρ2−ρ1∣=Kρ12+ρ22−2ρ1ρ2≤Kρ12+ρ22−2ρ1ρ2cosφ1−φ2=Ka+ρ1cosφ1,b+ρ1sinφ1−a+ρ2cosφ2,b+ρ2sinφ2=KX1−X2,for all X1,X2∈DC,R.Theorem 8.
Suppose that the mappingF:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(20)∣12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy∣≤KR3,where ∂C,R is the boundary of DC,R and ψ:0,2π→ℝ2 is its corresponding curve. Also, inequality (20) is sharp.Proof.
SinceF is Lipschitzian with respect to K>0 and the Euclidean norm on DC,R, then, we have(21)∣∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ−∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ∣≤∫02π∫0R∣Fa+ρcosφ,b+ρsinφ−Fa+Rcosφ,b+Rsinφ∣ρdρdφ≤K∫02π∫0Rρ−Rcosφ,ρ−Rsinφρdρdφ=K∫02π∫0Rρρ2+R2−2ρRdρdφ=K∫02π∫0RρR−ρdρdφ=KπR33.
Now, consider the constantR and the curve ψ:0,2π→ℝ2 defined by(22)ψφ:xφ=a+Rcosφ,yφ=b+Rsinφ,φ∈0,2π.
It is clear thatψ0,2π=∂C,R, and then by integrating, we obtain that(23)∫∂C,RFψdlψ=∫02πFxφ,yφ∂xφ∂φ2+∂yφ∂φ21/2dφ=R∫02πFa+Rcosφ,b+Rsinφdφ,where ∂xφ/∂φ2+∂yφ/∂φ2=R2sin2φ+R2cos2φ1/2=R, and ∂xφ/∂φ,∂yφ/∂φ are derivatives of “xφ” and “yφ” with respect to φ, respectively. So the fact that(24)∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ=R22∫02πFa+Rcosφ,b+Rsinφdφ,implies that(25)∫02π∫0RFa+Rcosφ,b+Rsinφρdρdφ=R2∫∂C,RFψdlψ.
Also, by the use of polar coordinates, we get to(26)∫∫DC,RFx,ydxdy=∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ.
Finally, by replacing (25) and (26) in (21) and then dividing the result with “πR2,” we deduce the desired result. To prove sharpness of (20), consider the function F:DC,R→ℝ defined by(27)Fa+ρcosφ,b+ρsinφ=KR−ρ,for fixed K>0 and all 0≤ρ≤R, 0≤φ≤2π. The function f is K-Lipschitzian by Lemma 7. It is not hard to see that Fa+ρcosφ,b+ρsinφ≥0 for all 0≤ρ≤R and also for the case that ρ=R, we have Fa+Rcosφ,b+Rsinφ=0. Now applying these results in (21) implies that(28)∣1πR2∫∫DC,RFx,ydxdy−12πR∫∂C,RFψdlψ∣=1πR2∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ=1πR2∫02π∫0RKR−ρρdρdφ=KR3.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:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(29)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤2KR3.Furthermore, inequality (29) is sharp.Proof.
Since the mappingF satisfies a Lipschitz condition with respect to a constant K>0 and the Euclidean norm on DC,R, we have(30)∣Fa+ρcosφ,b+ρsinφ−Fa,b∣≤Kρcosφ,ρsinφ=Kρ,for all ρ∈0,R and φ∈0,2π. It follows that(31)∣∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ−∫02π∫0RFa,bρdρdφ∣≤∫02π∫0R∣Fa+ρcosφ,b+ρsinφ−Fa,b∣ρdρdφ≤K∫02π∫0Rρ2dρdφ=2KπR33.
By the use of identity (26) in inequality (31), we obtain that(32)∣∫∫DC,RFx,ydxdy−πR2fC∣≤2KπR33.
Finally, it is enough to divide (32) with “πR2” to get the result. For the sharpness of (29), consider the function F:DC,R→ℝ defined by(33)Fa+ρcosφ,b+ρsinφ=Kρ,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 Fa+ρcosφ,b+ρsinφ≥0 and Fa,b=0. So, we have(34)∣1πR2∫∫DC,RFx,ydxdy−fC∣=1πR2∫02π∫0Rfa+ρcosφ,b+ρsinφρdρdφ=1πR2∫02π∫0RKρ2dρdφ=2KR3,showing that inequality (29) is sharp.Corollary 10.
Suppose thatU⊂ℝ2 is an open set with DC,R⊂U. If F is a convex function defined on U, then Theorem D of Section 41 in [26] implies that F satisfies a Lipschitz condition on DC,R with respect to a constant K>0 and so from inequalities (20) and (29) along with inequality (1), we have the following results:(35)0≤12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy≤KR3,0≤1πR2∫∫DC,RFx,ydxdy−FC≤2KR3.In the following example, for a given function, it is illustrated how we can obtain a Lipschitz constantK for a real valued bifunction defined on a disk.Example 11.
ConsiderFx,y=x−an+y−bn, x,y∈DC,R, n∈ℕ. We find a Lipschitz constant for F as follows:
ForX1,X2∈DC,R, consider the path η:0,1→DC,R from X2 to X1 in DC,R as(36)ηs=sX1+1−sX2,for s∈0,1. The fundamental theorem of calculus implies that(37)∣FX1−FX2∣=∣Fη1−Fη0∣=∫01dFηsdsds.
Also, the chain rule for differentiation implies that(38)dFηsds=∇Fηs.η′s=∇FηsX1−X2,where ∇f is the gradient vector of F. So,(39)∣∫01dFηsdsds∣=∣∫01∇FηsX1−X2ds∣≤X1−X2∫01∇Fηsds≤X1−X2supw∈DC,R∇Fw,which implies that(40)∣FX1−FX2∣≤X1−X2supw∈DC,R∇Fw.Now, we conclude thatK=supw∈DC,R∇Fw (if exists) is a Lipschitz constant for F. Therefore, for any x,y∈DC,R, we have(41)∇Fx,y=nx−an−1,ny−bn−1,and then by the use of polar transformation, we get(42)∇Fw=nx−an−12+ny−bn−12=nρ2cos2φn−1+ρ2sin2φn−1≤nρ2cos2φ+ρ2sin2φn−1=nρn−1≤nRn−1.So, we can chooseK=supw∈DC,R∇Fw=nRn−1 as a Lipschitz constant for F on DC,R.Remark 12.
According to the above example, if we have a functionF:DC,R→ℝ such that K=supw∈DC,R∇Fw<∞ with respect to the Euclidean norm ∙, then we can consider K 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 coordinatesρ,φ, the partial derivative of considered function with respect to the variable ρ is Lipschitzian in the Euclidean norm ∙.Theorem 13.
Consider a setI⊂ℝ2 with DC,R⊂I∘ and a mapping F:DC,R→ℝ such that ∂F/∂ρ (partial derivative of F with respect to the variable ρ in polar coordinates) is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(43)∣12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy∣≤KR24.Proof.
For any fixedφ∈0,2π, if we set(44)xρ=a+ρcosφ,yρ=b+ρsinφ,then we obtain that ∂xρ/∂ρ2+∂xρ/∂ρ21/2=sin2φ+cos2φ1/2=1, where ∂xρ/∂ρ, ∂yρ/∂ρ are the derivatives of xρ,yρ, respectively, with respect to the variable ρ in 0,R. By the above facts, using integration by parts and identities (25) and (26) obtained in Theorem 8, we get(45)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=R2∫02πFa+Rcosφ,b+Rsinφdφ−2∫02π∫0RFa+ρcosφ,b+ρsinφρdρdφ=R∫∂C,RFψdlψ−2∫∫DC,RFx,ydxdy.
On the other hand, we have(46)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ+∫π2π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφρ2dρdφ−∫0π∫0R∂F∂ρa−ρcosφ,b−ρsinφρ2dρdφ.
So from (45) and (46), we obtain that(47)∣R∫∂C,RFψdlψ−2∫∫DC,RFx,ydxdy∣≤∫0π∫0R∣∂F∂ρa+ρcosφ,b+ρsinφ−∂F∂ρa−ρcosφ,b−ρsinφ∣ρ2dρdφ≤K∫0π∫0R2ρcosφ,2ρsinφρ2dρdφ=πKR42.
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 Euclidean norm ∙.Theorem 14.
Consider a setI⊂ℝ2 with DC,R⊂I∘ and a mapping F:DC,R→ℝ such that ∂F/∂ρ (partial derivative of f with respect to the variable ρ in polar coordinates) is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then,(48)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤3KR24.Proof.
Using the description provided in the beginning of Theorem13, it is not hard to see that(49)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ=∫02πFa+Rcosφ,b+Rsinφdφ−2πFC.
Also by the use of (23) in Theorem 8, we have(50)∫02πFa+Rcosφ,b+Rsinφdφ=1R∫∂C,RFψdlψ.
On the other hand, we have(51)∫02π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφdρdφ−∫0π∫0R∂F∂ρa−ρcosφ,b−ρsinφdρdφ.
Then,(52)1R∫∂C,RFψdlψ−2πFC=∫0π∫0R∂F∂ρa+ρcosφ,b+ρsinφ−∂F∂ρa−ρcosφ,b−ρsinφdρdφ.
Since∂F/∂ρ is K-Lipschitzian, we have that(53)∣12πR∫∂C,RFψdlψ−FC∣≤KR22.
Now triangle inequality and inequality (43) imply that(54)∣1πR2∫∫DC,RFx,ydxdy−FC∣≤∣1πR2∫∫DC,RFx,ydxdy−12πR∫∂C,RFψdlψ∣+∣12πR∫∂C,RFψdlψ−FC∣≤KR24+KR22=3KR24.Here, we provide two examples in connection with results obtained in this subsection.Example 15.
Define a functionF:DC,R→ℝby(55)Fa+ρcosφ,b+ρsinφ=Mρ2,for M>0, 0≤ρ≤R, and 0≤φ≤2π. Since ∂F/∂ρ=2Mρ, then according to Remark 12, we can consider K=supw∈DC,R∇Fw=2M<∞ as a Lipschitz constant with respect to the Euclidean norm ∙. On the other hand,(56)FC=Fa,b=0,∫∫DC,RFx,ydxdy=MπR42,∫∂C,RFψdlψ=2MπR3.
So,(57)12πR∫∂C,RFψdlψ−1πR2∫∫DC,RFx,ydxdy=MR22,which shows that (43) is sharp. Also,(58)12πR∫∂C,RFψdlψ−FC=MR2,which implies that (53) is sharp.Example 16.
Considera,b>0, 0<R≤mina,b and 0≤ρ≤R. For n∈ℕ and polar function Fρ,φ=a−ρn+b−ρn which is defined on Da,b,R, by some calculations we can conclude that(59)∇∂F∂ρρ,φ=nn−1a−ρn−2+b−ρn−2,0,and then(60)K=sup0≤ρ≤R,0≤φ≤2π∇∂F∂ρρ,φ=nn−1an−2+bn−2,is a Lipschitz constant for F. Then, from inequality (53), we have that(61)∣Aa−Rn,b−Rn−Aan,bn∣≤nn−1Aan−2,bn−2R22,where Aa,b=a+b/2 is arithmetic mean of a and b. Also for the function F, by the use of (43) and (48), we can obtain other arithmetic mean type inequalities.


## 3. MappingsH and h

In this section, by the use of two mappingsHt:0,1→ℝ and ht:0,1→ℝ defined in [1], we give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian with respect to Euclidean norm ∙ on a disk DC,R:(62)Ht=1πR2∫∫DC,RFtC+1−tx,ydxdy,ht=12πtR∫∂C,tR,Fγdlγt,t∈0,1,FC,t=0.By the use of some properties for the mappingsh and H, we give some refinements for trapezoid and midpoint type inequalities obtained in previous sections for K-Lipschitzian mappings F:DC,R→ℝ.Theorem 17.
Suppose that the mappingF:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. Then, the mapping H is Lipschitzian with respect to “2KR/3” and the mapping h is Lipschitzian with respect to “KR.” The following inequalities also for all t∈0,1 hold.(63)∣1πR2∫∫DC,RFtC+1−tx,ydxdy−tFC−1−tπR2∫∫DC,RFx,ydxdy∣≤4KRt1−t3,(64)∣12πtR∫∂C,tRFγdlγt−1πR2∫∫DC,RFtC+1−tx,ydxdy∣≤2KRt3,(65)∣12πtR∫∂C,tRFγdlγt−t2πR∫∂C,RFγdlγ−1−tFC∣≤2KRt1−t.Proof.
Consider the following relations fort1,t2∈0,1, which prove the first part of this theorem:(66)∣Ht1−Ht2∣≤1πR2∫∫DC,R∣Ft1C+1−t1x,y−Ft2C+1−t2x,y∣dxdy≤K∣t1−t2∣πR2∫∫DC,Ra−x,b−ydxdy=K∣t1−t2∣πR2∫02π∫0Rρcosφ,ρsinφρdρdφ=K∣t1−t2∣πR2∫02π∫0Rρ2dρdφ=2KR3t1−t2.
Also,(67)∣ht1−ht2∣=∣12πt1R∫∂C,t1RFγdlγt1−12πt2R∫∂C,t2RFγdlγt2∣=1πR2∣∫02π∫0RFa+t1Rcosφ,b+t1Rsinφρdρdφ−∫02π∫0RFa+t2Rcosφ,b+t2Rsinφρdρdφ∣≤K∣t1−t2∣πR2∫02π∫0RRcosφ,Rsinφρdρdφ=KRt1−t2.
For inequality (63), we use the definition of H and the fact that F is K-Lipschitzian:(68)∣Ht−tH1−1−tH0∣≤t∣Ht−H1∣+1−t∣Ht−H0∣≤tπR2∫∫DC,R∣FtC+1−tx,y−FC∣dxdy+1−tπR2∫∫DC,R∣FtC+1−tx,y−Fx,y∣dxdy≤2Kt1−tπR2∫∫DC,Rx−a,y−bdxdy=2Kt1−tπR2∫02π∫0Rρ2dρdφ=4KRt1−t3.
To prove (64), if t∈0,1, we consider the following identity presented in [1],(69)Ht=1πR2t2∫∫DC,tRFx,ydxdy.
Now, consider transformation(70)xρ=a+tρcosφ;ρ∈0,R,φ∈0,2π,t∈0,1,yρ=b+tρsinφ.
This implies that(71)Ht=1πR2∫02π∫0RFa+tρcosφ,b+tρsinφρdρdφ.
Also, we have(72)ht=12πtR∫∂C,tRFγdlγt=12π∫02πFa+tRcosφ,b+tRsinφdφ=1πR2∫02π∫0RFa+tRcosφ,b+tRsinφρdρdφ.
So, we conclude that(73)∣ht−Ht∣≤1πR2∫02π∫0R∣Fa+tRcosφ,b+tRsinφ−Fa+tρcosφ,b+tρsinφ∣ρdρdφ≤KπR2∫02π∫0RtR−ρcosφ,tR−ρsinφρdρdφ=KtπR2∫02π∫0RR−ρρdρdφ=KtR3.
Inequality (65) is a consequence of the fact that(74)∣ht−th1−1−th0∣≤t∣ht−h1∣+1−tht−h0.
The details are omitted.The following results also are of interest:Theorem 18.
Suppose that the mappingF:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙. The following inequalities hold:(75)∣Ht−H0∣=∣1πR2∫∫DC,RFtC+1−tx,ydxdy−1πR2∫∫DC,RFx,ydxdy∣≤2KtR3,for all t∈0,1.(76)∣H1−Ht∣=∣FC−1πR2∫∫DC,RFtC+1−tx,ydxdy∣≤2K1−tR3,for all t∈0,1.(77)∣ht−h0∣=∣12πtR∫∂C,tRFγdlγt−FC∣≤KtR,for all t∈0,1, and(78)∣h1−ht∣=∣12πR∫∂C,RFγdlγ−12πtR∫∂C,tRFγdlγt∣≤K1−tR,for all t∈0,1.Proof.
It is enough to consider special cases fort1 and t2 in two inequalities (66) and (67) obtained in the proof of previous theorem.Remark 19.
(1)
For a convex functionF:DC,R→ℝwithK=supw∈DC,R∇Fw<∞, if we consider (1), some results obtained in [1], and the following inequality(79)FC≤1πR2t2∫∫DC,tRFx,ydxdy≤12πtR∫∂C,tRFγdlγt≤h1,t∈0,1,then, we deduce that (64) and (75)–(78) hold without using absolute value symbol.(2)
If we considert=1 in inequality (75) or consider t=0 in inequality (76), then, we recapture inequality (29) in Theorem 9. Also from inequality (77), we obtain this new inequality(80)∣12πR∫∂C,RFγdlγ−FC∣≤KR,where F:DC,R→ℝ is Lipschitzian with respect to a constant K>0 and the Euclidean norm ∙.



---

*Source: 6615626-2021-02-08.xml*