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

Some sharp trapezoid and midpoint type inequalities for Lipschitzian bifunctions de ﬁ ned 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 de ﬁ ned on a closed disk and its reverse are given. Furthermore, two mappings H ð t Þ and h ð t Þ are considered to give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian in Euclidean sense on a disk.


Introduction and Preliminaries
Consider that D ðC, RÞ is a closed disk in the plane centered at the point C = ða, bÞ having the radius R > 0. In [1] (see also [2]), the Hermite-Hadamard inequality for a convex function defined on D ðC, RÞ has been obtained as follows: Theorem 1. If the mapping F : D ðC, RÞ → R is convex on D ðC, RÞ, then one has the inequality where ∂ðC, RÞ is the circle centered at the point C = ða, bÞ 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 function F defined on a convex subset A ⊂ ℝ 2 , (1) if F is convex on A, then for any DðC, RÞ ⊂ A, we have where ∂ðC, RÞ is the boundary of D ðC, RÞ (2) if (2) holds for all DðC, RÞ ⊂ A, then F is convex Proof.
(1) Consider the change of coordinates M : ½a − R, a + R × ½0, 1 → DðC, RÞ defined as It follows that Now consider y = ± ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi such that Since F is continuous on A, then there exists R > 0 and a point C 0 = ða 0 , b 0 Þ in convex combination of X 1 and X 2 such that (5) holds on whole of DðC 0 , RÞ ⊂ A. Now, if we follow the proof of part (1) for F by the use of (5) on DðC 0 , RÞ and ∂ðC 0 , RÞ, then we have This contradiction proves the convexity of F on A.
We remind that the classic form of Hermite-Hadamard inequality (see [3][4][5]) for a real valued convex function f defined on ½a, b is the following: 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: where f : For more results about convex functions, related inequalities, and generalizations of (7)-(9), see [8][9][10][11][12][13][14][15][16][17][18][19][20][21][22][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 set I ⊂ ℝ 2 with DðC, RÞ ⊂ I ∘ . Suppose that the mapping F : DðC, RÞ → ℝ has continuous partial derivatives in the disk DðC, RÞ with respect to the variables ρ and φ in polar coordinates. If for any constant φ ∈ ½ 0, 2π, the function |∂F/∂ρ | is convex with respect to the variable ρ on ½0, R then 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 → ℝ be an M -Lipschitzian mapping on I and a, b ∈ I with a < b. Then, we have the inequalities 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, RÞ in 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 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: where M 1 , M 2 > 0, then we have the inequalities The main point is that the Euclidean Lipschitz condition used in this paper is a stronger condition than (13) in the case that M 1 = M 2 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.

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 ðρ, ϕÞ is Lipschitzian (in Euclidean sense).
Definition 6 [26]. A function F : I ⊂ ℝ 2 → ℝ is said to satisfy a Lipschitz condition (briefly K-Lipschitzian) on I with respect to a norm k•k, if there exists a constant K > 0 such that for any X 1 , X 2 ∈ I.
If F : DðC, RÞ → ℝ is Lipschitzian with respect to a constant K > 0 and the Euclidean norm k•k, then for any for any ρ 1 , ρ 2 ∈ ½0, R and φ 1 , φ 2 ∈ ½0, 2π. Also, it is obvious that if F : I ⊆ ℝ 2 → 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.

Journal of Function Spaces
Proof. Since F is Lipschitzian with respect to K > 0 and the Euclidean norm on DðC, RÞ, then, we have Now, consider the constant R and the curve ψ : It is clear that ψð½0, 2πÞ = ∂ðC, RÞ, and then by integrating, we obtain that where ½ð∂xðφÞÞ/∂φ 2 + ½ð∂yðφÞÞ/∂φ 2 = ðR 2 sin 2 φ + R 2 cos 2 φÞ 1/2 = R, and ð∂xðφÞÞ/∂φ, ð∂yðφÞÞ/∂φ are derivatives of "xðφÞ" and "yðφÞ" with respect to φ, respectively. So the fact that implies that Also, by the use of polar coordinates, we get to Finally, by replacing (25) and (26) in (21) and then dividing the result with "πR 2 ," we deduce the desired result. To prove sharpness of (20), consider the function F : DðC, R Þ → ℝ defined by 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 Fð a + ρ cos φ, b + ρsinφÞ ≥ 0 for all 0 ≤ ρ ≤ R and also for the case that ρ = R, we have Fða + R cos φ, b + R sin φÞ = 0. Now applying these results in (21) implies that The following result is the midpoint type inequality related to (1) for Lipschitzian functions defined on a closed disk.
Theorem 9. Suppose that the mapping F : DðC, RÞ → ℝ is Lipschitzian with respect to a constant K > 0 and the Euclidean norm k•k. Then, Furthermore, inequality (29) is sharp.
Proof. Since the mapping F satisfies a Lipschitz condition with respect to a constant K > 0 and the Euclidean norm on DðC, RÞ, we have Journal of Function Spaces for all ρ ∈ ½0, R and φ ∈ ½0, 2π. It follows that By the use of identity (26) in inequality (31), we obtain that 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 → ℝ defined by 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 + ρ cos φ, b + ρ sin φÞ ≥ 0 and Fða, bÞ = 0. So, we have showing that inequality (29) is sharp.

Corollary 10. Suppose that U ⊂ ℝ 2 is an open set with DðC
, RÞ ⊂ 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 DðC, RÞ with respect to a constant K > 0 and so from inequalities (20) and (29) along with inequality (1), we have the following results: In the following example, for a given function, it is illustrated how we can obtain a Lipschitz constant K for a real valued bifunction defined on a disk. Example 11. Consider Fðx, yÞ = ðx − aÞ n + ðy − bÞ n , ðx, yÞ ∈ DðC, RÞ, n ∈ ℕ. We find a Lipschitz constant for F as follows: For X 1 , X 2 ∈ DðC, RÞ, consider the path η : ½0, 1 → Dð C, RÞ from X 2 to X 1 in DðC, RÞ as for s ∈ ½0, 1. The fundamental theorem of calculus implies that Also, the chain rule for differentiation implies that where ∇f is the gradient vector of F. So, which implies that Now, we conclude that K = sup w∈DðC,RÞ k∇FðwÞk (if exists) is a Lipschitz constant for F. Therefore, for any ðx, y Þ ∈ DðC, RÞ, we have and then by the use of polar transformation, we get So, we can choose K = sup w∈DðC,RÞ k∇FðwÞk = nR n−1 as a Lipschitz constant for F on DðC, RÞ.
Remark 12. According to the above example, if we have a function F : DðC, RÞ → ℝ such that K = sup w∈DðC,RÞ k∇ FðwÞk < ∞ with respect to the Euclidean norm k•k, then 5 Journal of Function Spaces we can consider K as a Lipschitz constant and then obtain inequalities (20) and (29).

∂F/∂ρ Is Lipschitzian.
In this part, we investigate the trapezoid and midpoint type inequalities in the case that in polar coordinates ðρ, φÞ, the partial derivative of considered function with respect to the variable ρ is Lipschitzian in the Euclidean norm k•k.
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 k•k.

Theorem 14.
Consider a set I ⊂ ℝ 2 with DðC, RÞ ⊂ I ∘ and a mapping F : DðC, RÞ → 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 k•k. Then, Proof. Using the description provided in the beginning of Theorem 13, it is not hard to see that Also by the use of (23) in Theorem 8, we have On the other hand, we have Journal of Function Spaces Then, Since ∂F/∂ρ is K-Lipschitzian, we have that Now triangle inequality and inequality (43) imply that Here, we provide two examples in connection with results obtained in this subsection.
Example 16. Consider a, b > 0, 0 < R ≤ min fa, bg and 0 ≤ ρ ≤ R. For n ∈ ℕ and polar function Fðρ, φÞ = ða − ρÞ n + ðb − ρÞ n which is defined on Dðða, bÞ, RÞ, by some calculations we can conclude that and then is a Lipschitz constant for F. Then, from inequality (53), we have that where Aða, bÞ = ða + bÞ/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.

Mappings H and h
In this section, by the use of two mappings HðtÞ: ½0, 1 → ℝ and hðtÞ: ½0, 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 k•k on a disk DðC, RÞ: By the use of some properties for the mappings h and H, we give some refinements for trapezoid and midpoint type inequalities obtained in previous sections for K-Lipschitzian mappings F : DðC, RÞ → ℝ.
Theorem 17. Suppose that the mapping F : DðC, RÞ → ℝ is Lipschitzian with respect to a constant K > 0 and the Euclidean norm k•k. 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 ∈ ð 7 Journal of Function Spaces 0, 1Þ hold.
Proof. Consider the following relations for t 1 , t 2 ∈ ½0, 1, which prove the first part of this theorem: Also, For inequality (63), we use the definition of H and the fact that F is K-Lipschitzian: To prove (64), if t ∈ ð0, 1, we consider the following identity presented in [1], Now, consider transformation This implies that Also, we have The details are omitted.
The following results also are of interest: Theorem 18. Suppose that the mapping F : DðC, RÞ → ℝ is Lipschitzian with respect to a constant K > 0 and the Euclidean norm k•k. The following inequalities hold: for all t ∈ ð0, 1.
Proof. It is enough to consider special cases for t 1 and t 2 in two inequalities (66) and (67) obtained in the proof of previous theorem.
(2) If we consider t = 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 where F : DðC, RÞ → ℝ is Lipschitzian with respect to a constant K > 0 and the Euclidean norm k•k.

Data Availability
No data were used to support this study.