Abstract
This paper deals with some trapezoid and mid-point type inequalities on closed balls in \(\mathbb{R}^{3}\). Three kinds of functions are considered: convex, Lipschitz, and bounded functions. The spherical coordinates are used to obtain sharp inequalities. Also a reverse result is given for the right-hand side of Hermite–Hadamard’s inequality obtained on closed balls in \(\mathbb{R}^{3}\).
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1 Introduction and preliminaries
Consider the closed ball \(\bar{\mathcal{B}}(\mathcal{C},R)\) in the space \(\mathbb{R}^{3}\) with center \(\mathcal{C}=(a,b,c)\in \mathbb{R}^{3}\) and radius \(R>0\) defined as
Also consider \({\sigma }(\mathcal{C},R)\) as the boundary (the surface) of \(\bar{B}(\mathcal{C},R)\), i.e.,
The following result has been proved in [1], which is the Hermite–Hadamard’s inequality for convex functions defined on closed ball \(\bar{\mathcal{B}}(\mathcal{C},R)\).
Theorem 1.1
Let\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)be a convex mapping on the ball\(\bar{\mathcal{B}}(\mathcal{C},R)\). Then we have the inequality
where\(v(\bar{\mathcal{B}}(\mathcal{C},R))=\frac{4\pi R^{3}}{3}\)and\(\sigma (\bar{B}(\mathcal{C},R))=\frac{1}{4\pi R^{2}}\).
The main purpose of this paper is estimating two bounds \(\mathcal{B}_{1}\) and \(\mathcal{B}_{2}\) such that
and
Depending on the properties of the function f and the radius R, different values will be obtained for \(\mathcal{B}_{1}\) and \(\mathcal{B}_{2}\).
We call (2) a mid-point type inequality due to the following result obtained in [2] and interpretation of Fig. 1.
Theorem 1.2
Let\(f : I^{\circ }\subseteq \mathbb{R}\to \mathbb{R}\)be a differentiable mapping on\(I^{\circ }\), \(a, b\in I^{\circ }\)with\(a < b\). If\(\vert f' \vert \)is convex on\([a, b]\), then we have
According to (4), we have an estimate for the difference between the area under the graph of f, i.e., \(\int _{a}^{b}f(x)\,dx\), and the area of rectangle \(abcd\), i.e., \((b-a)f( \frac{a+b}{2}) \) (see Fig. 1).
Also we call (3) a trapezoid type inequality due to the following result and Fig. 2.
Theorem 1.3
([3])
Let\(f : I^{\circ }\subseteq \mathbb{R}\to \mathbb{R}\)be a differentiable mapping on\(I^{\circ }\), \(a, b\in I^{\circ }\)with\(a < b\). If\(\vert f' \vert \)is convex on\([a, b]\), then the following inequality holds:
According to (5), we can estimate the difference between the area of trapezoid \(abcd\), i.e., \((b-a)\frac{f(a)+f(b)}{2}\), and the area under the graph of f (see Fig. 2).
Note that to obtain (4) and (5), the absolute values of the derivative of f at boundary points of interval \([a,b]\) play a fundamental role. For more results about Hermite–Hadamard’s inequality, we refer an interested reader to [4–18] and the references therein.
Before presenting our main results, here we obtain a new representation of (1) and also give a reverse type theorem.
If we consider a convex function \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) and the change of coordinates
where \(\bar{D}((a,b),R)\) is a closed disk centered at the point \((a,b)\) having radius \(R>0\), then we obtain
Choosing \(z=\sqrt{R^{2}-x^{2}-y^{2}}\) in the latter integrals, the fact that \(\sqrt{1+(\frac{\partial z}{\partial x})^{2}+( \frac{\partial z}{\partial y})^{2}}= \frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}}=\frac{R}{z}\), and using the surface integral formula for \(\sigma (\mathcal{C},R)\) imply that
Inequality (8) gives another representation for (1).
In a special case for a convex function \(f:\bar{D}((a,b),R)\to \mathbb{R}\) we get
Now for a reverse type result, consider a continuous function f defined on a convex subset \(\mathcal{V}\subset \mathbb{R}^{3}\) such that (8) holds for all closed balls included in \(\mathcal{V}\). Then f is convex on \(\mathcal{V}\) because otherwise there would exist \(\mathcal{X},\mathcal{Y}\in \mathcal{V}\) and \(\lambda \in (0,1)\) such that
Since f is continuous on \(\mathcal{V}\), we can find \(R>0\) and a point \(\mathcal{Z}=(\bar{a},\bar{b},\bar{c})\) in a convex combination of \(\mathcal{X}\) and \(\mathcal{Y}\) such that (9) holds on the whole of \(\bar{\mathcal{B}}((\bar{a},\bar{b},\bar{c}),R)\subset \mathcal{V}\). So by the change of coordinates (6) and structure presented in (7) for \(\bar{D}((\bar{a},\bar{b}),R)\) and \(\bar{\mathcal{B}}((\bar{a},\bar{b},\bar{c}),R)\), we obtain that
which is a contradiction and this proves the convexity of f on \(\mathcal{V}\).
In the following sections we consider convex, Lipschitz, and bounded functions to obtain some trapezoid and mid-point type inequalities on a closed ball. We use the spherical coordinates in calculating the integrals.
2 Convex functions
In this section we obtain trapezoid and mid-point type inequalities for the case that the partial derivative absolute values of a considered function with respect to the radius in spherical coordinates is convex. We need the following lemma.
Lemma 2.1
For an integrable function\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\), we have
and
Proof
Consider the spherical transformation
It is obvious that the Jacobian of this transformation is \(J=\rho ^{2} \sin \varphi \). So we have (10).
For (11), consider the curve \(\eta : [0,\pi ] \times [0, 2\pi ]\to \mathbb{R}^{3}\) defined by
It is clear that \(\eta ([0,\pi ] \times [0, 2\pi ] ) =\sigma (C,R)\) and then by integrating with respect to the surface (arc length) we get
This proves (11). □
The following is a sharp trapezoid type inequality related to (1), where we consider a function with convex partial derivative (with respect to the radius ρ) absolute values defined on \(\bar{\mathcal{B}}(\mathcal{C},R)\).
Theorem 2.2
For\(\mathcal{V}\subset \mathbb{R}^{3}\), suppose that\(\bar{\mathcal{B}}(\mathcal{C},R)\subset \mathcal{V}^{\circ }\)where\(\mathcal{V}^{\circ }\)is the interior of\(\mathcal{V}\). Consider\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)which has continuous partial derivatives with respect to the variablesρ, φ, andθon\(\bar{\mathcal{B}}(\mathcal{C},R)\)in spherical coordinates. If\(\vert \frac{\partial f}{\partial \rho } \vert \)is convex on\(\bar{\mathcal{B}}(\mathcal{C},R)\), then
Furthermore, inequality (12) is sharp.
Proof
For fixed \(\varphi \in [0,\pi ]\) and \(\theta \in [0,2\pi ]\) and arbitrary \(\rho \in [0,R]\), since
by integration by parts we have
So integrating with respect to \(\varphi \in [0,\pi ]\) and \(\theta \in [0,2\pi ]\) in (13), along with (10) and (11) obtained in Lemma 2.1 and the convexity of \(\vert \frac{\partial f}{\partial \rho } \vert \) on \(\bar{\mathcal{B}}(\mathcal{C},R)\), implies that
By considering the left-hand side of (1) for \(\vert \frac{\partial f}{\partial \rho } \vert \) and applying it in (14), we have
By dividing (15) with \(4\pi R^{3}\), we obtain the desired result (12).
To show the sharpness of (12), consider the function \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) defined as
Using spherical coordinates, we have \(f(\rho ,\varphi ,\theta )=R-\rho \), for \(\rho \in [0,R]\), \(\varphi \in [0,\pi ]\) and \(\theta \in [0,2\pi ]\). With some calculations we obtain that
and
On the other hand, since \(\vert \frac{\partial f}{\partial \rho } \vert =1\),
From (16) and (17) we have the sharpness of (12). □
Now we obtain the midpoint type inequality related to (1), where the partial derivative absolute value of considered function defined on \(\bar{\mathcal{B}}(\mathcal{C},R)\) is convex.
Theorem 2.3
Suppose that\(\bar{\mathcal{B}}(\mathcal{C},R)\subset \mathcal{V}^{\circ }\), where\(\mathcal{V}\subset \mathbb{R}^{3}\). Consider\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)which has continuous partial derivatives with respect to the variablesρ, φ, andθon\(\bar{\mathcal{B}}(\mathcal{C},R)\)in spherical coordinates. If\(\vert \frac{\partial f}{\partial \rho } \vert \)is convex on\(\bar{\mathcal{B}}(\mathcal{C},R)\), then
Proof
Similar to the proof of Theorem 2.2, for fixed \(\varphi \in [0,\pi ]\) and \(\theta \in [0,2\pi ]\), we have
Integration with respect to the variables \(\varphi \in [0,\pi ]\) and \(\theta \in [0,2\pi ]\) in (19) implies that
So from the convexity of \(\vert \frac{\partial f}{\partial \rho } \vert \) we get
It follows from triangle inequality, (20), (12) and (1)(for \(\vert \frac{\partial f}{\partial \rho } \vert \)) that
which implies the desired result. □
Corollary 2.4
([17])
Consider a set\(I\subset \mathbb{R}^{2}\)with\(D(C,R)\subset I^{\circ }\). Suppose that the mapping\(f:D(C,R)\to \mathbb{R}\)has continuous partial derivatives in the disk\(D(C,R)\)with respect to the variablesrandθin polar coordinates. If for any constant\(\theta \in [0,2\pi ]\), the function\(\vert \frac{\partial f}{\partial r} \vert \)is convex with respect to the variableron\([0,R]\)then
Remark 2.5
In the proof of Theorem 2.3, we can find the following inequality:
Although (18) is not sharp, if we consider \(f(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}}\) for \(x,y,z\in \bar{\mathcal{B}}(\mathcal{C},R)\), we will find that inequality (21) is sharp.
Remark 2.6
If we drop out the convexity condition of \(\vert \frac{\partial f}{\partial \rho } \vert \) in Theorems 2.2, 2.3, and consider the condition
instead of that, then we get the following Ostrowski type inequalities (see [19, 20]) on a closed ball:
and
3 Lipschitz functions
In this section we consider Lipschitz functions with respect to the Euclidian norm to obtain some trapezoid and mid-point type inequalities on \(\bar{\mathcal{B}}(\mathcal{C},R)\).
Definition 3.1
([21])
A function \(f:\mathcal{V}\subset \mathbb{R}^{3}\to \mathbb{R}\) is said to satisfy a Lipschitz condition (briefly, f is \(\mathcal{L}\)-Lipschitz) on \(\mathcal{V}\) with respect to a norm \(\Vert \cdot \Vert \), if there exists a constant \(\mathcal{L}>0\) such that
for any \(x,y\in \mathcal{V}\).
If \(f: \bar{\mathcal{B}}(\mathcal{C},R)\) is Lipschitz with respect to the Euclidian norm with the constant \(\mathcal{L}>0\), then for any \(x=(a+\rho _{1} \cos \theta _{1} \sin \varphi _{1},b+\rho _{1} \sin \theta _{1} \sin \varphi _{1}, c+\rho _{1} \cos \varphi _{1})\) and \(y=(a+\rho _{2} \cos \theta _{2} \sin \varphi _{2},b+\rho _{2} \sin \theta _{2} \sin \varphi _{2}, c+\rho _{2} \cos \varphi _{2})\), with some calculations we obtain that
where \(M(\varphi _{1},\varphi _{2},\theta _{1},\theta _{2})= [\sin \varphi _{1} \sin \varphi _{2} \cos (\theta _{1}-\theta _{2})+\cos \varphi _{1} \cos \varphi _{2} ]\), \(\rho _{1},\rho _{2}\in [0,R]\), \(\theta _{1},\theta _{2}\in [0,2\pi ]\) and \(\varphi _{1},\varphi _{2}\in [0,\pi ]\). Also it is obvious that if \(f:\mathcal{V}\subseteq \mathbb{R}^{3}\to \mathbb{R}\) is Lipschitz with a constant \(\mathcal{L}>0\) on \(\mathcal{V}\), then it is continuous and so integrable on \(\mathcal{V}\). We need the following result.
Lemma 3.2
For any\(\varphi _{i}\in [0,\pi ]\)and\(\theta _{i}\in [0,2\pi ]\) (\(i\in \{1,2\}\)) we have
Proof
For any \(\theta _{1},\theta _{2}\in [0,2\pi ]\) it is obvious that \(\cos (\theta _{1}-\theta _{2})\leq 1\). On the other hand, since for any \(\varphi _{1},\varphi _{2}\in [0,\pi ]\), \(\sin \varphi _{1} \sin \varphi _{2}\) is nonnegative,
So
Similarly, we can prove that \(\mathcal{M}(\varphi _{1},\varphi _{2},\theta _{1},\theta _{2})\geq \cos (\varphi _{1} +\varphi _{2} )\geq -1\). □
The following trapezoid type inequality related to (1) for \(\mathcal{L}\)-Lipschitz functions on \(\bar{\mathcal{B}}(\mathcal{C},R)\) holds.
Theorem 3.3
Let\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)be an\(\mathcal{L}\)-Lipschitz function. Then
Inequality (22) is sharp.
Proof
Since f is Lipschitz with constant \(\mathcal{L}>0\) on \(\bar{\mathcal{B}}(\mathcal{C},R)\), we get
Now by replacing (10) and (11) in (23) and then dividing the result by \(\frac{4}{3}\pi R^{3}\), we deduce the desired result.
To prove the sharpness of (22), consider the function \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) defined by
for \(\mathcal{L}>0\), \(\rho \in [0,R]\), \(\varphi \in [0,\pi ]\), and \(\theta \in [0,2\pi ]\). The function f is Lipschitz with constant \(\mathcal{L}\). Consider \(x=(a+\rho _{1} \cos \theta _{1} \sin \varphi _{1},b+\rho _{1} \sin \theta _{1} \sin \varphi _{1}, c+\rho _{1} \cos \varphi _{1})\) and \(y=(a+\rho _{2} \cos \theta _{2} \sin \varphi _{2},b+\rho _{2} \sin \theta _{2} \sin \varphi _{2}, c+\rho _{2} \cos \varphi _{2})\), for \(\rho _{1},\rho _{2}\in [0,R]\), \(\varphi _{1}, \varphi _{2} \in [0,\pi ]\), \(\theta _{1},\theta _{2}\in [0,2\pi ]\). Then by Lemma 3.2 we have
It is not hard to see that \(f (a+\rho \cos \theta \sin \varphi , b+\rho \sin \theta \cos \varphi , c+\rho \cos \varphi )\geq 0\) for all \(0\leq \rho \leq R\), \(0\leq \varphi \leq \pi \), and \(0\leq \theta \leq 2\pi \). Also for the case \(\rho =R\), we have \(f (a+R \cos \theta \sin \varphi , b+R \sin \theta \cos \varphi , c+R \cos \varphi )=0\). So we have
□
For \(\mathcal{L}\)-Lipschitz functions we can obtain a mid-point type inequality as follows:
Theorem 3.4
Let\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)be an\(\mathcal{L}\)-Lipschitz function. Then
Inequality (24) is sharp.
Proof
Since the function f is \(\mathcal{L}\)-Lipschitz on \(\bar{\mathcal{B}}(\mathcal{C},R)\), we have
for all \(\rho \in [0,R]\), \(\varphi \in [0,\pi ]\), and \(\theta \in [0,2\pi ]\). It follows that
So we obtain that
which implies the desired result.
Now consider the function \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) defined by
for \(\mathcal{L}>0\), \(0\leq \rho \leq R\), \(0\leq \varphi \leq \pi \), and \(0\leq \theta \leq 2\pi \). It is obvious that \(f(\mathcal{C})=0\). By a similar method used in the proof of Theorem 3.3, the function f is \(\mathcal{L}\)-Lipschitz. So we have
showing that inequality (24) is sharp. □
Remark 3.5
Consider an open set \(\mathcal{V}\subset \mathbb{R}^{3}\) including \(\bar{\mathcal{B}}(\mathcal{C},R)\). For convex function f defined on \(\mathcal{V}\), from Theorem D of Sect. 41 in [21] we have that f is \(\mathcal{L}\)-Lipschitz on \(\bar{\mathcal{B}}(\mathcal{C},R)\) and so from inequalities (22) and (24), along with inequality (1), we get the following results:
and
In the following, as an example we obtain a Lipschitz constant \(\mathcal{L}\) for a real-valued function defined on a closed ball in \(\mathbb{R}^{3}\).
Example 3.6
Consider \(W=f(x,y,z)=(x-a)^{n}+(y-b)^{n}+(z-c)^{n}\), \(n\in \mathbb{N}\), \((x,y,z)\in \bar{\mathcal{B}}(\mathcal{C},R)\). To find a Lipschitz constant for f, we will do some calculations as follows.For \(A,B\in \bar{\mathcal{B}}(\mathcal{C},R)\), consider the path \(\psi :[0,1]\to \bar{\mathcal{B}}(\mathcal{C},R)\) from B to A in \(\bar{\mathcal{B}}(\mathcal{C},R)\) as
for \(t\in [0,1]\). Now using the fundamental theorem of calculus, we obtain that
On the other hand, from the chain rule for differentiation, we get
where ∇f is the gradient vector of f. So using the Euclidean norm \(\Vert \cdot \Vert \), we obtain
which implies
This shows that \(\mathcal{L}=\sup_{u\in \bar{\mathcal{B}}(\mathcal{C},R)} \Vert \nabla f(u) \Vert \) (if it exists) is a Lipschitz constant for f. Now for any \(w=(x,y,z)\in \bar{\mathcal{B}}(\mathcal{C},R)\), we have
and then
So we can choose \(\mathcal{L}=\sup_{u\in \bar{\mathcal{B}}(\mathcal{C},R)} \Vert \nabla f(u) \Vert =nR^{n-1}\) as a Lipschitz constant for f on \(\bar{\mathcal{B}}(\mathcal{C},R)\).
Using the above example, we have the following result:
Example 3.7
For \(n\in \mathbb{N}\setminus \{1\}\), consider the function \(f(\rho ,\varphi ,\theta )=(x_{0}-\rho )^{n}+(y_{0}-\rho )^{n}+(z_{0}- \rho )^{n}\) defined on \(\bar{\mathcal{B}}((x_{0},y_{0},z_{0}),R)\) such that \(x_{0},y_{0},z_{0}>0\), \(0< R\leq \min \{x_{0},y_{0},z_{0}\}\) and \(0\leq \rho \leq R\). It follows that
and then
is a Lipschitz constant for \(\nabla (\frac{\partial f}{\partial \rho } )\). On the other hand, it is not hard to prove that
So by (25), we have the following numerical inequality:
Remark 3.8
For any function \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\), we can apply the structure mentioned in the above example to obtain a Lipschitz constant \(\mathcal{L}=\sup_{z\in \bar{\mathcal{B}}(\mathcal{C},R)} \Vert \nabla f(z) \Vert \) with respect to the Euclidian norm \(\Vert \cdot \Vert \), provided that the gradient vector of f exists everywhere in \(\bar{\mathcal{B}}(\mathcal{C},R)\) and also \(\mathcal{L}<\infty \).
Remark 3.9
In Theorems 3.3 and 3.4, if we consider that \(\frac{\partial f}{\partial \rho }:\bar{\mathcal{B}}(\mathcal{C},R) \to \mathbb{R}\) is \(\mathcal{L}\)-Lipschitz and \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) is integrable, then by (13) and (19) we can obtain (the details are omitted)
and
4 Bounded functions
In the last section we investigate trapezoid and mid-point type inequalities where considered functions are bounded.
Theorem 4.1
Suppose that\(\mathcal{V}\subset \mathbb{R}^{3}\), \(\bar{\mathcal{B}}(\mathcal{C},R)\subset \mathcal{V}^{\circ }\)and\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)has continuous partial derivatives with respect to the variablesρ, φ, andθon\(\bar{\mathcal{B}}(\mathcal{C},R)\)in spherical coordinates. If\(\frac{\partial f}{\partial \rho }\)is bounded on\(\bar{\mathcal{B}}(\mathcal{C},R)\), then
where\(\mathcal{L}_{B}\)and\(\mathcal{U}_{B}\)are lower and upper bounds of\(\frac{\partial f}{\partial \rho }\)on\(\bar{\mathcal{B}}(\mathcal{C},R)\), respectively.
Proof
Consider \(\mathcal{U}_{B}\) and \(\mathcal{L}_{B}\) as the upper and lower bounds of an arbitrary function g defined on a set \(\mathcal{V}\subset \mathbb{R}^{3}\), respectively. Then for all \(x,y,z\in \mathcal{V}\), we have
which implies that
for all \(x,y,z\in \mathcal{V}\). On the other hand, from (13) we get
Now if in (27) we consider \(g=\frac{\partial f}{\partial \rho }\), \(\mathcal{V}=\bar{\mathcal{B}}(\mathcal{C},R)\), and utilize Lemma 2.1, then we obtain that
Finally, by the use of the triangle inequality and dividing the result by \(4\pi R^{3}\), we obtain inequality (26). □
Theorem 4.2
Suppose that\(\mathcal{V}\subset \mathbb{R}^{3}\), \(\bar{\mathcal{B}}(\mathcal{C},R)\subset \mathcal{V}^{\circ }\)and\(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\)has continuous partial derivatives with respect to the variablesρ, φ, andθon\(\bar{\mathcal{B}}(\mathcal{C},R)\)in spherical coordinates. If\(\frac{\partial f}{\partial \rho }\)is bounded on\(\bar{\mathcal{B}}(\mathcal{C},R)\), then
where\(\mathcal{L}_{B}\)and\(\mathcal{U}_{B}\)are lower and upper bounds of\(\frac{\partial f}{\partial \rho }\)on\(\bar{\mathcal{B}}(\mathcal{C},R)\), respectively.
Proof
Consider \(\mathcal{L}_{B}\) and \(\mathcal{U}_{B}\) as the upper and lower bounds of \(\frac{\partial f}{\partial \rho }\). By (19), the following relations hold:
This implies that
Finally, by using the triangle inequality, we get
□
Remark 4.3
If \(f:\bar{\mathcal{B}}(\mathcal{C},R)\to \mathbb{R}\) is a convex function and bounded from above on \(\bar{\mathcal{B}}(\mathcal{C},R)\) (\(\mathcal{U}_{B}\) exists), then f is bounded on \(\bar{\mathcal{B}}(\mathcal{C},R)\) because for an arbitrary \(X\in \bar{\mathcal{B}}(0,R)\) and \(\mathcal{C}=\frac{1}{2}(X+\mathcal{C})+\frac{1}{2}(-X+\mathcal{C})\), from the convexity of f we have \(2f(\mathcal{C})-f(-X+\mathcal{C})\leq f(X+\mathcal{C})\). This implies that \(2f(\mathcal{C})-\mathcal{U}_{B}\leq f(X+\mathcal{C})\) where \(X+\mathcal{C}\) and \(-X+\mathcal{C}\) belong to \(\bar{\mathcal{B}}(\mathcal{C},R)\). Now it is enough to set \(\mathcal{L}_{B}=2f(\mathcal{C})-\mathcal{U}_{B}\).
So if \(\frac{\partial f}{\partial \rho }:\bar{\mathcal{B}}( \mathcal{C},R)\to \mathbb{R}\) is convex and bounded from above, then by (26), (28), and (1), the following inequalities hold:
and
where \(\mathcal{L}_{B}\) and \(\mathcal{U}_{B}\) are lower and upper bounds of \(\frac{\partial f}{\partial \rho }\) on \(\bar{\mathcal{B}}(\mathcal{C},R)\), respectively.
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Rostamian Delavar, M. Sharp trapezoid and mid-point type inequalities on closed balls in \(\mathbb{R}^{3}\). J Inequal Appl 2020, 114 (2020). https://doi.org/10.1186/s13660-020-02377-x
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DOI: https://doi.org/10.1186/s13660-020-02377-x