Hermite-Hadamard's Type Inequalities on a Ball

TL;DR

This paper develops new inequalities related to Hermite-Hadamard for functions defined on a ball in space, extending classical results to a geometric setting.

Contribution

It introduces trapezoid and mid-point type inequalities for Hermite-Hadamard inequalities on a ball, expanding the scope of these inequalities.

Findings

Derived new trapezoid type inequalities

Established mid-point type inequalities

Extended Hermite-Hadamard inequalities to a geometric setting

Abstract

Some trapezoid and mid-point type inequalities related to the Hermite-Hadamard inequality for the mappings defined on a ball in the space are obtained.

Equations

$$\bar{B}(C,R)=\{(x,y,z)\in {\mathbb{R}}^3|(x-a{)}^2+(y-b{)}^2+(z-c{)}^2\le R^2\}.$$

$$\sigma (C,R)=\{(x,y,z)\in {\mathbb{R}}^3|(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=R^2\}.$$

$$f(a,b,c)\le \frac{3}{4\pi R^3}\int \int {\int }_{\bar{B}(C,R)}f(x,y,z)dv\le$$

$$\frac{1}{4\pi R^2}\int {\int }_{\sigma (C,R)}f(x,y,z)d\sigma .$$

$$\displaystyle\bigg{|}\frac{1}{4\pi R^{2}}\int\int_{\sigma(C,R)}f(x,y,z)d\sigma-\frac{1}{\frac{4}{3}\pi R^{3}}\int\int\int_{\bar{B}(C,R)}f(x,y,z)dv\bigg{|}\leq$$

$$\displaystyle\frac{1}{16\pi R}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma.$$

$$\int \int {\int }_{\bar{B}(C,R)}f(x,y,z)dv=$$

$${\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^{R}f(a+\rho cos\theta sin\phi ,b+\rho sin\theta cos\phi ,c+\rho cos\phi ){\rho }^{2}sin\phi d\rho d\phi d\theta .$$

$$\int {\int }_{\sigma (C,R)}f(x,y,z)d\sigma =$$

$${\int }_0^{2\pi }{\int }_0^{\pi }f(a+Rcos\theta sin\phi ,b+Rsin\theta cos\phi ,c+Rcos\phi )R^{2}sin\phi d\phi d\theta .$$

$${\int }_0^R\frac{\partial f}{\partial \rho }(a+\rho cos\theta sin\phi ,b+\rho sin\theta cos\phi ,c+\rho cos\phi ){\rho }^{3}sin\phi d\rho =$$

$$R^{3}f(a+Rcos\theta sin\phi ,b+Rsin\theta cos\phi ,c+Rcos\phi )-$$

$$-3{\int }_0^{R}f(a+\rho cos\theta sin\phi ,b+\rho sin\theta cos\phi ,c+\rho cos\phi ){\rho }^{2}sin\phi d\rho .$$

$$\displaystyle\bigg{|}R\int\int_{\sigma(C,R)}f(x,y,z)d\sigma-3\int\int\int_{\bar{B}(C,R)}f(x,y,z)dv\bigg{|}\leq$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a+\rho cos\theta sin\varphi,b+\rho sin\theta cos\varphi,c+\rho cos\varphi)\rho^{3}sin\varphi d\rho d\varphi d\theta\leq$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\big{|}\frac{\partial f}{\partial\rho}\big{|}\Big{(}(1-\frac{\rho}{R})(a,b,c)+\frac{\rho}{R}(a+Rcos\theta sin\varphi,b+Rsin\theta cos\varphi,c+Rcos\varphi\Big{)}\times$$

$${\rho }^{3}sin\phi d\rho d\phi d\theta \le$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\rho^{3}\big{(}1-\frac{\rho}{R}\big{)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a,b,c)sin\varphi d\rho d\varphi d\theta+$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\frac{\rho^{4}}{R}\big{|}\frac{\partial f}{\partial\rho}\big{|}\big{(}a+Rcos\theta sin\varphi,b+Rsin\theta cos\varphi,c+Rcos\varphi\big{)}sin\varphi d\rho d\varphi d\theta=$$

$$\displaystyle\frac{\pi R^{4}}{5}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a,b,c)+$$

$$\displaystyle\frac{R^{4}}{5}\int_{0}^{2\pi}\int_{0}^{\pi}\big{|}\frac{\partial f}{\partial\rho}\big{|}\big{(}a+Rcos\theta sin\varphi,b+Rsin\theta cos\varphi,c+Rcos\varphi\big{)}sin\varphi d\varphi d\theta=$$

$$\displaystyle\frac{\pi R^{4}}{5}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a,b,c)+\frac{R^{2}}{5}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma.$$

$$\displaystyle\bigg{|}R\int\int_{\sigma(C,R)}f(x,y,z)d\sigma-3\int\int\int_{\bar{B}(C,R)}f(x,y,z)dv\bigg{|}\leq$$

$$\displaystyle\frac{R^{2}}{20}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma+\frac{R^{2}}{5}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma=$$

$$\displaystyle\frac{R^{2}}{4}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma.$$

$$f(x,y,z)=R-\sqrt{(x-a{)}^2+(y-b{)}^2+(z-c{)}^2}.$$

$$\frac{1}{\frac{4}{3}\pi R^3}\int \int {\int }_{\bar{B}(C,R)}f(x,y,z)dv=\frac{1}{\frac{4}{3}\pi R^3}{\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^R(R-\rho ){\rho }^{2}sin\phi d\rho d\phi d\theta =\frac{R}{4}.$$

$$\int {\int }_{\sigma (C,R)}f(x,y,z)d\sigma =$$

$${\int }_0^{2\pi }{\int }_0^{\pi }f(a+Rcos\theta sin\phi ,b+Rsin\theta cos\phi ,c+Rcos\phi )R^{2}sin\phi d\phi d\theta =$$

$${\int }_0^{2\pi }{\int }_0^{\pi }(R-R)R^{2}sin\phi d\phi d\theta =0.$$

$$\displaystyle\frac{1}{16\pi R}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma=\frac{4\pi R^{2}}{16\pi R}=\frac{R}{4},$$

$$\displaystyle\bigg{|}\frac{1}{\frac{4}{3}\pi R^{3}}\int\int\int_{\bar{B}(C,R)}f(x,y,z)dv-f(a,b,c)\bigg{|}\leq\frac{5}{16\pi R}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma.$$

$${\int }_0^R\frac{\partial f}{\partial \rho }(a+\rho cos\theta sin\phi ,b+\rho sin\theta cos\phi ,c+\rho cos\phi )sin\phi d\rho =$$

$$f(a+Rcos\theta sin\phi ,b+Rsin\theta cos\phi ,c+Rcos\phi )sin\phi -f(a,b,c)sin\phi .$$

$${\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^R\frac{\partial f}{\partial \rho }(a+\rho cos\theta sin\phi ,b+\rho sin\theta cos\phi ,c+\rho cos\phi )sin\phi d\rho d\phi d\theta =$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}f(a+Rcos\theta sin\varphi,b+Rsin\theta cos\varphi,c+Rcos\varphi\Big{)}sin\varphi d\varphi d\theta-$$

$$\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}f(a,b,c)sin\varphi d\varphi d\theta=\frac{1}{R^{2}}\int\int_{\sigma(C,R)}\big{|}\frac{\partial f}{\partial\rho}\big{|}(x,y,z)d\sigma-4\pi f(a,b,c).$$

$$\displaystyle\bigg{|}\frac{1}{4\pi R^{2}}\int\int_{\sigma(C,R)}f(x,y,z)d\sigma-f(a,b,c)\bigg{|}\leq$$

$$\displaystyle\frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a+\rho cos\theta sin\varphi,b+\rho sin\theta cos\varphi,c+\rho cos\varphi)sin\varphi d\rho d\varphi d\theta\leq$$

$$\displaystyle\frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}(1-\frac{\rho}{R})\big{|}\frac{\partial f}{\partial\rho}\big{|}(a,b,c)sin\varphi d\rho d\varphi d\theta+$$

$$\displaystyle\frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\frac{\rho}{R}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a+Rcos\theta sin\varphi,b+Rsin\theta cos\varphi,c+Rcos\varphi)sin\varphi d\rho d\varphi d\theta=$$

$$\displaystyle\frac{R}{2}\big{|}\frac{\partial f}{\partial\rho}\big{|}(a,b,c)+\frac{1}{8\pi R}\int\int_{\sigma(C,R)}f(x,y,z)d\sigma.$$

Full text

1. Introduction

Consider the ball $\bar{B}(C,R)$ in the space where $C=(a,b,c)\in\mathbb{R}^{3}$, $R>0$ and

[TABLE]

Also consider $\sigma(C,R)$ as the boundary of $\bar{B}(C,R)$, i.e.

[TABLE]

The following result has proved in [1], which is the Hermite-Hadamard inequality for convex functions defined on a ball $\bar{B}(C,R)$.

Theorem 1.1**.**

Let $\bar{B}(C,R)\to\mathbb{R}$ be a convex mapping on the ball $\bar{B}(C,R)$. Then we have the inequality:

[TABLE]

Motivated by (1), we obtain some trapezoid and mid-point type inequalities related to the Hermite-Hadamard inequality for the mappings defined on a ball $\bar{B}(C,R)$ in the space. In this paper we use the spherical coordinates to prove our results.

2. Main Results

The following is trapezoid type inequalities related to the (1) for the mappings defined on $\bar{B}(C,R)$.

Theorem 2.1**.**

Suppose that $\bar{B}(C,R)\subset I^{\circ}$, where $I\subset R^{3}$ and consider $f:\bar{B}(C,R)\to\mathbb{R}$ which has continuous partial derivatives with respect to the variables $\rho$, $\varphi$ and $\theta$ on $I^{\circ}$ in spherical coordinates. If $|\frac{\partial f}{\partial\rho}|$ is convex on $\bar{B}(C,R)$, then

[TABLE]

Furthermore above inequality is sharp.

Proof.

First notice that

[TABLE]

Second notice that

[TABLE]

Now for fixed $\varphi\in[0,\pi]$ and $\theta\in[0,2\pi]$, we have

[TABLE]

So integrating with respect to $\varphi\in[0,\pi]$ and $\theta\in[0,2\pi]$ in (5) along with (3), (4) and the convexity of \big{|}\frac{\partial f}{\partial\rho}\big{|} imply that

[TABLE]

[TABLE]

Since \big{|}\frac{\partial f}{\partial\rho}\big{|} is convex, then from (1) and (6) we obtain that

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Dividing (7) with ”$4\pi R^{3}$” we obtain the desired result (2). For the sharpness of (2) consider the function $f:\bar{B}(C,R)\to\mathbb{R}$ defined as

[TABLE]

By the use of 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

[TABLE]

Also

[TABLE]

On the other hand since \big{|}\frac{\partial f}{\partial\rho}\big{|}=1, then

[TABLE]

which along with (8) and (9) show that (2) is sharp. ∎

The following is trapezoid type inequalities related to the (1) for the mappings defined on $\bar{B}(C,R)$.

Theorem 2.2**.**

Suppose that $\bar{B}(C,R)\subset I^{\circ}$, where $I\subset R^{3}$ and consider $f:\bar{B}(C,R)\to\mathbb{R}$ which has continuous partial derivatives with respect to the variables $\rho$, $\varphi$ and $\theta$ on $I^{\circ}$ in spherical coordinates. If $|\frac{\partial f}{\partial\rho}|$ is convex on $\bar{B}(C,R)$, then

[TABLE]

Proof.

For fixed $\varphi\in[0,\pi]$ and $\theta\in[0,2\pi]$ we have

[TABLE]

Integration with respect to the variables $\varphi\in[0,\pi]$ and $\theta\in[0,2\pi]$ in (10) implies that

[TABLE]

So from the convexity of \big{|}\frac{\partial f}{\partial\rho}\big{|} we get

[TABLE]

It follows from triangle inequality, (11), (1) and (2) that

[TABLE]

which implies the desired result. ∎