International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 909261 | https://doi.org/10.5402/2011/909261

Chaobang Gao, Jiajin Wen, "Theories and Inequalities on the Satellite System", International Scholarly Research Notices, vol. 2011, Article ID 909261, 22 pages, 2011. https://doi.org/10.5402/2011/909261

Theories and Inequalities on the Satellite System

Academic Editor: D. Han
Received18 Jan 2011
Accepted13 Feb 2011
Published03 May 2011

Abstract

We define the satellite system without any central, the satellite system with a central, and the satellite system of single point with a central. For the satellite system without any central 𝑆{Γ,Γ𝑁}, we establish the inequality: Γ𝑑𝑠Γ𝑁|𝑃𝑄|𝑑𝑠𝑄(𝑁|Γ|3/4𝜋2)sin(2𝜋/𝑁). For the satellite system with a central 𝑆{𝑂,Γ,Γ𝑁}, we establish the following inequality under the proper hypothesis: Γ𝑀𝑁[𝑡](𝑟)𝑑𝑠(|Γ|2/2𝜋)cos(𝜋/𝑁)(𝑡],2]). As an application, we get the inequalities Γ𝜌𝑃𝑑𝑠Γ𝑟𝑃𝑑𝑠2|𝐷(Γ)|, for the satellite system of single point with a central 𝑆{𝑂,Γ,𝑃}. For these results, there are generalized backgrounds in the fields of differential geometry and space science.

1. Introduction and the Main Results

In this paper, we will use the following symbols: 𝑅 and 𝑍 to denote the set of real numbers and the set of integers, respectively, 𝑎=(𝑎1,𝑎2,,𝑎𝑛), and 𝑅𝑛={𝑎|𝑎𝑖𝑅,1𝑖𝑛}[1, 2].

Definition 1.1. Let Γ=Γ(𝐴𝐵)𝑟=𝑟(𝑡)=𝑥(𝑡)𝑖+𝑦(𝑡)𝑗+𝑧(𝑡)𝑘(𝑎𝑡𝑏) be a smooth curve in three-space 𝑅3. 𝐴=𝑟(𝑎) and 𝐵=𝑟(𝑏) are called the initial point and terminal point of Γ, respectively. If 𝑎<𝑡1<𝑏, 𝑎𝑡2𝑏, 𝑡1𝑡2, and 𝑟(𝑡1)=𝑟(𝑡2), then 𝑟(𝑡1) is called a coincident point of Γ; smooth curve without any coincident point is called a simple curve; If a simple curve Γ satisfies 𝐴=𝐵, then Γ is said be a simple closed curve. Especially, we say Γ is a Jordan closed curve if Γ is a plane curve in 𝑅3. For a Jordan closed curve Γ, 𝐷(Γ) denotes the closed region bounded by Γ and its area is written as |𝐷(Γ)|, |Γ| denotes the length of Γ, and we have the following Jordan curve theorem.

Theorem 1.2 (Jordan Curve [3]). An arbitrary Jordan closed curve must divide a plane into two parts, where one part is bounded, and the other is unbounded. The bounded part is called interior and another is outside of the Jordan closed curve.

Definition 1.3 (see [4, 5]). Assume that Γ(𝐴𝐵) is a smooth space curve with end points 𝐴 and 𝐵, 𝑃 is a fixed point in 𝑅3, and 𝑄Γ(𝐴𝐵), where 𝑄 is a moving point, then the trail of the line segment 𝑃𝑄 is called a bounded cone surface, written as Ω[𝑃,Γ(𝐴𝐵)]. We Also say 𝑃𝑄 is the generating line, 𝑃 is the vertex, and Γ(𝐴𝐵) is the generating curve of the bounded cone surface. |Ω[𝑃,Γ(𝐴𝐵)]| denotes the area of Ω[𝑃,Γ(𝐴𝐵)].
If Γ is a given smooth curve in 𝑅3, then its length is written as |Γ|; Γ(𝐴𝐵) denotes the smooth curve segment with end points 𝐴 and 𝐵, and Γ(𝐴𝐵)Γ.

Definition 1.4. Let Γ be a fixed smooth closed curve in 𝑅3, the 𝑁-polygon Γ𝑁𝐴1𝐴2𝐴𝑁𝐴1(𝑁3) be inscribed in Γ, and, 𝐴𝑖Γ(𝐴𝑖1𝐴𝑖+1)(𝑖=1,2,,𝑁). If all vertices𝐴1,𝐴2,,𝐴𝑁 of Γ𝑁 move continuously along Γ and keep 𝑙𝑖=|Γ(𝐴𝑖𝐴𝑖+1)| invariable, where 0<𝑙𝑖<|Γ|/2,𝑖=1,2,,𝑁, we say that the set 𝑆{Γ,Γ𝑁}={Γ,Γ𝑁} is the satellite system without any central in three-space 𝑅3.
For 𝑆{Γ,Γ𝑁}, we write the set of the vertices of Γ𝑁 as 𝑉(Γ𝑁)={𝐴1,𝐴2,,𝐴𝑁} and define 𝐴𝑖=𝐴𝑗𝑖𝑗(mod𝑁), for all 𝑖,𝑗𝑍 [6].

Definition 1.5. For 𝑆{Γ,Γ𝑁}, if the following statements are valid:(i)Γis a fixed closed Jordan curve in 𝑅3,(ii)𝑂 is a fixed point in the interior of Γ, and(iii)Γ𝑁 is always a Jordan closed curve and Ω[𝑂,Γ𝑁]=𝐷(Γ𝑁) as all vertices of Γ𝑁 move continuously along Γ,then we say the set 𝑆{𝑂,Γ,Γ𝑁}:={𝑂,Γ,Γ𝑁} is the satellite system with a central and 𝑂 is its central.
For 𝑆{𝑂,Γ,Γ𝑁}, 𝑟𝑖=Dis(𝑂,𝐴𝑖𝐴𝑖+1) denotes the distance from 𝑂 to the line 𝐴𝑖𝐴𝑖+1, 𝑖=1,2,,𝑁. And we write 𝑟=(𝑟1,𝑟2,,𝑟𝑁).

Remark 1.6. The 𝑆{𝑂,Γ,Γ𝑁} may be explained as follows: the point 𝑂 denotes the central of earth, Γ denotes the trajectory on which satellites move, and the vertexes of Γ𝑁 are viewed as 𝑁 satellites moving on the same trajectory Γ. In order to avoid hitting, they must move by same curve velocity, that is, 𝑙𝑖=|Γ(𝐴𝑖𝐴𝑖+1)| is invariable and 0<𝑙𝑖<|Γ|/2,𝑖=1,2,,𝑁.

Definition 1.7. The tth power mean of the positive real numbers sets 𝑎=(𝑎1,𝑎2,,𝑎𝑁)(𝑁2), written as 𝑀𝑁[𝑡](𝑎), is defined by [715]: 𝑀𝑁[𝑡]1(𝑎)=𝑁𝑁𝑖=1𝑎𝑡𝑖1/𝑡,𝑡𝑅,0<|𝑡|<+𝑁𝑁𝑖=1𝑎𝑖,𝑡=0.(1.1)

Definition 1.8 (see [1, 7, 10, 11, 14]). Let Γ be a smooth curve in 𝑅3. For Riemann integrable function 𝑓Γ𝑅, 𝑀(𝑓,Γ)=(1/|Γ|)Γ𝑓𝑑𝑠 is called function means of 𝑓 on Γ. For the function 𝑓Γ]0,[ and the real number 𝑡, if 𝑓𝑡(𝑡𝑅,𝑡0) and ln𝑓 are Riemann integrable on Γ, then 𝑀[𝑡](1𝑓,Γ)=||Γ||Γ𝑓𝑡𝑑𝑠1/𝑡1,𝑡𝑅,𝑡0exp||Γ||Γln𝑓𝑑𝑠,𝑡=0(1.2) is called the tth power means of 𝑓 on Γ.

Now, we give our main results as follow.

Theorem 1.9. Let 𝑆{Γ,Γ𝑁} be a satellite system without any central. For any 𝑃𝑉(Γ𝑁)={𝐴1,𝐴2,,𝐴𝑁}, we have inequality Γ𝑑𝑠Γ𝑁||||𝑃𝑄𝑑𝑠𝑄𝑁||Γ||34𝜋2sin2𝜋𝑁,(1.3) that is, 𝑀||Ω𝑃,Γ𝑁||𝑁||Γ||,Γ28𝜋2sin2𝜋𝑁,(1.4) where 𝑃𝑄 is the generating line of the bounded cone surface Ω[𝑃,Γ𝑁]. A sufficient condition of equalities in (1.3) and (1.4) is that Γ is a circle and Γ𝑁 is always a regular polygon with 𝑁 sides.

Theorem 1.10. Let 𝑆{𝑂,Γ,Γ𝑁} be a satellite system with a central. If 𝑁=3 or 𝑁4, 0<𝑙𝑖|Γ|/4,𝑖=1,2,,𝑁, we have inequality, for any the real number 𝑡2, Γ𝑀𝑁[𝑡]||Γ||(𝑟)𝑑𝑠2𝜋2𝜋cos𝑁,(1.5) that is, 𝑀𝑀𝑁[𝑡]||Γ||(𝑟),Γ𝜋2𝜋cos𝑁.(1.6) The equalities of (1.5) and (1.6) occur if and only if Γ is a circle, 𝑂 is the central of the circle, and Γ𝑁 is a regular polygon with sides 𝑁.

In Section 5, we will give some applications of these results and theories.

2. Preliminaries

Lemma 2.1. For the quadrilateral 𝐴𝐵𝐶𝐷𝐴 in 𝑅3, writing 𝑎=|𝐴𝐵|,𝑏=|𝐵𝐶|,𝑐=|𝐶𝐷|,𝑑=|𝐷𝐴|, we obtain ||||+||||1Δ𝐵𝐴𝐷Δ𝐵𝐶𝐷4(𝑎+𝑑)2(𝑏𝑐)2(𝑏+𝑐)2(𝑎𝑑)2,(2.1) with equality if and only if 𝐵𝐴𝐷+𝐵𝐶𝐷=𝜋. If the quadrilateral 𝐴𝐵𝐶𝐷𝐴 is convex and in plane, equality holds if and only if the quadrilateral 𝐴𝐵𝐶𝐷𝐴 is inscribed in a circle.

Proof. Write 𝑠=|Δ𝐵𝐴𝐷|+|Δ𝐵𝐶𝐷|,𝐵𝐴𝐷=𝛼,𝐵𝐶𝐷=𝛽. By the area formula of triangle and cosine theorem, we get 1𝑠=2𝑎(𝑎𝑑sin𝛼+𝑏𝑐sin𝛽),(2.2)2+𝑑22𝑎𝑑cos𝛼=𝑏2+𝑐22𝑏𝑐cos𝛽,0𝛼,𝛽𝜋.(2.3) Consider 𝛽=𝛽(𝛼) as the implicit function with respect to 𝛼. Finding the derivatives of the two sides of (2.2) and (2.3), respectively, we have 𝑑𝑠=1𝑑𝛼2𝑎𝑑cos𝛼+𝑏𝑐cos𝛽𝑑𝛽,𝑑𝛼𝑑𝛽=𝑑𝛼𝑎𝑑sin𝛼𝑏𝑐sin𝛽0.(2.4) Therefore, (𝑑𝑠/𝑑𝛼)=(1/2)𝑎𝑑sin𝛼(cot𝛼+cot𝛽) and 𝛽=𝛽(𝛼) is increasing with respect to 𝛼. Since ||||||||||||||||||||||||𝑏𝑐𝐵𝐷𝑎+𝑑,𝑎𝑑𝐵𝐷𝑏+𝑐,𝑏𝑐𝑎+𝑑,𝑎𝑑𝑏+𝑐,(𝑎+𝑑)2(𝑏𝑐)20,(𝑎𝑑)2(𝑏+𝑐)20,2(𝑎𝑑+𝑏𝑐)𝑎2+𝑑2𝑏2𝑐2||||𝑎2(𝑎𝑑+𝑏𝑐),2+𝑑2𝑏2𝑐2||||2(𝑎𝑑+𝑏𝑐)1,(2.5) we have 𝑑𝑠𝑑𝛼=0𝛼+𝛽=𝜋,𝑎2+𝑑22𝑎𝑑cos𝛼=𝑏2+𝑐22𝑏𝑐cos𝛽,𝛼=𝛼0𝑎=arccos2+𝑑2𝑏2𝑐2𝛼2(𝑎𝑑+𝑏𝑐),𝛽=𝛽0=𝜋𝛼0,0𝛼<𝛼0𝛼0𝛽=𝛽(𝛼)<𝛽0=𝜋𝛼00𝛽<𝜋𝛼𝑑𝑠𝛼𝑑𝛼>0,0𝛼<𝛼𝜋𝜋𝛽=𝛽(𝛼)>𝛽0=𝜋𝛼0𝜋𝛽>𝜋𝛼𝑑𝑠𝑑𝛼<0.(2.6) Hence 𝑠 is max if and only if 𝛼+𝛽=𝜋. When 𝑠 is max, 𝛽=𝜋𝛼,𝛼=𝛼0. Thus, 1𝑠2𝑎𝑑sin𝛼0+𝑏𝑐sin𝛼0=12(𝑎𝑑+𝑏𝑐)𝑎12+𝑑2𝑏2𝑐22(𝑎𝑑+𝑏𝑐)2=14(𝑎+𝑑)2(𝑏𝑐)2(𝑏+𝑐)2(𝑎𝑑)2.(2.7) In other words, (2.1) holds. Equality holds if and only if 𝛼+𝛽=𝜋𝐵𝐴𝐷+𝐵𝐶𝐷=𝜋. This completes the proof.

Lemma 2.2. Let 𝑆{Γ,Γ𝑁} be a satellite system without any central. For any 𝑃𝑉(Γ𝑁), we have inequality: 12Γ𝑁||||𝑃𝑄𝑑𝑠𝑄=||Ω𝑃,Γ𝑁||||Γ𝑁||2𝜋4𝑁cot𝑁,(2.8) where 𝑃𝑄 is the generating line of the bounded cone surface Ω[𝑃,Γ𝑁]. The second equality occurs if Γ𝑁 is a regular polygon with sides 𝑁.

Proof. By the definition and geometric meaning of curve integral, we obtain 12Γ𝑁||||𝑃𝑄𝑑𝑠𝑄=||Ω𝑃,Γ𝑁||.(2.9) Now we prove the inequality in (2.8).
If 𝑁=3, (2.8) is known; if 𝑁=4, by Lemma 2.1, (2.8) holds. In the following, we suppose that 𝑁5. First, fix the value of |Γ𝑁|. Without loss of generality, we set 𝑃=𝐴1. By the theory of differential geometry, we know a bounded cone surface is a developable surface, which implies that Ω[𝑃,Γ𝑁] can be developed into a bounded cone surface Ω[𝐴1,Γ𝑁] in the plane, where Γ𝑁𝐴1𝐴2𝐴𝑁𝐴1(𝑃=𝐴1=𝐴1) is the developed graph of Γ𝑁 in the plane and it is a polygon with sides 𝑁 (may not be a closed Jordan curve) and satisfies Δ𝐴𝑖1𝐴1𝐴𝑖Δ𝐴𝑖𝐴1𝐴𝑖+1=𝐴1𝐴𝑖||Γ(𝑖=3,4,,𝑁1),𝑁||=||Γ𝑁||,||Ω𝑃,Γ𝑁||=||Ω𝐴1,Γ𝑁||=𝑁𝑖=3||Δ𝐴𝑖1𝐴1𝐴𝑖||.(2.10) When |Ω[𝐴1,Γ𝑁]| is max, for any 𝑖{2,3,,𝑁1}, the quadrilateral 𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑖1 must be convex and the polygon with 5 sides 𝐴1𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴1(𝐴𝑁+1=𝐴1) must be also convex by the plane geometry. Now, we prove the four points 𝐴𝑖1,𝐴𝑖,𝐴𝑖+1 and 𝐴𝑖+2 are on a common circle. Otherwise, there exists some 𝑖{2,3,,𝑁1} such that 𝐴𝑖1,𝐴𝑖,𝐴𝑖+1,𝐴𝑖+2 are not on a common circle. Therefore, fix the point 𝐴𝑗(𝑗{1,2,,𝑁}{𝑖,𝑖+1}) and modify 𝐴𝑖,𝐴𝑖+1 to 𝐴𝑖,𝐴𝑖+1 such that 𝐴𝑖1,𝐴𝑖,𝐴𝑖+1,𝐴𝑖+2 are on a common circle and |||𝐴𝑖1𝐴𝑖|||=|||𝐴𝑖1𝐴𝑖|||,|||𝐴𝑖𝐴𝑖+1|||=|||𝐴𝑖𝐴𝑖+1|||,|||𝐴𝑖+1𝐴𝑖+2|||=|||𝐴𝑖+1𝐴𝑖+2|||.(2.11) Hence, we obtain a new polygon with 𝑁 sides and write Γ𝑁𝐴1𝐴2𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑁𝐴1.(2.12) It follows that ||Ω𝐴1,Γ𝑁||=||Ω𝐴1,𝐴1𝐴2𝐴𝑖1𝐴𝑖+2𝐴𝑁𝐴1||+||𝐷𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑖1||,||Ω𝐴1,Γ𝑁||=||Ω𝐴1,𝐴1𝐴2𝐴𝑖1𝐴𝑖+2𝐴𝑁𝐴1||+||𝐷𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑖1||.(2.13) By Lemma 2.1, we have ||Ω𝐴1,Γ𝑁||||Ω𝐴1,Γ𝑁||=||𝐷𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑖1||||𝐷𝐴𝑖1𝐴𝑖𝐴𝑖+1𝐴𝑖+2𝐴𝑖1||>0.(2.14) This contradicts the greatest |Ω[𝐴1,Γ𝑁]|. Since three points confirm a unique circle and for any 𝑖{2,3,,𝑁1}, we know 𝐴1,𝐴2,,𝐴𝑁1,𝐴𝑁 are on a common circle if |Ω[𝐴1,Γ𝑁]| is greatest, and since Δ𝐴𝑖1𝐴1𝐴𝑖Δ𝐴𝑖𝐴1𝐴𝑖+1=𝐴1𝐴𝑖,𝑖=3,4,,𝑁1,(2.15)Γ𝑁 is inscribed in a circle and |Ω[𝑃,Γ𝑁]|=|Ω[𝐴1,Γ𝑁]|=|𝐷(Γ𝑁)|. Otherwise, there exists 𝑘{2,3,,𝑁1} such that Δ𝐴𝑘1𝐴1𝐴𝑘Δ𝐴𝑘𝐴1𝐴𝑘+1𝐴1𝐴𝑘. When the perimeter of a circle is a fixed value, the area of regular polygon with 𝑁 sides is greatest in all N-polygons inscribed in the circle [16]. Therefore, Γ𝑁 is a regular polygon with 𝑁 sides if |Ω[𝐴1,Γ𝑁]| is greatest.
Now let Γ𝑁 be a regular polygon with 𝑁 sides. By the plane geometry, we know||Ω𝐴1,Γ𝑁||=||𝐷Γ𝑁||=||Γ𝑁||2𝜋4𝑁cot𝑁=||Γ𝑁||2𝜋4𝑁cot𝑁.(2.16) Hence, the inequality holds in (2.8). Equality occurs if Γ𝑁 is a regular 𝑁-polygon. This completes the proof.

Lemma 2.3. For the quadrilateral 𝐴𝐵𝐶𝐷𝐴 in 𝑅3, we have ||||𝐴𝐶2+||||𝐵𝐷2||||𝐵𝐶2+||||𝐷𝐴2||||||||,+2𝐴𝐵𝐶𝐷(2.17) with equality if and only if (𝐴𝐵,𝐶𝐷)=𝜋.

Proof. Write (𝐴𝐵,𝐵𝐶,𝐶𝐷,𝐷𝐴,𝐴𝐶,𝐵𝐷)=(𝑎,𝑏,𝑐,𝑑,𝑒,𝑓), 𝑎𝑏=𝑎,𝑏 denotes the inner product of vectors 𝑎 and 𝑏, especially, 𝑎2=𝑎𝑎=|𝑎|2. Thus, (2.17) is expressed as 𝑒2+𝑓2𝑏2+𝑑2+2|𝑎||𝑐|.(2.18) Since 𝑎+𝑏=𝑒𝑐+𝑑=𝑒𝑏+𝑐=𝑓𝑑+𝑎=𝑓2𝑎𝑏=𝑒2𝑎2𝑏22𝑐𝑑=𝑒2𝑐2𝑑22𝑏𝑐=𝑓2𝑏2𝑐22𝑑𝑎=𝑓2𝑑2𝑎2(2.19) and 𝑎=𝑎+𝑏+𝑐+𝑑=0, which implies that 𝑎0=2=𝑎2+𝑏2+𝑐2+𝑑2𝑎+2(𝑎𝑏+𝑐𝑑+𝑏𝑐+𝑑𝑎)+2(𝑎𝑐+𝑏𝑑)=2+𝑏2+𝑐2+𝑑2𝑒+22+𝑓2𝑎+2(𝑎𝑐+𝑏𝑑)=22+𝑏2+𝑐2+𝑑2𝑒+22+𝑓2+(𝑎+𝑐)2+(𝑏+𝑑)2𝑎=22+𝑏2+𝑐2+𝑑2𝑒+22+𝑓2+2(𝑎+𝑐)2𝑏=22+𝑑2𝑒+22+𝑓2+4𝑎𝑐,(2.20) it follows that 𝑒2+𝑓2=𝑏2+𝑑22𝑎𝑐𝑏2+𝑑2+2|𝑎||𝑐|.(2.21)(2.18) is proved. Equality of (2.18) holds if and only if (𝐴𝐵,𝐶𝐷)=(𝑎,𝑐)=𝜋. This completes the proof.

Lemma 2.4. Let Γ𝑁(𝑁4) be a polygon with sides 𝑁 in 𝑅3. Setting 𝑆𝑘=sin𝑘𝜋𝑁,𝐿𝑘=𝑁(2𝑆𝑘)2𝑁1𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2,𝑘=1,2,,𝑁1,(2.22) we get the following inequality: 𝐿𝑘𝑆1𝑆𝑘2𝐿1+𝑆𝑘1𝑆𝑘+12𝑆2𝑘𝐿𝑘+1+𝐿𝑘1,(2.23) for 2𝑘𝑁2. A sufficient condition of equality is that Γ𝑁 is a plane regular 𝑁-polygon in 𝑅3.

Proof. Consider the quadrilateral 𝐴𝑖1𝐴𝑖𝐴𝑖1+𝑘𝐴𝑖+𝑘𝐴𝑖1. From 𝑆1𝑘1||𝐴𝑖𝐴𝑖1+𝑘||𝑆1𝑘+1||𝐴𝑖1𝐴𝑖+𝑘||20,(2.24) we obtain 2||𝐴𝑖𝐴𝑖1+𝑘||||𝐴𝑖1𝐴𝑖+𝑘||𝑆𝑘+1𝑆𝑘1||𝐴𝑖𝐴𝑖1+𝑘||2+𝑆𝑘1𝑆𝑘+1||𝐴𝑖1𝐴𝑖+𝑘||2.(2.25) It follows from Lemma 2.3 and (2.25) that ||𝐴𝑖𝐴𝑖+𝑘||2+||𝐴𝑖1𝐴𝑖1+𝑘||2||𝐴𝑖1𝐴𝑖||2+||𝐴𝑖1+𝑘𝐴𝑖+𝑘||2||𝐴+2𝑖𝐴𝑖1+𝑘||||𝐴𝑖1𝐴𝑖+𝑘||||𝐴𝑖1𝐴𝑖||2+||𝐴𝑖1+𝑘𝐴𝑖+𝑘||2+𝑆𝑘+1𝑆𝑘1||𝐴𝑖𝐴𝑖1+𝑘||2+𝑆𝑘1𝑆𝑘+1||𝐴𝑖1𝐴𝑖+𝑘||2,(2.26) which implies that 𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2+||𝐴𝑖1𝐴𝑖1+𝑘||2𝑁𝑖=1||𝐴𝑖1𝐴𝑖||2+||𝐴𝑖1+𝑘𝐴𝑖+𝑘||2+𝑆𝑘+1𝑆𝑘1||𝐴𝑖𝐴𝑖1+𝑘||2+𝑆𝑘1𝑆𝑘+1||𝐴𝑖1𝐴𝑖+𝑘||2.(2.27) Since 𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2=𝑁𝑖=1||𝐴𝑖1𝐴𝑖1+𝑘||2=4𝑁𝑆2𝑘𝐿𝑘,𝑁𝑖=1||𝐴𝑖1𝐴𝑖||2=𝑁𝑖=1||𝐴𝑖1+𝑘𝐴𝑖+𝑘||2=4𝑁𝑆21𝐿1,𝑁𝑖=1||𝐴𝑖𝐴𝑖1+𝑘||2=4𝑁𝑆2𝑘1𝐿𝑘1,𝑁𝑖=1||𝐴𝑖1𝐴𝑖+𝑘||2=4𝑁𝑆2𝑘+1𝐿𝑘+1,(2.28) the inequality (2.27) is equivalent to 8𝑁𝑆2𝑘𝐿𝑘8𝑁𝑆21𝐿1+4𝑁𝑆𝑘+1𝑆𝑘1𝐿𝑘1+4𝑁𝑆𝑘1𝑆𝑘+1𝐿𝑘+1𝐿𝑘𝑆1𝑆𝑘2𝐿1+𝑆𝑘1𝑆𝑘+12𝑆2𝑘𝐿𝑘+1+𝐿𝑘1.(2.29) Inequality (2.23) is proved. From this proof and Lemma 2.3, we know that a sufficient condition of equality is that Γ𝑁 is a regular polygon with 𝑁 sides in 𝑅3.

Remark 2.5. A sufficient condition of equality of (2.23) is that Γ𝑁 is a regular polygon with 𝑁 sides in 𝑅3. This condition is not necessary. For example, when 𝑁=4, the equality holds in (2.23) if and only if Γ4 is a parallelogram in 𝑅3.

Remark 2.6. If Γ𝑁 is a regular 𝑁-polygon, 𝐿𝑘 defined by Lemma 2.4 is equal to 𝑅20, where 𝑅0 denotes the circumradius of Γ𝑁. Namely, 𝐿𝑘=𝑅20,𝑘=1,2,,𝑁1.(2.30)

Lemma 2.7. Suppose 𝐿𝑘 is defined by Lemma 2.4, for any positive integer 𝑘,𝑗𝑘2,𝑘+𝑗𝑁1, there exist constants 𝐶𝑘+𝑗,𝑗,𝐶𝑘1,𝑗,𝐶1,𝑗 which is only related to 𝑘,𝑗,𝑁 such that 𝐿𝑘𝐶𝑘+𝑗,𝑗𝐿𝑘+𝑗+𝐶𝑘1,𝑗𝐿𝑘1+𝐶1,𝑗𝐿1,𝐶𝑘+𝑗,𝑗+𝐶𝑘1,𝑗+𝐶1,𝑗=1.(2.31) A sufficient condition of equality is that Γ𝑁 is a regular 𝑁-polygon in 𝑅3.

Proof. We prove it by mathematical induction with respect to 𝑗.
(i) When 𝑗=1, let 𝐶𝑘+1,1=𝐶𝑘1,1=𝑆𝑘1𝑆𝑘+1/(2𝑆2𝑘)(>0),𝐶1,1=(𝑆1/𝑆𝑘)2(>0) from Lemma 2.4, we have𝐿𝑘𝐶𝑘+1,1𝐿𝑘+1+𝐶𝑘1,1𝐿𝑘1+𝐶1,1𝐿1.(2.32) Let Γ𝑁 be a regular 𝑁-polygon. By Remark 2.6, we know 𝐿𝑘=𝐿𝑘+1=𝐿𝑘1=𝐿1=𝑅20>0; it follows from Lemma 2.4 that the equality of (2.32) holds, thus, 𝐶𝑘+1,1+𝐶𝑘1,1+𝐶1,1=1.
(ii) Assume that Lemma 2.7 holds for 𝑗=𝑛1. Now we want to prove that Lemma 2.7 holds for 𝑗=𝑛+1. By the hypothesis 𝑘,𝑛+1𝑘2,𝑘+𝑛+1𝑁1, we know that 𝑘,𝑛𝑘2,𝑘+𝑛𝑁2𝑁1. Thus, from the induction hypothesis, there exist constants 𝐶𝑘+𝑛,𝑛,𝐶𝑘1,𝑛,𝐶1,𝑛 such that𝐶𝑘+𝑛,𝑛+𝐶𝑘1,𝑛+𝐶1,𝑛𝐿=1,(2.33)𝑘𝐶𝑘+𝑛,𝑛𝐿𝑘+𝑛+𝐶𝑘1,𝑛𝐿𝑘1+𝐶1,𝑛𝐿1.(2.34) A sufficient condition of equality of (2.34) is that Γ𝑁 is a regular polygon with 𝑁 sides. Since 𝑘+1,𝑛𝑘+132,(𝑘+1)+𝑛𝑁1, by induction hypothesis, substitute 𝑘 by 𝑘+1 in (2.33) and (2.34), in other words, there exist constants 𝐶𝑘+1+𝑛,𝑛,𝐶𝑘,𝑛,𝐶1,𝑛 such that 𝐶𝑘+1+𝑛,𝑛+𝐶𝑘,𝑛+𝐶1,𝑛𝐿=1,(2.35)𝑘+1𝐶𝑘+1+𝑛,𝑛𝐿𝑘+1+𝑛+𝐶𝑘,𝑛𝐿𝑘+𝐶1,𝑛𝐿1.(2.36) Substituting (2.36) into (2.32), we get the following inequality: 𝐿𝑘𝐶𝑘+1,1𝐶𝑘+1+𝑛,𝑛𝐿𝑘+1+𝑛+𝐶𝑘,𝑛𝐿𝑘+𝐶1,𝑛𝐿1+𝐶𝑘1,1𝐿𝑘1+𝐶1,1𝐿1.(2.37) Notice 0<𝐶𝑘+1,1<1,0<𝐶𝑘,𝑛<1,1𝐶𝑘+1,1𝐶𝑘,𝑛>0. Solving inequality (2.37) with respect to 𝐿𝑘, we obtain 𝐿𝑘𝐶𝑘+𝑛+1,𝑛+1𝐿𝑘+𝑛+1+𝐶𝑘1,𝑛+1𝐿𝑘1+𝐶1,𝑛+1𝐿1,(2.38) where 𝐶𝑘+𝑛+1,𝑛+1=𝐶𝑘+1,1𝐶𝑘+𝑛+1,𝑛1𝐶𝑘+1,1𝐶𝑘,𝑛𝐶>0,𝑘1,𝑛+1=𝐶𝑘1,11𝐶𝑘+1,1𝐶𝑘,𝑛𝐶>0,1,𝑛+1=𝐶𝑘+1,1𝐶1,𝑛+𝐶1,11𝐶𝑘+1,1𝐶𝑘,𝑛>0.(2.39) Let Γ𝑁 be a regular 𝑁-polygon. From Remark 2.6, we know 𝐿𝑘=𝐿𝑘+𝑛+1=𝐿𝑘1=𝐿1=𝑅20>0. It follows from Lemma 2.4 and induction hypothesis that the equality of (2.38) holds. Thus, 𝐶𝑘+𝑛+1,𝑛+1+𝐶𝑘1,𝑛+1+𝐶1,𝑛+1=1, that is, Lemma 2.7 holds for 𝑗=𝑛+1. This completes the proof.

Lemma 2.8. Suppose 𝐿𝑘 is defined by Lemma 2.4, we have the inequality as follows: 𝐿𝑘𝐿1,𝑘=2,3,,𝑁2.(2.40) A sufficient condition of equality is that Γ𝑁 is a regular polygon with 𝑁 sides in 𝑅3.

Proof. Setting 𝑘+𝑗=𝑁1 in (2.31), we get 𝐿𝑘𝐶𝑁1,𝑁1𝑘𝐿𝑁1+𝐶𝑘1,𝑁1𝑘𝐿𝑘1+𝐶1,𝑁1𝑘𝐿1.(2.41) Since 𝐴𝑖=𝐴𝑗𝑖𝑗(mod𝑁), 𝐿𝑁1𝑁=2𝑆𝑁121𝑁1𝑖=1||𝐴𝑖𝐴𝑖+𝑁1||2=𝑁2𝑆121𝑁1𝑖=1||𝐴𝑖𝐴𝑖1||2=𝐿1.(2.42) It follows from (2.41) and (2.42) that there exist constants 𝐶𝑘1,𝐶1𝐶𝑘1+𝐶1=1 such that 𝐿𝑘𝐶𝑘1𝐿𝑘1+𝐶1𝐿1,(2.43) where 𝐶𝑘1=𝐶𝑘1,𝑁1𝑘>0,𝐶1=𝐶𝑁1,𝑁1𝑘+𝐶1,𝑁1𝑘>0, for any 𝑘{2,3,,𝑁2}. Using (2.43) repeatedly, we get 𝐿𝑘𝐶𝑘1𝐿𝑘1+𝐶1𝐿1𝐶𝑘1𝐶𝑘2𝐿𝑘2+𝐶1𝐿1+𝐶1𝐿1=𝐶𝑘2𝐿𝑘2+𝐶1𝐿1𝐶𝑘3𝐿𝑘3+𝐶1𝐿1𝐶𝐿1,(2.44) therefore, 𝐿𝑘𝐶𝐿1.(2.45) Set Γ𝑁 is a regular polygon with 𝑁 sides in 𝑅3. By Remark 2.6, we know 𝐿𝑘=𝐿1=𝑅20>0; from Lemma 2.7, we have the equality of (2.45) holds, which implies that 𝐶=1. A sufficient condition of equality of (2.40) is that Γ𝑁 is a regular 𝑁-polygon in 𝑅3. This completes the proof.

Lemma 2.9. Let Γ=Γ(𝐴𝐵)𝑟=𝑟(𝑡)=𝑥(𝑡)𝑖+𝑦(𝑡)𝑗+𝑧(𝑡)𝑘(𝑎𝑡𝑏) be a smooth curve with the end points 𝐴 and 𝐵 in 𝑅3. If the function 𝑓Γ(𝐴𝐵)𝑅 is Riemann integrable on Γ(𝐴𝐵), considering a partition of Γ(𝐴𝐵) by means of 𝑁1(𝑁3) points 𝐴=𝐴0,𝐴1,,𝐴𝑖1,𝐴𝑖,,𝐴𝑁=𝐵 such that |𝐴0𝐴1𝐴|=|1𝐴2𝐴|==|𝑖1𝐴𝑖𝐴|==|𝑁1𝐴𝑁|=|Γ𝑁|/𝑁, where Γ𝑁𝐴0𝐴1𝐴𝑁1𝐴𝑁 is a broken line, we have 1𝑀(𝑓,Γ(𝐴𝐵))=||Γ||(𝐴𝐵)Γ(𝐴𝐵)𝑓𝑑𝑠=lim𝑁1𝑁𝑁1𝑖=0𝑓𝐴𝑖.(2.46)

Proof. Since Γ(𝐴𝐵) is a smooth curve with the end points 𝐴 and 𝐵 in 𝑅3, ||||=Γ(𝐴𝐵)𝑏𝑎𝑥2(𝑡)+𝑦2(𝑡)+𝑧2(𝑡)𝑑𝑡(2.47) exists and |Γ(𝐴𝐵)|>0. And 𝑓Γ(𝐴𝐵)𝑅 is Riemann integrable on Γ(𝐴𝐵), it follows that 𝑓Γ(𝐴𝐵)𝑅 is bounded on Γ(𝐴𝐵), that is, there exists a constant 𝑀1>0 such that |𝑓(𝑃)|𝑀1 for any 𝑃Γ(𝐴𝐵). For any 𝑖{0,1,2,,𝑁1}, lim𝑁||Γ𝐴𝑖𝐴𝑖+1||=lim𝑁||𝐴𝑖𝐴𝑖+1||=lim𝑁||Γ𝑁||𝑁=||||Γ(𝐴𝐵)lim𝑁𝑁=0.(2.48) It follows from the definition of the curve integral Γ(𝐴𝐵)𝑓𝑑𝑠=lim𝑁𝑁1𝑖=0𝑓𝑃𝑖||Γ𝐴𝑖𝐴𝑖+1||,𝑃𝑖𝐴Γ𝑖𝐴𝑖+1(2.49) that Γ(𝐴𝐵)𝑓𝑑𝑠=lim𝑁𝑁1𝑖=0𝑓𝐴𝑖||Γ𝐴𝑖𝐴𝑖+1||.(2.50) Since lim𝑁(|Γ(𝐴𝑖𝐴𝑖+1𝐴)|/|𝑖𝐴𝑖+1|)=1(𝑖{0,1,2,,𝑁1}), for any 𝜀>0, there exists 𝑚𝑍(𝑚>1) such that 1|Γ(𝐴𝑖𝐴𝑖+1𝐴)|/|𝑖𝐴𝑖+1|<1+𝜀, for 𝑁>𝑚, therefore ||Γ𝐴0𝑖𝐴𝑖+1||||𝐴𝑖𝐴𝑖+1||||𝐴<𝜀𝑖𝐴𝑖+1||=𝜀||Γ𝑁||𝑁𝜀||||Γ(𝐴𝐵)𝑁.(2.51) Thus, |||||𝑁1𝑖=0𝑓𝐴𝑖||Γ𝐴𝑖𝐴𝑖+1||𝑁1𝑖=0𝑓𝐴𝑖||𝐴𝑖𝐴𝑖+1|||||||=|||||𝑁1𝑖=0𝑓𝐴𝑖||Γ𝐴𝑖𝐴𝑖+1||||𝐴𝑖𝐴𝑖+1|||||||𝑁1𝑖=0||𝑓𝐴𝑖||||Γ𝐴𝑖𝐴𝑖+1||||𝐴𝑖𝐴𝑖+1||𝑁1𝑖=0𝑀1𝜀||||Γ(𝐴𝐵)𝑁=𝑀1||||Γ(𝐴𝐵)𝜀.(2.52) It follows that lim𝑁𝑁1𝑖=0𝑓𝐴𝑖||Γ𝐴𝑖𝐴𝑖+1||=lim𝑁𝑁1𝑖=0𝑓𝐴𝑖||𝐴𝑖𝐴𝑖+1||.(2.53) In view of (2.50), we get 1𝑀(𝑓,Γ(𝐴𝐵))=||Γ||(𝐴𝐵)Γ1𝑓𝑑𝑠=||Γ||(𝐴𝐵)lim𝑁𝑁1𝑖=0𝑓𝐴𝑖||Γ𝐴𝑖𝐴𝑖+1||=1||||Γ(𝐴𝐵)lim𝑁𝑁1𝑖=0𝑓𝐴𝑖||𝐴𝑖𝐴𝑖+1||=lim𝑁1||||Γ(𝐴𝐵)𝑁1𝑖=0𝑓𝐴𝑖||Γ𝑁||𝑁=lim𝑁||Γ𝑁||||||Γ(𝐴𝐵)lim𝑁1𝑁𝑁1𝑖=0𝑓𝐴𝑖=lim𝑁1𝑁𝑁1𝑖=0𝑓𝐴𝑖.(2.54) This completes the proof of Lemma 2.9.

Lemma 2.10. Let Γ be a smooth closed curve in 𝑅3. If the points 𝐴,𝐵 move continuously on Γ and keep |Γ(𝐴𝐵)| invariable, where 0<𝑙=|Γ(𝐴𝐵)|<|Γ|/2, we obtain 𝑀[2]||||𝐴𝐵,Γ=1||Γ||Γ||||𝐴𝐵2||Γ||𝑑𝑠𝜋sin𝑙𝜋||Γ||.(2.55) A sufficient condition of equality is that Γ is a circle in 𝑅3.

Proof. By the theory of real number, there exists an increasing sequence of positive integers {𝑘𝑛} such that lim𝑛𝑘𝑛10𝑛=𝑙||Γ||.(2.56) Write 𝑘=𝑘𝑛,𝑁=10𝑛. Consider a partition of Γ by means of 𝑁(𝑁10) points 𝐴1,𝐴2,,𝐴𝑖1,𝐴𝑖,,𝐴𝑁 such that |𝐴1𝐴2𝐴|=|2𝐴3𝐴|==|𝑁1𝐴𝑁𝐴|=|𝑁𝐴1|=|Γ𝑁|/𝑁, where Γ𝑁𝐴1𝐴2𝐴𝑁1𝐴𝑁𝐴1 is a polygon with 𝑁 sides.
First, we give the following fact: if the point 𝐴 moves to some point 𝐴𝑖 along Γ, the point 𝐵 moves to 𝐴𝑖 along Γ, in other words, 𝐴=𝐴𝑖Γ,𝐵=𝐴𝑖Γ,|Γ(𝐴𝑖𝐴𝑖)|=𝑙, thenlim𝑛|||𝐴𝑖𝐴𝑖+𝑘|||=0.(2.57) In fact, from lim𝑛Γ𝑁(𝐴𝑖𝐴𝑖+𝑘)Γ,|Γ(𝐴𝑖𝐴𝑖)|=𝑙, we only need to prove lim𝑛||Γ𝑁𝐴𝑖𝐴𝑖+𝑘||=𝑙.(2.58) Since lim𝑛||Γ𝑁𝐴𝑖𝐴𝑖+𝑘||=lim𝑛𝑘||𝐴𝑖𝐴𝑖+1||=lim𝑛𝑘||Γ𝑁||𝑁=lim𝑛||Γ𝑁||lim𝑛𝑘𝑁=||Γ||𝑙||Γ||=𝑙,(2.59)(2.58) holds, it follows that (2.57) holds.
From (2.57), we know for any 𝜀>0, there exists 𝑚𝑍(𝑚>1), when 𝑁>𝑚, such that|||||||𝐴𝑖𝐴𝑖|||2||𝐴𝑖𝐴𝑖+𝑘||2|||||||||1<𝜀,𝑁𝑁𝑖=1|||𝐴𝑖𝐴𝑖|||21𝑁𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2|||||=1𝑁|||||𝑁𝑖=1|||𝐴𝑖𝐴𝑖|||2||𝐴𝑖𝐴𝑖+𝑘||2|||||1𝑁𝑁𝑖=1|||||||𝐴𝑖𝐴𝑖|||2||𝐴𝑖𝐴𝑖+𝑘||2||||<1𝑁𝑁𝑖=1𝜀=𝜀.(2.60) By Lemma 2.9, we get 1||Γ||Γ||||𝐴𝐵21𝑑𝑠=||Γ||Γ||||𝐴𝐵2𝑑𝑠𝐴=lim𝑛1𝑁𝑁𝑖=1|||𝐴𝑖𝐴𝑖|||2=lim𝑛1𝑁𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2.(2.61) It follows from inequality (2.40) that 1𝑁𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||21𝑁sin(𝑘𝜋/𝑁)sin(𝜋/𝑁)2𝑁𝑖=1||𝐴𝑖𝐴𝑖+1||2=1𝑁sin(𝑘𝜋/𝑁)sin(𝜋/𝑁)2𝑁𝑖=1||Γ𝑁||𝑁2=1𝑁sin(𝑘𝜋/𝑁)sin(𝜋/𝑁)2||Γ𝑁||2𝑁=sin(𝑘𝜋/𝑁)𝑁sin(𝜋/𝑁)2||Γ𝑁||2.(2.62) It follows from (2.61) and (2.62) that 1||Γ||Γ||||𝐴𝐵2𝑑𝑠=lim𝑛1𝑁𝑁𝑖=1||𝐴𝑖𝐴𝑖+𝑘||2lim𝑛sin(𝑘𝜋/𝑁)𝑁sin(𝜋/𝑁)2||Γ𝑁||2=sinlim𝑛(𝑘𝜋/𝑁)lim𝑛𝑁sin(𝜋/𝑁)2lim𝑛||Γ𝑁||2=||Γ||sin𝑙𝜋/𝜋2||Γ||2=||Γ||𝜋sin𝑙𝜋||Γ||2.(2.63) Namely, (2.55) holds. From Lemma 2.8, a sufficient condition of equality of (2.55) is that Γ is a circle in 𝑅3. This completes the proof.

Lemma 2.11 (Cauchy inequality [1, page 6]). Let Γ be a smooth closed curve in 𝑅3. If the function 𝑓Γ𝑅 and 𝑔Γ𝑅 are smooth, we have the inequality as follows: ||||Γ||||𝑓𝑔𝑑𝑠Γ𝑓2𝑑𝑠Γ𝑔2𝑑𝑠.(2.64)

Lemma 2.12. Assume Γ is a closed Jordan curve in 𝑅3 and 𝑂𝑅3, then 𝐷(Γ)Ω[𝑂,Γ].

Proof. Let 𝐴,𝐵𝑅3,𝐴𝐵, 𝐴𝐵 denote the straight line through the points 𝐴 and 𝐵, and let 𝐴𝐵 denote the oriented line segment with the initial point 𝐴 and terminal point 𝐵.
For any 𝑃𝐷(Γ), if 𝑃=𝑂, then 𝑃Ω[𝑂,Γ]; if 𝑃𝑂, then 𝑂𝑃𝐷(Γ)=𝑖1(mod2),1𝑖𝑘1𝑄𝑖𝑄𝑖+1,𝑄𝑖Γ,𝑖=1,2,,𝑘,𝑘2,𝑃𝑂𝑃𝐷(Γ)𝑖𝑖1(mod2),1𝑖𝑘1,𝑃𝑄𝑖𝑄𝑖+1.(2.65) If 𝑂𝑄𝑖𝑄𝑖+1, then 𝑃𝑄𝑖𝑄𝑖+1=𝑄𝑖𝑂𝑂𝑄𝑖+1𝑃𝑄𝑖[]𝑂Ω𝑂,Γ𝑃𝑂𝑄𝑖+1[][].Ω𝑂,Γ𝑃Ω𝑂,Γ(2.66) If 𝑂𝑄𝑖𝑄𝑖+1, without loss of generality, we set |𝑂𝑄𝑖||𝑂𝑄𝑖+1|, then 𝑃𝑄𝑖𝑄𝑖+1𝑂𝑄𝑖+1[][].Ω𝑂,Γ𝑃Ω𝑂,Γ(2.67) In all, for any 𝑃𝐷(Γ)𝑃Ω[𝑂,Γ], hence, 𝐷(Γ)Ω[𝑂,Γ]. This completes the proof.

Lemma 2.13. For 𝑆{𝑂,Γ,Γ𝑁}, we have the following inequality: 𝑀𝑁[2]1(𝑟)=𝑁𝑁𝑖=1𝑟𝑖21/212