Sharp lower bounds involving circuit layout system

Abstract

The circuit layout system $CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$ in a Euclidean space is defined. By means of algebraic,analytic, geometric and inequality theories, we obtain several sharp lowerbounds involving the circuit layout system.

MSC: 51K05, 26D15.

1 Introduction

We first introduce a circuit layout problem as follows. Let Γ be a rectangular(or polygon) courtyard (or street). Five light poles (or street lamp), with a fixedminimal distance apart from each other, are proposed to be erected on the boundaryof Γ, and straight underground pipes are planned to connect these poles (seeFigure 1). Assuming that the major cost of theconstruction project is the price of the pipes, it is then important to find out theminimal total lengths required for the project, its purpose is toestimate the installation costs.

Figure 1

The graph of $\mathbf{CLS}{\mathbf{\{}{\mathbf{\Gamma }}_{\mathbf{4}}\mathbf{(}\mathbf{A}\mathbf{)}\mathbf{,}{\mathbf{\Gamma }}_{\mathbf{5}}\mathbf{(}{\mathbf{A}}^{\mathbf{\ast }}\mathbf{)}\mathbf{,}\mathbf{2}\mathbf{\}}}_{{\mathbb{R}}^{\mathbf{2}}}$ .

We can easily illustrate this problem by means of Figure 1in a later Example 4.3 in which the corners of the courtyard Γ areindicated by the points $A_1$, $A_2$, $A_3$ and $A_4$, while the light poles are indicated by$A_1^{\ast }$, $A_2^{\ast }$, $A_3^{\ast }$, $A_4^{\ast }$ and $A_5^{\ast }$, respectively. The light poles are kept apart fromeach other for clear reasons so that we may assume the distances

$$\parallel A_2^{\ast }-A_1^{\ast }\parallel ,\parallel A_3^{\ast }-A_2^{\ast }\parallel ,\parallel A_4^{\ast }-A_3^{\ast }\parallel ,\parallel A_5^{\ast }-A_4^{\ast }\parallel ,\parallel A_5^{\ast }-A_1^{\ast }\parallel ⩾\delta >0.$$

We need to find among all possible locations of $A_1^{\ast },\dots ,A_5^{\ast }$ such that the total length

$$\parallel A_2^{\ast }-A_1^{\ast }\parallel +\parallel A_3^{\ast }-A_2^{\ast }\parallel +\parallel A_4^{\ast }-A_3^{\ast }\parallel +\parallel A_5^{\ast }-A_4^{\ast }\parallel +\parallel A_5^{\ast }-A_1^{\ast }\parallel$$

is the minimal one.

The above problem can easily be generalized. To this end, we need to recall somebasic concepts as follows.

Let be a Euclideanspace, and $\alpha ,\beta \in \mathbb{E}$. The inner product of α andβ is denoted by $〈\alpha ,\beta 〉$ and the norm of α is denoted by$\parallel \alpha \parallel$. The dimension $dim\mathbb{E}$ of satisfies$dim\mathbb{E}\ge n$ if and only if there exist n linearlyindependent vectors ${\mathit{\epsilon }}_1,{\mathit{\epsilon }}_2,\dots ,{\mathit{\epsilon }}_n$ in (see [1]).

Let B, C be points in , the closed, openand closed-open segments joining them will respectively be denoted by

$$[BC],(BC),[BC)\text{and}(BC]$$

and defined as usual by

$$\begin{array}{c}\{{\chi }_{B,C}(t)\mid t\in [0,1]\},\{{\chi }_{B,C}(t)\mid t\in (0,1)\},\hfill \\ \{{\chi }_{B,C}(t)|t\in [0,1)\}\text{and}\{{\chi }_{B,C}(t)\mid t\in (0,1]\},\hfill \end{array}$$

where

$${\chi }_{B,C}(t):=(1-t)B+tC.$$

Let $dim\mathbb{E}⩾2$, $\mathbf{A}=(A_1,A_2,\dots ,A_n)\in {\mathbb{E}}^n$, where

$$A_i\ne A_{i+1},i=1,2,\dots ,n,n⩾3,$$

be a sequence of points in and

$$A_i=A_j\iff i\equiv j(modn),i,j=0,\pm 1,\pm 2,\dots .$$

We call the set

$${\mathrm{\Gamma }}_n(\mathbf{A})=\bigcup _{i=1}^n[A_i{A}_{i+1})$$

an n-polygon, or a polygon if no confusion is caused. The angle^aat $A_i$ and the angle ∠A are defined as

$$\mathrm{\angle }{A}_i:=\mathrm{\angle }(A_i-A_{i-1},A_{i+1}-A_i),i=1,2,\dots ,n\text{and}\mathrm{\angle }A:=\underset{1⩽i⩽n}{min}\{\mathrm{\angle }{A}_i\}.$$

In case each $\mathrm{\angle }{A}_i$ is the same, we say that our polygon is equiangular.We will also denote the total length (or perimeter) of an n-gon by

$$|{\mathrm{\Gamma }}_n(\mathbf{A})|=\sum _{i=1}^n{a}_i=\sum _{i=1}^n\parallel A_{i+1}-A_i\parallel ,$$

where, and in the future,

$$a_i:=\parallel A_{i+1}-A_i\parallel ,i=1,2,\dots ,n.$$

Now we give the definition of the circuit layout system in a Euclidean space asfollows.

Definition 1.1 Let ${\mathrm{\Gamma }}_n(\mathbf{A})$ and ${\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })$, where $N⩾n⩾3$, be two polygons in with the dimension$dim\mathbb{E}⩾2$. We say that the set

$$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right),\delta \}}_{\mathbb{E}}:=\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right),\delta \}$$

is a circuit layout system (or CLS for short) if the following conditions aresatisfied:

(H1.1) $\mathrm{\angle }{A}_i\in (0,\pi )$, $i=1,2,\dots ,n$.

(H1.2) $A_j^{\ast }\in {\mathrm{\Gamma }}_n(\mathbf{A})$ for $j\in \{1,2,\dots ,N\}$ and $A_1^{\ast }\in [A_1{A}_2)$.

(H1.3) If $A_j^{\ast },A_{j+1}^{\ast }\in [A_i{A}_{i+1})$, then $A_{j+1}^{\ast }\in (A_j^{\ast }{A}_{i+1})$ for $i=1,2,\dots ,n$ and $j=1,2,\dots ,N$.

(H1.4) If $A_j^{\ast }\in [A_i{A}_{i+1})$ and $A_k^{\ast }\in [A_{i+1}{A}_{i+2})$ for $j,k\in \{1,2,\dots ,N\}$ and $i\in \{1,2,\dots ,n\}$, then $j<k$.

(H1.5) For any $i\in \{1,2,\dots ,n\}$, there exists $j\in \{1,2,\dots ,N\}$ such that $A_j^{\ast }\in [A_i{A}_{i+1})$.

(H1.6) For any $j\in \{1,2,\dots ,N\}$, there is $\delta >0$ such that

$$\parallel A_{j+1}^{\ast }-A_j^{\ast }\parallel ⩾\delta .$$

In this paper, we are concerned with the sharp lower bound (see [2–5]) of $|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|$, its purpose is to estimate the installation costs ofthe circuit layout problem. In other words, we will mainly be concerned with thefollowing problem.

Problem 1.1 (Circuit layout problem)

Let $CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$ be a CLS. How can we determine the lower bound of$|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|$ by means of n, N, δand ${\mathrm{\Gamma }}_n(\mathbf{A})$?

In this paper, by means of algebraic, analytic, geometric and inequality theories,several sharp lower bounds of $|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|$ in Problem 1.1 are obtained. As applications of ourresults, in Section 4, we calculate that $inf\{|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|\}$ for the special circuit layout system$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{{\mathbb{R}}^2}$ by means of three effective examples.

2 Preliminaries

We provide in this section some basic terminologies and results which are necessaryfor the investigation of Problem 1.1.

We first recall the concept of parallel vectors for later use. Two vectors xand y in are said to be inthe same (opposite) direction if (i) $\mathbf{x}=0$ or $\mathbf{y}=0$, or (ii) $\mathbf{x}\ne 0$ and $\mathbf{y}\ne 0$ and x is a positive (respectively negative)constant multiple of y. Two vectors x and y in the same(opposite) direction are indicated by $\mathbf{x}\uparrow \mathbf{y}$ (respectively $\mathbf{x}\downarrow \mathbf{y}$) (see [1]).

Next, we set that

$$\mathbf{x}={(x_1,x_2,\dots ,x_n)}^T,{\parallel \mathbf{x}\parallel }_p={(|x_1{|}^p+|x_2{|}^p+\cdots +|x_n{|}^p)}^{1/p},$$

where $p\in (1,\mathrm{\infty })$.

In order to study Problem 1.1, we need six lemmas as follows.

Lemma 2.1 (Minkowski’s inequality [6])

If$\mathbf{x},\mathbf{y}\in {\mathbb{R}}^n$and$p\in (1,\mathrm{\infty })$, then

$${\parallel \mathbf{x}+\mathbf{y}\parallel }_p⩽{\parallel \mathbf{x}\parallel }_p+{\parallel \mathbf{y}\parallel }_p.$$

(1)

Furthermore, the equality holds if and only if$\mathbf{x}\uparrow \mathbf{y}$.

According to Lemma 2.1 and the algebraic theory, we easily get the followinglemma.

Lemma 2.2 (Minkowski-type inequality [6])

Let$A\in {\mathbb{R}}^{n\times n}$. If$A^T=A$, $A⩾0$and$f(\mathbf{x})={\mathbf{x}}^{T}A\mathbf{x}$, then for any$\mathbf{x},\mathbf{y}\in {\mathbb{R}}^n$, we have

$$\sqrt{f(\mathbf{x}+\mathbf{y})}⩽\sqrt{f(\mathbf{x})}+\sqrt{f(\mathbf{y})}.$$

(2)

Furthermore, if$A>0$, then the equality in (2) holds if andonly if$\mathbf{x}\uparrow \mathbf{y}$.

Lemma 2.3 Let the function $\phi :[0,c]\to (0,\mathrm{\infty })$ be defined by

$$\phi (u)=\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }+u,$$

where$c,u,y\in [0,\mathrm{\infty })$, $\theta \in (0,\pi )$. Then the function φ is nondecreasing. If in addition, $y>0$, then φ is increasing.

Proof Note that

$$\begin{array}{rl}\frac{\partial \phi }{\partial u}& =-\frac{(c-u)-cos\theta y}{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }}+1\\ =\frac{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }-[(c-u)-cos\theta y]}{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }}\\ ⩾\frac{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }-|(c-u)-cos\theta y|}{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }}\\ =\frac{{[{(c-u)}^2+y^2-2(c-u)ycos\theta ]}^{-1/2}{sin}^2\theta y^2}{\sqrt{{(c-u)}^2+y^2-2(c-u)ycos\theta }+|(c-u)-cos\theta y|}\\ ⩾0.\end{array}$$

Thus, $\phi :[0,c]\to (0,\mathrm{\infty })$ is nondecreasing. Furthermore, if$y>0$, then $\phi :[0,c]\to (0,\mathrm{\infty })$ is a strictly increasing function. This ends theproof. □

Lemma 2.4 Let$B,C\in \mathbb{E}$. If$B\ne C$and$D\in [BC]$, then

$$\parallel C-B\parallel =\parallel C-D\parallel +\parallel D-B\parallel .$$

(3)

The result of Lemma 2.4 is well known.

By our assumptions (H1.2)-(H1.5), we may easily get the following result.

Lemma 2.5 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS. Then, for any$i\in \{1,2,\dots ,n\}$, there exist$\sigma (i)\in \{1,2,\dots ,N\}$and$\tau (i)\in \{0,1,2,\dots ,N-n\}$such that

$$A_{\sigma (i)+k}^{\ast }\in [A_i{A}_{i+1}),k=0,1,2,\dots ,\tau (i).$$

Furthermore,

$$\sum _{i=1}^n\tau (i)=N-n.$$

(4)

Lemma 2.6 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS. If the infimum of$|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|$can be attained, then for any

$$i\in \{1,2,\dots ,n\},k\in \{1,2,\dots ,\tau (i)\},\tau (i)⩾1,$$

we have

$$\parallel A_{\sigma (i)+k}^{\ast }-A_{\sigma (i)+k-1}^{\ast }\parallel =\delta ,$$

(5)

where$\sigma (i)$and$\tau (i)$are defined in Lemma  2.5.

Proof Suppose to the contrary that there exist $i\in \{1,2,\dots ,n\}$ and $k\in \{1,2,\dots ,\tau (i)\}$ such that

$$\parallel A_{\sigma (i)+k}^{\ast }-A_{\sigma (i)+k-1}^{\ast }\parallel >\delta .$$

(6)

We construct a new $CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast \ast }),\delta \}}_{\mathbb{E}}$ as follows: If

$$j\ne \sigma (i)+k,k=0,1,\dots ,\tau (i)-1,$$

then $A_j^{\ast \ast }=A_j^{\ast }$. If there exists $k\in \{0,1,\dots ,\tau (i)-1\}$ such that $j=\sigma (i)+k$, then

$$A_j^{\ast \ast }\in \left[A_i{A}_{\sigma (i)+\tau (i)}^{\ast }\right)$$

and

$$\parallel A_{\sigma (i)+1}^{\ast \ast }-A_{\sigma (i)}^{\ast \ast }\parallel =\parallel A_{\sigma (i)+2}^{\ast \ast }-A_{\sigma (i)+1}^{\ast \ast }\parallel =\cdots =\parallel A_{\sigma (i)+\tau (i)}^{\ast \ast }-A_{\sigma (i)+\tau (i)-1}^{\ast \ast }\parallel =\delta .$$

(7)

Now fix $A_{\sigma (i)+\tau (i)}^{\ast }\in [A_i{A}_{i+1})$. Denote

$$(c_i,u_i,y_i):=(\parallel A_{\sigma (i)+\tau (i)}^{\ast }-A_i\parallel ,\parallel A_{\sigma (i)+\tau (i)}^{\ast }-A_{\sigma (i)}^{\ast }\parallel ,\parallel A_{\sigma (i-1)+\tau (i-1)}^{\ast }-A_i\parallel ).$$

(8)

Without loss of generality, we can assume that $y_i>0$, $i=1,2,\dots ,n$. By condition (H1.3) and Lemma 2.4, we obtainthat

$$\parallel A_{\sigma (i)}^{\ast }-A_i\parallel =c_i-u_i,c_i⩾u_i=\parallel A_{\sigma (i)+\tau (i)}^{\ast }-A_{\sigma (i)}^{\ast }\parallel =\sum _{k=1}^{\tau (i)}\parallel A_{\sigma (i)+k}^{\ast }-A_{\sigma (i)+k-1}^{\ast }\parallel .$$

(9)

Since

$$\parallel \mathit{\alpha }-\mathit{\beta }\parallel =\sqrt{{\parallel \mathit{\alpha }\parallel }^2+{\parallel \mathit{\beta }\parallel }^2-2\parallel \mathit{\alpha }\parallel \cdot \parallel \mathit{\beta }\parallel cos\mathrm{\angle }(\mathit{\alpha },\mathit{\beta })},\mathrm{\forall }\mathit{\alpha },\mathit{\beta }\in \mathbb{E},$$

we see that

$$\begin{array}{rl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& =\sum _{j=1}^n(\parallel A_{\sigma (j)}^{\ast }-A_{\sigma (j-1)+\tau (j-1)}^{\ast }\parallel +\parallel A_{\sigma (j)+\tau (j)}^{\ast }-A_{\sigma (j)}^{\ast }\parallel )\\ =\sum _{j=1}^n[\parallel (A_{\sigma (j)}^{\ast }-A_i)-(A_{\sigma (j-1)+\tau (j-1)}^{\ast }-A_j)\parallel +\parallel A_{\sigma (j)+\tau (j)}^{\ast }-A_{\sigma (j)}^{\ast }\parallel ]\\ =\sum _{j=1}^n[\sqrt{{(c_j-u_j)}^2+y_j^2-2(c_j-u_j)y_{j}cos\mathrm{\angle }{A}_j}+u_j]\\ =\sum _{j=1}^n{\phi }_j(u_j)\\ =\sum _{1⩽j⩽n,j\ne i}{\phi }_j(u_j)+{\phi }_i(u_i),\end{array}$$

(10)

where

$${\phi }_j(u_j)=\sqrt{{(c_j-u_j)}^2+y_j^2-2(c_j-u_j)y_{j}cos\mathrm{\angle }{A}_j}+u_j,j=1,2,\dots ,n.$$

By condition (H1.6), (6) and (9), we see that

$$u_i=\sum _{k=1}^{\tau (i)}\parallel A_{\sigma (i)+k}^{\ast }-A_{\sigma (i)+k-1}^{\ast }\parallel >\tau (i)\delta .$$

(11)

According to Lemma 2.3, the function ${\phi }_i:[0,c_i]\to (0,\mathrm{\infty })$ is increasing. Thus, by (10) and (11), we have

$$\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|=\sum _{1⩽j⩽n,j\ne i}{\phi }_j(u_j)+{\phi }_i(u_i)>\sum _{1⩽j⩽n,j\ne i}{\phi }_j(u_j)+{\phi }_i(\tau (i)\delta )=\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast \ast }\right)\right|.$$

This is contrary to the minimality of $|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|$. The proof is completed. □

3 Study of Problem 1.1

3.1 The case where n is an odd number

We first study the case of Problem 1.1 where n is an odd number. In thissituation we have the following result.

Theorem 3.1 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS and n is an odd number. Then we have the following inequality:

$$\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|⩾|{\mathrm{\Gamma }}_n(\mathbf{A})|sin\frac{\mathrm{\angle }A}{2}+(1-sin\frac{\mathrm{\angle }A}{2})(N-n)\delta .$$

(12)

Proof We construct another $CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast \ast }),\delta \}}_{\mathbb{E}}$ such that (7) holds for any$i\in \{1,2,\dots ,n\}$. Set

$$\parallel A_{\sigma (j-1)+\tau (j-1)}^{\ast \ast }-A_j\parallel =y_j,\parallel A_{\sigma (j)}^{\ast \ast }-A_j\parallel =x_j.$$

By equality (7) and Lemma 2.4, we see that

$$\begin{array}{rl}{y}_{j+1}& =\parallel A_{j+1}-A_{\sigma (j)+\tau (j)}^{\ast \ast }\parallel \\ =\parallel (A_{j+1}-A_j)-(A_{\sigma (j)+\tau (j)}^{\ast \ast }-A_j)\parallel \\ =\parallel A_{j+1}-A_j\parallel -\parallel A_{\sigma (j)+\tau (j)}^{\ast \ast }-A_j\parallel \\ =a_j-x_j-\tau (j)\delta ,\end{array}$$

therefore,

$$x_j+y_{j+1}=a_j-\tau (j)\delta ,j\in \{1,2,\dots ,n\}.$$

(13)

Since

$$\sum _{i=1}^n{x}_{i+1}=\sum _{i=1}^n{x}_i,\sum _{i=1}^n{y}_{i+1}=\sum _{i=1}^n{y}_i,$$

by (13) and (14) we obtain that

$$\sum _{j=1}^n{x}_j+\sum _{j=1}^n{y}_j=\sum _{j=1}^n{a}_j-\delta \sum _{j=1}^n\tau (j)=|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n).$$

(14)

In view of Lemma 2.6 and equality (10), we see that

$$\begin{array}{rl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast \ast }\right)\right|\\ =\sum _{j=1}^n(\parallel A_{\sigma (j)}^{\ast \ast }-A_{\sigma (j-1)+\tau (j-1)}^{\ast \ast }\parallel +\parallel A_{\sigma (j)+\tau (j)}^{\ast \ast }-A_{\sigma (j)}^{\ast \ast }\parallel )\\ =\sum _{j=1}^n[\parallel (A_{\sigma (j)}^{\ast \ast }-A_i)-(A_{\sigma (j-1)+\tau (j-1)}^{\ast \ast }-A_j)\parallel +\parallel A_{\sigma (j)+\tau (j)}^{\ast \ast }-A_{\sigma (j)}^{\ast \ast }\parallel ]\\ =\sum _{j=1}^n[\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }{A}_j}+\tau (j)\delta ]\\ ⩾\sum _{j=1}^n[\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}+\tau (j)\delta ]\\ =\sum _{j=1}^n\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}+\sum _{j=1}^n\tau (j)\delta \\ =\sum _{j=1}^n\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}+\delta (N-n)\\ =\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\delta (N-n),\end{array}$$

i.e.,

$$\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|⩾\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\delta (N-n),$$

(15)

where

$$f{(x_j,y_j)}^T:=x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A=(x_j,y_j)\left[\begin{array}{cc}1& -cos\mathrm{\angle }A\\ -cos\mathrm{\angle }A& 1\end{array}\right]\left(\begin{array}{c}{x}_j\\ y_j\end{array}\right).$$

Note that condition (H1.1) implies $0<\mathrm{\angle }A<\pi$, thus, where the matrix is positive definite.According to inequality (15), Lemma 2.2 and equality (14), we obtain that

$$\begin{array}{rl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\delta (N-n)\\ =\frac{1}{2}[\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}]+\delta (N-n)\\ =\frac{1}{2}[\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\sum _{j=1}^n\sqrt{f{(y_j,x_j)}^T}]+\delta (N-n)\\ ⩾\frac{1}{2}\sqrt{f\left\{\sum _{j=1}^n[{(x_j,y_j)}^T+{(y_j,x_j)}^T]\right\}}+\delta (N-n)\\ =\frac{1}{2}\sqrt{f\left\{\left[\sum _{j=1}^n{(x_j+y_j,y_j+x_j)}^T\right]\right\}}+\delta (N-n)\\ =\frac{1}{2}\sqrt{f\left\{{[\sum _{j=1}^n(x_j+y_j),\sum _{j=1}^n(y_j+x_j)]}^T\right\}}+\delta (N-n)\\ =\frac{1}{2}\sqrt{f\left[(\sum _{j=1}^n{x}_j+\sum _{j=1}^n{y}_j){(1,1)}^T\right]}+\delta (N-n)\\ =\frac{1}{2}(\sum _{j=1}^n{x}_j+\sum _{j=1}^n{y}_j)\sqrt{f{(1,1)}^T}+\delta (N-n)\\ =\frac{1}{2}\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]\sqrt{2-2cos\mathrm{\angle }A}+\delta (N-n)\\ =\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]sin\frac{\mathrm{\angle }A}{2}+\delta (N-n)\\ =|{\mathrm{\Gamma }}_n(\mathbf{A})|sin\frac{\mathrm{\angle }A}{2}+(1-sin\frac{\mathrm{\angle }A}{2})(N-n)\delta .\end{array}$$

This means that inequality (12) holds.

In addition, from the above analysis we may easily see that the equality in (12)holds if

(H3.1) ${\mathrm{\Gamma }}_n(\mathbf{A})$ is an equiangular n-gon;

(H3.2) for any $i\in \{1,2,\dots ,n\}$, $\tau (i)⩾1$, equality (5) holds;

(H3.3) equality (4) holds; and

(H3.4) there exist $x,y\in {[0,\mathrm{\infty })}^n$ such that

$$\{\begin{array}{ll}{(x_j,y_j)}^T\uparrow {(x_1,y_1)}^T\uparrow {(y_j,x_j)}^T,& j=1,2,\dots ,n,\\ x_j+y_{j+1}=a_j-\tau (j)\delta ,& j=1,2,\dots ,n,\\ \sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}⩾\delta ,& j=1,2,\dots ,n,\end{array}$$

where

$${(x_i,y_i)}^T={(x_j,y_j)}^T\iff i\equiv j(modn),i,j=0,\pm 1,\pm 2,\dots .$$

The proof is completed. □

3.2 The case where n is an even number

Now, we consider the case of Problem 1.1 where n is an even number.

Theorem 3.2 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS, n is even, and let

$$\sum _{i=1}^n{(-1)}^{j+1}{a}_j⩾0.$$

Then we have the following two assertions:

• (I) If

  $$\delta (N-n)>\sum _{j=1}^n{(-1)}^{j+1}{a}_j,$$

then we have

$$\begin{array}{rcl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{min}^2\{\{\omega \},1-\{\omega \}\}\}}^{1/2}+\delta (N-n),\end{array}$$

(16)

where

$$\omega =\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta },\{\omega \}=\omega -[\omega ]\in [0,1),$$

and$[\omega ]$is the Gaussian function.

• (II) If

  $$\delta (N-n)⩽\sum _{j=1}^n{(-1)}^{j+1}{a}_j,$$

then we have

$$\begin{array}{rcl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+{cos}^2\frac{\mathrm{\angle }A}{2}{[\sum _{j=1}^n{(-1)}^{j+1}{a}_j-\delta (N-n)]}^2\}}^{1/2}+\delta (N-n).\end{array}$$

(17)

Proof First we consider the case where

$$\delta (N-n)>\sum _{j=1}^n{(-1)}^{j+1}{a}_j.$$

We show that for any $i\in \{1,2,\dots ,n\}$, equality (7) holds by constructing a new$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast \ast }),\delta \}}_{\mathbb{E}}$. Since n is an even number, by the proofof Theorem 3.1 and (13), we see that

$$\sum _{j=1}^{n/2}(x_{2j-1}+y_{2j})=\sum _{j=1}^{n/2}{a}_{2j-1}-\delta \sum _{i=1}^{n/2}\tau (2j-1)$$

(18)

and

$$\begin{array}{rl}\sum _{j=1}^{n/2}(y_{2j-1}+x_{2j})& =(y_1+x_2)+(y_3+x_4)+\cdots +(y_{n-1}+x_n)\\ =(x_2+y_3)+(x_4+y_5)+\cdots +(x_{n-2}+y_{n-1})+(x_n+y_1)\\ =\sum _{j=1}^{n/2}(x_{2j}+y_{2j+1})\\ =\sum _{j=1}^{n/2}{a}_{2j}-\delta \sum _{i=1}^{n/2}\tau (2j).\end{array}$$

(19)

Inequality (15) is still valid where

$$f{(x_j,y_j)}^T:=x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A=(x_j,y_j)\left[\begin{array}{cc}1& -cos\mathrm{\angle }A\\ -cos\mathrm{\angle }A& 1\end{array}\right]\left(\begin{array}{c}{x}_j\\ y_j\end{array}\right)$$

is a positive definite quadratic function. By means of inequality (15),Lemma 2.2, (18), (19) and (4), we then obtain that

$$\begin{array}{rl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast \ast }\right)\right|\\ ⩾\sum _{j=1}^n\sqrt{f{(x_j,y_j)}^T}+\delta (N-n)\\ =\sum _{j=1}^{n/2}\sqrt{f{(x_{2j-1},y_{2j-1})}^T}+\sum _{j=1}^{n/2}\sqrt{f{(x_{2j},y_{2j})}^T}+\delta (N-n)\\ =\sum _{j=1}^{n/2}\sqrt{f{(x_{2j-1},y_{2j-1})}^T}+\sum _{j=1}^{n/2}\sqrt{f{(y_{2j},x_{2j})}^T}+\delta (N-n)\\ ⩾\sqrt{f[\sum _{j=1}^{n/2}{(x_{2j-1},y_{2j-1})}^T+\sum _{j=1}^{n/2}{(y_{2j},x_{2j})}^T]}+\delta (N-n)\\ =\sqrt{f{[\sum _{j=1}^{n/2}(x_{2j-1}+y_{2j}),\sum _{j=1}^{n/2}(y_{2j-1}+x_{2j})]}^T}+\delta (N-n)\\ =\sqrt{f{[\sum _{j=1}^{n/2}{a}_{2j-1}-\delta \sum _{i=1}^{n/2}\tau (2j-1),\sum _{j=1}^{n/2}{a}_{2j}-\delta \sum _{i=1}^{n/2}\tau (2j)]}^T}+\delta (N-n)\\ =\sqrt{f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T}+\delta (N-n),\end{array}$$

i.e.,

$$\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|⩾\sqrt{f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T}+\delta (N-n),$$

(20)

where

$$P_n:=\sum _{j=1}^{n/2}{a}_{2j-1}\in (0,\mathrm{\infty }),Q_n:=\sum _{j=1}^{n/2}{a}_{2j}\in (0,\mathrm{\infty }),P_n⩾Q_n$$

and

$$\tau :=\sum _{i=1}^{n/2}\tau (2j-1)\in \{0,1,2,\dots ,N-n\}.$$

Note that

$$\begin{array}{c}f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T\hfill \\ ={(P_n-\delta \tau )}^2+{[Q_n-\delta (N-n-\tau )]}^2-2(P_n-\delta \tau )[Q_n-\delta (N-n-\tau )]cos\mathrm{\angle }A\hfill \\ ={[P_n+Q_n-\delta (N-n)]}^2-2(1+cos\mathrm{\angle }A)(P_n-\delta \tau )[Q_n-\delta (N-n-\tau )]\hfill \\ ={[P_n+Q_n-\delta (N-n)]}^2-4{cos}^2\frac{\mathrm{\angle }A}{2}(P_n-\delta \tau )[Q_n-\delta (N-n-\tau )]\hfill \\ ={sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{[\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta }-\tau ]}^2\hfill \\ ={sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{(\{\omega \}+[\omega ]-\tau )}^2,\hfill \end{array}$$

i.e.,

$$\begin{array}{c}f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T\hfill \\ ={sin}^2\frac{\mathrm{\angle }A}{2}{[|{\mathrm{\Gamma }}_n|-\delta (N-n)]}^2+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{(\{\omega \}+[\omega ]-\tau )}^2,\hfill \end{array}$$

(21)

where

$$\omega =\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta }.$$

(22)

Set

$${\tau }_0=\{\begin{array}{ll}[\omega ]& \text{if }0⩽\{\omega \}⩽\frac{1}{2},\\ [\omega ]+1& \text{if }\frac{1}{2}<\{\omega \}<1.\end{array}$$

(23)

Since

$$\delta (N-n)>\sum _{j=1}^n{(-1)}^{j+1}{a}_j,\tau \in \{0,1,\dots ,N-n\},$$

we see that

$$\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta }-(N-n)⩽\{\omega \}+[\omega ]-\tau ⩽\{\omega \}+[\omega ].$$

(24)

Thus, if

$$0⩽\{\omega \}⩽\frac{1}{2},$$

then

$$\begin{array}{c}f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T\hfill \\ ⩾{sin}^2\frac{\mathrm{\angle }A}{2}{[|{\mathrm{\Gamma }}_n|-\delta (N-n)]}^2+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{\{\omega \}}^2,\hfill \end{array}$$

(25)

where equality holds if and only if $\tau =[\omega ]={\tau }_0$, and if

$$\frac{1}{2}<\{\omega \}<1,$$

then

$$\begin{array}{c}f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T\hfill \\ ⩾{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{(1-\{\omega \})}^2.\hfill \end{array}$$

(26)

The equalities in (25) and (26) hold if and only if $\tau =[\omega ]+1={\tau }_0$. From (20), (21), (25) and (26), we see that

$$\begin{array}{rcl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾& \sqrt{f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T}+\delta (N-n)\\ ⩾& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{min}^2\{\{\omega \},1-\{\omega \}\}\}}^{1/2}+\delta (N-n).\end{array}$$

Thus, inequality (16) is proved.

Second, we consider the case

$$\delta (N-n)⩽\sum _{j=1}^n{(-1)}^{j+1}{a}_j.$$

Since

$$\delta (N-n)⩽\sum _{j=1}^n{(-1)}^{j+1}{a}_j,\tau \in \{0,1,\dots ,N-n\},$$

we see that

$$0⩽\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta }-(N-n)⩽\{\omega \}+[\omega ]-\tau ⩽\{\omega \}+[\omega ]$$

(27)

and

$$\begin{array}{rl}{(\{\omega \}+[\omega ]-\tau )}^2& ⩾{[\frac{{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j+\delta (N-n)}{2\delta }-(N-n)]}^2\\ =\frac{1}{4{\delta }^2}{[\sum _{j=1}^n{(-1)}^{j+1}{a}_j-\delta (N-n)]}^2,\end{array}$$

(28)

where equality holds in (28) if and only if $\tau =N-n$. By (20), (21) and (28), we have

$$\begin{array}{rcl}\left|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\right|& ⩾& \sqrt{f{[P_n-\delta \tau ,Q_n-\delta (N-n-\tau )]}^T}+\delta (N-n)\\ ⩾& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+{cos}^2\frac{\mathrm{\angle }A}{2}{[\sum _{j=1}^n{(-1)}^{j+1}{a}_j-\delta (N-n)]}^2\}}^{1/2}+\delta (N-n).\end{array}$$

Thus, inequality (16) is proved.

Finally, the conditions for the equality in (16) to hold are as follows:

(H3.5) The n-gon ${\mathrm{\Gamma }}_n(\mathbf{A})$ is an equiangular n-gon.

(H3.6) Equality (5) holds for any $i\in \{1,2,\dots ,n\}$, $\tau (i)⩾1$.

(H3.7) Equality (4) holds.

(H3.8) ${\sum }_{i=1}^{n/2}\tau (2j-1)={\tau }_0\in \{0,1,2,\dots ,N-n\}$, where ${\tau }_0$ is defined by (23).

(H3.9) There exist $x,y\in {[0,\mathrm{\infty })}^n$ such that

$$\{\begin{array}{ll}{(x_{2j-1},y_{2j-1})}^T\uparrow {(x_1,y_1)}^T\uparrow {(y_{2j},x_{2j})}^T,& j=1,2,\dots ,n/2,\\ x_j+y_{j+1}=a_j-\tau (j)\delta ,& j=1,2,\dots ,n,\\ \sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}⩾\delta ,& j=1,2,\dots ,n.\end{array}$$

While the conditions for the equality in (17) to hold are as follows:

(H3.10) The conditions (H3.5)-(H3.7) and (H3.9) hold.

(H3.11) ${\sum }_{i=1}^{n/2}\tau (2j-1)=N-n$.

Here,

$${(x_i,y_i)}^T={(x_j,y_j)}^T\iff i\equiv j(modn),i,j=0,\pm 1,\pm 2,\dots .$$

This completes the proof of this theorem. □

3.3 The case ${\mathrm{\Gamma }}_n(\mathbf{A})$is an equiangular n-gon

Equiangular polygon is a special kind of polygons. Regular polygon in${\mathbb{R}}^2$ is an equiangular polygon. If is a Euclideanspace with $dim\mathbb{E}⩾2$, then there is an equiangular 4-gon${\mathrm{\Gamma }}_4(\mathbf{A})$ in . Indeed, in , there exist atleast two linearly independent vectors α,β. Then, by the Gram-Schmidt orthogonalizationprocess, we may obtain two orthogonal unit vectors $\mathbf{i},\mathbf{j}\in \mathbb{E}$ from α, β. If we set

$$A_1=a(\mathbf{i}cos\frac{\pi }{3}+\mathbf{j}sin\frac{\pi }{3}),A_2=-b(\mathbf{i}cos\frac{\pi }{3}+\mathbf{j}sin\frac{\pi }{3}),A_3=-b\mathbf{i},A_4=a\mathbf{i},$$

where $a,b\in (0,\mathrm{\infty })$, then the 4-gon ${\mathrm{\Gamma }}_4(\mathbf{A})$ is an equiangular polygon in with

$$\mathrm{\angle }{A}_1=\mathrm{\angle }{A}_2=\mathrm{\angle }{A}_3=\mathrm{\angle }{A}_4=\frac{\pi }{3}.$$

Similarly, if $dim\mathbb{E}⩾3$, then the 8-gon

$${\mathrm{\Gamma }}_8(\mathbf{A}):={\mathrm{\Gamma }}_8(\mathbf{0},\mathbf{i},\mathbf{i}+\mathbf{j},\mathbf{j},\mathbf{j}+\mathbf{k},\mathbf{i}+\mathbf{j}+\mathbf{k},\mathbf{i}+\mathbf{k},\mathbf{k})$$

is equiangular with

$$\mathrm{\angle }{A}_1=\mathrm{\angle }{A}_2=\cdots =\mathrm{\angle }{A}_8=\frac{\pi }{2},$$

where i, j, k are three mutually orthogonal unit vectors in .

We now turn to the calculation of $inf\{|{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })|\}$.

Theorem 3.3 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS with n odd, and let${\mathrm{\Gamma }}_n(\mathbf{A})$be an equiangular n-gon. Suppose that there exist$\tau (i)\in \{0,1,\dots ,N-n\}$for each$i=1,2,\dots ,n$, such that:

(H3.12) ${\sum }_{i=1}^n\tau (i)=N-n$.

(H3.13) $\frac{1}{2}{\sum }_{j=1}^n[{(-1)}^{j+1}(a_j-\tau (j)\delta )]⩾\frac{1}{2}\delta csc\frac{\mathrm{\angle }A}{2}$.

(H3.14) ${(-1)}^k\{{\sum }_{j=1}^{k-1}[{(-1)}^{j+1}(a_j-\tau (j)\delta )]-\frac{1}{2}{\sum }_{j=1}^n[{(-1)}^{j+1}(a_j-\tau (j)\delta )]\}⩾\frac{1}{2}\delta csc\frac{\mathrm{\angle }A}{2}$, where$k=2,3,\dots ,n$.

Then

$$inf\left\{\right|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\left|\right\}=|{\mathrm{\Gamma }}_n(\mathbf{A})|sin\frac{\mathrm{\angle }A}{2}+(1-sin\frac{\mathrm{\angle }A}{2})(N-n)\delta .$$

(29)

Proof By the assumptions in Theorem 3.3, conditions (H3.1) and(H3.3) hold. Since $dim\mathbb{E}⩾2$, there exists ${\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast })\subset \mathbb{E}$ such that condition (H3.2) holds. Thus, we justneed to consider condition (H3.4).

From

$${(x_j,y_j)}^T\uparrow {(x_1,y_1)}^T\uparrow {(y_j,x_j)}^T,j=1,2,\dots ,n,$$

we have

$$x_j=y_j,j=1,2,\dots ,n.$$

(30)

By

$$x_j+y_{j+1}=a_j-\tau (j)\delta ,j=1,2,\dots ,n,$$

and (30), we get

$$y_j+y_{j+1}=a_j-\tau (j)\delta ,j=1,2,\dots ,n.$$

(31)

We can rewrite (31) as

$${(-1)}^{j+1}{y}_{j+1}-{(-1)}^j{y}_j={(-1)}^{j+1}(a_j-\tau (j)\delta ),j=1,2,\dots ,n.$$

(32)

By (30), (32), $y_{n+1}=y_1$ and n is odd number, we obtain that

$$\begin{array}{c}\sum _{j=1}^k[{(-1)}^{j+1}{y}_{j+1}-{(-1)}^j{y}_j]=\sum _{j=1}^k\left[{(-1)}^{j+1}(a_j-\tau (j)\delta )\right],\hfill \\ {(-1)}^{k+1}{y}_{k+1}+y_1=\sum _{j=1}^k\left[{(-1)}^{j+1}(a_j-\tau (j)\delta )\right],k=1,2,\dots ,n,\hfill \\ x_1=y_1=\frac{1}{2}\sum _{j=1}^n\left[{(-1)}^{j+1}(a_j-\tau (j)\delta )\right],\hfill \end{array}$$

(33)

$$x_k=y_k={(-1)}^k\{\sum _{j=1}^{k-1}\left[{(-1)}^{j+1}(a_j-\tau (j)\delta )\right]-\frac{1}{2}\sum _{j=1}^n\left[{(-1)}^{j+1}(a_j-\tau (j)\delta )\right]\},$$

(34)

where $k=2,3,\dots ,n$. By (30), $x,y\in {[0,\mathrm{\infty })}^n$ and

$$\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}⩾\delta ,j=1,2,\dots ,n,$$

we have

$$x_j=y_j⩾\frac{1}{2}\delta csc\frac{\mathrm{\angle }A}{2},j=1,2,\dots ,n.$$

(35)

According to the assumption in Theorem 3.3, and (33)-(35), condition (H3.4)holds. Consequently, by Theorem 3.1, Theorem 3.3 holds.

This completes the proof of Theorem 3.3. □

Theorem 3.4 Let$CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$be a CLS. Assume that${\mathrm{\Gamma }}_n(\mathbf{A})$is an equiangular n-gon where n is an even number, and

$$\sum _{j=1}^n{(-1)}^{j+1}{a}_j⩾0.$$

Suppose that there exist$\tau (i)\in \{0,1,\dots ,N-n\}$for each$i=1,2,\dots ,n$and free variable$y_n\in (0,\mathrm{\infty })$such that:

(H3.15) Condition (H3.12) holds.

(H3.16) The following inequalities hold:

$$y_j\sqrt{{\lambda }^{2{(-1)}^{j+1}}+1-{\lambda }^{{(-1)}^{j+1}}cos\mathrm{\angle }A}⩾\delta ,j=1,2,\dots ,n,$$

where

$$\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\\ ⋮\\ y_{n-2}\\ y_{n-1}\end{array}\right)={\left[\begin{array}{cccccc}\lambda & 1& 0& 0& 0& 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& ⋮\\ 0& \cdots & {\lambda }^{{(-1)}^{j+1}}& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& 0& 0& 0& {\lambda }^{-1}& 1\\ 0& 0& 0& 0& 0& \lambda \end{array}\right]}^{-1}\left(\begin{array}{c}{a}_1-\tau (1)\delta \\ ⋮\\ a_j-\tau (j)\delta \\ ⋮\\ a_{n-2}-\tau (j-2)\delta \\ a_{n-1}-\tau (n-1)\delta -y_n\end{array}\right),$$

(36)

$$\lambda =\frac{{\sum }_{j=1}^{n/2}[a_{2j-1}-\tau (2j-1)\delta ]}{{\sum }_{j=1}^{n/2}[a_{2j}-\tau (2j)\delta ]}.$$

(37)

(H3.17)

$$\sum _{j=1}^{n/2}\tau (2j-1)=\{\begin{array}{ll}[\omega ],& \delta (N-n)>{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j,0⩽\{\omega \}⩽\frac{1}{2},\\ [\omega ]+1,& \delta (N-n)>{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j,\frac{1}{2}<\{\omega \}<1,\\ N-n,& \delta (N-n)⩽{\sum }_{j=1}^n{(-1)}^{j+1}{a}_j,\end{array}$$

where ω is defined by (22).

Then we have the following two assertions:

• (I) If

  $$\delta (N-n)>\sum _{j=1}^n{(-1)}^{j+1}{a}_j,$$

then

$$\begin{array}{rcl}inf\left\{\right|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\left|\right\}& =& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+4{\delta }^2{cos}^2\frac{\mathrm{\angle }A}{2}{min}^2\{\{\omega \},1-\{\omega \}\}\}}^{1/2}+\delta (N-n).\end{array}$$

(38)

• (II) If

  $$\delta (N-n)⩽\sum _{j=1}^n{(-1)}^{j+1}{a}_j,$$

then

$$\begin{array}{rcl}inf\left\{\right|{\mathrm{\Gamma }}_N\left({\mathbf{A}}^{\ast }\right)\left|\right\}& =& \{{sin}^2\frac{\mathrm{\angle }A}{2}{\left[\right|{\mathrm{\Gamma }}_n(\mathbf{A})|-\delta (N-n)]}^2\\ {+{cos}^2\frac{\mathrm{\angle }A}{2}{[\sum _{j=1}^n{(-1)}^{j+1}{a}_j-\delta (N-n)]}^2\}}^{1/2}+\delta (N-n).\end{array}$$

(39)

Proof We first look for the conditions for equality in (16)-(17) tohold. The conditions are either (H3.5)-(H3.9) or (H3.10)-(H3.11). By theassumptions in Theorem 3.4 and $dim\mathbb{E}⩾2$, conditions (H3.5)-(H3.8) and (H3.11) hold. If(H3.5)-(H3.9) hold, then (H3.10) hold. Therefore we just need to show that(H3.9) holds.

Form (18)-(19) and

$${(x_{2j-1},y_{2j-1})}^T\uparrow {(x_1,y_1)}^T\uparrow {(y_{2j},x_{2j})}^T,j=1,2,\dots ,\frac{n}{2},$$

we see that

$$\frac{x_{2j-1}}{y_{2j-1}}=\frac{y_{2j}}{x_{2j}}=\lambda ,j=1,2,\dots ,\frac{n}{2},$$

(40)

where

$$\lambda =\frac{{\sum }_{j=1}^{n/2}(x_{2j-1}+y_{2j})}{{\sum }_{j=1}^{n/2}(y_{2j-1}+x_{2j})}=\frac{{\sum }_{j=1}^{n/2}[a_{2j-1}-\tau (2j-1)\delta ]}{{\sum }_{j=1}^{n/2}[a_{2j}-\tau (2j)\delta ]}.$$

(41)

Consequently,

$$x_j={\lambda }^{{(-1)}^{j+1}}{y}_j,j=1,2,\dots ,n.$$

(42)

By

$$x_j+y_{j+1}=a_j-\tau (j)\delta ,j=1,2,\dots ,n,$$

and (42), we have

$${\lambda }^{{(-1)}^{j+1}}{y}_j+y_{j+1}=a_j-\tau (j)\delta ,j=1,2,\dots ,n-1$$

(43)

and

$${\lambda }^{-1}{y}_n+y_1=a_n-\tau (n)\delta .$$

(44)

Equalities (40)-(43) imply that if $j=n$, then (43) holds, i.e., (44) holds. From(43) we have get (36), where $y_n$ are free variables. By (42),$x,y\in {[0,\mathrm{\infty })}^n$,

$$\sqrt{x_j^2+y_j^2-2x_j{y}_{j}cos\mathrm{\angle }A}⩾\delta ,j=1,2,\dots ,n,$$

we have

$$y_j\sqrt{{\lambda }^{2{(-1)}^{j+1}}+1-{\lambda }^{{(-1)}^{j+1}}cos\mathrm{\angle }A}⩾\delta ,j=1,2,\dots ,n.$$

(45)

This means that condition (H3.9) can be deduced from conditions (H3.15)-(H3.17).Thus Theorem 3.4 holds by applying Theorem 3.2.

This completes the proof of Theorem 3.4. □

4 Three effective examples

For a general $CLS{\{{\mathrm{\Gamma }}_n(\mathbf{A}),{\mathrm{\Gamma }}_N({\mathbf{A}}^{\ast }),\delta \}}_{\mathbb{E}}$, the equalities in (12), (16) and (17) may not hold,this is most probably because conditions (H3.1)-(H3.4) or conditions (H3.5)-(H3.11)cannot be met at the same time. We will discuss Problem 1.1 of a special CLS in${\mathbb{R}}^2$.

Example 4.1 Consider the $CLS{\{\mathrm{\Delta }ABC,\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast },\delta \}}_{{\mathbb{R}}^2}$, where

$$A^{\ast }\in [BC),B^{\ast }\in [CA),C^{\ast }\in [AB),0<\delta <min\{a,b,c\},$$

and a, b, c is the length of the sides of the triangle$\mathrm{\Delta }ABC$. We will calculate that $inf\{|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }|\}$.

By Theorem 3.1 we have

$$\left|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }\right|⩾|\mathrm{\Delta }ABC|sin\frac{min\{A,B,C\}}{2},$$

(46)

where A, B, C are three inner angles of the triangle$\mathrm{\Delta }ABC$. According to conditions (H3.1)-(H3.4), the equalityin (46) holds if and only if $\mathrm{\Delta }ABC$ is a normal triangle, and $A^{\ast }$, $B^{\ast }$, $C^{\ast }$ are the midpoints of line segment$[BC]$, $[CA]$, $[AB]$, respectively. Consequently, the equality ininequality (12) does not hold in general.

It is well known that $\mathrm{\Delta }ABC$ is an acute triangle if and only if

$$\left[A{A}^{\ast }\right]\perp [BC],\left[B{B}^{\ast }\right]\perp [CA],\left[C{C}^{\ast }\right]\perp [AB]$$

implies that $|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }|$ takes the minimum. By this we see that

$$inf\left\{\right|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }\left|\right\}=acosA+bcosB+ccosC.$$

(47)

If $\pi /2⩽A<\pi$ and $\delta =0$, then

$$B^{\ast }=C^{\ast }=A,\left[A{A}^{\ast }\right]\perp [BC]$$

is necessary and sufficient for $|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }|$ to take the minimum. By this we see that

$$inf\left\{\right|\mathrm{\Delta }{A}^{\ast }{B}^{\ast }{C}^{\ast }\left|\right\}=2\left|A{A}^{\ast }\right|=2bsinC=2csinB.$$

(48)

Example 4.2 Consider the $CLS{\{{\mathrm{\Gamma }}_3(\mathbf{A}),{\mathrm{\Gamma }}_4({\mathbf{A}}^{\ast }),1\}}_{{\mathbb{R}}^2}$ (see Figure 2), where

$$\parallel A_1-A_2\parallel =6,\parallel A_2-A_3\parallel =8,\parallel A_3-A_2\parallel =10$$

and

$$A_1^{\ast }\in [A_1{A}_2),A_2^{\ast }\in [A_2{A}_3),A_3^{\ast },A_4^{\ast }\in [A_3{A}_1).$$

We will calculate that $inf\{|{\mathrm{\Gamma }}_4({\mathbf{A}}^{\ast })|\}$.

Figure 2

The graph of $\mathbf{CLS}{\mathbf{\{}{\mathbf{\Gamma }}_{\mathbf{3}}\mathbf{(}\mathbf{A}\mathbf{)}\mathbf{,}{\mathbf{\Gamma }}_{\mathbf{4}}\mathbf{(}{\mathbf{A}}^{\mathbf{\ast }}\mathbf{)}\mathbf{,}\mathbf{1}\mathbf{\}}}_{{\mathbb{R}}^{\mathbf{2}}}$ .

Note that

$$cos\mathrm{\angle }{A}_1=\frac{3}{5},cos\mathrm{\angle }{A}_2=0,cos\mathrm{\angle }{A}_3=\frac{4}{5}.$$

Without loss of generality, we may assume that $\parallel A_4^{\ast }-A_3^{\ast }\parallel =1$. By Lemma 2.6, we have

$$\begin{array}{rcl}|{\mathrm{\Gamma }}_4^{\ast }(\mathbf{A})|& ⩾& \psi (x,y,z)\\ =& \sqrt{x^2+{(6-z)}^2}+\sqrt{{(8-x)}^2+y^2-\frac{8}{5}(8-x)y}\\ +\sqrt{{(9-y)}^2+z^2-\frac{6}{5}(9-y)z}+1,\end{array}$$

where $(x,y,z)\in [0,6]\times [0,9]\times [0,6]$, and

$$\{\begin{array}{l}\sqrt{x^2+{(6-z)}^2}⩾1,\\ \sqrt{{(8-x)}^2+y^2-\frac{8}{5}(8-x)y}⩾1,\\ \sqrt{{(9-y)}^2+z^2-\frac{6}{5}(9-y)z}⩾1.\end{array}$$

(49)

By the help of Mathematica software, we obtain that

$$\begin{array}{rcl}inf\left\{\right|{\mathrm{\Gamma }}_4\left({\mathbf{A}}^{\ast }\right)\left|\right\}& =& \psi (0.621684\dots ,5.90012\dots ,5.1705\dots )\\ =& 10.600001492661695\dots .\end{array}$$

We note that (49) and

$$\frac{\partial \psi (x,y,z)}{\partial x}=\frac{\partial \psi (x,y,z)}{\partial y}=\frac{\partial \psi (x,y,z)}{\partial z}=0$$

hold if

$$(x,y,z)=(0.621684\dots ,5.90012\dots ,5.1705\dots ).$$

Example 4.3 Consider the $CLS{\{{\mathrm{\Gamma }}_4(\mathbf{A}),{\mathrm{\Gamma }}_5({\mathbf{A}}^{\ast }),2\}}_{{\mathbb{R}}^2}$ (see Figure 1), where${\mathrm{\Gamma }}_4(\mathbf{A})$ is a rectangle, and

$$\parallel A_2-A_1\parallel =\parallel A_4-A_3\parallel =6,\parallel A_3-A_2\parallel =\parallel A_1-A_4\parallel =5$$

and

$$A_1^{\ast }\in [A_1{A}_2),A_2^{\ast }\in [A_2{A}_3),A_3^{\ast },A_4^{\ast }\in [A_3{A}_4),A_5^{\ast }\in [A_4{A}_1).$$

We will calculate that $inf\{|{\mathrm{\Gamma }}_5({\mathbf{A}}^{\ast })|\}$.

We may assume that $\parallel A_4^{\ast }-A_3^{\ast }\parallel =2$. By Lemma 2.6, we have

$$\begin{array}{rcl}\left|{\mathrm{\Gamma }}_5\left({\mathbf{A}}^{\ast }\right)\right|& ⩾& \phi (x,y,z,w)\\ =& \sqrt{x^2+{(5-w)}^2}+\sqrt{{(6-x)}^2+y^2}\\ +\sqrt{{(5-y)}^2+z^2}+\sqrt{{(4-z)}^2+w^2}+2,\end{array}$$

where $(x,y,z,w)\in [0,6]\times [0,5]\times [0,4]\times [0,5]$, and

$$\{\begin{array}{l}\sqrt{x^2+{(5-w)}^2}⩾2,\\ \sqrt{{(6-x)}^2+y^2}⩾2,\\ \sqrt{{(5-y)}^2+z^2}⩾2,\\ \sqrt{{(4-z)}^2+w^2}⩾2.\end{array}$$

(50)

By the help of Mathematica software, we obtain that

$$\begin{array}{rl}inf\{\phi (x,y,z,w)\}& =\phi (2.75296\dots ,3.24704\dots ,1.75296\dots ,2.24704\dots )\\ =16.1421356237309\dots \\ =10\sqrt{2}+2.\end{array}$$

(51)

On the other hand, by Theorem 3.2, we have

$$\begin{array}{rl}\left|{\mathrm{\Gamma }}_5\left({\mathbf{A}}^{\ast }\right)\right|& ⩾|{\mathrm{\Gamma }}_n(\mathbf{A})|sin\frac{\mathrm{\angle }A}{2}+(1-sin\frac{\mathrm{\angle }A}{2})(N-n)\delta \\ =10\sqrt{2}+2.\end{array}$$

It should be noted that the apparent error is caused by the computer. Therefore

$$inf\left\{\right|{\mathrm{\Gamma }}_5\left({\mathbf{A}}^{\ast }\right)\left|\right\}=10\sqrt{2}+2.$$

(52)

We note that (51) holds if

$$(x,y,z,w)=(2.75296\dots ,3.24704\dots ,1.75296\dots ,2.24704\dots ).$$

We can also give another intuitive proof of equation (52) as follows.

By Theorem 3.2, the equality in (51) holds if conditions (H3.5)-(H3.11) hold. Infact, there exist

$$(x,y,z,w)\in [0,6]\times [0,5]\times [0,4]\times [0,5],$$

such that (50) holds. From (H3.9) we get

$$\frac{x}{5-w}=\frac{z}{5-y}=\frac{6-x}{y}=\frac{4-z}{w}=\frac{x+z+(6-x)+(4-z)}{(5-w)+(5-y)+y+w}=1,$$

i.e.,

$$\{\begin{array}{l}x=5-w,\\ y=1+w,\\ z=4-w.\end{array}$$

(53)

Combined with (50), (53), we get

$$\sqrt{2}⩽w⩽4,$$

(54)

where w is a free variable. This means that (51) holds if and only if(53)-(54) hold. This proves equation (52).

In addition, we can also prove (52) by Theorem 3.4.

Example 4.3 is a geometry problem. However, we can see this example as a circuitlayout problem of a family. In addition, this example also means that the equalitiesin (12), (16) and (17) can hold.

Remark 4.1 Since a Euclidean space is an abstract space, we will find theapplications of CLS in theoretical fields such as statistics (see [1, 7]), matrix theory (see [8]), geometry (see [6, 9–12]) and space science (see [1, 9]), etc.

Remark 4.2 A large number of theories of algebra, analysis, geometry, computer(see [9, 12–14]) with inequality are used in this paper, which can be found in the latestliterature [1, 6–16].

Endnote

The angle between two nonzero vectors B and C is defined to be$\mathrm{\angle }(B,C)=:arccos(〈B,C〉/\parallel B\parallel \parallel C\parallel )\in [0,\pi ]$.