A novel computational framework for the expected number of real roots of stochastic functions on a given interval

TL;DR

This paper introduces a new computational framework to estimate the expected number of real roots of stochastic functions, bypassing the intractable joint density calculations of classical methods, and provides explicit formulas for specific distributions.

Contribution

A novel framework that avoids joint density calculations by using a cumulative expectation function, enabling analysis of stochastic functions' roots with broader applicability.

Findings

Explicit formulas for Gaussian and uniform distributions.

New analytical results for linear stochastic functions.

Broader applicability of root expectation analysis.

Abstract

We propose a new computational framework for the expected number of real roots of a stochastic function on a given interval. The classical Kac-Rice formula requires the joint density of the function and its derivative, which is often intractable. Our approach avoids this requirement entirely by introducing a cumulative expectation function. Through analysis of its absolute continuity and differential structure, we derive two complementary computational schemes: one expresses the expectation as a derivative of a variable-domain integral under weak conditions; the other yields an explicit integral representation without joint densities or variable-domain differentiation. We illustrate the method in detail for linear stochastic functions, obtaining explicit formulas for Gaussian and uniform distributions, together with several new analytical results. The framework substantially broadens the scope of problems amenable to rigorous analysis and provides a powerful tool for applications in stochastic analysis and beyond.

Equations

$$F(t;x)=0,t\in \Delta ,$$

$$E_n(\rho ,F,\Delta )={\int }_{{\mathbb{R}}^n}{N}_F(x,\Delta )\rho (x)dx,$$

$$F(t;x)=\sum _{i=1}^n{x}_i{t}^i,$$

$$F(t;x)=\sum _{i=1}^n{x}_{2i-1}\cos (it)+x_{2i}\sin (it),$$

$$F(t;x)=\sum _{i=1}^n{f}_i(t)x_i.$$

$$0\le \kappa <+\infty ,|\{x\in {\mathbb{R}}^n|N_F(x,\Delta )>\kappa \}|=0.$$

$$E_n(\rho ,F,\Delta )\le \kappa <+\infty ,$$

$$N_F(x,A\bigcup B)=N_F(x,A)+N_F(x,B)-N_F(x,A\bigcap B),$$

$$E_n(\rho ,F,B)=E_n(\rho ,F,A)+E_n(\rho ,F,B\backslash A)\ge E_n(\rho ,F,A).\text{qed}$$

$$S_F(A)=\{{\Lambda }_{N_F}^i(A){\}}_{i=1}^{\infty },$$

$${\Lambda }_{N_F}^i(A)=\{x\in {\mathbb{R}}^n|N_F(x,A)\ge i\}.$$

$${\Lambda }_{N_F}^0(A)={\mathbb{R}}^n,\forall A\subseteq \mathbb{R}.$$

$${\int }_{S_F(A)}f(x)dx=\sum _{i=1}^{\infty }{\int }_{{\Lambda }_{N_F}^i(A)}f(x)dx.$$

$$L_F(t^{\ast })=\{x\in {\mathbb{R}}^n|F(t^{\ast };x)=0\}.$$

$${\Lambda }_{N_F}^i(A\bigcup B)=\{x\in {\mathbb{R}}^n|N_F(x,A)+N_F(x,B)-N_F(x,A\bigcap B)\ge i\}.$$

$${\Lambda }_{N_F}^i(A\bigcup B)=\{x\in {\mathbb{R}}^n|N_F(x,A)+N_F(x,B)\ge i\}=\bigcup _{j,k\ge 0,j+k=i}({\Lambda }_{N_F}^j(A)\bigcap {\Lambda }_{N_F}^k(B)).$$

$${\Lambda }_{N_F}^i(A\bigcup B)=\bigcup _{j,k\ge 0,j+k=i}({\Lambda }_{N_F}^j(A)\bigcap {\Lambda }_{N_F}^k(B)).$$

$$N_F(x,A)+N_F(x,B\setminus A)\ge i,$$

$$x\in {\Lambda }_{N_F}^i(A\bigcup (B\setminus A))={\Lambda }_{N_F}^i(B).$$

$$E_n(\rho ,F,\Delta )=\sum _{i=1}^{\infty }{\int }_{{\Lambda }_{N_F}^i(\Delta )}\rho (x)dx={\int }_{S_F(\Delta )}\rho (x)dx$$

$${\int }_{{\mathbb{R}}^n}{N}_F(x,\Delta )\rho (x)dx={\int }_{\{x\in {\mathbb{R}}^n|N_F(x,\Delta )\ge 1\}}\sum _{i=1}^{\infty }\delta (x,i)\rho (x)dx$$

$$={\int }_{{\Lambda }_{N_F}^1(\Delta )}\sum _{i=1}^{\infty }\delta (x,i)\rho (x)dx=\sum _{i=1}^{\infty }{\int }_{{\Lambda }_{N_F}^i(\Delta )}\rho (x)dx$$

$$={\int }_{S_F(\Delta )}\rho (x)dx.\text{qed}$$

$$E_n(\rho ,F,\Delta )=0.$$

$$0\le |{\Lambda }_{N_F}^j(\Delta )|\le \sum _{i=1}^{+\infty }|L_F(t_i)|=0,\forall j\in \{1,2,\cdots \},$$

$$E_n(\rho ,F,\Delta )={\int }_{S_F(\Delta )}\rho (x)dx=\sum _{i=1}^{+\infty }{\int }_{{\Lambda }_{N_F}^i(\Delta )}\rho (x)dx=0.\text{qed}$$

$$\Delta =[l,u],$$

$$\underset{\epsilon \to 0_{+}}{\lim }{E}_n(\rho ,F,[t^{\ast },t^{\ast }+\epsilon ])=0$$

$$\underset{\epsilon \to 0_{-}}{\lim }{E}_n(\rho ,F,[t^{\ast }-\epsilon ,t^{\ast }])=0.$$

$$\underset{\epsilon \to 0_{+}}{\lim }{E}_n(\rho ,F,[t^{\ast },t^{\ast }+\epsilon ])\ne 0,$$

$$0\le E_n(\rho ,F,[t^{\ast },t^{\ast }+{\epsilon }_{k+1}])\le E_n(\rho ,F,[t^{\ast },t^{\ast }+{\epsilon }_k]),|[t^{\ast },t^{\ast }+{\epsilon }_k]|\to 0,$$

Full text

Yi Xu

School of Mathematics

Southeast University

2 Sipailou, Nanjing, 210096, China

1 Introduction

How many real zeros $t$ of

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with a parameter $x\in\mathbb{R}^{n}$, where $F:\Delta\mapsto\mathbb{R}$, $\Delta\subseteq\mathbb{R}$. It is obliviously that the number of the real zeros $t$ of (1) is different with different $x$. If $x$ is a random variable obeying a distribution with the density function $\rho(x)$, then the number of different real zeros $t$ of (1) is also a random variable. In this paper, we focus on getting its analytical expectation.

Definition 1.1**.**

Defining

[TABLE]

as the expectation of the number of real zeros of (1), where $N_{F}(x,\Delta)$ is the number of different real zeros $t\in\Delta$ of (1) with the parameter $x\in\mathbb{R}^{n}$, $\displaystyle\int dx$ is the Lebesgue integral.

This problem has been studied for about a hundred years, and there are a lot of results on this subject. Most of them discuss some specific equations $F(t;x)$ with some specific distributions, e.g, random polynomial equations[17, 18, 19, 7, 8, 9, 20, 3, 15, 16, 22, 23, 12, 13, 14, 5]:

[TABLE]

which parameter $x$ obeys the discrete distribution or the standard normal distribution; trigonometric equations[21, 10, 6, 11, 1, 2]:

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which parameter $x$ obeys the standard normal distribution, etc. Although, $E_{n}(\rho,F,\Delta)$ of above problems are all got, the methods used can’t extend to general cases. In [7], Edelman and Kostlan present a new view to review this problem, where $F$ is linear and homogeneous in $x$

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They find that this problem could be turned into computing the ratio of the area of a scanned area on a sup-sphere and the area of this sup-sphere. By this idea, they promote a general result to this problem, and get the same results of above problems got by other papers. However, this method still has two disadvantages. Firstly, this method only suits for the standard normal distribution; Secondly, (3) is still not general enough, since (3) is linear and homogeneous in $x$, it misses a constant term $f_{0}(t)$. Meanwhile, we consider that their method has great potential, so in this paper, we extend this method into more general cases.

The main results are presented here. We find that $E_{n}(\rho,F,\Delta)$ is just the integral of $\rho(x)$ on $\mathbb{R}^{n}$ with multiplicity, or in other words, the sum of integral of $\rho(x)$ on some subsets of $\mathbb{R}^{n}$. Then we get that $E_{n}(\rho,F,[l,t])$ is a continuous function in $t$. Under some assumptions, its derivative $H_{\rho,F}(t)$ exists also. So we could turn the problem that computing $E_{n}(\rho,F,\Delta)$ into computing $H_{\rho,F}(t)$. When $F(t;x)$ is linear in $x$ (not homogeneous necessary), we discuss two cases $n=2$ and $n\geq 4$ in subsection 3.1 and 3.2, and analytical expressions of $H_{\rho,F}(t)$ for different $n$ are presented, respectively. Furthermore, we discuss more concrete problems, which $x$ obeys some common distributions, e.g, the uniform distribution in a sup-sphere, the normal distribution, etc, then more concise results are got. We use our results to some classical problems just mentioned above, and find that our results is exactly same as the result existed, it tells us that all existed results with the proper density function are our corollaries.

The remainder of this paper is organized as follows. We close Section 1 by summarizing notations and symbols used throughout this paper, and then Section 2 introduces some properties of $E_{n}(\rho,F,\Delta)$. Then, in Section 3, we discuss the case that $F(t;x)$ is a linear function in $x$. In this section, we discuss two cases that $n=2$ and $n\geq 4$. Moreover, in Section 4, we present the analytic result when $x$ obeys the uniform distribution in a sup-sphere. Furthermore, we discuss other distributions in Section 5. Section 6 concludes this paper and presents some directions for future research. Finally, we present some proofs and lemmas used in this paper in Appendix.

In closing this section, we introduce some basic notations and symbols that will be used in our subsequent analysis.

Notations and Symbols : $\mathbb{R}$ and $\mathbb{R}^{n}$ are real Euclidean space and $n$ dimension real Euclidean space, respectively, $\mathbb{R}_{+}=\{x|x\geq 0\}$; $M_{:,i}$ is a column vector which is the $i$th column of matrix $M$; $\#\{\Omega\}$ is the number of elements in set $\Omega$; $|\Omega|$ is the Lebesgue measure of $\Omega$; $cl(\Omega)$, $int(\Omega)$, $\partial(\Omega)$ and $\bar{\Omega}$ are the closure, interior, boundary and complement of $\Omega$, respectively; $l(A,B)$ is the line passing through point $A$ and point $B$; $||x||$ is 2-norm of $x$; $B^{n}(x^{*},r)=\{x\in\mathbb{R}^{n}|||x-x^{*}||\leq r\}$;W_{f,g}(t)=\left(\begin{array}[]{cc}f(t)&g(t)\\ f^{\prime}(t)&g^{\prime}(t)\\ \end{array}\right); $T(A,B,C)\subset\mathbb{R}^{2}$ is a triangle with three vertices $A$, $B$ and $C$; If $x$ obeys the uniform distribution on $B^{n}(0,r),r>0$, we add an extra subscript $r$ to its density function $\rho_{r}(x)$ to show the radii of $B^{n}(0,r).$

Given a derivable convex(concave) function $G(x):\mathbb{R}\rightarrow\mathbb{R}$ and $x^{*}\in\mathbb{R}$, then the tangent line $y-G(x^{*})=G^{\prime}(x^{*})(x-x^{*})$ at $x^{*}$ separates $\mathbb{R}^{2}$ into two open half planes $\Omega$ and $\bar{\Omega}$, and a line $l(x^{*},G)=\{(x,y)|y-G(x^{*})=G^{\prime}(x^{*})(x-x^{*})\}$. Denoting $B(G)=\{(x,y)|y\geq(\leq)G(x)\}$. Supposing $\bar{\Omega}\bigcup l\supseteq B(G)$, then we define $A(x^{*},G)=\Omega\bigcup l$ as a closed half plane, $\bar{A}(x^{*},G)=\bar{\Omega}\bigcup l$, $C(x^{*},G)=(\bar{\Omega}\setminus int(B(G)))\bigcup l$.

Given $x\in\mathbb{R}^{n}$, where $n\geq 4$, we separate $x$ into $(y,\bar{x}),$ where $y\in\mathbb{R}^{2},\bar{x}\in\mathbb{R}^{n-2}$. For a $n(n\geq 4)$ dimension set $\Omega\subset\mathbb{R}^{n}$, $\{y\in\mathbb{R}^{2}|y=0\}\times\mathbb{P}(\Omega)$ is the projection $\Omega$ on the subspace $\{(y,\bar{x})\in\mathbb{R}^{n}|y=0\}$. Given $\bar{x}^{*}\in\mathbb{R}^{n-2}$, $\Upsilon(\Omega,\bar{x}^{*})=\{y\in\mathbb{R}^{2}|(y,\bar{x}^{*})\in\Omega\}.$

Given two hyperplanes $l_{1}$ and $l_{2}$ in $\mathbb{R}^{n}$. If $l_{1}\bigcap l_{2}\neq\emptyset,$ then $l_{1}$ and $l_{2}$ separate $\mathbb{R}^{n}$ into 4 separated parts $\Omega_{i},i=1,2,3,4$, where $\Omega_{1}$ and $\Omega_{2}$ are congruent with smaller facial angle, $\Omega_{3}$ and $\Omega_{4}$ are congruent with bigger facial angle. We define $\Phi(l_{1},l_{2})$ as the $cl(\Omega_{1}\bigcup\Omega_{2})$.

2 Some properties of $E_{n}(\rho,F,\Delta)$

In this section, we study the properties of $E_{n}(\rho,F,\Delta)$ firstly. There are some assumptions of $\rho(x)$ used in this paper.

Assumption 2.1**.**

• (a)

$P(||x||\geq r)\leq ae^{-br}$, where $a>0,b>0.$

• (b)

There exists a constant $Q\geq 0$ such that $|\{x\in\mathbb{R}^{n}|\rho(x)>Q\}|=0.$

• (c)

$\displaystyle\int_{\Omega}\rho(x)dx=0,\forall\Omega\subset\mathbb{R}^{n},|\Omega|=0.$**

• (d)

When $n\geq 4$, $\displaystyle\int_{\Omega}\rho(x)dx=\int_{\mathbb{P}(\Omega)}d\bar{x}\int_{\Upsilon(\Omega,\bar{x})}\rho(y,\bar{x})dy,$ $\forall\Omega\subset\mathbb{R}^{n}.$

A lot of distributions satisfy Assumption 2.1, e.g. the uniform distribution, the normal distribution, etc. For avoiding discussing the case that $E_{n}(\rho,F,\Delta)=+\infty$, we give the following assumption.

Assumption 2.2**.**

There exists a constant $\kappa$ such that

[TABLE]

Assumption 2.2 tells that for almost every $x\in\mathbb{R}^{n}$, (1) has $\kappa$ real zeros at most, so

[TABLE]

if Assumption 2.1(c) holds.

We present some basic properties of $E_{n}(\rho,F,A)$ here.

Lemma 2.3**.**

• (a)

$E_{n}(\rho,F,A)\geq 0,\forall A\subset\mathbb{R}$, $E_{n}(\rho,F,\emptyset)=0.$

• (b)

$E_{n}(\rho,F,A\bigcup B)=E_{n}(\rho,F,A)+E_{n}(\rho,F,B)-E_{n}(\rho,F,A\bigcap B).$**

• (c)

If $A\bigcap B=\emptyset$, then $E_{n}(\rho,F,A\bigcup B)=E_{n}(\rho,F,A)+E_{n}(\rho,F,B).$

• (d)

If $A\subseteq B$, then $E_{n}(\rho,F,A)\leq E_{n}(\rho,F,B).$

Proof.

(a) It is obviously that (a) is true.

(b) Since

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so the result is right.

(c) It is obviously true by (b).

(d) $B=A\bigcup(B\backslash A)$ and $A\bigcap(B\backslash A)=\emptyset$, then

[TABLE]

Definition 2.4 is the key of this paper, all works base on this definition.

Definition 2.4**.**

• (a)

Family $S_{F}(A)$ with ${N_{F}}(x,A)$: $S_{F}(A)$ is denoted to the family with ${N_{F}}(x,A)$

[TABLE]

where

[TABLE]

Furthermore, we define

[TABLE]

It is easy to see that $|\Lambda^{i}_{N_{F}}(A)|=0,\forall i>\kappa,$ if Assumption 2.2 holds.

• (b)

Integral on $S_{F}(A)$: Supposing that $f(x)$ is a Lebesgue integrable function on $\mathbb{R}^{n}$. Integrating $f(x)$ on $S_{F}(A)$ is defined as

[TABLE]

• (c)

Given $t^{*}\in\Delta$, we define

[TABLE]

Some properties of $S_{F}(A)$ and $\Lambda^{i}_{N_{F}}(A)$ are presented here.

Lemma 2.5**.**

• (a)

$\Lambda^{i+1}_{N_{F}}(A)\subseteq\Lambda^{i}_{N_{F}}(A)$.

• (b)

$S_{F}(A\bigcup B)=\{\Lambda^{i}_{N_{F}}(A\bigcup B)\}_{i=1}^{\infty},$* where*

[TABLE]

• (c)

If $A\bigcap B=\emptyset$, then $S_{F}(A\bigcup B)=\{\Lambda^{i}_{N_{F}}(A\bigcup B)\}_{i=1}^{\infty},$ where

[TABLE]

• (d)

If $A\subseteq B$, then $\Lambda_{N_{F}}^{i}(A)\subseteq\Lambda_{N_{F}}^{i}(B).$

• (e)

$\Lambda_{N_{F}}^{i}(A)\subseteq\displaystyle\bigcup_{t\in A}L_{F}(t),\forall i\geq 1.$**

Proof.

It is obviously that (a) and (b) are true.

(c) It is obviously true by (b). And it is easy to see that

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(d) Since $B=A\bigcup(B\setminus A)$, then $\forall x\in\Lambda_{N_{F}}^{i}(A)$, we have ${N_{F}}(x,A)\geq i$, so

[TABLE]

then

[TABLE]

(e) $\forall x\in\Lambda_{N_{F}}^{i}(A),i\geq 1$, then there exists $t^{*}\in A$ such that $F(t^{*};x)=0$, which means that $x\in L_{F}(t^{*})\in\displaystyle\bigcup_{t\in A}L_{F}(t).\qed$

Using the above definition, we could get the following theorem.

Theorem 2.6**.**

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Proof.

Let \delta(x,i)=\left\{\begin{array}[]{cc}1,&x\in\Lambda^{i}_{N_{F}}(\Delta)\\ 0,&x\not\in\Lambda^{i}_{N_{F}}(\Delta)\\ \end{array}\right.. It is clear that $N_{F}(x,\Delta)=\displaystyle\sum_{i=1}^{\infty}\delta(x,i)$, if $N_{F}(x,\Delta)\geq 1$, so

[TABLE]

For avoiding to discuss too strange functions, e.g. Hilbert Curve, we give the following assumption.

Assumption 2.7**.**

$|L_{F}(t)|=0,\forall t\in\Delta.$**

If $\Delta$ is countable, we have the following theorem.

Theorem 2.8**.**

Supposing that Assumption 2.1(c) and 2.7 hold. If $\Delta$ is countable, then

[TABLE]

Proof.

That $\Delta$ is countable means that $L_{F}(t),t\in\Delta$ is also countable, so we could sort $L_{F}(t)$ as $L_{F}(t_{i})$.

Since $\Lambda_{N_{F}}^{j}(\Delta)\subseteq\displaystyle\bigcup_{i=1}^{+\infty}L_{F}(t_{i}),\forall j\in\{1,2,\cdots\},$ then

[TABLE]

which means $|\Lambda_{N_{F}}^{j}(\Delta)|=0.$ Then

[TABLE]

Combining Lemma 2.3 and Theorem 2.8, it is easy to get the following proposition.

Proposition 2.9**.**

Supposing that Assumption 2.1(c) and 2.7 hold, then

• (a)

if $A\bigcap B$ is countable, then $E_{n}(\rho,F,A\bigcup B)=E_{n}(\rho,F,A)+E_{n}(\rho,F,B).$

• (b)

if $B$ is countable, then $E_{n}(\rho,F,A\bigcup B)=E_{n}(\rho,F,A).$

In the following paper, we always suppose that

[TABLE]

where $-\infty<l<u<+\infty$.

Theorem 2.10**.**

Supposing that Assumption 2.1(c) and 2.7 hold. Then $E_{n}(\rho,F,[l,t])$ is a continuous function in $t\geq l$.

Proof.

We just need to prove $\forall t^{*}\in[l,+\infty)$,

[TABLE]

and $\forall t^{*}\in(l,+\infty)$

[TABLE]

If there is $t^{*}\in[l,+\infty)$ such that

[TABLE]

then there exists $\vartheta_{0}>0,\varepsilon_{k}\searrow 0_{+}$ such that $E_{n}(\rho,F,[t^{*},t^{*}+\varepsilon_{k}])\geq\vartheta_{0},\forall k$. Noting that

[TABLE]

then $E_{n}(\rho,F,\{t^{*}\})\geq\vartheta_{0}$, which is contradict to Theorem 2.8. So

[TABLE]

By the same way, we could prove that $\forall t^{*}\in(l,+\infty)$,

[TABLE]

So $E_{n}(\rho,F,[l,t])$ is a continuous function in $t\geq l.\qed$

Theorem 2.11**.**

Supposing that Assumption 2.1(c) and 2.7 hold. For every $t^{*}\in int(\Delta)$. If

[TABLE]

exist and are same. Let

[TABLE]

then

[TABLE]

Proof.

We define

[TABLE]

It is easy to see that $\Pi(l)=0.$

From Theorem 2.10, $\Pi(t)$ is continuous in $t\in\Delta$. Let $t\in int(\Delta)$, then

[TABLE]

[TABLE]

It means that

[TABLE]

So

[TABLE]

Theorem 2.11 tells us if we want to compute $E_{n}(\rho,F,\Delta)$, we just need to compute $H_{\rho,F}(t).$

3 $F(t;x)$ is a linear function in $x$.

In the following paper, we always suppose that $F(t;x)$ is linear in $x$:

[TABLE]

where $f_{i}(t):\Delta\rightarrow\mathbb{R},i=0,\cdots,n.$

For simplifying the problem, we give the following theorem.

Theorem 3.1**.**

[TABLE]

where $\Xi=\{t\in\Delta|\displaystyle\sum_{i=1}^{n}f_{i}^{2}(t)=0\}.$

Proof.

Since $E_{n}(\rho,F,\Xi)=\#\{t\in\Xi|f_{0}(t)=0\}$, then

[TABLE]

Since $\#\{t\in\Xi|f_{0}(t)=0\}$ is easy to get, we only need to consider the case that the interval $\Delta$ satisfied

[TABLE]

In the following paper, we always suppose that (5) holds.

Now we give some assumptions of $f_{i}(t)$ here.

Assumption 3.2**.**

$f_{i}^{\prime\prime\prime}(t),i=0,1,\cdots,n$* exist, and are continuous functions on $\Delta$.*

Assumption 3.3**.**

$f_{1}(t)$* has finite number of real zeros in $\Delta$.*

Mark: (a) For convenience, in the following paper, we always suppose that Assumption 3.2 holds.

(b) If Assumption 3.2 holds, it is obviously that $f_{i}(t),f_{i}^{\prime}(t)$ and $f_{i}^{\prime\prime}(t)$ $i=0,1,\cdots,n$ are continuous derivable functions on $\Delta$.

(c) Since (4) and (5) hold, $L_{F}(t)$ is a hyperplane, so $|L_{F}(t)|=0,\forall t\in\Delta$, Assumption 2.7 always holds in the following paper.

3.1 How to compute $E_{2}(\rho,F,\Delta)$

In this subsection, we only consider the case $n=2$. Let

[TABLE]

If we consider $t$ as a parameter of (6), then (6) scans an area in $\mathbb{R}^{2}$ with a envelop curve. This envelop curve of the family of (6) is denoted as $H(x_{1},x_{2})=0$, which could be got by following equations

[TABLE]

For studying the envelop curve of (6), we give the following assumption.

Assumption 3.4**.**

W_{f_{1},f_{2}}(t)=\left(\begin{array}[]{cc}f_{1}(t)&f_{2}(t)\\ f^{\prime}_{1}(t)&f^{\prime}_{2}(t)\\ \end{array}\right)* is not singular on $\Delta$.*

If Assumption 3.4 holds, then we could solve (7) and get

[TABLE]

Bringing $x_{1}(t),x_{2}(t)$ back to $(r_{f}(x;t))_{t}^{(i)}$, and denoting it as $s_{f}^{i}(t),$

[TABLE]

Let $\Gamma_{f}(\Delta)\subseteq\Delta$ be the set of real zeros of

[TABLE]

Some properties are presented, and for the sake of brevity, we put the proofs of Lemma 3.6, 3.7, 3.8 and 3.9 into Appendix.

Proposition 3.5**.**

If Assumption 3.4 holds, then

• (a)

$x_{1}(t),x_{2}(t),x_{1}^{\prime}(t),x_{2}^{\prime}(t)$* and $s_{f}^{i}(t),i=0,1,2$ are continues derivable functions on $\Delta$. $x_{1}^{\prime\prime}(t),x_{2}^{\prime\prime}(t)$ and $s_{f}^{3}(t)$ are continues functions on $\Delta$.*

• (b)

$s_{f}^{0}(t)=s_{f}^{1}(t)\equiv 0,\forall t\in\Delta.$**

Lemma 3.6**.**

If Assumption 3.4 holds, then

• (a)

If $x^{\prime}_{1}(t)=0,x^{\prime}_{2}(t)=0,t\in int(\Delta)$, then $t\in\Gamma_{f}(\Delta).$

• (b)

$\forall t\in\Gamma_{f}(\Delta)$, $x_{1}^{\prime}(t)=0,x_{2}^{\prime}(t)=0$.

• (c)

If $f_{1}(t)\neq 0,x_{2}^{\prime}(t)=0,t\in int(\Delta)$, then $x_{1}^{\prime}(t)=0,t\in\Gamma_{f}(\Delta).$

• (d)

Furthermore, if $s_{f}^{3}(t)\neq 0,f_{1}(t)\neq 0,\forall t\in\Gamma_{f}(\Delta)$, then $\forall t\in\Gamma_{f}(\Delta)$ must be all the strict minimum point or maximum point of $x_{2}(t).$

• (e)

If $s^{3}_{f}(t)\neq 0,\forall t\in int(\Delta),f_{1}(t)\neq 0$ $\forall t\in\Delta$, then $s^{2}_{f}(t)$ has one zero in $\Delta$ at most.

Lemma 3.7**.**

If Assumption 3.4 holds. Supposing $\#\{\Gamma_{f}(\Delta)\}<+\infty,$ then

• (a)

$(x_{2}(t),x_{1}(t))|_{t\in\Delta}$* is the envelop curve $H(x_{1},x_{2})=0$ of (7) in $\mathbb{R}^{2}.$*

• (b)

If $f_{1}(t)\neq 0,\forall t\in\Delta$, there is not a interval $\hat{\Delta}\subseteq\Delta$ such that $int(\hat{\Delta})\neq\emptyset,x_{2}(t)\equiv C,\forall t\in\hat{\Delta}.$

Lemma 3.7 gives us the chance to discuss $\displaystyle\frac{dx_{1}}{dx_{2}}$ on $\Delta.$

Lemma 3.8**.**

If Assumption 3.4 holds. Supposing that $f_{1}(t)\neq 0,\forall t\in\Delta$, $\#\{\Gamma_{f}(\Delta)\}<+\infty,$ then

[TABLE]

[TABLE]

Lemma 3.9**.**

If Assumption 3.4 holds. Supposing that $f_{1}(t)\neq 0,\forall t\in\Delta$, $\#\{\Gamma_{f}(\Delta)\}<+\infty$, then $\forall t\in\Delta$, the tangent line of $x_{1}=H(x_{2})$ at $(x_{2}(t),x_{1}(t))$ is

[TABLE]

Mark:(1) If $f_{1}(t)\neq 0,\forall t\in\Delta$, the change of the concavity and convexity of $H(x_{1},x_{2})=0$ only happens at $t\in\Gamma_{f}(\Delta)$.

(2) if $f_{1}(t)\neq 0,s_{f}^{2}(t)\neq 0,\forall t\in\Delta$, $H(x_{1},x_{2})=0$ must be strict convex or concave.

(3) Every point of $H(x_{1},x_{2})=0$ has its tangent line not perpendicular to $x_{2}$ axis, if $f_{1}(t)\neq 0,\forall t\in\Delta$.

Lemma 3.6 and 3.8 tell us when $x_{1}=H(x_{2})$ is a convex(concave) function. Using this result, we could present the following important theorem to give the expression of $S_{F}(\Delta).$

Theorem 3.10**.**

If Assumption 3.4 holds. Supposing that $\Gamma_{f}(\Delta)=\emptyset$, $f_{1}(t)\neq 0,\forall t\in\Delta$, then

[TABLE]

where

[TABLE]

[TABLE]

$E^{i}_{f}(\Delta),i=1,2,3,4,5,6$* are defined in (3.1).*

Proof.

If $\exists t^{*}\in int(\Delta),x^{\prime}_{2}(t^{*})=0,$ since $f_{1}(t)\neq 0,\forall t\in\Delta$, then by Lemma 3.6, we have $t^{*}\in\Gamma_{f}(\Delta)$, which is against to $\Gamma_{f}(\Delta)=\emptyset,$ so $x^{\prime}_{2}(t)\neq 0,\forall t\in int(\Delta).$ Then $x_{2}(t)$ is a strict increasing or decreasing function on $\Delta$. Without general, we suppose that $x_{2}(t)$ is strict increasing on $\Delta$, which means $x_{2}(\Delta)=[x_{2}(l),x_{2}(u)].$

Then by Lemma 3.8, there exists a second order derivable function

[TABLE]

which is the envelop curve of $r_{f}(x;t)=0,t\in\Delta.$

$\Gamma_{f}(\Delta)=\emptyset$ means that $s^{2}_{f}(t)\neq 0,\forall t\in int(\Delta)$, then by Lemma 3.8, $x_{1}=H(x_{2})$ is a strict convex or concave function on $[x_{2}(l),x_{2}(u)]$. Without general, we suppose that $x_{1}=H(x_{2})$ is strict convex in $[x_{2}(l),x_{2}(u)].$

Since $x_{1}=H(x_{2})$ is strict convex, then $\displaystyle\frac{f_{2}(l)}{f_{1}(l)}\neq\frac{f_{2}(u)}{f_{1}(u)}$, which means

[TABLE]

We could extend the domain of $H(x)$ to $\mathbb{R}$, and make the extended one satisfy conditions in Lemma 7.2. Then $L_{F}(l)$, $L_{F}(u)$ and $x_{1}=H(x_{2})$ separate $\mathbb{R}^{2}$ into 6 parts.

[TABLE]

We will prove

[TABLE]

where

[TABLE]

Noting that $\forall t\in\Delta$ by Lemma 3.9,

[TABLE]

is the tangent line of $x_{1}=H(x_{2})$ at $(x_{2}(t),x_{1}(t)).$ Since $x_{1}=H(x_{2})$ is strict convex and derivable, then there is only one tangent line of $L_{F}(t)$ at $x_{2}\in[x_{2}(l),x_{2}(u)]$, which means for every tangent line of $x_{1}=H(x_{2})$ at $x_{2}\in[x_{2}(l),x_{2}(u)]$, there is only one $t\in\Delta$ corresponding to it.

(a) By Corollary 7.3, $\forall(x_{2},x_{1})\in E^{1}_{f}(\Delta)$, so

[TABLE]

Since $(x_{2},x_{1})\in int(A(x_{2}(l),H))$, there exists only one $z^{*}>x_{2}(l)$ such that $l(z^{*},H)$ passes through $(x_{2},x_{1})$. Since $(x_{2},x_{1})\in int(A(x_{2}(u),H))$, then $z^{*}>x_{2}(u).$ So $N_{F}((x_{1},x_{2}),\Delta)=0,\forall(x_{2},x_{1})\in E^{1}_{f}(\Delta)$.

(b) By Corollary 7.3, $\forall(x_{2},x_{1})\in E^{2}_{f}(\Delta)$, so

[TABLE]

If $(x_{2},x_{1})\in int(A(x_{2}(l),H))$, there exists only one $z^{*}>x_{2}(l)$ such that $l(z^{*},H)$ passes through $(x_{2},x_{1})$. Since $(x_{2},x_{1})\in C(x_{2}(u),H)$, then $z^{*}\leq x_{2}(u).$

If $(x_{2},x_{1})\in\partial(A(x_{2}(l),H))$, by the same discussion, besides $l(x_{2}(l),H)$, there exists $l(z^{*},H)$ passing through $(x_{2},x_{1})$ with $z^{*}$ such that $x_{2}(l)<z^{*}\leq x_{2}(u)$.

So $2\geq N_{F}((x_{1},x_{2}),\Delta)\geq 1,\forall(x_{2},x_{1})\in E^{2}_{f}(\Delta)$, $E^{2}_{f}(\Delta)\subset\Lambda^{1}_{N_{F}}(\Delta)$.

(c) By the same way, $E^{3}_{f}(\Delta)\subset\Lambda^{1}_{N_{F}}(\Delta)$.

(d) By Corollary 7.3, $\forall(x_{2},x_{1})\in E^{4}_{f}(\Delta)$, we have

[TABLE]

there must be only two tangent lines $l(z_{1}^{*},H)$ and $l(z_{2}^{*},H)$ with tangent points $z_{1}^{*},z^{*}_{2}$ passing through $(x_{2},x_{1})$, respectively, such that $x_{2}(l)<z_{1}^{*}<z_{2}^{*}<x_{2}(u)$. Then $N_{F}((x_{1},x_{2}),\Delta)=2,$ which means

[TABLE]

(e) $\forall(x_{2},x_{1})\in\{(x_{2},x_{1})|x_{1}=H(x_{2}),x_{2}\in[x_{2}(l),x_{2}(u)]\}$, since $x_{1}=H(x_{2})$ is strict convex and derivable on $\Delta$, there exists only one tangent line

[TABLE]

which means $\{(x_{2},x_{1})|x_{1}=H(x_{2}),x_{2}\in[x_{2}(l),x_{2}(u)]\}\subset\Lambda^{1}_{N_{F}}(\Delta).$ So it is easy to see that

[TABLE]

and

[TABLE]

(f) We define a new function $x_{1}=\bar{H}(x_{2})$ as

[TABLE]

Then $E^{6}_{f}(\Delta)=\{(x_{2},x_{1})|x_{1}>\bar{H}(x_{2})\}$ . So $\forall(x_{2},x_{1})\in E^{6}_{f}(\Delta)$,

[TABLE]

can’t be a tangent line of $x_{1}=H(x_{2})$, which means

[TABLE]

So we could get

[TABLE]

Furthermore, it is easy to see that

[TABLE]

So

[TABLE]

Corollary 3.11**.**

If Assumption 3.4 holds. Supposing that $\Gamma_{f}(\Delta)=\{t^{*}\}$, $f_{1}(t)\neq 0,\forall t\in\Delta$, then

[TABLE]

where

[TABLE]

Proof.

By Theorem 3.10,

[TABLE]

then

[TABLE]

Since $\Lambda^{i}_{N_{F}}(\Delta)=\displaystyle\bigcup_{j\geq 0,k\geq 0,j+k=i}(\Lambda^{k}_{N_{F}}([l,t^{*}])\bigcap\Lambda^{j}_{N_{F}}((t^{*},u])),$ then

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Theorem 3.12**.**

When $n=2$, if Assumption 2.1(b)(c), 3.3 and 3.4 hold. Given $t^{*}\in int(\Delta)$ satisfied the following conditions

• (a)

$f_{1}(t^{*})\neq 0$, $t^{*}\not\in\Gamma_{f}(\Delta)$.

• (b)

[TABLE]

exist and are same.

Then

[TABLE]

Proof.

Since Assumption 3.3 holds, there exists $\varepsilon_{1}>0$ small enough such that $f_{1}(t)\neq 0$, $\forall t\in[t^{*},t^{*}+\varepsilon_{1}]$. Since $t^{*}\not\in\Gamma_{f}(\Delta)$, there exists $\varepsilon_{2}>0$ small enough such that $s^{2}_{f}(t)\neq 0$ $\forall t\in[t^{*},t^{*}+\varepsilon_{2}]$, $\Gamma_{f}([t^{*},t^{*}+\varepsilon_{2}])=\emptyset$.

Let $0<\bar{\varepsilon}\leq\min\{\varepsilon_{1},\varepsilon_{2}\}$, then $x_{1}=H(x_{2})$ exists on $[x_{2}(t^{*}),x_{2}(t^{*}+\bar{\varepsilon})]$ or $[x_{2}(t^{*}+\bar{\varepsilon}),x_{2}(t^{*})]$ with $H^{\prime\prime}(x_{2})\neq 0$, and keeps its concavity and convexity. So $\displaystyle-\frac{f_{2}(t)}{f_{1}(t)}$ must be strict increasing or decreasing, which means $\forall t\in(t^{*},t^{*}+\bar{\varepsilon}],$

[TABLE]

So $\forall\varepsilon\in(0,\bar{\varepsilon}]$, the intersection point $(x_{2}^{*},x_{1}^{*})$ of $l(x_{2}(t^{*}),H)$ and $l(x_{2}(t^{*}+\varepsilon),H)$ exists and satisfies

[TABLE]

and

[TABLE]

By Corollary 7.3, it is easy to get that $x_{2}^{*}$ is between $x_{2}(t^{*})$ and $x_{2}(t^{*}+\varepsilon)$.

By Theorem 3.10, $\forall\varepsilon\in(0,\bar{\varepsilon}]$, we have

[TABLE]

where

[TABLE]

[TABLE]

$E^{i}_{f}(\cdot),i=1,\cdots,5$ are defined in (3.1).

Then $\displaystyle\int_{S_{F}([t^{*},t^{*}+\varepsilon])}\rho(x)dx=\sum_{i=1}^{2}\int_{\Lambda^{i}_{N_{F}}([t^{*},t^{*}+\varepsilon])}\rho(x)dx,$ and

[TABLE]

[TABLE]

We will prove $\displaystyle\int_{cl(E^{4}_{f}([t^{*},t^{*}+\varepsilon]))}\rho(x)dx=o(\varepsilon).$

Since $x_{1}=H(x_{2})$ is convex (concave) on $[x_{2}(t^{*}),x_{2}(t^{*}+\varepsilon)]([x_{2}(t^{*}+\varepsilon),x_{2}(t^{*})])$, it is easy to see that

[TABLE]

Then

[TABLE]

and the area of $\Psi$ is

[TABLE]

if $f_{0}(t^{*})\neq 0,f_{2}(t^{*})\neq 0$. It is easy to see that the above result holds too, if $f_{0}(t^{*})=0$ or $f_{2}(t^{*})=0.$

So

[TABLE]

Now we get

[TABLE]

exists.

It is easy to see that $E^{2}_{f}([t^{*},t^{*}+\varepsilon])\bigcup E^{3}_{f}([t^{*},t^{*}+\varepsilon])$ is just $\Phi(L_{F}(t^{*}),L_{F}(t^{*}+\varepsilon))$, when $\varepsilon$ is small enough, so

[TABLE]

By the same way, we could prove that

[TABLE]

So

[TABLE]

Assumption 3.13**.**

There are finite number of $t\in\Delta$ such that $W_{f_{1},f_{2}}(t)$ is singular.

Corollary 3.14**.**

When $n=2$, if Assumption 2.1(b)(c), 2.2, 3.3 and 3.13 hold. Supposing that

• (a)

$\#\{\Gamma_{f}(\Delta)\}<+\infty$*, *

• (b)

the number of $t\in int(\Delta)$ that does’t hold the condition(b) in Theorem 3.12 is finite,

then

[TABLE]

where $H_{\rho,F}(t)$ is defined in Theorem 3.12.

Proof.

Since Assumption 2.2 holds, $E_{2}(\rho,F,\Delta)$ exists. We call the point $t\in\Delta$ satisfied either $f_{1}(t)=0$ or that $W_{f_{1},f_{2}}(t)$ is singular or that condition(b) in Theorem 3.12 doesn’t hold or $s_{f}^{2}(t)=0$ as the bad points. By the conditions, the number of bad points is finite, we could separate $\Delta$ into several closed intervals $\Delta^{1}_{i},\Delta^{2}_{i}$ connected end to end, where $\Delta_{i}^{1}$ doesn’t have any bad point, $\displaystyle\bigcup_{i}\Delta_{i}^{2}$ includes all the bad points, while every $\Delta_{i}^{2}$ just has one bad point, and $|\Delta_{i}^{2}|=\varepsilon$ with a small enough $\varepsilon>0$. Except the case that $l$ or $u$ is the bad point, we all ask the bad point must be at the middle of $\Delta_{i}^{2}.$

Then

[TABLE]

so

[TABLE]

Mark: If $\Gamma_{f}(\Delta)$ is consisted of finite number of points and intervals, we could get same results of Corollary 3.14.

3.2 How to compute $E_{n}(\rho,F,\Delta)$ with $n\geq 4$

In this section, we consider the case: $n\geq 4$. Let

[TABLE]

then

[TABLE]

Denote

[TABLE]

for $\forall\bar{x}\in\mathbb{R}^{n-2},\Lambda^{i}_{G_{\bar{x}}}(A)\subseteq\mathbb{R}^{2},A\subseteq\mathbb{R}.$

The following theorem is one of main theorems of this paper.

Theorem 3.15**.**

When $n\geq 4$, if Assumption 2.1, 2.2, 3.3 and 3.4 hold. Given $t^{*}\in int(\Delta)$ satisfied

• (a)

$f_{1}(t^{*})\neq 0,$**

• (b)

\left(\begin{array}[]{c}h(t^{*})\\ u(t^{*})\\ \end{array}\right)* is full rank, where $h(t^{*})$ and $u(t^{*})$ are row vectors defined in (30) and (31), respectively,*

• (c)

[TABLE]

exist and are same.

Then

[TABLE]

Proof.

For $t^{*}\in int(\Delta)$, by Assumption 3.4, there exists $\varepsilon^{*}_{1}>0$ such that

[TABLE]

By Assumption 3.3, there exists $\varepsilon_{2}^{*}>0$, such that $f_{1}(t)\neq 0,\forall t\in[t^{*},t^{*}+\varepsilon^{*}_{2}].$

There exists $\varepsilon_{3}^{*}>0$ such that $\forall\varepsilon\in(0,\varepsilon_{3}^{*}]$, then \left(\begin{array}[]{cc}f_{1}(t^{*})&f_{2}(t^{*})\\ f_{1}(t^{*}+\varepsilon)&f_{2}(t^{*}+\varepsilon)\\ \end{array}\right) is not singular. If it is not true, then $\forall\bar{\varepsilon}>0,$ there exists $\hat{\varepsilon}\in(0,\bar{\varepsilon}]$ such that \left(\begin{array}[]{cc}f_{1}(t^{*})&f_{2}(t^{*})\\ f_{1}(t^{*}+\hat{\varepsilon})&f_{2}(t^{*}+\hat{\varepsilon})\\ \end{array}\right) is singular. Let $\bar{\varepsilon}\rightarrow 0$, then $\hat{\varepsilon}\rightarrow 0,$

[TABLE]

is singular, which is against to that $W_{f_{1},f_{2}}(t^{*})$ is not singular.

Let $\varepsilon^{*}=\min\{\varepsilon_{1}^{*},\varepsilon_{2}^{*},\varepsilon_{3}^{*},0.5\}.$

[TABLE]

where

[TABLE]

[TABLE]

Since $f^{(i)}_{j}(t),i=0,1,2,j=0,1,\cdots,n$ are continuous derivable functions on $\Delta$, $f^{{}^{\prime\prime\prime}}_{j}(t),j=0,1,\cdots,n$ are continuous functions on $\Delta$, and $f_{1}(t)f_{2}^{\prime}(t)-f_{1}^{\prime}(t)f_{2}(t)\neq 0,$ $\forall t\in[t^{*},t^{*}+\varepsilon^{*}]$, then $f_{0}^{\prime\prime}(t),f^{\prime\prime\prime}_{0}(t),h_{i}(t)$ and $u_{i}(t)$ are all Lip-continuous functions on $[t^{*},t^{*}+\varepsilon^{*}]$ with positive Lip constants $\alpha_{2,0},\alpha_{3,0},\beta_{2,i}$ and $\beta_{3,i}$ respectively.

Since $h(t^{*})\neq 0$, we always could take $0<\varepsilon<\varepsilon^{*}$ satisfied

[TABLE]

then we could separate $\mathbb{R}^{n}$ into following 4 parts

[TABLE]

It is easy to get $\Omega^{i}\bigcap\Omega^{j}=\emptyset,i\neq j$, $\mathbb{R}^{n}=\displaystyle\bigcup_{i=1}^{4}\Omega^{i}.$ Given $\varepsilon\in(0,\varepsilon^{*})$, we need to compute $\displaystyle\int_{S_{F}([t^{*},t^{*}+\varepsilon])}\rho(x)dx$.

Noting that

[TABLE]

we just need to compute $\displaystyle\int_{\Lambda_{F}^{i}([t^{*},t^{*}+\varepsilon])}\rho(x)dx.$

For every $\displaystyle\int_{\Lambda_{N_{F}}^{i}([t^{*},t^{*}+\varepsilon])}\rho(x)dx$, we have

[TABLE]

For convenience, denoting $\Omega^{i}_{j}=\Omega^{i}\bigcap\Lambda_{N_{F}}^{j}([t^{*},t^{*}+\varepsilon]).$ Noting that

[TABLE]

(1) $\displaystyle\int_{\Omega^{1}_{i}}\rho(x)dx.$

[TABLE]

Furthermore, by Assumption 2.2,

[TABLE]

(2) $\displaystyle\int_{\Omega^{2}_{i}}\rho(x)dx.$

Since \left(\begin{array}[]{c}h(t^{*})\\ u(t^{*})\\ \end{array}\right) is full rank, $h(t^{*})\neq 0$, by the definition of $\varepsilon$, we get $\Omega^{2}\neq\emptyset.$ $\forall\bar{x}\in\mathbb{P}(\Omega^{2}),\forall\bar{\varepsilon}\in(0,\varepsilon]$, we have that

[TABLE]

which means $\Gamma_{g_{\bar{x}}}([t^{*},t^{*}+\varepsilon])=\emptyset,\forall\bar{x}\in\mathbb{P}(\Omega^{2}).$

By Theorem 3.10, we have $\Lambda_{N_{G_{\bar{x}}}}^{i}([t^{*},t^{*}+\varepsilon])=\emptyset,i\geq 3,\forall\bar{x}\in\mathbb{P}(\Omega^{2}).$

(2.1) $\Lambda_{N_{G_{\bar{x}}}}^{i}([t^{*},t^{*}+\varepsilon])=\emptyset,i\geq 3,$ when $\bar{x}\in\mathbb{P}(\Omega^{2})$. So

[TABLE]

We just need to compute

[TABLE]

(2.2) $i=2.$

[TABLE]

Since $(g_{\bar{x}})_{1}(t^{*})=f_{1}(t^{*})\neq 0$, $\Gamma_{g_{\bar{x}}}([t^{*},t^{*}+\varepsilon])=\emptyset,$ $\forall(y,\bar{x})\in\Omega^{2}$, by the proof of Theorem 3.12, we know that

[TABLE]

where

[TABLE]

[TABLE]

So $x_{1}^{\prime}(t)=(p^{1}_{0})^{\prime}(t)+\displaystyle\sum_{i=3}^{n}(p^{1}_{i})^{\prime}(t)x_{i}$, $x^{\prime}_{2}(t)=(p^{2}_{0})^{\prime}(t)+\displaystyle\sum_{i=3}^{n}(p^{2}_{i})^{\prime}(t)x_{i}.$

$\forall\bar{x}\in\mathbb{P}(\Omega^{2})$,

[TABLE]

so

[TABLE]

$\forall\bar{x}\in\mathbb{P}(\Omega^{2}).$

Then

[TABLE]

(2.3) $i=1$, then

[TABLE]

(3) $\displaystyle\int_{\Omega^{3}_{i}}\rho(x)dx$

If $\Omega^{3}=\emptyset,$ then $\displaystyle\int_{\Omega^{3}_{i}}\rho(x)dx=0,\forall i,$ we only need to discuss the case $\Omega^{3}\neq\emptyset.$

By the same discussion in (2), $\forall\bar{\varepsilon}\in(0,\varepsilon],$

[TABLE]

Since $\bar{x}\in\mathbb{P}(\Omega^{3})$, then

[TABLE]

so $\forall\bar{\varepsilon}\in(0,\varepsilon]$, we have

[TABLE]

which means that $s^{3}_{g_{\bar{x}}}(t^{*}+\bar{\varepsilon})\neq 0,\forall\bar{\varepsilon}\in(0,\varepsilon]$. By Lemma 3.6(e), we get either

[TABLE]

If $\Gamma_{g_{\bar{x}}}([t^{*},t^{*}+\varepsilon])=\emptyset$, we could get the same result in (2).

If $\hat{t}\in(t^{*},t^{*}+\varepsilon)$ and $\bar{x}\in\mathbb{P}(\Omega^{3})$, then $|s^{2}_{g_{\bar{x}}}(t^{*})|<2(\alpha_{2,0}+\displaystyle\sqrt{n}\sum_{i=3}^{n}\beta_{2,i}(-\frac{2\ln\varepsilon}{b}))\varepsilon$ and $||\bar{x}||<\displaystyle-\frac{2\ln\varepsilon}{b}.$

(3.1) $i=1$.

[TABLE]

Now we discuss

[TABLE]

and

[TABLE]

Since $\Gamma_{g_{\bar{x}}}([t^{*},\hat{t}])=\emptyset$, by the same discussion in Theorem 3.12, \left(\begin{array}[]{cc}f_{1}(t^{*})&f_{2}(t^{*})\\ f_{1}(\hat{t})&f_{2}(\hat{t})\\ \end{array}\right) is not singular, then

[TABLE]

and

[TABLE]

must has a intersection point

[TABLE]

So

[TABLE]

where

[TABLE]

[TABLE]

Then

[TABLE]

So

[TABLE]

It is easy to see that

[TABLE]

where $\theta=|\arctan\displaystyle\frac{f_{1}(t^{*})f_{2}(\hat{t})-f_{2}(t^{*})f_{1}(\hat{t})}{f_{1}(t^{*})f_{1}(\hat{t})+f_{2}(t^{*})f_{2}(\hat{t})}|\leq O(\varepsilon)$.

So

[TABLE]

By the same way,

[TABLE]

So

[TABLE]

since $|\Lambda^{2}_{N_{G_{\bar{x}}}}([t^{*},\hat{t}])|=o(\varepsilon)$, $|\Lambda^{2}_{N_{G_{\bar{x}}}}([\hat{t},t^{*}+\varepsilon])|=o(\varepsilon),$ which could be proved just like in (2.2).

(3.2) $i=2,3,4,$

[TABLE]

(3.3) $i\geq 5.$ Whatever $\Gamma_{g_{\bar{x}}}([t^{*},t^{*}+\varepsilon])=\emptyset$ or $\Gamma_{g_{\bar{x}}}([t^{*},t^{*}+\varepsilon])=\{\hat{t}\},\forall(y,\bar{x})\in\Omega^{3}$, by Corollary 3.11,

[TABLE]

so

[TABLE]

(4) $\displaystyle\int_{\Omega^{4}_{i}}\rho(x)dx$

Since $h(t^{*})$ and $u(t^{*})$ are linear independent, then

[TABLE]

Furthermore, by Assumption 2.2,

[TABLE]

So

[TABLE]

[TABLE]

Then,

[TABLE]

Next we will prove

[TABLE]

Noting that given $\bar{x}\in\mathbb{P}(\Omega^{2})$, $s^{2}_{g_{\bar{x}}}(t)$ doesn’t change its sign on $[t^{*},t^{*}+\varepsilon]$, which means that there exists $x_{1}=H_{\bar{x}}(x_{2})$ which is a convex or concave function.

Since $0<\varepsilon<\varepsilon^{*}_{3}$, \left(\begin{array}[]{cc}f_{1}(t^{*})&f_{2}(t^{*})\\ f_{1}(t^{*}+\varepsilon)&f_{2}(t^{*}+\varepsilon)\\ \end{array}\right) is not singular, then

[TABLE]

and

[TABLE]

separate $\mathbb{R}^{2}$ into 2 sets $\Phi(L_{G_{\bar{x}}}(t^{*}),L_{G_{\bar{x}}}(t^{*}+\varepsilon))$ and $\mathbb{R}^{2}\setminus\Phi(L_{G_{\bar{x}}}(t^{*}),L_{G_{\bar{x}}}(t^{*}+\varepsilon)).$ Then $L_{G_{\bar{x}}}(t^{*})$ and $L_{G_{\bar{x}}}(t^{*}+\varepsilon)$ are the tangent lines of $x_{1}=H_{\bar{x}}(x_{2})$. Then the image of $x_{1}=H_{\bar{x}}(x_{2})$ must be in $cl(\mathbb{R}^{2}\setminus\Phi_{\bar{x}}(t^{*},t^{*}+\varepsilon)),$ so by Theorem 3.10,

[TABLE]

when $\varepsilon$ is small enough, so

[TABLE]

Noting that

[TABLE]

[TABLE]

If $\Omega^{3}=\emptyset,$

[TABLE]

If $\Omega^{3}\neq\emptyset,$

[TABLE]

where $\theta=|\arctan\displaystyle\frac{f_{1}(t^{*})f_{2}(t^{*}+\varepsilon)-f_{2}(t^{*})f_{1}(t^{*}+\varepsilon)}{f_{1}(t^{*})f_{1}(t^{*}+\varepsilon)+f_{2}(t^{*})f_{2}(t^{*}+\varepsilon)}|\leq O(\varepsilon),$ so

[TABLE]

Then

[TABLE]

Now we have

[TABLE]

By the same way, we could prove that

[TABLE]

when $\varepsilon>0$ is small enough.

Then if

[TABLE]

exist and are same,

[TABLE]

Mark: From above proof, it is easy to see that if we use the following assumption:

[TABLE]

to instead of Assumption 2.1(a), we could get all the same result.

We could use the following lemma to verify the condition(b) of Theorem 3.15.

Lemma 3.16**.**

If $W_{f_{3},f_{4}}(t)$ is not singular, then \left(\begin{array}[]{c}h(t)\\ u(t)\\ \end{array}\right) is full rank, where $h(t)$ and $u(t)$ are defined in (30) and (31), respectively.

Proof.

Since $W_{f_{3},f_{4}}(t)$ is not singular, then

[TABLE]

is full rank,

[TABLE]

is full rank too, where

[TABLE]

[TABLE]

So \left(\begin{array}[]{c}u(t)\\ h(t)\\ \end{array}\right) is full rank.∎

Corollary 3.17**.**

When $n\geq 4$, if Assumption 2.1, 2.2, 3.3 and 3.13 hold. Supposing that

• (a)

there are finite number of $t\in\Delta$ such that $W_{f_{1},f_{2}}(t)$ or $W_{f_{3},f_{4}}(t)$ is singular.

• (b)

$\forall t\in int(\Delta)$* such that $f_{1}(t)\neq 0,$ then condition(c) in Theorem 3.15 doesn’t hold at finite number of points at most.*

Then

[TABLE]

Proof.

Since Assumption 2.2 holds, $E_{n}(\rho,F,\Delta)$ exists. Since there are finite number of $t\in int(\Delta)$ such that either $W_{f_{1},f_{2}}(t)$ singular, or $W_{f_{3},f_{4}}(t)$ singular, or $f_{1}(t)=0$, or the condition(c) in Theorem 3.10 doesn’t hold, we call this points as bad points. We could separate $\Delta$ into closed intervals $\Delta_{i}^{1}$ and $\Delta_{i}^{2}$ connected end to end, where $\Delta_{i}^{1}$ doesn’t have any bad point, $\displaystyle\bigcup_{i}\Delta_{i}^{2}$ includes all the bad points, while every $\Delta_{i}^{2}$ just has one bad point, and $|\Delta_{i}^{2}|=\varepsilon$ with a small enough $\varepsilon>0$. Except the case that $l$ or $u$ is the bad point, we all ask the bad point must be at the middle of $\Delta_{i}^{2}.$

Since $W_{f_{3},f_{4}}(t)$ is not singular $\forall t\in\Delta_{i}^{1}$, then $u(t)$ and $h(t)$ defined in (30) and (31) are linear independent, for all $t\in\Delta_{i}^{1}.$

Then

[TABLE]

so

[TABLE]

4 When $x$ obeys the uniform distribution in a sup-sphere

Now we are at the position to consider a practical problem. It is well known that the Lebesgue measure of sup-sphere $B^{n}(0,r)\subset\mathbb{R}^{n}$ with radii $r$ is

[TABLE]

In this section, we discuss the case that $\rho(x)=\rho_{r}(x).$ It is easy to see that $\rho_{r}(x)$ satisfies Assumption 2.1. From Theorem 3.15, we just need to get $H_{\rho_{r},F}(t).$

Theorem 4.1**.**

When $n\geq 4$, if Assumption 2.2, 3.3 and 3.4 hold. Given $t^{*}\in int(\Delta),f_{1}(t^{*})\neq 0$, $W_{f_{1},f_{2}}(t^{*})$ and $W_{f_{3},f_{4}}(t^{*})$ are not singular, and

[TABLE]

Let

[TABLE]

[TABLE]

[TABLE]

• (a)

When $|a|<r,\chi^{2}+a^{2}<r^{2},$

[TABLE]

• (b)

When $|a|<r,$ $\chi<-\sqrt{r^{2}-a^{2}},$

[TABLE]

• (c)

When $|a|<r,$ $\chi>\sqrt{r^{2}-a^{2}},$

[TABLE]

• (d)

When $|a|>r,$

[TABLE]

Proof.

Since Assumption 2.1 holds for $\rho_{r}(x)$, by Theorem 3.15, we just need to compute

[TABLE]

Let $t\in int(\Delta)$, $|\varepsilon|$ be small enough, we consider $L_{F}(t^{*})$ and $L_{F}(t^{*}+\varepsilon)$. Denoting

[TABLE]

By QR decomposition, we have $Q\in\mathbb{R}^{n\times n}$ and $R\in\mathbb{R}^{n\times 2}$ such that $QR=B$, where $Q$ is a orthogonal matrix, $R$ is a upper triangular matrix with positive diagonal elements.

Let

[TABLE]

[TABLE]

with

[TABLE]

[TABLE]

When $|\varepsilon|$ is small enough, $C_{1,2}(\varepsilon)\neq 0,$ and keep its sign, since $\displaystyle\sum_{i=1}^{n}f^{2}_{i}(t^{*})\neq 0.$

We could rotate $L_{F}(t^{*})$ and $L_{F}(t^{*}+\varepsilon)$ into

[TABLE]

[TABLE]

Let

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By Theorem 3.10, since $W_{f_{1},f_{2}}(t^{*})$ and $W_{f_{3},f_{4}}(t^{*})$ are not singular, if

[TABLE]

exists, then

[TABLE]

exist and are same, then they are all $H_{\rho_{r},F}(t^{*})$, so we only need to discuss

[TABLE]

which is just

[TABLE]

We present some values needed here, where $\varepsilon>0$ is small enough.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

if $b^{\prime}(0)\neq 0.$

Basing on the relationship among the $l$, $l(\varepsilon)$ and $B^{n}(0,r)$, we just need to discuss 3 different cases.

• (a)

$f_{0}(t^{*})<0$

• (a.1)

$|a|<r$, $\chi^{2}+a^{2}<r^{2}$.

• (a.2)

$|a|<r$, $\chi<-\sqrt{r^{2}-a^{2}}.$

• (a.3)

$|a|<r$, $\chi>\sqrt{r^{2}-a^{2}}.$

• (a.4)

$|a|>r.$

• (b)

$f_{0}(t^{*})=0$

• (b.1)

$|a|<r$, $\chi^{2}+a^{2}<r^{2}$.

• (b.2)

$|a|<r$, $\chi<-\sqrt{r^{2}-a^{2}}.$

• (b.3)

$|a|<r$, $\chi>\sqrt{r^{2}-a^{2}}.$

• (c)

$f_{0}(t^{*})>0$

• (c.1)

$|a|<r$, $\chi^{2}+a^{2}<r^{2}$.

• (c.2)

$|a|<r$, $\chi<-\sqrt{r^{2}-a^{2}}.$

• (d.3)

$|a|<r$, $\chi>\sqrt{r^{2}-a^{2}}.$

• (c.4)

$|a|>r.$

In (a), there are 4 subcases. It is obviously that in (a.4) $H_{\rho,F}(t^{*})=0.$ In other three cases, we just prove (a.1) in detail, since the proofs of the left two cases are similar.

When $|\varepsilon|$ is small enough, $\displaystyle\lim_{\varepsilon\rightarrow 0}\displaystyle|\frac{c(\varepsilon)}{\sqrt{1+b^{2}(\varepsilon)}}|=|c(0)|=|a|<r,$ which means

[TABLE]

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

[TABLE]

since $b^{\prime}(0)\neq 0.$

[TABLE]

where

[TABLE]

[TABLE]

(a.1.1)

[TABLE]

where

[TABLE]

Since

[TABLE]

so

[TABLE]

(a.1.2)

[TABLE]

where

[TABLE]

Since

[TABLE]

so

[TABLE]

The proof of (a.1.3) and (a.1.4) are similar as (a.1.1) and (a.1.2), respectively. Then we get

(a.1.3)

[TABLE]

(a.1.4)

[TABLE]

Now we have

[TABLE]

(a.2) When $|\varepsilon|$ is small enough, $\displaystyle\lim_{\varepsilon\rightarrow 0}\displaystyle|\frac{c(\varepsilon)}{\sqrt{1+b^{2}(\varepsilon)}}|=|c(0)|=|a|<r,$ which means

[TABLE]

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By the same discussion in (a.1), we could get

[TABLE]

(a.3) When $|\varepsilon|$ is small enough, $\displaystyle\lim_{\varepsilon\rightarrow 0}\displaystyle|\frac{c(\varepsilon)}{\sqrt{1+b^{2}(\varepsilon)}}|=|c(0)|=|a|<r,$ which means

[TABLE]

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where

[TABLE]

[TABLE]

By the same discussion in (a.1), we could get

[TABLE]

For all other cases in (b) and (c), we all get the same result in (a). ∎

Mark: (a) There is only one case that Theorem 4.1 doesn’t mention, which is $|a|=r.$ If we assume that there are finite number of $t\in\Delta$ such that

[TABLE]

then it doesn’t affect to compute $E_{n}(\rho_{r},F,\Delta),$ since the continuity of $E_{n}(\rho,F,[l,t]).$

(b) If $f_{0}(t)\equiv 0,\forall t\in\Delta$ then

[TABLE]

Corollary 4.2**.**

$n\geq 2,f_{0}(t)=d,f_{2k-1}(t)=\cos(kt)$, $f_{2k}=\sin(kt),k=1,\cdots,n,$ $\Delta=[0,2\pi],$

[TABLE]

Proof.

It is easy to see that all conditions in Theorem 4.1 hold for this problem.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

if $\displaystyle\frac{|d|}{\sqrt{n}}<r,n\geq 2.$ So $E_{n}(\rho_{r},F,\Delta)=2\displaystyle\sqrt{\displaystyle\frac{(n+1)(2n+1)}{6}}\frac{(r^{2}-\frac{d^{2}}{n})^{n}}{r^{2n}}$, if $\displaystyle\frac{|d|}{\sqrt{n}}<r,n\geq 2.$

If $\displaystyle\frac{|d|}{\sqrt{n}}>r,n\geq 2$, $H_{\rho_{r},F}(t)=0,E_{n}(\rho_{r},F,\Delta)=0.$

If $\displaystyle\frac{|d|}{\sqrt{n}}=r,n\geq 2$, it is easy to see that $E_{n}(\rho_{r},F,\Delta)=0.\qed$

Corollary 4.3**.**

$n\geq 4,f_{0}(t)=d$, $|d|\sqrt{\varpi^{2}+1}<r,$ $\displaystyle\varpi=\frac{\sqrt{3}\,\left(n-1\right)}{\sqrt{n}\,\sqrt{n^{2}-1}},$ $f_{i}(t)=t^{i-1},i=1,\cdots,n,$ $\Delta=[l,u],$

[TABLE]

where $\chi,\alpha,\beta,a$ and $h(a,x_{2})$ are defined in Theorem 4.1.

When $d=0,$ we could get

[TABLE]

[TABLE]

Proof.

It is easy to see that all conditions in Theorem 4.1 hold for this problem.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

so only when $t=\pm 1$,

[TABLE]

Since

[TABLE]

[TABLE]

then

[TABLE]

where

[TABLE]

When $d=0,$ we could get

[TABLE]

[TABLE]

5 When $x$ does’t obey the uniform distribution

In this section, we discuss the case that $\rho(x)$ has a more general form.

Assumption 5.1**.**

There exists a measurable function $\phi(t)$ such that $\rho(x)=\displaystyle\int_{||x||}^{+\infty}\phi(t)dt$.

If Assumption 5.1 holds, $\phi(t)=-\varphi^{\prime}(t),$ where $\varphi(||x||)=\rho(x).$

Theorem 5.2**.**

If Assumption 5.1 holds, then

[TABLE]

Proof.

[TABLE]

Noting that

[TABLE]

is $E_{n}({\rho}_{t},F,\Delta)$, so we get

[TABLE]

Corollary 5.3**.**

$n\geq 2$, $f_{0}(t)=d,f_{2k-1}(t)=\cos(kt)$, $f_{2k}=\sin(kt),k=1,\cdots,n.$ $\Delta=[0,2\pi]$, $\rho(x)=\displaystyle\frac{1}{{(2\pi\sigma)}^{n}}e^{-\frac{||x||^{2}}{2\sigma}},$

[TABLE]

Proof.

Since $\rho(x)=\displaystyle\frac{1}{{(2\pi\sigma)}^{n}}e^{-\frac{||x||^{2}}{2\sigma}}$, $\phi(t)=\displaystyle\frac{t}{(2\pi)^{n}\sigma^{n+1}}e^{-\frac{t^{2}}{2\sigma}}.$

[TABLE]

It is easy to check

[TABLE]

with $u=d\sqrt{n},\sigma=1,$ which is just the main result of [10].

Corollary 5.4**.**

$n\geq 4$, $f_{0}(t)=0,f_{i}(t)=t^{i-1},i=1,\cdots,n,n\geq 4.$ $\rho(x)=\displaystyle\frac{1}{{(2\pi\sigma)}^{\frac{n}{2}}}e^{-\frac{||x||^{2}}{2\sigma}}.$ $\Delta=[l,u],$

[TABLE]

Proof.

Since $E_{n}(\rho_{t},F,[l,u])=\displaystyle\frac{1}{\pi}\int_{l}^{u}\frac{\sqrt{(t^{2n}-1)^{2}-n^{2}t^{2n-2}(t^{2}-1)^{2}}}{(t^{2}-1)(t^{2n}-1)}dt,$

[TABLE]

where $\phi(t)=\displaystyle\frac{t}{(2\pi)^{\frac{n}{2}}\sigma^{\frac{n}{2}+1}}e^{-\frac{t^{2}}{2\sigma}}.$

So

[TABLE]

which is just the result of [7].

6 Conclusion and Further works

We discuss how to get analytical expressions of $E_{n}(\rho,F,\Delta)$ in this paper. Firstly we discuss the prosperities of $E_{n}(\rho,F,\Delta)$ for general cases. Then we give some expressions of $E_{n}(\rho,F,\Delta)$ by calculate $H_{\rho,F}(t)$, when $F$ is linear in $x$. At last, for some practical problems, e.g. $x$ obeys the uniform distribution and the normal distribution, we give more concise results, which are same as the results existed.

Our result suits for every $n$, except $n=1,3.$ In the further work, we will go for it.

Acknowledgements: Yi Xu was supported in part by the National Natural Science Foundation of China (No. 11501100, 11571178, 11671082 and 11871149).

7 Appendix

7.1 Proofs of Lemma 3.6, 3.7, 3.8 and 3.9

The proof of Lemma 3.6:

Proof.

[TABLE]

(a) Since $t\in int(\Delta)$, $x^{\prime}_{1}(t)=x_{2}^{\prime}(t)=0$, then $s_{f}^{2}(t)=0$, which means $t\in\Gamma_{f}(\Delta).$

(b) Since $t\in\Gamma_{f}(\Delta)$, then

[TABLE]

And because

[TABLE]

Bringing (33) and (34) to (32), we get

[TABLE]

Since $W_{f_{1},f_{2}}(t)$ is not singular on $\Delta$, then $x^{\prime}_{1}(t)=x^{\prime}_{2}(t)=0.$

(c) Since $f_{1}(t)\neq 0,$ $t\in int(\Delta),$

[TABLE]

then $x^{\prime}_{1}(t)=0$, which means $t\in\Gamma_{f}(\Delta).$

(d) Furthermore,

[TABLE]

By the same discussion in (b), we could get $\forall t\in\Gamma_{f}(\Delta)$,

[TABLE]

[TABLE]

[TABLE]

Since $s^{3}_{f}(t)\neq 0,f_{1}(t)\neq 0,\forall t\in\Gamma_{f}(\Delta)$, then $x^{\prime\prime}_{2}(t)\neq 0,\forall t\in\Gamma_{f}(\Delta)$, which means that every $t\in\Gamma_{f}(\Delta)$ must be a strict minimum or maximum of $x_{2}(t).$

(e) If there are two different points $t_{1},t_{2}\in\Gamma_{f}(\Delta),t_{1}<t_{2}$, by (d), since

[TABLE]

then $t_{1}$ and $t_{2}$ are all the strict minimum or maximum of $x_{2}(t)$. Then there must be $t_{3}\in(t_{1},t_{2})$, such that $t_{3}$ is the strict maximum or minimum of $x_{2}(t)$, so $x_{2}^{\prime}(t_{3})=0.$ Then by (c), $t_{3}\in\Gamma_{f}(\Delta)$. Since $t_{3}$ is the strict maximum or minimum of $x_{2}(t)$, then by (d), $s_{f}^{3}(t_{3})$ has different sign from the sign of $s_{f}^{3}(t_{1})$, then $\exists t_{4}\in(t_{1},t_{3})$ such that $s^{3}_{f}(t_{4})=0$, which is against to $s^{3}_{f}(t)\neq 0,\forall t\in int(\Delta)$. So there is one point in $\Gamma_{f}(\Delta)$ at most.∎

The proof of Lemma 3.7:

Proof.

(a) If $(x_{2}(t),x_{1}(t))$ is a fix point for every $t\in\Delta,$ then $s^{2}_{f}(t)=(s^{1}_{f}(t))^{\prime}_{t}=0,\forall t\in int(\Delta)$, which means $\Gamma_{f}(\Delta)=int(\Delta)$, it is against to $\#\{\Gamma_{f}(\Delta)\}<+\infty.$

(b) If there exists a interval $\hat{\Delta}\subseteq\Delta$ such that $x_{2}(t)\equiv C,\forall t\in\hat{\Delta},$ which means $x^{\prime}_{2}(t)=0,\forall t\in int(\hat{\Delta}).$ By Lemma 3.6(c), $x_{1}^{\prime}(t)=0,\forall t\in int(\hat{\Delta}),$ so $x_{1}(t)\equiv\hat{C},\forall t\in\hat{\Delta},$ which means $s^{2}_{f}(t)=0,\forall t\in\hat{\Delta},$ $\hat{\Delta}\subseteq\Gamma_{f}(\Delta)$, it is against to $\#\{\Gamma_{f}(\Delta)\}<+\infty.$ ∎.

The proof of Lemma 3.8:

Proof.

[TABLE]

Since $s^{1}_{f}(t)=0,f_{1}(t)\neq 0$ and $s^{2}_{f}(t)\neq 0,\forall t\in\Delta\setminus\Gamma_{f}(\Delta),$ then

[TABLE]

[TABLE]

[TABLE]

Since $\#\{\Gamma_{f}(\Delta)\}<+\infty$, $\forall t^{*}\in\Gamma_{f}(\Delta)$, there is $\varepsilon>0$ such that $[t^{*}-\varepsilon,t^{*}+\varepsilon]\subset\Delta$ and $[t^{*}-\varepsilon,t^{*}+\varepsilon]\bigcap\Gamma_{f}(\Delta)=\{t^{*}\}$, then

[TABLE]

we still get \displaystyle\frac{dx_{1}}{dx_{2}}\Big{|}_{(x_{2}(t^{*}),x_{1}(t^{*}))}=-\frac{f_{2}(t^{*})}{f_{1}(t^{*})}. ∎

The proof of Lemma 3.9:

Proof.

By Lemma 3.8, the tangent line at $(x_{2}(t),x_{1}(t))$ is

[TABLE]

which is just

[TABLE]

7.2 Some properties of one variable convex functions

In this subsection, we present some prosperities of one variable convex functions used in this paper.

Lemma 7.1**.**

$G(x)$* is a continuous convex function in $x\in\mathbb{R}$. If there are three points $(x_{1},G(x_{1})),(x_{2},G(x_{2})),(x_{3},G(x_{3}))$ $(x_{1}<x_{2}<x_{3})$ on a line, then $G(x)\equiv kx+b,x_{1}\leq x\leq x_{3}.$*

Proof.

Since $(x_{1},G(x_{1})),(x_{2},G(x_{2})),(x_{3},G(x_{3}))$ $(x_{1}<x_{2}<x_{3})$ are on a line. It is obviously that this line is not perpendicular to $x$ axis. Supposing that this line is $y=kx+b$, now we consider $E(x)=G(x)-kx-b.$

It is clear that $E(x)$ is still a continuous convex function with $E(x_{1})=E(x_{2})=E(x_{3})=0$. Let $\Xi_{1}=\displaystyle\arg\min_{x\in[x_{1},x_{2}]}E(x)$, $\Xi_{2}=\displaystyle\arg\min_{x\in[x_{2},x_{3}]}F(x).$

• (a)

If $x_{2}\in\Xi_{1},x_{2}\in\Xi_{2}$ then $E(x)=0,\forall x\in[x_{1},x_{3}]$.

• (b)

If $x_{2}\in\Xi_{1},x_{2}\not\in\Xi_{2}$, then there exists $z_{2}\in\Xi_{2}\subset[x_{2},x_{3}],z_{2}\neq x_{2}$ such that $E(z_{2})<0.$ Then we have

[TABLE]

where $\displaystyle\lambda=\frac{x_{2}-x_{1}}{z_{2}-x_{1}}\neq 0.$ It is contradict to $E(x_{2})=0$.

• (c)

If $x_{2}\not\in\Xi_{1},x_{2}\in\Xi_{2}$, it is same as (b).

• (d)

If $x_{2}\not\in\Xi_{1},x_{2}\notin\Xi_{2}.$ Let $z_{1}\in\Xi_{1},z_{1}\neq x_{2},z_{2}\in\Xi_{2},z_{2}\neq x_{2}$, we have

[TABLE]

Then

[TABLE]

where $\lambda=\displaystyle\frac{x_{2}-z_{1}}{z_{2}-z_{1}}\neq 0$, which is contradict to $E(x_{2})=0.$

So $E(x)\equiv 0,\forall x\in[x_{1},x_{3}]$, which means $G(x)\equiv kx+b,\forall x\in[x_{1},x_{3}].\qed$

Lemma 7.2**.**

$y=G(x)$* is a continuous second order derivable convex function defined in $\mathbb{R}$. Supposing that $\displaystyle\lim_{x\rightarrow+\infty}\frac{G(x)}{x}$ and $\displaystyle\lim_{x\rightarrow-\infty}\frac{G(x)}{x}$ don’t exists, then $\forall(x^{*},y^{*})$, satisfied $y^{*}<G(x^{*})$, there are only two different tangent lines of $y=G(x)$ at $(x_{i},G(x_{i})),i=1,2$ passing through $(x^{*},y^{*}).$*

Proof.

Since $y=G(x)$ is a derivable convex function in $x\in\mathbb{R}$, then

[TABLE]

must has an unique optimal solution $(\hat{x},\hat{y})=(\hat{x},G(\hat{x})).$

Let $\displaystyle\theta^{*}=\tan^{-1}\frac{y^{*}-\hat{y}}{x^{*}-\hat{x}},\hat{\theta}=\tan^{-1}(G^{\prime}(\hat{x}))$, then $\hat{\theta}\neq\displaystyle\frac{\pi}{2}.$

If $\theta^{*}=\hat{\theta}$, then $\theta^{*}\neq\displaystyle\frac{\pi}{2}$. Since $(\hat{x},G(\hat{x}))$ is the projection of $(x^{*},y^{*})$ on $y=G(x)$, then

[TABLE]

[TABLE]

Then the tangent line of $G(x)$ at $(\hat{x},G(\hat{x}))$ must pass through $(x^{*},y^{*})$. By the differentiability of $G(x)$ at $\hat{x}$, there must be a point $(x_{1},G(x_{1}))\in\Upsilon$, which is contradict to the definition of $(\hat{x},G(\hat{x})).$ So $\theta^{*}\neq\hat{\theta},$ which means

[TABLE]

$\theta^{*}$ and $\hat{\theta}$ separate a circle into two parts. Without general, we consider $[\theta^{*},\hat{\theta}]$ (assuming $\theta^{*}<\hat{\theta}$).

Let

[TABLE]

It is easy to see $v(\hat{\theta})\geq\displaystyle\frac{|y^{*}-G(\hat{x})-G^{\prime}(\hat{x})(x^{*}-\hat{x})|}{\sqrt{1+G^{\prime 2}(\hat{x})}}>0,v(\theta^{*})=0$. Let $\tilde{\theta}=\sup\{\theta\in[\theta^{*},\hat{\theta}]|v(\theta)=0\}.$

Firstly, we prove some properties:

(1) $\tilde{\theta}\neq\displaystyle\frac{\pi}{2}$.

If $\tilde{\theta}=\displaystyle\frac{\pi}{2}$ and $v(\tilde{\theta})=0$, there are three cases:

(a) $x=x^{*}$ and $y=G(x)$ intersect at $(x^{*},G(x^{*}))$, and there are $(x_{1},G(x_{1}))$ and $(x_{2},G(x_{2}))$ such as $x_{1}<x^{*}<x_{2}$, then there must be a $\theta_{1}>\tilde{\theta}$ such that $v(\theta_{1})=0,$ which is against to the definition of $\tilde{\theta}$.

(b) $x=x^{*}$ and $y=G(x)$ intersect at $(x^{*},G(x^{*}))$, and there are not $(x_{1},G(x_{1}))$ and $(x_{2},G(x_{2}))$ such that $x_{1}<x^{*}<x_{2}$, then $y=G(x)$ belongs to only one side of $x=x^{*}$.

(c) there are not intersection points between $x=x^{*}$ and $y=G(x)$, which means $y=G(x)$ belongs to only one side of $x=x^{*}$. Either (b) or (c) is against to the definition of $G(x).$

If $\tilde{\theta}=\displaystyle\frac{\pi}{2}$ and $v(\tilde{\theta})\neq 0$, which means $y=G(x)$ belongs to only one side of $x=x^{*}$, which is also against to the definition of $G(x).$ So $\tilde{\theta}\neq\displaystyle\frac{\pi}{2}$.

(2) $\tilde{\theta}\neq\hat{\theta}$.

If $\tilde{\theta}=\hat{\theta}$, which means that $v(\theta)=0,\forall\theta\in[\theta^{*},\hat{\theta})$, while $v(\hat{\theta})\neq 0$. Then $\forall\tau>0$ such that $\displaystyle\frac{\pi}{2}\not\in[\hat{\theta}-\tau,\hat{\theta}]$, we have $v(\hat{\theta}-\tau)=0$. So $\forall\sigma>0$ small enough, there exists $(\tilde{x},G(\tilde{x}),\tilde{t})$ such that

[TABLE]

By solving

[TABLE]

we know that $\tilde{t}$ must be positive and

[TABLE]

Since $G^{\prime}(\hat{x})=\tan(\hat{\theta}),\hat{\theta}\neq\displaystyle\frac{\pi}{2},$ $\tilde{x}=x^{*}+\cos(\hat{\theta}-\tau)\tilde{t}$ must be $\infty$, when $\tau\rightarrow 0_{+}$.

Taken $\tau_{n}=\displaystyle\frac{1}{n}$, we could get $\{(\tilde{x}_{n},\tilde{t}_{n})\}$, then we select a subsequence $\{(\bar{x}_{n},\bar{t}_{n})\}$ of $\{(\tilde{x}_{n},\tilde{t}_{n})\}$, such that $\bar{x}_{n}\rightarrow+\infty$ or $\bar{x}_{n}\rightarrow-\infty$. Without general, we just discuss $\bar{x}_{n}\rightarrow+\infty$. Taken $\sigma_{n}=\displaystyle\frac{1}{n}$, we get

[TABLE]

[TABLE]

[TABLE]

[TABLE]

since $\tan(\hat{\theta})$ exists, then $\lim_{n\rightarrow\infty}\displaystyle\frac{G(\bar{x}_{n})}{\bar{x}_{n}}$ exists.

Since $G(x)$ is a second order derivable convex function, then $G^{\prime\prime}(x)\geq 0$, $\displaystyle(\frac{G(x)}{x})^{\prime}=\frac{G^{\prime}(x)x-G(x)}{x^{2}}.$ Then $(G^{\prime}(x)x-G(x))^{\prime}=xG^{\prime\prime}(x).$ Supposing $x>0$, then $G^{\prime}(x)x-G(x)$ is a nondecreasing function, which means either $G^{\prime}(x)x-G(x)$ is negative or there is $M>0$, when $x>M$, then $G^{\prime}(x)x-G(x)\geq 0$. So when $x>M$, $\displaystyle\frac{G(x)}{x}$ is monotone. Then $\displaystyle\lim_{x\rightarrow+\infty}\frac{G(x)}{x}$ exists, which is contradict to conditions of $G(x)$, it means $\hat{\theta}\neq\tilde{\theta}.$

(3) There exists $\theta_{l},\theta_{u}$ such that $[\theta_{l},\theta_{u}]\subset\{\theta|v(\theta)=0\},\theta_{l}<\theta_{u}$.

If it is not true, then $\{\theta^{*}\}=\{\theta|v(\theta)=0\}$. If $\theta^{*}=\displaystyle\frac{\pi}{2}$, then $\forall\tau>0$ small enough, $\tan(\theta^{*}+\tau)$ and $\tan(\theta^{*}-\tau)$ are well defined. If $\theta^{*}\neq\displaystyle\frac{\pi}{2}$, there must be $\forall\tau>0$ small enough such that $\displaystyle\frac{\pi}{2}\not\in[\theta^{*}-\tau,\theta^{*}+\tau]$, so $\tan(\theta^{*}+\tau)$ and $\tan(\theta^{*}-\tau)$ are well defined too. Then we have $v(\theta^{*}-\tau)>0$ and $v(\theta^{*}+\tau)>0$, $\forall\tau>0$ small enough. $y-y^{*}=\tan(\theta^{*}-\tau)(x-x^{*})$ and $y-y^{*}=\tan(\theta^{*}+\tau)(x-x^{*})$ separate $\mathbb{R}^{2}$ into $4$ parts. Since $v(\theta^{*}-\tau)>0$, $v(\theta^{*}+\tau)>0$, $\{(x,y)|y\geq G(x)\}$ must be in only one part only. When $\tau\rightarrow 0_{+}$, the domain of $y=G(x)$ is not $\mathbb{R}$, which is contradict to the definition of $G(x)$. So $int\{\theta|v(\theta)=0\}\neq\emptyset$.

Now we prove that $\tilde{\theta}$ is the tangent angle of a tangent line.

(1) If $\theta^{*}\neq\tilde{\theta}$.

At $\tilde{\theta}$, let $\alpha>0,\varepsilon>0$, we consider

[TABLE]

where $\theta\in[\tilde{\theta}-\varepsilon,\tilde{\theta}+\varepsilon]\subset[\theta^{*},\hat{\theta}]$ and $\displaystyle\frac{\pi}{2}\not\in[\tilde{\theta}-\varepsilon,\tilde{\theta}+\varepsilon]$

We will prove that there exists $\alpha^{*}>0,M^{*}>0$ such that

[TABLE]

If it is not true, then $\forall\alpha>0,\forall M>0$, there exists $\bar{\theta}\in[\tilde{\theta}-\varepsilon,\tilde{\theta}+\varepsilon],(\bar{x},\bar{t})\in\Omega_{\alpha}(\bar{\theta})$ such that $||(x,t)||>M$. Let $\alpha_{n}=\displaystyle\frac{1}{n},M_{n}=n$. Since $(\bar{x}_{n},\bar{t}_{n})\in\Omega_{\alpha_{n}}(\bar{{\theta}}_{n})$, then $(\bar{x}_{n}-x^{*}-\cos(\bar{\theta}_{n})\bar{t}_{n})^{2}\leq\alpha$, so $|\bar{x}_{n}|\rightarrow\infty$, while $n\rightarrow\infty$. Without general, we select the subsequence $\{(\tilde{x}_{n},\tilde{t}_{n},\tilde{\theta}_{n})\}$ from $\{(\bar{x}_{n},\bar{t}_{n},\bar{\theta}_{n})\}$ such that $\tilde{x}_{n}\rightarrow+\infty,$ and $\tilde{\theta}_{n}\rightarrow{\check{\theta}}$. Then

[TABLE]

[TABLE]

[TABLE]

[TABLE]

which means that $\displaystyle\lim_{x\rightarrow+\infty}\frac{G({x})}{{x}}$ exists, and it is contradict to conditions of $G(x)$. So there exists $\alpha^{*}>0,M^{*}>0$ such that

[TABLE]

Furthermore, since $H(x,t,\theta)$ and $H_{\theta}(x,t,\theta)$ are continuous at $(x,t,\theta)\in\mathbb{R}\times\mathbb{R}_{+}\times[\tilde{\theta}-\varepsilon,\tilde{\theta}+\varepsilon]$, then $v(\theta)$ is continuous at $\tilde{\theta}$ by [4], which means $v(\tilde{\theta})=0.$ Furthermore, since $\Omega_{\alpha^{*}}(\tilde{\theta})$ is bounded and closed, $\displaystyle\arg\inf_{t\geq 0,x}H(x,t,\tilde{\theta})$ is attached, and we could select a point $\displaystyle(x_{0},t_{0})\in\arg\inf_{t\geq 0,x}H(x,t,\tilde{\theta}).$

Now we will prove that $y-G(x_{0})=\tan(\tilde{\theta})(x-x_{0})$ is a tangent line at $(x_{0},G(x_{0}))$ and passes through $(x^{*},y^{*}).$

It is easy to see that $H(x_{0},t_{0},\tilde{\theta})=0$, which means

[TABLE]

so $(x_{0},G(x_{0})),(x^{*},y^{*})$ are all in the line $y-G(x_{0})=\tan(\tilde{\theta})(x-x_{0})$.

Let $\theta_{1}=\tilde{\theta}+\varepsilon$, where $\varepsilon>0$ is small enough. $\forall x_{1}\in\mathbb{R}$,

[TABLE]

By $v(\theta_{1})>0$, without general, we suppose $Q(x_{1},\theta_{1})>0$. So $\displaystyle\lim_{\varepsilon\rightarrow 0_{+}}Q(x_{1},\theta_{1})\geq 0$, $\forall x_{1}\in\mathbb{R}$, which means $\forall(x_{1},G(x_{1}))$ is at only one side of $y-y^{*}=\tan(\tilde{\theta})(x-x^{*})$, then $y-y^{*}=\tan(\tilde{\theta})(x-x^{*})$ must be the tangent line at $(x_{0},G(x_{0})).$

(2) If $\tilde{\theta}=\theta^{*}$. By the similar discussion, $y-y^{*}=\tan(\theta^{*})(x-x^{*})$ is also a tangent line of $y=G(x)$.

At last, we will prove there has just two tangent lines of $y=G(x)$ passing through $(x^{*},y^{*}).$

(1) There is only one tangent line of $y=G(x)$ passing through $(x^{*},y^{*}).$

By above discussion, we have know that $[\theta^{*},\hat{\theta}]$ and $[\hat{\theta},\theta^{*}]$ each have a tangent line. We just need to prove that they are different. The only chance that they are same is that $\theta^{*}=\sup\{\theta\in[\theta^{*},\hat{\theta}]|v(\theta)=0\}$ and $\theta^{*}=\inf\{\theta\in[\hat{\theta},\theta^{*}]|v(\theta)=0\},$ which is against that there exists $\theta_{l},\theta_{u}$ such that $\theta^{*}\in[\theta_{l},\theta_{u}]\subset\{\theta|v(\theta)=0\},\theta_{l}<\theta_{u}$.

(2) There are not more than 2 different tangent lines of $y=G(x)$ passing through $(x^{*},y^{*}).$

If there are 3 or more different tangent lines of $y=G(x)$ passing through $(x^{*},y^{*})$, we select 3 different tangent lines and denote their tangent points as $(x_{1},G(x_{1}))$, $(x_{2},G(x_{2}))$, $(x_{3},G(x_{3})),$ where $x_{1}<x_{2}<x_{3}.$

$l((x_{3},G(x_{3})),(x_{1},G(x_{1})))$ and $l((x_{2},G(x_{2})),(x^{*},y^{*}))$ must intersect at the point $(\hat{x},\hat{y})\in B(G)$. It is obviously that $(\hat{x},\hat{y})\in int(B(G))$ is impossible, then $\hat{y}=G(\hat{x})$, which means $\hat{x}=x_{2}$. Then $(x_{1},G(x_{1}))$, $(x_{2},G(x_{2}))$, $(x_{3},G(x_{3}))$ are in a line. By Lemma 7.1, $G(x)=kx+b,x\in[x_{1},x_{3}]$, then the tangent lines at $x_{1}$, and $x_{3}$ are same, which is against to that there are 3 different tangent lines. ∎

Corollary 7.3**.**

Supposing that all conditions in Lemma 7.2 hold. Given $x^{*}\in\mathbb{R}$, then

• (a)

if $(\hat{x},\hat{y})\in int(A(x^{*},G))$, then there are only two different tangent lines passing through $(\hat{x},\hat{y})$, with their tangent points $x_{1},x_{2}$ satisfied $x_{1}<x^{*}<x_{2}.$

• (b)

if $(\hat{x},\hat{y})\in int(C(x^{*},G))$, then there are only two different tangent lines passing through $(\hat{x},\hat{y})$, with their tangent points $x_{1},x_{2}$ are all at only one side of $x^{*},$ which means either $x_{1}<x_{2}<x^{*}$ or $x_{1}>x_{2}>x^{*}.$

Proof.

Either $(\hat{x},\hat{y})\in int(A(x^{*},G))$ or $(\hat{x},\hat{y})\in int(C(x^{*},G))$, from Lemma 7.2, there exist two tangent points $x_{1},x_{2}$ such that two different tangent lines at $x_{1},x_{2}$ of $G(x)$ passing through $(\hat{x},\hat{y})$. If $x_{1}=x^{*}$ or $x_{2}=x^{*}$, then by Lemma 7.2, $(\hat{x},\hat{y})$ must be at the tangent line of $G(x)$ at $(x^{*},G(x^{*}))$, which is against to definition of $(\hat{x},\hat{y}).$

(a) Supposing $x_{1}>x_{2}>x^{*}$, then $l((x_{1},G(x_{1})),(x^{*},G(x^{*})))$ and $l((\hat{x},\hat{y}),(x_{2},G(x_{2})))$ must intersect at $(\bar{x},\bar{y})$ with $\bar{y}\geq G(\bar{x})$.

If $\bar{y}=G(\bar{x})$, $(x_{1},G(x_{1}))$, $(\bar{x},G(\bar{x})$, $(x^{*},G(x^{*}))$ are all in a line. By Lemma 7.1, $G(x)=kx+b,\forall x\in[x^{*},x_{1}]$, then the tangent line at $x_{1}$, $x_{2}$ are same, which is contradict to Lemma 7.2.

If $\bar{y}>G(\bar{x})$, which means that $y-G(x_{2})=G^{\prime}(x_{2})(x-x_{2})$ intersects $int(B(G))$, it is contradict to the definition of $x_{2}.$

By the same discussion, $x_{1}<x_{2}<x^{*}$ is impossible too. So $x_{1}<x^{*}<x_{2}$.

(b) If $x_{1}<x^{*}<x_{2}$, $(x_{1},G(x_{1}))$ or $(x_{2},G(x_{2}))$ are at different sides of $x=x^{*}$, and at only one side of $y-y^{*}=G^{\prime}(x^{*})(x-x^{*})$, then the line $l((x_{1},G(x_{1})),(\hat{x},\hat{y}))$ or $l((\hat{x},\hat{y}),(x_{2},G(x_{2})))$ must intersect $y=G(x)$ twice at least. Denoting one of intersection points as $(\tilde{x},G(\tilde{x}))$.

If the image of $G(x)$ includes $l((\hat{x},\hat{y}),(\tilde{x},G(\tilde{x})))$, then the tangent line at $x^{*}$ is same as $l((\hat{x},\hat{y}),(\tilde{x},G(\tilde{x})))$, which means $(\hat{x},\hat{y})\in\{(x,y)|y-G(y^{*})=G^{\prime}(x^{*})(x-x^{*})\}$, it is contradict to the definition of $(\hat{x},\hat{y}).$

If the image of $y=G(x)$ does not include $l((\hat{x},\hat{y}),(\tilde{x},G(\tilde{x})))$, which means

[TABLE]

it is contradict to the definition of $x_{1},x_{2}$.

So either $x_{1}<x_{2}<x^{*}$ or $x_{1}>x_{2}>x^{*}.\qed$

Noting that Lemma 7.1, Lemma 7.2 and Corollary 7.3 also suit for the concave function of one variable, so if the concavity and convexity of a function of one variable does not change in $\mathbb{R}$, Lemma 7.1, Lemma 7.2 and Corollary 7.3 hold too.