Yingjun Ni, Yuli Fan^*
School of Mathematics and Physics, Southwest University of Science and Technology, Mianyang, China.
DOI: 10.4236/jamp.2024.127152   PDF    HTML   XML   198 Downloads   609 Views   Citations

Abstract

A function which is reflexive is called by pre-mean, a more generalized definition of a mean. In this paper, we define a new pre-mean and study its properties, and then using the given invariant curve we consider the problem of convergence of Gauss iteration of a kind of pre-mean type mappings generated by the exponential and logarithmic functions.

Keywords

Pre-Mean-Type Mapping, Invariant Equation, Invariant Curves

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1. Introduction

Mean value is an important fundamental concept in mathematics. For thousands of years, many forms of mean have been proposed, such as the arithmetic mean $A\mathrm{<mo>:</mo>}{ℝ}^p\to ℝ$ :

$A\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_p\right)\mathrm{<mo>=</mo>}\frac{x_1+\cdots +x_p}{p}\mathrm{,}$

the geometric mean $G\mathrm{<mo>:</mo>}{\left(\mathrm{0,}+\infty \right)}^p\to \left(\mathrm{0,}\infty \right)$ :

$G\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_p\right)=\sqrt[p]{x_1\cdots x_p}\mathrm{,}$

the harmonize mean $H\mathrm{<mo>:</mo>}{\left(\mathrm{0,}+\infty \right)}^p\to \left(\mathrm{0,}\infty \right)$ :

$H\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_p\right)=\frac{p}{\frac{1}{x_1}+\cdots +\frac{1}{x_p}}\mathrm{.}$

Although there are various mean forms, they all meet the following definitions ([1]):

Let $I\subset ℝ$ be an open interval. If the function $M\mathrm{<mo>:</mo>}{I}^p\to I$ satisfies

$\min \left\{x_1,\cdots ,x_p\right\}\le M\left(x,y\right)\le \max \left\{x_1,\cdots ,x_p\right\},x_1,\cdots ,x_p\in I,$ (1)

then M is called a mean on the interval I. When the above inequalities strictly hold, M is called strict mean.

Theorem 1. ([2] [3]) If the mean M and N of interval I are continuous, that is, any $x\mathrm{,}y\in I$ , $x\ne y$ , we have

$\max \left(M\left(x,y\right),N\left(x,y\right)\right)-\min \left(M\left(x,y\right),N\left(x,y\right)\right)<\max \left(x,y\right)-\min \left(x,y\right),$

then

1) For any $n\in \mathbb{N}$ , Gauss iteration ${\left(M\mathrm{,}N\right)}^n$ of the mapping $\left(M\mathrm{,}N\right)$ is the mean type map on $I^2$ .

2) There is a continuous mean $K\mathrm{<mo>:</mo>}{I}^2\to I$ , so that ${\left({\left(M\mathrm{,}N\right)}^n\right)}_{n=0}^{\infty }$ converges to $\left(K\mathrm{,}K\right)$ .

3) K is $\left(M\mathrm{,}N\right)$ -invariant mean, and is the unique $\left(M\mathrm{,}N\right)$ -invariant mean.

4) If $\left(M\mathrm{,}N\right)$ is a strict mean mapping, that is, M and N are a strict mean, then K is also a strict mean.

5) If M and N are (strictly) increasing for each variable, then K is also (strictly) increasing for each variable.

The above theorem provides the relationship between invariant equations and convergence of iteration of mean-type mappings. By studying the problem of invariant equations, we can study the dynamic behavior of mean-type mappings. There are many works to study kinds of invariant equations with different means ([4]-[8]). For the pre-mean-type mapping, the results have been still few. Next we will introduce some definitions for the pre-mean.

Obviously, it can be seen from the inequalities (1) that any mean satisfies the reflexivity, i.e.

$M\left(x\mathrm{,}\cdots \mathrm{,}x\right)=x\mathrm{,}x\in I\mathrm{.}$

However, not all the functions that satisfy the reflexivity are means. In other words, there are functions in the interval I that satisfy the reflexivity but not be means. In 2006, Matkowski ([9]) gave the definition of pre-mean as follows

If the function $M\mathrm{<mo>:</mo>}{I}^p\to I$ satisfies reflexivity, M is called a pre-mean.

This indicates that a mean must be a pre-mean, but a pre-mean is not necessarily a mean, and the pre-mean is a generalized function than the mean. For example, the function $M\mathrm{<mo>:</mo>}{ℝ}^2\to ℝ$ is defined as

$M\left(x\mathrm{,}y\right)=\ln \left(\frac{\frac{1}{3}{\text{e}}^{2x}+\frac{2}{3}{\text{e}}^{2y}}{\frac{1}{3}{\text{e}}^x+\frac{2}{3}{\text{e}}^y}\right)\mathrm{,}x\mathrm{,}y\in ℝ\mathrm{,}$

which satisfies reflexivity, but $M\left(\ln \frac{1}{2},0\right)=\ln \frac{27}{24}>\max \left\{\ln \frac{1}{2},0\right\}$ does not satisfy the definition of mean, so this function is a pre-mean but not a mean. But if a pre-mean is (strictly) increasing with respect to each variable, then it must be a (strict) mean ([9] [10]).

In this paper, we will consider the convergence of iterates of pre-mean type mappings of the form $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ with $\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ , $p\mathrm{,}q\in ℝ$ , $p\ne q$ , where $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{<mo>:</mo>}{ℝ}^2\to ℝ$ satisfies

$E_{\lambda ,\mu }^{\left[p,q\right]}\left(x,y\right)=\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}},x,y\in ℝ.$

2. Some Definitions and Auxiliary Results

Lemma 1 Suppose $p\mathrm{,}q\in ℝ$ , $p\ne q$ and $\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ , then

1) $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ is the pre-mean.

2) $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ is the mean if and only if $\lambda =\mu$ or $pq\le 0$ .

Proof. 1) If $\forall x\in ℝ$ , make $x=y$ , then $E_{\lambda ,\mu }^{\left[p\mathrm{,}q\right]}\left(x,x\right)=\ln {\left(\frac{{\text{e}}^{px}}{{\text{e}}^{qx}}\right)}^{\frac{1}{p-q}}=x$ , so it is a pre-mean.

2) Firstly, for the case $\lambda =\mu$ , $E_{\lambda ,\mu }^{\left[p\mathrm{,}q\right]}\left(x,x\right)=\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}}$ . Let $x<y$ , $p>q$ , then $\left(p-q\right)x<\left(p-q\right)y$ and

${\text{e}}^{px+qy}<{\text{e}}^{qx+py}.$

It can be obtained after doing a simple calculation on the above formula

$\left(\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}\right){\text{e}}^{qy}<\left(\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}\right){\text{e}}^{py},$

that is

$\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}}<y.$

Similarly, in the same way, we get $\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}}>x,$ i.e.

$x<\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}}<y.$

For the case of $p<q$ , it can be proved similarly. From the definition of the mean value, the $\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\lambda {\text{e}}^{qx}+\left(1-\lambda \right){\text{e}}^{qy}}\right)}^{\frac{1}{p-q}}\in \left(x,y\right),$ $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ is a mean at $\lambda =\mu$ .

Secondly, for the case $pq\le 0$ , it is obvious that the function $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ is increasing with respect to each of the variables, so it is a mean. The proof of the “if” part has been finished. To prove the “only if” result, assume that $\lambda =\mu$ or $pq\le 0$ and consider two possible cases.

Case $p>0$ and $q>0$ .

If $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ were a mean, there is

$E_{\lambda ,\mu }^{\left[p\mathrm{,}q\right]}\left(\ln \left(x\right),0\right)<0\text{and}{E}_{\lambda ,\mu }^{\left[p\mathrm{,}q\right]}\left(0,\ln \left(x\right)\right)<0,x\in \left(0,1\right)$

that is

${\left(\frac{\lambda x^p+\left(1-\lambda \right)}{\mu x^q+\left(1-\mu \right)}\right)}^{\frac{1}{p-q}}<1\text{and}{\left(\frac{\lambda +\left(1-\lambda \right)x^p}{\mu +\left(1-\mu \right)x^q}\right)}^{\frac{1}{p-q}}<1,x\in \left(0,1\right).$

Therefore, if $p-q>0$ , then

$\frac{\lambda x^p+\left(1-\lambda \right)}{\mu x^q+\left(1-\mu \right)}<1\text{and}\frac{\lambda +\left(1-\lambda \right)x^p}{\mu +\left(1-\mu \right)x^q}<1,x\in \left(0,1\right).$

Let $x\to 0$ , then

$\frac{\lambda }{1-\mu }\le 1\text{and}\frac{\lambda }{\mu }\le 1.$ (2)

If $p-q<0$ , then

$\frac{\lambda x^p+\left(1-\lambda \right)}{\mu x^q+\left(1-\mu \right)}>1\text{and}\frac{\lambda +\left(1-\lambda \right)x^p}{\mu +\left(1-\mu \right)x^q}>1,x\in \left(0,1\right).$

Let $x\to 0$ , then

$\frac{\lambda }{1-\mu }\ge 1\text{and}\frac{\lambda }{\mu }\ge 1.$ (3)

So, if and only if $\lambda =\mu$ , Equations (2) and (3) hold.

Case $p<0$ and $q<0$ .

Since

$\begin{array}{l}{E}_{\lambda ,\mu }^{\left[p\mathrm{,}q\right]}\left(x,y\right)+E_{1-\lambda ,1-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x,y\right)\\ =\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{qy}}\cdot \frac{\left(1-\mu \right){\text{e}}^{-qx}+\mu {\text{e}}^{-qy}}{\left(1-\lambda \right){\text{e}}^{-px}+\mu {\text{e}}^{-py}}\right)}^{\frac{1}{p-q}}\\ =\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{qy}}\cdot \frac{{\text{e}}^{p\left(x+y\right)}}{{\text{e}}^{q\left(x+y\right)}}\cdot \frac{\left(1-\mu \right){\text{e}}^{qy}+\mu {\text{e}}^{qx}}{\left(1-\lambda \right){\text{e}}^{py}+\lambda {\text{e}}^{px}}\right)}^{\frac{1}{p-q}}\\ =\frac{1}{p-q}\ln {\text{e}}^{\left(p-q\right)\left(x+y\right)}=x+y,\end{array}$

then $E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}y\right)=x+y-E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}y\right)$ . Therefore, if $E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ were a mean, that is, for all $x<y$ we have

$x\le E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}y\right)\le y\mathrm{,}$

then by $E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}y\right)=x+y-E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}y\right)$ , we have

$x\le E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}y\right)\le y\mathrm{,}$

that is, $E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}$ would be a mean, which contradicts the previous case. $\square$

Lemma 2 Let $p\mathrm{,}q\in ℝ$ , $p\ne q$ , $\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ , then the arithmetic mean A is invariant with respect to the pre-mean type mapping $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ , that is

$A\circ \left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)=A\mathrm{.}$

Proof. For all $x\mathrm{,}y\in ℝ$ , we have

$\begin{array}{l}A\circ \left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)\left(x,y\right)\\ =\frac{\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{qy}}\cdot \frac{\left(1-\mu \right){\text{e}}^{-qx}+\mu {\text{e}}^{-qy}}{\left(1-\lambda \right){\text{e}}^{-px}+\mu {\text{e}}^{-py}}\right)}^{\frac{1}{p-q}}}{2}\\ =\frac{\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{py}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{qy}}\cdot \frac{{\text{e}}^{p\left(x+y\right)}}{{\text{e}}^{q\left(x+y\right)}}\cdot \frac{\left(1-\mu \right){\text{e}}^{qy}+\mu {\text{e}}^{qx}}{\left(1-\lambda \right){\text{e}}^{py}+\lambda {\text{e}}^{px}}\right)}^{\frac{1}{p-q}}}{2}\\ =\frac{1}{2\left(p-q\right)}\ln {\text{e}}^{\left(p-q\right)\left(x+y\right)}=\frac{x+y}{2}=A\left(x,y\right).\end{array}$

$\square$

To consider the convergence of iterates of the pre-mean but not mean mappings, we now give the definition of invariant curves.

Let $M\mathrm{,}N\mathrm{<mo>:</mo>}{I}^2\to I$ , $K\mathrm{<mo>:</mo>}{I}^2\to ℝ$ be some functions, let $J\subset I$ , $f\mathrm{<mo>:</mo>}J\to I$ be a function on J. If there is

$f\left(M\left(x\mathrm{,}f\left(x\right)\right)\right)=N\left(x\mathrm{,}f\left(x\right)\right)\mathrm{,}x\in J\mathrm{,}$

we say that the graph of the function f is an invariant curve with respect to the map $\left(M\mathrm{,}N\right)$ , briefly, $\left(M\mathrm{,}N\right)$ -invariant curve ([2]).

To explore the connection between the pre-mean map and its invariant curve, Lemma 3 is introduced.

Lemma 3 Let $p\mathrm{,}q\in ℝ$ , $p\ne q$ , $\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ , then

(1) For every $a\in ℝ$ , the graph of the function $f\left(x\right)=2a-x$ ,$x\in ℝ$ , that is,

$H_a=\left\{\left(x\mathrm{,2}a-x\right)\mathrm{<mo>|</mo>}x\in ℝ\right\}\mathrm{,}$

is $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ -invariant curve, and in particular $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)\left(H_a\right)\subset H_a$ .

(2) The point $\left(a\mathrm{,}a\right)$ is the only fixed point of the mapping $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ in the set $H_a$ .

(3) The family of sets $\left\{H_a:a>0\right\}$ forms a partition of ${ℝ}^2$ , that is $H_a\cap H_b=\varnothing$ for all $a\mathrm{,}b\in ℝ$ and $a\ne b$ , and ${\displaystyle {\cup }_{a\in ℝ}{H}_a}={ℝ}^2$ .

Proof. Fix $a\in ℝ$ , we have

$A\left(x,f\left(x\right)\right)=\frac{x+f\left(x\right)}{2}=a,$

and using Lemma 2, we get

$A\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}f\left(x\right)\right)\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}f\left(x\right)\right)\right)=A\left(x\mathrm{,}f\left(x\right)\right)=a\mathrm{,}$

that is

$\frac{E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}f\left(x\right)\right)+E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}f\left(x\right)\right)}{2}=\frac{x+f\left(x\right)}{2}=a\mathrm{,}$

$E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}f\left(x\right)\right)=2a-E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}f\left(x\right)\right)\mathrm{,}$

therefore,

$f\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}f\left(x\right)\right)\right)=E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\left(x\mathrm{,}f\left(x\right)\right)\mathrm{,}$

According to the definition of invariant curves, the function $f\left(x\right)=2a-x$ is the invariant curve of $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ . Therefore, the first result is correct and the other two parts are obvious. $\square$

3. Convergence of Iteration of the Pre-Mean Mappings

Using Theorem 1 and Lemma 2, we can get the following:

Theorem 2 If $\lambda =\mu$ or $pq\le 0$ , the iterative sequence ${\left({\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^n\right)}_{n=0}^{\infty }$ is pointwise convergent on ${ℝ}^2$ and

$\underset{n\to \infty }{lim}{\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^n\left(x\mathrm{,}y\right)=\left(\frac{x+y}{2}\mathrm{,}\frac{x+y}{2}\right)\mathrm{,}x\mathrm{,}y\in ℝ\mathrm{.}$

It is difficult to consider the convergence problem for the case $\lambda \ne \mu$ and $pq>0$ . Lemma 3 tells us that, $H_a$ is the invariant curve of $\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)$ . We will study the convergence problem of the restrictions of this map to the invariant sets $H_a\mathrm{,}a\in ℝ$ .

Theorem 3 Let $p\mathrm{,}q\in ℝ$ , $p\ne q$ ,$\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ . For every $a\in ℝ$ and $n\in \mathbb{N}$ , we have for all $x\in ℝ$ it holds

${\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^n\left(x\mathrm{,2}a-x\right)=\left({\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)}^n\left(x\right)\mathrm{,2}a-{\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)}^n\left(x\right)\right)\mathrm{,}$ (4)

where the function $f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}$ is defined as

$f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\left(x\right)=E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,2}a-x\right)\mathrm{,}x\in ℝ\mathrm{,}$

that is

$f_{\left[\lambda ,\mu \right];a}^{\left[p,q\right]}\left(x\right)=\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{p\left(2a-x\right)}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{q\left(2a-x\right)}}\right)}^{\frac{1}{p-q}},x\in ℝ.$

Proof. For the convenience of writing, write $E=E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}$ , $f_a=f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}$ . Because

$E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\left(x\mathrm{,}y\right)+E_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}=x+y\mathrm{,}$

there is,

$\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)\left(x\mathrm{,}y\right)=\left(E\left(x\mathrm{,}y\right)\mathrm{,}x+y-E\left(x\mathrm{,}y\right)\right)\mathrm{,}x\mathrm{,}y\in ℝ\mathrm{.}$

When $n=1$ , the Equation (4) is correct, in fact

$\begin{array}{c}\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)\left(x\mathrm{,2}a-x\right)=\left(E\left(x\mathrm{,2}a-x\right)\mathrm{,2}a-E\left(x\mathrm{,2}a-x\right)\right)\\ =\left(f_a\left(x\right)\mathrm{,2}a-f_a\left(x\right)\right)\mathrm{.}\end{array}$

Suppose (4) holds for some $n=k\in \mathbb{N}$ , that is

${\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^k\left(x\mathrm{,2}a-x\right)=\left(f_a^k\left(x\right)\mathrm{,2}a-f_a^k\left(x\right)\right)\mathrm{,}$

then for all $x\in ℝ$

$\begin{array}{l}{\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^{k+1}\left(x\mathrm{,2}a-x\right)\\ ={\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^k\left(E\left(x\mathrm{,2}a-x\right)\mathrm{,2}a-E\left(x\mathrm{,2}a-x\right)\right)\\ =\left(f_a^k\left(E\left(x\mathrm{,2}a-x\right)\right)\mathrm{,2}a-f_a^k\left(E\left(x\mathrm{,2}a-x\right)\right)\right)\\ =\left(f_a^k\left(f_a\left(x\right)\right)\mathrm{,2}a-f_a^k\left(f_a\left(x\right)\right)\right)=\left(f_a^{k+1}\left(x\right)\mathrm{,2}a-f_a^{k+1}\left(x\right)\right)\mathrm{,}\end{array}$

that is to say, when $n=k+1$ , (4) is still established, the certificate is finished. $\square$

To further study the theorem 3, we obtain the following

Corollary 1 Let $p\mathrm{,}q\in ℝ$ , $p\ne q$ , $\lambda \mathrm{,}\mu \in \left(\mathrm{0,1}\right)$ , if

$\left|\frac{\left(2\lambda -1\right)p-\left(2\mu -1\right)q}{p-q}\right|<\mathrm{1,}$

then there is an open set $U\subset {ℝ}^2$ containing the diagonal $\Delta =\left(x\mathrm{,}x\right)\mathrm{<mo>|</mo>}x\in ℝ$ such that

$\underset{n\to \infty }{lim}{\left(E_{\lambda \mathrm{,}\mu }^{\left[p\mathrm{,}q\right]}\mathrm{,}{E}_{1-\lambda \mathrm{,1}-\mu }^{\left[-p\mathrm{,}-q\right]}\right)}^n\left(x\mathrm{,}y\right)=\left(\frac{x+y}{2}\mathrm{,}\frac{x+y}{2}\right)\mathrm{,}\left(x\mathrm{,}y\right)\in U\mathrm{.}$

Proof. According to theorem 3, we can obtain that

$f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\left(a\right)=E\left(a\mathrm{,}a\right)=a\mathrm{,}$

as well as

$\begin{array}{c}\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)\left(x\right)=\left[\ln {\left(\frac{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{2a-x}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{2a-x}}\right)}^{\frac{1}{p-q}}\right]\\ =\frac{1}{p-q}\left[p-\frac{2p\left(1-\lambda \right){\text{e}}^{p\left(2a-x\right)}}{\lambda {\text{e}}^{px}+\left(1-\lambda \right){\text{e}}^{p\left(2a-x\right)}}-q+\frac{2q\left(1-\mu \right)e^{q\left(2a-x\right)}}{\mu {\text{e}}^{qx}+\left(1-\mu \right){\text{e}}^{q\left(2a-x\right)}}\right].\end{array}$

So,

$\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)\left(a\right)=\frac{\left(2\lambda -1\right)p-\left(2\mu -1\right)q}{p-q}\mathrm{.}$

It can be seen that any real number a, a are the fixed point of $f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}$ and $\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)\left(a\right)$ is independent of the value of a.

Since $\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)\left(a\right)<1$ , according to the compression mapping principle,

$\underset{n\to \infty }{lim}{\left(f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\right)}^n\left(x\right)=f_{\left[\lambda \mathrm{,}\mu \right]\mathrm{<mo>;</mo>}a}^{\left[p\mathrm{,}q\right]}\left(a\right)=a=\frac{x+y}{2}\mathrm{,}$

which complete the proof. $\square$

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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