1942年,Menger在文献^[1]提出了概率度量空间的概念. 1972年, Sehgal和Bharucha-Reid在文献^[5]中建立了概率度量空间中的不动点理论. 众所周知, 度量空间中的压缩不动点理论(见文献^{[1, 2, 3, 4, 5, 6, 7, 8, 9]})为半序度量空间中的不动点理论的研究奠定了基础. 文献^{[9, 10, 11, 12, 13, 14, 15]}研究了半序度量空间中二元重合点与公共不动点.随后,Borcut在文献^{[16, 17]}中将半序度量空间中的二元重合点与公共不动点定理推广到了三元. 最近在文献^[15]中,胡新启和马晓燕在半序概率度量空间中得到可交换映射的二元重合点定理.本文在半序概率度量空间中建立了映射对$G:X\times X\times X\rightarrow X$与 $g:X\rightarrow X$的相容性概念. 在不需要可交换的条件下, 研究了在满足更一般压缩条件下的相容映射的三元重合点与三元不动点问题, 所得结果推广了文献^[18] 中的二元重合点与二元公共不动点定理. 最后,我们给出主要结果的一个具体应用.

先回顾如下概念.

设${\Bbb R}$表示全体实数所成集合,${{\Bbb R}^ + }$表示全体非负实数所成集合, ${\Bbb N}$表示全体自然数所成集合.

映像$f:{\Bbb R} \to {{\Bbb R}^ + }$称为分布函数,如果它是非减的、左连续的, 且满足条件 $$\mathop {\inf }\limits_{t \in R} f(t) = 0, \mathop {\sup }\limits_{t \in R} f(t) = 1.$$

用$D$表示一切分布函数所成集合,$H(t)$表示如下定义的特殊分布函数 $$H(t) = \left\{ {\matrix{ {0, t \le 0,} \cr {1, t0.} \cr } } \right.$$ 映射$\Delta :[0, 1] \times [0, 1] \to [0, 1]$称为三角范数(简称为$t$ -范数), 如果它满足如下条件: 对一切的$a,b,c,d \in [0, 1]$,有

($\Delta$-1) $\Delta(a,1)=a,$ $\Delta(0,0)=0;$

($\Delta$-2) $\Delta(a,b)=\Delta(b,a);$

($\Delta$-3) $a\geq b,c\geq d\Rightarrow \Delta(a,c)\geq \Delta(b,d);$

($\Delta$-4) $\Delta(a,\Delta(b,c))=\Delta(\Delta(a,b),c).$

其中$\Delta_p(a,b)=ab$与$\Delta_M(a,b)=\min(a,b)$都是典型的连续$t$ -范数.

$t$ -范数$\Delta$称为$H$ -型的,如果序列函数$\{\triangle^n(x)\}_{n\in N}$ 在$x=1$处是等度连续的,其中$\Delta^1(x)=\Delta(x,x)$,$\Delta^n(x)=\Delta(x, \Delta^{n-1}(x)),n \in N,x\in[0, 1]$.

定义2.1 ^{[1, 2]}Menger概率度量空间(简称为Menger PM -空间)是一三元组$(X,F,\Delta )$,其中$X$是一抽象集,$\Delta $为$t$ -范数, $F$是$X \times X$到$D$的映象(记分布函数$F(x,y)$为${F_{x,y}}$, 而${F_{x,y}}(t)$表示${F_{x,y}}$在$t \in R$的值),并且假定${F_{x,y}}, x,y \in X$满足下面的条件:

(PM-1)~ ${F_{x,y}}(t) = H(t),\forall t \in R,$ 当且仅当$x=y$;

(PM-2)~ 对任意的$x,y\in X,t\in R$,有${\rm{}}{F_{x,y}}(t) = {F_{y,x}}(t);$

(PM-3)~ 对任意的$x,y,z \in X,{t_1},{t_2}\geq 0$,有$ {F_{x,z}}({t_1} + {t_2}) \ge \Delta ({F_{x,y}}({t_1}),{F_{y,z}}({t_2})).$

定义 2.2 ^[18]分布函数$F_1$和$F_2$的代数和$F_1\oplus F_2$定义为 $$ (F_1\oplus F_2)(t)=\sup_{t_1+t_2=t}\min\{F_1(t_1),F_2(t_2)\}, \forall t\in R.$$

显然有 $$ (F_1\oplus F_2)(2t)\geq\min\{F_1(t),F_2(t)\}, \forall t>0.$$

注 2.1 Schweizer和Sclar在文献^{[2, 4]}中指出,如果Menger PM -空间$(X,F,\Delta )$ 中$t$ -范数$\Delta$满足条件$\sup\limits_{0 ＜t<1}\Delta(t,t)=1$, 则$(X,F,\Delta )$是$(\varepsilon,\lambda)$ -邻域系所诱导的拓扑$\tau$下的Hausdorff拓扑空间 $$ \{U_x(\varepsilon,\lambda):\varepsilon>0,\lambda\in(0,1]\} (x\in X), $$ 其中$U_x(\varepsilon,\lambda)=\{y\in X:F_{x,y}(\varepsilon)>1-\lambda\}$.

根据$\tau$拓扑的性质,在Menger PM -空间$(X,F,\Delta )$中,序列$\{ {x_n}\} $ 称为$\tau$ - 收敛于${x} \in X$,如果对任意的$t>0$, 有$\lim\limits_{n\rightarrow\infty}F_{x_n,x}(t)=1$. $\{ {x_n}\} $ 称为$X$中的$\tau$- Cauchy列,如果对任意的$\varepsilon> 0$和 $\lambda> 0$,存在正整数$N = N(\varepsilon ,\lambda )$, 当$m,n \ge N$时有${F_{{x_m},{x_n}}}(\varepsilon ) > 1 - \lambda $.Menger PM -空间$(X,F,\Delta )$称为$\tau$ -完备的, 如果$X$中的每一$\tau$- Cauchy列都$\tau$ -收敛于$X$中的某一点.

定义 2.3 ^[17]设$(X,\leq)$是一半序集, $G:X\times X\times X\rightarrow X,g:X\rightarrow X$. $G$称为具有$g$ -混合单调性质, 如果$G(x,y,z)$关于$x$ $g$ -单调不减,关于$y$ $g$ -单调不增,关于$z$ $g$ -单调不减, 即对任意的$x,y,z\in X$, $$ x_1,x_2\in X,g(x_1)\leq g(x_2)\Rightarrow G(x_1,y,z)\leq G(x_2,y,z); $$ $$ y_1,y_2\in X,g(y_1)\leq g(y_2)\Rightarrow G(x,y_1,z)\geq G(x,y_2,z); $$ $$ z_1,z_2\in X,g(z_1)\leq g(z_2)\Rightarrow G(x,y,z_1)\leq G(x,y,z_2).$$

定义 2.4 ^[17]元素$(x,y,z)\in X\times X\times X$称为$G$与$g$的三元重合点, 如果 $$G(x,y,z)=g(x),~~G(y,x,y)=g(y),~~G(z,y,x)=g(z).$$

定义 2.5 ^[16]设$(X,\leq)$是一半序集, $G:X\times X\times X\rightarrow X$. $G$称为具有混合单调性质, 如果$G(x,y,z)$关于$x$单调不减,关于$y$ 单调不增,关于$z$单调不减, 即对任意的$x,y,z\in X$, $$ x_1,x_2\in X,x_1\leq x_2\Rightarrow G(x_1,y,z)\leq G(x_2,y,z); $$ $$ y_1,y_2\in X,y_1\leq y_2\Rightarrow G(x,y_1,z)\geq G(x,y_2,z); $$ $$ z_1,z_2\in X,z_1\leq z_2\Rightarrow G(x,y,z_1)\leq G(x,y,z_2). $$

定义 2.6 ^[16]元素$(x,y,z)\in X\times X\times X$称为映射 $G:X\times X\times X\rightarrow X$的三元不动点,如果 $$ G(x,y,z)=x,~~G(y,x,y)=y,~~G(z,y,x)=z.$$

定义2.7 ^[12]设$(X,d)$是一度量空间, $G:X\times X\rightarrow X,g:X\rightarrow X$. 称$G$与$g$是相容的,如果 $$ \lim\limits_{n\rightarrow\infty}d(g(G(x_n,y_n)),G(g(x_n),g(y_n)))=0, $$ $$ \lim\limits_{n\rightarrow\infty}d(g(G(y_n,x_n)),G(g(y_n),g(x_n)))=0, $$ 其中$\{x_n\},\{y_n\}$是$X$中的序列,则对任意的$x,y\in X$,有 $$ \lim\limits_{n\rightarrow\infty}G(x_n,y_n)=\lim\limits_{n\rightarrow\infty}g(x_n)=x, $$ $$\lim\limits_{n\rightarrow\infty}G(y_n,x_n)=\lim\limits_{n\rightarrow\infty}g(y_n)=y.$$

定义$\Phi=\{\phi: {\Bbb R}^+\rightarrow {\Bbb R}^+\}$,其中$\phi\in\Phi$满足下列条件:

(a)~ $\phi$严格增;

(b)~ $\phi$右上半连续;

(c)~ $\sum\limits_{n=0}\limits^\infty \phi^n(t)＜+\infty,\forall t>0$, 其中$\phi^n(t)$是$\phi(t)$的第$n$次迭代.

易证如果$\phi\in\Phi$,则$\phi(t) ＜t,\forall t>0$.

引理 2.1 ^[9]设$\{y_n\}$是Menger PM -空间$(X,F,\Delta )$ 中的序列,其中$t$ -范数$\Delta$是$H$ -型的. 如果存在函数$\phi\in\Phi$,使得对任意的$t>0,n\in N$,有 $$ F_{y_n,y_{n+1}}(\phi(t))\geq\min\{F_{y_{n-1},y_n}(t),F_{y_n,y_{n+1}}(t)\}, $$ 则$\{y_n\}$是$X$中的Cauchy列.

引理2.2 ^[9]设$(X,F,\Delta )$是一个Menger PM -空间, 其中$t$ -范数$\Delta=\Delta_M,x,y\in X$. 如果存在$\phi\in\Phi$,使得对任意的$t>0$,有 $$ F_{x,y}(\phi(t)+0)\geq F_{x,y}(t), $$ 则$x=y$.

定义3.1  设$(X,\leq)$是一半序集, $(X,F,\Delta )$是一个完备的Menger PM -空间, $G:X\times X\times X\rightarrow X,$ $ g:X\rightarrow X$. 称$G$与$g$相容的, 如果对任意的$t>0$, $$ \lim\limits_{n\rightarrow\infty}F_{g(G(x_n,y_n,z_n)), G(g(x_n),g(y_n),g(z_n))}(t)=1, $$ $$ \lim\limits_{n\rightarrow\infty}F_{g(G(y_n,x_n,y_n)), G(g(y_n),g(x_n),g(y_n))}(t)=1, $$ $$ \lim\limits_{n\rightarrow\infty}F_{g(G(z_n,y_n,x_n)), G(g(z_n),g(y_n),g(x_n))}(t)=1, $$ 其中$\{x_n\},\{y_n\},\{z_n\}$是$X$中的序列,则对任意的$x,y,z\in X$,有 $$ \lim\limits_{n\rightarrow\infty}G(x_n,y_n,z_n)= \lim\limits_{n\rightarrow\infty}g(x_n)=x, $$ $$ \lim\limits_{n\rightarrow\infty}G(y_n,x_n,y_n)= \lim\limits_{n\rightarrow\infty}g(y_n)=y, $$ $$ \lim\limits_{n\rightarrow\infty}G(z_n,y_n,x_n)= \lim\limits_{n\rightarrow\infty}g(z_n)=z.$$

定理 3.1  设$(X,\leq)$是一半序集, $(X,F,\Delta )$是一个完备的Menger PM -空间,$t$ -范数$\triangle$是$H$ -型的. 设 $G:X\times X\times X\rightarrow X,g:X\rightarrow X$, 且$G$具有混合$g$ -单调性质. 设存在$\phi\in\Phi$, 使得对任意的$x,y,z,u,v,w\in X,t>0,\beta\in(0,1),\xi\in\{G(x,y,z),G(u,v,w)\}$,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t))&\geq& \min\Big\{F_{g(x),g(u)}(t),F_{g(x),G(x,y,z)}(t),F_{g(u),G(u,v,w)}(t), \\ && [F_{G(x,y,z),\xi}\oplus F_{\xi,G(u,v,w)}((2-\beta)t)]\Big\}, \tag{3.1}\end{eqnarray} 其中$g(x)\geq g(u),g(y)\leq g(v),g(z)\geq g(w)$ (或者$g(x)\leq g(u),g(y)\geq g(v),g(z)\leq g(w)$).

设$G(X\times X\times X)\subseteq g(X)$,$G$和$g$连续且是相容的, 并且$g$是单调不减的. 如果存在$x_0,y_0,z_0\in X$使得 $$ g(x_0)\leq G(x_0,y_0,z_0),~~g(y_0)\geq G(y_0,x_0,y_0),~~g(z_0)\leq G(z_0,y_0,x_0), $$ 则存在$x,y,z\in X$使得 $$ g(x)=G(x,y,z),~~g(y)=G(y,x,y),~~g(z)=G(z,y,x). $$ 即,$G$与$g$存在三元重合点.

证  根据题意,存在$x_0,y_0,z_0\in X$使得 \begin{equation} g(x_0)\leq G(x_0,y_0,z_0),~~g(y_0)\geq G(y_0,x_0,y_0),~~g(z_0)\leq G(z_0,y_0,x_0).\tag{3.2} \end{equation} 由于$G(X\times X\times X)\subseteq g(X)$, 故存在$x_1,y_1,z_1\in X$使得对任意的$n\geq0$,有 \begin{equation} g(x_1)= G(x_0,y_0,z_0),~~g(y_1)= G(y_0,x_0,y_0),~~g(z_1)=G(z_0,y_0,x_0).\tag{3.3} \end{equation} 依此类推,存在$\{x_n\},\{y_n\},\{z_n\}\in X$使得对任意的$n\geq0$,有 \begin{equation} g(x_{n+1})= G(x_n,y_n,z_n),~~g(y_{n+1})= G(y_n,x_n,y_n),~~ g(z_{n+1})= G(z_n,y_n,x_n).\tag{3.4} \end{equation}

下面用数学归纳法证明,对任意的$n\geq0$,有 \begin{equation}g(x_n)\leq g(x_{n+1}),~~g(y_n)\geq g(y_{n+1}),~~g(z_n)\leq g(z_{n+1}). \tag{3.5} \end{equation} 事实上,由(3.2)式和(3.3)式可知 $$ g(x_0)\leq g(x_1),~~g(y_0)\geq g(y_1),~~g(z_0)\leq g(z_1).$$ 即(3.5)式对$n=0$成立. 假设(3.5)式对某一$n\geq0$成立,即 $$g(x_n)\leq g(x_{n+1}),~~g(y_n)\geq g(y_{n+1}),~~g(z_n)\leq g(z_{n+1}), $$ 由$G$是$g$ -混合单调算子,可得 $$ g(x_{n+1})= G(x_n,y_n,z_n)\leq G(x_{n+1},y_n,z_n)\leq G(x_{n+1},y_{n+1},z_{n+1})=g(x_{n+2}), $$ $$ g(y_{n+1})= G(y_n,x_n,y_n)\geq G(y_{n+1},x_n,y_n)\geq G(y_{n+1},x_{n+1},y_{n+1})=g(y_{n+2}), $$ $$ g(z_{n+1})= G(z_n,y_n,x_n)\leq G(z_{n+1},y_n,x_n)\leq G(z_{n+1},y_{n+1},x_{n+1})=g(z_{n+2}). $$ 由数学归纳法,(3.5)式对任意的$n\geq0$成立,即有 $$g(x_0)\leq g(x_1)\leq g(x_2)\leq\cdots\leq g(x_n)\leq g(x_{n+1})\leq\cdots, $$ $$ g(y_0)\geq g(y_1)\geq g(y_2)\geq\cdots\geq g(y_n)\geq g(y_{n+1})\geq\cdots, $$ $$ g(z_0)\leq g(z_1)\leq g(z_2)\leq\cdots\leq g(z_n)\leq g(z_{n+1})\leq\cdots. $$

在(3.1)式中令$x=x_{n-1},y=y_{n-1},z=z_{n-1},u=x_n,v=y_n,w=z_n$,可得 \begin{eqnarray*} && F_{G(x_{n-1},y_{n-1},z_{n-1}),G(x_n,y_n,z_n)}(\phi(t))\\ &\geq& \min\{F_{g(x_{n-1}),g(x_n)}(t),F_{g(x_{n-1}),G(x_{n-1},y_{n-1},z_{n-1})}(t), F_{g(x_n),G(x_n,y_n,z_n)}(t),\\ &&[(F_{G(x_{n-1},y_{n-1},z_{n-1}),\xi} \oplus F_{\xi,G(x_n,y_n,z_n)})(t+(1-\beta)t)]\}.\end{eqnarray*} 由(3.4)式可得 \begin{eqnarray} F_{g(x_n),g(x_{n+1})}(\phi(t))&\geq&\min\{F_{g(x_{n-1}),g(x_n)}(t), F_{g(x_{n-1}),g(x_n)}(t),F_{g(x_n),g(x_{n+1})}(t), \\ &&F_{g(x_n),g(x_{n+1})}(t),F_{g(x_{n+1}),g(x_{n+1})}((1-\beta)t)\} \\ &=&\min\{F_{g(x_{n-1}),g(x_n)}(t),F_{g(x_n),g(x_{n+1})}(t),1\} \\ &=&\min\{F_{g(x_{n-1}),g(x_n)}(t),F_{g(x_n),g(x_{n+1})}(t)\}(n\in N,t>0).\tag{3.6}\end{eqnarray} 由引理2.1可知,$\{g(x_n)\}$是$X$中的Cauchy列.

同理可证,$\{g(y_n)\}$与$\{g(z_n)\}$也是$X$中的Cauchy列.

因为$X$完备,故存在$x,y,z\in X$使得 \begin{equation}\lim\limits_{n\rightarrow\infty}G(x_n,y_n,z_n)= \lim\limits_{n\rightarrow\infty}g(x_n)=x,\tag{3.7} \end{equation} \begin{equation}\lim\limits_{n\rightarrow\infty}G(y_n,x_n,y_n)= \lim\limits_{n\rightarrow\infty}g(y_n)=y,\tag{3.8} \end{equation} \begin{equation}\lim\limits_{n\rightarrow\infty}G(z_n,y_n,x_n)= \lim\limits_{n\rightarrow\infty}g(z_n)=z.\tag{3.9} \end{equation} 又由于$g$连续,由(3.7)-(3.9)式可得 \begin{equation}\lim\limits_{n\rightarrow\infty}g(G(x_n,y_n,z_n))= \lim\limits_{n\rightarrow\infty}g(g(x_{n+1}))=g(x),\tag{3.10} \end{equation} \begin{equation}\lim\limits_{n\rightarrow\infty}g(G(y_n,x_n,y_n))= \lim\limits_{n\rightarrow\infty}g(g(y_{n+1}))=g(y),\tag{3.11} \end{equation} \begin{equation}\lim\limits_{n\rightarrow\infty}g(G(z_n,y_n,x_n))= \lim\limits_{n\rightarrow\infty}g(g(z_{n+1}))=g(z).\tag{3.12} \end{equation} 由$G$与$g$的相容性和(3.7)-(3.9)式及定义3.1可得 \begin{equation} \lim\limits_{n\rightarrow\infty}F_{g(G(x_n,y_n,z_n)), G(g(x_n),g(y_n),g(z_n))}(t)=1,\forall t>0,\tag{3.13} \end{equation} \begin{equation} \lim\limits_{n\rightarrow\infty}F_{g(G(y_n,x_n,y_n)), G(g(y_n),g(x_n),g(y_n))}(t)=1,\forall t>0,\tag{3.14} \end{equation} \begin{equation} \lim\limits_{n\rightarrow\infty}F_{g(G(z_n,y_n,x_n)), G(g(z_n),g(y_n),g(x_n))}(t)=1,\forall t>0.\tag{3.15} \end{equation}

下证$g(x)=G(x,y,z),g(y)=G(y,x,y),g(z)=G(z,y,x)$.

因为$t-\phi(t)>0$,故由(PM-3)可知对任意的$n\geq0,t>0$,有 \begin{equation} F_{g(x),G(x,y,z)}(t)\geq \Delta\{F_{g(x), G(g(x_n),g(y_n),g(z_n))}(t-\phi(t)),F_{G(g(x_n),g(y_n),g(z_n)),G(x,y,z)}(\phi(t))\}, \tag{3.16} \end{equation} 又由(PM-3)可得 \begin{eqnarray} &&F_{g(x),G(g(x_n),g(y_n),g(z_n))}(t-\phi(t)) \\ &\geq& \Delta\Big\{F_{g(x),g(G(x_n,y_n,z_n))}(\frac{t-\phi(t)}{2}), F_{g(G(x_n,y_n,z_n)),G(g(x_n),g(y_n),g(z_n))} (\frac{t-\phi(t)}{2})\Big\}.\tag{3.17}\end{eqnarray} 由(3.10)式和拓扑$\tau$的收敛定义,有 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(x),g(G(x_n,y_n,z_n))} (\frac{t-\phi(t)}{2})=1.\tag{3.18} \end{equation} 因为$\frac{t-\phi(t)}{2}>0$,故由(3.13)式可得 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(G(x_n,y_n,z_n)), G(g(x_n),g(y_n),g(z_n))}(\frac{t-\phi(t)}{2})=1.\tag{3.19} \end{equation} 由$H$型$t$ -范数的性质可知 $$\lim\limits_{a\rightarrow1,b\rightarrow1}\Delta(a,b)=1.$$ 从而由上式和(3.17)-(3.19)式可得 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(x),G(g(x_n),g(y_n), g(z_n))}(t-\phi(t))=1,\forall t>0.\tag{3.20} \end{equation} 又由$G$的连续性和(3.7)-(3.9)式可得 $$ \lim\limits_{n\rightarrow\infty}G(g(x_n),g(y_n),g(z_n))= G(\lim\limits_{n\rightarrow\infty}g(x_n),\lim\limits_{n\rightarrow\infty}g(y_n), \lim\limits_{n\rightarrow\infty}g(z_n))=G(x,y,z).$$ 故由拓扑$\tau$的收敛定义,有 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{G(g(x_n),g(y_n),g(z_n)), G(x,y,z)}(\phi(t))=1.\tag{3.21} \end{equation} 由(3.16)式和(3.20)-(3.21)式及$\lim\limits_{a\rightarrow1,b\rightarrow1}\Delta(a,b)=1$可得 \begin{eqnarray} F_{g(x),G(x,y,z)}(t)&\geq& \lim\limits_{n\rightarrow\infty}\Delta\{F_{g(x), G(g(x_n),g(y_n),g(z_n))}(t-\phi(t)),F_{G(g(x_n),g(y_n),g(z_n)),G(x,y,z)}(\phi(t))\} \\ &=&1.\tag{3.22}\end{eqnarray} 故对任意的$t>0$,有$F_{g(x),G(x,y,z)}(t)=1$,由(PM-1)可得$g(x)=G(x,y,z)$.

同理可证,$F_{g(y),G(y,x,y)}(t)=1,$ $F_{g(z),G(z,y,x)}(t)=1,$ $\forall t>0$. 由(PM-1)可知$g(y)=G(y,x,y),$ $g(z)=G(z,y,x)$.

即$G$与$g$存在三元重合点.

推论 3.1  设$(X,\leq)$是一半序集, $(X,F,\Delta )$是一个完备的Menger PM -空间, $\Delta$是一连续$t$ -范数且$\Delta=\Delta_M$. 设 $G:X\times X\times X\rightarrow X,g:X\rightarrow X$,且$G$具有混合$g$ -单调性质. 设存在$\phi\in\Phi$,使得对任意的$x,y,z,u,v,w\in X,t>0,\beta\in(0,1), \xi\in\{g(x),g(u),G(x,y,z),G(u,v,w)\}$,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t)) &\geq& \min\{F_{g(x),g(u)}(t), F_{g(x),G(x,y,z)}(t),F_{g(u),G(u,v,w)}(t), \\ && [F_{G(x,y,z),\xi}\oplus F_{\xi,G(u,v,w)}((2-\beta)t)]\}, \tag{3.23}\end{eqnarray} 其中$g(x)\geq g(u),g(y)\leq g(v),g(z)\geq g(w)$ (或者$g(x)\leq g(u), g(y)\geq g(v),g(z)\leq g(w)$).

设$G(X\times X\times X)\subseteq g(X)$,$G$和$g$是相容的,$g$连续且单调不减, 并且$X$具有下列性质:

(i)~ 若不减序列$\{x_n\}\rightarrow x$,则对任意的$n\in N,x_n\leq x$;

(ii)~ 若不增序列$\{y_n\}\rightarrow y$,则对任意的$n\in N,y_n\geq y$.

如果存在$x_0,y_0,z_0\in X$使得 $$ g(x_0)\leq G(x_0,y_0,z_0),~~g(y_0)\geq G(y_0,x_0,y_0),~~ g(z_0)\leq G(z_0,y_0,x_0), $$ 则存在$x,y,z\in X$使得 $$ g(x)=G(x,y,z),~~g(y)=G(y,x,y),~~g(z)=G(z,y,x).$$ 即,$G$与$g$存在三元重合点.

在定理3.1中令$g=I$ ($I$为恒等映射),则下面推论成立.

推论 3.2  设$(X,\leq)$是一半序集, $(X,F,\Delta )$是一个完备的Menger PM -空间,$t$ -范数$\triangle$是$H$ -型的. 设 $G:X\times X\times X\rightarrow X$,$G$具有混合单调性质且$G$连续. 设存在$\phi\in\Phi$使得对任意的$x,y,z,u,v,w\in X,t>0,\beta\in(0,1), \xi\in\{G(x,y,z),G(u,v,w)\}$,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t))&\geq& \min\{F_{x,u}(t), F_{x,G(x,y,z)}(t),F_{u,G(u,v,w)}(t), \\ &&[F_{G(x,y,z),\xi}\oplus F_{\xi,G(u,v,w)}((2-\beta)t)]\}, \tag{3.24}\end{eqnarray} 如果存在$x_0,y_0,z_0\in X$使得 $$ x_0\leq G(x_0,y_0,z_0),~~y_0\geq G(y_0,x_0,y_0),~~z_0\leq G(z_0,y_0,x_0), $$ 则存在$x,y,z\in X$使得 $$ x=G(x,y,z),~~y=G(y,x,y),~~z=G(z,y,x). $$ 即,$G$存在三元不动点.

在定理3.1中令$\phi(t)=\kappa t$,则下面推论成立.

推论 3.3  设$(X,\leq)$是一半序集,$(X,F,\Delta )$是一个完备的 Menger PM -空间,$t$ -范数$\triangle$是$H$ -型的. 设 $G:X\times X\times X\rightarrow X,g:X\rightarrow X$,且$G$具有混合$g$ -单调性质, 并且存在$\kappa\in(0,1)$,使得对任意的$x,y,z,u,v,w\in X,t>0,\beta\in(0,1), \xi\in\{G(x,y,z),G(u,v,w)\}$,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\kappa t)&\geq& \min\{F_{g(x),g(u)}(t),F_{g(x), G(x,y,z)}(t),F_{g(u),G(u,v,w)}(t), \\ &&[F_{G(x,y,z),\xi}\oplus F_{\xi,G(u,v,w)}((2-\beta)t)]\}, \tag{3.25}\end{eqnarray} 其中$g(x)\geq g(u),g(y)\leq g(v),g(z)\geq g(w)$ (或者$g(x)\leq g(u),g(y)\geq g(v),g(z)\leq g(w)$).

设$G(X\times X\times X)\subseteq g(X)$,$G$和$g$连续且是相容的,并且$g$是单调不减的. 如果存在$x_0,y_0,z_0\in X$使得 $$ g(x_0)\leq G(x_0,y_0,z_0),~~g(y_0)\geq G(y_0,x_0,y_0),~~ g(z_0)\leq G(z_0,y_0,x_0), $$ 则存在$x,y,z\in X$使得 $$ g(x)=G(x,y,z),~~g(y)=G(y,x,y),~~g(z)=G(z,y,x).$$ 即,$G$与$g$存在三元重合点.

下面,通过一个例子来给出定理3.1的应用.

例4.1  令$X=[0, 1]$,$F_{x,y}(t)=\frac{t}{t+|x-y|}$,$\Delta=\Delta_M$,则$(X,F,\Delta )$是一个完备的Menger PM -空间.

定义$g:X\rightarrow X,G:X\times X\times X\rightarrow X$如下 $$g(x)=x^3,\forall x\in X,$$ $$G(x,y,z)= \left \{ \begin{array}{ll} \frac{x^3-y^3+z^3}{3},~~ & {x,y,z\in[0, 1],x\geq y\geq z,}\ 0.\end{array} \right.$$ 容易验证,$G(X\times X\times X)\subseteq g(X)$,$g$连续且单调不减, $G$具有混合$g$ -单调性质.

设$\phi(t)=\frac{2}{3}t,t\in[0,\infty).$ 并设$x_0=z_0=0,y_0=c\in[0, 1]$是$X$中三点,则 $$g(x_0)=g(0)=0=G(0,c,0)\leq G(x_0,y_0,z_0),$$ $$g(y_0)=g(c)=c^3\geq \frac{c^3-0^3+c^3}{3}=G(c,0,c)=G(y_0,x_0,y_0),$$ $$g(z_0)=g(0)=0=G(0,c,0)\leq G(z_0,y_0,x_0).$$

下证定理3.1中的(3.1)式成立. 取$x,y,z,u,v,w\in X$, 使$g(x)\leq g(u),$ $g(y)\geq g(v),$ $g(z)\leq g(w)$,也即,$x\leq u,y\geq v,z\leq w$.

下面分四种情况讨论:

情形1 当$G(x,y,z)=\frac{x^3-y^3+z^3}{3},G(u,v,w)=\frac{u^3-v^3+w^3}{3}$时, 也即,$x\geq y\geq z,u\geq v\geq w$时,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t))&=&F_{\frac{x^3-y^3+z^3}{3},\frac{u^3-v^3+w^3}{3}}(\frac{2}{3}t) \\ &=&\frac{\frac{2}{3}t}{\frac{2}{3}t+|\frac{x^3-y^3+z^3}{3}-\frac{u^3-v^3+w^3}{3}|} \\ &=&\frac{2t}{2t+|(u^3-x^3)+(y^3-v^3)+(w^3-z^3)|}.\tag{4.1}\end{eqnarray} 由$x\leq u,y\geq v,z\leq w$,可得 $$|(u^3-x^3)+(y^3-v^3)+(w^3-z^3)|=(u^3-x^3)+(y^3-v^3)+(w^3-z^3)\leq u^3-v^3+w^3,$$ 故由(4.1)式可得 \begin{equation}F_{G(x,y,z),G(u,v,w)}(\phi(t))\geq\frac{2t}{2t+(u^3-v^3+w^3)} \geq\frac{t}{t+|\frac{u^3-v^3+w^3}{3}-u^3|}=F_{g(u),G(u,v,w)}(t),\tag{4.2} \end{equation} 从而由(4.2)式可知,$G$与$g$满足(3.1)式.

情形2 当$G(x,y,z)=\frac{x^3-y^3+z^3}{3},G(u,v,w)=0$时, 即$x\geq y\geq z$时,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t))&=&F_{\frac{x^3-y^3+z^3}{3},0}(\frac{2}{3}t) \\ &=&\frac{\frac{2}{3}t}{\frac{2}{3}t+|\frac{x^3-y^3+z^3}{3}-0|} \\ &=&\frac{2t}{2t+|x^3-y^3+z^3|} \\ &\geq&\frac{t}{t+|\frac{x^3-y^3+z^3}{3}-x^3|} \\ &=&F_{g(x),G(x,y,z)}(t), \end{eqnarray} 从而$G$与$g$满足(3.1)式.

情形3 当$G(x,y,z)=0,G(u,v,w)=\frac{u^3-v^3+w^3}{3}$时, 即 $u\geq v\geq w$时,有 \begin{eqnarray} F_{G(x,y,z),G(u,v,w)}(\phi(t))&=&F_{0,\frac{u^3-v^3+w^3}{3}}(\frac{2}{3}t) \\ &=&\frac{\frac{2}{3}t}{\frac{2}{3}t+|0-\frac{u^3-v^3+w^3}{3}|} \\ &=&\frac{2t}{2t+|u^3-v^3+w^3|} \\ &\geq&\frac{t}{t+|\frac{u^3-v^3+w^3}{3}-u^3|} \\ &=&F_{g(u),G(u,v,w)}(t), \end{eqnarray} 从而$G$与$g$满足(3.1)式.

情形4 当$G(x,y,z)=0,G(u,v,w)=0$时,有 $$ F_{G(x,y,z),G(u,v,w)}(\phi(t))=F_{0,0}(\frac{2}{3}t)= \frac{\frac{2}{3}t}{\frac{2}{3}t+|0-0|}=1\geq F_{g(x),g(u)}(t), $$ 从而$G$与$g$满足(3.1)式.

设$\{x_n\},\{y_n\},\{z_n\}\in X$使得 \begin{equation}\lim\limits_{n\rightarrow\infty}G(x_n,y_n,z_n)=a,~~ \lim\limits_{n\rightarrow\infty}g(x_n)=a,\tag{4.3} \end{equation} \begin{equation} \lim\limits_{n\rightarrow\infty}G(y_n,x_n,y_n)=b,~~ \lim\limits_{n\rightarrow\infty}g(y_n)=b,\tag{4.4} \end{equation} \begin{equation} \lim\limits_{n\rightarrow\infty}G(z_n,y_n,x_n)=c,~~ \lim\limits_{n\rightarrow\infty}g(z_n)=c.\tag{4.5} \end{equation} 因为$G(x,y,z)=0,G(u,v,w)=0,\forall x,y,z,u,v,w\in X$, 故由(4.3)-(4.5)式可知$a=0,b=0,$ $c=0$,从而可得 $$\lim\limits_{n\rightarrow\infty}x_n=\lim\limits_{n\rightarrow\infty}y_n= \lim\limits_{n\rightarrow\infty}z_n= \lim\limits_{n\rightarrow\infty}\sqrt[3]{a}=\lim\limits_{n\rightarrow\infty} \sqrt[3]{b}=\lim\limits_{n\rightarrow\infty}\sqrt[3]{c}=0.$$ 因为$\lim\limits_{n\rightarrow\infty}x_n=\lim\limits_{n\rightarrow\infty}y_n= \lim\limits_{n\rightarrow\infty}z_n=0$及$G$与$g$是连续的,故有 $$\lim\limits_{n\rightarrow\infty}G(x_n,y_n,z_n)= \lim\limits_{n\rightarrow\infty}G(y_n,x_n,y_n)= \lim\limits_{n\rightarrow\infty}G(z_n,y_n,x_n)= G(0,0,0)=0, $$ $$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } G(g({x_n}),g({y_n}),g({z_n})) = \mathop {\lim }\limits_{n \to \infty } G(g({y_n}),g({x_n}),g({y_n})) \cr & = \mathop {\lim }\limits_{n \to \infty } G(g({z_n}),g({y_n}),g({x_n})) = G(0,0,0) = 0, \cr} \tag{4.6}$$ \begin{equation}\lim\limits_{n\rightarrow\infty}g(G(x_n,y_n,z_n))= \lim\limits_{n\rightarrow\infty}[G(x_n,y_n,z_n)]^3=[G(0,0,0)]^3=0.\tag{4.7} \end{equation} 由(4.6)式和(4.7)式可得 \begin{equation} \lim\limits_{n\rightarrow\infty}F_{G(g(x_n),g(y_n),g(z_n)),0}(t)=1, \lim\limits_{n\rightarrow\infty}F_{g(G(x_n,y_n,z_n)),0}(t)=1,\forall t>0.\tag{4.8} \end{equation} 由(PM-3)有 \begin{equation} F_{g(G(x_n,y_n,z_n)),G(g(x_n),g(y_n),g(z_n))}(t)\geq \Delta\Big\{F_{g(G(x_n,y_n,z_n)),0}(\frac{t}{2}), F_{G(g(x_n),g(y_n),g(z_n)),0}(\frac{t}{2})\Big\}.\tag{4.9} \end{equation} 由(4.9)式和(4.8)式及$\lim\limits_{a\rightarrow1,b\rightarrow1}\Delta(a,b)=1$可得 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(G(x_n,y_n,z_n)), G(g(x_n),g(y_n),g(z_n))}(t)=1,\forall t>0.\tag{4.10} \end{equation} 同理可证 \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(G(y_n,x_n,y_n)), G(g(y_n),g(x_n),g(y_n))}(t)=1,\forall t>0; \tag{4.11} \end{equation} \begin{equation}\lim\limits_{n\rightarrow\infty}F_{g(G(z_n,y_n,x_n)), G(g(z_n),g(y_n),g(x_n))}(t)=1,\forall t>0.\tag{4.12} \end{equation} 从而由(4.10)-(4.12)式可知,$G$与$g$是相容的. 综上,$G,g$,$\phi$满足定理3.1中的所有条件. 因为$G(0,0,0)=g(0)=0$,故$(0,0,0)$是$G$与$g$在$X$中的三元重合点.

下证$G$与$g$不是可交换的.$$g(G({x_n},{y_n},{z_n})) = \left\{ {\matrix{ {\matrix{ {{{({{x_n^3 - y_n^3 + z_n^3} \over 3})}^3},} \hfill & {{x_n},{y_n},{z_n} \in [0,1],{x_n} \ge {y_n} \ge {z_n},} \hfill & {} \hfill \cr } } \cr {0.} \cr } } \right.$$

设$x_n eq0,x_n>y_n\geq z_n$,有 \begin{eqnarray} g(G(z_n,y_n,x_n))&=&(\frac{x_n^3-y_n^3+z_n^3}{3})^3 eq \frac{(x_n^3)^3 -(y_n^3)^3+(z_n^3)^3}{3} \\ &=&\frac{(g(x_n))^3-(g(y_n))^3+(g(z_n))^3}{3}=G(g(x_n),g(y_n),g(z_n)). \end{eqnarray}