Banach压缩映射原理是不动点理论的一个基本结果.由于它在数学学科和工程方面的广泛应用, 许多学者对它进行了大量的研究并将它在不同的方向加以推广, 其中对压缩条件的推广是其重要方向之一.文献[1]在Hilbert空间中引进弱压缩映射的概念, 随后文献[2]在度量空间中研究了弱压缩映射的不动点问题并得到一些不动点定理.文献[3-9]研究了一些更一般的压缩条件.在研究单个映射的不动点问题的同时, 对于满足某种压缩条件的几个映射的公共不动点问题也得到人们的广泛关注和研究(参见文献[10-13]).人们还研究其它各类空间如锥度量空间, 偏度量空间, 概率度量空间等空间中的不动点问题.由于对赋予了一个偏序的空间, 压缩条件可不必对空间中的所有点成立, 文献[14-18]在赋予了一个偏序的度量空间或其它空间中研究映射的不动点或公共不动点.文献[19-23]研究了有理分式型压缩映射.

本文研究偏序偏度量空间中四个映射的公共不动点问题, 需要如下的概念和结果.

定义1.1^[3] 设函数$\psi:[0, +\infty)\rightarrow [0, +\infty)$满足下列条件

(ⅰ) $\psi$连续非减;

(ⅱ) $\psi(t)=0$当且仅当$t=0$.

则称$\psi$是改变距离函数(altering distance function).

设$f, g:X\rightarrow X$两个映射, 若$w=fx=gx$对$X$中某个$x$成立, 则称$x$是$f$和$g$的叠合点(coincidence point), 称$w$是叠合$f$和$g$的点(point of coincidence of $f$ and $g$).

定义1.2^[13] 称映射$f, g:X\rightarrow X$是弱相容的, 如果对每个$x\in X$, 只要$fx=gx$, 都有$fgx =gfx$成立, 即$f$和$g$在它们的每个叠合点可交换.也称$\{f, g\}$是一个弱相容映射对.

定义1.3^[6] 设$(X, \preceq)$是偏序集且$f, g, T:X\rightarrow X$是三个映射, 满足$f(X)\subseteq T(X), $ $g(X)\subseteq T(X)$.称$f$和$g$关于$T$弱增, 如果对所有$x\in X$, 有$fx\preceq gy$对每个$y\in T^{-1}(fx)$成立, 且$gx\preceq fy$对每个$y\in T^{-1}(gx)$成立.如果$f=g$, 称$f$关于$T$弱增.

定义1.4^[14] 设$(X, \preceq)$是偏序集且$f, g, T:X\rightarrow X$是三个映射使得$f(X)\subseteq T(X)$, 称$(f, g)$关于$T$部分弱增(partially weakly increasing), 如果对所有$x\in X$, 有$fx\preceq gy$对每个$y\in T^{-1}(fx)$成立.

定义1.5^[24] 设$X$非空集, $p:X\times X\rightarrow [0, +\infty)$是一个函数使得对所有$x, y, z\in X$,

(P1) $x=y$当且仅当$p(x, x)=p(x, y)=p(y, y);$

(P2) $p(x, x)\leq p(x, y)$;

(P3) $p(x, y)=p(y, x)$;

(P4) $p(x, z)+p(y, y)\leq p(x, y)+p(y, z)$.

则称$p$是$X$上的一个偏度量, 称$(X, P)$是偏度量空间(partial metric space).

注1.1 显然, 每个度量空间是偏度量空间, 反之不真.如果$p$是$X$上的一个偏度量, 且$p(x, y)=0$, 则由(P1) 和(P2), 有$x=y$.但$p(x, x)$未必为$0$.

对$X$上的一个偏度量$p$, 若令$d_{p}(x, y)=2p(x, y)-p(x, x)-P(y, y)$, 则$d_{p}:X\times X\rightarrow [0, \infty)$是$X$上的一个度量.

$X$上的每个偏度量$p$生成$X$上的一个$T_{0}$拓扑$\tau_{p}$, 所有开$p$-球所成的簇$B_{p}(x, \epsilon)$是该拓扑的一个基, 其中$B_{p}(x, \epsilon)=\{y\in X:p(x, y)＜p(x, x)+\epsilon\}$.

定义1.6^[20] 设$(X, p)$是一个偏度量空间, $\{x_{n}\}$是$X$中的一个序列, 称

(ⅰ) $\{x_{n}\}$收敛于$x\in X$如果$p(x, x)=\lim\limits_{n\rightarrow \infty}p(x, x_{n});$

(ⅱ) $\{x_{n}\}$是一个Cauchy序列如果$\lim\limits_{n, m\rightarrow \infty}p(x_{n}, x_{m})$存在(有限);

(ⅲ) $(X, p)$完备如果$X$中每个Cauchy序列$\{x_{n}\}$关于$\tau_{p}$收敛于$X$中一点$x$使得$p(x, x)=\lim\limits_{n, m\rightarrow \infty}p(x_{n}, x_{m})$.

显然, 如果$A\subset X$是偏度量空间$(X, p)$中的一个闭子集, $\{x_{n}\}\subset A$收敛于$x$, 则$x\in A$.

引理1.1^[24] 设$(X, p)$是一个偏度量空间.则

(ⅰ) $\{x_{n}\}$是$(X, p)$中的Cauchy序列当且仅当$\{x_{n}\}$是度量空间$(X, d_{p})$中的Cauchy序列.

(ⅱ) $(X, p)$完备当且仅当度量空间$(X, d_{p})$完备.进一步,

$\lim\limits_{n\rightarrow\infty}d_{p}(x, x_{n})=0\Leftrightarrow p(x, x)=\lim\limits_{n\rightarrow\infty}p(x, x_{n})= \lim\limits_{n, m\rightarrow\infty}p(x_{n}, x_{m}).$

一个偏序偏度量空间是一个赋予了一个偏序$"\preceq"$的偏度量空间, 记为$(X, p, \preceq)$.

定义1.7 设$(X, p, \preceq)$是偏序偏度量空间, $f:X\rightarrow X$是一个映射, 称$f$有性质(A$_{1})$如果$\{x_{n}\}$是$X$中任何非减收敛于某个$fx_{0}$的序列, 则对所有$n$都有$fx_{n}\preceq fx_{0}$.如果还有$fx_{0}\preceq ffx_{0}$, 则称$f$有性质(A), 其中, $x_{0}\in X$.

设$f, g, T, S:X\rightarrow X$是四个映射, $f(X)\subseteq T(X), g(X)\subseteq S(x)$. $x_{0}$是$X$中任意一点, 则存在$x_{1}, x_{2}\in X$使得$fx_{0}=Tx_{1}, gx_{1}=Sx_{2}$.归纳地, 我们可以得到两个序列$\{x_{n}\}, \{y_{n}\}\subseteq X$使得$y_{2n}=fx_{2n}=Tx_{2n+1}, y_{2n+1}=gx_{2n+1}=Sx_{2n+2}, n=0, 1, 2, \cdots .$称$\{y_{n}\}$是一个以$x_{0}$为初始点的$(f, g)$-$(T, S)$ -序列.

定理2.1 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f, g, S, T:X\rightarrow X$满足下列条件

(ⅰ)对所有使得$Sx$和$Ty$可比较, 且$Sx\neq Ty$或$fx\neq gy$的$x, y\in X$, 有

$\begin{eqnarray} \psi(p(fx, gy))\leq \psi (M(x, y))-\varphi(M(x, y)), \end{eqnarray}$

(2.1)

其中

$\begin{eqnarray*} M(x, y)&=&\max\bigg\{p(Sx, Ty), \frac{p(Sx, fx)p(Ty, gy)+p(Sx, gy)p(Ty, fx)} {3(p(fx, gy)+p(Sx, Ty))}, \\ &&\frac{p(Sx, fx)+p(Ty, gy)}{2}, \frac{p(Sx, gy)+p(Ty, fx)}{2}\bigg\}, \end{eqnarray*}$

$\psi$是改变距离函数, $\varphi:[0, +\infty)\rightarrow [0, +\infty)$是下半连续函数, $\varphi(t)=0$当且仅当$t=0$;

(ⅱ) $f(X)\subseteq T(X), g(X)\subseteq S(X)$, 且$T(X)$与$S(X)$都是$(X, p)$中的闭子集;

(ⅲ) $(f, g)$关于$T$部分弱增, $(g, f)$关于$S$部分弱增;

(ⅳ) $S$和$T$有性质(A$_{1}).$

则存在一点$w\in X$既是叠合$f$和$S$的点又是叠合$g$和$T$的点.

证 设$x_{0}$是$X$中任意一点, $\{y_{n}\}$是以$x_{0}$为初始点的$(f, g)$-$(T, S)$-序列.由于$x_{2n+1}\in T^{-1}(fx_{2n}), $ $x_{2n+2}\in S^{-1}(gx_{2n+1})$, 由条件(ⅲ), 有$y_{2n}=fx_{2n}\preceq gx_{2n+1}=y_{2n+1}$及$y_{2n+1}=gx_{2n+1}\preceq fx_{2n+2}=y_{2n+2}, n=0, 1, 2, \cdots .$

如果存在自然数$n_{0}$使得$y_{n_{0}-1}=y_{n_{0}}=y_{n_{0}+1}$, 当$n_{0}$是偶数, 即对某个$k_{0}$有$n_{0}=2k_{0}$时, 则$Sx_{2k_{0}}=fx_{2k_{0}}=gx_{2k_{0}+1}=Tx_{2k_{0}+1}$, 这表明$x_{2k_{0}}$是$f$和$S$的一个叠合点, $x_{2k_{0}+1}$是$g$和$T$的一个叠合点, 且$w=Sx_{2k_{0}}=fx_{2k_{0}}=gx_{2k_{0}+1}=Tx_{2k_{0}+1}$既是叠合$f$和$S$的点又是叠合$g$和$T$的点.当$n_{0}$是奇数也有同样的结论.下面假定对任意$n$, 都有$y_{n-1}\neq y_{n}$或$y_{n}\neq y_{n+1}$.

由于$Sx_{2n}$与$Tx_{2n+1}$可比较, 且$Sx_{2n}\neq Tx_{2n+1}$或$fx_{2n}\neq gx_{2n+1}$, 由(2.1) 式, 有

$\begin{eqnarray} \psi(p(y_{2n}, y_{2n+1}))&=&\psi(p(fx_{2n}, gx_{2n+1})) \leq\psi(M(x_{2n}, x_{2n+1}))-\varphi(M(x_{2n}, x_{2n+1})). \end{eqnarray}$

(2.2)

其中

$\begin{eqnarray*} &&M(x_{2n}, x_{2n+1})\\ &=&\max\bigg\{p(Sx_{2n}, Tx_{2n+1}), \\ &&\frac{p(Sx_{2n}, fx_{2n})p(Tx_{2n+1}, gx_{2n+1})+p(Sx_{2n}, gx_{2n+1})p(Tx_{2n+1}, fx_{2n})}{3(p(fx_{2n}, gx_{2n+1})+p(Sx_{2n}, Tx_{2n+1}))}, \\&&\frac{p(Sx_{2n}, fx_{2n})+p(Tx_{2n+1}, gx_{2n+1})}{2}, \frac{p(Sx_{2n}, gx_{2n+1})+p(Tx_{2n+1}, fx_{2n})}{2}\bigg\}\\ &=&\bigg\{p(y_{2n-1}, y_{2n}), \frac{p(y_{2n-1}, y_{2n})p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})p(y_{2n}, y_{2n})}{3(p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n}))}, \\&&\frac{p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1})}{2}, \frac{p(y_{2n-1}, y_{2n+1})+p(y_{2n}, y_{2n})}{2}\bigg\}. \end{eqnarray*}$

因为

$\begin{eqnarray*} &&\frac{p(y_{2n-1}, y_{2n})p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})p(y_{2n}, y_{2n})}{3(p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n}))}\\&\leq& \frac{p(y_{2n-1}, y_{2n})p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})p(y_{2n-1}, y_{2n})}{3(p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n}))} \\ &\leq& \frac{p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})}{3}\\ &\leq&\frac{p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})+p(y_{2n}, y_{2n})}{3}\\ &\leq& \frac{p(y_{2n-1}, y_{2n})+2p(y_{2n}, y_{2n+1})}{3};\\ &&\frac{p(y_{2n-1}, y_{2n+1})+p(y_{2n}, y_{2n})}{2}\leq \frac{p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1})}{2}. \end{eqnarray*}$

我们有

$\begin{eqnarray*} &&M(x_{2n}, x_{2n+1}) \\ &\leq& \max \bigg\{p(y_{2n-1}, y_{2n}), \frac{p(y_{2n-1}, y_{2n})+2p(y_{2n}, y_{2n+1})}{3}, \frac{p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1})}{2} \bigg\}. \end{eqnarray*}$

如果$p(y_{2n}, y_{2n+1})\geq p(y_{2n-1}, y_{2n})$, 则

$\frac{p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1})}{2}\leq p(y_{2n}, y_{2n+1}).$

$\frac{p(y_{2n-1}, y_{2n})+2p(y_{2n}, y_{2n+1})}{3}\leq p(y_{2n}, y_{2n+1}).$

从而$M(x_{2n}, x_{2n+1})\leq p(y_{2n}, y_{2n+1}).$由$\psi$的性质, 得$\psi(p(y_{2n}, y_{2n+1}))\leq \psi(p(y_{2n}, y_{2n+1}))-\varphi(M(x_{2n}, x_{2n_+1})), $这表明$\varphi(M(x_{2n}, x_{2n_+1}))=0$, 即$M(x_{2n}, x_{2n_+1})=0.$所以, $p(y_{2n-1}, y_{2n})=0, $ $p(y_{2n}, y_{2n+1})=0, $ $p(y_{2n-1}, y_{2n+1})=0, p(y_{2n}, y_{2n})=0, $因此$y_{2n-1}=y_{2n}=y_{2n+1}, $矛盾.

所以

$\begin{eqnarray} p(y_{2n}, y_{2n+1})＜ p(y_{2n-1}, y_{2n}). \end{eqnarray}$

(2.3)

再由$Sx_{2n+2}$与$Tx_{2n+1}$可比较, 且$Sx_{2n+2}\neq Tx_{2n+1}$或$fx_{2n+2}\neq gx_{2n+1}$, 同理可得

$\begin{eqnarray} p(y_{2n+2}, y_{2n+1})＜ p(y_{2n+1}, y_{2n}). \end{eqnarray}$

(2.4)

由(2.3) 和(2.4) 式得$\{p(y_{n}, y_{n+1})\}$是递减非负实数列, 设$\lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n+1})=r.$

如果$r\neq 0$, 则对所有$n$, $p(y_{n}, y_{n+1})\neq 0$, 从而(2.2) 式成立.因为

$\begin{eqnarray*}&&\limsup\limits_{n\rightarrow \infty}\frac{p(y_{2n-1}, y_{2n})p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n+1})p(y_{2n}, y_{2n})}{3(p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n}))}\\ &\leq&\limsup\limits_{n\rightarrow \infty}\frac{p(y_{2n-1}, y_{2n})p(y_{2n}, y_{2n+1})+(p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1}))p(y_{2n}, y_{2n+1})}{3(p(y_{2n}, y_{2n+1})+p(y_{2n-1}, y_{2n}))}\\ &=&\frac{r^{2}+2r^{2}}{3(r+r)}=\frac{r^{2}}{2r}＜r, \end{eqnarray*}$

$\begin{eqnarray*}&&\limsup\limits_{n\rightarrow \infty}\frac{p(y_{2n-1}, y_{2n+1})+p(y_{2n}, y_{2n})}{2}\leq \lim\limits_{n\rightarrow\infty}\frac{p(y_{2n-1}, y_{2n})+p(y_{2n}, y_{2n+1})}{2}=r.\end{eqnarray*}$

所以, $\limsup\limits_{n\rightarrow \infty}M(x_{2n}, x_{2n_+1})=r.$在(2.2) 式中取上极限得

$\limsup\limits_{n\rightarrow\infty}\psi(p(y_{2n}, y_{2n+1}))\leq \limsup\limits_{n\rightarrow\infty}\psi(M(x_{2n}, x_{2n+1}))- \liminf\limits_{n\rightarrow\infty}\varphi(M(x_{2n}, x_{2n+1})). $

由$\psi$的连续性和$\varphi$的下半连续性, 得$\psi(r)\leq\psi(r)-\varphi(r), $由此得$\varphi(r)=0, $即$r=0$, 矛盾.因此

$\begin{equation} \lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n+1})=0. \end{equation}$

(2.5)

同时由于$p(y_{n}, y_{n})\leq p(y_{n}, y_{n+1})$, 还有

$\begin{equation} \lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n})=0. \end{equation}$

(2.6)

从而

$\begin{equation} \lim\limits_{n\rightarrow\infty}d_{p}(y_{n}, y_{n+1})=\lim\limits_{n\rightarrow\infty}(2p(y_{n}, y_{n+1})-p(y_{n}, y_{n})-p(y_{n+1}, y_{n+1}))=0. \end{equation}$

(2.7)

下面证明$\{y_{n}\}$是度量空间$(X, d_{p})$中的Cauchy序列.根据(2.7) 式, 只需证明$\{y_{2n}\}$是Cauchy序列.事实上, 如果$\{y_{2n}\}$不是Cauchy序列, 则存在$\varepsilon_{0}＞0$, 对任何自然数$k$, 存在自然数$n_{k}, m_{k}$, $n_{k}＞m_{k}＞k$使得

$\begin{equation} d_{p}(y_{2n_{k}}, y_{2m_{k}})\geq\varepsilon_{0}. \end{equation}$

(2.8)

且$n_{k}$是符合这一条件的最小自然数, 那么

$\begin{equation} d_{p}(y_{2n_{k}-2}, y_{2m_{k}})＜\varepsilon_{0}. \end{equation}$

(2.9)

由(2.8), (2.9) 式及三角不等式, 有

$\begin{eqnarray} \varepsilon_{0}\leq d_{p}(y_{2n_{k}}, y_{2m_{k}})&\leq& d_{p}(y_{2n_{k}}, y_{2n_{k}-1})+d_{p}(y_{2n_{k}-1}, y_{2n_{k}-2})+d_{p}(y_{2n_{k}-2}, y_{2m_{k}})\nonumber\\ &＜&d_{p}(y_{2n_{k}}, y_{2n_{k}-1})+d_{p}(y_{2n_{k}-1}, y_{2n_{k}-2})+\varepsilon_{0}. \end{eqnarray}$

(2.10)

在(2.10) 式中令$k\rightarrow\infty$并利用(2.7) 式, 得到

$\begin{eqnarray} \lim _{k\rightarrow\infty}d_{p}(y_{2n_{k}}, y_{2m_{k}})=\varepsilon_{0}. \end{eqnarray}$

(2.11)

利用三角不等式, 还有

$d_{p}(y_{2n_{k}}, y_{2m_{k}+1})\leq d_{p}(y_{2n_{k}}, y_{2m_{k}})+d_{p}(y_{2m_{k}}, y_{2m_{k}+1});$

$d_{p}(y_{2n_{k}}, y_{2m_{k}})\leq d_{p}(y_{2n_{k}}, y_{2m_{k}+1})+d_{p}(y_{2m_{k}+1}, y_{2m_{k}});$

$d_{p}(y_{2n_{k}-1}, y_{2m_{k}})\leq d_{p}(y_{2n_{k}-1}, y_{2n_{k}})+d_{p}(y_{2n_{k}}, y_{2m_{k}});$

$d_{p}(y_{2n_{k}}, y_{2m_{k}})\leq d_{p}(y_{2n_{k}}, y_{2n_{k}-1})+d_{p}(y_{2n_{k}-1}, y_{2m_{k}});$

$d_{p}(y_{2n_{k}-1}, y_{2m_{k}+1})\leq d_{p}(y_{2n_{k}-1}, y_{2n_{k}})+d_{p}(y_{2n_{k}}, y_{2m_{k}})+d_{p}(y_{2m_{k}}, y_{2m_{k}+1});$

$d_{p}(y_{2n_{k}}, y_{2m_{k}})\leq d_{p}(y_{2n_{k}}, y_{2n_{k}-1})+d_{p}(y_{2n_{k}-1}, y_{2m_{k}+1})+d_{p}(y_{2m_{k}+1}, y_{2m_{k}}).$

在上面六个不等式中令$k\rightarrow\infty$并利用(2.7), (2.11) 式, 得$\lim\limits_{k\rightarrow\infty}d_{p}(y_{2n_{k}}, y_{2m_{k}+1})=\varepsilon_{0};$ $\lim\limits_{k\rightarrow\infty}d_{p}(y_{2n_{k}-1}, y_{2m_{k}})=\varepsilon_{0};$ $\lim\limits_{k\rightarrow\infty}d_{p}(y_{2n_{k}-1}, y_{2m_{k}+1})=\varepsilon_{0}.$从而

$\begin{equation} \lim\limits_{k\rightarrow\infty}p(y_{2n_{k}}, y_{2m_{k}})=\frac{\varepsilon_{0}}{2}; \end{equation}$

(2.12)

$\begin{equation} \lim\limits_{k\rightarrow\infty}p(y_{2n_{k}}, y_{2m_{k}+1})=\frac{\varepsilon_{0}}{2}; \end{equation}$

(2.13)

$\begin{equation} \lim\limits_{k\rightarrow\infty}p(y_{2n_{k}-1}, y_{2m_{k}})=\frac{\varepsilon_{0}}{2}; \end{equation}$

(2.14)

$\begin{equation} \lim\limits_{k\rightarrow\infty}p(y_{2n_{k}-1}, y_{2m_{k}+1})=\frac{\varepsilon_{0}}{2}. \end{equation}$

(2.15)

由于对任意$k$, $Sx_{2n_{k}}$与$Tx_{2m_{k}+1}$可比较, 且$Sx_{2n_{k}}\neq Tx_{2m_{k}+1}$或$fx_{2n_{k}}\neq gx_{2m_{k}+1}$, 有

$\begin{eqnarray} \psi(p(y_{2n_{k}}, y_{2m_{k}+1}))&=&\psi(p(fx_{2n_{k}}, gx_{2m_{k}+1}))\nonumber\\ &\leq &\psi(M(x_{2n_{k}}, x_{2m_{k}+1}))-\varphi(M(x_{2n_{k}}, x_{2m_{k}+1})), \end{eqnarray}$

(2.16)

其中

$\begin{eqnarray*} &&M(x_{2n_{k}}, x_{2m_{k}+1})\\ &=&\max\bigg\{p(Sx_{2n_{k}}, Tx_{2m_{k}+1}), \\ && \frac{p(Sx_{2n_{k}}, fx_{2n_{k}})p(Tx_{2m_{k}+1}, gx_{2m_{k}+1})+p(Sx_{2n_{k}}, gx_{2m_{k}+1})p(Tx_{2m_{k}+1}, fx_{2n_{k}})}{3(p(fx_{2n_{k}}, gx_{2m_{k}+1})+p(Sx_{2n_{k}}, Tx_{2m_{k}+1}))}, \\&&\frac{p(Sx_{2n_{k}}, fx_{2n_{k}})+p(Tx_{2m_{k}+1}, gx_{2m_{k}+1})}{2}, \frac{p(Sx_{2n_{k}}, gx_{2m_{k}+1})+p(Tx_{2m_{k}+1}, fx_{2n_{k}})}{2}\bigg\}\\ &=&\max\bigg\{p(y_{2n_{k}-1}, y_{2m_{k}}), \frac{p(y_{2n_{k}-1}, y_{2n_{k}})p(y_{2m_{k}}, y_{2m_{k}+1})+p(y_{2n_{k}-1}, y_{2m_{k}+1})p(y_{2m_{k}}, y_{2n_{k}})}{3(p(y_{2n_{k}}, y_{2m_{k}+1})+p(y_{2n_{k}-1}, y_{2m_{k}}))}, \\&&\frac{p(y_{2n_{k}-1}, y_{2n_{k}})+p(y_{2m_{k}}, y_{2m_{k}+1})}{2}, \frac{p(y_{2n_{k}-1}, y_{2m_{k}+1})+p(y_{2m_{k}}, y_{2n_{k}})}{2}\bigg\}. \end{eqnarray*}$

由(2.5) 和(2.12)-(2.15) 式得

$\begin{eqnarray} \lim\limits_{k\rightarrow\infty}M(x_{2n_{k}}, x_{2m_{k}+1})=\max \bigg\{\frac{\varepsilon_{0}}{2}, \frac{\varepsilon_{0}}{12}, 0, \frac{\varepsilon_{0}}{2} \bigg\}=\frac{\varepsilon_{0}}{2}. \end{eqnarray}$

(2.17)

在(2.16) 式中取上极限, 并由$\psi$的连续性和$\varphi$的下半连续性, 及(2.13) 和(2.17) 式, 得$\psi(\frac{\varepsilon_{0}}{2})\leq\psi(\frac{\varepsilon_{0}}{2})-\varphi(\frac{\varepsilon_{0}}{2}).$因此, $\varphi(\frac{\varepsilon_{0}}{2})=0, $即$\varepsilon_{0}=0, $矛盾.所以, $\{y_{n}\}$是度量空间$(X, d_{p})$中的Cauchy序列.因为$(X, p)$完备, 由引理1.1(ⅱ), $(X, d_{p})$也完备, 所以存在$w\in X$使$\lim\limits _{n\rightarrow\infty}d_{p}(y_{n}, w)=0. $再由引理1.1(ⅱ), 有

$\begin{eqnarray} \lim\limits_{n, m\rightarrow\infty}p(y_{n}, y_{m})=\lim\limits_{n\rightarrow\infty}p(y_{n}, w)=p(w, w), \end{eqnarray}$

(2.18)

这表明$\{y_{n}\}, \{y_{2n}\}$和$\{y_{2n+1}\}$在$(X, p)$中收敛于$w$.另外, 由(2.6) 式, 有$\lim\limits_{n\rightarrow\infty}d_{p}(y_{n}, w)=\lim \limits_{n\rightarrow\infty}(2p(y_{n}, w)-p(y_{n}, y_{n})-p(w, w))=p(w, w), $从而$p(w, w)=0$.

因为$f(X)\subseteq T(X)$, $g(X)\subseteq S(X)$, 且$S(X), T(X)$是$(X, p)$中闭子集, 及$\{y_{2n}\}$与$\{y_{2n+1}\}$都在$(X, p)$中收敛于$w$, 可得$w\in T(X)\cap S(X)$, 从而存在$z_{1}, z_{2}\in X$使$w=Tz_{1}=Sz_{2}, $由条件(ⅳ), $y_{2n}\preceq Tz_{1}, y_{2n+1}\preceq Sz_{2}.$

如果$fz_{2}\neq w$, 则$p(fz_{2}, w)\neq 0$.由于$Tx_{2n+1}$与$Sz_{2}=Tz_{1}$可比较, 且$Sz_{2}\neq Tx_{2n+1}$(否则, $w= y_{2n}$, 由于$y_{2n}\preceq y_{2n+1}\preceq y_{2n+2}\preceq w$, 我们有$y_{2n}= y_{2n+1}= y_{2n+2}$), 由(2.1) 式得

$\begin{eqnarray} \psi(p(fz_{2}, y_{2n+1}))&=&\psi(p(fz_{2}, gx_{2n+1}))\\ &\leq & \psi(M(z_{2}, x_{2n+1}))-\varphi(M(z_{2}, x_{2n+1})). \end{eqnarray}$

(2.19)

对$n=1, 2, \cdots $成立, 其中

$\begin{eqnarray*} &&M(z_{2}, x_{2n+1})\\ &=&\max\bigg\{p(Sz_{2}, Tx_{2n+1}), \frac{p(Sz_{2}, fz_{2})p(Tx_{2n+1}, gx_{2n+1})+p(Sz_{2}, gx_{2n+1})p(Tx_{2n+1}, fz_{2})}{3(p(fz_{2}, gx_{2n+1})+p(Sz_{2}, Tx_{2n+1}))}, \\&&\frac{p(Sz_{2}, fz_{2})+p(Tx_{2n+1}, gx_{2n+1})}{2}, \frac{p(Sz_{2}, gx_{2n+1})+p(Tx_{2n+1}, fz_{2})}{2}\bigg\}\\ &=&\max\bigg\{p(w, y_{2n}), \frac{p(w, fz_{2})p(y_{2n}, y_{2n+1})+p(w, y_{2n+1})p(y_{2n}, fz_{2})}{3(p(fz_{2}, y_{2n+1})+p(w, y_{2n}))}, \\&&\frac{p(w, fz_{2})+p(y_{2n}, y_{2n+1})}{2}, \frac{p(w, y_{2n+1})+p(y_{2n}, fz_{2})}{2}\bigg\}. \end{eqnarray*}$

由$\{y_{n}\}$在度量空间$(X, d_{p})$中收敛于$w$得$\lim\limits_{n\rightarrow\infty}d_{p}(fz_{2}, y_{n})=d_{p}(fz_{2}, w)$.而

$\begin{eqnarray*} \lim\limits_{n\rightarrow\infty}d_{p}(fz_{2}, y_{n})&=&2\lim\limits_{n\rightarrow\infty}p(fz_{2}, y_{n})-p(fz_{2}, fz_{2})-\lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n})\\ &=&2\lim\limits_{n\rightarrow\infty}p(fz_{2}, y_{n})-p(fz_{2}, fz_{2}). \end{eqnarray*}$

$ d_{p}(fz_{2}, w)=2p(fz_{2}, w)-p(fz_{2}, fz_{2})-p(w, w)=2p(fz_{2}, w)-p(fz_{2}, fz_{2}). $

我们有

$\begin{eqnarray} \lim\limits_{n\rightarrow\infty}p(fz_{2}, y_{n})=p(fz_{2}, w). \end{eqnarray}$

(2.20)

所以

$\begin{eqnarray*} &&\lim\limits_{n\rightarrow\infty}\frac{p(w, fz_{2})p(y_{2n}, y_{2n+1})+p(w, y_{2n+1})p(y_{2n}, fz_{2})}{3(p(fz_{2}, y_{2n+1})+p(w, y_{2n}))}=\frac{p(w, fz_{2})\times 0+0\times p(w, fz_{2})}{3(p(w, fz_{2})+0)}=0; \end{eqnarray*}$

$\lim\limits_{n\rightarrow\infty}\frac{p(w, fz_{2})+p(y_{2n}, y_{2n+1})}{2}=\frac{p(w, fz_{2})}{2};$

$\begin{eqnarray*} \lim\limits_{n\rightarrow\infty}\frac{p(w, y_{2n+1})+p(y_{2n}, fz_{2})}{2}=\frac{p(w, fz_{2})}{2}. \end{eqnarray*}$

从而

$\begin{eqnarray} \lim\limits_{n\rightarrow\infty}M(z_{2}, x_{2n+1})=\frac{p(w, fz_{2})}{2}. \end{eqnarray}$

(2.21)

在(2.19) 式中取上极限, 并由$\psi$的性质, 及(2.20) 和(2.21) 式, 得

$\psi(p(fz_{2}, w))\leq \psi(\frac{p(w, fz_{2})}{2})-\varphi(\frac{p(w, fz_{2})}{2})\leq\psi(p(fz_{2}, w))-\varphi(\frac{p(w, fz_{2})}{2}), $

因此$\varphi(\frac{p(w, fz_{2})}{2})=0, $即$p(w, fz_{2})=0, $矛盾.所以, $fz_{2}=w.$ $z_{2}$是$f$与$S$的一个叠合点.同理可得$z_{1}$是$g$与$T$的一个叠合点.因此, $w$既是叠合$f$和$S$的点又是叠合$g$和$T$的点.

定理2.2 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f, g, S, T:X\rightarrow X$满足定理2.1中的条件(ⅰ), (ⅱ), (ⅲ)及

(ⅳ)′ $S$和$T$有性质(A).

如果$\{f, S\}$和$\{g, T\}$都是弱相容映射对, 则$f, g, S, T$有一个公共不动点.

证 由定理2.1, 设$w$是同时叠合$f$和$S$与$g$和$T$的点.定理2.1的证明及条件(ⅳ)$' $表明, 存在$z_{1}, z_{2}\in X$使$fz_{2}=Sz_{2}=w=gz_{1}=Tz_{1}, $且$Tz_{1}\preceq TTz_{1}, Sz_{2}\preceq SSz_{2}.$因为$f$和$S$弱相容, 故$ffz_{2}=fSz_{2}=Sfz_{2}=SSz_{2}.$如果$Sw\neq w$, 则$Sw\neq Tz_{1}$.由于$w=Tz_{1}=Sz_{2}\preceq SSz_{2}=Sw, $我们有

$\begin{eqnarray} \psi(p(fw, w))=\psi(p(fw, gz_{1}))\leq \psi(M(w, z_{1}))-\varphi(M(w, z_{1})). \end{eqnarray}$

(2.22)

对$n=1, 2, \cdots $成立, 其中

$\begin{eqnarray*} M(w, z_{1})&=&\max\bigg\{p(Sw, Tz_{1}), \frac{p(Sw, fw)p(Tz_{1}, gz_{1})+p(Sw, gz_{1})p(Tz_{1}, fw)}{3(p(fw, gz_{1})+p(Sw, Tz_{1}))}, \\ &&\frac{p(Sw, fw)+p(Tz_{1}, gz_{1})}{2}, \frac{p(Sw, gz_{1})+p(Tz_{1}, fw)}{2}\bigg\}\\ &=&\max\bigg\{p(fw, w), \frac{p(fw, fw)p(w, w)+p(fw, w)p(fw, w)}{3(p(Sw, w)+p(fw, w))}, \\ &&\frac{p(fw, fw)+p(w, w)}{2}, \frac{p(fw, w)+p(fw, w)}{2}\bigg\}. \end{eqnarray*}$

由于$p(fw, fw)\leq p(fw, w)$, $p(w, w)\leq p(fw, w)$, $p(fw, w)=p(Sw, w)\neq 0$, 有

$\frac{p(fw, fw)p(w, w)+p(fw, w)p(fw, w)}{3(p(Sw, w)+p(fw, w))}\leq\frac{p(fw, w)}{3}, $

$\frac{p(fw, fw)+p(w, w)}{2}\leq p(fw.w), $

从而$M(w, z_{1})=p(fw, w)$.由(2.22) 式

$\psi(p(fw, w))\leq \psi(p(fw, w))-\varphi(p(fw, w)), $

得$\varphi(p(fw, w))=0$.因此$p(fw, w)=0$, 即$fw=w $, 从而$Sw=w$.同理, 由$g$和$T$弱相容, 可得$gw=Tw=w.$所以, $w$是$f, g, S, T$有一个公共不动点.

推论2.1 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f, g, H:X\rightarrow X$满足下列条件

(ⅰ)对所有使得$Hx$与$Hy$可比较, 且$Hx\neq Hy$或$fx\neq gy$的$x, y\in X$, 有

$ \psi(p(fx, gy))\leq \psi (M(x, y))-\varphi(M(x, y)), $

其中

$\begin{eqnarray*} M(x, y)&=&\max\bigg\{p(Hx, Hy), \frac{p(Hx, fx)p(Hy, gy)+p(Hx, gy)p(Hy, fx)}{3(p(fx, gy)+p(Hx, Hy))}, \\&&\frac{p(Hx, fx)+p(Hy, gy)}{2}, \frac{p(Hx, gy)+p(Hy, fx)}{2}\bigg\}. \end{eqnarray*}$

$\psi$是改变距离函数, $\varphi:[0, +\infty)\rightarrow [0, +\infty)$下半连续, $\varphi(t)=0$当且仅当$t=0$;

(ⅱ) $f(X)\subseteq H(X), g(X)\subseteq H(X)$且$H(X)$是$(X, p)$中的闭子集;

(ⅲ) $(f, g)$关于$H$弱增;

(ⅳ) $H$有性质(A).

如果$\{f, H\}$和$\{g, H\}$都是弱相容映射对, 则$f, g, H$有一个公共不动点.

证 在定理2.2中令$S=T=H$即可.

推论2.2 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f, S, T:X\rightarrow X$满足下列条件

(ⅰ)对所有使得$Sx$与$Ty$可比较, 且$Sx\neq Ty$或$fx\neq fy$的$x, y\in X$, 有

$\psi(p(fx, fy))\leq \psi (M(x, y))-\varphi(M(x, y)), $

其中

$\begin{eqnarray*} M(x, y)&=&\max\bigg\{p(Sx, Ty), \frac{p(Sx, fx)p(Ty, fy)+p(Sx, fy)p(Ty, fx)}{3(p(fx, fy)+p(Sx, Ty))}, \\&&\frac{p(Sx, fx)+p(Ty, fy)}{2}, \frac{p(Sx, fy)+p(Ty, fx)}{2} \bigg\}. \end{eqnarray*}$

$\psi$是改变距离函数, $\varphi:[0, +\infty)\rightarrow [0, +\infty)$下半连续, $\varphi(t)=0$当且仅当$t=0$;

(ⅱ) $f(X)\subseteq S(X), f(X)\subseteq T(X)$且$S(X)$和$T(X)$都是$(X, p)$中的闭子集;

(ⅲ) $f$关于$S$和$T$都是弱增;

(ⅳ) $S$和$T$有性质(A).

如果$\{f, S\}$和$\{f, T\}$都是弱相容映射对, 则$f, S, T$有一个公共不动点.

证 在定理2.2中令$f=g$即可.

推论2.3 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f:X\rightarrow X$满足下列条件

(ⅰ)对所有使得$x$与$y$可比较, 且$x\neq y$的$x, y\in X$, 有

$\psi(p(fx, fy))\leq \psi (M(x, y))-\varphi(M(x, y)), $

其中

$\begin{eqnarray*} M(x, y)&=&\max\bigg\{p(x, y), \frac{p(x, fx)p(y, fy)+p(x, fy)p(y, fx)}{3(p(fx, fy)+p(x, y))}, \\&&\frac{p(x, fx)+p(y, fy)}{2}, \frac{p(x, fy)+p(y, fx)}{2} \bigg\}. \end{eqnarray*}$

$\psi$是改变距离函数, $\varphi:[0, +\infty)\rightarrow [0, +\infty)$是下半连续函数, $\varphi(t)=0$当且仅当$t=0$;

(ⅱ)对所有$x\in X$, $fx\preceq ffx$;

(ⅲ)如果$\{x_{n}\}$非减收敛于$x$, 则$x_{n}\preceq x$.

则$f$有一个不动点.

证 在定理2.2中令$f=g$, $S=T=I_{X}$即可, 其中$I_{X}$是$X$上的恒同映射.

推论2.4 设映射$f, g, S, T:X\rightarrow X$满足定理2.2中除条件(ⅰ)的所有条件, 而条件(ⅰ)由下列条件代替:存在正的Lebesgue可积函数$\mu(t):[0, \infty)\rightarrow [0, \infty)$使得对每个$\varepsilon＞0$, $\int_{0}^{\varepsilon}\mu(t){\rm d}t ＞0$且

$\begin{eqnarray} \int_{0}^{\psi(p(fx, gy))}\mu(t){\rm d}t\leq \int_{0}^{\psi(M(x, y))}\mu(t){\rm d}t-\int_{0}^{\varphi(M(x, y))}\mu(t){\rm d}t. \end{eqnarray}$

(2.23)

其中$M(x, y)$如定理2.1所述.则$f, g, S, T$有一个公共不动点.

证 考虑函数

$\phi(u)=\int_{0}^{x}\mu(t){\rm d}t.$

令$\phi \circ \psi=\psi_{1}$, $\phi \circ \varphi=\varphi_{1}$.注意到$\psi_{1}$和$\varphi_{1}$具有定理2.1中所述性质, 而且(2.23) 式变为

$\psi_{1}(p(fx, gy))\leq \psi_{1}(M(x, y))-\varphi_{1}(M(x, y)).$

应用定理2.2即得该推论的证明.

例2.1 设$X=\{a, b, c, e\}$, $\preceq=\{(a, a), (b, b), (c, c), (e, e), (b, e), (a, b), (a, e)\}$, 则$(X, \preceq)$是偏序集.定义$p:X\times X\rightarrow [0, \infty)$如下

$p(a, a)=p(c, c)=\frac{1}{4}, p(b, b)=p(e, e)=0;$

$p(a, b)=p(b, a)=1, p(a, c)=p(c, a)=\frac{3}{2}, p(a, e)=p(e, a)=\frac{3}{4}, $

$ p(b, c)=p(c, b)=1, p(b, e)=p(e, b)=\frac{1}{2}, p(c, e)=p(e, c)=1.$

则$(X, p)$是完备偏度量空间.

函数$\psi, \varphi:[0, \infty)\rightarrow[0, \infty)$定义为对所有$t\in [0, \infty)$, $\psi(t)=2t$, $\varphi(t)=\frac{t}{1+t^{2}}$.则$\psi$是改变距离函数且$\varphi$连续, $\varphi(t)=0$当且仅当$t=0$.

映射$f, g, S, T:X\rightarrow X$定义如下

$ fx= e, x\in X.\qquad Tx= \left\{ \begin{array}{ll} c, x=a \ \mbox{或}\ c;\\ e, x=b\ \mbox{或}\ e. \end{array} \right. $

$ gx= \left\{ \begin{array}{ll} b, x=a\ \mbox{或}\ c;\\ e, x=b\ \mbox{或}\ e. \end{array} \right.\qquad Sx= \left\{ \begin{array}{ll} b, x=a\ \mbox{或}\ b;\\ c, x=c;\\ e, x=e. \end{array} \right. $

则$f(X)\subseteq T(X), g(X)\subseteq S(X)$, 且$\{f, T\}$和$\{g, S\}$都是弱相容行摄对.容易看到, $T$和$S$都有性质(A).

对任意$x\in X$和$y\in T^{-1}(fx)=T^{-1}(e)=\{b, e\}$, $fx=e\preceq e=gy$, 这表明$(f, g)$关于$T$部分弱增.同理$(g, f)$关于$S$部分弱增.

下面验证(2.1) 式对满足$Sx$和$Ty$可比较且$Sx\neq Ty$或$fx\neq gy$的所有$x, y\in X$都成立成立.事实上, 由于$Sc=c$和$Ta=c$可比较, 且$fc\neq ga$.

$\psi(p(fc, ga))=\psi(p(e, b))=\psi(\frac{1}{2})=1.$

$\begin{eqnarray*} M(c, a)&=&\max \bigg\{p(Sc, Ta), \frac{p(Sc, fc)p(Ta, ga)+p(Sc, ga)p(Ta, fc)}{3(p(fc, ga)+p(Sc, Ta))}, \\&&\frac{p(Sc, fc)+p(Ta, ga)}{2}, \frac{p(Sc, ga)+p(Ta, fc)}{2} \bigg\}\\ &=&\max \bigg\{p(c, c), \frac{p(c, e)p(c, b)+p(c, b)p(c, e)}{3(p(e, b)+p(c, c))}, \frac{p(c, e)+p(c, b)}{2}, \frac{p(c, b)+p(c, e)}{2} \bigg\}\\ &=&\max\bigg\{\frac{1}{4}, \frac{8}{9}, 1\bigg\}=1. \end{eqnarray*}$

因此$\psi(p(fc, ga))=1\leq\psi(M(c, a))-\varphi(M(c, a))=2-\frac{1}{2}=\frac{3}{2}, $所以, (2.1) 式对$c$和$a$成立.满足$Sx$和$Ty$可比较且$Sx\neq Ty$或$fx\neq gy$的$x, y\in X$还有$Sa\preceq Tb$ ($Sa\neq Tb$), $Sa\preceq Te$ ($Sa\neq Te$), $Sb\preceq Tb$ ($Sb\neq Tb$), and $Sc\preceq Tc$ ($fc\neq gc$), 不难验证(2.1) 式对各种情形都是成立的.由定理2.2, $f, g, S, T$有公共不动点$w=e$.

定理2.3 设$(X, p, \preceq)$是偏序完备偏度量空间.映射$f, g, S, T:X\rightarrow X$满足定理2.2的所有条件, 如果下列条件还满足

(ⅴ)对任意$x, y\in S(X)\cap T(X)$, 存在$z_{0}\in X$使得$x\preceq fz_{0}, y\preceq gz_{0}$, 且$\lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n}' )=0$, 其中$\{y_{n}\}$是以$z_{0}$为初始点的$(f, g)$-$(T, S)$-序列, $\{y_{n}' \}$是以$z_{0}$为初始点的$(g, f)$-$(S, T)$-序列.

则$f, g, S, T$有唯一公共不动点.

证 设$w_{1}$和$w_{2}$是$f, g, S, T$的两个公共不动点且$w_{1}\neq w_{2}, $则$p(w_{1}, w_{2})\neq 0.$如果$w_{1}$和$w_{2}$可比较, 即$Sw_{1}$和$Tw_{2}$可比较, 且$Sw_{1}\neq Tw_{2}$, 则

$\begin{eqnarray} \psi(p(w_{1}, w_{2}))=\psi(p(fw_{1}, gw_{2}))\leq \psi (M(w_{1}, w_{2}))-\varphi(M(w_{1}, w_{2})), \end{eqnarray}$

(2.24)

其中

$\begin{eqnarray*} M(w_{1}, w_{2})&=&\max \bigg\{p(Sw_{1}, Tw_{2}), \frac{p(Sw_{1}, fw_{1})p(Tw_{2}, gw_{2})+p(Sw_{1}, gw_{2})p(Tw_{2}, fw_{1})}{3(p(fw_{1}, gw_{2})+p(Sw_{1}, Tw_{2}))}, \\&&\frac{p(Sw_{1}, fw_{1})+p(Tw_{2}, gw_{2})}{2}, \frac{p(Sw_{1}, gw_{2})+p(Tw_{2}, fw_{1})}{2} \bigg\}\\ &=&\max\bigg\{p(w_{1}, w_{2}), \frac{p(w_{1}, w_{1})p(w_{2}, w_{2})+p(w_{1}, w_{2})p(w_{2}, w_{1})}{3(p(w_{1}, w_{2})+p(w_{1}, w_{2}))}, \\&&\frac{p(w_{1}, w_{1})+p(w_{2}, w_{2})}{2}, \frac{p(w_{1}, w_{2})+p(w_{2}, w_{1})}{2} \bigg\}. \end{eqnarray*}$

由于

$\begin{eqnarray*} \frac{p(w_{1}, w_{1})p(w_{2}, w_{2})+p(w_{1}, w_{2})p(w_{2}, w_{1})}{3(p(w_{1}, w_{2})+p(w_{1}, w_{2}))}&\leq& \frac{p(w_{1}, w_{2})p(w_{1}, w_{2})+p(w_{1}, w_{2})p(w_{2}, w_{1})}{6p(w_{1}, w_{2})}\\&=& \frac{p(w_{2}, w_{2})}{3}\leq p(w_{1}, w_{2}), \end{eqnarray*}$

$\frac{p(w_{1}, w_{1})+p(w_{2}, w_{2})}{2}\leq\frac{p(w_{1}, w_{2})+p(w_{1}, w_{2})}{2}=p(w_{1}, w_{2}), $

我们有$M(w_{1}, w_{2})=p(w_{1}, w_{2}).$由(2.24) 式, 得$\varphi(p(w_{1}, w_{2}))=0$, 即$p(w_{1}, w_{2})=0$, 矛盾, 所以, $w_{1}=w_{2}.$

如果$w_{1}$和$w_{2}$不可比较, 则存在$z_{0}\in X$使$w_{1}\preceq fz_{0}$, $w_{2}\preceq gz_{0}$.设$\{y_{n}\}$是以$z_{0}$为初始点的$(f, g)$-$(T, S)$-序列, $\{y_{n}' \}$是以$z_{0}$为初始点的$(g, f)$-$(S, T)$ -序列, 则存在两个序列$\{z_{n}\}, \{z_{n}' \}\subseteq X$, $z_{0}' =z_{0}.$使得

$y_{2n}=fz_{2n}=Tz_{2n+1}, y_{2n+1}=gz_{2n+1}=Sz_{2n+2}, n=0, 1, 2, \cdots .$

$y_{2n}' =gz_{2n}' =Sz_{2n+1}', y_{2n+1}' =fz_{2n+1}' =Tz' _{2n+2}, n=0, 1, 2, \cdots .$

$fz_{0}=y_{0}\preceq y_{n}, gz_{0}=y_{0}' \preceq y_{n}', n=1, 2, \cdots .$

如果存在正整数$n_{0}$使对所有$n>n_{0}$, 有$p(y_{n}, y_{n+1})=0$, 则$\lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n+1})=0.$如果存在无限多个正整数$n_{i}$使$p(y_{n_{i}}, y_{n_{i}+1})\neq0$, 类似于定理2.1的证明可得, 对每个$i$有$p(y_{n_{i}}, y_{n_{i}+1})＜p(y_{n_{i}-1}, y_{n_{i}})$, 这表明$p(y_{n_{i}-1}, y_{n_{i}})\neq 0, $从而$p(y_{n}, y_{n+1})\neq 0$, $n=1, 2, \cdots $, 同时$\{p(y_{n}, y_{n+1})\}$是递减数列.而且同样由定理2.1的证明, 有$\lim\limits_{n\rightarrow\infty}p(y_{n}, y_{n+1})=0.$同理也有$\lim\limits_{n\rightarrow\infty}p(y_{n}', y' _{n+1})=0.$

下面证明$p(w_{1}, y_{2n})\geq p(w_{1}, y_{2n+1})$, $n=1, 2, \cdots $.事实上, 如果对某个$n_{0}$, 有$p(w_{1}, y_{2n_{0}})＜ p(w_{1}, y_{2n_{0}+1})$.由于$Sw_{1}=w_{1}\preceq fz_{0}\preceq y_{2n_{0}}=Tz_{2n_{0}+1}$, 且$p(fw_{1}, gz_{2n_{0}+1})\neq 0$ (否则, 有$p(w_{1}, y_{2n_{0}})= p(w_{1}, y_{2n_{0}+1})$), 由(2.1) 式得

$\begin{eqnarray} \psi(p(w_{1}, y_{2n_{0}+1}))=\psi(p(fw_{1}, gz_{2n_{0}+1}))\leq \psi (M(w_{1}, z_{2n_{0}+1}))-\varphi(M(w_{1}, z_{2n_{0}+1})), \end{eqnarray}$

(2.25)

其中

$\begin{eqnarray*} &&M(w_{1}, z_{2n_{0}+1}) \\ &=&\max\bigg\{p(Sw_{1}, Tz_{2n_{0}+1}), \\ &&\frac{p(Sw_{1}, fw_{1})p(Tz_{2n_{0}+1}, gz_{2n_{0}+1}+p(Sw_{1}, gz_{2n_{0}+1})p(Tz_{2n_{0}+1}, fw_{1}))}{3(p(fw_{1}, gz_{2n_{0}+1})+p(Sw_{1}, Tz_{2n_{0}+1}))}, \\&&\frac{p(Sw_{1}, fw_{1})+p(Tz_{2n_{0}+1}, gz_{2n_{0}+1})}{2}, \frac{p(Sw_{1}, gz_{2n_{0}+1})+p(Tz_{2n_{0}+1}, fw_{1})}{2}\bigg\}\\ &=&\max\bigg\{p(w_{1}, y_{2n_{0}}), \frac{p(w_{1}, w_{1})p(y_{2n_{0}}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}+1})p(w_{1}, y_{2n_{0}})}{3(p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}}))}, \\&&\frac{p(w_{1}, w_{1})+p(y_{2n_{0}}, y_{2n_{0}+1})}{2}, \frac{p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}})}{2}\bigg\}. \end{eqnarray*}$

因为

$\frac{p(w_{1}, w_{1})+p(y_{2n_{0}}, y_{2n_{0}+1})}{2}\leq \frac{p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}})}{2}＜p(w_{1}, y_{2n_{0}+1}).$

$\begin{eqnarray*} &&\frac{p(w_{1}, w_{1})p(y_{2n_{0}}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}+1})p(w_{1}, y_{2n_{0}})}{3(p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}}))}\\&\leq & \frac{p(w_{1}, w_{1})(p(y_{2n_{0}}, w_{1})+p(w_{1}, y_{2n_{0}+1}))+p(w_{1}, y_{2n_{0}+1})p(w_{1}, y_{2n_{0}})}{3(p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}}))} \\&\leq&\frac{p(w_{1}, w_{1})(p(y_{2n_{0}}, w_{1})+p(w_{1}, y_{2n_{0}+1}))}{3(p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}}))}+\frac{p(w_{1}, y_{2n_{0}+1})p(w_{1}, y_{2n_{0}})}{3(p(w_{1}, y_{2n_{0}+1})+p(w_{1}, y_{2n_{0}}))} \\ &\leq & p(w_{1}, y_{2n_{0}+1}). \end{eqnarray*}$

由(2.25) 式得

$\psi(p(w_{1}, y_{2n_{0}+1}))\leq \psi (p(w_{1}, y_{2n_{0}+1}))-\varphi(M(w_{1}, z_{2n_{0}+1})), $

这表明$\varphi(M(w_{1}, z_{2n_{0}+1}))=0, $从而$M(w_{1}, z_{2n_{0}+1})=0$.因此$p(w_{1}, y_{2n_{0}})=0=p(w_{1}, y_{2n_{0}+1})$, 矛盾.所以, $p(w_{1}, y_{2n+1})\leq p(w_{1}, y_{2n}), $ $n=1, 2, \cdots $.类似可得$p(w_{1}, y_{2n+2})\leq p(w_{1}, y_{2n+1}), $ $n=1, 2, \cdots $.所以, $\{p(w_{1}, y_{n})\}$是非增数列, 设$\lim\limits_{n\rightarrow\infty}p(w_{1}, y_{n})=r\geq 0.$

如果$r＞0, $则$p(w_{1}, y_{n})>0$, $n=1, 2, \cdots $, 从而(2.25) 式对$n=1, 2, \cdots $成立.由于$\limsup\limits_{n\rightarrow\infty}M(w_{1}, z_{2n+1})=r, $在(2.25) 式取上极限得$\psi(r)\leq \psi(r)-\varphi(r).$所以, $\varphi(r)=0, $从而$r=0.$即$\lim\limits_{n\rightarrow\infty}p(w_{1}, y_{n})=0.$同理可得$\lim\limits_{n\rightarrow\infty}p(w_{2}, y_{n}' )=0.$因为

$\begin{eqnarray*} p(w_{1}, w_{2})&\leq&p(w_{1}, w_{2})+p(y_{n}, y_{n})+p(y_{n}', y_{n}' )\\&\leq&p(w_{1}, y_{n})+p(y_{n}, w_{2})+p(y_{n}', y_{n}' )\\ &\leq&p(w_{1}, y_{n})+p(y_{n}, y_{n}' )+p(y_{n}', w_{2}), \end{eqnarray*}$

在上面的不等式中取极限, 得$p(w_{1}, w_{2})=0$, 即$w_{1}=w_{2}$.

考虑如下同伦问题:

定理3.1 设$(X, p, \preceq)$是偏序完备偏度量空间, 对$X$中任何非减收敛于$x\in X$的序列$\{x_{n}\}$, 都有$x_{n}\preceq x$, $n=1, 2, \cdot, \cdot, \cdot$.又设$U$是$X$中的开集, $\varphi:[0, \infty]\rightarrow [0, \infty]$是非减连续函数使$\varphi(t)＜t$, $\varphi(t)=0$当且仅当$t=0$, 且$\phi(t)=t-\varphi(t)$非减.如果$H:X\times [0, 1]\rightarrow X $满足下列条件

(ⅰ)对每个$x \in X\setminus U $及$t\in [0, 1]$有$x \neq H(x, t)$;

(ⅱ)对$x, y\in X$, $x\neq y $及$t\in [0, 1]$, 有

$p(H(x, t), H(y, t))\leq \phi(p(x, y));$

(ⅲ)存在连续函数$\eta:[0, 1]\rightarrow {\Bbb R}$使对所有$t, s\in [0, 1]$及每个$x\in X $有

$p(H(x, t), H(x, s))\leq |\eta(t)-\eta(s)|;$

(ⅳ)对每个$x\in X$和$t\in [0, 1]$有$H(x, t)\preceq H(H(x, t), t)$.

则$H(\cdot, 0)$有不动点当且仅当$H(\cdot, 1)$有不动点.

证 设$H(\cdot, 0)$有不动点.记$W=\{t\in [0, 1]:\exists x\in U, x=H(x, t)\}$, 由条件(ⅰ), $H(\cdot, 0)$的不动点必属于$ U$, 故$0\in W$, 从而$W$非空.

Let $t_{0}\in W$, $x_{0}\in U$且$x_{0}=H(x_{0}, t_{0})$.由于$U$是$(X, p)$中的开集, 存在$r>0$使$B_{p}(x_{0}, r)\subseteq U$.记$\varepsilon =\varphi(p(x_{0}, x_{0})+r)$.由$\eta $的连续性, 存在$\delta >0$使对任意$t\in(t_{0}-\delta, t_{0}+\delta)$有$|\eta(t)-\eta(t_{0})|＜ \varepsilon$.设$t\in (t_{0}-\delta, t_{0}+\delta)$.对$x\in\overline{B_{p}(x_{0}, r)}=\{x\in X: p(x_{0}, x)\leq p(x_{0}, x_{0})+r\}, $如果$x=x_{0}$, 则$p(H(x, t), x_{0})=p(H(x_{0}, t), H(x_{0}, t_{0}))\leq |\eta(t)-\eta(t_{0})|＜ \varepsilon =\varphi(p(x_{0}, x_{0})+r)＜p(x_{0}, x_{0})+r$, 从而$H(x_{0}, t)\in \overline{B_{p}(x_{0}, r)}.$如果$x\neq x_{0}$, 由条件(ⅱ)和(ⅲ), 有

$\begin{eqnarray*} p(H(x, t), x_{0})&=&p(H(x_{0}, t), H(x_{0}, t_{0}))\\&\leq& p(H(x, t), H(x, t_{0}))+p(H(x, t_{0}), H(x_{0}, t_{0}))\\ &\leq&|\eta(t)-\eta(t_{0})|+\phi(p(x, x_{0}))\\ &\leq & \varepsilon + \phi(p(x_{0}, x_{0})+r)=p(x_{0}, x_{0})+r, \end{eqnarray*}$

这表明$H(x, t)\in \overline{B_{p}(x_{0}, r)}.$因此对每个固定的$t\in(t_{0}-\delta, t_{0}+\delta)$, $H(\cdot, t):\overline{B_{p}(x_{0}, r)}\rightarrow \overline{B_{p}(x_{0}, r)}.$

由条件(ⅱ)

$p(H(x, t), H(y, t))\leq \phi(p(x, y))\leq \phi(M(x, y))=M(x, y)-\varphi(M(x, y)), $

其中

$\begin{eqnarray*} M(x, y)&=&\max \bigg\{p(x, y), \frac{p(x, H(x, t))p(y, H(y, t))+p(x, H(y, t))p(y, H(x, t))}{3(p(H(x, t), H(y, t))+p(x, y))}, \\&&\frac{p(x, H(x, t))+p(y, H(y, t))}{2}, \frac{p(x, H(y, t))+p(y, H(x, t))}{2} \bigg\}. \end{eqnarray*}$

因此, 映射$f=H(\cdot, t)$和函数$\psi(t)=t.$及$\varphi$满足推论2.3的所有条件, $H(\cdot, t)$有一个不动点.由条件(ⅰ), 这个不动点必属于$U$, 即$t\in W$, 从而$(t_{0}-\delta, t_{0}+\delta)\subseteq W$, $W$是$[0, 1]$中的开集.

设$\{t_{n}\}$是$W$中的数列且$t_{n}\rightarrow t^{*}$ ($n\rightarrow \infty), $由$W$的定义, 对每个$n=1, 2, \cdots, $存在$x_{n}\in U$使$x_{n}=H(x_{n}, t_{n})$.由

$\begin{eqnarray*} p(x_{m}, x_{n})&=&p(H(x_{m}, t_{m}), H(x_{n}, t_{n}))\\&\leq& p(H(x_{m}, t_{m}), H(x_{m}, t_{n}))+p(H(x_{m}, t_{n}), H(x_{n}, t_{n}))\\ &\leq&|\eta(t_{m})-\eta(t_{n})|+\phi(p(x_{m}, x_{n})), \end{eqnarray*}$

得$\varphi(p(x_{m}, x_{n}))\leq |\eta(t_{m})-\eta(t_{n})|.$由于$\varphi$非减, $\varphi$和$\eta$连续且$\{t_{n}\}$收敛, 令$n, m\rightarrow\infty$, 得$\lim\limits_{n, m\rightarrow\infty}p(x_{m}, x_{n})=0$, 因此$\{x_{n}\}$是$(X, p)$中的Cauchy序列.由$(X, p)$完备性, 存在$x^{*}\in X$使$p(x^{*}, x^{*})=\lim\limits_{n\rightarrow\infty}p(x^{*}, x_{n})=\lim\limits_{n, m\rightarrow\infty}p(x_{m}, x_{n})=0.$另一方面, 有

$\begin{eqnarray*} p(x_{n}, H(x^{*}, t^{*}))&=&p(H(x_{n}, t_{n}), H(x^{*}, t^{*}))\\&\leq& p(H(x_{n}, t_{n}), H(x_{n}, t^{*}))+p(H(x_{n}, t^{*}), H(x^{*}, t^{*}))\\ &\leq&|\eta(t_{n})-\eta(t^{*})|+\phi(p(x_{n}, x^{*})), \end{eqnarray*}$

在上式中令$n\rightarrow\infty $, 得$\lim\limits_{n\rightarrow\infty}p(x_{n}, H(x^{*}, t^{*}))=0$, 从而

$p(x^{*}, H(x^{*}, t^{*})) =\lim\limits_{n\rightarrow\infty}p(x_{n}, H(x^{*}, t^{*}))=0, $

所以$x^{*}=H(x^{*}, t^{*})$, 由条件(ⅰ), $x^{*}\in U$.因此$t^{*}\in W$, $W$是$[0, 1]$中的闭集.

现在, $W$在$[0, 1]$中既开又闭, 由$[0, 1]$的连通性, 得$W=[0, 1]$, 这也就得到$H(\cdot, 1)$有不动点.同理可证, 如果$H(\cdot, 1)$有不动点, 则$H(\cdot, 0)$有不动点.