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Fixed point for set-valued mappings satisfying an implicit relation in partially ordered metric spaces. (English) Zbl 1176.54028
Let $(X,d;\le)$ be a complete partially ordered metric space and $F,G:B(x_0,r)\to C(X)$ be two maps with bounded values, fulfilling {\parindent=8mm\item{(i)} $T(D(Fx,Gy),d(x,y),d(x,Fx),d(y,Gy),d(x,Gy),d(y,Gx))\le 0$ for all $x,y\in B(x_0,r)$, where $T:\bbfR_+^6\to\bbfR_+$ satisfies some mild conditions, \item{(ii)} for each $x\in X$, there exists $y\in Lx$ with $x\le y$ such that $d(x,y)\le d(x,Lx)+\varepsilon$, where $L\in \{F,G\}$, \item{(iii)} if $(x_n)\subseteq B(x_0,r)$ fulfills $x_n\le x_{n+1}$ for all $n$ and $x_n\to x$ as $n\to \infty$, then $x_n\le x$ for all $n$. \par}In addition, assume that there exists a continuous strictly increasing function $\Phi:\bbfR_+\to \bbfR_+$ with $\Phi(t)< t$ for all $t> 0$, such that {\parindent=8mm\item{(iv)} $d(x_0,x_1)< r-\Phi(r)$ for some $x_1\in Fx_0$ with $x_0\le x_1$, \item{(v)} $\sum_n \Phi^n(r-\Phi(r))\le \Phi(r)$.\par} Then there exists in $B(x_0,r)$ a common fixed point for $F$ and $G$. Reviewer’s remark: The seminal 2004 fixed point result of {\it A. C. M. Ran} and {\it M. C. B. Reurings} [Proc. Am. Math. Soc. 132, No. 5, 1435--1443 (2004; Zbl 1060.47056)] was obtained in 1986 by the reviewer [J. Math. Anal. Appl. 117, 100--127 (1986; Zbl 0613.47037)].

54H25Fixed-point and coincidence theorems in topological spaces
54F05Linearly, generalized, and partial ordered topological spaces
54C60Set-valued maps (general topology)
Full Text: DOI
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