A Kannan-like contraction in partially ordered spaces (Q2860783)

Let \((X,\leq)\) be a partially ordered set; and \(d:X\times X\to [0,\infty)\) be a metric over \(X\) such that \((X,d)\) is complete. Further, let \(f:X\to X\) be a selfmap of \(X\) and set, for simplicity, \(A(x,y)=(1/2)[d(x,fx)+d(y,fy)]\), \(x,y\in X\). Finally, denote by \(S\) the class of all functions \(\beta:(0,\infty)\to [0,1)\) with [\(\beta(t_n)\to 1\) implies \(t_n\to 0\)].NEWLINENEWLINEThe following is the main result in this paper:NEWLINENEWLINE{Theorem.} Suppose that \(f\) is non-decreasing andNEWLINENEWLINE(K1)\ \(d(fx,fy)\leq \beta(A(x,y))A(x,y)\) for all comparable \(x,y\in X\);NEWLINENEWLINE(K2)\ \(x_0\leq fx_0\) for at least one \(x_0\in X\).NEWLINENEWLINEThen the following conclusions hold:NEWLINENEWLINE{(I)} If, in addition, one of the extra conditions holds: {(i)} \(f\) is continuous, or {(ii)} \((x_n)\)=non-decreasing and \(x_n\to x\) implies \(x_n\leq x\) for all \(n\), then \(f\) has a fixed point in \(X\).NEWLINENEWLINE{(II)} If, in addition, for each \(x,y\in X\) there exists \(z\in X\), comparable with \(x\) and \(y\), such that \(z\leq fz\), then \(f\) has a unique fixed point in \(X\).NEWLINENEWLINESome technical connections with other statements in the area are also discussed.