Diametrically contractive multivalued mappings (Q2472289)

Let \({\mathfrak I}(X)\) be the collection of nonempty closed subsets of \(X\). The authors prove the following theorems. Theorem 2.2 Let \(M\) be a weakly compact subset of a Banach space \(X\) and let \(T: M\to{\mathfrak I}(X)\), \(Tx\cap M\neq\varnothing\) for all \(x\in M\) and \(\delta(TA\cap A)< \delta(A)\) for all closed sets \(A\) with \(\delta(A)> 0\). Then \(T\) has a unique fixed point. Theorem 3.2. Let \((X, d)\) be a bounded hyperconvex metric space and let \(T: X\to E(X\)) be asymptotically regular, satisfying the following conditions: {\parindent7mm \begin{itemize}\item[(i)] there exists \(\varphi\in\Phi_1\) such that \(\varphi(x)\leq x\), \(\varphi(x+ y)\leq \varphi(x)+ \varphi(y)\), \(\varphi(x)= 0\) iff \(x= 0\), and \(H(Tx, Ty)\leq\varphi(d(x, y))\) for all \(x\), \(y\) in \(X\); \item[(ii)] \(H(Tx, Ty)< d(x, y)\) for all \(x,y\in X\) with \(x\neq y\). \end{itemize}} If \(\delta(T^n x)\to 0\) for each \(x\in X\), then \(T\) has a contractive fixed point. That is, there exists a unique point \(\xi\) in \(X\) such that, for each \(x\in X\), \(T^n x\to\{\xi\}= \text{Fix\,} T\).