نتایج جستجو برای: uniformly l lipschitzianasymptotically pseudocontractive mappings
تعداد نتایج: 669703 فیلتر نتایج به سال:
Let E be a real Banach space, {Ti}i=1 be a finite family of continuous pseudocontractive self mappings of E and G : E → E be a mapping which is both δ -strongly accretive and λ -strictly pseudocontractive of Browder-Petryshyn type such that δ + λ 1 . We propose a new implicit iteration scheme with perturbed mapping G for the approximation of common fixed points of {Ti}i=1 . For an arbitrary ini...
A extension of Nakajo and Takahashi’s modification of Mann’s iterative process to the Ishikawa iterative process is given. The strong convergence of a modified Ishikawa iterative scheme to a common fixed point of a finite family of Lipschitz pseudocontractive self-mappings on a closed convex subset of a Hilbert space is proved. Our theorem extends several known results. c © 2007 Elsevier Ltd. A...
*Correspondence: [email protected] 2Basic Experimental & Teaching Center, Henan University, Kaifeng, 475000, China Full list of author information is available at the end of the article Abstract In this article, a mean iterative algorithm is investigated for finding a common element in the solution set of generalized equilibrium problems and in the fixed point set of strictly pseudocontractive ma...
We propose an implicit iterative scheme and an explicit iterative scheme for finding a common element of the set of fixed point of infinitely many strict pseudocontractive mappings and the set of solutions of an equilibrium problem by the general iterative methods. In the setting of real Hilbert spaces, strong convergence theorems are proved. Our results improve and extend the corresponding res...
and Applied Analysis 3 We know that T is pseudocontractive if and only if T satisfies the condition Tx − Ty 2 ≤ x − y 2 + (I − T)x − (I − T)y 2 (15) for all x, y ∈ C. Since u ∈ Fix(T), we have from (15) that ‖Tx − u‖ 2 ≤ ‖x − u‖ 2 + ‖x − Tx‖ 2 , (16) for all x ∈ C. By using (13) and (16), we obtain Tyn − u 2 ≤ yn − u 2 + yn − Tyn 2 = (1 − γn)xn + γnT...
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