Sentence examples for by a nonexpansive mapping from inspiring English sources

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Mann [19] introduced an iterative scheme and employed it to approximate the solution of a fixed point problem defined by a nonexpansive mapping where the Picard iterative scheme fails to converge.

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By (i), is a nonexpansive mapping on the bounded closed convex set Thus has a fixed point, say that is, Thus Consequently, (3.9).

(1) is nonexpansive and for each ; (2)for each and for each positive integer, the limit exists; (3 the mapping define by (2.10). is a nonexpansive mapping satisfying and it is called the -mapping generated by and.

Hence, z ∈ M − 1 ( 0 ) by the maximality of M. On the other hand, from (3.14), we get that z ∈ F ( S ) by the demiclosedness of a nonexpansive mapping [4, 12].

One of the well-known results is proven by Reich [16] for a nonexpansive mapping T on C, which asserts the weak convergence of the sequence { x n } generated by (1.10) in a uniformly convex Banach space with a Frechet differentiable norm under the control condition ∑ n = 1 ∞ α n ( 1 − α n ) = ∞.

Since the mapping T is defined by (Tx=lim_{nrightarrowinfty}T_{n}x) for all (xin C), by Lemma 3, T is a nonexpansive mapping, and (operatorname {Fix}( T ) neqemptyset).

Define a nonexpansive mapping by.

By Lemma 3.4, each is a nonexpansive mapping satisfying.

This implicit iteration was introduced by Browder [2] for a nonexpansive mapping T in Hilbert space.

In 2007, Takahashi et al. [23] proved the following strong convergence theorem for a nonexpansive mapping by using the shrinking projection method in mathematical programming.

In 2006, Martinez-Yanes and Xu [1] obtained strong convergence theorems for finding a fixed point of a nonexpansive mapping by a new hybrid method in a Hilbert space.

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