Fixed Point Theory and Applications
Volume 2010 (2010), Article ID 805326, 2 pages
doi:10.1155/2010/805326
Letter to the Editor

A Note on Strong Convergence of a Modified Halpern's Iteration for Nonexpansive Mappings

Shuang Wang

Department of Mathematics, Hubei Normal University, Huangshi 435002, China

Received 25 September 2009; Accepted 18 January 2010

Academic Editor: Jerzy Jezierski

Copyright © 2010 Shuang Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In the paper by Hu in 2008, the author proved a strong convergence result for nonexpansive mappings using a modified Halpern's iteration algorithm. Unfortunately, the case l i m 𝑛 → ∞ 𝛽 𝑛 = 1 does not guarantee the strong convergence of the sequence { 𝑥 𝑛 } . In this note, we provide a counter-example to the theorem.

In [1], the author introduced a modified Halpern’s iteration. For any 𝑢 , 𝑥 0 ∈ 𝐶 , the sequence { 𝑥 𝑛 } is defined by 𝑥 𝑛 + 1 = 𝛼 𝑛 𝑢 + 𝛽 𝑛 𝑥 𝑛 + 𝛾 𝑛 𝑇 𝑥 𝑛 , 𝑛 ≥ 0 , ( ( I ) ) w here { 𝛼 𝑛 } , { 𝛽 𝑛 } , and { 𝛾 𝑛 } are three real sequences in ( 0 , 1 ) , satisfying 𝛼 𝑛 + 𝛽 𝑛 + 𝛾 𝑛 = 1 . The author proved the following strong convergence theorem.

Theorem 1 (see [1]). Let 𝐶 be a nonempty closed convex subset of a real Banach space 𝐸 which has a uniformly Gâteaux differentiable norm. Let 𝑇 ∶ 𝐶 → 𝐶 be a nonexpansive mapping with F i x ( 𝑇 ) ≠ ∅ . Assume that { 𝑧 𝑡 } converges strongly to a fixed point 𝑧 of 𝑇 as 𝑡 → 0 , where 𝑧 𝑡 is the unique element of 𝐶 which satisfies 𝑧 𝑡 = 𝑡 𝑢 + ( 1 − 𝑡 ) 𝑇 𝑧 𝑡   for any 𝑢 ∈ 𝐶 . Let { 𝛼 𝑛 } , { 𝛽 𝑛 } , and { 𝛾 𝑛 } be three real sequences in ( 0 , 1 ) which satisfy the following conditions: ( 𝐶 1 ) l i m 𝑛 → ∞ 𝛼 𝑛 = 0 and ∑ ( 𝐶 2 ) ∞ 𝑛 = 0 𝛼 𝑛 = + ∞ . For any 𝑥 0 ∈ 𝐶 , the sequence { 𝑥 𝑛 } is defined by the iteration in ( I ). Then the sequence { 𝑥 𝑛 } converges strongly to a fixed point of 𝑇 .

Counter Example
Let 𝐸 be a real Banach space whose norm is uniformly Gâteaux differentiable. Let 𝐶 be a nonempty closed and convex subset of 𝐸 , defined by [ ] 𝐶 = { 𝑥 ∈ 𝐸 ∶ 𝑥 = 𝜆 𝑦 , 𝜆 ∈ 0 , 3 } , ( 1 ) where 𝑦 ≠ 0 , with ‖ 𝑦 ‖ = 1 a fixed element of 𝐸 . Let 𝑇 ∶ 𝐶 → 𝐶 be a mapping defined by 𝑇 𝑥 = 0 for all 𝑥 ∈ 𝐶 . It is obvious that 𝑇 is a nonexpansive mapping and F i x ( 𝑇 ) = { 0 } . Take 𝛼 𝑛 = 1 / ( 𝑛 + 2 ) , 𝛽 𝑛 = 1 − 2 / ( 𝑛 + 2 ) , and 𝛾 𝑛 = 1 / ( 𝑛 + 2 ) for all 𝑛 ≥ 0 and 𝑥 0 = 𝑦 , 𝑢 = 3 𝑦 . We also can obtain that 𝑧 𝑡 = 3 𝑡 𝑦 → 0 ( 𝑡 → 0 ) . Observe that all conditions of Theorem 1 are satisfied. However, the iterative sequence { 𝑥 𝑛 } does not converge strongly to the fixed point 𝑧 = 0 of 𝑇 .

Claim 1. If ‖ 𝑥 𝑛 ‖ ≤ 1 , then ‖ 𝑥 𝑛 + 1 ‖ > ‖ 𝑥 𝑛 ‖ .

Proof. In fact, we have 𝑥 𝑛 + 1 = 1 2 𝑛 + 2 3 𝑦 + 1 − 𝑥 𝑛 + 2 𝑛 + 1 𝑛 + 2 𝑇 𝑥 𝑛 = 3 2 𝑛 + 2 𝑦 + 1 − 𝑥 𝑛 + 2 𝑛 = 3 2 𝑛 + 2 𝑦 + 1 − 𝜆 𝑛 + 2 𝑛 𝑦 , ( 2 ) where 𝑥 𝑛 can be denoted as 𝑥 𝑛 = 𝜆 𝑛 𝑦 . If ‖ 𝑥 𝑛 ‖ ≤ 1 , then 0 < 𝜆 𝑛 = ‖ 𝑥 𝑛 ‖ ≤ 1 . From the above equality we have ‖ ‖ 𝑥 𝑛 + 1 ‖ ‖ = ‖ ‖ ‖ 3 + 2 𝑛 + 2 1 − 𝜆 𝑛 + 2 𝑛 𝑦 ‖ ‖ ‖ = 3 + 2 𝑛 + 2 1 − 𝜆 𝑛 + 2 𝑛 = 2 𝑛 + 2 1 − 𝜆 𝑛 + 1 𝑛 + 2 + 𝜆 𝑛 > 𝜆 𝑛 = ‖ ‖ 𝑥 𝑛 ‖ ‖ . ( 3 ) Hence { 𝑥 𝑛 } does not converge strongly to 𝑧 = 0 .

Remark 1. Why does the proof of Theorem 1 fail? It is not difficult to check that the proof of Case  2 ( l i m 𝑛 → ∞ 𝛽 𝑛 = 1 ) is not suitable.

Acknowledgments

This work is supported by the National Science Foundation of China under Grant no. 10771175 and the Natural Science Foundational Committee of Hubei Province (D200722002).

References

  1. L.-G. Hu, “Strong convergence of a modified Halpern's iteration for nonexpansive mappings,” Fixed Point Theory and Applications, vol. 2008, Article ID 649162, 9 pages, 2008.