Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 717341, 6 pages
doi:10.1155/2010/717341
Research Article

An Iterative Algorithm of Solution for Quadratic Minimization Problem in Hilbert Spaces

1Department of Mathematics, Cangzhou Normal University, Hebei, Cangzhou 061001, China
2Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China

Received 24 October 2009; Accepted 28 December 2009

Academic Editor: Jong Kim

Copyright © 2010 Li Liu et al. 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

The purpose of this paper is to introduce an iterative algorithm for finding a solution of quadratic minimization problem in the set of fixed points of a nonexpansive mapping and to prove a strong convergence theorem of the solution for quadratic minimization problem. The result of this article improved and extended the result of G. Marino and H. K. Xu and some others.

1. Introduction and Preliminaries

Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [1, 2] and the references therein. A typical problem is to minimize a quadratic function over the set of fixed points of a nonexpansive mapping on a real Hilbert space 𝐻 :

m i n 𝑥 𝐶 1 2 𝐴 𝑥 , 𝑥 𝑥 , 𝑢 , ( 1 . 1 ) where 𝐶 is the fixed point set of a nonexpansive mapping 𝑇 defined on 𝐻 , and 𝑢 is a given point in 𝐻 . Let 𝐴 be a strongly positive operator defined on 𝐻 , that is, there is a constant 𝛾 > 0 with the property

𝐴 𝑥 , 𝑥 𝛾 𝑥 2 , 𝑥 𝐻 . ( 1 . 2 ) Then minimization (1.1) has a unique solution 𝑥 𝐶 which satisfies the optimality condition

𝐴 𝑥 𝑢 , 𝑥 𝑥 0 , 𝑥 𝐶 . ( 1 . 3 ) In [1, 2] it is proved that the sequence { 𝑥 𝑛 } generated by the following algorithm

𝑥 𝑛 + 1 = 𝐼 𝛼 𝑛 𝐴 𝑇 𝑥 𝑛 + 𝛼 𝑛 𝑢 , 𝑛 0 ( 1 . 4 ) converges in norm to the solution 𝑥 of (1.1) provided that the sequence { 𝛼 𝑛 } in ( 0 , 1 ) satisfies conditions

l i m 𝑛 𝛼 𝑛 = 0 , ( 𝐶 1 ) 𝑛 = 1 𝛼 𝑛 = , ( 𝐶 2 ) and additionally, either the condition 𝑛 = 1 | | 𝛼 𝑛 + 1 𝛼 𝑛 | | < , ( 𝐶 3 ) or the condition l i m 𝑛 | | 𝛼 𝑛 + 1 𝛼 𝑛 | | 𝛼 𝑛 + 1 = 0 . ( 𝐶 4 )

The purpose of this paper is to introduce the following iterative algorithm:

𝑥 𝑛 + 1 = 𝐼 𝛼 𝑛 𝐴 𝑦 𝑛 + 𝛼 𝑛 𝑦 𝑢 , 𝑛 = 𝛽 𝑛 𝑥 𝑛 + 1 𝛽 𝑛 𝑇 𝑥 𝑛 , ( 1 . 5 ) and to prove that the iterative sequence { 𝑥 𝑛 } defined by (1.5) converges strongly to the solution 𝑥 of (1.1) under the conditions ( 𝐶 1 ), ( 𝐶 2 ) and 0 < 𝑎 𝛽 𝑛 𝑏 < 1 for some constants 𝑎 , 𝑏 .

Lemma 1.1 (see [3, 4]). Let { 𝑥 𝑛 } and { 𝑦 𝑛 } be bounded sequences in a Banach space 𝑋 such that 𝑥 𝑛 + 1 = 𝜆 𝑛 𝑥 𝑛 + 1 𝜆 𝑛 𝑦 𝑛 , 𝑛 0 , ( 1 . 6 ) where { 𝜆 𝑛 } is a sequence in [ 0 , 1 ] such that 0 < l i m i n f 𝑛 𝜆 𝑛 l i m s u p 𝑛 𝜆 𝑛 < 1 . ( 1 . 7 ) Assume that l i m s u p 𝑛 𝑦 𝑛 + 1 𝑦 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 0 . ( 1 . 8 ) Then l i m 𝑛 𝑦 𝑛 𝑥 𝑛 = 0 .

Lemma 1.2 (see [1]). Assume that 𝐴 is a strongly positive linear bounded operator on a real Hilbert space 𝐻 with coefficient 𝛾 > 0 and 0 < 𝛼 𝐴 1 . Then 𝐼 𝛼 𝐴 ( 1 𝛼 𝛾 ) .

Lemma 1.3 (see [5]). Let 𝐻 be a Hilbert space, 𝐾 a closed convex subset of 𝐻 , and 𝑇 𝐾 𝐾 a nonexpansive mapping with nonempty fixed point set 𝐹 ( 𝑇 ) . If { 𝑥 𝑛 } is a sequence in 𝐾 weakly converging to 𝑥 and if 𝑥 𝑛 𝑇 𝑥 𝑛 converges strongly to 0 , then 𝑥 = 𝑇 𝑥 .

Lemma 1.4 (see [6]). Assume that { 𝑎 𝑛 } is a sequence of nonnegative real numbers such that 𝑎 𝑛 + 1 1 𝛾 𝑛 𝑎 𝑛 + 𝛾 𝑛 𝛿 𝑛 , ( 1 . 9 ) where { 𝛾 𝑛 } is a sequence in ( 0 , 1 ) and { 𝛿 𝑛 } is a real sequence such that (i) 𝑛 = 1 𝛾 𝑛 = ,(ii) l i m s u p 𝑛 𝛿 𝑛 0 or 𝑛 = 1 𝛾 𝑛 | 𝛿 𝑛 | < . Then l i m 𝑛 𝑎 𝑛 = 0 .

2. Main Results

Theorem 2.1. Suppose that 𝐴 is strongly positive operator with coefficient 𝛾 > 0 as given in (1.2). Suppose that the sequences { 𝛼 𝑛 } , { 𝛽 𝑛 } satisfy the conditions ( 𝐶 1 ), ( 𝐶 2 ) and 0 < 𝑎 𝛽 𝑛 𝑏 < 1 for some constants 𝑎 , 𝑏 . Then the sequence { 𝑥 𝑛 } generated by algorithm (1.5) converges strongly to the unique solution 𝑥 of the minimization problem (1.1).

Proof. First we show that { 𝑥 𝑛 } is bounded. As a matter of fact, take 𝑝 𝐹 ( 𝑇 ) and use Lemma 1.2 to obtain 𝑥 𝑛 + 1 = 𝑝 𝐼 𝛼 𝑛 𝐴 𝛽 𝑛 𝑥 𝑛 + 1 𝛽 𝑛 𝑇 𝑥 𝑛 𝑝 + 𝛼 𝑛 ( 𝑢 𝐴 𝑝 ) 1 𝛾 𝛼 𝑛 𝑥 𝑛 𝑝 + 𝛼 𝑛 𝑢 𝐴 𝑝 . ( 2 . 1 ) By induction we can get 𝑥 𝑛 𝑥 𝑝 m a x 0 , 1 𝑝 𝛾 𝑢 𝐴 𝑝 , 𝑛 0 . ( 2 . 2 ) Hence, { 𝑥 𝑛 } is bounded and so is { 𝑦 𝑛 } . Next rewrite 𝑥 𝑛 + 1 in the form 𝑥 𝑛 + 1 = 1 𝜆 𝑛 𝑥 𝑛 + 𝜆 𝑛 𝑧 𝑛 , ( 2 . 3 ) where 𝜆 𝑛 = 1 1 𝛼 𝑛 𝛽 𝑛 𝑧 , ( 2 . 4 ) 𝑛 = 𝛼 𝑛 𝛽 𝑛 𝜆 𝑛 ( 𝐼 𝐴 ) 𝑥 𝑛 + 1 𝛽 𝑛 𝜆 𝑛 𝐼 𝛼 𝑛 𝐴 𝑇 𝑥 𝑛 + 𝛼 𝑛 𝜆 𝑛 𝑢 . ( 2 . 5 ) Since 𝛼 𝑛 0 and 0 < 𝑎 𝛽 𝑛 𝑏 < 1 , then 0 < l i m i n f 𝑛 𝜆 𝑛 l i m s u p 𝑛 𝜆 𝑛 < 1 . ( 2 . 6 )
Next some manipulations give us that 𝑧 𝑛 + 1 𝑧 𝑛 = 𝛽 𝑛 + 1 𝛼 𝑛 + 1 𝜆 𝑛 + 1 ( 𝐼 𝐴 ) 𝑥 𝑛 + 1 𝛽 𝑛 𝛼 𝑛 𝜆 𝑛 ( 𝐼 𝐴 ) 𝑥 𝑛 + 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝛼 𝑛 𝜆 𝑛 𝑢 + 1 𝛽 𝑛 + 1 𝜆 𝑛 + 1 𝑇 𝑥 𝑛 + 1 𝑇 𝑥 𝑛 1 𝛽 𝑛 + 1 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝐴 𝑇 𝑥 𝑛 + 1 𝑇 𝑥 𝑛 + 1 𝛽 𝑛 + 1 𝜆 𝑛 + 1 1 𝛽 𝑛 𝜆 𝑛 𝑇 𝑥 𝑛 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝛼 𝑛 𝜆 𝑛 1 𝛽 𝑛 𝐴 𝑇 𝑥 𝑛 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝛽 𝑛 𝛽 𝑛 + 1 𝐴 𝑇 𝑥 𝑛 . ( 2 . 7 ) Therefore, 𝑧 𝑛 + 1 𝑧 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 𝛽 𝑛 + 1 𝛼 𝑛 + 1 𝜆 𝑛 + 1 ( 𝐼 𝐴 ) 𝑥 𝑛 + 1 + 𝛽 𝑛 𝛼 𝑛 𝜆 𝑛 ( 𝐼 𝐴 ) 𝑥 𝑛 + | | | | 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝛼 𝑛 𝜆 𝑛 | | | | + 𝑢 1 𝛽 𝑛 + 1 𝜆 𝑛 + 1 𝑥 1 𝑛 + 1 𝑥 𝑛 + 1 𝛽 𝑛 + 1 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝑥 𝐴 𝑛 + 1 𝑥 𝑛 + | | | | 1 𝛽 𝑛 + 1 𝜆 𝑛 + 1 1 𝛽 𝑛 𝜆 𝑛 | | | | 𝑇 𝑥 𝑛 + | | | | 𝛼 𝑛 + 1 𝜆 𝑛 + 1 𝛼 𝑛 𝜆 𝑛 | | | | 1 𝛽 𝑛 𝐴 𝑇 𝑥 𝑛 + 𝛼 𝑛 + 1 𝜆 𝑛 + 1 | | 𝛽 𝑛 𝛽 𝑛 + 1 | | 𝐴 𝑇 𝑥 𝑛 . ( 2 . 8 ) Since 𝜆 𝑛 = 1 ( 1 𝛼 𝑛 ) 𝛽 𝑛 and 𝛼 𝑛 0 , then l i m 𝑛 1 𝛽 𝑛 𝜆 𝑛 = l i m 𝑛 𝛼 1 𝑛 𝛽 𝑛 𝜆 𝑛 = 1 . ( 2 . 9 ) Then last inequality implies that l i m s u p 𝑛 𝑧 𝑛 + 1 𝑧 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 0 , ( 2 . 1 0 ) and so an application of Lemma 1.1 asserts that l i m 𝑛 𝑧 𝑛 𝑥 𝑛 = 0 . ( 2 . 1 1 ) By (2.5) we have that 𝑧 𝑛 𝑇 𝑥 𝑛 = 𝛼 𝑛 𝛽 𝑛 𝜆 𝑛 ( 𝐼 𝐴 ) 𝑥 𝑛 + 1 𝛽 𝑛 𝜆 𝑛 1 𝑇 𝑥 𝑛 1 𝛽 𝑛 𝛼 𝑛 𝜆 𝑛 𝐴 𝑇 𝑥 𝑛 + 𝛼 𝑛 𝜆 𝑛 𝑢 . ( 2 . 1 2 ) Again since 𝛼 𝑛 0 , { 𝑥 𝑛 } is bounded, and 𝜆 𝑛 = 1 ( 1 𝛼 𝑛 ) 𝛽 𝑛 , then we deduce from (2.12) that l i m 𝑛 𝑧 𝑛 𝑇 𝑥 𝑛 = 0 . ( 2 . 1 3 ) This together with (2.11) yields l i m 𝑛 𝑥 𝑛 𝑇 𝑥 𝑛 = 0 . ( 2 . 1 4 ) By using Lemma 1.3, we obtain 𝜔 𝑤 ( 𝑥 𝑛 ) 𝐹 ( 𝑇 ) , where 𝜔 𝑤 ( 𝑥 𝑛 ) = { 𝑧 𝑥 𝑛 𝑘 𝑧 } is the set of weak 𝜔 - limit points of sequence { 𝑥 𝑛 } .
Let 𝑥 be the unique solution to the minimization (1.1). Then by the definition of algorithm (1.5), we can write 𝑥 𝑛 + 1 𝑥 = 𝐼 𝛼 𝑛 𝐴 𝛽 𝑛 𝑥 𝑛 + 1 𝛽 𝑛 𝑇 𝑥 𝑛 𝑥 + 𝛼 𝑛 𝑢 𝐴 𝑥 . ( 2 . 1 5 ) Since 𝐻 is a Hilbert space, then we have that 𝑥 𝑛 + 1 𝑥 2 ( 𝐼 𝛼 𝑛 𝐴 ) ( 𝛽 𝑛 𝑥 𝑛 + 1 𝛽 𝑛 𝑇 𝑥 𝑛 𝑥 ) 2 + 2 𝛼 𝑛 𝑢 𝐴 𝑥 , 𝑥 𝑛 + 1 𝑥 1 𝛾 𝛼 𝑛 𝑥 𝑛 𝑥 + 2 𝛼 𝑛 𝑢 𝐴 𝑥 , 𝑥 𝑛 + 1 𝑥 . ( 2 . 1 6 ) However, we can take a subsequence { 𝑥 𝑛 𝑘 } of { 𝑥 𝑛 } such that l i m s u p 𝑛 𝑢 𝐴 𝑥 , 𝑥 𝑛 𝑥 = l i m 𝑘 𝑢 𝐴 𝑥 , 𝑥 𝑛 𝑘 𝑥 , ( 2 . 1 7 ) and also { 𝑥 𝑛 𝑘 } converges weakly to a fixed point 𝑝 𝐹 ( 𝑇 ) . It follows from optimality condition (1.3) that l i m s u p 𝑛 𝑢 𝐴 𝑥 , 𝑥 𝑛 𝑥 = 𝑢 𝐴 𝑥 , 𝑝 𝑥 0 . ( 2 . 1 8 ) Therefore, by using Lemma 1.4 and noticing (2.18), we conclude that 𝑥 𝑛 𝑥 . This completes the proof.

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