Open Access

A new explicit iteration method for variational inequalities on the set of common fixed points for a finite family of nonexpansive mappings

Journal of Inequalities and Applications20132013:419

https://doi.org/10.1186/1029-242X-2013-419

Received: 8 April 2013

Accepted: 18 August 2013

Published: 3 September 2013

Abstract

In this paper, we introduce a new explicit iteration method based on the steepest descent method and Krasnoselskii-Mann type method for finding a solution of a variational inequality involving a Lipschitz continuous and strongly monotone mapping on the set of common fixed points for a finite family of nonexpansive mappings in a real Hilbert space.

MSC:41A65, 47H17, 47H20.

Keywords

contraction common fixed points hybrid steepest descent method nonexpansive mappings monotone mappings

1 Introduction and preliminaries

Let C be a nonempty closed and convex subset of a real Hilbert space H with the inner product , and the norm , and let F : H H be a nonlinear mapping. The variational inequality problem is to find a point p C such that
F ( p ) , p p 0 , p C .
(1.1)

Variational inequalities were initially studied by Kinderlehrer and Stampacchia in [1], and since then have been widely investigated. They cover partial differential equations, optimal control, optimization, mathematical programming, mechanics, and finance (see [13]).

It is well known that if F is an L-Lipschitz continuous and η-strongly monotone, i.e., F satisfies the following conditions:
F ( x ) F ( y ) L x y , F ( x ) F ( y ) , x y η x y 2 ,
where L and η are fixed positive numbers, then (1.1) has a unique solution. It is also known that (1.1) is equivalent to the fixed point equation
p = P C ( p μ F ( p ) ) ,
(1.2)

where P C denotes the metric projection from x H onto C and μ is an arbitrarily positive constant.

The fixed point formulation (1.2) involves the metric projection P C . To overcome the complexity caused by P C , Yamada [4] introduced a hybrid steepest descent method for solving (1.1). His idea is stated as follows. Assume that C = i = 1 N Fix ( T i ) , the set of common fixed points of a finite family of nonexpansive mappings T i on H with an integer N 1 .

Recall that T : H H is nonexpansive if
T x T y x y , x , y H
and
Fix ( T ) = { x H : x = T x }
denotes the fixed point set of T. Yamada proposed the following algorithm in [4]
u k + 1 = T [ k + 1 ] u k λ k + 1 μ F ( T [ k + 1 ] u k ) ,
(1.3)

where T [ n ] = T n mod N , for integer n 1 , with the mod-function taking values in the set { 1 , 2 , , N } , μ ( 0 , 2 η / L 2 ) and { λ k } ( 0 , 1 ) , and proved that the sequence { u k } in (1.3) converges strongly to p under the following conditions:

(L1) lim λ k = 0 ;

(L2) λ k = ;

(L3) | λ k λ k + N | < .

Further, Zeng and Yao [5] proved the same result with (L3) replaced by

(L4) lim ( λ k λ k + N ) / λ k + N = 0 .

Theorem 1.1 [5]

Let H be a real Hilbert space, and let F : H H be an L-Lipschitz continuous and η-strongly monotone mapping for some constants L , η > 0 . Let { T i } i = 1 N be N nonexpansive self-maps of H such that C = i = 1 N Fix ( T i ) , μ ( 0 , 2 η / L 2 ) , and let conditions (L1), (L2), (L4) be satisfied. Assume in addition that
C = i = 1 N Fix ( T i ) = Fix ( T 1 T 2 T N ) = Fix ( T N T 1 T 2 T N 1 ) = = Fix ( T 2 T 3 T N T 1 ) .
(1.4)

Then the sequence { u k } defined by (1.3) converges strongly to the unique element p in (1.1).

It is not difficult to show that (L3) implies (L4) if lim λ k / λ k + N exists. However, in general, conditions (L3) and (L4) are not comparable, i.e., neither one of them implies the other (see [6] for details).

Recently, Zeng et al. [7] proposed the following iterative scheme:
u k + 1 = T [ k + 1 ] u k λ k + 1 μ k + 1 F ( T [ k + 1 ] u k ) ,
(1.5)

and proved the following result.

Theorem 1.2 [7]

Let H be a real Hilbert space, and let F : H H be an L-Lipschitz continuous and η-strongly monotone mapping for some constants L , η > 0 . Let { T i } i = 1 N be N nonexpansive self-maps of H such that
C = i = 1 N Fix ( T i ) ,
and let μ k ( 0 , 2 η / L 2 ) . Assume that the following conditions hold:
  1. (i)

    λ k = , where { λ k } ( 0 , 1 ) ;

     
  2. (ii)

    | μ k η / L 2 | η 2 c L 2 / L 2 for some c ( 0 , η 2 / L 2 ) ;

     
  3. (iii)

    lim ( μ k + N ( λ k / λ k + N ) μ k ) = 0 .

     
Assume in addition that (1.4) holds. If
lim sup k T [ k + N ] T [ k + 1 ] u k u k + N , T [ k + N ] T [ k + 1 ] u k u k 0 ,
(1.6)

then the sequence { u k } defined by (1.5) converges strongly to the unique element p in (1.1).

They also showed that conditions (L1), (L2) and (L4) are sufficient for { u k } to be bounded and
lim k u k T [ k + N ] T [ k + 1 ] u k = 0 .

So, (1.6) is satisfied. They did not give another sufficient condition different from (L1), (L2) and (L4).

Let F x = A x u , where A is a self-adjoint bounded linear mapping such that A is strongly positive, i.e.,
A x , x η x 2 , x H
and u is some fixed element in H. Xu [6] introduced the following iteration process:
u k + 1 = ( I λ k + 1 A ) T k + 1 u k + λ k + 1 u ,
(1.7)

where I is the identity mapping of H, and proved the following result.

Theorem 1.3 [6]

Let conditions (L1), (L2) and (L3) or (L4) be satisfied. Assume in addition that (1.4) holds. Then the sequence { u k } generated by algorithm (1.7) converges strongly to the unique solution of (1.1) with F x = A x u .

Very recently, Liu and Cui [8] showed that the condition
C = i = 1 N Fix ( T i ) = Fix ( T 1 T 2 T N )
(1.8)

is sufficient for (1.4) if C .

In this paper, we introduce a new algorithm based on a combination of the steepest descent method for variational inequalities with the Krasnoselskii-Mann method for fixed point problems to solve (1.1) with C = i = 1 N Fix ( T i ) , where T i is a nonexpansive mapping on H for each i.

Given a starting point x 1 H , the iteration is defined by
{ x 1 H , y k 0 = ( I λ k μ F ) x k , y k i = ( 1 β k i ) x k + β k i T i y k i 1 , i = 1 , , N , x k + 1 = y k N ,
(1.9)
and the sequences of parameters { λ k } and { β k i } ( 0 , 1 ) satisfy the following conditions:
{ λ k 0 as  k and λ k = ; β k i β i , 0 < β i < 1 , i = 1 , , N 1 ; 0 < lim inf k β k N lim sup k β k N < 1 .
(1.10)

In Section 2, we prove the strong convergence theorem for (1.9)-(1.10) without conditions (L3), (L4) and (1.8). An application to the case that T i is a γ i -strictly pseudocontractive mapping is given in Section 3.

2 Main results

We need the following lemmas for the proof of our main result.

Lemma 2.1 [9]

  1. (i)

    x + y 2 x 2 + 2 y , x + y , x , y H .

     
  2. (ii)

    ( 1 t ) x + t y 2 = ( 1 t ) x 2 + t y 2 ( 1 t ) t x y 2 , x , y H , and for any fixed t [ 0 , 1 ] .

     

From [4], we have the following lemma.

Lemma 2.2 [4]

T λ x T λ y ( 1 λ τ ) x y , x , y H and for a fixed number μ ( 0 , 2 η / L 2 ) , λ ( 0 , 1 ) , where τ = 1 1 μ ( 2 η μ L 2 ) ( 0 , 1 ) and T λ x = ( I λ μ F ) x for x H .

Lemma 2.3 [10, 11]

Assume that T is a nonexpansive self-map of a closed convex subset K of a real Hilbert space H. If T has a fixed point, then I T is demiclosed; that is, whenever { x k } is a sequence in K weakly converging to some x K and the sequence { ( I T ) x k } strongly converges to some y, it follows that ( I T ) x = y .

Lemma 2.4 [12]

Let { x k } and { z k } be bounded sequences in a Banach space E such that
x k + 1 = ( 1 β k ) x k + β k z k
for k 1 , where { β k } is in [ 0 , 1 ] such that
0 < lim inf k β k lim sup k β k < 1 .
Assume that
lim sup k z k + 1 z k x k + 1 x k 0 .

Then lim k x k z k = 0 .

Lemma 2.5 [6]

Let { a k } be a sequence of nonnegative real numbers satisfying the condition
a k + 1 ( 1 b k ) a k + b k c k ,
where { b k } and { c k } are sequences of real numbers such that
  1. (i)

    b k [ 0 , 1 ] and k = 1 b k = ;

     
  2. (ii)

    lim sup k c k 0 .

     

Then lim k a k = 0 .

Now, we are in a position to prove the following main result.

Theorem 2.6 Let H be a real Hilbert space, and let F : H H be an L-Lipschitz continuous and η-strongly monotone mapping for some constants L , η > 0 . Let { T i } i = 1 N be N nonexpansive self-maps of H such that
C = i = 1 N Fix ( T i ) .

Then the sequence { x k } defined by (1.9)-(1.10) converges strongly to the unique element p in (1.1).

Proof First, we prove that { x k } is bounded. By Lemma 2.2, we have, for any p C , from (1.9) that
y k 0 p = ( I λ k μ F ) x k p = ( I λ k μ F ) x k ( I λ k μ F ) p λ k μ F ( p ) ( 1 λ k τ ) x k p + λ k μ F ( p ) .
Put M p = max { x 1 p , μ τ F ( p ) } . Then x 1 p M p . So, if x k p M p , then y k 0 p M p . This conclusion has a place for { y k i } with i = 1 , , N 1 . Indeed,
y k i p = ( 1 β k i ) ( x k p ) + β k i ( T i y k i 1 T i p ) ( 1 β k i ) x k p + β k i y k i 1 p ( 1 β k i ) M p + β k i M p = M p .
Then
x k + 1 p = ( 1 β k N ) ( x k p ) + β k N ( T i y k i 1 T i p ) ( 1 β k N ) x k p + β k i N y k N 1 p ( 1 β k i ) M p + β k i M p = M p .

Therefore, the sequence { x k } is bounded. So, the sequences { F ( x k ) } , { y k i } , and { T i y k i 1 } ( i = 1 , 2 , , N ) are also bounded. Without loss of generality, we assume that they are bounded by a positive constant M 1 .

Let z k = T N y k N 1 . Then we have from (1.9) that
x k + 1 = ( 1 β k N ) x k + β k N z k
and
z k + 1 z k = T N y k + 1 N 1 T N y k N 1 y k + 1 N 1 y k N 1 = ( 1 β k + 1 N 1 ) x k + 1 + β k + 1 N 1 T N 1 y k + 1 N 2 [ ( 1 β k N 1 ) x k + β k N 1 T N 1 y k N 2 ] ( 1 β k + 1 N 1 ) x k + 1 x k + 2 M 1 | β k + 1 N 1 β k N 1 | + β k + 1 N 1 T N 1 y k + 1 N 2 T N 1 y k N 2 ( 1 β k + 1 N 1 ) x k + 1 x k + 2 M 1 | β k + 1 N 1 β k N 1 | + β k + 1 N 1 y k + 1 N 2 y k N 2 ( 1 β k + 1 N 1 ) x k + 1 x k + 2 M 1 | β k + 1 N 1 β k N 1 | + β k + 1 N 1 [ ( 1 β k + 1 N 2 ) x k + 1 x k + 2 M 1 | β k + 1 N 2 β k N 2 | + β k + 1 N 2 y k + 1 N 3 y k N 3 ] ( 1 β k + 1 N 1 β k + 1 N 2 ) x k + 1 x k + 2 M 1 ( | β k + 1 N 1 β k N 1 | + | β k + 1 N 2 β k N 2 | ) + β k + 1 N 1 β k + 1 N 2 y k + 1 N 3 y k N 3 ( 1 i = 1 N 1 β k + 1 i ) x k + 1 x k + i = 1 N 1 β k + 1 i y k + 1 0 y k 0 + 2 M 1 i = 1 N 1 | β k + 1 i β k i | .
On the other hand,
y k + 1 0 y k 0 = ( I λ k + 1 μ F ) x k + 1 ( I λ k μ F ) x k x k + 1 x k + M 1 ( λ k + 1 + λ k ) .
So, we obtain that
z k + 1 z k x k + 1 x k M 1 ( λ k + 1 + λ k ) i = 1 N 1 β k + 1 i + 2 M 1 i = 1 N 1 | β k + 1 i β k i | .
Since λ k 0 and β k i β i , i = 1 , , N 1 , we have
lim sup k z k + 1 z k x k + 1 x k 0 .

By Lemma 2.4, x k z k 0 as k , i.e., x k T N y k N 1 0 .

Now, we prove that x k T i y k i 1 0 for i = 1 , , N 1 . First, we prove x k T N 1 y k N 2 0 . Let { x k n } be a subsequence of { x k } such that
lim sup k x k T N 1 y k N 2 = lim n x k n T N 1 y k n N 2
and let { x k j } be a subsequence of { x k n } such that
lim sup n x k n p = lim j x k j p .
Further,
x k j p x k j T N y k j N 1 + T N y k j N 1 T N p x k j T N y k j N 1 + y k j N 1 p x k j T N y k j N 1 + ( 1 β k N 1 ) x k j p + β k j N 1 T N 1 y k j N 2 T N 1 p x k j T N y k j N 1 + ( 1 β k N 1 ) x k j p + β k j N 1 y k j N 2 p x k j T N y k j N 1 + ( 1 β k N 1 ) x k j p + β k j N 1 [ ( 1 β k N 2 ) x k j p + β k j N 2 T N 2 y k j N 3 T N 2 p ] x k j T N y k j N 1 + ( 1 β k N 1 β k j N 2 ) x k j p + β k j N 1 β k j N 2 y k j N 2 p x k j T N y k j N 1 + ( 1 i = 1 N 1 β k j i ) x k j p + i = 1 N 1 β k j i y k j 0 p .
Since
y k j 0 p ( 1 λ k j τ ) x k j p + λ k j μ F ( p ) ,
we have
x k j p x k j T N y k j N 1 + y k j N 1 p x k j T N y k j N 1 + x k j p + i = 1 N 1 β k j i λ k j μ F ( p ) .
Therefore,
lim j x k j p = lim j y k j N 1 p .
(2.1)
Next, by Lemma 2.1 we have
y k j N 1 p 2 = ( 1 β k j N 1 ) x k j p 2 + β k j N 1 T N 1 y k j N 2 p 2 ( 1 β k j N 1 ) β k j N 1 x k j T N 1 y k j N 2 2 ( 1 β k j N 1 ) x k j p 2 + β k j N 1 y k j N 2 p 2 ( 1 β k j N 1 ) β k j N 1 x k j T N 1 y k j N 2 2 ( 1 β k j N 1 ) x k j p 2 + β k j N 1 ( 1 β k j N 2 ) ( x k j p ) + β k j N 2 ( T N 2 y k j N 3 p ) 2 ( 1 β k j N 1 ) β k j N 1 x k j T N 1 y k j N 2 2 ( 1 β k j N 1 β k j N 2 ) x k j p 2 + β k j N 1 β k j N 2 y k j N 3 p 2 ( 1 β k j N 1 ) β k j N 1 x k j T N 1 y k j N 2 2 ( 1 i = 1 N 1 β k j i ) x k j p 2 + i = 1 N 1 β k j i y k j 0 p 2 ( 1 β k j N 1 ) β k j N 1 x k j T N 1 y k j N 2 2 .
On the other hand, by Lemma 2.1 we get
y k 0 p 2 = ( I λ k μ F ) x k p 2 = x k p ( I λ k μ F ) x k x k p 2 2 λ k μ F ( x k ) , x k p x k p 2 + 2 λ k μ M 1 M p .
Without loss of generality, assume that α β k i β , k 1 for i = 1 , , N 1 and some α , β ( 0 , 1 ) . Then we obtain that
α ( 1 β ) x k j T N 1 y k j N 2 2 x k j p 2 y k j N 1 p 2 + 2 λ k j μ M 1 M p i = 1 N 1 β k j i ,

which with λ k 0 and (2.1) implies that x k j T N 1 y k j N 2 0 as j . So, x k T N 1 y k N 2 0 as k .

Similarly, we obtain that x k T N 2 y k N 3 0 , …, x k T 1 y k 0 0 as k .

Further, we prove that x k T i x k 0 as k for i = 1 , , N . First, note that y k 0 x k = λ k F ( x k ) 0 as k because λ k 0 and F ( x k ) M 1 , and y k i x k = β k i x k T i y k i 1 0 because x k T i y k i 1 0 , for i = 1 , , N . Now, from
x k T i x k x k T i y k i 1 + T i y k i 1 T i x k x k T i y k i 1 + y k i 1 x k

and x k T i y k i 1 , y k i 1 x k 0 , it follows that x k T i x k 0 for i = 1 , , N .

Further, we have
lim sup k F ( p ) , p x k 0 .
(2.2)
Indeed, let { x k j } be a subsequence of { x k } that converges weakly to p ˜ such that
lim sup k F ( p ) , p x k = lim j F ( p ) , p x k j .

Then x k j T i x k j 0 . So, by Lemma 2.3, p ˜ C . Therefore, from (1.1) it implies (2.2).

Finally, we estimate the value
x k + 1 p 2 ( 1 β k N ) x k p 2 + β k N T N y k N 1 T N p 2 ( 1 β k N ) x k p 2 + β k N y k N 1 T N p 2 ( 1 β k N ) x k p 2 + β k N [ ( 1 β k N 1 ) x k p 2 + β k N 1 T N 1 y k N 2 T N 1 p 2 ] ( 1 β k N β k N 1 ) x k p 2 + β k N β k N 1 y k N 1 T N p 2 ( 1 β k N β k N 1 ) x k p 2 + β k N β k N 1 [ ( 1 β k N 2 ) x k p 2 + β k N 2 T N 2 y k N 3 T N 2 p 2 ] ( 1 i = 1 N β k i ) x k p 2 + i = 1 N β k i y k 0 p 2 .
On the other hand, since
y k 0 p 2 = ( I λ k μ F ) x k p 2 = ( I λ k μ F ) ( x k p ) λ k μ F ( p ) 2 ( 1 λ k τ ) x k p 2 2 λ k μ F ( x k ) , y k 0 p ,
we have
x k + 1 p 2 ( 1 i = 1 N β k i ) x k p 2 + i = 1 N β k i ( 1 λ k τ ) x k p 2 + [ 2 λ k μ F ( p ) , p x k + 2 λ k μ F ( p ) , x k y k 0 ] i = 1 N β k i ( 1 λ k τ i = 1 N β k i ) x k p 2 + λ k τ i = 1 N β k i [ 2 μ τ F ( p ) , p x k + 2 μ τ F ( p ) , x k y k 0 ] .
Using Lemma 2.5 with
a k = x k p , b k = λ k τ i = 1 N β k i , c k = 2 μ τ F ( p ) , p x k + 2 μ τ F ( p ) , x k y k 0 ,

x k y k 0 0 and (2.2), we have that x k p 0 . This completes the proof. □

3 Application

Recall that a mapping S : H H is called γ-strictly pseudocontractive if there exists a constant γ [ 0 , 1 ) such that
S x S y 2 x y 2 + γ ( I S ) x ( I S ) y 2 , x , y H .

It is well known [9] that a mapping T : H H by T x = α x + ( 1 α ) S x with a fixed α [ γ , 1 ) for all x H is a nonexpansive mapping and Fix ( T ) = Fix ( S ) .

Using this fact, we can extend our result to the case C = i = 1 N Fix ( S i ) , where S i is γ i -strictly pseudocontractive as follows.

Let α i [ γ i , 1 ) be fixed numbers. Then C = i = 1 N Fix ( T ˜ i ) with
T ˜ i y = α i y + ( 1 α i ) S i y ,
(3.1)

a nonexpansive mapping, for each i = 1 , , N . So, we have the following result.

Theorem 3.1 Let H be a real Hilbert space, and let F : H H be an L-Lipschitzian and η-strongly monotone mapping for some constants L , η > 0 . Let { S i } i = 1 N be N γ i -strictly pseudocontractive self-maps of H such that
C = i = 1 N Fix ( S i ) .

Let α i [ γ i , 1 ) , μ ( 0 , 2 η / L 2 ) . Assume that { λ k } , { β k i } ( 0 , 1 ) satisfy (1.10). Then the sequence { x k } defined by (1.9) with T i replaced by T ˜ i of (3.1) converges strongly to the unique element p of (1.1).

4 Numerical example

Consider the following optimization problem: find an element
p C : φ ( p ) = min C φ ( x ) ,
where φ ( x ) = x 2 / 2 , x = ( x 1 , x 2 ) E 2 , Euclid space, and C = C 1 C 2 , defined by
C 1 = { ( x 1 , x 2 ) E 2 : x 1 2 x 2 + 1 0 } , C 2 = { ( x 1 , x 2 ) E 2 : 4 x 1 x 2 3 0 } .
Clearly, the above problem possesses a unique solution p = ( 1 ; 1 ) and F, the Fréchet derivative of φ, is 1-Lipschitz continuous and ( 1 / 2 ) -strongly monotone. Starting with the point x 0 = ( x 1 0 ; x 2 0 ) = ( 4 ; 15 ) , μ = 1 / 10 ( 0 ; 2 η / L 2 ) and λ k = ( k + 1 ) 1 / 2 , we obtained the result in Table 1.
Table 1

Iterations of scheme ( 1.9 ), where starting point x 0 = ( 4 ; 15 )

k th

x k by algorithm (1.9)

0

(4.00000;15.00000)

10

(4.10755;13.434117)

20

(3.89572; 12.58288)

1,000

(1.52757; 3.110295)

1,500

(1.28179; 2.12718)

2,000

(1.13269; 1.53076)

2,500

(1.03367; 1.13468)

3,000

(0.99933; 0.99738)

3,500

(0.99938; 0.99517)

4,000

(0.99942; 0.99768)

4,500

(0.99945; 0.99781)

5,000

(0.99948; 0.99792)

Declarations

Acknowledgements

This work was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2012R1A1A2042138).

Authors’ Affiliations

(1)
Department of Mathematics Education, Kyungnam University
(2)
Vietnamese Academy of Science and Technology, Institute of Information Technology

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© Kim and Buong; licensee Springer. 2013

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