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Hierarchical convergence of an implicit double-net algorithm for nonexpansive semigroups and variational inequality problems Fixed Point Theory and Applications 2011, 2011:101 doi:10.118

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Hierarchical convergence of an implicit double-net algorithm for nonexpansive

semigroups and variational inequality problems

Fixed Point Theory and Applications 2011, 2011:101 doi:10.1186/1687-1812-2011-101

Yonghong Yao (yaoyonghong@yahoo.cn) Yeol Je Cho (yjcho@gsnu.ac.kr) Yeong-Cheng Liou (simplex_liou@hotmail.com)

ISSN 1687-1812

Article type Research

Submission date 3 November 2010

Acceptance date 20 December 2011

Publication date 20 December 2011

Article URL http://www.fixedpointtheoryandapplications.com/content/2011/1/101

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Fixed Point Theory and

Applications

Hierarchical convergence of an implicit

double-net algorithm for nonexpansive

semigroups and variational inequality problems

Yonghong Yao1, Yeol Je Cho∗2 and Yeong-Cheng Liou3

1Department of Mathematics, Tianjin Polytechnic University,

Tianjin 300160, People’s Republic of China

2Department of Mathematics Education and the RINS,

Gyeongsang National University, Chinju 660-701, Republic of Korea

3Department of Information Management, Cheng Shiu University,

Kaohsiung 833, Taiwan

∗Corresponding author: yjcho@gsnu.ac.kr

E-mail addresses:

YY:yaoyonghong@yahoo.cn Y-CL:simplex liou@hotmail.com

Abstract

In this paper, we show the hierarchical convergence of the following implicit

double-net algorithm:

x s,t = s[tf (x s,t ) + (1 − t)(x s,t − µAx s,t )] + (1 − s)1

λ s

Z λ s

0

T (ν)x s,t dν, ∀s, t ∈ (0, 1),

where f is a ρ-contraction on a real Hilbert space H, A : H → H is an α-inverse strongly monotone mapping and S = {T (s)} s≥0 : H → H is a nonexpansive

semi-group with the common fixed points set F ix(S) 6= ∅, where F ix(S) denotes the set

of fixed points of the mapping S, and, for each fixed t ∈ (0, 1), the net {x s,t }

con-verges in norm as s → 0 to a common fixed point x t ∈ F ix(S) of {T (s)} s≥0and, as

t → 0, the net {x t } converges in norm to the solution x ∗ of the following variational

inequality:











x ∗ ∈ F ix(S);

hAx ∗ , x − x ∗ i ≥ 0, ∀x ∈ F ix(S).

Keywords: fixed point; variational inequality; double-net algorithm; hierarchical convergence; Hilbert space.

MSC(2000): 49J40; 47J20; 47H09; 65J15.

In nonlinear analysis, a common approach to solving a problem with multiple solutions is

to replace it by a family of perturbed problems admitting a unique solution and to obtain

a particular solution as the limit of these perturbed solutions when the perturbation vanishes

In this paper, we introduce a more general approach which consists in finding a particular part of the solution set of a given fixed point problem, i.e., fixed points which

solve a variational inequality More precisely, the goal of this paper is to present a method

for finding hierarchically a fixed point of a nonexpansive semigroup S = {T (s)} s≥0 with

respect to another monotone operator A, namely,

Find x ∗ ∈ F ix(S) such that

This is an interesting topic due to the fact that it is closely related to convex

pro-This paper is devoted to solve the problem (1.1) For this purpose, we propose a

double-net algorithm which generates a net {x s,t } and prove that the net {x s,t }

hierar-chically converges to the solution of the problem (1.1), that is, for each fixed t ∈ (0, 1), the net {x s,t } converges in norm as s → 0 to a common fixed point x t ∈ F ix(S) of the

nonexpansive semigroup {T (s)} s≥0 and, as t → 0, the net {x t } converges in norm to the

unique solution x ∗ of the problem (1.1)

Let H be a real Hilbert space with inner product h·, ·i and norm k · k, respectively Recall

a mapping f : H → H is called a contraction if there exists ρ ∈ [0, 1) such that

kf (x) − f (y)k ≤ ρkx − yk, ∀x, y ∈ H.

A mapping T : C → C is said to be nonexpansive if

kT x − T yk ≤ kx − yk, ∀x, y ∈ H.

gramming problems For the related works, refer to [1–19]

Denote the set of fixed points of the mapping T by F ix(T ).

Recall also that a family S := {T (s)} s≥0 of mappings of H into itself is called a

nonexpansive semigroup if it satisfies the following conditions:

(S1) T (0)x = x for all x ∈ H;

(S2) T (s + t) = T (s)T (t) for all s, t ≥ 0;

(S3) kT (s)x − T (s)yk ≤ kx − yk for all x, y ∈ H and s ≥ 0;

(S4) for all x ∈ H, s → T (s)x is continuous.

We denote by F ix(T (s)) the set of fixed points of T (s) and by F ix(S) the set of all common fixed points of S, i.e., F ix(S) = Ts≥0 F ix(T (s)) It is known that F ix(S) is

closed and convex ([20], Lemma 1)

A mapping A of H into itself is said to be monotone if

hAu − Av, u − vi ≥ 0, ∀u, v ∈ H,

and A : C → H is said to be α-inverse strongly monotone if there exists a positive real number α such that

hAu − Av, u − vi ≥ αkAu − Avk2, ∀u, v ∈ H.

It is obvious that any α-inverse strongly monotone mapping A is monotone and

1

α-Lipschitz continuous

Now, we introduce some lemmas for our main results in this paper

Lemma 2.1 [21] Let H be a real Hilbert space Let the mapping A : H → H be α-inverse

strongly monotone and µ > 0 be a constant Then, we have

k(I − µA)x − (I − µA)yk2≤ kx − yk2+ µ(µ − 2α)kAx − Ayk2, ∀x, y ∈ H.

In particular, if 0 ≤ µ ≤ 2α, then I − µA is nonexpansive.

Lemma 2.2 [22] Let C be a nonempty bounded closed convex subset of a Hilbert space

H and {T (s)} s≥0 be a nonexpansive semigroup on C Then, for all h ≥ 0,

lim

t→∞sup

x∈C

°

°

°1

t

Z t

0

T (s)xds − T (h)1

t

Z t

0

T (s)xds

°

°

° = 0.

Lemma 2.3 [23] (Demiclosedness Principle for Nonexpansive Mappings) Let C be a

nonempty closed convex subset of a real Hilbert space H and T : C → C be a nonexpansive mapping with F ix(T ) 6= ∅ If {x n } is a sequence in C converging weakly to a point x ∈ C and {(I − T )x n } converges strongly to a point y ∈ C, then (I − T )x = y In particular,

if y = 0, then x ∈ F ix(T ).

Lemma 2.4 Let H be a real Hilbert space Let f : H → H be a ρ-contraction with

coefficient ρ ∈ [0, 1) and A : H → H be an α-inverse strongly monotone mapping Let

µ ∈ (0, 2α) and t ∈ (0, 1) Then, the variational inequality











x ∗ ∈ F ix(S);

htf (z) + (1 − t)(I − µA)z − z, x ∗ − zi ≥ 0, ∀z ∈ F ix(S),

(2.1)

is equivalent to its dual variational inequality











x ∗ ∈ F ix(S);

htf (x ∗ ) + (1 − t)(I − µA)x ∗ − x ∗ , x ∗ − zi ≥ 0, ∀z ∈ F ix(S).

(2.2)

Proof Assume that x ∗ ∈ F ix(S) solves the problem (2.1) For all y ∈ F ix(S), set

x = x ∗ + s(y − x ∗ ) ∈ F ix(S), ∀s ∈ (0, 1).

We note that

htf (x) + (1 − t)(I − µA)x − x, x ∗ − xi ≥ 0.

Hence, we have

htf (x ∗ + s(y − x ∗ )) + (1 − t)(I − µA)(x ∗ + s(y − x ∗ )) − x ∗ − s(y − x ∗ ), s(x ∗ − y)i ≥ 0,

which implies that

htf (x ∗ + s(y − x ∗ )) + (1 − t)(I − µA)(x ∗ + s(y − x ∗ )) − x ∗ − s(y − x ∗ ), x ∗ − yi ≥ 0.

Letting s → 0, we have

htf (x ∗ ) + (1 − t)(I − µA)(x ∗ ) − x ∗ , x ∗ − yi ≥ 0,

which implies the point x ∗ ∈ F ix(S) is a solution of the problem (2.2).

Conversely, assume that the point x ∗ ∈ F ix(S) solves the problem (2.2) Then, we

have

htf (x ∗ ) + (1 − t)(I − µA)x ∗ − x ∗ , x ∗ − zi ≥ 0.

Noting that I − f and A are monotone, we have

h(I − f )z − (I − f )x ∗ , z − x ∗ i ≥ 0

and

hAz − Ax ∗ , z − x ∗ i ≥ 0.

Thus, it follows that

th(I − f )z − (I − f )x ∗ , z − x ∗ i + (1 − t)µhAz − Ax ∗ , z − x ∗ i ≥ 0,

which implies that

htf (z) + (1 − t)(I − µA)z − z, x ∗ − zi

≥ htf (x ∗ ) + (1 − t)(I − µA)x ∗ − x ∗ , x ∗ − zi

≥ 0.

This implies that the point x ∗ ∈ F ix(S) solves the problem (2.1) This completes the

proof

In this section, we first introduce our double-net algorithm and then prove a strong convergence theorem for this algorithm

Throughout, we assume that

(C1) H is a real Hilbert space;

(C2) f : H → H is a ρ-contraction with coefficient ρ ∈ [0, 1), A : H → H is an

α-inverse strongly monotone mapping, and S = {T (s)} s≥0 : H → H is a nonexpansive semigroup with F ix(S) 6= ∅;

(C3) the solution set Ω of the problem (1.1) is nonempty;

(C4) µ ∈ (0, 2α) is a constant, and {λ s } 0<s<1 is a continuous net of positive real numbers satisfying lims→0 λ s = +∞.

For any s, t ∈ (0, 1), we define the following mapping

x 7→ W s,t x := s[tf (x) + (1 − t)(x − µAx)] + (1 − s) 1

λ s

Z λ s

0

T (ν)xdν.

We note that the mapping W s,t is a contraction In fact, we have

kW s,t x − W s,t yk =

°

°

°s[tf (x) + (1 − t)(x − µAx)] + (1 − s)1

λ s

Z λ s

0

T (ν)xdν

−s[tf (y) + (1 − t)(y − µAy)] − (1 − s) 1

λ s

Z λ s

0

T (ν)ydν

°

°

°

≤ st

°

°f (x) − f (y)k + s(1 − t)k(x − µAx) − (y − µAy)k

+(1 − s)k 1

λ s

Z λ s

0

(T (ν)x − T (ν)y)dν

°

°

≤ stρkx − yk + s(1 − t)kx − yk + (1 − s)kx − yk

= [1 − (1 − ρ)st]kx − yk,

which implies that W s,t is a contraction Hence, by Banach’s Contraction Principle, W s,t has a unique fixed point, which is denoted x s,t ∈ H, that is, x s,t is the unique solution

in H of the fixed point equation

x s,t = s[tf (x s,t ) + (1 − t)(x s,t − µAx s,t)]

+ (1 − s)1

λ s

Z λ s

0

T (ν)x s,t dν, ∀s, t ∈ (0, 1).

(3.1)

Now, we give some lemmas for our main result

Lemma 3.1 For each fixed t ∈ (0, 1), the net {x s,t } defined by (3.1) is bounded.

Proof Taking any z ∈ F ix(S), since I − µA is nonexpansive (by Lemma 2.1), it follows

from (3.1) that

kx s,t − zk

=

°

°s[tf (x s,t ) + (1 − t)(I − µA)x s,t ] + (1 − s) 1

λ s

Z λ s

0

T (ν)x s,t dν − z

°

°

≤ sktf (x s,t ) + (1 − t)(I − µA)x s,t − zk + (1 − s)

°

°

° 1

λ s

Z λ s

0

T (ν)x s,t dν − z

°

°

°

≤ s

h

tkf (x s,t ) − f (z)k + tkf (z) − zk + (1 − t)k(I − µA)x s,t − (I − µA)zk

+(1 − t)k(I − µA)z − zk

i

+ (1 − s)kx s,t − zk

≤ s[tρkx s,t − zk + tkf (z) − zk + (1 − t)kx s,t − zk + (1 − t)µkAzk]

+(1 − s)kx s,t − zk

= [1 − (1 − ρ)st]kx s,t − zk + stkf (z) − zk + s(1 − t)µkAzk.

This implies that

(1 − ρ)t (tkf (z) − zk + (1 − t)µkAzk)

(1 − ρ)t max{kf (z) − zk, µkAzk}.

Thus, it follows that, for each fixed t ∈ (0, 1), {x s,t } is bounded and so are the nets {f (x s,t )} and {(I − µA)x s,t } This completes the proof.

Lemma 3.2 x s,t → x t ∈ F ix(S) as s → 0.

Proof For each fixed t ∈ (0, 1), we set R t := 1

(1−ρ)t max{kf (z) − zk, µkAzk} It is clear that, for each fixed t ∈ (0, 1), {x s,t } ⊂ B(p, R t ), where B(p, R t) denotes a closed ball

with the center p and radius R t Notice that

°

° 1

λ s

Z λ s

0

T (ν)x s,t dν − p

°

° ≤ kx s,t − pk ≤ R t

Moreover, we observe that if x ∈ B(p, R t), then

kT (s)x − pk ≤ kT (s)x − T (s)pk ≤ kx − pk ≤ R t ,

that is, B(p, R t ) is T (s)-invariant for all s ∈ (0, 1) From (3.1), it follows that

kT (τ )x s,t − x s,t k ≤

°

°

°T (τ )x s,t − T (τ ) 1

λ s

Z λ s

0

T (ν)x s,t dν

°

°

° +

°

°T (τ )1

λ s

Z λ s

0

T (ν)x s,t dν − 1

λ s

Z λ s

0

T (ν)x s,t dν

°

° +

°

°

° 1

λ s

Z λ s

0

T (ν)x s,t dν − x s,t

°

°

°

≤

°

°

°T (τ )1

λ s

Z λ s

0

T (ν)x s,t dν − 1

λ s

Z λ s

0

T (ν)x s,t dν

°

°

° +2

°

°x s,t − 1

λ s

Z λ s

0

T (ν)x s,t dν

°

°

≤ 2s

°

°

°tf (x s,t ) + (1 − t)(x s,t − µAx s,t ) − 1

λ s

Z λ s

0

T (ν)x s,t dν

°

°

° +

°

°

°T (τ )1

λ s

Z λ s

0

T (ν)x s,t dν − 1

λ s

Z λ s

0

T (ν)x s,t dν

°

°

°.

By Lemma 2.2, for all 0 ≤ τ < ∞ and fixed t ∈ (0, 1), we deduce

lim

Next, we show that, for each fixed t ∈ (0, 1), the net {x s,t } is relatively norm-compact

as s → 0 In fact, from Lemma 2.1, it follows that

kx s,t − µAx s,t − (z − µAz)k2 ≤ kx s,t − zk2+ µ(µ − 2α)kAx s,t − Azk2. (3.3)

By (3.1), we have

kx s,t − zk2

= sthf (x s,t ) − f (z), x s,t − zi + sthf (z) − z, x s,t − zi

+s(1 − t)h(I − µA)x s,t − (I − µA)z, x s,t − zi

+s(1 − t)h(I − µA)z − z, x s,t − zi

+(1 − s)D 1

λ s

Z λ s

0

T (ν)x s,t dν − z, x s,t − z

E

≤ stkf (x s,t ) − f (z)kkx s,t − zk + sthf (z) − z, x s,t − zi

+s(1 − t)k(I − µA)x s,t − (I − µA)zkkx s,t − zk − s(1 − t)µhAz, x s,t − zi

+(1 − s)

°

°

°1

λ s

Z λ s

0

T (ν)x s,t dν − zkkx s,t − z

°

°

°

≤ stρkx s,t − zk2+ sthf (z) − z, x s,t − zi − s(1 − t)µhAz, x s,t − zi

+s(1 − t)k(I − µA)x s,t − (I − µA)zkkx s,t − zk + (1 − s)kx s,t − zk2

≤ stρkx s,t − zk2+ sthf (z) − z, x s,t − zi − s(1 − t)µhAz, x s,t − zi

+s(1 − t)

2 (k(I − µA)x s,t − (I − µA)zk

2+ kx s,t − zk2) + (1 − s)kx s,t − zk2.

This together with (3.3) imply that

kx s,t − zk2

≤ stρkx s,t − zk2+ sthf (z) − z, x s,t − zi − s(1 − t)µhAz, x s,t − zi

+s(1 − t)

2 (kx s,t − zk

2+ µ(µ − 2α)kAx s,t − Azk2+ kx s,t − zk2) + (1 − s)kx s,t − zk2

≤ [1 − (1 − ρ)st]kx s,t − zk2+ sthf (z) − z, x s,t − zi

−s(1 − t)µhAz, x s,t − zi,

which implies that

kx s,t − zk2

(1 − ρ)t htf (z) + (1 − t)(I − µA)z − z, x s,t − zi, ∀z ∈ F ix(S).

(3.4)

Assume that {s n } ⊂ (0, 1) is such that s n → 0 as n → ∞ By (3.4), we obtain

immedi-ately that

kx s n ,t − zk2

(1 − ρ)t htf (z) + (1 − t)(I − µA)z − z, x s n ,t − zi, ∀z ∈ F ix(S).

(3.5)

Since {x s n ,t } is bounded, without loss of generality, we may assume that, as s n → 0, {x s n ,t } converges weakly to a point x t From (3.2) and Lemma 2.3, we get x t ∈ F ix(S).

Further, if we substitute x t for z in (3.5), then it follows that

kx s n ,t − x t k2 ≤ 1

(1 − ρ)t htf (x t ) + (1 − t)(I − µA)x t − x t , x s n ,t − x t i.

Therefore, the weak convergence of {x s n ,t } to x t actually implies that x s n ,t → x tstrongly

This has proved the relative norm-compactness of the net {x s,t } as s → 0.

Now, if we take the limit as n → ∞ in (3.5), we have

kx t − zk2

(1 − ρ)t htf (z) + (1 − t)(I − µA)z − z, x t − zi, ∀z ∈ F ix(S).

In particular, x t solves the following variational inequality:













x t ∈ F ix(S);

htf (z) + (1 − t)(I − µA)z − z, x t − zi ≥ 0, ∀z ∈ F ix(S),

or the equivalent dual variational inequality (see Lemma 2.4):















x t ∈ F ix(S);

htf (x t ) + (1 − t)(I − µA)x t − x t , x t − zi ≥ 0, ∀z ∈ F ix(S).

(3.6)

Notice that (3.6) is equivalent to the fact that x t = P F ix(S) (tf + (1 − t)(I − µA))x t,

that is, x t is the unique element in F ix(S) of the contraction P F ix(S) (tf +(1−t)(I −µA)) Clearly, it is sufficient to conclude that the entire net {x s,t } converges in norm to x t ∈

F ix(S) as s → 0 This completes the proof.

Lemma 3.3 The net {x t } is bounded.

Proof In (3.6), if we take any y ∈ Ω, then we have

htf (x t ) + (1 − t)(I − µA)x t − x t , x t − yi ≥ 0. (3.7)

By virtue of the monotonicity of A and the fact that y ∈ Ω, we have

h(I − µA)x t − x t , x t − yi ≤ h(I − µA)y − y, x t − yi ≤ 0. (3.8) Thus, it follows from (3.7) and (3.8) that

and hence

kx t − yk2 ≤ hx t − y, x t − yi + hf (x t ) − x t , x t − yi

= hf (x t ) − f (y), x t − yi + hf (y) − y, x t − yi

≤ ρkx t − yk2+ hf (y) − y, x t − yi.

Therefore, we have

kx t − yk2 ≤ 1

1 − ρ hf (y) − y, x t − yi, ∀y ∈ Ω. (3.10)

In particular,

kx t − yk ≤ 1

1 − ρ kf (y) − yk, ∀t ∈ (0, 1), which implies that {x t } is bounded This completes the proof.

Lemma 3.4 If the net {x t } converges in norm to a point x ∗ ∈ Ω, then the point solves the variational inequality

Proof First, we note that the solution of the variational inequality (3.11) is unique.

Next, we prove that ω w (x t ) ⊂ Ω, that is, if (t n ) is a null sequence in (0, 1) such that

x t n → x 0 weakly as n → ∞, then x 0 ∈ Ω To see this, we use (3.6) to get

hµAx t , z − x t i ≥ t

1 − t h(I − f )x t , x t − zi, ∀z ∈ F ix(S).

However, since A is monotone, we have

hAz, z − x t i ≥ hAx t , z − x t i.

Combining the last two relations yields that

hµAz, z − x t i ≥ t

1 − t h(I − f )x t , x t − zi, ∀z ∈ F ix(S). (3.12) Letting t = t n → 0 as n → ∞ in (3.12), we get

hAz, z − x 0 i ≥ 0, ∀z ∈ F ix(S),