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Iterative algorithms for quasi-variational inclusions and fixed point problems of pseudocontractions

Abstract

In this paper, quasi-variational inclusions and fixed point problems of pseudocontractions are considered. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

MSC:49J40, 47J20, 47H09, 65J15.

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Let A:CH be a single-valued nonlinear mapping and B:H 2 H be a multi-valued mapping. The ‘so called’ quasi-variational inclusion problem is to find an u 2 H such that

0Au+Bu.
(1.1)

The set of solutions of (1.1) is denoted by ( A + B ) 1 (0). A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see for instance [14]. For related work, see [510]. The problem (1.1) includes many problems as special cases.

  1. (1)

    If B=ϕ:H 2 H , where ϕ:HR+ is a proper convex lower semi-continuous function and ∂ϕ is the subdifferential of ϕ, then the variational inclusion problem (1.1) is equivalent to finding uH such that

    Au,yu+ϕ(y)ϕ(u)0,yH,

    which is called the mixed quasi-variational inequality (see [11]).

  2. (2)

    If B= δ C , where C is a nonempty closed convex subset of H and δ C :H[0,] is the indicator function of C, i.e.,

    δ C ={ 0 , x C , + , x C ,

then the variational inclusion problem (1.1) is equivalent to finding uC such that

Au,vu0,vC.

This problem is called the Hartman-Stampacchia variational inequality (see [12]).

Let T:CC be a nonlinear mapping. The iterative scheme of Mann’s type for approximating fixed points of T is the following: x 0 C and

x n + 1 = α n x n +(1 α n )T x n ,

for all n1, where { α n } is a sequence in [0,1]; see [13]. For two nonlinear mappings S and T, Takahashi and Tamura [14] considered the following iteration procedure: x 0 C and

x n + 1 = α n x n +(1 α n )S ( β n x n + ( 1 β n ) T x n ) ,

for all n1, where { α n } and { β n } are two sequences in [0,1]. Algorithms for finding the fixed points of nonlinear mappings or for finding the zero points of maximal monotone operators have been studied by many authors. The reader can refer to [1519]. Especially, Takahashi et al. [20] recently gave the following convergence result.

Theorem 1.1 Let C be a closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let J λ B = ( I + λ B ) 1 be the resolvent of B for λ>0 and let T be a nonexpansive mapping of C into itself, such that F(T) ( A + B ) 1 0. Let x 1 =xC and let { x n }C be a sequence generated by

x n + 1 = β n x n +(1 β n )T ( α n x + ( 1 α n ) J λ n B ( x n λ n A x n ) ) ,

for all n0, where { λ n }(0,2α), { α n }(0,1) and { β n }(0,1) satisfy

0 < a λ n b < 2 α , 0 < c β n d < 1 , lim n ( λ n + 1 λ n ) = 0 , lim n α n = 0 and n α n = .

Then { x n } converges strongly to a point of F(T) ( A + B ) 1 0.

Recently, Zhang et al. [21] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [22] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping.

Motivated and inspired by the works in this field, the purpose of this paper is to consider the quasi-variational inclusions and fixed point problems of pseudocontractions. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

2 Notations and lemmas

Let H be a real Hilbert space with inner product , and norm , respectively. Let C be a nonempty closed convex subset of H. It is well known that in a real Hilbert space H, the following equality holds:

t x + ( 1 t ) y 2 =t x 2 +(1t) y 2 t(1t) x y 2
(2.1)

for all x,yH and t[0,1].

Recall that a mapping T:CC is called

(D1) L-Lipschitzian there exists L>0 such that TxTyLxy for all x,yC; in the case of L=1, T is said to be nonexpansive;

(D2) Firmly nonexpansive T x T y 2 x y 2 ( I T ) x ( I T ) y 2 T x T y 2 TxTy,xy for all x,yC;

(D3) Pseudocontractive TxTy,xy x y 2 T x T y 2 x y 2 + ( I T ) x ( I T ) y 2 for all x,yC;

(D4) Strongly monotone there exists a positive constant γ ˜ such that TxTy,xy γ ˜ xy for all x,yC;

(D5) Inverse strongly monotone TxTy,xyα T x T y 2 for some α>0 and for all x,yC.

Let B be a mapping of H into 2 H . The effective domain of B is denoted by dom(B), that is, dom(B)={xH:Bx}. A multi-valued mapping B is said to be a monotone operator on H iff

xy,uv0

for all x,ydom(B), uBx, and vBy. A monotone operator B on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone operator on H. Let B be a maximal monotone operator on H and let B 1 0={xH:0Bx}.

For a maximal monotone operator B on H and λ>0, we may define a single-valued operator J λ B = ( I + λ B ) 1 :Hdom(B), which is called the resolvent of B for λ. It is known that the resolvent J λ B is firmly nonexpansive, i.e.,

J λ B x J λ B y 2 J λ B x J λ B y , x y

for all x,yC and B 1 0=Fix( J λ B ) for all λ>0.

Usually, the convergence of fixed point algorithms requires some additional smoothness properties of the mapping T such as demi-closedness.

Recall that a mapping T is said to be demiclosed if, for any sequence { x n } which weakly converges to x ˜ , and if the sequence {T( x n )} strongly converges to z, then T( x ˜ )=z. For the pseudocontractions, the following demiclosed principle is well known.

Lemma 2.1 ([23])

Let H be a real Hilbert space, C a closed convex subset of H. Let U:CC be a continuous pseudo-contractive mapping. Then

  1. (i)

    Fix(U) is a closed convex subset of C,

  2. (ii)

    (IU) is demiclosed at zero.

Lemma 2.2 ([24])

Let { r n } be a sequence of real numbers. Assume { r n } does not decrease at infinity, that is, there exists at least a subsequence { r n k } of { r n } such that r n k r n k + 1 for all k0. For every nN, define an integer sequence {τ(n)} as

τ(n)=max{in: r n i < r n i + 1 }.

Then τ(n) as n, and for all nN

max{ r τ ( n ) , r n } r τ ( n ) + 1 .

Lemma 2.3 ([25])

Assume { a n } is a sequence of nonnegative real numbers such that

a n + 1 (1 γ n ) a n + δ n γ n ,

where { γ n } is a sequence in (0,1) and { δ n } is a sequence such that

  1. (1)

    n = 1 γ n =;

  2. (2)

    lim sup n δ n 0 or n = 1 | δ n γ n |<.

Then lim n a n =0.

In the sequel we shall use the following notations:

  1. 1.

    ω w ( u n )={x: u n j x weakly} denote the weak ω-limit set of { u n };

  2. 2.

    u n x stands for the weak convergence of { u n } to x;

  3. 3.

    u n x stands for the strong convergence of { u n } to x;

  4. 4.

    Fix(T) stands for the set of fixed points of T.

3 Main results

In this section, we consider a strong convergence theorem for quasi-variational inclusions and fixed point problems of pseudocontractive mappings in a Hilbert space.

Algorithm 3.1 Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let J λ B = ( I + λ B ) 1 be the resolvent of B for λ. Let F:CH be an L 1 -Lipschitzian and ς strongly monotone mapping and f:CC be a ρ-contraction such that ρ<max{1,ς/2}. Let T:CC be an L 2 (>1)-Lipschitzian pseudocontraction. For x 0 C, define a sequence { x n } as follows:

{ z n = J λ B ( I λ A ) x n , y n = ν z n + ( 1 ν ) T ( ( 1 ζ ) z n + ζ T z n ) , x n + 1 = α n x n + ( 1 α n ) ( β n f ( x n ) + ( I β n F ) y n ) ,
(3.1)

for all nN, where λ, ν and ζ are three constants, { α n } and { β n } are two sequences in [0,1].

Now, we demonstrate the convergence analysis of the algorithm (3.1).

Theorem 3.2 Suppose Γ:=Fix(T) ( A + B ) 1 (0). Assume the following conditions are satisfied:

(C1) α n [a,b](0,1);

(C2) lim n β n =0 and n = 1 β n =;

(C3) λ(0,2α) and 0<1νζ< 1 1 + L 2 2 + 1 .

Then the sequence { x n } defined by (3.1) converges strongly to u= P Γ (IF+f)u.

Proof Let x Fix(T) ( A + B ) 1 (0). Then, we get x = J λ B (IλA) x =T x . From (3.1), we have

z n x 2 = J λ B ( I λ A ) x n J λ B ( I λ A ) x 2 x n x λ ( A x n A x ) 2 = x n x 2 2 λ A x n A x , x n x + λ 2 A x n A x 2 x n x 2 2 λ α A x n A x 2 + λ 2 A x n A x 2 = x n x 2 λ ( 2 α λ ) A x n A x 2 x n x 2 .
(3.2)

It follows that

z n x x n x .
(3.3)

Since x Fix(T), we have from (D3) that

T x x 2 x x 2 + T x x 2 ,
(3.4)

for all xC.

Thus,

T ( ( 1 ζ ) I + ζ T ) z n x 2 ( 1 ζ ) ( z n x ) + ζ ( T z n x ) 2 + ( ( 1 ζ ) I + ζ T ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 .
(3.5)

By (3.4), (3.5), and (2.1), we obtain

T ( ( 1 ζ ) I + ζ T ) z n x 2 ( 1 ζ ) ( z n x ) + ζ ( T z n x ) 2 + ( ( 1 ζ ) I + ζ T ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 = ( 1 ζ ) ( z n T ( ( 1 ζ ) I + ζ T ) z n ) + ζ ( T z n T ( ( 1 ζ ) I + ζ T ) z n ) 2 + ( 1 ζ ) ( z n x ) + ζ ( T z n x ) 2 = ( 1 ζ ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 + ζ T z n T ( ( 1 ζ ) I + ζ T ) z n 2 ζ ( 1 ζ ) z n T z n 2 + ( 1 ζ ) z n x 2 + ζ T z n x 2 ζ ( 1 ζ ) z n T z n 2 ( 1 ζ ) z n x 2 + ζ ( z n x 2 + z n T z n 2 ) 2 ζ ( 1 ζ ) z n T z n 2 + ( 1 ζ ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 + ζ T z n T ( ( 1 ζ ) I + ζ T ) z n 2 .

Noting that T is L 2 -Lipschitzian and z n ((1ζ)I+ζT) z n =ζ( z n T z n ), we have

T ( ( 1 ζ ) I + ζ T ) z n x 2 ( 1 ζ ) z n x 2 + ζ ( z n x 2 + z n T z n 2 ) 2 ζ ( 1 ζ ) z n T z n 2 + ( 1 ζ ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 + ζ 3 L 2 2 z n T z n 2 = z n x 2 + ( 1 ζ ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 ζ ( 1 2 ζ ζ 2 L 2 2 ) z n T z n 2 .
(3.6)

Since ζ< 1 1 + L 2 2 + 1 , we have 12ζ ζ 2 L 2 2 >0. From (3.6), we can deduce

T ( ( 1 ζ ) I + ζ T ) z n x 2 z n x 2 +(1ζ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 .
(3.7)

Hence,

y n x 2 = ν z n + ( 1 ν ) T ( ( 1 ζ ) I + ζ T ) z n x 2 = ν ( z n x ) + ( 1 ν ) ( T ( ( 1 ζ ) I + ζ T ) z n x ) 2 = ν z n x 2 + ( 1 ν ) T ( ( 1 ζ ) I + ζ T ) z n x 2 ν ( 1 ν ) T ( ( 1 ζ ) I + ζ T ) z n z n 2 ν z n x 2 + ( 1 ν ) [ z n x 2 + ( 1 ζ ) z n T ( ( 1 ζ ) I + ζ T ) z n 2 ] ν ( 1 ν ) T ( ( 1 ζ ) I + ζ T ) z n z n 2 = z n x 2 + ( 1 ν ) ( 1 ζ ν ) T ( ( 1 ζ ) I + ζ T ) z n z n 2 .
(3.8)

By (C3) and (3.8), we obtain

y n x z n x .
(3.9)

Let u n = β n f( x n )+(I β n F) y n for all n0. Then, we have

u n x = β n f ( x n ) + ( I β n F ) y n x β n f ( x n ) F x + ( I β n F ) y n ( I β n F ) x β n f ( x n ) f ( x ) + β n f ( x ) F x + ( I β n F ) y n ( I β n F ) x β n ρ x n x + β n f ( x ) F x + ( I β n F ) y n ( I β n F ) x .
(3.10)

Since F is L 1 -Lipschitzian and ς strongly monotone, we have

( I β n F ) y n ( I β n F ) x 2 = ( y n x ) β n ( F y n F x ) 2 = y n x 2 2 β n F y n F x , y n x + β n 2 F y n F x 2 y n x 2 2 β n ς y n x 2 + β n 2 L 1 2 y n x 2 = ( 1 2 β n ς + β n 2 L 1 2 ) y n x 2 .
(3.11)

Noting that L 1 ς and lim n β n =0, without loss of generality, we assume that β n < ς L 1 2 ς / 4 for all n0. Thus, 12 β n ς+ β n 2 L 1 2 ( 1 β n ς 2 ) 2 . So,

( I β n F ) y n ( I β n F ) x ( 1 β n ς 2 ) y n x .
(3.12)

We have from (3.9), (3.10), and (3.12)

u n x β n ρ x n x + β n f ( x ) F x + ( 1 β n ς 2 ) x n x = [ 1 ( ς 2 ρ ) β n ] x n x + β n f ( x ) F x .
(3.13)

From (3.1) and (3.13), we have

x n + 1 x = α n ( x n x ) + ( 1 α n ) ( u n x ) ( 1 α n ) ( [ 1 ( ς 2 ρ ) β n ] x n x + β n f ( x ) F x ) + α n x n x = [ 1 ( ς 2 ρ ) ( 1 α n ) β n ] x n x + ( 1 α n ) β n f ( x ) F x .
(3.14)

By the definition of x n , we have

x n + 1 x n = α n x n + ( 1 α n ) ( β n f ( x n ) + ( I β n F ) y n ) x n = ( 1 α n ) [ β n f ( x n ) β n F y n + y n x n ] .
(3.15)

Hence,

x n + 1 x n , x n x = ( 1 α n ) [ β n f ( x n ) β n F y n + y n x n ] , x n x = ( 1 α n ) β n f ( x n ) , x n x ( 1 α n ) β n F y n , x n x + ( 1 α n ) y n x n , x n x .
(3.16)

Since 2 x n + 1 x n , x n x = x n + 1 x 2 x n x 2 x n + 1 x n 2 and 2 y n x n , x n x = y n x 2 x n x 2 y n x n 2 , it follows from (3.16), (3.3), and (3.9) that

x n + 1 x 2 x n x 2 x n + 1 x n 2 = 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x + ( 1 α n ) [ y n x 2 x n x 2 y n x n 2 ] 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x ( 1 α n ) y n x n 2 .
(3.17)

By (3.15), we obtain

x n + 1 x n 2 ( 1 α n ) 2 [ β n f ( x n ) F y n + y n x n ] 2 = ( 1 α n ) 2 [ β n 2 f ( x n ) F y n 2 + y n x n 2 + 2 β n f ( x n ) F y n y n x n ] .
(3.18)

Combining (3.17) and (3.18) to deduce

x n + 1 x 2 x n x 2 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x ( 1 α n ) y n x n 2 + ( 1 α n ) 2 [ β n 2 f ( x n ) F y n 2 + y n x n 2 + 2 β n f ( x n ) F y n y n x n ] 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x ( 1 α n ) α n y n x n 2 + ( 1 α n ) 2 [ β n 2 f ( x n ) F y n 2 + 2 β n f ( x n ) F y n y n x n ] .

Hence, we obtain

x n + 1 x 2 x n x 2 + ( 1 α n ) α n y n x n 2 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x + ( 1 α n ) 2 [ β n 2 f ( x n ) F y n 2 + 2 β n f ( x n ) F y n y n x n ] .

It follows that, hence, we obtain

( 1 α n ) α n y n x n 2 x n x 2 x n + 1 x 2 + 2 ( 1 α n ) β n f ( x n ) , x n x 2 ( 1 α n ) β n F y n , x n x + ( 1 α n ) 2 [ β n 2 f ( x n ) F y n 2 + 2 β n f ( x n ) F y n y n x n ] .
(3.19)

Next we divide our proof into two possible cases.

Case 1. There exists an integer number m such that x n + 1 x x n x for all nm. In this case, we have lim n x n x exists. Since α n [a,b](0,1) and lim n β n =0, by (3.19), we derive

lim n y n x n =0.
(3.20)

This together with (3.18) implies that

lim n x n + 1 x n =0.
(3.21)

Note that

u n y n = β n f ( x n ) + ( I β n F ) y n y n β n f ( x n ) F y n .

So,

lim n u n y n =0.
(3.22)

By (3.20) and (3.22), we obtain

lim n u n x n =0.
(3.23)

From (3.2) and (3.9), we have

y n x 2 z n x 2 x n x 2 λ(2αλ) A x n A x 2 .

Hence,

λ ( 2 α λ ) A x n A x 2 x n x 2 y n x 2 x n y n ( x n x + y n x ) .

Therefore,

lim n A x n A x =0.
(3.24)

Since J λ B is firmly nonexpansive and A is monotone, we have

z n x 2 = J λ B ( I λ A ) x n J λ B ( I λ A ) x 2 ( I λ A ) x n ( I λ A ) x , z n x = z n x , x n x λ z n x , A x n A x = 1 2 ( z n x 2 + x n x 2 z n x n 2 ) λ x n x , A x n A x λ z n x n , A x n A x 1 2 ( z n x 2 + x n x 2 z n x n 2 ) + λ z n x n A x n A x .

It follows that

z n x 2 x n x 2 z n x n 2 +2λ z n x n A x n A x .
(3.25)

By (3.25) and (3.9), we deduce

y n x 2 z n x 2 x n x 2 z n x n 2 +2λ z n x n A x n A x .

Therefore,

z n x n 2 x n x 2 y n x 2 + 2 λ z n x n A x n A x x n y n ( x n x + y n x ) + 2 λ z n x n A x n A x .
(3.26)

Equations (3.20), (3.24), and (3.26) imply that

lim n z n x n =0.
(3.27)

Notice that Ff is (ςρ) strongly monotone. Thus, the variational inequality of finding yΓ such that (Ff)y,xy0 for all xΓ has a unique solution, denoted by x , that is, x = P Γ (IV+F)( x ). Next, we prove that

lim sup n ( f F ) x , u n x 0.

Since u n is bounded, without loss of generality, we assume that there exists a subsequence { z n i } of { u n } such that u n i x ˜ for some x ˜ H and

lim sup n ( f F ) x , u n x = lim sup i ( f F ) x , u n i x .

Thus, we have that x n i x ˜ and

lim i J λ B ( I λ A ) x n i x n i =0.

Therefore, x ˜ Fix( J λ B (IλA))= ( A + B ) 1 (0).

Next we show that x ˜ Fix(T). First, we show that Fix(T)=Fix(T((1ζ)I+ζT)). As a matter of fact, Fix(T)Fix(T((1ζ)I+ζT)) is obvious. Next, we show that Fix(T((1ζ)I+ζT))Fix(T).

Take any x Fix(T((1ζ)I+ζT)). We have T((1ζ)I+ζT) x = x . Set S=(1ζ)I+ζT. We have TS x = x . Write S x = y . Then, T y = x . Now we show x = y . In fact,

x y = T y S x = T y ( 1 ζ ) x ζ T x = ζ T y T x ζ L 2 y x .

Since, ζ< 1 1 + L 2 2 + 1 < 1 L 2 , we deduce y = x Fix(S)=Fix(T). Thus, x Fix(T). Hence, Fix(T((1ζ)I+ζT))Fix(T). Therefore, Fix(T((1ζ)I+ζT))=Fix(T).

By (3.1), (3.20), and (3.27), we deduce

lim n T ( ( 1 ζ ) I + ζ T ) x n x n =0.
(3.28)

Next we prove that T((1ζ)I+ζT)I is demiclosed at 0. Let the sequence { w n } H 2 satisfying w n x and w n T((1ζ)I+ζT) w n 0. Next, we will show that x Fix(T((1ζ)I+ζT))=Fix(T).

Since T is L 2 -Lipschizian, we have

w n T w n w n T ( ( 1 ζ ) I + ζ T ) w n + T ( ( 1 ζ ) I + ζ T ) w n T w n w n T ( ( 1 ζ ) I + ζ T ) w n + ζ L w n T w n .

It follows that

w n T w n 1 1 ζ L w n T ( ( 1 ζ ) I + ζ T ) w n .

Hence,

lim n w n T w n =0.

Since TI is demiclosed at 0 by Lemma 2.1, we immediately deduce x Fix(T)=Fix(T((1ζ)I+ζT)). Therefore, T((1ζ)I+ζT)I is demiclosed at 0. By (3.28), we deduce x ˜ Fix(T). Hence, x ˜ Γ. So,

lim sup n ( f F ) x , u n x = lim sup i ( f F ) x , u n i x = ( f F ) x , x ˜ x 0 .
(3.29)

Note that

u n x 2 = β n ( f ( x n ) f ( x ) ) + β n ( f ( x ) F x ) + ( I β n F ) ( y n x ) 2 ( I β n F ) ( y n x ) 2 + 2 β n f ( x n ) f ( x ) , u n x + 2 β n f ( x ) F x , u n x ( 1 β n ς 2 ) 2 x n x 2 + 2 β n ρ x n x u n x + 2 β n f ( x ) F x , u n x ( 1 β n ς 2 ) 2 x n x 2 + 2 β n ρ x n x 2 + 1 2 u n x 2 + 2 β n f ( x ) F x , u n x .

It follows that

u n x 2 [ 1 2 ( ς 2 ρ ) β n ] x n x 2 + β n 2 ς 4 2 x n x 2 + 4 β n f ( x ) F x , u n x .

So,

x n + 1 x 2 = α n ( x n x ) + ( 1 α n ) ( u n x ) 2 α n x n x 2 + ( 1 α n ) u n x 2 [ 1 2 ( ς 2 ρ ) ( 1 α n ) β n ] x n x 2 + ( 1 α n ) β n 2 ς 2 4 x n x 2 + 4 ( 1 α n ) β n f ( x ) F x , u n x = [ 1 ( ς 2 ρ ) ( 1 α n ) β n ] x n x 2 + ( ς 2 ρ ) ( 1 α n ) β n { β n ς 2 4 ( ς 2 ρ ) x n x 2 + 4 ς 2 ρ f ( x ) F x , u n x } .
(3.30)

Applying Lemma 2.3 to (3.30) we deduce x n x .

Case 2. Assume there exists an integer n 0 such that x n 0 x x n 0 + 1 x . In this case, we set ω n ={ x n x }. Then, we have ω n 0 ω n 0 + 1 . Define an integer sequence { τ n } for all n n 0 as follows:

τ(n)=max{lN| n 0 ln, ω l ω l + 1 }.

It is clear that τ(n) is a non-decreasing sequence satisfying

lim n τ(n)=

and

ω τ ( n ) ω τ ( n ) + 1 ,

for all n n 0 . From (3.19), we get

( 1 α τ ( n ) ) α τ ( n ) y τ ( n ) x τ ( n ) 2 x τ ( n ) x 2 x τ ( n ) + 1 x 2 + 2 ( 1 α τ ( n ) ) β τ ( n ) f ( x τ ( n ) ) , x τ ( n ) x 2 ( 1 α τ ( n ) ) β τ ( n ) F y τ ( n ) , x τ ( n ) x + ( 1 α τ ( n ) ) 2 [ β τ ( n ) 2 f ( x τ ( n ) ) F y τ ( n ) 2 + 2 β τ ( n ) f ( x τ ( n ) ) F y τ ( n ) y τ ( n ) x τ ( n ) ] .
(3.31)

It follows that

lim n y τ ( n ) x τ ( n ) =0.

By a similar argument to that of (3.29) and (3.30), we can prove that

lim sup n ( f F ) x , u τ ( n ) x 0,
(3.32)

and

ω τ ( n ) + 1 2 [ 1 2 ( ς 2 ρ ) ( 1 α τ ( n ) ) β τ ( n ) ] ω τ ( n ) 2 + ( 1 α τ ( n ) ) β τ ( n ) 2 ς 2 4 ω τ ( n ) 2 + 4 ( 1 α τ ( n ) ) β τ ( n ) f ( x ) F x , u τ ( n ) x .
(3.33)

Since ω τ ( n ) ω τ ( n ) + 1 , we have from (3.33)

ω τ ( n ) 2 16 4 ( ς 2 2 ρ ) ς 2 β τ ( n ) f ( x ) F x , u τ ( n ) x .
(3.34)

Combining (3.33) and (3.34), we have

lim sup n ω τ ( n ) 0,

and hence

lim n ω τ ( n ) =0.
(3.35)

From (3.33), we also obtain

lim sup n ω τ ( n ) + 1 lim sup n ω τ ( n ) .

This together with (3.35) imply that

lim n ω τ ( n ) + 1 =0.

Applying Lemma 2.2 to get

0 ω n max{ ω τ ( n ) , ω τ ( n ) + 1 }.

Therefore, ω n 0. That is, x n x . This completes the proof. □

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Acknowledgements

Yonghong Yao was supported in part by NSFC 71161001-G0105. Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.

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Correspondence to Yeong-Cheng Liou.

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All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Yao, Y., Agarwal, R.P. & Liou, Y. Iterative algorithms for quasi-variational inclusions and fixed point problems of pseudocontractions. Fixed Point Theory Appl 2014, 82 (2014). https://doi.org/10.1186/1687-1812-2014-82

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Keywords

  • quasi-variational inclusions
  • fixed point problem
  • pseudocontractions
  • maximal monotone
  • firmly nonexpansive mappings