Let H be a real Hilbert space and C be a nonempty closed convex subset of H. The paper is concerned with a method for finding solutions to equilibrium problems, stated as follows:

(1.1)Find  x*∈C  such that f(x*,y)≥0  ∀y∈C,

where f:H×H→H, is a function such that f(x,.) is convex and subdifferentiable on H for every fixed x ∈ C. From now on, we denote the solution set of Problem (1.1) by Sol(C,f). Problem (1.1) is a general model in the sense that it unifies, in a simple form, numerous known models of optimization problems, nonlinear complementary problems, and variational inequalites [3, 10, 12, 13]. Equilibrium problems have many direct applications in other fields, such as transportation, electricity markets, and network problems [7, 13, 15, 19, 24, 26, 27]. This may explain why the problem has become an attractive field and has received a lot of attention by many authors. Some notable methods for solving it have been proposed, to highlight a few, see [1, 2, 3, 4, 6, 7, 17, 28, 29, 30, 32, 33, 36].

In the special case, f(x,y)=⟨F(x),y−x⟩, where F:C→H, Problem (1.1) is equivalent to the following variational inequality problem:

(1.2)Find  x*∈C  such that ⟨F(x*),x−x*⟩≥0  ∀y∈C.

In order to solve the variational inequality problems, many iterative methods have been proposed. If F is strongly monotone, the problems can be solved by the Newton method [34] or by single projection methods [9, 16]. Under the assumption that F is monotone and L-Lipschitz continuous, the extragradient method for solving Problem (1.2) is introduced by Korpelevich in [20]. In this method, two projections onto the feasible set C are used at each iteration:

x0∈C,yk=PrC(xk−λkF(xk)),xk+1=PrC(xk−λkF(yk)),

where λk∈(0,1L) and Pr_C denotes Euclidean projection onto C. Korpelevich showed that the iterative sequence generated by the algorithm is convergent in Euclidean spaces. His extragradient method has since been expanded and improved by many mathematicians in different ways [11, 22]. Recently, in [33], the authors introduced a single projection method which combines a projection method with the Halpern iteration technique. This method requires only one projection onto the feasible set C at each iteration and the iterative process is given by

(1.3)x0∈C,yk=PrC(xk−λkF(xk)),dk:=xk−yk−λk(F(xk)−F(yk)),xk+1=αkx0+(1−αk)(xk−γρkdk),

where γ∈(0,2), αk∈(0,1), limk→∞αk=0,∑ k=0∞αk=+∞, λk∈(0,∞) and

ρk=⟨xk−yk,dk⟩‖dk‖2 if dk≠00 if dk=0.

Recall that the strong convergence result of the iterative sequence generated by the proposed method is only established in real Hilbert spaces when F is pseudomonotone and L-Lipschitz-continuous. When F is a multivalued mapping from C to H, then Problem (1.1) becomes the following multivalued variational inequality problem

(1.4)Find (x*,w*)∈C×F(x*) such that ⟨w*,x−x*⟩≥0  ∀x∈C.

Recall that the Hausdorff distance ρ(A,B) between two subsets A and B of H is defined by:

ρ(A,B):=max{d(A,B),d(B,A)},

where d(A,B):=supa∈Ainfb∈B‖a−b‖. A multivalued mapping is said to be Lipschitz continuous on C with constant L if

ρ(F(x),F(y))≤L‖x−y‖2,∀x, y∈C.

In paper [8], by replacing a projection onto C in 1.3 at each iteration by an approximate projection or a proximal operator, we improved the method in [33] and proposed a new algorithm for solving Problem (1.4). We proved that the algorithm is strongly convergent under the assumption of the pseudomonotonicity and Lipschitz continuity of cost mappings.

In this paper, we extend the single projection method to pseudomonotone variational inequality in [33] (Algorithm 1.3) for solving Problem (1.1) in a real Hilbert space. Under the assumptions that the equilibrium bifunction f(x,y) is pseudomonotone and

(1.5)ρ∂2f(x,⋅)(y),∂2f(y,⋅)(y)≤L‖x−y‖, ∀x∈H,y∈C,

we have proved that the sequence {xk} generated by the algorithm is strongly convergent to a solution of the problem. Note that in many other methods, the assumption of Lipschitz-type continuity with the constants c₁, c₂ of bifunction f(x,y) is necessary for obtaining the convergence theorem of the algorithm [1, 17, 18, 30]. In our algorithm, the condition (1.5) is considered to be an alternative to the condition that f(x,y) is Lipschitz-type continuous. On the basis of the given algorithm for equilibrium problem we developed a new algorithm for finding a common solution of Problem (1.1) and of fixed point problems. The strong convergence of the algorithm is established under mild assumptions.

The paper is organized as follows. In Section 2 we review some necessary concepts and lemmas that will be used in proving the main results of the paper. Im Section 3 we give then Halpern subgradient method for solving Problem (1.1) and prove its convergence. In section 4, we develop the algorithm forom Section 3 for problem of finding a common solution of Problem (1.1) and fixed point problems. In the last section, several fundamental experiments are provided to illustrate the convergence of the algorithm from Section 3, and to compare it to other algorithms.

Throughout this paper, unless otherwise mentioned, let H denote a Hilbert space with inner product ⟨.,.⟩ and

the induced norm ‖.‖.

Definition 2.1. Let C be a nonempty closed convex subset in H. The metric projection from H onto C is denoted by Pr_C and

PrC(x)=argmin{‖x−y‖: y∈C},∀x∈H.

It is well known that the metric projection PrC(⋅) has the following basic property:

⟨x−PrC(x),y−PrC(x)⟩≤0, ∀x∈H, y∈C.

Definition 2.2. Let C be a nonempty closed convex subset of a real Hilbert space H. A bifunction f:C×C→H is called

• (i) β-strongly monotone on C, if f(x,y)+f(y,x)≤−β‖x−y‖2  ∀x,y∈C;

• (ii) monotone on C, if f(x,y)+f(y,x)≤0  ∀x,y∈C;

• (iii) pseudomonotone on C, if f(x,y)≥0⇒f(y,x)≤0 ∀x,y∈C.

Definition 2.3. Let C⊂H be a nonempty subset. An operator S:C→H is called

• (i) β-demicontractive on C, if Fix(S) is nonempty and there exists β∈[0,1) such that (2.1)‖Sx−p‖2≤‖x−p‖2+β‖x−Sx‖2 ∀x∈C, ∀p∈Fix(S);

• (ii) demiclosed, if for any sequence {xk}⊂C, xk⇀z∈C, (I−S)(xk)⇀0 implies z∈Fix(S).

It is well known that if S is β-demicontractive on C then S is demiclosed and (2.1) is equivalent to (see [25])

(2.2)⟨x−Sx,x−p⟩≥12(1−β)‖x−Sx‖2 ∀x∈C, ∀p∈Fix(S).

To prove the main result in Section 3 and 4, we shall use the following lemmas in the sequel.

Lemma 2.4. For every x, y∈H, we have the following assertions.

• (i) ‖x+y‖2=‖x‖2+2⟨x,y⟩+‖y‖2;

• (ii) ‖x+y‖2≤‖x‖2+2⟨y,x+y⟩.

Lemma 2.5. Let ak be a sequence of nonnegative real numbers satisfying the following condition:

ak+1≤(1−αk)ak+αkδk+βk, ∀k≥1,

where αk⊂[0,1], ∑ k=0∞αk=+∞, limsupk→∞δk≤0 and βk≥0, ∑ n=1∞βk<∞.

Then, limk→∞ak=0.

The subdifferential of a convex function g:C→R∪{+∞} is defined by

∂g(x)={u∈H:⟨u,y−x⟩≤g(y)−g(x)  ∀y∈C}.

In convex programming, we have the following result.

Lemma 2.6. ([31])

Let C be a convex subset of a real Hilbert space H and g:C→R∪{+∞} be subdifferentiable.

Then, x^* is a solution to the following convex problem:

min{g(x):x∈C}

if and only if 0∈∂g(x*)+NC(x*), where ∂g denotes the subdifferential of g and NC(x*) is the outer normal cone of C at x*∈C, that is, NC(x*)={u∈H:⟨u,y−x*⟩≤0  ∀y∈C}.

3.1. Assumption

In this article, in order to find a point in Sol(f,C), we assume that the bifunction f:H×H→H satisfies the following conditions:

• A₁. f(x,x)=0 for all x ∈ C, f is pseudomonotone and weakly continuous on H,

  i.e., xk⇀x¯ and yk⇀y¯⇒f(xk,yk)→f(x¯,y¯);

• A₂. There exists a real nonnegative number L such that ρ∂2f(x,⋅)(y),∂2f(y,⋅)(y)≤L‖x−y‖, ∀x∈H,y∈C;

• A₃. Sol(C,f) is nonempty.

• A₄. f(x,⋅) is convex and subdifferentiable on H.

Remark 3.1. (a) Let g:H→H be a convex subdifferentiable and weakly continuous on H. Clearly, f(x,y):=g(y)−g(x) satisfies conditions A1−A2. We well known that the following optimization problem

ming(x) such that x∈C,

is equivalent to Problem (1.1).

(b) Let F(x) be a Lipschitz continuous and weakly continuous function on H, i.e., xk⇀x¯⇒F(xk)→F(x¯).

Setting f(x,y):=⟨F(x),y−x⟩, it is easy to see that f(x,y) satisfies conditions A₁ and A₂.

3.2. Algorithm

Algorithm 3.2. Choose starting point x0∈H, L>L, sequences αk,{λk} and ϵk such that

(3.1)αk⊂(0,1),limk→∞αk=0,∑ k=0∞αk=+∞,0<ρkϵk≤αk3,∑ k=0∞(ρkϵk)12<∞,{λk}⊂[a,b]⊂0,1L⊂(0,∞).

Step 1. (k=0,1,...) Find yk∈H such that

yk=argminλkf(xk,y)+12‖y−xk‖2:y∈C.

If xk−yk=0 then STOP.

Step 2. Take uk∈∂2f(xk,yk) satisfying ⟨xk−yk−λkuk,yk−x⟩≥−ϵk, ∀x∈C and

vk∈B(uk,L‖xk−yk‖)∩∂2f(yk,yk),

where Buk,L‖xk−yk‖:={x∈H:‖x−uk‖≤L‖xk−yk‖}.

Step 3. Compute xk+1=αkx0+(1−αk)zk, where zk:=xk−ρkdk and

dk:=xk−yk−λk(uk−vk), ρk=⟨xk−yk,dk⟩‖dk‖2.

Step 4. Set k:= k+1, and go to Step 1.

Remark 3.3. (a) By Step 1 and Lemma 2.6, we have

0∈λk∂2f(xk,yk)+yk−xk+NC(yk),

it follows that there is a uk∈∂2f(xk,yk) such that

⟨xk−yk−λkuk,yk−x⟩≥0, ∀x∈C,

i.e., the u^k in Step 2 always exists.

(b) If d^k=0 then

‖xk−yk‖=λk‖uk−vk‖≤λkL‖xk−yk‖.

Since λk≤1L for all k we have

(1−λkL)‖xk−yk‖≤0⇒xk=yk.

Thus, we observe that d^k ≠ 0 and ρ_k of Step 3 is defined.

3.3. Convergence analysis of algorithm

In this section, we show that the algorithm proposed is strongly convergent if assumptions A₁-A₄ satisfy.

Lemma 3.4. Let sequence {xk} be generated by Algorithm 3.2 and x*∈Sol(C,f). Then,

• (i) ‖zk−x*‖2≤‖xk−x*‖2−‖zk−xk‖2+2ρkϵk;

• (ii) sequences {xk} and {zk} are bounded.

Proof. Using Step 2, we have

⟨xk−yk−λkuk,yk−x⟩≥−ϵk ∀x∈C.

Replace x by x*∈C in the last inequality, we have

⟨xk−yk−λkuk,yk−x*⟩≥−ϵk.

Using x*∈Sol(C,f), f(yk,yk)=0, vk∈∂2f(yk,yk) and the pseudomonotone assumption of f, we get

λk⟨vk,yk−x*⟩≥λk[f(yk,yk)−f(yk,x*)]≥0.

Adding two last inequalities, it follows that

−ϵk≤⟨yk−x*,xk−yk−λkuk+λkvk⟩=⟨yk−x*,dk⟩.

Using the above inequality and the definition of z^k, we have

‖zk−x*‖2=‖xk−ρkdk−x*‖2=‖xk−x*‖2−2ρk⟨xk−x*,dk⟩+ρk2‖dk‖2≤‖xk−x*‖2−2ρk⟨xk−yk,dk⟩+ρk2‖dk‖2+2ρkϵk=‖xk−x*‖2−2ρk⟨xk−yk,dk⟩+ρk⟨xk−yk,dk⟩+2ρkϵk==‖xk−x*‖2−ρk⟨xk−yk,dk⟩+2ρkϵk,

and ρk⟨xk−yk,dk⟩=‖zk−xk‖2. Therefore

‖zk−x*‖2≤‖xk−x*‖2−ρk⟨xk−yk,dk⟩+2ρkϵk=‖xk−x*‖2−‖zk−xk‖2+2ρkϵk.

This follows (i). We now prove (ii). From (i) and condition (3.1), it follows that

(3.2)‖xk+1−x*‖=‖αkx0+(1−αk)zk−x*‖≤αk‖x0−x*‖+(1−αk)‖zk−x*‖≤αk‖x0−x*‖+(1−αk)‖xk−x*‖2+2ρkϵk≤αk‖x0−x*‖+(1−αk)(‖xk−x*‖+2ρkϵk)≤max‖x0−x*‖,‖xk−x*‖+2ρkϵk...≤max‖x0−x*‖,‖x0−x*‖+2∑ i=0 kρiϵi≤‖x0−x*‖+2∑ i=0∞ρiϵi<+∞.

It implies that {xk} is bounded. Using again (i), we have that the sequence {zk} is bounded.

Lemma 3.5. Let x*∈Sol(C,f). Set ak=‖xk−x*‖2, bk=2⟨x0−x*,xk+1−x*⟩ and βk=2ρkϵk. Then,

• (i) ak+1≤(1−αk)ak+αkbk+βk;

• (ii) βk≥0, ∑ k=1∞βk<∞;

• (iii) limk→∞βkαk=0;

• (iv) −1≤limsupk→∞bk<∞.

Proof. Using Lemma 2.4 (ii), we get

‖xk+1−x*‖2=‖αk(x0−x*)+(1−αk)(zk−x*)‖2     ≤(1− α k )2‖zk−x*‖2+2αk(1−αk)⟨x0−x*,xk+1−x*⟩     ≤(1−αk)‖zk−x*‖2+2αk⟨x0−x*,xk+1−x*⟩.

Combining the last inequality and Lemma 3.4 (i), we get

‖xk+1−x*‖2≤(1−αk)‖xk−x*‖2+2αk⟨x0−x*,xk+1−x*⟩+(1−αk)2ρkϵk    ≤(1−αk)‖xk−x*‖2+2αk⟨x0−x*,xk+1−x*⟩+2ρkϵk.

This follows (i). Assumptions (ii) and (iii) are directly inferred from condition (3.1).

Since {xk} is bounded, we have bk≤2‖x0−x*‖‖xk+1−x*‖<∞, and so limsupk→∞bk<∞. We now assume contradiction that limsupk→∞bk<−1. There exists k0∈N such that bk<−1 for all k≥k0. It follows from (i) that

ak+1≤(1−αk)ak+αkbk+βk  <(1−αk)ak−αk+βk  =ak−(ak+1)αk+βk  ≤ak−αk+βk.  ⋯  ≤ak0−∑i=k0kαi+∑i=k0kβi  ∀k≥k0.

Using this and the result (ii), one have

limsupk→∞ak≤ak0−∑ i= k 0 +∞αi+∑ i= k 0 +∞βi=−∞.

This contradicts the fact that ak≥0 for all k∈N.

Therefore, limsupk→∞bk≥−1.

Theorem 3.6. Let bifunction f:H×H→H be satisfying the assumptions A₁-A₄. Then, the sequence {xk} generated by Algorithm 3.2 converges strongly to a solution z∈Sol(C,f), where z=PrSol(C,f)(x0).

Proof. Set ak:=‖xk−z‖. In oder to prove this theorem, we consider two following cases.

Case 1. Suppose that there exists k0∈N such that ak+1≤ak for all k≥k0. Then, there exists the limit limk→∞ak∈[0,∞). From Step 3, Lemma 3.4 (i) and Lemma 2.4 (ii), it follows that

‖xk+1−z‖2=‖(1−αk)(zk−z)+αk(x0−z)‖2    ≤(1− α k )2‖zk−z‖2+2αk⟨x0−z,xk+1−z⟩    ≤‖zk−z‖2+2αk⟨x0−z,xk+1−z⟩    ≤‖xk−z‖2−‖zk−xk‖2+2αk⟨x0−z,xk+1−z⟩+2ρkϵk    ≤‖xk−z‖2−‖zk−xk‖2+2ρkϵk+αkQ0,

where Q0:=sup{2⟨x0−z,xk+1−z⟩:k=0,1,...}<∞. This implies that

(3.3)ak+1−ak+‖zk−xk‖2≤+2ρkϵk+αkQ0 ∀k≥0.

Taking k→∞ in the last inequality and using the assumptions limk→∞αk=0, limk→∞2ρkϵk=0,

we have limk→∞‖zk−xk‖=0. From vk∈B(uk,L‖xk−yk‖) for all k, it follows

(3.4)⟨xk−yk,dk⟩=‖xk−yk‖2−λk⟨xk−yk,uk−vk⟩      ≥‖xk−yk‖2−λk‖xk−yk‖.‖uk−vk‖      ≥(1−bL)‖xk−yk‖2,

and

(3.5)‖dk‖=‖xk−yk−λk(uk−vk)‖  ≤‖xk−yk‖+λk‖uk−vk‖  ≤(1+λkL)‖xk−yk‖  ≤(1+bL)‖xk−yk‖.

Using Step 3, (3.4) and (3.5), we get ρk≥1−bL(1+bL)2 and

‖xk−yk‖2≤1(1−bL)⟨xk−yk,dk⟩    =1(1−bL)ρk‖zk−xk‖2    ≤(1+bL)2(1−bL)2‖zk−xk‖2.

From the above inequality and limk→∞‖zk−xk‖=0, it follows that

limk→∞‖xk−yk‖=0.

Using this and limk→∞‖zk−xk‖=0, we obtain limk→∞‖zk−yk‖=0.

By the definition of xk+1 and Lemma 3.4 (ii), we have

‖xk+1−zk‖=αk‖x0−zk‖≤αkQ1→0 as k→∞,

where Q1=sup{‖x0−zk‖: k=0,1,...}<+∞. This together with limk→∞‖zk−xk‖=0 implies that

‖xk+1−xk‖≤‖xk+1−zk‖+‖zk−xk‖→0 as k→∞.

Since sequence {xk} is bounded, so there exists a subsequence {xki+1} such that xki+1⇀p as i→∞, and

(3.6)limsupk→∞⟨x0−z,xk+1−z⟩=limi→∞⟨x0−z,xki+1−z⟩.

We will show that p∈Sol(C,f). Indeed, by Step 2, one has

⟨xki+1−yki−λki+1uki+1,yki+1−x⟩≥0 ∀x∈C.

Combining this inequality and uki+1∈∂2f(xki+1,yki+1), we get

⟨xki+1−yki+1,x−yki+1⟩≤λki+1⟨uki+1,x−yki+1⟩≤λki+1[f(xki+1,x)−f(xki+1,yki+1)].

Since ‖xk−yk‖→0 as k→∞, {xki+1} is bounded and converges weakly to p as i→∞, {yki+1} also is bounded and yki+1⇀p. For each fixed point x ∈ C, take the limit as i→∞, using limi→∞‖xki+1−yki+1‖=0 and weakly continuity of bifunction f(x,y), we get

f(p,x)≥0  ∀x∈C.

Using p∈Sol(C,f) and (3.6), we have

limsupk→∞bk=limsupk→∞⟨x0−z,xk+1−z⟩,    =2limk→∞⟨x0−z,xki+1−z⟩    =2⟨x0−z,p−z⟩≤0.

Applying Lemma 2.5 for Lemma 3.5 (i) and using the last inequality, we deduce

limk→∞‖xk−z‖=0.

Thus, {xk} converges strongly to the solution z=PrSol(C,f)(x0).

Case 2. We now assume that there is not k¯∈N such that {ak}k=k¯∞ is monotonically decreasing. So, there exists an integer k0≥k¯ such that ak0≤ak0+1. Then, there exists a subsequence {aτ(k)}of {ak} such that (see Remark 4.4, [21])

0≤ak≤aτ(k)+1,aτ(k)≤aτ(k)+1  ∀k≥k0,

where τ(k)=maxi∈N:k0≤i≤k,ai≤ai+1. Using aτ(k)≤aτ(k)+1, ∀k≥k0 and (3.3), one has

0≤‖wτ(k)−xτ(k)‖≤aτ(k)+1−aτ(k)+‖wτ(k)−xτ(k)‖≤ατ(k)Q0+2ρkϵk→0 as k→∞,

and so limk→∞‖wτ(k)−xτ(k)‖=0. By a similar way as in Case 1, we can show that

(3.7)limn→∞‖xτ(k)+1−xτ(k)‖=limn→∞‖xτ(k)−yτ(k)‖=limn→∞‖wτ(k)−yτ(k)‖=0.

Since {xτ(k)} is bounded, there exists a subsequence of {xτ(k)}, still denoted by {xτ(k)}, which converges weakly to p.

Arguing similarly as in Case 1, we can prove that p∈Sol(C,f) and

(3.8)limsupk→∞bτ(k)≤0.

From Lemma 3.5 (i) and aτ(k)≤aτ(k)+1, ∀k≥k0, it follows that

ατ(k)aτ(k)≤aτ(k)−aτ(k)+1+ατ(k)bτ(k)+βτ(k)≤ατ(k)bτ(k)+βτ(k).

It is equivalent to

aτ(k)≤bτ(k)+βτ(k)ατ(k).

By Lemma 3.5 (iii), (3.8) and the last inequality, we get

limsupk→∞aτ(k)≤limsupk→∞bτ(k)≤0.

Therefore, limk→∞aτ(k)=0. As a consequence, we get

aτ(k)+1=‖xτ(k)+1−z‖    ≤‖xτ(k)+1−xτ(k)‖+aτ(k)→0, k→∞.

It follows that limk→∞aτ(k)+1=0. Furthermore, 0≤ak≤aτ(k)+1 for all k≥k0. Hence, limk→∞ak=0, i.e., xk→z, as  k→∞.

Let a finite system of mappings Sj (j∈J:={1,2,...,r}) of H into itself. Denote the fixed point set of S_j by

Fix(Sj):={x∈H:Sjx=x}.

We consider the problem which finds a common element of the solution set of Problem (1.1) and the set of fixed

points of a finite system of mappings Sj (j∈J), namely:

 Find x*∈∩j∈JFix(Sj)∩Sol(C,f).

4.1. Assumption

In this section, we assume that the bifunction f:H×H→H and the mappings Sj (j∈J) satisfy the following conditions:

B₁. f(x,x)=0 for all x∈ C, f is pseudomonotone and weakly continuous on H and f(x,⋅) is convex and subdifferentiable on H;

B₂. There exists a real nonnegative number L such that

ρ∂2f(x,⋅)(y),∂2f(y,⋅)(y)≤L‖x−y‖, ∀x∈H,y∈C;

B₃. ∩j∈JFix(Sj)∩Sol(C,f)≠∅;

B₄. Sj:H→H is β_j-demicontractive for every j∈ J.

4.2. Algorithm

Algorithm 4.1. Choose starting point x0∈H, L>L, sequences αk,{λk} and ϵk such that

(4.1)αk⊂(0,1),limk→∞αk=0,∑ k=0∞αk=+∞,0<ρkϵk≤αk3,∑ k=0∞(ρkϵk)12<∞,{λk}⊂[a,b]⊂0,1L⊂(0,∞).

Step 1*. (k=0,1,...) Find yk∈H such that

yk=argminλkf(xk,y)+12‖y−xk‖2:y∈C.

If xk−yk=0 then STOP.

Step 2*. Take uk∈∂2f(xk,yk) satisfying ⟨xk−yk−λkuk,yk−x⟩≥−ϵk, ∀x∈C and

vk∈B(uk,L‖xk−yk‖)∩∂2f(yk,yk),

where Buk,L‖xk−yk‖:={x∈H:‖x−uk‖≤L‖xk−yk‖}. Set dk:=xk−yk−λk(uk−vk) and zk:=xk−ρkdk with

dk:=xk−yk−λk(uk−vk), ρk=⟨xk−yk,dk⟩‖dk‖2.

Step 3*. Compute

pk=αkx0+(1−αk)zk,qjk=(1−ω)pk+ωSjpk, 0<ω<1−βj2 ∀j∈J, (4.2)xk+1=qj0k, j0=argmax{||qjk−pk||, j∈J}, k≥1.

Step 4*. Set k:= k+1, and go to Step 1*.

Lemma 4.2. The sequences {pk},{xk},{zk} and {yk} are bounded.}

Proof. Let x*∈∩j∈JFix(Sj)∩Sol(C,f). Using Step 3* and the βj demicontractive assumption of Sj, j=1,2,..., we get

(4.3)|xk+1−x*|2=|(1−ω)pk+ωSj0pk−x*|2    =|(pk−x*)+ω(Sj0pk−pk)|2    ≤pk−x*|2+2ω⟨pk−x*,Sj0pk−pk⟩+ω2Sj0pk−pk|2    ≤pk−x*|2+ω(ω+βj0−1)Sj0pk−pk|2    ≤|pk−x*|2.

From Lemma 3.4 (i) and the last inequality, it follows that

(4.4)||zk+1−x*||≤||pk−x*||+2ρk+1ϵk+1.

Using Step 3*, condition (4.1) and (4.4), we have

‖pk+1−x*‖=||αk+1(x0−x*)+(1−αk+1)(zk+1−x*)||    ≤αk+1||x0−x*||+(1−αk+1)||zk+1−x*||    ≤αk+1||x0−x*||+(1−αk+1)(||pk−x*||+2ρk+1ϵk+1)    ≤max{||pk−x*||+2ρk+1ϵk+1}...    ≤max{||p0−x*||+∑ i=1 k+12ρk+1ϵk+1, ||x0−x*||}<+∞.

So, the sequence {pk} is bounded. From (4.3) and (4.4), it follows that the sequences {xk} and {zk} are bounded.

Lemma 4.3. Let x*∈∩j∈JFix(Sj)∩Sol(C,f). Set ak=‖xk−x*‖2, βk=2ρkϵk

and bk=2⟨x0−x*,pk−x*⟩. Then,

• (i) ak+1≤(1−αk)ak+αkbk+βk;

• (ii) βk≥0, ∑ n=1∞βk<∞;

• (iii) limk→∞βkαk=0;

• (iv) −1≤limsupn→∞bk<∞.

Proof. Using Lemma 2.4 (ii), Lemma 3.4 (i) and Step 3*, we get

(4.5)|pk−x*|2=|αk(x0−x*)+(1−αk)(zk−x*)|2    ≤(1−αk)|zk−x*|2+2αk⟨x0−x*,pk−x*⟩    ≤(1−αk)|xk−x*|2+2αk⟨x0−x*,pk−x*⟩+2ρkϵk(1−αk)    ≤(1−αk)|xk−x*|2+2αk⟨x0−x*,pk−x*⟩+2ρkϵk.

Using last inequality and (4.3), we have

||xk+1−x*||2≤(1−αk)||xk−x*||+2αk⟨x0−x*,pk−x*⟩+2ρkϵk.

We have (i). By arguing similarly as in the proof of Lemma 3.5, we obtain (ii), (iii) and (iv).

Theorem 4.4. Suppose that conditions B₁-B₄ are satisfied. Let {xk} be a sequence generated by Algorithm 4.1. Then, the sequence {xk} converges strongly to a solution

z∈∩j∈JFix(Sj)∩Sol(C,f),

where z=Pr∩j∈JFix(Sj)∩Sol(C,f)(x0).

Proof. Set ak:=‖xk−z‖. In oder to prove this theorem, we consider two following cases.

Case 1. Suppose that there exists k0∈N such that ak+1≤ak for all k≥k0. Then, there exists the limit limk→∞ak∈[0,∞).

Using Step 3*, Lemma 2.1 (ii), Lemma 3.4 (i) and (2.2), we obtain

(4.6)‖xk+1−z‖2=‖(1−ω)pk+ωSj0pk−z‖2    =‖pk−z‖2−2ω⟨pk−z,pk−Sj0pk⟩+ω2‖pk−Sj0pk‖2    ≤‖pk−z‖2−ω(1−βj0−ω)‖pk−Sj0pk‖2    =||αk(x0−z)+(1−αk)(zk−z)||2−1ω(1−βj0−ω)‖xk+1−pk‖2    ≤(1−αk)||zk−z||2+2αk⟨x0−z,pk−z⟩−‖xk+1−pk‖2    ≤||zk−z||2+2αk⟨x0−z,pk−z⟩−‖xk+1−pk‖2    ≤‖xk−z‖2−‖zk−xk‖2+2αk⟨x0−z,pk−z⟩−‖xk+1−pk‖2+2ρkϵk    ≤‖xk−z‖2−‖zk−xk‖2+αkM0−‖xk+1−pk‖2+2ρkϵk,

where M0:=sup{2⟨x0−z,pk−z⟩:k=0,1,...}<∞. It follows that

(4.7)ak+1−ak+‖zk−xk‖2+‖xk+1−pk‖2≤αkM0+2ρkϵk  ∀k≥0.

Passing the limit as k→∞ and using the assumptions

limk→∞αk=0,limk→∞2ρkϵk=0,

we have

limk→∞‖zk−xk‖=0, limk→∞‖xk+1−pk‖=0.

By a similar way as in the proof of Theorem 3.6, we can show that

limk→∞‖xk−yk‖=0.

It follows that

‖zk−yk‖≤‖zk−xk‖+‖xk−yk‖→0,  as  k→∞.

Using Step 3*, we have

‖pk−zk‖=αk‖x0−zk‖≤αkM1→0, as k→∞,

where M1=sup{‖x0−zk‖: k=0,1,...}0<+∞.

Therefore,

‖xk+1−xk‖≤‖xk+1−pk‖+‖pk−zk‖+‖zk−xk‖→0  as  k→∞.

From this and

‖xk−pk‖≤‖xk+1−xk‖+‖xk+1−pk‖,

it follows that limk→∞‖xk−pk‖=0. Since sequence {xk} is bounded, there exists a subsequence {xki} such that xki⇀p∈H, pki⇀p and

(4.8)limsupk→∞⟨x0−z,pk−z⟩=limi→∞⟨x0−z,pki−z⟩.

Now, we will show that

p∈∩j∈JFix(Sj)∩Sol(C,f).

By a similar way as in the proof of Theorem 3.6,

we can prove that p∈Sol(C,f). For each j ∈ J, using Step 3^*, we have

||pk−Sjpk||=1ω||pk−qjk||≤1ω||pk−qj0k||=1ω||xk+1−pk||.

By limk→∞‖xk+1−pk‖=0 and the last inequality, we get

||pk−Sjpk||→0, k→∞.

From limk→∞‖xk−pk‖=0 and xki⇀p, it follows that pki⇀p. Using this, limk→∞||pk−Sjpk||=0 and the demiclosedness of S_j, we have p∈Fix(Sj). Therefore,

p∈∩j∈JFix(Sj)∩Sol(C,f).

This together with (4.8) implies that

limsupk→∞bk=2limi→∞⟨x0−z,pki−z⟩=2⟨x0−z,p−z⟩≤0.

Using this, Lemma 2.5 and Lemma 4.3, we obtain limk→∞‖xk−z‖=0.

Case 2. We now assume that there is not k¯∈N such that {ak}k=k¯∞ is monotonically decreasing. So, there exists an integer k0≥k¯ such that ak0≤ak0+1. Then, there exists a subsequence {aτ(k)} of {ak}such that (see Remark 4.4, [21])

0≤ak≤aτ(k)+1,aτ(k)≤aτ(k)+1  ∀k≥k0,

where τ(k)=maxi∈N:k0≤i≤k,ai≤ai+1. Using aτ(k)≤aτ(k)+1, for all k≥k0 and (3.3), we get

‖zτ(k)−xτ(k)‖→0,‖xτ(k)+1−pτ(k)‖→0, k→∞.

By a similar way as in case 1, we can show that

(4.9)limk→∞‖xτ(k)−pτ(k)‖=limk→∞‖xτ(k)−yτ(k)‖=limk→∞‖zτ(k)−yτ(k)‖=0.

Since {xτ(k)} is bounded, there exists a subsequence of {xτ(k)}, still denoted by {xτ(k)}, which converges weakly to p∈H. By a similar way as in case 1, we can prove that p∈∩j∈JFix(Sj)∩Sol(C,f) and

limsupk→∞bτ(k)≤0.

Using Lemma 4.3 (i) and aτ(k)≤aτ(k)+1, ∀k≥k0, we have

ατ(k)aτ(k)≤aτ(k)−aτ(k)+1+ατ(k)bτ(k)+βτ(k)≤ατ(k)bτ(k)+βτ(k).

Since ατ(k)>0, we get

aτ(k)≤bτ(k)+βτ(k)ατ(k).

From Lemma 4.3 (iii) and last inequality,

it follows that

limsupk→∞aτ(k)≤limsupk→∞bτ(k)≤0.

Hence, limk→∞aτ(k)=0. It follows that

aτ(k)+1=‖xτ(k)+1−z‖2   ≤(‖xτ(k)+1−xτ(k)‖+‖xτ(k)−z‖)2→0, k→∞.

Using 0≤ak≤aτ(k)+1 for all k≥k0, we get limn→∞ak=0. Hence, xk→z as  k→∞.