A Tseng-Type Algorithm with Self-Adaptive Techniques for Solving the Split Problem of Fixed Points and Pseudomonotone Variational Inequalities in Hilbert Spaces

Abstract

In this paper, we survey the split problem of fixed points of two pseudocontractive operators and variational inequalities of two pseudomonotone operators in Hilbert spaces. We present a Tseng-type iterative algorithm for solving the split problem by using self-adaptive techniques. Under certain assumptions, we show that the proposed algorithm converges weakly to a solution of the split problem. An application is included.

1. Introduction

In this paper, we survey the variational inequality (in short, VI$(C,\varphi )$) of seeking an element $p^{†}\in C$ such that

$$\begin{array}{c}\hfill ⟨\varphi \left(p^{†}\right),x-p^{†}⟩,\mathrm{for}\mathrm{all}x\in C,\end{array}$$

(1)

where C is a nonempty closed convex set in a real Hilbert space H, $⟨·,·⟩$ means the inner product of H, and $\varphi :H\to H$ is a nonlinear operator. Denote by $\mathrm{Sol}(C,\varphi )$ the solution set of variational inequality (1).

A host of problems such as optimization problem, saddle point, equilibrium problem, fixed point problem can be converted into the form of variational inequality (1), see [1,2,3,4,5,6,7,8,9,10,11,12]. Many numerical algorithms have been proposed and developed for solving (1) and related problems, see [13,14,15,16,17,18,19,20,21,22,23,24,25] and the references therein. Generally speaking, $\varphi$ should satisfy the following assumptions

• $\varphi$ is strongly monotone, i.e., there exists a positive constant $\gamma$ such that

  $$\begin{array}{c}\hfill ⟨\varphi \left(u\right)-\varphi \left(\widehat{u}\right),u-\widehat{u}⟩\ge \gamma {\parallel u-\widehat{u}\parallel }^2,\mathrm{for}\mathrm{all}u,\widehat{u}\in H.\end{array}$$

  (2)
• $\varphi$ is Lipschitz continuous, i.e., there exists a positive constant $\kappa$ such that

  $$\begin{array}{c}\hfill \parallel \varphi \left(u\right)-\varphi \left(\widehat{u}\right)\parallel \le \kappa \parallel u-\widehat{u}\parallel ,\mathrm{for}\mathrm{all}u,\widehat{u}\in H.\end{array}$$

  (3)

In order to abate the restriction (2), Korpelevich’s extragradient algorithm ([26]) was proposed in 1976

$$\left\{\begin{array}{c}{y}^k={\mathrm{proj}}_C[x^k-\tau \varphi \left(x^k\right)],\hfill \\ x^{k+1}={\mathrm{proj}}_C[x^k-\tau \varphi \left(y^k\right)],\hfill \end{array}\right.$$

(4)

where ${\mathrm{proj}}_C$ denotes the orthogonal projection from H onto C and the step-size $\tau$ is in $(0,\frac{1}{\kappa })$.

Extragradient algorithm (4) guarantees the convergence of the sequence $\left\{x^k\right\}$ provided $\varphi$ is monotone. Extragradient algorithm and its variant have been investigated extensively, see [27,28,29,30,31]. However, we have to compute (i) twice ${\mathrm{proj}}_C$ at two different points and (ii) two values $\varphi \left(x^k\right)$ and $\varphi \left(y^k\right)$. Two important modifications of extragradient algorithm have been made. One is proposed in [32] by Censor, Gibali and Reich and another is the following remarkable algorithm proposed in [33] by Tseng

$$\left\{\begin{array}{c}{y}^k={\mathrm{proj}}_C[x^k-\gamma \varphi \left(x^k\right)],\hfill \\ x^{k+1}=y^k+\gamma [\varphi \left(x^k\right)-\varphi \left(y^k\right)],\hfill \end{array}\right.$$

(5)

where $\gamma \in (0,\frac{1}{\kappa })$.

On the other hand, if $\varphi$ is not Lipschitz continuous or its Lipschitz constant $\kappa$ is difficult to estimate, then algorithms (4) and (5) are invalid. To avoid this obstacle, Iusem [34] used a self-adaptive technique without prior knowledge of Lipschitz constant $\kappa$ of $\varphi$ for solving (1). Some related works on self-adaptive methods for solving (1), please refer to [35,36,37,38].

Let $H_1$ and $H_2$ be two real Hilbert spaces. Let C and Q be two nonempty closed and convex subsets of $H_1$ and $H_2$, respectively. Let $S:C\to C$, $T:Q\to Q$, $f:H_1\to H_1$ and $g:H_2\to H_2$ be four nonlinear operators. We consider the classical split problem which is to find a point $x^{\ast }\in C$ such that

$$x^{\ast }\in \mathrm{Fix}\left(S\right)\cap \mathrm{Sol}(C,f)\mathrm{and}A{x}^{\ast }\in \mathrm{Fix}\left(T\right)\cap \mathrm{Sol}(Q,g),$$

(6)

where $\mathrm{Fix}\left(S\right):=\left\{u^{†}\right|u^{†}=S{u}^{†}\}$ and $\mathrm{Fix}\left(T\right):=\left\{v^{†}\right|v^{†}=T{v}^{†}\}$ are the fixed point sets of S and T, respectively.

The solution set of (6) is denoted by $\Gamma$, i.e.,

$$\Gamma =\{x^{\ast }\in \mathrm{Fix}\left(S\right)\cap \mathrm{Sol}(C,f),A{x}^{\ast }\in \mathrm{Fix}\left(T\right)\cap \mathrm{Sol}(Q,g)\}.$$

Let f and g be the null operators in C and Q, respectively. Then, the split problem (6) becomes to the split fixed point problem studied in [39,40] which is to find an element point $x^{\ast }\in C$ such that

$$x^{\ast }\in \mathrm{Fix}\left(S\right)\mathrm{and}A{x}^{\ast }\in \mathrm{Fix}\left(T\right).$$

(7)

Let S and T be the identity operators in C and Q, respectively. Then, the split problem (6) becomes to the split variational inequality problem studied in [41] which is to find an element $x^{\ast }\in C$ such that

$$x^{\ast }\in \mathrm{Sol}(C,f)\mathrm{and}A{x}^{\ast }\in \mathrm{Sol}(Q,g).$$

(8)

The solution set of (8) is denoted by ${\Gamma }_1$, i.e.,

$$\Gamma =\{x^{\ast }\in \mathrm{Sol}(C,f),A{x}^{\ast }\in \mathrm{Sol}(Q,g)\}.$$

The split problems (6)–(8) have a common prototype that is the split feasibility ([42]) problem of finding a point $x^{\ast }$ such that

$$x^{\ast }\in C\mathrm{and}A{x}^{\ast }\in Q.$$

(9)

The split problems have emerged their powerful applications in image recovery and signal processing, control theory, biomedical engineering and geophysics. Some iterative algorithms for solving the split problems have been studied and extended by many scholars, see [43,44,45,46,47].

Motivated by the work in this direction, in this paper, we further survey the split problem (6) in which S and T are two pseudocontractive operators and f and g are two pseudomonotone operators. We present a Tseng-type iterative algorithm for solving the split problem (6) by using self-adaptive techniques. Under certain conditions, we show that the proposed algorithm converges weakly to a solution of the split problem (6).

2. Preliminaries

Let H be a real Hilbert space equipped with inner product $⟨·,·⟩$ and the induced norm defined by $x\mapsto \parallel x\parallel :=\sqrt{(x,x)}$. For any $x,x^{†}\in H$ and constant $\eta \in \mathbb{R}$, we have

$$\parallel \eta x+(1-\eta )x^{†}{\parallel }^2={\eta \parallel x\parallel }^2+(1-\eta )\parallel x^{†}{\parallel }^2-\eta (1-\eta ){\parallel x-x^{†}\parallel }^2.$$

(10)

The symbol $“\rightharpoonup ”$ denotes the weak convergence and the symbol $“\to ”$ denotes the strong convergence. Use ${\omega }_w\left(u^k\right)$ to denote the set of all weak cluster points of the sequence $\left\{u^k\right\}$, namely, ${\omega }_w\left(u^k\right)=\{u^{†}:\mathrm{there}\mathrm{exists}\left\{u^{k_i}\right\}\subset \left\{u^k\right\}\mathrm{such}\mathrm{that}{u}^{k_i}\rightharpoonup u^{†}\mathrm{as}i\to \infty \}$.

Recall that an operator $\varphi :H\to H$ is said to be

• Pseudomonotone, if

  $$\begin{array}{c}\hfill ⟨\varphi \left(\tilde{x}\right),x-\tilde{x}⟩\ge 0\mathrm{implies}⟨\varphi \left(x\right),x-\tilde{x}⟩\ge 0,\forall x,\tilde{x}\in H.\end{array}$$
• Weakly sequentially continuous, if $H\ni u^k\rightharpoonup \tilde{u}$ implies that $\varphi \left(u^k\right)\rightharpoonup \varphi \left(\tilde{u}\right)$.

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that an operator $S:C\to C$ is said to be pseudocontractive if

$$\parallel S\left(x\right)-S\left(x^{†}\right){\parallel }^2\le \parallel x-x^{†}{\parallel }^2+{\parallel (I-S)x-(I-S)x^{†}\parallel }^2,\mathrm{for}\mathrm{all}x,x^{†}\in C.$$

For given $u^{†}\in H$, there exists a unique point in C, denoted by ${\mathrm{proj}}_C\left[u^{†}\right]$ such that

$$\begin{array}{c}\hfill \parallel u^{†}-{\mathrm{proj}}_C\left[u^{†}\right]\parallel \le \parallel x-u^{†}\parallel ,\mathrm{for}\mathrm{all}x\in C.\end{array}$$

It is known that ${\mathrm{proj}}_C$ is firmly nonexpansive, that is, ${\mathrm{proj}}_C$ satisfies

$$\begin{array}{c}\hfill \parallel {\mathrm{proj}}_C\left[x^{\ast }\right]-{\mathrm{proj}}_C\left[x^{†}\right]{\parallel }^2\le ⟨{\mathrm{proj}}_C\left[x^{\ast }\right]-{\mathrm{proj}}_C\left[x^{†}\right],x^{\ast }-x^{†}⟩,\mathrm{for}\mathrm{all}{x}^{\ast },x^{†}\in H.\end{array}$$

It is obvious that ${\mathrm{proj}}_C$ is nonexpansive, i.e., $\parallel {\mathrm{proj}}_C\left[x^{\ast }\right]-{\mathrm{proj}}_C\left[x^{†}\right]\parallel \le \parallel x^{\ast }-x^{†}\parallel$ for all $x^{\ast },x^{†}\in H$. Moreover, ${\mathrm{proj}}_C$ satisfies the following inequality ([48])

$$\begin{array}{c}\hfill ⟨x^{\ast }-{\mathrm{proj}}_C\left[x^{\ast }\right],x^{†}-{\mathrm{proj}}_C\left[x^{\ast }\right]⟩\le 0,x^{\ast }\in H\mathrm{and}\mathrm{for}\mathrm{all}{x}^{†}\in C.\end{array}$$

(11)

Lemma 1 ([49]).

Let C be a nonempty, convex and closed subset of a Hilbert space H. Assume that the operator $S:C\to C$ is pseudocontractive and κ-Lipschitz continuous. Then, for all $\tilde{u}\in C$ and $u^{†}\in \mathrm{Fix}\left(S\right)$, we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel u^{†}-S((1-\alpha )\tilde{u}+\alpha S\left(\tilde{u}\right)){\parallel }^2\le \parallel \tilde{u}-u^{†}{\parallel }^2+(1-\alpha ){\parallel \tilde{u}-S((1-\alpha )\tilde{u}+\alpha S\left(\tilde{u}\right))\parallel }^2,\end{array}\end{array}$$

where α is a constant in $(0,\frac{1}{\sqrt{1+{\kappa }^2}+1})$.

Lemma 2 ([50]).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $f:C\to H$ be a continuous and pseudomonotone operator. Then $p^{†}\in \mathrm{Sol}(C,f)$ iff $p^{†}$ solves the following variational inequality

$$\begin{array}{c}\hfill ⟨f\left(u\right),u-p^{†}⟩\ge 0,\mathrm{for}\mathrm{all}u\in C.\end{array}$$

Lemma 3 ([51]).

Let C be a nonempty, convex and closed subset of a Hilbert space H. Let the operator $S:C\to C$ be continuous pseudocontractive. Then, S is demiclosed, i.e., $u^k\rightharpoonup \tilde{u}$ and $S\left(u^k\right)\to u^{†}$ as $k\to +\infty$ imply that $S\left(\tilde{u}\right)=u^{†}$.

Lemma 4 ([52]).

Let Γ be a nonempty closed convex subset of a real Hilbert space H. Let $\left\{x^k\right\}\subset H$ be a sequence. If the following assumptions are satisfied

$\left(i\right)$

$\forall x^{\ast }\in \Gamma$, ${lim}_{k\to +\infty }\parallel x^k-x^{\ast }\parallel$ exists;

$\left(ii\right)$

$w_{\omega }\left(x^k\right)\subset \Gamma$.

Then the sequence $\left\{x^k\right\}$ converges weakly to some point in Γ.

3. Main Results

In this section, we present our main results.

Let $H_1$ and $H_2$ be two real Hilbert spaces. Let $C\subset H_1$ and $Q\subset H_2$ be two nonempty closed convex sets. Let $S:C\to C$, $T:Q\to Q$, $f:H_1\to H_1$ and $g:H_2\to H_2$ be four nonlinear operators. Let $A:C\to Q$ be a bounded linear operator with its adjoint $A^{\ast }$.

Let $\left\{{\alpha }_k\right\}$, $\left\{{\beta }_k\right\}$, $\left\{{\zeta }_k\right\}$ and $\left\{{\lambda }_k\right\}$ be four real number sequences. Let $\vartheta$, $\delta$, $\omega$, $\mu$ and $\epsilon$ be five constants. Let ${\gamma }_0$ and ${\tau }_0$ be two positive constants.

Next, we introduce an iterative algorithm for solving the split problem (6).

In order to demonstrate the convergence analysis of Algorithm 1, we add some conditions on the operators and the parameters.

Algorithm 1: Select an initial point $x^0\in C$. Set $k=0$.

Step 1. Assume that the present iterate $x^k$ and the step-sizes ${\gamma }_k$ and ${\tau }_k$ are given.   Compute $$\begin{array}{c}\hfill \left\{\begin{array}{}\mathrm{(12)}& \hfill v^k=(1-{\beta }_k)x^k+{\beta }_{k}S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)],\mathrm{(13)}& y^k={\mathrm{proj}}_C[v^k-{\gamma }_{k}f\left(v^k\right)],\hfill \mathrm{(14)}& u^k=(1-\vartheta )v^k+\vartheta y^k+\vartheta {\gamma }_k[f\left(v^k\right)-f\left(y^k\right)],\hfill \mathrm{(15)}& w^k={\mathrm{proj}}_Q[A{u}^k-{\tau }_{k}g\left(A{u}^k\right)],\hfill \mathrm{(16)}& t^k=(1-\delta )A{u}^k+\delta w^k+\delta {\tau }_k[g\left(A{u}^k\right)-g\left(w^k\right)],\hfill \mathrm{(17)}& q^k=(1-{\zeta }_k)t^k+{\zeta }_{k}T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)].\hfill \end{array}\right.\end{array}$$ Step 2. Compute the next iterate $x^{k+1}$ by the following form $$\begin{array}{cc}\hfill x^{k+1}& ={\mathrm{proj}}_C[u^k+\epsilon A^{\ast }(q^k-A{u}^k)].\hfill \end{array}$$ (18) Step 3. Increase k by 1 and go back to Step 1. Meanwhile, update $${\gamma }_{k+1}=\left\{\begin{array}{cc}min\left\{{\gamma }_k,\frac{\omega \parallel y^k-v^k\parallel }{\parallel f\left(y^k\right)-f\left(v^k\right)\parallel }\right\},\hfill & f\left(y^k\right)\ne f\left(v^k\right),\hfill \\ {\gamma }_k,\hfill & else.\hfill \end{array}\right.$$ (19) and $$\begin{array}{c}\hfill {\tau }_{k+1}=\left\{\begin{array}{cc}min\left\{{\tau }_k,\frac{\mu \parallel w^k-A{u}^k\parallel }{\parallel g\left(w^k\right)-g\left(A{u}^k\right)\parallel }\right\},\hfill & g\left(w^k\right)\ne g\left(A{u}^k\right),\hfill \\ {\tau }_k,\hfill & else.\hfill \end{array}\right.\end{array}$$ (20)

Suppose that

(c1):

S and T are two pseudocontractive operators with Lipschitz constants $L_1$ and $L_2$, respectively;

(c2):

the operator f is pseudomonotone on $H_1$, weakly sequentially continuous and ${\kappa }_1$-Lipschitz continuous on C;

(c3):

the operator g is pseudomonotone on $H_2$, weakly sequentially continuous and ${\kappa }_2$-Lipschitz continuous on Q.

(r1):

$L_1>1$ and $0<\underline{\beta }<{\beta }_k<\overline{\beta }<{\alpha }_k<\overline{\alpha }<\frac{1}{\sqrt{1+L_1^2}+1}(\forall k\ge 0)$;

(r2):

$L_2>1$ and $0<\underline{\zeta }<{\zeta }_k<\overline{\zeta }<{\lambda }_k<\overline{\lambda }<\frac{1}{\sqrt{1+L_2^2}+1}(\forall k\ge 0)$;

(r3):

$\vartheta \in (0,1]$, $\delta \in (0,1]$, $\omega \in (0,1)$, $\mu \in (0,1)$ and $\epsilon \in (0,1/\parallel A{\parallel }^2)$.

Remark 1.

According to (19), the sequence $\left\{{\gamma }_k\right\}$ is monotonically decreasing. Moreover, by the ${\kappa }_1$-Lipschitz continuity of f, we obtain that $\frac{\omega \parallel y^k-v^k\parallel }{\parallel f\left(y^k\right)-f\left(v^k\right)\parallel }\ge \frac{\omega }{{\kappa }_1}$. Thus, $\left\{{\gamma }_k\right\}$ has a lower bound $min\{{\gamma }_0,\frac{\omega }{{\kappa }_1}\}$. Therefore, the limit ${lim}_{k\to +\infty }{\gamma }_k$ exists. Similarly, the sequence $\left\{{\tau }_k\right\}$ is monotonically decreasing and has a lower bound $min\{{\tau }_0,\frac{\mu }{{\kappa }_2}\}$. So, the limit ${lim}_{k\to +\infty }{\tau }_k$ exists.

Now, we prove our main theorem.

Theorem 1.

Suppose that $\Gamma \ne \varnothing$. Then the sequence $\left\{x^k\right\}$ generated by Algorithm 1 converges weakly to some point $p\in \Gamma$.

Proof. 

Let $x^{\ast }\in \Gamma$. Then, $x^{\ast }\in \mathrm{Fix}\left(S\right)\cap \mathrm{Sol}(C,f)$ and $A{x}^{\ast }\in \mathrm{Fix}\left(T\right)\cap \mathrm{Sol}(Q,g)$. By (10) and (12), we have

$$\begin{array}{cc}\hfill \parallel v^k-x^{\ast }{\parallel }^2& =\parallel (1-{\beta }_k)(x^k-x^{\ast })+{\beta }_k(S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^{\ast }){\parallel }^2\hfill \\ & =(1-{\beta }_k)\parallel x^k-x^{\ast }{\parallel }^2+{\beta }_k{\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^{\ast }\parallel }^2\hfill \\ & -(1-{\beta }_k){\beta }_k{\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2.\hfill \end{array}$$

(21)

Using Lemma 1, we obtain

$$\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^{\ast }{\parallel }^2\le \parallel x^k-x^{\ast }{\parallel }^2+(1-{\alpha }_k){\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2.$$

(22)

Combining (21) and (22), we derive

$$\begin{array}{cc}\hfill \parallel v^k-x^{\ast }{\parallel }^2& \le (1-{\beta }_k)\parallel x^k-x^{\ast }{\parallel }^2+{\beta }_k(1-{\alpha }_k){\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2\hfill \\ & +{\beta }_k\parallel x^k-x^{\ast }{\parallel }^2-(1-{\beta }_k){\beta }_k{\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2\hfill \\ & =\parallel x^k-x^{\ast }{\parallel }^2-{\beta }_k({\alpha }_k-{\beta }_k){\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2\hfill \\ \hfill \left(\mathrm{by}\mathrm{condition}\right(\mathrm{r}1\left)\right)& \le \parallel x^k-x^{\ast }{\parallel }^2.\hfill \end{array}$$

(23)

Similarly, according to (10), Lemma 1 and (17), we have the following estimate

$$\begin{array}{cc}\hfill \parallel q^k-A{x}^{\ast }{\parallel }^2& \le \parallel t^k-A{x}^{\ast }{\parallel }^2-({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2\hfill \\ \hfill \left(\mathrm{by}\mathrm{condition}\right(\mathrm{r}2\left)\right)& \le \parallel t^k-A{x}^{\ast }{\parallel }^2.\hfill \end{array}$$

(24)

Applying the inequality (11) to (13), we obtain

$$\begin{array}{c}\hfill ⟨y^k-v^k+{\gamma }_{k}f\left(v^k\right),y^k-x^{\ast }⟩\le 0.\end{array}$$

(25)

Since $x^{\ast }\in \mathrm{Sol}(C,f)$ and $y^k\in C$, $⟨f\left(x^{\ast }\right),y^k-x^{\ast }⟩\ge 0$. This together with the pseudomonotonicity of f implies that

$$\begin{array}{c}\hfill ⟨f\left(y^k\right),y^k-x^{\ast }⟩\ge 0.\end{array}$$

(26)

Based on (25) and (26), we get

$$\begin{array}{c}\hfill ⟨y^k-v^k,y^k-x^{\ast }⟩+{\gamma }_k⟨f\left(v^k\right)-f\left(y^k\right),y^k-x^{\ast }⟩\le 0.\end{array}$$

It follows that

$$\begin{array}{c}\hfill \frac{1}{2}(\parallel y^k-v^k{\parallel }^2+\parallel y^k-x^{\ast }{\parallel }^2-\parallel v^k-x^{\ast }{\parallel }^2)+{\gamma }_k⟨f\left(v^k\right)-f\left(y^k\right),y^k-x^{\ast }⟩\le 0,\end{array}$$

which yields

$$\parallel y^k-x^{\ast }{\parallel }^2\le \parallel v^k-x^{\ast }{\parallel }^2-2{\gamma }_k⟨f\left(v^k\right)-f\left(y^k\right),y^k-x^{\ast }⟩-{\parallel y^k-v^k\parallel }^2.$$

(27)

By (14), we have

$$\begin{array}{cc}\hfill \parallel u^k-x^{\ast }{\parallel }^2& =\parallel (1-\vartheta )(v^k-x^{\ast })+\vartheta (y^k-x^{\ast })+\vartheta {\gamma }_k[f\left(v^k\right)-f\left(y^k\right)]{\parallel }^2\hfill \\ \hfill & =\parallel (1-\vartheta )(v^k-x^{\ast })+\vartheta (y^k-x^{\ast }){\parallel }^2+{\vartheta }^2{\gamma }_k^2{\parallel f\left(v^k\right)-f\left(y^k\right)\parallel }^2\hfill \\ \hfill & +2\vartheta (1-\vartheta ){\gamma }_k⟨v^k-x^{\ast },f\left(v^k\right)-f\left(y^k\right)⟩\hfill \\ \hfill & +2{\vartheta }^2{\gamma }_k⟨y^k-x^{\ast },f\left(v^k\right)-f\left(y^k\right)⟩.\hfill \end{array}$$

(28)

From (10), we obtain

$$\begin{array}{cc}\hfill \parallel (1-\vartheta )(v^k-x^{\ast })+\vartheta (y^k-x^{\ast }){\parallel }^2& =(1-\vartheta )\parallel v^k-x^{\ast }{\parallel }^2+\vartheta {\parallel y^k-x^{\ast }\parallel }^2\hfill \\ \hfill & -(1-\vartheta )\vartheta \parallel v^k-y^k{\parallel }^2.\hfill \end{array}$$

(29)

Substituting (27) and (29) into (28), we deduce

$$\begin{array}{cc}\hfill \parallel u^k-x^{\ast }{\parallel }^2& \le (1-\vartheta )\parallel v^k-x^{\ast }{\parallel }^2+\vartheta \parallel v^k-x^{\ast }{\parallel }^2-2\vartheta {\gamma }_k⟨f\left(v^k\right)-f\left(y^k\right),y^k-x^{\ast }⟩-\vartheta {\parallel y^k-v^k\parallel }^2\hfill \\ \hfill & -(1-\vartheta )\vartheta \parallel v^k-y^k{\parallel }^2+{\vartheta }^2{\gamma }_k^2{\parallel f\left(v^k\right)-f\left(y^k\right)\parallel }^2+2\vartheta (1-\vartheta ){\gamma }_k⟨v^k-x^{\ast },f\left(v^k\right)-f\left(y^k\right)⟩\hfill \\ \hfill & +2{\vartheta }^2{\gamma }_k⟨y^k-x^{\ast },f\left(v^k\right)-f\left(y^k\right)⟩\hfill \\ \hfill & =\parallel v^k-x^{\ast }{\parallel }^2-(2-\vartheta )\vartheta \parallel v^k-y^k{\parallel }^2+{\vartheta }^2{\gamma }_k^2{\parallel f\left(v^k\right)-f\left(y^k\right)\parallel }^2\hfill \\ \hfill & -2\vartheta (1-\vartheta ){\gamma }_k⟨f\left(v^k\right)-f\left(y^k\right),y^k-v^k⟩\hfill \\ \hfill & \le \parallel v^k-x^{\ast }{\parallel }^2-(2-\vartheta )\vartheta \parallel v^k-y^k{\parallel }^2+{\vartheta }^2{\gamma }_k^2{\parallel f\left(v^k\right)-f\left(y^k\right)\parallel }^2\hfill \\ \hfill & +2\vartheta (1-\vartheta ){\gamma }_k\parallel f\left(v^k\right)-f\left(y^k\right)\parallel \parallel y^k-v^k\parallel .\hfill \end{array}$$

(30)

Thanks to (19), $\parallel f\left(v^k\right)-f\left(y^k\right)\parallel \le \frac{\omega \parallel y^k-v^k\parallel }{{\gamma }_{k+1}}$. It follows from (30) that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel u^k-x^{\ast }{\parallel }^2& \le \parallel v^k-x^{\ast }{\parallel }^2-(2-\vartheta )\vartheta \parallel v^k-y^k{\parallel }^2+{\vartheta }^2{\omega }^2\frac{{\gamma }_k^2}{{\gamma }_{k+1}^2}\parallel y^k-v^k\parallel \hfill \\ & +2\vartheta (1-\vartheta )\omega \frac{{\gamma }_k}{{\gamma }_{k+1}}{\parallel y^k-v^k\parallel }^2\hfill \\ & =\parallel v^k-x^{\ast }{\parallel }^2-\vartheta \left[2-\vartheta -\vartheta {\omega }^2\frac{{\gamma }_k^2}{{\gamma }_{k+1}^2}-2(1-\vartheta )\omega \frac{{\gamma }_k}{{\gamma }_{k+1}}\right]{\parallel y^k-v^k\parallel }^2.\hfill \end{array}\end{array}$$

(31)

By Remark 1, ${lim}_{k\to +\infty }\frac{{\gamma }_k}{{\gamma }_{k+1}}=1$. So,

$$\begin{array}{cc}\hfill & \underset{k\to +\infty }{lim}\left[2-\vartheta -\vartheta {\omega }^2\frac{{\gamma }_k^2}{{\gamma }_{k+1}^2}-2(1-\vartheta )\omega \frac{{\gamma }_k}{{\gamma }_{k+1}}\right]=2-\vartheta -\vartheta {\omega }^2-2(1-\vartheta )\omega =-\vartheta (\omega -1)(\omega +\frac{2-\vartheta }{\vartheta })>0.\end{array}$$

Then, there exists $\sigma >0$ and $m_1$ such that $2-\vartheta -\vartheta {\omega }^2\frac{{\gamma }_k^2}{{\gamma }_{k+1}^2}-2(1-\vartheta )\omega \frac{{\gamma }_k}{{\gamma }_{k+1}}\ge \sigma$ when $k\ge m_1$. In combination with (31), we get

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel u^k-x^{\ast }{\parallel }^2\le \parallel v^k-x^{\ast }{\parallel }^2-\sigma \vartheta {\parallel y^k-v^k\parallel }^2.\end{array}\end{array}$$

This together with (23) implies that

$$\begin{array}{cc}\hfill \parallel u^k-x^{\ast }{\parallel }^2& \le \parallel x^k-x^{\ast }{\parallel }^2-{\beta }_k({\alpha }_k-{\beta }_k){\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel }^2\hfill \\ \hfill & -\sigma \vartheta \parallel y^k-v^k{\parallel }^2.\hfill \end{array}$$

(32)

By the property (11) of ${\mathrm{proj}}_Q$ and (15), we have

$$\begin{array}{c}\hfill ⟨w^k-A{u}^k+{\tau }_{k}g\left(A{u}^k\right),w^k-A{x}^{\ast }⟩\le 0.\end{array}$$

(33)

Since $A{x}^{\ast }\in \mathrm{Sol}(Q,g)$ and $w^k\in Q$, $⟨g\left(A{x}^{\ast }\right),w^k-A{x}^{\ast }⟩\ge 0$. By the pseudomonotonicity of g, we obtain

$$⟨g\left(w^k\right),w^k-A{x}^{\ast }⟩\ge 0.$$

(34)

Taking into account (33) and (34), we obtain

$$⟨w^k-A{u}^k,w^k-A{x}^{\ast }⟩+{\tau }_k⟨g\left(A{u}^k\right)-g\left(w^k\right),w^k-A{x}^{\ast }⟩\le 0,$$

which is equivalent to

$$\frac{1}{2}(\parallel w^k-A{u}^k{\parallel }^2+\parallel w^k-A{x}^{\ast }{\parallel }^2-\parallel A{u}^k-A{x}^{\ast }{\parallel }^2)+{\tau }_k⟨g\left(A{u}^k\right)-g\left(w^k\right),w^k-A{x}^{\ast }⟩\le 0.$$

It follows that

$$\parallel w^k-A{x}^{\ast }{\parallel }^2\le \parallel A{u}^k-A{x}^{\ast }{\parallel }^2-2{\tau }_k⟨g\left(A{u}^k\right)-g\left(w^k\right),w^k-A{x}^{\ast }⟩-{\parallel w^k-A{u}^k\parallel }^2.$$

(35)

From (14), we receive

$$\begin{array}{cc}\hfill \parallel t^k-A{x}^{\ast }{\parallel }^2& =\parallel (1-\delta )(A{u}^k-A{x}^{\ast })+\delta (w^k-A{x}^{\ast })+\delta {\tau }_k[g\left(A{u}^k\right)-g\left(w^k\right)]{\parallel }^2\hfill \\ \hfill & =\parallel (1-\delta )(A{u}^k-A{x}^{\ast })+\delta (w^k-A{x}^{\ast }){\parallel }^2+{\delta }^2{\tau }_k^2{\parallel g\left(A{u}^k\right)-g\left(w^k\right)\parallel }^2\hfill \\ \hfill & +2\delta (1-\delta ){\tau }_k⟨A{u}^k-A{x}^{\ast },g\left(A{u}^k\right)-g\left(w^k\right)⟩\hfill \\ \hfill & +2{\delta }^2{\tau }_k⟨w^k-A{x}^{\ast },g\left(A{u}^k\right)-g\left(w^k\right)⟩.\hfill \end{array}$$

(36)

By virtue of (10), we achieve

$$\begin{array}{cc}\hfill \parallel (1-\delta )(A{u}^k-A{x}^{\ast })+\delta (w^k-A{x}^{\ast }){\parallel }^2& =(1-\delta )\parallel A{u}^k-A{x}^{\ast }{\parallel }^2+\delta {\parallel w^k-A{x}^{\ast }\parallel }^2\hfill \\ \hfill & -(1-\delta )\delta \parallel A{u}^k-w^k{\parallel }^2.\hfill \end{array}$$

(37)

Substituting (35) and (37) into (36), we obtain

$$\begin{array}{cc}\hfill \parallel t^k-A{x}^{\ast }{\parallel }^2& \le \parallel A{u}^k-A{x}^{\ast }{\parallel }^2-(2-\delta )\delta \parallel A{u}^k-w^k{\parallel }^2+{\delta }^2{\tau }_k^2{\parallel g\left(A{u}^k\right)-g\left(w^k\right)\parallel }^2\hfill \\ \hfill & -2\delta (1-\delta ){\tau }_k⟨w^k-A{u}^k,g\left(A{u}^k\right)-g\left(w^k\right)⟩\hfill \\ \hfill & \le \parallel A{u}^k-A{x}^{\ast }{\parallel }^2-(2-\delta )\delta \parallel A{u}^k-w^k{\parallel }^2+{\delta }^2{\tau }_k^2{\parallel g\left(A{u}^k\right)-g\left(w^k\right)\parallel }^2\hfill \\ \hfill & +2\delta (1-\delta ){\tau }_k\parallel w^k-A{u}^k\parallel \parallel g\left(A{u}^k\right)-g\left(w^k\right)\parallel .\hfill \end{array}$$

(38)

Duo to (20), we have

$$\parallel g\left(A{u}^k\right)-g\left(w^k\right)\parallel \le \frac{\mu \parallel A{u}^k-w^k\parallel }{{\tau }_{k+1}}.$$

This together with (38) implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel t^k-A{x}^{\ast }{\parallel }^2& \le \parallel A{u}^k-A{x}^{\ast }\parallel -(2-\delta )\delta \parallel A{u}^k-w^k{\parallel }^2+{\delta }^2{\mu }^2\frac{{\tau }_k^2}{{\tau }_{k+1}^2}\parallel w^k-A{u}^k\parallel \hfill \\ & +2\delta (1-\delta )\mu \frac{{\tau }_k}{{\tau }_{k+1}}{\parallel A{u}^k-w^k\parallel }^2\hfill \\ & =\parallel A{u}^k-A{x}^{\ast }\parallel -\delta \left[2-\delta -\delta {\mu }^2\frac{{\tau }_k^2}{{\tau }_{k+1}^2}-2(1-\delta )\mu \frac{{\tau }_k}{{\tau }_{k+1}}\right]{\parallel A{u}^k-w^k\parallel }^2.\hfill \end{array}\end{array}$$

(39)

By Remark 1, ${lim}_{k\to +\infty }\frac{{\tau }_k}{{\tau }_{k+1}}=1$ and hence

$$\begin{array}{c}\hfill \underset{k\to \infty }{lim}\left[2-\delta -\delta {\mu }^2\frac{{\tau }_k^2}{{\tau }_{k+1}^2}-2(1-\delta )\mu \frac{{\tau }_k}{{\tau }_{k+1}}\right]=2-\delta -\delta {\mu }^2-2(1-\delta )\mu >0.\end{array}$$

So, there exists $\varrho >0$ and $m_2$ such that

$$2-\delta -\delta {\mu }^2\frac{{\tau }_k^2}{{\tau }_{k+1}^2}-2(1-\delta )\mu \frac{{\tau }_k}{{\tau }_{k+1}}\ge \varrho ,$$

when $k\ge m_2$.

In the light of (39), we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel t^k-A{x}^{\ast }{\parallel }^2\le \parallel A{u}^k-A{x}^{\ast }\parallel -\varrho \delta \parallel w^k-A{u}^k{\parallel }^2.\end{array}\end{array}$$

(40)

Owing to (24) and (40), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel q^k-A{x}^{\ast }{\parallel }^2& \le \parallel A{u}^k-A{x}^{\ast }\parallel -({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2\hfill \\ & -\varrho \delta \parallel w^k-A{u}^k{\parallel }^2.\hfill \end{array}\end{array}$$

(41)

Observe that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨u^k-x^{\ast },A^{\ast }(q^k-A{u}^k)⟩& =⟨A{u}^k-A{x}^{\ast },q^k-A{u}^k⟩\hfill \\ & =\frac{1}{2}[\parallel q^k-A{x}^{\ast }{\parallel }^2-\parallel A{u}^k-A{x}^{\ast }{\parallel }^2]-\frac{1}{2}{\parallel q^k-A{u}^k\parallel }^2.\hfill \end{array}\end{array}$$

(42)

Combining (41) and (42), we acquire

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨u^k-x^{\ast },A^{\ast }(q^k-A{u}^k)⟩& \le -\frac{1}{2}\varrho \delta \parallel w^k-A{u}^k{\parallel }^2-\frac{1}{2}{\parallel q^k-A{u}^k\parallel }^2\hfill \\ & -\frac{1}{2}({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2.\hfill \end{array}\end{array}$$

(43)

In view of (18), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x^{k+1}-x^{\ast }{\parallel }^2& =\parallel {\mathrm{proj}}_C[u^k+\epsilon A^{\ast }(q^k-A{u}^k)]-{\mathrm{proj}}_C\left[x^{\ast }\right]{\parallel }^2\hfill \\ & \le \parallel u^k-x^{\ast }+\epsilon A^{\ast }(q^k-A{u}^k){\parallel }^2\hfill \\ & =\parallel u^k-x^{\ast }{\parallel }^2+{\parallel \epsilon A^{\ast }(q^k-A{u}^k)\parallel }^2+2\epsilon ⟨A^{\ast }(q^k-A{u}^k),u^k-x^{\ast }⟩.\hfill \end{array}\end{array}$$

It follows from (32) and (43) that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x^{k+1}-x^{\ast }{\parallel }^2& \le \parallel u^k-x^{\ast }{\parallel }^2+{\epsilon }^2{\parallel A\parallel }^2\parallel q^k-A{u}^k{\parallel }^2-\epsilon \varrho \delta {\parallel w^k-A{u}^k\parallel }^2\hfill \\ & -\epsilon \parallel q^k-A{u}^k{\parallel }^2-\epsilon ({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2\hfill \\ & =\parallel u^k-x^{\ast }{\parallel }^2{-\epsilon (1-\epsilon \parallel A\parallel }^2)\parallel q^k-A{u}^k{\parallel }^2-\epsilon \varrho \delta {\parallel w^k-A{u}^k\parallel }^2\hfill \\ & -\epsilon ({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2\hfill \\ & \le \parallel x^k-x^{\ast }{\parallel }^2-\epsilon \varrho \delta \parallel w^k-A{u}^k{\parallel }^2{-\epsilon (1-\epsilon \parallel A\parallel }^2)\parallel q^k-A{u}^k{\parallel }^2\hfill \\ & -{\beta }_k({\alpha }_k-{\beta }_k)\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k{\parallel }^2-\sigma \vartheta {\parallel y^k-v^k\parallel }^2\hfill \\ & -\epsilon ({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T\left(t^k\right)]-t^k\parallel }^2\hfill \\ & \le \parallel x^k-x^{\ast }{\parallel }^2,\hfill \end{array}\end{array}$$

(44)

which implies that ${lim}_{k\to +\infty }\parallel x^k-x^{\ast }\parallel$ exists. Since $\parallel x^{k+1}-x^{\ast }\parallel \le \parallel u^k-x^{\ast }\parallel \le \parallel v^k-x^{\ast }\parallel \le \parallel x^k-x^{\ast }\parallel$, we deduce

$$\begin{array}{c}\hfill \underset{k\to +\infty }{lim}\parallel u^k-x^{\ast }\parallel =\underset{k\to +\infty }{lim}\parallel v^k-x^{\ast }\parallel =\underset{k\to +\infty }{lim}\parallel x^k-x^{\ast }\parallel .\end{array}$$

(45)

So, the sequences $\left\{x^k\right\}$, $\left\{u^k\right\}$ and $\left\{v^k\right\}$ are all bounded.

By virtue of (44), we derive

$$\begin{array}{c}\hfill \begin{array}{cc}& {\beta }_k({\alpha }_k-{\beta }_k)\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S{x}^k]-x^k{\parallel }^2+\sigma \vartheta \parallel y^k-v^k{\parallel }^2+\epsilon \varrho \delta {\parallel w^k-A{u}^k\parallel }^2\hfill \\ & {+\epsilon (1-\epsilon \parallel A\parallel }^2)\parallel q^k-A{u}^k{\parallel }^2+\epsilon ({\lambda }_k-{\zeta }_k){\zeta }_k{\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T{t}^k]-t^k\parallel }^2\hfill \\ & \le \parallel x^k-x^{\ast }{\parallel }^2-{\parallel x^{k+1}-x^{\ast }\parallel }^2\to 0,\hfill \end{array}\end{array}$$

which implies that

$$\begin{array}{c}\hfill \left\{\begin{array}{}\mathrm{(46)}& \underset{k\to +\infty }{lim}\parallel q^k-A{u}^k\parallel =0,\hfill \mathrm{(47)}& \underset{k\to +\infty }{lim}\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S(x^k)]-x^k\parallel =0,\hfill \mathrm{(48)}& \underset{k\to +\infty }{lim}\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)]-t^k\parallel =0,\hfill \mathrm{(49)}& \underset{k\to +\infty }{lim}\parallel y^k-v^k\parallel =0,\hfill \mathrm{(50)}& \underset{k\to +\infty }{lim}\parallel w^k-A{u}^k\parallel =0.\hfill \end{array}\right.\end{array}$$

By the $L_1$-Lipschitz continuity of S, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel S\left(x^k\right)-x^k\parallel & \le \parallel S\left(x^k\right)-S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]\parallel +\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel \hfill \\ & \le L_1{\alpha }_k\parallel S\left(x^k\right)-x^k\parallel +\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel .\hfill \end{array}\end{array}$$

It follows that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel S\left(x^k\right)-x^k\parallel \le \frac{1}{1-L_1{\alpha }_k}\parallel S[(1-{\alpha }_k)x^k+{\alpha }_{k}S\left(x^k\right)]-x^k\parallel .\end{array}\end{array}$$

This together with (47) implies that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \underset{k\to +\infty }{lim}\parallel S\left(x^k\right)-x^k\parallel =0.\end{array}\end{array}$$

(51)

From (12) and (47), we conclude that $\parallel x^k-v^k\parallel \to 0$.

Next, we show that ${\omega }_w\left(x^k\right)\subset \Gamma$. Pick any $p^{†}\in {\omega }_w\left(x^k\right)$. Then, there exists a subsequence $\left\{x^{k_i}\right\}$ of $\left\{x^k\right\}$ such that $x^{k_i}\rightharpoonup p^{†}$ as $i\to +\infty$. In addition, $y^{k_i}\rightharpoonup p^{†}$ and $v^{k_i}\rightharpoonup p^{†}$ as $i\to +\infty$.

First, we prove that $p^{†}\in \mathrm{Sol}(C,f)$. In view of (11) and $y^{k_i}={\mathrm{proj}}_C[v^{k_i}-{\gamma }_{k_i}f\left(v^{k_i}\right)]$, we achieve

$$\begin{array}{c}\hfill ⟨y^{k_i}-v^{k_i}+{\gamma }_{k_i}f\left(v^{k_i}\right),y^{k_i}-u⟩\le 0,\mathrm{for}\mathrm{all}u\in C.\end{array}$$

It follows that

$$\frac{1}{{\gamma }_{k_i}}⟨v^{k_i}-y^{k_i},u-y^{k_i}⟩+⟨f\left(v^{k_i}\right),y^{k_i}-v^{k_i}⟩\le ⟨f\left(v^{k_i}\right),u-v^{k_i}⟩,\mathrm{for}\mathrm{all}u\in C.$$

(52)

Noting that from (49), we have ${lim}_{i\to +\infty }\parallel v^{k_i}-y^{k_i}\parallel =0$. Meanwhile, $\left\{y^{k_i}\right\}$ and $\left\{f\left(v^{k_i}\right)\right\}$ are bounded. Then, by (52), we deduce

$$\begin{array}{c}\hfill \underset{i\to +\infty }{lim\; inf}⟨f\left(v^{k_i}\right),u-v^{k_i}⟩\ge 0,\mathrm{for}\mathrm{all}u\in C.\end{array}$$

(53)

Let $\left\{{ϵ}_j\right\}$ be a positive real numbers sequence satisfying ${lim}_{j\to +\infty }{ϵ}_j=0$. On account of (53), for each ${ϵ}_j$, there exists the smallest positive integer $n_j$ such that

$$\begin{array}{c}\hfill ⟨f(v^{k_{i_j}}),u-v^{k_{i_j}}⟩+{ϵ}_j\ge 0,\mathrm{for}\mathrm{all}j\ge n_j.\end{array}$$

(54)

Moreover, for each $j>0$, $f(v^{k_{i_j}})\ne 0$. Setting $\phi (v^{k_{i_j}})=\frac{f(v^{k_{i_j}})}{\parallel f(v^{k_{i_j}}){\parallel }^2}$, we have $⟨f(v^{k_{i_j}}),\phi (v^{k_{i_j}})⟩=1$. From (54), we have

$$\begin{array}{c}\hfill ⟨f(v^{k_{i_j}}),u+{ϵ}_j\phi (v^{k_{i_j}})-v^{k_{i_j}}⟩\ge 0.\end{array}$$

By the pseudomonotonicity of f, we get

$$\begin{array}{c}\hfill ⟨f(u+{ϵ}_j\phi (v^{k_{i_j}})),u+{ϵ}_j\phi (v^{k_{i_j}})-v^{k_{i_j}}⟩\ge 0,\end{array}$$

which implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨f(u),u-v^{k_{i_j}}⟩& \ge ⟨f(u)-f(u+{ϵ}_j\phi (v^{k_{i_j}})),u+{ϵ}_j\phi (v^{k_{i_j}})-v^{k_{i_j}}⟩\hfill \\ & +⟨f(u),-{ϵ}_j\phi (v^{k_{i_j}})⟩.\hfill \end{array}\end{array}$$

(55)

Because of $f(v^{k_{i_j}})\rightharpoonup f(p^{†})$, we have

$$\begin{array}{c}\hfill \underset{j\to +\infty }{lim\; inf}\parallel f(v^{k_{i_j}})\parallel \ge \parallel f(p^{†})\parallel >0.\end{array}$$

Then,

$$\begin{array}{c}\hfill \underset{j\to +\infty }{lim}\parallel {ϵ}_j\phi (v^{k_{i_j}})\parallel =\underset{j\to +\infty }{lim}\frac{{ϵ}_j}{\parallel f(v^{k_{i_j}})\parallel }=0.\end{array}$$

This together with (55) implies that

$$\begin{array}{c}\hfill ⟨f(u),u-p^{†}⟩\ge 0.\end{array}$$

(56)

By Lemma 2 and (56), we conclude that $p^{†}\in \mathrm{Sol}(C,f)$.

On the other hand, by (51), $\parallel S{x}^{k_i}-x^{k_i}\parallel \to 0$ as $i\to +\infty$. This together with $x^{k_i}\rightharpoonup p^{†}$ and Lemma 3 implies that $p^{†}\in \mathrm{Fix}(S)$. Therefore, $p^{†}\in \mathrm{Fix}(S)\cap \mathrm{Sol}(C,f)$.

Next, we show that $A{p}^{†}\in \mathrm{Fix}(T)\cap \mathrm{Sol}(Q,g)$. Observe that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel T(t^k)-t^k\parallel & \le \parallel T(t^k)-T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)]\parallel +\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)]-t^k\parallel \hfill \\ & \le L_2{\lambda }_k\parallel T(t^k)-t^k\parallel +\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)]-t^k\parallel .\hfill \end{array}\end{array}$$

It follows that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel T(t^k)-t^k\parallel \le \frac{1}{1-L_2{\lambda }_k}\parallel T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)]-t^k\parallel .\end{array}\end{array}$$

This together with (48) implies that

$$\begin{array}{c}\hfill \underset{k\to +\infty }{lim}\parallel T(t^k)-t^k\parallel =0.\end{array}$$

(57)

From (14), $u^{k_i}\rightharpoonup p^{†}$ as $i\to +\infty$. Thanks to (17) and (48), we have $q^{k_i}-t^{k_i}\to 0$ as $i\to +\infty$. Combining with (46), we deduce that $t^{k_i}\rightharpoonup A{p}^{†}$. Applying Lemma 3 to (57), we obtain that $A{p}^{†}\in \mathrm{Fix}(T)$.

Next, we show that $A{p}^{†}\in \mathrm{Sol}(Q,g)$. In view of (10) and $w^{k_i}=pro{j}_Q[A{u}^{k_i}-{\tau }_{k_i}g(A{u}^{k_i})]$, we achieve

$$\begin{array}{c}\hfill ⟨w^{k_i}-A{u}^{k_i}+{\tau }_{k_i}g(A{u}^{k_i}),w^{k_i}-v⟩\le 0,\mathrm{for}\mathrm{all}v\in Q.\end{array}$$

It follows that

$$\frac{1}{{\tau }_{k_i}}⟨w^{k_i}-A{u}^{k_i},w^{k_i}-v⟩+⟨g(A{u}^{k_i}),w^{k_i}-A{u}^{k_i}⟩\le ⟨g(A{u}^{k_i}),v-A{u}^{k_i}⟩,\mathrm{for}\mathrm{all}v\in Q.$$

(58)

Noting that from (r3), we have ${lim}_{i\to +\infty }\parallel w^{k_i}-A{u}^{k_i}\parallel =0$. Then, by (58), we deduce

$$\begin{array}{c}\hfill \underset{i\to +\infty }{lim\; inf}⟨g(A{u}^{k_i}),v-A{u}^{k_i}⟩\ge 0,\mathrm{for}\mathrm{all}v\in Q.\end{array}$$

(59)

Choose a positive real numbers sequence $\left\{{\upsilon }_j\right\}$ such that ${lim}_{j\to +\infty }{\upsilon }_j=0$. In terms of (59), for each ${\upsilon }_j$, there exists the smallest positive integer $m_j$ such that

$$\begin{array}{c}\hfill ⟨g(A{u}^{k_i}),v-A{u}^{k_i}⟩+{\upsilon }_j\ge 0,\mathrm{for}\mathrm{all}j\ge m_j.\end{array}$$

(60)

Moreover, for each $j>0$, $g(A{u}^{k_i})\ne 0$. Setting $\psi (u^{k_i})=\frac{g(A{u}^{k_i})}{\parallel g(A{u}^{k_i}){\parallel }^2}$, we have $⟨g(A{u}^{k_i}),\psi (u^{k_i})⟩=1$. From (60), we have

$$\begin{array}{c}\hfill ⟨g(A(u^{k_{i_j}})),v+{\upsilon }_j\psi (u^{k_{i_j}})-A{u}^{k_{i_j}}⟩\ge 0.\end{array}$$

By the pseudomonotonicity of g, we get

$$\begin{array}{c}\hfill ⟨g(v+{\upsilon }_j\psi (u^{k_{i_j}})),v+{\upsilon }_j\psi (u^{k_{i_j}})-A{u}^{k_{i_j}}⟩\ge 0,\end{array}$$

which implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨g(v),v-A{u}^{k_{i_j}}⟩& \ge ⟨g(v)-g(v+{\upsilon }_j\psi (u^{k_{i_j}})),v+{\upsilon }_j\psi (u^{k_{i_j}})-A{u}^{k_{i_j}}⟩\hfill \\ & +⟨g(v),-{\upsilon }_j\psi (u^{k_{i_j}})⟩.\hfill \end{array}\end{array}$$

(61)

Because of $g(A(u^{k_{i_j}}))\rightharpoonup g(A{p}^{†})$, we have

$$\begin{array}{c}\hfill \underset{j\to +\infty }{lim\; inf}\parallel g(A(u^{k_{i_j}}))\parallel \ge \parallel g(A{p}^{†})\parallel >0.\end{array}$$

Then,

$$\begin{array}{c}\hfill \underset{j\to +\infty }{lim}\parallel {\upsilon }_j\psi (u^{k_{i_j}}))\parallel =\underset{j\to \infty }{lim}\frac{{\upsilon }_j}{\parallel g(A{u}^{k_i})\parallel }=0.\end{array}$$

This together with (61) implies that

$$\begin{array}{c}\hfill ⟨g(v),v-A{p}^{†}⟩\ge 0,\mathrm{for}\mathrm{all}v\in Q.\end{array}$$

(62)

By Lemma 2 and (62), we conclude that $A{p}^{†}\in \mathrm{Sol}(Q,g)$. So, $p\in \Gamma$ and ${\omega }_w(x^k)\subset \Gamma$.

Finally, we show that the entire sequence $\left\{x^k\right\}$ converges weakly to $p^{†}$. As a matter of fact, we have the following facts:

(i)

$\forall x^{\ast }\in \Gamma$, ${lim}_{k\to +\infty }\parallel x^k-x^{\ast }\parallel$ exists;

(ii)

$w_{\omega }(x^k)\subset \Gamma$;

(iii)

$p^{†}\in w_{\omega }(x^k)$.

Thus, by Lemma 4, we deduce that the sequence $\left\{x^k\right\}$ weakly converges to $p^{†}\in \Gamma$. This completes the proof. □

Corollary 1.

Suppose that ${\Gamma }_1\ne \varnothing$. Then the sequence $\left\{x^k\right\}$ generated by Algorithm 2 converges weakly to some point $p_1\in {\Gamma }_1$.

Algorithm 2: Select an initial point $x^0\in C$. Set $k=0$.

Step 1. Assume that the present iterate $x^k$ and the step-sizes ${\gamma }_k$ and ${\tau }_k$ are given.   Compute $$\begin{array}{c}\hfill \left\{\begin{array}{}\mathrm{(63)}& y^k={\mathrm{proj}}_C[x^k-{\beta }_{k}f(x^k)],\hfill \mathrm{(64)}& u^k=(1-\vartheta )x^k+\vartheta y^k+\vartheta {\gamma }_k[f(x^k)-f(y^k)],\hfill \mathrm{(65)}& w^k={\mathrm{proj}}_Q[A{u}^k-{\tau }_{k}g(A{u}^k)],\hfill \mathrm{(66)}& t^k=(1-\delta )A{u}^k+\delta w^k+\delta {\tau }_k[g(A{u}^k)-g(w^k)],\hfill \end{array}\right.\end{array}$$ Step 2. Compute the next iterate $x^{k+1}$ by the following form $$\begin{array}{cc}\hfill x^{k+1}& ={\mathrm{proj}}_C[u^k+\epsilon A^{\ast }(t^k-A{u}^k)].\hfill \end{array}$$ (67) Step 3. Increase k by 1 and go back to Step 1. Meanwhile, update $$\begin{array}{c}\hfill {\gamma }_{k+1}=\left\{\begin{array}{cc}min\left\{{\gamma }_k,\frac{\omega \parallel y^k-x^k\parallel }{\parallel f(y^k)-f(x^k)\parallel }\right\},\hfill & f(y^k)\ne f(x^k),\hfill \\ {\gamma }_k,\hfill & else.\hfill \end{array}\right.\end{array}$$ (68) $$\begin{array}{c}\hfill {\tau }_{k+1}=\left\{\begin{array}{cc}min\left\{{\tau }_k,\frac{\mu \parallel w^k-A{u}^k\parallel }{\parallel g(w^k)-g(A{u}^k)\parallel }\right\},\hfill & g(w^k)\ne g(A{u}^k),\hfill \\ {\tau }_k,\hfill & else.\hfill \end{array}\right.\end{array}$$ (69)

4. Application to Split Pseudoconvex Optimization Problems and Fixed Point Problems

In this section, we apply Algorithm 1 to solve split pseudoconvex optimization problems and fixed point problems.

Let ${\mathbb{R}}^n$ be the Euclidean space. Let C be a closed convex set in ${\mathbb{R}}^n$. Recall that a differentiable function $F:{\mathbb{R}}^n\to \mathbb{R}$ is said to be pseudoconvex on C if for every pair of distinct points $x,y\in C$,

$$\nabla F{(x)}^T(y-x)\ge 0\mathrm{implies}F(y)\ge F(x).$$

Now, we consider the following optimization problem

$$\begin{array}{c}\hfill minF(x)\mathrm{subject}\mathrm{to}x\in C,\end{array}$$

(70)

where $F(x)$ is pseudoconvex and twice continuously differentiable.

Denote by $\mathrm{SOP}(C,F)$ the solution set of optimization problem (70).

The following lemma reveals the relationship between the variational inequality and the pseudoconvex optimization problem.

Lemma 5

([53]). Suppose that $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable and pseudoconvex on C. Then $x^{\ast }\in C$ satisfies

$$\nabla F{(x^{\ast })}^T(x-x^{\ast })\ge 0\mathrm{for}\mathrm{all}x\in C$$

if and only if $x^{\ast }$ is a minimum of $F(x)$ in C.

Let ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$ be two Euclidean spaces. Let $C\subset {\mathbb{R}}^n$ and $Q\subset {\mathbb{R}}^m$ be two nonempty closed convex sets. Let A be a given $m\times n$ real matrix. Let $S:C\to C$ and $T:Q\to Q$ be two pseudocontractive operators with Lipschitz constants $L_1$ and $L_2$, respectively. Let $F:{\mathbb{R}}^n\to \mathbb{R}$ be a differentiable function with ${\kappa }_1$-Lipschitz continuous gradient which is also pseudoconvex on C. Let $G:{\mathbb{R}}^m\to \mathbb{R}$ be a differentiable function with ${\kappa }_2$-Lipschitz continuous gradient which is also pseudoconvex on Q.

Consider the following split problem of finding a point $x^{\ast }\in C$ such that

$$x^{\ast }\in \mathrm{Fix}(S)\cap \mathrm{SOP}(C,F)\mathrm{and}A{x}^{\ast }\in \mathrm{Fix}(T)\cap \mathrm{SOP}(Q,G).$$

(71)

The solution set of (71) is denoted by ${\Gamma }_2$, i.e.,

$${\Gamma }_2=\{x^{\ast }\in \mathrm{Fix}(S)\cap \mathrm{SOP}(C,F),A{x}^{\ast }\in \mathrm{Fix}(T)\cap \mathrm{SOP}(Q,G)\}.$$

Next, we introduce an iterative algorithm for solving the split problem (71).

Let $\left\{{\alpha }_k\right\}$, $\left\{{\beta }_k\right\}$, $\left\{{\zeta }_k\right\}$ and $\left\{{\lambda }_k\right\}$ be four real number sequences. Let $\vartheta$, $\delta$, $\omega$, $\mu$ and $\epsilon$ be five constants. Let ${\gamma }_0$ and ${\tau }_0$ be two positive constants.

Theorem 2.

Suppose that ${\Gamma }_2\ne \varnothing$ and the conditions (r1)–(r3) hold. Then the sequence $\left\{x^k\right\}$ generated by Algorithm 3 converges to some point $p\in {\Gamma }_2$.

Algorithm 3: Select an initial point $x^0\in C$. Set $k=0$.

Step 1. Assume that the present iterate $x^k$ and the step-sizes ${\gamma }_k$ and ${\tau }_k$ are given.   Compute $$\begin{array}{c}\hfill \left\{\begin{array}{c}{v}^k=(1-{\beta }_k)x^k+{\beta }_{k}S[(1-{\alpha }_k)x^k+{\alpha }_{k}S(x^k)],\hfill \\ y^k={\mathrm{proj}}_C[v^k-{\gamma }_k\nabla F(v^k)],\hfill \\ u^k=(1-\vartheta )v^k+\vartheta y^k+\vartheta {\gamma }_k[\nabla F(v^k)-\nabla F(y^k)],\hfill \\ w^k={\mathrm{proj}}_Q[A{u}^k-{\tau }_k\nabla G(A{u}^k)],\hfill \\ t^k=(1-\delta )A{u}^k+\delta w^k+\delta {\tau }_k[\nabla G(A{u}^k)-\nabla G(w^k)],\hfill \\ q^k=(1-{\zeta }_k)t^k+{\zeta }_{k}T[(1-{\lambda }_k)t^k+{\lambda }_{k}T(t^k)].\hfill \end{array}\right.\end{array}$$ Step 2. Compute the next iterate $x^{k+1}$ by the following form $$\begin{array}{cc}\hfill x^{k+1}& ={\mathrm{proj}}_C[u^k+\epsilon A^{\ast }(q^k-A{u}^k)].\hfill \end{array}$$ Step 3. Increase k by 1 and go back to Step 1. Meanwhile, update $$\begin{array}{c}\hfill {\gamma }_{k+1}=\left\{\begin{array}{cc}min\left\{{\gamma }_k,\frac{\omega \parallel y^k-v^k\parallel }{\parallel \nabla F(y^k)-\nabla F(v^k)\parallel }\right\},\hfill & \nabla F(y^k)\ne \nabla F(v^k),\hfill \\ {\gamma }_k,\hfill & else.\hfill \end{array}\right.\end{array}$$ and $$\begin{array}{c}\hfill {\tau }_{k+1}=\left\{\begin{array}{cc}min\left\{{\tau }_k,\frac{\mu \parallel w^k-A{u}^k\parallel }{\parallel \nabla G(w^k)-\nabla G(A{u}^k)\parallel }\right\},\hfill & \nabla G(w^k)\ne \nabla G(A{u}^k),\hfill \\ {\tau }_k,\hfill & else.\hfill \end{array}\right.\end{array}$$

5. Concluding Remarks

In this paper, we survey iterative methods for solving the split problem of fixed points of two pseudocontractive operators and variational inequalities of two pseudomonotone operators in Hilbert spaces. By using self-adaptive techniques, we construct a Tseng-type iterative algorithm for solving this split problem. We prove that the proposed Tseng-type iterative algorithm converges weakly to a solution of the split problem under some additional conditions imposed the operators and the parameters. Finally, we apply our algorithm to solve split pseudoconvex optimization problems and fixed point problems.