0 Introduction

H is presumed to be a real Hilbert interspace, which possesses a given non-null, closed convex subset C. In this article, variational inequality problem is mainly considered, which is hereinafter abbreviated as VIP. It can be represented as seeking a dot x^*∈C that satisfies the below expression:

$ \left\langle\boldsymbol{x}-\boldsymbol{x}^*, F\left(\boldsymbol{x}^*\right)\right\rangle \geqslant 0, \forall \boldsymbol{x} \in \boldsymbol{C} $

(1)

where 〈·, ·〉 is known as transvection and ‖·‖ is referred to as norm. The answer set of (1) is written as sol(F, C). As a crucial part and extension of many problems in nonlinear analysis, VIP has appeared in diverse physical models, economic science, transportation, engineering, optimal control and other problems^[1-4]. Because of its significance and applications, considerable interest has been attached to research methods for approximating solutions of VIP. In the last few years, the projection method has been widely noticed by some authors^[5-16], who have proposed a series of variant forms about this method and obtained the convergence of these forms.

Fixed point problem (FPP), a well-explored subject in functional analysis, is another major and related problem.It is of great importance to study the common solutions of FPP and VIP due to the fact that constraint conditions of some mathematical models can be converted into fixed point and variational inequality problems. These examples can be discovered in various practical matters such as image recovery and signal processing^[17-18]. This article focuses on solving FPP and VIP, in other words, seeking a dot x^* to satisfy Eq.(2).

$ \boldsymbol{x}^* \in \operatorname{Fix}(U) \cap \operatorname{sol}(F, \boldsymbol{C}) $

(2)

where U: H→H and Fix(U)= {x∈H: Ux=x}. In recent years, lots of methods to solve problem (2) have been proposed and discussed ^[19-26].

On the basis of extragradient method^[9], Nadezhkina and Takahashi^[20] put forward an iterative approach to solve problem (2) and confirmed the theorem of weak convergence. The most distinguishing characteristic of this approach is the projection in each iteration, which is based on practicable set C and demands to be carried out twice. However, when C is equipped with a comparatively complicated structure, the calculation cost will be expensive. After that, Censor et al.^[10] presented a modified approach. The target is to substitute the second projection onto C for a specific semi-space which can be computed explicitly. In addition, Kraikaew and Saejung^[22] proposed another improved method for solving problem (2) by combining Halpern iterative method with subgradient extergradient method, and the strong convergence result was testified.

Looking from another aspect, inertial technique was applied by Alvarez and Attouch^[27] to acquire the inertial proximal point algorithm. Meanwhile, the term θ_k(x_k－x_k－1) was introduced, which is known as the inertia. Since inertial type algorithm can be viewed as a procedure, which is conducive to promote the convergence rate of sequences, fast iterative algorithms based on inertial technique have been extensively studied by some authors^{[15, 21, 23-24]}.

Inspired by the above research results^{[15-16, 24-25]}, an innovative algorithm is presented in this paper for resolving quasi-nonexpansive fixed point and pseudomonotone variational inequality problem. Main contributions of this algorithm are as follows. A rule of non-monotonic step size is employed which demands no information of Lipschitz constant. In addition, the algorithm is constructed by using the subgradient extragradient method which is highly applied, and the inertial term θ_k(x_k－x_k－1) is utilized which is beneficial to speeding up the convergence and the viscosity method which ensures the strong convergence. Comparing the proposed algorithm in this paper with those in Refs. [16] and [25], pseudomonotone variational inequalities are considered, and under the weaker condition, it is proved that the proposed algorithm is provided with the property of strong convergence. Through the analysis of algorithms^[11-12], it can be found that the step size composition of the new algorithm is more concise and does not make use of linesearch, which means that no calculation of additional projections is needed. Furthermore, numerical experiments show that the new algorithm has preferable runtime and number of iterations than the presented ones.

The synopsis of this article is arranged as shown below. Several correlative and fundamental results are given in Section 1. Algorithm A and the proof of convergence constitute Section 2. Ultimately, some classical numerical problems and comparisons are provided in Section 3.

1 Preliminaries

A small number of elementary concepts and lemmas are expounded in this portion, which can be directly applied later.Following this, strong convergence and weak convergence of {ξ_n} for dot ξ are expressed as ξ_n→ξ and ξ_n$\rightharpoonup $ξ, respectively.

Definition 1^[7]   Given ξ, η∈H, the mapping F is called as follows:

(ⅰ) Monotone, if

$ \langle F(\boldsymbol{\xi})-F(\boldsymbol{\eta}), \boldsymbol{\xi}-\boldsymbol{\eta})\rangle \geqslant 0 $

(ⅱ)Pseudomonotone, if

$ \langle F(\boldsymbol{\eta}), \boldsymbol{\xi}-\boldsymbol{\eta})\rangle \geqslant 0 \Rightarrow\langle F(\boldsymbol{\xi}), \boldsymbol{\xi}-\boldsymbol{\eta})\rangle \geqslant 0 $

(ⅲ) L-Lipschitz continuous, if

$ \exists L>0, \text { s.t. }\|F(\boldsymbol{\xi})-F(\boldsymbol{\eta})\| \leqslant L\|\boldsymbol{\xi}-\boldsymbol{\eta}\| $

(ⅳ) Sequentially weakly continuous, if

$ \forall\left\{\boldsymbol{\xi}_n\right\} \subseteq \boldsymbol{H}, \boldsymbol{\xi}_n \rightharpoonup \boldsymbol{\xi} \Rightarrow F\left(\boldsymbol{\xi}_n\right) \rightharpoonup F(\boldsymbol{\xi}) $

Remark 1    For each monotone mapping, according to Definition 1, it is not difficult to observe that it is pseudomonotone. However, the opposite implication is not true.

Definition 2^[28]   Let U: H→H and Fix(U)≠Ø.

(ⅰ) U is referred to as a contraction, if

$ \begin{gathered} \exists l \in(0, 1), \text { s.t. }\|U \boldsymbol{\xi}-U \boldsymbol{\eta}\| \leqslant l\|\boldsymbol{\xi}-\boldsymbol{\eta}\|, \\ \forall \boldsymbol{\xi}, \boldsymbol{\eta} \in \boldsymbol{H} \end{gathered} $

If l=1, U is nonexpansive.

(ⅱ) U is quasi-nonexpansive, if

$ \|U \boldsymbol{\xi}-\boldsymbol{\eta}\| \leqslant\|\boldsymbol{\xi}-\boldsymbol{\eta}\|, \quad \forall \boldsymbol{\xi} \in \boldsymbol{H}, \boldsymbol{\eta} \in \operatorname{Fix}(U) $

(ⅲ) I－U is said to be demiclosed at zero, if

$ \begin{gathered} \forall\left\{\xi_n\right\} \subseteq \boldsymbol{H}, \xi_n \rightharpoonup \xi, \\ (I-U) \xi_n \rightarrow 0 \Rightarrow \xi \in \operatorname{Fix}(U) \end{gathered} $

Remark 2   With respect to each nonexpansive mapping, it is common knowledge that it is quasi-nonexpansive only when the disaggregation of fixed points is non-null.

Lemma 1^[28]   C is supposed to be a known non-null closed convex subset. For any dot ξ∈H, the below expressions are true:

(ⅰ)$\boldsymbol{\zeta}=P_C \boldsymbol{\xi} \Leftrightarrow\langle\boldsymbol{\zeta}-\boldsymbol{\xi}, \boldsymbol{\zeta}-\boldsymbol{\eta}\rangle \leqslant 0, \forall \boldsymbol{\eta}, \boldsymbol{\zeta} \in \boldsymbol{C}$

(ⅱ) $\|\boldsymbol{\xi}+\boldsymbol{\eta}\|^2 \leqslant\|\boldsymbol{\xi}\|^2+2\langle\boldsymbol{\eta}, \boldsymbol{\xi}+\boldsymbol{\eta}\rangle, \forall \boldsymbol{\eta} \in \boldsymbol{H}$

(ⅲ) $\left\|P_C(\boldsymbol{\xi})-\boldsymbol{\eta}\right\|^2+\left\|\boldsymbol{\xi}-P_C(\boldsymbol{\xi})\right\|^2 \leqslant$$\|\boldsymbol{\xi}-\boldsymbol{\eta}\|^2, \forall \boldsymbol{\eta} \in \boldsymbol{C}$

(ⅳ) $\left\langle\boldsymbol{\xi}-P_C(\boldsymbol{\xi}), \boldsymbol{\eta}-P_C(\boldsymbol{\xi})\right\rangle \leqslant 0, \forall \boldsymbol{\eta} \in \boldsymbol{C}$

Lemma 2^[29]   F: C→H is presumed to be a pseudomonotone succession, when x^*∈C is an answer of inequality (1), it is equivalent to seeking dot x^* that satisfies the below formula:

$ \left\langle\boldsymbol{x}-\boldsymbol{x}^*, F(\boldsymbol{x})\right\rangle \geqslant 0, \forall \boldsymbol{x} \in \boldsymbol{C} $

Lemma 3^[30]   Select two nonnegative real successions {s_n} and {v_n}, where s_n∈(0, 1) and $\sum\limits_{n = 0}^\infty {{s_n}} = \infty $. Assume that v_n+1≤v_n(1－s_n)+s_nt_n for all n≥1. If $\mathop {\lim }\limits_{k \to \infty } \sup {t_{{n_k}}} \le 0$ for each subsequence {v_nk}of {v_n} satisfies $\lim\limits _{k \rightarrow \infty} \inf \left(v_{n_k+1}-v_{n_k}\right) \geqslant 0$, there is $\lim\limits _{n \rightarrow \infty} v_n=0$.

2 Algorithm and Convergence Analysis

Novel Algorithm A is stated and studied in this part. First, it is assumed that the presented algorithm satisfies the below conditions.

(C1) C is a known non-null, closed convex aggregation and Fix(U)∩sol(F, C)≠Ø;

(C2) F is sequentially weakly continuous over C, pseudomonotone and L-Lipschitz continuous in H;

(C3) I－U is demiclosed at zero and U is quasi-nonexpansive;

(C4) f is a contraction on H such that contraction coefficient l∈(0, 1);

(C5){q_n}, {t_n} and {s_n} are successions obtained from (0, 1) and these successions satisfy q_n+t_n+s_n=1;

(C6){θ_n} is assumed to be a nonnegative succession which satisfies $\mathop {\lim }\limits_{n \to \infty } \left({{\theta _n}/{s_n}} \right) = 0$, together with $\sum\limits_{n = 1}^\infty {{s_n}} = \infty \; {\rm{and}}\; \mathop {\lim }\limits_{{\rm{n}} \to \infty } {s_n} = 0$.

At present, a clear and accurate description of the algorithm is given. The form is listed below:

Algorithm A

Step 0: Preset starting points x₀, x₁∈H and parameters λ₁>0, μ∈(0, 1). Take a succession of nonnegative real numbers {p_n} to satisfy $\sum\limits_{n = 1}^\infty {{p_n}} < + \infty $.

Step 1: Define u_n=τ_n(x_n－x_n-1)+x_n, and figure up

$ \boldsymbol{y}_n=P_C\left(\boldsymbol{u}_n-\lambda_n F\left(\boldsymbol{u}_n\right)\right) $

Step 2: With T_n={x∈H: 〈y_n－x, u_n－y_n－λ_nF(u_n)〉≥0}, reckon up

$ \boldsymbol{z}_n=P_{T_n}\left(\boldsymbol{u}_n-\lambda_n F\left(\boldsymbol{y}_n\right)\right) $

Step 3: Set x_n+1=q_nUz_n+t_nz_n+s_nf(x_n), and update

$ \lambda_{n+1}=\left\{\begin{array}{l} \min \left\{\lambda_n+p_n, \frac{\mu\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|}{\left\|F\left(\boldsymbol{y}_n\right)-F\left(\boldsymbol{u}_n\right)\right\|}\right\} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text { if } F\left(\boldsymbol{u}_n\right)-F\left(\boldsymbol{y}_n\right) \neq 0 \\ \lambda_n+p_n, \quad \text { else } \end{array}\right. $

Put n=n+1 and turn to Step 1.

Remark 3   According to conditions listed above, expression (3) is acquired:

$ \lim \limits_{n \rightarrow \infty} \frac{\tau_n}{s_n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_n\right\|=0 $

(3)

Indeed, choose parameter τ_n to be

$ \tau_n=\left\{\begin{aligned} \min \left\{\frac{1}{n+1}, \right. & \left.\frac{\theta_n}{\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_n\right\|}\right\}, \\ & \text { if } \boldsymbol{x}_{n-1}-\boldsymbol{x}_n \neq 0 \\ (n+1)^{-1}, \text { else } \end{aligned}\right. $

Consequently, with respect to any positive integer n and the selection of parameter τ_n mentioned above, it is known that τ_n‖x_n－1－x_n‖≤θ_n. By the condition (C6), the following formula can be procurable:

$ \lim \limits_{n \rightarrow \infty} \frac{\tau_n}{s_n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_n\right\| \leqslant \lim \limits_{n \rightarrow \infty} \frac{\theta_n}{s_n}=0 $

Next, some lemmas are proposed, which lay a preliminary foundation for the convergence property of Algorithm A.

Lemma 4    Succession step size {λ_n} obtained from Algorithm A satisfies expression min{λ₁, μ/L}≤λ≤λ₁+P, where $\lambda=\lim \limits_{n \rightarrow \infty} \lambda_n$ and $P=\sum\limits_{n=1}^{\infty} p_n$.

Proof  F is known to be L-Lipschitz continuous, when F(u_n)－F(y_n)≠0, it can be derived that

$ \frac{\mu\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|}{\left\|F\left(\boldsymbol{y}_n\right)-F\left(\boldsymbol{u}_n\right)\right\|} \geqslant \frac{\mu\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|}{L\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|}=\frac{\mu}{L} $

Through the notion of λ_n+1 and induction, it can be attained that the lower bound and upper bound of sequence {λ_n} are min{λ₁, μ/L} and λ₁+P, respectively. Bringing h_n+1=λ_n+1－λ_n into the equation, it also can be seen from the concept of λ_n+1 that

$ \sum\limits_{n=1}^{\infty} p_n \geqslant \sum\limits_{n=1}^{\infty} h_{n+1}^{+}, \sum\limits_{n=1}^{\infty} p_n <+\infty $

(4)

where $h_{n+1}^{-}=\max \left\{0, -h_{n+1}\right\}$ and $h_{n+1}^{+}=\max \left\{0, h_{n+1}\right\}$. From Eq.(4), it is clear to know that the positive series $\sum\limits_{n=1}^{\infty} h_{n+1}^{+}$ converge. Next, we prove that the positive series $\sum\limits_{n=1}^{\infty} h_{n+1}^{-}$ has the same result. Assume $\sum\limits_{n=1}^{\infty} h_{n+1}^{-}=-\infty$. Since the expression $h_{n+1}=h_{n+1}^{+}-h_{n+1}^{-}$ holds, the formula below can be found:

$ \lambda_{m+1}-\lambda_1=\sum\limits_{n=1}^m h_{n+1}=\sum\limits_{n=1}^m h_{n+1}^{+}-\sum\limits_{n=1}^m h_{n+1}^{-} $

(5)

Taking the limit as m→+∞ in Eq.(5), so λ_m→+∞, which is not consistent with the fact that {λ_n} is bounded, so $\sum\limits_{n=1}^{\infty} h_{n+1}^{-}$ converges. From the above analysis, it can be seen that when m inclines to positive infinity in Eq.(5), it is evident that $\lim \limits_{n \rightarrow \infty} \lambda_n=\lambda$. By the boundedness of {λ_n}, min{λ₁, μ/L}≤λ≤λ₁+P is obtained.

Lemma 5   Successions {x_n}, {y_n} and {u_n} are provided by Algorithm A. Assume that $\lim \limits_{k \rightarrow \infty} \| \boldsymbol{x}_{n_k}- \boldsymbol{u}_{n_k}\left\|=\lim \limits_{k \rightarrow \infty}\right\| \boldsymbol{y}_{n_k}-\boldsymbol{z}_{n_k}\left\|=\lim \limits_{k \rightarrow \infty}\right\| \boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k} \|= \mathop {\lim }\limits_{k \to \infty } \left\| {U{\mathit{\boldsymbol{z}}_{{n_k}}} - {\mathit{\boldsymbol{z}}_{{n_k}}}} \right\| = 0$. If sequence {x_nk} weakly converges to p∈H, it can be obtained that p∈Fix(U)∩sol(F, C).

Proof  Since $\boldsymbol{x}_{n_k} \rightharpoonup \boldsymbol{p} \in \boldsymbol{H}$ and $\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_k}-\boldsymbol{u}_{n_k}\right\|=\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}\right\|=0$, we can easily get $\boldsymbol{u}_{n_k} \rightharpoonup \boldsymbol{p}$ and $\boldsymbol{y}_{n_k} \rightharpoonup \boldsymbol{p}$. From {y_n}⊆C and closed convex property of set C, there is p∈C.

For each y∈C, p∈sol(F, C) can be testified. Indeed, according to Lemma 1(ⅳ) and the notion of y_nk, it can be derived that

$ \left\langle\boldsymbol{u}_{n_k}-\lambda_{n_k} F\left(\boldsymbol{u}_{n_k}\right)-\boldsymbol{y}_{n_k}, \boldsymbol{y}-\boldsymbol{y}_{n_k}\right\rangle \leqslant 0 $

This signifies that

$ \begin{gathered} \left\langle F\left(\boldsymbol{u}_{n_k}\right)-F\left(\boldsymbol{y}_{n_k}\right), \boldsymbol{y}-\boldsymbol{y}_{n_k}\right\rangle+\left\langle F\left(\boldsymbol{y}_{n_k}\right), \boldsymbol{y}-\boldsymbol{y}_{n_k}\right\rangle \geqslant \\ \lambda_{n_k}^{-1}\left\langle\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}, \boldsymbol{y}-\boldsymbol{y}_{n_k}\right\rangle \end{gathered} $

(6)

Using L-Lipschitz continuity of F, $\lim \limits_{n \rightarrow \infty} \lambda_n=\lambda>0$, $\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}\right\|=0$ and letting k→∞ in inequality (6), the following inequality can be obtained

$ \lim \limits_{k \rightarrow \infty} \inf \left\langle F\left(\boldsymbol{y}_{n_k}\right), \boldsymbol{y}-\boldsymbol{y}_{n_k}\right\rangle \geqslant 0 $

(7)

Select a decreasing positive sequence{r_k} such that $\lim \limits_{k \rightarrow \infty} r_k=0$. According to the property of limit inferior, for each r_k, use N_k to represent the smallest positive integer which satisfies the following inequality:

$ r_k+\left\langle F\left(\boldsymbol{y}_{n_j}\right), \boldsymbol{y}-\boldsymbol{y}_{n_j}\right\rangle \geqslant 0, \forall j \geqslant N_k $

(8)

Since{r_k} decreases, it is easy to get {N_k} is increasing.

For any k, putting

$ \boldsymbol{d}_{N_k}=\frac{F\left(\boldsymbol{y}_{N_k}\right)}{\left\|F\left(\boldsymbol{y}_{N_k}\right)\right\|^2} $

there is $\left\langle\boldsymbol{d}_{N_k}, F\left(\boldsymbol{y}_{N_k}\right)\right\rangle=1$. By Eq.(8), it is deduced that

$ \left\langle F\left(\boldsymbol{y}_{N_k}\right), r_k \boldsymbol{d}_{N_k}+\boldsymbol{y}-\boldsymbol{y}_{N_k}\right\rangle \geqslant 0 $

Through utilizing the pseudomonotonicity of F, the formula below can be attained:

$ \left\langle F\left(r_k \boldsymbol{d}_{N_k}+y\right), r_k \boldsymbol{d}_{N_k}+\boldsymbol{y}-\boldsymbol{y}_{N_k}\right\rangle \geqslant 0 $

(9)

Furthermore, according to the concept of sequentially weakly continuity and $\boldsymbol{y}_{n_k} \rightharpoonup \boldsymbol{p}$, it is derived that $F\left(\boldsymbol{y}_{n_k}\right) \rightharpoonup F(\boldsymbol{p})$. Assume F(p)≠0 (if not, p is a solution). Apparently, applying lower semicontinuity of norm mapping, it is easy to obtain:

$ \lim \limits_{k \rightarrow \infty} \inf \left\|F\left(\boldsymbol{y}_{n_k}\right)\right\| \geqslant\|F(\boldsymbol{p})\|>0 $

Combining{y_Nk}⊆{y_nk} and $\lim \limits_{k \rightarrow \infty} r_k=0$, there is

$ \begin{aligned} 0 \leqslant & \lim \limits_{k \rightarrow \infty} \sup \left\|r_k \boldsymbol{d}_{N_k}\right\|=\lim \limits_{k \rightarrow \infty} \sup \left(\frac{r_k}{\left\|F\left(\boldsymbol{y}_{N_k}\right)\right\|}\right) \leqslant \\ & \frac{\lim \limits_{k \rightarrow \infty} \sup r_k}{\liminf \limits_{k \rightarrow \infty} \inf \left\|\left(\boldsymbol{y}_{N_k}\right)\right\|}=0 \end{aligned} $

This implies that $\lim \limits_{k \rightarrow \infty} r_k \boldsymbol{d}_{N_k}=0$. Taking k→∞ in Eq.(9), 〈F(y), y－p〉≥0. From Lemma 2, p∈sol(F, C) can be obtained.

Now, with expressions $\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{y}_{n_k}-\boldsymbol{z}_{n_k}\right\|=0$ and $\boldsymbol{y}_{n_k} \rightharpoonup \boldsymbol{p}$, there is z_nk$\rightharpoonup$p. Since $\mathop {\lim }\limits_{k \to \infty } \left\| {U{\mathit{\boldsymbol{z}}_{{n_k}}} - {\mathit{\boldsymbol{z}}_{{n_k}}}} \right\| = 0$ and because of the demiclosedness of I－U, p∈Fix(U) can be inferred. Accordingly, combined with the above proof, Lemma 5 can be obtained.

Lemma 6    Suppose that successions {z_n}, {y_n} and {u_n} are produced by Algorithm A, the expression given below holds:

$ \begin{gathered} \left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right) \cdot \\ \left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2-\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{u}_n-\boldsymbol{y}_n\right\|^2 \end{gathered} $

Proof  Take v∈sol(F, C). It follows from z_n=P_Tn(u_n－λ_nF(y_n)) and Lemma 1(ⅲ) that

$ \left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2=\left\|P_{T_n}\left(\boldsymbol{u}_n-\lambda_n F\left(\boldsymbol{y}_n\right)\right)-\boldsymbol{v}\right\|^2 \leqslant\\ \left\|\boldsymbol{u}_n-\lambda_n F\left(\boldsymbol{y}_n\right)-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{u}_n-\lambda_n F\left(\boldsymbol{y}_n\right)-\boldsymbol{z}_n\right\|^2=\\ \left\|\boldsymbol{v}-\boldsymbol{u}_n+\lambda_n F\left(\boldsymbol{y}_n\right)\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{u}_n+\lambda_n F\left(\boldsymbol{y}_n\right)\right\|^2=\\ \left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2+2 \lambda_n\left\langle\boldsymbol{v}-\boldsymbol{u}_n, F\left(\boldsymbol{y}_n\right)\right\rangle-\left\|\boldsymbol{z}_n-\boldsymbol{u}_n\right\|^2-\\ 2 \lambda_n\left\langle\boldsymbol{z}_n-\boldsymbol{u}_n, F\left(\boldsymbol{y}_n\right)\right\rangle=\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{u}_n\right\|^2+\\ 2 \lambda_n\left\langle\boldsymbol{v}-\boldsymbol{z}_n, F\left(\boldsymbol{y}_n\right)\right\rangle=\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{u}_n\right\|^2+\\ 2 \lambda_n\left\langle\boldsymbol{y}_n-\boldsymbol{z}_n, F\left(\boldsymbol{y}_n\right)\right\rangle+2 \lambda_n\left\langle\boldsymbol{v}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_n\right)\right\rangle $

By y_n∈C and v∈sol(F, C), it can be deduced that 〈y_n－v, F(v)〉≥0. From the pseudomonotonicity of F, the formula given below can be attained:

$ \left\langle\boldsymbol{y}_n-\boldsymbol{v}, F\left(\boldsymbol{y}_n\right)\right\rangle \geqslant 0 $

(10)

According to z_n∈T_n, there is

$ \left\langle \boldsymbol{z}_n-\boldsymbol{y}_n, \boldsymbol{u}_n-\boldsymbol{y}_n-\lambda_n F\left(\boldsymbol{u}_n\right)\right\rangle \leqslant 0 $

Using the above inequality, Eq.(10) and Cauchy-Schwarz inequality, the expression below can be attained:

$ \begin{aligned} &\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{u}_n\right\|^2+ \\ & 2 \lambda_n\left\langle\boldsymbol{y}_n-\boldsymbol{z}_n, F\left(\boldsymbol{y}_n\right)\right\rangle=\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2- \\ & \left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+2\left\langle\boldsymbol{z}_n-\boldsymbol{y}_n, \boldsymbol{u}_n-\boldsymbol{y}_n-\lambda_n F\left(\boldsymbol{y}_n\right)\right\rangle= \\ & \left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2-\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+ \\ & 2 \lambda_n\left\langle\boldsymbol{z}_n-\boldsymbol{y}_n, F\left(\boldsymbol{u}_n\right)-F\left(\boldsymbol{y}_n\right)\right\rangle+2\left\langle\boldsymbol{z}_n-\boldsymbol{y}_n, \boldsymbol{u}_n-\right. \\ & \left.\boldsymbol{y}_n-\lambda_n F\left(\boldsymbol{u}_n\right)\right\rangle \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2- \\ & \left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+2 \boldsymbol{\lambda}_n\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\| \| F\left(\boldsymbol{u}_n\right)- \\ & F\left(\boldsymbol{y}_n\right) \| \end{aligned} $

(11)

Combining the definition of λ_n+1 and Eq.(11), there is

$ \begin{aligned} & \left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2- \\ & \left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+2 \lambda_{n+1} \frac{\lambda_n}{\lambda_{n+1}}\left\|z_n-\boldsymbol{y}_n\right\| \| F\left(\boldsymbol{u}_n\right)- \\ & F\left(\boldsymbol{y}_n\right)\|\leqslant\| \boldsymbol{v}-\boldsymbol{u}_n\left\|^2-\right\| \boldsymbol{z}_n-\boldsymbol{y}_n \|^2- \\ & \left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+2 \mu \frac{\lambda_n}{\lambda_{n+1}}\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|\left\|\boldsymbol{u}_n-\boldsymbol{y}_n\right\| \leqslant \\ & \left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2-(1- \\ & \left.\mu \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2 \end{aligned} $

Lemma 7   Succession {x_n} is provided by Algorithm A. If the presumptions (C1)-(C6) are given, it can be inferred that {x_n} is bounded.

Proof   Applying $\lim \limits_{n \rightarrow \infty} \lambda_n=\lambda>0$, there is $\mathop {\lim }\limits_{n \to \infty } \left({1 - \mu \left({{\lambda _n}/{\lambda _{n + 1}}} \right)} \right) = 1 - \mu > 0$. Therefore, for any n≥N, where N>0, there is

$ 1-\mu\left(\lambda_n / \lambda_{n+1}\right)>0 $

Using Lemma 6, we get

$ \begin{gathered} \left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2-\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2- \\ \quad\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{u}_n-\boldsymbol{y}_n\right\|^2 \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2 \end{gathered} $

(12)

Using the notion of u_n, it can be deduced that

$ \begin{gathered} \left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|=\left\|\boldsymbol{v}-\left(\boldsymbol{\tau}_n\left(\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right)+\boldsymbol{x}_n\right)\right\| \leqslant \\ \tau_n\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|+\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|= \\ s_n \frac{\tau_n}{s_n}\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|+\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\| \end{gathered} $

(13)

Utilizing Eq.(3), the existence of positive number M₁ can be inferred, together with

$ s_n \frac{\tau_n}{s_n}\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\| \leqslant s_n M_1 $

(14)

From Eqs.(12), (13) and (14), the expression below is obtained:

$ \left\|z_n-\boldsymbol{v}\right\| \leqslant\left\|\boldsymbol{v}-\boldsymbol{u}_n\right\| \leqslant\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|+s_n M_1, \forall n \geqslant N $

(15)

Combining Eq.(15) and the formula of x_n+1, there is

$ \begin{aligned} & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|=\left\|q_n U \boldsymbol{z}_n+t_n z_n+s_n f\left(\boldsymbol{x}_n\right)-\boldsymbol{v}\right\| \leqslant \\ & q_n\left\|U \boldsymbol{z}_n-\boldsymbol{v}\right\|+t_n\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|+s_n\left(\left\|f\left(\boldsymbol{x}_n\right)-f(\boldsymbol{v})\right\|+\right. \\ & \|f(\boldsymbol{v})-\boldsymbol{v}\|) \leqslant\left(t_n+q_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|+s_n\left(l\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|+\right. \\ & \|f(\boldsymbol{v})-\boldsymbol{v}\|) \leqslant\left(1-s_n\right)\left(\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|+s_n M_1\right)+ \\ & s_n\left(l\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|+\|f(\boldsymbol{v})-\boldsymbol{v}\|\right) \leqslant s_n(\|f(\boldsymbol{v})-\boldsymbol{v}\|+ \\ & \left.M_1\right)+\left(1-(1-l) s_n\right)\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|=(1- \\ & l) s_n \frac{\|f(\boldsymbol{v})-\boldsymbol{v}\|+M_1}{1-l}+\left(1-(1-l) s_n\right) \| \boldsymbol{v}- \\ & \boldsymbol{x}_n \| \leqslant \max \left\{\frac{\|f(\boldsymbol{v})-\boldsymbol{v}\|+M_1}{1-l}, \left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|\right\} \leqslant \\ & \cdots \leqslant \max \left\{\frac{\|f(\boldsymbol{v})-\boldsymbol{v}\|+M_1}{1-l}, \left\|\boldsymbol{v}-\boldsymbol{x}_N\right\|\right\} \end{aligned} $

Therefore, it is concluded that succession{x_n} is bounded. Meanwhile, this signifies that {y_n}, {u_n}, {f(x_n)}, {z_n} and {Uz_n} have the same property.

Theorem 1   Succession {x_n} can be provided by Algorithm A. If conditions (C1)- (C6) are given, {x_n} to dot v must be strong convergence, where $\mathit{\boldsymbol{v}} = {P_{{\rm{Fix }}(U) \cap {\mathop{\rm sol}\nolimits} (F, \mathit{\boldsymbol{C}})}}f(\mathit{\boldsymbol{v}})$ and v∈Fix(U)∩sol(F, C).

Proof   Three parts listed below are the process of this proof.

Part 1   The expression below can be illustrated:

$ \begin{gathered} t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+\left(1-s_n\right)\left(1-\mu\left(\lambda_n / \lambda_{n+1}\right)\right) \cdot \\ \left(\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2\right) \leqslant \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+s_n M_4 \end{gathered} $

Indeed, using inequality (15), there is

$ \begin{aligned} & \left\|\boldsymbol{v}-\boldsymbol{u}_n\right\|^2 \leqslant\left(\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|+s_n M_1\right)^2= \\ & \left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|^2+s_n\left(s_n M_1^2+2 M_1\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|\right)= \\ & \left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|^2+s_n M_2 \end{aligned} $

(16)

for some M₂>0.

On the other hand, from the expression of x_n+1, it can be deduced that

$ \begin{aligned} & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2=\left\|q_n U \boldsymbol{z}_n+t_n \boldsymbol{z}_n+s_n f\left(\boldsymbol{x}_n\right)-\boldsymbol{v}\right\|^2= \\ & \left\|q_n\left(U \boldsymbol{z}_n-\boldsymbol{v}\right)+t_n\left(\boldsymbol{z}_n-\boldsymbol{v}\right)+s_n\left(f\left(\boldsymbol{x}_n\right)-\boldsymbol{v}\right)\right\|^2 \leqslant \\ & q_n\left\|U \boldsymbol{z}_n-\boldsymbol{v}\right\|^2+t_n\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2+s_n\left\|f\left(\boldsymbol{x}_n\right)-\boldsymbol{v}\right\|^2- \\ & t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2 \leqslant\left(t_n+q_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2+ \\ & s_n\left(\left\|f\left(\boldsymbol{x}_n\right)-f(\boldsymbol{v})\right\|+\|f(\boldsymbol{v})-\boldsymbol{v}\|\right)^2-t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2 \leqslant \\ & \left(1-s_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2+s_n\left(l\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|+\right. \\ & \|f(\boldsymbol{v})-\boldsymbol{v}\|)^2-t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2 \leqslant \\ & \left(1-s_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2+s_n\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|^2- \\ & t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+s_n\left(\|f(\boldsymbol{v})-\boldsymbol{v}\|^2+\right. \\ & \left.2\|f(\boldsymbol{v})-\boldsymbol{v}\|\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|\right) \leqslant s_n\left\|\boldsymbol{v}-\boldsymbol{x}_n\right\|^2+ \\ & \left(1-s_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2-t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+ \\ & s_n M_3 \end{aligned} $

(17)

for some M₃>0.

Combining formulas (16), (17) and Lemma 6, the expression below can be derived:

$ \begin{aligned} & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant s_n\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2-t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+ \\ & s_n M_3+\left(1-s_n\right)\left(\left\|\boldsymbol{u}_n-\boldsymbol{v}\right\|^2-\left(1-\mu \frac{\lambda_n}{\lambda_{n+1}}\right) \cdot\right. \\ & \left.\left(\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2\right)\right) \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2- \\ & t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+s_n M_2+s_n M_3-\left(1-s_n\right)(1- \\ & \left.\quad \boldsymbol{\lambda _ { n }}\right)\left(\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2\right) \end{aligned} $

(18)

That is

$ \begin{aligned} & t_n q_n\left\|U \boldsymbol{z}_n-\boldsymbol{z}_n\right\|^2+\left(1-s_n\right)\left(1-\mu\left(\lambda_n / \lambda_{n+1}\right)\right) \cdot \\ & \quad\left(\left\|\boldsymbol{y}_n-\boldsymbol{u}_n\right\|^2+\left\|\boldsymbol{z}_n-\boldsymbol{y}_n\right\|^2\right) \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2- \\ & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+s_n M_4 \end{aligned} $

(19)

where M₄ ∶ = M₂+M₃.

Part 2   The expression below can be attained:

$ \begin{aligned} & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant\left(1-(1-l) s_n\right)\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+ \\ & (1-l) \cdot s_n\left(\frac{2}{1-l}\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle+\right. \\ & \left.\frac{3 M}{1-l} \cdot \frac{\boldsymbol{\tau}_n}{s_n}\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|\right) \end{aligned} $

Indeed, using the formula of u_n, there is

$ \begin{gathered} \left\|\boldsymbol{u}_n-\boldsymbol{v}\right\|^2=\left\|\boldsymbol{\tau}_n\left(\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right)+\boldsymbol{x}_n-\boldsymbol{v}\right\|^2 \leqslant \\ \left(\tau_n\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|+\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|\right)^2= \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+3 M \tau_n\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\| \end{gathered} $

(20)

where $M: =\sup\limits _{n \geqslant 1}\left\{\tau_n\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|, \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|\right\}>0$.

According to Lemma 1(ⅱ), inequality(12) and (20), it can be concluded

$ \begin{aligned} & \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2=\| q_n\left(U \boldsymbol{z}_n-\boldsymbol{v}\right)+t_n\left(\boldsymbol{z}_n-\boldsymbol{v}\right)+ \\ & \boldsymbol{s}_n\left(f\left(\boldsymbol{x}_n\right)-f(\boldsymbol{v})\right)+\boldsymbol{s}_n(f(\boldsymbol{v})-\boldsymbol{v})\left\|^2 \leqslant\right\| q_n\left(U \boldsymbol{z}_n-\boldsymbol{v}\right)+ \\ & t_n\left(z_n-\boldsymbol{v}\right)+\boldsymbol{s}_n\left(f\left(\boldsymbol{x}_n\right)-f(\boldsymbol{v})\right) \|^2+2 \boldsymbol{s}_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \leqslant \\ & \left(1-\boldsymbol{s}_n\right)\left\|\boldsymbol{z}_n-\boldsymbol{v}\right\|^2+\boldsymbol{s}_n\left\|f\left(\boldsymbol{x}_n\right)-f(\boldsymbol{v})\right\|^2+ \\ & 2 \boldsymbol{s}_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \leqslant\left(1-\boldsymbol{s}_n\right)\left\|\boldsymbol{u}_n-\boldsymbol{v}\right\|^2+ \\ & \boldsymbol{s}_n l\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+2 \boldsymbol{s}_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \leqslant(1- \\ & \left.\boldsymbol{s}_n\right)\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+\boldsymbol{s}_n l\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+2 \boldsymbol{s}_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, \right. \\ & f(\boldsymbol{v})-\boldsymbol{v}\rangle+3 M \tau_n\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|=2 \boldsymbol{s}_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, \right. \\ & f(\boldsymbol{v})-\boldsymbol{v}\rangle+\left(1-(1-l) \boldsymbol{s}_n\right)\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+3 M \tau_n \| \boldsymbol{x}_n- \\ & \boldsymbol{x}_{n-1}\left\|=\left(1-(1-l) \boldsymbol{s}_n\right)\right\| \boldsymbol{x}_n-\boldsymbol{v} \|^2+(1-l) \boldsymbol{s}_n \cdot \\ & \left(\frac{3 M}{1-l} \cdot \frac{\boldsymbol{\tau}_n}{\boldsymbol{s}_n} \cdot\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n-1}\right\|+\frac{2}{1-l}\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle\right) \end{aligned} $

(21)

Part 3   It can be proved that the succession {‖x_n－v‖} strongly converges to zero.

Indeed, the existence of a subsequence {‖x_nk－v‖} can be presumed, and this subsequence satisfies inequality $\mathop {\lim}\limits _{k \to \infty }\inf \left({\left\| {{\mathit{\boldsymbol{x}}_{{n_k} + 1}} - \mathit{\boldsymbol{v}}} \right\| - } \right.\left. {\left\| {{\mathit{\boldsymbol{x}}_{{n_k}}} - \mathit{\boldsymbol{v}}} \right\|} \right) \ge 0$. The expression below can be attained

$ \left. {\mathop {\lim }\limits_{k \to \infty } \inf {{\left\| {{\mathit{\boldsymbol{x}}_{{n_k} + 1}} - \mathit{\boldsymbol{v}}} \right\|}^2} - {{\left\| {{\mathit{\boldsymbol{x}}_{{n_k}}} - \mathit{\boldsymbol{v}}} \right\|}^2}} \right) \ge 0 $

(22)

Combining Part 1 and inequality (22), it can be infered that

$ \begin{aligned} & \lim \limits_{k \rightarrow \infty} \sup \left(t_{n_k} q_{n_k}\left\|U \boldsymbol{z}_{n_k}-\boldsymbol{z}_{n_k}\right\|^2+\left(1-s_{n_k}\right) \cdot\right. \\ & \left.\quad\left(1-\mu \frac{\lambda_{n_k}}{\lambda_{n_k+1}}\right)\left(\left\|\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}\right\|^2+\left\|\boldsymbol{y}_{n_k}-\boldsymbol{z}_{n_k}\right\|^2\right)\right) \leqslant \\ & \lim \limits_{k \rightarrow \infty} \sup \left(\left\|\boldsymbol{x}_{n_k}-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{x}_{n_k+1}-\boldsymbol{v}\right\|^2+s_{n_k} M_4\right) \leqslant \\ & \lim \limits_{k \rightarrow \infty} \sup \left(\left\|\boldsymbol{x}_{n_k}-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{x}_{n_k+1}-\boldsymbol{v}\right\|^2\right)+ \\ & \lim \limits_{k \rightarrow \infty} \sup s_{n_k} M_4=-\lim \limits_{k \rightarrow \infty} \inf \left(\left\|\boldsymbol{x}_{n_k+1}-\boldsymbol{v}\right\|^2-\right. \\ & \left.\left\|\boldsymbol{x}_{n_k}-\boldsymbol{v}\right\|^2\right) \leqslant 0 \end{aligned} $

(23)

This implies that

$ \left\|\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}\right\| \rightarrow 0(k \rightarrow \infty) $

(24)

$ \left\|\boldsymbol{y}_{n_k}-ct {z}_{n_k}\right\| \rightarrow 0(k \rightarrow \infty) $

(25)

$ \left\|U \boldsymbol{z}_{n_k}-\boldsymbol{z}_{n_k}\right\| \rightarrow 0(k \rightarrow \infty) $

(26)

By the definition of u_n, there is

$ \left\|\boldsymbol{u}_{n_k}-\boldsymbol{x}_{n_k}\right\|=s_{n_k} \frac{\tau_{n_k}}{s_{n_k}}\left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}_{n_k-1}\right\| \rightarrow 0, (k \rightarrow \infty) $

(27)

Combining Eqs.(24)-(27), the following expressions are derived:

$ \begin{aligned} & \left\|\boldsymbol{x}_{n_k}-\boldsymbol{y}_{n_k}\right\| \leqslant\left\|\boldsymbol{x}_{n_k}-\boldsymbol{u}_{n_k}\right\|+\left\|\boldsymbol{u}_{n_k}-\boldsymbol{y}_{n_k}\right\| \rightarrow 0, \\ & (k \rightarrow \infty) \end{aligned} $

(28)

$ \begin{gathered} \left\|\boldsymbol{x}_{n_k+1}-\boldsymbol{z}_{n_k}\right\| \leqslant s_{n_k}\left\|f\left(\boldsymbol{x}_{n_k}\right)-\boldsymbol{z}_{n_k}\right\|+t_{n_k} \| \boldsymbol{z}_{n_k}- \\ \boldsymbol{z}_{n_k}\left\|+q_{n_k}\right\| U \boldsymbol{z}_{n_k}-\boldsymbol{z}_{n_k} \| \rightarrow 0, (k \rightarrow \infty) \end{gathered} $

(29)

$ \begin{aligned} & \left\|\boldsymbol{z}_{n_k}-\boldsymbol{u}_{n_k}\right\| \leqslant\left\|\boldsymbol{z}_{n_k}-\boldsymbol{y}_{n_k}\right\|+\left\|\boldsymbol{y}_{n_k}-\boldsymbol{u}_{n_k}\right\| \rightarrow 0, \\ & (k \rightarrow \infty) \end{aligned} $

(30)

$ \begin{gathered} \left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}_{n_k+1}\right\| \leqslant\left\|\boldsymbol{x}_{n_k}-\boldsymbol{u}_{n_k}\right\|+\left\|\boldsymbol{u}_{n_k}-\boldsymbol{z}_{n_k}\right\|+ \\ \left\|\boldsymbol{z}_{n_k}-\boldsymbol{x}_{n_k+1}\right\| \rightarrow 0, (k \rightarrow \infty) \end{gathered} $

(31)

For{x_nk} is bounded, it is not hard to infer that succession $\left\{\boldsymbol{x}_{n_{k_j}}\right\} \subseteq\left\{\boldsymbol{x}_{n_k}\right\}$ and point x^*∈H exist, together with $\boldsymbol{x}_{n_{k_j}} \rightharpoonup \boldsymbol{x}^*$. Then there is

$ \begin{aligned} & \lim \limits_{k \rightarrow \infty} \sup \left\langle\boldsymbol{x}_{n_k}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle= \\ & \lim \limits_{k \rightarrow \infty}\left\langle\boldsymbol{x}_{n_{k_j}}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle=\left\langle\boldsymbol{x}^*-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \end{aligned} $

(32)

According to Lemma 5, it is derived that x^*∈Fix(U)∩sol(F, C). Since v=P_{Fix(U)∩sol(F, C)} f(v), Eq.(33) can be obtained:

$ \lim \limits_{k \rightarrow \infty} \sup \left\langle\boldsymbol{x}_{n_k}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle=\left\langle\boldsymbol{x}^*-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \leqslant 0 $

(33)

Combining Eqs.(31) and (33), the result below is deduced:

$ \begin{aligned} & \lim \limits_{k \rightarrow \infty} \sup \left\langle\boldsymbol{x}_{n_k+1}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle=\lim \limits_{k \rightarrow \infty} \sup \left(\left\langle\boldsymbol{x}_{n_k+1}-\right.\right. \\ & \left.\left.\boldsymbol{x}_{n_k} f(\boldsymbol{v})-\boldsymbol{v}\right\rangle+\left\langle\boldsymbol{x}_{n_k}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle\right) \leqslant \\ & \lim \limits_{k \rightarrow \infty} \sup \left\|\boldsymbol{x}_{n_k+1}-\boldsymbol{x}_{n_k}\right\|\|f(\boldsymbol{v})-\boldsymbol{v}\|+ \\ & \lim \limits_{h \rightarrow \infty} \sup \left\langle\boldsymbol{x}_{n_k}-\boldsymbol{v}, f(\boldsymbol{v})-\boldsymbol{v}\right\rangle \leqslant 0 \end{aligned} $

(34)

Therefore, using Lemma 3, Eq.(3), Part 2 and Eq.(34), it can be derived that $\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|=0$.

3 Numerical Experiments

On this portion, the validation of Algorithm A is verified by several numerical experiments and comparisons. In the numerical implementations, the first two inquire into the test problem of finite dimensional spaces, and the last one is about infinite dimensional spaces. In experimental results described in the tables below, the runtime (time) and the number of iterations (ite.) are written down in seconds. In addition, ‖x_n－y_n‖≤ε is used as stopping condition for all algorithms, where ε is the iteration accuracy. With respect to Algorithm A, the parameters listed below are taken:

$ \begin{aligned} \mu & =0.2, \lambda_1=0.09, f(x)=0.15 \boldsymbol{x} \\ \theta_n & =6 /(n+1)^4, p_n=(n+1)^{-2} \\ t_n & =(n+1)^{-1}, s_n=\left(n^3+1\right)^{-1} \end{aligned} $

Problem 1   The first problem is a classical problem which was considered in Refs. [24] and [25]. Let H=R. Presume that U: H→H is expressed as Ux=(x/2)sinx and F: H→H is expressed as Fx=x+sinx. In this problem, preset C=[－5, 5] and start point x₀=x₁=1. F, by simple calculation, can be verified to be monotone and consequently pseudomonotone on C. Meanwhile, U is not nonexpansive but quasi-nonexpansive on C. For Algorithm 3.3^[11], set σ=0.9, γ=0.9, θ_n=0.2 and α_n=(n³+1)^-1. For Algorithm 3.1^[25], take μ=0.2, τ₁=0.09, f(x)=0.15x, α_n=(n³+1)^-1 and β_n=(n+1)^-1. Different choices of ε are given, and the results of the experiment are illustrated in Table 1.

Problem 2   The second problem was applied in Refs. [12] and [13]. Suppose that H=R². Let F: R²→R² be represented as

$ \boldsymbol{F}\left(x_1, x_2\right)=\left[\begin{array}{c} \left(x_1^2+\left(x_2-1\right)^2\left(1+x_2\right)\right. \\ -x_1\left(x_2-1\right)^2-x_1^3 \end{array}\right] $

and U: R²→R² be represented as Ux=x－x³. Let the initial point be x₀=x₁ and feasible set C be R₂⁺, where P_C=max(0, x). It follows from the Monte Carlo approach^[8] that F is not monotone but pseudomonotone on C. Furthermore, it can be shown that U is not nonexpansive but quasi-nonexpansive on C. For Algorithm 2^[12], choose μ=0.2, γ=3 and l=0.7. For Algorithm 3.1^[16], take γ=0.2, λ₁=0.09 and f(x)=0.15x. For all tests, set ε=10^-3 and consider the different choices of initial point. Table 2 is a record of the experimental results.

Problem 3   In the third experiment, a problem in an infinite dimension space is given, which was used for numerical experiment in Refs. [12] and [24]. Assume that H=L²([0, 1]) with the transvection $\langle\boldsymbol{\eta}, \xi\rangle=\int_0^1 \boldsymbol{\eta}(t) \xi(t) \mathrm{d} t$ and the induced norm $\|\xi\|_2=\left(\int_0^1(|\xi(t)|)^2 \mathrm{~d} t\right)^{\frac{1}{2}}$. Presume F : H→His described as

$ (F x)(t)=e^{-\|x\|_2} \int_0^t x(v) \mathrm{d} v, t \in[0, 1] $

and U: H→H is described as $(U \boldsymbol{x})(t)=\int_0^1 t \boldsymbol{x}(v) \mathrm{d} v$, t∈[0, 1]. Consider practicable set C={x∈H: ‖x‖₂≤2}. It is known that F is pseudomonotone but not monotone on C. Moreover, U is nonexpansive and consequently quasi-nonexpansive on C. Let the initial point be x₀=x₁. For Algorithm 3.1^[25], take μ=0.2, τ₁=0.09, f(x)=0.15x, α_n=(n³+1)^-1 and β_n=(n+1)^-1. For all tests, set ε=10^－2 and consider the different choices of initial point. Table 3 records the experimental results.

4 Conclusions

Based on several classical approaches, a novel algorithm is presented. In particular, the suggested algorithm adopts a non-monotonic step size rule so that its convergence has no requirement to reckon Lipschitz constant. When parameters satisfy certain conditions, strong convergence is confirmed. Eventually, the testification of algorithm validation is undertaken by some numerical problems.