0 Introduction

For real Hilbert space H possessing transvection 〈·, ·〉 and norm ‖·‖, strong and weak converge of succession {x_n} to a dot x are distinguished and expressed as x_n→x and x_n⇀x. With regard to a preset non-null, closed convex set C⊆H, variational inequality problem (VI (P)) need to be considered. A dot x^′∈C is sought and the below formula holds:

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

(1)

As is known, x^′ is a solution of inequality (1) which equals x^′=P_C(x^′ －λF(x^′)) for all λ>0.

Variational inequality is a particularly practical model. Quite a few problems can be expressed as VI (P), for instance, nonlinear equation, complementarity matters, barrier questions, and many others^[1-2]. In recent decades, some projection algorithms to solve VI (P) have been raised and developed by many researchers^[3-11].

The most classical algorithm is projection gradient method^[12-13]:

$ \boldsymbol{x}_{n+1}=P_C\left(\boldsymbol{x}_n-\lambda F\left(\boldsymbol{x}_n\right)\right) $

Only when the mapped F is a(n) (inverse) strong monotone L-Lipschitz consecutive mapping and the step length satisfies λ∈(0, (2η)/L²), the convergence of algorithm can be obtained, where L>0 and η>0 represent Lipschitz and strong monotonicity constants, respectively. To solve the monotone L-Lipschitz continuous variational inequality, Korpelevich^[14] proposed an extragradient algorithm. This method has attracted the attention of many researchers and presents a variety of improved methods^[15-19]. Later, Censor et al.^[5] introduced an algorithm: the second projection is converted from C to a specific semi-space, which is called subgradient extragradient algorithm. Its form is listed as follows:

$ \boldsymbol{z}_{n+1}=P_{T_n}\left(\boldsymbol{z}_n-\lambda F \boldsymbol{x}_n\right), \boldsymbol{x}_{n+1}=P_C\left(\boldsymbol{z}_{n+1}-\lambda F \boldsymbol{z}_{n+1}\right) $

where

$ \lambda \in\left(0, \frac{1}{L}\right), \boldsymbol{T}_n=\left\{\boldsymbol{x} \in \boldsymbol{H}:\left\langle\boldsymbol{x}_n-\boldsymbol{x}, \boldsymbol{x}_n+\lambda F z_n-z_n\right\rangle \leqslant 0\right\} $

In the subgradient extragradient method, two F values need to be calculated per iteration. Recently, Popov^[20] and Malitsky and Semenov^[3] studied an algorithm that merely needs one F value per iteration, which is known as modified subgradient extragradient algorithm, but it requires knowledge of the L-Lipschitz constant or estimated value in advance. In 2018, Yang et al.^[21] proposed a modified subgradient extragradient method. When strict conditions are satisfied, the algorithm is proved to be at least R-linearly convergent. On this basis, this paper weakens its condition and solves some more general VI (P) under a non-monotonic stepsize strategy.

The other part of this article is arranged as follows. Section 1 reviews several relevant knowledge topics for later use. Section 2 states a subgradient extragradient method with a non-monotonic stepsize strategy. Section 3 includes two parts: the weak convergence and R-linearly convergence rate under the hypothesis of pseudomonotonicity and strong pseudomonotonicity respectively. In the last section, two classical examples are given to explain the advantages of the proposed algorithm.

1 Preliminaries

Several notations and lemmas are provided for later use.

Definition 1.1^[22-23]  For any α, β ∈ H, assume F on H is

(a) strongly monotone

$ \exists \gamma>0 \text {, s.t. } \gamma\|\boldsymbol{\alpha}-\boldsymbol{\beta}\|^2 \leqslant\langle\boldsymbol{\alpha}-\boldsymbol{\beta}, F \boldsymbol{\alpha}-F \boldsymbol{\beta}\rangle $

(b) monotone

$ \langle\boldsymbol{\alpha}-\boldsymbol{\beta}, F \boldsymbol{\beta}\rangle \leqslant\langle\boldsymbol{\alpha}-\boldsymbol{\beta}, F \boldsymbol{\alpha}\rangle $

(c) strongly pseudomonotone

$ \begin{array}{r} \exists \gamma>0 \text {, s.t. } 0 \leqslant\langle\boldsymbol{\beta}-\boldsymbol{\alpha}, F \boldsymbol{\alpha}\rangle \Rightarrow \\ \gamma\|\boldsymbol{\beta}-\boldsymbol{\alpha}\|^2 \leqslant\langle\boldsymbol{\beta}-\boldsymbol{\alpha}, F \boldsymbol{\beta}\rangle \end{array} $

(d) pseudomonotone

$ 0 \leqslant\langle\boldsymbol{\alpha}-\boldsymbol{\beta}, F \boldsymbol{\beta}\rangle \Rightarrow 0 \leqslant\langle\boldsymbol{\alpha}-\boldsymbol{\beta}, F \boldsymbol{\alpha}\rangle $

(e) L-Lipschitz continuous

$ \exists L>0 \text {, s.t. }\|F \boldsymbol{\alpha}-F \boldsymbol{\beta}\| \leqslant L\|\boldsymbol{\alpha}-\boldsymbol{\beta}\| $

(f) sequentially weakly continuous: if random succession {α_n} satisfies α_n⇀α, there is Fα_n⇀Fα.

Remark 1^[22]  The deduced process of (a)⇒(b) and (c)⇒(d) are obvious. Furthermore, (a) and (c) ensure that VI (P) has at most one solution.

Lemma 1.1^[24]  For a non-null closed convex set C⊆H, any α∈H, β ∈ C satisfies

(a) 〈P_Cα － α, β －P_Cα 〉≥0

(b) ‖P_Cα－α ‖²+‖P_C α－β ‖²≤‖ α－β ‖²

Lemma 1.2^[21]  (Peter-Paul inequality) When a and b are real numbers, and ε>0, the following inequality below can be obtained:

$ 2 a b \leqslant \frac{a^2}{\bar{\varepsilon}}+\bar{\varepsilon} b^2 $

Lemma 1.3^[25]  For two nonnegative real successions {a_n} and {b_n}, the existence of N>0 makes a_n+1≤a_n－b_n, ∀n≥N hold. It can be deduced that {a_n} is convergent and $\lim\limits_{n \rightarrow \infty} b_n=0$.

Lemma 1.4^[26]  For VI (P), C⊆H is a non-null enclosed convex and F: C → H is pseudo-monotone successive. α^′ as an answer to VI (P) is equal to

$ \left\langle\boldsymbol{\alpha}-\boldsymbol{\alpha}^{\prime}, F \boldsymbol{\alpha}\right\rangle \geqslant 0, \quad \forall \boldsymbol{\alpha} \in \boldsymbol{C} $

Lemma 1.5 (Opial)  Consider succession {α_n}⊆H. If α_n⇀α is true, there is

$ \liminf\limits_{n \rightarrow \infty}\left\|\boldsymbol{\alpha}_n-\boldsymbol{\alpha}\right\|<\liminf\limits_{n \rightarrow \infty}\left\|\boldsymbol{\alpha}_n-\overline{\boldsymbol{\alpha}}\right\|, \forall \overline{\boldsymbol{\alpha}} \neq \boldsymbol{\alpha} $

2 Algorithm

The proposed subgradient extragradient algorithm combines a non-monotonic stepsize strategy, which will be illustrated in this section.

Algorithm A

Step 1   Take

$ \boldsymbol{x}_0=\boldsymbol{y}_0 \in \boldsymbol{H}, \lambda_0 \geqslant \lambda_1>0, \boldsymbol{\alpha} \in(0, 2-\sqrt{2}) $

Choose a non-negative real number succession {p_n} which satisfies $\sum\limits_{n=1}^{\infty} p_n<+\infty$. Calculate

$ \boldsymbol{x}_1=P_C\left(\boldsymbol{x}_0-\lambda_0 F\left(\boldsymbol{y}_0\right)\right), \boldsymbol{y}_1=P_C\left(\boldsymbol{x}_1-\lambda_1 F\left(\boldsymbol{y}_0\right)\right) $

Step 2  Knowing the present iterates y_n-1, x_n, and y_n, construct

$ \boldsymbol{T}_n=\left\{\boldsymbol{x} \in \boldsymbol{H}:\left\langle\boldsymbol{y}_n-\boldsymbol{x}, \boldsymbol{x}_n-\lambda_n F\left(\boldsymbol{y}_{n-1}\right)-\boldsymbol{y}_n\right\rangle \geqslant 0\right\} $

and figure out

$ \boldsymbol{x}_{n+1}=P_{\boldsymbol{T}_n}\left(\boldsymbol{x}_n-\lambda_n F\left(\boldsymbol{y}_n\right)\right) $

Step 3  Calculate a stepsize

$ \lambda_{n+1}= \\ \left\{\begin{array}{l} \min \left\{\frac{\boldsymbol{\alpha}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|^2+\frac{\boldsymbol{\alpha}}{2}\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2}{2\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle}\right\} , \\ \left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle>0 \\ \lambda_n+p_n, \text { otherwise } \end{array}\right. $

Step 4  Compute

$ \boldsymbol{y}_{n+1}=P_C\left(\boldsymbol{x}_{n+1}-\lambda_{n+1} F\left(\boldsymbol{y}_n\right)\right) $

If x_n+1=x_n and y_n= y_n－1 (or y_n+1= x_n+1= y_n), stop and y_n is a solution. Otherwise, let n: =n+1, and go to Step 2.

Lemma 2.1  Algorithm A can produce step lengths order {λ_n}, and it satisfies $\lim\limits_{n \rightarrow \infty} \lambda_n=\lambda$ and $\min \left\{\frac{\boldsymbol{\alpha}}{\sqrt{2} L}, \lambda_1\right\} \leqslant \lambda \leqslant \lambda_1+P, \text { where } P=\sum\limits_{n=1}^{\infty} p_n$.

Proof  Because F is L-Lipschitz continuous mapping, if 〈x_n+1－y_n, F(y_n－1)－F(y_n)〉>0, then

$ \begin{aligned} &\frac{\boldsymbol{\alpha}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|^2+\frac{\boldsymbol{\alpha}}{2}\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2}{2\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle} \geqslant\\ &\frac{\sqrt{2} \boldsymbol{\alpha}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|}{2\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|\left\|F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\|} \geqslant\\ &\frac{\boldsymbol{\alpha}\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|}{\sqrt{2} L\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|}=\frac{\boldsymbol{\alpha}}{\sqrt{2} L} \end{aligned} $

Through derivation of λ_n+1 and mathematical induction, it is known that the highest and lowest values of step succession {λ_n} correspond to λ₁+P and $\min \left\{\frac{\boldsymbol{\alpha}}{\sqrt{2} L}, \lambda_1\right\}$, respectively. From the expression of λ_n, order t_n+1=λ_n+1－λ_n, and the following inequalities are obtained:

$ t_{n+1}^{+} \leqslant p_n, \sum\limits_{n=1}^{\infty} p_n<+\infty $

Here $t_{n+1}^{+}=\max \left\{0, t_{n+1}\right\}, t_{n+1}^{-}=\max \left\{0, -t_{n+1}\right\}$. The positive term $\sum\limits_{n=1}^{\infty} t_{n+1}^{+}$ is astringent in the aforesaid inequality. The subsequent task is to derive that $\sum\limits_{n=1}^{\infty} t_{n+1}^{-}$ also converges. Otherwise, assuming $\sum\limits_{n=1}^{\infty} t_{n+1}^{-}=+\infty$, there is

$ t_{n+1}=t_{n+1}^{+}-t_{n+1}^{-} $

which implies that

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

As m→+∞, λ_m→－∞, the above equation is in conflict with the previous conclusion. Because $\sum\limits_{n=1}^{\infty} t_{n+1}^{+}$ and $\sum\limits_{n=1}^{\infty} t_{n+1}^{-}$ are convergent, let m→+∞, then $\lim\limits_{m \rightarrow \infty} \lambda_m$=λ in the above formula. Further, since {λ_n} has the lowest value $\min \left\{\frac{\boldsymbol{\alpha}}{\sqrt{2} L}, \lambda_1\right\}$ and the highest value λ₁+P, there must be $\min \left\{\frac{\boldsymbol{\alpha}}{\sqrt{2} L}, \lambda_1\right\} \leqslant \lambda \leqslant \lambda_1+P$.

3 Convergence Analysis

The relevant convergence is given in this part. Firstly, some sequences weakly converging to the solution were analyzed and the R-linear convergence rate was obtained. Moreover, the following hypotheses is made about VI (P):

(A1) The disaggregation S of inequality (1) is non-null;

(A2) F is pseudomonotone over H and weakly sequentially continuous over C;

(A3) F is L-Lipschitz consecutive over H.

Convergence of the algorithm is discussed below. The following lemma is essential to gain the main theorem.

Lemma 3.1  Algorithm A can produce two successions {x_n} and {y_n}, which are bounded.

Proof  For v ∈ S, y_n ∈ C naturally has

$ \left\langle\boldsymbol{y}_n-\boldsymbol{v}, F(\boldsymbol{v})\right\rangle \geqslant 0, \quad \forall n \geqslant 1 $

Then 〈y_n－v, F(y_n)〉≥0 in consideration of pseudomonotonicity, which in turn is equivalent to

$ \left\langle\boldsymbol{y}_n-\boldsymbol{x}_{n+1}, F\left(\boldsymbol{y}_n\right)\right\rangle \geqslant\left\langle\boldsymbol{v}-\boldsymbol{x}_{n+1}, F\left(\boldsymbol{y}_n\right)\right\rangle $

(2)

The proving process is analogue to Lemma 3.3 of Ref. [23], it can be obtained that

$ \begin{aligned} &\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1) \| \boldsymbol{x}_n- \\ &\boldsymbol{y}_{n-1}\left\|^2-\left(1-\frac{1}{\sqrt{2}}\right)\right\| \boldsymbol{y}_{n-1}-\boldsymbol{y}_n\left\|^2-\right\| \boldsymbol{y}_n-\boldsymbol{x}_{n+1} \|^2+ \\ &2 \lambda_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle \end{aligned} $

(3)

In accordance with the definition of λ_n, inequality (3) yields that

$ \begin{aligned} &\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \leqslant \\ &\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2- \\ &\left(1-\frac{1}{\sqrt{2}}\right)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2-(2-\sqrt{2})\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2+ \\ &\frac{\lambda_n}{\lambda_{n+1}}\left(\boldsymbol{\alpha}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|^2+\frac{\boldsymbol{\alpha}}{2}\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2\right)= \\ &\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2- \\ &\left(1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2- \\ &\left(2-\sqrt{2}-\boldsymbol{\alpha} \frac{\lambda_n}{\lambda_{n+1}}\right)\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \end{aligned} $

(4)

Taking the limit n→∞, due to $\boldsymbol{\alpha} \in(0, 2-\sqrt{2})$,

$ \begin{aligned} &1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \cdot \frac{\lambda_n}{\lambda_{n+1}} \rightarrow 1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2}>0 \\ &2-\sqrt{2}-\boldsymbol{\alpha} \frac{\lambda_n}{\lambda_{n+1}} \rightarrow 2-\sqrt{2}-\boldsymbol{\alpha}>0 \end{aligned} $

N₁>0 and G₁>0, such that for any n>N₁, there is

$ 1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \cdot \frac{\lambda_n}{\lambda_{n+1}}>G_1, 2-\sqrt{2}-\boldsymbol{\alpha} \frac{\lambda_n}{\lambda_{n+1}}>G_1 $

Inequality (4) can be rewritten by

$ \begin{gathered} \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \leqslant \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2 \\ G_1\left(\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2+\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2\right) \end{gathered} $

(5)

For n>N₁, set

$ \begin{gathered} a_n=\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2 \\ b_n=G_1\left(\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2+\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2\right) \end{gathered} $

Therefore, inequality (5) can be rewritten as a_n+1≤a_n－b_n, ∀n≥N₁. According to Lemma 1.3, {a_n} is convergent and $\lim\limits_{n \rightarrow \infty} b_n=0 . \quad\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2 \leqslant a_n$, which means that {x_n} is bounded and

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

By means of trigonometric inequality, there is x_n→y_n and {y_n} is limited as well.

Theorem 3.1  Suppose that successions {x_n} and {y_n} are produced by Algorithm A, it can be deduced that they weakly converge to the same pot x^′∈ S.

Proof  Lemma 3.1 yields the result that there is a sub-series {x_nk}⊆{ x_n} that converges weakly to a dot x^′. Obviously there is {y_nk} weakly converging to x^′∈C.

Since y_nk+1=P_C(x_nk+1－λ_nk+1F(y_nk)), applying Lemma 1.1(a), there is

$ \left\langle z-\boldsymbol{y}_{n_k+1}, \boldsymbol{y}_{n_k+1}-\boldsymbol{x}_{n_k+1}+\lambda_{n_k+1} F\left(\boldsymbol{y}_{n_k}\right)\right\rangle \geqslant 0, \forall z \in \boldsymbol{C} $

or equivalently

$ \begin{aligned} 0 \leqslant &\left\langle\boldsymbol{z}-\boldsymbol{y}_{n_k+1}, \boldsymbol{y}_{n_k+1}-\boldsymbol{x}_{n_k+1}\right\rangle+\lambda_{n_k+1}\left\langle\boldsymbol{z}-\boldsymbol{y}_{n_k+1}\right.\\ &\left.F\left(\boldsymbol{y}_{n_k}\right)\right\rangle=\left\langle\boldsymbol{z}-\boldsymbol{y}_{n_k+1}, \boldsymbol{y}_{n_k+1}-\boldsymbol{x}_{n_k+1}\right\rangle+\lambda_{n_k+1}\langle\boldsymbol{z}-\\ &\left.\boldsymbol{y}_{n_k}, F\left(\boldsymbol{y}_{n_k}\right)\right\rangle+\lambda_{n_k+1}\left\langle\boldsymbol{y}_{n_k}-\boldsymbol{y}_{n_k+1}, F\left(\boldsymbol{y}_{n_k}\right)\right\rangle \end{aligned} $

Taking k→∞ in the above-mentioned expression, since λ_nk+1>0, the following is derived:

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

(6)

There is a positive succession {ε_k}, which decreases and has a limit of zero. A rigorously increasing sequence of positive integers {N_k} can be found, thus the inequality below is obtained:

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

(7)

In light of inequality (6), it can be concluded that N_k exists. In addition, for each k, let

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

There is 〈F(y_Nk), w_Nk〉=1. It follows from inequality (7) that for each k, there is

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

Furthermore, F is pseudomonotone over H, so there is

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

which corresponds to

$ \begin{gathered} \left\langle \boldsymbol{z}+\varepsilon_k \boldsymbol{w}_{N_k}-\boldsymbol{y}_{N_k}, F(\boldsymbol{z})-F\left(\boldsymbol{z}+\varepsilon_k \boldsymbol{w}_{N_k}\right)\right\rangle- \\ \varepsilon_k\left\langle\boldsymbol{w}_{N_k}, F(\boldsymbol{\boldsymbol{z}})\right\rangle \leqslant\left\langle {z}-\boldsymbol{y}_{N_k}, F(\boldsymbol{z})\right\rangle \end{gathered} $

(8)

$\lim\limits_{k \rightarrow \infty} \varepsilon_k \boldsymbol{w}_{N_k}=0$ is known. According to the definition of weak sequential continuity, it is evident that {F(y_nk)} converges weakly to {F(x^′)}. Assume that F(x^′)≠0 (or else x^′ is the solution), because of the weak sequential lower continuity of norm mapping, the following result is obtained:

$ 0<\left\|F\left(\boldsymbol{x}^{\prime}\right)\right\| \leqslant \lim\limits_{k \rightarrow \infty} \inf \left\|F\left(\boldsymbol{y}_{n_k}\right)\right\| $

Due to {y_Nk}⊂{ y_nk} and $\lim\limits_{k \rightarrow \infty} \varepsilon_k=0$, the following expression is deduced:

$ \begin{gathered} 0 \leqslant \lim\limits_{k \rightarrow \infty} \sup \left\|\varepsilon_k \boldsymbol{w}_{N_k}\right\|=\lim\limits_{k \rightarrow \infty} \sup \left(\frac{\varepsilon_k}{\left\|F\left(\boldsymbol{y}_{N_k}\right)\right\|}\right) \leqslant \\ \frac{\lim\limits_{k \rightarrow \infty} \sup \varepsilon_k}{\lim\limits_{k \rightarrow \infty} \inf \left\|F\left(\boldsymbol{y}_{N_k}\right)\right\|}=0 \end{gathered} $

which implies that $\lim\limits_{k \rightarrow \infty} \varepsilon_k \boldsymbol{w}_{N_k}=0$. When k→∞, the right limit of inequality (8) is zero. It is not difficult to figure out $\lim\limits_{k \rightarrow \infty} \inf \left\langle \boldsymbol{z}-\boldsymbol{y}_{N_k}, F(\boldsymbol{z})\right\rangle \geqslant 0$. Hence,

$ \begin{gathered} \left\langle \boldsymbol{z}-\boldsymbol{x}^{\prime}, F(\boldsymbol{z})\right\rangle=\lim\limits_{k \rightarrow \infty}\left\langle \boldsymbol{z}-\boldsymbol{y}_{N_k}, F(\boldsymbol{z})\right\rangle= \\ \lim\limits_{k \rightarrow \infty} \inf \left\langle \boldsymbol{z}-\boldsymbol{y}_{N_k}, F(\boldsymbol{z})\right\rangle \geqslant 0, \forall \boldsymbol{z} \in \boldsymbol{z} \end{gathered} $

Invoking Lemma 1.4 yields x^′∈S.

Finally, the conclusion that {x_n} converges weakly to x^′ is gained. Suppose that it has at any rate two weak accumulation dots x^″∈S and x^′∈S. At this time, there is x^″≠x^′. Let the limit of {x_ni} be x^″, from Lemma 1.5 the following can be known:

$ \begin{aligned} &\lim\limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_n-\boldsymbol{x}^{\prime \prime}\right\|=\lim\limits_{i \rightarrow \infty}\left\|\boldsymbol{x}_{n_i}-\boldsymbol{x}^{\prime \prime}\right\|= \\ &\quad \lim\limits_{i \rightarrow \infty} \inf \left\|\boldsymbol{x}_{n_i}-\boldsymbol{x}^{\prime \prime}\right\|<\lim\limits_{i \rightarrow \infty} \inf \left\|\boldsymbol{x}_{n_i}-\boldsymbol{x}^{\prime}\right\| \\ &\quad \lim\limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_n-\boldsymbol{x}^{\prime}\right\|=\lim\limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}^{\prime}\right\|= \\ &\quad \lim\limits_{k \rightarrow \infty} \inf \left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}^{\prime}\right\|<\lim\limits_{k \rightarrow \infty} \inf \left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}^{\prime \prime}\right\| \\ &\lim\limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_k}-\boldsymbol{x}^{\prime \prime}\right\|=\lim\limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_n-\boldsymbol{x}^{\prime \prime}\right\| \end{aligned} $

From this contradiction, x_n⇀x^′ is obtained. Since $\lim_{n \rightarrow \infty}\left\|\boldsymbol{x}_n-\boldsymbol{y}_n\right\|=0$, there is y_n⇀x^′. The proof is complete.

Subsequently, (A2) was replaced with (A2^*) to prove that Algorithm A has R-linear convergence rate.

(A2^*) F is strongly pseudomonotone over H.

Theorem 3.2  If the supposition (A1), (A2^*), and (A3) comes into existence, Algorithm A can produce succession {x_n}. It can be derived that {x_n} at least R-linearly strongly converges to the only resolution u of inequality (1).

Proof  〈y_n－v, F(v)〉≥0, ∀ v ∈ S is true. According to the hypothesis that F is strongly pseudo-monotone, the following formula is true:

$ \gamma\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2 \leqslant\left\langle\boldsymbol{y}_n-\boldsymbol{v}, F\left(\boldsymbol{y}_n\right)\right\rangle $

It could also be described as

$ \begin{aligned} \gamma &\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2 \leqslant\left\langle\boldsymbol{y}_n-\boldsymbol{x}_{n+1}, F\left(\boldsymbol{y}_n\right)\right\rangle-\\ &\left\langle\boldsymbol{v}-\boldsymbol{x}_{n+1}, F\left(\boldsymbol{y}_n\right)\right\rangle \end{aligned} $

(9)

Therefore, invoking the regulation of T_n in the Algorithm A yields that

$ \begin{aligned} &\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, \boldsymbol{x}_n-\lambda_n F\left(\boldsymbol{y}_n\right)-\boldsymbol{y}_n\right\rangle \leqslant \\ &\lambda_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle \end{aligned} $

(10)

Combining Lemma 1.1 (b) and inequality (9) with inequality (10), the following can be derived:

$ \begin{gathered} \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{x}_n-\boldsymbol{x}_{n+1}\right\|^2+ \\ 2 \lambda_n\left\langle\boldsymbol{v}-\boldsymbol{x}_{n+1}, F\left(\boldsymbol{y}_n\right)\right\rangle \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2- \\ \left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|^2-\left\|\boldsymbol{y}_n-\boldsymbol{x}_n\right\|^2+2\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, \right. \\ \left.\boldsymbol{x}_n-\boldsymbol{y}_n-\lambda_n F\left(\boldsymbol{y}_n\right)\right\rangle-2 \gamma \lambda_n\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2 \leqslant \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2-\left\|\boldsymbol{x}_n-\boldsymbol{y}_{n-1}\right\|^2-\left\|\boldsymbol{y}_{\mathrm{n}-1}-\boldsymbol{y}_n\right\|^2+ \\ 2\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|\left\|\boldsymbol{x}_n-\boldsymbol{y}_{n-1}\right\|-\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2+ \\ 2 \lambda_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-F\left(\boldsymbol{y}_n\right)\right\rangle-2 \gamma \lambda_n\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2 \end{gathered} $

(11)

From Lemma 1.2, substituting $\bar{\varepsilon}=\sqrt{2}$ into inequality (11) gives us inequality (12), namely,

$ \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1) \cdot \\ \begin{aligned} &\left\|\boldsymbol{x}_n-\boldsymbol{y}_{n-1}\right\|^2-\left(1-\frac{1}{\sqrt{2}}\right)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2- \\ &\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2+2 \lambda_n\left\langle\boldsymbol{x}_{n+1}-\boldsymbol{y}_n, F\left(\boldsymbol{y}_{n-1}\right)-\right. \\ &\left.F\left(\boldsymbol{y}_n\right)\right\rangle-2 \gamma \lambda_n\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2 \end{aligned} $

(12)

From the form defined by λ_n, ∃N₂>0, λ>0, ∀n>N₂ s.t. λ_n>λ, ∀n>N₂ is obtained. The above formula can be arranged as

$ \begin{array}{r} \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \leqslant \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2- \\ \left(1-\frac{1}{\sqrt{2}}\right)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2-(2-\sqrt{2}) \cdot \\ \left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2+\frac{\lambda_n}{\lambda_{n+1}}\left(\boldsymbol{\alpha}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{y}_n\right\|^2+\right. \\ \left.\frac{\boldsymbol{\alpha}}{2}\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2\right)-2 \gamma \lambda_n\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2= \\ \left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2- \\ 2 \gamma \lambda\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2-\left(1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \cdot \frac{\lambda_n}{\lambda_{n+1}}\right) \cdot \end{array} $

$ \begin{aligned} &\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2-\left(2-\sqrt{2}-\boldsymbol{\alpha} \cdot \frac{\lambda_n}{\lambda_{n+1}}\right) \cdot \\ &\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \end{aligned} $

(13)

In short, denote

$ \begin{aligned} \varepsilon &=\frac{1}{4} \cdot\left(4-2 \sqrt{2}+\frac{2-\sqrt{2}-\boldsymbol{\alpha}}{\gamma \lambda}+\right.\\ & {\left.\left[\left(4-2 \sqrt{2}+\frac{2-\sqrt{2}-\boldsymbol{\alpha}}{\gamma \lambda}\right)^2+8(2 \sqrt{2}-2)\right]^{1 / 2}\right) } \end{aligned} $

which gives the fact that

$ \varepsilon>\frac{4-2 \sqrt{2}+\sqrt{(4-2 \sqrt{2})^2+16(\sqrt{2}-1)}}{4}=1 $

Invoking Lemma 1.2 can deduce

$ \begin{aligned} &-\left\|\boldsymbol{y}_n-\boldsymbol{v}\right\|^2=-\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2-\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2- \\ &2\left\langle\boldsymbol{y}_n-\boldsymbol{x}_{n+1}, \boldsymbol{x}_{n+1}-\boldsymbol{v}\right\rangle \leqslant(\varepsilon-1) \cdot \\ &\quad\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2+\left(\frac{1}{\varepsilon}-1\right) \cdot\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \end{aligned} $

(14)

Using inequality (14), the following formula is obtained by rearranging inequality (13) and shifting items

$ \begin{aligned} &\left(1-2 \gamma \lambda\left(\frac{1}{\varepsilon}-1\right)\right)\left(\left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2+(\sqrt{2}-1) \cdot\right. \\ &\left.\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2\right) \leqslant\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1) \cdot \\ &\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2-\left(1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \cdot \frac{\lambda_n}{\lambda_{n+1}}\right) \cdot \\ &\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2-\left(2-\sqrt{2}-\boldsymbol{\alpha} \cdot \frac{\lambda_n}{\lambda_{n+1}}+(4-\right. \\ &\left.\left.2 \sqrt{2}-2 \varepsilon+\frac{2 \sqrt{2}-2}{\varepsilon}\right) \gamma \lambda\right) \cdot\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2 \end{aligned} $

(15)

Similar to the processing method in inequality (4), there is N₃>N₂>0, G₂>0, such that for any n>N₃, there is

$ \begin{aligned} &1-\frac{1}{\sqrt{2}}-\frac{\boldsymbol{\alpha}}{2} \cdot \frac{\lambda_n}{\lambda_{n+1}}>G_2 \\ &2-\sqrt{2}-\boldsymbol{\alpha} \cdot \frac{\lambda_n}{\lambda_{n+1}}+ \\ &\left(4-2 \sqrt{2}-2 \varepsilon+\frac{2 \sqrt{2}-2}{\varepsilon}\right) \gamma \lambda>G_2 \end{aligned} $

Set $\rho=-2 \gamma \lambda\left(\frac{1}{\varepsilon}-1\right)$. Obviously there is ρ>0. For ∀n>N₁, set

$ \begin{gathered} a_n=\left\|\boldsymbol{x}_n-\boldsymbol{v}\right\|^2+(\sqrt{2}-1)\left\|\boldsymbol{y}_{n-1}-\boldsymbol{x}_n\right\|^2 \\ b_n=G_2\left(\left\|\boldsymbol{y}_{n-1}-\boldsymbol{y}_n\right\|^2+\left\|\boldsymbol{y}_n-\boldsymbol{x}_{n+1}\right\|^2\right) \end{gathered} $

Then inequality (15) is equivalent to

$ (1+\rho) a_{n+1} \leqslant a_n-b_n, \forall n>N_3 $

From induction, it may be deduced that for ∀n>N₃ there is

$ \begin{gathered} \left\|\boldsymbol{x}_{n+1}-\boldsymbol{v}\right\|^2 \leqslant a_{n+1} \leqslant \frac{1}{1+\rho} a_n \leqslant \cdots \leqslant \\ \left(\frac{1}{1+\rho}\right)^{n+1-N_3} a_{N_3} \end{gathered} $

which completes the proof.

Remark 2  Compared with Theorem 2.1 in Ref. [23], the above theorem has three major advantages:

1) Fewer parameters are used;

2) A non-monotonic stepsize strategy is adopted instead of the monotonic decreasing stepsize;

3) The previous assumptions are weakened, namely, monotonicity are replaced with pseudomonotonicity and weak sequential continuity of F, and strong monotonicity is replaced with strong pseudomonotonicity.

4 Numerical Results

By comparing Algorithm A with Algorithm 1^[21] in numerical experiments, the effectiveness of the proposed algorithm is demonstrated.

The iteration times (ite.) and calculation time (t) conducted in all tests were recorded in seconds. The condition for terminating the algorithm was ‖x_n+1－x_n‖≤ε. For the tolerance, ε=10^－6 was set in all experiments. The following parameters were used in the algorithm mentioned below:

$ \boldsymbol{x}_0=\boldsymbol{y}_0, \lambda_0=\lambda_1=0.3, \mu=0.2, \theta=\frac{1}{1000} $

Problem 1  In the first group of numerical experiments (known as HpHard problem, also considered in Refs. [25] and [27]), let G(x)=d+Ax with d∈Rⁿ and A = BB^T+ C + D, where B, C, D∈R^n×n. Each item of B and skew-symmetric array C is consistently produced from (－5, 5). Each term of d and diagonal D is consistently produced from (－500, 0) and (0, 0.3), respectively. The practicable set is R_n⁺, where P_C=x₊=max(0, x). For all tests, x₀=y₀=(1, 1, ..., 1) is taken. $\boldsymbol{\alpha}=2-\sqrt{2}-\frac{1}{1000}$ is taken for Algorithm A and three stochastic values are produced based on unlike options of A and d. The results in Table 1 indicate the effect of the proposed algorithm.

From Table 1, p_n controls the non-monotonicity of the stepsize. When p_n=0, the stepsize decreases monotonically. At this time, the iteration will stop because the stepsize is too small to find the best point. Non-monotonic stepsize increases or decreases to quickly find the best advantage and make the iteration stop. It can also be seen that the non-monotone stepsize has less iterations and faster time than monotone stepsize.

Problem 2  The second question was applied to Refs. [28] and [29], where

$ \begin{gathered} \boldsymbol{H}(\boldsymbol{x})=\left(h_1(\boldsymbol{x}), h_2(\boldsymbol{x}), \ldots, h_m(\boldsymbol{x})\right) \\ h_i(\boldsymbol{x})=x_i^2+x_{i-1}^2+x_{i+1} x_i+x_i x_{i-1}+x_{i+1}+4 x_i- \\ 2 x_{i-1}-1 \\ x_0=x_{m+1}=0, i=1, 2, \ldots, m \end{gathered} $

The practicable set is C = R_m⁺ and P_C=x₊=max(0, x). Initial point is taken as x₀=y₀=(0, 0, ..., 0). Table 2 collects the results.

As shown in Table 2, it can be observed that Algorithm A is more advanced than Algorithm 1 in terms of iterations and calculation time. Moreover, when the value of α is larger, the effect of the new algorithm is better.

5 Conclusions

The paper presents the convergence analysis of Lipschitz continuous pseudomonotone VI (P). When the value of Lipschitz constant is unclear, weak convergence of the sequence is proved by using non-monotonically updating stepsize strategy. Furthermore, the R-linear convergence rate of sequence was obtained and the performance of the scheme was demonstrated on some numerical examples to show the effectiveness of the proposed algorithms. References