Subgradient Extragradient Methods for Equilibrium Problems and Fixed Point Problems in Hilbert Space

Author NameAffiliation
Lulu YinSchool of Mathematics and Statistics, Xidian University, Xi′an 710126, China
Hongwei LiuSchool of Mathematics and Statistics, Xidian University, Xi′an 710126, China

Abstract:
Inspired by inertial methods and extragradient algorithms, two algorithms were proposed to investigate fixed point problem of quasinonexpansive mapping and pseudomonotone equilibrium problem in this study. In order to enhance the speed of the convergence and reduce computational cost, the algorithms used a new step size and a cutting hyperplane. The first algorithm was proved to be weak convergence, while the second algorithm used a modified version of Halpern iteration to obtain strong convergence. Finally, numerical experiments on several specific problems and comparisons with other algorithms verified the superiority of the proposed algorithms.
Key words:subgradient extragradient methodsinertial methodspseudomonotone equilibrium problemsfixed point problemsLipschitz-type condition
DOI：10.11916/j.issn.1005-9113.2020029
Clc Number:O224
Fund:

Lulu Yin, Hongwei Liu. Subgradient extragradient methods for equilibrium problems and fixed point problems in Hilbert space[J]. Journal of Harbin Institute of Technology (New Series), 2022, 29(1): 15-23. DOI: 10.11916/j.issn.1005-9113.2020029
Corresponding author Lulu Yin, E-mail: yinluluxidian@163.com Article history Received: 2020-06-16



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Subgradient extragradient methods for equilibrium problems and fixed point problems in Hilbert space
Lulu Yin, Hongwei Liu
School of Mathematics and Statistics, Xidian University, Xi'an 710126, China
Received: 2020-06-16
Corresponding author: Lulu Yin, E-mail: yinluluxidian@163.com.

Abstract: Inspired by inertial methods and extragradient algorithms, two algorithms were proposed to investigate fixed point problem of quasinonexpansive mapping and pseudomonotone equilibrium problem in this study. In order to enhance the speed of the convergence and reduce computational cost, the algorithms used a new step size and a cutting hyperplane. The first algorithm was proved to be weak convergence, while the second algorithm used a modified version of Halpern iteration to obtain strong convergence. Finally, numerical experiments on several specific problems and comparisons with other algorithms verified the superiority of the proposed algorithms.
Keywords: subgradient extragradient methodsinertial methodspseudomonotone equilibrium problemsfixed point problemsLipschitz-type condition
0 Introduction Let H equip with associated norm ‖·‖ and inner product 〈·, ·〉. With regard to a preset non-null, closed convex set X?H and bifunction f: H×H→R contenting f(z, z)=0, z∈X. The equilibrium problem (EP) related to f lies in seeking out $\breve{\boldsymbol{x}} \in \boldsymbol{X} $ such that
$f(\breve{\boldsymbol{x}}, \boldsymbol{z}) \geqslant 0, \quad \forall \boldsymbol{z} \in \boldsymbol{X}$ (1)
which is also called Ky Fan inequality^[1]. Here, its disaggregation is denoted by EP(X, f). Interestingly, fixed point problem, Nash equilibrium problem, variational inequality, optimization problem, and many other models can be transformed into EP^[2-4]. In recent years, numerous measures of seeking the approximate solution of problem (1) have been discussed^[5-12].
A common measure is extragradient method^[7-8], which tackles two strongly convex programming models in every iteration. Evaluation of the subprograms of the algorithm may be extremely expensive for involute structure of bifunctions and/or feasible sets. In recent years, there is a vast amount of literature concerning the study and improvement of this algorithm, among which Refs. [9-13] are representative. To combat this drawback, Van Hieu^[10] put forward an modified way to settle fixed point problem and equilibrium problem by using Halpern iteration method and subgradient extragradient, and established the strong convergence result. It is clear that the second strongly convex programming problems is performed in half space. Furthermore, to speed up the convergence of the algorithm, Rehman et al.^[14] recently designed a new algorithm by applying the inertial technique^[15-17], and obtained the weak convergence results under appropriate assumptions.
Motivated and inspired by the several above-mentioned advantages in Refs.[10] and [14], this paper introduces two new algorithms to analyze fixed point problem with quasinonexpansive mapping and pseudomonotone equilibrium problem. Moreover, the weak convergence results of one algorithm and the strong convergence results of another algorithm were obtained. Among the two proposed algorithms, the new step size avoids Lipschitz constants of bifunction. The experimental results reflect the numerical behavior.
The paper is structured as follows. Some preliminary information is reviewed in Section 1. Section 2 is the research of the convergence results. The similar application to variational inequalities is elaborated in Section 3. Eventually, two experiments of Section 4 show that the proposed algorithms possess perfect efficiency.
1 Preliminaries In this paper, several notations and lemmas were introduced for later use. Stipulate s⁺=max{0, s}, s^-=max{0, -s}, ?s∈R. Presume that a, b, c∈H and τ∈R are known. Then
$\begin{array}{l}\|\tau \boldsymbol{b}+(1-\tau) \boldsymbol{c}\|^{2}=\tau\|\boldsymbol{b}\|^{2}+(1-\tau)\|\boldsymbol{c}\|^{2}- \\\ \ \ \ \ \ \ \ \ \ \ \ \tau(1-\tau)\|\boldsymbol{c}-\boldsymbol{b}\|^{2}\end{array}$ (2)
$\begin{array}{l}2\langle\boldsymbol{b}-\boldsymbol{c}, \boldsymbol{b}-\boldsymbol{a}\rangle=\|\boldsymbol{b}-\boldsymbol{c}\|^{2}+\|\boldsymbol{a}-\boldsymbol{b}\|^{2}- \\\ \ \ \ \ \ \ \ \|\boldsymbol{c}-\boldsymbol{a}\|^{2}\end{array}$ (3)
Evidently, the projection P_X possesses the following feature:
$\mathit{\boldsymbol{c}} = {\mathit{\boldsymbol{P}}_X}\mathit{\boldsymbol{b}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \Leftrightarrow {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \langle \mathit{\boldsymbol{b}} - \mathit{\boldsymbol{c}}, \mathit{\boldsymbol{a}} - \mathit{\boldsymbol{c}}\rangle \leqslant 0, \forall \mathit{\boldsymbol{a}} \in \mathit{\boldsymbol{X}}$
Definition 1.1?? Bifunction f: H×H→R is
●??Pseudomonotone in X:
$f(\boldsymbol{b}, \boldsymbol{c}) \geqslant 0 \Rightarrow f(\boldsymbol{c}, \boldsymbol{b}) \leqslant 0, \forall \boldsymbol{a}, \boldsymbol{b} \in \boldsymbol{X}$
●??Lipschitz-type condition in X:
$\begin{aligned}&\exists l_{1}, l_{2}>0 \\&\text { s.t. }f(\boldsymbol{c}, \boldsymbol{a})-f(\boldsymbol{c}, \boldsymbol{b})-f(\boldsymbol{b}, \boldsymbol{a}) \leqslant \\&\qquad\qquad l_{1}\|\boldsymbol{b}-\boldsymbol{a}\|-l_{2}\|\boldsymbol{c}-\boldsymbol{b}\| , \\&\qquad\qquad\quad \forall \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \in \boldsymbol{X}\end{aligned}$
Further, given c∈X and function h: X→(-∞, ∞], its subdifferential is described by
$\begin{aligned}\partial h(\boldsymbol{c})=&\{\xi \in \boldsymbol{H}: h(\boldsymbol{b})-h(\boldsymbol{c}) \geqslant\\&\langle\xi, \boldsymbol{b}-\boldsymbol{c}\rangle, \forall \boldsymbol{b} \in \boldsymbol{X}\}\end{aligned}$
and its normal cone of X is
$\boldsymbol{N}_{X}(\boldsymbol{c})=\{\eta \in \boldsymbol{H}:\langle\eta, \boldsymbol{b}-\boldsymbol{c}\rangle \leqslant 0, \boldsymbol{b} \in \boldsymbol{X}\}$
Lemma 1.1^[18]?? Suppose h: X→(-∞, +∞] is subdifferentiable, lower hemicontinuous, and convex. Suppose that there is a point on X that makes h continuous, or h is finite at some interior point on X. Then c^*=arg min{h(c): c∈X} is equivalent to 0∈?h(c^*)+N_X(c^*).
Definition 1.2?? Fix(T)≠? and T: H→H are known, then
1) T is known as quasi-nonexpansive:
$\forall \boldsymbol{a} \in \boldsymbol{H}, \boldsymbol{b} \in \operatorname{Fix}(T), \|T \boldsymbol{a}-\boldsymbol{b}\| \leqslant\|\boldsymbol{a}-\boldsymbol{b}\|$
2) T-I is known as demiclosed at zero:
$\forall\left\{c_{n}\right\} \subset \boldsymbol{H}, \text { s.t. } \boldsymbol{c}_{n} \rightharpoonup \boldsymbol{z}, $
$T \boldsymbol{c}_{n}-\boldsymbol{c}_{n} \rightarrow 0 \Rightarrow \boldsymbol{c} \in \operatorname{Fix}(T)$
The information of the proximal operator is recalled which is the basic tool for the proposed algorithm. Assume that a function h: H→(-∞, +∞] is lower hemicontinuous, proper, and convex. Given λ < 0 and c∈H, the proximal operator of h is described as
$\operatorname{prox}_{\lambda h}(\boldsymbol{c})=\arg \min \left\{\lambda h(\boldsymbol{b})+\frac{1}{2}\|\boldsymbol{c}-\boldsymbol{b}\|^{2}: \boldsymbol{b} \in \boldsymbol{X}\right\}$
Remark 1.1?? For ease of notation for lower hemicontinuous and proper convex function h: X→(-∞, +∞], symbol prox_λh^X(x) represents proximal operator of h with λ on X.
Then a crucial attribute of proximal operators is recommended.
Lemma 1.2^[19]?? Let c∈H, b∈X, and λ > 0. Then
$\begin{aligned}&\lambda\left\{h(\boldsymbol{b})-h\left(\operatorname{prox}_{\lambda h}^{X}(\boldsymbol{c})\right)\right\} \geqslant \\&\quad\left\langle\boldsymbol{c}-\operatorname{prox}_{\lambda h}^{X}(\boldsymbol{c}), \boldsymbol{b}-\operatorname{prox}_{\lambda h}^{X}(\boldsymbol{c})\right\rangle\end{aligned}$ (4)
Lemma 1.3 (Peter-Paul inequality)?? Given ε > 0 and a₁, a₂∈R, the following property holds:
$2 a_{1} a_{2} \leqslant \frac{a_{1}^{2}}{\varepsilon}+\varepsilon a_{2}^{2}$ (5)
Lemma 1.4 (Opial) Given progression?? {c_n}?H, suppose $\boldsymbol{c}_{n} \rightharpoonup \boldsymbol{c} $. Then
$\liminf \limits_{n \rightarrow \infty}\left\|\boldsymbol{c}_{n}-\boldsymbol{c}\right\|<\liminf \limits_{n \rightarrow \infty}\left\|\boldsymbol{c}_{n}-\boldsymbol{b}\right\|, \forall \boldsymbol{c} \neq \boldsymbol{b}$ (6)
Lemma 1.5^[16]?? Sequences $\left\{\vartheta_{n}\right\}, \quad\left\{\alpha_{n}\right\} $, and {ι_n} in [0, +∞) satisfy
$\iota_{n+1} \leqslant \iota_{n}+\alpha_{n}\left(\iota_{n}-\iota_{n-1}\right)+\vartheta_{n}, \sum\limits_{n=1}^{+\infty} \vartheta_{n}<+\infty$
Suppose that real number α exists and has 0≤α_n≤α < 1, ?n∈N. Thus, the following can be obtained:
1) $\sum\limits_{n=1}^{+\infty}\left(\iota_{n}-\iota_{n-1}\right)^{+}<+\infty $;
2) There is ι^*∈[0, +∞) such that $\lim \limits_{n \rightarrow+\infty} \iota_{n}=\iota^{*} $.
Lemma 1.6^[20]?? Given progressions {a_n}, {b_n}, and {α_n}, where a_n≥0, α_n∈(0, 1), and $\sum\limits_{n=0}^{\infty} \alpha_{n}=\infty $, suppose a_n+1≤ α_nb_n+(1-α_n)a_n for all n∈N. If for each subsequence {a_nk} of {a_n}, $\limsup \limits_{k \rightarrow \infty} b_{n_{k}} \leqslant 0 $ is established when
$\limsup \limits_{n \rightarrow+\infty}\left(a_{n_{k}+1}-a_{n_{k}}\right) \geqslant 0$
then there is $\lim \limits_{n \rightarrow+\infty} a_{n}=0 $.
2 Algorithms and Convergence Analysis Two methods are proposed and studied in this section. To prove the algorithm's convergence, the following assumptions are proposed.
Condition (A):
(A1) f is pseudomonotone over X;
(A2) f(z, ·) has convexity and subdifferentiability over X for any z∈H;
(A3) $\mathop {\lim \sup }\limits_{n \to \infty } f\left( {{\mathit{\boldsymbol{z}}_n}, \mathit{\boldsymbol{e}}} \right) \leqslant f(\mathit{\boldsymbol{z}}, \mathit{\boldsymbol{e}}) $ from e∈X and {z_n}?X with $\boldsymbol{z}_{n} \rightharpoonup \boldsymbol{z} $;
(A4) f has Lipschitz-type condition over H with l₁, l₂ > 0.
Remark 2.1?? It is explicit when Condition (A) is satisfied. The EP(f) of problem (1) is convex and closed^{[7, 10]}. In addition, Fix(T)?H is also convex and closed under the condition that T is quasi-nonexpansive^[21]. Moreover, the symbol Λ=EP(X, f)∩Fix(T) is taken for convenience.
2.1 Weak Convergence First, inspired by the works of Ref.[14], the algorithm's weak convergence is derived. In addition, the algorithm's step size is specially selected, which makes it unnecessary for the algorithm to know Lipschitz constants in advance. The first algorithm has the following form:
Algorithm 2.1
Step 0?? Select x₀, x₁∈X, μ∈(0, 1), λ₁ > 0. Choose a non-negative real sequence {p_n} satisfying $\sum\limits_{n=0}^{\infty} p_{n}<+\infty $.
Step 1?? Assume that x_n-1 and x_n are known. Compute
$\boldsymbol{w}_{n}=\boldsymbol{x}_{n}+\alpha_{n}\left(\boldsymbol{x}_{n}-\boldsymbol{x}_{n-1}\right)$
$\begin{aligned}\boldsymbol{y}_{n}=&\arg \min \left\{\boldsymbol{\lambda}_{n} f\left(\boldsymbol{w}_{n}, \boldsymbol{y}\right)+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{y}\right\|^{2}, \boldsymbol{y} \in\right. \\&\boldsymbol{X}\}=\operatorname{prox}_{\lambda_{n} f\left(\boldsymbol{w}_{n}, \cdot\right)}^{X}\left(\boldsymbol{w}_{n}\right)\end{aligned}$
Step 2?? Select v_n∈?₂f(w_n, y_n) such that
$\boldsymbol{w}_{n}-\lambda_{n} \boldsymbol{v}_{n}-\boldsymbol{y}_{n} \in \boldsymbol{N}_{X}\left(\boldsymbol{y}_{n}\right)$
Compute
$\begin{aligned}\boldsymbol{z}_{n}=& \arg \min \left\{\lambda_{n} f\left(\boldsymbol{y}_{n}, \boldsymbol{z}\right)+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{z}\right\|^{2}, \boldsymbol{z} \in \boldsymbol{T}_{n}\right\}=\\& \operatorname{prox}_{\lambda_{n} f\left(\boldsymbol{y}_{n}, \cdot\right)}^{T_{n}}\left(\boldsymbol{w}_{n}\right)\end{aligned}$
where
$\boldsymbol{T}_{n}=\left\{\boldsymbol{x} \in \boldsymbol{H}:\left\langle\boldsymbol{w}_{n}-\lambda_{n} \boldsymbol{v}_{n}-\boldsymbol{y}_{n}, \boldsymbol{y}_{n}-\boldsymbol{x}\right\rangle \geqslant 0\right\}$
Step 3?? Define x_n+1=(1-β_n)w_n+β_nTz_n and
$\begin{aligned}&\lambda_{n+1}= \\&\left\{\begin{array}{l}\min \left\{\frac{\mu\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)}{2\left(f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right)}, \lambda_{n}+p_{n}\right\}, \\\qquad\quad \text { if } f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)>0 \\\lambda_{n}+p_{n}, \text { otherwise }\end{array}\right.\end{aligned}$
If w_n = y_n = x_n+1, then stop, wn ∈ Λ.
Take n: =n+1 and revert to Step 1.
Remark 2.2?? It is easy to confirm the existence of v_n and X?T_n from Algorithm 2.1. Please refer to Ref.[22] for detailed proof.
Remark 2.3?? The existence of the parameter μ in Algorithm 2.1 is necessary for the subsequent proof that the proposed algorithm is convergent.
Lemma 2.1?? Algorithm 2.1 generates progression {λ_n}, which satisfies the limit λ existence of {λ_n} and $ \min \left\{\frac{\mu}{2 \max \left\{l_{1}, l_{2}\right\}}, \lambda_{0}\right\} \leqslant \lambda \leqslant \lambda_{0}+P$ hold where $ P=\sum\limits_{n=0}^{\infty} p_{n}$.
Proof?? When
$f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)>0$
the Lipschitz condition of f is engendered:
$\begin{aligned}&\frac{\mu\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)}{2\left(f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right)} \geqslant \\&\ \ \ \ \ \ \ \ \frac{\mu\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)}{2\left(l_{1}\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+l_{2}\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)} \geqslant \\&\ \ \ \ \ \ \ \ \frac{\mu}{2 \max \left\{l_{1}, l_{2}\right\}}\end{aligned}$
Using the above inequality and induction, it is obvious that the following expression can be derived:
$\min \left\{\frac{\mu}{2 \max \left\{l_{1}, l_{2}\right\}}, \lambda_{0}\right\} \leqslant \lambda_{n} \leqslant \lambda_{0}+P$
Taking s_n+1=λ_n+1-λ_n, definition of {λ_n} leads to
$\sum\limits_{n=0}^{\infty} s_{n+1}^{+} \leqslant \sum\limits_{n=0}^{\infty} p_{n}<+\infty$ (7)
Therefore, $\sum\limits_{n=0}^{\infty} s_{n+1}^{+} $ converges. Then, the next task is to find evidence that $\sum\limits_{n=0}^{\infty} s_{n+1}^{-} $ has a limit. Suppose $\sum\limits_{n=0}^{\infty} s_{n+1}^{-}=-\infty $. Since $s_{n+1}=s_{n+1}^{+}-s_{n+1}^{-} $, the following equation is obtained:
$\lambda_{k+1}-\lambda_{0}=\sum\limits_{n=0}^{k} s_{n+1}^{+}-\sum\limits_{n=0}^{k} s_{n+1}^{-}$ (8)
In Eq.(8), let k→+∞, there is λ_k→-∞, which is impossible. Because $\sum\limits_{n=0}^{\infty} s_{n+1}^{+} $ and $\sum\limits_{n=0}^{\infty} s_{n+1}^{-} $ are convergent, let k→+∞, then $\lim \limits_{n \rightarrow \infty} \lambda_{n}=\lambda $ is deduced. By boundedness of {λ_n}, $\min \left\{\frac{\mu}{2 \max \left\{l_{1}, l_{2}\right\}}, \lambda_{0}\right\} $≤λ≤λ₀+P is obtained.
Remark 2.4?? Algorithm 2.1 updates progression {λ_n} which is not monotonically decreasing, thus the dependence on the initial step size λ₀ is reduced. Furthermore, it is evident that $\lim \limits_{n\rightarrow \infty} \frac{\lambda_{n}}{\lambda_{n+1}}=1 $ can be obtained by the rule of four arithmetic operations of the convergent sequence.
Now, a lemma is introduced to pave the way for proof of convergence result.
Lemma 2.2?? Suppose progression {z_n} is updated by Algorithm 2.1. For $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\in \mathrm{ }\!\!\boldsymbol{\varLambda}\!\!\text{ } $, it can deduce
$\begin{aligned}&\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \leqslant\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}- \\&\quad\left(1-\frac{\lambda_{n}}{\lambda_{n+1}} \mu\right)\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{y}_{n}-\boldsymbol{z}_{n}\right\|^{2}\right)\end{aligned}$
Proof?? From Lemma 1.2 and$\boldsymbol{z}_{n}=\operatorname{prox}_{\lambda_{n}f\left(\boldsymbol{y}_{n}, \cdot\right)}^{\boldsymbol{T}_{n}}\left(\boldsymbol{w}_{n}\right) $, the following expression is deduced:
$\begin{aligned}\lambda_{n}(&\left.f\left(\boldsymbol{y}_{n}, \boldsymbol{z}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right) \geqslant \\&\left\langle\boldsymbol{w}_{n}-\boldsymbol{z}_{n}, \boldsymbol{z}-\boldsymbol{z}_{n}\right\rangle, \forall z \in \boldsymbol{T}_{n}\end{aligned}$ (9)
Note that Λ?EP(X, f)?X?T_n. Given $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $∈Λ, employing z=$\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $ in Eq.(8) leads to
$\lambda_{n}\left(f\left(\boldsymbol{y}_{n}, \mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right) \geqslant\left\langle\boldsymbol{w}_{n}-\boldsymbol{z}_{n}, \mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}-\boldsymbol{z}_{n}\right\rangle$ (10)
For any n≥0, y_n∈X and $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $∈EP(X, f) can export f(y_n, $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $)≤0. Because f is pseudo-monotone, f(y_n, $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $)≤0 can be deduced. Hence, from inequality (9) and λ_n > 0, it is found that
$-\lambda_{n} f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right) \geqslant\left\langle\boldsymbol{w}_{n}-\boldsymbol{z}_{n}, \mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}-\boldsymbol{z}_{n}\right\rangle$ (11)
Using the concept of the subdifferential, z_n∈T_n?H and v_n∈?₂f(w_n, y_n), the following expression is derived:
$f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right) \geqslant\left\langle\boldsymbol{v}_{n}, \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle$ (12)
Owing to the given form of T_n, it is found that
$\left\langle\boldsymbol{w}_{n}-\lambda_{n} \boldsymbol{v}_{n}-\boldsymbol{y}_{n}, \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle \leqslant 0$
Hence, there is
$\lambda_{n}\left(f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)\right) \geqslant\left\langle\boldsymbol{y}_{n}-\boldsymbol{w}_{n}, \boldsymbol{y}_{n}-\boldsymbol{z}_{n}\right\rangle$ (13)
Applying inequalities (11) and (13) yields
$\begin{array}{c}2 \lambda_{n}\left(f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right) \geqslant 2\left\langle\boldsymbol{y}_{n}-\right. \\\left.\boldsymbol{w}_{n}, \boldsymbol{y}_{n}-\boldsymbol{z}_{n}\right\rangle+2\left\langle\boldsymbol{w}_{n}-\boldsymbol{z}_{n}, \mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}-\boldsymbol{z}_{n}\right\rangle=\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}+ \\\left\|\boldsymbol{y}_{n}-\boldsymbol{w}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}\end{array}$ (14)
Through the representation of λ_n, the following is achieved:
$\begin{array}{c}\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \leqslant\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\left(\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\right. \\\left.\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)+\frac{\lambda_{n}}{\lambda_{n+1}} \mu\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\right. \\\left.\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)=\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-(1- \\\left.\frac{\lambda_{n}}{\lambda_{n+1}} \mu\right)\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)\end{array}$ (15)
Lemma 2.3?? Algorithm 2.1 formulates progressions {x_n}, {w_n}, and {y_n}. Suppose
$\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_{k}}-\boldsymbol{w}_{n_{k}}\right\|=0, \lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{w}_{n_{k}}-\boldsymbol{y}_{n_{k}}\right\|=0 $
$\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{w}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|=0, \lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|=0$
$\lim \limits_{n \rightarrow \infty}\left\|T \boldsymbol{z}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|=0$
If the sequence {x_nk} to x"∈H is weakly convergent, then x"∈Λ.
Proof?? Apparently, there is $ \boldsymbol{w}_{n_{k}} \rightharpoonup \boldsymbol{x}^{\prime \prime}, \boldsymbol{y}_{n_{k}} \rightharpoonup \boldsymbol{x}^{\prime \prime}, \boldsymbol{z}_{n_{k}} \rightharpoonup \boldsymbol{x}^{\prime \prime}$, and x^"∈X. From the relation of Eq.(8), the following expression is deduced:
$\begin{aligned}&\lambda_{n_{k}}\left(f\left(\boldsymbol{y}_{n_{k}}, \boldsymbol{z}\right)-f\left(\boldsymbol{y}_{n_{k}}, \boldsymbol{z}_{n_{k}}\right)\right) \geqslant \\&\ \ \left\langle\boldsymbol{w}_{n_{k}}-\boldsymbol{z}_{n_{k}}, \boldsymbol{z}-\boldsymbol{z}_{n_{k}}\right\rangle, \quad \forall \boldsymbol{z} \in \boldsymbol{T}_{n}\end{aligned}$ (16)
The Lipschitz-type condition of f on X yields
$\begin{gathered}\lambda_{n_{k}} f\left(\boldsymbol{y}_{n_{k}}, \boldsymbol{z}_{n_{k}}\right) \geqslant \lambda_{n_{k}}\left(f\left(\boldsymbol{w}_{n_{k}}, \boldsymbol{z}_{n_{k}}\right)-f\left(\boldsymbol{w}_{n_{k}}, \boldsymbol{y}_{n_{k}}\right)\right)- \\\lambda_{n_{k}} l_{1}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{w}_{n_{k}}\right\|^{2}-\lambda_{n_{k}} l_{2}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|^{2}\end{gathered}$ (17)
According to inequalities (13) and (17), the following expression can be obtained:
$\begin{gathered}\lambda_{n_{k}} f\left(\boldsymbol{y}_{n_{k}}, \boldsymbol{z}_{n_{k}}\right) \geqslant\left\langle\boldsymbol{w}_{n_{k}}-\boldsymbol{y}_{n_{k}}, \boldsymbol{z}_{n_{k}}-\boldsymbol{y}_{n_{k}}\right\rangle- \\\lambda_{n_{k}} l_{1}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{w}_{n_{k}}\right\|^{2}-\lambda_{n_{k}} l_{2}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|^{2}\end{gathered}$ (18)
Combining inequalities (17) and (18), and X?T_n, it can be deduced for all z∈X that
$\begin{aligned}f\left(\boldsymbol{y}_{n_{k}}, \boldsymbol{z}\right) \geqslant & \frac{1}{\lambda_{n_{k}}}\left\langle\boldsymbol{w}_{n_{k}}-\boldsymbol{z}_{n_{k}}, \boldsymbol{z}-\boldsymbol{z}_{n_{k}}\right\rangle+\\& \frac{1}{\lambda_{n_{k}}}\left\langle\boldsymbol{w}_{n_{k}}-\boldsymbol{y}_{n_{k}}, \boldsymbol{z}_{n_{k}}-\boldsymbol{y}_{n_{k}}\right\rangle-\\& l_{1}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{w}_{n_{k}}\right\|^{2}-l_{2}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|^{2}\end{aligned}$
Taking the limit in the last inequality and using
$\begin{gathered}\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{w}_{n_{k}}-\boldsymbol{y}_{n_{k}}\right\|=\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{w}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|= \\\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{y}_{n_{k}}-\boldsymbol{z}_{n_{k}}\right\|=0 \end{gathered}$
$\lim \limits_{n \rightarrow \infty} \lambda_{n}=\lambda>0$
f(x^", z)≥0, ?z∈X is deduced from the assumption (A3). In other words, x^"∈EP(X, f). Moreover, since $\boldsymbol{z}_{n_{k}} \rightharpoonup \boldsymbol{x}^{\prime \prime} $ and demiclosedness of zero of I-T, x^"∈Fix(T) is discovered. Then, x^"∈Λ.
Theorem 2.1?? In progressions {α_n} and {β_n}, 0≤α_n≤α_n+1≤α < $\frac{1}{3} $ and 0 < β≤β_n≤$\frac{1}{2} $. Suppose Condition (A) holds and Λ≠?, then each weak converge limit point of progression {x_n} formed by Algorithm 2.1 belongs to Λ.
Proof?? $ 1-\lambda_{n} \frac{\mu}{\lambda_{n+1}}>0, \forall n \geqslant N$ is derived from the limit $\lim \limits_{n \rightarrow \infty}\left(1-\frac{\lambda_{n}}{\lambda_{n+1}} \mu\right)=1-\mu>0 $. Using Lemma 2.2, it is derived that
$\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\| \leqslant\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|, \forall n \geqslant N$ (19)
By invoking quasi-nonexpansion of T, β_n≤1/2 and x_n+1=(1-β_n)w_n+β_nT z_n, the following expression is obtained:
$\begin{gathered}\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}=\left(1-\beta_{n}\right)\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}+\beta_{n}\left\|T \boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}- \\\beta_{n}\left(1-\beta_{n}\right)\left\|T \boldsymbol{z}_{n}-\boldsymbol{w}_{n}\right\|^{2} \leqslant\left(1-\beta_{n}\right)\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}+ \\\beta_{n}\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}-\frac{1-\beta_{n}}{\beta_{n}}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{w}_{n}\right\|^{2}= \\\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} \right\|^{2}-\left\|\boldsymbol{x}_{n+1}-\boldsymbol{w}_{n}\right\|^{2}\end{gathered}$ (20)
Furthermore, by utilizing the definition of w_n and inequality (2), it is derived
$\begin{aligned}&\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}=\left(1+\alpha_{n}\right)\left\|\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}_{n}-\boldsymbol{x}\right\|^{2}- \\&\ \ \alpha_{n}\left\|\boldsymbol{x}_{n-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{{x}}}\right\|^{2}+\alpha_{n}\left(1+\alpha_{n}\right)\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}\end{aligned}$ (21)
and
$\begin{gathered}\left\|\boldsymbol{x}_{n+1}-\boldsymbol{w}_{n}\right\|^{2}=\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}+\alpha_{n}^{2}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}- \\2 \alpha_{n}\left\langle\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}, \boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\rangle \geqslant\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}+ \\\alpha_{n}^{2}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}-\alpha_{n}\left(\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|+\right. \\\left.\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|\right)=\left(1-\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}- \\\left(\alpha_{n}-\alpha_{n}^{2}\right)\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}\end{gathered}$ (22)
Applying inequalities (20), (22), and Eq. (21), the following expression can be deduced:
$\begin{gathered}\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \leqslant\left(1+\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}- \\\alpha_{n}\left\|\boldsymbol{x}_{n-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+2 \alpha_{n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}- \\\left(1-\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}\end{gathered}$ (23)
The non-decreasing property of {α_n} leads to
$\begin{gathered}\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\alpha_{n+1}\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+ \\2 \alpha_{n+1}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2} \leqslant\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}- \\\alpha_{n}\left\|\boldsymbol{x}_{n-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+\left(2 \alpha_{n+1}-1+\right. \\\left.\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}+2 \alpha_{n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}\end{gathered}$ (24)
Let
$\begin{aligned}&\iota_{n}=\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\alpha_{n}\left\|\boldsymbol{x}_{n-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+ \\&\ \ \ \ 2 \alpha_{n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}\end{aligned}$
then (23) is equivalent to
$\iota_{n+1}-\iota_{n} \leqslant\left(2 \alpha_{n+1}-1+\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}$ (25)
Since $ 0 \leqslant \alpha_{n} \leqslant \alpha<\frac{1}{3}$, it holds that
$2 \alpha_{n+1}-1+\alpha_{n} \leqslant 3 \alpha-1<0$
In inequality (25), $-\vartheta=3 \alpha-1 $ leads to
$0 \geqslant-\vartheta\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2} \geqslant \iota_{n+1}-\iota_{n}$ (26)
In other words, progression {ι_n} does not increase. Moreover, utilizing the form of {ι_n}, the following expression is deduced:
$\iota_{j} \geqslant\left\|\boldsymbol{x}_{j}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\alpha_{j}\left\|\boldsymbol{x}_{j-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}$ (27)
Also, considering the form of ι_n+1, the following formula is obtained:
$\alpha_{j+1}\left\|\boldsymbol{x}_{j}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \geqslant-\iota_{j+1}$ (28)
Hence, inequalities (27) and (28) indicate that
$\begin{aligned}-\iota_{j+1} & \leqslant \alpha\left(\iota_{j}+\alpha\left\|\boldsymbol{x}_{j-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}\right) \leqslant \cdots \leqslant \\& \alpha\left(\alpha^{j-N}\left\|\boldsymbol{x}_{N}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+\iota_{N}\left(1+\alpha+\cdots+\alpha^{j-N-1}\right)\right) \leqslant \\& \alpha^{j-N+1}\left\|\boldsymbol{x}_{N}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+\frac{\alpha \iota_{N}}{1-\alpha}\end{aligned}$ (29)
By making use of inequalities (26) and (29), there is
$\begin{gathered}-\vartheta \sum\limits_{n=N}^{j}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2} \leqslant \iota_{N}-\iota_{j+1} \leqslant \iota_{N}+ \\\alpha^{j-N+1}\left\|\boldsymbol{x}_{N}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+\frac{\alpha \iota_{N}}{1-\alpha}\end{gathered}$ (30)
Set k→∞ from inequality (30), the following expression is derived:
$\sum\limits_{n=N}^{\infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|^{2}<+\infty$ (31)
which indicates
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|=0$ (32)
By using α_n≤α, there is
$\begin{array}{c}\left\|\boldsymbol{w}_{n}-\boldsymbol{x}_{n+1}\right\| \leqslant\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|+\alpha_{n}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\| \leqslant \\\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|+\alpha\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|\end{array}$ (33)
Therefore, by Eq.(32) and inequality (33), the following expression is obtained:
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{w}_{n}-\boldsymbol{x}_{n+1}\right\|=0$ (34)
Combining Eqs.(32) and (34) yields
$\begin{gathered}\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{w}_{n}\right\| \leqslant \lim \limits_{n \rightarrow \infty}\left(\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n+1}\right\|+\right. \\\left.\left\|\boldsymbol{w}_{n}-\boldsymbol{x}_{n+1}\right\|\right)=0\end{gathered}$
So
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{w}_{n}\right\|=0$ (35)
From expression (22), for n≥N, the following expression is obtained:
$\begin{aligned}&\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \leqslant\left(1+\alpha_{n}\right)\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}- \\&\ \ \ \ \ \ \ \ \alpha_{n}\left\|\boldsymbol{x}_{n-1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+2 \alpha\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|^{2}\end{aligned}$ (36)
By inequality(31) and (36), and invoking Lemma 1.5, there is
$\left\|\boldsymbol{x}_{n}-\boldsymbol{x}^{\prime}\right\|^{2} \rightarrow \sigma$ (37)
Eq.(35) leads to
$\left\|\boldsymbol{w}_{n}-\boldsymbol{x}^{\prime}\right\|^{2} \rightarrow {\sigma}$ (38)
Because of the relationship provided in inequality (20), the following is obtained:
$\begin{gathered}\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{{x}}}\right\|^{2} \leqslant\left(1-\beta_{n}\right)\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+ \\\beta_{n}\left\|z_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}\end{gathered}$
which means
$\begin{gathered}\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \geqslant\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}+\\\frac{\left\|\boldsymbol{x}_{n+1}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}}{\beta_{n}}\end{gathered}$ (39)
Applying expressions (37), (38), inequality (39), and $0<\beta \leqslant \beta_{n} \leqslant \frac{1}{2} $, there is
$\liminf \limits_{n \rightarrow \infty}\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \geqslant \lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}=\sigma$
Exploiting Lemma 2.2, the following expression is derived:
$\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\| \leqslant\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\| \rightarrow \sqrt{\sigma}$
Therefore,
$\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2} \rightarrow \sigma$ (40)
Lemma 2.2 is invoked to obtain
$\begin{gathered}\left(1-\frac{\lambda_{n}}{\lambda_{n+1}} \mu\right)\left(\left\|\boldsymbol{y}_{n}-\boldsymbol{w}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right) \leqslant \\\left\|\boldsymbol{w}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}-\left\|\boldsymbol{z}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}, \forall n \geqslant N\end{gathered}$
The last expression implies that
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{y}_{n}-\boldsymbol{w}_{n}\right\|=\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{y}_{n}-\boldsymbol{z}_{n}\right\|=0$
So it is found that
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{w}_{n}-\boldsymbol{z}_{n}\right\|=0$
By modality of x_n+1, inequality (29), and $0<\beta \leqslant \beta_{n} \leqslant \frac{1}{2} $, $\lim \limits_{n \rightarrow \infty}\left\|T \boldsymbol{z}_{n}-\boldsymbol{w}_{n}\right\|=0 $ is obtained. Since $ \lim \limits_{n \rightarrow \infty}\left\|T \boldsymbol{z}_{n}-\boldsymbol{z}_{n}\right\| \leqslant \lim \limits_{n \rightarrow \infty}\left(\left\|T \boldsymbol{z}_{n}-\boldsymbol{w}_{n}\right\|+\left\|\boldsymbol{w}_{n}-\boldsymbol{z}_{n}\right\|\right)=0$ there is
$\lim \limits_{n \rightarrow \infty}\left\|T \boldsymbol{z}_{n}-\boldsymbol{z}_{n}\right\|=0$
Overall, the conclusion is drawn that progressions {x_n}, {w_n}, {y_n}, and {z_n} are bounded and $ \lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}}\right\|^{2}$ exists. Boundedness of progression {x_n} means that succession {x_nk}?{x_n} weakly converges to some x^"∈H. Utilizing Lemma 2.3, it leads to x^"∈Λ.
The next task is to verify that $\boldsymbol{x}_{n} \rightharpoonup \boldsymbol{x}^{\prime \prime} $. Suppose ?x^", $\hat{\boldsymbol{x}} $∈Λ with x^"≠$\hat{\boldsymbol{x}} $. Given that progression {x_ni} satisfies the weak convergence of x_ni to $\hat{\boldsymbol{x}} $(i→∞), it is worth noting that the following expression holds:
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}^{\prime}\right\|=\sqrt{\sigma} \in R, \quad \forall \boldsymbol{x}^{\prime} \in \boldsymbol{\varLambda}$ (41)
Applying Lemma 1.4, it is deduced that
$\begin{aligned}&\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\hat{\boldsymbol{x}}\right\|=\lim \limits_{i \rightarrow \infty}\left\|\boldsymbol{x}_{n_{i}}-\hat{\boldsymbol{x}}\right\|= \\&\ \ \ \ \liminf \limits_{i \rightarrow \infty}\left\|\boldsymbol{x}_{n_{i}}-\hat{x}\right\|<\liminf \limits_{i \rightarrow \infty} \left\|\boldsymbol{x}_{n_i}-\boldsymbol{x}^{\prime \prime}\right\|= \\&\ \ \ \ \lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}^{\prime \prime}\right\|=\lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_{k}}-\boldsymbol{x}^{\prime \prime}\right\|= \\&\ \ \ \ \liminf \limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_{k}}-\boldsymbol{x}^{\prime \prime}\right\|<\liminf \limits_{i \rightarrow \infty} \left\|\boldsymbol{x}_{n_k}-\hat{\boldsymbol{x}}\right\|= \\&\ \ \ \ \lim \limits_{k \rightarrow \infty}\left\|\boldsymbol{x}_{n_{k}}-\hat{\boldsymbol{x}}\right\|=\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\hat{\boldsymbol{x}}\right\|\end{aligned}$ (42)
There is a contradiction in Eq.(42). As a result, $ \boldsymbol{x_{n}} \rightharpoonup \boldsymbol{x}^{\prime \prime}$. Given
$\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{y}_{n}\right\|=\lim \limits_{n \rightarrow \infty}\left\|\boldsymbol{x}_{n}-\boldsymbol{z}_{n}\right\|=0$
there is $\boldsymbol{y}_{n} \rightharpoonup \boldsymbol{x}^{\prime \prime} $ and $\boldsymbol{x}_{n} \rightharpoonup \boldsymbol{x}^{\prime \prime} $.
2.2 Strong Convergence In this framework, inspired by Refs.[10], [12] and [23], Algorithm 2.1 was improved by using the modified version of Halpern iteration. To study the algorithhm's strong convergence, some assumptions are added.
Condition (B):
(B1) β_n∈(0, 1), $ \lim \limits_{n \rightarrow \infty} \beta_{n}=0, \text { and } \sum\limits_{n=0}^{\infty} \beta_{n}=\infty$;
(B2) γ_n∈[a, b]?(0, 1);
(B3) ε_n=o(β_n), i.e., $ \lim \limits_{n \rightarrow \infty}\left(\varepsilon_{n} / \beta_{n}\right)=0$.
Next, the form of algorithm is described in detail.
Algorithm 2.2
Step 0?? Let x₀, x₁∈X, μ∈(0, 1), λ₁ > 0. Select a non-negative real sequence {p_n} satisfying $\sum\limits_{n=0}^{\infty} p_{n}<+\infty $.
Step 1?? With x_n-1 and x_n known, α_n with 0≤α_n≤α^′_n is selected and
$\alpha^{\prime}{ }_{n}=\left\{\begin{array}{l}\min \left\{\frac{\varepsilon_{n}}{\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|}, \alpha\right\}, \text { if } \boldsymbol{x}_{n-1}-\boldsymbol{x}_{n} \neq 0 \\\alpha, \quad \text { else }\end{array}\right.$
Step 2?? Take w_n=x_n-α_n(x_n-1-x_n) and calculate
$\begin{aligned}\boldsymbol{y}_{n}=& \arg \min \left\{\boldsymbol{\lambda}_{n} f\left(\boldsymbol{w}_{n}, \boldsymbol{y}\right)+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{y}\right\|^{2}, \boldsymbol{y} \in \boldsymbol{X}\right\}=\\& \operatorname{prox}_{\lambda_{n} f\left(\boldsymbol{w}_{n}, \cdot\right)}^{X}\left(\boldsymbol{w}_{n}\right)\end{aligned}$
Step 3?? Pick v_n∈?₂ f(w_n, y_n) such that
$\boldsymbol{w}_{n}-\lambda_{n} \boldsymbol{v}_{n}-\boldsymbol{y}_{n} \in \boldsymbol{N}_{X}\left(\boldsymbol{y}_{n}\right)$
compute
$\begin{aligned}z_{n}=& \arg \min \left\{\lambda_{n} f\left(\boldsymbol{y}_{n}, \boldsymbol{z}\right)+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{z}\right\|^{2}, \boldsymbol{z} \in \boldsymbol{T}_{n}\right\}=\\& \operatorname{prox}_{\lambda_{n} f\left(\boldsymbol{y}_{n}, \cdot\right)}^{T_{n}}\left(\boldsymbol{w}_{n}\right)\end{aligned}$
where
$\boldsymbol{T}_{n}=\left\{\boldsymbol{x} \in \boldsymbol{H} \mid\left\langle\boldsymbol{w}_{n}-\lambda_{n} \boldsymbol{v}_{n}-\boldsymbol{y}_{n}, \boldsymbol{y}_{n}-\boldsymbol{x}\right\rangle \geqslant 0\right\}$
Step 4?? Formulate
$\boldsymbol{x}_{n+1}=\gamma_{n} T\left(\beta_{n} \boldsymbol{x}_{0}+\left(1-\beta_{n}\right) \boldsymbol{z}_{n}\right)+\left(1-\gamma_{n}\right) \boldsymbol{x}_{n}$
and
$\begin{aligned}&\lambda_{n+1}= \\&\left\{\begin{array}{l}\min \left\{\frac{\mu\left(\left\|\boldsymbol{w}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{y}_{n}-\boldsymbol{z}_{n}\right\|^{2}\right)}{2\left(f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)\right)}, \lambda_{n}+p_{n}\right\}, \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text { if } f\left(\boldsymbol{w}_{n}, \boldsymbol{z}_{n}\right)-f\left(\boldsymbol{w}_{n}, \boldsymbol{y}_{n}\right)-f\left(\boldsymbol{y}_{n}, \boldsymbol{z}_{n}\right)>0 \\\lambda_{n}+p_{n}, \quad \text { otherwise }\end{array}\right.\end{aligned}$
Take n: =n+1 and transfer to Step 1.
Remark 2.2?? Apparently, it holds that
$\lim \limits_{n \rightarrow \infty} \frac{\alpha_{n}}{\beta_{n}}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|=0$
Actually, the following expression can be easily obtained:
$\alpha_{n}\left\|\boldsymbol{x}_{n}-\boldsymbol{x}_{n-1}\right\| \leqslant \varepsilon_{n}$
From the above formula and hypothesis (B3), it is directly deduced that
$\frac{\alpha_{n}}{\beta_{n}}\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\| \leqslant \frac{\varepsilon_{n}}{\beta_{n}} \rightarrow 0$
Theorem 2.2?? Let conditions (A) and (B) hold and Λ≠?. Algorithm 2.2 forms progression {x_n}. Then {x_n} to $\mathrm{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\boldsymbol{x}}} $∈P_Λ(x₀) is strongly convergent.
Proof?? Please refer to Refs.[12] and [22] for the detailed proof of the theorem.
3 Research on Variational Inequalities This portion elaborates applications of the outcomes to fixed point problem and variational inequality. Define f(a, b)=〈F(a), b-a〉 for each b, a∈X, where F: X→X is a nonlinear operator. This special case of EP is simplified to variational inequality problem (VIP) which is elaborated as seeking out $\breve{\boldsymbol{x}}$∈X such that
$\langle F(\breve{\boldsymbol{x}}), \boldsymbol{z}-\breve{\boldsymbol{x}}\rangle \geqslant 0, \forall \boldsymbol{z} \in \boldsymbol{X}$ (43)
where VI(X, F) represents the disaggregation of problem (43). Here f is called
● pseudo-monotone over X:
$\langle \boldsymbol{F}(\boldsymbol{d}), \boldsymbol{e}-\boldsymbol{d}\rangle \geqslant 0\Rightarrow \langle \boldsymbol{F}(\boldsymbol{e}), \boldsymbol{e}-\boldsymbol{d}\rangle \geqslant 0, \quad \forall \boldsymbol{d}, \boldsymbol{e}\in \boldsymbol{X}$
● L-Lipschitz continuous over X:
$\begin{align} & \exists L>0, \text{ s}\text{.t}\text{. }\left\| F(\boldsymbol{d})-F(\boldsymbol{e}) \right\|\leqslant L\left\| \boldsymbol{d}-\boldsymbol{e} \right\|, \\ & \forall \boldsymbol{d}, \boldsymbol{e}\in \boldsymbol{X} \\ \end{align}$
Make the following presumptions about VIP:
(B1) f is pseudo-monotone over X;
(B2) f is weakly sequentially continuous over X for any progression {z_n}: {z_n} weakly converging to z, which means {F(z_n)} weakly converges to F(z).
(B3) f possess L-Lipschitz successive over X.
Obviously, f(y, z)=〈F(y), z-y〉 makes conditions (A1)-(A3) hold. Condition (A4) holds in situations of Lipschitz-type constant $l_{1}=l_{2}=\frac{L}{2} $ on f. According to given forms of y_n and f, the following is derived:
$\begin{aligned}&\boldsymbol{y}_{n}=\arg \min \left\{\lambda_{n} f\left(\boldsymbol{w}_{n}, \boldsymbol{y}\right)+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{y}\right\|^{2}, \boldsymbol{y} \in \boldsymbol{X}\right\}= \\&\quad \arg \min \left\{\lambda_{n}\left\langle F\left(\boldsymbol{w}_{n}\right), \boldsymbol{y}-\boldsymbol{w}_{n}\right\rangle+\frac{1}{2}\left\|\boldsymbol{w}_{n}-\boldsymbol{y}\right\|^{2}, \boldsymbol{y} \in \boldsymbol{X}\right\}= \\&\quad \arg \min \left\{\frac{1}{2}\left\|\boldsymbol{y}-\left(\boldsymbol{w}_{n}-\lambda_{n} F\left(\boldsymbol{w}_{n}\right)\right)\right\|^{2}, \boldsymbol{y} \in \boldsymbol{X}\right\}- \\&\quad \frac{\lambda_{n}^{2}}{2}\left\|F\left(\boldsymbol{w}_{n}\right)\right\|^{2}=P_{X}\left(\boldsymbol{w}_{n}-\lambda_{n} F\left(\boldsymbol{w}_{n}\right)\right)\end{aligned}$
Similarly, z_n in the proposed algorithms reduces to
$\boldsymbol{z}_{n}=\boldsymbol{P}_{T_{n}}\left(\boldsymbol{w}_{n}-\lambda_{n} F\left(\boldsymbol{y}_{n}\right)\right)$
Firstly, the following conclusions are obtained by observing the proof about Algorithm 2 for Ref. [23].
Theorem 3.1?? Progressions {α_n} and {β_n} satisfying 0≤α_n≤α_n+1≤α < $\frac{1}{3} $ and 0 < β≤β_n≤$\frac{1}{2} $ is known. Suppose that Condition (A) holds and Fix(T)∩VI(X, F)≠?. Take μ∈(0, 1) and λ₁ > 0. Choose a non-negative real sequence {p_n} such that $\sum\limits_{n=0}^{\infty} p_{n}<+\infty $. Select arbitrary points x₀= x₁∈X and clarify a progression {x_n}?H:
$\boldsymbol{d}_{n}=\boldsymbol{x}_{n}-\alpha_{n}\left(\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right)$
$\boldsymbol{y}_{n}=P_{X}\left(\boldsymbol{d}_{n}-\lambda_{n} F\left(\boldsymbol{d}_{n}\right)\right)$
$\boldsymbol{z}_{n}=P_{\boldsymbol{T}_{n}}\left(\boldsymbol{d}_{n}-\lambda_{n} F\left(\boldsymbol{y}_{n}\right)\right)$
$\boldsymbol{T}_{n}=\left\{\boldsymbol{x} \in \boldsymbol{H} \mid\left\langle\lambda_{n} F\left(\boldsymbol{d}_{n}\right)+\boldsymbol{y}_{n}-\boldsymbol{d}_{n}, \boldsymbol{y}_{n}-\boldsymbol{x}\right\rangle \leqslant 0\right\}$
$\boldsymbol{x}_{n+1}=\left(1-\beta_{n}\right) \boldsymbol{d}_{n}+\beta_{n} T \boldsymbol{z}_{n}$
$\lambda_{n+1}=\left\{\begin{array}{l}\min \left\{\frac{\mu\left(\left\|\boldsymbol{d}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)}{2\left\langle F\left(\boldsymbol{d}_{n}\right)-F\left(\boldsymbol{y}_{n}\right), \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle}, \lambda_{n}+p_{n}\right\}, \\\ \ \ \ \ \ \ \ \ \ \text { if }\ \ \left\langle F\left(\boldsymbol{d}_{n}\right)-F\left(\boldsymbol{y}_{n}\right), \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle>0 \\\lambda_{n}+p_{n}, \text { otherwise }\end{array}\right.$
Then progression {x_n} weakly converges to a certain spot of Fix(T)∩VI(X, F).
Then, a strong convergence schemeis presented. Its proof is similar to that in Ref.[24] and is thus omitted.
Theorem 3.2?? Let conditions (A), (B) hold and Fix(T)∩VI(X, F)≠?. Let μ∈(0, 1) and λ₁ > 0. Choose a non-negative real sequence {p_n} such that $\sum\limits_{n=0}^{\infty} p_{n}<+\infty $. Take arbitrary points x₀=x₁∈X. A progression {x_n}?H is formed as follows:
Choose α_n with 0≤α_n≤α_n, where
$\bar{\alpha}_{n}=\left\{\begin{array}{l}\min \left\{\frac{\varepsilon_{n}}{\left\|\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right\|}, \alpha\right\}, \text { if } \boldsymbol{x}_{n-1} \neq \boldsymbol{x}_{n} \\\alpha, \text { otherwise }\end{array}\right.$
$\boldsymbol{d}_{n}=\boldsymbol{x}_{n}-\alpha_{n}\left(\boldsymbol{x}_{n-1}-\boldsymbol{x}_{n}\right)$
$\boldsymbol{y}_{n}=P_{X}\left(\boldsymbol{d}_{n}-\lambda_{n} F\left(\boldsymbol{w}_{n}\right)\right)$
$\boldsymbol{z}_{n}=P_{T_{n}}\left(\boldsymbol{d}_{n}-\lambda_{n} F\left(\boldsymbol{y}_{n}\right)\right)$
$\boldsymbol{T}_{n}=\left\{\boldsymbol{x} \in H \mid\left\langle\lambda_{n} F\left(\boldsymbol{d}_{n}\right)+\boldsymbol{y}_{n}-\boldsymbol{d}_{n}, \boldsymbol{y}_{n}-\boldsymbol{x}\right\rangle \leqslant 0\right\}$
$\boldsymbol{x}_{n+1}=\left(1-\gamma_{n}\right) \boldsymbol{z}_{n}+\gamma_{n} T\left(\left(1-\beta_{n}\right) \boldsymbol{z}_{n}+\beta_{n} \boldsymbol{x}_{0}\right)$
$\begin{aligned}&\lambda_{n+1}= \\&\left\{\begin{array}{l}\min \left\{\frac{\mu\left(\left\|\boldsymbol{d}_{n}-\boldsymbol{y}_{n}\right\|^{2}+\left\|\boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\|^{2}\right)}{2\left\langle F\left(\boldsymbol{d}_{n}\right)-F\left(\boldsymbol{y}_{n}\right), \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle}, \lambda_{n}+p_{n}\right\}, \\\ \ \ \ \ \ \ \ \ \ \ \ \text { if }\ \ \left\langle F\left(\boldsymbol{d}_{n}\right)-F\left(\boldsymbol{y}_{n}\right), \boldsymbol{z}_{n}-\boldsymbol{y}_{n}\right\rangle>0 \\\lambda_{n}+p_{n}, \quad \text { otherwise }\end{array}\right.\end{aligned}$
Then progression {x_n} strongly converges to a dot x^′, where x^′∈P_{Fix(T)∩VI(X, F)}(x₀).
Remark 3.1?? For convenience, the algorithms proposed by Theorem 3.1 and Theorem 3.2 are denoted as Algorithm 3.1 and Algorithm 3.2, respectively. They are direct applications of Algorithm 2.1 and Algorithm 2.2 to variational inequalities. Moreover, it is worth noting that the formula of calculating shadow onto a closed convex set T_n^[25] is stated as
${P_{{\mathit{\boldsymbol{T}}_n}}}(\mathit{\boldsymbol{d}}) = \mathit{\boldsymbol{d}} - \max \left\{ {\frac{{\langle \mathit{\boldsymbol{e}}, \mathit{\boldsymbol{d}} - \mathit{\boldsymbol{x}}\rangle }}{{{{\left\| \mathit{\boldsymbol{e}} \right\|}^2}}}, 0} \right\}\mathit{\boldsymbol{e}}$
where e=w_n-λ_nF(w_n)-y_n and x=y_n.
4 Numerical Experiments The proposed algorithms are compared with other algorithms in numerical experiments, and their effectiveness is illustrated in this section. For Algorithm 2.1 and Algorithm 3.1, take μ=0.6, λ₁=0.12, α_n=0.32, β_n=0.5, and p_n=1/(n+1)¹⁰. For all the tests, record the number of iterations (Iter.) and calculation time (Time) implemented in the passing seconds. Particularly, ‖x_n-x_n+1‖≤ε and ε=10^-3 are adopted as terminate principle.
Problem Ⅰ?? The first experiment is focused on equilibrium problem. f: H×H→R is known and make f(y, z)=〈z-y, Ky+Mz+d〉 hold. Each item of d∈R^m is casually created from [-5, 5]. It holds that M and K-M are symmetric positive semi-definite matrices of m×m. The practicable set is
$\boldsymbol{X}=\left\{\boldsymbol{x} \in {\bf{R}}^{m}:-2 \leqslant x_{i} \leqslant 2, i=1, \cdots, m\right\}$
For Algorithm 2.1, take Tx=x/2. For Algorithm 3.1 in Ref.[14], λ=1/2l₁, $\alpha_{n}=0.99(\sqrt{5}-2) $, and β_n=0.5 are used. In Algorithm 1 of Ref.[26], λ=1/2l₁ is chosen. x₀=x₁=(1, …, 1) is taken for any algorithms. As can be seen from Table 1, three sets of stochastic values are produced on account of distinct options of K, d and M.
表 1
Table 1 Numerical results of problem Ⅰ with starting dot (1, 1, ..., 1) m Algorithm 2.1 Algorithm 3.1^[14] Algorithm 1^[26]
Iter. Time Iter. Time Iter. Time
10 11 0.094 36 0.546 22 0.156
13 0.156 46 0.562 30 0.187
50 17 0.515 111 1.888 77 1.466
17 0.484 75 1.716 48 0.764
500 22 57.908 123 94.303 86 30.997
22 56.472 137 123.568 94 36.348

Table 1 Numerical results of problem Ⅰ with starting dot (1, 1, ..., 1)


Problem Ⅱ?? Consider the HpHard problem, where the feasible set is X=R_n⁺. Let G(x)=Ex+d possess d=0 and E=BB^T+J+K, where B, J, K∈R^n×n. Each item of matrix B and skew-symmetric matrix J is stochastically selected in [-2, 2], and each diagonal term of diagonal K is consistently given by (0, 2). For all tests, x₀=x₁=(1, …, 1) is taken. For Algorithm 2.1, Tx=-x/2 is used. In Algorithm 4.3 in Ref.[27], γ=1.99, λ=0.9/‖M‖, α_n=1/(13t+2) are chosen. For Algorithm 3.1 in Ref.[28], μ=0.5, λ₁=0.7, and α_n=0.15 are taken. The results described in Table 2 reflect the effect of the proposed algorithms.
表 2
Table 2 Numerical results of problem Ⅱ with starting dot (1, 1, ..., 1) m Algorithm 3.1 Algorithm 4.3^[27] Algorithm 3.1^[28]
Iter. Time Iter. Time Iter. Time
10 9 0 55 0.016 250 0
9 0 29 0 180 0.016
50 15 0.078 92 0 1034 0.187
14 0 86 0.062 874 0.156
500 34 0.062 149 0.187 2891 0.936
34 0.078 140 0.218 2690 1.186

Table 2 Numerical results of problem Ⅱ with starting dot (1, 1, ..., 1)


In conclusion, compared with Algorithm 3.1 of Ref.[27] and Algorithm 4.3 of Ref.[28], this algorithm has better performance.
5 Conclusions With regard to the fixed point problem and equilibrium problem, two new algorithms were proposed. Under appropriate circumstances, the convergence of algorithms was discussed. In particular, the variational inequalities were also studied. The performance of the proposed algorithms was demonstrated by observing the numerical results, which shows that the algorithms are effective.

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