0 Introduction

Now we take it as a rule that H is considered as a real Hilbert space. In addition, H has a non-null, closed convex subset named C. In accordance with the definition of variational inequality problem, or VIP is elaborated as seeking i^*∈C such that

$ \left\langle F i^*, v-i^*\right\rangle \geqslant 0, \forall v \in C $

(1)

where 〈·, ·〉 represents the transvection and F is an operator defined as F: H→H. The solution set of VIP is written as Sol(C, F).The related theory of variational inequality has appeared in diverse models of economics, physics, engineering, mathematical programming and other problems^[1-4]. Because variational inequality problems as a crucial tool to solve these problems, many numerical methods have been proposed to solve it and some related optimization problems^[5-7].

In recent years, we know that the projection method has been widely popular, and some authors have proposed many variant forms of this method, and the proof of convergence was given^[8-15].

In order to avoid the strong limitation that F is monotonous, Korpelevich^[16] presented the extragradient algorithm (EGM)

$ \begin{gathered} v^k=P_C\left(u^k-\rlap{-} \lambda F u^k\right) \\ u^{k+1}=P_C\left(u^k-\rlap{-}\lambda F v^k\right) \end{gathered} $

where $ \rlap{-} \lambda $∈(0, 1/L). It is well known that the (EGM) has been noticed and some achievements have been made in real Hilbert spaces recently. In fact, in this approach, two predictions for the feasibility set C need to be calculated for each iteration. This, of course, has serious implications for efficiency. To overcome this disadvantage and avoid this inefficient situation, Thong and Hieu^[17] proposed a self adaptive strategy called the Tseng's extragradient algorithm. What's more, inertial-type algorithm has also been further studied^[18-20] and proved to be a very efficient two-step iterative strategy. The main idea is to derive the next iteration from the previous two iterations. Since the algorithm can promote the convergence rate of sequences, many scholars have studied it. For example, Fan and Qin^[20] introduced a new iterative format that combined Armijo-like step size to deal with (VIP) problems^[20]:

$ \left\{\begin{array}{l} h^k=u^k+\sigma^k\left(u^k-u^{k-1}\right) \\ v^k=P_C\left(h^k-\rlap{-} \lambda_k F h^k\right) \\ r^k=v^k-\rlap{-} \lambda^k\left(F v^k-F h^k\right) \\ u^{k+1}=d^k Y\left(u^k\right)+g^k r^k+m^k \chi_k \end{array}\right. $

where Y refers to a contractive mapping and F is a monotone operator.

In addition, Duong Viet Thong employs the special self adaptive inertial subgradient extragradient algorithms^[21]. The step size here is monotonic and the iterative form is

$ \begin{gathered} \rlap{-} \lambda_{k+1}=\left\{\begin{array}{l} \min \left\{\frac{\sigma\left\|h^k-v^k\right\|}{\left\|F h^k-F v^k\right\|}, \rlap{-} \lambda_k\right\}, {\text { if }}\; F h^k-F v^k \neq 0 \\ \rlap{-} \lambda_k, {\text { else }} \end{array}\right. \\ u^{k+1}=d^k Y\left(r^k\right)+\left(1-d^k\right) r^k \end{gathered} $

Now, the research results of inertial type algorithms are becoming more and more mature, and more algorithms have been developed.

Taking inspiration from the above research results^{[14, 15, 21-22]}, this paper proposes two different methods to solve the Lipschitz pseudomonotone VIP. Under the previous assumption, we obtain weak and strong convergence theorems for these two algorithms, respectively. Relative to algorithms in Refs.[13-14], it can be seen that the two new algorithms do not make use of linear search, which means that no additional projections need to be calculated. In addition, we designed three different comparative experiments, and the results show that for our two new algorithms, the number of iterations and runtime are lower.

The organizational structure of this article is arranged as follows: Some of the relevant definitions and preliminary lemma will be mentioned in Section 1. The two new algorithms are elaborated and weak-strong convergence is proved in Section 2. Then some related numerical problems and comparisons are designed in Section 3. Finally, we have carried on a brief but comprehensive summary in Section 4.

1 Preliminaries

Definition 1^{[20, 23]}   First, we illustrate some definitions and lemmas that will be used later. The F: H→H is defined as

(1) Monotone:

$ \langle F(\gamma)-F(\iota), \gamma-\iota\rangle \geqslant 0, \forall \gamma, \iota \in H $

(2) Pseudomonotone:

$ \begin{gathered} \langle F(\iota), \gamma-\iota\rangle \geqslant 0 \Rightarrow\langle F(\gamma), \gamma-\iota\rangle \geqslant 0, \\ \forall \gamma, \iota \in H \end{gathered} $

(3) L-Lipschitz continuous:

$ \begin{aligned} \exists L> & 0 {\text {, s.t. }}\|F(\gamma)-F(\iota)\| \leqslant L\|\gamma-\iota\|, \\ & \forall \gamma, \iota \in H \end{aligned} $

(4) Sequentially weakly continuous:

Suppose for all {u^k}, we get it weakly converges to the same u, then call Fu^k converges weakly to Fu.

Lemma 1^[23]   Let C be a nonempty closed convex subset of a real Hilbert space H. Given u∈H and ι∈C. We have

(1) ι=P_Cu⇔〈u－ι, ι－a〉≥0, ∀a∈C.

(2)〈u－P_C(u), a－P_C(u)〉≤0, ∀a∈C.

Lemma 2^[24]   Let {ξ^k}, {χ^k} and {${\ell ^k}$} be sequences in [0, +∞) such that

$ \begin{aligned} & \xi^{k+1} \leqslant \xi^k+\ell^k\left(\xi^k-\xi^{k-1}\right)+\chi^k, \\ & \forall k \geqslant 1, \sum\limits_{k=1}^{+\infty} \chi^k<+\infty \\ & \end{aligned} $

and there has a real number $ \ell $ with 0≤${\ell ^k}$≤ $ \ell $ < 1 for all k ∈N.

(1) Define [p]₊∶ =max{p, 0}, then we have $ \sum\limits_{k=1}^{+\infty}\left[\xi^k-\xi^{k-1}\right]_{+}<+\infty $;

(2) There must exist ξ^*∈[0, +∞) such that $ \lim\limits _{k \rightarrow+\infty} \xi^k=\xi^* $.

Lemma 3^[25]   If C⊆H is nonempty and {u^k}∈H satisfies the following conditions:

1) ∀u∈C, $ \lim\limits _{k \rightarrow \infty}\left\|u^k-u\right\| $ exists; 2) Every sequential weak cluster point of {u^k} is in C.

Then {u^k} converges weakly to some point in C.

Lemma 4^[26]   If {a^k} is a sequence of nonnegative numbers and fulfilling:

$ a^{k+1} \leqslant b^k a^k+c^k, \forall k \in N $

The elements in {b^k}, {c^k} are greater than or equal to zero, {b^k}⊂[1, +∞), $ \sum\limits_{k=1}^{\infty}\left(b^k-1\right)<\infty$ and $ \sum\limits_{k=1}^{\infty} c^k<\infty $. Then $ \lim \limits_{k \rightarrow \infty} a^k $ exists.

Lemma 5^[27]   For the (VIP), suppose F: C→H is pseudomonotone and continuous and C being a nonempty, closed, convex subset of a real Hilbert space H. We have the following conclusion: i is a solution to the problem we discussed above if and only if

$ \langle F u, u-i\rangle \geqslant 0, \forall u \in C $

Lemma 6^[28]   Suppose {s^k}, {t^k}, {λ^k} are all sequences of real numbers, and {s^k} is required to be nonnegative. λ^k∈(0, 1] satisfies $\sum\limits_{k=1}^{\infty} \lambda^k=\infty $. If lim sup_j→∞t^k_j≤0 and s^k+1≤(1－λ^k)s^k+λ^kt^k, ∀k≥1, for every subsequence {s^k_j} of {s^k}, having $\liminf\limits _{j \rightarrow \infty}\left(s^{k_j+1}-s^{k_j}\right) \geqslant 0 $, we can get $ \lim\limits _{k \rightarrow \infty} s^k=0 $.

2 Algorithms and Convergence Analysis 2.1 Weak Convergence

First, we illustrate the first of the two new algorithms mentioned above and prove its weak convergence. Until then, assume the following is true:

(C1) The given C has closed and convex properties. In addition, C and Sol(C, F) have non-zero elements.

(C2) The operator F: C→H in the algorithm satisfies: pseudomonotone, sequentially weakly continuous, and L-Lipschitz continuous on C.

Algorithm 1   is detailed in the following four steps:

Step 0: Let starting point u⁰, u¹∈H be arbitrary and given parameters $\bar{\omega} $>0, $\rlap{-} \lambda$>0, σ∈(0, 1/3).Take a succession of nonnegation real numbers {χ^k} to content χ^k∈[a, b]⊂(0, 1).

Step 1: Given the u^k－1, u^k. Then define h^k=u^k+ $ \bar{\omega}^{k} $(u^k－u^k－1),

where

$ \bar{\omega}^k=\left\{\begin{array}{l} \min \left\{\bar{\omega}, \frac{1}{\left\|u^k-u^{k+1}\right\|^2 \cdot k^2}\right\}, {\text { if } }\;u^k-u^{k+1} \neq 0 \\ \bar{\omega}, \text { else } \end{array}\right. $

(2)

Step 2: Compute

$ v^k=P_C\left(h^k-\rlap{-} \lambda_k F h^k\right) $

If h^k=v^k or Av^k=0, then stop, because v^k∈Sol(C, F)

Step 3: Set u^k+1=χ^kr^k+(1－χ^k)h^k.

Take r^k in the formula above as the following form,

$ r^k=v^k-\rlap{-}\lambda_k\left(F v^k-F h^k\right) $

Next update

$ \rlap{-}\lambda_{{\mathrm{k}}+1}=\left\{\begin{array}{l} \min \left\{\varPhi_k \rlap{-}\lambda_k+\varGamma_k, \frac{\sigma\left\|h^k-v^k\right\|}{\left\|F h^k-F v^k\right\|}\right\}, {\text { if }}\; F h^k-F v^k \neq 0 \\ \varPhi_k \rlap{-}\lambda_k+\varGamma_k, {\text { else }} \end{array}\right. $

(3)

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

Remark 1   According to Eq.(2), we can get $ \bar{\omega}^k $‖u^k－u^k－1‖²≤1/k², ∀k∈N. Where 0≤ $ \bar{\omega}^k $≤ $ \bar{\omega} $, ∀k∈N, that is

$ \sum\limits_{k=1}^{\infty} \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2 \leqslant+\infty $

Remark 1 is useful for our later proof.

Lemma 7   For {$ \rlap{-}\lambda_k $} derived from the above Algorithm 1, we can obtain $ \lim\limits _{k \rightarrow \infty} \rlap{-}\lambda_k=\rlap{-}\lambda $ and $ \rlap{-}\lambda$>min{ $ \rlap{-}\lambda_1 $, σ/L}>0.

Proof   When Fh^k－Fv^k≠0, by the continuity requirements of F in Condition 2) we can obtain ‖Fh^k－Fv^k‖≤L‖h^k－v^k‖, which yields that

$ \frac{\sigma\left\|h^k-v^k\right\|}{\left\|F h^k-F v^k\right\|} \geqslant \frac{\sigma\left\|h^k-v^k\right\|}{L\left\|h^k-v^k\right\|}=\frac{\sigma}{L} $

That means:

$ \begin{aligned} \rlap{-}\lambda_{k+1} & =\min \left\{\varPhi_k \rlap{-}\lambda_k+\varGamma_k, \frac{\sigma\left\|h^k-v^k\right\|}{\left\|F h^k-F v^k\right\|}\right\} \geqslant \\ & \min \left\{\rlap{-}\lambda_k, \frac{\sigma}{L}\right\} \end{aligned} $

where Γ_k≥0 and Φ_k≥1.

By the proof, we can attain that the {$ \rlap{-}\lambda_k $} be the succession step size by above Algorithm 1 has a lower bound min{ $ \rlap{-}\lambda_1 $, σ/L}.

When k=1, $ \rlap{-}\lambda_1 $≥min{$ \rlap{-}\lambda_1 $, σ/L}.

When k=m(m≥1), the formula $ \rlap{-}\lambda_m $≥min{$ \rlap{-}\lambda_1 $, σ/L} holds.

When k=m+1, $ \rlap{-}\lambda_{m+1} $≥min{$ \rlap{-}\lambda_{m} $, σ/L}≥ min{$ \rlap{-}\lambda_{1} $, σ/L}.

By induction, we get the results we wanted.

From Eq.(3), it is clear that

$ \rlap{-}\lambda_{k+1} \leqslant \varPhi_k \chi_k+\varGamma_k $

Thanks to Lemma 4, we clearly know that $ \lim\limits _{_k \rightarrow \infty} \rlap{-}\lambda_k $ exist. Now, we denote $ \lim\limits _{k \rightarrow \infty} \rlap{-}\lambda_k=\rlap{-}\lambda $. From the above analysis, it can be seen that sequence {$ \rlap{-}\lambda_k $} has the lower bound min {$ \rlap{-}\lambda_1$, σ/L}. Apparently, $ \rlap{-}\lambda $>0.

Lemma 8   Succession {h^k} is provided by Algorithm 1, if the presumptions Conditions 1)-2) are given and there exists a subsequence {h^k_j} convergent weakly to some ι, $ \lim\limits _{j \rightarrow \infty}\left\|h^{k_j}-v^{k_j}\right\|=0 $, we have ι∈Sol(C, F).

Proof : Since

$ \begin{gathered} v^k=P_C\left(h^k-\rlap{-}\lambda_k F h^k\right) \\ \left\langle h^{k_j}-\rlap{-}\lambda_k F h^{k_j}-v^{k_j}, e-v^{k_j}\right\rangle \leqslant 0, \quad \forall e \in C \end{gathered} $

or

$ \frac{1}{\rlap{-}\lambda_{k_j}}\left\langle h^{k_j}-v^{k_j}, e-v^{k_j}\right\rangle \leqslant\left\langle F h^{k_j}, e-v^{k_j}\right\rangle, \forall e \in C $

This implies

$ \begin{aligned} & \frac{1}{\rlap{-}\lambda_{k_j}}\left\langle h^{k_j}-v^{k_j}, e-v^{k_j}\right\rangle+\left\langle F h^{k_j}, v^{k_j}-h^{k_j}\right\rangle \leqslant \\ & \quad\left\langle F h^{k_j}, e-h^{k_j}\right\rangle, \forall e \in C \end{aligned} $

(4)

For {h^k_j}, we know that {h^k_j} is bounded because of its weak convergence. On the other hand, according to the nature of F, we get {Fh^k_j} is also bounded. Let ‖h^k_j－v^k_j‖→0, one sees {v^k_j} is bounded.

By Lemma 7, $ \rlap{-}\lambda_{k_j} \geqslant \min \left\{\rlap{-}\lambda_1, \sigma / L\right\} $. Now, passing inequality (4) to let as j→∞, we have

$ \liminf\limits _{j \rightarrow \infty}\left\langle F h^{k_j}, e-h^{k_j}\right\rangle \geqslant 0, \forall e \in C $

(5)

Moreover,

$ \begin{gathered} \left\langle F v^{k_j}, e-v^{k_j}\right\rangle=\left\langle F v^{k_j}-F h^{k_j}, e-h^{k_j}\right\rangle+ \\ \left\langle F h^{k_j}, e-h^{k_j}\right\rangle+\left\langle F v^{k_j}, h^{k_j}-v^{k_j}\right\rangle \end{gathered} $

(6)

From the above assumption that F is L-Lipschitz continuous and $\lim\limits _{j \rightarrow \infty}\left\|h^{k_j}-v^{k_j}\right\|=0 $, one sees

$ \lim\limits _{j \rightarrow \infty}\left\|F h^{k_j}-F v^{k_j}\right\|=0 $

which combining inequality (5) and Eq.(6), we have

$ \liminf\limits _{j \rightarrow \infty}\left\langle e-v^{k_j}, F v^{k_j}\right\rangle \geqslant 0 $

Let's prove concretely that ι∈Sol(C, F). First, select a decreasing positive sequence {τ_j} to satisfy $\lim\limits _{j \rightarrow \infty} \tau_j=0 $. For each j, K_j represents the smallest positive integer such that

$ \left\langle\boldsymbol{F} v^{k_i}, e-v^{k_i}\right\rangle+\tau_k \geqslant 0, \quad \forall i \geqslant K_j $

(7)

Since {τ_j} decrease, we can get {K_j} is obviously increasing.

For any j, hypothesis Fv^K_j≠0 (when Fv^K_j=0, v^K_j is a solution). Putting

$ \vartheta^{K_j}=\frac{F v^{K_j}}{\left\|F v^{K_j}\right\|^2} $

Combined with the knowledge of norm, one obtains 〈Fv^K_j, $ \vartheta^{K_j} $〉=1. By inequality (7), we can deduce that

$ \left\langle F v^{K_j}, e+\tau_j \vartheta^{K_j}-v^{K_j}\right\rangle \geqslant 0 $

Through (2) of Definition 1, one gets

$ \left\langle F\left(e+\tau_j \vartheta^{K_j}\right), e+\tau_j \vartheta^{K_j}-v^{K_j}\right\rangle \geqslant 0 $

To transform the upper formula, we have

$ \begin{gathered} \left\langle F e, e-v^{K_j}\right\rangle \geqslant\left\langle F e-F\left(e+\tau_j \vartheta^{K_j}\right), e+\right. \\ \left.\tau_j \vartheta^{K_j}-v^{K_j}\right\rangle-\tau_j\left\langle F e, \vartheta^{K_j}\right\rangle \end{gathered} $

(8)

Since h^k_j⇀ι and $\lim\limits _{j \rightarrow \infty}\left\|h^{k_j}-v^{k_j}\right\|=0 $, when j→∞ one has v^k_j⇀ι. Furthermore, thanks to F has the property of sequentially weakly continuity on C, one sees {Fv^k_j} converges weakly to Fι. Now, assume Fι≠0 (if not, ι is a solution). Apparently, applying lower semicontinuity of the norm mapping, one gets

$ 0<\|F \iota\| \leqslant \liminf\limits _{j \rightarrow \infty}\left\|F v^{k_j}\right\| $

According to the previous certificate, it is easy to obtain

$ 0 \leqslant \limsup\limits _{j \rightarrow \infty}\left\|\tau_j \vartheta^{K_j}\right\|=\limsup\limits _{j \rightarrow \infty}\left(\frac{\tau_j}{\left\|Fv^{k_j}\right\|}\right) \leqslant\\\frac{\limsup\limits _{j \rightarrow \infty} \tau_j}{\liminf \limits_{j \rightarrow \infty}\left\|F v^{K_j}\right\|}=0 $

which indicates that $ \lim\limits _{j \rightarrow \infty} \tau_j \vartheta^{K_j}=0 $.

In addition, {h^K_j}, {$ \vartheta^{K_j} $} are bounded, F is uniformly continuous and $ \lim\limits _{j \rightarrow \infty} \tau_j \vartheta^{K_j}=0 $, so the right hand side of inequality (8) tend to 0 as j→∞.

Hence,

$ \liminf\limits _{j \rightarrow \infty}\left\langle F e, e-v^{K_j}\right\rangle \geqslant 0 $

For all e∈C,

$ \begin{gathered} \langle F e, e-\iota\rangle=\lim\limits _{j \rightarrow \infty}\left\langle F e, e-v^{K_j}\right\rangle= \\ \liminf\limits _{j \rightarrow \infty}\left\langle F e, e-v^{K_j}\right\rangle \geqslant 0 \end{gathered} $

Using Lemma 5, we can get ι∈Sol(C, F).

Lemma 9   If {u^k} be a sequence generated by above Algorithm 1. Then

$ \begin{gathered} \left\|u^{k+1}-i\right\|^2 \leqslant\left\|h^k-i\right\|^2-\chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right) \cdot\\ \left\|v^k-h^k\right\|^2, \forall i \in {\rm{Sol}}(C, F) \end{gathered} $

Proof : Through the notion of u^k+1,

$ \begin{array}{l} \left\|u^{k+1}-i\right\|^2=\left\|\left(1-\chi^k\right) h^k+\chi^k r^k-i\right\|^2= \\ \left\|\chi^k\left(r^k-i\right)+\left(1-\chi^k\right)\left(h^k-i\right)\right\|^2= \\ \chi^k\left\|r^k-i\right\|^2+\left(1-\chi^k\right)\left\|h^k-i\right\|^2- \\ \chi^k\left(1-\chi^k\right)\left\|h^k-r^k\right\|^2 \end{array} $

In view of r^k=v^k－ $\rlap{-}\lambda_{k} $(Fv^k－Fh^k), we have

$ \begin{array}{l} \left\|r^k-i\right\|^2=\left\|v^k-i\right\|^2+\rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2- \\ 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k-F h^k\right\rangle=\| v^k-h^k+h^k- \\ i\left\|^2+\rlap{-}\lambda_k^2\right\| F v^k-F h^k \|^2-2 \rlap{-}\lambda_k\left\langle v^k-i, \right. \\ \left.F v^k-F h^k\right\rangle=\left\|h^k-i\right\|^2+\left\|v^k-h^k\right\|^2+ \\ \rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2+2\left\langle v^k-h^k, h^k-i\right\rangle- \\ 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k-F h^k\right\rangle=\left\|h^k-i\right\|^2+ \\ \left\|v^k-h^k\right\|^2+\rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2-2\left\langle v^k-\right. \\ \left.h^k, v^k-h^k\right\rangle+2\left\langle v^k-h^k, v^k-i\right\rangle-2 \rlap{-}\lambda_k\left\langle v^k-\right. \\ \left.i, F v^k-F h^k\right\rangle=\left\|h^k-i\right\|^2-\left\|v^k-h^k\right\|^2+ \\ \rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2+2\left\langle v^k-h^k, v^k-i\right\rangle- \\ 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k-F h^k\right\rangle \end{array} $

(9)

From v^k=P_C(h^k－ $ \rlap{-}\lambda_k $Fh^k), one arrives at

$ \begin{aligned} & \left\langle v^k-h^k+\rlap{-}\lambda_k F h^k, v^k-i\right\rangle= \\ & \quad\left\langle v^k-h^k, v^k-i\right\rangle+\left\langle\rlap{-}\lambda_k F h^k, v^k-i\right\rangle \leqslant 0 \end{aligned} $

Hence,

$ \left\langle v^k-h^k, v^k-i\right\rangle \leqslant-\rlap{-}\lambda_k\left\langle F h^k, v^k-i\right\rangle $

(10)

Combining Eqs.(9), (10) and Lemma 5, one sees

$ \begin{array}{l} \left\|u^{k+1}-i\right\|^2 \leqslant\left(1-\chi^k\right)\left\|h^k-i\right\|^2+ \\ \chi^k\left\|r^k-i\right\|^2 \leqslant\left\|h^k-i\right\|^2+\chi^k \rlap{-}\lambda_k^2\cdot \\ \left\|F v^k-F h^k\right\|^2-\chi^k\left\|h^k-v^k\right\|^2-\\ 2 \chi^k \rlap{-}\lambda_k\left\langle F h^k, v^k-i\right\rangle-2 \chi^k \rlap{-}\lambda_k\left\langle v^k-i, F v^k-\right. \\ \left.F h^k\right\rangle \leqslant\left\|h^k-i\right\|^2+\chi^k \sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\left\|h^k-v^k\right\|^2- \\ \chi^k\left\|h^k-v^k\right\|^2-2 \chi^k \rlap{-}\lambda_k\left\langle v^k-i, F v^k\right\rangle \leqslant \\ \left\|h^k-i\right\|^2-\chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|h^k-v^k\right\|^2 \end{array} $

Theorem 1    Under the premise that the previously mentioned Conditions 1)-2) are true, {u^k} converges weakly to some i, where i∈Sol(C, F).

Proof : By the conclusion of Lemma 9 we have

$ \begin{aligned} & \left\|u^{k+1}-i\right\|^2 \leqslant\left\|h^k-i\right\|^2- \\ & \quad \chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|v^k-h^k\right\|^2 \end{aligned} $

and

$ \begin{aligned} & \left\|h^k-i\right\|^2=\left\|\bar{\omega}^k\left(u^k-u^{k-1}\right)+u^k-i\right\|^2= \\ & \left\|\left(\bar{\omega}^k+1\right)\left(u^k-i\right)-\bar{\omega}^k\left(u^{k-1}-i\right)\right\|^2= \\ & \left(\bar{\omega}^k+1\right)\left\|u^k-i\right\|^2-\bar{\omega}^k\left\|u^{k-1}-i\right\|^2+ \\ & \left(\bar{\omega}^k+1\right) \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2 \end{aligned} $

Combine the above two formulas, one sees

$ \begin{gathered} \left\|u^{k+1}-i\right\|^2 \leqslant\left(\bar{\omega}^k+1\right)\left\|u^k-i\right\|^2- \\ \bar{\omega}^k\left\|u^{k-1}-i\right\|^2+\left(\bar{\omega}^k+1\right) \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2- \\ \chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|v^k-h^k\right\|^2 \leqslant\left\|u^k-i\right\|^2+ \\ \bar{\omega}^k\left(\left\|u^k-i\right\|^2-\left\|u^{k-1}-i\right\|^2\right)+(\bar{\omega}+1) \bar{\omega}^k \cdot \\ \left\|u^k-u^{k-1}\right\|^2-\chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|v^k-h^k\right\|^2 \end{gathered} $

Since

$ \lim\limits _{n \rightarrow \infty}\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)=1-\sigma^2>0 $

(11)

there

$ \exists k^0 \in N {\text { s.t. }}\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)>0, \forall k \geqslant k_0 $

Through the above analysis, one has

$ \begin{aligned} & \left\|u^{k+1}-i\right\|^2 \leqslant\left\|u^k-i\right\|^2+\bar{\omega}^k\left(\left\|u^k-i\right\|^2-\right. \\ & \left.\left\|u^{k-1}-i\right\|^2\right)+(\bar{\omega}+1) \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2, \\ & \forall k \geqslant k_0 \end{aligned} $

(12)

Applying Lemma 2 with

$ \xi^k=\left\|u^k-i\right\|^2 $

and

$ \chi_n=(\bar{\omega}+1) \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2 $

Using Lemma 2, we get $ \lim\limits _{k \rightarrow \infty}\left\|u^k-i\right\|^2 $ exists and

$ \sum\limits_{k=1}^{\infty}\left[\left\|u^k-i\right\|^2-\left\|u^{k-1}-i\right\|^2\right]_{+}<+\infty $

where [p]₊=max {p, 0}.

Hence,

$ \lim\limits _{k \rightarrow \infty}\left[\left\|u^k-i\right\|^2-\left\|u^{k-1}-i\right\|^2\right]_{+}=0 $

Thanks to inequality (12),

$ \begin{aligned} & \chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|v^k-h^k\right\|^2 \leqslant\left\|u^k-i\right\|^2- \\ & \left\|u^{k+1}-i\right\|^2+\bar{\omega}^k\left(\left\|u^k-i\right\|^2-\left\|u^{k-1}-i\right\|^2\right)+ \\ & \bar{\omega}^k(\bar{\omega}+1)\left\|u^k-u^{k-1}\right\|^2 \leqslant\left\|u^k-i\right\|^2- \\ & \left\|u^{k+1}-i\right\|^2+\bar{\omega}^k\left[\left\|u^k-i\right\|^2-\left\|u^{k-1}-i\right\|^2\right]_{+}+ \\ & \bar{\omega}^k(\bar{\omega}+1)\left\|u^k-u^{k-1}\right\|^2, \forall k \geqslant k_0 \end{aligned} $

That means further

$ \lim \limits_{k \rightarrow \infty}\left\|v^k-h^k\right\|=0 $

Using the notion of h^k, we get

$ \begin{gathered} \left\|h^k-u^k\right\|^2=\left(\bar{\omega}^k\right)^2\left\|u^k-u^{k-1}\right\|^2 \leqslant \\ \bar{\omega} \cdot \bar{\omega}^k\left\|u^k-u^{k-1}\right\|^2 \end{gathered} $

So,

$ \lim\limits _{k \rightarrow \infty}\left\|h^k-u^k\right\|=0 $

(13)

From $ \lim\limits _{k \rightarrow \infty}\left\|u^k-i\right\| $ exists, the sequence {u^k} is bounded.

We select {u^k_j} of {u^k} such that u^k_j→u^*. Thanks to Eq.(13), h^k_j⇀u^*.

Thanks to Lemma 8, one has u^*∈Sol(C, F).

So, we prove the following two conclusions:

∀i ∈ Sol(C, F), from the above analysis, $ \lim\limits _{k \rightarrow \infty}\left\|u^k-i\right\| $ exists;

(2) For every u^*, where u^* is the sequential weak cluster point of produced by {u^k}, it is in Sol(C, F).

Using Lemma 3, one concludes that {u^k} converges weakly to some u^* of the solution set Sol(C, F).

Theorem 2   Suppose F is β-strongly pseudomonotone and the previous assumption is still valid. If 0 < r < 1 and

$ \bar{\omega}^k=\left\{\begin{array}{l} 0, k=2 k \\ \min \left\{\bar{\omega}^k, \theta\right\}, k=2 k-1 \end{array}\right. $

(14)

Then the sequence {u^k} generated by Algorithm 1 is R-linearly convergent to the unique solution of (VIP) problem.

Proof:

Due to β-strongly pseudomonotonicity of F, the VIP problem has a unique solution. Let's call it i^*. By using v^k=P_C(h^k－ $ \rlap{-}\lambda_{k} $Fh^k) and (2) in Lemma 1, the following formula is established.

$ \begin{array}{l} \left\langle h^k-v^k, i^*-v^k\right\rangle=\left\langle h^k-\rlap{-}\lambda_k F h^k-v^k, i^*-v^k\right\rangle+ \\ \rlap{-}\lambda_k\left\langle F h^k, i^*-v^k\right\rangle \leqslant \rlap{-}\lambda_k\left\langle F h^k, i^*-v^k\right\rangle= \\ \rlap{-}\lambda_k\left\langle F v^k, i^*-v^k\right\rangle+\rlap{-}\lambda_k\left\langle F h^k-F v^k, i^*-v^k\right\rangle \end{array} $

Because F is β-strongly pseudomonotone on C, we know〈Fv^k, v^k－i^*〉≥β‖v^k－i^*‖².

So

$ \begin{array}{l} \left\langle h^k-v^k, i^*-v^k\right\rangle \leqslant-\beta \rlap{-}\lambda_k\left\|i^*-v^k\right\|^2+ \\ \rlap{-}\lambda_k\left\langle F h^k-F v^k, i^*-v^k\right\rangle \leqslant-\beta \rlap{-}\lambda_k \| i^*- \\ v^k\left\|^2+\sigma \frac{\rlap{-}\lambda_k}{\rlap{-}\lambda_{k+1}}\right\| h^k-v^k\|\cdot\| i^*-v^k \| \end{array} $

That means

$ \begin{aligned} & \beta \rlap{-}\lambda_k\left\|i^*-v^k\right\|^2 \leqslant-\left\langle h^k-v^k, i^*-v^k\right\rangle+ \\ & \sigma \frac{\rlap{-}\lambda_k}{\rlap{-}\lambda_{k+1}}\left\|h^k-v^k\right\| \cdot\left\|i^*-v^k\right\| \leqslant\left\|h^k-v^k\right\| \cdot \\ & \quad\left\|i^*-v^k\right\|+\sigma \frac{\rlap{-}\lambda_k}{\rlap{-}\lambda_{k+1}}\left\|h^k-v^k\right\| \cdot\left\|i^*-v^k\right\|= \\ & \quad\left(1+\sigma \frac{\rlap{-}\lambda_k}{\rlap{-}\lambda_{k+1}}\right)\left\|h^k-v^k\right\| \cdot\left\|i^*-v^k\right\| \end{aligned} $

Then we can get

$ \left\|i^*-v^k\right\|^2 \leqslant\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\left\|h^k-v^k\right\| $

Therefore, in combination with the previous derivation, we can see that

$ \begin{gathered} \left\|h^k-i^*\right\| \leqslant\left\|i^*-v^k\right\|+\left\|h^k-v^k\right\| \leqslant \\ {\left[1+\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\right]\left\|h^k-v^k\right\|} \end{gathered} $

That is

$ \left\|h^k-v^k\right\| \geqslant\left[1+\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\right]^{-1}\left\|h^k-i^*\right\| $

In addition, combining the above formula with the conclusion of Lemma 9, one gets

$ \begin{aligned} & \left\|u^{k+1}-i^*\right\|^2 \leqslant\left\|h^k-i^*\right\|^2-\chi^k(1- \\ & \left.\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left\|v^k-h^k\right\|^2 \leqslant\left[1-\chi^k(1-\right. \\ & \left.\left.\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left[1+\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\right]^{-2}\right] . \\ & \left\|h^k-i^*\right\|^2 \end{aligned} $

(15)

If assume

$ \begin{gathered} y=\liminf\limits _{k \rightarrow \infty}\left[1+\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\right]^{-2} \\ x=\liminf\limits _{k \rightarrow \infty} \chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)>0 \end{gathered} $

By Lemma 7, there must be y>0. Then $\exists $K>0, s.t.

$ \begin{gathered} \chi^k\left(1-\frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2} \sigma^2\right)\left[1+\left(\frac{1}{\beta \rlap{-}\lambda_k}+\frac{\sigma}{\beta \rlap{-}\lambda_{k+1}}\right)\right]^{-2} \geqslant \\ x y>0, \forall k \geqslant K \end{gathered} $

Define r²=1－xy, there is obviously r² < 1.Combined with inequality (15), we can obtain

$ \left\|u^{k+1}-i^*\right\|^2 \leqslant r^2\left\|h^k-i^*\right\|^2, \quad \forall k \geqslant K $

(16)

Now we consider the convergence results in two cases.

When k=2k+1, using the definition of h^k and Eq.(14), we know

$ h^{2 k}=u^{2 k}+\bar \omega^{2 k}\left(u^{2 k}-u^{2 k-1}\right)=u^{2 k} $

Then

$ \begin{gathered} \left\|u^{2 k+1}-i^*\right\|^2 \leqslant r^2\left\|h^{2 k}-i^*\right\|^2= \\ r^2\left\|u^{2 k}-i^*\right\|^2 \end{gathered} $

(17)

That means

$ \begin{gathered} \left\|u^{2 k+1}-i^*\right\| \leqslant r\left\|u^{2 k}-i^*\right\| \leqslant \cdots \leqslant \\ r^{k-K+1}\left\|u^{2 K}-i^*\right\|, \quad \forall k \geqslant K \end{gathered} $

(18)

Let's consider next, when k=2k+2, using the definition of h^k and Eq.(14), one gets

$ \begin{gathered} \left\|h^{2 k+1}-i^*\right\|^2=\| u^{2 k+1}+\bar{\omega}^{2 k+1}\left(u^{2 k+1}-u^{2 k}\right)- \\ i^*\left\|^2=\right\|\left(\bar{\omega}^{2 k+1}+1\right)\left(u^{2 k+1}-i^*\right)- \\ \bar{\omega}^{2 k+1}\left(u^{2 k}-i^*\right) \|^2 \end{gathered} $

By inequality (16), we know

$ \begin{aligned} & \left\|u^{2 k+2}-i^*\right\|^2 \leqslant r^2\left\|h^{2 k+1}-i^*\right\|^2= \\ & r^2 \|\left(\bar{\omega}^{2 k+1}+1\right)\left(u^{2 k+1}-i^*\right)-\bar{\omega}^{2 k+1}\left(u^{2 k}-\right. \\ & \left.i^*\right) \|^2=r^2\left[\left(\bar{\omega}^{2 k+1}+1\right)\left\|u^{2 k+1}-i^*\right\|^2-\right. \\ & \bar{\omega}^{2 k+1}\left\|u^{2 k}-i^*\right\|^2+\bar{\omega}^{2 k+1}\left(\bar{\omega}^{2 k+1}+1\right) \cdot \\ & \left.\left\|u^{2 k+1}-u^{2 k}\right\|^2\right] \end{aligned} $

Combined with Eqs.(14) and (17), the following formula can be obtained

$ \begin{gathered} \left\|u^{2 k+2}-i^*\right\|^2 \leqslant r^2\left[r^2\left(\bar{\omega}^{2 k+1}+1\right)\left\|u^{2 k}-i^*\right\|^2-\right. \\ \bar{\omega}^{2 k+1}\left\|u^{2 k}-i^*\right\|^2+\bar{\omega}^{2 k+1}\left(\bar{\omega}^{2 k+1}+1\right) \cdot \\ \left.\left(\left\|u^{2 k+1}-i^*\right\|+\left\|u^{2 k}-i^*\right\|\right)^2\right] \leqslant \\ r^2\left[r^2\left(\bar{\omega}^{2 k+1}+1\right)\left\|u^{2 k}-i^*\right\|^2-\right. \\ \bar{\omega}^{2 k+1}\left\|u^{2 k}-i^*\right\|^2+\bar{\omega}^{2 k+1}\left(\bar{\omega}^{2 k+1}+1\right)\left(r^2+1\right) \cdot \\ \left.\left\|u^{2 k}-i^*\right\|^2\right]=r^2\left[r^2\left(\bar{\omega}^{2 k+1}+1\right)-\right. \\ \left.\bar{\omega}^{2 k+1}+\left(r^2+1\right) \bar{\omega}^{2 k+1}\left(\bar{\omega}^{2 k+1}+1\right)\right] \cdot \\ \left\|u^{2 k}-i^*\right\|^2 \end{gathered} $

(19)

Now if we want

$ r^2\left(\bar{\omega}^{2 k+1}+1\right)-\bar{\omega}^{2 k+1}+\left(r^2+1\right)\left(\bar{\omega}^{2 k+1}+1\right) \bar{\omega}^{2 k+1} \leqslant 1 $ we obviously need to make sure that $ \bar{\omega}^{2 k+1} \leqslant \frac{1-r^2}{(1+r)^2}= \frac{1-r}{1+r}$, when we define it. So, let's take $ \theta=\frac{1-r}{1+r} $ in Eq.(14).

Then inequality (19) is transformed into

$ \left\|u^{2 k+2}-i^*\right\|^2 \leqslant r^2\left\|u^{2 k}-i^*\right\|^2 $

That means

$ \begin{gathered} \left\|u^{2 k+2}-i^*\right\| \leqslant r\left\|u^{2 k}-i^*\right\| \leqslant \cdots \leqslant \\ r^{k-K+1}\left\|u^{2 K}-i^*\right\|, \quad \forall k \geqslant K \end{gathered} $

(20)

By the inequalities (18) and (20), we can obtain the sequence {u^k} converges to i^* with an R linear rate.

2.2 Strong Convergence

Now we modify the Algorithm 1 and obtain strong convergence.

Until then, let's assume the following six conditions hold true.

(C1) F is sequentially weakly continuous over C, pseudomonotone and L-Lipschitz continuous in H.

(C2) We define operator Y is some contraction on H such that contraction coefficient κ∈(0, 1).

(C3) $\left\{\bar{\omega}^k\right\} \subset[0, 1) $.

(C4) {m^k}, {d^k}, {g^k} are three sequences in 0, 1), and $\sum\limits_{k=1}^{\infty} d^k=\infty $.

(C5) On the basis of (C4), m^k+d^k+g^k=1.

(C6) {ξ^k} is a positive sequence such that $\lim\limits _{k \rightarrow \infty} \frac{\xi^k}{d^k}=0 $, together with $ \lim\limits _{k \rightarrow \infty} d^k=0$.

Remark 2 According to

$ \bar{\omega}^k=\left\{\begin{array}{l} \min \left\{\bar{\omega}, \frac{\xi^k}{\left\|u^k-u^{k-1}\right\|^2}\right\} \\ \bar{\omega}, {\text { else }} \end{array}\right. $, if u^k－u^k－1≠0 one sees

$ \lim\limits _{n \rightarrow \infty} \frac{\bar{\omega}^k}{d^k}\left\|u^k-u^{k-1}\right\|=0 $

The truth is, $\bar{\omega}^k\left\|u^k-u^{k-1}\right\| \leqslant \xi^k $, ∀k>0 and $ \lim\limits _{k \rightarrow \infty} \frac{\xi^k}{d^k}=0 $, which suggests

$ \lim\limits _{k \rightarrow \infty} \frac{\bar{\omega}^k}{d^k}\left\|u^k-u^{k-1}\right\| \leqslant \lim\limits _{k \rightarrow \infty} \frac{\xi^k}{d^k}=0 $

Theorem 3 Succession {u^k} can be provided by Algorithm 2. If the conditions (C1) - (C6) hold, {u^k} converges strongly to i∈Sol(C, F), here we define i=P_{Sol(C, F)}·Y(i).

Algorithm 2

Step 0: Let starting point u⁰, u¹∈H be arbitrary and given parameters $ \overline{\overline \omega } $>0, $ \rlap{-}\lambda_{1} $>0, σ∈(0, 1).

Step 1: When k≥1, through the previously u^k－1 and u^k, define h^k=u^k+$ \overline{\overline \omega }^{k} $(u^k－u^k－1).

where

$ \overline{\bar{\omega}}^k=\left\{\begin{array}{l} \min \left\{\overline{\bar{\omega}}, \frac{\xi^k}{\left\|u^k-u^{k-1}\right\|^2}\right\} \\ \overline{\bar{\omega}}, {\text { else }} \end{array}\right. $, if u^k－u^k－1≠0

Step 2: Compute

$ v^k=P_C\left(h^k-\rlap{-}\lambda_k F h^k\right) $

If h^k=v^k or Fv^k=0, end all iteration steps and v^k∈Sol(C, F).

Step 3: Set u^k+1=d^kY(u^k)+g^kr^k+m^kh^k,

where

$ r^k=v^k-\rlap{-}\lambda_k\left(F v^k-F h^k\right) $

Then update this iteration step size

$ \rlap{-}\lambda_{k+1}=\left\{\begin{array}{l} \min \left\{\varPhi_k \rlap{-}\lambda_k+\varGamma_k, \frac{\sigma\left\|h^k-v^k\right\|}{\left\|F h^k-F v^k\right\|}\right\}, {\text { if }}\; F h^k-F v^k \neq 0 \\ \varPhi_k \rlap{-}\lambda_k+\varGamma_k, {\text { else }} \end{array}\right. $

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

Proof   For clarity, our derivation process is roughly broken down into four small parts.

Part 1   Prove that {u^k} is bounded.

Let i=P_{Sol(C, F)}·Y(i). By Eqs.(9), (10) in the Lemma 9 and P_{Sol(C, F)}·Y is a compressed map,

we can get

$ \begin{aligned} \left\|r^k-i\right\|^2 & =\left\|h^k-i\right\|^2-\left\|h^k-v^k\right\|^2+ \\ & 2\left\langle v^k-h^k, v^k-i\right\rangle+ \\ & \rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2- \\ & 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k-F h^k\right\rangle \end{aligned} $

and

$ \left\langle v^k-h^k, v^k-i\right\rangle \leqslant-\rlap{-}\lambda_k\left\langle F h^k, v^k-i\right\rangle $

That is

$ \begin{aligned} & \left\|r^k-i\right\|^2 \leqslant\left\|h^k-i\right\|^2-\left\|h^k-v^k\right\|^2+ \\ & \rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2-2 \rlap{-}\lambda_k\left\langle F h^k, v^k-i\right\rangle- \\ & 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k-F h^k\right\rangle=\left\|h^k-i\right\|^2- \\ & \left\|h^k-v^k\right\|^2+\rlap{-}\lambda_k^2\left\|F v^k-F h^k\right\|^2- \\ & 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k\right\rangle \leqslant\left\|h^k-i\right\|^2- \\ & \left\|h^k-v^k\right\|^2+\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\left\|h^k-v^k\right\|^2- \\ & 2 \rlap{-}\lambda_k\left\langle v^k-i, F v^k\right\rangle=\left\|h^k-i\right\|^2-(1- \\ & \left.\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right)\left\|h^k-v^k\right\|^2-2 \rlap{-}\lambda_k\left\langle v^k-i, \right. \\ & \left.F v^k\right\rangle \leqslant\left\|h^k-i\right\|^2-\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right) \\ & \left\|h^k-v^k\right\|^2 \\ & \end{aligned} $

(21)

Applying Eq.(11), we have

$ 1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}>0 $

It implies that ∀k≥k₀,

$ \left\|r^k-i\right\| \leqslant\left\|h^k-i\right\| $

(22)

Using the notion of h^k, we can deduce that

$ \begin{aligned} & \left\|h^k-i\right\|=\left\|\overline{\bar{\omega}}^k\left(u^k-u^{k-1}\right)+u^k-i\right\| \leqslant \\ & \overline{\bar{\omega}}^k\left\|u^k-u^{k-1}\right\|+\left\|u^k-i\right\|= \\ & \left\|u^k-i\right\|+d^k \cdot \frac{\overline{\bar{\omega}}^k}{d^k}\left\|u^k-u^{k-1}\right\| \\ & \end{aligned} $

(23)

From Remark 2, we attain

$ \frac{\overline{\bar{\omega}}^k}{d^k}\left\|u^k-u^{k-1}\right\| \rightarrow 0 $

(24)

Then $ \exists $Q₁>0, such that

$ \frac{\overline{\bar{\omega}}^k}{d^k}\left\|u^k-u^{k-1}\right\| \leqslant Q_1, \forall k \geqslant 1 $

(25)

Combing inequalities (23), (24) and (25), we attain

$ \left\|r^k-i\right\| \leqslant\left\|h^k-i\right\| \leqslant\left\|u^k-i\right\|+d^k Q_1 $

(26)

Using the notion of u^k+1, one can see that

$ \begin{aligned} & \left\|u^{k+1}-i\right\| \leqslant m^k\left\|h^k-i\right\|+d^k\left\|Y\left(u^k\right)-i\right\| \\ & \quad+g^k\left\|r^k-i\right\| \leqslant m^k\left\|h^k-i\right\|+d^k \| Y\left(u^k\right)- \\ & Y(i)\left\|+d^k\right\| Y(i)-i\left\|+g^k\right\| r^k-i \| \leqslant \\ & m^k\left\|h^k-i\right\|+d^k \kappa\left\|u^k-i\right\|+d^k\|Y(i)-i\|+ \\ & g^k\left\|r^k-i\right\| \leqslant m^k\left\|u^k-i\right\|+d^k \kappa\left\|u^k-i\right\|+ \\ & d^k\|Y(i)-i\|+g^k\left\|r^k-i\right\|+2 d^k Q_1 \leqslant\left(d^k \kappa+\right. \\ & \left.g^k+m^k\right) \cdot\left\|u^k-i\right\|+d^k\|Y(i)-i\|+2 d^k Q_1= \\ & \left(1-(1-\kappa) d^k\right)\left\|u^k-i\right\|+(1-\kappa) d^k . \\ & \quad \frac{\|Y(i)-i\|+2 Q_1}{1-\kappa} \end{aligned} $

Further transformation, we can get the following set up:

$ \begin{aligned} \left\|u^{k+1}-i\right\| & =(1-\kappa) \cdot d^k \cdot \frac{\|Y(i)-i\|+2 Q_1}{1-\kappa}+ \\ & \left(1-(1-\kappa) d^k\right)\left\|u^k-i\right\| \leqslant \\ & \max \left\{\frac{\|Y(i)-i\|+2 Q_1}{1-\kappa}, \left\|u^k-i\right\|\right\} \leqslant \cdots \\ & \leqslant \max \left\{\frac{\|Y(i)-i\|+2 Q_1}{1-\kappa}, \right. \\ & \left.\left\|u^{k_0}-i\right\|\right\} \end{aligned} $

Therefore, the succession {u^k} is bounded. In addition, {h^k}, {r^k} and Y(u^k) have the same property.

Part 2

$ \begin{gathered} g^k\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right)\left\|h^k-v^k\right\|^2 \leqslant\left\|u^k-i\right\|^2- \\ \left\|u^{k+1}-i\right\|^2+d^k Q_4 \end{gathered} $

where Q₄>0.

Indeed, we have

$ \begin{array}{l} \left\|u^{k+1}-i\right\|^2 \leqslant m^k\left\|h^k-i\right\|^2+d^k \cdot\left\|Y\left(u^k\right)-i\right\|^2+ \\ g^k\left\|r^k-i\right\|^2 \leqslant m^k\left\|h^k-i\right\|^2+d^k \| Y\left(u^k\right)-Y(i)+ \\ Y(i)-i\left\|^2+g^k\right\| r^k-i\left\|^2 \leqslant m^k\right\| h^k-i \|^2+ \\ d^k\left(\kappa\left\|u^k-i\right\|+\|Y(i)-i\|\right)^2+g^k\left\|r^k-i\right\|^2 \leqslant \\ m^k\left\|h^k-i\right\|^2+d^k\left(\left\|u^k-i\right\|+\|Y(i)-i\|\right)^2+ \\ g^k\left\|r^k-i\right\|^2 \leqslant d^k\left\|u^k-i\right\|^2+d^k\left(\|Y(i)-i\|^2+\right. \\ \left.2\left\|u^k-i\right\| \cdot\|Y(i)-i\|\right)+\\ m^k\left\|h^k-i\right\|^2+g^k\left\|r^k-i\right\|^2 \leqslant \\ d^k\left\|u^k-i\right\|^2+m^k\left\|h^k-i\right\|^2+ \\ g^k\left\|r^k-i\right\|^2+d^k Q_2 \end{array} $

For some Q₂>0.

Combining formulas (21) and (27), the expression below can be derived

$ \begin{gathered} \left\|u^{k+1}-i\right\|^2 \leqslant d^k\left\|u^k-i\right\|^2+\left(1-d^k\right) \cdot \\ \left\|h^k-i\right\|^2-g^k\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right) \cdot \\ \left\|h^k-v^k\right\|^2+d^k Q_2 \end{gathered} $

(28)

Using inequality (21), one sees

$ \begin{aligned} & \left\|h^k-i\right\|^2 \leqslant\left(\left\|u^k-i\right\|+d^k Q_1\right)^2=\left\|u^k-i\right\|^2+ \\ & \quad d^k\left(2 Q_1\left\|u^k-i\right\|+d^k Q_1^2\right) \leqslant \\ & \quad\left\|u^k-i\right\|^2+d^k Q_3 \end{aligned} $

(29)

for some Q₃>0.

From inequalities (28) and (29), we can see

$ \begin{array}{r} \left\|u^{k+1}-i\right\|^2 \leqslant\left(1-d^k\right)\left\|u^k-i\right\|^2+ \\ d^k\left\|u^k-i\right\|^2-g^k\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right) \cdot \\ \left\|h^k-v^k\right\|^2+d^k Q_3+d^k Q_2= \\ \left\|u^k-i\right\|^2-g^k\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right) \cdot \\ \left\|h^k-v^k\right\|^2+d^k Q_3+d^k Q_2 \end{array} $

That is

$ \begin{gathered} g^k\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right)\left\|h^k-v^k\right\|^2 \leqslant\left\|u^k-i\right\|^2- \\ \left\|u^{k+1}-i\right\|^2+d^k Q_4 \end{gathered} $

(30)

define Q₄∶ =Q₂+Q₃.

Part 3

$ \begin{array}{r} \left\|u^{k+1}-i\right\|^2 \leqslant\left(1-(1-\kappa) d^k\right)\left\|u^k-i\right\|^2+ \\ \quad(1-\kappa) d^k \cdot\left[\frac{2\left\langle Y(i)-i, u^{k+1}-i\right\rangle}{1-\kappa}+\right. \\ \left.\frac{3 Q_5}{1-\kappa} \cdot \frac{\overline{\bar{\omega}}^k}{d^k} \cdot\left\|u^k-u^{k-1}\right\|\right], \forall k \geqslant k_0 \end{array} $

where $Q_5:=\sup _{k \in N}\left\{\left\|u^k-i\right\|, \overline{\bar{\omega}}\left\|u^k-u^{k-1}\right\|\right\}>0 $.

Indeed, we have

$ \begin{aligned} & \left\|h^k-i\right\|^2=\left\|\overline{\overline{\boldsymbol{\omega}}}^k\left(u^k-u^{k-1}\right)+u^k-i\right\|^2= \\ & \left(\overline{\overline{\boldsymbol{\omega}}}^k\right)^2\left\|u^k-u^{k-1}\right\|^2+2 \overline{\overline{\boldsymbol{\omega}}}^k\left\langle u^k-i, u^k-u^{k-1}\right\rangle+ \\ & \left\|u^k-i\right\|^2 \leqslant\left(\overline{\overline{{\omega}}}^k\right)^2\left\|u^k-u^{k-1}\right\|^2+2 \overline{\overline{\boldsymbol{\omega}}}^k \| u^k- \\ & i\|\cdot\| u^k-u^{k-1}\|+\| u^k-i \|^2 \end{aligned} $

(31)

Using the formula ‖γ+i‖²≤‖γ‖²+2〈i, γ+i〉, one sees

$ \begin{aligned} & \left\|u^{k+1}-i\right\|^2 \leqslant \| d^k\left(Y\left(u^k\right)-Y(i)\right)+g^k\left(r^k-\right. \\ & i)+m^k\left(h^k-i\right) \|^2+2 d^k\langle Y(i)- \\ & \left.i, u^{k+1}-i\right\rangle \leqslant d^k\left\|Y\left(u^k\right)-Y(i)\right\|^2+ \\ & \quad g^k\left\|r^k-i\right\|^2+m^k\left\|h^k-i\right\|^2+ \\ & \quad 2 d^k\left\langle Y(i)-i, u^{k+1}-i\right\rangle \end{aligned} $

According to the properties of Y in (C2), the following is immediately obtained:

$ \begin{gathered} \left\|u^{k+1}-i\right\|^2 \leqslant d^k \kappa^2\left\|u^k-i\right\|^2+g^k\left\|r^k-i\right\|^2+ \\ m^k\left\|h^k-i\right\|^2+2 d^k\left\langle Y(i)-i, u^{k+1}-i\right\rangle \leqslant \\ \quad\left(1-d^k\right)\left\|h^k-i\right\|^2+d^k \kappa\left\|u^k-i\right\|^2+ \\ 2 d^k\left\langle Y(i)-i, u^{k+1}-i\right\rangle \end{gathered} $

(32)

Substituting (31) to (32), one gets

$ \begin{aligned} \| u^{k+1}- & i\left\|^2=\left(1-(1-\kappa) d^k\right)\right\| u^k-i \|^2+ \\ & (1-\kappa) d^k \cdot \frac{2\left\langle Y(i)-i, u^{k+1}-i\right\rangle}{1-\kappa}+ \\ & \overline{\bar{\omega}}^k\left\|u^k-u^{k-1}\right\|\left(2\left\|u^k-i\right\|+\bar{\omega} \| u^k-\right. \\ & \left.u^{k-1} \|\right) \leqslant\left(1-(1-\kappa) d^k\right)\left\|u^k-i\right\|^2+ \\ & (1-\kappa) d^k\left[\frac{2\left\langle Y(i)-i, u^{k+1}-i\right\rangle}{1-\kappa}+\frac{3 Q_5}{1-\kappa} \cdot\right. \\ & - \\ & \left.\frac{\eta_n}{d_n} \cdot\left\|u^k-u^{k-1}\right\|\right], \forall k \geqslant k_0 \end{aligned} $

where $Q_5:=\sup _{k \in N}\left\{\left\|u^k-i\right\|, \overline{\bar{\omega}}\left\|u^k-u^{k-1}\right\|\right\}>0 $.

Part 4   We prove {u^k} strongly converges to i. In fact, in order to use Lemma 6 to derive the desired result, it needs to satisfy the condition.

For this, consider a subset {‖u^k_j－i‖} of {‖u^k－ i‖} satisfies

$ \liminf\limits _{j \rightarrow \infty}\left(\left\|u^{k_j+1}-i\right\|-\left\|u^{k_j}-i\right\|\right) \geqslant 0 $

Then, we have

$ \begin{aligned} & \liminf\limits _{j \rightarrow \infty}\left(\left\|u^{k_j+1}-i\right\|^2-\left\|u^{k_j}-i\right\|^2\right)= \\ & \quad \liminf\limits _{j \rightarrow \infty}\left[\left(\left\|u^{k_j+1}-i\right\|+\right.\right. \\ & \left.\left\|u^{k_j}-i\right\|\right)\left(\left\|u^{k_j+1}-i\right\|-\right. \\ & \left.\left.\quad\left\|u^{k_j}-i\right\|\right)\right] \geqslant 0 \end{aligned} $

According to Part 2, one sees

$ \begin{aligned} & \limsup\limits _{j \rightarrow \infty} g^{k_j}\left(1-\sigma^2 \frac{\rlap{-}\lambda_k^2}{\rlap{-}\lambda_{k+1}^2}\right)\left\|h^{k_j}-v^{k_j}\right\|^2 \leqslant \\ & \limsup\limits _{j \rightarrow \infty}\left[\left\|u^{k_j}-i\right\|^2-\left\|u^{k_j+1}-i\right\|^2+\right. \\ & \left.d^{k_j} Q_4\right] \leqslant \limsup\limits _{j \rightarrow \infty}\left[\left\|u^{k_j}-i\right\|^2-\right. \\ & \left.\left\|u^{k_j+1}-i\right\|^2\right]+\underset{k \rightarrow \infty}{\rm{lmsup}} d^{k_j} Q_4= \\ & -\liminf\limits _{j \rightarrow \infty}\left[\left\|u^{k_j+1}-i\right\|^2-\right. \\ & \left.\left\|u^{k_j}-i\right\|^2\right] \leqslant 0 \\ & \end{aligned} $

This implies that

$ \lim\limits _{j \rightarrow \infty}\left\|h^{k_j}-v^{k_j}\right\|=0 $

(33)

Because of r^k=v^k－ $ \rlap{-}\lambda_{k}$(Fv^k－Fh^k) and the renewal formula of $ \rlap{-}\lambda_{k}$, we can obtain

$ \left\|r^{k_j}-h^{k_j}\right\|=\left\|v^{k_j}-\rlap{-}\lambda_{k_i}\left(F v^{k_j}-F h^{k_j}\right)-h^{k_j}\right\| \leqslant\\ \left\|v^{k_j}-h^{k_j}\right\|+\rlap{-}\lambda_{k_j}\left\|F v^{k_j}-F h^{k_j}\right\| \leqslant \\ \left\|v^{k_j}-h^{k_j}\right\|+\sigma \frac{\rlap{-}\lambda_{k_j}}{\rlap{-}\lambda_{k_j+1}}\left\|v^{k_j}-h^{k_j}\right\| \leqslant \\ \left(\sigma \frac{\rlap{-}\lambda_{k_j}}{\rlap{-}\lambda_{k_j+1}}+1\right)\left\|v^{k_j}-h^{k_j}\right\| \rightarrow 0 $

(34)

According to (C5) and the definition of u^k, combining inequality (34) we get

$ \begin{array}{r} \left\|u^{k_j+1}-\rlap{-}\lambda_{k_j}\right\|=\| d^{k_j} Y\left(u^{k_j}\right)+\left(g^{k_j}-1\right) r^{k_j}+ \\ m^{k_j} h^{k_j}\|=\| d^{k_j}\left(Y\left(u^{k_j}\right)-r^{k_j}\right)+m^{k_j}\left(h^{k_j}-\right. \\ \left.r^{k_j}\right)\left\|\leqslant d^{k_j}\right\| Y\left(u^{k_j}\right)-r^{k_j}\left\|+m^{k_j}\right\| h^{k_j}- \\ r^{k_j} \| \rightarrow 0 \end{array} $

(35)

By the notion of h^k, we also get

$ \begin{gathered} \left\|u^{k_j}-h^{k_j}\right\|=\overline{\overline{\omega}}^{k_j}\left\|u^{k_j}-u^{k_j-1}\right\|= \\ d^{k_j} \cdot \frac{\overline{\overline{\omega}}^{k_j}}{d^{k_j}}\left\|u^{k_j}-u^{k_j-1}\right\| \rightarrow 0 \end{gathered} $

(36)

Using Eqs.(34), (35) and (36), one derives that

$ \begin{aligned} & \left\|u^{k_j+1}-u^{k_j}\right\| \leqslant\left\|u^{k_j+1}-r^{k_j}\right\|+ \\ & \left\|r^{k_j}-h^{k_j}\right\|+\left\|h^{k_j}-u^{k_j}\right\| \rightarrow 0 \end{aligned} $

(37)

According to {u^k_j} is bounded, now we assume u^k_ji⇀ι, {u^k_ji} of {u^k_j}.

One sees

$ \begin{gathered} \limsup\limits _{j \rightarrow \infty}\left\langle Y(i)-i, u^{k_j}-i\right\rangle= \\ \lim\limits _{i \rightarrow \infty}\left\langle Y(i)-i, u^{k_{j_i}}-i\right\rangle= \\ \langle Y(i)-i, \iota-i\rangle \end{gathered} $

(38)

Thanks to Eq.(36), we have

$ h^{k_j} \rightharpoonup \iota, j \rightarrow \infty $

Using Lemma 8 and Eq.(33), one obtains ι∈Sol(C, F).

Due to the preceding description of i and ι∈Sol(C, F), one derives that

$ \langle Y(i)-i, \iota-i\rangle \leqslant 0 $

By using inequality (37) and the inequality 〈Y(i)－i, ι－i〉≤0 we have

$ \begin{aligned} & \limsup\limits _{k \rightarrow \infty}\left\langle Y(i)-i, u^{k_j+1}-i\right\rangle \leqslant \\ & \quad \limsup\limits _{k \rightarrow \infty}\left\langle Y(i)-i, u^{k_j}-i\right\rangle \leqslant 0 \end{aligned} $

(39)

Note {‖u^k－i‖} above is the {s^k} in Lemma 6, and obviously {s^k} is nonnegative.

Thus, using inequality (39), part 3, and

$ \begin{gathered} \liminf\limits _{j \rightarrow \infty}\left(\left\|u^{k_j+1}-i\right\|^2-\left\|u^{k_j}-i\right\|^2\right) \geqslant 0 \\ \lim\limits _{k \rightarrow \infty} \frac{\omega^k}{d^k}\left\|u^k-u^{k-1}\right\|=0 \end{gathered} $

the condition of Lemma 6 is satisfied. So, we obtain $ \lim \limits_{k \rightarrow \infty}\left\|u^k-i\right\|=0 $ from Lemma 6.

At this point, we have completed the proof of strong convergence.

3 Numerical Experiments

We designed several numerical experiments and comparisons in order to illustrate the validity of the two algorithms in part three. First, we illustrate the effectiveness of Algorithm 1 using Problems 1 and Problem 2 in finite-dimensional space. Then for Algorithm 2, we discuss the validity of it using Problem 3 in infinite-dimensional space. For convenience, in the final result, ite. and Time represent the number and time of iterations, respectively.In all the numerical experiments, we choose u¹=u⁰ and ‖u^k－v^k‖≤ε as the stopping condition for the iteration, where ε is the iteration accuracy.

3.1 Experiments for Algorithm 1

Problem 1   First, let H=R, C=[－5, 5] and F: H→H is defined as Fu=sinu. So obviously F is monotone and consequently pseudomonotone on C. In this problem, the Algorithm 1 is compared with Algorithm 3.1 in Ref.[18] and Algorithm 3.3 in Ref.[13], and the numerical results are obtained. Here we denote the latter two algorithms as Algorithm A and Algorithm B, respectively. For Algorithm A, we take μ=0.1, τ₁=0.09, f(u)=0.15u, Uu=(u/2)sinu, α_k=(k³+1)^－1 and β_k=(k+1)^－1. For Algorithm B, we choose σ=0.9, γ=0.9, θ_k=0.2 and α_k=(k³+1)^－1.

We give ε=10^－3 and different choices of start point.

Problem 2   The second problem is discussed in Refs. [14] and ^[15]. Let H=R² and F: R²→R² be represented as

$ \begin{aligned} F(u, \iota) & =\left[\left(u^2+(\iota-1)^2\right)(1+\iota)-\right. \\ & \left.u(\iota-1)^2-u^3\right] \end{aligned} $

Now let the feasible set C be R₂⁺, where P_C=max(0, u). This shows that F is not monotone but pseudomonotone on C. In this problem, the Algorithm 1 is compared with Algorithm 3.1 in Ref.[5] and Algorithm 2 in Ref.[14], and the numerical results are obtained. Here we denote the latter two algorithms as Algorithm C and Algorithm D, respectively. For Algorithm C, set γ=0.2, λ₁=0.09 and f(u)=0.15u. For Algorithm D, we choose μ=0.2, γ=3 and l=0.7. For the test, we choose ε=10^－3 and consider the different choices of initial point.

In our Algorithm 1, we select the appropriate parameters as shown in Table 1, and the comparison results are presented in Table 2 and Table 3, respectively.

3.2 Experiments for Algorithm 2

Problem 3   In the last experiment, we consider problem in a given infinite dimensional space, which was used for numerical experiment in Refs. [14] and ^[29]. Given that H=L²([0, 1]) with the transvection $ \langle\eta, \xi\rangle=\int_0^1 \eta(y) \xi(y) \mathrm{d} y $ and the induced norm $\|\xi\|_2=\left(\int_0^1\left(|\xi(y)|^2\right) \mathrm{d} y\right)^{\frac{1}{2}} $. Assume F: H→H is described as

$ (F u)(y)=e^{-\|u\|_2} \int_0^y u(\iota) {\mathrm{d}} \iota, y \in[0, 1] $

Now we consider set C={u∈H: ‖u‖₂≤2}. we can see that F is pseudomonotone but not monotone on C. In this problem, we compare Algorithm 1 with Algorithm 3.1 in Ref.[18], and still mark Algorithm 3.1 in Ref.[18] as Algorithm A. For Algorithm A, we take μ=0.9, τ₁=0.09, and f(x)=0.15x, α_n=(n³+1)^－1, β_n=(n+1)^－1. Let (Tu)(y)= $ \int_0^1 y u(\iota) {\mathrm{d}} \iota $, y∈[0, 1], by our definition, T is nonexpansive and consequently quasi-nonexpansive on C.For all test, let ε=10^－2 and consider the different choices of initial point. We select (Yu)(y)= $ \int_0^1 y u(\iota) {\mathrm{d}} \iota $, y∈[0, 1] and the appropriate parameters as shown in Table 4.

Finally, we put the resulting data in Table 5.

4 Conclusions

Based on the previous classical algorithms, we propose two new extragradient algorithms in Hilbert space. In particular, the two algorithms adopt non-monotonic step size rule.So, we do not need the Lipschitz constant to get its convergence. In addition, under certain conditions, we prove that the Algorithm 1 has R-linear convergence rate. Finally, we prove that our two new algorithms are very efficient by comparing the numerical results with those of other algorithms.