1 Introduction

We focus on the following classical variational inequality (VI) Fichera (1963, 1964) which consists in finding a point \(x^*\in C\) such that

$$\begin{aligned} \langle Fx^*,x-x^*\rangle \ge 0 \quad \forall x\in C, \end{aligned}$$
(1)

where C is a nonempty closed convex subset in a real Hilbert space H,  and is a single-valued mapping \(F: C\rightarrow H\). As commonly done, we denote by \(\mathrm{Sol}(C,F)\) the solution set of VI (1). Variational inequalities are fundamental in a broad range of mathematical and applied sciences; the theoretical and algorithmic foundations as well as the applications of variational inequalities have been extensively studied in the literature and continue to attract intensive research. For the current state of the art in finite-dimensional setting, see for instance Facchinei and Pang (2003) and the extensive list of references therein.

Many authors have proposed and analyzed several iterative methods for solving the variational inequality (1). The simplest one is the following projection method, which can be seen an extension of the projected gradient method introduced for solving optimization problems. However, the convergence of this method is quite restrictive and assume that F is L-Lipschitz continuous and \(\alpha \)-strongly monotone.

In a way to weaken the convergence assumptions of the gradient method, Korpelevich Korpelevich (1976) (also independently by Antipin (1976)) proposed a double projection method known as the extragradient method in Euclidean space when F is monotone and L-Lipschitz continuous. In recent years, the extragradient method was further extended to infinite-dimensional spaces in various ways, see, e.g., Censor et al. (2011a, 2011b, 2011c, 2012), Gibali (2015), Khobotov (1987), Malitsky (2015), Malitsky and Semenov (2015), Marcotte (1991), Solodov and Svaiter (1999), Gibali and Shehu (2019), Shehu et al. (2020), Shehu and Iyiola (2020) and the references therein.

However, in the case the inequality variational mapping is not Lipschitz continuous or its Lipschitz constant is difficult to compute/approximate, Korpelevich’s method fails since the step-size depends on this. Hence, the method extragradient method cannot be applied. Motivated by this problem it would be of interest to propose an iterative method for solving VI (1) with the underline cost function F is uniformly continuous on bounded subsets of C but not Lipschitz continuous on C. In Iusem (1994), Iusem proposed the algorithm so that it can overcome this barrier. Recently, Cai et al. (2021) introduced the modification of the subgradient extragradient method which it uses a different Armijo-type line-search and the mapping F is only assumed to be pseudo-monotone on C in the sense of Karamardian (1976) and Vuong (2018). In this work, motivated by the works of Cai et al., we propose a new extragradient method for solving variational inequality problems of pseudo-monotone and non-Lipschitz continuous operator in real Hilbert spaces. We present strong convergence and convergence rate of the proposed algorithm under suitable conditions. Our results improve some recent contributions in the literature.

The paper is organized as follows. We first recall some basic definitions and results in Sect. 2. Our algorithms are presented and analyzed in Sect. 3. In Sect. 4, we present some numerical experiments to demonstrate the performance of the proposed algorithm.

2 Preliminaries

    Let H be a real Hilbert space and C be a nonempty, closed and convex subset of H. The weak convergence of \(\{x_n\}_{n=1}^{\infty }\) to x is denoted by \(x_{n}\rightharpoonup x\) as \(n\rightarrow \infty \), while the strong convergence of \(\{x_n\}_{n=1}^{\infty }\) to x is written as \(x_n\rightarrow x\) as \(n\rightarrow \infty .\) For each \(x,y,z\in H\), we have

$$\begin{aligned} \Vert (1-\beta )x+\beta y\Vert ^2= & {} (1-\beta )\Vert x\Vert ^2+\beta \Vert y\Vert ^2-\beta (1-\beta )\Vert x-y\Vert ^2\quad \forall \beta \in \mathbb {R}. \end{aligned}$$
(2)
$$\begin{aligned} \Vert x+y\Vert ^2\le & {} \Vert x\Vert ^2+2\langle y,x+y\rangle . \end{aligned}$$
(3)
$$\begin{aligned} \Vert \alpha x+\beta y+\gamma z\Vert ^2= & {} \alpha \Vert x\Vert ^2 + \beta \Vert y\Vert ^2 + \gamma \Vert z\Vert ^2- \alpha \beta \Vert x-y\Vert ^2 \nonumber \\&- \alpha \gamma \Vert x-z\Vert ^2 - \beta \gamma \Vert y-z\Vert ^2 \end{aligned}$$
(4)

for all \(\alpha , \beta , \gamma \in [0; 1]\) with \(\alpha + \beta + \gamma = 1.\)

Definition 2.1

Let \(T:H\rightarrow H\) be an operator.

  1. 1.

    The operator T is called L-Lipschitz continuous with \(L>0\) if

    $$\begin{aligned} \Vert Tx-Ty\Vert \le L \Vert x-y\Vert \quad \forall x,y \in H. \end{aligned}$$
  2. 2.

    The operator T is called monotone if

    $$\begin{aligned} \langle Tx-Ty,x-y \rangle \ge 0 \quad \forall x,y \in H. \end{aligned}$$
  3. 3.

    The operator T is called pseudo-monotone if

    $$\begin{aligned} \langle Tx,y-x \rangle \ge 0 \Longrightarrow \langle Ty,y-x \rangle \ge 0 \quad \forall x,y \in H. \end{aligned}$$
  4. 4.

    The operator T is called \(\delta \)-strongly pseudo-monotone if there is \(\delta > 0\) such that

    $$\begin{aligned} \langle Tx,y-x \rangle \ge 0 \Longrightarrow \langle Ty,y-x \rangle \ge \delta \Vert y - x\Vert ^2. \end{aligned}$$
  5. 5.

    The operator T is called \(\alpha \)-strongly monotone if there exists a constant \(\alpha >0\) such that

    $$\begin{aligned} \langle Tx-Ty,x-y\rangle \ge \alpha \Vert x-y\Vert ^2 \quad \forall x,y\in H. \end{aligned}$$
  6. 6.

    The operator T is called sequentially weakly continuous if for each sequence \(\{x_n\}\) we have: \(\{x_n\}\) converges weakly to x implies \({Tx_n}\) converges weakly to Tx.

It is easy to see that very monotone operator is pseudo-monotone but the converse is not true.

For every point \(x\in H\), there exists a unique nearest point in C, denoted by \(P_Cx\) such that \(\Vert x-P_Cx\Vert \le \Vert x-y\Vert \ \forall y\in C\). \(P_C\) is called the metric projection of H onto C. It is known that \(P_C\) is nonexpansive.

Lemma 2.1

(Goebel and Reich 1984) Let C be a nonempty closed convex subset of a real Hilbert space H. Given \(x\in H\) and \(z\in C\). Then, \(z=P_Cx\Longleftrightarrow \langle x-z,z-y\rangle \ge 0 \ \ \forall y\in C.\)

Lemma 2.2

(Goebel and Reich 1984) Let C be a closed and convex subset in a real Hilbert space H,  \(x\in H\). Then,

  1. (i)

    \(\Vert P_Cx-P_Cy\Vert ^2\le \langle P_C x-P_C y,x-y\rangle \ \forall y\in C\);

  2. (ii)

    \(\Vert P_C x-y\Vert ^2\le \Vert x-y\Vert ^2-\Vert x-P_Cx\Vert ^2 \ \forall y\in C;\)

  3. (iii)

    \(\langle (I-P_C)x-(I-P_C)y,x-y\rangle \ge \Vert (I-P_C)x-(I-P_C)y\Vert ^2 \ \forall y\in C.\)

For properties of the metric projection, the interested reader could be referred to Goebel and Reich (1984), Section 3.

The following Lemmas are useful for the convergence of our proposed methods.

Lemma 2.3

(Denisov et al. 2015) For \(x\in H\) and \(\alpha \ge \beta >0\), the following inequalities hold:

$$\begin{aligned} \dfrac{\Vert x-P_{C}(x-\alpha Fx)\Vert }{\alpha }\le \dfrac{\Vert x-P_C(x-\beta Fx)\Vert }{\beta },\\ \Vert x-P_{C}(x-\beta Fx)\Vert \le \Vert x-P_C(x-\alpha Fx)\Vert , \end{aligned}$$

where \(F:C\rightarrow H\) is a nonlinear mapping.

Lemma 2.4

(Iusem and Nasri 2011) Let \(H_1\) and \(H_2\) be two real Hilbert spaces. Suppose \(F : H_1 \rightarrow H_2\) is uniformly continuous on bounded subsets of \(H_1\) and M is a bounded subset of \(H_1\). Then, F(M) (the image of M under F) is bounded.

Lemma 2.5

(Cottle and Yao 1992) Consider the problem \(\mathrm{Sol}(C, F)\) with C being a nonempty, closed, convex subset of a real Hilbert space H and \(F : C \rightarrow H\) being pseudo-monotone and continuous. Then, \(x^*\) is a solution of \(\mathrm{Sol}(C, F)\) if and only if

$$\begin{aligned} \langle Fx, x - x^*\rangle \ge 0 \quad \forall x \in C. \end{aligned}$$

Definition 2.2

(Ortega and Rheinboldt 1993) Let \(\{x_n\}\) be a sequence in H.

  1. (i)

    \(\{x_n\}\) is said to converge R-linearly to \(x^*\) with rate \(\rho \in [0, 1)\) if there is a constant \(c>0\) such that

    $$\begin{aligned} \Vert x_n-x^*\Vert \le c\rho ^n \quad \forall n\in \mathbb {N}. \end{aligned}$$
  2. (ii)

    \(\{x_n\}\) is said to converge Q-linearly to \(x^*\) with rate \(\rho \in [0, 1)\) if

    $$\begin{aligned} \Vert x_{n+1}-x^*\Vert \le \rho \Vert x_n-x^*\Vert \quad \forall n\in \mathbb {N}. \end{aligned}$$

Lemma 2.6

(Saejung and Yotkaew 2012) Let \(\{a_n\}\) be a sequence of nonnegative real numbers, \(\{\gamma _n\}\) be a sequence of real numbers in (0, 1) with \(\sum _{n=1}^\infty \gamma _n=\infty \) and \(\{b_n\}\) be a sequence of real numbers. Assume that

$$\begin{aligned} a_{n+1}\le (1-\gamma _n)a_n+\gamma _n b_n,\, \, \, \forall n\ge 1. \end{aligned}$$

If \(\limsup _{k\rightarrow \infty } b_{n_k} \le 0\) for every subsequence \(\{a_{n_k}\}\) of \(\{a_n\}\) satisfying

$$\begin{aligned} \liminf _{k\rightarrow \infty }(a_{n_k+1}-a_{n_k})\ge 0, \end{aligned}$$

then \(\lim _{n\rightarrow \infty }{a_n} = 0\).

3 Main results

In this section, we present the strong convergence and convergence rate of the sequence generated by the proposed method under the suitable assumptions. The proposed algorithm is of the form:

figure a

Remark 3.1

(5) can rewritten as follows

$$\begin{aligned} \gamma l^m \langle Fy_n-Fv_n,y_n-w_n\rangle \le \dfrac{\mu }{2}(\Vert v_n-y_n\Vert ^2+\Vert y_n-w_n\Vert ^2). \end{aligned}$$

3.1 Strong convergence

To prove the strong convergence, we need the following conditions:

Condition 1

The feasible set C of the VI (1) is a nonempty, closed, and convex subset of the real Hilbert space H.

Condition 2

The mapping \(F:C\rightarrow H\) is a pseudo-monotone, uniformly continuous on bounded subsets of C.

Condition 3

The solution set of the VI (1) is nonempty, that is \(\mathrm{Sol}(C,F)\ne \emptyset \).

We start the algorithm’s convergence analysis by proving that (5) terminates after a finite number of steps.

Lemma 3.7

Assume that the mapping \(F:C\rightarrow H\) is uniformly continuous on bounded subsets of C. The Armijo line-search rule (5) is well defined. In addition, we have \(\tau _n\le \gamma .\)

Proof

If \(v_n\in \mathrm{Sol}(C,F)\), then \(v_n=P_{C}(v_n-\gamma Fv_n)\) and \(m_n=0\). We consider the situation \(v_n\notin \mathrm{Sol}(C,F)\) and assume the contrary that for all m, we have

$$\begin{aligned}&\gamma l^{m} \langle F P_C(v_n-\gamma l^{m} Fv_n)-Fv_n,P_C(v_n-\gamma l^{m} Fv_n)-P_C(v_n-\gamma l^{m}FP_C(v_n-\gamma l^{m}Fv_n)\rangle \nonumber \\&\quad > \dfrac{\mu }{2}(\Vert v_n-P_C(v_n-\gamma l^{m} Fv_n)\Vert ^2+\Vert P_C(v_n-\gamma l^{m} Fv_n)\nonumber \\&\qquad -P_C(v_n-\gamma l^{m}P_C(v_n-\gamma l^{m} Fv_n))\Vert ^2). \end{aligned}$$
(6)

By Cauchy–Schwartz inequality, we have

$$\begin{aligned}&\gamma l^{m} \langle F P_C(v_n-\gamma l^{m} Fv_n)-Fv_n,P_C(v_n-\gamma l^{m} Fv_n)-P_C(v_n-\gamma l^{m}FP_C(v_n-\gamma l^{m}Fv_n)\rangle \nonumber \\&\quad \le \gamma l^m \Vert F P_C(v_n-\gamma l^{m} Fv_n)-Fv_n\Vert \Vert P_C(v_n-\gamma l^{m} Fv_n)\nonumber \\&\qquad -P_C(v_n-\gamma l^{m}FP_C(v_n-\gamma l^{m}Fv_n)\Vert , \end{aligned}$$
(7)

and

$$\begin{aligned}&(\Vert v_n-P_C(v_n-\gamma l^{m} Fv_n)\Vert ^2+\Vert P_C(v_n-\gamma l^{m} Fv_n)-P_C(v_n-\gamma l^{m}P_C(v_n-\gamma l^{m} Fv_n))\Vert ^2)\nonumber \\&\quad \ge 2\Vert v_n-P_C(v_n-\gamma l^{m} Fv_n)\Vert \Vert P_C(v_n-\gamma l^{m} Fv_n)-P_C(v_n-\gamma l^{m}P_C(v_n-\gamma l^{m} Fv_n))\Vert . \end{aligned}$$
(8)

Combining (6) and (7) and (8), we find

$$\begin{aligned} \gamma l^m\Vert F P_C(v_n-\gamma l^{m} Fv_n)-Fv_n\Vert > \mu \Vert P_C(v_n-\gamma l^m Fv_n)-v_n\Vert . \end{aligned}$$
(9)

This implies that

$$\begin{aligned} \Vert Fv_n-FP_C(v_n-\gamma l^m Fv_n)\Vert > \mu \dfrac{\Vert v_n-P_C(v_n-\gamma l^m Fv_n)\Vert }{\gamma l^m}. \end{aligned}$$
(10)

Since \(v_n\in C\) for all n and \(P_C\) is continuous, we have \(\lim _{m\rightarrow \infty }\Vert v_n-P_C(v_n-\gamma l^m Fv_n)\Vert =0.\) From the uniform continuity of the operator F on bounded subsets of C, it implies that

$$\begin{aligned} \lim _{m\rightarrow \infty }\Vert Fv_n-FP_C(v_n-\gamma l^m Fv_n)\Vert =0. \end{aligned}$$
(11)

Combining (10) and (11), we get

$$\begin{aligned} \lim _{m\rightarrow \infty }\dfrac{\Vert v_n-P_C(v_n-\gamma l^m Fv_n)\Vert }{\gamma l^m}=0. \end{aligned}$$
(12)

Assume that \(z_m=P_C(v_n-\gamma l^m Fv_n)\), we have

$$\begin{aligned} \langle z_m-v_n+\gamma l^m Fv_n,x-z_m\rangle \ge 0 \quad \forall x\in C. \end{aligned}$$

This implies that

$$\begin{aligned} \left\langle \dfrac{ z_m-v_n}{\gamma l^m},x-z_m\right\rangle + \langle Fv_n,x-z_m\rangle \ge 0 \quad \forall x\in C. \end{aligned}$$
(13)

Taking the limit \(m\rightarrow \infty \) in (13) and using (12), we obtain

$$\begin{aligned} \langle Fv_n,x-v_n\rangle \ge 0 \quad \forall x\in C, \end{aligned}$$

which implies that \(v_n\in \mathrm{Sol}(C,F)\) this is a contraction. \(\square \)

Lemma 3.8

Assume that Conditions 1–3 hold and let \(\{v_n\}\) be any sequence generated by Algorithm 3.1. Moreover, we assume that the mapping \(F:C\rightarrow H\) satisfies the following condition:

$$\begin{aligned} \text { whenever } \{x_n\} \subset C, x_n \rightharpoonup z, \text { one has } \Vert Fz\Vert \le \liminf _{n\rightarrow \infty }\Vert Fx_n\Vert . \end{aligned}$$
(14)

If there exists a subsequence \(\{v_{n_k}\}\) of \(\{v_n\}\) such that \(\{v_{n_k}\}\) converges weakly to \(z\in C\) and \(\lim _{k\rightarrow \infty }\Vert v_{n_k}-y_{n_k}\Vert =0\) then \(z\in \mathrm{Sol}(C,F).\)

Proof

We have \(y_{n_k}=P_C(v_{n_k}-\tau _{n_k}Fv_{n_k})\), thus

$$\begin{aligned} \langle v_{n_k}-\tau _{n_k}Fv_{n_k}-y_{n_k},x-y_{n_k}\rangle \le 0 \quad \forall x\in C. \end{aligned}$$

or equivalently

$$\begin{aligned} \dfrac{1}{\tau _{n_k}}\langle v_{n_k}-y_{n_k},x-y_{n_k}\rangle \le \langle Fv_{n_k},x-y_{n_k}\rangle \quad \forall x\in C. \end{aligned}$$

This implies that

$$\begin{aligned} \dfrac{1}{\tau _{n_k}}\langle v_{n_k}-y_{n_k},x-y_{n_k}\rangle +\langle Fv_{n_k},y_{n_k}-v_{n_k}\rangle \le \langle Fv_{n_k},x-v_{n_k}\rangle \quad \forall x\in C. \end{aligned}$$
(15)

Now, we show that

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fv_{n_k},x-v_{n_k}\rangle \ge 0. \end{aligned}$$
(16)

For showing this, we consider two possible cases. Suppose first that \(\liminf _{k\rightarrow \infty }\tau _{n_k}>0\). We have \(\{v_{n_k}\}\) is a bounded sequence, F is uniformly continuous on bounded subsets of C. By Lemma 2.4, we get that \(\{Fv_{n_k}\}\) is bounded. Taking \(k\rightarrow \infty \) in (15) since \(\Vert v_{n_k}-y_{n_k}\Vert \rightarrow 0\), we get

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fv_{n_k},x-v_{n_k}\rangle \ge 0. \end{aligned}$$

Now, we assume that \(\liminf _{k\rightarrow \infty }\tau _{n_k}=0\). Assume \(z_{n_k}=P_{C}(v_{n_k}-\tau _{n_k}.l^{-1}Fv_{n_k})\), we have \(\tau _{n_k}l^{-1}>\tau _{n_k}\). Applying Lemma 2.3, we obtain

$$\begin{aligned} \Vert v_{n_k}-z_{n_k}\Vert \le \dfrac{1}{l}\Vert v_{n_k}-y_{n_k}\Vert \rightarrow 0 \quad \text { as } k\rightarrow \infty . \end{aligned}$$

Consequently, \(z_{n_k}\rightharpoonup z\in C\), this implies that \(\{z_{n_k}\}\) is bounded, which the uniformly continuity of the operator F on bounded subsets of C, it follows that

$$\begin{aligned} \Vert Fv_{n_k}-Fz_{n_k}\Vert \rightarrow 0\quad \text { as } k\rightarrow \infty . \end{aligned}$$
(17)

By the Armijo line-search rule (5) and the proof is similar to the inequality (9), we have

$$\begin{aligned} \tau _{n_k}.l^{-1}\Vert Fv_{n_k}-FP_C(v_{n_k}-\tau _{n_k}l^{-1} Fv_{n_k})\Vert > \mu \Vert v_{n_k}-P_C(v_{n_k}-\tau _{n_k}l^{-1} Fv_{n_k})\Vert . \end{aligned}$$

That is,

$$\begin{aligned} \dfrac{1}{\mu }\Vert Fv_{n_k}-Fz_{n_k}\Vert >\dfrac{\Vert v_{n_k}-z_{n_k}\Vert }{\tau _{n_k}l^{-1}}. \end{aligned}$$
(18)

Combining (17) and (18), we obtain

$$\begin{aligned} \lim _{k\rightarrow \infty }\dfrac{\Vert v_{n_k}-z_{n_k}\Vert }{\tau _{n_k}l^{-1}}=0. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \langle v_{n_k}-\tau _{n_k} l^{-1}Fv_{n_k}-z_{n_k},x-z_{n_k}\rangle \le 0 \quad \forall x\in C. \end{aligned}$$

This implies that

$$\begin{aligned} \dfrac{1}{\tau _{n_k}l^{-1}}\langle v_{n_k}-z_{n_k},x-z_{n_k}\rangle +\langle Fv_{n_k},z_{n_k}-v_{n_k}\rangle \le \langle Fv_{n_k},x-v_{n_k}\rangle \quad \forall x\in C. \end{aligned}$$
(19)

Taking the limit \(k\rightarrow \infty \) in (19), we get

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fv_{n_k},x-v_{n_k}\rangle \ge 0. \end{aligned}$$

Therefore, the inequality (16) is proved. Next, we show that \(z\in \mathrm{Sol}(C,F)\).

Now, we choose a sequence \(\{\varepsilon _k\}\) of positive numbers decreasing and tending to 0. For each k, we denote by \(N_k\) the smallest positive integer such that

$$\begin{aligned} \langle Fv_{n_j},x-v_{n_j}\rangle +\varepsilon _k \ge 0 \quad \forall j\ge N_k, \end{aligned}$$
(20)

where the existence of \(N_k\) follows from (16). Since \(\{ \varepsilon _k\}\) is decreasing, it is easy to see that the sequence \(\{N_k\}\) is increasing. Furthermore, for each k, since \(\{v_{N_k}\}\subset C\), we have \(Fv_{N_k}\ne 0\) and, setting

$$\begin{aligned} u_{N_k} = \dfrac{Fv_{N_k}}{\Vert Fv_{N_k}\Vert ^2 }, \end{aligned}$$

we have \(\langle Fv_{N_k}, u_{N_k}\rangle = 1\) for each k. Now, we can deduce from (20) that for each k

$$\begin{aligned} \langle Fv_{N_k}, x+\varepsilon _k u_{N_k}-v_{N_k}\rangle \ge 0. \end{aligned}$$

Since the fact that F is pseudo-monotone, we get

$$\begin{aligned} \langle F(x+\varepsilon _k u_{N_k}), x+\varepsilon _k u_{N_k}-v_{N_k}\rangle \ge 0. \end{aligned}$$

This implies that

$$\begin{aligned} \langle Fx, x-v_{N_k}\rangle \ge \langle Fx-F(x+\varepsilon _k u_{N_k}), x+\varepsilon _k u_{N_k}-v_{N_k} \rangle -\varepsilon _k \langle Fx, u_{N_k}\rangle . \end{aligned}$$
(21)

Now, we show that \(\lim _{k\rightarrow \infty }\varepsilon _k u_{N_k}=0\). Indeed, we have \(v_{n_k}\rightharpoonup z \text { as } k \rightarrow \infty \). Since F satisfies the condition (14). We have

$$\begin{aligned} 0 < \Vert Fz\Vert \le \liminf _{k\rightarrow \infty }\Vert Fv_{n_k}\Vert \text { (note that } Fz \ne 0\text { otherwise, } z\text { is a solution)}. \end{aligned}$$

Since \(\{v_{N_k}\}\subset \{v_{n_k}\}\) and \(\varepsilon _k \rightarrow 0\) as \(k \rightarrow \infty \), we obtain

$$\begin{aligned} 0 \le \limsup _{k\rightarrow \infty } \Vert \varepsilon _k u_{N_k} \Vert = \limsup _{k\rightarrow \infty } \left( \dfrac{\varepsilon _k}{\Vert Fv_{n_k}\Vert }\right) \le \dfrac{\limsup _{k\rightarrow \infty }\varepsilon _k }{\liminf _{k\rightarrow \infty }\Vert Fv_{n_k}\Vert }=0, \end{aligned}$$

which implies that \(\lim _{k\rightarrow \infty } \varepsilon _k u_{N_k} = 0.\)

Now, letting \(k\rightarrow \infty \), then the right hand side of (21) tends to zero by F is uniformly continuous, \(\{v_{N_k}\}, \{u_{N_k}\}\) are bounded and \(\lim _{k\rightarrow \infty }\varepsilon _k u_{N_k}=0\). Thus, we get

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fx,x-v_{N_k}\rangle \ge 0. \end{aligned}$$

Hence, for all \(x\in C\), we have

$$\begin{aligned} \langle Fx, x-z\rangle =\lim _{k\rightarrow \infty } \langle Fx, x-v_{N_k}\rangle =\liminf _{k\rightarrow \infty } \langle Fx, x-v_{N_k}\rangle \ge 0. \end{aligned}$$

By Lemma 2.5, we obtain \(z\in \mathrm{Sol}(C,F)\) and the proof is complete. \(\square \)

Remark 3.2

We should here emphasize that if F is sequentially weakly continuous, then F satisfies condition (14) but the converse is not true, hence condition (14) is strictly weaker than the sequential weak continuity of the operator F. This is the assumption which has frequently been used in recent articles on pseudo-monotone variational inequality problems, e.g., Vuong (2018), Thong et al. (2020) and Vuong and Shehu (2019). Moreover, when F is monotone, we can remove condition (14) (Denisov et al. 2015; Vuong 2018).

Theorem 3.1

Assume that Conditions 1–3 hold. Moreover, we assume that the mapping \(F:C\rightarrow H\) satisfies the following condition:

$$\begin{aligned} \text { whenever } \{x_n\} \subset C, x_n \rightharpoonup z, \text { one has } \Vert Fz\Vert \le \liminf _{n\rightarrow \infty }\Vert Fx_n\Vert . \end{aligned}$$

Then, any sequence \(\{v_n\}\) generated by Algorithm 3.1 converges strongly to an element \(u^*\) \(\mathrm{Sol}(C,F)\), where

$$\begin{aligned} \Vert u^*\Vert =\min \{ \Vert z\Vert : z\in \mathrm{Sol}(C,F)\}. \end{aligned}$$

Proof

Let \(u^*\) be a certain solution of (VIP). We divide the proof into two claims.

Claim 1.

$$\begin{aligned} \Vert w_n-u^*\Vert ^2\le \Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2-(1-\mu )\Vert w_n-y_n\Vert ^2. \end{aligned}$$
(22)

Indeed, from \(u^*\in \mathrm {Sol}(C,F)\subset C\), we have

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&=\Vert P_C(v_n-\tau _n Fy_n)-P_Cu^*\Vert ^2\le \langle w_n-u^*,v_n-\tau _n Fy_n-u^*\rangle \\&=\dfrac{1}{2}\Vert w_n-u^*\Vert ^2+\dfrac{1}{2}\Vert v_n-\tau _n Fy_n-u^*\Vert ^2-\dfrac{1}{2}\Vert w_n-v_n+\tau _n Fy_n\Vert ^2\\&= \dfrac{1}{2}\Vert w_n-u^*\Vert ^2+\dfrac{1}{2}\Vert v_n-u^*\Vert ^2+\dfrac{1}{2}\tau ^2_n \Vert Fy_n\Vert ^2- \langle v_n-u^*,\tau _n Fy_n \rangle \\&\quad - \dfrac{1}{2}\Vert w_n-v_n\Vert ^2-\dfrac{1}{2}\tau ^2_n \Vert Fy_n\Vert ^2-\langle w_n-v_n, \tau _n Fy_n\rangle \\&=\dfrac{1}{2}\Vert w_n-u^*\Vert ^2+\dfrac{1}{2}\Vert v_n-u^*\Vert ^2 -\dfrac{1}{2}\Vert w_n-v_n\Vert ^2-\langle w_n-u^*, \tau _n Fy_n\rangle . \end{aligned}$$

This implies that

$$\begin{aligned} \Vert w_n-u^*\Vert ^2\le \Vert v_n-u^*\Vert ^2 -\Vert w_n-v_n\Vert ^2-2\langle w_n-u^*, \tau _n Fy_n\rangle . \end{aligned}$$
(23)

Since \(u^*\) is a solution of the problem (VI), we have \(\langle Fu^*,x-u^*\rangle \ge 0\) for all \(x\in C\). By the pseudo-monotonicity of F on C, we have \(\langle Fx,x-u^*\rangle \ge 0\) for all \(x\in C\). Taking \(x:=y_n\in C\), we get

$$\begin{aligned} \langle Fy_n,u^*-y_n\rangle \le 0. \end{aligned}$$

Thus, we have

$$\begin{aligned} \langle Fy_n,u^*-w_n\rangle =&\langle Fy_n,u^*-y_{n}\rangle +\langle Fy_n,y_n-w_n\rangle \le \langle Fy_n,y_n-w_n\rangle . \end{aligned}$$
(24)

From (23) and (24), we obtain

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-\Vert w_n-v_n\Vert ^2+2\tau _n\langle Fy_n,y_n-w_n\rangle \nonumber \\&=\Vert v_n-u^*\Vert ^2-\Vert w_n-y_n\Vert ^2-\Vert y_{n}-v_n\Vert ^2-2\langle w_n-y_n,y_{n}-v_n \rangle \nonumber \\&\quad +2\tau _n\langle Fy_n,y_n-w_n\rangle \nonumber \\&=\Vert v_n-u^*\Vert ^2-\Vert w_n-y_n\Vert ^2-\Vert y_{n}-v_n\Vert ^2 +2\langle v_n-\tau _n Fy_n-y_n,w_n-y_n\rangle . \end{aligned}$$
(25)

Since \(y_n=P_{C}(v_n-\tau _n Fv_n)\) and \(w_n\in C\), we have

$$\begin{aligned} 2\langle v_n{-}\tau _n Fy_n{-}y_n,w_n-y_n\rangle&=2\langle v_n-\tau _n Fv_n-y_n,w_n-y_n\rangle +2\tau _n \langle Fv_n-Fy_n,w_n-y_n\rangle \nonumber \\&\le 2\tau _n \langle Fv_n-Fy_n,w_n-y_n\rangle . \end{aligned}$$
(26)

Combining (5) and (26), we obtain

$$\begin{aligned} 2\langle v_n-\tau _n Fy_n-y_n,w_n-y_n\rangle \le \mu \Vert v_n-y_n\Vert ^2+\mu \Vert y_n-w_n\Vert ^2. \end{aligned}$$
(27)

Substituting (27) into (25), we obtain

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2-(1-\mu )\Vert w_n-y_n\Vert ^2\\&\le \Vert v_n-u^*\Vert ^2. \end{aligned}$$

Claim 2. The sequence \(\{v_n\}\) is bounded. Indeed, we have

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert&=\Vert (1-\alpha _n-\beta _n)v_n+\beta _n w_n- u^*\Vert \\&=\Vert (1-\alpha _n-\beta _n)(v_n-u^*)+\beta _n(w_n-u^*)-\alpha _n u^*\Vert \\&\le \Vert (1-\alpha _n-\beta _n)(v_n-u^*)+\beta _n(w_n-u^*)\Vert +\alpha _n \Vert u^*\Vert \\&\le (1-\alpha _n-\beta _n)\Vert v_n-u^*\Vert +\beta _n\Vert w_n-u^*\Vert +\alpha _n \Vert u^*\Vert \\&\le (1-\alpha _n-\beta _n)\Vert v_n-u^*\Vert +\beta _n\Vert v_n-u^*\Vert +\alpha _n \Vert u^*\Vert \\&= (1-\alpha _n) \Vert v_n-u^*\Vert +\alpha _n\Vert u^*\Vert \\&\le \max \{\Vert v_n-u^*\Vert ,\Vert u^*\Vert \}| \\&\le \cdots \\&\le \max \{\Vert v_0-u^*\Vert ,\Vert u^*\Vert \}. \end{aligned}$$

That is, the sequence \(\{v_n\}\) is bounded and \(\{w_n\}\) is also.

Claim 3.

$$\begin{aligned} a(1-&\mu )\Vert v_n-y_n\Vert ^2 +a(1-\mu )\Vert y_n-w_n\Vert ^2\le \Vert v_n-u^*\Vert ^2-\Vert v_{n+1}-u^*\Vert ^2 +\alpha _n \Vert u^*\Vert ^2. \end{aligned}$$

Indeed, using (4), we have

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&=\Vert (1-\alpha _n-\beta _n)v_n+\beta _n w_n- u^*\Vert ^2\nonumber \\&=\Vert (1-\alpha _n-\beta _n)(v_n-u^*)+\beta _n (w_n- u^*)+\alpha _n (-u^*)\Vert ^2\nonumber \\&=(1-\alpha _n-\beta _n)\Vert v_n-u^*\Vert ^2+\beta _n \Vert w_n- u^*\Vert ^2+\alpha _n \Vert u^*\Vert ^2\nonumber \\&\quad -\beta _n (1-\alpha _n-\beta _n)\Vert v_n-w_n\Vert ^2\nonumber \\&\quad -\alpha _n(1-\alpha _n-\beta _n) \Vert v_n\Vert ^2-\alpha _n \beta _n \Vert w_n\Vert ^2\nonumber \\&\le (1-\alpha _n-\beta _n)\Vert v_n-u^*\Vert ^2+\beta _n \Vert w_n- u^*\Vert ^2+\alpha _n \Vert u^*\Vert ^2. \end{aligned}$$
(28)

Substituting (22) into the above inequality, we get

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&\le (1-\alpha _n-\beta _n)\Vert v_n-u^*\Vert ^2+\beta _n \Vert v_n- u^*\Vert ^2-\beta _n(1-\mu )\Vert v_n-y_n\Vert ^2\\&\quad -\beta _n(1-\mu )\Vert y_n-w_n\Vert ^2 +\alpha _n \Vert u^*\Vert ^2\\&=(1-\alpha _n)\Vert v_n-u^*\Vert ^2-\beta _n(1-\mu )\Vert v_n-y_n\Vert ^2\\&\quad -\beta _n(1-\mu )\Vert y_n-w_n\Vert ^2+\alpha _n \Vert u^*\Vert ^2\\&\le \Vert v_n-u^*\Vert ^2-\beta _n(1-\mu )\Vert v_n-y_n\Vert ^2 -\beta _n(1-\mu )\Vert y_n-w_n\Vert ^2+\alpha _n \Vert u^*\Vert ^2. \end{aligned}$$

Thus, we get

$$\begin{aligned} \beta _n(1-\mu )\Vert v_n-y_n\Vert ^2 +\beta _n(1-\mu )\Vert y_n-w_n\Vert ^2\le \Vert v_n-u^*\Vert ^2-\Vert v_{n+1}-u^*\Vert ^2 +\alpha _n \Vert u^*\Vert ^2. \end{aligned}$$

Moreover, since \(\beta _n\ge a\) for all \(n\ge 1\), we obtain

$$\begin{aligned} a(1-\mu )\Vert v_n-y_n\Vert ^2 +a(1-\mu )\Vert y_n-w_n\Vert ^2\le \Vert v_n-u^*\Vert ^2-\Vert v_{n+1}-u^*\Vert ^2 +\alpha _n \Vert u^*\Vert ^2. \end{aligned}$$

Claim 4.

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2\le (1-\alpha _n)\Vert v_n-u^*\Vert ^2{+}\alpha _n[2\beta _n\Vert v_n{-}w_n\Vert \Vert v_{n+1}{-}u^*\Vert +2\langle u^*,u^*-v_{n+1}\rangle ]. \end{aligned}$$

Indeed, setting \(t_n=(1-\beta _n)v_n+\beta _n w_n\) for each \(n\ge 1\), we have

$$\begin{aligned} \Vert t_n-u^*\Vert&=\Vert (1-\beta _n)(v_n-u^*)+\beta _n(w_n-u^*)\Vert \le (1-\beta _n)\Vert v_n-u^*\Vert +\beta _n\Vert w_n-u^*\Vert \nonumber \\&\le (1-\beta _n)\Vert v_n-u^*\Vert +\beta _n\Vert v_n-u^*\Vert =\Vert v_n-u^*\Vert , \end{aligned}$$
(29)

and

$$\begin{aligned} \Vert t_n-v_n\Vert =\beta _n \Vert v_n-w_n\Vert . \end{aligned}$$
(30)

Using (29) and (30), we get

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&=\Vert (1-\alpha _n-\beta _n)v_n+\beta _n w_n-u^*\Vert ^2\nonumber \\&=\Vert (1-\beta _n)v_n+\beta _n w_n-\alpha _n v_n-u^*\Vert ^2\nonumber \\&=\Vert (1-\alpha _n)(t_n-u^*)-\alpha _n(v_n-t_n)-\alpha _n u^*\Vert ^2. \end{aligned}$$
(31)

Now, using the inequality (3), we get

$$\begin{aligned}&\Vert (1-\alpha _n)(t_n-u^*)-\alpha _n(v_n-t_n)-\alpha _n u^*\Vert ^2\le (1-\alpha _n)^2\Vert t_n-u^*\Vert ^2 \nonumber \\&\quad -2\langle \alpha _n(v_n-t_n)+\alpha _n u^*,v_{n+1}-u^*\rangle . \end{aligned}$$
(32)

Combining (31) and (32), we obtain

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&\le (1-\alpha _n)^2\Vert t_n-u^*\Vert ^2 +2 \alpha _n\langle v_n-t_n,u^*-v_{n+1}\rangle +2\alpha _n \langle u^*,u^*-v_{n+1}\rangle \\&\le (1-\alpha _n)\Vert t_n-u^*\Vert ^2 + 2\alpha _n \Vert v_n-t_n\Vert \Vert v_{n+1}-u^*\Vert +2\alpha _n \langle u^*,u^*-v_{n+1}\rangle \\&\le (1-\alpha _n)\Vert v_n-u^*\Vert ^2 + \alpha _n [2\beta _n\Vert v_n-w_n\Vert \Vert v_{n+1}-u^*\Vert +2 \langle u^*,u^*-v_{n+1}\rangle ]. \end{aligned}$$

Claim 5. \(\{\Vert v_n-u^*\Vert ^2\}\) converges to zero. Indeed, for each \(n\ge 0\), set

$$\begin{aligned} a_n:=\Vert v_n-u^*\Vert ^2 \quad \text { and }\quad b_n:=2\beta _n\Vert v_n-w_n\Vert \Vert v_{n+1}-u^*\Vert +2 \langle u^*,u^*-v_{n+1}\rangle . \end{aligned}$$

Then, Claim 4 can be rewritten as follows:

$$\begin{aligned} a_{n+1}\le (1-\alpha _n)a_n+\alpha _n b_n. \end{aligned}$$

By Lemma 2.6, it is sufficient to show that \(\limsup _{k\rightarrow \infty }b_{n_k}\le 0\) for every subsequence \(\{a_{n_k}\}\) of \(\{a_n\}\) satisfying

$$\begin{aligned} \liminf _{k\rightarrow \infty }(a_{n_k+1}-a_{n_k})\ge 0. \end{aligned}$$

This is equivalently to that we need to show \(\limsup _{k\rightarrow \infty }\langle u^*, u^*-v_{n_k+1}\rangle \le 0\) and \(\limsup _{k\rightarrow \infty }\Vert v_{n_k}-w_{n_k}\Vert \le 0\) for every subsequence \(\{\Vert v_{n_k}-u^*\Vert \}\) of \(\{\Vert v_n-u^*\Vert \}\) satisfying

$$\begin{aligned} \liminf _{k\rightarrow \infty }(\Vert v_{n_k+1}-u^*\Vert -\Vert v_{n_k}-u^*\Vert )\ge 0. \end{aligned}$$

Suppose that \(\{\Vert v_{n_k}-u^*\Vert \}\) is a subsequence of \(\{\Vert v_n-u^*\Vert \}\) such that

$$\begin{aligned} \liminf _{k\rightarrow \infty }(\Vert v_{n_k+1}-u^*\Vert -\Vert v_{n_k}-u^*\Vert )\ge 0. \end{aligned}$$

Then, we have

$$\begin{aligned}&\liminf _{k\rightarrow \infty }(\Vert v_{n_k+1}-u^*\Vert ^2-\Vert v_{n_k}-u^*\Vert ^2)\\&\quad =\liminf _{k\rightarrow \infty }[(\Vert v_{n_k+1}-u^*\Vert -\Vert v_{n_k}-u^*\Vert )(\Vert v_{n_k+1}-u^*\Vert +\Vert v_{n_k}-u^*\Vert )]\\&\quad \ge 0. \end{aligned}$$

By Claim 3, we obtain

$$\begin{aligned}&\limsup _{k\rightarrow \infty }[a(1-\mu )\Vert v_{n_k}-y_{n_k}\Vert ^2 +a(1-\mu )\Vert y_{n_k}-w_{n_k}\Vert ^2]\\&\quad \le \limsup _{k\rightarrow \infty }[\Vert v_{n_k}-u^*\Vert ^2-\Vert v_{{n_k}+1}-u^*\Vert ^2 +\alpha _{n_k} \Vert u^*\Vert ^2]\\&\quad \le \limsup _{k\rightarrow \infty }[ \Vert v_{n_k}-u^*\Vert ^2-\Vert v_{n_k+1}-u^*\Vert ^2]+\limsup _{k\rightarrow \infty }\alpha _{n_k} \Vert u^*\Vert ^2\\&\quad = - \liminf _{k\rightarrow \infty }[ \Vert v_{n_k+1}-u^*\Vert ^2-\Vert v_{_{n_k}}-u^*\Vert ^2]\\&\quad \le 0. \end{aligned}$$

This implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert v_{n_k}-y_{n_k}\Vert =0,\quad \lim _{k\rightarrow \infty }\Vert y_{n_k}-w_{n_k}\Vert =0. \end{aligned}$$

Thus, we have

$$\begin{aligned} \Vert v_{n_k}-w_{n_k}\Vert \le \Vert v_{n_k}-y_{n_k}\Vert +\Vert y_{n_k}-w_{n_k}\Vert \rightarrow 0 \quad \text { as }\ k\rightarrow \infty . \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \Vert v_{{n_k}+1}-v_{n_k}\Vert \le \alpha _{n_k} \Vert v_{n_k}\Vert +\beta _{n_k}\Vert v_{n_k}-w_{n_k}\Vert \rightarrow 0 \quad \text{ as } \ k\rightarrow \infty . \end{aligned}$$
(33)

Since the sequence \(\{v_{n_k}\}\) is bounded, it follows that there exists a subsequence \(\{v_{n_{k_j}}\}\) of \(\{v_{n_k}\}\), which converges weakly to some \(z\in H\), such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle u^*,u^*-v_{n_k}\rangle =\lim _{j\rightarrow \infty }\langle u^*,u^*-v_{n_{k_j}}\rangle =\langle u^*,u^*-z\rangle . \end{aligned}$$
(34)

From \( \lim _{k\rightarrow \infty }\Vert v_{n_k}-y_{n_k}\Vert =0\) and Lemma 3.8, we have \(z\in \mathrm{Sol}(C,F)\) and, from (34) and the definition of \(u^*=P_{\mathrm{Sol}(C,F)} 0\), we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle u^*,u^*-v_{n_k}\rangle =\langle u^*,u^*-z\rangle \le 0. \end{aligned}$$
(35)

Combining (33) and (35), we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle u^*,u^*-v_{n_k+1}\rangle \le \limsup _{k\rightarrow \infty }\langle u^*,u^*-v_{n_k}\rangle =\langle u^*,u^*-z\rangle \le 0. \end{aligned}$$
(36)

Hence, it follows from (36), \(\lim _{k\rightarrow \infty }\Vert v_{n_k}-w_{n_k}\Vert =0\), Claim 4 and Lemma 2.6 that

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert v_n-u^*\Vert =0. \end{aligned}$$

This completes the proof. \(\square \)

Remark 3.3

Our result generalizes some related results in the literature and hence might be applied to a wider class of nonlinear mappings. For example, in the next section, we present the advantages of our method compared with the recent results (Cai et al. 2021, Theorem 3.1) as follows:

  1. (1)

      In Theorem 3.1, we replaced the sequentially weakly continuity of F by condition (14), which is strictly weaker than the sequential weak continuity of the operator F.

  2. (2)

      We also obtained the strong convergence without using the viscosity technique.

3.2 Convergence rate

In this section, we provide a result on the convergence rate of the iterative sequence generated by Algorithm 3.1 when \(\alpha _n=0\) and \(\beta _n=\beta \in (0,1)\) for all n and the mapping \(F: C\rightarrow H\) is L-Lipschitz continuous and \(\delta \)-strongly pseudo-monotone on C. In this case, Algorithm 3.1 is of the form:

figure b

Theorem 3.2

Assume that \(F: C\rightarrow H\) is L-Lipschitz continuous and \(\delta \)-strongly pseudo-monotone on C. Then, the sequence \(\{v_n\}\) generated by Algorithm 3.2 converges strongly to \(u^*\) with a Q-linear rate, where \(u^*\) is a unique solution of VI.

Proof

We have \(\langle Fu^*,y_n-u^*\rangle \ge 0\), by the strong pseudo-monotonicity of F on C, we get

$$\begin{aligned} \langle Fy_n,y_n-u^*\rangle \ge \delta \Vert y_n-u^*\Vert ^2. \end{aligned}$$
(37)

Using (37), we have

$$\begin{aligned} \langle Fy_n,u^*-w_n\rangle =&\langle Fy_n,u^*-y_{n}\rangle +\langle Fy_n,y_n-w_n\rangle \le -\delta \Vert y_n-u^*\Vert ^2 + \langle Fy_n,y_n-w_n\rangle . \end{aligned}$$
(38)

Substituting (38) into (23), we obtain

$$\begin{aligned} \Vert w_n-u^*\Vert ^2\le \Vert v_n-u^*\Vert ^2 -\Vert w_n-v_n\Vert ^2-2\tau _n\delta \Vert y_n-u^*\Vert ^2 +2\tau _n \langle Fy_n,y_n-w_n\rangle . \end{aligned}$$

Thus,

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-\Vert w_n-v_n\Vert ^2-2\tau _n\delta \Vert y_n-u^*\Vert ^2 +2\tau _n\langle Fy_n,y_n-w_n\rangle \nonumber \\&=\Vert v_n-u^*\Vert ^2-\Vert w_n-y_n\Vert ^2-\Vert y_{n}-v_n\Vert ^2-2\langle w_n-y_n,y_{n}-v_n \rangle \nonumber \\&\quad -2\tau _n\delta \Vert y_n-u^*\Vert ^2 +2\tau _n\langle Fy_n,y_n-w_n\rangle \nonumber \\&=\Vert v_n-u^*\Vert ^2-\Vert w_n-y_n\Vert ^2-\Vert y_{n}-v_n\Vert ^2-2\tau _n\delta \Vert y_n-u^*\Vert ^2 \nonumber \\&\quad +2\langle v_n-\tau _n Fy_n-y_n,w_n-y_n\rangle . \end{aligned}$$
(39)

Substituting (27) into (39), we obtain

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2-(1-\mu )\Vert w_n-y_n\Vert ^2-2\tau _n\delta \Vert y_n-u^*\Vert ^2. \end{aligned}$$
(40)

Now we show that \(\tau _n> \dfrac{\mu l}{L}\) for all n. Indeed, by the search rule (5), we know that \(\dfrac{\tau _n}{l}\) must violate inequality (5), i.e.,

$$\begin{aligned} \langle Fy_n-Fv_n,y_n-w_n\rangle > \dfrac{\mu }{2\frac{\tau _n}{l}}(\Vert v_n-y_n\Vert ^2+\Vert y_n-w_n\Vert ^2). \end{aligned}$$

This follows that

$$\begin{aligned} L\Vert y_n-v_n\Vert \Vert y_n-w_n\Vert > \dfrac{\mu }{2\frac{\tau _n}{l}}(\Vert v_n-y_n\Vert ^2+\Vert y_n-w_n\Vert ^2). \end{aligned}$$

Thus,

$$\begin{aligned} \dfrac{L}{2}(\Vert y_n-v_n\Vert ^2+\Vert y_n-w_n\Vert ^2)> \dfrac{\mu }{2\frac{\tau _n}{l}}(\Vert v_n-y_n\Vert ^2+\Vert y_n-w_n\Vert ^2). \end{aligned}$$

This implies that

$$\begin{aligned} \tau _n>\dfrac{\mu l}{L}\quad \forall n. \end{aligned}$$
(41)

Combining (40) and (41), we get

$$\begin{aligned} \Vert w_n-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2-(1-\mu )\Vert w_n-y_n\Vert ^2-2\dfrac{\mu l}{L}\delta \Vert y_n-u^*\Vert ^2\nonumber \\&\le \Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2-2\dfrac{\mu l}{L}\delta \Vert y_n-u^*\Vert ^2. \end{aligned}$$
(42)

On the other hand, using (2), we have

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&=\Vert (1-\beta )v_n+\beta w_n-u^*\Vert ^2\\&= \Vert (1-\beta )(v_n-u^*)+\beta (w_n-u^*)\Vert ^2\\&= (1-\beta )\Vert v_n-u^*\Vert ^2+\beta \Vert w_n-u^*\Vert ^2-\beta (1-\beta )\Vert v_n-w_n\Vert ^2\\&\le (1-\beta )\Vert v_n-u^*\Vert ^2+\beta [\Vert v_n-u^*\Vert ^2-(1-\mu )\Vert y_n-v_n\Vert ^2\\&\quad -2\dfrac{\mu l}{L}\delta \Vert y_n-u^*\Vert ^2]-\dfrac{1-\beta }{\beta }\Vert v_{n+1}-v_n\Vert ^2\\&\le \Vert v_n-u^*\Vert ^2+\beta (1-\mu )\Vert y_n-v_n\Vert ^2-2\beta \dfrac{\mu l}{L}\delta \Vert y_n-u^*\Vert ^2. \end{aligned}$$

Let \(\xi :=\dfrac{1}{2}\min \bigg \{\beta (1-\mu ),2\beta \dfrac{\mu l}{L}\delta \bigg \}\in (0,1)\). From (42), we get

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert ^2&\le \Vert v_n-u^*\Vert ^2-2\xi \Vert y_n-v_n\Vert ^2-2\xi \Vert y_n-u^*\Vert ^2\\&\le \Vert v_n-u^*\Vert ^2-\xi \Vert v_n-u^*\Vert ^2\\&=(1-\xi )\Vert v_n-u^*\Vert ^2. \end{aligned}$$

This implies that

$$\begin{aligned} \Vert v_{n+1}-u^*\Vert \le \sqrt{(1-\xi )}\Vert v_n-u^*\Vert . \end{aligned}$$

Since \(\sqrt{(1-\xi )}\in (0,1)\), it implies that the sequence \(\{v_n\}\) converges strongly to \(u^*\) with a Q-linear rate.

4 Numerical illustrations

In this section, we perform four numerical experiments to show the behaviors of Algorithm 3.2, and compare them with other algorithms. In the numerical results listed in the following tables, ‘Iter.’ denotes the total number of internal and external circulations, and the number in brackets denotes the number of total iterations of finding suitable \(\tau _n.\) The followings are the experiments in detail.

Example 4.1

Shehu et al. (2019) Consider \(C:=\{x \in H:\Vert x\Vert \le 2\}\). Let \(g: C\rightarrow \mathbb {R}\) be defined by

$$\begin{aligned} g(u):=\frac{1}{1+\Vert u\Vert ^2}. \end{aligned}$$

Observe that g is Lipschitz continuous with constant \(L_g=\frac{16}{25}\) and \(\frac{1}{5}\le g(u)\le 1, \quad \forall \, u \in C\). Define the Volterra integral operator \(A:L^2([0,1])\rightarrow L^2([0,1])\) by

$$\begin{aligned} A(u)(t):=\int _0^t u(s)\mathrm{d}s,\quad \forall u \in L^2([0,1]), t \in [0,1]. \end{aligned}$$

Then, A is bounded linear monotone and \(\Vert A\Vert =\frac{2}{\pi }\). Now, define \(F:C\rightarrow L^2([0,1])\) by

$$\begin{aligned} F(u)(t):=g(u)A(u)(t),\quad \forall u \in C, t \in [0,1]. \end{aligned}$$

Hence F is not monotone on C but pseudo-monotone and Lipschitz continuous with constant \(L = 82/\pi \). The parameters are chosen as follows:

Algorithm 3.2: \(\gamma =1.5,\,\mu =0.99,\,l=0.5,\,a=0.0001,\,\alpha _n=\frac{1}{n+2},\,\beta _n=a+0.99(1-a-\alpha _n)\).

Algorithm 3.1 in Cai et al. (2021): \(f(x)=0,\,\gamma =1.5,\,\mu =0.6,\,l=0.5,\,\beta _n=\frac{1}{n+2}\).

Algorithm 3 in Thong et al. (2020): \(f(x)=0,\,\gamma =1.1,\,\mu =0.5,\,l=0.5,\,\alpha _n=\frac{1}{n+2}\).

Algorithm 3.1 in Thong and Gibali (2019): \(\gamma =1.72,\,\mu =0.5,\,l=0.5,\,\alpha _n=\frac{1}{n+2}\).

We now make comparisons of four algorithms with the initial function \(x_0=x_1=t^2\) and \(\Vert x_{n}-y_n\Vert <\mathrm{Error}\) as stopping condition.

The results in Fig. 1 illustrate that Algorithm 3.2 behaves better than Algorithm 3.1 in Cai et al. (2021), Algorithm 3 in Thong et al. (2020) and Algorithm 3.1 in Thong and Gibali (2019).

Fig. 1
figure 1

Comparison of four algorithms in Example 4.1

Full size image

Example 4.2

(Harker and Pang 1990) Let \(F(x): = M x + q\) where

$$\begin{aligned} M= BB^T+ C + D, \end{aligned}$$

and B is an \(m\times m\) matrix, C is an \(m\times m\) skew-symmetric matrix, D is an \(m\times m\) diagonal matrix, whose diagonal entries are nonnegative (so M is positive semidefinite), q is a vector in \(\mathbb {R}^m\). The feasible set \(C\subset \mathbb {R}^m\) is a closed and convex subset defined by \(C: = \{x\in \mathbb {R}^m: Qx\le b\}\), where Q is an \(d\times m\) matrix and b is a nonnegative vector. It is clear that F is monotone and Lipschitz continuous with constant \(L =\Vert M\Vert \). Let \( q= 0\). Then, we obtain the solution set \(\varGamma = \{0\}\). The parameters are chosen as follows:

Algorithm 3.2: \(\gamma =0.01,\,\mu =0.99,\,l=0.55,\,\alpha _n=\frac{1}{n+2},\,\beta _n=a+0.99(1-a-\alpha _n)\).

Algorithm 3.2 in Gibali et al. (2019): \(f(x)=0,\,\lambda =0.01,\,\gamma =1,\,\mu =0.5,\,l=0.5,\,\alpha _n=\frac{1}{n+2}\).

We now make comparisons of two algorithms with \(d=20\), \(m=20\) and \(\Vert v_n\Vert <\mathrm{Error}\) as stopping condition. From Fig. 2 and Table 1, it is observed that the performance of Algorithm 3.2 is better than Algorithm 3.2 in Gibali et al. (2019).

Fig. 2
figure 2

Comparison of four algorithms in Example 4.2

Full size image

Example 4.3

(Shehu et al. 2019) Next, let us consider Sol(C, F) with

$$\begin{aligned} F(x)= \left( \begin{array}{c} (x_1^2+(x_2-1)^2)(1+x_2) \\ -x_1^3-x_1(x_2-1)^2 \\ \end{array} \right) \end{aligned}$$

and \(C:=\{x \in \mathbb {R}^2:-10 \le x_i \le 10, i=1,2\}.\) This problem has unique solution \(x^* =(0,-1)^T\). It is clear that F is not a monotone map on C. However, using the Monte Carlo approach (see Shehu et al. 2019), it can be shown that F is pseudo-monotone on C. The parameters are chosen as follows:

Table 1 Comparisons of Algorithm 3.2 and Algorithm 3.2 in Gibali et al. (2019) for Example 4.1
Full size table

Algorithm 3.2: \(\gamma =0.24,\,\mu =0.99,\,l=0.58,\,a=0.0001,\,\alpha _n=\frac{1}{n+2},\,\beta _n=a+0.99(1-a-\alpha _n)\).

Algorithm 3 in Thong et al. (2020): \(f(x)=0,\,\gamma =1.8,\,\mu =0.5,\,l=0.22,\,\alpha _n=\frac{1}{n+2}\).

Algorithm 3.1 in Thong and Gibali (2019): \(\gamma =0.84,\,\mu =0.01,\,l=0.99,\,\alpha _n=\frac{1}{n+2}\).

Algorithm 3.2 in Thong et al. (2020): \(\tau _1=0.43,\,\mu =0.99,\,\alpha =0,\,\beta _n=\frac{1}{n+2}\).

We now make comparisons of four algorithms with the initial function \(v_0=v_1=[-0.3;-0.3]\) and \(\Vert v_n-x^*\Vert <\mathrm{Error}\) as stopping condition. Figure 3 and Table 2 demonstrate that Algorithm 3.2 performs better than Algorithm 3 in Thong et al. (2020), Algorithm 1 in Thong and Gibali (2019) and Algorithm 3.2 in Thong et al. (2020).

Fig. 3
figure 3

Comparison of four algorithms in Example 4.3

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Table 2 Comparisons of Algorithm 3.2, Algorithm 3 in Thong et al. (2020), Algorithm 3.1 in Thong and Gibali (2019) and Algorithm 3.2 in Thong et al. (2020) for Example 4.3
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Example 4.4

Cai et al. (2021) Suppose that \(H = L^2([0,1])\) with norm \(\Vert x\Vert :=\Big (\int _0^1 |x(t)|^2\mathrm{d}t\Big )^{\frac{1}{2}}\) and inner product \(\langle x,y\rangle := \int _0^1 x(t)y(t)\mathrm{d}t,\quad x,y \in H\). Let \(C:=\{x \in H:\Vert x\Vert \le 1\}\) be the unit ball. Define an operator \(F:C\rightarrow H\) by

$$\begin{aligned} Fx(t)=\int _0^1 [x(t)-A(t,s)x(s)]\mathrm{d}s,\quad x \in C, t \in [0,1], \end{aligned}$$

where

$$\begin{aligned} A(t,s)=\frac{4}{\mathrm{e}^2-1}ts\mathrm{e}^{t+s}:=cts\mathrm{e}^{t+s}. \end{aligned}$$

It was shown in Cai et al. (2021) that F is sequentially weakly continuous on C, monotone and 2-Lipschitz continuous.

Table 3 Comparisons of Algorithm 3.2, Algorithm 3.1 in Cai et al. (2021), Algorithm 3.2 in Thong et al. (2020), Algorithm 1 in Yang and Hongwei (2019) for Example 4.4
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The parameters are chosen as follows:

Algorithm 3.2: \(\gamma =0.01,\,\mu =0.99,\,l=0.5,\,a=0.0001,\,\alpha _n=\frac{1}{n+2}, \,\beta _n=a+0.99(1-\alpha _n-a)\).

Algorithm 3.1 in Cai et al. (2021): \(f(x)=0,\,\gamma =0.02,\,\mu =0.8,\,l=0.5,\,\beta _n=\frac{1}{n+2}\).

Algorithm 3.2 in Thong et al. (2020): \(f(x)=0,\,\tau _1=0.4,\,\mu =0.03,\,\alpha _n=0,\,\beta _n=\frac{1}{n+2}\).

Algorithm 1 in Yang and Hongwei (2019): \(f(x)=0,\,\lambda _0=0.02,\,\mu =0.3,\,\alpha _n=\frac{1}{n+2}\).

We now make comparisons of four algorithms with the initial functions \(x_0=x_1=(0.5,0.5,\dots ,0.5)\) and the stopping condition \(\Vert x_n-y_n\Vert <Error\). The result in Table 3 illustrates that Algorithm 3.2 behaves better than Algorithm 3.1 in Cai et al. (2021), Algorithm 3.2 in Thong et al. (2020) and Algorithm 1 in Yang and Hongwei (2019) when the error is smaller.

5 Conclusions

In this paper, we introduce a new version of extragradient method for solving non-Lipschitzian pseudo-monotone variational inequalities in real Hilbert spaces. Strong convergence and convergence rate of the proposed are presented. Our work extends and generalizes some existing results in the literature. Finally, several numerical experiments are reported to illustrate the performance of the proposed algorithm.