1 Introduction

Let \({\mathcal {H}}\) be a real Hilbert space, and let \(A:{\mathcal {H}}\rightarrow 2^{{\mathcal {H}}}\) and \(B:{\mathcal {H}}\rightarrow {{\mathcal {H}}}\) be two monotone operators. The problem of finding a zero of the sum of A and B also known as the monotone inclusion problem is defined as

$$\begin{aligned} \text{ find } ~{\widehat{x}}\in {\mathcal {H}}~\text{ such } \text{ that }~0\in (A+B){\widehat{x}}. \end{aligned}$$
(1.1)

We denote the solution set of this problem by \((A+B)^{-1}(0)\).

The monotone inclusion is an important problem in optimization as well as in signal processing, image recovery, and machine learning. For instance, consider the minimization problem:

$$\begin{aligned} \min _{{\widehat{x}}\in {\mathcal {H}}} \{f({\widehat{x}})+g({\widehat{x}})\}, \end{aligned}$$
(1.2)

where \(f:{\mathcal {H}}\rightarrow (-\infty ,~+\infty ]\) is proper convex and lower semicontinuous, and \(g:{\mathcal {H}}\rightarrow {\mathbb {R}}\) is convex and continuously differentiable. The solutions to (1.2) are solutions to the problem:

$$\begin{aligned} \text{ find } ~{\widehat{x}}\in {\mathcal {H}}~\text{ such } \text{ that }~0\in (\partial f+ \nabla g){\widehat{x}}, \end{aligned}$$

where \(\partial f\) denotes the subdifferential of f and \(\nabla g\) is the gradient of g. Thus, the minimization problem of the sum of two convex functions is a special case of the monotone inclusion problem (1.1). Problems of the form (1.1) are often solved by splitting algorithms which involve evaluating A and B separately by means of a forward evaluation of B and a backward evaluation of A rather than their sum (\(A+B\)). These algorithms have undergone tremendous study which has been motivated by the desire to devise faster, computationally inexpensive and much more applicable methods. Among these splitting algorithms is the following forward–backward splitting algorithm (Lions and Mercier 1979; Passty 1979):

$$\begin{aligned} x_{n+1}=(I_{{\mathcal {H}}}+\lambda A)^{-1}(x_n-\lambda Bx_n),~n\ge 1, \end{aligned}$$
(1.3)

where \(I_{{\mathcal {H}}}\) is the identity operator on \({\mathcal {H}}\) and \(\lambda \) is a positive constant. This algorithm involves only one forward evaluation of B and one backward evaluation of A per iteration, and it is known to converge weakly to a solution of the inclusion problem (1.1) when B is \(L^{-1}\)-cocoercive (which is strict), \(\lambda \in (0, 2L^{-1})\), A is maximal monotone and \((A+B)^{-1}(0)\) is nonempty. Strongly convergent variants of Algorithm (1.3) have been studied under the strict cocoercive assumption on B in Takahashi et al. (2010); Wang and Wang (2018).

The strict cocoercive assumption on B in Algorithm (1.3) (and its strong convergent variants) was removed in Tseng (2000), where the author proposed the following forward–backward–forward splitting algorithm (also known as Tseng’s splitting method):

$$\begin{aligned} {\left\{ \begin{array}{ll} y_n=(I_{{\mathcal {H}}}+\lambda A)^{-1}(x_n-\lambda Bx_n),\\ x_{n+1}=y_n-\lambda By_n+\lambda Bx_n,~n\ge 1. \end{array}\right. } \end{aligned}$$
(1.4)

The main advantage of this algorithm is that it converges weakly to a solution of (1.1) under the much more weaker assumption that B is monotone and L-Lipschitz continuous, \(\lambda \in (0, L^{-1})\), A is maximal monotone and \((A+B)^{-1}(0)\ne \emptyset \). However, its main disadvantage is that it involves two forward evaluations of B, and this might affect the efficiency of the algorithm especially when this algorithm is applied to optimization arising from large-scale problems and optimal control theory, where computations of pertinent operators are often very expensive (see (Lions 1971)). Strongly convergent variants of Algorithm (1.4) were studied in Gibali and Thong (2018); Thong and Cholamjiak (2019), and they also have the disadvantage of requiring two forward evaluations of B.

This disadvantage was recently overcome by Malitsky and Tam (2020); they proposed the following forward–reflected–backward splitting algorithm:

$$\begin{aligned} x_{n+1}=(I_{{\mathcal {H}}}+\lambda A)^{-1}(x_n-2\lambda Bx_n+\lambda Bx_{n-1}),~n\ge 1,~\lambda \in \left( 0, \frac{1}{2}L^{-1}\right) . \end{aligned}$$
(1.5)

The main advantage of Algorithm (1.5) is that it requires only one forward evaluation of B even when B is monotone and Lipschitz continuous. The reflexive Banach space variant of Algorithm (1.5) was studied in Izuchukwu et al. (2022), and when B is linear, Algorithm (1.5) has the same structure as the reflected–forward–backward splitting algorithm proposed and studied in Cevher and Vũ (2021). However, due to the computational structure of Algorithm (1.5) (unlike Algorithm (1.3) and Algorithm (1.4)), its strongly convergent variants are very rare in the literature despite that in infinite-dimensional spaces, strong convergence results are much more desirable than weak convergence results.

Recently, fast convergent algorithms for solving optimization problems have been of great interest to many researchers. On one hand, the anchored extrapolation step is known to be one of the most important ingredients for improving the convergence rate of optimization algorithms (see (Qi and Xu 2021; Yoon and Ryu 2021) for details). On the other hand, the inertial technique which is based upon a discrete analog of a second-order dissipative dynamical system is also known for its efficiency in improving the convergence speed of iterative algorithms. The one-step inertial extrapolation \(x_n+{\vartheta }(x_n-x_{n-1})\) is the most commonly used technique for this purpose. It originates from the heavy ball method of the second-order dynamical system for minimizing a smooth convex function:

$$\begin{aligned} \frac{d^2 x}{\textrm{d}t^2}(t)+\gamma \frac{\textrm{d} x}{\textrm{d}t}(t)+\nabla f(x(t))=0, \end{aligned}$$

where \(\gamma >0\) is a damping or friction parameter. Polyak (1964) was the first author to propose the heavy ball method, Alvarez and Attouch (2001) extended it to the setting of a general maximal monotone operator. In Bing and Cho (2021), the authors proposed the following one-step inertial viscosity-type forward–backward–forward splitting algorithm (Bing and Cho 2021, Algorithm 3.4):

$$\begin{aligned} {\left\{ \begin{array}{ll} w_n=x_n + \vartheta _n(x_n-x_{n-1}),\\ y_n=(I_{{\mathcal {H}}}+\lambda A)^{-1}(w_n-\lambda Bw_n),\\ z_n=y_n-\lambda By_n+\lambda Bx_n,\\ x_{n+1}=\alpha _n fx_n+(1-\alpha _n)z_n,~n\ge 1, \end{array}\right. } \end{aligned}$$
(1.6)

where f is a contraction mapping and \(\vartheta _n\) is the inertial parameter. They proved that the sequence \(\{x_n\}\) generated by Algorithm (1.6) converges strongly to a solution of the monotone inclusion problem (1.1) (see also (Bing and Cho (2021), Theorem 3.3), a version of Algorithm (1.6) without the contraction mapping). In (Suparatulatorn and Chaichana (2022), Algorithm 1), Suparatulatorn and Chaichana proposed a one-step inertial parallel shrinking-type algorithm for solving a finite family of monotone inclusion problems. Also, in Alakoya et al. (2022); Liu et al. (2021); Tan et al. (2022); Taiwo and Mewomo (2022), the authors proposed and studied one-step inertial algorithms that tackle the particular case, where A is the normal cone of some nonempty, closed and convex set (i.e., one-step inertial algorithms that solve variational inequality problems).

However, it was discussed in (Poon and Liang 2019, Section 3) that one-step inertial term may fail to provide acceleration to ADMM. Let \(H_1, H_2 \subset {\mathbb {R}}^2\) be two subspaces such that \(H_1 \cap H_2 \ne \emptyset \). Consider the feasibility problem:

$$\begin{aligned} \text{ find } ~{\widehat{x}}\in {\mathbb {R}}^2~\text{ such } \text{ that }~{\widehat{x}}\in H_1 \cap H_2. \end{aligned}$$
(1.7)

It was shown in [25, Section 4] that for problem (1.7), the two-step inertial fixed point iteration

$$\begin{aligned} x_{n+1}=T(x_n + \theta (x_n - x_{n-1})+\delta (x_{n-1} - x_{n-2})), \end{aligned}$$

where \(T:=\frac{1}{2}\Big (I+(2P_{H_1}-I)(2P_{H_2}-I)\Big )\), converges faster in terms of both the number of iterations and the CPU time, than the one-step inertial fixed point iteration

$$\begin{aligned} x_{n+1}=T(x_n + \theta (x_n - x_{n-1})). \end{aligned}$$
(1.8)

Furthermore, it was shown, using problem (1.7), that the sequences generated by the one-step inertial fixed point iteration (1.8) converge more slowly than those generated by its non-inertial version. This shows that the one-step inertial fixed point iteration (1.8) may fail to provide acceleration. But as discussed in (Liang 2016, Chapter 4), employing the multi-step inertial, like the two-step inertial term \(x_n+{\vartheta }(x_n-x_{n-1})+{\beta }(x_{n-1}-x_{n-2})\), \({\vartheta } > 0\), \({\beta }< 0\), could resolve the issue of not providing acceleration. Thus in Combettes and Glaudin (2017); Dong et al. (2019); Iyiola and Shehu (2022); Li et al. (2022); Polyak (1987), the authors recently studied multi-step inertial algorithms and showed that multi-step inertial terms (e.g., the two-step inertial term) enhances the speed of optimization algorithms.

In this paper, we investigate the strong convergence of a two-step inertial anchored variant of the forward-reflected-backward splitting algorithm (1.5). In other words, we propose a two-step inertial forward–reflected–anchored–backward splitting algorithm and prove that it converges strongly to a solution of Problem (1.1). The proposed algorithm involves only one forward evaluation of the monotone Lipschitz continuous operator B and one backward evaluation of the maximal monotone operator A at each iteration; a feature that is absent in many other available strongly convergent inertial splitting algorithms in the literature. Furthermore, we perform numerical experiments for problems emanating from image restoration, and these experiments confirm that our proposed algorithm is efficient and faster than other related strongly convergent splitting algorithms in the literature.

2 Preliminaries

The operator \(B: {\mathcal {H}}\rightarrow {\mathcal {H}}\) is called L-cocoercive (or inverse strongly monotone) if there exists \(L>0\) such that

$$\begin{aligned} \langle B{x}-B{y},{x}-{y}\rangle \ge L\Vert B{x}-B{y}\Vert ^2~ \forall x,y \in {\mathcal {H}}, \end{aligned}$$

and monotone if

$$\begin{aligned} \langle Bx-By,x-y\rangle \ge 0 \; ~ \forall x,y \in {\mathcal {H}}. \end{aligned}$$

The operator B is called L-Lipschitz continuous if there exists \(L>0\) such that

$$\begin{aligned} \Vert Bx-By\Vert \le L\Vert x-y\Vert ~ \forall x,y\in {\mathcal {H}}. \end{aligned}$$

Let A be a set-valued operator \(A:{\mathcal {H}}\rightarrow 2^{{\mathcal {H}}}\), then A is said to be monotone if

$$\begin{aligned} \langle {\widehat{x}}-{\widehat{y}}, x-y\rangle \ge 0 \; ~\forall x,y \in {\mathcal {H}},~{\widehat{x}}\in Ay,~{\widehat{y}}\in Ay. \end{aligned}$$

The monotone operator A is called maximal if the graph \({\mathcal {G}}(A)\) of A, defined by

$$\begin{aligned} {\mathcal {G}}(A):=\{(x, {\widehat{x}})\in {\mathcal {H}}\times {\mathcal {H}}: {\widehat{x}}\in Ax\}, \end{aligned}$$

is not properly contained in the graph of any other monotone operator. In other words, A is called a maximal monotone operator if for \((x,{\widehat{x}})\in {\mathcal {H}}\times {\mathcal {H}}\), we have that \(\langle {\widehat{x}}-{\widehat{y}}, x-y\rangle \ge 0\) for all \((y,{\widehat{y}})\in {\mathcal {G}}(A)\) implies \({\widehat{x}}\in Ax\).

For a set-valued operator A, the resolvent associated with it is the mapping \(J_{\lambda }^{A}: {\mathcal {H}}\rightarrow 2^{{\mathcal {H}}}\) defined by

$$\begin{aligned} J_{\lambda }^{A}(x):= (I_{{\mathcal {H}}}+\lambda A)^{-1}(x), ~~x\in {\mathcal {H}},~\lambda >0. \end{aligned}$$

If A is maximal monotone and B is single-valued, then both \(J_{\lambda }^{A}\) and \(J_{\lambda }^{A}(I_{{\mathcal {H}}}-\lambda B)\) are single-valued and everywhere defined on \({\mathcal {H}}\). Furthermore, the resolvent \(J_{\lambda }^{A}\) is nonexpansive.

The following hold in a real Hilbert space \({\mathcal {H}}\):

$$\begin{aligned} 2\langle {\widehat{x}}, {\widehat{y}}\rangle =\Vert {\widehat{x}}\Vert ^{2}+\Vert {\widehat{y}}\Vert ^{2}-\Vert {\widehat{x}}-{\widehat{y}}\Vert ^{2}=\Vert {\widehat{x}}+{\widehat{y}}\Vert ^{2}-\Vert {\widehat{x}}\Vert ^{2}-\Vert {\widehat{y}}\Vert ^{2}~~~\forall ~ {\widehat{x}},{\widehat{y}}\in {\mathcal {H}} \end{aligned}$$
(2.1)

and

$$\begin{aligned} \Vert (1+{t}){\widehat{x}}-({t}-{s}){\widehat{y}}-s {\widehat{z}}\Vert ^2= & {} (1+t)\Vert {\widehat{x}}\Vert ^2-(t-s)\Vert {\widehat{y}}\Vert ^2-s\Vert {\widehat{z}}\Vert ^2 +(1+t)(t-s)\Vert {\widehat{x}}-{\widehat{y}}\Vert ^2\nonumber \\{} & {} +s(1+t)\Vert {\widehat{x}}-{\widehat{z}}\Vert ^2-s(t-s)\Vert {\widehat{y}}-{\widehat{z}}\Vert ^2~~\forall ~ {\widehat{x}},{\widehat{y}},{\widehat{z}} \in {\mathcal {H}},~ t,s \in {\mathbb {R}}.\nonumber \\ \end{aligned}$$
(2.2)

Lemma 2.1

Saejung and Yotkaew (2012) Suppose that \(\{p_{n}\}\) is a sequence of nonnegative real numbers, \(\{\alpha _{n}\}\) is a sequence of real numbers in (0, 1) satisfying \(\sum _{n=1}^\infty \alpha _n=\infty \), and \(\{q_n\}\) is a sequence of real numbers such that

$$\begin{aligned} p_{n+1}\le (1-\alpha _{n})p_{n}+\alpha _{n}q_{n},~n\ge 1. \end{aligned}$$

If \(\limsup \limits _{i\rightarrow \infty } q_{n_i}\le 0\) for each subsequence \(\{p_{n_i}\}\) of \(\{p_{n}\}\) satisfying \(\liminf \limits _{i\rightarrow \infty }\left( p_{n_i+1}-p_{n_i}\right) \ge 0\), then \(\lim \limits _{n\rightarrow \infty } p_n=0\).

Lemma 2.2

Lemaire (1997) Suppose that \(A:{\mathcal {H}}\rightarrow 2^{{\mathcal {H}}}\) is maximal monotone and \(B:{\mathcal {H}}\rightarrow {\mathcal {H}}\) is monotone Lipschitz continuous, then \((A+B):{\mathcal {H}}\rightarrow 2^{{\mathcal {H}}}\) is maximal monotone.

Lemma 2.3

Maingé (2007) Suppose that \(\{p_{n}\}\) and \(\{r_{n}\}\) are sequences of nonnegative real numbers such that

$$\begin{aligned} p_{n+1}\le (1-\alpha _{n})p_{n}+s_{n}+r_{n},\ \ n\ge 1, \end{aligned}$$

where \(\{\alpha _{n}\}\) is a sequence in (0, 1) and \(\{s_{n}\}\) is a real sequence. Let \(\sum _{n=1}^{\infty }r_{n}<\infty \) and \(s_{n}\le \alpha _{n}M\) for some \(M\ge 0\). Then, \(\{p_{n}\}\) is bounded.

3 Two-step inertial forward–reflected–anchored–backward splitting algorithm

In this section, we first propose and then study the convergence analysis of the following algorithm.

Algorithm 3.1

Let \(\lambda _0,\lambda _1>0\), \(\vartheta \in [0, 1)\), \(\beta \le 0\), \(\delta \in \Big (t,~\frac{1-2t}{2}\Big )\) with \(t\in (0, \frac{1}{4})\), and choose sequences \(\{\alpha _n\}\) in (0, 1) and \(\{e_n\}\) in \([0, \infty )\) such that \(\sum _{n=1}^{\infty } e_n< \infty \). For arbitrary \({\hat{v}}, x_{-1},x_0,x_1 \in {\mathcal {H}}\), let the sequence \(\{x_n\}\) be generated by

$$\begin{aligned} x_{n+1}= & {} J^{A}_{\lambda _n}\left( \alpha _n {\hat{v}}+(1-\alpha _n)\left[ x_n+\vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})\right] -\lambda _nBx_n \right. \\{} & {} \quad \left. -\lambda _{n-1}(1-\alpha _n)(Bx_{n}-Bx_{n-1})\right) , \end{aligned}$$

for all \(n\ge 1,\) where

$$\begin{aligned} \lambda _{n+1}={\left\{ \begin{array}{ll} \min \left\{ \frac{\delta \Vert x_n-x_{n+1}\Vert }{\Vert Bx_n-Bx_{n+1}\Vert },~\lambda _{n}+e_{n}\right\} , &{} \text{ if }~ Bx_n\ne Bx_{n+1},\\ \lambda _{n}+e_n,&{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$
(3.1)

Algorithm 3.1 is called a two-step inertial forward–reflected–anchored–backward splitting algorithms since it involves an anchor \({\hat{v}}\), an anchoring coefficient \(\alpha _n\), a two-step inertial term and the forward–reflected–backward splitting algorithm (1.5). This algorithm can also be viewed as a two-step inertial Halpern-type forward–reflected–backward method since it is based on the Halpern iteration. For more information on the convergence of Halpern-type methods for solving optimization problems, see, for example, Qi and Xu (2021); Yoon and Ryu (2021).

Assumption 3.2

  1. (a)

    A is maximal monotone,

  2. (b)

    B is monotone and Lipschitz continuous with constant \({L}>0\),

  3. (c)

    \((A+B)^{-1}(0)\) is nonempty,

  4. (d)

    \(\vartheta \) and \(\beta \) satisfy \(0\le \vartheta<\frac{1}{3}\left( 1-2(\frac{1}{2}-t)\right) , ~~\frac{1}{3+4\vartheta }\left( 3\vartheta -1+2(\frac{1}{2}-t)\right) <\beta \le 0.\)

Remark 3.3

By (3.1), \(\lim \nolimits _{n\rightarrow \infty }\lambda _n={\lambda }\), where \({\lambda }\in [\min \{\delta {L}^{-1}, \lambda _1\}, ~\lambda _1+e]\) with \(e=\sum _{n=1}^\infty e_n\) (see (Liu and Yang 2020)). If \(e_n=0\), then the step size \(\lambda _n\) in (3.1) is similar to the one in Bing and Cho (2021), which is derived from the paper (Yang and Liu 2019) for solving variational inequalities.

Lemma 3.4

Let \(\{x_n\}\) be generated by Algorithm 3.1 when Assumption 3.2 holds. If \(\lim \limits _{n\rightarrow \infty }\alpha _n=0\), then the sequence \(\{x_n\}\) is bounded.

Proof

Let \({\widehat{x}}\in (A+B)^{-1}(0)\) and \(u_n:=\alpha _n{\hat{v}}+(1-\alpha _n) v_n\), where \(v_n=x_n+\vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})\). Then

$$\begin{aligned} -\lambda _n B {\widehat{x}} \in \lambda _n A {\widehat{x}} \end{aligned}$$
(3.2)

and

$$\begin{aligned} u_n-\lambda _nBx_n-\lambda _{n-1}(1-\alpha _n)(Bx_{n}-Bx_{n-1})-x_{n+1}\in \lambda _n A x_{n+1}. \end{aligned}$$
(3.3)

Since A is monotone, we get from (3.2) and (3.3) that

$$\begin{aligned} \left\langle u_n-\lambda _nBx_n-\lambda _{n-1}(1-\alpha _n)(Bx_{n}-Bx_{n-1})-x_{n+1}+\lambda _n B{\widehat{x}},~ x_{n+1}-{\widehat{x}}\right\rangle \ge 0. \end{aligned}$$

This implies

$$\begin{aligned} 0\le & {} 2\left\langle x_{n+1}- u_n+\lambda _nBx_n+\lambda _{n-1}(1-\alpha _n)(Bx_{n}-Bx_{n-1})-\lambda _n B{\widehat{x}},~ {\widehat{x}}-x_{n+1}\right\rangle \nonumber \\= & {} 2\langle x_{n+1}-u_n,{\widehat{x}}-x_{n+1}\rangle +2\lambda _{n}\langle Bx_n-B{\widehat{x}},{\widehat{x}}-x_{n+1}\rangle +2\lambda _{n-1}(1-\alpha _n)\nonumber \\{} & {} \langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},x_{n}-x_{n+1}\rangle \nonumber \\= & {} \Vert u_n-{\widehat{x}}\Vert ^2-\Vert x_{n+1}-{\widehat{x}}\Vert ^2-\Vert x_{n+1}-u_n\Vert ^2 +2\lambda _{n}\langle Bx_n-B{\widehat{x}},{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},x_{n}-x_{n+1}\rangle ,\nonumber \\ \end{aligned}$$
(3.4)

where the last equation follows from (2.1).

Now, using the monotonicity of B, we get

$$\begin{aligned} \langle Bx_n-B{\widehat{x}}, {\widehat{x}}-x_{n+1}\rangle \le \langle Bx_n-Bx_{n+1}, {\widehat{x}}-x_{n+1}\rangle . \end{aligned}$$
(3.5)

Also, using (3.1), we have

$$\begin{aligned} 2\lambda _{n-1}\langle Bx_n-Bx_{n-1},x_n-x_{n+1}\rangle&\le 2\lambda _{n-1} \Vert Bx_n-Bx_{n-1}\Vert \Vert x_n-x_{n+1}\Vert \\&\le \frac{2\lambda _{n-1}}{\lambda _n}{\delta }\Vert x_n-x_{n-1}\Vert \Vert x_n-x_{n+1}\Vert \nonumber \\&\le \frac{\lambda _{n-1}}{\lambda _n}{\delta }\Big (\Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\Big ). \end{aligned}$$

Using Remark 3.3 and noting that \(\delta \in \Big (t,~\frac{1-2t}{2}\Big ),\) we see that \(\lim _{n\rightarrow \infty }\frac{\lambda _{n-1}}{\lambda _{n}}\delta =\delta <\frac{1}{2}-t\). Hence, there exists \(n_0\ge 1\) such that \(\frac{\lambda _{n-1}}{\lambda _{n}}\delta <\frac{1}{2}-t~\forall n\ge n_0\). Thus, we obtain that

$$\begin{aligned} 2\lambda _{n-1}\langle Bx_n-Bx_{n-1},x_n-x_{n+1}\rangle&\le \left( \frac{1}{2}-t\right) \Big (\Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\Big ). \end{aligned}$$
(3.6)

Putting (3.5) and (3.6) in (3.4), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} \le \Vert u_n-{\widehat{x}}\Vert ^2-\Vert x_{n+1}-u_n\Vert ^2 +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} +(1-\alpha _n)\big (\frac{1}{2}-t\big )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) ~\forall n\ge n_0. \end{aligned}$$
(3.7)

From (2.1), we get

$$\begin{aligned} \Vert u_n-{\widehat{x}}\Vert ^2= & {} \Vert (v_n-{\widehat{x}})-\alpha _n(v_n-{\hat{v}})\Vert ^2\nonumber \\= & {} \Vert v_n-{\widehat{x}}\Vert ^2+\alpha _n^2\Vert v_n-{\hat{v}}\Vert ^2-2\alpha _n\langle v_n-{\widehat{x}}, v_n-{\hat{v}}\rangle \nonumber \\= & {} \Vert v_n-{\widehat{x}}\Vert ^2+\alpha _n^2\Vert v_n-{\hat{v}}\Vert ^2-\alpha _n\Vert v_n-{\hat{v}}\Vert ^2-\alpha _n\Vert v_n-{\widehat{x}}\Vert ^2+\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2.\nonumber \\ \end{aligned}$$
(3.8)

Replacing \({\widehat{x}}\) by \(x_{n+1}\) in (3.8), we get

$$\begin{aligned} \Vert u_n-x_{n+1}\Vert ^2= & {} \Vert v_n-x_{n+1}\Vert ^2+\alpha _n^2\Vert v_n-{\hat{v}}\Vert ^2-\alpha _n\Vert v_n-{\hat{v}}\Vert ^2-\alpha _n\Vert v_n-x_{n+1}\Vert ^2\nonumber \\{} & {} +\alpha _n\Vert {\hat{v}}-x_{n+1}\Vert ^2. \end{aligned}$$
(3.9)

Now, subtracting (3.9) from (3.8), we obtain

$$\begin{aligned}{} & {} \Vert u_n-{\widehat{x}}\Vert ^2-\Vert u_n-x_{n+1}\Vert ^2\nonumber \\= & {} (1-\alpha _n)\Vert v_n-{\widehat{x}}\Vert ^2+\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2-(1-\alpha _n)\Vert x_{n+1}-v_n\Vert ^2-\alpha _n\Vert x_{n+1}-{\hat{v}}\Vert ^2.\nonumber \\ \end{aligned}$$
(3.10)

Using (3.10) in (3.7), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} \le (1-\alpha _n)\Vert v_n-{\widehat{x}}\Vert ^2+\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2-(1-\alpha _n)\Vert x_{n+1}-v_n\Vert ^2-\alpha _n\Vert x_{n+1}-{\hat{v}}\Vert ^2 \nonumber \\{} & {} +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} +(1-\alpha _n)\big (\frac{1}{2}-t\big )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) ~\forall n\ge n_0. \end{aligned}$$
(3.11)

From (2.2), we get

$$\begin{aligned} \Vert v_n-{\widehat{x}}\Vert ^2= & {} \Vert x_n+\vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})-{\widehat{x}}\Vert ^2\nonumber \\= & {} \Vert (1+\vartheta )(x_n-{\widehat{x}})-(\vartheta -\beta )(x_{n-1}-{\widehat{x}})-\beta (x_{n-2}-{\widehat{x}})\Vert ^2 \nonumber \\= & {} (1+\vartheta )\Vert x_n-{\widehat{x}}\Vert ^2-(\vartheta -\beta )\Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} +(1+\vartheta )(\vartheta -\beta )\Vert x_n-x_{n-1}\Vert ^2+\beta (1+\vartheta )\Vert x_n-x_{n-2}\Vert ^2\nonumber \\{} & {} -\beta (\vartheta -\beta )\Vert x_{n-1}-x_{n-2}\Vert ^2. \end{aligned}$$
(3.12)

Also, from (2.1), we get

$$\begin{aligned} \Vert x_{n+1}-v_n\Vert ^2= & {} \Vert x_{n+1}-x_n\Vert ^2-2\langle x_{n+1}-x_n, \vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})\rangle \nonumber \\{} & {} + \Vert \vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})\Vert ^2\nonumber \\{} & {} =\Vert x_{n+1}-x_n\Vert ^2-2\vartheta \langle x_{n+1}-x_n, x_n-x_{n-1}\rangle \nonumber \\{} & {} -2\beta \langle x_{n+1}-x_n,x_{n-1}-x_{n-2} \rangle +\vartheta ^2\Vert x_n-x_{n-1}\Vert ^2\nonumber \\{} & {} +2\beta \vartheta \langle x_n-x_{n-1},x_{n-1}-x_{n-2}\rangle +\beta ^2\Vert x_{n-1}-x_{n-2}\Vert ^2. \end{aligned}$$
(3.13)

Now, observe that

$$\begin{aligned} -2\vartheta \langle x_{n+1}-x_n, x_{n}-x_{n-1}\rangle\ge & {} -2\vartheta \Vert x_{n+1}-x_n\Vert \Vert x_{n}-x_{n-1}\Vert \nonumber \\\ge & {} -\vartheta \Vert x_{n+1}-x_n\Vert ^2-\vartheta \Vert x_{n}-x_{n-1}\Vert ^2, \end{aligned}$$
(3.14)
$$\begin{aligned} -2\beta \langle x_{n+1}-x_n, x_{n-1}-x_{n-2}\rangle\ge & {} -2|\beta |\Vert x_{n+1}-x_n\Vert \Vert x_{n-1}-x_{n-2}\Vert \nonumber \\\ge & {} -|\beta |\Vert x_{n+1}-x_n\Vert ^2-|\beta |\Vert x_{n-1}-x_{n-2}\Vert ^2, \end{aligned}$$
(3.15)
$$\begin{aligned} 2\beta \vartheta \langle x_{n}-x_{n-1}, x_{n-1}-x_{n-2}\rangle\ge & {} -2|\beta |\vartheta \Vert x_{n}-x_{n-1}\Vert \Vert x_{n-1}-x_{n-2}\Vert \nonumber \\\ge & {} -|\beta |\vartheta \Vert x_{n}-x_{n-1}\Vert ^2-|\beta |\vartheta \Vert x_{n-1}-x_{n-2}\Vert ^2.\nonumber \\ \end{aligned}$$
(3.16)

Putting (3.14), (3.15) and (3.16) in (3.13), we obtain

$$\begin{aligned} \Vert x_{n+1}-v_n\Vert ^2\ge & {} \Vert x_{n+1}-x_n\Vert ^2-\vartheta \Vert x_{n+1}-x_n\Vert ^2-\vartheta \Vert x_{n}-x_{n-1}\Vert ^2 \nonumber \\{} & {} -|\beta |\Vert x_{n+1}-x_n\Vert ^2-|\beta |\Vert x_{n-1}-x_{n-2}\Vert ^2+\vartheta ^2\Vert x_n-x_{n-1}\Vert ^2\nonumber \\{} & {} -|\beta |\vartheta \Vert x_{n}-x_{n-1}\Vert ^2-|\beta |\vartheta \Vert x_{n-1}-x_{n-2}\Vert ^2+\beta ^2\Vert x_{n-1}-x_{n-2}\Vert ^2\nonumber \\{} & {} =(1-\vartheta -|\beta |)\Vert x_{n+1}-x_n\Vert ^2+(\vartheta ^2-\vartheta -\vartheta |\beta |)\Vert x_{n}-x_{n-1}\Vert ^2\nonumber \\{} & {} +(\beta ^2-|\beta |-\vartheta |\beta |)\Vert x_{n-1}-x_{n-2}\Vert ^2. \end{aligned}$$
(3.17)

Now, putting (3.12) and (3.17) in (3.11), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} \le (1-\alpha _n)\Big [(1+\vartheta )\Vert x_n-{\widehat{x}}\Vert ^2-(\vartheta -\beta )\Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} +(1+\vartheta )(\vartheta -\beta )\Vert x_n-x_{n-1}\Vert ^2+\beta (1+\vartheta )\Vert x_n-x_{n-2}\Vert ^2-\beta (\vartheta -\beta )\Vert x_{n-1}-x_{n-2}\Vert ^2\Big ]\nonumber \\{} & {} +\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2-(1-\alpha _n)\Big [(1-\vartheta -|\beta |)\Vert x_{n+1}-x_n\Vert ^2+(\vartheta ^2-\vartheta -\vartheta |\beta |)\Vert x_{n}-x_{n-1}\Vert ^2\nonumber \\{} & {} +(\beta ^2-|\beta |-\vartheta |\beta |)\Vert x_{n-1}-x_{n-2}\Vert ^2\Big ]-\alpha _n\Vert x_{n+1}-{\hat{v}}\Vert ^2 +2\lambda _{n-1}(1-\alpha _n) \nonumber \\{} & {} \times \langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle +(1-\alpha _n)\bigg (\frac{1}{2}-t\bigg )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) \\{} & {} \le (1-\alpha _n)\Big [(1+\vartheta )\Vert x_n-{\widehat{x}}\Vert ^2-(\vartheta -\beta )\Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} +(2\vartheta -\beta -\vartheta \beta +\vartheta |\beta |)\Vert x_n-x_{n-1}\Vert ^2+(|\beta |+|\beta |\vartheta -\beta \vartheta )\Vert x_{n-1}-x_{n-2}\Vert ^2\\{} & {} -(1-\vartheta -|\beta |)\Vert x_{n+1}-x_n\Vert ^2 +2\lambda _{n-1} \langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \Big ]\nonumber \\{} & {} +\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+(1-\alpha _n)\bigg (\frac{1}{2}-t\bigg )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) ~\forall n\ge n_0. \end{aligned}$$

This implies that

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2-\vartheta \Vert x_n-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_{n+1}-x_n\Vert ^2 \le (1-\alpha _n)\Big [\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} -\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\Big ]+\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} +(1-\alpha _n)\left[ 2(\frac{1}{2}-t)+3\vartheta -1+(1+\vartheta )(|\beta |-\beta )\right] \Vert x_n-x_{n-1}\Vert ^2\nonumber \\{} & {} +(1-\alpha _n)\left[ |\beta |+|\beta |\vartheta -\beta \vartheta \right] \Vert x_{n-1}-x_{n-2}\Vert ^2 = (1-\alpha _n)\nonumber \\{} & {} \times \Big [\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\nonumber \\{} & {} -\left( 2(\frac{1}{2}-t)+3\vartheta -1+(1+\vartheta )(|\beta |-\beta )\right) \left( \Vert x_{n-1}-x_{n-2}\Vert ^2-\Vert x_n-x_{n-1}\Vert ^2\right) \Big ]+\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} -(1-\alpha _n)\left[ -\left( 2(\frac{1}{2}-t)+3\vartheta -1+(1+\vartheta )(|\beta |-\beta )\right) -\left( |\beta |+|\beta |\vartheta -\beta \vartheta \right) \right] \Vert x_{n-1}\nonumber \\{} & {} \quad -x_{n-2}\Vert ^2~\forall n\ge n_0. \end{aligned}$$
(3.18)

Set \(c_1:=-\left( 2(\frac{1}{2}-t)+3\vartheta -1+(1+\vartheta )(|\beta |-\beta )\right) \), \(c_2:=1-3\vartheta -2(\frac{1}{2}-t)-2|\beta |-2\vartheta |\beta |+\beta +2\vartheta \beta \) and \(p_n:=\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2+c_1\Vert x_{n-1}-x_{n-2}\Vert ^2\). Then, (3.18) becomes

$$\begin{aligned} p_{n+1}\le & {} (1-\alpha _n)p_n +\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2-(1-\alpha _n)c_2\Vert x_{n-1}-x_{n-2}\Vert ^2 ~\forall n\ge n_0. \end{aligned}$$
(3.19)

Next, we show that \(c_1\), \(c_2\) are positive and \(p_n\) is nonnegative. From Assumption 3.2 (d), \(3\vartheta -1+2(\frac{1}{2}-t)<0\). Thus, \(\frac{1}{2+2\vartheta }\left( 3\vartheta -1+2(\frac{1}{2}-t)\right)<\frac{1}{3+4\vartheta }\left( 3\vartheta -1+2(\frac{1}{2}-t)\right) <\beta \), which implies that \(3\vartheta -1+2\left( \frac{1}{2}-t\right) -2\beta -2\vartheta \beta <0\).

Since \(|\beta |=-\beta \), we obtain

$$\begin{aligned} 3\vartheta -1+2\left( \frac{1}{2}-t\right) +|\beta |-\beta +\vartheta |\beta |-\vartheta \beta <0, \end{aligned}$$
(3.20)

which implies that \(c_1>0\).

Now, using \(\frac{1}{3+4\vartheta }\left( 3\vartheta -1+2(\frac{1}{2}-t)\right) <\beta \), we obtain that \(1-3\vartheta -2(\frac{1}{2}-t)+3\beta +4\beta \vartheta >0.\) Since \(|\beta |=-\beta \), we get

$$\begin{aligned} 1-3\vartheta -2\left( \frac{1}{2}-t\right) -2|\beta |+\beta -2|\beta |\vartheta +2\beta \vartheta >0. \end{aligned}$$
(3.21)

Hence, \(c_2>0\).

On the other hand, since \(\beta \le 0\) and \(c_1>0\), we get for all \(n\ge n_0\), that

$$\begin{aligned} p_n\ge & {} \Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} \quad +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\nonumber \\\ge & {} \Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2 -\frac{\lambda _{n-1}}{\lambda _{n}} \delta \left( \Vert x_n-x_{n-1}\Vert ^2+ \Vert x_{n}-{\widehat{x}}\Vert ^2\right) +(1-|\beta |-\vartheta \nonumber \\{} & {} \quad -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\nonumber \\\ge & {} \Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \left( 2\Vert x_n-x_{n-1}\Vert ^2+2\Vert x_{n}-{\widehat{x}}\Vert ^2\right) \nonumber \\{} & {} -\left( \frac{1}{2}-t\right) \left( \Vert x_n-x_{n-1}\Vert ^2+ \Vert x_{n}-{\widehat{x}}\Vert ^2\right) +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\nonumber \\= & {} \left( 1-2\vartheta -\left( \frac{1}{2}-t\right) \right) \Vert x_n-{\widehat{x}}\Vert ^2+(1-|\beta |-3\vartheta -2(\frac{1}{2}-t))\Vert x_n-x_{n-1}\Vert ^2\nonumber \\\ge & {} \left( 1-3\vartheta -\left( \frac{1}{2}-t\right) \right) \Vert x_n-{\widehat{x}}\Vert ^2+\left( 1-|\beta |-3\vartheta -2\left( \frac{1}{2}-t\right) \right) \Vert x_n-x_{n-1}\Vert ^2.\nonumber \\ \end{aligned}$$
(3.22)

Since \(\vartheta <\frac{1}{3}\left( 1-2(\frac{1}{2}-t)\right) \), we get \(3\vartheta -1+2(\frac{1}{2}-t)<0\). This imply that \(1-3\vartheta -(\frac{1}{2}-t)>0\), and \( 3\vartheta -1+2(\frac{1}{2}-t)<\frac{1}{3+4\theta }(3\vartheta -1+2(\frac{1}{2}-t))<\beta \). Hence, \(-|\beta |-3\vartheta +1-2(\frac{1}{2}-t))>0\). Therefore, we get from (3.22) that \(p_n\ge 0\) for all \(n\ge n_0\). Using these facts in (3.19), we obtain that \(\{p_n\}\) is bounded. It then follows from (3.22) that the sequence \(\{x_n\}\) is indeed bounded, as claimed. \(\square \)

We now state and prove the convergence theorem of this paper.

Theorem 3.5

Let \(\{x_n\}\) be generated by Algorithm 3.1 when Assumption 3.2 holds. If \(\lim _{n\rightarrow \infty }\alpha _n=0\) and \(\sum _{n=1}^\infty \alpha _n=\infty \), then \(\{x_n\}\) converges strongly to \(P_{(A+B)^{-1}(0)}{\hat{v}}\).

Proof

Let \({\widehat{x}}=P_{(A+B)^{-1}(0)}{\hat{v}}\). Then by (2.1), we get

$$\begin{aligned} \Vert u_n-{\widehat{x}}\Vert ^2= & {} \Vert \alpha _n({\hat{v}}-{\widehat{x}})+(1-\alpha _n)(v_n-{\widehat{x}})\Vert ^2\nonumber \\= & {} \alpha _n^2\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+(1-\alpha _n)^2\Vert v_n-{\widehat{x}}\Vert ^2+2\alpha _n(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle \nonumber \\ \end{aligned}$$
(3.23)

Again, using (2.1), we get

$$\begin{aligned} \Vert u_n-x_{n+1}\Vert ^2= & {} \alpha _n^2\Vert {\hat{v}}-x_{n+1}\Vert ^2+(1-\alpha _n)^2\Vert v_n-x_{n+1}\Vert ^2+2\alpha _n(1-\alpha _n)\langle {\hat{v}}-x_{n+1}, v_n-x_{n+1}\rangle \nonumber \\\ge & {} \alpha _n^2\Vert x_{n+1}-{\hat{v}}\Vert ^2+(1-\alpha _n)^2\Vert x_{n+1}-v_n\Vert ^2-2\alpha _n(1-\alpha _n)\Vert x_{n+1}-{\hat{v}}\Vert \Vert x_{n+1}-v_n\Vert \nonumber \\\ge & {} \alpha _n^2\Vert x_{n+1}-{\hat{v}}\Vert ^2+(1-\alpha _n)^2\Vert x_{n+1}-v_n\Vert ^2-2\alpha _n(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert , \end{aligned}$$
(3.24)

where \(M:=\sup \limits _{n\ge 1}\Vert x_{n+1}-{\hat{v}}\Vert \) which exists due to the boundedness of \(\{x_n\}\) proved in Lemma 3.4.

Now, using (3.23) and (3.24) in (3.7), we see that

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} \quad \le \alpha _n^2\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+(1-\alpha _n)^2\Vert v_n-{\widehat{x}}\Vert ^2+2\alpha _n(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle \nonumber \\{} & {} \qquad -\left( \alpha _n^2\Vert x_{n+1}-{\hat{v}}\Vert ^2+(1-\alpha _n)^2\Vert x_{n+1}-v_n\Vert ^2-2\alpha _n(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \right) \nonumber \\{} & {} \qquad +2\lambda _{n-1}(1-\alpha _n)\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \nonumber \\{} & {} \qquad +(1-\alpha _n)\big (\frac{1}{2}-t\big )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) \nonumber \\{} & {} \quad \le (1-\alpha _n)\left( \Vert v_n-{\widehat{x}}\Vert ^2+2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \right) \nonumber \\{} & {} \qquad +\alpha _n\left( \alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+2(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle +2(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \right) \nonumber \\{} & {} \qquad -(1-\alpha _n)^2\Vert x_{n+1}-v_n\Vert ^2 +(1-\alpha _n)\big (\frac{1}{2}-t\big )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) ~\forall n\ge n_0.\nonumber \\ \end{aligned}$$
(3.25)

Next, putting (3.12) and (3.17) in (3.25), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \nonumber \\{} & {} \quad \le (1-\alpha _n)\Big [(1+\vartheta )\Vert x_n-{\widehat{x}}\Vert ^2-(\vartheta -\beta )\Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2\nonumber \\{} & {} \qquad +(1+\vartheta )(\vartheta -\beta )\Vert x_n-x_{n-1}\Vert ^2+\beta (1+\vartheta )\Vert x_n-x_{n-2}\Vert ^2\nonumber \\{} & {} \qquad -\beta (\vartheta -\beta )\Vert x_{n-1}-x_{n-2}\Vert ^2+2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \Big ]\nonumber \\{} & {} \qquad +\alpha _n\left( \alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+2(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle +2(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \right) \nonumber \\{} & {} \qquad -(1-\alpha _n)^2\Big [(1-\vartheta -|\beta |)\Vert x_{n+1}-x_n\Vert ^2+(\vartheta ^2-\vartheta -\vartheta |\beta |)\Vert x_{n}-x_{n-1}\Vert ^2\nonumber \\{} & {} \qquad +(\beta ^2-|\beta |-\vartheta |\beta |)\Vert x_{n-1}-x_{n-2}\Vert ^2\Big ]\\{} & {} \qquad +(1-\alpha _n)\big (\frac{1}{2}-t\big )\left( \Vert x_n-x_{n-1}\Vert ^2+\Vert x_{n+1}-x_n\Vert ^2\right) . \end{aligned}$$

This implies that

$$\begin{aligned}{} & {} \Vert x_{n+1}-{\widehat{x}}\Vert ^2-\vartheta \Vert x_n-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-1}-{\widehat{x}}\Vert ^2 +2\lambda _{n}\langle Bx_{n+1}-Bx_n,{\widehat{x}}-x_{n+1}\rangle \\{} & {} \qquad +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_{n+1}-x_n\Vert ^2 \\{} & {} \quad \le (1-\alpha _n)\Big [\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \\{} & {} \qquad +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\Big ]\\{} & {} \qquad +\alpha _n\left( \alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+2(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle +2(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \right) \\{} & {} \qquad +(1-\alpha _n)\left[ 2(\frac{1}{2}-t)+2\vartheta -\beta -\vartheta \beta -1+|\beta |+\vartheta ^2-(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)\right] \\{} & {} \Vert x_n-x_{n-1}\Vert ^2 +(1-\alpha _n)\left[ \beta ^2-\beta \vartheta -(1-\alpha _n)(\beta ^2-|\beta |-\vartheta |\beta |\right] \Vert x_{n-1}-x_{n-2}\Vert ^2\\{} & {} \quad = (1-\alpha _n)\Big [\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle \\{} & {} \qquad +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2\\{} & {} -\left( 2(\frac{1}{2}-t)+2\vartheta -\beta -\vartheta \beta -1+|\beta |+\vartheta ^2-(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)\right) \\{} & {} \quad \left( \Vert x_{n-1}-x_{n-2}\Vert ^2-\Vert x_n-x_{n-1}\Vert ^2\right) \Big ]\\{} & {} \qquad +\alpha _n\left( \alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+2(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle +2(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \right) \\{} & {} \qquad -(1-\alpha _n)\Big [1-2\vartheta -2(\frac{1}{2}-t)+\beta +2\vartheta \beta -|\beta |-\beta ^2-\vartheta ^2+(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)\\{} & {} \qquad +(1-\alpha _n)(\beta ^2-|\beta |-\vartheta |\beta |)\Big ]\Vert x_{n-1}-x_{n-2}\Vert ^2~\forall n\ge n_0. \end{aligned}$$

That is,

$$\begin{aligned} t_{n+1}\le & {} (1-\alpha _n) t_n +\alpha _n q_n -(1-\alpha _n) d_n\Vert x_{n-1}-x_{n-2}\Vert ^2 ~\forall n\ge n_0, \end{aligned}$$
(3.26)

where \(d_n=1-2\vartheta -2(\frac{1}{2}-t)+\beta +2\vartheta \beta -|\beta |-\beta ^2-\vartheta ^2+(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)+(1-\alpha _n)(\beta ^2-|\beta |-\vartheta |\beta |)\), \(t_n=\Vert x_n-{\widehat{x}}\Vert ^2-\vartheta \Vert x_{n-1}-{\widehat{x}}\Vert ^2-\beta \Vert x_{n-2}-{\widehat{x}}\Vert ^2 +2\lambda _{n-1}\langle Bx_n-Bx_{n-1},{\widehat{x}}-x_{n}\rangle +(1-|\beta |-\vartheta -\frac{1}{2}+t)\Vert x_n-x_{n-1}\Vert ^2+c_n\Vert x_{n-1}-x_{n-2}\Vert ^2\), \(c_n=-\left[ 2(\frac{1}{2}-t)+2\vartheta -\beta -\vartheta \beta -1+|\beta |+\vartheta ^2-(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)\right] \) and \(q_n=\alpha _n\Vert {\hat{v}}-{\widehat{x}}\Vert ^2+2(1-\alpha _n)\langle {\hat{v}}-{\widehat{x}}, v_n-{\widehat{x}}\rangle +2(1-\alpha _n)M \Vert x_{n+1}-v_n\Vert \).

From (3.20), we have \(1-3\vartheta -2(\frac{1}{2}-t)-|\beta |+\beta -\vartheta |\beta |+\vartheta \beta >0,\) which implies that

$$\begin{aligned} \lim _{n\rightarrow \infty } c_n\!=\! & {} \lim _{n\rightarrow \infty } \!-\!\left[ 2\left( \frac{1}{2}-t\right) +2\vartheta -\beta -\vartheta \beta -1+|\beta |+\vartheta ^2-(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)\right] \nonumber \\= & {} 1-3\vartheta -2 \left( \frac{1}{2}-t\right) -|\beta |+\beta -\vartheta |\beta |+\vartheta \beta >0. \end{aligned}$$

Thus, there exists \(n_1\ge n_0\) such that \(c_n>0\) for all \(n\ge n_1\). Also, from (3.21), we have \(1-3\vartheta -2\left( \frac{1}{2}-t\right) -2|\beta |+\beta -2|\beta |\vartheta +2\beta \vartheta >0\), which implies

$$\begin{aligned} \lim _{n\rightarrow \infty } d_n= & {} \lim _{n\rightarrow \infty } \Big [1-2\vartheta -2\left( \frac{1}{2}-t\right) +\beta +2\vartheta \beta -|\beta |-\beta ^2-\vartheta ^2\\{} & {} +(1-\alpha _n)(\vartheta ^2-\vartheta -\vartheta |\beta |)+(1-\alpha _n)(\beta ^2-|\beta |-\vartheta |\beta |)\Big ]\nonumber \\= & {} 1-3\vartheta -2\left( \frac{1}{2}-t\right) -2|\beta |+\beta -2|\beta |\vartheta +2\beta \vartheta >0. \end{aligned}$$

There exists \(n_2\ge n_0\) such that \(d_n>0\) for all \(n\ge n_2\). Therefore,

$$\begin{aligned} t_{n+1}\le & {} (1-\alpha _n)t_n+\alpha _n q_n,~\forall n\ge n_2. \end{aligned}$$
(3.27)

Let \(\{t_{n_i}\}\) be a subsequence of \(\{t_n\}\) such that \(\liminf _{i\rightarrow \infty }\left( t_{n_i+1}-t_{n_i}\right) \ge 0\). Then, it follows from (3.26) that

$$\begin{aligned}{} & {} \limsup _{i\rightarrow \infty }\left[ (1-\alpha _{n_i})d_{n_i}\Vert x_{{{n_i}}-1}-x_{{{n_i}}-2}\Vert ^2\right] \\{} & {} \quad \le \limsup _{i\rightarrow \infty }\left[ (t_{n_i}-t_{{n_i}+1})+\alpha _{n_i} (q_{n_i}-t_{n_i})\right] \\{} & {} \quad \le -\liminf _{i\rightarrow \infty }(t_{{n_i}+1}-t_{n_i})\le 0. \end{aligned}$$

Since \(\lim _{i\rightarrow \infty }(1-\alpha _{n_i})d_{n_i}>0\), we get

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert x_{{{n_i}}-1}-x_{{{n_i}}-2}\Vert =0=\lim _{i\rightarrow \infty }\Vert x_{{n_i}+1}-x_{n_i}\Vert . \end{aligned}$$
(3.28)

Thus,

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert v_{{n_i}}-x_{n_i}\Vert =\lim _{i\rightarrow \infty }\Vert \vartheta (x_{n_i}-x_{n_i-1})+\beta (x_{n_i-1}-x_{n_i-2})\Vert =0. \end{aligned}$$
(3.29)

Using (3.28) and (3.29), we obtain

$$\begin{aligned} \lim _{i\rightarrow \infty } \Vert x_{n_i+1}-v_{n_i}\Vert =0. \end{aligned}$$
(3.30)

Since \(\lim _{n\rightarrow \infty }\alpha _n=0\), we get

$$\begin{aligned} \lim _{i\rightarrow \infty } \Vert u_{n_i}-v_{n_i}\Vert =\lim _{i\rightarrow \infty }\alpha _{n_i}\Vert {\hat{v}}-v_{n_i}\Vert =0. \end{aligned}$$
(3.31)

Using (3.30) and (3.31), we obtain

$$\begin{aligned} \lim _{i\rightarrow \infty } \Vert u_{n_i}-x_{n_i+1}\Vert =0. \end{aligned}$$
(3.32)

From (3.28) and the Lipschitz continuity of B, we find that

$$\begin{aligned} \lim _{i\rightarrow \infty }\Vert Bx_{{n_i}+1}-Bx_{n_i}\Vert =0. \end{aligned}$$
(3.33)

In the light of Lemma 3.4, we see that \(\{x_{n_i}\}\) is bounded. Thus, we can choose a subsequence \(\{x_{n_{i_j}}\}\) of \(\{x_{n_i}\}\) such that \(\{x_{n_{i_j}}\}\) converges weakly to some \(x^*\in {\mathcal {H}}\), and

$$\begin{aligned} \limsup _{i\rightarrow \infty }\langle {\hat{v}}-{\widehat{x}}, x_{n_i}-{\widehat{x}}\rangle =\lim _{j\rightarrow \infty }\langle {\hat{v}}-{\widehat{x}}, x_{{n_i}_j}-{\widehat{x}}\rangle =\langle {\hat{v}}-{\widehat{x}}, x^*-{\widehat{x}}\rangle . \end{aligned}$$
(3.34)

Now, consider \((u, v)\in {\mathcal {G}}(A+B)\). Then \(\lambda _{n_{i_j}}(v-Bu)\in \lambda _{n_{i_j}}Au\). Using this, (3.3) and the monotonicity of A, we find that

$$\begin{aligned} \langle \lambda _{n_{i_j}}(v-Bu){} & {} -u_{n_{i_j}}+\lambda _{n_{i_j}}Bx_{n_{i_j}}+\lambda _{{n_{i_j}}-1}(1-\alpha _{n_{i_j}})(Bx_{n_{i_j}}-Bx_{{n_{i_j}}-1})+x_{{n_{i_j}}+1}, u \nonumber \\{} & {} - x_{{n_{i_j}}+1}\rangle \ge 0. \end{aligned}$$

Thus, using the monotonicity of B, we obtain

$$\begin{aligned}{} & {} \langle v, u-x_{{n_{i_j}}+1}\rangle \ge \frac{1}{ \lambda _{n_{i_j}}} \langle \lambda _{n_{i_j}}Bu+u_{n_{i_j}}-\lambda _{n_{i_j}}Bx_{n_{i_j}}-\lambda _{{n_{i_j}}-1}(1-\alpha _{n_{i_j}})\nonumber \\{} & {} (Bx_{n_{i_j}}-Bx_{{n_{i_j}}-1})-x_{{n_{i_j}}+1}, u - x_{n_j+1}\rangle \nonumber \\{} & {} \quad = \langle Bu-Bx_{{n_{i_j}}+1}, u - x_{{n_{i_j}}+1}\rangle + \langle Bx_{{n_{i_j}}+1}-Bx_{n_{i_j}}, u - x_{{n_{i_j}}+1}\rangle \nonumber \\{} & {} \qquad +\frac{\lambda _{{n_{i_j}}-1}}{ \lambda _{n_{i_j}}}(1-\alpha _{n_{i_j}})\langle Bx_{{n_{i_j}}-1}-Bx_{n_{i_j}}, u - x_{{n_{i_j}}+1}\rangle + \frac{1}{ \lambda _{n_{i_j}}}\langle u_{n_{i_j}}-x_{{n_{i_j}}+1}, u - x_{{n_{i_j}}+1}\rangle \nonumber \\{} & {} \quad \ge \langle Bx_{{n_{i_j}}+1}-Bx_{n_{i_j}}, u - x_{{n_{i_j}}+1}\rangle +\frac{\lambda _{{n_{i_j}}-1}}{ \lambda _{n_{i_j}}}(1-\alpha _{n_{i_j}})\langle Bx_{{n_{i_j}}-1}-Bx_{n_{i_j}}, u - x_{{n_{i_j}}+1}\rangle \nonumber \\{} & {} \qquad + \frac{1}{ \lambda _{n_{i_j}}}\langle u_{n_{i_j}}-x_{{n_{i_j}}+1}, u - x_{{n_{i_j}}+1}\rangle . \end{aligned}$$
(3.35)

As \(j\rightarrow \infty \) in (3.35), we obtain, using (3.32) and (3.33), that \(\langle v, u-x^*\rangle \ge 0\). By Lemma 2.2, \(A+B\) is maximal monotone. Hence, we get that \(x^*\in (A+B)^{-1}(0)\).

Since \({\widehat{x}}=P_{(A+B)^{-1}(0)}{\hat{v}}\), it follows from (3.34) and the characterization of the metric projection that

$$\begin{aligned} \limsup _{i\rightarrow \infty }\langle {\hat{v}}-{\widehat{x}}, x_{n_i}-{\widehat{x}}\rangle =\langle {\hat{v}}-{\widehat{x}}, x^*-{\widehat{x}}\rangle \le 0. \end{aligned}$$
(3.36)

Using (3.29), (3.30) and (3.36), we obtain that \(\limsup _{i\rightarrow \infty }q_{n_{i}}\le 0\). Thus, in view of the condition \(\sum _{n=1}^\infty \alpha _n=\infty \), Lemma 2.1 and (3.27), we see that \(\lim _{n\rightarrow \infty }t_n=0\). This together with (3.22) imply that \(\{x_n\}\) converges strongly to \({\widehat{x}}=P_{(A+B)^{-1}(0)}{\hat{v}}\), as asserted. \(\square \)

The step size defined in (3.1) makes it possible for Algorithm 3.1 to be applied in practice even when the Lipschitz constant L of B is not known. However, when this constant is known or can be calculated, we simply adopt the following variant of Algorithm 3.1:

Algorithm 3.6

Let \(\lambda \in \Big (0,~\frac{1}{2\,L}\Big )\), \(\vartheta \in [0, 1)\), \(\beta \le 0\), and choose the sequence \(\{\alpha _n\}\) in (0, 1). For arbitrary \({\hat{v}}, x_{-1}, x_0,x_1 \in {\mathcal {H}}\), let the sequence \(\{x_n\}\) be generated by

$$\begin{aligned} x_{n+1}= & {} J^{A}_{\lambda }\big (\alpha _n {\hat{v}}+(1-\alpha _n) \left[ x_n+\vartheta (x_n-x_{n-1})+\beta (x_{n-1}-x_{n-2})\right] \\{} & {} \quad -\lambda Bx_n-\lambda (1-\alpha _n)(Bx_{n}-Bx_{n-1})\big ),~ n\ge 1. \end{aligned}$$

Remark 3.7

Using arguments similar to those in Lemma 3.4 and Theorem 3.5, we can establish that the sequence \(\{x_n\}\) generated by Algorithm 3.6 converges strongly to \(P_{(A+B)^{-1}(0)}{\hat{v}}\).

Remark 3.8

  1. (a)

    We obtained strong convergence results for Algorithm 3.1 without assuming that either A or B is strongly monotone (a condition that is quite restrictive). Rather, we modified the forward–reflected–backward splitting algorithm in Malitsky and Tam (2020) appropriately to obtain our strong convergent results.

  2. (b)

    Compared to Izuchukwu et al. (2023), we proved the strong convergence of Algorithm 3.1 using the double inertial technique.

Remark 3.9

A more careful examination of Algorithm 3.1 and it convergence analysis can help us to relax the Lipschitz continuity on B to uniform continuity (see, for example (Thong et al. (2023), page 1114). In a finite-dimensional space, B can even be continuous (see (Izuchukwu and Shehu 2021, Section 3)). However, as seen in these papers, this relaxation may be achieved with the cost of having strict restrictions on the stepsize \(\{\lambda _n\}\) (e.g., through some linesearch techniques). Therefore, we intend to investigate these restrictions in detail in a different project in the future.

4 Numerical illustrations

In this section, using test examples which originate from image restoration problem, as well as an academic example, we compare Algorithm 3.1 with other strongly convergent algorithms (Alakoya et al. 2022, Algorithm 3.2), (Bing and Cho (2021), Algorithm 3.3 and Algorithm 3.4) and (Tan et al. 2022, Algorithm 3.4).

Example 4.1

We consider the image restoration problem:

$$\begin{aligned} \min _{{\hat{x}}\in {\mathbb {R}}^m} ~\{||{\mathcal {D}}{\hat{x}}-{\hat{c}}||^2_2 +\lambda ||{\hat{x}}||_1\}, \end{aligned}$$
(4.1)

where \(\lambda >0\) (we take \(\lambda =1\)), \({\hat{x}}\in {\mathbb {R}}^m\) is the original image to be restored, \({\hat{c}}\in {\mathbb {R}}^N\) is the observed image and \({\mathcal {D}}: {\mathbb {R}}^m\rightarrow {\mathbb {R}}^N\) is the blurring operator. The quality of the restored image is measured by

$$\begin{aligned} \text{ SNR } = 20 \times \log _{10}\left( \frac{\Vert {\hat{x}}\Vert _2}{\Vert {\hat{x}}-x^*\Vert _2}\right) , \end{aligned}$$

where SNR means signal-to-noise ratio, and \(x^*\) is the recovered image.

For the implementation, we take \(x_0 = {\textbf {0}} \in {\mathbb {R}}^{m\times m}\) and \(x_{-1}=x_1 = {\textbf {1}} \in {\mathbb {R}}^{m\times N}\), and use the following image found in the MATLAB Image Processing Toolbox:

  1. (a)

    Tire Image of size \(205 \times 232\). To create the blurred and noisy image (observed image), we use the Gaussian blur of size \(9\times 9\) and standard deviation \(\sigma =4\).

  2. (b)

    Cameraman Image of size \(256 \times 256\). We use the Gaussian blur of size \(7\times 7\) and standard deviation \(\sigma =4\).

  3. (c)

    Medical Resonance Imaging (MRI) of size \(128 \times 128\). We use the Gaussian blur of size \(7\times 7\) and standard deviation \(\sigma =4\).

Table 1 Numerical results for Example 4.1
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Table 2 Numerical results for Example 4.2 with \(\epsilon =10^{-7}\)
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Fig. 1
figure 1

Numerical results for Tire

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Example 4.2

Let \({\mathcal {H}}={\mathfrak {l}}_2({\mathbb {R}}):=\{x=(x_1, x_2,...,x_i,...),~~x_i \in {\mathbb {R}}:\sum \nolimits _{i=1}^{\infty }|x_i|^2<\infty \}\) and \(||x||_{\mathfrak {l}_2}:= \left( \sum \nolimits _{i=1}^\infty |x_i|^2 \right) ^{\frac{1}{2}} \; ~~\forall x \in \mathfrak {l}_2({\mathbb {R}}).\)

Define \(A,B:\mathfrak {l}_2\rightarrow \mathfrak {l}_2\) by

$$\begin{aligned} Ax:= \left( 2x_1, 2x_2,...,2x_i,...\right) , ~\forall x\in \mathfrak {l}_2 \end{aligned}$$

and

$$\begin{aligned} Bx:= \left( \frac{x_1+|x_1|}{2}, \frac{x_2+|x_2|}{2},...,\frac{x_i+|x_i|}{2},...\right) , ~\forall x\in \mathfrak {l}_2. \end{aligned}$$

Then, A is maximal monotone and B is Lipschitz continuous and monotone with Lipschitz constant \({L}=1\).

Fig. 2
figure 2

Numerical results for Cameraman

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Fig. 3
figure 3

Numerical results for MRI

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Fig. 4
figure 4

The behavior of \(\text{ TOL}_n\) for Example 4.2 with \(\epsilon =10^{-7}\): Top Left: Case 1; Top Right: Case 2; Bottom left: Case 3; Bottom Right: Case 4

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For the implementation, we take the starting points:

Case 1: \(x_0=(\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \cdots )\), \(x_{-1}=x_1= (\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \cdots ) \).

Case 2: \(x_0=(\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \cdots )\), \(x_{-1}=x_1= (\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \cdots )\).

Case 3: \(x_0=(1, \frac{1}{2}, \frac{1}{4}, \cdots )\), \(x_{-1}=x_1= (\frac{4}{5}, \frac{16}{25}, \frac{64}{125}, \cdots )\).

Case 4: \(x_0= (1, \frac{1}{4}, \frac{1}{9}, \cdots )\), \(x_{-1}=x_1= (\frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \cdots ).\)

During the implementation, we make use of the following:

  • Algorithm 3.1: \(\lambda _0 = 0.1, \lambda _1 = 0.3, \delta =0.25, \alpha _n=\frac{0.005}{3n+25000}, e_n = \frac{16}{(n+1)^{1.1}}, t=0.2, \vartheta = 0.12, \beta =\{0, -0.01\}\).

  • Alakoya et al. (2022) (Alg. 3.2): \(\theta =0.12, \lambda _1=0.3,\mu =0.25,\alpha _n=\frac{0.005}{3n+25000}, \rho _n = \frac{16}{(n+1)^{1.1}}, \xi _n=\frac{1}{(2n+1)^4}, f(x)=\frac{1}{3}x\).

  • Bing and Cho (2021) (Alg. 3.3): \(\theta =0.12, \lambda _1=0.3,\mu =0.25,\alpha _n=\frac{0.005}{3n+25000}, \beta _n=0.5(1-\alpha _n), \epsilon _n=\frac{1}{(2n+1)^4}\).

  • Bing and Cho (2021) (Alg. 3.4): \(\theta =0.12, \lambda _1=0.3,\mu =0.25,\alpha _n=\frac{0.005}{3n+25000}, \epsilon _n=\frac{1}{(2n+1)^4}, f(x)=\frac{1}{3}x\).

  • Tan et al. (2022) (Alg. 3.4): \(\tau =0.12, \mu =0.25, \vartheta _1=0.3, \theta =1.5, \sigma _n=\frac{0.005}{3n+25000}, \varphi _n=0.5(1-\sigma _n), \epsilon _n=\frac{100}{(n+1)^2}, \xi _n=\frac{16}{(n+1)^{1.1}}\).

We then use the stopping criterion; \(\text{ TOL}_n:=0.5\Vert x_n - J^A(x_n - Bx_n)\Vert ^2<\epsilon \) for all algorithms, where \(\epsilon \) is the predetermined error.

All the computations are performed using Matlab 2016 (b) which is running on a personal computer with an Intel(R) Core(TM) i5-2600 CPU at 2.30GHz and 8.00 Gb-RAM.

In Tables 1 and 2, “Iter” and “CPU” mean the CPU time in seconds and the number of iterations, respectively.

Remark 4.3

Figures 1, 2, 3 can be seen clearly (or understood better) by looking at the graphs of “SNR” vs “Iteration number (n)”, and Table 1. Note that the larger the SNR, the better the quality of the restored image.

5 Conclusion

We have proposed a two-step inertial forward–reflected–anchored–backward splitting algorithm for solving the monotone inclusion problem (1.1) in a real Hilbert space. We have also proved that the sequence generated by this algorithm converges strongly to a solution of the monotone inclusion problem. This algorithm inherits the attractive features of the forward–reflected–backward splitting algorithm (1.5), namely it involves only one forward evaluation of B even when B is not required to be cocoercive. However, unlike the forward–reflected–backward splitting algorithm (1.5), our algorithm converges strongly. Numerical results show that the proposed algorithm is efficient and faster than other related strongly convergent splitting algorithms in the literature. We remark that our proposed algorithm involves a restrictive condition on \(\{e_n\}\); for example, the sequence \(\{\frac{1}{n}\}\) does not satisfy this condition. Therefore, we intend to relax the restriction on \(\{e_n\}\) in our ongoing projects.