An inertially constructed forward–backward splitting algorithm in Hilbert spaces

Abstract

In this paper, we develop an iterative algorithm whose architecture comprises a modified version of the forward–backward splitting algorithm and the hybrid shrinking projection algorithm. We provide theoretical results concerning weak and strong convergence of the proposed algorithm towards a common solution of the fixed point problem associated to a finite family of demicontractive operators, the split equilibrium problem and the monotone inclusion problem in Hilbert spaces. Moreover, we compute a numerical experiment to show the efficiency of the proposed algorithm. As a consequence, our results improve various existing results in the current literature.

1 Introduction

The theory of mathematical optimization provides a quantitative optimal solution associated with various real-world problems emerging in the fields of engineering, medicine, economics, management, and industry and other branches of the sciences. One of the main advantages of mathematical optimization is to provide effective iterative algorithms and the corresponding analysis of these iterative algorithms. Moreover, the viability of such iterative algorithms is evaluated in terms of computational performance and complexity. As a consequence, the theory of mathematical optimization has not only emerged as an independent subject to solve real-world problems but also serve as an interdisciplinary bridge between various branches of sciences.

Monotone operator theory is a fascinating field of research in nonlinear functional analysis and found valuable applications in the field of convex optimization, subgradients, partial differential equations, variational inequalities, signal and image processing, evolution equations and inclusions; see, for instance, [4, 12, 14, 30] and the references cited therein. It is noted that the convex optimization problem can be translated into finding a zero of a maximal monotone operator defined on a Hilbert space. On the other hand, the problem of finding a zero of the sum of two (maximal-) monotone operators is of fundamental importance in convex optimization and variational analysis [23, 27, 33]. The forward–backward algorithm is prominent among various splitting algorithms to find a zero of the sum of two maximal monotone operators [23]. The class of splitting algorithms has parallel computing architectures and thus reducing the complexity of the problems under consideration. On the other hand, the forward–backward algorithm efficiently tackle the situation for smooth and/or nonsmooth functions. It is worth mentioning that the forward–backward algorithm has been modified by employing the heavy ball method [28] for convex optimization problems.

Fixed point theory has been studied extensively in the current literature owing to its rich abstract structures. These structures and subsequent tools elegantly manipulate various mathematical problems from the areas such as control theory, game theory, mathematical economics, image recovery signal processing and image processing. In 2015, the problem of finding a common solution of the zero point problem and fixed point problem was studied by Takahashi et al. [32]. It is well known that the class of demicontractive operators [15] includes various classes of nonlinear operators and comparatively exhibits powerful applications. Therefore, it is natural to study the fixed point problems associated with the class of demicontractive operators.

The theory of equilibrium problems is a systematic approach to the study of a diverse range of problems arising in the field of physics, optimization, variational inequalities, transportation, economics, network and noncooperative games; see, for example, [5, 11–13] and the references cited therein. The classical equilibrium problem theory has been generalized in several interesting ways to solve real-world problems. In 2012, Censor et al. [9] proposed a theory regarding split variational inequality problem (SVIP) which aims to solve a pair of variational inequality problems in such a way that the solution of a variational inequality problem, under a given bounded linear operator, solves another variational inequality.

In 2011, Moudafi [26] suggested the concept of split monotone variational inclusions (SMVIP) which includes, as a special case, split variational inequality problem, split common fixed point problem, split zeros problem, split equilibrium problem (SEP) and split feasibility problem. These problems have already been studied and successfully employed as a model in intensity-modulated radiation therapy treatment planning; see [6, 8]. This formalism is also at the core of modeling of many inverse problems arising for phase retrieval and other real-world problems; for instance, in sensor networks in computerized tomography and data compression; see, for example, [10, 12]. Some methods have been proposed and analyzed to solve SEP and generalized SEP in Hilbert spaces; see, for example, [2, 3, 16–22] and the references cited therein.

Inspired and motivated by the above-mentioned results and the ongoing research in this direction, we aim to employ the modified inertial forward–backward algorithm to find a common solution of fixed point problem associated to a finite family of demicontractive operators, SEP and monotone inclusion problem in Hilbert spaces. The rest of the paper is organized as follows: Section 2 contains preliminary concepts and results regarding fixed point theory, equilibrium problem theory and monotone operator theory. Section 3 comprises weak and strong convergence results of the proposed algorithm. Section 4 deals with the efficiency of the proposed algorithm by a numerical experiment together with theoretical applications to the split feasibility problem, the split variational inequality problem and the split minimization problem.

2 Preliminaries

In this section, we recall concepts and results regarding fixed point theory, equilibrium problem theory and monotone operator theory. Throughout this paper, let $\mathcal{H}_{1}$ be a real Hilbert space with the inner product and the associated norm $\langle \cdot , \cdot \rangle$ and $\Vert \cdot \Vert$, respectively. The symbols ⇀ and → denotes weak and strong convergence.

An operator $P_{C}$ is said to be metric projection of $\mathcal{H}_{1}$ onto nonempty, closed and convex subset C, if for every $x \in \mathcal{H}_{1}$, there exists a unique nearest point in C denoted by $P_{C}x$ such that

$$ \Vert x-P_{C}x \Vert \leq \Vert x-z \Vert , \quad \text{for all } z\in C. $$

It is noted that $P_{C}$ is a firmly nonexpansive operator and $P_{C}x$ is characterized by the following property:

$$ \langle x-P_{C}x,P_{C}x-y\rangle \geq 0, \quad \text{for all } x \in \mathcal{H}_{1}\text{ and } y \in C. $$

Next, we recall the definitions of nonexpansive and related operators.

Definition 1

([4])

Let C be a nonempty subset of $\mathcal{H}_{1}$, for an operator $T:C \rightarrow \mathcal{H}_{1}$, we denote by $Fix(T)$ the set of fixed points of the operator T, that is, $Fix(T)=\{x \in \mathcal{H}_{1}| x=Tx\}$. The operator T is considered as:

• 1. nonexpansive if

  $$ \Vert Tx-Ty \Vert \leq \Vert x-y \Vert , \quad \forall x,y \in C; $$

• 2. firmly nonexpansive if

  $$ \Vert Tx-Ty \Vert ^{2}\leq \Vert x-y \Vert ^{2}- \bigl\Vert (Id-T)x-(Id-T)y \bigr\Vert ^{2}, \quad \forall x,y \in C; $$

• 3. quasi-nonexpansive if $Fix(T)\neq \emptyset$ such that

  $$ \Vert Tx-y \Vert \leq \Vert x-y \Vert , \quad \forall x \in C, y \in Fix(T); $$

• 4. -demicontractive if $Fix(T)\neq \emptyset$ and there exists such that

  

It follows immediately that a firmly nonexpansive operator is a nonexpansive operator.

We now define the concept of SEP. Let $\hbar :\mathcal{H}_{1}\rightarrow \mathcal{H}_{2}$ be a bounded linear operator. Let $F_{1}:C\times C\rightarrow \mathbb{R}$ and $F_{2}:Q\times Q\rightarrow \mathbb{R}$ be two bifunctions, then SEP is to find:

$$ x^{\ast }\in C\quad \text{such that}\quad F_{1} \bigl( x^{\ast },x \bigr) \geq 0\quad \text{for all }x\in C $$

(1)

and

$$ y^{\ast }=\hbar x^{\ast }\in Q\quad \text{such that}\quad F_{2} \bigl( y^{ \ast },y \bigr) \geq 0\quad \text{for all } y\in Q. $$

(2)

The solution set of the SEP (1) and (2) is denoted by

$$ \Omega := \bigl\{ x^{\ast } \in C: x^{\ast } \in EP(F_{1}) \text{ and } \hbar x^{\ast } \in EP(F_{2}) \bigr\} . $$

(3)

Now, we recall some important concepts related to monotone operator theory [4].

Let $\mathcal{A}: \mathcal{H}_{1} \rightarrow 2^{\mathcal{H}_{1}}$ be a set-valued operator. We denote its domain, range, graph and zeros by $Dom \mathcal{A}=\{x \in \mathcal{H}_{1}| \mathcal{A}x \neq 0\}$, $Ran\mathcal{A}=\{ u \in \mathcal{H}_{1}| (\exists x \in \mathcal{H}_{1})u \in \mathcal{A}x\}$, $Gra\mathcal{A}=\{(x,u) \in \mathcal{H}_{1}\times \mathcal{H}_{1}| u \in \mathcal{A}x\}$ and $Zer\mathcal{A}=\{x \in \mathcal{H}_{1}| 0 \in \mathcal{A}x\}$, respectively. Let the set-valued operator $\mathcal{A}$ is said to be monotone, if

$$ \langle x-y,u-v\rangle \geq 0, \quad \forall (x,u),(y,v) \in Gra \mathcal{A}. $$

Moreover, $\mathcal{A}$ is said to be maximal monotone if its graph is not strictly contained in the graph of any other monotone operator on $\mathcal{H}_{1}$. A well-known example of a maximal monotone operator is the subgradient operator of a proper, lower semicontinuous convex function $f:\mathcal{H}_{1} \rightarrow (-\infty ,+\infty ]$ defined by

$$ \partial f:\mathcal{H}_{1} \rightarrow 2^{\mathcal{H}_{1}}:x \mapsto \bigl\{ u \in \mathcal{H}_{1}| f(y) \geq f(x)+\langle u, y-x\rangle , \forall y \in \mathcal{H}_{1} \bigr\} . $$

For a maximal monotone operator, the associated resolvent operator with index $m > 0$ is defined as

$$ J_{m}=(Id+m\mathcal{A})^{-1}, $$

where Id denotes the identity operator.

It is well known that the resolvent operator $J_{m}$ is well-defined everywhere on Hilbert space $\mathcal{H}_{1}$. Furthermore, $J_{m}$ is single-valued and satisfies the firmly nonexpansiveness. Furthermore, $x \in \mathcal{A}^{-1}(0)$ if and only if $x=J_{m}(x)$.

Let $f:\mathcal{H}_{1}\rightarrow \mathbb{R}\cup \{ +\infty \}$ be a proper, convex and lower semicontinuous function and let $g:\mathcal{H}_{1}\rightarrow \mathbb{R}$ be a convex, differentiable and Lipschitz continuous function, then the convex minimization problem for f and g is defined as

$$ \min_{x\in \mathcal{H}_{1}} \bigl\{ f (x )+g ( x ) \bigr\} . $$

(4)

Definition 2

([4])

Let $\mathcal{B}:\mathcal{H}_{1} \rightarrow \mathcal{H}_{1}$ be a nonlinear operator. For $\gamma > 0$, the operator $\mathcal{B}$ is said to be γ-inverse strongly monotone (γ-ism) if

$$ \langle x-y,\mathcal{B}x-\mathcal{B}y\rangle \geq \gamma \Vert \mathcal{B}x- \mathcal{B}y \Vert ^{2}, \quad \forall x,y \in \mathcal{H}_{1}. $$

The γ-ism is also coined as γ-cocoercive operator. Moreover, γ-ism is $\frac{1}{\gamma }$-Lipschitz continuous. In connection with the problem (4), the monotone inclusion problem with respect to a maximally monotone operator $\mathcal{A}$ and an arbitrary operator $\mathcal{B}$ is to find:

$$ x^{\ast }\in C\quad \text{such that}\quad 0\in \mathcal{A}x^{\ast }+ \mathcal{B}x^{\ast }. $$

(5)

In the sequel, we list some important results in the form of lemmas for the convergence analysis.

Lemma 2.1

([4])

Let C be a nonempty, closed and convex subset of a real Hilbert space $\mathcal{H}_{1}$. Let $T: C \rightarrow C$ be an operator. Then the operator T is said to be demiclosed at zero, if for any sequence $(x_{k})$ in C that converges weakly to x and $(Id-T)x_{k}$ converges strongly to zero, then $x \in Fix(T)$.

Lemma 2.2

Let $x,y \in \mathcal{H}_{1}$ and $\beta \in \mathbb{R}$, then the following relations hold:

• $\Vert x+y \Vert ^{2} \leq \Vert x \Vert ^{2}+2 \langle y, x+y \rangle$;

• $\Vert \beta x+(1-\beta )y \Vert ^{2}=\beta \Vert x \Vert ^{2}+(1-\beta ) \Vert x \Vert ^{2}- \beta (1-\beta ) \Vert x-y \Vert ^{2}$.

Lemma 2.3

([31])

Let E be a Banach space satisfying Opial’s condition and let $\{x_{n}\}$ be a sequence in E. Let $l,m\in E$ be such that $\lim_{n\rightarrow \infty } \Vert x_{n}-l \Vert$ and $\lim_{n\rightarrow \infty } \Vert x_{n}-m \Vert$ exist. If $\{x_{n_{k}}\}$ and $\{x_{m_{k}}\}$ are subsequences of $\{x_{n}\}$ which converge weakly to l and m, respectively, then $l=m$.

Lemma 2.4

([24])

Let E be a Banach space. Let $\mathcal{A}:E\rightarrow 2^{E}$ be an m-accretive operator and let $\mathcal{B}:E\rightarrow E$ be an α-inverse strongly accretive operator. Then we have

• a) For $r > 0$, $Fix(T^{\mathcal{A},\mathcal{B}}_{r})=(\mathcal{A}+\mathcal{B})^{-1}(0)$,

• b) for $0 < s \leq r$ and $x \in E$, $\Vert x-T^{\mathcal{A},\mathcal{B}}_{s}x \Vert \leq 2 \Vert x-T^{\mathcal{A},\mathcal{B}}_{r} \Vert$.

Lemma 2.5

([24])

Let E be a uniformly convex and q-uniformly smooth Banach space for some $q\in (0,2]$. Let $\mathcal{A}:E\rightarrow 2^{E}$ be an m-accretive operator and let $\mathcal{B}:E\rightarrow E$ be an α-inverse strongly accretive operator. Then, given $r>0$, there exists a continuous, strictly increasing and convex function $\varphi _{q}:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ with $\varphi _{q}(0)=0$ such that for all $x,y\in B_{r}$

$$\begin{aligned} \bigl\Vert T_{r}^{\mathcal{A},\mathcal{B}}x-T_{r}^{\mathcal{A},\mathcal{B}}y \bigr\Vert ^{q} \leq & \Vert x-y \Vert ^{q}-r \bigl( \alpha q-r^{q-1}k_{q} \bigr) \Vert \mathcal{A}x-\mathcal{A}y \Vert ^{q} \\ & {} -\varphi _{q} \bigl( \bigl\Vert \bigl(Id-J_{r}^{\mathcal{B}} \bigr) (Id-r\mathcal{A})x- \bigl(Id-J_{r}^{ \mathcal{B}} \bigr) (Id-r \mathcal{A})y \bigr\Vert \bigr), \end{aligned}$$

where $k_{q}$ is the q-uniform smoothness coefficient of E.

Lemma 2.6

([1])

Let $\{\xi _{n}\}$, $\{\eta _{n}\}$ and $\{\alpha _{n}\}$ be sequences in $[0,+\infty )$ satisfying

$$ \xi _{n+1}\leq \xi _{n}+\alpha _{n}(\xi _{n}-\xi _{n-1})+\eta _{n}, \quad \textit{for all } n\geq 1, $$

provided that $\sum_{n=1}^{\infty }\eta _{n}<+\infty$ and $0\leq \alpha _{n}\leq \alpha <1$ for all $n\geq 1$. Then the following two relations hold:

$$\begin{aligned}& \textstyle\begin{cases} \textit{a)} & \sum_{n \geq 1}[\xi _{n} - \xi _{n-1}]_{+} < +\infty , \quad \textit{where } [t]_{+}=\max \{t,0\}; \\ \textit{b)} & \textit{there exists}\quad \xi ^{*} \in [0,+\infty ) \quad \textit{such that} \quad \lim_{n \rightarrow +\infty } \xi _{n} = \xi ^{*}.\end{cases}\displaystyle \end{aligned}$$

Lemma 2.7

([25])

Let C be a nonempty, closed and convex subset of a real Hilbert space $\mathcal{H}_{1}$. For every $x,y\in \mathcal{H}_{1}$ and $a\in \mathbb{R}$, the set

$$ D= \bigl\{ v\in C: \Vert y-v \Vert ^{2}\leq \Vert x-v \Vert ^{2}+\langle z,v \rangle +a \bigr\} $$

is closed and convex.

Assumption 2.8

Let C be a nonempty, closed and convex subset of a Hilbert space $\mathcal{H}_{1}$. Let $F_{1}:C\times C\rightarrow \mathbb{R}$ be a bifunction satisfying the following conditions:

• (A1): $F_{1}(x,x)=0$ for all $x\in C$;

• (A2): $F_{1}$ is monotone, i.e., $F_{1}(x,y)+F_{1}(y,x)\leq 0$ for all $x,y \in C$;

• (A3): for each $x,y,z\in C$, $\limsup_{t\rightarrow 0}F_{1}(tz+(1-t)x,y)\leq F_{1}(x,y)$;

• (A4): for each $x\in C$, $y\mapsto F_{1}(x,y)$ is convex and lower semicontinuous.

Lemma 2.9

([11])

Let C be a nonempty, closed and convex subset of a real Hilbert space $\mathcal{H}_{1}$ and let $F_{1}:C\times C\rightarrow \mathbb{R}$ be a bifunction satisfying Assumption 2.8. For $r>0$ and $x\in \mathcal{H}_{1}$, there exists $z\in C$ such that

$$ F_{1}(z,y)+\frac{1}{r}\langle y-z,z-x\rangle \geq 0,\quad \textit{for all }y \in C. $$

Moreover, define an operator $T_{r}^{F}:\mathcal{H}_{1}\rightarrow C$ by

$$ T_{r}^{F_{1}}(x)= \biggl\{ z\in C:F_{1}(z,y)+ \frac{1}{r}\langle y-z,z-x\rangle \geq 0\textit{, for all }y\in C \biggr\} , $$

for all $x\in \mathcal{H}_{1}$. Then we have the following observations:

• (1): for each $x \in \mathcal{H}_{1}$, $T_{r}^{F_{1}}(x)\neq \emptyset$;

• (2): $T_{r}^{F_{1}}$ is single-valued;

• (3): $T_{r}^{F_{1}}$ is firmly nonexpansive;

• (4): $Fix(T_{r}^{F_{1}})=EP(F_{1})$;

• (5): $EP(F_{1})$ is closed and convex.

It is noted that if $F_{2}:Q\times Q\rightarrow \mathbb{R}$ is a bifunction satisfying Assumption 2.8, where Q is a nonempty, closed and convex subset of a Hilbert space $\mathcal{H}_{2}$. Then, for each $s>0$ and $w \in \mathcal{H}_{2}$, we define the operator

$$ T_{s}^{F_{2}}(w)= \biggl\{ d\in C:F_{2}(d,e)+ \frac{1}{s}\langle e-d,d-w \rangle \geq 0\text{, for all }e\in Q \biggr\} . $$

Similarly, we have the following relations:

• (1): for each $w \in \mathcal{H}_{2}$, $T_{s}^{F_{2}}(w)\neq \emptyset$;

• (2): $T_{s}^{F_{2}}$ is single-valued;

• (3): $T_{s}^{F_{2}}$ is firmly nonexpansive;

• (4): $Fix(T_{s}^{F_{2}})=EP(F_{2})$;

• (5): $EP(F_{2})$ is closed and convex.

3 Algorithm and convergence analysis

In this section, we present an approach to the convergence analysis of inertial forward–backward splitting method for solving the fixed point problem associated to a finite family of demicontractive operators, SEP and monotone inclusion problem in Hilbert spaces. First, we set the following hypotheses required in the sequel: Let $\mathcal{H}_{1}$, $\mathcal{H}_{2}$ be two real Hilbert spaces and let $C \subseteq \mathcal{H}_{1}$, $Q \subseteq \mathcal{H}_{2}$ be nonempty, closed and convex subsets of Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. We consider the following hypotheses:

• (H1) Let $F_{1}: C \times C \rightarrow \mathbb{R}$ and $F_{2}: Q \times Q \rightarrow \mathbb{R}$ be two bifunctions satisfying Assumption 2.8 such that $F_{2}$ is upper semicontinuous;

• (H2) let $\hbar : \mathcal{H}_{1} \rightarrow \mathcal{H}_{2}$ be a bounded linear operator;

• (H3) let $\mathcal{A}:\mathcal{H}_{1} \rightarrow 2^{\mathcal{H}_{1}}$ be a maximal monotone operator and let $\mathcal{B}:\mathcal{H}_{1} \rightarrow \mathcal{H}_{1}$ be a γ-ism operator;

• (H4) for $i \in \{1,2,\ldots ,N\}$, let $S_{i}:\mathcal{H}_{1} \rightarrow \mathcal{H}_{1}$ be a finite family of -demicontractive operators;

• (H5) suppose that $\Gamma := zer(\mathcal{A}+\mathcal{B})\cap \Omega \cap \bigcap^{i=1}_{N}Fix(S_{i})$.

Theorem 3.1

If $\Gamma \neq \emptyset$ with hypotheses (H1)–(H5), then the sequence $(x_{k})$ generated by Algorithm 1 converges weakly to an element $\bar{x} \in \Gamma$, provided the following conditions hold:

• (C1) $\sum^{\infty }_{k=1}\Theta _{k} \Vert x_{k}-x_{k-1} \Vert <\infty$;

• (C2) $0 < a^{\ast } \leq \beta _{k}$, $\lambda _{k} \leq b^{\ast } < 1$ and ;

• (C3) $0 < \liminf_{k \rightarrow \infty } \lambda _{k} \leq \limsup_{k \rightarrow \infty }\lambda _{k} < 1$;

• (C4) $\liminf_{k \rightarrow \infty }u_{k} > 0$;

• (C5) $0 < \liminf_{k \rightarrow \infty } m_{k} \leq \limsup_{k \rightarrow \infty } m_{k} < 2\gamma$.

Algorithm 1

An inertially constructed forward–backward splitting algorithm

Proof

First we show that $\hbar ^{\ast }(Id-T_{u_{k}}^{F_{2}})\hbar$ is an $\frac{1}{L}$-ism operator. For this, we utilize the firmly nonexpansiveness of $T_{u_{k}}^{F_{2}}$, which implies that $(Id-T_{u_{k}}^{F_{2}})$ is an 1-ism operator. Now, observe that

$$\begin{aligned} \bigl\Vert \hbar ^{\ast } \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) \hbar x-\hbar ^{\ast } \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) \hbar y \bigr\Vert ^{2} =& \bigl\langle \hbar ^{\ast } \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) ( \hbar x-\hbar y),\hbar ^{\ast } \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) (\hbar x-\hbar y) \bigr\rangle \\ =& \bigl\langle \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) (\hbar x-\hbar y),\hbar ^{\ast } \hbar \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) ( \hbar x-\hbar y) \bigr\rangle \\ \leq & L \bigl\langle \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) (\hbar x- \hbar y), \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) ( \hbar x-\hbar y) \bigr\rangle \\ =&L \bigl\Vert \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) (\hbar x-\hbar y) \bigr\Vert ^{2} \\ \leq &L \bigl\langle x-y,\hbar ^{\ast } \bigl(Id-T_{u_{k}}^{F_{2}} \bigr) (\hbar x- \hbar y) \bigr\rangle , \end{aligned}$$

for all $x,y\in \mathcal{H}_{1}$. So, we observe that $\hbar ^{\ast }(Id-T_{u_{k}}^{F_{2}})\hbar$ is an $\frac{1}{L}$-ism. Moreover, $Id-\delta \hbar ^{\ast }(Id-T_{u_{k}}^{F_{2}})h$ is nonexpansive provided $\delta \in (0,\frac{1}{L})$. Now, we divided the rest of the proof into the following three steps:

Step 1. Show that $\lim_{k\rightarrow \infty } \Vert x_{k}-\hat{x} \Vert$ exists for every $\hat{x}\in \Gamma$.

For any $\hat{x}\in \Gamma$, we get

$$\begin{aligned} \Vert b_{k}-\hat{x} \Vert ^{2} =& \bigl\Vert x_{k}+\Theta _{k}(x_{k}-x_{k-1})- \hat{x} \bigr\Vert ^{2} \\ \leq & \Vert x_{k}-\hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2} \\ & {} +2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle . \end{aligned}$$

(6)

Since $T^{F_{1}}_{u_{k}} \hat{x}=\hat{x}$ and using (6), we have

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} =& \bigl\Vert T^{F_{1}}_{u_{k}} \bigl(Id-\delta \hbar ^{ \ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar \bigr)b_{k}- \hat{x} \bigr\Vert ^{2} \\ \leq & \bigl\Vert b_{k}-\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k}- \hat{x} \bigr\Vert ^{2} \\ \leq & \Vert b_{k}-\hat{x} \Vert ^{2}+\delta ^{2} \bigl\Vert \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\Vert ^{2} \\ & {} +2\delta \bigl\langle \hat{x}-b_{k},\hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\rangle . \end{aligned}$$

(7)

Thus, we have

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} \leq & \Vert b_{k}-\hat{x} \Vert ^{2}+\delta ^{2} \bigl\langle \hbar b_{k}-T^{F_{2}}_{u_{k}}\hbar b_{k},\hbar ^{\ast }\hbar \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\rangle \\ & {} +2\delta \bigl\langle \hat{x}-b_{k},\hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\rangle . \end{aligned}$$

(8)

Moreover, we have

$$\begin{aligned} \delta ^{2} \bigl\langle \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k},\hbar ^{ \ast }\hbar \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k} \bigr\rangle \leq & L\delta ^{2} \bigl\langle \hbar b_{k}-T^{F_{2}}_{u_{k}}\hbar b_{k},\hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\rangle \\ =&L\delta ^{2} \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2}. \end{aligned}$$

(9)

Note that

$$\begin{aligned} &2\delta \bigl\langle \hat{x}-b_{k},\hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\rangle \\ &\quad = 2\delta \bigl\langle \hbar (\hat{x}-b_{k}) , \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\rangle \\ &\quad = 2\delta \bigl[ \bigl\langle \hbar \hat{x}-T^{F_{2}}_{u_{k}} \hbar b_{k} , \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\rangle - \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2} \bigr] \\ & \quad \leq 2\delta \biggl[\frac{1}{2} \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2}- \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2} \biggr] \\ & \quad = - \delta \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2}. \end{aligned}$$

(10)

Utilizing (8)–(10), we have

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} \leq & \Vert b_{k}-\hat{x} \Vert ^{2}+L\delta ^{2} \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}}\hbar b_{k} \bigr\Vert ^{2}- \delta \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2} \\ \leq & \Vert b_{k}-\hat{x} \Vert ^{2}+\delta (L\delta -1) \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2}. \end{aligned}$$

(11)

Since $\delta \in (0 , \frac{1}{L})$, the estimate (11) implies that

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} \leq \Vert x_{k}-\hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2}+2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle . \end{aligned}$$

(12)

Furthermore, by using (6), (12) and (C2), we have

(13)

Moreover, it follows from (6), (12), (13) and Lemma 2.5 that

$$\begin{aligned} \Vert x_{k+1}-\hat{x} \Vert ^{2} =& \bigl\Vert \lambda _{k}w_{k}+(1- \lambda _{k})J_{k}w_{k}-\hat{x} \bigr\Vert ^{2} \\ \leq &\lambda _{k} \Vert w_{k}-\hat{x} \Vert ^{2}+(1-\lambda _{k}) \Vert J_{k}w_{k}- \hat{x} \Vert ^{2} \\ = & \Vert w_{k}-\hat{x} \Vert ^{2} \\ \leq & \Vert x_{k}-\hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2}+2 \Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle . \end{aligned}$$

(14)

From Lemma 2.6 and (C1), we conclude from the estimate (14) that $\lim_{k\rightarrow \infty } \Vert x_{k}-\hat{x} \Vert$ exists.

Step 2. Show that $x_{k}\rightharpoonup \bar{x}\in (\mathcal{A}+\mathcal{B})^{-1}(0)$.

Since $\hat{x}=J_{k}\hat{x}$, therefore it follows from Lemma 2.2 and Lemma 2.5 that

(15)

As $\lim_{k\rightarrow \infty } \Vert x_{k}-\hat{x} \Vert$ exists, therefore utilizing, (C1), (C4), (C5) and (15), we get

$$ \lim_{k\rightarrow \infty }(1-\lambda _{k})m_{k}(2\gamma -m_{k}) \Vert \mathcal{A}w_{k}-\mathcal{A}\hat{x} \Vert =0. $$

(16)

Also from (15), we get

$$ \lim_{k\rightarrow \infty } \Vert w_{k}-m_{k} \mathcal{A}w_{k}-J_{k}w_{k}+m_{k} \mathcal{A}\hat{x} \Vert =0. $$

(17)

Using (16), (17) and the triangle inequality

$$ \Vert w_{k}-m_{k}\mathcal{A}w_{k}-J_{k}w_{k}+m_{k} \mathcal{A} \hat{x} \Vert \leq \Vert w_{k}-J_{k}w_{k} \Vert +m_{k} \Vert \mathcal{A}w_{k}-\mathcal{A}\hat{x} \Vert , $$

we get

$$ \lim_{n\rightarrow \infty } \Vert J_{k}w_{k}-w_{k} \Vert =0. $$

(18)

Since $\liminf_{k\rightarrow \infty }m_{k}>0$ there exists $m>0$ such that $m_{k}\geq m$ for all $k\geq 0$. It follows from Lemma 2.4(b) that

$$ \bigl\Vert T_{m}^{\mathcal{A},\mathcal{B}}w_{k}-w_{k} \bigr\Vert \leq 2 \Vert J_{k}w_{k}-w_{k} \Vert . $$

Now utilizing (18), the above estimate implies that

$$ \lim_{k\rightarrow \infty } \bigl\Vert T_{m}^{\mathcal{A},\mathcal{B}}w_{k}-w_{k} \bigr\Vert =0. $$

(19)

As a consequence, we have

$$ \lim_{k\rightarrow \infty } \Vert x_{k+1}-w_{k} \Vert = \lim_{k \rightarrow \infty } \bigl(1-a^{\ast } \bigr) \Vert J_{k}w_{k}-w_{k} \Vert =0. $$

(20)

Again, from (15), we have

Rearranging the above estimate and using (C1), (C2), we get

$$ \lim_{k \rightarrow \infty } \bigl\Vert (Id-S_{i})\ell _{k} \bigr\Vert = 0. $$

(21)

This implies that

$$ \lim_{k \rightarrow \infty } \Vert w_{k}-\ell _{k} \Vert = \lim_{k \rightarrow \infty }b^{\ast } \bigl\Vert (Id-S_{i})\ell _{k} \bigr\Vert = 0. $$

(22)

Again, by Lemma 2.2, Lemma 2.6 and (11), we have

$$\begin{aligned} \Vert x_{k+1}-\hat{x} \Vert ^{2} \leq & \Vert x_{k}-\hat{x} \Vert ^{2}+2 \Theta _{k}\langle x_{k}-x_{k-1},b_{k}-\hat{x}\rangle + \delta (L\delta -1) \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2}. \end{aligned}$$

Rearranging the above estimate, we have

$$\begin{aligned} -\delta (L\delta -1) \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert ^{2} \leq & \Vert x_{k}-\hat{x} \Vert ^{2}- \Vert x_{k+1}-\hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2} \\ & {} +2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle . \end{aligned}$$

(23)

Since $\delta (L\delta -1)<0$, it follows from (C1) and (23) that

$$ \lim_{k \rightarrow \infty } \bigl\Vert \hbar b_{k}-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr\Vert =0. $$

(24)

Note that $T^{F_{1}}_{u_{k}}$ is firmly nonexpansive and $Id-\delta \hbar ^{\ast }(Id-T^{F_{2}}_{u_{k}})\hbar$ is nonexpansive, therefore we have

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} =& \bigl\Vert T^{F_{1}}_{u_{k}} \bigl(b_{k}-\delta \hbar ^{ \ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k} \bigr)-T^{F_{1}}_{u_{k}}\hat{x} \bigr\Vert ^{2} \\ \leq & \bigl\langle T^{F_{1}}_{u_{k}} \bigl(b_{k}- \delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr)-T^{F_{1}}_{u_{k}}\hat{x} , b_{k}- \delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k}-\hat{x} \bigr\rangle \\ =& \bigl\langle \ell _{k}-\hat{x} , b_{k}-\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k}-\hat{x} \bigr\rangle \\ =& \frac{1}{2} \bigl\{ \Vert \ell _{k}-\hat{x} \Vert ^{2}+ \bigl\Vert b_{k}-\delta \hbar ^{ \ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k}-\hat{x} \bigr\Vert ^{2} \\ & {} - \bigl\Vert \ell _{k}-b_{k}+\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k} \bigr\Vert ^{2} \bigr\} \\ \leq & \frac{1}{2} \bigl\{ \Vert \ell _{k}-\hat{x} \Vert ^{2}+ \Vert b_{k}-\hat{x} \Vert ^{2}- \bigl\Vert \ell _{k}-b_{k}+\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr)\hbar b_{k} \bigr\Vert ^{2} \bigr\} \\ =&\frac{1}{2} \bigl\{ \Vert \ell _{k}-\hat{x} \Vert ^{2}+ \Vert b_{k}-\hat{x} \Vert ^{2}- \bigl( \Vert \ell _{k}-b_{k} \Vert ^{2}+\delta ^{2} \bigl\Vert \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{b_{k}} \bigr)\hbar b_{k} \bigr\Vert ^{2} \\ & {} -2\delta \bigl\langle \ell _{k}-b_{k} , \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\rangle \bigr) \bigr\} . \end{aligned}$$

So, we have

$$\begin{aligned} \Vert \ell _{k}-\hat{x} \Vert ^{2} \leq & \Vert b_{k}-\hat{x} \Vert ^{2}- \Vert \ell _{k}-b_{k} \Vert ^{2}+2\delta \bigl\langle \ell _{k}-b_{k}, \hbar ^{\ast } \bigl(Id-T^{F_{\ast }}_{u_{k}} \bigr)\hbar b_{k} \bigr\rangle . \end{aligned}$$

(25)

Therefore, we have

$$\begin{aligned} \Vert \ell _{k}-b_{k} \Vert ^{2} \leq & \Vert b_{k}-\hat{x} \Vert ^{2}- \Vert \ell _{k}-\hat{x} \Vert ^{2}+2\delta \Vert \ell _{k}-b_{k} \Vert \bigl\Vert \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \bigr) \hbar b_{k} \bigr\Vert . \end{aligned}$$

(26)

Utilizing (24) and (C2), we have

$$ \lim_{k \rightarrow \infty } \Vert \ell _{k}-b_{k} \Vert =0. $$

(27)

From the definition of $(b_{k})$ and (27), we have

$$ \lim_{k \rightarrow \infty } \Vert \ell _{k}-x_{k} \Vert =0. $$

(28)

By the definition of $(b_{k})$ and (C1), we have

$$ \lim_{k\rightarrow \infty } \Vert b_{k}-x_{k} \Vert = \lim_{k \rightarrow \infty }\Theta _{k} \Vert x_{k}-x_{k-1} \Vert =0. $$

(29)

Since $(x_{k})$ is bounded and $\mathcal{H}_{1}$ is reflexive, $\nu _{\omega }(x_{k})=\{x\in \mathcal{H}_{1}:x_{k_{n}} \rightharpoonup x,(x_{k_{n}})\subset (x_{k})\}$ is nonempty. Let $\bar{x}\in \nu _{\omega }(x_{{k}})$ be an arbitrary element. Then there exists a subsequence $(x_{k_{n}})\subset (x_{k})$ converging weakly to x̄. Let $\hat{x}\in \nu _{\omega }(x_{k})$ and $(x_{k_{m}})\subset (x_{k})$ be such that $x_{k_{m}}\rightharpoonup \hat{x}$. From (24), we also have $\ell _{k_{n}}\rightharpoonup \bar{x}$ and $\ell _{k_{m}}\rightharpoonup \hat{x}$. Since $T_{m}^{\mathcal{A},\mathcal{B}}$ is nonexpansive, from (19) and Lemma 2.1, we have $\hat{x},\bar{x}\in (\mathcal{A}+\mathcal{B})^{-1}(0)$. By applying Lemma 2.3, we obtain $\hat{x}=\bar{x}$.

Step 3. Show that $\bar{x} \in \Omega$.

Let $\bar{x} \in EP(F_{1})$. For any $y\in \mathcal{H}_{1}$, we have

$$ F_{1}(\ell _{k},y)+\frac{1}{u_{k}} \bigl\langle y-\ell _{k},\ell _{k}-x_{k}- \delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}}\hbar b_{k} \bigr) \bigr\rangle \geq 0. $$

This implies that

$$ F_{1}(\ell _{k},y)+\frac{1}{u_{k}}\langle y-\ell _{k},\ell _{k}-x_{k} \rangle -\frac{1}{u_{k}} \bigl\langle y-\ell _{k},\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr) \bigr\rangle \geq 0. $$

From Assumption 2.8(A2), we have

$$\begin{aligned} \frac{1}{u_{k}}\langle y-\ell _{k},\ell _{k}-x_{k} \rangle - \frac{1}{u_{k}} \bigl\langle y-\ell _{k},\delta \hbar ^{\ast } \bigl(Id-T^{F_{2}}_{u_{k}} \hbar b_{k} \bigr) \bigr\rangle \geq -F_{1}(\ell _{k},y) \geq F_{1}(y,\ell _{k}). \end{aligned}$$

So, we have

$$\begin{aligned} \frac{1}{u_{k_{n}}}\langle y-\ell _{k_{n}},\ell _{k_{n}}-x_{k_{n}} \rangle -\frac{1}{u_{k_{n}}} \bigl\langle y-\ell _{k_{n}},\delta \hbar ^{ \ast } \bigl(Id-T^{F_{2}}_{u_{k}}\hbar b_{k_{n}} \bigr) \bigr\rangle \geq F(y,\ell _{k_{n}}). \end{aligned}$$

(30)

Utilizing (28) and (C2), we get $\ell _{k_{n}} \rightharpoonup \bar{x}$. Moreover, from (24) and Assumption 2.8(A4), we get

$$ F(y , \bar{x}) \leq 0, \quad \text{for all } y \in \mathcal{H}_{1}. $$

Let $y_{t}=ty+(1-t)\bar{x}$ for some $1 \geq t > 0$ and $y \in \mathcal{H}_{1}$. Since $\bar{x} \in \mathcal{H}_{1}$, consequently, $y_{t} \in \mathcal{H}_{1}$ and hence $F_{1}(y_{t},\bar{x})\leq 0$. Using Assumption 2.8((A1) and (A4)), it follows that

$$\begin{aligned} 0 =&F_{1}(y_{t},y_{t}) \\ \leq & t F_{1}(y_{t},y)+(1-t)F_{1}(y_{t}, \bar{x}) \\ \leq & t \bigl(F_{1}(y_{t},y) \bigr). \end{aligned}$$

This implies that

$$ F_{1}(y_{t},y) \geq 0, \quad \text{for all } y \in C. $$

Letting $t \rightarrow 0$, we have

$$ F_{1}(\bar{x},y) \geq 0, \quad \text{for all } y \in C. $$

Thus, $\bar{x} \in EP(F_{1})$. Similarly, we can show that $\bar{x} \in EP (F_{2})$. Since ħ is a bounded linear operator, we have $\hbar x_{k_{n}} \rightharpoonup \hbar \bar{x}$. It follows from (26) that

$$ T^{F_{2}}_{u_{k_{n}}}\hbar b_{k_{n}} \rightharpoonup \hbar \bar{x} \quad \text{as } n \rightarrow \infty . $$

(31)

Now, from Lemma 2.7 we have

$$\begin{aligned} F_{2} \bigl(T^{F_{2}}_{u_{k_{n}}}\hbar b_{k_{n}},y \bigr)+\frac{1}{u_{k_{n}}} \bigl\langle y-T^{F_{2}}_{u_{k_{n}}}\hbar b_{k_{n}},T^{F_{2}}_{u_{k_{n}}} \hbar b_{k_{n}}-\hbar b_{k_{n}} \bigr\rangle \geq 0, \end{aligned}$$

for all $y \in \mathcal{H}_{1}$. Since $F_{2}$ is upper semicontinuous in the first argument and from (31), we have

$$ F_{2}(\hbar \bar{x},y) \geq 0, $$

for all $y \in \mathcal{H}_{1}$. This implies that $\hbar \bar{x} \in EP(F_{2})$. Therefore, $\bar{x} \in \Omega$.

Step 4. From (21) and by using the demiclosed principle for $S_{i}$ (it is evident that $x_{k_{n}} \rightharpoonup \bar{x}$ and $\lim_{k \rightarrow \infty } \Vert (Id-S_{i})x_{k_{n}} \Vert =0$), we have $\bar{x} \in \bigcap^{i=1}_{N}Fix(S_{i})$ and hence $\bar{x} \in \Gamma$. This completes the proof. □

Now, we establish strong convergence results of Algorithm 2.

Algorithm 2

An inertially constructed forward–backward splitting algorithm

Theorem 3.2

If $\Gamma \neq \emptyset$ with hypotheses (H1)–(H5), then the sequence $(x_{k})$ generated by Algorithm 2 converges strongly to an element $\bar{x} \in P_{\Gamma }x_{1}$, provided the conditions (C1)–(C5) hold.

Proof

The proof is divided into the following steps:

Step 1. Show that the sequence $\{ x_{k} \}$ defined in Algorithm 2 is well-defined.

We know that $(\mathcal{A}+\mathcal{B})^{-1}(0)$, Ω and $Fix(S_{i})$ are closed and convex by Lemma 2.4 and Lemma 2.9. Moreover, from Lemma 2.7 we see that $C_{k+1}$ is closed and convex for each $k\geq 1$. Hence the projection $P_{C_{k+1}}x_{1}$ is well-defined. For any $\hat{x}\in \Gamma$, it follows from Algorithm 2 and the estimates (6), (12) and (13) that

(32)

It follows from the estimate (32) that $\Gamma \subset C_{k+1}$. Summing up these facts, we conclude that $C_{k+1}$ is nonempty, closed and convex for all $k \geq 1$, and hence the sequence $(x_{k})$ is well-defined.

Step 2. Show that $\lim_{k\rightarrow \infty } \Vert x_{k}-x_{1} \Vert$ exists.

Since Γ is nonempty closed and convex subset of $\mathcal{H}_{1}$, there exists a unique $x^{\ast }\in \Gamma$ such that $x^{\ast }=P_{\Gamma }x_{1}$. From $x_{k+1}=P_{C_{k+1}}x_{1}$, we have $\Vert x_{k+1}-x_{1} \Vert \leq \Vert x^{\ast }-x_{1} \Vert$, for all $\bar{x}\in \Gamma \subset C_{k+1}$. In particular $\Vert x_{k+1}-x_{1} \Vert \leq \Vert P_{\Gamma }x_{1}-x_{1} \Vert$. This proves that the sequence $(x_{k})$ is bounded. On the other hand, from $x_{k}=P_{C_{k}}x_{1}$ and $x_{k+1}=P_{C_{k+1}}x_{1}\in C_{k+1}$, we get

$$ \Vert x_{k}-x_{1} \Vert \leq \Vert x_{k+1}-x_{1} \Vert . $$

This implies that $(x_{k})$ is nondecreasing and hence

$$ \lim_{k\rightarrow \infty } \Vert x_{k}-x_{1} \Vert \quad \text{exists.} $$

(33)

Step 3. Show that $\bar{x}\in (\mathcal{A}+\mathcal{B})^{-1}(0)$.

In order to proceed, we first calculate the following estimates which are required in the sequel:

$$\begin{aligned} \Vert x_{k+1}-x_{k} \Vert ^{2} =& \Vert x_{k+1}-x_{1}+x_{1}-x_{k} \Vert ^{2} \\ =& \Vert x_{k+1}-x_{1} \Vert ^{2}+ \Vert x_{k}-x_{1} \Vert ^{2}-2 \langle x_{k}-x_{1},x_{k+1}-x_{1} \rangle \\ =& \Vert x_{k+1}-x_{1} \Vert ^{2}+ \Vert x_{k}-x_{1} \Vert ^{2}-2 \langle x_{k}-x_{1},x_{k+1}-x_{k}+x_{k}-x_{1} \rangle \\ =& \Vert x_{k+1}-x_{1} \Vert ^{2}- \Vert x_{k}-x_{1} \Vert ^{2}-2 \langle x_{k}-x_{1},x_{k+1}-x_{k} \rangle \\ \leq & \Vert x_{k+1}-x_{1} \Vert ^{2}- \Vert x_{k}-x_{1} \Vert ^{2}. \end{aligned}$$

Taking limsup on both sides of the above estimate and utilizing (33), we have $\limsup_{k\rightarrow \infty } \Vert x_{k+1}-x_{k} \Vert ^{2}=0$. That is,

$$ \lim_{k\rightarrow \infty } \Vert x_{k+1}-x_{k} \Vert =0. $$

(34)

Note that $x_{k+1}\in C_{k+1}$, therefore we have

$$ \Vert y_{k}-x_{k+1} \Vert \leq \Vert x_{k}-x_{k+1} \Vert +2\Theta _{k} \Vert x_{k}-x_{k-1} \Vert -2\Theta _{k}\langle x_{k}-x_{k+1},x_{k-1}-x_{k} \rangle . $$

Utilizing (34) and (C1), the above estimate implies that

$$ \lim_{k\rightarrow \infty } \Vert y_{k}-x_{k+1} \Vert =0. $$

(35)

From (34), (35) and the triangular inequality

$$ \Vert y_{k}-x_{k} \Vert \leq \Vert y_{k}-x_{k+1} \Vert + \Vert x_{k+1}-x_{k} \Vert , $$

we get

$$ \lim_{k\rightarrow \infty } \Vert y_{k}-x_{k} \Vert =0. $$

(36)

Also, from Lemma 2.2 and (21), we have

Rearranging the above estimate, we have

$$\begin{aligned} a^{\ast } \bigl(1-b^{\ast } \bigr) \Vert J_{k}w_{k}-w_{k} \Vert ^{2} \leq & \Vert x_{k}-\hat{x} \Vert ^{2}- \Vert y_{k}-\hat{x} \Vert ^{2}+2 \Theta _{k}\langle x_{k}-x_{k-1},b_{k}- \hat{x}\rangle \\ \leq & \bigl( \Vert x_{k}-\hat{x} \Vert + \Vert y_{k}-\hat{x} \Vert \bigr) \Vert x_{k}-y_{k} \Vert +2\Theta _{k}\langle x_{k}-x_{k-1},b_{k}- \hat{x}\rangle . \end{aligned}$$

The above estimate, by using (C1) and (36), implies that

$$ \lim_{k\rightarrow \infty } \Vert J_{k}w_{k}-w_{k} \Vert =0. $$

(37)

Making use of (37), we have the following estimate:

$$ \lim_{k\rightarrow \infty } \Vert y_{k}-w_{k} \Vert = \lim_{k \rightarrow \infty } \bigl(1-a^{\ast } \bigr) \Vert J_{k}w_{k}-w_{k} \Vert =0. $$

(38)

Reasoning as above, utilizing the estimate (37), the estimate (38) implies that

$$ \lim_{k\rightarrow \infty } \Vert w_{k}-x_{k} \Vert =0. $$

(39)

In a similar fashion, we have

$$ \lim_{k\rightarrow \infty } \bigl\Vert T_{m}^{\mathcal{A},\mathcal{B}}w_{k}-w_{k} \bigr\Vert =0. $$

(40)

Reasoning as above (Theorem 3.1, Step 2), we have the desired result.

Step 4. Show that $\bar{x}\in \Omega$.

See proof of Step 3 in Theorem 3.1.

Step 5. Show that $\bar{x}\in \bigcap^{i=1}_{N}Fix(S_{i})$.

See proof of Step 4 in Theorem 3.1.

Step 6. Show that $\bar{x}=P_{\Gamma }x_{1}$.

Let $x=P_{\Gamma }x_{1}$ imply that $x=P_{\Gamma }x_{1}\in C_{k+1}$. Since $x_{k+1}=P_{C_{k+1}}x_{1}\in C_{k+1}$, we have

$$ \Vert x_{k+1}-x_{1} \Vert \leq \Vert x-x_{1} \Vert . $$

On the other hand, we have

$$\begin{aligned} \Vert x-x_{1} \Vert \leq & \Vert \bar{x}-x_{1} \Vert \\ \leq &\liminf_{k\rightarrow \infty } \Vert x_{k}-x_{1} \Vert \\ \leq &\limsup_{k\rightarrow \infty } \Vert x_{k}-x_{1} \Vert \\ \leq & \Vert x-x_{1} \Vert . \end{aligned}$$

That is,

$$ \Vert \bar{x}-x_{1} \Vert =\lim_{k\rightarrow \infty } \Vert x_{k}-x_{1} \Vert = \Vert x-x_{1} \Vert . $$

Therefore, we conclude that $\lim_{k\rightarrow \infty }x_{k}=\bar{x}=P_{\Gamma }x_{1}$. This completes the proof. □

The following remark gives us a stopping criterion of Algorithm 2.

Remark 3.3

We remark here that the condition (C1) is easily implemented in numerical computation since the value of $\Vert x_{k}-x_{k-1} \Vert$ is known before choosing $\Theta _{k}$. The parameter $\Theta _{k}$ can be taken as $0 \leq \Theta _{k} \leq \widehat{\Theta _{k}}$,

$$ \widehat{\Theta _{k}} = \textstyle\begin{cases} \min \{\frac{\nu _{k}}{ \Vert x_{k}-x_{k-1} \Vert }, \Theta \} &\text{if }x_{k} \neq x_{k-1}; \\ \Theta &\text{otherwise}, \end{cases} $$

where $\{ \nu _{k}\}$ is a positive sequence such that $\sum^{\infty }_{k = 1}\nu _{k} < \infty$ and $\Theta _{k} = \Theta \in [0,1)$.

4 Numerical experiment and results

This section shows the effectiveness of Algorithm 2 by the following given example.

Example 4.1

Let $\mathcal{H}_{1} = \mathcal{H}_{2} = \mathbb{R}$ be the set of all real numbers, with the inner product defined by $\langle x , y\rangle = xy$, for all $x , y \in \mathbb{R}$ and the usual induced norm $\vert \cdot \vert$. Let $F_{1}:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a bifunction defined as $F_{1}(x,y)=2x(y-x)$ and let $F_{2}:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a bifunction defined as $F_{2}(p,q)=p(q-p)$. For all $x \in \mathbb{R}$, let the operators $\hbar ,\mathcal{A},\mathcal{B} : \mathbb{R}\rightarrow \mathbb{R}$ be defined as $\hbar (x)=3x$, $\mathcal{A}x=4x$ and $\mathcal{B}x=3x$, respectively. Let $S_{i}: \mathbb{R} \rightarrow \mathbb{R}$ be a finite family of demicontractive operators defined by

$$ S_{i}(x)= \textstyle\begin{cases} -\frac{x}{i}, &x \in [0,\infty ); \\ x, &x \in (-\infty ,0). \end{cases} $$

Then the sequence $(x_{k})$ generated by Algorithm 2 strongly converges to a point in Γ.

Proof

It is easy to prove that the bifunctions $F_{1}$ and $F_{2}$ satisfy Assumptions 2.8 and $F_{2}$ is upper semicontinuous with $\Omega = 0$. Moreover, ħ is bounded linear operator on $\mathbb{R}$ with the adjoint operator $\hbar ^{\ast }$ such that $\Vert \hbar \Vert = \Vert \hbar ^{\ast } \Vert =3$, $\mathcal{A}$ is a maximal monotone operator and $\mathcal{B}$ is a monotone and γ-Lipschitz operator for some $\gamma > 0$ with $(\mathcal{A} + \mathcal{B})^{-1}(0)=\{0\}$. Note that $S_{i}$ is a finite family of $\frac{1-i^{2}}{(1+i)^{2}}$-demicontractive operators with $\bigcap^{i=1}_{N}Fix(S_{i})=\{0\}$. Hence $\Gamma = (\mathcal{A} + \mathcal{B})^{-1}(0)\cap \Omega \cap \bigcap^{i=1}_{\infty }Fix(S_{i}) = 0$. Choose $\Theta = 0.5$, $u_{k}=\frac{k}{5k+1}$, $\beta _{k}=\frac{1}{100k+1}$, $\lambda _{k} =\frac{1}{100k+1}$, $\delta =0.04$, $L=3$ and $m=0.01$. Since

$$ \textstyle\begin{cases} \min \{\frac{1}{k^{2} \Vert x_{k}-x_{k-1} \Vert },0.5\}&\text{if } x_{k}\neq x_{k-1}; \\ 0.5&\text{otherwise}.\end{cases} $$

For the rest of the numerical experiment, we proceed as follows:

Step 1. Find $z \in F_{2}$ such that $F_{2}(z,y)+\frac{1}{u}\langle y - z,z - \hbar x \rangle \geq 0$ for all $y \in F_{2}$. We write

$$\begin{aligned} F_{2}(z,y)+\frac{1}{u}\langle y - z,z - \hbar x \rangle \geq 0\quad \Leftrightarrow & \quad z(y-z)+\frac{1}{u}\langle y - z,z - \hbar x \rangle \geq 0 \\ \Leftrightarrow & \quad uz(y-z)+(y-z) (z-\hbar x)\geq 0 \\ \Leftrightarrow & \quad (y-z) \bigl((1+u)z-\hbar x \bigr)\geq 0, \end{aligned}$$

for all $y \in F_{2}$. Thus, by Lemma 2.9(2), we know that $T^{F_{2}}_{u}\hbar x$ is single-valued for each $x \in F_{1}$. Hence $z=\frac{\hbar x}{1+u}$.

Step 2. Find $g \in F_{1}$ such that $g = x-\delta \hbar ^{\ast }(I-T_{r}^{F_{2}})\hbar x$. From Step 1, we get

$$\begin{aligned} g=x-\delta \hbar ^{\ast } \bigl(I-T_{u}^{F_{2}} \bigr) \hbar x = & x-\delta \hbar ^{ \ast } \bigl(I-T_{u}^{F_{2}} \bigr)\hbar x \\ = & x - \delta \biggl(3x-\frac{3(\hbar x)}{1+u} \biggr) \\ = & (1-3\delta )x+\frac{3\delta }{1+u}(\hbar x). \end{aligned}$$

Step 3. Find $p \in F_{1}$ such that $F_{1}(p,q)+\frac{1}{u}\langle p-q,p-g\rangle \geq 0$ for all $q \in F_{1}$. From Step 2, we have

$$\begin{aligned} F(p,q)+\frac{1}{u}\langle q-p,p-g\rangle \geq 0 \quad \Leftrightarrow &\quad (2p) (q - p)+\frac{1}{u}\langle q-p , p-g\rangle \geq 0 \\ \Leftrightarrow &\quad u(2p) (q-p)+(q-p) (p-g)\geq 0 \\ \Leftrightarrow &\quad (q-p) \bigl((1+2u)p-g \bigr)\geq 0, \end{aligned}$$

for all $q \in F_{1}$. Similarly, by Lemma 2.9(2), we obtain $p=\frac{g}{1+2u}=\frac{(1-3\delta )x}{1+2u}+ \frac{3\delta \hbar x}{(1+u)(1+2u)}$.

Step 4. Compute the numerical results for $x_{k+1}$.

We provide a numerical test of a comparison between our Inertial Forward–Backward Splitting Algorithm (IFBSA) defined in Algorithm 2 (i.e., $\Theta _{k}\neq 0$) and the Forward–Backward Splitting Algorithm (FBSA) (i.e., $\Theta _{k}=0$). The stopping criterion is defined as $\mathrm{Error}= E_{k}= \Vert x_{k+1}-x_{k} \Vert <10^{-6}$. The different choices of $x_{0}$ and $x_{1}$ are given in tables and figures.

The error plotting $E_{k}$ and $(x_{k})$ against $\Theta _{k} \neq 0$ and $\Theta _{k}=0$ for each choice in Table 1 is shown in Fig. 1.

Figure 1

Graph of IFBSA and FBSA plotted for Choice 1 with $N=20$

Table 1 Numerical results for Example 4.1

 □

We can see from Table 1 and Figs. 1 and 2 that IFBSA performs better as compared to FBSA. Elaborating the behavior of this algorithm with respect to Table 1, the error analysis is depicted in Figs. 1 and 2 whereas the number of iterations required to converge the sequence $(x_{k})$ towards the common solution are expressed in Figs. 3 and 4. Summarizing these facts, we say that the IFBSA exhibits a reduction in the error, time and the number of iterations of the function as compared to the FBSA.

Figure 2

Graph of IFBSA and FBSA plotted for Choice 2 with $N=20$

Figure 3

Comparison between trajectories of Algorithm 2 for Choice 1 with $N=20$ and $N=4$

Figure 4

Comparison between trajectories of Algorithm 2 for Choice 2 with $N=20$ and $N=4$

5 Applications

In this section, we illustrate the theoretical results which we have obtained in the previous section.

5.1 Split feasibility problems

Let $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ be two real Hilbert spaces and $\hbar : \mathcal{H}_{1} \rightarrow \mathcal{H}_{2}$ be a bounded linear operator. Let C and Q be closed, convex and nonempty subsets of $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. The split feasibility problem aims to find $\bar{x} \in C$ such that $S\bar{x} \in Q$. We represent the solution sets by $\omega := C \cap \hbar ^{-1}(Q) = \{\bar{y} \in C: \hbar \bar{y} \in Q\}$. Censor and Elfving [7] introduced it to solve inverse problems and their application to medical image reconstruction and radiation therapy in a finite dimensional Hilbert space. For the set C, recall the function

$$\begin{aligned}& b_{C}(\bar{x}):= \textstyle\begin{cases} 0, &\bar{x} \in C; \\ \infty , &\text{otherwise}. \end{cases}\displaystyle \end{aligned}$$

The proximal operator of $b_{C}$ is the metric projection on C,

$$\begin{aligned} prox_{b_{C}} =&\arg \min_{\bar{p} \in C} \Vert \bar{p}-\bar{x} \Vert \\ =&P_{C}(\bar{x}). \end{aligned}$$

Let $P_{Q}$ be the projection of $\mathcal{H}_{2}$ onto a nonempty, convex and closed subset Q. Take: $f(\bar{x})=\frac{1}{2} \Vert \hbar \bar{x}-P_{Q}\hbar \bar{x} \Vert ^{2}$ and $g(\bar{x})=b_{C}(\bar{x})$. Then we compute the split feasibility problem from the following result.

Corollary 5.1

Let $\mathcal{H}_{1}$, $\mathcal{H}_{2}$ be two Hilbert spaces and let $C \subseteq \mathcal{H}_{1}$ and $Q \subseteq \mathcal{H}_{2}$ be nonempty, closed and convex subsets of Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. Assume that $\Gamma =\omega \cap \Omega \cap \bigcap^{i=1}_{N}Fix(S_{i}) \neq \emptyset$ with hypotheses (H1)–(H4). Let $\Theta _{k}$ be a bounded real sequence and $m_{k} \in (0 , \frac{2}{ \Vert \hbar \Vert ^{2}})$. For given $x_{0},x_{1} \in \mathcal{H}_{1}$, let the iterative sequences $(x_{k})$, $(b_{k})$, $(\ell _{k})$, $(w_{k})$, $(y_{k})$ and $(x_{k+1})$ be generated by

$$\begin{aligned} \textstyle\begin{cases} b_{k}=x_{k}+\Theta _{k}(x_{k}-x_{k-1}); \\ \ell _{k}=T^{F_{1}}_{u_{k}}(I-\delta _{k}\hbar ^{\ast }(I-T^{F_{2}}_{u_{k}}) \hbar )b_{k}; \\ w_{k}=(1-\beta _{k})\ell _{k}+\beta _{k}S_{k}\ell _{k}; \\ y_{k}=\lambda _{k}w_{k}+(1-\lambda _{k})J_{k}w_{k}; \\ C_{k+1}=\{z \in C_{k}: \Vert y_{k}-z \Vert ^{2} \leq \Vert x_{k}- \hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2} \cdots \\ \hphantom{C_{k+1}=}{} +2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle \}; \\ x_{k+1}=P_{C_{k+1}}x_{1},\quad \forall k \geq 1, \end{cases}\displaystyle \end{aligned}$$

(41)

where $J_{k}=P_{C}(Id-m_{k}\hbar ^{\ast }(I-P_{Q})\hbar )$. Assume that the conditions (C1)–(C4) hold, then the sequence $(x_{k})$ generated by (41) converges strongly to an element $\bar{x}=P_{\Gamma }x_{1}$.

5.2 Monotone variational inequality problems

Let $\mathcal{H}_{1}$ be a Hilbert space and C be a nonempty, closed and convex subset of $\mathcal{H}_{1}$. Let $\mathcal{B}:C \rightarrow \mathcal{H}_{1}$ be a nonlinear monotone operator. The variational inequality problem aims to find a point $\bar{x} \in C$ such that

$$ \langle \mathcal{B}\bar{x},\bar{y}-\bar{x} \rangle \geq 0 \quad \forall \bar{y} \in C. $$

(42)

The solution set of the above problem is denoted by ω and assume that $\omega \neq \emptyset$. [29], The resolvent operator acts as a projection operator $P_{C}$. Then we compute monotone variational inequality problems from the following result.

Corollary 5.2

Let $\mathcal{H}_{1}$, $\mathcal{H}_{2}$ be two Hilbert spaces and let $C \subseteq \mathcal{H}_{1}$ and $Q \subseteq \mathcal{H}_{2}$ be nonempty, closed and convex subsets of Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. Assume that $\Gamma =\omega \cap \Omega \cap \bigcap^{i=1}_{N}Fix(S_{i}) \neq \emptyset$ with hypotheses (H1)–(H4). Let $\Theta _{k}$ be a bounded real sequence and $m_{k} \in (0 , \frac{2}{ \Vert \hbar \Vert ^{2}})$. For given $x_{0},x_{1} \in \mathcal{H}_{1}$, let the iterative sequences $(x_{k})$, $(b_{k})$, $(\ell _{k})$, $(w_{k})$, $(y_{k})$ and $(x_{k+1})$ be generated by

$$\begin{aligned} \textstyle\begin{cases} b_{k}=x_{k}+\Theta _{k}(x_{k}-x_{k-1}); \\ \ell _{k}=T^{F_{1}}_{u_{k}}(I-\delta _{k}\hbar ^{\ast }(I-T^{F_{2}}_{u_{k}}) \hbar )b_{k}; \\ w_{k}=(1-\beta _{k})\ell _{k}+\beta _{k}S_{k}\ell _{k}; \\ y_{k}=\lambda _{k}w_{k}+(1-\lambda _{k})J_{k}w_{k}; \\ C_{k+1}=\{z \in C_{k}: \Vert y_{k}-z \Vert ^{2} \leq \Vert x_{k}- \hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2} \cdots \\ \hphantom{C_{k+1}=}{} +2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle \}; \\ x_{k+1}=P_{C_{k+1}}x_{1}, \quad \forall k \geq 1, \end{cases}\displaystyle \end{aligned}$$

(43)

where $J_{k}=P_{C}(Id-m_{k}\mathcal{B})$. Assume that the conditions (C1)–(C4) hold, then the sequence $(x_{k})$ generated by (43) converges strongly to an element $\bar{x}=P_{\Gamma }x_{1}$.

5.3 Convex minimization problems

Let $f: \mathcal{H}_{1} \rightarrow \mathbb{R}$ and $g: \mathcal{H}_{1} \rightarrow \mathbb{R}$ be two convex, proper and lower semicontinuous functions. In Algorithm 2, set that $\mathcal{A} := \partial f$ and $\mathcal{B} := \nabla g$. Assume that ω is the set of solutions of problem (4) and $\omega \neq \emptyset$. Then we compute the convex minimization problem from the following result.

Corollary 5.3

Let $\mathcal{H}_{1}$, $\mathcal{H}_{2}$ be two Hilbert spaces and let $C \subseteq \mathcal{H}_{1}$ and $Q \subseteq \mathcal{H}_{2}$ be nonempty, closed and convex subsets of Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, respectively. Assume that $\Gamma =\omega \cap \Omega \cap \bigcap^{i=1}_{N}Fix(S_{i}) \neq \emptyset$ with hypotheses (H1)–(H4). Let $\Theta _{k}$ be a bounded real sequence and $m_{k} \in (0,\frac{2}{ \Vert \hbar \Vert ^{2}})$. Let $f , g:\mathcal{H}_{1} \rightarrow \mathbb{R}$ be two convex, proper and lower semicontinuous functions, such that f is sub-differential function and g is differentiable with γ-Lipschitz continuous gradient. For given $x_{0},x_{1} \in \mathcal{H}_{1}$, let the iterative sequences $(x_{k})$, $(b_{k})$, $(\ell _{k})$, $(w_{k})$, $(y_{k})$ and $(x_{k+1})$ be generated by

$$\begin{aligned} \textstyle\begin{cases} b_{k}=x_{k}+\Theta _{k}(x_{k}-x_{k-1}); \\ \ell _{k}=T^{F_{1}}_{u_{k}}(I-\delta _{k}\hbar ^{\ast }(I-T^{F_{2}}_{u_{k}}) \hbar )b_{k}; \\ w_{k}=(1-\beta _{k})\ell _{k}+\beta _{k}S_{k}\ell _{k}; \\ y_{k}=\lambda _{k}w_{k}+(1-\lambda _{k})J_{k}w_{k}; \\ C_{k+1}=\{z \in C_{k}: \Vert y_{k}-z \Vert ^{2} \leq \Vert x_{k}- \hat{x} \Vert ^{2}+\Theta ^{2}_{k} \Vert x_{k}-x_{k-1} \Vert ^{2} \cdots \\ \hphantom{C_{k+1}=}{} +2\Theta _{k}\langle x_{k}-\hat{x}, x_{k}-x_{k-1} \rangle \}; \\ x_{k+1}=P_{C_{k+1}}x_{1}, \quad \forall k \geq 1, \end{cases}\displaystyle \end{aligned}$$

(44)

where $J_{k}=J^{\partial f}_{m_{k}}(Id-m_{k}\nabla g)$. Assume that the conditions (C1)–(C4) hold, then the sequence $(x_{k})$ generated by (44) converges strongly to an element $\bar{x}=P_{\Gamma }x_{1}$.

6 Conclusions

In this paper, we have devised an inertially constructed forward–backward splitting algorithm for computing a common solution of the finite family of demicontractive operators, SEP and monotone inclusion problem in Hilbert spaces. The theoretical framework of the algorithm has been strengthened with an appropriate numerical example. Moreover, this framework has also been implemented to various instances of the monotone inclusion problems. We would like to emphasize that the above mentioned problems occur naturally in many applications, therefore, iterative algorithms are inevitable in this field of investigation. As a consequence, our theoretical framework constitutes an important topic of future research.