New Inertial Forward-Backward Mid-Point Methods for Sum of Infinitely Many Accretive Mappings, Variational Inequalities, and Applications

Abstract

Some new inertial forward-backward projection iterative algorithms are designed in a real Hilbert space. Under mild assumptions, some strong convergence theorems for common zero points of the sum of two kinds of infinitely many accretive mappings are proved. New projection sets are constructed which provide multiple choices of the iterative sequences. Some already existing iterative algorithms are demonstrated to be special cases of ours. Some inequalities of metric projection and real number sequences are widely used in the proof of the main results. The iterative algorithms have also been modified and extended from pure discussion on the sum of accretive mappings or pure study on variational inequalities to that for both, which complements the previous work. Moreover, the applications of the abstract results on nonlinear capillarity systems are exemplified.

1. Introduction and Preliminaries

Suppose H is a real Hilbert space with norm $\parallel ·\parallel$ and inner product $\langle ·,·\rangle$. Let K be the non-empty closed and convex subset of $H.$ We use → and ⇀ to denote the strong and weak convergence in $H,$ respectively.

We know that Hilbert space H satisfies Opial’s condition in the sense that $limin{f}_{n\to \infty }\parallel x_n-z\parallel <limin{f}_{n\to \infty }\parallel x_n-y\parallel$ for $\{x_n\}\subset H$ with $x_n\rightharpoonup z$ and $y\ne z$ (see [1]).

The inclusion problem for finding $u\in H$ such that

$$0\in Su+Tu$$

(1)

is studied intensively, where $S:H\to H$ is a mapping and $T:H\to 2^H$ is a multi-valued mapping. This is mainly because many problems appear in convex programming, variational inequalities, split feasibility problems, minimization problem, inverse problem and image processing can be modeled by (1).

A mapping $T:D(T)\subset H\to 2^H$ is said to be an accretive mapping (see [2]) if for each $x,y\in D(T),$ there exist $u\in Tx$ and $v\in Ty$ such that $\langle x-y,u-v\rangle \ge 0.$ An accretive mapping $T:D(T)\subset H\to 2^H$ is said to be m-accretive if $R(I+kT)=H,$ for $k>0.$

A mapping $S:D(S)\subset H\to H$ is said to be $\mu$-inversely strongly accretive mapping (see [3]) if for each $x,y\in D(S)$ and $\mu >0,$ $\langle x-y,Sx-Sy\rangle \ge {\mu \parallel Sx-Sy\parallel }^2.$

For a mapping $W:D(W)\subset H\to H,$ a point $x\in D(W)$ is called a zero point of W if $Wx=0.$ The set of zero points of W is denoted by $W^{-1}0.$ If $x\in D(W)\subset H$ satisfies that $Wx=x,$ then x is called a fixed point of W. The set of fixed points of W is denoted by $Fix(W).$

The study of the special case of inclusion problem (1), where T is accretive and S is $\mu$-inversely strongly accretive, has been a hot topic during the past few years. In particular, the constructions of the iterative algorithms for approximating the zero point of the sum of T and S are focused, see [3,4,5,6,7,8,9,10,11,12] and the references therein. The inertial forward-backward splitting method is one of the important iterative algorithms studied by some authors, see [7,8,9,13,14].

In 2015, Lorenz and Pock [9] proposed the following inertial forward-backward algorithm for approximating zero points of $T+S$, where $T:H\to 2^H$ is m-accretive and $S:H\to H$ is $\mu$-inversely strongly accretive:

$$\left\{\begin{array}{c}{u}_0,u_1\in H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ v_n=u_n+{\theta }_n(u_n-u_{n-1}),\hfill \\ u_{n+1}={(I+r_{n}T)}^{-1}(v_n-r_{n}S{v}_n),n\in \mathbb{N}.\hfill \end{array}\right.$$

(2)

In addition, the result that $u_n\rightharpoonup p\in {(T+S)}^{-1}0,$ as $n\to \infty ,$ is proved under some conditions.

To get strong convergence, Dong et al. proposed the following inertial forward-backward projection algorithm in Hilbert spaces in [14]:

$$\left\{\begin{array}{c}{u}_0,u_1\in H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ v_n=u_n+{\alpha }_n(u_n-u_{n-1}),\hfill \\ w_n={(I+r_{n}T)}^{-1}(v_n-r_{n}S{v}_n),\hfill \\ C_n=\{p\in H:\parallel w_n{-p\parallel }^2\le {\parallel u_n-p\parallel }^2-2{\alpha }_n\langle u_n-p,u_{n-1}-u_n\rangle \hfill \\ +{\alpha }_n^2\parallel u_{n-1}-u_n{\parallel }^2\},\hfill \\ Q_n=\{p\in H:\langle u_n-p,u_n-u_0\rangle \le 0\},\hfill \\ u_{n+1}=P_{C_n\bigcap Q_n}(u_0),n\in \mathbb{N},\hfill \end{array}\right.$$

(3)

where T and S are the same as those in (2) and $P_{C_n\bigcap Q_n}$ is the metric projection whose meaning can be seen in Definition 1. The projection sets $C_n$ and $Q_n$ play an important role in the iterative construction to ensure the strong convergence. The result that $u_n\to P_{{(T+S)}^{-1}0}(u_0),$ as $n\to \infty ,$ is proved under some conditions.

In 2018, Khan et al. proposed the following one in which the projection set $Q_n$ is deleted (see [7]):

$$\left\{\begin{array}{c}{u}_0,u_1\in H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ v_n=u_n+{\theta }_n(u_n-u_{n-1}),\hfill \\ w_n={\alpha }_n{u}_n+(1-{\alpha }_n){(I+r_{n}T)}^{-1}(v_n-r_{n}S{v}_n),\hfill \\ C_1:=H,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le {\parallel u_n-p\parallel }^2\hfill \\ -2{\theta }_n(1-{\alpha }_n)\langle u_n-p,u_{n-1}-u_n\rangle +2{\theta }_n^2\parallel u_{n-1}-u_n{\parallel }^2\},\hfill \\ u_{n+1}=P_{C_{n+1}}(u_0),n\in \mathbb{N},\hfill \end{array}\right.$$

(4)

where T and S are the same as those in (3). The strong convergence that $u_n\to P_{{(T+S)}^{-1}0}(u_0),$ as $n\to \infty ,$ is also obtained under some conditions.

On the other hand, the inclusion problem (1) is extended to the system of inclusion problems:

$$0\in S_{i}u+T_{i}u,$$

(5)

where $T_i$ is m-accretive and $S_i$ is ${\mu }_i$-inversely strongly accretive for $i\in \{1,2,\cdots ,m\}$ or $i\in \mathbb{N}.$ In addition, some iterative algorithms for approximating common zero points of $T_i+S_i$ are constructed in [3,15,16,17]. In particular, Wei et al. proposed the following implicit mid-point forward-backward projection algorithm in [17]:

$$\left\{\begin{array}{c}{u}_1\in H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ w_n={\alpha }_n\eta f(u_n)+(I-{\alpha }_{n}F)u_n,\hfill \\ v_n={\beta }_n{w}_n+{\delta }_n{\sum }_{i=1}^{\infty }{c}_{n,i}{(I+r_{n,i}{T}_i)}^{-1}[\frac{w_n+v_n}{2}-r_{n,i}{S}_i(\frac{w_n+v_n}{2})]+{\lambda }_n{e}_n,\hfill \\ C_1=H,\hfill \\ C_{n+1}=\{p\in C_n:\parallel v_n{-p\parallel }^2\le \frac{1+{\beta }_n-{\lambda }_n}{2-{\delta }_n}\parallel w_n{-p\parallel }^2+\frac{2{\lambda }_n}{2-{\delta }_n}\parallel e_n-p{\parallel }^2\},\hfill \\ u_{n+1}=P_{C_{n+1}}(u_1),n\in \mathbb{N},\hfill \end{array}\right.$$

(6)

where $f:H\to H$ is a contraction, $F:H\to H$ is strongly positive linear bounded mapping, $e_n$ is the computational error, $T_i$ is m-accretive and $S_i$ is ${\mu }_i$-inversely strongly accretive for $i\in \mathbb{N}.$ The result that $u_n\to P_{{\bigcap }_{i=1}^{\infty }{(T_i+S_i)}^{-1}0}(u_1),$ as $n\to \infty ,$ is proved under some conditions.

Recall that $f:H\to H$ is called a contraction (see [17]) if there exists a constant $l\in (0,1)$ such that $\parallel f(x)-f(y)\parallel \le l\parallel x-y\parallel$ for $x,y\in H.$

A mapping $F:H\to H$ is called strongly positive (see [17]) if there exists $\xi >0$ such that $\langle x,Fx\rangle \ge {\xi \parallel x\parallel }^2$ for $x\in H.$ In this case,

$$\parallel aI-bF\parallel =su{p}_{\parallel x\parallel \le 1}|\langle (aI-bF)x,x\rangle |,$$

where I is the identity mapping, $a\in [0,1]$ and $b\in [-1,1].$

A mapping $U:H\to H$ is said to be non-expansive (see [17]) if for each $x,y\in H,$ $\parallel Ux-Uy\parallel \le \parallel x-y\parallel .$

In 2018, Wei et al. proposed some new hybrid iterative algorithms to approximate the common element of the set of zero points of infinitely many m-accretive mappings $T_i:H\to H$ and the set of fixed points of infinitely many non-expansive mappings $B_i:H\to H$. A special case (see Corollary 3.6 in [18]) is presented as follows:

$$\left\{\begin{array}{c}{u}_1\in H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ y_n={\alpha }_n{u}_n+(1-{\alpha }_n){\sum }_{i=1}^{\infty }{c}_{n,i}{(I+r_{n,i}{T}_i)}^{-1}(u_n+e_n),\hfill \\ z_n={\beta }_n{u}_n+(1-{\beta }_n){\sum }_{i=1}^{\infty }{b}_i{B}_i{y}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel y_n{-p\parallel }^2\le {\alpha }_n\parallel u_n{-p\parallel }^2+(1-{\alpha }_n){\parallel u_n+e_n-p\parallel }^2,\hfill \\ \parallel z_n{-p\parallel }^2\le {\beta }_n\parallel u_n{-p\parallel }^2+(1-{\beta }_n)\parallel y_n-p{\parallel }^2\},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel u_1-P_{C_{n+1}}(u_1){\parallel }^2+{\lambda }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(7)

The result that $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in ({\bigcap }_{i=1}^{\infty }{T}_i^{-1}0)\bigcap ({\bigcap }_{i=1}^{\infty }Fix(B_i)),$ as $n\to \infty ,$ is proved under some conditions. We may notice that infinite choices of $\{u_n\}$ can be made, which is totally different from traditional projection iterative algorithms, e.g., (3).

In 2016, Wei et al. proposed an implicit forward-backward mid-point iterative algorithm for approximating common zero points of $T_i+S_i,$ where $T_i$ is m-accretive and $S_i$ is ${\mu }_i$-inversely strongly accretive, for $i\in \mathbb{N}.$ A special case in [19] in the frame of Hilbert space is presented as follows:

$$\left\{\begin{array}{c}{u}_1\in K\subset H\mathrm{chosen}\mathrm{arbitrarily},\hfill \\ y_n=P_K[(1-{\alpha }_n)(u_n+e_n^{\prime })],\hfill \\ z_n={\delta }_n{y}_n+{\beta }_n{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{T}_i)}^{-1}[(\frac{y_n+z_n}{2}-r_{n,i}{S}_i(\frac{y_n+z_n}{2})]+{\xi }_n{e}_n^{″},\hfill \\ u_{n+1}={\gamma }_n\eta f(u_n)+(I-{\gamma }_{n}F)z_n+e_n^{‴},n\in \mathbb{N},\hfill \end{array}\right.$$

(8)

where f and F are the same as those in (6), $\{e_n^{\prime }\}$, $\{e_n^{″}\}$ and $\{e_n^{‴}\}$ are the error sequences. Under some conditions, $\{u_n\}$ is proved to be convergent strongly to $u_0\in {\bigcap }_{i=1}^{\infty }{(T_i+S_i)}^{-1}0,$ which also satisfies the following variational inequality:

$$\langle F{u}_0-\eta f(u_0),u_0-z\rangle \le 0,\forall z\in \bigcap _{i=1}^{\infty }{(T_i+S_i)}^{-1}0.$$

(9)

We may notice that the connection between the common element of ${(T_i+S_i)}^{-1}0$ for $i\in \mathbb{N}$ and the solution of one kind variational inequality is set up in [19].

In this paper, our main purpose is formulated as follows: (1) obtain strong convergence theorems instead of weak ones; (2) construct new projection sets, which ensure that infinitely many iterative sequences can be generated compared to traditional projection iterative algorithms (3), (4) and (6); (3) inject the idea of inertial forward-backward algorithm into the iterative construction, compared to iterative algorithms (6)–(8); (4) set up the connection between the common zero point of the sum of two kinds of infinitely many accretive mappings and the solution of one kind variational inequality, which complements the corresponding work since rare studies of the projection iterative algorithms (e.g., (3)–(7)) have mentioned that; (5) provide the application of the abstract result to capillarity systems.

To begin our study, we need some preliminaries.

Definition 1.

(see [2]) For the Hilbert space H and its non-empty closed and convex subset K, there exists a unique point $x_0\in K$ such that $\parallel x-x_0\parallel =inf\{\parallel x-y\parallel :y\in K\},$ for each $x\in H.$ In this case, the metric projection mapping $P_K:H\to K$ is defined by $P_{K}x=x_0$, for $\forall x\in H.$

Definition 2.

(see [20]) Let $\{K_n\}$ be a sequence of non-empty closed and convex subsets of H. Then

(1) 

the strong lower limit of $\{K_n\}$, $s-liminf{K}_n,$ is defined as the set of all $x\in H$ such that there exists $x_n\in K_n$ for almost all n and it tends to x as $n\to \infty$ in the norm;

(2) 

the weak upper limit of $\{K_n\},$ $w-limsup{K}_n,$ is defined as the set of all $x\in H$ such that there exists a subsequence $\{K_{n_m}\}$ of $\{K_n\}$ and $x_{n_m}\in K_{n_m}$ for every $n_m$ and it tends to x as $n_m\to \infty$ in the weak topology;

(3) 

the limit of $\{K_n\},$ $lim{K}_n,$ is the common value when $s-liminf{K}_n=w-limsup{K}_n$.

Lemma 1.

(see [20]) Let $\{K_n\}$ be a decreasing sequence of closed and convex subsets of H, i.e., $K_n\subset K_m$ if $n\ge m.$ Then $\{K_n\}$ converges in H and $lim{K}_n={\bigcap }_{n=1}^{\infty }{K}_n.$

Lemma 2.

(see [21]) Suppose H is a real Hilbert space. If $lim{K}_n$ exists and is not empty, then $P_{K_n}x\to P_{lim{K}_n}x$ for every $x\in H,$ as $n\to \infty .$

Lemma 3.

(see [2,19]) If $B:H\to H$ is accretive, then ${(I+rB)}^{-1}:H\to H$ is non-expansive.

Lemma 4.

(see [22]) If H is a real Hilbert space with K its non-empty closed and convex subset, $T_i:K\to K$ is non-expansive for $i\in \mathbb{N}$ and ${\sum }_{i=1}^{\infty }{a}_i=1$ for $\{a_i\}\subset (0,1),$ then ${\sum }_{i=1}^{\infty }{a}_i{T}_i$ is non-expansive with $Fix({\sum }_{i=1}^{\infty }{a}_i{T}_i)={\bigcap }_{i=1}^{\infty }Fix(T_i)$ under the assumption that ${\bigcap }_{i=1}^{\infty }Fix(T_i)\ne \varnothing .$

Lemma 5.

(see [19]) If H is a real Hilbert space with K its non-empty closed and convex subset, $S:K\to H$ is a single-valued mapping and $T:H\to 2^H$ is an m-accretive mapping, then

$$Fix({(I+rT)}^{-1}(I-rS))={(T+S)}^{-1}0,$$

for $\forall r>0.$

Lemma 6.

(see [23]) Let H be a real Hilbert space and $r\in (0,+\infty )$. Then there exists a continuous, strictly increasing and convex function $g:[0,2r]\to [0,+\infty )$ with $g(0)=0$ such that ${\parallel kx+(1-k)y\parallel }^2\le {k\parallel x\parallel }^2+{(1-k)\parallel y\parallel }^2-k(1-k)g(\parallel x-y\parallel ),$ for $k\in [0,1],x,y\in H$ with $\parallel x\parallel \le r$ and $\parallel y\parallel \le r$.

Lemma 7.

(see [24]) Let K be the non-empty closed and convex subset of Hilbert space H and $P_K:H\to K$ be the metric projection. Then

(1) 

for $\forall x\in H$ and $\forall y\in K,$ $\parallel P_K{x-y\parallel }^2+\parallel P_K{x-x\parallel }^2\le {\parallel y-x\parallel }^2.$

(2) 

$y=P_{K}x$ if and only if there holds the following inequality $\langle x-y,y-z\rangle \ge 0,$ for $\forall z\in K.$

Lemma 8.

(see [25]) If $f:H\to H$ is a contraction, then there is a unique element $x\in H$ such that $f(x)=x.$

2. Some Inertial Forward-Backward Algorithms

In this section, unless otherwise stated, we always assume that:

(1)

H is a real Hilbert space;

(2)

$A_i:H\to H$ is ${\mu }_i$-inversely strongly accretive and $B_i:H\to H$ is m-accretive, for each $i\in \mathbb{N}$. In addition, ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\ne \varnothing ;$

(3)

$\{e_n\}\subset H$ is the computational error;

(4)

$\{{\sigma }_n\}$, $\{s_{n,i}\}$ and $\{{\mu }_i\}$ are three real number sequences in $(0,+\infty )$ for $i,n\in \mathbb{N}$;

(5)

$\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are three real number sequences in $(0,1)$ with ${\alpha }_n+{\beta }_n+{\gamma }_n\equiv 1,$ for $n\in \mathbb{N};$

(6)

$\{{\omega }_{n,i}\}$ is a real number sequence in $(0,1)$ with ${\sum }_{i=1}^{\infty }{\omega }_{n,i}=1,$ for $n\in \mathbb{N};$

(7)

$\{k_n\}$ is a real number sequence in $[0,k]$ for some $k\in [0,1).$

2.1. New Inertial Forward-Backward Projection Algorithms

Theorem 1.

Let $\{u_n\}$ be generated by the following iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ v_n=u_n+k_n(u_n-u_{n-1}),\hfill \\ w_n={\alpha }_n{u}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le (1-{\gamma }_n)\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\parallel u_n-u_{n-1}{\parallel }^2-2{\beta }_n{k}_n\langle u_n-p,u_{n-1}-u_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(10)

Under the assumptions that: $(i)$ $s_{n,i}\le {\mu }_i$ for $i,n\in \mathbb{N};$ $(ii)$ ${\sigma }_n\to 0,{\gamma }_n\to 0,$ as $n\to \infty ;$ $(iii)$ $0<in{f}_n{\beta }_n<1$; $(iv)$ there exists $M>0$ such that $\parallel e_n\parallel \le M,$ for $n\in \mathbb{N},$ we have: $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$ as $n\to \infty .$

Proof. 

We split the proof into nine steps.

Step 1. ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i):H\to H$ is non-expansive, for $n\in \mathbb{N}.$

The proof of Step 1 is essentially from that of Step 1 in Theorem 2.1 of [17]. For the sake of completeness, we present it as follows.

Since $s_{n,i}\le 2{\mu }_i,$ then for each $x,y\in H,$

$$\begin{array}{c}\parallel (I-s_{n,i}{A}_i)x-(I-s_{n,i}{A}_i){y\parallel }^2={\parallel (x-y)-s_{n,i}(A_{i}x-A_{i}y)\parallel }^2\hfill \\ ={\parallel x-y\parallel }^2-2s_{n,i}\langle x-y,A_{i}x-A_{i}y\rangle +s_{n,i}^2{\parallel A_{i}x-A_{i}y\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2+s_{n,i}(s_{n,i}-2{\mu }_i)\parallel A_{i}x-A_i{y\parallel }^2\le {\parallel x-y\parallel }^2.\hfill \end{array}$$

Thus, $(I-s_{n,i}{A}_i):H\to H$ is non-expansive, for $i,n\in \mathbb{N}.$ It then follows from Lemmas 3 and 4 that ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i):H\to H$ is non-expansive, for $n\in \mathbb{N}.$ Combining with Lemma 5, ${\bigcap }_{i=1}^{\infty }Fix({(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i))={\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

Step 2. ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset C_n,$ for $n\in \mathbb{N}.$

If $n=1,$ it is obvious that ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset C_1.$

Now, $\forall q\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$ suppose the result is true for $n=m+1,$ then if $n=m+2,$ it follows from (10) that

$$\begin{array}{c}\parallel v_{m+1}{-q\parallel }^2\hfill \\ =\parallel u_{m+1}{-q\parallel }^2+2k_{m+1}\langle u_{m+1}-q,u_{m+1}-u_m\rangle +k_{m+1}^2{\parallel u_{m+1}-u_m\parallel }^2.\hfill \end{array}$$

(11)

Using Step 1, we have:

$$\begin{array}{c}\parallel w_{m+1}{-q\parallel }^2\hfill \\ \le {\alpha }_{m+1}\parallel u_{m+1}{-q\parallel }^2+{\beta }_{m+1}\parallel v_{m+1}{-q\parallel }^2+{\gamma }_{m+1}{\parallel e_{m+1}-q\parallel }^2.\hfill \end{array}$$

(12)

Combining (11) and (12),

$$\begin{array}{c}\parallel w_{m+1}{-q\parallel }^2\le ({\alpha }_{m+1}+{\beta }_{m+1}){\parallel u_{m+1}-q\parallel }^2\hfill \\ +{\gamma }_{m+1}\parallel e_{m+1}{-q\parallel }^2+k_{m+1}^2{\parallel u_{m+1}-u_m\parallel }^2-2k_{m+1}{\beta }_{m+1}\langle u_{m+1}-q,u_m-u_{m+1}\rangle ,\hfill \end{array}$$

which ensures that $q\in C_{m+2}.$ Then by induction, $q\in C_n,$ for $n\in \mathbb{N}.$

Step 3. $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$

It is not difficult to see that

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\le (1-{\gamma }_n)\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2{\parallel u_n-u_{n-1}\parallel }^2-2{\beta }_n{k}_n\langle u_n-p,u_{n-1}-u_n\rangle \hfill \\ ⟺\parallel w_n{\parallel }^2-(1-{\gamma }_n)\parallel u_n{\parallel }^2-{\gamma }_n\parallel e_n{\parallel }^2-k_n^2{\parallel u_n-u_{n-1}\parallel }^2+2{\beta }_n{k}_n\langle u_n,u_{n-1}-u_n\rangle \hfill \\ \le 2\langle p,w_n\rangle -2(1-{\gamma }_n)\langle p,u_n\rangle -2{\gamma }_n\langle p,e_n\rangle +2{\beta }_n{k}_n\langle p,u_{n-1}-u_n\rangle .\hfill \end{array}$$

Then $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$

Step 4. $Q_n$ is non-empty for each $n\in \mathbb{N},$ which ensures that $\{u_n\}$ is well-defined.

From Step 3 and the definition of metric projection, for ${\sigma }_{n+1},$ there exists ${\delta }_{n+1}\in C_{n+1}$ such that $\parallel u_1-{\delta }_{n+1}{\parallel }^2\le (in{f}_{z\in C_{n+1}}\parallel u_1-{z\parallel )}^2+{\sigma }_{n+1}={\parallel P_{C_{n+1}}(u_1)-u_1\parallel }^2+{\sigma }_{n+1}.$ Thus, $Q_{n+1}\ne \varnothing ,$ for $n\in \mathbb{N}.$ And then $\{u_n\}$ is well-defined.

Step 5. $P_{C_{n+1}}(u_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

The proof of Step 5 is similar to Step 2 of Theorem 3.1 in [18]. It follows from Lemma 1 that $lim{C}_n$ exists and $lim{C}_n={\bigcap }_{n=1}^{\infty }{C}_n\ne \varnothing$. Then Lemma 2 implies that $P_{C_{n+1}}(u_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 6. $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $u_{n+1}\in Q_{n+1}\subset C_{n+1}$ and $C_n$ is a convex subset of H, then for $\forall t\in (0,1),$ $t{P}_{C_{n+1}}(u_1)+(1-t)u_{n+1}\in C_{n+1},$ which implies that

$$\parallel P_{C_{n+1}}(u_1)-u_1\parallel \le \parallel t{P}_{C_{n+1}}(x_1)+(1-t)u_{n+1}-u_1\parallel .$$

(13)

Using Lemma 6, we have

$$\begin{array}{c}\parallel t{P}_{C_{n+1}}(u_1)+(1-t)u_{n+1}-u_1{\parallel }^2={\parallel t(P_{C_{n+1}}(u_1)-u_1)+(1-t)(u_{n+1}-u_1)\parallel }^2\hfill \\ \le t\parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+(1-t)\parallel u_{n+1}-u_1{\parallel }^2-t(1-t)g(\parallel P_{C_{n+1}}(u_1)-u_{n+1}\parallel ).\hfill \end{array}$$

(14)

Then (13) and (14) ensure that $tg(\parallel P_{C_{n+1}}(u_1)-u_{n+1}\parallel )\le \parallel u_{n+1}-u_1{\parallel }^2-{\parallel P_{C_{n+1}}(u_1)-u_1\parallel }^2\le {\sigma }_{n+1}.$ Letting $t\to 1$ first and then $n\to \infty$, we know that $P_{C_{n+1}}(u_1)-u_{n+1}\to 0$ as $n\to \infty .$ From Step 5, $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 7. $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$ and $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $u_{n+1}-u_n\to 0$ and $u_{n+1}\in Q_{n+1}\subset C_{n+1},$ then from (10), $\parallel u_{n+1}-v_{n+1}\parallel =k_{n+1}\parallel u_{n+1}-u_n\parallel \to 0,$ as $n\to \infty$. Thus, $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$. Since $u_{n+1}\in Q_{n+1}\subset C_{n+1}$, then

$$\begin{array}{c}\parallel w_n-u_{n+1}{\parallel }^2\le (1-{\gamma }_n)\parallel u_{n+1}-u_n{\parallel }^2+{\gamma }_n{\parallel e_n-u_{n+1}\parallel }^2\hfill \\ +k_n^2{\parallel u_n-u_{n-1}\parallel }^2-2{\beta }_n{k}_n\langle u_n-u_{n+1},u_{n-1}-u_n\rangle \hfill \\ \le (1-{\gamma }_n)\parallel u_{n+1}-u_n{\parallel }^2+{\gamma }_n\parallel e_n-u_{n+1}{\parallel }^2+k_n^2\parallel u_n-u_{n-1}{\parallel }^2+2\parallel u_{n+1}-u_n\parallel \parallel u_{n-1}-u_n\parallel \to 0,\hfill \end{array}$$

as $n\to \infty$. Thus, $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 8. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

In fact, if, otherwise, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\overline{\in }{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$ Then $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\ne {\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$

Since $w_n={\alpha }_n{u}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n+{\gamma }_n{e}_n,$ then ${\beta }_n[{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n-w_n]={\alpha }_n(w_n-u_n)+{\gamma }_n(w_n-e_n)\to 0,$ as $n\to \infty .$

Since $in{f}_n{\beta }_n>0,$ then there exists a subsequence of $\{n\}$, which is still denoted by $\{n\}$ such that ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n-w_n\to 0,$ as $n\to \infty .$

Thus, ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty .$

Since H satisfies Opial’s condition, then

$$\begin{array}{c}limin{f}_{n\to \infty }\parallel v_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ <limin{f}_{n\to \infty }\parallel v_n-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel v_n-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)v_n\parallel \hfill \\ +limin{f}_{n\to \infty }\parallel v_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel v_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel ,\hfill \end{array}$$

which makes a contradiction! Thus, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

Step 9. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1).$

From Step 8, $\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$

On the other hand, since ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset {\bigcap }_{m=1}^{\infty }{C}_m$, then $\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel .$ Thus,

$$\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel =\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$$

Using Lemma 7, we have

$$\begin{array}{c}\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1){\parallel }^2+{\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2\hfill \\ \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1{\parallel }^2={\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2,\hfill \end{array}$$

which implies that $P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)=P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$ □

Remark 1.

Compared to (3) and (4), infinitely many m-accretive mappings and infinitely many ${\mu }_i$-inversely strongly accretive mappings are considered in (10). Compared to (6), the idea of inertial forward-backward algorithm is embodied in (10). Compared to (3), (4) and (6), infinite choices of the iterative sequences $\{u_n\}$ are defined.

Remark 2.

The traditional idea for choosing the unique iterative element $u_{n+1}$ as the metric projection of the initial element in iterative algorithms (3), (4) and (6) is contained in the ideas of (10) in our paper.

In fact, if take $u_{n+1}=P_{C_{n+1}}(u_1),$ we can easily see that

$$\parallel u_{n+1}-u_1{\parallel }^2\le {\parallel P_{C_{n+1}}(u_1)-u_1\parallel }^2+{\sigma }_{n+1}.$$

Thus, $u_{n+1}=P_{C_{n+1}}(u_1)\in Q_{n+1},$ which means that this $u_{n+1}$ is a kind of choice of (10).

Corollary 1.

If $k_n\equiv 0,$ then (10) in Theorem 1 becomes to the traditional forward-backward iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ w_n={\alpha }_n{u}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)u_n+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le (1-{\gamma }_n)\parallel u_n{-p\parallel }^2+{\gamma }_n\parallel e_n-p{\parallel }^2\},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

Corollary 2.

If $i\equiv 1,$ then (10) in Theorem 1 becomes to the following iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ v_n=u_n+k_n(u_n-u_{n-1}),\hfill \\ w_n={\alpha }_n{u}_n+{\beta }_n{(I+s_{n}B)}^{-1}(I-s_{n}A)v_n+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le (1-{\gamma }_n)\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\parallel u_n-u_{n-1}{\parallel }^2-2{\beta }_n{k}_n\langle u_n-p,u_{n-1}-u_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(15)

Remark 3.

Let $e_n\equiv 0,{\alpha }_n+{\beta }_n\equiv 1,$ for $n\in \mathbb{N}.$ After taking $u_{n+1}=P_{C_{n+1}}(u_1)$ in (15), we may see that $Q_{n+1}$ can be deleted which implies that (15) reduces to (4). However, the strong assumption that ${\sum }_{n=1}^{\infty }{k}_n\parallel u_n-u_{n-1}\parallel <+\infty$ in [7] is no longer needed in our paper.

2.2. New Mid-Point Inertial Forward-Backward Projection Algorithms

Theorem 2.

Suppose $f:H\to H$ is a contraction with $k\in (0,1)$ and $F:H\to H$ is a strongly positive linear bounded operator with coefficient $\xi >0.$ Let $\{u_n\}$ be generated by the following iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ z_0=u_0,\hfill \\ z_n={\delta }_n\lambda f(u_n)+(I-{\delta }_{n}F)u_n,\hfill \\ v_n=z_n+k_n(z_n-z_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-p\parallel }^2\hfill \\ +\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{k}_n^2\parallel z_n-z_{n-1}{\parallel }^2-2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle z_n-p,z_{n-1}-z_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(16)

Under the assumptions of $(i)-(iv)$ in Theorem 1 and $(v)$ $\lambda >0$ and $(vi)$ ${\delta }_n\to 0,$ as $n\to \infty ,$ we have: $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$ as $n\to \infty .$

Proof. 

We split the proof into ten steps.

Step 1. ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i):H\to H$ is non-expansive, for $n\in \mathbb{N}.$

Copy Step 1 in Theorem 1.

Step 2. $\{w_n\}$ is well-defined.

Define $U:H\to H$ by

$$Ux=ty+sT(\frac{x+y}{2})+(1-t-s)v,$$

where $T:H\to H$ is non-expansive and $t,s\in (0,1).$

It is easy to check that U is a contraction since

$$\begin{array}{c}\parallel Ux-Uz\parallel \le s\parallel \frac{x+y}{2}-\frac{z+y}{2}\parallel =\frac{s}{2}\parallel x-z\parallel <\parallel x-z\parallel ,\hfill \end{array}$$

for $\forall x,z\in H.$

In view of Lemma 8, there exists a unique element $x\in H$ such that $x=Ux.$ Combining with the result of Step 1, $\{w_n\}$ is well-defined.

Step 3. ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset C_n,$ for $n\in \mathbb{N}.$

If $n=1,$ it is obvious that ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset C_1.$

Now, $\forall q\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$ suppose the result is true for $n=m+1,$ then if $n=m+2,$ in view of Lemma 4, we have:

$$\begin{array}{c}\parallel w_{m+1}{-q\parallel }^2\le {\alpha }_{m+1}\parallel v_{m+1}{-q\parallel }^2+{\beta }_{m+1}\parallel \frac{v_{m+1}+w_{m+1}}{2}{-q\parallel }^2+{\gamma }_{m+1}{\parallel e_{m+1}-q\parallel }^2,\hfill \end{array}$$

which ensures that

$$\begin{array}{c}\parallel w_{m+1}{-q\parallel }^2\le \frac{2{\alpha }_{m+1}+{\beta }_{m+1}}{2-{\beta }_{m+1}}\parallel v_{m+1}{-q\parallel }^2+\frac{2{\gamma }_{m+1}}{2-{\beta }_{m+1}}{\parallel e_{m+1}-q\parallel }^2.\hfill \end{array}$$

(17)

It follows from (16) that

$$\begin{array}{c}\parallel v_{m+1}{-q\parallel }^2\hfill \\ =\parallel z_{m+1}{-q\parallel }^2+2k_{m+1}\langle z_{m+1}-q,z_{m+1}-z_m\rangle +k_{m+1}^2{\parallel z_{m+1}-z_m\parallel }^2.\hfill \end{array}$$

(18)

Combining (17) and (18),

$$\begin{array}{c}\parallel w_{m+1}{-q\parallel }^2\le \frac{2{\alpha }_{m+1}+{\beta }_{m+1}}{2-{\beta }_{m+1}}\parallel z_{m+1}{-q\parallel }^2+\frac{2{\gamma }_{m+1}}{2-{\beta }_{m+1}}{\parallel e_{m+1}-q\parallel }^2\hfill \\ +\frac{2{\alpha }_{m+1}+{\beta }_{m+1}}{2-{\beta }_{m+1}}{k}_{m+1}^2{\parallel z_{m+1}-z_m\parallel }^2-2k_{m+1}\frac{2{\alpha }_{m+1}+{\beta }_{m+1}}{2-{\beta }_{m+1}}\langle z_{m+1}-q,z_m-z_{m+1}\rangle .\hfill \end{array}$$

Thus, $q\in C_{m+2}.$ Then by induction, $q\in C_n,$ for $n\in \mathbb{N}.$

Step 4. $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$ It is not difficult to see that

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{\parallel z_n-z_{n-1}\parallel }^2-2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle z_n-p,z_{n-1}-z_n\rangle \hfill \\ ⟺\parallel w_n{\parallel }^2-\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{\parallel }^2-\frac{2{\gamma }_n}{2-{\beta }_n}\parallel e_n{\parallel }^2-k_n^2\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{\parallel z_n-z_{n-1}\parallel }^2+2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle z_n,z_{n-1}-z_n\rangle \hfill \\ \le 2\langle p,w_n\rangle -2\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle p,z_n\rangle -\frac{4{\gamma }_n}{2-{\beta }_n}\langle p,e_n\rangle +2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle p,z_{n-1}-z_n\rangle .\hfill \end{array}$$

Then $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$

Step 5. $Q_n$ is non-empty for each $n\in \mathbb{N},$ which ensures that $\{u_n\}$ is well-defined.

Copy Step 4 in Theorem 1.

Step 6. $P_{C_{n+1}}(u_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Copy Step 5 in Theorem 1.

Step 7. $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Copy Step 6 in Theorem 1.

Step 8. $z_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$, $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$ and $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $z_n-u_n={\delta }_n(\lambda f(u_n)-F{u}_n)$ and ${\delta }_n\to 0,$ then it is easy to see that $z_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$, as $n\to \infty$.

Since $v_n=z_n+k_n(z_n-z_{n-1}),$ then $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $u_{n+1}-u_n\to 0$ and $u_{n+1}\in Q_{n+1}\subset C_{n+1},$ then

$$\begin{array}{c}\parallel w_n-u_{n+1}{\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n-u_{n+1}{\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-u_{n+1}\parallel }^2\hfill \\ +k_n^2\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{\parallel z_n-z_{n-1}\parallel }^2-2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle z_n-u_{n+1},z_{n-1}-z_n\rangle \hfill \\ \le \parallel z_n-u_{n+1}{\parallel }^2+2{\gamma }_n\parallel e_n-z_{n+1}{\parallel }^2+k_n^2\parallel z_n-z_{n-1}{\parallel }^2+2\parallel u_{n+1}-z_n\parallel \parallel z_{n-1}-z_n\parallel \to 0.\hfill \end{array}$$

Thus, $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 9. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

In fact, if, otherwise, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\overline{\in }{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$ Then $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\ne {\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$

Since $w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})+{\gamma }_n{e}_n,$ then ${\beta }_n[{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})-w_n]={\alpha }_n(w_n-v_n)+{\gamma }_n(w_n-e_n)\to 0,$ as $n\to \infty .$

Since $in{f}_n{\beta }_n>0,$ then there exists a subsequence of $\{n\}$, which is still denoted by $\{n\}$ such that ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})-w_n\to 0,$ as $n\to \infty .$

Thus, ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty .$

Since H satisfies Opial’s condition, then

$$\begin{array}{c}limin{f}_{n\to \infty }\parallel \frac{v_n+w_n}{2}-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ <limin{f}_{n\to \infty }\parallel \frac{v_n+w_n}{2}-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel \frac{v_n+w_n}{2}-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)\frac{v_n+w_n}{2}\parallel \hfill \\ +limin{f}_{n\to \infty }\parallel \frac{v_n+w_n}{2}-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel \frac{v_n+w_n}{2}-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel ,\hfill \end{array}$$

which makes a contradiction. Thus, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

Step 10. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1).$

From Step 9, $\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$

On the other hand, since ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\subset {\bigcap }_{m=1}^{\infty }{C}_m$, then $\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel .$ Thus,

$$\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1\parallel =\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$$

Using Lemma 7, we have

$$\begin{array}{c}\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1){\parallel }^2+{\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2\hfill \\ \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-u_1{\parallel }^2={\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2,\hfill \end{array}$$

which implies that $P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)=P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$ □

Remark 4.

Similar to Remark 2, $u_{n+1}=P_{C_{n+1}}(u_1)$ is also a possible choice of $u_{n+1}\in Q_{n+1}$ in Theorem 2.

Corollary 3.

If ${\delta }_n\equiv 0$, then (16) becomes to the following traditional mid-point inertial forward-backward projection iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ v_n=u_n+k_n(u_n-u_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel u_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-p\parallel }^2\hfill \\ +\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{k}_n^2\parallel u_n-u_{n-1}{\parallel }^2-2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle u_n-p,u_{n-1}-u_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(19)

If, moreover, $k_n\equiv 0$ in (19), then it becomes to the following traditional forward-backward mid-point iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ w_n={\alpha }_n{u}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{u_n+w_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel u_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}\parallel e_n-p{\parallel }^2\},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(20)

Corollary 4.

If $k_n\equiv 0$ in (15), then it becomes to the following one:

$$\left\{\begin{array}{c}{u}_0,u_1\in Hchosenarbitrarily,e_1\in H,\hfill \\ z_n={\delta }_n\lambda f(u_n)+(I-{\delta }_{n}F)u_n,\hfill \\ w_n={\alpha }_n{z}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{w_n+z_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}\parallel e_n-p{\parallel }^2\},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(21)

If, moreover, take $u_{n+1}=P_{C_{n+1}}(u_1)$ in (21), then it becomes to (6).

2.3. Relationship with Variational Inequalities

A lot work has been done on designing iterative algorithms to approximate solution of variational inequalities due to their wide applications (e.g., [26,27]). A classical variational inequality is to find $u\in K$ such that for $\forall v\in K,$

$$\langle v-u,Tu\rangle \ge 0,$$

(22)

where $T:K\to H$ is a nonlinear mapping. The symbol $VI(K,T)$ denotes the solution of the above variational inequality.

2.3.1. The First Kind Iteration Theorems

Definition 3.

Let H be a real Hilbert space with K being its non-empty closed and convex subset. $T:K\to H$ is called a τ-Lipschitz continuous mapping if $\parallel Tx-Ty\parallel \le \tau \parallel x-y\parallel ,$ for $x,y\in K.$

Theorem 3.

Let H be a real Hilbert space with K being its non-empty closed and convex subset. Suppose $A_i:K\to H$ is ${\mu }_i$-inversely strongly accretive and $B_i:K\to H$ is m-accretive, for each $i\in \mathbb{N}$. Suppose $T:K\to H$ is an accretive and τ-Lipschitz continuous mapping. Let $\{u_n\}$ be generated by the following iterative algorithm:

$$\left\{\begin{array}{c}{u}_0\in K,u_1\in Kchosenarbitrarily,e_1\in H,\hfill \\ y_0=P_K(u_0-{\lambda }_{0}T{u}_0),\hfill \\ y_n=P_K(u_n-{\lambda }_{n}T{u}_n),\hfill \\ v_n=y_n+k_n(y_n-y_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{y}_n)+{\gamma }_n{e}_n,\hfill \\ C_1=K=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le {\alpha }_n\parallel y_n{-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\parallel y_n-y_{n-1}{\parallel }^2-2{\alpha }_n{k}_n\langle y_n-p,y_{n-1}-y_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

(23)

Under the assumptions of (i)–(iv) in Theorem 1 and the additional assumptions $(v)$ ${\lambda }_n\to 0,$ $(vi)$ ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)\ne \varnothing ,$ we have: $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1),$ as $n\to \infty .$

Proof. 

We split the proof into nine steps.

Step 1. ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i):K\to K$ is non-expansive, for $n\in \mathbb{N}.$

Copy the proof of Step 1 in Theorem 1.

Step 2. ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)\subset C_n,$ for $n\in \mathbb{N}.$

We first show that $\forall q\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T),$

$$\parallel P_K(u_n-{\lambda }_{n}T{y}_n)-q\parallel \le \parallel u_n-q\parallel .$$

(24)

In fact, in view of Lemma 7, we have:

$$\begin{array}{c}\parallel P_K(u_n-{\lambda }_{n}T{y}_n){-q\parallel }^2\hfill \\ \le \parallel u_n-q-{\lambda }_{n}T{y}_n{\parallel }^2-{\parallel u_n-P_K(u_n-{\lambda }_{n}T{y}_n)-{\lambda }_{n}T{y}_n\parallel }^2\hfill \\ =\parallel u_n{-q\parallel }^2-{\parallel u_n-P_K(u_n-{\lambda }_{n}T{y}_n)\parallel }^2\hfill \\ +2{\lambda }_n[\langle T{y}_n-Tq,q-y_n\rangle +\langle Tq,q-y_n\rangle +\langle T{y}_n,y_n-P_K(u_n-{\lambda }_{n}T{y}_n)\rangle ]\hfill \\ \le \parallel u_n{-q\parallel }^2-{\parallel u_n-P_K(u_n-{\lambda }_{n}T{y}_n)\parallel }^2+2{\lambda }_n\langle T{y}_n,y_n-P_K(u_n-{\lambda }_{n}T{y}_n)\rangle \hfill \\ =\parallel u_n{-q\parallel }^2-(\parallel u_n-y_n{\parallel }^2+\parallel y_n-P_K(u_n-{\lambda }_{n}T{y}_n){\parallel }^2)+2{\lambda }_n\langle T{u}_n-T{y}_n,P_K(u_n-{\lambda }_{n}T{y}_n)-y_n\rangle \hfill \\ +2\langle u_n-y_n-{\lambda }_{n}T{u}_n,P_K(u_n-{\lambda }_{n}T{y}_n)-y_n\rangle \hfill \\ \le \parallel u_n{-q\parallel }^2-(\parallel u_n-y_n{\parallel }^2+\parallel y_n-P_K(u_n-{\lambda }_{n}T{y}_n){\parallel }^2)+2{\lambda }_n\tau \parallel u_n-y_n\parallel \parallel y_n-P_K(u_n-{\lambda }_{n}T{y}_n)\parallel \hfill \\ \le \parallel u_n{-q\parallel }^2+({\lambda }_n^2{\tau }^2-1){\parallel u_n-y_n\parallel }^2\hfill \\ \le \parallel u_n{-q\parallel }^2,\hfill \end{array}$$

which implies that (24) is true.

Next, we can easily check the following by noticing the result of Step 1 and (24):

$$\begin{array}{c}\parallel w_n{-q\parallel }^2\le {\alpha }_n\parallel v_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+{\gamma }_n{\parallel e_n-q\parallel }^2\hfill \\ \le {\alpha }_n\parallel y_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+{\gamma }_n\parallel e_n{-q\parallel }^2+k_n^2{\parallel y_n-y_{n-1}\parallel }^2-2{\alpha }_n{k}_n\langle y_n-q,y_{n-1}-y_n\rangle .\hfill \end{array}$$

Thus, by induction as that in Theorem 1, $q\in C_n$, for $n\in \mathbb{N}$.

Step 3. $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$

It is not difficult to see that

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\le {\alpha }_n\parallel y_n{-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2{\parallel y_n-y_{n-1}\parallel }^2-2{\alpha }_n{k}_n\langle y_n-p,y_{n-1}-y_n\rangle \hfill \\ ⟺\parallel w_n{\parallel }^2-{\alpha }_n\parallel y_n{\parallel }^2-{\beta }_n\parallel u_n{\parallel }^2-{\gamma }_n\parallel e_n{\parallel }^2-k_n^2{\parallel y_n-y_{n-1}\parallel }^2+2{\alpha }_n{k}_n\langle y_n,y_{n-1}-y_n\rangle \hfill \\ \le 2\langle p,w_n\rangle -2{\alpha }_n\langle p,y_n\rangle -2{\beta }_n\langle p,u_n\rangle -2{\gamma }_n\langle p,e_n\rangle +2{\alpha }_n{k}_n\langle p,y_{n-1}-y_n\rangle .\hfill \end{array}$$

Then $C_n$ is a closed and convex subset of $H,$ for each $n\in \mathbb{N}.$

Copy the results of Steps 4–6 in Theorem 1, we have:

Step 4. $Q_n$ is non-empty for each $n\in \mathbb{N},$ which ensures that $\{u_n\}$ is well-defined.

Step 5. $P_{C_{n+1}}(u_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 6. $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 7. $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$, $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$ and $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

It is easy to see that $Q_n$ is a closed subset of K. Then $u_{n+1}\in Q_{n+1}\subset K$ and $u_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)$ imply that $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in K.$ Therefore,

$$\begin{array}{c}\parallel y_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \le \parallel u_n-{\lambda }_{n}T{u}_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le \parallel u_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel +{\lambda }_n\parallel T{u}_n\parallel \to 0,\hfill \end{array}$$

as $n\to \infty$. Thus, $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $y_{n+1}-y_n\to 0$ and $v_n=y_n+k_n(y_n-y_{n-1}),$ then $v_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Since $u_{n+1}\in Q_{n+1}\subset C_{n+1},$ then from (23), $\parallel w_n-u_{n+1}{\parallel }^2\le {\alpha }_n\parallel y_n-u_{n+1}{\parallel }^2+{\beta }_n\parallel u_{n+1}-u_n{\parallel }^2+{\gamma }_n\parallel e_n-u_{n+1}{\parallel }^2+k_n^2{\parallel y_n-y_{n-1}\parallel }^2-2{\alpha }_n{k}_n\langle y_n-u_{n+1},y_{n-1}-y_n\rangle \to 0,$ as $n\to \infty$. Thus, $w_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty$.

Step 8. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T).$

We shall first show that $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in VI(K,T).$

For this, define

$$Bv=\left\{\begin{array}{c}Tv+N_{K}v,ifv\in K,\hfill \\ \varnothing ,ifv\overline{\in }K,\hfill \end{array}\right.$$

where $N_{K}v=\{w\in H:\langle v-u,w\rangle \ge 0,\forall u\in K\}$ is the normal cone to K at $v\in K.$ It is well-known that $B:H\to H$ is m-accretive and $0\in Bv$ if and only if $v\in VI(K,T)$ [28].

Let $z\in Bv=Tv+N_{K}v,$ then $z-Tv\in N_{K}v.$ From the definition of the normal cone, we have

$$\langle v-y_n,z-Tv\rangle \ge 0.$$

(25)

From Lemma 7, we have:

$$\langle u_n-{\lambda }_{n}T{u}_n-P_K(u_n-{\lambda }_{n}T{u}_n),P_K(u_n-{\lambda }_{n}T{u}_n)-v\rangle \ge 0,\forall v\in K,$$

which implies that

$$\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),\frac{P_K(u_n-{\lambda }_{n}T{u}_n)-u_n}{{\lambda }_n}+T{u}_n\rangle \ge 0,\forall v\in K.$$

(26)

In view of (25) and (26), we know that

$$\begin{array}{c}\langle v-y_n,z\rangle \ge \langle v-y_n,Tv\rangle \hfill \\ =\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),Tv\rangle \hfill \\ \ge \langle v-P_K(u_n-{\lambda }_{n}T{u}_n),Tv\rangle -\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),\frac{P_K(u_n-{\lambda }_{n}T{u}_n)-u_n}{{\lambda }_n}+T{u}_n\rangle \hfill \\ =\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),Tv-T{P}_K(u_n-{\lambda }_{n}T{u}_n)\rangle \hfill \\ +\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),T{P}_K(u_n-{\lambda }_{n}T{u}_n)-T{u}_n\rangle \hfill \\ -\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),\frac{P_K(u_n-{\lambda }_{n}T{u}_n)-u_n}{{\lambda }_n}\rangle \hfill \\ \ge \langle v-P_K(u_n-{\lambda }_{n}T{u}_n),T{P}_K(u_n-{\lambda }_{n}T{u}_n)-T{u}_n\rangle \hfill \\ -\langle v-P_K(u_n-{\lambda }_{n}T{u}_n),\frac{P_K(u_n-{\lambda }_{n}T{u}_n)-u_n}{{\lambda }_n}\rangle \hfill \\ =\langle v-y_n,T{y}_n-T{u}_n\rangle -\langle v-y_n,\frac{y_n-u_n}{{\lambda }_n}\rangle .\hfill \end{array}$$

Taking limit on both sides of the above inequality, we have: $\langle v-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),z\rangle \ge 0,$ which implies from the fact B is m-accretive that $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in B^{-1}0,$ and then $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in VI(K,T).$

Next, we shall show that $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

In fact, if, otherwise, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\overline{\in }{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$ Then $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\ne {\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$

Since $w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{y}_n)+{\gamma }_n{e}_n,$ then ${\beta }_n[{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{y}_n)-w_n]={\alpha }_n(w_n-v_n)+{\gamma }_n(w_n-e_n)\to 0,$ as $n\to \infty .$

Since $in{f}_n{\beta }_n>0,$ then there exists a subsequence of $\{n\}$, which is still denoted by $\{n\}$ such that ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{y}_n)-w_n\to 0,$ as $n\to \infty .$

Thus, ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{y}_n)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty .$ From Step 1 and $u_n-y_n\to 0$, we can also know that ${\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{u}_n)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1),$ as $n\to \infty .$

Since H satisfies Opial’s condition, ${\lambda }_n\to 0$ and $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in K$, then

$$\begin{array}{c}limin{f}_{n\to \infty }\parallel u_n-{\lambda }_{n}T{u}_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ <limin{f}_{n\to \infty }\parallel u_n-{\lambda }_{n}T{u}_n-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel u_n-{\lambda }_{n}T{u}_n-{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{u}_n)\parallel \hfill \\ +limin{f}_{n\to \infty }\parallel {\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_K(u_n-{\lambda }_{n}T{u}_n)\hfill \\ -{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel \hfill \\ \le limin{f}_{n\to \infty }\parallel u_n-{\lambda }_{n}T{u}_n-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\parallel ,\hfill \end{array}$$

which makes a contradiction! Thus, $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

Step 9. $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1).$

From Step 8, $\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$

On the other hand, since ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)\subset {\bigcap }_{m=1}^{\infty }{C}_m$, then

$$\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)-u_1\parallel .$$

Thus

$$\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)-u_1\parallel =\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel .$$

Using Lemma 7, we have

$$\begin{array}{c}\parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1){\parallel }^2+{\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2\hfill \\ \le \parallel P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)-u_1{\parallel }^2={\parallel P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1)-u_1\parallel }^2,\hfill \end{array}$$

which implies that $P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0\bigcap VI(K,T)}(u_1)=P_{{\bigcap }_{m=1}^{\infty }{C}_m}(u_1).$

This completes the proof. □

2.3.2. The Second Kind Iteration Theorems

The following result is a special result of Lemma 10 in [19]:

Theorem 4.

Let H be a real Hilbert space and K be a non-empty closed and convex subset of H. Suppose $f:H\to H$ is a contraction with $k\in (0,1),$ $F:H\to H$ is a strongly positive linear bounded operator with coefficient ξ and $U:H\to H$ is non-expansive mapping. If $0<\lambda <\frac{\xi }{2k}$, then there exists $x_t$ which satisfies $x_t=t\lambda f(x_t)+(I-tF)U{x}_t,$ for $0<t\le {\parallel F\parallel }^{-1}.$ Moreover, $x_t\to p_0$ as $t\to 0,$ and $p_0$ satisfies the following variational inequality: for $\forall z\in Fix(U),$

$$\langle (F-\lambda f)p_0,p_0-z\rangle \le 0.$$

(27)

In Lemma 10 of [19], we can also know that the solution of the variational inequality (27) is unique.

Theorem 5.

Under the assumptions of Theorem 2, $\{u_n\}$ generated by (16) converges strongly to $P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)$. Set $\tilde{x}=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1).$ If $\tilde{x}=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}[\lambda f(\tilde{x})-F(\tilde{x})+\tilde{x}],$ then $\tilde{x}$ satisfies the following variational inequality: $\forall z\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$

$$\langle (F-\lambda f)\tilde{x},\tilde{x}-z\rangle \le 0.$$

(28)

Proof. 

It follows from Lemma 7 that $\langle (F-\lambda f)\tilde{x},\tilde{x}-z\rangle \le 0,$ $\forall z\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$ Since Theorem 4 tells us that (28) has a unique solution, then we know that $\{u_n\}$ generated by (16) converges strongly to the unique solution of variational inequality (28). □

Remark 5.

The assumption that $\tilde{x}=P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}[\lambda f(\tilde{x})-F(\tilde{x})+\tilde{x}]$ is reasonable. For example, we may take $f(x)=\frac{x}{2}$ and $F(x)=\frac{\lambda x}{2},$ for $x\in H.$

Remark 6.

For projection iterative algorithms such as (16), rare work can be found to show that the limit of the iterative sequences is also the solution of a kind of variational inequalities.

3. Applications

3.1. Preparation for Discussion of Capillarity Systems

To present some examples in this section, we need some basic definitions in Banach spaces.

Let E be a real Banach space with $E^{*}$ being its dual space and let $\langle ·,·\rangle$ denote the generalized duality pairing between E and $E^{*}$.

Definition 4.

(see [29]) Recall that $J:E\to 2^{E^{*}}$ is called the normalized duality mapping if $\forall x\in E,$

$$Jx=\{f\in E^{*}:\langle x,f\rangle =\parallel x{\parallel }^2=\parallel f{\parallel }^2\}.$$

Definition 5.

(see [29]) Recall that $\mathcal{A}:E\to E^{*}$ is said to be a monotone mapping if $\forall x_i\in D(\mathcal{A}),i=1,2,$ one has

$$\langle x_1-x_2,\mathcal{A}{x}_1-\mathcal{A}{x}_2\rangle \ge 0.$$

A monotone mapping $\mathcal{A}:E\to E^{*}$ is said to be maximal monotone if $R(J+r\mathcal{A})=E^{*}$, $\forall r>0$.

Definition 6.

(see [29]) Recall that a mapping $\mathcal{B}:E\to E^{*}$ is said to be coercive if $\{x_n\}\subset D(\mathcal{B})$ with ${lim}_{n\to \infty }\parallel x_n\parallel =+\infty ,$ then ${lim}_{n\to \infty }\frac{\langle x_n,\mathcal{B}{x}_n\rangle }{\parallel x_n\parallel }=+\infty .$

Definition 7.

(see [29]) Recall that a mapping $\mathcal{B}:D(\mathcal{B})=E\to E^{*}$ is said to be a hemi-continuous mapping if $\mathcal{B}(x+ty)\rightharpoonup \mathcal{B}x$, as $t\to 0,$ for $\forall x,y\in E.$

Definition 8.

(see [29]) $\psi :E\to (-\infty ,+\infty ]$ is said to be a proper convex functional if there exists $u_0\in E$ such that $\psi (u_0)<+\infty$ and $\psi ((1-\lambda )u+\lambda v)\le (1-\lambda )\psi (u)+\lambda \psi (v),$ for $\forall u,v\in E$ and $\lambda \in [0,1].$ $\psi :E\to (-\infty ,+\infty ]$ is said to be lower-semi-continuous: ${lim\; inf}_{y\to x}\psi (y)\ge \psi (x),$ for $\forall x\in E.$ The subdifferential of ψ, $\partial \psi :E\to E^{*},$ is defined by:

$$\partial \psi (u)=\{w\in E^{*}:\psi (u)-\psi (v)\le \langle u-v,w\rangle ,\forall v\in E\},\forall u\in E.$$

3.2. Applications to Capillarity Elliptic Systems

Example 1.

Suppose Ω is bounded conical domain in $R^n$ $(n\in \mathbb{N})$ with $\mathsf{\Gamma }\in C^1,$ϑ is the exterior normal derivative of Γ, ${\epsilon }_i$ is a non-negative constant, ${\lambda }_i$ is a positive number, $f_i(x)\in L^{p_i}(\mathsf{\Omega })$ is a given function, for $i=1,2,\cdots ,M$. In addition, ${\beta }_x:R\to R$ is the subdifferential of ${\phi }_x$, where ${\phi }_x=\phi (x,·):R\to R,$ for each $x\in \mathsf{\Gamma }$. Suppose $\frac{2n}{n+1}<p_i<+\infty ,$ $i=1,2,\cdots ,M$. If $p_i\ge n$ then $1\le q_i,r_i<+\infty$ and if $p_i<n$ then $1\le q_i,r_i\le \frac{n{p}_i}{n-p_i}$ for $i=1,2,\cdots ,M.$

The following capillarity system is studied in [30]:

$$\left\{\begin{array}{c}-div[(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}})|\nabla u^{(i)}{|}^{p_i-2}\nabla u^{(i)}]+{\lambda }_i(|u^{(i)}{|}^{q_i-2}{u}^{(i)}+|u^{(i)}{|}^{r_i-2}{u}^{(i)})\hfill \\ +{\epsilon }_i{g}_i(x,\nabla u^{(i)},u^{(i)})=f_i(x),x\in \mathsf{\Omega }\hfill \\ -<\vartheta ,(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}}){|\nabla u^{(i)}|}^{p_i-2}\nabla u^{(i)}>\in {\beta }_x(u^{(i)}(x)),x\in \mathsf{\Gamma },i=1,2,\cdots ,M,\hfill \end{array}\right.$$

(29)

where $|·|$ and $<·,·>$ denote the norm and inner product in $R^n,$ respectively.

The study on (29) in [30] is based on the following assumptions.

(1)

$\forall x\in \mathsf{\Gamma },{\phi }_x=\phi (x,·):R\to R$ is a proper convex and lower-semi-continuous mapping with ${\phi }_x(0)=0$.

(2)

$0\in {\beta }_x(0)$, $\forall t\in R,$$x\in \mathsf{\Gamma }\to {(I+\lambda {\beta }_x)}^{-1}(t)\in R$ is measurable for $\lambda >0$.

(3)

For each $i\in \{1,2,\cdots ,M\},$ $g_i:\mathsf{\Omega }\times R^n\times R\to R$ satisfies Caratheodory’s conditions and satisfies that

$$|g_i(x,r_1,r_2,\cdots ,r_{n+1})-g_i(x,s_1,s_2,\cdots ,s_{n+1})|\le b_i|r_{n+1}-s_{n+1}|,$$

for $\forall (r_1,r_2,\cdots ,r_{n+1}),(s_1,s_2,\cdots ,s_{n+1})\in R^{n+1}.$

By using splitting method, the sufficient condition that (29) has non-trivial solution is obtained:

Theorem 6.

(see [30]) If $u^{(i)}\in L^{p_i}(\mathsf{\Omega })$ satisfies that

$${\int }_{\mathsf{\Omega }}[g_i(x,\nabla u^{(i)},u^{(i)})-f_i]{|u^{(i)}|}^{p_i-2}{u}^{(i)}dx\ge 0,$$

for $i=1,2,\cdots ,M,$ then $u=(u^{(1)},u^{(2)},\cdots ,u^{(M)})$ is the non-trivial solution of capillarity system (29).

Based on Example 1, we present the following example:

Example 2.

Suppose Ω, Γ and ϑ are the same as those in Example 1. Suppose ${\lambda }_i>0$, $\frac{2n}{n+1}<p_i<+\infty$. If $p_i\ge n,$ then suppose $1\le q_i,r_i<+\infty$ and if $p_i<n,$ then suppose $1\le q_i,r_i\le \frac{n{p}_i}{n-p_i},$ for $i\in \mathbb{N}.$

Now, we will discuss the following capillarity systems.

$$\left\{\begin{array}{c}-div[(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}})|\nabla u^{(i)}{|}^{p_i-2}\nabla u^{(i)}]\hfill \\ +{\lambda }_i(|u^{(i)}{|}^{q_i-2}{u}^{(i)}+|u^{(i)}{|}^{r_i-2}{u}^{(i)})+u^{(i)}(x)=f_i(x),x\in \mathsf{\Omega }\hfill \\ -<\vartheta ,(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}}){|\nabla u^{(i)}|}^{p_i-2}\nabla u^{(i)}>=0,x\in \mathsf{\Gamma },i\in \mathbb{N}.\hfill \end{array}\right.$$

(30)

Please note that (30) is the extension from the finite case of (29) to that for infinite case. However, both the capillarity equations and the boundary conditions are the special case of (29) in the sense that ${\epsilon }_i\equiv 1$ and $g_i(x,\nabla u^{(i)},u^{(i)})\equiv u^{(i)}$ for $i\in \mathbb{N}$ and ${\beta }_x\equiv 0,$ for $x\in \mathsf{\Gamma }.$

Lemma 9.

(see [30]) The mapping $U_i:W^{1,p_i}(\mathsf{\Omega })\to {(W^{1,p_i}(\mathsf{\Omega }))}^{*}$ defined by

$$\langle w,U_{i}u\rangle ={\int }_{\mathsf{\Omega }}<(1+\frac{{|\nabla u|}^{p_i}}{\sqrt{1+{|\nabla u|}^{2p_i}}}){|\nabla u|}^{p_i-2}\nabla u,$$

$$\nabla v>dx+{\lambda }_i{\int }_{\mathsf{\Omega }}{|u(x)|}^{q_i-2}u(x)v(x)dx+{\lambda }_i{\int }_{\mathsf{\Omega }}{|u(x)|}^{r_i-2}u(x)v(x)dx,$$

for $\forall u,w\in W^{1,p_i}(\mathsf{\Omega }),$ is everywhere defined, hemi-continuous, monotone and coercive, for each $i\in \mathbb{N}$.

Lemma 10.

(see [30]) Define $B_i:L^2(\mathsf{\Omega })\to L^2(\mathsf{\Omega })$ by

$$D(B_i)=\{u\in L^2(\mathsf{\Omega })|\exists f\in L^2(\mathsf{\Omega })suchthatf\in U_{i}u\}.$$

For $u\in D(B_i)$, $B_{i}u=\{f\in L^2(\mathsf{\Omega })|f\in U_{i}u\}$. Then $B_i:L^2(\mathsf{\Omega })\to L^2(\mathsf{\Omega })$ is m-accretive, for each $i\in \mathbb{N}$.

Lemma 11.

(see [30]) The mapping $A_i:D(A_i)\subset L^2(\mathsf{\Omega })\to L^2(\mathsf{\Omega })$ defined by

$$(A_{i}u)(x)=u(x)-f_i(x),\forall u(x)\in D(A_i),$$

is ${\mu }_i$-inversely strongly accretive, for ${\mu }_i\in (0,1]$, for $i\in \mathbb{N}.$

Theorem 7.

If $f_i(x)\equiv {\lambda }_i{(|k|}^{q_i-1}+{|k|}^{r_i-1})sgnk+k,$ then $\{u^{(i)}\equiv k:i\in \mathbb{N}\}$ is the solution of capillarity system (30). Moreover, $\{k\}={\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

Proof. 

It is easy to see that $\{u^{(i)}\equiv k:i\in \mathbb{N}\}$ is the solution of capillarity system (30) and $\{k\}\subset {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$ Now, we shall show that $\{k\}\supset {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0.$

In fact, if $A_{i}u+B_{i}u=0$ and $A_{i}v+B_{i}v=0,$ then $u+B_{i}u=v+B_{i}v,$ which implies that

$$0\le \langle u-v,B_{i}u-B_{i}v\rangle =\langle u-v,v-u\rangle \le 0.$$

Thus, $u=v$ and then ${\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0$ is a singleton. Since $k\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$ then the result follows. □

Theorem 8.

Let $f_i(x)\equiv {\lambda }_i{(|k|}^{q_i-1}+{|k|}^{r_i-1})sgnk+k,$ for $i\in \mathbb{N}.$ Suppose $A_i$ and $B_i$ are the same as those in Lemmas 10 and 11, respectively. Let $F:L^2(\mathsf{\Omega })\to L^2(\mathsf{\Omega })$ be any strongly positive linear bounded operator with coefficient $\xi >0$ and $f:L^2(\mathsf{\Omega })\to L^2(\mathsf{\Omega })$ be a contraction with coefficient $k\in (0,1).$ Constructing the following iterative algorithm:

$$\left\{\begin{array}{c}{u}_0,u_1\in L^2(\mathsf{\Omega })chosenarbitrarily,e_1\in L^2(\mathsf{\Omega }),\hfill \\ z_0=u_0,\hfill \\ z_n={\delta }_n\lambda f(u_n)+(I-{\delta }_{n}F)u_n,\hfill \\ v_n=z_n+k_n(z_n-z_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n{\sum }_{i=1}^{\infty }{\omega }_{n,i}{(I+s_{n,i}{B}_i)}^{-1}(I-s_{n,i}{A}_i)(\frac{v_n+w_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=L^2(\mathsf{\Omega })=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-p\parallel }^2\hfill \\ +\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{k}_n^2\parallel z_n-z_{n-1}{\parallel }^2-2k_n\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\langle z_n-p,z_{n-1}-z_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(u_1)-u_1{\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in \mathbb{N}.\hfill \end{array}\right.$$

Under the assumptions of Theorem 2, using the result of Theorem 5, one has $u_n(x)\to P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1),$ which is the unique solution of capillarity system (30) and satisfies the following variational inequality: For $\forall z(x)\in {\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0,$

$$\langle (F-\lambda f)P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1),P_{{\bigcap }_{i=1}^{\infty }{(A_i+B_i)}^{-1}0}(u_1)-z)\rangle \le 0.$$

Remark 7.

From Theorem 8 we can easily see the relationship among the solution of capillarity system, the solution of variational inequality and the zero of sum of infinitely many m-accretive mappings and infinitely many ${\mu }_i$-inversely strongly accretive mappings.