Variational Inequalities Approaches to Minimization Problems with Constraints of Generalized Mixed Equilibria and Variational Inclusions

Abstract

Multistep composite implicit and explicit extragradient-like schemes are presented for solving the minimization problem with the constraints of variational inclusions and generalized mixed equilibrium problems. Strong convergence results of introduced schemes are given under suitable control conditions.

1. Introduction

Let H be a real Hilbert space and $\varnothing \ne C\subset H$ be a closed convex set. Let $\mathcal{A}:C\to H$ be a nonlinear operator. Let us consider the variational inequality problem (VIP) which aims to find $x^{*}\in C$ verifying

$$\langle \mathcal{A}{x}^{*},x-x^{*}\rangle \ge 0,\forall x\in C.$$

(1)

We denote the solution set of VIP (1) by $\mathrm{VI}(C,\mathcal{A})$.

In [1], Korpelevich suggested an extragradient method for solving VIP (1). Korpelevich’s method has been studied by many authors, see e.g., references [2,3,4,5,6,7,8,9,10,11,12,13] and references therein. In 2011, Ceng, Ansari and Yao [14] considered the following algorithm

$$u_{k+1}=P_C[{\alpha }_k\gamma V{u}_k+(I-{\alpha }_k\mu F)T{u}_k],\forall k\ge 0.$$

(2)

They showed that $\{u_k\}$ strongly converges to $x^{†}\in \mathrm{Fix}(T)$ which solves the following VIP

$$\langle (\mu F-\gamma V)x^{†},x^{†}-\tilde{x}\rangle \le 0,\forall \tilde{x}\in \mathrm{Fix}(T).$$

(3)

Ceng, Guu and Yao [15] presented a composite implicit scheme

$$x_t=(I-{\theta }_t\mathcal{B})T{x}_t+{\theta }_t[T{x}_t-t(\mu FT{x}_t-\gamma V{x}_t)],$$

(4)

and another composite explicit scheme

$$\left\{\begin{array}{c}{y}_n=(I-{\alpha }_n\mu F)T{x}_n+{\alpha }_n\gamma V{x}_n,\hfill \\ x_{n+1}=(I-{\beta }_n\mathcal{B})T{x}_n+{\beta }_n{y}_n,\forall n\ge 0.\hfill \end{array}\right.$$

(5)

Ceng, Guu and Yao proved that $\{x_t\}$ and $\{x_n\}$ converge strongly to the same point $x^{†}\in \mathrm{Fix}(T)$, which solves the following VIP

$$\langle (\mathcal{B}-I)x^{†},x^{†}-\tilde{x}\rangle \le 0,\forall \tilde{x}\in \mathrm{Fix}(T).$$

(6)

In [2], Peng and Yao considered the generalized mixed equilibrium problem (GMEP) which aims to find $x^{†}\in C$ verifying

$$\phi (y)-\phi (x^{†})+\Theta (x^{†},y)+\langle \mathcal{A}{x}^{†},y-x^{†}\rangle \ge 0,\forall y\in C,$$

(7)

where $\phi :C\to \mathbf{R}$ is a real-valued function and $\Theta :C\times C\to \mathbf{R}$ is a bifunction. The set of solutions of GMEP (7) is denoted by $\mathrm{GMEP}(\Theta ,\phi ,\mathcal{A})$.

It is clear that GMEP (7) includes optimization problems, VIP and Nash equilibrium problems as special cases.

Special cases:

(i) Letting $\phi =0$, GMEP (7) reduces to find $x^{†}\in C$ verifying

$$(GEP):\Theta (x^{†},y)+\langle \mathcal{A}{x}^{†},y-x^{†}\rangle \ge 0,\forall y\in C,$$

which was studied in [13,16].

(ii) Letting $\mathcal{A}=0$, GMEP (7) reduces to find $x^{†}\in C$ verifying

$$(MEP):\Theta (x^{†},y)+\phi (y)-\phi (x^{†})\ge 0,\forall y\in C,$$

which was considered in [17].

(iii) Letting $\phi =0$ and $\mathcal{A}=0$, GMEP (7) reduces to find $x^{†}\in C$ verifying

$$(EP):\Theta (x^{†},y)\ge 0,\forall y\in C,$$

which was discussed in [18,19].

GMEP (7) has been discussed extensively in the literature; see e.g., [6,7,20,21,22,23,24,25,26,27,28,29,30,31,32].

Now, we consider the minimization problem (CMP):

$$\underset{x\in C}{min}f(x),$$

(8)

where $f:C\to \mathbf{R}$ is a convex and continuously Fréchet differentiable functional. Use $\Xi$ to denote the set of minimizers of CMP (8).

For solving (8), an efficient algorithm is the gradient-projection algorithm (GPA) which is defined by: for given the initial guess $x_0$,

$$z_{k+1}:=P_C(z_k-\mu \nabla f(z_k)),\forall k\ge 0,$$

(9)

or a general form,

$$z_{n+1}:=P_C(z_k-{\mu }_k\nabla f(z_k)),\forall k\ge 0.$$

(10)

It is known that $S:=P_C(I-\lambda \nabla f)$ is a contractive operator if $\nabla f$ is $\alpha$-strongly monotone and L-Lipschitz and $0<\lambda <\frac{2\alpha }{L^2}$. In this case, GPA (9) has strong convergence.

Recall that the variational inclusion problem is to find a point $x\in C$ verifying

$$0\in Bx+Rx.$$

(11)

where $B:C\to H$ is a single-valued mapping and R is a multivalued mapping with $D(R)=C$. Use $\mathrm{I}(B,R)$ to denote the solution set of (11).

Taking $B=0$ in (11), it reduces to the problem introduced by Rockafellar [33]. Let $R:D(R)\subset H\to 2^H$ be a maximal monotone operator. The resolvent operator $J_{R,\lambda }:H\to \overline{D(R)}$ is defined by $J_{R,\lambda }={(I+\lambda R)}^{-1},\forall x\in H.$ In [34], Huang considered problem (11) under the assumptions that B is strongly monotone Lipschitz operator and R is maximal monotone. Zeng, Guu and Yao [35] discussed problem (11) under more general cases. Related work, please refer to [36,37] and the references therein.

In the present paper, we introduce two composite schemes for finding a solution of the CMP (8) with the constraints of finitely many GMEPs and finitely many variational inclusions for maximal monotone and inverse strongly monotone mappings in a real Hilbert space H. Strong convergence of the suggested algorithms are given. Our theorems complement, develop and extend the results obtained in [6,15,38], having as background [39,40,41,42,43,44,45,46].

2. Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that the metric projection $P_C:H\to C$ is defined by $\parallel x-P_{C}x\parallel ={inf}_{y\in C}\parallel x-y\parallel =:d(x,C)$. A mapping $A:H\to H$ is called strongly positive if $\langle Ax,x\rangle \ge \overline{\gamma }{\parallel x\parallel }^2,\forall x\in H$ for $\overline{\gamma }>0$. A mapping $F:C\to H$ is called Lipschitz if $\parallel Fx-Fy\parallel \le L\parallel x-y\parallel ,\forall x,y\in C$ for some $L\ge 0$. F is called nonexpansive if $L=1$ and it is called contractive provided $L\in [0,1)$.

Proposition 1.

Let $x\in H$ and $z\in C$. Then, the following hold

• $z=P_{C}x\iff \langle x-z,z-x^{†}\rangle \ge 0,\forall x^{†}\in C$;
• $z=P_{C}x\iff {\parallel x-z\parallel }^2+\parallel x^{†}{-z\parallel }^2\le {\parallel x-x^{†}\parallel }^2,\forall x^{†}\in C$;
• $\langle P_{C}x-P_C{x}^{†},x-x^{†}\rangle \ge {\parallel P_{C}x-P_C{x}^{†}\parallel }^2,\forall x^{†}\in H$.

Definition 1.

A mapping $F:C\to H$ is called

• monotone if $\langle Fx-F{x}^{†},x-x^{†}\rangle \ge 0,\forall x,x^{†}\in C$;
• η-strongly monotone if $\langle Fx-F{x}^{†},x-x^{†}\rangle \ge \eta {\parallel x-x^{†}\parallel }^2,\forall x,y\in C$ for some $\eta >0$;
• α-inverse-strongly monotone (ism) if $\langle Fx-F{x}^{†},x-x^{†}\rangle \ge \alpha {\parallel Fx-F{x}^{†}\parallel }^2,\forall x,x^{†}\in C$ for some $\alpha >0$. In this case, we have for all $u,u^{†}\in C$,

  $$\begin{array}{c}\hfill \parallel (I-\mu F)u-(I-\mu F)u^{†}{\parallel }^2\le \parallel u-u^{†}{\parallel }^2+\mu (\mu -2\alpha ){\parallel Fu-F{u}^{†}\parallel }^2,\end{array}$$

  (12)

  where $\mu >0$ is a constant.

Definition 2.

A mapping $T:H\to H$ is called firmly nonexpansive if $2T-I$ is nonexpansive. Thus, T is firmly nonexpansive iff $T=\frac{1}{2}(I+S)$, where $S:H\to H$ is nonexpansive.

Let $\Theta :C\times C\to \mathbf{R}$ be a bifunction satisfying the following assumptions:

(A1)

$\Theta (x,x)=0$ for all $x\in C$;

(A2)

$\Theta (x,x^{†})+\Theta (x^{†},x)\le 0$ for all $x,x^{†}\in C$;

(A3)

for $x,x^{†},y\in C$, ${lim\; sup}_{t\to 0^{+}}\Theta (ty+(1-t)x,x^{†})\le \Theta (x,x^{†})$;

(A4)

$\Theta (x,·)$ is convex and lower semicontinuous for each $x\in C$.

Let $\phi :C\to \mathbf{R}$ be a lower semicontinuous and convex function satisfying the following conditions:

(B1)

for each $x\in H$ and $r>0$, there exists a bounded subset $D_x\subset C$ and $y_x\in C$ such that for any $z\in C\backslash D_x$,

$$\Theta (z,y_x)+\phi (y_x)-\phi (z)+\frac{1}{r}\langle y_x-z,z-x\rangle <0;$$

(B2)

C is a bounded set.

Let $r>0$. Define a mapping $T_r^{(\Theta ,\phi )}:H\to C$ by the following form: for any $x\in H$,

$$T_r^{(\Theta ,\phi )}(x):=\{x^{†}\in C:\Theta (x^{†},y)+\phi (y)-\phi (x^{†})+\frac{1}{r}\langle x^{†}-x,y-x^{†}\rangle \ge 0,\forall y\in C\}.$$

If $\phi \equiv 0$, then $T_r^{\Theta }(x):=\{x^{†}\in C:\Theta (x^{†},y)+\frac{1}{r}\langle x^{†}-x,y-x^{†}\rangle \ge 0,\forall y\in C\}$.

Proposition 2

([17]). Let a bifunction $\Theta :C\times C\to \mathbf{R}$ satisfying assumptions (A1)–(A4). Let a convex function $\phi :C\to \mathbf{R}$ be a proper lower semi-continuous. Assume conditions (B1) or (B2) holds. Then, we have

• $\forall x\in H$, $T_r^{(\Theta ,\phi )}(x)\ne \varnothing$ is single-valued;
• $T_r^{(\Theta ,\phi )}$ is firmly nonexpansive;
• $\mathrm{Fix}(T_r^{(\Theta ,\phi )})=\mathrm{MEP}(\Theta ,\phi )$;
• $\mathrm{MEP}(\Theta ,\phi )$ is convex and closed;
• $\forall s,t>0$, $\parallel T_s^{(\Theta ,\phi )}{z}^{†}-T_t^{(\Theta ,\phi )}{z}^{†}{\parallel }^2\le \frac{s-t}{s}\langle T_s^{(\Theta ,\phi )}{z}^{†}-T_t^{(\Theta ,\phi )}{z}^{†},T_s^{(\Theta ,\phi )}{z}^{†}-z^{†}\rangle$ for $z^{†}\in H$.

Proposition 3

([41]). We have the following conclusions:

• A mapping T is nonexpansive iff the complement $I-T$ is $\frac{1}{2}$-ism.
• If a mapping T is α-ism, then $\gamma T$ is $\frac{\alpha }{\gamma }$-ism where $\gamma >0$.
• A mapping T is averaged iff the complement $I-T$ is α-ism for some $\alpha >1/2$.

Lemma 1.

Let E be a real inner product space. Then,

$$\parallel x+x^{†}{\parallel }^2\le {\parallel x\parallel }^2+2\langle x^{†},x+x^{†}\rangle ,\forall x,x^{†}\in E.$$

Lemma 2.

In a real Hilbert space H, we have the following results

• $\parallel x-x^{†}{\parallel }^2={\parallel x\parallel }^2-2\langle x-x^{†},x^{†}\rangle -{\parallel x^{†}\parallel }^2$, $\forall x,x^{†}\in H$;
• $\parallel \lambda x+\gamma x^{†}{\parallel }^2={\lambda \parallel x\parallel }^2+\gamma \parallel x^{†}{\parallel }^2-\lambda \gamma {\parallel x-x^{†}\parallel }^2$, $\forall x,x^{†}\in H$ and $\lambda ,\gamma \in [0,1]$ with $\lambda +\gamma =1$;
• $\{u_n\}\subset H$ is a sequence satisfying $u_n\rightharpoonup x$. Then,

  $$\underset{n\to \infty }{lim\; sup}\parallel u_n-x^{†}{\parallel }^2=\underset{n\to \infty }{lim\; sup}\parallel u_n{-x\parallel }^2+{\parallel x-x^{†}\parallel }^2,\forall x^{†}\in H.$$

Lemma 3

([45]). Let H be a real Hilbert space and $\varnothing \ne C\subset H$ be a closed convex set. Let $T:C\to C$ be a k-strictly pseudocontractive mapping. Then,

• $\parallel Tx-Ty\parallel \le \frac{1+k}{1-k}\parallel x-y\parallel$,$\forall x,y\in C$.
• $I-T$ is demiclosed at 0.
• $\mathrm{Fix}(T)$ of T is closed and convex.

Lemma 4

([39]). Let H be a real Hilbert space and $\varnothing \ne C\subset H$ be a closed convex set. Let $T:C\to C$ be a k-strictly pseudocontractive mapping. Then,

$$\parallel \gamma (x-x^{†})+\mu (Tx-T{x}^{†})\parallel \le (\gamma +\mu )\parallel x-x^{†}\parallel ,\forall x,x^{†}\in C,$$

where $\gamma \ge 0$ and $\mu \ge 0$ with $(\gamma +\mu )k\le \gamma$.

Let $T:C\to C$ be a nonexpansive operator and the operator $F:C\to H$ be $\kappa$-Lipschitzian and $\eta$-strongly monotone. Define an operator $T^{\lambda }:C\to H$ by $T^{\lambda }x:=Tx-\lambda \mu F(Tx),\forall x\in C$, where $\lambda \in (0,1]$ and $\mu >0$ are two constants.

Lemma 5

([42]). Let $0<\mu <\frac{2\eta }{{\kappa }^2}$. Then, for $\forall x,y\in C$, we have $\parallel T^{\lambda }x-T^{\lambda }y\parallel \le (1-\lambda \tau )\parallel x-y\parallel$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$.

Lemma 6

([46]). Let $\{a_n\}\subset [0,+\infty )$, $\{{\omega }_n\}\subset [0,1]$, $\{{\delta }_n\}$ and $\{r_n\}\subset [0,+\infty )$ be four sequences. If

(i) 

${\sum }_{n=0}^{\infty }{\omega }_n=\infty$ and ${\sum }_{n=1}^{\infty }{r}_n<\infty$;

(ii) 

either ${lim\; sup}_{n\to \infty }{\delta }_n\le 0$ or ${\sum }_{n=0}^{\infty }{\omega }_n|{\delta }_n|<\infty$;

(iii) 

$a_{n+1}\le (1-{\omega }_n)a_n+{\omega }_n{\delta }_n+r_n$, $\forall n\ge 0$.

Then, ${lim}_{n\to \infty }{a}_n=0$.

Lemma 7

([44]). Let bounded linear operator $A:H\to H$ be $\overline{\gamma }$-strongly positive. Then, $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$ provided $0<\rho \le {\parallel A\parallel }^{-1}$.

Let $\mathrm{LIM}$ be a Banach limit. Then, we have following properties:

• $a_n\le c_n,n\ge 1$⇒${\mathrm{LIM}}_n{a}_n\le {\mathrm{LIM}}_n{c}_n$;
• ${\mathrm{LIM}}_n{a}_{n+N}={\mathrm{LIM}}_n{a}_n$, where N is a fixed positive integer;
• ${lim\; inf}_{n\to \infty }{a}_n\le {\mathrm{LIM}}_n{a}_n\le {lim\; sup}_{n\to \infty }{a}_n$, $\forall \{a_n\}\in l^{\infty }$.

Lemma 8

([40]). Assume that the sequence $\{s_n\}\in l^{\infty }$ satisfy ${\mathrm{LIM}}_n{s}_n\le M$, where M is a constant. If ${lim\; sup}_{n\to \infty }(s_{n+1}-s_n)\le 0$, then ${lim\; sup}_{n\to \infty }{s}_n\le M$.

Let the operator $R:D(R)\subset H\to 2^H$ be maximal monotone. Let $\lambda ,\mu >0$ be two constants.

Lemma 9

([43]). There holds the resolvent identity

$$J_{R,\lambda }x=J_{R,\mu }(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{R,\lambda }x),\forall x\in H.$$

Remark 1.

The resolvent has the following property

$$\parallel J_{R,\lambda }x-J_{R,\mu }{x}^{†}\parallel \le \parallel x-x^{†}\parallel +|\lambda -\mu |(\frac{1}{\lambda }\parallel J_{R,\lambda }x-x^{†}\parallel +\frac{1}{\mu }\parallel x-J_{R,\mu }{x}^{†}\parallel ),\forall x,x^{†}\in H.$$

(13)

Lemma 10

([34,35]). $J_{R,\lambda }$ satisfies

$$\langle J_{R,\lambda }x-J_{R,\lambda }y,x-y\rangle \ge {\parallel J_{R,\lambda }x-J_{R,\lambda }y\parallel }^2,\forall x,y\in H.$$

3. Main Results

Let H be a real Hilbert space and $\varnothing \ne C\subset H$ be a closed convex set. In what follows, we assume:

$f:C\to \mathbf{R}$ is a convex functional with gradient $\nabla f$ being L-Lipschitz, the operator $F:C\to H$ is $\eta$-strongly monotone and $\kappa$-Lipschitzian, $R_i:C\to 2^H$ is a maximal monotone mapping and $B_i:C\to H(i=1,\dots ,N)$ is ${\eta }_i$-ism, ${\Theta }_j:C\times C\to \mathbf{R}$ satisfies (A1)–(A4), ${\phi }_j:C\to \mathbf{R}\cup \{+\infty \}$ is a proper convex and lower semicontinuous satisfying (B1) or (B2), and $A_j:C\to H$ is ${\mu }_j$-ism for each $j=1,\dots ,M$;

$V:C\to H$ is a nonexpansive mapping and $A:H\to H$ is a $\overline{\gamma }$-strongly positive bounded linear operator such $1<\overline{\gamma }<2$, $0<\mu <\frac{2\eta }{{\kappa }^2}$ and $0\le \gamma <\tau$ with $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$;

$P_C(I-{\lambda }_t\nabla f)=s_{t}I+(1-s_t)T_t$ where $T_t$ is nonexpansive, $s_t=\frac{2-{\lambda }_{t}L}{4}\in (0,\frac{1}{2})$ and ${\lambda }_t:(0,1)\to (0,\frac{2}{L})$ with ${lim}_{t\to 0}{\lambda }_t=\frac{2}{L}$;

The operator ${\Lambda }_t^N:C\to C$ is defined by ${\Lambda }_t^{N}x=J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t}{B}_N)\cdots J_{R_1,{\lambda }_{1,t}}(I-{\lambda }_{1,t}{B}_1)x,t\in (0,1)$, for $\{{\lambda }_{i,t}\}\subset [a_i,b_i]\subset (0,2{\eta }_i),i=1,\dots ,N$;

The operator ${\Lambda }_n^N:C\to C$ is defined by ${\Lambda }_n^{N}x=J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N)\cdots J_{R_1,{\lambda }_{1,n}}(I-{\lambda }_{1,n}{B}_1)x$ with $\{{\lambda }_{i,n}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$ and ${lim}_{n\to \infty }{\lambda }_{i,n}={\lambda }_i$, for each $i=1,\dots ,N$;

The operator ${\Delta }_t^M:C\to C$ is defined by ${\Delta }_t^{M}x=T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{A}_M)\cdots T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{A}_1)x,t\in (0,1)$, for $\{r_{j,t}\}\subset [c_j,d_j]\subset (0,2{\mu }_j),j=1,\dots ,M$;

The operator ${\Delta }_n^M:C\to C$ is defined by ${\Delta }_n^{M}x=T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M)\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{A}_1)x$ with $\{r_{j,n}\}\subset [c_j,d_j]\subset (0,2{\mu }_j)$ and ${lim}_{n\to \infty }{r}_{j,n}=r_j$, for each $j=1,\dots ,M$;

$\Omega :={\cap }_{j=1}^M\mathrm{GMEP}({\Theta }_j,{\phi }_j,A_j)\cap {\cap }_{i=1}^N\mathrm{I}(B_i,R_i)\cap \Xi \ne \varnothing$;

$\{{\alpha }_n\}\subset [0,1],\{s_n\}\subset (0,min\{\frac{1}{2}{,\parallel A\parallel }^{-1}\})$ and ${\{s_t\}}_{t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma }\})}\subset (0,min\{\frac{1}{2}{,\parallel A\parallel }^{-1}\})$.

Next, set

$${\Lambda }_t^i=J_{R_i,{\lambda }_{i,t}}(I-{\lambda }_{i,t}{B}_i)J_{R_{i-1},{\lambda }_{i-1,t}}(I-{\lambda }_{i-1,t}{B}_{i-1})\cdots J_{R_1,{\lambda }_{1,t}}(I-{\lambda }_{1,t}{B}_1),\forall t\in (0,1),$$

$${\Lambda }_n^i=J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i)J_{R_{i-1},{\lambda }_{i-1,n}}(I-{\lambda }_{i-1,n}{B}_{i-1})\cdots J_{R_1,{\lambda }_{1,n}}(I-{\lambda }_{1,n}{B}_1),\forall n\ge 0,$$

$${\Delta }_t^j=T_{r_{j,t}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,t}{A}_j)T_{r_{j-1,t}}^{({\Theta }_{j-1},{\phi }_{j-1})}(I-r_{j-1,t}{A}_{j-1})\cdots T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{A}_1),\forall t\in (0,1),$$

and

$${\Delta }_n^j=T_{r_{j,n}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,n}{A}_j)T_{r_{j-1,n}}^{({\Theta }_{j-1},{\phi }_{j-1})}(I-r_{j-1,n}{A}_{j-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{A}_1),\forall n\ge 0,$$

for $j=1,\dots ,M$ and $i=1,\dots ,N$, ${\Lambda }_t^0={\Lambda }_n^0=I$ and ${\Delta }_t^0={\Delta }_n^0=I$.

By Proposition 3, $\lambda \nabla f$ is $\frac{1}{\lambda L}$-ism for $\lambda >0$. In addition, hence, $I-\lambda \nabla f$ is $\frac{\lambda L}{2}$-averaged. It is clear that $P_C(I-{\lambda }_t\nabla f)$ is $\frac{2+{\lambda }_{t}L}{4}$-averaged for each ${\lambda }_t\in (0,\frac{2}{L})$. Thus, $P_C(I-{\lambda }_t\nabla f)=s_{t}I+(1-s_t)T_t$, where $T_t$ is nonexpansive and $s_t:=s_t({\lambda }_t)=\frac{2-{\lambda }_{t}L}{4}\in (0,\frac{1}{2})$ for each ${\lambda }_t\in (0,\frac{2}{L})$. Similarly, for each $n\ge 0$, $P_C(I-{\lambda }_n\nabla f)$ is $\frac{2+{\lambda }_{n}L}{4}$-averaged for each ${\lambda }_n\in (0,\frac{2}{L})$ and $P_C(I-{\lambda }_n\nabla f)=s_{n}I+(1-s_n)T_n$. Please note that $\mathrm{Fix}(T_t)=\mathrm{Fix}(T_n)=\Xi$. Since $\{{\lambda }_{i,t}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$, by (12) and Lemma 10, we deduce

$$\begin{array}{c}\hfill \parallel {\Lambda }_t^{N}x-{\Lambda }_t^{N}y\parallel =\parallel J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t}{B}_N){\Lambda }_t^{N-1}x-J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t}{B}_N){\Lambda }_t^{N-1}y\parallel \le \parallel x-y\parallel .\end{array}$$

On the other hand, since $\{r_{i,t}\}\subset [c_i,d_i]\subset (0,2{\mu }_i)$, according to (12) and Proposition 2, we get

$$\begin{array}{cc}\hfill \parallel {\Delta }_t^{M}x-{\Delta }_t^{M}y\parallel & =\parallel T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{A}_M){\Delta }_t^{M-1}x-T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{A}_M){\Delta }_t^{M-1}y\parallel \hfill \\ & \le \parallel (I-r_{M,t}{A}_M){\Delta }_t^{M-1}x-(I-r_{M,t}{A}_M){\Delta }_t^{M-1}y\parallel \hfill \\ & \le \parallel x-y\parallel .\hfill \end{array}$$

Next, we present the following net ${\{x_t\}}_{t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma }\})}$:

$$\left\{\begin{array}{c}{u}_t=T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{A}_M)T_{r_{M-1,t}}^{({\Theta }_{M-1},{\phi }_{M-1})}(I-r_{M-1,t}{A}_{M-1})\cdots T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{A}_1)x_t,\hfill \\ v_t=J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t}{B}_N)J_{R_{N-1},{\lambda }_{N-1,t}}(I-{\lambda }_{N-1,t}{B}_{N-1})\cdots J_{R_1,{\lambda }_{1,t}}(I-{\lambda }_{1,t}{B}_1)u_t,\hfill \\ x_t=P_C[(I-s_{t}A)T_t{v}_t+s_t[V{x}_t-t(\mu FV{x}_t-\gamma T_t{v}_t)].\hfill \end{array}\right.$$

(14)

We prove the strong convergence of $\{x_t\}$ as $t\to 0$ to a point $\tilde{x}\in \Omega$ which solves

$$\langle (A-V)\tilde{x},p-\tilde{x}\rangle \ge 0,\forall p\in \Omega .$$

(15)

Let $x_0\in C$, define a sequence $\{x_n\}$ as follows

$$\left\{\begin{array}{c}{u}_n=T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M)T_{r_{M-1,n}}^{({\Theta }_{M-1},{\phi }_{M-1})}(I-r_{M-1,n}{A}_{M-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{A}_1)x_n,\hfill \\ v_n=J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N)J_{R_{N-1},{\lambda }_{N-1,n}}(I-{\lambda }_{N-1,n}{B}_{N-1})\cdots J_{R_1,{\lambda }_{1,n}}(I-{\lambda }_{1,n}{B}_1)u_n,\hfill \\ y_n={\alpha }_n\gamma T_n{v}_n+(I-{\alpha }_n\mu F)V{x}_n,\hfill \\ x_{n+1}=P_C[(I-s_{n}A)T_n{v}_n+s_n{y}_n],\forall n\ge 0.\hfill \end{array}\right.$$

(16)

We will show the convergence of $\{x_n\}$ as $n\to \infty$ to $\tilde{x}\in \Omega$, which solves (15).

Now, for $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma }\})$, and $s_t\in (0,min\{\frac{1}{2}{,\parallel A\parallel }^{-1}\})$, let $Q_t:C\to C$ be defined by $Q_{t}x=P_C[(I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^{M}x+s_t[Vx-t(\mu FVx-\gamma T_t{\Lambda }_t^N{\Delta }_t^{M}x)],\forall x\in C$. By Lemmas 5 and 7, we have

$$\begin{array}{cc}\hfill \parallel Q_{t}x-Q_{t}y\parallel & \le \parallel (I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^{M}x-(I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^{M}y\parallel \hfill \\ & +s_t\parallel ((I-t\mu F)Vx+t\gamma T_t{\Lambda }_t^N{\Delta }_t^{M}x)-((I-t\mu F)Vy+t\gamma T_t{\Lambda }_t^N{\Delta }_t^{M}y)\parallel \hfill \\ & \le (1-s_t\overline{\gamma })\parallel x-y\parallel +s_t[(1-t\tau )\parallel x-y\parallel +t\gamma \parallel x-y\parallel ]\hfill \\ & =[1-s_t(\overline{\gamma }-1+t(\tau -\gamma ))]\parallel x-y\parallel .\hfill \end{array}$$

It is easy to check that $0<1-s_t(\overline{\gamma }-1+t(\tau -\gamma ))<1$. Therefore, $Q_t:C\to C$ is contraction and $Q_t$ has a unique fixed point, denoted by $x_t$.

Proposition 4.

Let $\{x_t\}$ be defined via (14). Then

(i) 

$\{x_t\}$ is bounded for $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma }\})$;

(ii) 

${lim}_{t\to 0}\parallel x_t-T_t{x}_t\parallel =0,{lim}_{t\to 0}\parallel x_t-{\Lambda }_t^N{x}_t\parallel =0$ and ${lim}_{t\to 0}\parallel x_t-{\Delta }_t^M{x}_t\parallel =0$ provided ${lim}_{t\to 0}{\lambda }_t=\frac{2}{L}$;

(iii) 

$x_t,s_t,{\lambda }_{i,t},r_{j,t}$ are locally Lipschitzian.

Proof. 

(i) Pick up $p\in \Omega$. Noting that $\mathrm{Fix}(T_t)=\Xi ,{\Lambda }_t^{i}p=p$ and ${\Delta }_t^{j}p=p$ for all $i\in \{1,\dots ,N\}$ and $j\in \{1,\dots ,M\}$, by the nonexpansivity of $T_t,{\Lambda }_t^i$ and ${\Delta }_t^j$ and Lemmas 5 and 7 we get

$$\begin{array}{cc}\hfill \parallel x_t-p\parallel & \le \parallel (I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^M{x}_t-(I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^{M}p\parallel +s_t\parallel (I-t\mu F)V{x}_t+t\gamma T_t{\Lambda }_t^N{\Delta }_t^M{x}_t-Ap\parallel \hfill \\ & \le s_t\parallel (I-t\mu F)V{x}_t-(I-t\mu F)Vp+t(\gamma T_t{\Lambda }_t^N{\Delta }_t^M{x}_t-\mu FVp)+Vp-Ap\parallel \hfill \\ & +(1-s_t\overline{\gamma })\parallel T_t{\Lambda }_t^N{\Delta }_t^M{x}_t-T_t{\Lambda }_t^N{\Delta }_t^{M}p\parallel \hfill \\ & \le [1-s_t(\overline{\gamma }-1+t(\tau -\gamma ))]\parallel x_t-p\parallel +s_t(t\parallel (\gamma I-\mu FV)p\parallel +\parallel (V-A)p\parallel ).\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill \parallel x_t-p\parallel \le \frac{\parallel (V-A)p\parallel +\parallel (\gamma I-\mu FV)p\parallel }{\overline{\gamma }-1}.\end{array}$$

Hence $\{x_t\}$ is bounded and so are $\{V{x}_t\},\{u_t\},\{v_t\},\{T_t{v}_t\}$ and $\{FV{x}_t\}$.

(ii) From (14), we obtain

$$\begin{array}{cc}\hfill \parallel x_t-T_t{v}_t\parallel & =\parallel P_C[(I-s_{t}A)T_t{v}_t+s_t((I-t\mu F)V{x}_t+t\gamma T_t{v}_t)]-P_C{T}_t{v}_t\parallel \hfill \\ & \le s_t\parallel V{x}_t-A{T}_t{v}_t\parallel +t\parallel \gamma T_t{v}_t-\mu FV{x}_t\parallel \to 0\mathrm{as}t\to 0.\hfill \end{array}$$

(17)

From (12), we have

$$\begin{array}{cc}\hfill \parallel v_t{-p\parallel }^2& \le \parallel {\Lambda }_t^i{u}_t{-p\parallel }^2\hfill \\ & =\parallel J_{R_i,{\lambda }_{i,t}}(I-{\lambda }_{i,t}{B}_i){\Lambda }_t^{i-1}{u}_t-J_{R_i,{\lambda }_{i,t}}(I-{\lambda }_{i,t}{B}_i){p\parallel }^2\hfill \\ & \le \parallel x_t{-p\parallel }^2+{\lambda }_{i,t}({\lambda }_{i,t}-2{\eta }_i){\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel }^2,\hfill \end{array}$$

(18)

and

$$\begin{array}{cc}\hfill \parallel u_t{-p\parallel }^2& \le \parallel {\Delta }_t^j{x}_t{-p\parallel }^2\hfill \\ & =\parallel T_{r_{j,t}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,t}{A}_j){\Delta }_t^{j-1}{x}_t-T_{r_{j,t}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,t}{A}_j){p\parallel }^2\hfill \\ & \le \parallel x_t{-p\parallel }^2+r_{j,t}(r_{j,t}-2{\mu }_j){\parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel }^2.\hfill \end{array}$$

(19)

Simple calculations show that

$$\begin{array}{cc}\hfill x_t-p& =x_t-w_t+(I-s_{t}A)(T_t{v}_t-T_{t}p)+s_t[(I-t\mu F)V{x}_t-(I-t\mu F)Vp\hfill \\ & +t\gamma (T_t{v}_t-T_{t}p)+t(\gamma I-\mu FV)p]+s_t(V-A)p,\hfill \end{array}$$

(20)

where $w_t=(I-s_{t}A)T_t{v}_t+s_t(t\gamma T_t{v}_t+(I-t\mu F)V{x}_t)$. Then, by the nonexpansivity of $T_t$, Proposition 1, and Lemmas 5 and 7, from (18)–(20) we obtain that

$$\begin{array}{cc}\hfill \parallel x_t{-p\parallel }^2& \le \langle (I-s_{t}A)(T_t{v}_t-T_{t}p),x_t-p\rangle +s_t[\langle (I-t\mu F)V{x}_t-(I-t\mu F)Vp,x_t-p\rangle \hfill \\ & +t\gamma \langle (T_t{v}_t-T_{t}p),x_t-p\rangle +t\langle (\gamma I-\mu FV)p,x_t-p\rangle ]+s_t\langle (V-A)p,x_t-p\rangle \hfill \\ & \le (1-s_t(\overline{\gamma }-t\gamma ))\frac{1}{2}[\parallel x_t{-p\parallel }^2+r_{j,t}(r_{j,t}-2{\mu }_j){\parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel }^2\hfill \\ & +{\lambda }_{i,t}({\lambda }_{i,t}-2{\eta }_i)\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_i{p\parallel }^2+\parallel x_t{-p\parallel }^2]+s_t[(1-t\tau ){\parallel x_t-p\parallel }^2\hfill \\ & +t\parallel (\gamma I-\mu FV)p\parallel \parallel x_t-p\parallel ]+s_t\parallel (V-A)p\parallel \parallel x_t-p\parallel \hfill \\ & \le \parallel x_t{-p\parallel }^2-\frac{1-s_t(\overline{\gamma }-t\gamma )}{2}[r_{j,t}(2{\mu }_j-r_{j,t}){\parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel }^2+{\lambda }_{i,t}(2{\eta }_i-{\lambda }_{i,t})\hfill \\ & \times \parallel B_i{\Lambda }_t^{i-1}{u}_t-B_i{p\parallel }^2]+s_t(t\parallel (\gamma I-\mu FV)p\parallel +\parallel (V-A)p\parallel )\parallel x_t-p\parallel ,\hfill \end{array}$$

(21)

which together with $\{{\lambda }_{i,t}\}\subset [a_i,b_i]\subset (0,2{\eta }_i),i=1,\dots ,N$ and $\{r_{j,t}\}\subset [c_j,d_j]\subset (0,2{\mu }_j),j=1,\dots ,M$, implies that

$$\begin{array}{c}\frac{1-s_t(\overline{\gamma }-t\gamma )}{2}[c_j(2{\mu }_j-d_j)\parallel A_j{\Delta }_t^{j-1}{x}_t-A_j{p\parallel }^2+a_i(2{\eta }_i-b_i)\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p{\parallel }^2]\hfill \\ \le \frac{1-s_t(\overline{\gamma }-t\gamma )}{2}[r_{j,t}(2{\mu }_j-r_{j,t})\parallel A_j{\Delta }_t^{j-1}{x}_t-A_j{p\parallel }^2+{\lambda }_{i,t}(2{\eta }_i-{\lambda }_{i,t})\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p{\parallel }^2]\hfill \\ \le s_t(t\parallel (\gamma I-\mu FV)p\parallel +\parallel (V-A)p\parallel )\parallel x_t-p\parallel .\hfill \end{array}$$

Since ${lim}_{t\to 0}{s}_t=0$ and $\{x_t\}$ is bounded, we have

$$\underset{t\to 0}{lim}\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel =0\mathrm{and}\underset{t\to 0}{lim}\parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel =0(i\in \{1,\dots ,N\},j\in \{1,\dots ,M\}).$$

(22)

According to Proposition 2, we have

$$\begin{array}{cc}\hfill \parallel {\Delta }_t^j{x}_t{-p\parallel }^2& \le \langle (I-r_{j,t}{A}_j){\Delta }_t^{j-1}{x}_t-(I-r_{j,t}{A}_j)p,{\Delta }_t^j{x}_t-p\rangle \hfill \\ & \le \frac{1}{2}(\parallel {\Delta }_t^{j-1}{x}_t{-p\parallel }^2+\parallel {\Delta }_t^j{x}_t{-p\parallel }^2-\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t-r_{j,t}(A_j{\Delta }_t^{j-1}{x}_t-A_{j}p){\parallel }^2),\hfill \end{array}$$

which implies that

$$\begin{array}{cc}\hfill \parallel {\Delta }_t^j{x}_t{-p\parallel }^2& \le \parallel {\Delta }_t^{j-1}{x}_t{-p\parallel }^2-{\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t-r_{j,t}(A_j{\Delta }_t^{j-1}{x}_t-A_{j}p)\parallel }^2\hfill \\ & \le \parallel x_t{-p\parallel }^2-\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t{\parallel }^2+2r_{j,t}\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel \parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel .\hfill \end{array}$$

(23)

Also, by Lemma 10, we obtain that for each $i\in \{1,\dots ,N\}$

$$\begin{array}{cc}\hfill \parallel {\Lambda }_t^i{u}_t{-p\parallel }^2& \le \langle (I-{\lambda }_{i,t}{B}_i){\Lambda }_t^{i-1}{u}_t-(I-{\lambda }_{i,t}{B}_i)p,{\Lambda }_t^i{u}_t-p\rangle \hfill \\ & =\frac{1}{2}(\parallel (I-{\lambda }_{i,t}{B}_i){\Lambda }_t^{i-1}{u}_t-(I-{\lambda }_{i,t}{B}_i){p\parallel }^2+{\parallel {\Lambda }_t^i{u}_t-p\parallel }^2\hfill \\ & -\parallel (I-{\lambda }_{i,t}{B}_i){\Lambda }_t^{i-1}{u}_t-(I-{\lambda }_{i,t}{B}_i)p-({\Lambda }_t^i{u}_t-p){\parallel }^2)\hfill \\ & \le \frac{1}{2}(\parallel u_t{-p\parallel }^2+\parallel {\Lambda }_t^i{u}_t{-p\parallel }^2-\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t-{\lambda }_{i,t}(B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p){\parallel }^2),\hfill \end{array}$$

which implies

$$\begin{array}{cc}\hfill \parallel {\Lambda }_t^i{u}_t{-p\parallel }^2& \le \parallel u_t{-p\parallel }^2-{\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t-{\lambda }_{i,t}(B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p)\parallel }^2\hfill \\ & \le \parallel u_t{-p\parallel }^2-\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t{\parallel }^2+2{\lambda }_{i,t}\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t\parallel \parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel .\hfill \end{array}$$

(24)

Thus, utilizing Lemma 1, from (21), (23) and (24) we have

$$\begin{array}{cc}\hfill \parallel x_t{-p\parallel }^2& \le (1-s_t(\overline{\gamma }-t\gamma ))\frac{1}{2}(\parallel {\Lambda }_t^i{u}_t{-p\parallel }^2+\parallel x_t{-p\parallel }^2)+s_t\parallel (V-A)p\parallel \parallel x_t-p\parallel \hfill \\ & +s_t[(1-t\tau )\parallel x_t{-p\parallel }^2+t\parallel (\gamma I-\mu FV)p\parallel \parallel x_t-p\parallel ]\hfill \\ & \le (1-s_t(\overline{\gamma }-t\gamma ))\frac{1}{2}[\parallel x_t{-p\parallel }^2-{\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel }^2\hfill \\ & +2r_{j,t}\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel \parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel -\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t{\parallel }^2\hfill \\ & +2{\lambda }_{i,t}\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t\parallel \parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel +\parallel x_t-p{\parallel }^2]\hfill \\ & +s_t[(1-t\tau )\parallel x_t{-p\parallel }^2+t\parallel (\gamma I-\mu FV)p\parallel \parallel x_t-p\parallel ]+s_t\parallel (V-A)p\parallel \parallel x_t-p\parallel \hfill \\ & \le \parallel x_t{-p\parallel }^2-\frac{1-s_t(\overline{\gamma }-t\gamma )}{2}(\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t{\parallel }^2+\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t{\parallel }^2)\hfill \\ & +r_{j,t}\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel \parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel +{\lambda }_{i,t}\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t\parallel \parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel \hfill \\ & +s_t(t\parallel (\gamma I-\mu FV)p\parallel +\parallel (V-A)p\parallel )\parallel x_t-p\parallel ,\hfill \end{array}$$

which together with $\{{\lambda }_{i,t}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$ and $\{r_{j,t}\}\subset [c_j,d_j]\subset (0,2{\mu }_j)$, leads to

$$\begin{array}{c}\frac{1-s_t(\overline{\gamma }-t\gamma )}{2}(\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t{\parallel }^2+\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t{\parallel }^2)\hfill \\ \le d_j\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel \parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel +b_i\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t\parallel \parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel \hfill \\ +s_t(t\parallel (\gamma I-\mu FV)p\parallel +\parallel (V-A)p\parallel )\parallel x_t-p\parallel .\hfill \end{array}$$

Since ${lim}_{t\to 0}{s}_t=0,{lim}_{t\to 0}\parallel A_j{\Delta }_t^{j-1}{x}_t-A_{j}p\parallel =0$ and ${lim}_{t\to 0}\parallel B_i{\Lambda }_t^{i-1}{u}_t-B_{i}p\parallel =0$ (due to (22)), we deduce from the boundedness of $\{x_t\},\{{\Lambda }_t^i{u}_t\}$ and $\{{\Delta }_t^j{x}_t\}$ that

$$\underset{t\to 0}{lim}\parallel {\Lambda }_t^{i-1}{u}_t-{\Lambda }_t^i{u}_t\parallel =0\mathrm{and}\underset{t\to 0}{lim}\parallel {\Delta }_t^{j-1}{x}_t-{\Delta }_t^j{x}_t\parallel =0.$$

(25)

Hence we get

$$\begin{array}{cc}\hfill \parallel x_t-u_t\parallel & \le \parallel {\Delta }_t^0{x}_t-{\Delta }_t^1{x}_t\parallel +\parallel {\Delta }_t^1{x}_t-{\Delta }_t^2{x}_t\parallel +\cdots +\parallel {\Delta }_t^{M-1}{x}_t-{\Delta }_t^M{x}_t\parallel \hfill \\ & \to 0\mathrm{as}t\to 0,\hfill \end{array}$$

(26)

and

$$\begin{array}{cc}\hfill \parallel u_t-v_t\parallel & \le \parallel {\Lambda }_t^0{u}_t-{\Lambda }_t^1{u}_t\parallel +\parallel {\Lambda }_t^1{u}_t-{\Lambda }_t^2{u}_t\parallel +\cdots +\parallel {\Lambda }_t^{N-1}{u}_t-{\Lambda }_t^N{u}_t\parallel \hfill \\ & \to 0\mathrm{as}t\to 0.\hfill \end{array}$$

(27)

So, taking into account that $\parallel x_t-v_t\parallel \le \parallel x_t-u_t\parallel +\parallel u_t-v_t\parallel$, we have

$$\underset{t\to 0}{lim}\parallel x_t-v_t\parallel =0.$$

(28)

In the meantime, from the nonexpansivity of $T_t$ and ${\Lambda }_t^N$, it is easy to see that

$$\begin{array}{cc}\hfill \parallel x_t-T_t{x}_t\parallel & \le \parallel x_t-T_t{v}_t\parallel +\parallel T_t{v}_t-T_t{x}_t\parallel \hfill \\ & \le \parallel x_t-T_t{v}_t\parallel +\parallel v_t-x_t\parallel ,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \parallel x_t-{\Lambda }_t^N{x}_t\parallel & \le \parallel x_t-{\Lambda }_t^N{u}_t\parallel +\parallel {\Lambda }_t^N{u}_t-{\Lambda }_t^N{x}_t\parallel \hfill \\ & \le \parallel x_t-v_t\parallel +\parallel u_t-x_t\parallel .\hfill \end{array}$$

Consequently, from (17), (26) and (28) we immediately deduce that

$$\underset{t\to 0}{lim}\parallel x_t-T_t{x}_t\parallel =0\mathrm{and}\underset{t\to 0}{lim}\parallel x_t-{\Lambda }_t^N{x}_t\parallel =0.$$

(29)

(iii) Since $\nabla f$ is $\frac{1}{L}$-ism, $P_C(I-{\lambda }_t\nabla f)$ is nonexpansive for ${\lambda }_t\in (0,\frac{2}{L})$. Then,

$$\begin{array}{cc}\hfill \parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}\parallel & \le \parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}-P_C(I-{\lambda }_t\nabla f)p\parallel +\parallel p\parallel \hfill \\ & \le \parallel v_{t_0}\parallel +2\parallel p\parallel .\hfill \end{array}$$

This implies that $\{P_C(I-{\lambda }_t\nabla f)v_{t_0}\}$ is bounded. Also, observe that

$$\begin{array}{cc}\hfill \parallel T_t{v}_{t_0}-T_{t_0}{v}_{t_0}\parallel & \le \frac{4L|{\lambda }_{t_0}-{\lambda }_t|\parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}\parallel }{(2+{\lambda }_{t}L)(2+{\lambda }_{t_0}L)}+\frac{4L|{\lambda }_t-{\lambda }_{t_0}|}{(2+{\lambda }_{t}L)(2+{\lambda }_{t_0}L)}\parallel v_{t_0}\parallel \hfill \\ & +\frac{4(2+{\lambda }_{t}L)\parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}-P_C(I-{\lambda }_{t_0}\nabla f)v_{t_0}\parallel }{(2+{\lambda }_{t}L)(2+{\lambda }_{t_0}L)}\hfill \\ & \le |{\lambda }_t-{\lambda }_{t_0}|[L\parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}\parallel +4\parallel \nabla f(v_{t_0})\parallel +L\parallel v_{t_0}\parallel ]\hfill \\ & \le \tilde{M}|{\lambda }_t-{\lambda }_{t_0}|,\hfill \end{array}$$

(30)

where ${sup}_t\{L\parallel P_C(I-{\lambda }_t\nabla f)v_{t_0}\parallel +4\parallel \nabla f(v_{t_0})\parallel +L\parallel v_{t_0}\parallel \}\le \tilde{M}$ for some $\tilde{M}>0$. So, by (30), we have that

$$\begin{array}{cc}\hfill \parallel T_t{v}_t-T_{t_0}{v}_{t_0}\parallel & \le \parallel T_t{v}_t-T_t{v}_{t_0}\parallel +\parallel T_t{v}_{t_0}-T_{t_0}{v}_{t_0}\parallel \hfill \\ & \le \parallel v_t-v_{t_0}\parallel +\tilde{M}|{\lambda }_t-{\lambda }_{t_0}|\hfill \\ & \le \parallel v_t-v_{t_0}\parallel +\frac{4\tilde{M}}{L}|s_t-s_{t_0}|.\hfill \end{array}$$

(31)

Utilizing (12) and (13), we obtain that

$$\begin{array}{cc}\hfill \parallel v_t-v_{t_0}\parallel & \le \parallel J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t}{B}_N){\Lambda }_t^{N-1}{u}_t-J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t\parallel \hfill \\ & +\parallel J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t-J_{R_N,{\lambda }_{N,t_0}}(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_{t_0}^{N-1}{u}_{t_0}\parallel \hfill \\ & \le \parallel (I-{\lambda }_{N,t}{B}_N){\Lambda }_t^{N-1}{u}_t-(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t\parallel \hfill \\ & +\parallel (I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t-(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_{t_0}^{N-1}{u}_{t_0}\parallel +|{\lambda }_{N,t}-{\lambda }_{N,t_0}|\hfill \\ & \times (\frac{1}{{\lambda }_{N,t}}\parallel J_{R_N,{\lambda }_{N,t}}(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t-(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_{t_0}^{N-1}{u}_{t_0}\parallel \hfill \\ & +\frac{1}{{\lambda }_{N,t_0}}\parallel (I-{\lambda }_{N,t_0}{B}_N){\Lambda }_t^{N-1}{u}_t-J_{R_N,{\lambda }_{N,t_0}}(I-{\lambda }_{N,t_0}{B}_N){\Lambda }_{t_0}^{N-1}{u}_{t_0}\parallel )\hfill \\ & \le |{\lambda }_{N,t}-{\lambda }_{N,t_0}|(\parallel B_N{\Lambda }_t^{N-1}{u}_t\parallel +\widehat{M})+\parallel {\Lambda }_t^{N-1}{u}_t-{\Lambda }_{t_0}^{N-1}{u}_{t_0}\parallel \hfill \\ & \le {\tilde{M}}_0{\displaystyle \sum _{i=1}^N}|{\lambda }_{i,t}-{\lambda }_{i,t_0}|+\parallel u_t-u_{t_0}\parallel ,\hfill \end{array}$$

(32)

where

$$\begin{array}{c}{\displaystyle \underset{t}{sup}}\{\frac{1}{{\lambda }_{i,t}}\parallel J_{R_i,{\lambda }_{i,t}}(I-{\lambda }_{i,t_0}{B}_i){\Lambda }_t^{i-1}{u}_t-(I-{\lambda }_{i,t_0}{B}_i){\Lambda }_{t_0}^{i-1}{u}_{t_0}\parallel \hfill \\ +\frac{1}{{\lambda }_{i,t_0}}\parallel (I-{\lambda }_{i,t_0}{B}_i){\Lambda }_t^{i-1}{u}_t-J_{R_i,{\lambda }_{i,t_0}}(I-{\lambda }_{i,t_0}{B}_i){\Lambda }_{t_0}^{i-1}{u}_{t_0}\parallel \}\le \widehat{M},\hfill \end{array}$$

for some $\widehat{M}>0$ and ${sup}_t\{{\sum }_{i=1}^N\parallel B_i{\Lambda }_t^{i-1}{u}_t\parallel +\widehat{M}\}\le {\tilde{M}}_0$ for some ${\tilde{M}}_0>0$. Also, utilizing Proposition 2, we deduce that

$$\begin{array}{cc}\hfill \parallel u_t-u_{t_0}\parallel & \le \parallel T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{B}_M){\Delta }_t^{M-1}{x}_t-T_{r_{M,t_0}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{B}_M){\Delta }_t^{M-1}{x}_t\parallel \hfill \\ & +\parallel T_{r_{M,t_0}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{B}_M){\Delta }_t^{M-1}{x}_t-T_{r_{M,t_0}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t_0}{B}_M){\Delta }_t^{M-1}{x}_t\parallel \hfill \\ & +\parallel (I-r_{M,t_0}{B}_M){\Delta }_t^{M-1}{x}_t-(I-r_{M,t_0}{B}_M){\Delta }_{t_0}^{M-1}{x}_{t_0}\parallel \hfill \\ & \le |r_{M,t}-r_{M,t_0}|[\parallel B_M{\Delta }_t^{M-1}{x}_t\parallel +\frac{1}{r_{M,t}}\parallel T_{r_{M,t}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,t}{B}_M){\Delta }_t^{M-1}{x}_t\hfill \\ & -(I-r_{M,t}{B}_M){\Delta }_t^{M-1}{x}_t\parallel ]+\cdots +\parallel {\Delta }_t^0{x}_t-{\Delta }_{t_0}^0{x}_{t_0}\parallel \hfill \\ & +|r_{1,t}-r_{1,t_0}|[\parallel B_1{\Delta }_t^0{x}_t\parallel +\frac{1}{r_{1,t}}\parallel T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{B}_1){\Delta }_t^0{x}_t-(I-r_{1,t}{B}_1){\Delta }_t^0{x}_t\parallel ]\hfill \\ & {\displaystyle \le {\tilde{M}}_1\sum _{j=1}^M|r_{j,t}-r_{j,t_0}|+\parallel x_t-x_{t_0}\parallel ,}\hfill \end{array}$$

(33)

where ${\tilde{M}}_1>0$ is a constant and

$$\sum _{j=1}^M[\parallel B_j{\Delta }_t^{j-1}{x}_t\parallel +\frac{1}{r_{j,t}}\parallel T_{r_{j,t}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,t}{B}_j){\Delta }_t^{j-1}{x}_t-(I-r_{j,t}{B}_j){\Delta }_t^{j-1}{x}_t\parallel ]\le {\tilde{M}}_1.$$

In terms of (31)–(33), we calculate

$$\begin{array}{c}\hfill {\displaystyle \parallel T_t{v}_t-T_{t_0}{v}_{t_0}\parallel \le \parallel x_t-x_{t_0}\parallel +{\tilde{M}}_1\sum _{j=1}^M|r_{j,t}-r_{j,t_0}|+{\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,t}-{\lambda }_{i,t_0}|+\frac{4\tilde{M}}{L}|s_t-s_{t_0}|,}\end{array}$$

and hence

$$\begin{array}{cc}\hfill \parallel x_t-x_{t_0}\parallel & \le \parallel (I-s_{t}A)T_t{v}_t+s_t((I-t\mu F)V{x}_t+t\gamma T_t{v}_t)-(I-s_{t_0}A)T_{t_0}{v}_{t_0}\hfill \\ & -s_{t_0}((I-t_0\mu F)V{x}_{t_0}+t_0\gamma T_{t_0}{v}_{t_0})\parallel \hfill \\ & \le |s_t-s_{t_0}|\parallel A\parallel \parallel T_t{v}_t\parallel +(1-s_{t_0}\overline{\gamma })\parallel T_t{v}_t-T_{t_0}{v}_{t_0}\parallel +t_0\gamma \parallel T_t{v}_t-T_{t_0}{v}_{t_0}\parallel \hfill \\ & +|s_t-s_{t_0}|[\parallel V{x}_t\parallel +t(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )]+s_{t_0}[(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )|t-t_0|\hfill \\ & +(1-t_0\tau )\parallel x_t-x_{t_0}\parallel ]\hfill \\ & \le (1-s_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma )))(\parallel x_t-x_{t_0}\parallel +{\tilde{M}}_1{\displaystyle \sum _{j=1}^M}|r_{j,t}-r_{j,t_0}|+{\tilde{M}}_0{\displaystyle \sum _{i=1}^N}|{\lambda }_{i,t}-{\lambda }_{i,t_0}|\hfill \\ & +\frac{4\tilde{M}}{L}|s_t-s_{t_0}|)+|s_t-s_{t_0}|[\parallel A\parallel \parallel T_t{v}_t\parallel +\parallel V{x}_t\parallel +t(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )]\hfill \\ & +s_{t_0}(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )|t-t_0|.\hfill \end{array}$$

This immediately implies that

$$\begin{array}{cc}\hfill \parallel x_t-x_{t_0}\parallel & \le \frac{\parallel A\parallel \parallel T_t{v}_t\parallel +\parallel V{x}_t\parallel +t(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )+\frac{4\tilde{M}}{L}}{s_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma ))}|s_t-s_{t_0}|\hfill \\ & +\frac{\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel }{\overline{\gamma }-1+t_0(\tau -\gamma )}|t-t_0|+\frac{{\tilde{M}}_1}{s_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma l))}{\displaystyle \sum _{j=1}^M}|r_{j,t}-r_{j,t_0}|\hfill \\ & +\frac{{\tilde{M}}_0}{s_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma l))}{\displaystyle \sum _{i=1}^N}|{\lambda }_{i,t}-{\lambda }_{i,t_0}|.\hfill \end{array}$$

Since $s_t,{\lambda }_{i,t},r_{j,t}$ are locally Lipschitzian, we conclude that $x_t$ is locally Lipschitzian. □

Theorem 1.

Assume that ${lim}_{t\to 0}{s}_t=0$. Then, $x_t$ defined by (14) converges strongly to $\tilde{x}\in \Omega$ as $t\to 0$, which solves (15).

Proof. 

Let $\tilde{x}$ be the unique solution of (15). Let $p\in \Omega$. Then, we have

$$\begin{array}{cc}\hfill x_t-p& =x_t-w_t+s_t[(I-t\mu F)V{x}_t-(I-t\mu F)Vp+t\gamma (T_t{v}_t-T_{t}p)+t(\gamma I-\mu FV)p]\hfill \\ & +(I-s_{t}A)(T_t{v}_t-T_{t}p)+s_t(V-A)p,\hfill \end{array}$$

where $w_t=(I-s_{t}A)T_t{v}_t+s_t((I-t\mu F)V{x}_t+t\gamma T_t{v}_t)$. Then, by Proposition 1 and the nonexpansivity of $T_t$, we obtain from (18) that

$$\begin{array}{cc}\hfill \parallel x_t{-p\parallel }^2& \le \langle (I-s_{t}A)(T_t{v}_t-T_{t}p),x_t-p\rangle +s_t[\langle (I-t\mu F)V{x}_t-(I-t\mu F)Vp,x_t-p\rangle \hfill \\ & +t\gamma \langle T_t{v}_t-T_{t}p,x_t-p\rangle +t\langle (\gamma I-\mu FV)p,x_t-p\rangle ]+s_t\langle (V-A)p,x_t-p\rangle \hfill \\ & \le (1-s_t\overline{\gamma })\parallel v_t-p\parallel \parallel x_t-p\parallel +s_t[(1-t\tau ){\parallel x_t-p\parallel }^2\hfill \\ & +t\gamma \parallel v_t-p\parallel \parallel x_t-p\parallel +t\langle (\gamma I-\mu FV)p,x_t-p\rangle ]+s_t\langle (V-A)p,x_t-p\rangle \hfill \\ & \le [1-s_t((\overline{\gamma }-1)+t(\tau -\gamma ))]{\parallel x_t-p\parallel }^2+\langle (V-A)p,x_t-p\rangle +s_t(t\langle (\gamma I-\mu FV)p,x_t-p\rangle ).\hfill \end{array}$$

Hence,

$$\parallel x_t{-p\parallel }^2\le \frac{1}{\overline{\gamma }-1+t(\tau -\gamma )}(\langle (V-A)p,x_t-p\rangle +t\langle (\gamma I-\mu FV)p,x_t-p\rangle ).$$

(34)

It follows that $x_{t_n}\to x^{*}$, where $\{t_n\}\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma }\})$ such that $t_n\to 0$. By Proposition 4, we get ${lim}_{n\to \infty }\parallel x_{t_n}-T_{t_n}{x}_{t_n}\parallel =0$. Observe that

$$\begin{array}{cc}\hfill \parallel P_C(I-{\lambda }_{t_n}\nabla f)x_{t_n}-x_{t_n}\parallel & =\parallel s_{t_n}{x}_{t_n}+(1-s_{t_n})T_{t_n}{x}_{t_n}-x_{t_n}\parallel \hfill \\ & \le \parallel T_{t_n}{x}_{t_n}-x_{t_n}\parallel ,\hfill \end{array}$$

where $s_{t_n}=\frac{2-{\lambda }_{t_n}L}{4}\in (0,\frac{1}{2})$ for ${\lambda }_{t_n}\in (0,\frac{2}{L})$. Hence we have

$$\begin{array}{cc}\hfill \parallel P_C(I-\frac{2}{L}\nabla f)x_{t_n}-x_{t_n}\parallel & \le \parallel (I-\frac{2}{L}\nabla f)x_{t_n}-(I-{\lambda }_{t_n}\nabla f)x_{t_n}\parallel +\parallel P_C(I-{\lambda }_{t_n}\nabla f)x_{t_n}-x_{t_n}\parallel \hfill \\ & \le (\frac{2}{L}-{\lambda }_{t_n})\parallel \nabla f(x_{t_n})\parallel +\parallel T_{t_n}{x}_{t_n}-x_{t_n}\parallel .\hfill \end{array}$$

By the boundedness of $\{x_{t_n}\},s_{t_n}\to 0$ ($\iff {\lambda }_{t_n}\to \frac{2}{L}$) and $\parallel T_{t_n}{x}_{t_n}-x_{t_n}\parallel \to 0$, we deduce

$$\parallel x^{*}-P_C(I-\frac{2}{L}\nabla f)x^{*}\parallel =\underset{n\to \infty }{lim}\parallel x_{t_n}-P_C(I-\frac{2}{L}\nabla f)x_{t_n}\parallel =0.$$

Therefore, $x^{*}\in \mathrm{VI}(C,\nabla f)=\Xi$.

Furthermore, from (25), (26) and (28), we have that ${\Delta }_{t_n}^j{x}_{t_n}\to x^{*},{\Lambda }_{t_n}^m{u}_{t_n}\to x^{*},u_{t_n}\to x^{*}$ and $v_{t_n}\to x^{*}$. First, we prove that $x^{*}\in {\cap }_{m=1}^N\mathrm{I}(B_m,R_m)$. Please note that $R_m+B_m$ is maximal monotone. Let $(v,g)\in G(R_m+B_m)$, i.e., $g-B_{m}v\in R_{m}v$. Noting that ${\Lambda }_{t_n}^m{u}_{t_n}=J_{R_m,{\lambda }_{m,t_n}}(I-{\lambda }_{m,t_n}{B}_m){\Lambda }_{t_n}^{m-1}{u}_{t_n},m\in \{1,\dots ,N\}$, we have

$${\Lambda }_{t_n}^{m-1}{u}_{t_n}-{\lambda }_{m,t_n}{B}_m{\Lambda }_{t_n}^{m-1}{u}_{t_n}\in (I+{\lambda }_{m,t_n}{R}_m){\Lambda }_{t_n}^m{u}_{t_n},$$

that is,

$$\frac{1}{{\lambda }_{m,t_n}}({\Lambda }_{t_n}^{m-1}{u}_{t_n}-{\Lambda }_{t_n}^m{u}_{t_n}-{\lambda }_{m,t_n}{B}_m{\Lambda }_{t_n}^{m-1}{u}_{t_n})\in R_m{\Lambda }_{t_n}^m{u}_{t_n}.$$

According to the monotonicity of $R_m$, we have

$$\langle v-{\Lambda }_{t_n}^m{u}_{t_n},g-B_{m}v-\frac{1}{{\lambda }_{m,t_n}}({\Lambda }_{t_n}^{m-1}{u}_{t_n}-{\Lambda }_{t_n}^m{u}_{t_n}-{\lambda }_{m,t_n}{B}_m{\Lambda }_{t_n}^{m-1}{u}_{t_n})\rangle \ge 0$$

and hence

$$\begin{array}{cc}\hfill \langle v-{\Lambda }_{t_n}^m{u}_{t_n},g\rangle & \ge \langle v-{\Lambda }_{t_n}^m{u}_{t_n},B_{m}v+\frac{1}{{\lambda }_{m,t_n}}({\Lambda }_{t_n}^{m-1}{u}_{t_n}-{\Lambda }_{t_n}^m{u}_{t_n}-{\lambda }_{m,t_n}{B}_m{\Lambda }_{t_n}^{m-1}{u}_{t_n})\rangle \hfill \\ & \ge \langle v-{\Lambda }_{t_n}^m{u}_{t_n},B_m{\Lambda }_{t_n}^m{u}_{t_n}-B_m{\Lambda }_{t_n}^{m-1}{u}_{t_n}\rangle +\langle v-{\Lambda }_{t_n}^m{u}_{t_n},\frac{1}{{\lambda }_{m,t_n}}({\Lambda }_{t_n}^{m-1}{u}_{t_n}-{\Lambda }_{t_n}^m{u}_{t_n})\rangle .\hfill \end{array}$$

Since $\parallel {\Lambda }_{t_n}^m{u}_{t_n}-{\Lambda }_{t_n}^{m-1}{u}_{t_n}\parallel \to 0$ (due to (25)) and $\parallel B_m{\Lambda }_{t_n}^m{u}_{t_n}-B_m{\Lambda }_{t_n}^{m-1}{u}_{t_n}\parallel \to 0$, we deduce from ${\Lambda }_{t_n}^m{u}_{t_n}\to x^{*}$ and $\{{\lambda }_{m,t_n}\}\subset [a_m,b_m]\subset (0,2{\eta }_m)$ that

$$\underset{n\to \infty }{lim}\langle v-{\Lambda }_{t_n}^m{u}_{t_n},g\rangle =\langle v-x^{*},g\rangle \ge 0.$$

By the maximal monotonicity of $B_m+R_m$, we derive $0\in (R_m+B_m)x^{*}$, i.e., $x^{*}\in \mathrm{I}(B_m,R_m)$. Thus, $x^{*}\in {\bigcap }_{m=1}^N\mathrm{I}(B_m,R_m)$. Next we prove that $x^{*}\in {\cap }_{j=1}^M\mathrm{GMEP}({\Theta }_j,{\phi }_j,A_j)$. Since ${\Delta }_{t_n}^j{x}_{t_n}=T_{r_{j,t_n}}^{({\Theta }_j,{\phi }_j)}(I-r_{j,t_n}{A}_j){\Delta }_{t_n}^{j-1}{x}_{t_n},j\in \{1,\dots ,M\}$, we have

$${\Theta }_j({\Delta }_{t_n}^j{x}_{t_n},y)+{\phi }_j(y)-{\phi }_j({\Delta }_{t_n}^j{x}_{t_n})+\langle A_j{\Delta }_{t_n}^{j-1}{x}_{t_n},y-{\Delta }_{t_n}^j{x}_{t_n}\rangle +\langle y-{\Delta }_{t_n}^j{x}_{t_n},\frac{{\Delta }_{t_n}^j{x}_{t_n}-{\Delta }_{t_n}^{j-1}{x}_{t_n}}{r_{j,t_n}}\rangle \ge 0.$$

By (A2), we have

$${\phi }_j(y)-{\phi }_j({\Delta }_{t_n}^j{x}_{t_n})+\langle A_j{\Delta }_{t_n}^{j-1}{x}_{t_n},y-{\Delta }_{t_n}^j{x}_{t_n}\rangle +\langle y-{\Delta }_{t_n}^j{x}_{t_n},\frac{{\Delta }_{t_n}^j{x}_{t_n}-{\Delta }_{t_n}^{j-1}{x}_{t_n}}{r_{j,t_n}}\rangle \ge {\Theta }_j(y,{\Delta }_{t_n}^j{x}_{t_n}).$$

Let $y\in C$ and $t\in (0,1]$. Set $z_t=ty+(1-t)x^{*}$. Thus, $z_t\in C$. Hence,

$$\begin{array}{cc}\hfill \langle z_t-{\Delta }_{t_n}^j{x}_{t_n},A_j{z}_t\rangle & \ge {\phi }_j({\Delta }_{t_n}^j{x}_{t_n})-{\phi }_j(z_t)+\langle z_t-{\Delta }_{t_n}^j{x}_{t_n},A_j{z}_t-A_j{\Delta }_{t_n}^j{x}_{t_n}\rangle +{\Theta }_j(z_t,{\Delta }_{t_n}^j{x}_{t_n})\hfill \\ & +\langle z_t-{\Delta }_{t_n}^j{x}_{t_n},A_j{\Delta }_{t_n}^j{x}_{t_n}-A_j{\Delta }_{t_n}^{j-1}{x}_{t_n}\rangle -\langle z_t-{\Delta }_{t_n}^j{x}_{t_n},\frac{{\Delta }_{t_n}^j{x}_{t_n}-{\Delta }_{t_n}^{j-1}{x}_{t_n}}{r_{j,t_n}}\rangle .\hfill \end{array}$$

(35)

By (25), we have $\parallel A_j{\Delta }_{t_n}^j{x}_{t_n}-A_j{\Delta }_{t_n}^{j-1}{x}_{t_n}\parallel \to 0$ as $n\to \infty$. Please note that $\langle z_t-{\Delta }_{t_n}^j{x}_{t_n},A_j{z}_t-A_j{\Delta }_{t_n}^j{x}_{t_n}\rangle \ge 0$. Then,

$$\langle z_t-x^{*},A_j{z}_t\rangle \ge {\phi }_j(x^{*})-{\phi }_j(z_t)+{\Theta }_j(z_t,x^{*}).$$

(36)

Applying (A1), (A4) and (36), we obtain

$$\begin{array}{cc}\hfill 0& ={\Theta }_j(z_t,z_t)+{\phi }_j(z_t)-{\phi }_j(z_t)\hfill \\ & \le t[{\Theta }_j(z_t,y)+{\phi }_j(y)-{\phi }_j(z_t)]+(1-t)t\langle y-x^{*},A_j{z}_t\rangle ,\hfill \end{array}$$

and hence

$$0\le {\Theta }_j(z_t,y)+{\phi }_j(y)-{\phi }_j(z_t)+(1-t)\langle y-x^{*},A_j{z}_t\rangle .$$

Thus,

$$0\le {\Theta }_j(x^{*},y)+{\phi }_j(y)-{\phi }_j(x^{*})+\langle y-x^{*},A_j{x}^{*}\rangle .$$

So, $x^{*}\in \mathrm{GMEP}({\Theta }_j,{\phi }_j,A_j)$ and $x^{*}\in {\cap }_{j=1}^M\mathrm{GMEP}({\Theta }_j,{\phi }_j,A_j)$. Therefore, $x^{*}\in {\cap }_{j=1}^M\mathrm{GMEP}({\Theta }_j,{\phi }_j,A_j)\cap {\cap }_{m=1}^N\mathrm{I}(B_m,R_m)\cap \Xi =:\Omega$.

Next, we prove that $x_t\to \tilde{x}$ as $t\to 0$. First, let us assert that $x^{*}$ is a solution of the VIP (15). As a matter of fact, since $x_t=x_t-w_t+(I-s_{t}A)T_t{\Lambda }_t^N{\Delta }_t^M{x}_t+s_t((I-t\mu F)V{x}_t+t\gamma T_t{\Lambda }_t^N{\Delta }_t^M{x}_t)$, we have

$$x_t-T_t{\Lambda }_t^N{\Delta }_t^M{x}_t=x_t-w_t+s_t(V-A)T_t{\Lambda }_t^N{\Delta }_t^M{x}_t+s_t(V{x}_t-V{T}_t{\Lambda }_t^N{\Delta }_t^M{x}_t+t(\gamma T_t{\Lambda }_t^N{\Delta }_t^M{x}_t-\mu FV{x}_t)).$$

Since $T_t,{\Lambda }_t^N$ and ${\Delta }_t^M$ are nonexpansive mappings, $I-T_t{\Lambda }_t^N{\Delta }_t^M$ is monotone. By the monotonicity of $I-T_t{\Lambda }_t^N{\Delta }_t^M$, we have

$$\begin{array}{cc}\hfill 0& \le \langle (I-T_t{\Lambda }_t^N{\Delta }_t^M)x_t-(I-T_t{\Lambda }_t^N{\Delta }_t^M)p,x_t-p\rangle \hfill \\ & \le s_t\langle (V-A)T_t{v}_t-(V-A)x_t,x_t-p\rangle +s_t\langle V{x}_t-V{T}_t{v}_t,x_t-p\rangle \hfill \\ & +s_t\langle (V-A)x_t,x_t-p\rangle +s_{t}t\langle (\gamma T_t{v}_t-\mu FV{x}_t),x_t-p\rangle .\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \langle (A-V)x_t,x_t-p\rangle & \le \parallel (V-A)T_t{v}_t-(V-A)x_t\parallel \parallel x_t-p\parallel +\parallel V{x}_t-V{T}_t{v}_t\parallel \parallel x_t-p\parallel \hfill \\ & +t\parallel \gamma T_t{v}_t-\mu FV{x}_t\parallel \parallel x_t-p\parallel \hfill \\ & \le (2+\parallel A\parallel )\parallel T_t{v}_t-x_t\parallel \parallel x_t-p\parallel +t(\gamma \parallel T_t{v}_t\parallel +\mu \parallel FV{x}_t\parallel )\parallel x_t-p\parallel .\hfill \end{array}$$

(37)

Now, replacing t in (37) with $t_n$ and noticing the boundedness of $\{\gamma \parallel T_{t_n}{v}_{t_n}\parallel +\mu \parallel FV{x}_{t_n}\parallel \}$ and the fact that $(V-A)T_{t_n}{v}_{t_n}-(V-A)x_{t_n}\to 0$, we deduce

$$\langle (A-V)x^{*},x^{*}-p\rangle \le 0.$$

Thus, $x^{*}\in \Omega$ is a solution of (15); hence $x^{*}=\tilde{x}$ by uniqueness. Consequently, $x_t\to \tilde{x}$ as $t\to 0$. □

Theorem 2.

Assume that the sequences $\{{\alpha }_n\}\subset [0,1]$ and $\{s_n\}\subset (0,\frac{1}{2})$ satisfy: $(C1)$${lim}_{n\to \infty }{\alpha }_n=0$ and ${lim}_{n\to \infty }{s}_n=0$. Let the sequence $\{x_n\}$ be defined by (16). Then ${\mathrm{LIM}}_n\langle (A-V)\tilde{x},\tilde{x}-x_n\rangle \le 0$, where $\tilde{x}={lim}_{t\to 0^{+}}{x}_t$ and $x_t$ is defined by

$$x_t=P_C[(I-s_{t}A)T{\Lambda }^N{\Delta }^M{x}_t+s_t(V{x}_t-t(\mu FV{x}_t-\gamma T{\Lambda }^N{\Delta }^M{x}_t))],$$

(38)

where $T,{\Lambda }^N,{\Delta }^M:C\to C$ are defined by $Tx=P_C(I-\frac{2}{L}\nabla f)x,{\Lambda }^{N}x=J_{R_N,{\lambda }_N}(I-{\lambda }_N{B}_N)\cdots J_{R_1,{\lambda }_1}(I-{\lambda }_1{B}_1)x$ and ${\Delta }^{M}x=T_{r_M}^{({\Theta }_M,{\phi }_M)}(I-r_M{A}_M)\cdots T_{r_1}^{({\Theta }_1,{\phi }_1)}(I-r_1{A}_1)x$ for ${\lambda }_i\in [a_i,b_i]\subset (0,2{\eta }_i),i=1,\dots ,N$ and $r_j\in [c_j,d_j]\subset (0,2{\mu }_j),j=1,\dots ,M$.

Proof. 

We assume, without loss of generality, that $0<s_n\le {\parallel A\parallel }^{-1}$ for all $n\ge 0$. Let ${lim}_{t\to 0}{x}_t=\tilde{x}$ and $\tilde{x}$ is the unique solution of (15). By Proposition 4, we deduce that the nets $\{x_t\}$, $\{V{x}_t\},\{{\Delta }^M{x}_t\},\{{\Lambda }^N{\Delta }^M{x}_t\}$ and $\{FV{x}_t\}$ are bounded.

Let $p\in \Omega$. Then $T_{n}p=p,{\Lambda }_n^{N}p=p$ and ${\Delta }_n^{M}p=p$. Applying Lemmas 5 and 7, we obtain that

$$\begin{array}{cc}\hfill \parallel x_{n+1}-p\parallel & \le s_n[(1-{\alpha }_n\tau )\parallel x_n-p\parallel +{\alpha }_n\gamma \parallel v_n-p\parallel +{\alpha }_n\parallel (\gamma I-\mu FV)p\parallel ]\hfill \\ & +(1-s_n\overline{\gamma })\parallel v_n-p\parallel +s_n\parallel (V-A)p\parallel \hfill \\ & \le s_n[(1-{\alpha }_n\tau )\parallel x_n-p\parallel +{\alpha }_n\gamma \parallel x_n-p\parallel +{\alpha }_n\parallel (\gamma I-\mu FV)p\parallel ]\hfill \\ & +s_n\parallel (V-A)p\parallel +(1-s_n\overline{\gamma })\parallel x_n-p\parallel \hfill \\ & \le max\{\frac{\parallel (V-A)p\parallel +\parallel (\gamma I-\mu FV)p\parallel }{\overline{\gamma }-1},\parallel x_n-p\parallel \}.\hfill \end{array}$$

By induction

$$\parallel x_n-p\parallel \le max\{\frac{\parallel (V-A)p\parallel +\parallel (\gamma I-\mu FV)p\parallel }{\overline{\gamma }-1},\parallel x_0-p\parallel \},\forall n\ge 0.$$

Thus, $\{x_n\}$, $\{u_n\},\{v_n\},\{T_n{v}_n\},\{FV{x}_n\},\{V{x}_n\}$ and $\{y_n\}$ are all bounded. By (C1), we obtain

$$\begin{array}{c}\hfill \parallel x_{n+1}-T_n{v}_n\parallel \le \parallel (I-s_{n}A)T_n{v}_n+s_n{y}_n-T_n{v}_n\parallel =s_n\parallel y_n-A{T}_n{v}_n\parallel \to 0\mathrm{as}n\to \infty .\end{array}$$

By (31), we also have

$$\parallel T{\Lambda }^N{\Delta }^M{x}_n-T_n{\Lambda }_n^N{\Delta }_n^M{x}_n\parallel \le \parallel {\Lambda }^N{\Delta }^M{x}_n-{\Lambda }_n^N{\Delta }_n^M{x}_n\parallel +\widehat{M}|\frac{2}{L}-{\lambda }_n|,$$

(39)

where ${sup}_{n\ge 0}\{L\parallel P_C(I-\frac{2}{L}\nabla f)v_n\parallel +4\parallel \nabla f(v_n)\parallel +L\parallel v_n\parallel \}\le \widehat{M}$ for some $\widehat{M}>0$. According to (32), we have

$$\parallel {\Lambda }^N{\Delta }^M{x}_n-{\Lambda }_n^N{\Delta }_n^M{x}_n\parallel \le {\widehat{M}}_0{\displaystyle \sum _{i=1}^N}|{\lambda }_i-{\lambda }_{i,n}|+\parallel {\Delta }^M{x}_n-{\Delta }_n^M{x}_n\parallel ,$$

(40)

where

$$\begin{array}{c}{\displaystyle \underset{n\ge 0}{sup}}\{\frac{1}{{\lambda }_i}\parallel J_{R_i,{\lambda }_i}(I-{\lambda }_{i,n}{B}_i){\Lambda }^{i-1}{\Delta }^M{x}_n-(I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n\parallel \hfill \\ +\frac{1}{{\lambda }_{i,n}}\parallel (I-{\lambda }_{i,n}{B}_i){\Lambda }^{i-1}{\Delta }^M{x}_n-J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n\parallel \}\le \widehat{N},\hfill \end{array}$$

for some $\widehat{N}>0$ and ${sup}_{n\ge 0}\{{\sum }_{i=1}^N\parallel B_i{\Lambda }^{i-1}{\Delta }^M{x}_n\parallel +\widehat{N}\}\le {\widehat{M}}_0$ for some ${\widehat{M}}_0>0$. In terms of (33), we have

$${\displaystyle \parallel {\Delta }^M{x}_n-{\Delta }_n^M{x}_n\parallel \le {\widehat{M}}_1\sum _{j=1}^M|r_j-r_{j,n}|,}$$

(41)

where ${sup}_{n\ge 0}\{{\sum }_{j=1}^M[\parallel A_j{\Delta }^{j-1}{x}_n\parallel +\frac{1}{r_j}\parallel T_{r_j}^{({\Theta }_j,{\phi }_j)}(I-r_j{A}_j){\Delta }^{j-1}{x}_n-(I-r_j{A}_j){\Delta }^{j-1}{x}_n\parallel ]\}\le {\widehat{M}}_1$ for some ${\widehat{M}}_1>0$. In terms of (39)–(41) we calculate

$$\begin{array}{c}\hfill {\displaystyle \parallel T{\Lambda }^N{\Delta }^M{x}_n-T_n{\Lambda }_n^N{\Delta }_n^M{x}_n\parallel \le {\widehat{M}}_1\sum _{j=1}^M|r_j-r_{j,n}|+{\widehat{M}}_0\sum _{i=1}^N|{\lambda }_i-{\lambda }_{i,n}|+\widehat{M}|\frac{2}{L}-{\lambda }_n|.}\end{array}$$

It is clear that

$$\begin{array}{cc}\hfill \parallel T{\Lambda }^N{\Delta }^M{x}_t-x_{n+1}\parallel & \le \parallel T{\Lambda }^N{\Delta }^M{x}_t-T{\Lambda }^N{\Delta }^M{x}_n\parallel +\parallel T{\Lambda }^N{\Delta }^M{x}_n-T_n{\Lambda }_n^N{\Delta }_n^M{x}_n\parallel \hfill \\ & +\parallel T_n{\Lambda }_n^N{\Delta }_n^M{x}_n-x_{n+1}\parallel \hfill \\ & \le \parallel x_t-x_n\parallel +{\widehat{M}}_1{\displaystyle \sum _{j=1}^M}|r_j-r_{j,n}|+{\widehat{M}}_0{\displaystyle \sum _{i=1}^N}|{\lambda }_i-{\lambda }_{i,n}|+\widehat{M}|\frac{2}{L}-{\lambda }_n|\hfill \\ & +\parallel T_n{v}_n-x_{n+1}\parallel \hfill \\ & =\parallel x_t-x_n\parallel +{ϵ}_n,\hfill \end{array}$$

(42)

where ${ϵ}_n={\widehat{M}}_1{\sum }_{j=1}^M|r_j-r_{j,n}|+{\widehat{M}}_0{\sum }_{i=1}^N|{\lambda }_i-{\lambda }_{i,n}|+\widehat{M}|\frac{2}{L}-{\lambda }_n|+\parallel T_n{v}_n-x_{n+1}\parallel \to 0$ as $n\to \infty$. Please note that

$$\langle A{x}_t-A{x}_n,x_t-x_n\rangle \ge \overline{\gamma }{\parallel x_t-x_n\parallel }^2.$$

(43)

Furthermore, for simplicity, we write $w_t=(I-s_{t}A)T{\Lambda }^N{\Delta }^M{x}_t+s_t((I-t\mu F)V{x}_t+t\gamma T{\Lambda }^N{\Delta }^M{x}_t)$. By (38), we get $x_t=P_C{w}_t$ and

$$\begin{array}{cc}\hfill x_t-x_{n+1}& =(I-s_{t}A)T{\Lambda }^N{\Delta }^M{x}_t-(I-s_{t}A)x_{n+1}+s_t[(I-t\mu F)V{x}_t-(I-t\mu F)V{x}_{n+1}\hfill \\ & +t(\gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1})+(V-A)x_{n+1}]+x_t-w_t.\hfill \end{array}$$

Applying Lemma 1 and Proposition 1, we have

$$\begin{array}{cc}\hfill \parallel x_t-x_{n+1}{\parallel }^2& \le 2s_{t}t\langle \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}),x_t-x_{n+1}\rangle +2s_t\langle (V-A)x_{n+1},x_t-x_{n+1}\rangle \hfill \\ & +2s_t\langle (I-t\mu F)V{x}_t-(I-t\mu F)V{x}_{n+1},x_t-x_{n+1}\rangle \hfill \\ & +\parallel (I-s_{t}A)T{\Lambda }^N{\Delta }^M{x}_t-(I-s_{t}A)x_{n+1}{\parallel }^2\hfill \\ & \le {(1-s_t\overline{\gamma })}^2{\parallel T{\Lambda }^N{\Delta }^M{x}_t-x_{n+1}\parallel }^2+2s_t\langle V{x}_t-V{x}_{n+1},x_t-x_{n+1}\rangle \hfill \\ & -2s_{t}t\mu \langle FV{x}_t-FV{x}_{n+1},x_t-x_{n+1}\rangle +2s_t\langle (V-A)x_{n+1},x_t-x_{n+1}\rangle \hfill \\ & +2s_{t}t\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel \parallel x_t-x_{n+1}\parallel \hfill \\ & \le {(1-s_t\overline{\gamma })}^2{\parallel T{\Lambda }^N{\Delta }^M{x}_t-x_{n+1}\parallel }^2+2s_t\langle V{x}_t-V{x}_{n+1},x_t-x_{n+1}\rangle \hfill \\ & +2s_{t}t(\mu \parallel FV{x}_t-FV{x}_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +2s_t\langle (V-A)x_{n+1},x_t-x_{n+1}\rangle .\hfill \end{array}$$

(44)

Using (42) and (43) in (44), we obtain

$$\begin{array}{cc}\hfill \parallel x_t-x_{n+1}{\parallel }^2& \le {(1-s_t\overline{\gamma })}^2(\parallel x_t-x_n{\parallel +{ϵ}_n)}^2+2s_t\langle V{x}_t-V{x}_{n+1},x_t-x_{n+1}\rangle \hfill \\ & +2s_{t}t(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +2s_t\langle (V-A)x_{n+1},x_t-x_{n+1}\rangle \hfill \\ & \le s_t^2\overline{\gamma }\langle A(x_t-x_n),x_t-x_n\rangle +\parallel x_t-x_n{\parallel }^2+{(1-s_t\overline{\gamma })}^2(2\parallel x_t-x_n\parallel {ϵ}_n+{ϵ}_n^2)\hfill \\ & +2s_{t}t(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +2s_t[\langle (V-A)x_t,x_t-x_{n+1}\rangle +\langle A(x_t-x_{n+1}),x_t-x_{n+1}\rangle \hfill \\ & -\langle A(x_t-x_n),x_t-x_n\rangle ].\hfill \end{array}$$

(45)

Applying the Banach limit $\mathrm{LIM}$ to (45), we have

$$\begin{array}{cc}\hfill {\mathrm{LIM}}_n{\parallel x_t-x_{n+1}\parallel }^2& \le s_t^2\overline{\gamma }{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle +{\mathrm{LIM}}_n{\parallel x_t-x_n\parallel }^2\hfill \\ & +2s_{t}t{\mathrm{LIM}}_n(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +2s_t[{\mathrm{LIM}}_n\langle (V-A)x_t,x_t-x_{n+1}\rangle +{\mathrm{LIM}}_n\langle A(x_t-x_{n+1}),x_t-x_{n+1}\rangle \hfill \\ & -{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle ].\hfill \end{array}$$

(46)

Using the property ${\mathrm{LIM}}_n{a}_n={\mathrm{LIM}}_n{a}_{n+1}$, we have

$$\begin{array}{cc}\hfill {\mathrm{LIM}}_n& \langle (A-V)x_t,x_t-x_n\rangle ={\mathrm{LIM}}_n\langle (A-V)x_t,x_t-x_{n+1}\rangle \hfill \\ & \le \frac{s_t\overline{\gamma }}{2}{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle +\frac{1}{2s_t}[{\mathrm{LIM}}_n\parallel x_t-x_n{\parallel }^2-{\mathrm{LIM}}_n\parallel x_t-x_{n+1}{\parallel }^2]\hfill \\ & +t{\mathrm{LIM}}_n(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +{\mathrm{LIM}}_n\langle A(x_t-x_{n+1}),x_t-x_{n+1}\rangle -{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle \hfill \\ & =\frac{s_t\overline{\gamma }}{2}{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle \hfill \\ & +t{\mathrm{LIM}}_n(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel .\hfill \end{array}$$

(47)

Since

$$s_t\langle A(x_t-x_n),x_t-x_n\rangle \le s_t\parallel A\parallel \parallel x_t-x_n{\parallel }^2\le s_t\parallel A\parallel K^2\to 0\mathrm{as}t\to 0,$$

(48)

where $\parallel x_t-x_n\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_n\parallel \le K$,

$$t\parallel x_t-x_{n+1}{\parallel }^2\to 0\mathrm{and}t\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel \parallel x_t-x_{n+1}\parallel \to 0\mathrm{as}t\to 0,$$

(49)

we conclude from (47)–(49) that

$$\begin{array}{cc}\hfill {\mathrm{LIM}}_n\langle (A-V)\tilde{x},\tilde{x}-x_n\rangle & \le {\displaystyle \underset{t\to 0}{lim\; sup}}{\mathrm{LIM}}_n\langle (A-V)x_t,x_t-x_n\rangle \hfill \\ & \le {\displaystyle \underset{t\to 0}{lim\; sup}}t{\mathrm{LIM}}_n(\mu \kappa \parallel x_t-x_{n+1}\parallel +\parallel \gamma T{\Lambda }^N{\Delta }^M{x}_t-\mu FV{x}_{n+1}\parallel )\parallel x_t-x_{n+1}\parallel \hfill \\ & +{\displaystyle \underset{t\to 0}{lim\; sup}}\frac{s_t\overline{\gamma }}{2}{\mathrm{LIM}}_n\langle A(x_t-x_n),x_t-x_n\rangle \hfill \\ & =0.\hfill \end{array}$$

This completes the proof. □

Theorem 3.

Let the sequences $\{{\alpha }_n\}\subset [0,1]$ and $\{s_n\}\subset (0,\frac{1}{2})$ satisfy: (C1) ${lim}_{n\to \infty }{\alpha }_n=0,{lim}_{n\to \infty }{s}_n=0$ and (C2) ${\sum }_{n=0}^{\infty }{s}_n=\infty$. Let the sequence $\{x_n\}$ be defined by (16). If $x_{n+1}-x_n\rightharpoonup 0$, then $x_n$ converges strongly to $\tilde{x}\in \Omega$, which solves (15).

Proof. 

We assume, without loss of generality, that ${\alpha }_n\tau <1$ and $\frac{2s_n(\overline{\gamma }-1)}{1-s_n}<1$ for all $n\ge 0$. Let $x_t$ be defined by (38). Then, ${lim}_{t\to 0}{x}_t:=\tilde{x}\in \Omega$ (due to Theorem 1). We divide the rest of the proof into several steps.

Step 1. It is easy to show that

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel (V-A)p\parallel +\parallel (\gamma I-\mu FV)p\parallel }{\overline{\gamma }-1}\},\forall n\ge 0.$$

Hence $\{x_n\}$, $\{u_n\},\{v_n\}$, $\{T_n{v}_n\},\{FV{x}_n\},\{V{x}_n\}$ and $\{y_n\}$ are all bounded.

Step 2. We show that ${lim\; sup}_{n\to \infty }\langle (V-A)\tilde{x},x_n-\tilde{x}\rangle \le 0$. To this end, let

$$a_n:=\langle (A-V)\tilde{x},\tilde{x}-x_n\rangle ,\forall n\ge 0.$$

According to Theorem 2, we deduce ${\mathrm{LIM}}_n{a}_n\le 0$. Choose a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that

$$\underset{n\to \infty }{lim\; sup}(a_{n+1}-a_n)=\underset{j\to \infty }{lim\; sup}(a_{n_j+1}-a_{n_j})$$

and $x_{n_j}\rightharpoonup v\in H$. This indicates that $x_{n_j+1}\rightharpoonup v$. Hence,

$$w-\underset{j\to \infty }{lim}(\tilde{x}-x_{n_j+1})=w-\underset{j\to \infty }{lim}(\tilde{x}-x_{n_j})=(\tilde{x}-v),$$

and therefore

$$\underset{n\to \infty }{lim\; sup}(a_{n+1}-a_n)=\underset{j\to \infty }{lim}\langle (A-V)\tilde{x},(\tilde{x}-x_{n_j+1})-(\tilde{x}-x_{n_j})\rangle =0.$$

Then, by Lemma 8, we derive

$$\underset{n\to \infty }{lim\; sup}\langle (V-A)\tilde{x},x_n-\tilde{x}\rangle =\underset{n\to \infty }{lim\; sup}\langle (A-V)\tilde{x},\tilde{x}-x_n\rangle \le 0.$$

Step 3. We show that ${lim}_{n\to \infty }\parallel x_n-\tilde{x}\parallel =0$. Set $w_n=(I-s_{n}A)T_n{v}_n+s_n{y}_n$ for all $n\ge 0$. Then $x_{n+1}=P_C{w}_n$. Utilizing (16) and $T_n{\Lambda }_n^N{\Delta }_n^M\tilde{x}=T_n\tilde{x}=\tilde{x}$, we have

$$\begin{array}{cc}\hfill x_{n+1}-\tilde{x}& =s_n[(I-{\alpha }_n\mu F)V{x}_n-(I-{\alpha }_n\mu F)V\tilde{x}+{\alpha }_n\gamma (T_n{v}_n-T_n\tilde{x})+{\alpha }_n(\gamma I-\mu FV)\tilde{x}]\hfill \\ & +(I-s_{n}A)(T_n{v}_n-T_n\tilde{x})+s_n(V-A)\tilde{x}+x_{n+1}-w_n.\hfill \end{array}$$

Thus, utilizing Proposition 1 and Lemma 1, we get

$$\begin{array}{cc}\hfill \parallel x_{n+1}-\tilde{x}{\parallel }^2& \le [(1-s_n\overline{\gamma })\parallel T_n{v}_n-T_n\tilde{x}\parallel +s_n((1-{\alpha }_n\tau )\parallel x_n-\tilde{x}\parallel +{\alpha }_n\gamma \parallel T_n{v}_n-T_n\tilde{x}{\parallel )]}^2\hfill \\ & +2s_n[{\alpha }_n\parallel (\gamma I-\mu FV)\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +\langle (A-V)\tilde{x},\tilde{x}-x_{n+1}\rangle ]\hfill \\ & \le [1-s_n(\overline{\gamma }-1+{\alpha }_n(\tau -\gamma ))]{\parallel x_n-\tilde{x}\parallel }^2\hfill \\ & +2s_n[{\alpha }_n\parallel (\gamma I-\mu FV)\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +\langle (A-V)\tilde{x},\tilde{x}-x_{n+1}\rangle ]\hfill \\ & =[1-s_n(\overline{\gamma }-1+{\alpha }_n(\tau -\gamma ))]{\parallel x_n-\tilde{x}\parallel }^2+s_n(\overline{\gamma }-1+{\alpha }_n(\tau -\gamma ))\hfill \\ & \times \frac{2}{\overline{\gamma }-1+{\alpha }_n(\tau -\gamma )}[{\alpha }_n\parallel (\gamma I-\mu FV)\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +\langle (A-V)\tilde{x},\tilde{x}-x_{n+1}\rangle ]\hfill \\ & =(1-{\omega }_n){\parallel x_n-\tilde{x}\parallel }^2+{\omega }_n{\delta }_n,\hfill \end{array}$$

where ${\omega }_n=s_n(\overline{\gamma }-1+{\alpha }_n(\tau -\gamma ))$ and

$${\delta }_n=\frac{2}{\overline{\gamma }-1+{\alpha }_n(\tau -\gamma )}[{\alpha }_n\parallel (\gamma I-\mu FV)\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +\langle (A-V)\tilde{x},\tilde{x}-x_{n+1}\rangle ].$$

It is easy to check that ${\omega }_n\to 0,{\sum }_{n=0}^{\infty }{\omega }_n=\infty$ and ${lim\; sup}_{n\to \infty }{\delta }_n\le 0$. By Lemma 6 with $r_n=0$, we deduce that ${lim}_{n\to \infty }\parallel x_n-\tilde{x}\parallel =0$. This completes the proof. □