Theorem 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H, A:C\to H be an αinversestrongly monotone mapping, S:C\to C be a quasinonexpansive mapping such that IS is demiclosed at zero, and B be a maximal monotone operator on H such that the domain of B is included in C. Assume that \mathcal{F}=F(S)\cap {(A+B)}^{1}(0)\ne \mathrm{\varnothing}. Let \{{\lambda}_{n}\} be a positive real number sequence. Let \{{\alpha}_{n}\} be a real number sequence in [0,1]. Let \{{x}_{n}\} be a sequence in C generated in the following iterative process:
\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {C}_{1}=C,\hfill \\ {y}_{n}={\alpha}_{n}{x}_{n}+(1{\alpha}_{n})S{J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}),\hfill \\ {C}_{n+1}=\{z\in {C}_{n}:\parallel {y}_{n}z\parallel \le \parallel {x}_{n}z\parallel \},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 1,\hfill \end{array}
where {J}_{{\lambda}_{n}}={(I+{\lambda}_{n}B)}^{1}. Suppose that the sequences \{{\alpha}_{n}\} and \{{\lambda}_{n}\} satisfy the following restrictions:

(a)
0\le {\alpha}_{n}\le a<1;

(b)
0<b\le {\lambda}_{n}\le c<2\alpha.
Then the sequence \{{x}_{n}\} converges strongly to {P}_{\mathcal{F}}{x}_{1}.
Proof First, we show that {C}_{n} is closed and convex. Notice that {C}_{1}=C is closed and convex. Suppose that {C}_{i} is closed and convex for some i\ge 1. We show that {C}_{i+1} is closed and convex for the same i. Indeed, for any v\in {C}_{i}, we see that
\parallel {y}_{i}z\parallel \le \parallel {x}_{i}z\parallel
is equivalent to
{\parallel {y}_{i}\parallel}^{2}{\parallel {x}_{i}\parallel}^{2}2\u3008z,{y}_{i}{x}_{i}\u3009\ge 0.
Thus {C}_{i+1} is closed and convex. This shows that {C}_{n} is closed and convex.
Next, we prove that I{\lambda}_{n}A is a nonexpansive mapping. Indeed, we have
In view of the restriction (b), we obtain that I{\lambda}_{n}A is nonexpansive. Next, we show that \mathcal{F}\subset {C}_{n} for each n\ge 1. From the assumption, we see that \mathcal{F}\subset C={C}_{1}. Assume that \mathcal{F}\subset {C}_{i} for some i\ge 1. For any z\in \mathcal{F}\subset {C}_{i}, we find from Lemma that
z=Sz={J}_{{\lambda}_{i}}(z{\lambda}_{i}Az).
Put {z}_{n}={J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}). Since {J}_{{\lambda}_{n}} and I{\lambda}_{n}A are nonexpansive, we have
\begin{array}{rcl}\parallel {z}_{n}p\parallel & \le & \parallel ({x}_{n}{\lambda}_{n}A{x}_{n})(p{\lambda}_{n}Ap)\parallel \\ \le & \parallel {x}_{n}p\parallel .\end{array}
(2.1)
It follows from (2.1) that
\begin{array}{rl}\parallel {y}_{i}z\parallel & =\parallel {\alpha}_{i}{x}_{i}+(1{\alpha}_{i})S{z}_{i}z\parallel \\ \le {\alpha}_{i}\parallel {x}_{i}z\parallel +(1{\alpha}_{i})\parallel {z}_{i}z\parallel \\ \le \parallel {x}_{i}z\parallel .\end{array}
This shows that z\in {C}_{i+1}. This proves that \mathcal{F}\subset {C}_{n}. Notice that {x}_{n}={P}_{{C}_{n}}{x}_{1}. For every z\in \mathcal{F}\subset {C}_{n}, we have
\parallel {x}_{1}{x}_{n}\parallel \le \parallel {x}_{1}z\parallel .
In particular, we have
\parallel {x}_{1}{x}_{n}\parallel \le \parallel {x}_{1}{P}_{\mathcal{F}}{x}_{1}\parallel .
This implies that \{{x}_{n}\} is bounded. Since {x}_{n}={P}_{{C}_{n}}{x}_{1} and {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}\subset {C}_{n}, we arrive at
\begin{array}{rl}0& \le \u3008{x}_{1}{x}_{n},{x}_{n}{x}_{n+1}\u3009\\ \le {\parallel {x}_{1}{x}_{n}\parallel}^{2}+\parallel {x}_{1}{x}_{n}\parallel \parallel {x}_{1}{x}_{n+1}\parallel .\end{array}
It follows that
\parallel {x}_{n}{x}_{1}\parallel \le \parallel {x}_{n+1}{x}_{1}\parallel .
This implies that {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{x}_{1}\parallel exists. On the other hand, we have
It follows that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}{x}_{n+1}\parallel =0.
(2.2)
Notice that {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}. It follows that
\parallel {y}_{n}{x}_{n+1}\parallel \le \parallel {x}_{n}{x}_{n+1}\parallel .
This in turn implies that
\parallel {y}_{n}{x}_{n}\parallel \le \parallel {y}_{n}{x}_{n+1}\parallel +\parallel {x}_{n}{x}_{n+1}\parallel \le 2\parallel {x}_{n}{x}_{n+1}\parallel .
In view of (2.2), we obtain that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}{y}_{n}\parallel =0.
(2.3)
On the other hand, we have
\parallel {x}_{n}{y}_{n}\parallel =(1{\alpha}_{n})\parallel {x}_{n}S{z}_{n}\parallel .
It follows from (2.3) that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}S{z}_{n}\parallel =0.
(2.4)
For any p\in \mathcal{F}, we see that
\begin{array}{rl}{\parallel {z}_{n}p\parallel}^{2}& ={\parallel {J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}){J}_{{\lambda}_{n}}(p{\lambda}_{n}Ap)\parallel}^{2}\\ \le {\parallel {x}_{n}p\parallel}^{2}2\u3008{x}_{n}p,A{x}_{n}Ap\u3009+{\lambda}_{n}^{2}{\parallel A{x}_{n}Ap\parallel}^{2}\\ \le {\parallel {x}_{n}p\parallel}^{2}{\lambda}_{n}(2\alpha {\lambda}_{n}){\parallel A{x}_{n}Ap\parallel}^{2}.\end{array}
(2.5)
Notice that
\begin{array}{rl}{\parallel {y}_{n}p\parallel}^{2}& \le {\alpha}_{n}{\parallel {x}_{n}p\parallel}^{2}+(1{\alpha}_{n}){\parallel S{z}_{n}p\parallel}^{2}\\ \le {\alpha}_{n}{\parallel {x}_{n}p\parallel}^{2}+(1{\alpha}_{n}){\parallel {z}_{n}p\parallel}^{2}.\end{array}
(2.6)
Substituting (2.5) into (2.6), we see that
{\parallel {y}_{n}p\parallel}^{2}\le {\parallel {x}_{n}p\parallel}^{2}(1{\alpha}_{n}){\lambda}_{n}(2\alpha {\lambda}_{n}){\parallel A{x}_{n}Ap\parallel}^{2}.
It follows that
\begin{array}{rl}(1{\alpha}_{n}){\lambda}_{n}(2\alpha {\lambda}_{n}){\parallel A{x}_{n}Ap\parallel}^{2}& \le {\parallel {x}_{n}p\parallel}^{2}{\parallel {y}_{n}p\parallel}^{2}\\ \le (\parallel {x}_{n}p\parallel +\parallel {y}_{n}p\parallel )\parallel {x}_{n}{y}_{n}\parallel .\end{array}
This implies from (2.3) that
\underset{n\to \mathrm{\infty}}{lim}\parallel A{x}_{n}Ap\parallel =0.
(2.7)
On the other hand, we have
\begin{array}{rcl}{\parallel {z}_{n}p\parallel}^{2}& =& {\parallel {J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}){J}_{{\lambda}_{n}}(p{\lambda}_{n}Ap)\parallel}^{2}\\ \le & \u3008({x}_{n}{\lambda}_{n}A{x}_{n})(p{\lambda}_{n}Ap),{z}_{n}p\u3009\\ =& \frac{1}{2}({\parallel ({x}_{n}{\lambda}_{n}A{x}_{n})(p{\lambda}_{n}Ap)\parallel}^{2}+{\parallel {z}_{n}p\parallel}^{2}\\ {\parallel ({x}_{n}{\lambda}_{n}A{x}_{n})(p{\lambda}_{n}Ap)({z}_{n}p)\parallel}^{2})\\ \le & \frac{1}{2}({\parallel {x}_{n}p\parallel}^{2}+{\parallel {z}_{n}p\parallel}^{2}{\parallel {x}_{n}{z}_{n}{\lambda}_{n}(A{x}_{n}Ap)\parallel}^{2})\\ \le & \frac{1}{2}({\parallel {x}_{n}p\parallel}^{2}+{\parallel {z}_{n}p\parallel}^{2}{\parallel {x}_{n}{z}_{n}\parallel}^{2}{\lambda}_{n}^{2}{\parallel A{x}_{n}Ap\parallel}^{2}\\ +2{\lambda}_{n}\parallel {x}_{n}{z}_{n}\parallel \parallel A{x}_{n}Ap\parallel )\\ \le & \frac{1}{2}({\parallel {x}_{n}p\parallel}^{2}+{\parallel {z}_{n}p\parallel}^{2}{\parallel {x}_{n}{z}_{n}\parallel}^{2}+2{\lambda}_{n}\parallel {x}_{n}{z}_{n}\parallel \parallel A{x}_{n}Ap\parallel ).\end{array}
It follows that
{\parallel {z}_{n}p\parallel}^{2}\le {\parallel {x}_{n}p\parallel}^{2}{\parallel {x}_{n}{z}_{n}\parallel}^{2}+2{\lambda}_{n}\parallel {x}_{n}{z}_{n}\parallel \parallel A{x}_{n}Ap\parallel .
(2.8)
Substituting (2.8) into (2.6), we see that
{\parallel {y}_{n}p\parallel}^{2}\le {\parallel {x}_{n}p\parallel}^{2}(1{\alpha}_{n}){\parallel {x}_{n}{z}_{n}\parallel}^{2}+2(1{\alpha}_{n}){\lambda}_{n}\parallel {x}_{n}{z}_{n}\parallel \parallel A{x}_{n}Ap\parallel .
It follows that
In view of the restriction (a), we obtain from (2.7) that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}{z}_{n}\parallel =0.
(2.9)
Since \{{x}_{n}\} is bounded, we may assume that there is a subsequence \{{x}_{{n}_{i}}\} of \{{x}_{n}\} converging weakly to some point {x}^{\ast}. It follows from (2.9) that {z}_{{n}_{i}} converges weakly to {x}^{\ast}. Notice that
\parallel S{z}_{n}{z}_{n}\parallel \le \parallel S{z}_{n}{x}_{n}\parallel +\parallel {x}_{n}{z}_{n}\parallel .
It follows from (2.4) and (2.9) that
\underset{n\to \mathrm{\infty}}{lim}\parallel S{z}_{n}{z}_{n}\parallel =0.
In view of the assumption that S is demiclosed at zero, we see that {x}^{\ast}\in F(S).
Next, we show that {x}^{\ast}\in {(A+B)}^{1}(0). Notice that {z}_{n}={J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}). This implies that
{x}_{n}{\lambda}_{n}A{x}_{n}\in (I+{\lambda}_{n}B){z}_{n}.
That is,
\frac{{x}_{n}{z}_{n}}{{\lambda}_{n}}A{x}_{n}\in B{z}_{n}.
Since B is monotone, we get for any (u,v)\in B, that
\u3008{z}_{n}u,\frac{{x}_{n}{z}_{n}}{{\lambda}_{n}}A{x}_{n}v\u3009\ge 0.
(2.10)
Replacing n by {n}_{i} and letting i\to \mathrm{\infty}, we obtain from (2.10) that
\u3008\omega u,A\omega v\u3009\le 0.
This means A\omega \in B\omega, that is, 0\in (A+B)(\omega ). Hence, we get \omega \in {(A+B)}^{1}(0). This completes the proof that {x}^{\ast}\in \mathcal{F}.
Notice that {P}_{\mathcal{F}}{x}_{1}\subset {C}_{n+1} and {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}, we have
\parallel {x}_{1}{x}_{n+1}\parallel \le \parallel {x}_{1}{P}_{\mathcal{F}}{x}_{1}\parallel .
On the other hand, we have
\begin{array}{rl}\parallel {x}_{1}{P}_{\mathcal{F}}{x}_{1}\parallel & \le \parallel {x}_{1}{x}^{\ast}\parallel \\ \le \underset{i\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\parallel {x}_{1}{x}_{{n}_{i}}\parallel \\ \le \underset{i\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\parallel {x}_{1}{x}_{{n}_{i}}\parallel \\ \le \parallel {x}_{1}{P}_{\mathcal{F}}{x}_{1}\parallel .\end{array}
We, therefore, obtain that
\parallel {x}_{1}{x}^{\ast}\parallel =\underset{i\to \mathrm{\infty}}{lim}\parallel {x}_{1}{x}_{{n}_{i}}\parallel =\parallel {x}_{1}{P}_{\mathcal{F}}{x}_{1}\parallel .
This implies {x}_{{n}_{i}}\to {x}^{\ast}={P}_{\mathcal{F}}{x}_{1}. Since \{{x}_{{n}_{i}}\} is an arbitrary subsequence of \{{x}_{n}\}, we obtain that {x}_{n}\to {P}_{\mathcal{F}}{x}_{1} as n\to \mathrm{\infty}. This completes the proof. □
From Theorem 2.1, we have the following results immediately.
Corollary 2.2 Let C be a nonempty closed convex subset of a real Hilbert space H, A:C\to H be an αinversestrongly monotone mapping, and B be a maximal monotone operator on H such that the domain of B is included in C. Assume that {(A+B)}^{1}(0)\ne \mathrm{\varnothing}. Let \{{\lambda}_{n}\} be a positive real number sequence. Let \{{\alpha}_{n}\} be a real number sequence in [0,1]. Let \{{x}_{n}\} be a sequence in C generated in the following iterative process:
\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {C}_{1}=C,\hfill \\ {y}_{n}={\alpha}_{n}{x}_{n}+(1{\alpha}_{n}){J}_{{\lambda}_{n}}({x}_{n}{\lambda}_{n}A{x}_{n}),\hfill \\ {C}_{n+1}=\{z\in {C}_{n}:\parallel {y}_{n}z\parallel \le \parallel {x}_{n}z\parallel \},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 1,\hfill \end{array}
where {J}_{{\lambda}_{n}}={(I+{\lambda}_{n}B)}^{1}. Suppose that the sequences \{{\alpha}_{n}\} and \{{\lambda}_{n}\} satisfy the following restrictions:

(a)
0\le {\alpha}_{n}\le a<1;

(b)
0<b\le {\lambda}_{n}\le c<2\alpha.
Then the sequence \{{x}_{n}\} converges strongly to {P}_{{(A+B)}^{1}(0)}{x}_{1}.
Let f:H\to (\mathrm{\infty},\mathrm{\infty}] be a proper lower semicontinuous convex function. Define the subdifferential
\partial f(x)=\{z\in H:f(x)+\u3008yx,z\u3009\le f(y),\mathrm{\forall}y\in H\}
for all x\in H. Then ∂f is a maximal monotone operator of H into itself; see [23] for more details. Let C be a nonempty closed convex subset of H and {i}_{C} be the indicator function of C, that is,
{i}_{C}x=\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ \mathrm{\infty},\hfill & x\notin C.\hfill \end{array}
Furthermore, we define the normal cone {N}_{C}(v) of C at v as follows:
{N}_{C}v=\{z\in H:\u3008z,yv\u3009\le 0,\mathrm{\forall}y\in H\}
for any v\in C. Then {i}_{C}:H\to (\mathrm{\infty},\mathrm{\infty}] is a proper lower semicontinuous convex function on H and \partial {i}_{C} is a maximal monotone operator. Let {J}_{\lambda}x={(I+\lambda \partial {i}_{C})}^{1}x for any \lambda >0 and x\in H. From \partial {i}_{C}x={N}_{C}x and x\in C, we get
\begin{array}{rcl}v={J}_{\lambda}x\phantom{\rule{1em}{0ex}}& \iff & \phantom{\rule{1em}{0ex}}x\in v+\lambda {N}_{C}v\\ \iff & \phantom{\rule{1em}{0ex}}\u3008xv,yv\u3009\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in C,\\ \iff & \phantom{\rule{1em}{0ex}}v={P}_{C}x,\end{array}
where {P}_{C} is the metric projection from H into C. Similarly, we can get that x\in {(A+\partial {i}_{C})}^{1}(0)\iff x\in \mathit{VI}(A,C). Putting B=\partial {i}_{C} in Theorem 2.1, we can see {J}_{{\lambda}_{n}}={P}_{C}. The following is not hard to derive.
Corollary 2.3 Let C be a nonempty closed convex subset of a real Hilbert space H, A:C\to H be an αinversestrongly monotone mapping, and S:C\to C be a quasinonexpansive mapping such that IS is demiclosed at zero. Assume that \mathcal{F}=F(S)\cap \mathit{VI}(C,A)\ne \mathrm{\varnothing}. Let \{{\lambda}_{n}\} be a positive real number sequence. Let \{{\alpha}_{n}\} be a real number sequence in [0,1]. Let \{{x}_{n}\} be a sequence in C generated in the following iterative process:
\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {C}_{1}=C,\hfill \\ {y}_{n}={\alpha}_{n}{x}_{n}+(1{\alpha}_{n})S{P}_{C}({x}_{n}{\lambda}_{n}A{x}_{n}),\hfill \\ {C}_{n+1}=\{z\in {C}_{n}:\parallel {y}_{n}z\parallel \le \parallel {x}_{n}z\parallel \},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 1.\hfill \end{array}
Suppose that the sequences \{{\alpha}_{n}\} and \{{\lambda}_{n}\} satisfy the following restrictions:

(a)
0\le {\alpha}_{n}\le a<1;

(b)
0<b\le {\lambda}_{n}\le c<2\alpha.
Then the sequence \{{x}_{n}\} converges strongly to {P}_{\mathcal{F}}{x}_{1}.
In view of Corollary 2.3, we have the following corollary on variational inequalities.
Corollary 2.4 Let C be a nonempty closed convex subset of a real Hilbert space H and A:C\to H be an αinversestrongly monotone mapping. Assume that \mathcal{F}=\mathit{VI}(C,A)\ne \mathrm{\varnothing}. Let \{{\lambda}_{n}\} be a positive real number sequence. Let \{{\alpha}_{n}\} be a real number sequence in [0,1]. Let \{{x}_{n}\} be a sequence in C generated in the following iterative process:
\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {C}_{1}=C,\hfill \\ {y}_{n}={\alpha}_{n}{x}_{n}+(1{\alpha}_{n}){P}_{C}({x}_{n}{\lambda}_{n}A{x}_{n}),\hfill \\ {C}_{n+1}=\{z\in {C}_{n}:\parallel {y}_{n}z\parallel \le \parallel {x}_{n}z\parallel \},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 1.\hfill \end{array}
Suppose that the sequences \{{\alpha}_{n}\} and \{{\lambda}_{n}\} satisfy the following restrictions:

(a)
0\le {\alpha}_{n}\le a<1;

(b)
0<b\le {\lambda}_{n}\le c<2\alpha.
Then the sequence \{{x}_{n}\} converges strongly to {P}_{\mathit{VI}(C,A)}{x}_{1}.