## 1 Introduction

Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces and let $C\subset {H}_{1}$ and $Q\subset {H}_{2}$ be two nonempty, closed, and convex sets. Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint ${A}^{\ast }$. Let $S:{H}_{2}\to {H}_{2}$ and $T:{H}_{1}\to {H}_{1}$ be two nonlinear mappings.

The purpose of this paper is to study the following split feasibility problem and fixed point problem:

(1.1)

Special cases:

(i) Finding a point ${x}^{\ast }$ which satisfies

${x}^{\ast }\in C\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}A{x}^{\ast }\in Q.$
(1.2)

This problem, referred to as the split feasibility problem, was introduced by Censor and Elfving [1], modeling phase retrieval and other image restoration problems, and further studied by many researchers; see, for instance, [27].

(ii) Find a point ${x}^{\ast }$ with the property

${x}^{\ast }\in Fix\left(T\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}A{x}^{\ast }\in Fix\left(S\right).$
(1.3)

This problem, referred to as the split common fixed point problem, was first introduced by Censor and Segal [8].

Next, we recall some existing algorithms for solving (1.1)-(1.3) in the literature.

In order to solve (1.2), Censor and Elfving [1] introduced the following algorithm:

${x}_{n+1}={A}^{-1}{P}_{Q}\left({P}_{A\left(C\right)}\left(A{x}_{n}\right)\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$
(1.4)

where C and Q are closed and convex sets in ${\mathbb{R}}^{n}$, A is a full rank $n×n$ matrix and $A\left(C\right)=\left\{y\in {\mathbb{R}}^{n}\mid y=Ax,x\in C\right\}$.

Now (1.4) is not popular because it involves the computation of the inverse ${A}^{-1}$.

A more popular algorithm that solves (1.4) seems to be the CQ algorithm presented by Byrne [5, 7]:

${x}_{n+1}={P}_{C}\left({x}_{n}-\tau {A}^{\ast }\left(I-{P}_{Q}\right)A{x}_{n}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$
(1.5)

where $\tau \in \left(0,\frac{2}{L}\right)$, with L being the largest eigenvalue of the matrix ${A}^{\ast }A$.

Note that ${x}^{\ast }$ solves (1.2) if and only if ${x}^{\ast }$ solves the fixed point equation

${x}^{\ast }={P}_{C}\left(I-\lambda {A}^{\ast }\left(I-{P}_{Q}\right)A\right){x}^{\ast }.$
(1.6)

The above equivalence relation (1.6) reminds us to use fixed point method to solve (1.2). Many authors have given a continuation of the study on the CQ algorithm and its variant form. For related work, please refer to [916]. Especially, the following regularized method was presented by Xu [6]:

${x}_{n+1}={P}_{C}\left(\left(1-{\alpha }_{n}{\gamma }_{n}\right){x}_{n}-{\gamma }_{n}{A}^{\ast }\left(I-{P}_{Q}\right)A{x}_{n}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$
(1.7)

It should be pointed out that (1.7) can be used to find the minimum norm solution of (1.2).

For solving (1.3), Censor and Segal [8] invented an algorithm which generates a sequence $\left\{{x}_{n}\right\}$ according to the iterative procedure:

${x}_{n+1}=T\left({x}_{n}-\gamma {A}^{\ast }\left(I-S\right)A{x}_{n}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$
(1.8)

Note that (1.8) is more general than (1.5). Some further generations of this algorithm were studied by Moudafi [17] and Wang and Xu [18] and others; see, for example, [1922].

Motivated by the results in this direction, the purpose of this paper is to study the split feasibility problem and the fixed point problem involved in the pseudocontractive mappings. We construct an iterative algorithm and prove its strong convergence.

## 2 Preliminaries

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty, closed, and convex subset of H.

Recall that a mapping $T:C\to C$ is called pseudocontractive if

$〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^{2}$

for all $x,y\in C$. It is well known that T is pseudocontractive if and only if

${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$
(2.1)

for all $x,y\in C$. A mapping $T:C\to C$ is called L-Lipschitzian if there exists $L>0$ such that

$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel$

for all $x,y\in C$. If $L=1$, we call T nonexpansive.

We will use $Fix\left(T\right)$ to denote the set of fixed points of T, that is,

$Fix\left(T\right)=\left\{x\in C:x=Tx\right\}.$

We know that the metric projection ${P}_{C}:H\to C$ satisfies

$\parallel x-{P}_{C}\left(x\right)\parallel =inf\left\{\parallel x-y\parallel :y\in C\right\}.$

It is well known that the metric projection ${P}_{C}:H\to C$ is firmly nonexpansive, that is,

$\begin{array}{r}〈x-y,{P}_{C}\left(x\right)-{P}_{C}\left(y\right)〉\ge {\parallel {P}_{C}\left(x\right)-{P}_{C}\left(y\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}{\parallel {P}_{C}\left(x\right)-{P}_{C}\left(y\right)\parallel }^{2}\le {\parallel x-y\parallel }^{2}-{\parallel \left(I-{P}_{C}\right)x-\left(I-{P}_{C}\right)y\parallel }^{2}\end{array}$
(2.2)

for all $x,y\in H$.

For all $x,y\in H$, the following conclusions hold:

${\parallel tx+\left(1-t\right)y\parallel }^{2}=t{\parallel x\parallel }^{2}+\left(1-t\right){\parallel y\parallel }^{2}-t\left(1-t\right){\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],$
(2.3)
${\parallel x+y\parallel }^{2}={\parallel x\parallel }^{2}+2〈x,y〉+{\parallel y\parallel }^{2},$
(2.4)

and

${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,x+y〉.$
(2.5)

Lemma 2.1 ([23])

Let H be a real Hilbert space, C a closed convex subset of H. Let $T:C\to C$ be a continuous pseudocontractive mapping. Then

1. (i)

$Fix\left(T\right)$ is a closed convex subset of C,

2. (ii)

$\left(I-T\right)$ is demiclosed at zero.

Lemma 2.2 ([24])

Assume that $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that

${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n},\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$

where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that

1. (1)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (2)

${lim sup}_{n\to \mathrm{\infty }}\frac{{\delta }_{n}}{{\gamma }_{n}}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }$.

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

Lemma 2.3 ([25])

Let $\left\{{w}_{n}\right\}$ be a sequence of real numbers. Assume $\left\{{w}_{n}\right\}$ does not decrease at infinity, that is, there exists at least a subsequence $\left\{{w}_{{n}_{k}}\right\}$ of $\left\{{w}_{n}\right\}$ such that ${w}_{{n}_{k}}\le {w}_{{n}_{k}+1}$ for all $k\ge 0$. For every $n\ge {N}_{0}$, define an integer sequence $\left\{\tau \left(n\right)\right\}$ as

$\tau \left(n\right)=max\left\{i\le n:{w}_{{n}_{i}}<{w}_{{n}_{i}+1}\right\}.$

Then $\tau \left(n\right)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and for all $n\ge {N}_{0}$

$max\left\{{w}_{\tau \left(n\right)},{w}_{n}\right\}\le {w}_{\tau \left(n\right)+1}.$

## 3 Main results

Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces and let $C\subset {H}_{1}$ and $Q\subset {H}_{2}$ be two nonempty closed convex sets. Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint ${A}^{\ast }$. Let $S:{H}_{2}\to {H}_{2}$ nonexpansive mapping and let $T:{H}_{1}\to {H}_{1}$ be an L-Lipschitzian pseudocontractive mapping with $L>1$.

We use Γ to denote the set of solutions of (1.1), that is,

$\mathrm{\Gamma }=\left\{{x}^{\ast }\mid {x}^{\ast }\in C\cap Fix\left(T\right),A{x}^{\ast }\in Q\cap Fix\left(S\right)\right\}.$

In the sequel, we assume $\mathrm{\Gamma }\ne \mathrm{\varnothing }$.

Now, we present our algorithm for finding ${x}^{\ast }\in \mathrm{\Gamma }$.

Algorithm 3.1 For fixed $u\in {H}_{1}$ and ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

$\left\{\begin{array}{c}{u}_{n}={P}_{C}\left[{\alpha }_{n}u+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}\right)\right],\hfill \\ {x}_{n+1}=\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right),\hfill & n\in \mathbb{N},\hfill \end{array}$
(3.1)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$, ${\left\{{\beta }_{n}\right\}}_{n\in \mathbb{N}}$, and ${\left\{{\gamma }_{n}\right\}}_{n\in \mathbb{N}}$ are three real number sequences in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Theorem 3.2 Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

(C3) $0.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.1) converges strongly to the point ${x}^{\ast }$, given by ${x}^{\ast }={P}_{\mathrm{\Gamma }}\left(u\right)$.

Proof Set ${x}^{\ast }={P}_{\mathrm{\Gamma }}\left(u\right)$. Then we have ${x}^{\ast }\in C\cap Fix\left(T\right)$ and $A{x}^{\ast }\in Q\cap Fix\left(S\right)$. Set ${z}_{n}={P}_{Q}A{x}_{n}$ and ${y}_{n}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}\right)$ for all $n\in \mathbb{N}$. Thus ${u}_{n}={P}_{C}{y}_{n}$ for all $n\in \mathbb{N}$.

Since ${P}_{C}$ and ${P}_{Q}$ are nonexpansive, we have

$\parallel {z}_{n}-A{x}^{\ast }\parallel =\parallel {P}_{Q}A{x}_{n}-{P}_{Q}A{x}^{\ast }\parallel \le \parallel A{x}_{n}-A{x}^{\ast }\parallel$
(3.2)

and

$\parallel {u}_{n}-{x}^{\ast }\parallel =\parallel {P}_{C}{y}_{n}-{P}_{C}{x}^{\ast }\parallel \le \parallel {y}_{n}-{x}^{\ast }\parallel .$
(3.3)

By (2.2), we get

$\begin{array}{rl}{\parallel S{z}_{n}-A{x}^{\ast }\parallel }^{2}& ={\parallel S{P}_{Q}A{x}_{n}-S{P}_{Q}A{x}^{\ast }\parallel }^{2}\\ \le {\parallel {P}_{Q}A{x}_{n}-{P}_{Q}A{x}^{\ast }\parallel }^{2}\\ \le {\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-A{x}_{n}\parallel }^{2}.\end{array}$
(3.4)

From (2.1), we have

${\parallel T{u}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel T{u}_{n}-{u}_{n}\parallel }^{2}$
(3.5)

and

$\begin{array}{r}{\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel \left(1-{\gamma }_{n}\right)\left({u}_{n}-{x}^{\ast }\right)+{\gamma }_{n}\left(T{u}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\parallel \left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}.\end{array}$
(3.6)

Applying equality (2.3), we have

$\begin{array}{r}{\parallel \left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left(1-{\gamma }_{n}\right)\left({u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\right)+{\gamma }_{n}\left(T{u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}+{\gamma }_{n}{\parallel T{u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}.\end{array}$
(3.7)

Since T is L-Lipschitzian and ${u}_{n}-\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)={\gamma }_{n}\left({u}_{n}-T{u}_{n}\right)$, by (3.7), we get

$\begin{array}{r}{\parallel \left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}+{\gamma }_{n}^{3}{L}^{2}{\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}+\left({\gamma }_{n}^{3}{L}^{2}+{\gamma }_{n}^{2}-{\gamma }_{n}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}.\end{array}$
(3.8)

By (2.3) and (3.5), we have

$\begin{array}{r}{\parallel \left(1-{\gamma }_{n}\right)\left({u}_{n}-{x}^{\ast }\right)+{\gamma }_{n}\left(T{u}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{\parallel T{u}_{n}-{x}^{\ast }\parallel }^{2}-{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\gamma }_{n}\right){\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}\left({\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}-{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}^{2}{\parallel {u}_{n}-T{u}_{n}\parallel }^{2}.\end{array}$
(3.9)

From (3.6), (3.8), and (3.9), we deduce

$\begin{array}{r}{\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-{\gamma }_{n}\left(1-2{\gamma }_{n}-{\gamma }_{n}^{2}{L}^{2}\right){\parallel {u}_{n}-T{u}_{n}\parallel }^{2}.\end{array}$
(3.10)

Since ${\gamma }_{n}, we derive that

$1-2{\gamma }_{n}-{\gamma }_{n}^{2}{L}^{2}>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$

This together with (3.10) implies that

$\begin{array}{r}{\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}.\end{array}$
(3.11)

By (2.3), (3.1), (3.11), and (C3), we have

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}=& {\parallel \left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ =& \left(1-{\beta }_{n}\right){\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ -{\beta }_{n}\left(1-{\beta }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \le & {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}-{\beta }_{n}\left({\gamma }_{n}-{\beta }_{n}\right){\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \le & {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}.\end{array}$
(3.12)

By the convexity of the norm and by using (2.4), we get

$\begin{array}{rl}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}=& {\parallel {\alpha }_{n}\left(u-{x}^{\ast }\right)+\left(1-{\alpha }_{n}\right)\left({x}_{n}-{x}^{\ast }+\delta {A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)\right)\parallel }^{2}\\ \le & \left(1-{\alpha }_{n}\right){\parallel \left({x}_{n}-{x}^{\ast }+\delta {A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)\right)\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}\\ =& \left(1-{\alpha }_{n}\right)\left[\parallel {x}_{n}-{x}^{\ast }\parallel +{\delta }^{2}{\parallel {A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)\parallel }^{2}\\ +2\delta 〈{x}_{n}-{x}^{\ast },{A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)〉\right]+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}.\end{array}$
(3.13)

Since A is a linear operator with its adjoint ${A}^{\ast }$, we have

$\begin{array}{r}〈{x}_{n}-{x}^{\ast },{A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)〉\\ \phantom{\rule{1em}{0ex}}=〈A\left({x}_{n}-{x}^{\ast }\right),S{z}_{n}-A{x}_{n}〉\\ \phantom{\rule{1em}{0ex}}=〈A{x}_{n}-A{x}^{\ast }+S{z}_{n}-A{x}_{n}-\left(S{z}_{n}-A{x}_{n}\right),S{z}_{n}-A{x}_{n}〉\\ \phantom{\rule{1em}{0ex}}=〈S{z}_{n}-A{x}^{\ast },S{z}_{n}-A{x}_{n}〉-{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}.\end{array}$
(3.14)

Again using (2.4), we obtain

$〈S{z}_{n}-A{x}^{\ast },S{z}_{n}-A{x}_{n}〉=\frac{1}{2}\left({\parallel S{z}_{n}-A{x}^{\ast }\parallel }^{2}+{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}-{\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\right).$
(3.15)

From (3.4), (3.14), and (3.15), we get

$\begin{array}{rl}〈{x}_{n}-{x}^{\ast },{A}^{\ast }\left(S{z}_{n}-A{x}_{n}\right)〉=& \frac{1}{2}\left({\parallel S{z}_{n}-A{x}^{\ast }\parallel }^{2}+{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}-{\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\right)\\ -{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ \le & \frac{1}{2}\left({\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-A{x}_{n}\parallel }^{2}+{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ -{\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\right)-{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ =& -\frac{1}{2}{\parallel {z}_{n}-A{x}_{n}\parallel }^{2}-\frac{1}{2}{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}.\end{array}$
(3.16)

Substituting (3.16) into (3.13) we deduce

$\begin{array}{rl}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\le & \left(1-{\alpha }_{n}\right)\left[{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\delta }^{2}{\parallel A\parallel }^{2}{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ +2\delta \left(-\frac{1}{2}{\parallel {z}_{n}-A{x}_{n}\parallel }^{2}-\frac{1}{2}{\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\right)\right]+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}\\ =& \left(1-{\alpha }_{n}\right)\left[{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\left({\delta }^{2}{\parallel A\parallel }^{2}-\delta \right){\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ -\delta {\parallel {z}_{n}-A{x}_{n}\parallel }^{2}\right]+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}\\ \le & \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}.\end{array}$
(3.17)

From (3.3), (3.12), and (3.17), we get

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& \le {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ \le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}\\ \le max\left\{{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2},{\parallel u-{x}^{\ast }\parallel }^{2}\right\}.\end{array}$

The boundedness of the sequence $\left\{{x}_{n}\right\}$ yields our result.

Using the firmly nonexpansivenessity of ${P}_{C}$ (2.2), we have

$\begin{array}{rl}{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}& ={\parallel {P}_{C}{y}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {P}_{C}{y}_{n}-{y}_{n}\parallel }^{2}\\ ={\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{y}_{n}\parallel }^{2}.\end{array}$
(3.18)

Thus

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& \le {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}\\ \le {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{y}_{n}\parallel }^{2}\\ \le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{y}_{n}\parallel }^{2}.\end{array}$

It follows that

${\parallel {u}_{n}-{y}_{n}\parallel }^{2}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}.$
(3.19)

Next, we consider two possible cases.

Case 1. Assume there exists some integer $m>0$ such that $\left\{\parallel {x}_{n}-{x}^{\ast }\parallel \right\}$ is decreasing for all $n\ge m$. In this case, we know that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}^{\ast }\parallel$ exists. From (3.19), we deduce

$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{y}_{n}\parallel =0.$
(3.20)

Returning to (3.17), we have

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le & {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ \le & \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({\delta }^{2}{\parallel A\parallel }^{2}-\delta \right){\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}\\ -\left(1-{\alpha }_{n}\right)\delta {\parallel {z}_{n}-A{x}_{n}\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}.\end{array}$

Hence,

$\begin{array}{r}\left(1-{\alpha }_{n}\right)\left(\delta -{\delta }^{2}{\parallel A\parallel }^{2}\right){\parallel S{z}_{n}-A{x}_{n}\parallel }^{2}+\left(1-{\alpha }_{n}\right)\delta {\parallel {z}_{n}-A{x}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2},\end{array}$

which implies that

$\underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_{n}-A{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-A{x}_{n}\parallel =0.$
(3.21)

So,

$\underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_{n}-{z}_{n}\parallel =0.$
(3.22)

Note that

$\begin{array}{rl}\parallel {y}_{n}-{x}_{n}\parallel & =\parallel \delta {A}^{\ast }\left(S{P}_{Q}-I\right)A{x}_{n}+{\alpha }_{n}\left({x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}-u\right)\parallel \\ \le \delta \parallel A\parallel \parallel S{z}_{n}-A{x}_{n}\parallel +{\alpha }_{n}\parallel {x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}-u\parallel .\end{array}$

It follows from (3.21) that

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{y}_{n}\parallel =0.$
(3.23)

From (3.3), (3.12), and (3.17), we deduce

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le & {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}-{\beta }_{n}\left({\gamma }_{n}-{\beta }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}\\ -{\beta }_{n}\left({\gamma }_{n}-{\beta }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}.\end{array}$

It follows that

$\begin{array}{r}{\beta }_{n}\left({\gamma }_{n}-{\beta }_{n}\right){\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}.\end{array}$

Therefore,

$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel =0.$
(3.24)

Observe that

$\begin{array}{rcl}\parallel {u}_{n}-T{u}_{n}\parallel & \le & \parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel +\parallel T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)-T{u}_{n}\parallel \\ \le & \parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel +L{\gamma }_{n}\parallel {u}_{n}-T{u}_{n}\parallel .\end{array}$

Thus,

$\parallel {u}_{n}-T{u}_{n}\parallel \le \frac{1}{1-L{\gamma }_{n}}\parallel {u}_{n}-T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right)\parallel .$

This together with (3.24) implies that

$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-T{u}_{n}\parallel =0.$
(3.25)

Now, we show that

$\underset{n\to \mathrm{\infty }}{lim sup}〈u-{x}^{\ast },{y}_{n}-{x}^{\ast }〉\le 0.$

Choose a subsequence $\left\{{y}_{{n}_{i}}\right\}$ of $\left\{{y}_{n}\right\}$ such that

$\underset{n\to \mathrm{\infty }}{lim sup}〈u-{x}^{\ast },{y}_{n}-{x}^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈u-{x}^{\ast },{y}_{{n}_{i}}-{x}^{\ast }〉.$
(3.26)

Since the sequence $\left\{{y}_{{n}_{i}}\right\}$ is bounded, we can choose a subsequence $\left\{{y}_{{n}_{{i}_{j}}}\right\}$ of $\left\{{y}_{{n}_{i}}\right\}$ such that ${y}_{{n}_{{i}_{j}}}⇀z$. For the sake of convenience, we assume (without loss of generality) that ${y}_{{n}_{i}}⇀z$. Consequently, we derive from the above conclusions that

${x}_{{n}_{i}}⇀z,\phantom{\rule{2em}{0ex}}{u}_{{n}_{i}}⇀z,\phantom{\rule{2em}{0ex}}A{x}_{{n}_{i}}⇀Az\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{{n}_{i}}⇀Az.$
(3.27)

Applying Lemma 2.1, we deduce

$z\in Fix\left(T\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Az\in Fix\left(S\right).$

Note that ${u}_{{n}_{i}}={P}_{C}{y}_{{n}_{i}}\in C$ and ${z}_{{n}_{i}}={P}_{Q}A{x}_{{n}_{i}}\in Q$. From (3.27), we deduce

$z\in C\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Az\in Q.$

To this end, we deduce

$z\in C\cap Fix\left(T\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Az\in Q\cap Fix\left(S\right).$

That is to say, $z\in \mathrm{\Gamma }$.

Therefore,

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim sup}〈u-{x}^{\ast },{y}_{n}-{x}^{\ast }〉& =\underset{i\to \mathrm{\infty }}{lim}〈u-{x}^{\ast },{y}_{{n}_{i}}-{x}^{\ast }〉\\ =\underset{i\to \mathrm{\infty }}{lim}〈u-{x}^{\ast },z-{x}^{\ast }〉\\ \le 0.\end{array}$
(3.28)

Using (2.5), we have

$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& \le {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ ={\parallel \left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}-{x}^{\ast }\right)+{\alpha }_{n}\left(u-{x}^{\ast }\right)\parallel }^{2}\\ \le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}-{x}^{\ast }\parallel }^{2}+2{\alpha }_{n}〈u-{x}^{\ast },{y}_{n}-{x}^{\ast }〉\\ \le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\alpha }_{n}〈u-{x}^{\ast },{y}_{n}-{x}^{\ast }〉.\end{array}$
(3.29)

Applying Lemma 2.2 and (3.28) to (3.29), we deduce ${x}_{n}\to {x}^{\ast }$.

Case 2. Assume there exists an integer ${n}_{0}$ such that

$\parallel {x}_{{n}_{0}}-{x}^{\ast }\parallel \le \parallel {x}_{{n}_{0}+1}-{x}^{\ast }\parallel .$

Set ${\omega }_{n}=\left\{\parallel {x}_{n}-{x}^{\ast }\parallel \right\}$. Then we have

${\omega }_{{n}_{0}}\le {\omega }_{{n}_{0}+1}.$

Define an integer sequence $\left\{{\tau }_{n}\right\}$ for all $n\ge {n}_{0}$ as follows:

$\tau \left(n\right)=max\left\{l\in \mathbb{N}\mid {n}_{0}\le l\le n,{\omega }_{l}\le {\omega }_{l+1}\right\}.$

It is clear that $\tau \left(n\right)$ is a non-decreasing sequence satisfying

$\underset{n\to \mathrm{\infty }}{lim}\tau \left(n\right)=\mathrm{\infty }$

and

${\omega }_{\tau \left(n\right)}\le {\omega }_{\tau \left(n\right)+1},$

for all $n\ge {n}_{0}$.

By a similar argument to that of Case 1, we can obtain

$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{\tau \left(n\right)}-{y}_{\tau \left(n\right)}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{\tau \left(n\right)}-{y}_{\tau \left(n\right)}\parallel =0,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_{\tau \left(n\right)}-A{x}_{\tau \left(n\right)}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{\tau \left(n\right)}-A{x}_{\tau \left(n\right)}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_{\tau \left(n\right)}-{z}_{\tau \left(n\right)}\parallel =0,\hfill \end{array}$

and

$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{\tau \left(n\right)}-T{u}_{\tau \left(n\right)}\parallel =0.$

This implies that

${\omega }_{w}\left({y}_{\tau \left(n\right)}\right)\subset \mathrm{\Gamma }.$

Thus, we obtain

$\underset{n\to \mathrm{\infty }}{lim sup}〈u-{x}^{\ast },{y}_{\tau \left(n\right)}-{x}^{\ast }〉\le 0.$
(3.30)

Since ${\omega }_{\tau \left(n\right)}\le {\omega }_{\tau \left(n\right)+1}$, we have from (3.29) that

${\omega }_{\tau \left(n\right)}^{2}\le {\omega }_{\tau \left(n\right)+1}^{2}\le \left(1-{\alpha }_{\tau \left(n\right)}\right){\omega }_{\tau \left(n\right)}^{2}+2{\alpha }_{\tau \left(n\right)}〈u-{x}^{\ast },{y}_{\tau \left(n\right)}-{x}^{\ast }〉.$
(3.31)

It follows that

${\omega }_{\tau \left(n\right)}^{2}\le 2〈u-{x}^{\ast },{y}_{\tau \left(n\right)}-{x}^{\ast }〉.$
(3.32)

Combining (3.30) and (3.32), we have

$\underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)}\le 0,$

and hence

$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau \left(n\right)}=0.$
(3.33)

By (3.31), we obtain

$\underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)+1}^{2}\le \underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)}^{2}.$

This together with (3.33) implies that

$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau \left(n\right)+1}=0.$

Applying Lemma 2.3 to get

$0\le {\omega }_{n}\le max\left\{{\omega }_{\tau \left(n\right)},{\omega }_{\tau \left(n\right)+1}\right\}.$

Therefore, ${\omega }_{n}\to 0$. That is, ${x}_{n}\to {x}^{\ast }$. This completes the proof. □

Algorithm 3.3 For ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

$\left\{\begin{array}{c}{u}_{n}={P}_{C}\left[\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S{P}_{Q}\right)A{x}_{n}\right)\right],\hfill \\ {x}_{n+1}=\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right),\hfill & n\in \mathbb{N},\hfill \end{array}$
(3.34)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$, ${\left\{{\beta }_{n}\right\}}_{n\in \mathbb{N}}$, and ${\left\{{\gamma }_{n}\right\}}_{n\in \mathbb{N}}$ are three real number sequences in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Corollary 3.4 Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

(C3) $0.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.34) converges strongly to ${x}^{\ast }={P}_{\mathrm{\Gamma }}\left(0\right)$, which is the minimum norm in Γ.

Algorithm 3.5 For fixed $u\in {H}_{1}$ and ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

${x}_{n+1}={P}_{C}\left[{\alpha }_{n}u+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-{P}_{Q}\right)A{x}_{n}\right)\right],\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$
(3.35)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$ is a real number sequence in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Corollary 3.6 Suppose ${\mathrm{\Gamma }}_{1}$, the set of the solutions of (1.2), is nonempty. Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.35) converges strongly to ${x}^{\ast }={P}_{{\mathrm{\Gamma }}_{1}}\left(u\right)$.

Algorithm 3.7 For ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

${x}_{n+1}={P}_{C}\left[\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-{P}_{Q}\right)A{x}_{n}\right)\right],\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$
(3.36)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$ is a real number sequence in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Corollary 3.8 Suppose ${\mathrm{\Gamma }}_{1}$, the set of the solutions of (1.2), is nonempty. Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.36) converges strongly to ${x}^{\ast }={P}_{{\mathrm{\Gamma }}_{1}}\left(0\right)$, which is the minimum norm in ${\mathrm{\Gamma }}_{1}$.

Algorithm 3.9 For fixed $u\in {H}_{1}$ and ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

$\left\{\begin{array}{c}{u}_{n}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S\right)A{x}_{n}\right),\hfill \\ {x}_{n+1}=\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right),\hfill & n\in \mathbb{N},\hfill \end{array}$
(3.37)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$, ${\left\{{\beta }_{n}\right\}}_{n\in \mathbb{N}}$, and ${\left\{{\gamma }_{n}\right\}}_{n\in \mathbb{N}}$ are three real number sequences in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Corollary 3.10 Suppose ${\mathrm{\Gamma }}_{2}$, the set of the solutions of (1.3), is nonempty. Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

(C3) $0.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.37) converges strongly to ${x}^{\ast }={P}_{{\mathrm{\Gamma }}_{2}}\left(u\right)$.

Algorithm 3.11 For and ${x}_{0}\in {H}_{1}$ arbitrarily, let $\left\{{x}_{n}\right\}$ be a sequence defined by

$\left\{\begin{array}{c}{u}_{n}=\left(1-{\alpha }_{n}\right)\left({x}_{n}-\delta {A}^{\ast }\left(I-S\right)A{x}_{n}\right),\hfill \\ {x}_{n+1}=\left(1-{\beta }_{n}\right){u}_{n}+{\beta }_{n}T\left(\left(1-{\gamma }_{n}\right){u}_{n}+{\gamma }_{n}T{u}_{n}\right),\hfill & n\in \mathbb{N},\hfill \end{array}$
(3.38)

where ${\left\{{\alpha }_{n}\right\}}_{n\in \mathbb{N}}$, ${\left\{{\beta }_{n}\right\}}_{n\in \mathbb{N}}$, and ${\left\{{\gamma }_{n}\right\}}_{n\in \mathbb{N}}$ are three real number sequences in $\left(0,1\right)$ and δ is a constant in $\left(0,\frac{1}{{\parallel A\parallel }^{2}}\right)$.

Corollary 3.12 Suppose ${\mathrm{\Gamma }}_{2}$, the set of the solutions of (1.3), is nonempty. Assume the following conditions are satisfied:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

(C3) $0.

Then the sequence $\left\{{x}_{n}\right\}$ generated by algorithm (3.38) converges strongly to ${x}^{\ast }={P}_{{\mathrm{\Gamma }}_{2}}\left(0\right)$ which is the minimum norm in ${\mathrm{\Gamma }}_{2}$.

Example 3.13 Let ${H}_{1}={H}_{2}=\mathbb{R}$ with the inner product defined by $〈x,y〉=xy$ for all $x,y\in \mathbb{R}$ and the standard norm $|\cdot |$. Let $C=\left[0,\mathrm{\infty }\right)$ and $Q=\mathbb{R}$. Let $Sx=\frac{x}{2}-1$ for all $x\in Q$ and let $Tx=x-1+\frac{4}{x+1}$ for all $x\in C$. Let $Ax=-\frac{2x}{3}$ for all $x\in \mathbb{R}$. Then A is a bounded linear operator with its adjoint ${A}^{\ast }=A$. Observe that $Fix\left(T\right)=3$ and $Fix\left(S\right)=-2$. It is easy to see that

$\begin{array}{rl}〈Tx-Ty,x-y〉& =〈x-1+\frac{4}{x+1}-y+1-\frac{4}{y+1},x-y〉\\ \le \left[1-\frac{4}{\left(x+1\right)\left(y+1\right)}\right]{|x-y|}^{2}\\ \le {|x-y|}^{2},\end{array}$

and

$\begin{array}{rcl}|Tx-Ty|& \le & |x-1+\frac{4}{x+1}-y+1-\frac{4}{y+1}|\\ \le & |1-\frac{4}{\left(x+1\right)\left(y+1\right)}||x-y|\\ \le & 5|x-y|,\end{array}$

for all $x,y\in C$.

But

$|T\left(\frac{1}{4}\right)-T\left(0\right)|=\frac{11}{20}>\frac{1}{4}.$

Hence, T is a Lipschitzian pseudocontractive mapping but not a nonexpansive one.

Note that $\parallel A\parallel =\parallel {A}^{\ast }\parallel =\frac{2}{3}$. Let $u=3$ and $\delta =\frac{9}{8}$. Then we have

$\begin{array}{rcl}{u}_{n}& =& {P}_{C}\left[3{\alpha }_{n}+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\frac{9}{8}×\left(-\frac{2I}{3}\right)×\left(\frac{I}{2}+1\right)×\left(\frac{-2{x}_{n}}{3}\right)\right)\right]\\ =& {P}_{C}\left[3{\alpha }_{n}+\left(1-{\alpha }_{n}\right)\left(\frac{3}{4}{x}_{n}+\frac{3}{4}\right)\right].\end{array}$

Let ${\beta }_{n}=\frac{1}{8}$ and ${\gamma }_{n}=\frac{1}{7}$ for all n. It is not hard to compute that

$\begin{array}{rl}|{x}_{n+1}-3|& \le |{u}_{n}-3|\\ \le \left(1-{\alpha }_{n}\right)|\frac{3}{4}{x}_{n}+\frac{3}{4}-3|\\ \le \frac{3}{4}|{x}_{n}-3|\\ \le \cdots \\ \le {\left(\frac{3}{4}\right)}^{n}|{x}_{1}-3|,\end{array}$

which shows ${x}_{n}\to 3\in \mathrm{\Gamma }$.