Non-convex hybrid algorithm for a family of countable quasi-Lipschitz mappings and application

Abstract

The purpose of this article is to establish a kind of non-convex hybrid iteration algorithms and to prove relevant strong convergence theorems of common fixed points for a uniformly closed asymptotically family of countable quasi-Lipschitz mappings in Hilbert spaces. Meanwhile, the main result is applied to get the common fixed points of finite family of quasi-asymptotically nonexpansive mappings. It is worth pointing out that a non-convex hybrid iteration algorithm is first presented in this article, a new technique is applied in our process of proof. Finally, an example is given which is a uniformly closed asymptotically family of countable quasi-Lipschitz mappings. The results presented in this article are interesting extensions of some current results.

1 Introduction

Construction of fixed points of nonexpansive mappings (and asymptotically nonexpansive mappings) is an important subject in the theory of nonexpansive mappings and finds application in a number of applied areas. Recently, a great deal of literature on iteration algorithms for approximating fixed points of nonexpansive mappings has been published since one has a variety of applications in inverse problem, image recovery, and signal processing; see [1–8]. Mann’s iteration process [1] is often used to approximate a fixed point of the operators, but it has only weak convergence (see [3] for an example). However, strong convergence is often much more desirable than weak convergence in many problems that arise in infinite dimensional spaces (see [7] and references therein). So, attempts have been made to modify Mann’s iteration process so that strong convergence is guaranteed (see [9–24] and references therein).

In 2003, Nakajo and Takahashi [25] proposed a modification of Mann iteration method for a single nonexpansive mapping in a Hilbert space. In 2006, Kim and Xu [26] proposed a modification of Mann iteration method for asymptotically nonexpansive mapping T in a Hilbert space. They also proposed a modification of the Mann iteration method for asymptotically nonexpansive semigroup in a Hilbert space. In 2006, Martinez-Yanes and Xu [27] proposed a modification of the Ishikawa iteration method for nonexpansive mapping in a Hilbert space. Martinez-Yanes and Xu [27] proposed also a modification of the Halpern iteration method for nonexpansive mapping in a Hilbert space. In 2008, Su and Qin [28] proposed first a monotone hybrid iteration method for nonexpansive mapping in a Hilbert space. In 2015, Dong and Lu [29] proposed a new iteration method for nonexpansive mapping in a Hilbert space. In 2015, Liu et al. [30] proposed a new iteration method for a finite family of quasi-asymptotically pseudocontractive mappings in a Hilbert spaces.

Throughout this paper, let H be a real Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and norm $\|\cdot\|$. We write $x_{n} \rightarrow x$ to indicate that the sequence $\{x_{n}\}$ converges strongly to x. We write $x_{n} \rightharpoonup x$ to indicate that the sequence $\{x_{n}\}$ converges weakly to x. Let C be a nonempty, closed, and convex subset of H, we denote by $P_{C}(\cdot)$ the metric projection onto C. It is well known that $z = P_{C}(x)$ is equivalent to that $z\in C$ and $\langle z-y,x- z\rangle\geq0$ for every $y\in C$. Recall that $T:C\rightarrow C$ is nonexpansive if $\|Tx-Ty\|\leq \|x-y\|$ for all $x,y \in C$. A point $x\in C$ is a fixed point of T provided $Tx=x$. Denote by $F(T)$ the set of fixed points of T, that is, $F(T)=\{x\in C:Tx=x\}$. It is well known that $F(T)$ is closed and convex. A mapping $T: C\rightarrow C$ is said to be quasi-Lipschitz, if the following conditions hold:

• (1) the fixed point set $F(T)$ is nonempty;

• (2) $\|Tx-p\|\leq L\|x-p\|$ for all $x \in C$, $p \in F(T)$,

where $1\leq L<+\infty$ is a constant. T is said to be quasi-nonexpansive, if $L=1$.

Recall that a mapping $T:C\rightarrow C$ is said to be closed if $x_{n}\rightarrow x$ and $\|Tx_{n}-x_{n}\|\rightarrow0$ as $n\rightarrow\infty$ implies $Tx=x$. A mapping $T:C\rightarrow C$ is said to be weak closed if $x_{n}\rightharpoonup x$ and $\|Tx_{n}-x_{n}\|\rightarrow0$ as $n\rightarrow\infty$ implies $Tx=x$. It is obvious that a weak closed mapping must be a closed mapping, the inverse is not true.

Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let $\{T_{n}\}$ be sequence of mappings from C into itself with a nonempty common fixed point set F. $\{T_{n}\}$ is said to be uniformly closed if for any convergent sequence $\{z_{n}\} \subset C$ such that $\|T_{n}z_{n}-z_{n}\|\rightarrow0$ as $n\rightarrow\infty$, the limit of $\{z_{n}\}$ belongs to F.

The purpose of this article is to establish a kind of non-convex hybrid iteration algorithms and to prove relevant strong convergence theorems of common fixed points for a uniformly closed asymptotically family of countable quasi-Lipschitz mappings in Hilbert spaces. Meanwhile, the main result was applied to get the common fixed points of finite family of quasi-asymptotically nonexpansive mappings. It is worth pointing out that a non-convex hybrid iteration algorithm was first presented in this article, a new technique has been applied in our process of proof. Finally, an example has been given which is a uniformly closed asymptotically family of countable quasi-Lipschitz mappings. The results presented in this article are interesting extensions of some current results.

2 Main results

The following lemma is well known and is useful for our conclusions.

Lemma 2.1

Let C be a nonempty, closed, and convex subset of real Hilbert space H. Given $x \in H$ and $z\in C$. Then $z = P_{C}x$ if and only if we have the relation

$$\langle x-z, z-y \rangle\geq0 $$

for all $y \in C$.

Definition 2.2

Let H be a Hilbert space, let C be a closed convex subset of E, and let $\{T_{n}\}$ be a family of countable quasi-$L_{n}$-Lipschitz mappings from C into itself, $\{T_{n}\}$ is said to be asymptotically, if $\lim_{n\rightarrow\infty}L_{n}=1$.

Lemma 2.3

Let H be a Hilbert space, let C be a closed convex subset of E, and let $\{T_{n}\}$ be a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings from C into itself. Then the common fixed point set F is closed and convex.

Proof

Let $p_{n} \in F$ and $p_{n}\rightarrow p$ as $n\rightarrow\infty$, we have

$$\|T_{n}p_{n}-p_{n}\|=0\rightarrow0, \quad p_{n}\rightarrow p $$

as $n\rightarrow\infty$. Since $\{T_{n}\}$ is uniformly closed, we know that $p \in F$, therefore F is closed. Next we show that F is also convex. For any $x,y \in F$, let $z=tx+(1-t)y$ for any $t \in(0,1)$, we have

$$\begin{aligned} \Vert T_{n}z-z\Vert ^{2} =& \langle T_{n}z-z,T_{n}z-z\rangle \\ =& \Vert T_{n}z\Vert ^{2}-2 \langle T_{n}z,z \rangle+ \Vert z\Vert ^{2} \\ =& \Vert T_{n}z\Vert ^{2}-2 \bigl\langle T_{n}z, tx+(1-t)y\bigr\rangle + \Vert z\Vert ^{2} \\ =& \Vert T_{n}z\Vert ^{2}-2 t\langle T_{n}z, x \rangle+2(1-t)\langle T_{n}z, y\rangle + \Vert z\Vert ^{2} \\ =&t \Vert T_{n}z-x\Vert ^{2}+(1-t)\Vert T_{n}z-y\Vert ^{2}-t\Vert x\Vert ^{2}-(1-t) \Vert y\Vert ^{2}+\Vert z\Vert ^{2} \\ \leq& t L_{n}^{2}\Vert z-x\Vert ^{2}+(1-t)L_{n}^{2} \Vert z-y\Vert ^{2}-t\Vert x\Vert ^{2}-(1-t)\Vert y \Vert ^{2}+\Vert z\Vert ^{2} \\ =& t \Vert z-x\Vert ^{2}+(1-t)\Vert z-y\Vert ^{2}-t\Vert x\Vert ^{2}-(1-t)\Vert y\Vert ^{2}+\Vert z\Vert ^{2} \\ &{} +t\bigl(L_{n}^{2}-1\bigr)\Vert z-x\Vert ^{2}+(1-t) \bigl(L_{n}^{2}-1\bigr)\Vert y-x \Vert ^{2} \\ =& \Vert z\Vert ^{2}- 2\langle z, z\rangle+\Vert z\Vert ^{2} \\ &{}+t\bigl(L_{n}^{2}-1\bigr)\Vert z-x\Vert ^{2}+(1-t) \bigl(L_{n}^{2}-1\bigr)\Vert y-x \Vert ^{2}\rightarrow0 \end{aligned}$$

as $n\rightarrow\infty$. Since $z\rightarrow z$, and $\{T_{n}\}$ is uniformly closed, $z\in F$. Therefore F is convex. This completes the proof. □

The following conclusion is well known.

Lemma 2.4

Let C be a closed convex subset of a Hilbert space H, for any given $x_{0} \in H$, we have

$$p=P_{C}x_{0}\quad \Leftrightarrow\quad \langle p-z, x_{0}-p\rangle\geq0, \quad \forall z \in C. $$

Theorem 2.5

Let C be a closed convex subset of a Hilbert space H, and let $\{T_{n}\} : C\rightarrow C$ be a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings from C into itself. Assume that $\alpha_{n} \in(a,1]$ holds for some $a \in(0,1)$. Then $\{x_{n}\}$ generated by

$$\textstyle\begin{cases} x_{0} \in C=Q_{0} \quad \textit{chosen arbitrarily}, \\ y_{n} =(1-\alpha_{n})x_{n}+ \alpha_{n}T_{n}x_{n},\quad n\geq0, \\ C_{n}=\{z\in C: \|y_{n}-z\|\leq(1+(L_{n}-1)\alpha_{n}) \|x_{n}-z\|\}\cap A,\quad n\geq0 , \\ Q_{n}=\{ z \in Q_{n-1}: \langle x_{n}-z, x_{0}-x_{n}\rangle\geq0 \}, \quad n\geq1, \\ x_{n+1}=P_{\overline{\operatorname{co}}\, C_{n}\cap Q_{n}}x_{0}, \end{cases} $$

converges strongly to $P_{F}x_{0}$, where $\overline{\operatorname{co}}\, C_{n}$ denotes the closed convex closure of $C_{n}$ for all $n\geq1$, $A=\{z \in H: \|z-P_{F}x_{0}\|\leq1\}$.

Proof

We split the proof into seven steps.

Step 1. It is obvious that $\overline{\operatorname{co}}\, C_{n}$, $Q_{n}$ are closed and convex for all $n\geq0$. Next, we show that $F\cap A\subset\overline{\operatorname{co}}\, C_{n}$ for all $n\geq0$. Indeed, for each $p\in F\cap A$, we have

$$\begin{aligned} \Vert y_{n}-p\Vert & =\bigl\Vert (1-\alpha_{n})x_{n}+ \alpha_{n}T_{n}x_{n}-p\bigr\Vert \\ &=\bigl\Vert \alpha_{n}(x_{n}-p)+(1-\alpha_{n}) (T_{n}x_{n}-p)\bigr\Vert \\ & \leq(1-\alpha_{n})\Vert x_{n}-p\Vert + \alpha_{n}L_{n}\Vert x_{n}-p\Vert \\ & = \bigl(1+(L_{n}-1)\alpha_{n}\bigr) \Vert x_{n}-z\Vert \end{aligned}$$

and $p\in A$, so $p\in C_{n}$ which implies that $F\cap A\subset C_{n}$ for all $n\geq0$. Therefore, $F\cap A\subset\overline{\operatorname{co}}\, C_{n}$ for all $n\geq0$.

Step 2. We show that $F\cap A \subset \overline{\operatorname{co}}\, C_{n}\cap Q_{n}$ for all $n\geq0$. It suffices to show that $F\cap A \subset Q_{n}$, for all $n\geq0$. We prove this by mathematical induction. For $n=0$, we have $F\cap A \subset C= Q_{0}$. Assume that $F\cap A \subset Q_{n}$. Since $x_{n+1}$ is the projection of $x_{0}$ onto $\overline{\operatorname{co}}\, C_{n}\cap Q_{n}$, from Lemma 2.1, we have

$$\langle x_{n+1}-z,x_{n+1}-x_{0}\rangle\leq0, \quad \forall z\in\overline {\operatorname{co}}\, C_{n}\cap Q_{n} $$

as $F\cap A \subset\overline{\operatorname{co}}\, C_{n}\cap Q_{n}$, the last inequality holds, in particular, for all $z\in F\cap A$. This together with the definition of $Q_{n+1}$ implies that $F\cap A\subset Q_{n+1}$. Hence the $F\cap A\subset\overline{\operatorname{co}}\, C_{n}\cap Q_{n}$ holds for all $n\geq0$.

Step 3. We prove $\{x_{n}\}$ is bounded. Since F is a nonempty, closed, and convex subset of C, there exists a unique element $z_{0}\in F$ such that $z_{0}=P_{F}x_{0}$. From $x_{n+1}=P_{\overline{\operatorname{co}}\, C_{n}\cap Q_{n}}x_{0}$, we have

$$\|x_{n+1}-x_{0}\|\leq\|z-x_{0}\| $$

for every $z\in\overline{\operatorname{co}}\, C_{n}\cap Q_{n}$. As $z_{0}\in F \cap A \subset\overline{\operatorname{co}}\, C_{n}\cap Q_{n}$, we get

$$\|x_{n+1}-x_{0}\|\leq\|z_{0}-x_{0}\| $$

for each $n\geq0$. This implies that $\{x_{n}\}$ is bounded.

Step 4. We show that $\{x_{n}\}$ converges strongly to a point of C (we show that $\{x_{n}\}$ is a Cauchy sequence). As $x_{n+1}=P_{\overline{\operatorname{co}}\, C_{n}\cap Q_{n}}x_{0}\subset Q_{n}$ and $x_{n}=P_{Q_{n}}x_{0}$ (Lemma 2.4), we have

$$\|x_{n+1}-x_{0}\|\geq\|x_{n}-x_{0}\| $$

for every $n\geq0$, which together with the boundedness of $\|x_{n}-x_{0}\|$ implies that there exists the limit of $\|x_{n} -x_{0}\|$. On the other hand, from $x_{n+m}\in Q_{n}$, we have $\langle x_{n}-x_{n+m}, x_{n}-x_{0}\rangle\leq0$ and hence

$$\begin{aligned} \begin{aligned} \|x_{n+m}-x_{n}\|^{2}&=\bigl\Vert (x_{n+m}-x_{0})-(x_{n}-x_{0})\bigr\Vert ^{2} \\ &\leq\|x_{n+m}-x_{0}\|^{2}-\|x_{n}-x_{0} \|^{2}-2\langle x_{n+m}-x_{n}, x_{n}-x_{0} \rangle \\ &\leq\|x_{n+m}-x_{0}\|^{2}-\|x_{n}-x_{0} \|^{2}\rightarrow0 ,\quad n\rightarrow\infty \end{aligned} \end{aligned}$$

for any $m\geq1$. Therefore $\{x_{n}\}$ is a Cauchy sequence in C, then there exists a point $q\in C$ such that $\lim_{n\rightarrow\infty}x_{n}=q$.

Step 5. We show that $y_{n}\rightarrow q$, as $n\rightarrow\infty$. Let

$$D_{n}=\bigl\{ z \in C: \|y_{n}-z\|^{2}\leq \|x_{n}-z\|^{2}+4(L_{n}-1) (L_{n}+1) \bigr\} . $$

From the definition of $D_{n}$, we have

$$\begin{aligned} D_{n} =&\bigl\{ z\in C: \langle y_{n}-z,y_{n}-z \rangle\leq\langle x_{n}-z,x_{n}-z \rangle +(L_{n}-1) (L_{n}+1)2 \bigr\} \\ =&\bigl\{ z\in C: \|y_{n}\|^{2}-2 \langle y_{n},z \rangle+\|z\|^{2} \leq\|x_{n}\|^{2}-2\langle x_{n}, z\rangle+\|z \|^{2}+4(L_{n}-1) (L_{n}+1) \bigr\} \\ =&\bigl\{ z\in C: 2 \langle x_{n}- y_{n},z\rangle \leq\|x_{n}\|^{2}-\|y_{n}\|^{2}+4(L_{n}-1) (L_{n}+1) \bigr\} . \end{aligned}$$

This implies that $D_{n}$ is closed and convex, for all $n\geq0$. Next, we show that

$$C_{n}\subset D_{n}, \quad n\geq0. $$

In fact, for any $z \in C_{n}$, we have

$$\begin{aligned} \|y_{n}-z\|^{2}&\leq\bigl(1+(L_{n}-1) \alpha_{n}\bigr)^{2} \|x_{n}-z\|^{2} \\ &=\|x_{n}-z\|^{2}+\bigl[2(L_{n}-1) \alpha_{n}+(L_{n}-1)^{2}\alpha_{n}^{2} \bigr]\|x_{n}-z\|^{2} \\ &\leq\|x_{n}-z\|^{2}+\bigl[2(L_{n}-1)+(L_{n}-1)^{2} \bigr]\|x_{n}-z\|^{2} \\ &= \|x_{n}-z\|^{2}+(L_{n}-1) (L_{n}+1) \|x_{n}-z\|^{2}. \end{aligned}$$

From

$$C_{n}=\bigl\{ z\in C: \|y_{n}-z\|\leq\bigl(1+(L_{n}-1) \alpha_{n}\bigr) \|x_{n}-z\|\bigr\} \cap A,\quad n\geq0, $$

we have $C_{n}\subset A$, $n\geq0$. Since A is convex, we also have $\overline{\operatorname{co}}\, C_{n}\subset A$, $n\geq0$. Consider $x_{n} \in \overline{\operatorname{co}}\, C_{n-1}$, we know that

$$\begin{aligned} \|y_{n}-z\|^{2} & \leq\|x_{n}-z \|^{2}+(L_{n}-1) (L_{n}+1)\|x_{n}-z \|^{2} \\ & \leq\|x_{n}-z\|^{2}+4(L_{n}-1) (L_{n}+1). \end{aligned}$$

This implies that $z \in D_{n}$ and hence $C_{n}\subset D_{n}$, $n\geq0$. Since $D_{n}$ is convex, we have $\overline{\operatorname{co}} (C_{n})\subset D_{n}$, $n\geq0$. Therefore

$$\|y_{n}-x_{n+1}\|^{2} \leq\|x_{n}-x_{n+1} \|^{2}+4(L_{n}-1) (L_{n}+1)\rightarrow0 $$

as $n\rightarrow\infty$. That is, $y_{n}\rightarrow q$ as $n\rightarrow\infty$.

Step 6. We show that $q\in F$. From the definition of $y_{n}$, we have

$$\alpha_{n}\|T_{n}x_{n}-x_{n}\|= \|y_{n}-x_{n}\|\rightarrow0 $$

as $n\rightarrow\infty$. Since $\alpha_{n} \in(a,1]\subset[0,1]$, from the above limit we have

$$\lim_{n\rightarrow\infty}\|T_{n}x_{n}-x_{n}\|= 0. $$

Since $\{T_{n}\}$ is uniformly closed and $x_{n}\rightarrow q$, we have $q\in F$.

Step 7. We claim that $q=z_{0}=P_{F}x_{0}$, if not, we have that $\|x_{0}-p\|>\|x_{0}-z_{0}\|$. There must exist a positive integer N, if $n>N$ then $\|x_{0}-x_{n}\|>\|x_{0}-z_{0}\|$, which leads to

$$ \|z_{0}-x_{0}\|^{2}=\|z_{0}-x_{n}+x_{n}-x_{0} \|^{2} =\|z_{0}-x_{n}\|^{2}+\|x_{n}-x_{0} \|^{2}+2\langle z_{0}-x_{n}, x_{n}-x_{0} \rangle. $$

It follows that $\langle z_{0}-x_{n}, x_{n}-x_{0}\rangle<0$ which implies that $z_{0} \, \overline{\in}\, Q_{n}$, so that $z_{0}\, \overline{\in}\, F$, this is a contradiction. This completes the proof. □

Next, we give an example of $C_{n}$ not involving a convex subset.

Example 2.6

Let $H=R^{2}$, $T_{n}: R^{2}\rightarrow R^{2}$ be a sequence of mappings defined by

$$T_{n}: (t_{1},t_{2})\mapsto\biggl(t_{1}, \frac{1}{8}t_{2}\biggr), \quad \forall (t_{1},t_{2}) \in R^{2}, \forall n\geq0. $$

It is obvious that $\{T_{n}\}$ is a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings with the common fixed point set $F=\{(t_{1},0): t_{1} \in(-\infty,+\infty)\}$. Take $x_{0}=(4,0)$, $\alpha_{0}=\frac{6}{7}$, we have

$$y_{0}=\frac{1}{7}x_{0}+\frac{6}{7}T_{0}x_{0}= \biggl(4\times\frac{1}{7}+\frac {4}{8}\times\frac{6}{7}, 0 \biggr)=(1,0). $$

Take $1+(L_{0}-1)\alpha_{0}=\sqrt{\frac{5}{2}}$, we have

$$C_{0}=\biggl\{ z \in R^{2}: \|y_{0}-z\|\leq\sqrt{ \frac{5}{2}}\|x_{0}-z\| \biggr\} . $$

It is easy to show that $z_{1}=(1,3), z_{2}=(-1,3) \in C_{0}$. But

$$z'=\frac{1}{2}z_{1}+\frac{1}{2}z_{2}=(0,3) \, \overline{\in}\, C_{0}, $$

since $\|y_{0}-z'\|=2$, $\|x_{0}-z'\|=1$. Therefore $C_{0}$ is not convex.

Corollary 2.7

Let C be a closed convex subset of a Hilbert space H, and let $T : C\rightarrow C$ be a closed quasi-nonexpansive mapping from C into itself. Assume that $\alpha_{n} \in(a,1]$ holds for some $a \in(0,1)$. Then $\{x_{n}\}$ generated by

$$\textstyle\begin{cases} x_{0} \in C=Q_{0} \quad \textit{chosen arbitrarily}, \\ y_{n} =(1-\alpha_{n})x_{n}+ \alpha_{n}Tx_{n}, \quad n\geq0, \\ C_{n}=\{z\in C: \|y_{n}-z\|\leq \|x_{n}-z\|\}\cap A,\quad n\geq0 , \\ Q_{n}=\{ z \in Q_{n-1}: \langle x_{n}-z, x_{0}-x_{n}\rangle\geq0 \}, \quad n\geq1, \\ x_{n+1}=P_{ C_{n}\cap Q_{n}}x_{0}, \end{cases} $$

converges strongly to $P_{F(T)}x_{0}$, where $A=\{z \in H: \|z-P_{F}x_{0}\| \leq1\}$.

Proof

Take $T_{n}\equiv T$, $L_{n}\equiv1$ in Theorem 2.5, in this case, $C_{n}$ is closed and convex, for all $n\geq0$, by using Theorem 2.5, we obtain Corollary 2.7. □

Since a nonexpansive mapping must be a closed quasi-nonexpansive mapping, from Corollary 2.7, we obtain the following result.

Corollary 2.8

Let C be a closed convex subset of a Hilbert space H, and let $T : C\rightarrow C$ be a nonexpansive mapping from C into itself. Assume that $\alpha_{n} \in(a,1]$ holds for some $a \in(0,1)$. Then $\{x_{n}\}$ generated by

$$\textstyle\begin{cases} x_{0} \in C=Q_{0} \quad \textit{chosen arbitrarily}, \\ y_{n} =(1-\alpha_{n})x_{n}+ \alpha_{n}Tx_{n}, \quad n\geq0, \\ C_{n}=\{z\in C: \|y_{n}-z\|\leq \|x_{n}-z\|\}\cap A, \quad n\geq0 , \\ Q_{n}=\{ z \in Q_{n-1}: \langle x_{n}-z, x_{0}-x_{n}\rangle\geq0 \}, \quad n\geq1, \\ x_{n+1}=P_{ C_{n}\cap Q_{n}}x_{0}, \end{cases} $$

converges strongly to $P_{F(T)}x_{0}$, where $A=\{z \in H: \|z-P_{F}x_{0}\| \leq1\}$.

3 Application to family of quasi-asymptotically nonexpansive mappings

In this section, we will apply the above result to study the following finite family of asymptotically quasi-nonexpansive mappings $\{T_{n}\}_{n=0}^{N-1}$. Let

$$\bigl\Vert T_{i}^{j}x-p\bigr\Vert \leq k_{i,j} \|x-p\|, \quad \forall x \in C, p \in F, $$

where F denotes the common fixed point set of $\{T_{n}\}_{n=0}^{N-1}$, $\lim_{j\rightarrow\infty}k_{i,j}=1$ for all $0\leq i \leq N-1$. The finite family of asymptotically quasi-nonexpansive mappings $\{T_{n}\}_{n=0}^{N-1}$ is said to be uniformly L-Lipschitz, if

$$\bigl\Vert T_{i}^{j}x-T_{i}^{j}y \bigr\Vert \leq L\|x-y\|, \quad \forall x, y \in C $$

for all $i=0,1,2,\ldots,N-1$, $j\geq1$, where $L\geq1$.

Theorem 3.1

Let C be a closed convex subset of a Hilbert space H, and let $\{T_{n}\}_{n=0}^{N-1} : C\rightarrow C$ be a uniformly L-Lipschitz finite family of asymptotically quasi-nonexpansive mappings with nonempty common fixed point set F. Assume that $\alpha_{n} \in(a,1]$ holds for some $a \in(0,1)$. Then $\{x_{n}\}$ generated by

$$\textstyle\begin{cases} x_{0} \in C=Q_{0} \quad \textit{chosen arbitrarily}, \\ y_{n} =(1-\alpha_{n})x_{n}+ \alpha_{n}T_{i(n)}^{j(n)}x_{n}, \quad n\geq0, \\ C_{n}=\{z\in C: \|y_{n}-z\|\leq(1+(k_{i(n),j(n)}-1)\alpha_{n}) \|x_{n}-z\|\}\cap A, \quad n\geq 0 , \\ Q_{n}=\{ z \in Q_{n-1}: \langle x_{n}-z, x_{0}-x_{n}\rangle\geq0 \}, \quad n\geq1, \\ x_{n+1}=P_{\overline{\operatorname{co}}\, C_{n}\cap Q_{n}}x_{0}, \end{cases} $$

converges strongly to $P_{F}x_{0}$, where $\overline{\operatorname{co}}\, C_{n}$ denotes the closed convex closure of $C_{n}$ for all $n\geq1$, $n=(j(n)-1)N+i(n)$ for all $n\geq0$, $A=\{z \in H: \|z-P_{F}x_{0}\|\leq1\}$.

Proof

It is sufficient to prove the following two conclusions.

Conclusion 1

$\{T_{i(n)}^{j(n)}\}_{n=0}^{\infty}$ is a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings from C into itself.

Conclusion 2

$F=\bigcap_{n=0}^{N}F(T_{n})=\bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})$, where $F(T)$ denotes the fixed point set of the mapping T.

Proof of Conclusion 1

Let

$$\bigl\Vert T_{i(n)}^{j(n)}x_{n}-x_{n}\bigr\Vert \rightarrow0, \quad x_{n}\rightarrow p $$

as $n\rightarrow\infty$. Observe that

$$\begin{aligned} \Vert T_{i(n)}x_{n}-x_{n}\Vert \leq&\bigl\Vert T_{i(n)}^{j(n)}x_{n}-x_{n}\bigr\Vert + \bigl\Vert T_{i(n)}^{j(n)}x_{n}-T_{i(n)}x_{n} \bigr\Vert \\ \leq&\bigl\Vert T_{i(n)}^{j(n)}x_{n}-x_{n} \bigr\Vert + L\bigl\Vert T_{i(n)}^{j(n)-1}x_{n}-x_{n} \bigr\Vert \\ \leq&\bigl\Vert T_{i(n)}^{j(n)}x_{n}-x_{n} \bigr\Vert + L\bigl\Vert T_{i(n)}^{j(n-N)}x_{n}-T_{i(n)}^{j(n-N)}x_{n-N} \bigr\Vert \\ &{} + L\bigl\Vert T_{i(n-N)}^{j(n-N)}x_{n-N}-x_{n-N} \bigr\Vert +L\Vert x_{n-N}-x_{n}\Vert \\ \leq& \bigl\Vert T_{i(n)}^{j(n)}x_{n}-x_{n} \bigr\Vert +\bigl(L+L^{2}\bigr)\Vert x_{n-N}-x_{n} \Vert \\ &{} +L\bigl\Vert T_{i(n-N)}^{j(n-N)}x_{n-N}-x_{n-N} \bigr\Vert \end{aligned}$$

from which it turns out that $\|T_{i(n)}x_{n}-x_{n}\|\rightarrow0$ as $n\rightarrow\infty$. This implies there exists subsequence $\{n_{k}\}\subset\{x_{n}\}$ such that

$$\Vert T_{i}x_{n_{k}}-x_{n_{k}}\Vert \rightarrow0, \quad i=0,1,2, \ldots, N-1 $$

as $k\rightarrow\infty$. That is, $p \in F=\bigcap_{n=0}^{N-1}F(T_{n})$. Therefore, $p \in \bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})$, hence $\{T_{i(n)}^{j(n)}\}$ is uniformly closed. On the other hand, we have

$$\bigl\Vert T_{i(n)}^{j(n)}x-p\bigr\Vert \leq k_{i(n),j(n)} \|x-p\|, \quad \forall x \in C, p\in\bigcap_{n=0}^{\infty}F \bigl(T_{i(n)}^{j(n)}\bigr), $$

and $\lim_{n\rightarrow\infty}k_{i(n),j(n)}=1$. So, $\{T_{i(n)}^{j(n)}\}$ is a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings from C into itself with $L_{n}=k_{i(n),j(n)}$. □

Proof of Conclusion 2

It is obvious that

$$\bigcap_{n=0}^{N-1}F(T_{n})\subset \bigcap_{n=0}^{\infty}F\bigl(T_{i(n)}^{j(n)} \bigr). $$

On the other hand, for any $p\in \bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})$, let $n=0,1,2, \ldots, N-1$, we obtain

$$p\in F(T_{0}), \qquad p\in F(T_{1}),\qquad p\in F(T_{2}),\qquad \ldots,\qquad p\in F(T_{n-1}), $$

which implies that

$$\bigcap_{n=0}^{N-1}F(T_{n})\supset \bigcap_{n=0}^{\infty}F\bigl(T_{i(n)}^{j(n)} \bigr). $$

Hence

$$\bigcap_{n=0}^{N-1}F(T_{n})= \bigcap_{n=0}^{\infty}F\bigl(T_{i(n)}^{j(n)} \bigr). $$

 □

By using Theorem 2.5, the iterative sequence $\{x_{n}\}$ converges strongly to $P_{\bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})}x_{0}=P_{F}x_{0}$. This completes the proof of Theorem 3.1.  □

Corollary 3.2

Let C be a closed convex subset of a Hilbert space H, and let $T : C\rightarrow C$ be a L-Lipschitz asymptotically quasi-nonexpansive mappings with nonempty fixed point set F. Assume that $\alpha_{n} \in(a,1]$ holds for some $a \in(0,1)$. Then $\{x_{n}\}$ generated by

$$\textstyle\begin{cases} x_{0} \in C=Q_{0} \quad \textit{chosen arbitrarily}, \\ y_{n} =(1-\alpha_{n})x_{n}+ \alpha_{n}T^{n}x_{n},\quad n\geq0, \\ C_{n}=\{z\in C: \|y_{n}-z\|\leq(1+(k_{n}-1)\alpha_{n}) \|x_{n}-z\|\}\cap A, \quad n\geq0 , \\ Q_{n}=\{ z \in Q_{n-1}: \langle x_{n}-z, x_{0}-x_{n}\rangle\geq0 \}, \quad n\geq1, \\ x_{n+1}=P_{\overline{\operatorname{co}}\, C_{n}\cap Q_{n}}x_{0}, \end{cases} $$

converges strongly to $P_{F(T)}x_{0}$, where $\overline{\operatorname{co}}\, C_{n}$ denotes the closed convex closure of $C_{n}$ for all $n\geq1$, $A=\{z \in H: \|z-P_{F}x_{0}\|\leq1\}$.

Proof

Take $T_{n}\equiv T$ in Theorem 3.1, we obtain Corollary 3.2. □

Since a nonexpansive mapping must be a Lipschitz asymptotically quasi-nonexpansive mapping, from Corollary 3.2, we can obtain Corollary 2.8.

4 Example

Conclusion 4.1

Let H be a Hilbert space, $\{x_{n}\}_{n=1}^{\infty}\subset H$ be a sequence such that it converges weakly to a non-zero element $x_{0}$ and $\|x_{i}-x_{j}\|\geq 1$ for any $i\neq j$. Define a sequence of mappings $T_{n}: H\rightarrow H$ as follows:

$$T_{n}(x)= \textstyle\begin{cases} L_{n}x_{n} & \textit{if } x=x_{n}\ (\exists n\geq 1) , \\ -x & \textit{if } x\neq x_{n}\ (\forall n\geq1), \end{cases} $$

where $\{L_{n}\}_{n=1}^{\infty}$ is a sequence of number such that $L_{n}>1$, $\lim_{n\rightarrow\infty}L_{n}=1$. Then $\{T_{n}\}$ is a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings with the common fixed point set $F=\{0\}$.

Proof

It is obvious that $\{T_{n}\}$ has a unique common fixed point 0. Next, we prove that $\{T_{n}\}$ is uniformly closed. In fact, for any strong convergent sequence $\{z_{n}\}\subset E$ such that $z_{n}\rightarrow z_{0}$ and $\|z_{n}-T_{n}z_{n}\|\rightarrow0$ as $n\rightarrow\infty$, there exists sufficiently large natural number N such that $z_{n}\neq x_{m}$, for any $n, m >N$. Then $T_{n}z_{n}=-z_{n}$ for $n>N$, it follows from $\|z_{n}-T_{n}z_{n}\|\rightarrow0$ that $2z_{n}\rightarrow 0$ and hence $z_{0} \in F$. Finally, from the definition of $\{T_{n}\}$, we have

$$\|T_{n}x-0\|= \|T_{n}x\|\leq\|L_{n}x \|=L_{n}\|x-0\|, \quad \forall x \in H, $$

so that $\{T_{n}\}$ is a uniformly closed asymptotically family of countable quasi-$L_{n}$-Lipschitz mappings. □

Remark

In the result of Liu et al. [30], the boundedness of C was assumed and the hybrid iterative process was complex. In our hybrid iterative process, $C_{n}$ was constructed as a non-convex set can makes it more simple, meanwhile, the boundedness of C can be removed. Of course, a new technique has been applied in our process of proof.