Hybrid iterative algorithms for nonexpansive and nonspreading mappings in Hilbert spaces

Abstract

Recently, Iemoto and Takahashi considered a weak convergence iterative scheme for a nonspreading mapping and a nonexpansive mapping in Hilbert spaces. In this paper, we suggest two hybrid iterative algorithms by modifying Iemoto and Takahashi’s iterative scheme for a countable family of nonspreading mappings and a nonexpansive mapping in Hilbert spaces.

MSC:47H05, 47H09.

1 Introduction and preliminaries

Let H be a Hilbert space and C be a nonempty closed convex subset of H. Let T be a nonlinear mapping of C into itself. We use $F(T)$ and $P_C$ to denote the set of fixed points of T and the metric projection from H onto C, respectively.

Recall that T is said to be nonexpansive if

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

(1.1)

for all $x,y\in C$.

For approximating the fixed point of a nonexpansive mapping in a Hilbert space, Mann [1] in 1953 introduced the famous iterative scheme as follows:

$$\mathrm{\forall }{x}_1\in C,x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,\mathrm{\forall }n\ge 1,$$

(1.2)

where T is a nonexpansive mapping of C into itself and $\{{\alpha }_n\}$ is a sequence in $(0,1)$. It is well known that $\{x_n\}$ defined in (1.2) converges weakly to a fixed point of T.

Attempts to modify the normal Mann iteration method (1.2) for nonexpansive mappings so that strong convergence is guaranteed have recently been made; see, e.g., [2–9].

Let T be a mapping from C into itself. Then T is called nonspreading [3] if

$$2{\parallel Tx-Ty\parallel }^2\le {\parallel Tx-y\parallel }^2+{\parallel x-Ty\parallel }^2$$

for all $x,y\in C$. A mapping $T:C\to C$ is called quasi-nonexpansive if $F(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in C$ and $y\in F(T)$. If T is a nonspreading mapping from C into itself and $F(T)$ is nonempty, then T is quasi-nonexpansive. Further, we know that the set of fixed points of each quasi-nonexpansive mapping is closed and convex; see [10].

In [11], by using Moudafi’s iterative scheme [12], Iemoto and Takahashi considered the following weak convergence theorem.

Theorem IT ([11])

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a nonspreading mapping of C into itself, and let T be a nonexpansive mapping of C into itself such that $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Define a sequence $\{x_n\}$ as follows:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n\{{\beta }_{n}S{x}_n+(1-{\beta }_n)T{x}_n\}\hfill \end{array}$$

(1.3)

for all $n\in N$, where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Then the following hold:

• (i) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\beta }_n)<\mathrm{\infty }$, then $\{x_n\}$ converges weakly to $v\in F(S)$;

• (ii) If ${\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$, then $\{x_n\}$ converges weakly to $v\in F(T)$;

• (iii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$, then $\{x_n\}$ converges weakly to $v\in F(S)\cap F(T)$.

In this paper, we modify (1.1) by a hybrid iterative scheme and obtain the strong convergence theorems for a family of nonspreading mappings and a nonexpansive mapping in a Hilbert space.

Let E be a Banach space and K be a nonempty closed convex subset of E. Let $\{T_n\}:K\to K$ be a family of mappings. Then $\{T_n\}$ is said to satisfy the AKTT-condition [13] if for each bounded subset B of K, one has

$$\sum _{n=1}^{\mathrm{\infty }}sup\{\parallel T_{n+1}z-T_{n}z\parallel :z\in B\}<\mathrm{\infty }.$$

The following is an important result on a family of mappings ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ satisfying the AKTT-condition.

Lemma 1.1 ([13])

Let K be a nonempty and closed subset of a Banach space E, and let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a family of mappings of K into itself which satisfies the AKTT-condition. Then, for each $x\in K$, $\{T_{n}x\}$ converges strongly to a point in K. Moreover, let the mapping $T:K\to K$ be defined by

$$Tx=\underset{n\to \mathrm{\infty }}{lim}{T}_{n}x,\mathrm{\forall }x\in K.$$

Then, for each bounded subset B of K,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\parallel Tz-T_{n}z\parallel :z\in B\}=0.$$

Obviously, if a family of mappings ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ satisfies the AKTT-condition and $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$ for each $x\in K$, then it is unnecessary that $F(T)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. To show this, see the following example.

Example 1.1 Let $E=\mathbb{R}$ and $K=[0,2]$. Define a family of mappings ${\{T_n\}}_{n=1}^{\mathrm{\infty }}:K\to K$ by

$$T_{1}x=0,T_n=\frac{1}{n}(1+x),n\ge 2.$$

Then ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ satisfy the AKTT-condition. It is easy to see that for each $x\in K$, ${lim}_{n\to \mathrm{\infty }}{T}_{n}x=0$. Define the mapping $T:K\to K$ by $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$. That is, $Tx=0$ for all $x\in K$. But $F(T)\ne {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$.

In this paper, we call that $\{T_n,T\}$ satisfy the AKTT-condition if ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ satisfy the AKTT-condition with $F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$.

Lemma 1.2 ([11])

Let C be a nonempty closed subset of a Hilbert space H. Then a mapping $T:C\to C$ is nonspreading if and only if

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

for all $x,y\in C$.

By using Lemma 1.2, we get the following simple but important result.

Lemma 1.3 Let H be a Hilbert space and C be a nonempty subset of H. Let $\{T_n\}$ be a family of nonspreading mappings of C into itself, and assume that ${lim}_{n\to \mathrm{\infty }}{T}_{n}x$ exists for each $x\in C$. Define the mapping $T:C\to C$ by $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$. Then the mapping T is a nonspreading mapping.

Proof In fact, for all $x,y\in C$, we have

$$\begin{array}{rcl}{\parallel Tx-Ty\parallel }^2& =& {\parallel \underset{n\to \mathrm{\infty }}{lim}{T}_{n}x-\underset{n\to \mathrm{\infty }}{lim}{T}_{n}y\parallel }^2\\ =& \underset{n\to \mathrm{\infty }}{lim}{\parallel T_{n}x-T_{n}y\parallel }^2\\ \le & \underset{n\to \mathrm{\infty }}{lim}[{\parallel x-y\parallel }^2+2〈x-T_{n}x,y-T_{n}y〉]\\ =& {\parallel x-y\parallel }^2+2〈x-\underset{n\to \mathrm{\infty }}{lim}{T}_{n}x,y-\underset{n\to \mathrm{\infty }}{lim}{T}_{n}y〉\\ =& {\parallel x-y\parallel }^2+2〈x-Tx,y-Ty〉.\end{array}$$

Lemma 1.2 shows that the mapping T is a nonspreading mapping. □

Lemma 1.4 Let C be a closed convex subset of a real Hilbert space H, and let $P_C$ be the metric projection from H onto C (i.e., for $x\in H$, $P_{C}x$ is the only point in C such that $\parallel x-P_{C}x\parallel =inf\{\parallel x-z\parallel :z\in C\}$). Given $x\in H$ and $z\in C$. Then $z=P_{C}x$ if and only if the following relation holds:

$$〈x-z,y-z〉\le 0,\mathrm{\forall }y\in C.$$

Lemma 1.5 ([14])

Let H be a real Hilbert space. Then the following equation holds:

$${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2,\mathrm{\forall }x\in C\mathit{\text{and}}\mathrm{\forall }t\in [0,1].$$

2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of nonspreading mappings such that $F=F(S)\cap [{\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)]\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_n[{\beta }_{n}S{x}_n+{\sum }_{i=1}^n({\beta }_{i-1}-{\beta }_i)T_i{x}_n],\hfill \\ C_n=\{v\in C:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ D_n={\bigcap }_{j=1}^n{C}_j,\hfill \\ x_{n+1}=P_{D_n}x,n\ge 1,\hfill \end{array}$$

(2.1)

where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Assume that $\{{\beta }_n\}$ is strictly decreasing and ${\beta }_0=1$. Then the following hold:

• (i) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, then $\{x_n\}$ strongly converges to $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$;

• (ii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, then $\{x_n\}$ converges strongly to $q\in F$.

Proof Obviously, each $C_n$ is closed and convex and hence $D_n$ is closed and convex. Next, we show that $F\subset D_n$ for all $n\ge 1$. To end this, we need to prove that $F\subset C_n$ for all $n\ge 1$. Indeed, for each $p\in F$, we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & (1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n[{\beta }_n\parallel S{x}_n-p\parallel +\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)\parallel T_i{x}_n-p\parallel ]\\ \le & (1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n[{\beta }_n\parallel x_n-p\parallel +\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)\parallel x_n-p\parallel ]\\ =& \parallel x_n-p\parallel .\end{array}$$

(2.2)

This implies that

$$p\in C_n\text{for all }n\ge 1.$$

Therefore, $F\subset C_n$ and hence $C_n$ is nonempty for all $n\ge 1$. On the other hand, from the definition of $D_n$, we see that $F\subset D_n={\bigcap }_{i=1}^n{C}_j$ for all $n\ge 1$.

From $x_{n+1}=P_{D_n}x$, we have

$$\parallel x_{n+1}-x\parallel \le \parallel v-x\parallel ,\mathrm{\forall }v\in D_n,n\ge 1.$$

Since $P_{F}x\in F\subset D_n$, one has

$$\parallel x_{n+1}-x\parallel \le \parallel P_{F}x-x\parallel ,n\ge 1.$$

(2.3)

This implies that $\{x_n\}$ is bounded and hence $\{y_n\}$ is bounded.

On the other hand, since $D_{n+1}\subset D_n$ for all $n\ge 1$, we have

$$x_{n+2}=P_{D_{n+1}}x\in D_{n+1}\subset D_n$$

for all $n\ge 1$. From $x_{n+1}=P_{D_n}x$ one has

$$\parallel x_{n+1}-x\parallel \le \parallel x_{n+2}-x\parallel$$

(2.4)

for all $n\ge 1$. It follows from (2.3) and (2.4) that the limit of $\{x_n-x\}$ exists.

Since $D_m\subset D_n$ and $x_{m+1}=P_{D_m}x\in D_m\subset D_n$ for all $m\ge n$ and $x_{n+1}=P_{D_n}x$, by Lemma 1.4 one has

$$〈x_{n+1}-x,x_{m+1}-x_{n+1}〉\ge 0.$$

(2.5)

It follows from (2.5) that

$$\begin{array}{c}{\parallel x_{m+1}-x_{n+1}\parallel }^2\hfill \\ ={\parallel x_{m+1}-x-(x_{n+1}-x)\parallel }^2\hfill \\ ={\parallel x_{m+1}-x\parallel }^2+{\parallel x_{n+1}-x\parallel }^2-2〈x_{n+1}-x,x_{m+1}-x〉\hfill \\ ={\parallel x_{m+1}-x\parallel }^2+{\parallel x_{n+1}-x\parallel }^2-2〈x_{n+1}-x,x_{m+1}-x_{n+1}+x_{n+1}-x〉\hfill \\ ={\parallel x_{m+1}-x\parallel }^2-{\parallel x_{n+1}-x\parallel }^2-2〈x_{n+1}-x,x_{m+1}-x_{n+1}〉\hfill \\ \le {\parallel x_{m+1}-x\parallel }^2-{\parallel x_{n+1}-x\parallel }^2.\hfill \end{array}$$

(2.6)

Since the limit of $\parallel x_n-x\parallel$ exists, we get

$$\underset{m,n\to \mathrm{\infty }}{lim}\parallel x_m-x_n\parallel =0.$$

It follows that $\{x_n\}$ is a Cauchy sequence. Since H is a Hilbert space and C is closed and convex, there exists $q\in C$ such that

$$x_n\to q,\text{as }n\to \mathrm{\infty }.$$

(2.7)

By taking $m=n+1$ in (2.6), one arrives at

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+2}-x_{n+1}\parallel =0,$$

i.e.,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(2.8)

Noticing that $x_{n+1}=P_{D_n}x\in D_n\subset C_n$, we get

$$\parallel y_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel \to 0,$$

and hence

$$\parallel y_n-x_n\parallel \le \parallel y_n-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel \to 0.$$

(2.9)

From (2.7) and (2.9) it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel =\parallel q-p\parallel ,\mathrm{\forall }p\in F.$$

(2.10)

Now we prove (i). Note that

$$\begin{array}{rcl}{y}_n& =& (1-{\alpha }_n)x_n+{\alpha }_n[{\beta }_{n}S{x}_n+\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)(T_i{x}_n-x_n)]+{\alpha }_n(1-{\beta }_n)x_n\\ =& (1-{\alpha }_n{\beta }_n)x_n+{\alpha }_n{\beta }_{n}S{x}_n+{\alpha }_n\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)(T_i{x}_n-x_n).\end{array}$$

Hence,

$${\alpha }_n\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)(T_i{x}_n-x_n)=(1-{\alpha }_n{\beta }_n)(y_n-x_n)+{\alpha }_n{\beta }_n(y_n-S{x}_n).$$

(2.11)

On the other hand, for any $p\in F$, from Lemma 1.2 we have

$$\begin{array}{rcl}{\parallel x_n-p\parallel }^2& =& 2〈x_n-T_i{x}_n,p-T_{i}p〉+{\parallel x_n-p\parallel }^2\\ \ge & {\parallel T_i{x}_n-T_{i}p\parallel }^2={\parallel T_i{x}_n-p\parallel }^2={\parallel T_i{x}_n-x_n+(x_n-p)\parallel }^2\\ =& {\parallel T_i{x}_n-x_n\parallel }^2+{\parallel x_n-p\parallel }^2+2〈T_i{x}_n-x_n,x_n-p〉,\end{array}$$

and hence

$${\parallel T_i{x}_n-x_n\parallel }^2\le 2〈x_n-T_i{x}_n,x_n-p〉,\mathrm{\forall }i\in \mathbb{N}.$$

(2.12)

Note that $\{{\beta }_n\}$ is strictly decreasing. Hence from (2.11) and (2.12) we get

$$\begin{array}{rcl}{\parallel T_i{x}_n-x_n\parallel }^2& \le & \frac{1}{2{\alpha }_n({\beta }_{i-1}-{\beta }_i)}[(1-{\alpha }_n{\beta }_n)〈y_n-x_n,x_n-T_{i}p〉\\ +{\alpha }_n{\beta }_n〈y_n-S{x}_n,x_n-p〉],i\ge 1.\end{array}$$

(2.13)

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, from (2.9) and (2.13) it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i{x}_n-x_n\parallel =0,\mathrm{\forall }i\in \mathbb{N}.$$

(2.14)

Since each $T_i$ is a nonspreading mapping, by Lemma 1.2, (2.7) and (2.10), we have

$${\parallel T_{i}q-T_i{x}_n\parallel }^2\le {\parallel x_n-q\parallel }^2+2〈q-T_{i}q,x_n-T_i{x}_n〉\to 0,\mathrm{\forall }i\in \mathbb{N}.$$

(2.15)

Further, one has

$$\parallel q-T_{i}q\parallel \le \parallel q-x_n\parallel +\parallel x_n-T_i{x}_n\parallel +\parallel T_i{x}_n-T_{i}q\parallel \to 0,\mathrm{\forall }i\in \mathbb{N}.$$

(2.16)

So, we have $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$.

To prove (ii), first we show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel =0$. For any $p\in F$, we have

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\beta }_n[(1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p]+\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)[(1-{\alpha }_n)x_n+{\alpha }_n{T}_i{x}_n-p]\parallel }^2\\ \le & {\beta }_n{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2+\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i){\parallel (1-{\alpha }_n)x_n+{\alpha }_n{T}_i{x}_n-p\parallel }^2\\ \le & {\beta }_n{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2\\ +\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i)[(1-{\alpha }_n){\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel T_i{x}_n-p\parallel }^2]\\ \le & {\beta }_n{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2+\sum _{i=1}^n({\beta }_{i-1}-{\beta }_i){\parallel x_n-p\parallel }^2\\ =& {\beta }_n{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2+(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2,\end{array}$$

and hence by (2.10) we get

$$\begin{array}{rcl}0& \le & {\parallel x_n-p\parallel }^2-{\beta }_n{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2-(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ =& {\beta }_n[{\parallel x_n-p\parallel }^2-{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2]\\ \le & {\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2\to 0.\end{array}$$

(2.17)

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, it follows from (2.17) that

$$\underset{n\to \mathrm{\infty }}{lim}({\parallel x_n-p\parallel }^2-{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2)=0.$$

(2.18)

From (2.18) and

$${\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2=(1-{\alpha }_n){\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel S{x}_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel x_n-S{x}_n\parallel }^2,$$

we get

$$\begin{array}{c}{\alpha }_n(1-{\alpha }_n){\parallel x_n-S{x}_n\parallel }^2\hfill \\ =({\parallel x_n-p\parallel }^2-{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2)-{\alpha }_n{\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel S{x}_n-p\parallel }^2\hfill \\ \le ({\parallel x_n-p\parallel }^2-{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2)-{\alpha }_n{\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel x_n-p\parallel }^2\hfill \\ ={\parallel x_n-p\parallel }^2-{\parallel (1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n-p\parallel }^2\to 0.\hfill \end{array}$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S{x}_n\parallel =0.$$

(2.19)

Now, using (2.19), (2.7) and

$$\parallel q-Sq\parallel \le \parallel q-x_n\parallel +\parallel x_n-S{x}_n\parallel +\parallel S{x}_n-Sq\parallel \le 2\parallel q-x_n\parallel +\parallel x_n-S{x}_n\parallel \to 0,$$

which implies that $q\in F(S)$.

Note that (2.9) and (2.19) imply that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-S{x}_n\parallel =0$. Then, repeating (2.11) to (2.16), we get $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. So, $q\in F$. This completes the proof. □

Theorem 2.2 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of nonspreading mappings such that $F=F(S)\cap [{\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)]\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_n[{\beta }_{n}S{x}_n+(1-{\beta }_n)T_n{x}_n],\hfill \\ C_n=\{v\in C:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ D_n={\bigcap }_{j=1}^n{C}_j,\hfill \\ x_{n+1}=P_{D_n}x,n\ge 1,\hfill \end{array}$$

(2.20)

where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Assume that $\{T_n,T\}$ satisfies the AKTT-condition. Then the following hold:

• (i) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, then $\{x_n\}$ strongly converges to $v\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$;

• (ii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, then $\{x_n\}$ converges strongly to $z\in F$.

Proof By a process similar to the proof of Theorem 2.1, we can conclude that $\{x_n\}$ converges strongly to some $q\in C$ and

$$x_n-y_n\to 0.$$

We first prove (i). From (2.20) we have

$$T_n{x}_n-x_n=\frac{1}{{\alpha }_n(1-{\beta }_n)}(y_n-x_n)-\frac{{\beta }_n}{1-{\beta }_n}(S{x}_n-x_n),$$

and hence

$$\parallel T_n{x}_n-x_n\parallel \le \frac{1}{{\alpha }_n(1-{\beta }_n)}\parallel y_n-x_n\parallel +\frac{{\beta }_n}{1-{\beta }_n}\parallel S{x}_n-x_n\parallel .$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_n{x}_n-x_n\parallel =0.$$

(2.21)

Further, by Lemma 1.1 and (2.21), we have

$$\begin{array}{rcl}\parallel x_n-T{x}_n\parallel & \le & \parallel x_n-T_n{x}_n\parallel +\parallel T_n{x}_n-T{x}_n\parallel \\ \le & \parallel x_n-T_n{x}_n\parallel +sup\{\parallel T_{n}z-Tz\parallel :z\in \{x_n\}\}\to 0.\end{array}$$

(2.22)

Since each $T_n$ is a nonspreading mapping, Lemma 1.3 shows that T is a nonspreading mapping. Further, by using Lemma 1.2, we have

$${\parallel Tq-T{x}_n\parallel }^2\le {\parallel x_n-q\parallel }^2+2〈q-Tq,x_n-T{x}_n〉\to 0,\mathrm{\forall }i\in \mathbb{N}.$$

(2.23)

From (2.21) and (2.23) it follows that

$$\parallel q-Tq\parallel \le \parallel q-x_n\parallel +\parallel x_n-T{x}_n\parallel +\parallel T{x}_n-Tq\parallel \to 0.$$

(2.24)

It follows that $q\in F(T)$. Since $(\{T_n\},T)$ satisfies the AKTT-condition, one has $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)=F(T)$. This completes (i).

Next we show (ii). By a process similar to the proof of Theorem 2.1 and from (2.22) to (2.24), we can get that

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S{x}_n\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_n{x}_n\parallel =0,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\parallel T{x}_n-Tq\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T{x}_n\parallel =0.\hfill \end{array}$$

Finally, by

$$\parallel q-Sq\parallel \le \parallel q-x_n\parallel +\parallel x_n-S{x}_n\parallel +\parallel S{x}_n-Sq\parallel \le 2\parallel x_n-q\parallel +\parallel x_n-S{x}_n\parallel \to 0$$

and

$$\parallel q-Tq\parallel \le \parallel q-x_n\parallel +\parallel x_n-T{x}_n\parallel +\parallel T{x}_n-Tq\parallel \to 0,$$

we get $q\in F(S)\cap F(T)$. Since $(\{T_n\},T)$ satisfies the AKTT-condition, we conclude that $q\in F$. This completes (ii). □

Letting $T_i=T$ for all $i\in \mathbb{N}$ in Theorem 2.1 and Theorem 2.2, we get the following corollary.

Corollary 2.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping and $T:C\to C$ be a nonspreading mapping such that $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_n[{\beta }_{n}S{x}_n+(1-{\beta }_n)T{x}_n],\hfill \\ C_n=\{v\in C:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ D_n={\bigcap }_{j=1}^n{C}_j,\hfill \\ x_{n+1}=P_{D_n}x,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Then the following hold:

• (i) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, then $\{x_n\}$ strongly converges to $x^{\prime }\in F(T)$;

• (ii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, then $\{x_n\}$ converges strongly to $q\in F(S)\cap F(T)$ with $q=P_{F}x$.

Letting $S=I$ in Theorems 2.1 and 2.2, we get the following corollary.

Corollary 2.2 Let C be a nonempty closed convex subset of a Hilbert space H. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of nonspreading mappings such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n(1-{\beta }_n))x_n+{\alpha }_n{\sum }_{i=1}^n({\beta }_{i-1}-{\beta }_i)T_i{x}_n,\hfill \\ C_n=\{v\in C:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ D_n={\bigcap }_{j=1}^n{C}_j,\hfill \\ x_{n+1}=P_{D_n}x,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Assume that $\{{\beta }_n\}$ is strictly decreasing and ${\beta }_0=1$. Then if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, then $\{x_n\}$ strongly converges to $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$.

Corollary 2.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of nonspreading mappings such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n(1-{\beta }_n))x_n+{\alpha }_n(1-{\beta }_n)T_n{x}_n,\hfill \\ C_n=\{v\in C:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ D_n={\bigcap }_{j=1}^n{C}_j,\hfill \\ x_{n+1}=P_{D_n}x,n\ge 1,\hfill \end{array}$$

where $\{{\gamma }_n\}\subset [0,1]$. Assume that $(\{T_n,T\})$ satisfies the AKTT-condition. Then if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, then $\{x_n\}$ strongly converges to $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$.