Implicit iteration scheme for two phi-hemicontractive operators in arbitrary Banach spaces

Abstract

The purpose of this paper is to characterize conditions for the convergence of the implicit Ishikawa iterative scheme with errors in the sense of Agarwal et al. (J. Math. Anal. Appl. 272:435-447, 2002) to a common fixed point of two ϕ-hemicontractive mappings in a nonempty convex subset of an arbitrary Banach space.

MSC:47H09, 47J25.

1 Introduction and preliminaries

Let K be a nonempty subset of an arbitrary Banach space E and $E^{\ast }$ be its dual space. The symbols T and $F(T)$ stand for the self-map of K and the set of fixed points of T. We denote by J the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

$$J(x)=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^2={\parallel f^{\ast }\parallel }^2\}.$$

Definition 1.1 [1–4]

• (i) T is said to be strongly pseudocontractive if there exists a constant $t>1$ such that for each $x,y\in K$, there exists $j(x-y)\in J(x-y)$ satisfying

  $$〈Tx-Ty,j(x-y)〉\le \frac{1}{t}{\parallel x-y\parallel }^2.$$

• (ii) T is said to be strictly hemicontractive if $F(T)\ne \mathrm{\varnothing }$ and if there exists a constant $t>1$ such that for each $x\in K$ and $q\in F(T)$, there exists $j(x-q)\in J(x-q)$ satisfying

  $$〈Tx-Tq,j(x-q)〉\le \frac{1}{t}{\parallel x-q\parallel }^2.$$

• (iii) T is said to be ϕ-strongly pseudocontractive if there exists a strictly increasing function $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $\varphi (0)=0$ such that for each $x,y\in K$, there exists $j(x-y)\in J(x-y)$ satisfying

  $$〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2-\varphi (\parallel x-y\parallel ).$$

• (iv) T is said to be ϕ-hemicontractive if $F(T)\ne \mathrm{\varnothing }$ and if there exists a strictly increasing function $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $\varphi (0)=0$ such that for each $x\in K$ and $q\in F(T)$, there exists $j(x-q)\in J(x-q)$ satisfying

  $$〈Tx-Tq,j(x-q)〉\le {\parallel x-q\parallel }^2-\varphi (\parallel x-q\parallel ).$$

Clearly, each strictly hemicontractive operator is ϕ-hemicontractive.

Chidume [1] established that the Mann iteration sequence converges strongly to the unique fixed point of T in the case that T is a Lipschitz strongly pseudo-contractive mapping from a bounded, closed, convex subset of $L_p$ (or $l_p$) into itself. Afterwards, several authors have generalized this result of Chidume in various directions [4–12].

In 2001, Xu and Ori [13] introduced the following implicit iteration process for a finite family of nonexpansive mappings $\{T_i:i\in I\}$ (here $I=\{1,2,\dots ,N\}$), with $\{{\alpha }_n\}$ a real sequence in $(0,1)$, and an initial point $x_0\in K$:

$$\begin{array}{c}{x}_1=(1-{\alpha }_1)x_0+{\alpha }_1{T}_1{x}_1,\hfill \\ x_2=(1-{\alpha }_2)x_1+{\alpha }_2{T}_2{x}_2,\hfill \\ ⋮\hfill \\ x_N=(1-{\alpha }_N)x_{N-1}+{\alpha }_N{T}_N{x}_N,\hfill \\ x_{N+1}=(1-{\alpha }_{N+1})x_N+{\alpha }_{N+1}{T}_{N+1}{x}_{N+1},\hfill \\ ⋮\hfill \end{array}$$

which can be written in the following compact form:

$$x_n=(1-{\alpha }_n)x_{n-1}+{\alpha }_n{T}_n{x}_n\text{for all }n\ge 1,$$

(XO)

where $T_n=T_{n(modN)}$ (here the modN function takes values in I). Xu and Ori [13] proved that the process converges weakly to a common fixed point of a finite family in a Hilbert space. They remarked further that it is yet unclear what assumptions on the mappings and the parameters $\{{\alpha }_n\}$ are sufficient to guarantee the strong convergence of the sequence $\{x_n\}$.

In [14], Osilike proved the following results.

Theorem 1.2 [[14], Theorem 2]

Let E be a real Banach space and K be a nonempty closed convex subset of E. Let $\{T_i:i\in I\}$ be N strictly pseudocontractive self-mappings of K with $F={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be a real sequence satisfying the conditions:

$$\begin{array}{c}(\text{i})0<{\alpha }_n<1,\hfill \\ (\text{ii})\sum _{n=1}^{\mathrm{\infty }}(1-{\alpha }_n)=\mathrm{\infty },\hfill \\ (\text{iii})\sum _{n=1}^{\mathrm{\infty }}{(1-{\alpha }_n)}^2<\mathrm{\infty }.\hfill \end{array}$$

For arbitrary $x_0\in K$, define the sequence $\{x_n\}$ by the implicit iteration process (XO). Then $\{x_n\}$ converges strongly to a common fixed point of the mappings $\{T_i:i\in I\}$ if and only if ${lim}_{n\to \mathrm{\infty }}infd(x_n,F)=0$.

It is well known that ${\alpha }_n=1-\frac{1}{n^{\frac{1}{2}}}$, $\sum {(1-{\alpha }_n)}^2=\mathrm{\infty }$. Hence the results of Osilike [14] need to be improved.

The purpose of this paper is to characterize conditions for the convergence of the implicit Ishikawa iterative scheme with errors in the sense of Agarwal et al. [15] to a common fixed point of two ϕ-hemicontractive mappings in a nonempty convex subset of an arbitrary Banach space. Our studying results improve and generalize most results in recent literature [4, 5, 7–9, 12].

2 Main results

The following results are now well known.

Lemma 2.1 [16]

For all $x,y\in E$ and $j(x+y)\in J(x+y)$,

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

Lemma 2.2 [17]

Let $\{{\theta }_n\}$ be a sequence of nonnegative real numbers, and let $\{{\lambda }_n\}$ be a real sequence satisfying

$$0\le {\lambda }_n\le 1,\sum _{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }.$$

Suppose that there exists a strictly increasing function $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $\varphi (0)=0$. If there exists a positive integer $n_0$ such that

$${\theta }_{n+1}^2\le {\theta }_n^2-{\lambda }_n\varphi ({\theta }_{n+1})+{\sigma }_n+{\gamma }_n$$

for all $n\ge n_0$, with ${\sigma }_n\ge 0$, $\mathrm{\forall }n\in \mathbb{N}$, ${\sigma }_n=0({\lambda }_n)$ and $\sum _{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Now we prove our main results.

Theorem 2.3 Let K be a nonempty convex subset of an arbitrary Banach space E, and let $T,S:K\to K$ be two uniformly continuous with $F(T)\cap F(S)\ne \mathrm{\varnothing }$ and ϕ-hemicontractive mappings. Suppose that ${\{u_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{v_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded sequences in K and ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{a_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ and ${\{c_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions

• (i) $a_n+b_n+c_n=a_n^{\prime }+b_n^{\prime }+c_n^{\prime }=1$,

• (ii) ${lim}_{n\to \mathrm{\infty }}{b}_n={lim}_{n\to \mathrm{\infty }}{c}_n={lim}_{n\to \mathrm{\infty }}{b}_n^{\prime }=0$,

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n^{\prime }<\mathrm{\infty }$, and

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n^{\prime }=\mathrm{\infty }$.

For any $x_0\in K$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ inductively as follows:

$$\begin{array}{r}{y}_n=a_n{x}_{n-1}+b_{n}S{x}_n+c_n{u}_n,\\ x_n=a_n^{\prime }{x}_{n-1}+b_n^{\prime }T{y}_n+c_n^{\prime }{v}_n,n\ge 1.\end{array}$$

(2.1)

Then the following conditions are equivalent:

• (a) ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S,

• (b) ${lim}_{n\to \mathrm{\infty }}T{y}_n=q$,

• (c) ${\{T{y}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

Proof Since T and S are ϕ-hemicontractive, then the common fixed point of $F(T)\cap F(S)$ is unique. Suppose that p and q are all common fixed points of T and S, then

$${\parallel p-q\parallel }^2=〈p-q,j(p-q)〉=〈Tp-Tq,j(p-q)〉\le {\parallel p-q\parallel }^2-\varphi (\parallel p-q\parallel )<{\parallel p-q\parallel }^2,$$

which is a contradiction. So, we denote the unique fixed point q.

Suppose that ${lim}_{n\to \mathrm{\infty }}{x}_n=q$. Then (ii) and the uniform continuity of T and S yield that

$$\underset{n\to \mathrm{\infty }}{lim}{y}_n=\underset{n\to \mathrm{\infty }}{lim}[a_n{x}_{n-1}+b_{n}S{x}_n+c_n{u}_n]=q,$$

which implies that ${lim}_{n\to \mathrm{\infty }}T{y}_n=q$. Therefore ${\{T{y}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

Put

$$M_1=\parallel x_0-q\parallel +\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +\underset{n\ge 1}{sup}\parallel v_n-q\parallel .$$

Obviously, $M_1<\mathrm{\infty }$. It is clear that $\parallel x_0-p\parallel \le M_1$. Let $\parallel x_{n-1}-p\parallel \le M_1$. Next we prove that $\parallel x_n-p\parallel \le M_1$.

Consider

$$\begin{array}{rcl}\parallel x_n-q\parallel & =& \parallel a_n^{\prime }{x}_{n-1}+b_n^{\prime }T{y}_n+c_n^{\prime }{v}_n-q\parallel \\ =& \parallel a_n^{\prime }(x_{n-1}-q)+b_n^{\prime }(T{y}_n-q)+c_n^{\prime }(v_n-q)\parallel \\ \le & (1-b_n^{\prime })\parallel x_{n-1}-q\parallel +b_n^{\prime }\parallel T{y}_n-q\parallel +c_n^{\prime }\parallel v_n-q\parallel \\ \le & (1-b_n^{\prime })M_1+b_n^{\prime }\parallel T{y}_n-q\parallel +c_n^{\prime }\parallel v_n-q\parallel \\ =& (1-b_n^{\prime })[\parallel x_0-q\parallel +\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +\underset{n\ge 1}{sup}\parallel v_n-q\parallel ]\\ +b_n^{\prime }\parallel T{y}_n-q\parallel +c_n^{\prime }\parallel v_n-q\parallel \\ \le & \parallel x_0-q\parallel +((1-b_n^{\prime })\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel +b_n^{\prime }\parallel T{y}_n-q\parallel )\\ +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +((1-b_n^{\prime })\underset{n\ge 1}{sup}\parallel v_n-q\parallel +b_n^{\prime }\parallel v_n-q\parallel )\\ \le & \parallel x_0-q\parallel +((1-b_n^{\prime })\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel +b_n^{\prime }\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel )\\ +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +((1-b_n^{\prime })\underset{n\ge 1}{sup}\parallel v_n-q\parallel +b_n^{\prime }\underset{n\ge 1}{sup}\parallel v_n-q\parallel )\\ =& \parallel x_0-q\parallel +\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +\underset{n\ge 1}{sup}\parallel v_n-q\parallel \\ =& M_1.\end{array}$$

So, from the above discussion, we can conclude that the sequence ${\{x_n-q\}}_{n\ge 0}$ is bounded. Since S is uniformly continuous, so ${\{\parallel S{x}_n-q\parallel \}}_{n=1}^{\mathrm{\infty }}$ is also bounded. Thus there is a constant $M_2>0$ satisfying

$$\begin{array}{rcl}{M}_2& =& \underset{n\ge 1}{sup}\parallel x_n-q\parallel +\underset{n\ge 1}{sup}\parallel S{x}_n-q\parallel +\underset{n\ge 1}{sup}\parallel T{y}_n-q\parallel \\ +\underset{n\ge 1}{sup}\parallel u_n-q\parallel +\underset{n\ge 1}{sup}\parallel v_n-q\parallel .\end{array}$$

(2.2)

Denote $M=M_1+M_2$. Obviously, $M<\mathrm{\infty }$. Let $w_n=\parallel T{y}_n-T{x}_n\parallel$ for each $n\ge 1$. The uniform continuity of T ensures that

$$\underset{n\to \mathrm{\infty }}{lim}{w}_n=0.$$

(2.3)

Because

$$\begin{array}{rcl}\parallel y_n-x_n\parallel & =& \parallel b_n(S{x}_n-x_{n-1})+b_n^{\prime }(x_{n-1}-T{y}_n)+c_n(u_n-x_{n-1})-c_n^{\prime }(v_n-x_{n-1})\parallel \\ \le & b_n\parallel S{x}_n-x_{n-1}\parallel +b_n^{\prime }\parallel x_{n-1}-T{y}_n\parallel +c_n\parallel u_n-x_{n-1}\parallel +c_n^{\prime }\parallel v_n-x_{n-1}\parallel \\ \le & 2M_2(b_n+c_n+b_n^{\prime }+c_n^{\prime })\\ \to & 0\end{array}$$

as $n\to \mathrm{\infty }$.

By virtue of Lemma 2.1 and (2.1), we infer that

$$\begin{array}{rcl}{\parallel x_n-q\parallel }^2& =& {\parallel a_n^{\prime }{x}_{n-1}+b_n^{\prime }T{y}_n+c_n^{\prime }{v}_n-q\parallel }^2\\ =& {\parallel a_n^{\prime }(x_{n-1}-q)+b_n^{\prime }(T{y}_n-q)+c_n^{\prime }(v_n-q)\parallel }^2\\ \le & {(1-b_n^{\prime })}^2{\parallel x_{n-1}-q\parallel }^2+2b_n^{\prime }〈T{y}_n-q,j(x_n-q)〉\\ +2c_n^{\prime }〈v_n-q,j(x_n-q)〉\\ \le & {(1-b_n^{\prime })}^2{\parallel x_{n-1}-q\parallel }^2+2b_n^{\prime }〈T{y}_n-T{x}_n,j(x_n-q)〉\\ +2b_n^{\prime }〈T{x}_n-q,j(x_n-q)〉+2c_n^{\prime }\parallel v_n-q\parallel \parallel x_n-q\parallel \\ \le & {(1-b_n^{\prime })}^2{\parallel x_{n-1}-q\parallel }^2+2b_n^{\prime }\parallel T{y}_n-T{x}_n\parallel \parallel x_n-q\parallel \\ +2b_n^{\prime }{\parallel x_n-q\parallel }^2-2b_n^{\prime }\varphi (\parallel x_n-q\parallel )+2M^2{c}_n^{\prime }\\ \le & {(1-b_n^{\prime })}^2{\parallel x_{n-1}-q\parallel }^2+2M{b}_n^{\prime }{w}_n+2b_n^{\prime }{\parallel x_n-q\parallel }^2\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )+2M^2{c}_n^{\prime }.\end{array}$$

(2.4)

Consider

$$\begin{array}{rcl}{\parallel x_n-q\parallel }^2& =& {\parallel a_n^{\prime }{x}_{n-1}+b_n^{\prime }T{y}_n+c_n^{\prime }{v}_n-q\parallel }^2\\ =& {\parallel a_n^{\prime }(x_{n-1}-q)+b_n^{\prime }(T{y}_n-q)+c_n^{\prime }(v_n-q)\parallel }^2\\ \le & a_n^{\prime }{\parallel x_{n-1}-q\parallel }^2+b_n^{\prime }{\parallel T{y}_n-q\parallel }^2+c_n^{\prime }{\parallel v_n-q\parallel }^2\\ \le & {\parallel x_{n-1}-q\parallel }^2+M^2(b_n^{\prime }+c_n^{\prime }),\end{array}$$

(2.5)

where the first inequality holds by the convexity of ${\parallel \cdot \parallel }^2$.

Substituting (2.5) in (2.4), we get

$$\begin{array}{r}{\parallel x_n-q\parallel }^2\\ \le [{(1-b_n^{\prime })}^2+2b_n^{\prime }]{\parallel x_{n-1}-q\parallel }^2+2M{b}_n^{\prime }(w_n+M(b_n^{\prime }+c_n^{\prime }))+2M^2{c}_n^{\prime }\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\\ =(1+b_n^{{\prime }^2}){\parallel x_{n-1}-q\parallel }^2+2M{b}_n^{\prime }(w_n+M(b_n^{\prime }+c_n^{\prime }))+2M^2{c}_n^{\prime }\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\\ \le {\parallel x_{n-1}-q\parallel }^2+M{b}_n^{\prime }(3M{b}_n^{\prime }+2w_n+2M{b}_n^{\prime })+2M^2{c}_n^{\prime }\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\\ ={\parallel x_{n-1}-q\parallel }^2+b_n^{\prime }{l}_n+2M^2{c}_n^{\prime }-2b_n^{\prime }\varphi (\parallel x_n-q\parallel ),\end{array}$$

(2.6)

where

$$l_n=M(3M{b}_n^{\prime }+2w_n+2M{b}_n^{\prime })\to 0$$

(2.7)

as $n\to \mathrm{\infty }$.

Denote

$$\begin{array}{c}{\theta }_n=\parallel x_n-q\parallel ,\hfill \\ {\lambda }_n=2b_n^{\prime },\hfill \\ {\sigma }_n=b_n^{\prime }{l}_n,\hfill \\ {\gamma }_n=2M^2{c}_n^{\prime }.\hfill \end{array}$$

Condition (i) assures the existence of a rank $n_0\in \mathbb{N}$ such that ${\lambda }_n=2b_n^{\prime }\le 1$ for all $n\ge n_0$. Now, with the help of conditions (ii), (iii), (2.3), (2.7) and Lemma 2.2, we obtain from (2.6) that

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

completing the proof. □

Using the method of proof in Theorem 2.3, we have the following result.

Corollary 2.4 Let E, K, T, S, ${\{u_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ be as in Theorem  2.3. Suppose that ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{a_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ and ${\{c_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions (i), (ii), (iv) and $c_n^{\prime }=o(b_n^{\prime })$. Then the conclusions of Theorem  2.3 hold.

Proof From the condition $c_n^{\prime }=o(b_n^{\prime })$, set $c_n^{\prime }=b_n^{\prime }{t}_n$, where ${lim}_{n\to \mathrm{\infty }}{t}_n=0$. Substituting (2.5) in (2.4), we get

$$\begin{array}{r}{\parallel x_n-q\parallel }^2\\ \le [{(1-b_n^{\prime })}^2+2b_n^{\prime }]{\parallel x_{n-1}-q\parallel }^2+2M{b}_n^{\prime }(w_n+M(b_n^{\prime }+c_n^{\prime }+t_n))\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\parallel x_n-q\parallel \\ =(1+b_n^{{\prime }^2}){\parallel x_{n-1}-q\parallel }^2+2M{b}_n^{\prime }(w_n+M(b_n^{\prime }+c_n^{\prime }+t_n))\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\\ \le {\parallel x_{n-1}-q\parallel }^2+M{b}_n^{\prime }(2w_n+M(3b_n^{\prime }+2c_n^{\prime }+2t_n))\\ -2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\\ ={\parallel x_{n-1}-q\parallel }^2-2b_n^{\prime }\varphi (\parallel x_n-q\parallel )\parallel x_n-q\parallel +b_n^{\prime }{l}_n^{\prime },\end{array}$$

(2.8)

where

$$l_n^{\prime }=M(2w_n+M(3b_n^{\prime }+2c_n^{\prime }+2t_n))\to 0$$

(2.9)

as $n\to \mathrm{\infty }$.

It follows from Lemma 2.2 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$. □

Corollary 2.5 Let K be a nonempty convex subset of an arbitrary Banach space E, and let $T:K\to K$ be a uniformly continuous and ϕ-hemicontractive mapping. Suppose that ${\{u_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{v_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded sequences in K and ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{a_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ and ${\{c_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions

• (i) $a_n+b_n+c_n=a_n^{\prime }+b_n^{\prime }+c_n^{\prime }=1$,

• (ii) ${lim}_{n\to \mathrm{\infty }}{b}_n={lim}_{n\to \mathrm{\infty }}{c}_n={lim}_{n\to \mathrm{\infty }}{b}_n^{\prime }=0$,

• (iii) $c_n^{\prime }=o(b_n^{\prime })$, and

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n^{\prime }=\mathrm{\infty }$.

For any $x_0\in K$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ inductively as follows:

$$\begin{array}{c}{y}_n=a_n{x}_{n-1}+b_{n}T{x}_n+c_n{u}_n,\hfill \\ x_n=a_n^{\prime }{x}_{n-1}+b_n^{\prime }T{y}_n+c_n^{\prime }{v}_n,n\ge 1.\hfill \end{array}$$

Then the following conditions are equivalent:

• (a) ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the unique fixed point q of T,

• (b) ${lim}_{n\to \mathrm{\infty }}T{y}_n=q$,

• (c) ${\{T{y}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

Corollary 2.6 Let E, K, T, ${\{u_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ be as in Corollary  2.5. Suppose that ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{a_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ and ${\{c_n^{\prime }\}}_{n=1}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions (i), (ii), (iv) and ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n^{\prime }<\mathrm{\infty }$. Then the conclusions of Corollary  2.5 hold.

Corollary 2.7 Let K be a nonempty convex subset of an arbitrary Banach space E, and let $T,S:K\to K$ be two uniformly continuous and ϕ-hemicontractive mappings. Suppose that ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ are any sequences in $[0,1]$ satisfying

• (i) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0={lim}_{n\to \mathrm{\infty }}{\alpha }_n$,

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

For any $x_0\in K$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ inductively as follows:

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_{n-1}+{\beta }_{n}S{x}_n,\\ x_n=(1-{\alpha }_n)x_{n-1}+{\alpha }_{n}T{y}_n,n\ge 1.\end{array}$$

Then the following conditions are equivalent:

• (a) ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S,

• (b) ${lim}_{n\to \mathrm{\infty }}T{y}_n=q$,

• (c) ${\{T{y}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

Corollary 2.8 Let K be a nonempty convex subset of an arbitrary Banach space E, and let $T:K\to K$ be a uniformly continuous and ϕ-hemicontractive mapping. Suppose that ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ are any sequences in $[0,1]$ satisfying

• (i) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0={lim}_{n\to \mathrm{\infty }}{\alpha }_n$,

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

For any $x_0\in K$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ inductively as follows:

$$\begin{array}{c}{y}_n=(1-{\beta }_n)x_{n-1}+{\beta }_{n}T{x}_n,\hfill \\ x_n=(1-{\alpha }_n)x_{n-1}+{\alpha }_{n}T{y}_n,n\ge 1.\hfill \end{array}$$

Then the following conditions are equivalent:

• (a) ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the unique fixed point q of T,

• (b) ${lim}_{n\to \mathrm{\infty }}T{y}_n=q$,

• (c) ${\{T{y}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

Corollary 2.9 Let K be a nonempty convex subset of an arbitrary Banach space E, and let $T:K\to K$ be a uniformly continuous and ϕ-hemicontractive mapping. Suppose that ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is any sequence in $[0,1]$ satisfying

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$,

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

For any $x_0\in K$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ inductively as follows:

$$x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T{x}_n,n\ge 1.$$

Then the following conditions are equivalent:

• (a) ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the unique fixed point q of T,

• (b) ${lim}_{n\to \mathrm{\infty }}T{x}_n=q$,

• (c) ${\{T{x}_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

All of the above results are also valid for Lipschitz ϕ-hemicontractive mappings.