Almost stability of the Mann type iteration method with error term involving strictly hemicontractive mappings in smooth Banach spaces

Abstract

Let K be a nonempty closed bounded convex subset of an arbitrary smooth Banach space X and $T:K\to K$ be a continuous strictly hemicontractive mapping. Under some conditions, we obtain that the Mann iteration method with error term converges strongly to a unique fixed point of T and is almost T-stable on K. As an application of our results, we establish strong convergence of a multi-step iteration process.

1 Introduction

Chidume [1] established that the Mann iteration sequence converges strongly to the unique fixed point of T in case T is a Lipschitz strongly pseudo-contractive mapping from a bounded closed convex subset of $L_p$ (or $l_p$) into itself. Schu [2] generalized the result in [1] to both uniformly continuous strongly pseudo-contractive mappings and real smooth Banach spaces. Park [3] extended the result in [1] to both strongly pseudocontractive mappings and certain smooth Banach spaces. Rhoades [4] proved that the Mann and Ishikawa iteration methods may exhibit different behavior for different classes of nonlinear mappings. Harder and Hicks [5, 6] revealed the importance of investigating the stability of various iteration procedures for various classes of nonlinear mappings. Harder [7] established applications of stability results to first-order differential equations. Afterwords, several generalizations have been made in various directions (see, for example, [2, 4, 8–21].

Let K be a nonempty closed bounded convex subset of an arbitrary smooth Banach space X and $T:K\to K$ be a continuous strictly hemicontractive mapping. Under some conditions, we obtain that the Mann iteration method with error term converges strongly to a unique fixed point of T and is almost T-stable on K. As an application, we shall also establish strong convergence of a multi-step iteration process. The results presented here generalize the corresponding results in [2–4, 10, 11, 22].

2 Preliminaries

Let K be a nonempty subset of an arbitrary Banach space X and $X^{\ast }$ be its dual space. The symbols $D(T)$, $R(T)$ and $F(T)$ stand for the domain, the range and the set of fixed points of $T:X\to X$ respectively (x is called a fixed point of T iff $T(x)=x$). We denote by J the normalized duality mapping from X to $2^{X^{\ast }}$ defined by

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

Let T be a self-mapping of K.

Definition 1 The mapping T is called Lipshitzian if there exists $L>0$ such that

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

for all $x,y\in K$. If $L=1$, then T is called non-expansive and if $0⩽L<1$, T is called contraction.

Definition 2 [10, 22]

• 1. The mapping T is said to be pseudocontractive if the inequality

  $$\parallel x-y\parallel ⩽\parallel x-y+t[(I-T)x-(I-T)y]\parallel$$

  (2.1)

holds for each $x,y\in K$ and for all $t>0$.

• 2. T is said to be strongly pseudocontractive if there exists $t>1$ such that

  $$\parallel x-y\parallel \le \parallel (1+r)(x-y)-rt(Tx-Ty)\parallel$$

  (2.2)

for all $x,y\in D(T)$ and $r>0$.

• 3. T is said to be local strongly pseudocontractive if for each $x\in D(T)$, there exists $t_x>1$ such that

  $$\parallel x-y\parallel \le \parallel (1+r)(x-y)-r{t}_x(Tx-Ty)\parallel$$

  (2.3)

for all $y\in D(T)$ and $r>0$.

• 4. T is said to be strictly hemicontractive if $F(T)\ne \mathrm{\varnothing }$ and if there exists $t>1$ such that

  $$\parallel x-q\parallel \le \parallel (1+r)(x-q)-rt(Tx-q)\parallel$$

  (2.4)

for all $x\in D(T)$, $q\in F(T)$ and $r>0$.

Clearly, each strongly pseudocontractive operator is local strongly pseudocontractive.

Definition 3 [5–7]

Let K be a nonempty convex subset of X and $T:K\to K$ be an operator. Assume that $x_o\in K$ and $x_{n+1}=f(T,x_n)$ defines an iteration scheme which produces a sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}\subset K$. Suppose, furthermore, that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to $q\in F(T)\ne \mathrm{\varnothing }$. Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any bounded sequence in K and put ${\epsilon }_n=\parallel y_{n+1}-f(T,y_n)\parallel$.

• (1) The iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ defined by $x_{n+1}=f(T,x_n)$ is said to be T-stable on K if ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$.

• (2) The iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ defined by $x_{n+1}=f(T,x_n)$ is said to be almost T-stable on K if ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$.

It is easy to verify that an iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ which is T-stable on K is almost T-stable on K.

Lemma 4 [3]

Let X be a smooth Banach space. Suppose one of the following holds:

• (1) J is uniformly continuous on any bounded subsets of X,

• (2) $〈x-y,j(x)-j(y)〉\le {\parallel x-y\parallel }^2$ for all x, y in X,

• (3) for any bounded subset D of X, there is a $c:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ such that

  $$Re〈x-y,j(x)-j(y)〉\le c(\parallel x-y\parallel ),$$

for all $x,y\in D$, where c satisfies

$$\underset{t\to 0^{+}}{lim}\frac{c(t)}{t}=0.$$

(2.5)

Then for any $ϵ>0$ and any bounded subset K, there exists $\delta >0$ such that

$${\parallel sx+(1-s)y\parallel }^2\le (1-2s){\parallel y\parallel }^2+2sRe〈x,j(y)〉+2sϵ$$

(2.6)

for all $x,y\in K$ and $s\in [0,\delta ]$.

Lemma 5 [10]

Let $T:D(T)\subseteq X\to X$ be an operator with $F(T)\ne \phi$. Then T is strictly hemicontractive if and only if there exists $t>1$ such that for all $x\in D(T)$ and $q\in F(T)$, there exists $j(x-q)\in J(x-q)$ satisfying

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

(2.7)

Lemma 6 [4]

Let X be an arbitrary normed linear space and $T:D(T)\subseteq X\to X$ be an operator.

• (1) If T is a local strongly pseudocontractive operator and $F(T)\ne \mathrm{\varnothing }$, then $F(T)$ is a singleton and T is strictly hemicontractive.

• (2) If T is strictly hemicontractive, then $F(T)$is a singleton.

3 Main results

We now prove our main results.

Lemma 7 Let ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=0}^{\mathrm{\infty }}$ be nonnegative real sequences, and let ${ϵ}^{\mathrm{\prime }}>0$ be a constant satisfying

$${\beta }_{n+1}\le (1-{\alpha }_n^l){\beta }_n+{ϵ}^{\mathrm{\prime }}{\alpha }_n+{\gamma }_n;l\ge 1,n\ge 0,$$

where ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n^l=\mathrm{\infty }$, ${\alpha }_n\le 1$ for all $n\ge 0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$. Then, ${lim}_{n\to \mathrm{\infty }}sup{\beta }_n\le {ϵ}^{\mathrm{\prime }}$.

Proof By a straightforward argument, for $n\ge k\ge 0$,

$${\beta }_{n+1}\le {\beta }_k\prod _{j=k}^n(1-{\alpha }_j^l)+{ϵ}^{\mathrm{\prime }}\sum _{j=k}^n{\alpha }_j\prod _{i=j+1}^n(1-{\alpha }_i^l)+\sum _{j=k}^n{\gamma }_j\prod _{i=j+1}^n(1-{\alpha }_i^l),$$

(3.1)

where we put ${\prod }_{i=n+1}^n(1-{\alpha }_i^l)=1$. Note that ${\sum }_{j=k}^n{\alpha }_j{\prod }_{i=j+1}^n(1-{\alpha }_i^l)\le 1$. It follows from (3.1) that

$${\beta }_{n+1}\le exp(-\sum _{j=k}^n{\alpha }_j^l){\beta }_k+{ϵ}^{\mathrm{\prime }}+\sum _{j=k}^n{\gamma }_j.$$

(3.2)

For a given $\delta >0$, there exists a positive integer k such that ${\sum }_{j=k}^{\mathrm{\infty }}{\gamma }_j<\delta$. Thus (3.2) ensures that

$$\underset{n\to \mathrm{\infty }}{lim}sup{\beta }_n\le {ϵ}^{\mathrm{\prime }}+\delta .$$

Letting $\delta \to 0^{+}$ yields ${lim}_{n\to \mathrm{\infty }}sup{\beta }_n\le {ϵ}^{\mathrm{\prime }}$. □

Remark 8

• (i) If ${\gamma }_n=0$ for each $n\ge 0$, then Lemma 7 reduces to Lemma 1 of Park [3].

• (ii) If $l=1$, then Lemma 7 reduces to Lemma 2.1 of Liu et al. [4].

Theorem 9 Let Xbe a smooth Banach space satisfying any one of the Axioms (1)-(3) of Lemma 4. Let K be a nonempty closed bounded convex subset of X and $T:K\to K$ be a continuous strictly hemicontractive mapping. Suppose that ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$ is an arbitrary sequence in K and ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ and ${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ are any sequences in $[0,1]$ satisfying conditions (i) $a_n^{\mathrm{\prime }}+b_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}=1$, (ii) $c_n^{\mathrm{\prime }}=o(b_n^{\mathrm{\prime }})$, (iii) ${lim}_{n\to \mathrm{\infty }}{b}_n^{\mathrm{\prime }}=0$ and (iv) ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n^{\mathrm{\prime }}=\mathrm{\infty }$.

For a sequence ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$ in K, suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is the sequence generated from an arbitrary $x_0\in K$ by

$$x_{n+1}=a_n^{\mathrm{\prime }}{x}_n+b_n^{\mathrm{\prime }}T{v}_n+c_n^{\mathrm{\prime }}{u}_n,n\ge 0,$$

(3.3)

and satisfying ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$.

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

(3.4)

where $p_n=a_n^{\mathrm{\prime }}{y}_n+b_n^{\mathrm{\prime }}T{v}_n+c_n^{\mathrm{\prime }}{u}_n$, such that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$.

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to a unique fixed point q of T,

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost T-stable on K,

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Proof From (ii), we have $c_n^{\mathrm{\prime }}=t_n{b}_n^{\mathrm{\prime }}$, where $t_n\to 0$ as $n\to \mathrm{\infty }$.

It follows from Lemma 6 that $F(T)$ is a singleton. That is, $F(T)=\{q\}$ for some $q\in K$.

Set $M=1+diamK$. For all $n\ge 0$, it is easy to verify that

$$\begin{array}{rcl}M& =& \underset{n\ge 0}{sup}\parallel x_n-q\parallel +\underset{n\ge 0}{sup}\parallel T{v}_n-q\parallel +\underset{n\ge 0}{sup}\parallel u_n-q\parallel \\ +\underset{n\ge 0}{sup}\{\parallel p_n-q\parallel \}+\underset{n\ge 0}{sup}\parallel y_n-q\parallel .\end{array}$$

(3.5)

For given any $ϵ>0$ and the bounded subset K, there exists a $\delta >0$ satisfying (2.6). Note that (ii), (iii), ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$ and the continuity of T ensure that there exists an N such that

$$b_n^{\mathrm{\prime }}<min\{\delta ,\frac{1}{2(1-k)}\},t_n\le \frac{ϵ}{16M^2},\parallel T{v}_n-T{x}_n\parallel \le \frac{ϵ}{4M},n\ge N,$$

(3.6)

where $k=\frac{1}{t}$ and t satisfies (2.7). Using (3.3) and Lemma 4, we infer that

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\parallel (1-b_n^{\mathrm{\prime }})(x_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)+c_n^{\mathrm{\prime }}(u_n-x_n)\parallel }^2\\ \le & {(\parallel (1-b_n^{\mathrm{\prime }})(x_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)\parallel +2M{c}_n^{\mathrm{\prime }})}^2\\ \le & {\parallel (1-b_n^{\mathrm{\prime }})(x_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)\parallel }^2+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2b_n^{\mathrm{\prime }}){\parallel x_n-q\parallel }^2+2b_n^{\mathrm{\prime }}Re(T{v}_n-q,j(x_n-q))+2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ =& (1-2b_n^{\mathrm{\prime }}){\parallel x_n-q\parallel }^2+2b_n^{\mathrm{\prime }}Re(T{x}_n-q,j(x_n-q))\\ +2b_n^{\mathrm{\prime }}Re(T{v}_n-T{x}_n,j(x_n-q))+2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2b_n^{\mathrm{\prime }}){\parallel x_n-q\parallel }^2+2k{b}_n^{\mathrm{\prime }}{\parallel x_n-q\parallel }^2\\ +2b_n^{\mathrm{\prime }}\parallel T{v}_n-T{x}_n\parallel \parallel x_n-q\parallel +2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2(1-k)b_n^{\mathrm{\prime }}){\parallel x_n-q\parallel }^2\\ +2M{b}_n^{\mathrm{\prime }}\parallel T{v}_n-T{x}_n\parallel +2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2(1-k)b_n^{\mathrm{\prime }}){\parallel x_n-q\parallel }^2+3ϵb_n^{\mathrm{\prime }},\end{array}$$

(3.7)

for all $n\ge N$.

Put

we have from (3.7)

$${\beta }_{n+1}\le (1-{\alpha }_n){\beta }_n+{ϵ}^{\mathrm{\prime }}{\alpha }_n+{\gamma }_n,n\ge 0.$$

Observe that ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, ${\alpha }_n<1$ for all $n\ge 0$. It follows from Lemma 7 that

$$\underset{n\to \mathrm{\infty }}{lim}sup{\parallel x_n-q\parallel }^2\le {ϵ}^{\mathrm{\prime }}.$$

Letting ${ϵ}^{\mathrm{\prime }}\to 0^{+}$, we obtain that ${lim}_{n\to \mathrm{\infty }}sup{\parallel x_n-q\parallel }^2=0$, which implies that $x_n\to q$ as $n\to \mathrm{\infty }$.

On the same lines, we obtain

$$\begin{array}{rcl}{\parallel p_n-q\parallel }^2& =& {\parallel (1-b_n^{\mathrm{\prime }})(y_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)+c_n^{\mathrm{\prime }}(u_n-y_n)\parallel }^2\\ \le & {(\parallel (1-b_n^{\mathrm{\prime }})(y_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)\parallel +2M{c}_n^{\mathrm{\prime }})}^2\\ \le & {\parallel (1-b_n^{\mathrm{\prime }})(y_n-q)+b_n^{\mathrm{\prime }}(T{v}_n-q)\parallel }^2+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2b_n^{\mathrm{\prime }}){\parallel y_n-q\parallel }^2+2b_n^{\mathrm{\prime }}Re(T{v}_n-q,j(y_n-q))+2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ =& (1-2b_n^{\mathrm{\prime }}){\parallel y_n-q\parallel }^2+2b_n^{\mathrm{\prime }}Re(T{y}_n-q,j(y_n-q))\\ +2b_n^{\mathrm{\prime }}Re(T{v}_n-T{y}_n,j(y_n-q))+2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2b_n^{\mathrm{\prime }}){\parallel y_n-q\parallel }^2+2k{b}_n^{\mathrm{\prime }}{\parallel y_n-q\parallel }^2\\ +2b_n^{\mathrm{\prime }}\parallel T{v}_n-T{y}_n\parallel \parallel y_n-q\parallel +2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2(1-k)b_n^{\mathrm{\prime }}){\parallel y_n-q\parallel }^2\\ +2M{b}_n^{\mathrm{\prime }}\parallel T{v}_n-T{y}_n\parallel +2ϵb_n^{\mathrm{\prime }}+8M^2{c}_n^{\mathrm{\prime }}\\ \le & (1-2(1-k)b_n^{\mathrm{\prime }}){\parallel y_n-q\parallel }^2+3ϵb_n^{\mathrm{\prime }},\end{array}$$

(3.8)

for all $n\ge N$.

Suppose that ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$. In view of (3.4) and (3.8), we infer that

$$\begin{array}{rcl}{\parallel y_{n+1}-q\parallel }^2& \le & {(\parallel y_{n+1}-p_n\parallel +\parallel p_n-q\parallel )}^2\\ \le & {\parallel p_n-q\parallel }^2+3M{\epsilon }_n\\ \le & [1-2b_n^{\mathrm{\prime }}(1-k)]{\parallel y_n-q\parallel }^2+3ϵb_n^{\mathrm{\prime }}+3M{\epsilon }_n,\end{array}$$

(3.9)

for all $n\ge N$.

Now, put

and we have from (3.9)

$${\beta }_{n+1}\le (1-{\alpha }_n){\beta }_n+{ϵ}^{\mathrm{\prime }}{\alpha }_n+{\gamma }_n,n\ge 0.$$

Observe that ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, ${\alpha }_n<1$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$ for all $n\ge 0$. It follows from Lemma 7 that

$$\underset{n\to \mathrm{\infty }}{lim}sup{\parallel y_n-q\parallel }^2\le {ϵ}^{\mathrm{\prime }}.$$

Letting ${ϵ}^{\mathrm{\prime }}\to 0^{+}$, we obtain that ${lim}_{n\to \mathrm{\infty }}sup{\parallel y_n-q\parallel }^2=0$, which implies that $y_n\to q$ as $n\to \mathrm{\infty }$.

Conversely, suppose that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, then (iii) and (3.8) imply that

$$\begin{array}{rcl}{\epsilon }_n& \le & \parallel y_{n+1}-q\parallel +\parallel p_n-q\parallel \\ \le & \parallel y_{n+1}-q\parallel +{[[1-2(1-k)b_n^{\mathrm{\prime }}]{\parallel y_n-q\parallel }^2+3ϵb_n^{\mathrm{\prime }}]}^{\frac{1}{2}}\\ \to & 0,\end{array}$$

as $n\to \mathrm{\infty }$, that is, ${\epsilon }_n\to 0$ as $n\to \mathrm{\infty }$. □

Using the methods of the proof of Theorem 9, we can easily prove the following.

Theorem 10 Let X, K, T and ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$, be as in Theorem 9. Suppose that ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ and ${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions (i), (iii)-(iv) and

$$\sum _{n=0}^{\mathrm{\infty }}{c}_n^{\mathrm{\prime }}<\mathrm{\infty }.$$

If ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{p_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ are as in Theorem 9, then the conclusions of Theorem 9 hold.

Corollary 11 Let X be a smooth Banach space satisfying any one of the Axioms (1)-(3) of Lemma 4. Let K be a nonempty closed bounded convex subset of X and $T:K\to K$ be a Lipschitz strictly hemicontractive mapping. Suppose that ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$ is an arbitrary sequence in K and ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ and ${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ are any sequences in $[0,1]$ satisfying conditions (i) $a_n^{\mathrm{\prime }}+b_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}=1$, (ii) $c_n^{\mathrm{\prime }}=o(b_n^{\mathrm{\prime }})$, (iii) ${lim}_{n\to \mathrm{\infty }}{b}_n^{\mathrm{\prime }}=0$ and (iv) ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n^{\mathrm{\prime }}=\mathrm{\infty }$.

For a sequence ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$ in K, suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is the sequence generated from an arbitrary $x_0\in K$ by

$$x_{n+1}=a_n^{\mathrm{\prime }}{x}_n+b_n^{\mathrm{\prime }}T{v}_n+c_n^{\mathrm{\prime }}{u}_n,n\ge 0,$$

and satisfying ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$.

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

where $p_n=a_n^{\mathrm{\prime }}{y}_n+b_n^{\mathrm{\prime }}T{v}_n+c_n^{\mathrm{\prime }}{u}_n$, such that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$.

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to a unique fixed point q of T,

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost T-stable on K,

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Corollary 12 Let X, K, T and ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$ be as in Corollary 11. Suppose that ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ and ${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$ are sequences in $[0,1]$ satisfying conditions (i), (iii)-(iv) and

$$\sum _{n=0}^{\mathrm{\infty }}{c}_n^{\mathrm{\prime }}<\mathrm{\infty }.$$

If ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{p_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ are as in Corollary 11, then the conclusions of Corollary 11 hold.

Corollary 13 Let X be a smooth Banach space satisfying any one of the Axioms (1)-(3) of Lemma 4. Let K be a nonempty closed bounded convex subset of X and $T:K\to K$ be a continuous strictly hemicontractive mapping. Suppose that ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$ is a sequence in $[0,1]$ satisfying conditions (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

For a sequence ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$ in K, suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is the sequence generated from an arbitrary $x_0\in K$ by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T{v}_n,n\ge 0,$$

and satisfying ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$.

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

where $p_n={\alpha }_n{y}_n+(1-{\alpha }_n)T{v}_n$, such that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$.

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to a unique fixed point q of T,

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost T-stable on K,

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Corollary 14 Let X be a smooth Banach space satisfying any one of the Axioms (1)-(3) of Lemma 4. Let K be a nonempty closed bounded convex subset of X and $T:K\to K$ be a Lipschitz strictly hemicontractive mapping. Suppose that ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$ is a sequence in $[0,1]$ satisfying conditions (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

For a sequence ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$ in K, suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is the sequence generated from an arbitrary $x_0\in K$ by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T{v}_n,n\ge 0,$$

and satisfying ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$.

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

where $p_n={\alpha }_n{y}_n+(1-{\alpha }_n)T{v}_n$, such that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$.

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to a unique fixed point q of T,

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost T-stable on K,

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

4 Applications to a multi-step iteration process

Khan et al. [23] have introduced and studied a multi-step iteration process for a finite family of selfmappings. We now introduce a modified multi-step process as follows:

Let K be a nonempty closed convex subset of a real normed space E and $T_1,T_2,\dots ,T_p:K\to K$ ($p\ge 2$) be a family of selfmappings.

Algorithm 1 For a given $x_0\in K$, compute the sequence ${\{x_n\}}_{n\ge 0}$ by the iteration process of arbitrary fixed order $p\ge 2$,

(4.1)

which is called the modified multi-step iteration process, where ${\{{\alpha }_n\}}_{n\ge 0},{\{{\beta }_n^i\}}_{n\ge 0}\subset [0,1]$, $i=1,2,\dots ,p-1$.

For $p=3$, we obtain the following three-step iteration process:

Algorithm 2 For a given $x_0\in K$, compute the sequence ${\{x_n\}}_{n\ge 0}$ by the iteration process:

(4.2)

where ${\{{\alpha }_n\}}_{n\ge 0}$, ${\{{\beta }_n^1\}}_{n\ge 0}$ and ${\{{\beta }_n^2\}}_{n\ge 0}$ are three real sequences in $[0,1]$.

For $p=2$, we obtain the Ishikawa [24] iteration process:

Algorithm 3 For a given $x_0\in K$, compute the sequence ${\{x_n\}}_{n\ge 0}$ by the iteration process

$$\begin{array}{r}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{T}_1{y}_n^1,\\ y_n^1=(1-{\beta }_n^1)x_n+{\beta }_n^1{T}_2{x}_n,n\ge 0,\end{array}$$

(4.3)

where ${\{{\alpha }_n\}}_{n\ge 0}$ and ${\{{\beta }_n^1\}}_{n\ge 0}$ are two real sequences in $[0,1]$.

If $T_1=T$, $T_2=I$, ${\beta }_n^1=0$ in (4.3), we obtain the Mann iteration process [14]:

Algorithm 4 For any given $x_0\in K$, compute the sequence ${\{x_n\}}_{n\ge 0}$ by the iteration process

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,n\ge 0,$$

(4.4)

where $\{{\alpha }_n\}$ is a real sequence in $[0,1]$.

Theorem 15 Let K be a nonempty closed bounded convex subset of a smooth Banach space X and $T_1,T_2,\dots ,T_p$ ($p\ge 2$) be selfmappings of K. Let $T_1$ be a continuous strictly hemicontractive mapping. Let ${\{{\alpha }_n\}}_{n\ge 0},{\{{\beta }_n^i\}}_{n\ge 0}\subset [0,1]$, $i=1,2,\dots ,p-1$ be real sequences in $[0,1]$ satisfying ${\sum }_{n\ge 0}{\alpha }_n=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n^1=0$. For arbitrary $x_0\in K$, define the sequence ${\{x_n\}}_{n\ge 0}$ by (4.1). Then ${\{x_n\}}_{n\ge 0}$ converges strongly to a point in ${\bigcap }_{i=1}^{p}F(T_i)\ne \mathrm{\varnothing }$.

Proof By applying Corollary 13 under assumption that $T_1$ is continuous strictly hemicontractive mapping, we obtain Theorem 15 which proves strong convergence of the iteration process defined by (4.1). We will check only the condition ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$ by taking $T_1=T$ and $v_n=y_n^1$,

$$\begin{array}{rcl}\parallel v_n-x_n\parallel & =& \parallel y_n^1-x_n\parallel \\ =& \parallel (1-{\beta }_n^1)x_n+{\beta }_n^1{T}_2{y}_n^2-x_n\parallel \\ =& {\beta }_n^1\parallel T_2{y}_n^2-x_n\parallel \\ \le & 2M{\beta }_n^1.\end{array}$$

Now, from the condition ${lim}_{n\to \mathrm{\infty }}{\beta }_n^1=0$, it can be easily seen that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$. □

Corollary 16 Let K be a nonempty closed bounded convex subset of a smooth Banach space X and $T_1,T_2,\dots ,T_p$ ($p\ge 2$) be selfmappings of K. Let $T_1$ be a Lipschitz strictly hemicontractive mapping. Let ${\{{\alpha }_n\}}_{n\ge 0},{\{{\beta }_n^i\}}_{n\ge 0}\subset [0,1]$, $i=1,2,\dots ,p-1$ be real sequences in $[0,1]$ satisfying ${\sum }_{n\ge 0}{\alpha }_n=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n^1=0$. For arbitrary $x_0\in K$, define the sequence ${\{x_n\}}_{n\ge 0}$ by (4.1). Then ${\{x_n\}}_{n\ge 0}$ converges strongly to a point in ${\bigcap }_{i=1}^{p}F(T_i)\ne \mathrm{\varnothing }$.

Remark 17 Similar results can be found for the iteration processes with error terms, we omit the details.