The convergence of the modified Mann and Ishikawa iterations in Banach spaces

Abstract

In this paper, under the new condition we show that the convergence of the modified Mann and Ishikawa iterations is equivalent for uniformly L-Lipschitz asymptotically pseudocontractive mappings in real Banach spaces. Our results extend and improve the corresponding results of Zeng (Acta Math. Sin. 47:219-228, 2004).

MSC:47H09, 47H10.

1 Introduction and preliminaries

Throughout the paper, we assume that E is an arbitrary real Banach space, D is a nonempty closed convex subset of E, $T:D\to D$ is a self-mapping and $F(T)$ is the fixed point set of T, i.e., $F(T)=\{x\in D:Tx=x\}$. Let J denote the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

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

(1.1)

where $E^{\ast }$ denotes the dual space of E and $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. The single-valued normalized duality mapping is denoted by j.

Definition 1.1 (see [1])

• (1) A mapping T is said to be uniformly L-Lipschitz if there exists a constant $L>0$ such that, for all $x,y\in D$,

  $$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }n\ge 1.$$

  (1.2)

• (2) The mapping T is said to be asymptotically nonexpansive with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ if, for all $x,y\in D$,

  $$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel ,\mathrm{\forall }n\ge 1.$$

  (1.3)

• (3) The mapping T is said to be asymptotically pseudocontractive with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ if, for all $x,y\in D$, there exists $j(x-y)\in J(x-y)$ such that

  $$〈T^{n}x-T^{n}y,j(x-y)〉\le k_n{\parallel x-y\parallel }^2,\mathrm{\forall }n\ge 1.$$

  (1.4)

Obviously, an asymptotically nonexpansive mapping is both asymptotically pseudocontractive and uniformly L-Lipschitz, but the converse is not true in general. For more details on uniformly L-Lipschitz asymptotically nonexpansive and asymptotically pseudocontractive mappings, see [2–6] and [7–11].

Definition 1.2 (see [1])

For any $u_1,x_1\in D$, the sequences $\{u_n\}$ and $\{x_n\}$ in D defined by

$$u_{n+1}=(1-a_n)u_n+a_n{T}^n{u}_n,\mathrm{\forall }n\ge 1,$$

(1.5)

and

$$\{\begin{array}{c}{y}_n=(1-b_n)x_n+b_n{T}^n{x}_n,\hfill \\ x_{n+1}=(1-a_n)x_n+a_n{T}^n{y}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(1.6)

are called the modified Mann and Ishikawa iterations, respectively, where $\{a_n\}$, $\{b_n\}$ are two real sequences in $[0,1]$ satisfying some conditions. For more details on the Mann and Ishikawa iterations, see [4, 12] and [11].

In 2001, Chidume and Mutangadura [13] constructed an example for every nontrivial Mann iteration failing to converge while Ishikawa iteration converges. Therefore, there exist some differences between convergence of two kinds of the iterative sequences. Since then, many authors have shown that the Mann (modified Mann) and Ishikawa (modified Ishikawa) iterations (with errors) converge strongly to fixed points of pseudocontractive mappings and others under appropriate conditions.

Especially, Chang [1] proved the following.

Theorem 1.3 [[1], Theorem 2.1]

Let D be a nonempty closed convex subset of E and $T:D\to D$ be a uniformly L-Lipschitzian asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and $L\ge 1$. Let $\{a_n\}$ and $\{b_n\}$ be two real sequences in $[0,1]$ satisfying the following conditions:

• (a) $a_n,b_n\to 0$ as $n\to \mathrm{\infty }$;

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$.

For any $x_0\in D$, let $\{x_n\}$ be the modified Ishikawa iteration defined by (1.2). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that

$$〈T^n{x}_{n+1}-q,j(x_{n+1}-q)〉\le k_n{\parallel x_{n+1}-q\parallel }^2-\mathrm{\Phi }(\parallel x_{n+1}-q\parallel ),\mathrm{\forall }n\ge 0,$$

where $j(x_{n+1}-q)\in J(x_{n+1}-q)$, then $\{x_n\}$ converges strongly to a fixed point q of T.

Theorem 1.4 [[1], Theorem 2.3]

Let D be a nonempty closed convex subset of E and $T:D\to D$ be a uniformly L-Lipschitzian asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and $L\ge 1$. Let $\{a_n\}$ be the real sequence in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$;

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$.

For any $u_0\in D$, let $\{u_n\}$ be the modified Mann iteration defined by (1.1). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ satisfying the condition (2.1) of [[1], Theorem 2.1], then $\{u_n\}$ converges strongly to a fixed point q of T.

Motivated by Theorems 1.3 and 1.4, Zeng [14] gave another interesting results as follows.

Theorem 1.5 [[14], Theorem 2.1]

Let D be a nonempty closed convex subset of E and $T:D\to D$ be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and $L\ge 1$. Let $\{a_n\}$ and $\{b_n\}$ be two real sequences in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$;

• (c) ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n<\mathrm{\infty }$.

For arbitrary $x_0\in D$, let $\{x_n\}$ be the modified Ishikawa iteration defined by (1.2). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that

$$〈T^n{x}_{n+1}-q,j(x_{n+1}-q)〉\le k_n{\parallel x_{n+1}-q\parallel }^2-\mathrm{\Phi }(\parallel x_{n+1}-q\parallel ),\mathrm{\forall }n\ge 0,$$

where $j(x_{n+1}-q)\in J(x_{n+1}-q)$, then $\{x_n\}$ converges strongly to the fixed point q of T.

Theorem 1.6 [[14], Theorem 2.3]

Let D be a nonempty closed convex subset of E and $T:D\to D$ be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and $L\ge 1$. Let $\{a_n\}$ be a real sequence in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$.

For arbitrary $u_0\in D$, let $\{u_n\}$ be the modified Mann iteration defined by (1.1). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ satisfying the condition (2.1) of [[1], Theorem 2.1], then $\{u_n\}$ converges strongly to the fixed point q of T.

It is worth mentioning that the result of Chang [1] is different from that of Zeng [14]. This can be seen from the following example.

Example 1.7 Set

$$a_n=\{\begin{array}{cc}0,\hfill & n=2i,\hfill \\ \frac{1}{n},\hfill & n=2i-1,\hfill \end{array}{b}_n=\{\begin{array}{cc}\frac{1}{2},\hfill & n=2i,\hfill \\ \frac{1}{n},\hfill & n=2i-1,\hfill \end{array}{k}_n=1+\frac{1}{n},\mathrm{\forall }i\ge 1,n\ge 1.$$

Then $a_n\to 0$ as $n\to \mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n{b}_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$, but $b_n\to 0$ as $n\to \mathrm{\infty }$ does not hold. On the other hand, let

$$a_n=\{\begin{array}{cc}0,\hfill & n=2i,\hfill \\ \frac{1}{\sqrt{n}},\hfill & n=2i-1,\hfill \end{array}{b}_n=\{\begin{array}{cc}0,\hfill & n=2i,\hfill \\ \frac{1}{\sqrt{n}},\hfill & n=2i-1,\hfill \end{array}{k}_n=1+\frac{1}{\sqrt{n}},\mathrm{\forall }i\ge 1,n\ge 1.$$

Then $a_n,b_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$, but ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2=\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n{b}_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n(k_n-1)=\mathrm{\infty }$.

The aim of this paper is to extend and improve Theorem 1.5 and Theorem 1.6.

For this, we need to use the following lemmas.

Lemma 1.8 [1]

Let E be a real Banach space and $J:E\to 2^{E^{\ast }}$ be a normalized duality mapping. Then, for all $x,y\in E$ and $j(x+y)\in J(x+y)$,

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

(1.7)

Lemma 1.9 [14]

Let $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ be three nonnegative real sequences satisfying

$$a_{n+1}\le (1+b_n)a_n+c_n,\mathrm{\forall }n\ge 0.$$

(1.8)

If ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

2 Main results

Now, we give the main results in this paper.

Theorem 2.1 Let D be a nonempty closed convex subset of E and $T:D\to D$ be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and $L\ge 1$. Let $\{a_n\}$ be a real sequence in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$.

For arbitrary $u_1\in D$, let $\{u_n\}$ be the modified Mann iteration defined by (1.5). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing continuous function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that

$$〈T^n{u}_{n+1}-q,j(u_{n+1}-q)〉\le k_n{\parallel u_{n+1}-q\parallel }^2-\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2},\mathrm{\forall }n\ge 1,$$

where $j(u_{n+1}-q)\in J(u_{n+1}-q)$, then $\{u_n\}$ converges strongly to a fixed point q of T.

Proof Applying (1.5) and Lemma 1.8, we have

$$\begin{array}{rcl}{\parallel u_{n+1}-q\parallel }^2& =& {\parallel (1-a_n)(u_n-q)+a_n(T^n{u}_n-q)\parallel }^2\\ \le & {(1-a_n)}^2{\parallel u_n-q\parallel }^2+2a_n〈T^n{u}_n-q,j(u_{n+1}-q)〉\\ =& {(1-a_n)}^2{\parallel u_n-q\parallel }^2+2a_n〈T^n{u}_n-T^n{u}_{n+1},j(u_{n+1}-q)〉\\ +2a_n〈T^n{u}_{n+1}-T^{n}q,j(u_{n+1}-q)〉\\ \le & {(1-a_n)}^2{\parallel u_n-q\parallel }^2+2a_{n}L\parallel u_n-u_{n+1}\parallel \cdot \parallel u_{n+1}-q\parallel \\ +2a_n[k_n{\parallel u_{n+1}-q\parallel }^2-\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2}].\end{array}$$

(2.1)

Observe that

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & =& \parallel a_n(T^n{u}_n-u_n)\parallel \\ \le & a_n\parallel T^n{u}_n-T^{n}q+q-u_n\parallel \\ \le & a_n(1+L)\parallel u_n-q\parallel .\end{array}$$

(2.2)

Substituting (2.2) into (2.1), we obtain

$$\begin{array}{rcl}{\parallel u_{n+1}-q\parallel }^2& \le & {(1-a_n)}^2{\parallel u_n-q\parallel }^2+a_n^{2}L(1+L)({\parallel u_n-q\parallel }^2+{\parallel u_{n+1}-q\parallel }^2)\\ +2a_n[k_n{\parallel u_{n+1}-q\parallel }^2-\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2}].\end{array}$$

(2.3)

Since $a_n\to 0$ and $k_n\to 1$ as $n\to \mathrm{\infty }$, without loss of generality, we assume that

$$\frac{1}{2}<1-a^{2}L(1+L)-2a_n{k}_n<1,\mathrm{\forall }n\ge 1.$$

Then (2.3) implies that

(2.4)

Since ${\sum }_{n=1}^{\mathrm{\infty }}\{4a_n(k_n-1)+2a_n^2[1+2L(1+L)]\}<\mathrm{\infty }$, by Lemma 1.9, ${lim}_{n\to \mathrm{\infty }}\parallel u_n-q\parallel$ exists. Denote $M={sup}_{n\ge 1}\{\parallel u_n-q\parallel \}$.

On the other hand, from (2.4), we have

(2.5)

Let ${inf}_{n\ge 1}\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2}=\delta$. Then $\delta =0$. Assume $\delta >0$. Then we have

$$\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2}\ge \delta ,\mathrm{\forall }n\ge 1.$$

It follows from (2.5) that

$$2a_n\delta \le {\parallel u_n-q\parallel }^2-{\parallel u_{n+1}-q\parallel }^2+[4a_n(k_n-1)+2a_n^2(1+2L(1+L))]M^2,$$

(2.6)

which implies that

$$2\delta \sum _{n=1}^{\mathrm{\infty }}{a}_n\le {\parallel u_1-q\parallel }^2+\sum _{n=1}^{\mathrm{\infty }}[4a_n(k_n-1)+2a_n^2(1+2L(1+L))]M^2<\mathrm{\infty },$$

(2.7)

which is a contradiction, and so $\delta =0$. Thus, there exists a subsequence

$$\left\{\frac{\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )+{\parallel u_{n_i+1}-q\parallel }^2}\right\}$$

of

$$\left\{\frac{\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n+1}-q\parallel )+{\parallel u_{n+1}-q\parallel }^2}\right\}$$

such that

$$\underset{i\to \mathrm{\infty }}{lim}\frac{\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )+{\parallel u_{n_i+1}-q\parallel }^2}=0.$$

Since $0\le \parallel u_n-q\parallel \le M$, it follows that

$$0\le \frac{\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )}{1+\mathrm{\Phi }(M)+M^2}\le \frac{\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )+{\parallel u_{n_i+1}-q\parallel }^2}.$$

Thus, ${lim}_{i\to \mathrm{\infty }}\mathrm{\Phi }(\parallel u_{n_i+1}-q\parallel )=0$. By the strictly increasing continuous function Φ, we obtain that ${lim}_{i\to \mathrm{\infty }}\parallel u_{n_i+1}-q\parallel =0$ and so ${lim}_{n\to \mathrm{\infty }}\parallel u_n-q\parallel =0$. This completes the proof. □

Theorem 2.2 Let E be a real Banach space and D be a nonempty closed convex subset of E. Let $T:D\to D$ be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$. Suppose that $\{a_n\}$ and $\{b_n\}$ are two real sequences in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$;

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n{b}_n<\mathrm{\infty }$.

For any $u_1,x_1\in D$, let $\{u_n\}$ and $\{x_n\}$ be the modified Mann and Ishikawa iterations defined by (1.5) and (1.6), respectively. If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing continuous function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that

where $j(x_{n+1}-u_{n+1})\in J(x_{n+1}-u_{n+1})$. Then the following two assertions are equivalent:

• (1) $\{u_n\}$ converges strongly to the fixed point q of T.

• (2) $\{x_n\}$ converges strongly to the fixed point q of T.

Proof If the iteration (1.6) converges to a fixed point q, then, by putting $b_n=0$, we can get the convergence of the iteration (1.5).

Conversely, we only need to prove that the iteration (1.5) ⇒ the iteration (1.6), i.e., $\parallel u_n-q\parallel \to 0$ as $n\to \mathrm{\infty }\Rightarrow \parallel x_n-q\parallel \to 0$ as $n\to \mathrm{\infty }$. Here, without loss of generality, let $\parallel u_n-q\parallel \le 1$. Then $\parallel T^n{u}_n-u_n\parallel \le (1+L)$.

Applying the iterations (1.5), (1.6) and Lemma 1.8, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-u_{n+1}\parallel }^2& =& {\parallel (1-a_n)(x_n-u_n)+a_n(T^n{y}_n-T^n{u}_n)\parallel }^2\\ \le & {(1-a_n)}^2{\parallel x_n-u_n\parallel }^2+2a_n〈T^n{y}_n-T^n{u}_n,j(x_{n+1}-u_{n+1})〉\\ =& {(1-a_n)}^2{\parallel x_n-u_n\parallel }^2+2a_n〈T^n{y}_n-T^n{x}_{n+1},j(x_{n+1}-u_{n+1})〉\\ +2a_n〈T^n{x}_{n+1}-T^n{u}_{n+1},j(x_{n+1}-u_{n+1})〉\\ +2a_n〈T^n{u}_{n+1}-T^n{u}_n,j(x_{n+1}-u_{n+1})〉\\ \le & {(1-a_n)}^2{\parallel x_n-u_n\parallel }^2+2a_{n}L\parallel y_n-x_{n+1}\parallel \cdot \parallel x_{n+1}-u_{n+1}\parallel \\ +2a_n[k_n{\parallel x_{n+1}-u_{n+1}\parallel }^2-\frac{\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )+{\parallel x_{n+1}-u_{n+1}\parallel }^2}]\\ +2a_{n}L\parallel u_{n+1}-u_n\parallel \cdot \parallel x_{n+1}-u_{n+1}\parallel .\end{array}$$

(2.8)

Observe that

(2.9)

(2.10)

(2.11)

where $A_n=b_n(L+1)+a_n[1+L(1+b_{n}L)]\to 0$ and $B_n=b_n(L+1)+a_n(L+1)(1+L{b}_n)\to 0$ as $n\to \mathrm{\infty }$. Substituting (2.9), (2.11) into (2.8), we obtain

(2.12)

Since $a_n,b_n,A_n,B_n,k_n-1\to 0$ as $n\to \mathrm{\infty }$, without loss of generality, we assume that

$$\frac{1}{2}<1-a_{n}L{A}_n-a_{n}L{B}_n-a_n^{2}L(1+L)-2a_n{k}_n<1,\mathrm{\forall }n\ge 1.$$

Then (2.12) implies that

(2.13)

Since

$$\sum _{n=1}^{\mathrm{\infty }}[4a_{n}L{A}_n+2a_{n}L{B}_n+2a_n^2(1+L+L^2)+4a_n(k_n-1)]<\mathrm{\infty }$$

and ${\sum }_{n=1}^{\mathrm{\infty }}[2a_{n}L{B}_n+2a_n^{2}L(1+L)]<\mathrm{\infty }$, by Lemma 1.9, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel$ exists. Denote $M_0={sup}_{n\ge 1}\{\parallel x_n-u_n\parallel \}$.

On the other hand, from (2.13), it follows that

(2.14)

Let ${inf}_{n\ge 1}\frac{\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )+{\parallel x_{n+1}-u_{n+1}\parallel }^2}=\delta$. Then $\delta =0$. Assume $\delta >0$. Then we have

$$\frac{\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n+1}-u_{n+1}\parallel )+{\parallel x_{n+1}-u_{n+1}\parallel }^2}\ge \delta ,\mathrm{\forall }n\ge 1.$$

It follows from (2.14) that

$$\begin{array}{rcl}2a_n\delta & \le & {\parallel x_n-u_n\parallel }^2-{\parallel x_{n+1}-u_{n+1}\parallel }^2\\ +[4a_{n}L{A}_n+2a_{n}L{B}_n+2a_n^2(1+L+L^2)+4a_n(k_n-1)]M^2\\ +2a_{n}L{B}_n+2a_n^{2}L(1+L),\end{array}$$

(2.15)

which implies that

$$\begin{array}{rcl}2\delta \sum _{n=1}^{\mathrm{\infty }}{a}_n& \le & {\parallel x_1-u_1\parallel }^2+\sum _{n=1}^{\mathrm{\infty }}[4a_{n}L{A}_n+2a_{n}L{B}_n+2a_n^2(1+L)+4a_n(k_n-1)]M^2\\ +\sum _{n=1}^{\mathrm{\infty }}[2a_{n}L{B}_n+2a_n^{2}L(1+L)]<\mathrm{\infty },\end{array}$$

(2.16)

which is a contradiction, and so $\delta =0$. Thus, there exists a subsequence

$$\left\{\frac{\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )+{\parallel x_{n_i+1}-u_{n_i+1}\parallel }^2}\right\}$$

of

$$\left\{\frac{\mathrm{\Phi }(\parallel x_{n+1}-u_{n_i+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n+1}-u_{n_i+1}\parallel )+{\parallel x_{n+1}-u_{n+1}\parallel }^2}\right\}$$

such that

$$\underset{i\to \mathrm{\infty }}{lim}\frac{\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )+{\parallel x_{n_i+1}-u_{n_i+1}\parallel }^2}=0.$$

Since $0\le \parallel x_n-u_n\parallel \le M$, it follows that

$$0\le \frac{\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )}{1+\mathrm{\Phi }(M)+M^2}\le \frac{\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )}{1+\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )+{\parallel x_{n_i+1}-u_{n_i+1}\parallel }^2}.$$

Thus, ${lim}_{i\to \mathrm{\infty }}\mathrm{\Phi }(\parallel x_{n_i+1}-u_{n_i+1}\parallel )=0$. By the strictly increasing continuous function Φ, we obtain that ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_i+1}-u_{n_i+1}\parallel =0$ and so ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$. Using the inequality $\parallel x_n-q\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-q\parallel \to 0$ as $n\to \mathrm{\infty }$, we know that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Theorem 2.3 Let E be a real Banach space and D be a nonempty closed convex subset of E. Let $T:D\to D$ be a uniformly L-Lipschitz asymptotically pseudocontractive mapping with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$. Suppose that $\{a_n\}$ and $\{b_n\}$ are two real sequences in $[0,1]$ satisfying the following conditions:

• (a) $a_n\to 0$ as $n\to \mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n(k_n-1)<\mathrm{\infty }$;

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n{b}_n<\mathrm{\infty }$.

For any $x_1\in D$, let $\{x_n\}$ be the modified Ishikawa iteration defined in (1.6). If $F(T)\ne \mathrm{\varnothing }$, $q\in F(T)$ and there exists a strictly increasing continuous function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that

$$〈T^n{x}_{n+1}-q,j(x_{n+1}-q)〉\le k_n{\parallel x_{n+1}-q\parallel }^2-\frac{\mathrm{\Phi }(\parallel x_{n+1}-q\parallel )}{1+\mathrm{\Phi }(\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2},\mathrm{\forall }n\ge 1,$$

where $j(x_{n+1}-q)\in J(x_{n+1}-q)$. Then $\{x_n\}$ converges strongly to the fixed point q of T.

Proof By Theorem 2.1 and Theorem 2.2, we obtain the proof of Theorem 2.3. □

Remark 2.4 Since the condition $〈T^{n}x-q,j(x-q)〉\le k_n{\parallel x-q\parallel }^2-\frac{\mathrm{\Phi }(\parallel x-q\parallel )}{1+\mathrm{\Phi }(\parallel x-q\parallel )+{\parallel x-q\parallel }^2}$ is weaker than $〈T^{n}x-q,j(x-q)〉\le k_n{\parallel x-q\parallel }^2-\mathrm{\Phi }(\parallel x-q\parallel )$, Theorem 2.1 and Theorem 2.3 generalize the corresponding results of Zeng [14]. Further, our proof methods are different from those of Zeng [14].

For the sake of convenience, we give the following definitions.

Definition 2.5 A mapping $T:D\to E$ is said to be weak generalized asymptotically φ-hemi-contractive with a sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ if there exists a strictly increasing continuous function $\phi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\phi (0)=0$ such that, for any $x\in D$ and $y\in F(T)$, there exists $j(x-y)\in J(x-y)$ such that

$$〈T^{n}x-T^{n}y,j(x-y)〉\le k_n{\parallel x-y\parallel }^2-\frac{\phi (\parallel x-y\parallel )}{1+\phi (\parallel x-y\parallel )+{\parallel x-y\parallel }^2},\mathrm{\forall }n\ge 1.$$

(2.17)

If the condition (2.17) is replaced by the following inequality:

$$〈T^{n}x-T^{n}y,j(x-y)〉\le k_n{\parallel x-y\parallel }^2-\phi (\parallel x-y\parallel ),\mathrm{\forall }n\ge 1,$$

(2.18)

then T is called a generalized asymptotically φ-hemi-contractive mapping. Clearly, if T is a generalized asymptotically asymptotically φ-hemi-contractive, then T must be a weak generalized asymptotically asymptotically φ-hemi-contractive mapping. However, the converse is not true in general. This can be seen from the following examples.

Example 2.6 Let $E=R$ be the set of real numbers with the usual norm $|\cdot |$ and $D=[0,+\mathrm{\infty })$. Define a mapping $T:D\to D$ by

$$Tx=\frac{2x^3}{1+2x^2},\mathrm{\forall }x\in D.$$

Then T is a monotonically increasing function with a fixed point $q=0\in D$. Define two functions $\mathrm{\Phi },\phi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by $\mathrm{\Phi }(t)=\frac{t^2}{1+2t^2}$ and $\phi (t)=t^2$, respectively. Then Φ and φ are two strictly increasing continuous functions with $\mathrm{\Phi }(0)=\phi (0)=0$. For all $x\in D$ and $q\in F(T)$, let $k_n=1$. Then we obtain that

(2.19)

(2.20)

Then T is a generalized asymptotically Φ-hemi-contraction and a weak generalized asymptotically φ-hemi-contraction.

Example 2.7 Let $E=R$ be the set of real numbers with the usual norm and $R^{+}=[0,+\mathrm{\infty })$. Define a mapping $T:R^{+}\to R$ by

$$Tx=\frac{x+x^3+x^{5/2}-x^{1/2}}{1+x^{3/2}+x^2},\mathrm{\forall }x\in R^{+}.$$

Then T has a fixed point $q=0\in R^{+}$. Define a function $\phi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by $\phi (t)=t^{3/2}$. Then φ is a strictly increasing continuous function with $\phi (0)=0$. For all $x\in R^{+}$ and $q\in F(T)$, let $n=1$ and $k_n=1$. Then we have

$$\begin{array}{rcl}〈Tx-Tq,j(x-q)〉& =& 〈\frac{x+x^3+x^{5/2}-x^{1/2}}{1+x^{3/2}+x^2}-0,j(x-0)〉\\ =& 〈\frac{x+x^3+x^{5/2}-x^{1/2}}{1+x^{3/2}+x^2},x〉\\ =& \frac{x^2+x^4+x^{7/2}-x^{3/2}}{1+x^{3/2}+x^2}\\ =& x^2-\frac{x^{3/2}}{1+x^{3/2}+x^2}\\ =& {|x-q|}^2-\frac{{|x-q|}^{3/2}}{1+{|x-q|}^{3/2}+{|x-q|}^2}\\ =& {|x-q|}^2-\frac{\phi (|x-q|)}{1+\phi (|x-q|)+{|x-q|}^2}.\end{array}$$

(2.21)

Then T is a weak generalized asymptotically φ-hemi-contraction, but not a generalized asymptotically Φ-hemi-contraction with $n=1$.