The convergence of implicit Mann and Ishikawa iterations for weak generalized φ-hemicontractive mappings in real Banach spaces

Abstract

Let E be a real Banach space and let D be a nonempty closed convex subset of E, let $T:D\to D$ be a continuous weak generalized φ-hemicontractive mapping. The existence theorem of a fixed point of a weak generalized φ-pseudocontractive mapping is obtained. And we also prove that implicit Mann and Ishikawa iterations converge strongly to the unique fixed point of T. Our results extend the corresponding results of Xiang (Nonlinear Anal. 70(6): 2277-2279, 2009).

MSC:47H09, 47J25.

1 Introduction

Throughout the paper we assume that E is an arbitrary real Banach space and $E^{\ast }$ is its dual space. Let D be a nonempty closed convex subset of E and let $F(T)=\{x\in D:Tx=x\}\ne \mathrm{\varnothing }$ be a fixed point set of T. We denote that the normalized duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

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

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. We denote the single-valued normalized duality mapping by j.

Definition 1.1 [1]

Let $T:D\to D$ be a mapping.

T is said to be strongly pseudocontractive if there exists a constant $k\in (0,1)$ such that for any $x,y\in D$, there exists $j(x-y)\in J(x-y)$ satisfying

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

(1.1)

T is called ϕ-strongly pseudocontractive if there exists a strictly increasing continuous function $\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\varphi (0)=0$ such that for any $x,y\in D$, there exists $j(x-y)\in J(x-y)$ satisfying

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

(1.2)

T is called generalized Φ-pseudocontractive if there exists a strictly increasing continuous function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that for any $x,y\in D$, there exists $j(x-y)\in J(x-y)$ satisfying

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

(1.3)

Furthermore, if the inequalities (1.1), (1.2) and (1.3) hold for any $x\in D$ and $y\in F(T)$, then the corresponding mapping T is called strongly hemicontractive, ϕ-strongly hemicontractive and generalized Φ-hemicontractive, respectively. Clearly, the generalized Φ-hemicontractive mappings not only include strongly hemicontractive and ϕ-strongly hemicontractive mappings, but also strongly pseudocontractive, ϕ-strongly pseudocontractive and Φ-pseudocontractive mappings. Thus, the class of generalized Φ-hemicontractive mappings is the most general in the class of above pseudocontractive mappings, i.e., {strongly hemicontractive mappings set} ⊂ {ϕ-strongly hemicontractive mappings set} ⊂ {generalized Φ-hemicontractive mappings set}. The converse is not true in general. The counterexamples are as follows. (See [2].)

Example 1.2 Let $E=R$ be a real numbers space with the usual norm and $D=[0,+\mathrm{\infty })$. Define $T:D\to D$ by

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

Observe that T has a fixed point $q=0\in D$. Define $\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by $\varphi (t)=\frac{t}{1+t}$. And ϕ is a strictly increasing function with $\varphi (0)=0$. Then T is a ϕ-strongly hemicontractive mapping. Indeed, for all $x\in D$, $q\in F(T)$, we have

$$\begin{array}{c}〈Tx-Tq,j(x-q)〉\hfill \\ =〈\frac{x^2}{1+x}-0,j(x-0)〉\hfill \\ =〈\frac{x^2}{1+x},x〉\hfill \\ =\frac{x^3}{1+x}\hfill \\ ={|x-q|}^2-\frac{|x-q|}{1+|x-q|}\cdot |x-q|\hfill \\ ={|x-q|}^2-\varphi (|x-q|)\cdot |x-q|.\hfill \end{array}$$

(1.4)

Hence, T is a ϕ-strongly hemicontractive mapping. But T is not a strongly hemicontractive mapping.

Example 1.3 Let $E=R$ be a real numbers space with the usual norm and $D=[0,+\mathrm{\infty })$. Define $T:D\to D$ by

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

Then T has a fixed point $q=0\in D$. Define $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by $\mathrm{\Phi }(t)=\frac{t^2}{1+t^2}$. Then Φ is a strictly increasing function with $\mathrm{\Phi }(0)=0$. For all $x\in D$, $q\in F(T)$, we obtain that

$$\begin{array}{c}〈Tx-Tq,j(x-q)〉\hfill \\ =〈\frac{x^3}{1+x^2}-0,j(x-0)〉\hfill \\ =〈\frac{x^3}{1+x^2},x〉\hfill \\ =\frac{x^4}{1+x^2}\hfill \\ ={|x-q|}^2-\frac{{|x-q|}^2}{1+{|x-q|}^2}\hfill \\ ={|x-q|}^2-\mathrm{\Phi }(|x-q|).\hfill \end{array}$$

(1.5)

Therefore, T is a generalized Φ-hemicontractive mapping. However, T is not a ϕ-strongly hemicontractive mapping. If it is not the case, then there exists a strictly increasing function $\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\varphi (0)=0$ such that

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

i.e., $\varphi (x)\le \frac{x}{1+x^2}$ for all $x\in [0,+\mathrm{\infty })$. So, ${lim}_{x\to +\mathrm{\infty }}\varphi (x)=0$. This is a contradiction with a strictly increasing function ϕ. Hence it holds.

Recently, Xiang [1] discussed the relationship between generalized Φ-pseudocontractive mappings and ϕ-strongly pseudocontractive mappings. The results are as follows.

Theorem 1.4 [[1], Proposition 1.1]

Let C be a bounded subset of E and let $T:C\to E$ be a mapping. Then T is generalized Φ-pseudocontractive if and only if T is Ψ-strongly pseudocontractive.

Theorem 1.5 [[1], Proposition 1.2]

Suppose that C is an unbounded subset of E and $T:C\to E$ is a generalized Φ-pseudocontractive mapping. Then T is ϕ-strongly pseudocontractive if and only if there exists a strictly increasing function $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\mathrm{\Phi }(0)=0$ such that (1.1) holds and ${lim}_{s\to \mathrm{\infty }}inf\frac{\mathrm{\Phi }(s)}{s}=\sigma >0$.

At the same time, Xiang [1] also proved the following existence theorem.

Theorem 1.6 [[1], Theorem 2.1]

Let E be real Banach space, let C be a nonempty closed convex subset of E, and let $T:C\to C$ be a continuous generalized Φ-pseudocontractive mapping. Then T has a unique fixed point in C.

In this paper, we extend the results of Xiang [1] and give the convergence of other iterative methods. For this, we need to introduce the following lemmas.

Lemma 1.7 [[3], Corollary 1]

Let D be a nonempty closed convex subset of E, and let $T:D\to D$ be a continuous strongly pseudocontractive mapping. Then T has the unique fixed point in D.

Lemma 1.8 [4]

Let E be a real Banach space, and let $J:E\to 2^{E^{\ast }}$ be a normalized duality mapping. Then

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

for all $x,y\in E$ and each $j(x+y)\in J(x+y)$.

2 Main results

In the sequel, we give the main results.

Definition 2.1 The map $T:D\to D$ is called weak generalized φ-pseudocontractive if there exists a strictly increasing continuous function $\phi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\phi (0)=0$ such that for any $x,y\in D$, there exists $j(x-y)\in J(x-y)$ satisfying

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

(2.1)

In Definition 2.1, if for any $x\in D$, $y\in F(T)$ such that (2.1) holds, then T is called a weak generalized φ-hemicontractive mapping. (See [5, 6].)

Remark 2.2 If T is generalized φ-hemicontractive, then T must be weak generalized φ-hemicontractive. That is,

$$\begin{array}{rl}〈Tx-Tq,j(x-q)〉& \le {\parallel x-q\parallel }^2-\phi (\parallel x-q\parallel )\\ \le {\parallel x-q\parallel }^2-\frac{\phi (\parallel x-q\parallel )}{1+\phi (\parallel x-q\parallel )+{\parallel x-q\parallel }^2}.\end{array}$$

However, the converse is not true in general. See the following example.

Counterexample 2.3 Let $E=R$ be a real numbers space with the usual norm and $D=R^{+}=[0,+\mathrm{\infty })$. Define $T:R^{+}\to R^{+}$ by

$$Tx=\{\begin{array}{ll}\frac{2}{3}x,& x\in [0,1];\\ \frac{x+x^3+x^2\sqrt{x}-\sqrt{x}}{1+x\sqrt{x}+x^2},& x\in (1,+\mathrm{\infty }).\end{array}$$

Then T has a fixed point $q=0\in R^{+}$. Set $\mathrm{\Phi }:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ by

$$\mathrm{\Phi }(t)=\{\begin{array}{ll}{t}^4,& t\in [0,1];\\ t^{3/2},& t\in (1,+\mathrm{\infty }).\end{array}$$

Then Φ is a strictly increasing continuous function with $\mathrm{\Phi }(0)=0$. And for any $x\in [0,1]$, $q\in F(T)$, we obtain that

$$\begin{array}{c}〈Tx-Tq,j(x-q)〉\hfill \\ =〈\frac{2}{3}x-0,j(x-0)〉\hfill \\ =\frac{2}{3}{x}^2\hfill \\ \le x^2-\frac{x^4}{1+x^4+x^2}\hfill \\ ={|x-q|}^2-\frac{{|x-q|}^4}{1+{|x-q|}^4+{|x-q|}^2}\hfill \\ ={|x-q|}^2-\frac{\mathrm{\Phi }(|x-q|)}{1+\mathrm{\Phi }(|x-q|)+{|x-q|}^2}.\hfill \end{array}$$

(2.2)

For any $x\in (1,+\mathrm{\infty })$, $q\in F(T)$, we have

$$\begin{array}{c}〈Tx-Tq,j(x-q)〉\hfill \\ =〈\frac{x+x^3+x^2\sqrt{x}-\sqrt{x}}{1+x\sqrt{x}+x^2}-0,j(x-0)〉\hfill \\ =〈\frac{x+x^3+x^2\sqrt{x}-\sqrt{x}}{1+x\sqrt{x}+x^2},x〉\hfill \\ =\frac{x^2+x^4+x^3\sqrt{x}-x\sqrt{x}}{1+x\sqrt{x}+x^2}\hfill \\ =x^2-\frac{x^{3/2}}{1+x^{3/2}+x^2}\hfill \\ ={|x-q|}^2-\frac{{|x-q|}^{3/2}}{1+{|x-q|}^{3/2}+{|x-q|}^2}\hfill \\ ={|x-q|}^2-\frac{\mathrm{\Phi }(|x-q|)}{1+\mathrm{\Phi }(|x-q|)+{|x-q|}^2}.\hfill \end{array}$$

(2.3)

Then T is a weak generalized Φ-hemicontractive mapping. But T is not a generalized φ-hemicontractive mapping. Therefore, it has more practical significance to research of the class of mappings in fixed point theory and applications. For this, we firstly give the existence theorem.

Theorem 2.4 Let E be a real Banach space, let D be a nonempty closed convex subset of E, and let $T:D\to D$ be a continuous weak generalized φ-pseudocontractive mapping. Then T has a unique fixed point in D.

Proof Similar, using the proof method of Xiang [1].

Step I. Construct the sequence $\{x_n\}$.

For any given $x_0\in D$, the mapping $S_1:D\to D$ is defined by $S_{1}x=\frac{1}{2}{x}_0+\frac{1}{2}Tx$ for all $x\in D$, then $S_1$ is a continuous strongly pseudocontractive mapping. So, there exists $x_1\in D$ such that $S_1{x}_1=x_1$, i.e., $x_1=\frac{1}{2}{x}_0+\frac{1}{2}T{x}_1$. The mapping $S_2:D\to D$ is defined by $S_{2}x=\frac{1}{2}{x}_1+\frac{1}{2}Tx$ for all $x\in D$, then $S_2$ is a continuous strongly pseudocontractive mapping. So, there exists $x_2\in D$ such that $S_2{x}_2=x_2$, i.e., $x_2=\frac{1}{2}{x}_1+\frac{1}{2}T{x}_2,\dots$ , we obtain the sequence $\{x_n\}$ by $x_{n+1}=\frac{1}{2}{x}_n+\frac{1}{2}T{x}_{n+1}$.

Step II. Show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n-1}\parallel =0$.

From the above sequence $\{x_n\}$, we notice that

$$x_{n+1}=x_n-x_{n+1}+T{x}_{n+1},x_n=x_{n-1}-x_n+T{x}_n.$$

Using the equalities above and Lemma 1.8, we have

$$\begin{array}{c}{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ ={\parallel (x_n-x_{n-1})-(x_{n+1}-x_n)+(T{x}_{n+1}-T{x}_n)\parallel }^2\hfill \\ \le {\parallel x_n-x_{n-1}\parallel }^2-2{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ +2[{\parallel x_{n+1}-x_n\parallel }^2-\frac{\phi (\parallel x_{n+1}-x_n\parallel )}{1+\phi (\parallel x_{n+1}-x_n\parallel )+{\parallel x_{n+1}-x_n\parallel }^2}]\hfill \\ \le {\parallel x_n-x_{n-1}\parallel }^2-\frac{\phi (\parallel x_{n+1}-x_n\parallel )}{1+\phi (\parallel x_{n+1}-x_n\parallel )+{\parallel x_{n+1}-x_n\parallel }^2}\hfill \\ \le {\parallel x_n-x_{n-1}\parallel }^2.\hfill \end{array}$$

(2.4)

Based on the monotone bounded principle, then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n-1}\parallel$ exists. And

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n-1}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel T{x}_n-x_n\parallel =A.$$

Denote $M={sup}_n\{\parallel x_n-x_{n-1}\parallel \}$. From (2.4), we have

$$\begin{array}{c}{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ \le {\parallel x_n-x_{n-1}\parallel }^2-\frac{\phi (\parallel x_{n+1}-x_n\parallel )}{1+\phi (\parallel x_{n+1}-x_n\parallel )+{\parallel x_{n+1}-x_n\parallel }^2}.\hfill \end{array}$$

(2.5)

Let ${inf}_{n\ge 0}\frac{\phi (\parallel x_{n+1}-x_n\parallel )}{1+\phi (\parallel x_{n+1}-x_n\parallel )+{\parallel x_{n+1}-x_n\parallel }^2}=\delta$, then $\delta =0$. If this is not the case, then $\delta >0$. We have

$$\frac{\phi (\parallel x_{n+1}-x_n\parallel )}{1+\phi (\parallel x_{n+1}-x_n\parallel )+{\parallel x_{n+1}-x_n\parallel }^2}\ge \delta$$

for all $n\ge 0$. It follows from (2.5) that

$$\delta \le {\parallel x_n-x_{n-1}\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2,$$

(2.6)

which implies that ${\sum }_{n=1}^{\mathrm{\infty }}\delta \le {\parallel x_1-x_0\parallel }^2<\mathrm{\infty }$, which is a contradiction. Then $\delta =0$. Thus there exists an infinite subsequence $\{\frac{\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )}{1+\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )+{\parallel x_{n_i+1}-x_{n_i}\parallel }^2}\}$ such that

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

Since $0\le \frac{\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )}{1+\phi (M)+M^2}\le \frac{\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )}{1+\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )+{\parallel x_{n_i+1}-x_{n_i}\parallel }^2}$, then ${lim}_{i\to \mathrm{\infty }}\phi (\parallel x_{n_i+1}-x_{n_i}\parallel )=0$. It leads to ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_i+1}-x_{n_i}\parallel =0$ by the strict increase and continuity of φ. Hence $A=0$.

Step III. $\{x_n\}$ is a Cauchy sequence.

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. For $\mathrm{\forall }ϵ\in (0,1)$, ∃N such that

$$\parallel x_{n+1}-x_n\parallel <ϵ,\parallel T{x}_n-x_n\parallel ,\parallel T{x}_m-x_m\parallel <\frac{\phi (ϵ)}{2[1+\phi (2ϵ)+4{ϵ}^2](1+2ϵ)}$$

for all $m,n\ge N$. By the induction method, we prove that $\parallel x_m-x_n\parallel <ϵ$ for all $m,n\ge N$. If $m=n+1$, then $\parallel x_{n+1}-x_n\parallel <ϵ$. Suppose that $\parallel x_m-x_n\parallel <ϵ$ holds for some $m\ge N$, then $\parallel x_{m+1}-x_n\parallel \le \parallel x_{m+1}-x_m\parallel +\parallel x_m-x_n\parallel <2ϵ$ (*). Next we want to show that $\parallel x_{m+1}-x_n\parallel <ϵ$. Since T is a weak generalized φ pseudocontractive mapping, then

$$〈T{x}_{m+1}-T{x}_n,j(x_{m+1}-x_n)〉\le {\parallel x_{m+1}-x_n\parallel }^2-\frac{\phi (\parallel x_{m+1}-x_n\parallel )}{1+\phi (\parallel x_{m+1}-x_n\parallel )+{\parallel x_{m+1}-x_n\parallel }^2},$$

i.e., $\frac{\phi (\parallel x_{m+1}-x_n\parallel )}{1+\phi (\parallel x_{m+1}-x_n\parallel )+{\parallel x_{m+1}-x_n\parallel }^2}$ ≤ ${\parallel x_{m+1}-x_n\parallel }^2-〈T{x}_{m+1}-T{x}_n,j(x_{m+1}-x_n)〉$ ≤ $[\parallel x_{m+1}-T{x}_{m+1}\parallel +\parallel x_n-T{x}_n\parallel ]\cdot \parallel x_{m+1}-x_n\parallel$. By the above inequalities, we have

$$\frac{\phi (\parallel x_{m+1}-x_n\parallel )}{1+\phi (2ϵ)+4{ϵ}^2}<\frac{2ϵ\phi (ϵ)}{[1+\phi (2ϵ)+4{ϵ}^2](1+2ϵ)}<\frac{\phi (ϵ)}{1+\phi (2ϵ)+4{ϵ}^2},$$

which implies that $\parallel x_{m+1}-x_n\parallel <ϵ$ by the strict increase of φ. Therefore $\{x_n\}$ is a Cauchy sequence. Since D is closed in Banach space E, then D is complete. Hence, there exists a point $q\in D$ such that $x_n\to q$ as $n\to \mathrm{\infty }$. Since T is continuous, then $q=Tq$. The uniqueness is obvious. □

3 Applications of the weak generalized φ-hemicontractive mappings

Now that the weak generalized φ-hemicontractive mappings are much more general mappings. Hence it is of interest to study the convergence of an iteration process of fixed points of the class mappings.

Definition 3.1 Let $T:D\to D$ be a mapping. For any given $u_1\in D$, define the sequence ${\{u_n\}}_{n=1}^{\mathrm{\infty }}\subset D$ by the iterative scheme

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

(3.1)

which is called the Mann iterative process, where ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ is a real sequence in $[0,1]$ satisfying certain conditions. Further, we assume that there exists ${(I-tT)}^{-1}$ for all $t\in (0,1)$. For any given $x_1\in D$, define the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}\subset D$ by the iterative scheme [3]

$$x_{n+1}=(1-a_n)x_n+a_{n}T{x}_{n+1},n\ge 1,$$

(3.2)

which is called the implicit Mann iterative process.

Definition 3.2 Let $T:D\to D$ be a mapping. For any given $u_1\in D$, define the sequence ${\{w_n\}}_{n=1}^{\mathrm{\infty }}\subset D$ by the iterative scheme

$$\{\begin{array}{l}{w}_1\in D,\\ v_n=(1-b_n)w_n+b_{n}T{w}_n,n\ge 1,\\ w_{n+1}=(1-a_n)w_n+a_{n}T{v}_n,n\ge 1,\end{array}$$

(3.3)

which is called the Ishikawa iterative process, where $\{a_n\}$ and $\{b_n\}$ are two real sequences in $[0,1]$ satisfying certain conditions. And for any given $z_1\in D$, define the sequence ${\{z_n\}}_{n=1}^{\mathrm{\infty }}\subset D$ by the iterative scheme

$$\{\begin{array}{l}{z}_1\in D,\\ y_n=(1-b_n)z_n+b_{n}T{y}_n,n\ge 1,\\ z_{n+1}=(1-a_n)y_n+a_{n}T{z}_{n+1},n\ge 1,\end{array}$$

(3.4)

which is called the implicit Ishikawa iterative process. Especially, if $b_n=0$, then the corresponding iterations (3.3) and (3.4) reduce to (3.1) and (3.2), respectively.

Lemma 3.3 [1]

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

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

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.

In the following, we study the convergence of implicit Mann and Ishikawa iterative processes for weak generalized φ-hemicontractive mappings in general real Banach spaces.

Theorem 3.4 Let E be a real Banach space and let D be a nonempty closed convex subset of E, let $T:D\to D$ be a weak generalized φ-hemicontractive mapping. Suppose that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is defined by (3.2) with the iteration parameter ${\{a_n\}}_{n=1}^{\mathrm{\infty }}\subset [0,\frac{1}{2})$ satisfying: $a_n\to 0$ as $n\to \mathrm{\infty }$; ${\sum }_{n=1}^{\mathrm{\infty }}\frac{a_n}{1-2a_n}=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}\frac{a_n^2}{1-2a_n}<\mathrm{\infty }$. Then the implicit Mann iteration ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the unique fixed point of T.

Proof Let $q\in F(T)$. Applying Lemma 1.8 and (3.4), we have

$$\begin{array}{c}{\parallel x_{n+1}-q\parallel }^2\hfill \\ ={\parallel (1-a_n)(x_n-q)+a_n(T{x}_{n+1}-Tq)\parallel }^2\hfill \\ \le {(1-a_n)}^2{\parallel x_n-q\parallel }^2+2a_n〈T{x}_{n+1}-Tq,j(x_{n+1}-q)〉\hfill \\ \le {(1-a_n)}^2{\parallel x_n-q\parallel }^2+2a_n[{\parallel x_{n+1}-q\parallel }^2-\frac{\phi (\parallel x_{n+1}-q\parallel )}{1+\phi (\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2}],\hfill \end{array}$$

(3.5)

which implies that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& \le \frac{{(1-a_n)}^2}{1-2a_n}{\parallel x_n-q\parallel }^2-\frac{2a_n}{1-2a_n}\cdot \frac{\phi (\parallel x_{n+1}-q\parallel )}{1+\phi (\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2}\\ =(1+\frac{a_n^2}{1-2a_n}){\parallel x_n-q\parallel }^2-\frac{2a_n}{1-2a_n}\cdot \frac{\phi (\parallel x_{n+1}-q\parallel )}{1+\phi (\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2}\\ \le (1+\frac{a_n^2}{1-2a_n}){\parallel x_n-q\parallel }^2.\end{array}$$

(3.6)

By Lemma 3.3, we obtain that ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-q\parallel }^2$ exists. Let $M={sup}_{n\ge 1}\{\parallel x_n-q\parallel \}$.

Set ${inf}_{n\ge 1}\frac{\phi (\parallel x_{n+1}-q\parallel )}{1+\phi (\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2}=\lambda$, then $\lambda =0$. If this is not the case, we assume that $\lambda >0$, then $\frac{\phi (\parallel x_{n+1}-q\parallel )}{1+\phi (\parallel x_{n+1}-q\parallel )+{\parallel x_{n+1}-q\parallel }^2}\ge \lambda$ for any n. From (3.6), we get

$${\parallel x_{n+1}-q\parallel }^2\le {\parallel x_n-q\parallel }^2+\frac{a_n^2}{1-2a_n}{M}^2-\frac{2\lambda a_n}{1-2a_n},$$

(3.7)

which implies that

$$\sum _{n=1}^{\mathrm{\infty }}\frac{2\lambda a_n}{1-2a_n}\le {\parallel x_1-q\parallel }^2+\sum _{n=1}^{\mathrm{\infty }}\frac{a_n^2}{1-2a_n}{M}^2<\mathrm{\infty },$$

(3.8)

which is a contradiction, and so $\lambda =0$. Consequently, there exists an infinite subsequence such that $\frac{\phi (\parallel x_{n_i+1}-q\parallel )}{1+\phi (\parallel x_{n_i+1}-q\parallel )+{\parallel x_{n_i+1}-q\parallel }^2}\to 0$ as $i\to \mathrm{\infty }$. Then we have

$$0\le \frac{\phi (\parallel x_{n_i+1}-q\parallel )}{1+\phi (M)+M^2}\le \frac{\phi (\parallel x_{n_i+1}-q\parallel )}{1+\phi (\parallel x_{n_i+1}-q\parallel )+{\parallel x_{n_i+1}-q\parallel }^2},$$

which implies that $\phi (\parallel x_{n_i+1}-q\parallel )\to 0$ as $i\to \mathrm{\infty }$. It leads to $\parallel x_{n_i+1}-q\parallel \to 0$ as $i\to \mathrm{\infty }$ by the strict increase and continuity of φ. Thus, we obtain that $\parallel x_n-q\parallel \to 0$ as $n\to \mathrm{\infty }$. This completes the proof. □

Theorem 3.5 Let E be a real Banach space and let D be a nonempty closed convex subset of E, let $T:D\to D$ be a weak generalized φ-hemicontractive mapping. Suppose that ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ is defined by (3.4) with the iteration parameters $a_n,b_n\in [0,\frac{1}{2})$ satisfying the conditions:

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

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}\frac{a_n}{1-2a_n}=\mathrm{\infty }$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}\frac{a_n^2}{1-2a_n}<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}\frac{b_n^2}{1-2b_n}<\mathrm{\infty }$.

Then the implicit Ishikawa iteration ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to the unique fixed point of T.

Proof By the definition of a weak generalized φ-hemicontractive mapping, we know that the fixed point of T is unique. Denote q. And for any $x\in D$, we have

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

(3.9)

Applying Lemma 1.8 and (3.2), we have

$$\begin{array}{c}{\parallel z_{n+1}-q\parallel }^2\hfill \\ ={\parallel (1-a_n)(y_n-q)+a_n(T{z}_{n+1}-Tq)\parallel }^2\hfill \\ \le {(1-a_n)}^2{\parallel y_n-q\parallel }^2+2a_n〈T{z}_{n+1}-Tq,j(z_{n+1}-q)〉\hfill \\ \le {(1-a_n)}^2{\parallel y_n-q\parallel }^2+2a_n[{\parallel z_{n+1}-q\parallel }^2-\frac{\phi (\parallel z_{n+1}-q\parallel )}{1+\phi (\parallel z_{n+1}-q\parallel )+{\parallel z_{n+1}-q\parallel }^2}],\hfill \end{array}$$

(3.10)

$$\begin{array}{c}{\parallel y_n-q\parallel }^2\hfill \\ ={\parallel (1-b_n)(z_n-q)+b_n(T{y}_n-Tq)\parallel }^2\hfill \\ \le {(1-b_n)}^2{\parallel z_n-q\parallel }^2+2b_n〈T{y}_n-Tq,j(y_n-q)〉\hfill \\ \le {(1-b_n)}^2{\parallel z_n-q\parallel }^2+2b_n[{\parallel y_n-q\parallel }^2-\frac{\phi (\parallel y_n-q\parallel )}{1+\phi (\parallel y_n-q\parallel )+{\parallel y_n-q\parallel }^2}],\hfill \end{array}$$

(3.11)

which implies that

$${\parallel y_n-q\parallel }^2\le \frac{{(1-b_n)}^2}{1-2b_n}{\parallel z_n-q\parallel }^2.$$

(3.12)

Substituting (3.14) into (3.12), we obtain that

$$\begin{array}{c}{\parallel z_{n+1}-q\parallel }^2\hfill \\ \le \frac{{(1-a_n)}^2{(1-b_n)}^2}{(1-2a_n)(1-2b_n)}{\parallel z_n-q\parallel }^2-\frac{2a_n}{1-2a_n}\cdot \frac{\phi (\parallel z_{n+1}-q\parallel )}{1+\phi (\parallel z_{n+1}-q\parallel )+{\parallel z_{n+1}-q\parallel }^2}\hfill \\ \le (1+\frac{a_n^2}{1-2a_n}+\frac{b_n^2}{1-2b_n}){\parallel z_n-q\parallel }^2-\frac{2a_n}{1-2a_n}\cdot \frac{\phi (\parallel z_{n+1}-q\parallel )}{1+\phi (\parallel z_{n+1}-q\parallel )+{\parallel z_{n+1}-q\parallel }^2}\hfill \\ \le (1+\frac{a_n^2}{1-2a_n}+\frac{b_n^2}{1-2b_n}){\parallel z_n-q\parallel }^2.\hfill \end{array}$$

(3.13)

By Lemma 3.3, we obtain that ${lim}_{n\to \mathrm{\infty }}{\parallel z_n-q\parallel }^2$ exists. Let $M_1={sup}_{n\ge 1}\{\parallel z_n-q\parallel \}$.

Set ${inf}_{n\ge 1}\frac{\phi (\parallel z_{n+1}-q\parallel )}{1+\phi (\parallel z_{n+1}-q\parallel )+{\parallel z_{n+1}-q\parallel }^2}=\delta$, then $\delta =0$. If this is not the case, we assume that $\delta >0$, then $\frac{\phi (\parallel z_{n+1}-q\parallel )}{1+\phi (\parallel z_{n+1}-q\parallel )+{\parallel z_{n+1}-q\parallel }^2}\ge \delta$ for any n. From (3.13), we get

$${\parallel z_{n+1}-q\parallel }^2\le {\parallel z_n-q\parallel }^2+(\frac{a_n^2}{1-2a_n}+\frac{b_n^2}{1-2b_n})M_1^2-\frac{2\delta a_n}{1-2a_n},$$

(3.14)

which implies that

$$\sum _{n=1}^{\mathrm{\infty }}\frac{2\delta a_n}{1-2a_n}\le {\parallel z_1-q\parallel }^2+\sum _{n=1}^{\mathrm{\infty }}(\frac{a_n^2}{1-2a_n}+\frac{b_n^2}{1-2b_n})M_1^2<\mathrm{\infty },$$

(3.15)

which is a contradiction, and so $\delta =0$. Consequently, there exists an infinite subsequence such that $\frac{\phi (\parallel z_{n_i+1}-q\parallel )}{1+\phi (\parallel z_{n_i+1}-q\parallel )+{\parallel z_{n_i+1}-q\parallel }^2}\to 0$ as $i\to \mathrm{\infty }$. Then we have

$$0\le \frac{\phi (\parallel z_{n_i+1}-q\parallel )}{1+\phi (M_1)+M_1^2}\le \frac{\phi (\parallel z_{n_i+1}-q\parallel )}{1+\phi (\parallel z_{n_i+1}-q\parallel )+{\parallel z_{n_i+1}-q\parallel }^2},$$

which implies that $\phi (\parallel z_{n_i+1}-q\parallel )\to 0$ as $i\to \mathrm{\infty }$. It leads to $\parallel z_{n_i+1}-q\parallel \to 0$ as $i\to \mathrm{\infty }$ by the strict increase and continuity of φ. Thus, we obtain that $\parallel z_n-q\parallel \to 0$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 3.6 Theorem 2.4 shows that the implicit iteration $\{x_n\}$ by $x_{n+1}=\frac{1}{2}{x}_n+\frac{1}{2}T{x}_{n+1}$ is convergent, and it converges strongly to the fixed point of T. And Theorem 3.4 and Theorem 3.5 also yield that the implicit Mann iteration and the implicit Ishikawa iteration converge strongly to the fixed point of T, respectively.