Convergence theorems for a generalized Φ-pseudo-contractive type mapping in real normal linear spaces

Abstract

In this paper, we first give a new notion of generalized Φ-pseudo-contractive type mapping, and then we consider some convergence theorems for a fixed point of the mapping. Our results improve and extend the corresponding results due to (Chidume and Chidume in J. Math. Anal. Appl. 302:545-554, 2005) and other papers.

1 Introduction and statement of results

Let E be a real normed linear space and $E^{\ast }$ be its dual space. The normalized duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

$$Jx=\{f\in E^{\ast }:〈x,f〉=\parallel x\parallel \cdot \parallel f\parallel ,\parallel x\parallel =\parallel f\parallel \},$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing.

Definition 1.1 [1, 2]

Let $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be a function for which $\varphi (0)=0$, $\mathrm{\forall }{r}_0>0$, ${lim\hspace{0.17em}inf}_{r\to r_0}\varphi (r)>0$. A mapping $T:D(T)\subset E\to E$ is called ϕ-strongly accretive if for each $x,y\in D(T)$, there exists $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉\ge \varphi (\parallel x-y\parallel )\parallel x-y\parallel .$$

We also say that $T:D(T)\subset E\to E$ is ϕ-strongly pseudo-contractive if $I-T$ is ϕ-strongly accretive.

Definition 1.2 Let $\mathrm{\Phi }:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be a function for which $\mathrm{\Phi }(0)=0$, $\mathrm{\forall }{r}_0>0$, ${lim\hspace{0.17em}inf}_{r\to r_0}\mathrm{\Phi }(r)>0$. A mapping $T:D(T)\subset E\to E$ is called generalized Φ-accretive if there exists $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉\ge \mathrm{\Phi }(\parallel x-y\parallel ),\mathrm{\forall }x,y\in D(T).$$

We also say that $T:D(T)\subset E\to E$ is generalized Φ-pseudo-contractive if $I-T$ is generalized ϕ-accretive.

Remark 1.3 Definition 1.1 and Definition 1.2 do not assume that $\varphi (r)$ ($\mathrm{\Phi }(r)$) is strictly increasing. Clearly, ϕ-strongly accretive maps (ϕ-strongly pseudo-contractive maps) are generalized by generalized ϕ-accretive maps (generalized Φ-pseudo-contractive maps) with $\mathrm{\Phi }(r)=r\varphi (r)$.

Definition 1.4 $T:D(T)\subset E\to E$ is called a generalized Φ-accretive type mapping if there exists $x^{\ast }\in D(T)$ such that for all $x\in D(T)$, there exists $j(x-x^{\ast })\in J(x-x^{\ast })$ such that

$$〈Tx-T{x}^{\ast },j(x-x^{\ast })〉\ge \mathrm{\Phi }\left(\parallel x-x^{\ast }\parallel \right),$$

where Φ is as in Definition 1.2. T is called a generalized Φ-pseudo-contractive type mapping if $I-T$ is a generalized Φ-accretive type mapping.

Recently, Chidume and Chidume proved the following theorems by using the conclusion that a uniformly continuous mapping on K is bounded.

Theorem CC1 [3]

Let E be a real normed linear space, K be a nonempty subset of E and $T:K\to E$ be a uniformly continuous generalized Φ-hemi-contractive mapping, i.e., there exist $x^{\ast }\in K$ and a strictly increasing function $\mathrm{\Phi }:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $\mathrm{\Phi }(0)=0$ such that for all $x\in K$, there exists $j(x-x^{\ast })\in J(x-x^{\ast })$ such that

$$〈Tx-T{x}^{\ast },j(x-x^{\ast })〉\le {\parallel x-x^{\ast }\parallel }^2-\mathrm{\Phi }\left(\parallel x-x^{\ast }\parallel \right).$$

• (a) If $y^{\ast }\in K$ is a fixed point of T, then $y^{\ast }=x^{\ast }$ and so T has at most one fixed point in K.

• (b) Suppose that there exists $x_0\in K$ such that the sequence $\{x_n\}$ defined by

  $$x_{n+1}=a_n{x}_n+b_{n}T{x}_n+c_n{u}_n,\mathrm{\forall }n\ge 0,$$

is contained in K, where $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ are real sequences in [0,1] satisfying the following conditions:

• (i) $a_n+b_n+c_n=1$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}(b_n+c_n)=\mathrm{\infty }$;

• (iii) ${\sum }_{n=0}^{\mathrm{\infty }}{(b_n+c_n)}^2<\mathrm{\infty }$;

• (iv) ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$; and $\{u_n\}$ is a bounded sequence in K.

Then $\{x_n\}$ converges strongly to $x^{\ast }$. In particular, if $y^{\ast }$ is a fixed point of T in K, then $\{x_n\}$ converges strongly to $y^{\ast }$.

Theorem CC2 [3]

Let E be a real normed linear space, $A:E\to E$ be a uniformly continuous generalized Φ-quasi-contractive mapping, i.e., there exists $x^{\ast }\in D(A)$ such that for all $x\in E$ , there exist $j(x-x^{\ast })\in J(x-x^{\ast })$ and a strictly increasing function $\mathrm{\Phi }:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $\mathrm{\Phi }(0)=0$ such that

$$〈Ax-A{x}^{\ast },j(x-x^{\ast })〉\ge \mathrm{\Phi }\left(\parallel x-x^{\ast }\parallel \right).$$

For arbitrary $x_0\in D(A)$, define the sequence $\{x_n\}$ iteratively by

$$x_{n+1}=a_n{x}_n+b_{n}S{x}_n+c_n{u}_n,\mathrm{\forall }n\ge 0,$$

where $S:E\to E$ is defined by $Sx:=x-Ax$ for all $x\in E$; and $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ are real sequences in $[0,1]$ satisfying the following conditions:

• (i) $a_n+b_n+c_n=1$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}(b_n+c_n)=\mathrm{\infty }$;

• (iii) ${\sum }_{n=0}^{\mathrm{\infty }}{(b_n+c_n)}^2<\mathrm{\infty }$;

• (iv) ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$; and $\{u_n\}$ is a bounded sequence in K.

Then $\{x_n\}$ converges strongly to $x^{\ast }$.

Remark 1.5 In Theorem CC1 and Theorem CC2, the condition that K is convex is needed. Since $K\subset E$ is a nonempty subset without assuming that K is convex, then a uniformly continuous mapping T on K is not necessarily bounded. See the following example.

Let $\{e_n\}$ be an orthonormal set of $l^2$, $K=\{x\in l^2\mid x=t{e}_n+(1-t)e_{n+1},t\in [0,1]\}$. Let $T:K\to l^2$ be a mapping defined by

$$Tx=(n+t)e_n+(n+1-t)e_{n+1},\text{where }x=t{e}_n+(1-t)e_{n+1}\in K.$$

Then T is uniformly continuous on a bounded and nonconvex set K. But T is not bounded.

Proof Clearly K is bounded and nonconvex. Let $x_m,y_m\in K$ such that $\parallel x_m-y_m\parallel \to 0$ ($m\to \mathrm{\infty }$). Then this implies that there exist $n_0\in N$ and $t_m,t_m^{\prime }\in [0,1]$ such that

$$\begin{array}{c}{x}_m=t_m{e}_{n_0}+(1-t_m)e_{n_0+1},\hfill \\ y_m=t_m^{\prime }{e}_{n_0}+(1-t_m^{\prime })e_{n_0+1},\hfill \\ \parallel t_m-t_m^{\prime }\parallel \to 0.\hfill \end{array}$$

So,

$$\begin{array}{rcl}\parallel T{x}_m-T{y}_m\parallel & =& \parallel (n_0+t_m)e_{n_0}+(n_0+1-t_m)e_{n_0+1}-(n_0+t_m^{\prime })e_{n_0}-(n_0+1-t_m^{\prime })e_{n_0+1}\parallel \\ =& |t_m-t_m^{\prime }|\parallel e_{n_0}+e_{n_0+1}\parallel \\ =& \sqrt{2}|t_m-t_m^{\prime }|\to 0(m\to \mathrm{\infty }).\end{array}$$

Hence T is uniformly continuous.

Let $x\in K$, then

$$\begin{array}{rcl}\parallel Tx\parallel & =& \parallel (n+t)e_n+(n+1-t)e_{n+1}\parallel \\ =& {({(n+t)}^2+{(n+1-t)}^2)}^{\frac{1}{2}}\to \mathrm{\infty }(n\to \mathrm{\infty }).\end{array}$$

This says that T is unbounded and completes the proof. □

In 1999, Morales and Chidume proved the following theorem.

Theorem MC [1]

Let E be a uniformly smooth Banach space, and let $A:E\to E$ be a bounded demicontinuous ϕ-strongly accretive mapping for some $x_0\in E$, ${lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\varphi (r)>\parallel A{x}_0\parallel$. Let $\{c_n\}$ be a real sequence in $[0,1]$ satisfying the following conditions: (i) ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n=\mathrm{\infty }$; (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{c}_{n}b(c_n)<\mathrm{\infty }$. Let $\{x_n\}$ be a sequence generated by

$$x_{n+1}=x_n-c_{n}A{x}_n,\mathrm{\forall }n\ge 0.$$

Then there exists a constant $r_0>0$ such that when $c_n<r_0$ ($\mathrm{\forall }n\ge 0$), the sequence $\{x_n\}$ converges strongly to the unique zero of A.

Inspired and motivated by these facts, we will give convergence theorems for a fixed point of the generalized Φ-pseudo-contractive type mapping. Our result generalizes the corresponding results in [1–9].

2 Main results

Let $F(T)=\{x\in K:Tx=x\}$, $N(A)=\{x\in D(A):Ax=0\}$.

We shall make use of the following well-known inequality.

Lemma 2.1 Let E be a real normed linear space. Then the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,j(x+y)〉,\mathrm{\forall }x,y\in E,\mathrm{\forall }j(x+y)\in J(x+y).$$

Theorem 2.2 Let E be a real normed linear space, K be a nonempty subset of E and $T:K\to E$ be a uniformly continuous generalized Φ-pseudo-contractive type mapping, i.e., there exist $x^{\ast }\in K$ and a function $\mathrm{\Phi }:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $\mathrm{\Phi }(0)=0$ such that for all $x\in K$, there exists $j(x-x^{\ast })\in J(x-x^{\ast })$ such that

$$〈Tx-T{x}^{\ast },j(x-x^{\ast })〉\le {\parallel x-x^{\ast }\parallel }^2-\mathrm{\Phi }\left(\parallel x-x^{\ast }\parallel \right).$$

(2.1)

• (a) If $y^{\ast }\in K$ is a fixed point of T, then $y^{\ast }=x^{\ast }$ and so T has at most one fixed point in K.

• (b) Let the above $x^{\star }\in F(T)$, $x_0\in K$, $T{x}_0\ne x_0$, $x_0\ne x^{\ast }$. Suppose that the sequence $\{x_n\}$ defined by

  $$x_{n+1}=a_n{x}_n+b_{n}T{x}_n+c_n{u}_n,\mathrm{\forall }n\ge 0,$$

  (2.2)

is contained in K, where $\{u_n\}$ is a bounded sequence in K and $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ are real sequences in $[0,1]$ satisfying the following conditions:

• (i) $a_n+b_n+c_n=1$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}(b_n+c_n)=\mathrm{\infty }$;

• (iii) $b_n+c_n\to 0$ as $n\to \mathrm{\infty }$;

• (iv) $c_n\le b_n^2$.

If ${lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\frac{\mathrm{\Phi }(r)}{1+r}>\parallel x_0-T{x}_0\parallel$ and $\{x_n-T{x}_n\}$ is bounded, then there exists a constant $d_0>0$ such that when $0<b_n+c_n\le d_0$, the sequence $\{x_n\}$ converges strongly to $x^{\ast }$.

Proof The proof of (a) is the same as the proof of Theorem CC1 [3].

(b) Define $a=sup\{r\in R^{+}:\frac{\mathrm{\Phi }(r)}{1+r}\le \parallel x_0-T{x}_0\parallel \}$. Then, by $\mathrm{\Phi }(0)=0$ and $\parallel x_0-T{x}_0\parallel >0$, we have $a>0$. We show that $a\ne \mathrm{\infty }$. If $a=\mathrm{\infty }$, then there exists $\{r_n\}\subset [0,\mathrm{\infty })$, $r_n\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, $\frac{\mathrm{\Phi }(r_n)}{1+r_n}\le \parallel x_0-T{x}_0\parallel$, and hence $\parallel x_0-T{x}_0\parallel <{lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\frac{\mathrm{\Phi }(r)}{1+r}\le \parallel x_0-T{x}_0\parallel$, a contradiction. Therefore, $a<\mathrm{\infty }$.

Let $N^{\ast }={sup}_n\parallel u_n-x^{\ast }\parallel$ and $M={sup}_n\parallel x_n-T{x}_n\parallel +N^{\ast }$. Since T is uniformly continuous on K, for $ϵ=\frac{\parallel x_0-T{x}_0\parallel }{6a}$, there exists $\delta >0$ such that $x,y\in K$ implies $\parallel Tx-Ty\parallel <ϵ$.

Let

$$d_0=\frac{1}{2(a+M)}min\{\delta ,a,\frac{\parallel x_0-T{x}_0\parallel }{24a}\}.$$

(2.3)

Claim 1 $\{x_n\}$ is bounded, i.e.,

$$\parallel x_n-x^{\ast }\parallel \le 2a,\mathrm{\forall }n\ge 0.$$

(2.4)

We show this by induction. By (2.1),

$$\frac{\mathrm{\Phi }(\parallel x_0-x^{\ast }\parallel )}{1+\parallel x_0-x^{\ast }\parallel }\le \parallel x_0-T{x}_0\parallel .$$

Therefore,$\parallel x_0-x^{\ast }\parallel \le a<2a$. Suppose $\parallel x_n-x^{\ast }\parallel \le 2a$, we show that $\parallel x_{n+1}-x^{\ast }\parallel \le 2a$. Suppose not, then $\parallel x_{n+1}-x^{\ast }\parallel >2a>a$ and from the definition of a, we have

$$\frac{\mathrm{\Phi }(\parallel x_{n+1}-x^{\ast }\parallel )}{1+\parallel x_{n+1}-x^{\ast }\parallel }>\parallel x_0-T{x}_0\parallel ,$$

and hence

$$\mathrm{\Phi }\left(\parallel x_{n+1}-x^{\ast }\parallel \right)>\parallel x_0-T{x}_0\parallel .$$

(2.5)

Set ${\alpha }_n=b_n+c_n$. Then Eq. (2.2) becomes

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

(2.6)

where $U_n=u_n-T{x}_n$. Observe that

$$\parallel U_n\parallel \le \parallel u_n-x^{\ast }\parallel +\parallel x_n-x^{\ast }\parallel +\parallel x_n-T{x}_n\parallel \le 2a+M.$$

(2.7)

Furthermore,

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & \le & \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel x_n-T{x}_n\parallel +c_n\parallel U_n\parallel \\ \le & 2a+d_0(2a+2M)\le 3a.\end{array}$$

(2.8)

Also,

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & {\alpha }_n\{\parallel x_n-T{x}_n\parallel +\parallel U_n\parallel \}\\ \le & {\alpha }_n(2a+2M)<d_0(2a+2M)\le \delta ,\end{array}$$

(2.9)

so that $\parallel T{x}_{n+1}-T{x}_n\parallel <ϵ$. Using Lemma 2.1, (2.1), (2.3), (2.5), (2.7)-(2.9) and recursion formula (2.6), we now obtain the following estimates:

$$\begin{array}{c}{\parallel x_{n+1}-x^{\ast }\parallel }^2={\parallel x_n-x^{\ast }-{\alpha }_n(x_n-T{x}_n)+c_n{U}_n\parallel }^2\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2}\le {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n〈x_n-T{x}_n,j(x_{n+1}-x^{\ast })〉+2c_n\parallel U_n\parallel \cdot \parallel x_{n+1}-x^{\ast }\parallel \hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2}\le {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n〈x_{n+1}-T{x}_{n+1}-x_{n+1}+T{x}_{n+1}+x_n-T{x}_n,j(x_{n+1}-x^{\ast })〉\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2\le }+6c_n(2a+M)a\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2}\le {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n\mathrm{\Phi }\left(\parallel x_{n+1}-x^{\ast }\parallel \right)+2{\alpha }_n\parallel x_{n+1}-x_n\parallel \cdot \parallel x_{n+1}-x^{\ast }\parallel \hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2\le }+2{\alpha }_n\parallel T{x}_{n+1}-T{x}_n\parallel \cdot \parallel x_{n+1}-x^{\ast }\parallel +6{\alpha }_n^2(2a+M)a\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2}\le {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n\parallel x_0-T{x}_0\parallel +2{\alpha }_n^2(2a+2M)\cdot 3a+2{\alpha }_n\cdot \frac{3a\parallel x_0-T{x}_0\parallel }{6a}\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2\le }+6{\alpha }_n^2(2a+M)a\hfill \\ \phantom{{\parallel x_{n+1}-x^{\ast }\parallel }^2}\le {\parallel x_n-x^{\ast }\parallel }^2-\frac{{\alpha }_n}{2}\parallel x_0-T{x}_0\parallel <{\parallel x_n-x^{\ast }\parallel }^2,\hfill \end{array}$$

and hence $\parallel x_{n+1}-x^{\ast }\parallel <2a$, a contraction. Hence $\{x_n\}$ is bounded.

Claim 2 ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$.

Suppose this is not true. Let ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =\sigma >0$. Then there exists an integer $N_0$ such that

$$\parallel x_n-x^{\ast }\parallel \ge \frac{\sigma }{2},\mathrm{\forall }n\ge N_0.$$

(2.10)

Since, for any $r_0>0$, ${lim\hspace{0.17em}inf}_{r\to r_0}\mathrm{\Phi }(r)>0$, then ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\mathrm{\Phi }(\parallel x_n-x^{\ast }\parallel )\triangleq \beta >0$. Hence there exists an integer $N_1>N_0$ such that

$$\mathrm{\Phi }\left(\parallel x_n-x^{\ast }\parallel \right)\ge \frac{\beta }{2},\mathrm{\forall }n\ge N_1.$$

(2.11)

Since $\{x_n-T{x}_n\}$, $\{u_n\}$ and $\{x_n\}$ are bounded,

$$\parallel x_{n+1}-x_n\parallel \le {\alpha }_n\parallel x_n-T{x}_n\parallel +c_n\parallel u_n-T{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Therefore, there exists an integer $N_2>N_1$ such that

$$\parallel x_{n+1}-x_n\parallel <\frac{\beta }{16a},\mathrm{\forall }n>N_2.$$

(2.12)

Since T is uniformly continuous, then there exists an integer $N_3>N_2$ such that

$$\parallel T{x}_{n+1}-T{x}_n\parallel <\frac{\beta }{16a},\mathrm{\forall }n>N_3.$$

(2.13)

Also, since ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$, there exists an integer $N_4>N_3$ such that

$${\alpha }_n<\frac{\beta }{16a(2a+M)},\mathrm{\forall }n>N_4.$$

(2.14)

By Lemma and (2.11)-(2.14), we obtain the following estimates:

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n〈x_n-T{x}_n,j(x_{n+1}-x^{\ast })〉+2c_n〈U_n,j(x_{n+1}-x^{\ast })〉\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n\mathrm{\Phi }\left(\parallel x_{n+1}-x^{\ast }\parallel \right)+2{\alpha }_n\parallel x_{n+1}-x_n\parallel \cdot \parallel x_{n+1}-x^{\ast }\parallel \\ +2{\alpha }_n\parallel T{x}_{n+1}-T{x}_n\parallel \cdot \parallel x_{n+1}-x^{\ast }\parallel +2{\alpha }_n^2(2a+M)\parallel x_{n+1}-x^{\ast }\parallel \\ \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n\cdot \frac{\beta }{2}+2{\alpha }_n\cdot \frac{\beta }{16a}\cdot 2a+2{\alpha }_n\cdot \frac{\beta }{16a}\cdot 2a\\ +2{\alpha }_n\cdot \frac{\beta }{16a(2a+M)}\cdot (2a+M)\cdot 2a\\ =& {\parallel x_n-x^{\ast }\parallel }^2-\frac{1}{4}{\alpha }_n\beta \end{array}$$

(2.15)

for all $n\ge N_4$, and this implies ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, a contraction to condition (ii) of Theorem 2.2. Hence Claim 2 holds.

Thus, there exists a subsequence $\{x_{n_j}\}$ such that $x_{n_j}\to x^{\ast }$ as $n\to \mathrm{\infty }$, i.e., for any $ϵ>0$, there exists some integer $n_{j_0}$ such that $\parallel x_{n_{j_0}}-x^{\ast }\parallel <ϵ$.

Claim 3 $\parallel x_{n_{j_0}+m}-x^{\ast }\parallel <ϵ$, $m=1,2,\dots$ .

Let $r_0=inf\{\mathrm{\Phi }(r):r\ge ϵ\}$, then $r_0>0$.

Since $\parallel x_{n+1}-x_n\parallel \to 0$, $\parallel T{x}_{n+1}-T{x}_n\parallel \to 0$ and ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$, then there exists an integer $N>0$ such that for all $n\ge N$, the following inequalities hold:

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \le \frac{r_0}{16a},\hfill \\ \parallel T{x}_{n+1}-T{x}_n\parallel \le \frac{r_0}{16a},\hfill \\ {\alpha }_n<\frac{r_0}{4a(2a+M)}.\hfill \end{array}$$

If $\parallel x_{n_{j_0}+1}-x^{\ast }\parallel \ge ϵ$, then $\mathrm{\Phi }(\parallel x_{n_{j_0}+1}-x^{\ast }\parallel )\ge r_0$. Using recursion formula (2.15), we obtain the following estimate:

$$\begin{array}{rcl}{\parallel x_{n_{j_0}+1}-x^{\ast }\parallel }^2& \le & {\parallel x_{n_{j_0}}-x^{\ast }\parallel }^2-2{\alpha }_n{r}_0+2{\alpha }_n\cdot \frac{r_0}{16a}\cdot 2a+2{\alpha }_n\cdot \frac{r_0}{16a}\cdot 2a\\ +2{\alpha }_n\cdot \frac{r_0}{4a(2a+M)}\cdot (2a+M)\cdot 2a\\ =& {\parallel x_{n_{j_0}}-x^{\ast }\parallel }^2-{\alpha }_n{r}_0+\frac{1}{2}{\alpha }_n{r}_0\\ =& {\parallel x_{n_{j_0}}-x^{\ast }\parallel }^2-\frac{1}{2}{\alpha }_n{r}_0<{\parallel x_{n_{j_0}}-x^{\ast }\parallel }^2<ϵ,\end{array}$$

a contradiction. Hence Claim 3 holds for $m=1$. Assume now that it holds for $m=k$. From the above argument, one easily proves that it holds for $m=k+1$. Hence, Claim 3 holds. This shows that $\{x_n\}$ converges strongly to $x^{\ast }$ as $n\to \mathrm{\infty }$, completing the proof of Theorem 2.2.  □

Theorem 2.3 Let E be a real normed linear space, and let $A:D(A)\subset E\to E$ be a uniformly continuous generalized Φ-accretive type mapping, i.e., there exists $x^{\ast }\in N(A)$ such that for all $x\in E$ , there exist $j(x-x^{\ast })\in J(x-x^{\ast })$ and a function $\mathrm{\Phi }:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $\mathrm{\Phi }(0)=0$ such that

$$〈Ax-A{x}^{\ast },j(x-x^{\ast })〉\ge \mathrm{\Phi }\left(\parallel x-x^{\ast }\parallel \right).$$

For arbitrary $x_0\in D(A)$, define the sequence $\{x_n\}$ iteratively by

$$x_{n+1}=a_n{x}_n+b_{n}S{x}_n+c_n{u}_n,\mathrm{\forall }n\ge 0,$$

where $S:E\to E$ is defined by $Sx:=x-Ax$ for all $x\in D(A)$; and $\{u_n\}$ is a bounded sequence in E, $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ are real sequences in $[0,1]$ satisfying the following conditions:

• (i) $a_n+b_n+c_n=1$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}(b_n+c_n)=\mathrm{\infty }$;

• (iii) $b_n+c_n\to 0$ as $n\to \mathrm{\infty }$;

• (iv) $c_n\le b_n^2$.

If ${lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\frac{\mathrm{\Phi }(r)}{1+r}>\parallel A{x}_0\parallel$ and $\{A{x}_n\}$ is bounded, then there exists a constant $d_0>0$ such that when $0<b_n+c_n\le d_0$, the sequence $\{x_n\}$ converges strongly to $x^{\ast }$.

Proof We simply observe that S is a uniformly continuous and generalized Φ-pseudo-contractive type mapping of $D(A)$ into E. The result can follow from Theorem 2.2. □

Remark 2.4 (1) Our theorems extend and improve Theorem CC1 and Theorem CC2 in the following ways:

• (i) Our theorems do not assume that $\mathrm{\Phi }(t)$ is a strictly increasing function.

• (ii) The conditions ${\sum }_{n=0}^{\mathrm{\infty }}{(b_n+c_n)}^2<\mathrm{\infty }$, ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$ are replaced by $b_n+c_n\to 0$ as $n\to \mathrm{\infty }$, $c_n\le b_n^2$, respectively. Our theorems enlarge the range of $b_n$ and $c_n$ values.

• (iii) We do not need the condition that K is convex. We added the condition that $\{x_n-T{x}_n\}$ is bounded. It is readily seen that $\{x_n\}$ converges strongly to $x^{\ast }$ if and only if $\{x_n-T{x}_n\}$($\{A{x}_n\}$) is bounded under the assumptions of Theorem 2.2 (Theorem 2.3).

(2) Since the class of generalized Φ-accretive maps (generalized Φ-pseudo-contractive maps) includes the class of ϕ-strongly accretive maps (ϕ-strongly pseudo-contractive maps), our results unify and extend many known results. In particular, since ${lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\varphi (r)>\parallel A{x}_0\parallel$ in Theorem MC implies ${lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\frac{\mathrm{\Phi }(r)}{1+r}={lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\frac{\varphi (r)r}{1+r}={lim\hspace{0.17em}inf}_{r\to \mathrm{\infty }}\varphi (r)>\parallel A{x}_0\parallel$, our Theorem 2.3 extends Theorem MC from uniformly smooth Banach spaces to arbitrary normed linear spaces.

(3) Our results also improve and extend the corresponding results in [2, 4–9].