Viscosity iteration algorithm for a ϱ-strictly pseudononspreading mapping in a Hilbert space

Abstract

In this paper, we discuss the strong convergence of the viscosity approximation method in Hilbert spaces relatively to the computation of fixed points of an operator in ϱ-strictly pseudononspreading. Under suitable conditions, some strong convergence theorems are proved. Our work improves previous results for nonspreading mappings.

1 Introduction

Throughout this paper, we always assume that H is a real Hilbert space endowed with an inner product and its induced norm denoted by $〈\cdot ,\cdot 〉$ and $|\cdot |$, respectively. Let C be a nonempty, closed and convex subset of H and let $A:C\to H$ be a nonlinear mapping.

Definition 1.1 A is said to be

• (i) monotone if

$$〈Ax-Ay,x-y〉\ge 0,\mathrm{\forall }x,y\in C;$$

• (ii) strongly monotone if there exists a constant $\alpha >0$ such that

  $$〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

For such a case, A is said to be α-strongly-monotone;

• (iii) inverse-strongly monotone if there exists a constant $\alpha >0$ such that

  $$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

For such a case, A is said to be α-inverse-strongly-monotone;

• (iv) k-Lipschitz continuous if there exists a constant $k\ge 0$ such that

  $$\parallel Ax-Ay\parallel \le k\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

Remark 1.2 Let $F=\mu B-\gamma f$, where B is a θ-Lipschitz and η-strongly monotone operator on H with $\theta >0$ and f is a Lipschitz mapping on H with coefficient $L>0$, $0<\gamma \le \frac{\mu \eta }{L}$. It is a simple matter to see that the operator F is $(\mu \eta -\gamma L)$-strongly monotone over H, i.e.,

$$〈Fx-Fy,x-y〉\ge (\mu \eta -\gamma L){\parallel x-y\parallel }^2,\mathrm{\forall }(x,y)\in H\times H.$$

The classical variational inequality, which is denoted by $VI(A,C)$, is to find $x\in C$ such that

$$〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(1.1)

The variational inequality has been extensively studied in literature (see [1–7] and the references therein).

A mapping $T:C\to C$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

A mapping T is said to be firmly nonexpansive if

$${\parallel Tx-Ty\parallel }^2\le 〈x-Tx,y-Ty〉,\mathrm{\forall }x,y\in C;$$

see, for instance, [8–11]. It is known that a mapping $T:C\to C$ is firmly nonexpansive if and only if

$${\parallel Tx-Ty\parallel }^2+{\parallel (I-T)x-(I-T)y\parallel }^2\le {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

T is said to be nonspreading in [12] if

$${\parallel Tx-Ty\parallel }^2\le {\parallel Tx-y\parallel }^2+{\parallel Ty-x\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.2)

It is shown in [13] that (1.2) is equivalent to

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+2〈x-Tx,y-Ty〉,\mathrm{\forall }x,y\in C.$$

These mappings are generalization of a firmly nonexpansive mapping in a Hilbert space. $T:C\to C$ is said to be firmly nonexpansive if

$${\parallel Tx-Ty\parallel }^2\le 〈x-y,Tx-Ty〉,\mathrm{\forall }x,y\in C.$$

See [14–17] for more information on firmly nonexpansive mappings.

Definition 1.3 $T:H\to H$ is called demicontractive on H if there exists a constant $\alpha <1$ such that

$${\parallel Tx-q\parallel }^2\le {\parallel x-q\parallel }^2+\alpha {\parallel x-Tx\parallel }^2,\mathrm{\forall }(x,q)\in H\times F_{ix}(T).$$

(1.3)

Definition 1.4 [18]

$T:D(T)\subseteq H\to H$ is ϱ-strictly pseudononspreading if there exists $\varrho \in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+\varrho {\parallel x-Tx-(y-Ty)\parallel }^2+2〈x-Tx,y-Ty〉,$$

(1.4)

for all $x,y\in D(T)$.

Remark 1.5 It is easy to claim that firmly nonexpansive mapping ⇒ nonspreading mapping ⇒ ϱ-strictly pseudononspreading mapping.

Indeed, from the definition of those mappings, $\mathrm{\forall }x,y\in C$, we obtain

Clearly, every nonspreading mapping is ϱ-strictly pseudononspreading. The following example shows that the class of ϱ-strictly pseudononspreading mappings is more general than the class of nonspreading mappings. Let us give an example of a ϱ-strictly pseudononspreading mapping satisfying the condition of Definition 1.4.

Example 1.6 Let $X=l^2$ with the norm $\parallel \cdot \parallel$ defined by

$$\parallel x\parallel =\sqrt{\sum _{i=1}^{\mathrm{\infty }}{x}_i^2},\mathrm{\forall }x=(x_1,x_2,\dots ,x_n,\dots )\in X,$$

$C=\{x=(x_1,x_2,\dots ,x_n,\dots )|x_i\in R^1,i=1,2,\dots \}$, and let C be an orthogonal subspace of X (i.e., $\mathrm{\forall }x,y\in C$, we have $〈x,y〉=0$). Then it is obvious that C is a nonempty closed convex subset of X. Now, for any $x=(x_1,x_2,\dots ,x_n,\dots )\in C$, define a mapping $T:C\to C$ as follows:

$$Tx=\{\begin{array}{cc}(x_1,x_2,\dots ,x_n,\dots ),\hfill & {\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0,\hfill \\ (-2x_1,-2x_2,\dots ,-2x_n,\dots ),\hfill & {\prod }_{i=1}^{\mathrm{\infty }}{x}_i\ge 0.\hfill \end{array}$$

(1.5)

To see that T is $\frac{1}{3}$-strictly pseudononspreading, we break the process of proof into three cases. $\mathrm{\forall }x,y\in C$,

Case 1: ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i<0$, observe that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2+2〈x-Tx,y-Ty〉,\frac{1}{3}\in [0,1),$$

since ${\parallel Tx-Ty\parallel }^2={\parallel x-y\parallel }^2$ and $\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2=2〈x-Tx,y-Ty〉=0$.

Case 2: ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i\le 0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i\ge 0$, we obtain ${\parallel Tx-Ty\parallel }^2={\parallel x+2y\parallel }^2={\parallel x\parallel }^2+4〈x,y〉+4{\parallel y\parallel }^2$, $2〈x-Tx,y-Ty〉=0$ and $\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2=3{\parallel y\parallel }^2$.

Hence,

Case 3: ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i\ge 0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i\ge 0$, we have ${\parallel Tx-Ty\parallel }^2=4{\parallel x-y\parallel }^2$, ${\parallel x-Tx-(y-Ty)\parallel }^2=9{\parallel x-y\parallel }^2$ and $2〈x-Tx,y-Ty〉=18〈x,y〉=0$. Thus

$$\begin{array}{rcl}{\parallel Tx-Ty\parallel }^2& =& 4{\parallel x-y\parallel }^2\\ =& {\parallel x-y\parallel }^2+\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2\\ \le & {\parallel x-y\parallel }^2+\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2+2〈x-Tx,y-Ty〉.\end{array}$$

From (1), (2) and (3), we obtain that T is $\frac{1}{3}$-strictly pseudononspreading, i.e.,

$${\parallel Tx-Ty\parallel }^2={\parallel x-y\parallel }^2+\frac{1}{3}{\parallel x-Tx-(y-Ty)\parallel }^2+2〈x-Tx,y-Ty〉,\mathrm{\forall }x,y\in R.$$

We can easily know that $F_{ix}(T)=\{(x_1,x_2,\dots ,x_n,\dots ),{\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0\}\cup \{0\}$, where $F_{ix}(T)$ is defined by the set of fixed points of T.

T is not nonspreading, since for $x=\{0,0,\dots ,0,\dots \}$, $y=\{1,0,\dots ,0,\dots \}$, we have ${\parallel Tx-Ty\parallel }^2=4$, ${\parallel x-y\parallel }^2=1$ and $2〈x-Tx,y-Ty〉=0$, we obtain

$${\parallel Tx-Ty\parallel }^2=4>1={\parallel x-y\parallel }^2+2〈x-Tx,y-Ty〉.$$

Since our class of maps contains the class of nonspreading mappings, it also contains the class of firmly nonexpansive mappings.

Remark 1.7 [19]

Let T be an α-demicontractive mapping on H with $F_{ix}(T)\ne \mathrm{\varnothing }$ and $T_{\omega }=(1-\omega )I+\omega T$ for $\omega \in (0,\mathrm{\infty })$:

(A1) T α-demicontractive is equivalent to

$$〈x-T_{\omega }x,x-q〉\ge \frac{\omega }{2}{\parallel x-Tx\parallel }^2,\mathrm{\forall }(x,q)\in H\times F_{ix}(T).$$

(A2) $F_{ix}(T)=F_{ix}(T_{\omega })$ if $\omega \ne 0$.

Remark 1.8 Observe that if T is ϱ-strictly pseudononspreading and $F_{ix}(T)\ne \mathrm{\varnothing }$, then $\mathrm{\forall }x\in D(T)$ and $\mathrm{\forall }p\in F_{ix}(T)$, we obtain

$${\parallel Tx-p\parallel }^2\le {\parallel x-p\parallel }^2+\varrho {\parallel x-Tx\parallel }^2.$$

Thus, every ϱ-strictly pseudononspreading mapping with a nonempty fixed point set $F_{ix}(T)$ is demicontractive (see [20, 21]).

Remark 1.9 According to Remark 1.7(A1) and the fact that the ϱ-strictly pseudononspreading mapping of T is demicontractive, let $I-T_{\omega }=\omega (I-T)$. Then we obtain

$$〈x-T_{\omega }x,x-q〉\ge \frac{\omega (1-\varrho )}{2}{\parallel x-Tx\parallel }^2,\mathrm{\forall }(x,q)\in H\times F_{ix}(T).$$

(1.6)

In 2011, Osilike and Isiogugu [12] introduced the following propositions and proved a strong convergence theorem somewhat related to a Halpern-type iteration algorithm for a ϱ-strictly pseudononspreading mapping in Hilbert spaces.

Proposition 1.10 [12]

Let C be a nonempty closed convex subset of a real Hilbert space H and let $T:C\to C$ be a ϱ-strictly pseudononspreading mapping. If $F_{ix}(T)\ne \mathrm{\varnothing }$, then it is closed and convex.

Proposition 1.11 [12]

Let C be a nonempty closed convex subset of a real Hilbert space H and let $T:C\to C$ be a ϱ-strictly pseudononspreading mapping. Then $(I-T)$ is demiclosed at 0.

Theorem 1.12 [12]

Let C be a nonempty closed convex subset of a real Hilbert space H and let $T:C\to C$ be a ϱ-strictly pseudononspreading mapping with a nonempty fixed point set $F_{ix}(T)$. Let $\alpha \in [\varrho ,1)$ and let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be a real sequence in $[0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$. Let $u\in C$, $\{x_n\}$ and $\{z_n\}$ be sequences in C generated from an arbitrary $x_1\in C$ by

$$\{\begin{array}{c}{x}_{n+1}={\alpha }_{n}u+(I-{\alpha }_n)z_n,n>0,\hfill \\ z_n=\frac{1}{n}{\sum }_{k=1}^{n-1}{T}_{\alpha }^k{x}_n,n\ge 1.\hfill \end{array}$$

(1.7)

Then $\{x_n\}$ and $\{z_n\}$ converge strongly to $P_{F_{ix}(T)}u$, where $P_{F_{ix}(T)}:H\to F_{ix}(T)$ is a metric projection of H onto $F_{ix}(T)$.

In 2010, Tian [22] introduced the following theorem for finding an element of a set of solutions to the fixed point of a nonexpansive mapping in a Hilbert space.

Theorem 1.13 [22]

Let f be a contraction on a real Hilbert space H and T be a nonexpansive mapping on H. Starting with an arbitrary initial $x_0\in H$, define a sequence $\{x_n\}$ generated by

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

(1.8)

where B is a θ-Lipschitz and η-strongly monotone operator on H with $\theta >0$, $\eta >0$ and $0<\mu <2\eta /{\theta }^2$. Assume also that a sequence $\{{\alpha }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

(c1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

(c2) ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$ or ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n+1}/{\alpha }_n=1$.

Then the sequence $\{x_n\}$ generated by (1.8) converges strongly to the unique solution $x^{\ast }\in F_{ix}(T)$ of the variational inequality

$$〈(\gamma f-\mu B)x^{\ast },x-x^{\ast }〉\le 0,\mathrm{\forall }x\in F_{ix}(T).$$

(1.9)

In this paper, we combine Theorem 1.12 and Theorem 1.13 and introduce the following general iterative algorithm for a ϱ-strictly pseudononspreading mapping T.

Algorithm 1.14 Let $x_0\in H$ be arbitrary

$$\{\begin{array}{c}{x}_{n+1}={\alpha }_n\gamma f(x_n)+(I-\mu {\alpha }_{n}B)z_n,n>0,\hfill \\ z_n=\frac{1}{n}{\sum }_{k=1}^n{T}_{\alpha }^k{x}_n,n\ge 1,\hfill \end{array}$$

where $B:H\to H$ is η-strongly monotone and boundedly Lipschitzian, f is an L-Lipschitz mapping on H with coefficient $L>0$ and $T_{\alpha }^k=(1-\alpha )I+\alpha T^k$, $\alpha \in ({\varrho }_k,\frac{1}{2})$.

Under suitable conditions, some strong convergence theorems are proved in the following chapter.

2 Preliminaries

Throughout this paper, we write $x_n\rightharpoonup x$ to indicate that the sequence $\{x_n\}$ converges weakly to x. $x_n\to x$ implies that $\{x_n\}$ converges strongly to x. The following lemmas are useful for main results.

Definition 2.1 A mapping T is said to be demiclosed if for any sequence $\{x_n\}$ which weakly converges to y, and if the sequence $\{T{x}_n\}$ strongly converges to z, then $T(y)=z$.

Lemma 2.2 [3]

Assume $\{{\alpha }_n\}$ is a sequence of nonnegative real numbers such that

$${\alpha }_{n+1}\le (1-{\gamma }_n){\alpha }_n+{\delta }_n,n\ge 0,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence in ℝ such that

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$,

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}=0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$.

Lemma 2.3 [1]

Let $\{{\mathcal{T}}_n\}$ be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence ${\{{\mathcal{T}}_{n_j}\}}_{j\ge 0}$ of $\{{\mathcal{T}}_n\}$ which satisfies ${\mathcal{T}}_{n_j}<{\mathcal{T}}_{n_j+1}$ for all $j\ge 0$. Also, consider the sequence of integers ${\{\delta (n)\}}_{n\ge n_0}$ defined by

$$\delta (n)=max\{k\le n|{\mathcal{T}}_k<{\mathcal{T}}_{k+1}\}.$$

(2.1)

Then ${\{\delta (n)\}}_{n\ge n_0}$ is a nondecreasing sequence verifying ${lim}_{n\to \mathrm{\infty }}\delta (n)=\mathrm{\infty }$, $\mathrm{\forall }n\ge n_0$. It holds that $T_{\delta (n)}<{\mathcal{T}}_{\delta (n)+1}$, and we have

$${\mathcal{T}}_n<{\mathcal{T}}_{\delta (n)+1}.$$

Lemma 2.4 Let K be a closed convex subset of a real Hilbert space H given $x\in H$ and $y\in K$. Then $y=P_{K}x$ if and only if the following inequality holds:

$$〈x-y,y-z〉\ge 0,\mathrm{\forall }z\in K.$$

3 Main results

Let C be a nonempty closed convex subset of a real Hilbert space H and let $T^k:C\to C$ be a ${\varrho }_k$-strictly pseudononspreading mapping with a common nonempty fixed point set ${\bigcap }_k^n{F}_{ix}(T^k)$. Let f be an L-Lipschitz mapping on H with coefficient $L>0$. Assume the set ${\bigcap }_k^n{F}_{ix}(T^k)$ is nonempty. Since ${\bigcap }_k^n{F}_{ix}(T^k)$ is closed and convex, the nearest point projection from C onto ${\bigcap }_k^n{F}_{ix}(T^k)$ is well defined. Recall $B:H\to H$ is η-strongly monotone and θ-Lipschitzian on H with $\theta >0$, $\eta >0$. Let $0<\mu <2\eta /{\theta }^2$, $0<\gamma <\mu (\eta -\frac{\mu {\theta }^2}{2})/L=\tau /L$, consider the following sequence $\{x_n\}$ defined by

$$\{\begin{array}{c}{x}_{n+1}={\alpha }_n\gamma f(x_n)+(I-\mu {\alpha }_{n}B)z_n,n>0,\hfill \\ z_n=\frac{1}{n}{\sum }_{k=1}^n{T}_{\alpha }^k{x}_n,n\ge 1,\hfill \end{array}$$

(3.1)

where $T_{\alpha }^k=(1-\alpha )I+\alpha T^k$, $\alpha \in ({\varrho }_k,\frac{1}{2})$, $k=\{1,2,\dots ,n\}$, and $\{{\alpha }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

(c1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$,

(c2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ or ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n+1}/{\alpha }_n=1$.

Remark 3.1 [23]

Let H be a real Hilbert space. Let B be a θ-Lipschitzian and η-strongly monotone operator on H with $\theta >0$, $\eta >0$. Let $0<\mu <2\eta /{\theta }^2$, let $S=(I-t\mu B)$ and $\mu (\eta -\frac{\mu {\theta }^2}{2})=\tau$, then for $t\in (0,min\{1,\frac{1}{\tau }\})$, S is a contraction with a constant $1-t\tau$.

Before stating our main result, we introduce some lemmas for algorithm (3.1) as follows.

Lemma 3.2 The sequence $\{x_n\}$ is generated by (3.1) with $T^k$ being a ϱ-strictly pseudononspreading mapping on H and $\{{\alpha }_n\}\subset (0,1)$. Then $\{x_n\}$ is bounded.

Proof Let $T_{\alpha }^{k}x=(1-\alpha )x+\alpha T^{k}x$ and $0<{\varrho }_k<\alpha <\frac{1}{2}$. Then $\mathrm{\forall }x,y\in C$, we have

$$\begin{array}{rcl}{\parallel T_{\alpha }^{k}x-T_{\alpha }^{k}y\parallel }^2& =& \alpha {\parallel x-y\parallel }^2+(1-\alpha ){\parallel T^{k}x-T^{k}y\parallel }^2-\alpha (1-\alpha ){\parallel x-T^{k}x-(y-T^{k}y)\parallel }^2\\ \le & \alpha {\parallel x-y\parallel }^2+(1-\alpha )[{\parallel x-y\parallel }^2+{\varrho }_k{\parallel x-T^{k}x-(y-T^{k}y)\parallel }^2\\ +2〈x-T^{k}x,y-T^{k}y〉]-\alpha (1-\alpha ){\parallel x-T^{k}x-(y-T^{k}y)\parallel }^2\\ =& {\parallel x-y\parallel }^2+2(1-\alpha )〈x-T^{k}x,y-T^{k}y〉\\ -(1-\alpha )(\alpha -{\varrho }_k){\parallel x-T^{k}x-(y-T^{k}y)\parallel }^2\\ \le & {\parallel x-y\parallel }^2+2(1-\alpha )〈x-T^{k}x,y-T^{k}y〉\\ =& {\parallel x-y\parallel }^2+\frac{2(1-\alpha )}{{\alpha }^2}〈x-T_{\alpha }^{k}x,y-T_{\alpha }^{k}y〉.\end{array}$$

(3.2)

From $p\in {\bigcap }_k^n{F}_{ix}(T^k)$ and (3.2), we also have

$$\parallel T_{\alpha }^k{x}_n-p\parallel \le \parallel x_n-p\parallel .$$

(3.3)

According to (3.3), (3.1) and Remark 3.1, we obtain

$$\parallel z_n-p\parallel =\parallel \frac{1}{n}\sum _{k=1}^n{T}_{\alpha }^k{x}_n-p\parallel \le \frac{1}{n}\sum _{k=1}^n\parallel T_{\alpha }^k{x}_n-p\parallel \le \frac{1}{n}\sum _{k=1}^n\parallel x_n-p\parallel =\parallel x_n-p\parallel .$$

(3.4)

Thus,

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n\gamma (f(x_n)-f(p))+{\alpha }_n(\gamma f(p)-\mu Bp)+(I-\mu {\alpha }_{n}B)(z_n-p)\parallel \\ \le & {\alpha }_n\gamma \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \gamma f(p)-\mu Bp\parallel +(1-{\alpha }_n\tau )\parallel z_n-p\parallel \\ \le & {\alpha }_n\gamma \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \gamma f(p)-\mu Bp\parallel +(1-{\alpha }_n\tau )\parallel x_n-p\parallel ,\end{array}$$

(3.5)

which combined with $\parallel f(x_n)-f(p)\parallel \le L\parallel x_n-p\parallel$ amounts to

$$\parallel x_{n+1}-p\parallel \le (1-{\alpha }_n(\tau -\gamma L))\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f(p)-\mu Bp\parallel .$$

(3.6)

Putting $M_1=max\{\parallel x_0-p\parallel ,\parallel \gamma f(p)-\mu Bp\parallel \}$, we clearly obtain $\parallel x_n-p\parallel \le M_1$. Hence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded. From (3.3), we have that ${\{T_{\alpha }^k{x}_n\}}_{n=1}^{\mathrm{\infty }}$ is also bounded. □

Now we are in a position to claim the main result.

Theorem 3.3 Assume C is a nonempty closed convex subset of a real Hilbert space H and let $T^k:C\to C$ be a ${\varrho }_k$-strictly pseudononspreading mapping with a common nonempty fixed point set ${\bigcap }_k^n{F}_{ix}(T^k)$. Let f be an L-Lipschitz mapping on H with coefficient $L>0$ and $B:H\to H$ be η-strongly monotone and θ-Lipschitzian on H with $\theta >0$, $\eta >0$. Let $0<\mu <2\eta /{\theta }^2$, $0<\gamma <\mu (\eta -\frac{\mu {\theta }^2}{2})/L=\tau /L$. Consider the sequences ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ to be sequences in C generated from an arbitrary $x_1\in C$ by (3.1), where $T_{\alpha }^k=(1-\alpha )I+\alpha T^k$, $\alpha \in ({\varrho }_k,\frac{1}{2})$, $k=\{1,2,\dots ,n\}$, ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}\in [0,1)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$. Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to the unique element $x^{\ast }$ in ${\bigcap }_k^n{F}_{ix}(T^k)$ verifying

$$P_{{\bigcap }_k^n{F}_{ix}(T^k)}(I-\mu B+\gamma f)x^{\ast }=x^{\ast },$$

(3.7)

which equivalently solves the following variational inequality problem:

$$x^{\ast }\in \bigcap _k^n{F}_{ix}\left(T^k\right),〈(\gamma f-\mu B)x^{\ast },v-x^{\ast }〉\le 0,\mathrm{\forall }v\in \bigcap _k^n{F}_{ix}\left(T^k\right).$$

(3.8)

Proof According to Lemma 3.2, it is simple to know that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{T_{\alpha }^k{x}_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded. Thus, for $\mathrm{\forall }y\in C$ and $\mathrm{\forall }k=0,1,2,\dots ,n-1$ and according to (3.2) and (3.1), we have

$$\begin{array}{rcl}{\parallel T_{\alpha }^{k+1}{x}_n-T_{\alpha }y\parallel }^2& =& {\parallel T_{\alpha }\left(T_{\alpha }^k{x}_n\right)-T_{\alpha }y\parallel }^2\\ \le & {\parallel T_{\alpha }^k{x}_n-y\parallel }^2+\frac{2}{(1-\alpha )}〈T_{\alpha }^k{x}_n-T_{\alpha }^{k+1}{x}_n,y-T_{\alpha }y〉\\ =& {\parallel T_{\alpha }^k{x}_n-T_{\alpha }y\parallel }^2+{\parallel T_{\alpha }y-y\parallel }^2+2〈T_{\alpha }^k{x}_n-T_{\alpha }y,T_{\alpha }y-y〉\\ +\frac{2}{(1-\alpha )}〈T_{\alpha }^k{x}_n-T_{\alpha }^{k+1}{x}_n,y-T_{\alpha }y〉.\end{array}$$

(3.9)

Summing (3.9) from $k=0$ to n and dividing by n, we obtain

$$\begin{array}{rcl}\frac{1}{n}\parallel T_{\alpha }^{k+1}{x}_n-T_{\alpha }y\parallel & \le & \frac{1}{n}{\parallel x_n-T_{\alpha }y\parallel }^2+{\parallel T_{\alpha }y-y\parallel }^2+2〈z_n-T_{\alpha }y,T_{\alpha }y-y〉\\ +\frac{2}{n(1-\alpha )}〈x_n-T_{\alpha }^n{x}_n,T_{\alpha }y-y〉.\end{array}$$

(3.10)

Since ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded, then there exists a subsequence ${\{z_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ of ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ which converges weakly to $\omega \in C$. Replacing n by $n_j$ in (3.10), we obtain

(3.11)

Since ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{T_{\alpha }^n{x}_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded, letting $j\to \mathrm{\infty }$ in (3.11) yields

$$0\le {\parallel T_{\alpha }y-y\parallel }^2+2〈\omega -T_{\alpha }y,T_{\alpha }y-y〉.$$

(3.12)

Let $y=\omega$ in (3.12). We obtain that $\omega \in F_{ix}(T_{\alpha })=F_{ix}(T)$.

Observe that since ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ are bounded, and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, then

$$\begin{array}{rcl}\parallel x_{n+1}-z_n\parallel & =& {\alpha }_n\parallel \gamma f(x_n)-\mu B{z}_n\parallel \\ \le & {\alpha }_n\gamma \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \gamma f(p)-\mu Bp\parallel +{\alpha }_n\parallel \mu B(z_n-p)\parallel \\ \le & {\alpha }_n\gamma L\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f(p)-\mu Bp\parallel +{\alpha }_n\tau \parallel z_n-p\parallel ,\end{array}$$

then

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-z_n\parallel =0.$$

(3.13)

We next show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-\mu B)z,x_{n+1}-z〉\le 0.$$

(3.14)

Indeed, take ${\{x_{n_j+1}\}}_{n=1}^{\mathrm{\infty }}$ of ${\{x_{n+1}\}}_{n=1}^{\mathrm{\infty }}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-\mu B)x^{\ast },x_{n+1}-x^{\ast }〉=\underset{j\to \mathrm{\infty }}{lim}〈(\gamma f-\mu B)x^{\ast },x_{n_j+1}-x^{\ast }〉,$$

where $x^{\ast }$ is obtained in (3.7). We may assume that $x_{n_j+1}\rightharpoonup z$ as $j\to \mathrm{\infty }$. From (3.13), we have $z_{n_j}\rightharpoonup z$ as $j\to \mathrm{\infty }$, then to arbitrary bounded linear functional g on H, we have

$$\begin{array}{rcl}\parallel g(z_{n_j})-g(z)\parallel & \le & \parallel g(z_{n_j})-g(x_{n_j+1})\parallel +\parallel g(x_{n_j+1})-g(z)\parallel \\ \le & \parallel g\parallel \parallel z_{n_j}-x_{n_j+1}\parallel +\parallel g(x_{n_j+1})-g(z)\parallel \\ \to & 0,\text{as }j\to 0.\end{array}$$

Thus, we obtain $z_{n_j}\to z$ as $j\to 0$, and $z\in F_{ix}(T)$. Hence, we have

$$\underset{j\to \mathrm{\infty }}{lim}〈(\gamma f-\mu B)x^{\ast },x_{n_j+1}-x^{\ast }〉=〈(\gamma f-\mu B)x^{\ast },z-x^{\ast }〉\le 0.$$

(3.15)

Moreover, from (3.1), (3.13) and (3.14), we have

(3.16)

As required, finally we show that $x_n\to x^{\ast }$ and $z_n\to x^{\ast }$.

According to (3.1), (3.4) and (3.16), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& {\parallel {\alpha }_n(\gamma f(x_n)-\mu B{x}^{\ast })+(I-\mu {\alpha }_{n}B)z_n-(I-\mu {\alpha }_{n}B)x^{\ast }\parallel }^2\\ =& {\alpha }_n^2{\parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel }^2+{\parallel (I-\mu {\alpha }_{n}B)z_n-(I-\mu {\alpha }_{n}B)x^{\ast }\parallel }^2\\ +2{\alpha }_n〈(I-\mu {\alpha }_{n}B)z_n-(I-\mu {\alpha }_{n}B)x^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉\\ \le & {\alpha }_n^2{\parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel }^2+{(1-{\alpha }_n\tau )}^2{\parallel z_n-x^{\ast }\parallel }^2\\ +2{\alpha }_n[〈z_n-x^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉-\mu {\alpha }_n〈B{z}_n-B{x}^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉]\\ \le & [{(1-{\alpha }_n\tau )}^2+2{\alpha }_n\gamma L]{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n[2〈z_n-x^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉\\ +{\alpha }_n{\parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel }^2+2\mu {\alpha }_n\parallel B{z}_n-B{x}^{\ast }\parallel \parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel ]\\ \le & [1-2{\alpha }_n(\tau -\gamma L)]{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n[2〈x_n-x^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉\\ +{\alpha }_n{\parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel }^2+2\mu {\alpha }_n\parallel B{z}_n-B{x}^{\ast }\parallel \parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel \\ +{\alpha }_n{\tau }^2{\parallel x_n-x^{\ast }\parallel }^2]\\ =& (1-\overline{{\alpha }_n}){\parallel x_n-x^{\ast }\parallel }^2+\overline{{\alpha }_n}\overline{{\beta }_n},\end{array}$$

where $\overline{{\alpha }_n}=2{\alpha }_n(\tau -\gamma L)$,

$$\begin{array}{rcl}\overline{{\beta }_n}& =& \frac{1}{2(\tau -\gamma L)}[2〈x_n-x^{\ast },\gamma f(x_n)-\mu B{x}^{\ast }〉+{\alpha }_n{\parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel }^2\\ +2\mu {\alpha }_n\parallel B{z}_n-B{x}^{\ast }\parallel \parallel \gamma f(x_n)-\mu B{x}^{\ast }\parallel +{\alpha }_n{\tau }^2{\parallel x_n-x^{\ast }\parallel }^2].\end{array}$$

It is easily seen that ${lim}_{n\to \mathrm{\infty }}\overline{{\alpha }_n}$, $\sum \overline{{\alpha }_n}=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\overline{{\beta }_n}\le 0$. By Lemma 2.2, we conclude that $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$, and $z_n$ also converges strongly to the unique element $x^{\ast }$ in $F_{ix}(T)$. In addition, the variational inequality (3.15) can be written as

$$〈(I-\mu B+\gamma f)x^{\ast }-x^{\ast },z-x^{\ast }〉\ge 0,z\in \bigcap _k^n{F}_{ix}\left(T^k\right).$$

So, by Lemma 2.4, it is equivalent to the fixed point equation

$$P_{{\bigcap }_k^n{F}_{ix}(T^k)}(I-\mu B+\gamma f)x^{\ast }=x^{\ast }.$$

 □

Remark 3.4 For a nonspreading mapping T, we have $\varrho =0$ in Theorem 3.3 to obtain the following corollary.

Corollary 3.5 Assume C is a nonempty closed convex subset of a real Hilbert space H and let $T^k:C\to C$ be a nonspreading mapping with a common nonempty fixed point set ${\bigcap }_k^n{F}_{ix}(T^k)$. Let f be an L-Lipschitz mapping on H with coefficient $L>0$ and $B:H\to H$ be η-strongly monotone and θ-Lipschitzian on H with $\theta >0$, $\eta >0$. Let $0<\mu <2\eta /{\theta }^2$, $0<\gamma <\mu (\eta -\frac{\mu {\theta }^2}{2})/L=\tau /L$, consider the sequences ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ to be sequences in C generated from an arbitrary $x_1\in C$ by

$$\{\begin{array}{c}{x}_{n+1}={\alpha }_n\gamma f(x_n)+(I-\mu {\alpha }_{n}B)z_n,n>0,\hfill \\ z_n=\frac{1}{n}{\sum }_{k=1}^n{T}_{\alpha }^k{x}_n,n\ge 1,\hfill \end{array}$$

where $T_{\alpha }^k=(1-\alpha )I+\alpha T^k$, $\alpha \in (0,\frac{1}{2})$, ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}\in [0,1)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$. Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to the unique element $x^{\ast }$ in ${\bigcap }_k^n{F}_{ix}(T^k)$ verifying

$$P_{{\bigcap }_k^n{F}_{ix}(T^k)}(I-\mu B+\gamma f)x^{\ast }=x^{\ast },$$

which equivalently solves the following variational inequality problem:

$$x^{\ast }\in \bigcap _k^n{F}_{ix}\left(T^k\right),〈(\gamma f-\mu B)x^{\ast },v-x^{\ast }〉\le 0,\mathrm{\forall }v\in \bigcap _k^n{F}_{ix}\left(T^k\right).$$