On generalized quasi-ϕ-nonexpansive mappings and their projection algorithms

Abstract

A fixed point problem of a generalized asymptotically quasi-ϕ-nonexpansive mapping and an equilibrium problem are investigated. A strong convergence theorem for solutions of the fixed point problem and the equilibrium problem is established in a Banach space.

1 Introduction and preliminaries

Let E be a real Banach space, and let $E^{\ast }$ be the dual space of E. We denote by J the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

$$Jx=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^2={\parallel f^{\ast }\parallel }^2\},$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. A Banach space E is said to be strictly convex if $\parallel \frac{x+y}{2}\parallel <1$ for all $x,y\in E$ with $\parallel x\parallel =\parallel y\parallel =1$ and $x\ne y$. It is said to be uniformly convex if ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$ for any two sequences $\{x_n\}$ and $\{y_n\}$ in E such that $\parallel x_n\parallel =\parallel y_n\parallel =1$ and ${lim}_{n\to \mathrm{\infty }}\parallel \frac{x_n+y_n}{2}\parallel =1$. Let $U_E=\{x\in E:\parallel x\parallel =1\}$ be the unit sphere of E. Then the Banach space E is said to be smooth provided

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists for each $x,y\in U_E$. It is also said to be uniformly smooth if the above limit is attained uniformly for $x,y\in U_E$. It is well known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E. It is also well known that E is uniformly smooth if and only if $E^{\ast }$ is uniformly convex.

Recall that a Banach space E enjoys the Kadec-Klee property if for any sequence $\{x_n\}\subset E$, and $x\in E$ with $x_n\rightharpoonup x$, and $\parallel x_n\parallel \to \parallel x\parallel$, then $\parallel x_n-x\parallel \to 0$ as $n\to \mathrm{\infty }$. For more details on the Kadec-Klee property, the readers can refer to [1] and the references therein. It is well known that if E is a uniformly convex Banach space, then E enjoys the Kadec-Klee property.

Let C be a nonempty subset of E. Let f be a bifunction from $C\times C$ to ℝ, where ℝ denotes the set of real numbers. In this paper, we investigate the following equilibrium problem. Find $p\in C$ such that

$$f(p,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.1)

We use $EP(f)$ to denote the solution set of the equilibrium problem (1.1). That is,

$$EP(f)=\{p\in C:f(p,y)\ge 0,\mathrm{\forall }y\in C\}.$$

Given a mapping $Q:C\to E^{\ast }$, let

$$f(x,y)=〈Qx,y-x〉,\mathrm{\forall }x,y\in C.$$

Then $p\in EP(f)$ iff p is a solution of the following variational inequality. Find p such that

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

(1.2)

In order to study the solution problem of the equilibrium problem (1.1), we assume that f satisfies the following conditions:

• (A1) $f(x,x)=0$, $\mathrm{\forall }x\in C$;

• (A2) f is monotone, i.e., $f(x,y)+f(y,x)\le 0$, $\mathrm{\forall }x,y\in C$;

• (A3)
  $$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}f(tz+(1-t)x,y)\le f(x,y),\mathrm{\forall }x,y,z\in C;$$

• (A4) for each $x\in C$, $y\mapsto f(x,y)$ is convex and weakly lower semi-continuous.

As we all know, if C is a nonempty closed convex subset of a Hilbert space H and $P_C:H\to C$ is the metric projection of H onto C, then $P_C$ is nonexpansive. This fact actually characterizes Hilbert spaces and, consequently, it is not available in more general Banach spaces. In this connection, Alber [2] recently introduced a generalized projection operator ${\mathrm{\Pi }}_C$ in a Banach space E, which is an analogue of the metric projection $P_C$ in Hilbert spaces.

Next, we assume that E is a smooth Banach space. Consider the functional defined by

$$\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

Observe that in a Hilbert space H, the equality is reduced to $\varphi (x,y)={\parallel x-y\parallel }^2$, $x,y\in H$. The generalized projection ${\mathrm{\Pi }}_C:E\to C$ is a map that assigns to an arbitrary point $x\in E$ the minimum point of the functional $\varphi (x,y)$, that is, ${\mathrm{\Pi }}_{C}x=\overline{x}$, where $\overline{x}$ is the solution to the minimization problem

$$\varphi (\overline{x},x)=\underset{y\in C}{min}\varphi (y,x).$$

Existence and uniqueness of the operator ${\mathrm{\Pi }}_C$ follows from the properties of the functional $\varphi (x,y)$ and strict monotonicity of the mapping J; see, for example, [1] and [2]. In Hilbert spaces, ${\mathrm{\Pi }}_C=P_C$. It is obvious from the definition of function ϕ that

$${(\parallel x\parallel -\parallel y\parallel )}^2\le \varphi (x,y)\le {(\parallel y\parallel +\parallel x\parallel )}^2,\mathrm{\forall }x,y\in E$$

(1.3)

and

$$\varphi (x,y)=\varphi (x,z)+\varphi (z,y)+2〈x-z,Jz-Jy〉,\mathrm{\forall }x,y,z\in E.$$

(1.4)

Remark 1.1 If E is a reflexive, strictly convex and smooth Banach space, then $\varphi (x,y)=0$ if and only if $x=y$; for more details, see [1] and [3].

Let $T:C\to C$ be a mapping. In this paper, we use $F(T)$ to denote the fixed point set of T. T is said to be asymptotically regular on C if, for any bounded subset K of C, ${lim}_{n\to \mathrm{\infty }}{sup}_{x\in K}\parallel T^{n+1}x-T^{n}x\parallel =0$. T is said to be closed if, for any sequence $\{x_n\}\subset C$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x_0$ and ${lim}_{n\to \mathrm{\infty }}T{x}_n=y_0$, $T{x}_0=y_0$. In this paper, we use → and ⇀ to denote the strong convergence and weak convergence, respectively.

A point p in C is said to be an asymptotic fixed point of T [3] iff C contains a sequence $\{x_n\}$ which converges weakly to p such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. The set of asymptotic fixed points of T will be denoted by $\tilde{F}(T)$. T is said to be relatively nonexpansive [4, 5] iff $\tilde{F}(T)=F(T)\ne \mathrm{\varnothing }$ and $\varphi (p,Tx)\le \varphi (p,x)$ for all $x\in C$ and $p\in F(T)$. T is said to be relatively asymptotically nonexpansive [6, 7] iff $\tilde{F}(T)=F(T)\ne \mathrm{\varnothing }$ and there exists a sequence $\{{\mu }_n\}\subset [1,\mathrm{\infty })$ with ${\mu }_n\to 1$ as $n\to \mathrm{\infty }$ such that $\varphi (p,Tx)\le {\mu }_n\varphi (p,x)$ for all $x\in C$, $p\in F(T)$ and $n\ge 1$. T is said to be quasi-ϕ-nonexpansive [8, 9] iff $F(T)\ne \mathrm{\varnothing }$ and $\varphi (p,Tx)\le \varphi (p,x)$ for all $x\in C$ and $p\in F(T)$. T is said to be asymptotically quasi-ϕ-nonexpansive [10–12] iff $F(T)\ne \mathrm{\varnothing }$ and there exists a sequence $\{{\mu }_n\}\subset [1,\mathrm{\infty })$ with ${\mu }_n\to 1$ as $n\to \mathrm{\infty }$ such that $\varphi (p,Tx)\le {\mu }_n\varphi (p,x)$ for all $x\in C$, $p\in F(T)$ and $n\ge 1$.

Remark 1.2 The class of asymptotically quasi-ϕ-nonexpansive mappings is more general than the class of relatively asymptotically nonexpansive mappings which requires the restriction $F(T)=\tilde{F}(T)$.

Remark 1.3 The classes of asymptotically quasi-ϕ-nonexpansive mappings and quasi-ϕ-nonexpansive mappings are the generalizations of asymptotically quasi-nonexpansive mappings and quasi-nonexpansive mappings in Hilbert spaces.

Recently, Qin et al. [13] introduced a class of generalized asymptotically quasi-ϕ-nonexpansive mappings. Recall that a mapping T is said to be generalized asymptotically quasi-ϕ-nonexpansive iff $F(T)\ne \mathrm{\varnothing }$ and there exist a sequence $\{{\mu }_n\}\subset [1,\mathrm{\infty })$ with ${\mu }_n\to 1$ as $n\to \mathrm{\infty }$ and a sequence $\{{\nu }_n\}\subset [0,\mathrm{\infty })$ with ${\nu }_n\to 0$ as $n\to \mathrm{\infty }$ such that $\varphi (p,Tx)\le {\mu }_n\varphi (p,x)+{\nu }_n$ for all $x\in C$, $p\in F(T)$ and $n\ge 1$.

Remark 1.4 The class of generalized asymptotically quasi-ϕ-nonexpansive mappings is a generalization of the class of generalized asymptotically quasi-nonexpansive mappings which was studied in [14].

Recently, fixed point and equilibrium problems (1.1) have been intensively investigated based on iterative methods; see [15–28]. The projection method which grants strong convergence of the iterative sequences is one of efficient methods for the problems. In this paper, we investigate the equilibrium problem (1.1) and a fixed point problem of the generalized quasi-ϕ-nonexpansive mapping based on a projection method. A strong convergence theorem for solutions of the equilibrium and the fixed point problem is established in a Banach space.

In order to state our main results, we need the following lemmas.

Lemma 1.5 [2]

Let E be a reflexive, strictly convex, and smooth Banach space, let C be a nonempty, closed, and convex subset of E, and $x\in E$. Then

$$\varphi (y,{\mathrm{\Pi }}_{C}x)+\varphi ({\mathrm{\Pi }}_{C}x,x)\le \varphi (y,x),\mathrm{\forall }y\in C.$$

Lemma 1.6 [2]

Let C be a nonempty, closed, and convex subset of a smooth Banach space E, and $x\in E$. Then $x_0={\mathrm{\Pi }}_{C}x$ if and only if

$$〈x_0-y,Jx-J{x}_0〉\ge 0,\mathrm{\forall }y\in C.$$

Lemma 1.7 [11]

Let E be a reflexive, strictly convex, and smooth Banach space such that both E and $E^{\ast }$ have the Kadec-Klee property. Let C be a nonempty closed and convex subset of E. Let $T:C\to C$ be a closed asymptotically quasi-ϕ-nonexpansive mapping. Then $F(T)$ is a closed convex subset of C.

Lemma 1.8 [29, 30]

Let C be a closed convex subset of a smooth, strictly convex, and reflexive Banach space E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Then there exists $z\in C$ such that $f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0$, $\mathrm{\forall }y\in C$. Define a mapping $S_r:E\to C$ by $S_{r}x=\{z\in C:f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C\}$. Then the following conclusions hold:

• (1) $S_r$ is a single-valued and firmly nonexpansive-type mapping, i.e., for all $x,y\in E$,

  $$〈S_{r}x-S_{r}y,J{S}_{r}x-J{S}_{r}y〉\le 〈S_{r}x-S_{r}y,Jx-Jy〉;$$

• (2) $F(S_r)=EP(f)$ is closed and convex;

• (3) $S_r$ is quasi-ϕ-nonexpansive;

• (4) $\varphi (q,S_{r}x)+\varphi (S_{r}x,x)\le \varphi (q,x)$, $\mathrm{\forall }q\in F(S_r)$.

Lemma 1.9 [31]

Let E be a smooth and uniformly convex Banach space, and let $r>0$. Then there exists a strictly increasing, continuous and convex function $g:[0,2r]\to R$ such that $g(0)=0$ and

$${\parallel tx+(1-t)y\parallel }^2\le t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t)g(\parallel x-y\parallel )$$

for all $x,y\in B_r=\{x\in E:\parallel x\parallel \le r\}$ and $t\in [0,1]$.

2 Main results

Theorem 2.1 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property. Let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let $T:C\to C$ be a closed generalized asymptotically quasi-ϕ-nonexpansive mapping. Assume that T is asymptotically regular on C and that $\mathcal{F}=F(T)\cap EP(f)$ is nonempty and bounded. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}^n{x}_n),\hfill \\ u_n\in C\mathit{\text{such that}}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+({\mu }_n-1)M_n+{\nu }_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1,\hfill \end{array}$$

where $M_n=sup\{\varphi (z,x_n):z\in \mathcal{F}\}$, $\{{\alpha }_n\}$ is a real sequence in $[0,1]$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and $\{r_n\}$ is a real sequence in $[a,\mathrm{\infty })$, where a is some positive real number. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathcal{F}}{x}_1$.

Proof In view of Lemma 1.7 and Lemma 1.8, we find that ℱ is closed and convex, so that ${\mathrm{\Pi }}_{\mathcal{F}}x$ is well defined for any $x\in C$. Next, we show that $C_n$ is closed and convex. It is obvious that $C_1=C$ is closed and convex. Suppose that $C_m$ is closed and convex for some $m\in \mathbb{N}$. We now show that $C_{m+1}$ is also closed and convex. For $z_1,z_2\in C_{m+1}$, we see that $z_1,z_2\in C_m$. It follows that $z=t{z}_1+(1-t)z_2\in C_m$, where $t\in (0,1)$. Notice that

$$\varphi (z_1,u_m)\le \varphi (z_1,x_m)+({\mu }_m-1)M_m+{\nu }_m$$

and

$$\varphi (z_1,u_h)\le \varphi (z_1,x_m)+({\mu }_m-1)M_m+{\nu }_m.$$

The above inequalities are equivalent to

$$2〈z_1,J{x}_m-J{u}_m〉\le {\parallel x_m\parallel }^2-{\parallel u_m\parallel }^2+({\mu }_m-1)M_m+{\nu }_m$$

(2.1)

and

$$2〈z_2,J{x}_m-J{u}_m〉\le {\parallel x_m\parallel }^2-{\parallel u_m\parallel }^2+({\mu }_m-1)M_m+{\nu }_m.$$

(2.2)

Multiplying t and $(1-t)$ on both sides of (2.1) and (2.2), respectively, yields that

$$2〈z,J{x}_m-J{u}_m〉\le {\parallel x_m\parallel }^2-{\parallel y_m\parallel }^2+({\mu }_m-1)M_m+{\nu }_m.$$

That is,

$$\varphi (z,u_m)\le \varphi (z,x_m)+({\mu }_m-1)M_m+{\nu }_m,$$

(2.3)

where $z\in C_m$. This gives that $C_{m+1}$ is closed and convex. Then $C_n$ is closed and convex. This shows that ${\mathrm{\Pi }}_{C_{n+1}}{x}_1$ is well defined.

Next, we show that $\mathcal{F}\subset C_n$. $\mathcal{F}\subset C_1=C$ is obvious. Suppose that $\mathcal{F}\subset C_m$ for some $m\in \mathbb{N}$. Fix $w\in \mathcal{F}\subset C_m$. It follows that

$$\begin{array}{rcl}\varphi (w,u_m)& =& \varphi (w,S_{r_m}{y}_m)\\ \le & \varphi (w,y_m)\\ =& \varphi (w,J^{-1}({\alpha }_{m}J{x}_m+(1-{\alpha }_m)J{T}^m{x}_m))\\ =& {\parallel w\parallel }^2-2〈w,{\alpha }_{m}J{x}_m+(1-{\alpha }_m)J{T}^m{x}_m〉+{\parallel {\alpha }_{m}J{x}_m+(1-{\alpha }_m)J{T}^m{x}_m\parallel }^2\\ \le & {\parallel w\parallel }^2-2{\alpha }_m〈w,J{x}_m〉-2(1-{\alpha }_m)〈w,J{T}^m{x}_m〉+{\alpha }_m{\parallel x_m\parallel }^2+(1-{\alpha }_m){\parallel T^m{x}_m\parallel }^2\\ =& {\alpha }_m\varphi (w,x_m)+(1-{\alpha }_m)\varphi (w,T^m{x}_m)\\ \le & {\alpha }_m\varphi (w,x_m)+(1-{\alpha }_m){\mu }_m\varphi (w,x_m)+(1-{\alpha }_m){\nu }_m\\ =& \varphi (w,x_m)-(1-{\alpha }_m)\varphi (w,x_m)+(1-{\alpha }_m){\mu }_m\varphi (w,x_m)+(1-{\alpha }_m){\nu }_m\\ \le & \varphi (w,x_m)+(1-{\alpha }_m)({\mu }_m-1)\varphi (w,x_m)+{\nu }_m\\ \le & \varphi (w,x_m)+({\mu }_m-1)M_m+{\nu }_m,\end{array}$$

(2.4)

which shows that $w\in C_{m+1}$. This implies that $\mathcal{F}\subset C_n$ for each $n\ge 1$. In view of $x_n={\mathrm{\Pi }}_{C_n}{x}_1$, from Lemma 1.6 we find that $〈x_n-z,J{x}_1-J{x}_n〉\ge 0$ for any $z\in C_n$. It follows from $\mathcal{F}\subset C_n$ that

$$〈x_n-w,J{x}_1-J{x}_n〉\ge 0,\mathrm{\forall }w\in \mathcal{F}.$$

(2.5)

It follows from Lemma 1.5 that

$$\begin{array}{rl}\varphi (x_n,x_1)& =\varphi ({\mathrm{\Pi }}_{C_n}{x}_1,x_1)\\ \le \varphi ({\mathrm{\Pi }}_{\mathcal{F}}{x}_1,x_1)-\varphi ({\mathrm{\Pi }}_{\mathcal{F}}{x}_1,x_n)\\ \le \varphi ({\mathrm{\Pi }}_{\mathcal{F}}{x}_1,x_1).\end{array}$$

This implies that the sequence $\{\varphi (x_n,x_1)\}$ is bounded. It follows from (1.3) that the sequence $\{x_n\}$ is also bounded. Since the space is reflexive, we may assume that $x_n\rightharpoonup \overline{x}$. Since $C_n$ is closed and convex, we find that $\overline{x}\in C_n$. On the other hand, we see from the weak lower semicontinuity of the norm that

$$\begin{array}{rl}\varphi (\overline{x},x_1)& ={\parallel \overline{x}\parallel }^2-2〈\overline{x},J{x}_1〉+{\parallel x_1\parallel }^2\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\parallel x_n\parallel }^2-2〈x_n,J{x}_1〉+{\parallel x_1\parallel }^2)\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_n,x_1)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (x_n,x_1)\\ \le \varphi (\overline{x},x_1),\end{array}$$

which implies that ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_1)=\varphi (\overline{x},x_1)$. Hence, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =\parallel \overline{x}\parallel$. In view of the Kadec-Klee property of E, we find that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$.

Now, we are in a position to prove that $\overline{x}\in F(T)$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_1$ and $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\in C_{n+1}\subset C_n$, we find that $\varphi (x_n,x_1)\le \varphi (x_{n+1},x_1)$. This shows that $\{\varphi (x_n,x_1)\}$ is nondecreasing. It follows from the boundedness that ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_1)$ exists. In view of the construction of $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\in C_{n+1}\subset C_n$, we arrive at

$$\begin{array}{rl}\varphi (x_{n+1},x_n)& =\varphi (x_{n+1},{\mathrm{\Pi }}_{C_n}{x}_1)\\ \le \varphi (x_{n+1},x_1)-\varphi ({\mathrm{\Pi }}_{C_n}{x}_1,x_1)\\ =\varphi (x_{n+1},x_1)-\varphi (x_n,x_1).\end{array}$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},x_n)=0.$$

(2.6)

In light of $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\in C_{n+1}$, we find that

$$\varphi (x_{n+1},u_n)\le \varphi (x_{n+1},x_n)+({\mu }_n-1)M_n+{\nu }_n.$$

Thanks to (2.6), we find that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},u_n)=0.$$

(2.7)

In view of (1.3), we see that ${lim}_{n\to \mathrm{\infty }}(\parallel x_{n+1}\parallel -\parallel u_n\parallel )=0$. It follows that ${lim}_{n\to \mathrm{\infty }}\parallel u_n\parallel =\parallel \overline{x}\parallel$. This is equivalent to

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{u}_n\parallel =\parallel J\overline{x}\parallel .$$

(2.8)

This implies that $\{J{u}_n\}$ is bounded. Note that both E and $E^{\ast }$ are reflexive. We may assume that $J{u}_n\rightharpoonup u^{\ast }\in E^{\ast }$. In view of the reflexivity of E, we see that $J(E)=E^{\ast }$. This shows that there exists an element $u\in E$ such that $Ju=u^{\ast }$. It follows that

$$\begin{array}{rl}\varphi (x_{n+1},u_n)& ={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel u_n\parallel }^2\\ ={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel J{u}_n\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ on both sides of the equality above yields that

$$\begin{array}{rl}0& \ge {\parallel \overline{x}\parallel }^2-2〈\overline{x},u^{\ast }〉+{\parallel u^{\ast }\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Ju〉+{\parallel Ju\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Ju〉+{\parallel u\parallel }^2\\ =\varphi (\overline{x},u).\end{array}$$

That is, $\overline{x}=u$, which in turn implies that $u^{\ast }=J\overline{x}$. It follows that $J{u}_n\rightharpoonup J\overline{x}\in E^{\ast }$. Since $E^{\ast }$ enjoys the Kadec-Klee property, we obtain from (2.8) that

$$\underset{n\to \mathrm{\infty }}{lim}J{u}_n=J\overline{x}.$$

Since E enjoys the Kadec-Klee property, we obtain that $u_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Note that $\parallel x_n-u_n\parallel \le \parallel x_n-\overline{x}\parallel +\parallel \overline{x}-u_n\parallel$. It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel =0.$$

(2.9)

This gives that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{u}_n\parallel =0.$$

(2.10)

Notice that

$$\begin{array}{rl}\varphi (w,x_n)-\varphi (w,u_n)& ={\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2-2〈w,J{x}_n-J{u}_n〉\\ \le \parallel x_n-u_n\parallel (\parallel x_n\parallel +\parallel u_n\parallel )+2\parallel w\parallel \parallel J{x}_n-J{u}_n\parallel .\end{array}$$

It follows from (2.9) and (2.10) that

$$\underset{n\to \mathrm{\infty }}{lim}(\varphi (w,x_n)-\varphi (w,u_n))=0.$$

(2.11)

Since E is uniformly smooth, we know that $E^{\ast }$ is uniformly convex. In view of Lemma 1.9, we see that

$$\begin{array}{r}\varphi (w,u_n)\\ =\varphi (w,S_{r_n}{y}_n)\\ \le \varphi (w,y_n)\\ =\varphi (w,J^{-1}[{\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}^n{x}_n])\\ ={\parallel w\parallel }^2-2〈w,{\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}^n{x}_n〉+{\parallel {\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}^n{x}_n\parallel }^2\\ \le {\parallel w\parallel }^2-2{\alpha }_n〈w,J{x}_n〉-2(1-{\alpha }_n)〈w,J{T}^n{x}_n〉+{\alpha }_n{\parallel x_n\parallel }^2+(1-{\alpha }_n){\parallel T^n{x}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n)g\left(\parallel J{x}_n-J\left(T^n{x}_n\right)\parallel \right)\\ ={\alpha }_n\varphi (w,x_n)+(1-{\alpha }_n)\varphi (w,T^n{x}_n)-{\alpha }_n(1-{\alpha }_n)g\left(\parallel J{x}_n-J\left(T^n{x}_n\right)\parallel \right)\\ \le {\alpha }_n\varphi (w,x_n)+(1-{\alpha }_n){\mu }_n\varphi (w,x_n)+{\nu }_n-{\alpha }_n(1-{\alpha }_n)g\left(\parallel J{x}_n-J\left(T^n{x}_n\right)\parallel \right)\\ \le \varphi (w,x_n)+(1-{\alpha }_n)({\mu }_n-1)\varphi (w,x_n)+{\nu }_n-{\alpha }_n(1-{\alpha }_n)g\left(\parallel J{x}_n-J\left(T^n{x}_n\right)\parallel \right).\end{array}$$

This implies that

$${\alpha }_n(1-{\alpha }_n)g\left(\parallel J{x}_n-J\left(T^n{x}_n\right)\parallel \right)\le \varphi (w,x_n)-\varphi (w,u_n)+(1-{\alpha }_n)({\mu }_n-1)\varphi (w,x_n)+{\nu }_n.$$

In view of the restrictions on the sequence $\{{\alpha }_n\}$, we find from (2.11) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J\left(T^n{x}_n\right)-J{x}_n\parallel =0.$$

(2.12)

Notice that

$$\parallel J\left(T^n{x}_n\right)-J\overline{x}\parallel \le \parallel J\left(T^n{x}_n\right)-J{x}_n\parallel +\parallel J{x}_n-J\overline{x}\parallel .$$

It follows from (2.12) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J\left(T^n{x}_n\right)-J\overline{x}\parallel =0.$$

(2.13)

The demicontinuity of $J^{-1}:E^{\ast }\to E$ implies that $T^n{x}_n\rightharpoonup \overline{x}$. Note that

$$|\parallel T^n{x}_n\parallel -\parallel \overline{x}\parallel |=|\parallel J\left(T^n{x}_n\right)\parallel -\parallel J\overline{x}\parallel |\le \parallel J\left(T^n{x}_n\right)-J\overline{x}\parallel .$$

This implies from (2.13) that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n\parallel =\parallel \overline{x}\parallel$. Since E has the Kadec-Klee property, we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-\overline{x}\parallel =0$. Since

$$\parallel T^{n+1}{x}_n-\overline{x}\parallel \le \parallel T^{n+1}{x}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-\overline{x}\parallel ,$$

we find from the asymptotic regularity of T that ${lim}_{n\to \mathrm{\infty }}\parallel T^{n+1}{x}_n-\overline{x}\parallel =0$, that is, $T{T}^n{x}_n-\overline{x}\to 0$ as $n\to \mathrm{\infty }$. It follows from the closedness of T that $T\overline{x}=\overline{x}$.

Next, we show that $\overline{x}\in EP(f)$. In view of Lemma 1.8, we find from (2.4) that

$$\begin{array}{rl}\varphi (u_n,y_n)& \le \varphi (w,y_n)-\varphi (w,u_n)\\ \le \varphi (w,x_n)+({\mu }_n-1)M_n+{\nu }_n-\varphi (w,u_n)\\ =\varphi (w,x_n)-\varphi (w,u_n)+(k_n-1)M_n.\end{array}$$

(2.14)

It follows from (2.11) that ${lim}_{n\to \mathrm{\infty }}\varphi (u_n,y_n)=0$. This implies that ${lim}_{n\to \mathrm{\infty }}(\parallel u_n\parallel -\parallel y_n\parallel )=0$. In view of (2.9), we see that $u_n\to \overline{x}$ as $n\to \mathrm{\infty }$. This implies that $\parallel y_n\parallel -\parallel \overline{x}\parallel \to 0$ as $n\to \mathrm{\infty }$. It follows that ${lim}_{n\to \mathrm{\infty }}\parallel J{y}_n\parallel =\parallel J\overline{x}\parallel$. Since $E^{\ast }$ is reflexive, we may assume that $J{y}_n\rightharpoonup r^{\ast }\in E^{\ast }$. In view of $J(E)=E^{\ast }$, we see that there exists $r\in E$ such that $Jr=r^{\ast }$. It follows that

$$\begin{array}{rl}\varphi (u_n,y_n)& ={\parallel u_n\parallel }^2-2〈u_n,J{y}_n〉+{\parallel y_n\parallel }^2\\ ={\parallel u_n\parallel }^2-2〈u_n,J{y}_n〉+{\parallel J{y}_n\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ on both sides of the equality above yields that

$$\begin{array}{rl}0& \ge {\parallel \overline{x}\parallel }^2-2〈\overline{x},r^{\ast }〉+{\parallel r^{\ast }\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Jr〉+{\parallel Jr\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Jr〉+{\parallel r\parallel }^2\\ =\varphi (\overline{x},r).\end{array}$$

That is, $\overline{x}=r$, which in turn implies that $r^{\ast }=J\overline{x}$. It follows that $J{y}_n\rightharpoonup J\overline{x}\in E^{\ast }$. Since $E^{\ast }$ enjoys the Kadec-Klee property, we obtain that $J{y}_n-J\overline{x}\to 0$ as $n\to \mathrm{\infty }$. Note that $J^{-1}:E^{\ast }\to E$ is demicontinuous. It follows that $y_n\rightharpoonup \overline{x}$. Since E enjoys the Kadec-Klee property, we obtain that $y_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Note that

$$\parallel u_n-y_n\parallel \le \parallel u_n-\overline{x}\parallel +\parallel \overline{x}-y_n\parallel .$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-y_n\parallel =0.$$

(2.15)

Since J is uniformly norm-to-norm continuous on any bounded sets, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{u}_n-J{y}_n\parallel =0.$$

From the assumption $r_n\ge a$, we see that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel J{u}_n-J{y}_n\parallel }{r_n}=0.$$

(2.16)

In view of $u_n=S_{r_n}{y}_n$, we see that

$$f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from (A2) that

$$\parallel y-u_n\parallel \frac{\parallel J{u}_n-J{y}_n\parallel }{r_n}\ge \frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge -f(u_n,y)\ge f(y,u_n),\mathrm{\forall }y\in C.$$

By taking the limit as $n\to \mathrm{\infty }$ in the above inequality, from (A4) and (2.16) we obtain that

$$f(y,\overline{x})\le 0,\mathrm{\forall }y\in C.$$

For $0<t<1$ and $y\in C$, define $y_t=ty+(1-t)\overline{x}$. It follows that $y_t\in C$, which yields that $f(y_t,\overline{x})\le 0$. It follows from (A1) and (A4) that

$$0=f(y_t,y_t)\le tf(y_t,y)+(1-t)f(y_t,\overline{x})\le tf(y_t,y).$$

That is,

$$f(y_t,y)\ge 0.$$

Letting $t\downarrow 0$, we obtain from (A3) that $f(\overline{x},y)\ge 0$, $\mathrm{\forall }y\in C$. This implies that $\overline{x}\in EP(f)$. This shows that $\overline{x}\in \mathcal{F}=EP(f)\cap F(T)$.

Finally, we prove that $\overline{x}={\mathrm{\Pi }}_{\mathcal{F}}{x}_1$. Letting $n\to \mathrm{\infty }$ in (2.5), we see that

$$〈\overline{x}-w,J{x}_1-J\overline{x}〉\ge 0,\mathrm{\forall }w\in \mathcal{F}.$$

In view of Lemma 1.6, we find that $\overline{x}={\mathrm{\Pi }}_{\mathcal{F}}{x}_1$. This completes the proof. □

If T is asymptotically quasi-ϕ-nonexpansive, then Theorem 2.1 is reduced to the following.

Corollary 2.2 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property. Let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let $T:C\to C$ be a closed asymptotically quasi-ϕ-nonexpansive mapping. Assume that T is asymptotically regular on C and that $\mathcal{F}=F(T)\cap EP(f)$ is nonempty and bounded. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}^n{x}_n),\hfill \\ u_n\in C\mathit{\text{such that}}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+({\mu }_n-1)M_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1,\hfill \end{array}$$

where $M_n=sup\{\varphi (z,x_n):z\in \mathcal{F}\}$, $\{{\alpha }_n\}$ is a real sequence in $[0,1]$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and $\{r_n\}$ is a real sequence in $[a,\mathrm{\infty })$, where a is some positive real number. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathcal{F}}{x}_1$.

If T is quasi-ϕ-nonexpansive, then Theorem 2.1 is reduced to the following.

Corollary 2.3 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property. Let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let $T:C\to C$ be a closed quasi-ϕ-nonexpansive mapping. Assume that $\mathcal{F}=F(T)\cap EP(f)$ is nonempty. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)JT{x}_n),\hfill \\ u_n\in C\mathit{\text{such that}}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ is a real sequence in $[0,1]$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and $\{r_n\}$ is a real sequence in $[a,\mathrm{\infty })$, where a is some positive real number. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathcal{F}}{x}_1$.

If T is the identity, then Theorem 2.1 is reduced to the following.

Corollary 2.4 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property. Let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4). Assume that $EP(f)$ is nonempty. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ u_n\in C\mathit{\text{such that}}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{x}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+({\mu }_n-1)M_n+{\nu }_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1,\hfill \end{array}$$

where $M_n=sup\{\varphi (z,x_n):z\in \mathcal{F}\}$, $\{{\alpha }_n\}$ is a real sequence in $[0,1]$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and $\{r_n\}$ is a real sequence in $[a,\mathrm{\infty })$, where a is some positive real number. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{EP(f)}{x}_1$.