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On generalized asymptotically quasiϕnonexpansive mappings and a Ky Fan inequality
Fixed Point Theory and Applications volume 2013, Article number: 237 (2013)
Abstract
Generalized asymptotically quasiϕnonexpansive mappings and a Ky Fan inequality are investigated. A strong convergence theorem for common solutions to a fixed point problem of generalized asymptotically quasiϕnonexpansive mappings and a Ky Fan inequality is established in a Banach space.
MSC:47H05, 47J25, 90C33.
1 Introduction
Iterative algorithms have been studied by many authors. The applications of iterative algorithms are found in a wide range of areas, including economics, image recovery and signal processing. Many wellknown problems can be studied by using algorithms which are iterative in their nature; see [1–14] and the references therein. As an example, in computer tomography with limited data, each piece of information implies the existence of a convex set, in which the required solution lies. The problem of finding a point in the intersection of these convex subsets is then of crucial interest, and it cannot be usually solved directly. Therefore, an iterative algorithm must be used to approximate such a point.
Mann iteration, introduced by Mann [15], is an efficient tool to study fixed point problems of asymptotical nonexpansive mappings. However, Mann iteration is only weak convergence in infinitedimensional spaces; see [10] and the references therein. The importance of strong convergence is underlined in [16], where a convex function f is minimized via the proximalpoint algorithm: it is shown that the rate of convergence of the value sequence \{f({x}_{n})\} is better when \{{x}_{n}\} converges strongly than when it converges weakly. Such properties have a direct impact when the process is executed directly in the underlying infinitedimensional space. To obtain strong convergence of Mann iteration, projection methods, which were first introduced by Haugazeau [17], have been considered for modifying Mann iteration to obtain strong convergence. The advantage of projection methods is that strong convergence of iterative sequences can be guaranteed without any compact assumptions.
The organization of this paper is as follows. In Section 2, we provide some necessary concepts and lemmas. In Section 3, fixed point problems of generalized asymptotically quasiϕnonexpansive mappings and solutions of a Ky Fan inequality are investigated. A strong convergence theorem is established in a Banach space.
2 Preliminaries
Recall that the normalized duality mapping J from E to {2}^{{E}^{\ast}} is defined by
where \u3008\cdot ,\cdot \u3009 denotes the generalized duality pairing. 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 iff
exists for each x,y\in {U}_{E}. It is also said to be uniformly smooth iff 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 normtonorm 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 E is said to be strictly convex iff \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 iff {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. Recall that E enjoys the KadecKlee 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 KadecKlee property, readers can refer to [18] and the references therein. It is well known that if E is a uniformly convex Banach space, then E enjoys the KadecKlee property.
Next, we assume that E is a smooth Banach space. Consider the functional defined by
Observe that in a Hilbert space H, the equality is reduced to \varphi (x,y)={\parallel xy\parallel}^{2}, x,y\in H. 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 [19] 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. Recall that 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
Existence and uniqueness of the operator {\mathrm{\Pi}}_{C} follow from the properties of the functional \varphi (x,y) and strict monotonicity of the mapping J; see, for example, [18]. In Hilbert spaces, {\mathrm{\Pi}}_{C}={P}_{C}. It is obvious from the definition of a function ϕ that
and
Remark 2.1 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 [18] and the reference therein.
Let C be a nonempty subset of E and 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,
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}, then T{x}_{0}={y}_{0}. In this paper, we use → and ⇀ to denote strong convergence and weak convergence, respectively. Recall that a point p in C is said to be an asymptotic fixed point of T iff C contains a sequence \{{x}_{n}\} which converges weakly to p so 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).
Recall that T is said to be relatively nonexpansive iff
Recall that T is said to be relatively asymptotically nonexpansive iff
where \{{\mu}_{n}\}\subset [0,\mathrm{\infty}) is a sequence such that {\mu}_{n}\to 0 as n\to \mathrm{\infty}.
Recall that a mapping T is said to be quasiϕnonexpansive iff
Recall that a mapping T is said to be asymptotically quasiϕnonexpansive iff there exists a sequence \{{\mu}_{n}\}\subset [0,\mathrm{\infty}) with {\mu}_{n}\to 0 as n\to \mathrm{\infty} such that
Remark 2.2 The class of asymptotically quasiϕnonexpansive mappings was considered in Zhou et al. [20] and Qin et al. [21]; see also [22] and [23].
Remark 2.3 The class of quasiϕnonexpansive mappings and the class of asymptotically quasiϕnonexpansive mappings are more general than the class of relatively nonexpansive mappings and the class of relatively asymptotically nonexpansive mappings [24]. Quasiϕnonexpansive mappings and asymptotically quasiϕnonexpansive mappings do not require the restriction F(T)=\tilde{F}(T).
Remark 2.4 The class of quasiϕnonexpansive mappings and the class of asymptotically quasiϕnonexpansive mappings are generalizations of the class of quasinonexpansive mappings and the class of asymptotically quasinonexpansive mappings in Banach spaces.
Recall that T is said to be generalized asymptotically quasiϕnonexpansive if 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} 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 2.5 The class of generalized asymptotically quasiϕnonexpansive mappings was considered in Qin et al. [25]; see also [26].
Let f be a bifunction from C\times C to ℝ, where ℝ denotes the set of real numbers, and let A:C\to {E}^{\ast} be a mapping. Consider the following Ky Fan inequality which is known as a generalized equilibrium problem. Find p\in C such that
We use S(f,A) to denote the solution set of inequality (2.3). That is,
If A=0, then problem (2.3) is reduced to the following Ky Fan inequality which is known as an equilibrium problem. Find p\in C such that
We use S(f) to denote the solution set of inequality (2.4). That is,
If f=0, then problem (2.3) is reduced to the classical variational inequality. Find p\in C such that
We use VI(C,A) to denote the solution set of inequality (2.5). That is,
Recall that a mapping A:C\to {E}^{\ast} is said to be αinversestrongly monotone if there exists \alpha >0 such that
For solving problem (2.3), let us assume that the nonlinear mapping A:C\to {E}^{\ast} is αinversestrongly monotone and the bifunction f:C\times C\to \mathbb{R} 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+(1t)x,y)\le F(x,y),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y,z\in C;

(A4)
for each x\in C, y\mapsto F(x,y) is convex and weakly lower semicontinuous.
Recently, many authors investigated the solutions of problems (2.3), (2.4) and (2.5) based on iterative methods; see [27–37]. In this paper, we investigate generalized asymptotically quasiϕnonexpansive mappings and problem (2.3). A strong convergence theorem for common solutions to a fixed point problem of generalized asymptotically quasiϕnonexpansive mappings and problem (2.3) is established in a Banach space.
In order to state our main results, we need the following lemmas, which play an import role in the paper.
Lemma 2.6 [28]
Let E be a smooth, strictly convex and reflexive Banach space and C be a nonempty closed convex subset of E. Let A:C\to {E}^{\ast} be an αinversestrongly monotone mapping and f be a bifunction satisfying conditions (A1)(A4). Let r>0 be any given number and x\in E be any given point. Then there exists p\in C such that
Lemma 2.7 [28]
Let E be a smooth, strictly convex and reflexive Banach space and C be a nonempty closed convex subset of E. Let A:C\to {E}^{\ast} be an αinversestrongly monotone mapping and f be a bifunction satisfying conditions (A1)(A4). Let r>0 be any given number and x\in E define a mapping {K}_{r}:C\to C as follows: for any x\in C,
Then the following conclusions hold:

(1)
{K}_{r} is singlevalued;

(2)
{K}_{r} is a firmly nonexpansivetype mapping, i.e., for all x,y\in E,
\u3008{K}_{r}x{K}_{r}y,J{K}_{r}xJ{K}_{r}y\u3009\le \u3008{S}_{r}x{S}_{r}y,JxJy\u3009; 
(3)
F({K}_{r})=S(f,A);

(4)
{K}_{r} is quasiϕnonexpansive;

(5)
\varphi (q,{K}_{r}x)+\varphi ({K}_{r}x,x)\le \varphi (q,x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}q\in F({K}_{r});

(6)
S(f,A) is closed and convex.
Lemma 2.8 [19]
Let C be a nonempty closed convex subset of a smooth Banach space E and x\in E. Then {x}_{0}={\mathrm{\Pi}}_{C}x if and only if
Lemma 2.9 [19]
Let E be a reflexive, strictly convex and smooth Banach space, C be a nonempty closed convex subset of E and x\in E. Then
Lemma 2.10 [25]
Let E be a uniformly smooth and strictly convex Banach space which also enjoys the KadecKlee property and C be a nonempty closed convex subset of E. Let T:C\to C be a closed generalized asymptotically quasiϕnonexpansive mapping. Then F(T) is closed and convex.
Lemma 2.11 [38]
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
for all x,y\in {B}_{r}=\{x\in E:\parallel x\parallel \le r\} and t\in [0,1].
3 Main results
Theorem 3.1 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the KadecKlee property and C be a nonempty closed convex subset of E. Let T:C\to C be a generalized asymptotically quasiϕnonexpansive mapping. Let f be a bifunction from C\times C to ℝ satisfying (A1)(A4) and A:C\to {E}^{\ast} be an αinversestrongly monotone mapping. Assume that T is closed and asymptotically regular on C, and F(T)\cap S(f,A) is nonempty and bounded. Let \{{x}_{n}\} be a sequence generated in the following manner:
where {W}_{n}=sup\{\varphi (p,{x}_{n}):p\in F(T)\cap S(f,A)\}, \{{\alpha}_{n}\} is a real number 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 number sequence such that {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{r}_{n}>0. Then the sequence \{{x}_{n}\} converges strongly to {\mathrm{\Pi}}_{F(T)\cap S(f,A)}{x}_{1}, where {\mathrm{\Pi}}_{F(T)\cap S(f,A)} is the generalized projection from E onto F(T)\cap S(f,A).
Proof First, we prove {C}_{n} is closed and convex so that the projection is well defined. We see that {C}_{1}=C is closed and convex. Assume that {C}_{m} is closed and convex for some positive integer m. For k\in {C}_{m}, we find that
which is equivalent to
It is easy to see that {C}_{m+1} is closed and convex. This proves that {C}_{n} is closed and convex so that {\mathrm{\Pi}}_{{C}_{n+1}}{x}_{1} is well defined. Set {u}_{n}={K}_{{r}_{n}}{y}_{n}. It follows from Lemma 2.7 that {K}_{{r}_{n}} is quasiϕnonexpansive.
Now, we are in a position to prove that F(T)\cap S(f,A)\subset {C}_{n}. Indeed, F(T)\cap S(f,A)\subset {C}_{1}=C is obvious. Assume that F(T)\cap S(f,A)\subset {C}_{m} for some positive integer m. Then, for \mathrm{\forall}e\in F(T)\cap S(f,A)\subset {C}_{m}, we have
which proves that e\in {C}_{m+1}. This implies that F(T)\cap S(f,A)\subset {C}_{n}. Notice that {x}_{n}={\mathrm{\Pi}}_{{C}_{n}}{x}_{1}. We find from Lemma 2.8 that \u3008{x}_{n}z,J{x}_{1}J{x}_{n}\u3009\ge 0 for any z\in {C}_{n}. Since F(T)\cap S(f,A)\subset {C}_{n}, we therefore find that
It follows from Lemma 2.9 that
This implies that the sequence \{\varphi ({x}_{n},{x}_{1})\} is bounded. This in turn implies that the sequence \{{x}_{n}\} is bounded. Since E is a uniform space, we find that E is reflexive. We may assume, without loss of generality, that {x}_{n}\rightharpoonup \stackrel{\u02c6}{x}. Next, we prove that \stackrel{\u02c6}{x}\in F(T)\cap S(f,A). Since {C}_{n} is closed and convex, we find that \stackrel{\u02c6}{x}\in {C}_{n}. This implies from {x}_{n}={\mathrm{\Pi}}_{{C}_{n}}{x}_{1} that \varphi ({x}_{n},{x}_{1})\le \varphi (\stackrel{\u02c6}{x},{x}_{1}). On the other hand, we see from the weakly lower semicontinuity of the norm \parallel \cdot \parallel that
which implies that {lim}_{n\to \mathrm{\infty}}\varphi ({x}_{n},{x}_{1})=\varphi (\stackrel{\u02c6}{x},{x}_{1}). Hence, we have {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}\parallel =\parallel \stackrel{\u02c6}{x}\parallel. Since E enjoys the KadecKlee property, we find that {x}_{n}\to \stackrel{\u02c6}{x} as n\to \mathrm{\infty}. In the light of {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. We obtain that {lim}_{n\to \mathrm{\infty}}\varphi ({x}_{n},{x}_{1}) exists. It follows that
This implies that {lim}_{n\to \mathrm{\infty}}\varphi ({x}_{n+1},{x}_{n})=0. In view of {x}_{n+1}={\mathrm{\Pi}}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}, we find that
It follows that
In view of (2.2), we see that {lim}_{n\to \mathrm{\infty}}(\parallel {x}_{n+1}\parallel \parallel {u}_{n}\parallel )=0. This implies that {lim}_{n\to \mathrm{\infty}}\parallel {u}_{n}\parallel =\parallel \stackrel{\u02c6}{x}\parallel. That is,
This implies that \{J{u}_{n}\} is bounded. Since both E and {E}^{\ast} are uniform, we find that both E and {E}^{\ast} are reflexive. We may assume, without loss of generality, 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
It follows that
That is, \stackrel{\u02c6}{x}=u, which in turn implies that {u}^{\ast}=J\stackrel{\u02c6}{x}. It follows that J{u}_{n}\rightharpoonup J\stackrel{\u02c6}{x}\in {E}^{\ast}. Since E is uniformly smooth, we know that {E}^{\ast} is uniformly convex. Therefore, {E}^{\ast} enjoys the KadecKlee property, we obtain that {lim}_{n\to \mathrm{\infty}}J{u}_{n}=J\stackrel{\u02c6}{x}. Since {J}^{1}:{E}^{\ast}\to E is demicontinuous and E enjoys the KadecKlee property, we obtain that {u}_{n}\to \stackrel{\u02c6}{x} as n\to \mathrm{\infty}. Note that
It follows that
Since E is uniformly smooth, we know that {E}^{\ast} is uniformly convex. In the light of Lemma 2.11, we find that
It follows that
Notice that
We find from (3.4) that
In view of the restriction on the sequences, we find from (3.5) that {lim}_{n\to \mathrm{\infty}}g(\parallel J{x}_{n}J{T}^{n}{x}_{n}\parallel )=0. Notice that
It follows that
The demicontinuity of {J}^{1}:{E}^{\ast}\to E implies that {T}_{i}^{n}{x}_{n}\rightharpoonup \stackrel{\u02c6}{x}. Note that
This implies that {lim}_{n\to \mathrm{\infty}}\parallel {T}^{n}{x}_{n}\parallel =\parallel \stackrel{\u02c6}{x}\parallel. Since E has the KadecKlee property, we obtain that {lim}_{n\to \mathrm{\infty}}\parallel {T}^{n}{x}_{n}\stackrel{\u02c6}{x}\parallel =0. Notice that
It follows from the uniformly asymptotic regularity of T that
That is, T{T}^{n}{x}_{n}\to \stackrel{\u02c6}{x}. From the closedness of T, we find \stackrel{\u02c6}{x}=T\stackrel{\u02c6}{x}. This proves \stackrel{\u02c6}{x}\in F(T). Next, we show that \stackrel{\u02c6}{x}\in S(f,A). It follows from Lemma 2.9 and (3.1) that
This yields that {lim}_{n\to \mathrm{\infty}}\varphi ({u}_{n},{y}_{n})=0. This implies from (2.2) that {lim}_{n\to \mathrm{\infty}}(\parallel {u}_{n}\parallel \parallel {y}_{n}\parallel )=0. It follows that
We, therefore, find that
This shows that \{J{y}_{n}\} is bounded. Since {E}^{\ast} is reflexive, we may assume that J{y}_{n}\rightharpoonup {y}^{\ast}\in {E}^{\ast}. In view of JE={E}^{\ast}, we see that there exists y\in E such that Jy={y}^{\ast}. It follows that
Taking {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}} on both sides of the equality above yields that
That is, \stackrel{\u02c6}{x}=y, which in turn implies that {y}^{\ast}=J\stackrel{\u02c6}{x}. It follows that J{y}_{n}\rightharpoonup J\stackrel{\u02c6}{x}\in {E}^{\ast}. Since {E}^{\ast} enjoys the KadecKlee property, we obtain that J{y}_{n}J\stackrel{\u02c6}{x}\to 0 as n\to \mathrm{\infty}. Note that {J}^{1}:{E}^{\ast}\to E is demicontinuous. It follows that {y}_{n}\rightharpoonup \stackrel{\u02c6}{x}. Since E enjoys the KadecKlee property, we obtain that {y}_{n}\to \stackrel{\u02c6}{x} as n\to \mathrm{\infty}. Note that
This implies that {lim}_{n\to \mathrm{\infty}}\parallel {u}_{n}{y}_{n}\parallel =0. Since J is uniformly normtonorm continuous on any bounded sets, we have {lim}_{n\to \mathrm{\infty}}\parallel J{u}_{n}J{y}_{n}\parallel =0. In view of the restriction {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{r}_{n}>0, we see that
Since {u}_{n}={K}_{{r}_{n}}{y}_{n}, we find that
where
It follows from (A2) that
In view of (A4), we find that
For 0<t<1 and q\in C, define {q}_{t}=tq+(1t)\stackrel{\u02c6}{x}. It follows that {q}_{t}\in C, which yields that F({q}_{t},\stackrel{\u02c6}{x})\le 0. It follows from (A1) and (A4) that
That is,
Letting t\downarrow 0, we obtain from (A3) that F(\stackrel{\u02c6}{x},q)\ge 0, \mathrm{\forall}q\in C. This implies that \stackrel{\u02c6}{x}\in S(f,A). This completes the proof \stackrel{\u02c6}{x}\in F(T)\cap S(f,A).
Finally, what we need to prove is \stackrel{\u02c6}{x}={\mathrm{\Pi}}_{F(T)\cap S(f,A)}{x}_{1}.
Letting n\to \mathrm{\infty} in (3.2), we obtain that
From Lemma 2.8, we immediately find that \stackrel{\u02c6}{x}={\mathrm{\Pi}}_{F(T)\cap S(f,A)}{x}_{1}. This completes the whole proof. □
Remark 3.2 Since the class of generalized asymptotically quasiϕnonexpansive mappings is a generalization of the class of asymptotically quasiϕnonexpansive mappings, Theorem 3.1 includes Kim’s [36] results as a special case.
Remark 3.3 Notice that every uniformly smooth and uniformly convex space is a uniformly smooth and strictly convex Banach space which also enjoys the KadecKlee property, and every uniformly convex Banach space enjoys the KadecKlee property. We find that Theorem 3.1 is still valid in the framework of every uniformly smooth and uniformly convex space.
Next, we consider the solution of problem (2.4).
If the mapping T is closed quasiϕnonexpansive, which is more general than relatively nonexpansive mappings, we have the following.
Corollary 3.4 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the KadecKlee property and C be a nonempty closed convex subset of E. Let T:C\to C be a quasiϕnonexpansive mapping and f be a bifunction from C\times C to ℝ satisfying (A1)(A4). Assume that T is closed and F(T)\cap S(f) is nonempty. Let \{{x}_{n}\} be a sequence generated in the following manner:
where \{{\alpha}_{n}\} is a real number 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 number sequence such that {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{r}_{n}>0. Then the sequence \{{x}_{n}\} converges strongly to {\mathrm{\Pi}}_{F(T)\cap S(f)}{x}_{1}, where {\mathrm{\Pi}}_{F(T)\cap S(f)} is the generalized projection from E onto F(T)\cap S(f).
In the framework of Hilbert spaces, we find from Theorem 3.1 the following.
Theorem 3.5 Let E be a Hilbert space and C be a nonempty closed convex subset of E. Let T:C\to C be a generalized asymptotically quasinonexpansive mapping. Let f be a bifunction from C\times C to ℝ satisfying (A1)(A4), and let A:C\to E be an αinversestrongly monotone mapping. Assume that T is closed and asymptotically regular on C, and F(T)\cap S(f,A) is nonempty and bounded. Let \{{x}_{n}\} be a sequence generated in the following manner:
where {W}_{n}=sup\{{\parallel p{x}_{n}\parallel}^{2}:p\in F(T)\cap S(f,A)\}, \{{\alpha}_{n}\} is a real number 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 number sequence such that {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{r}_{n}>0. Then the sequence \{{x}_{n}\} converges strongly to {P}_{F(T)\cap S(f,A)}{x}_{1}, where {P}_{F(T)\cap S(f,A)} is the metric projection from E onto F(T)\cap S(f,A).
Proof In the framework of Hilbert spaces, we see that \varphi (x,y)={\parallel xy\parallel}^{2} and the mapping J is reduced to the identity mapping. The desired conclusion can be immediately drawn from Theorem 3.1. □
For problem (2.4), we have the following result.
Corollary 3.6 Let E be Hilbert space and C be a nonempty closed convex subset of E. Let T:C\to C be a generalized asymptotically quasinonexpansive mapping. Let f be a bifunction from C\times C to ℝ satisfying (A1)(A4). Assume that T is closed and asymptotically regular on C, and F(T)\cap S(f) is nonempty and bounded. Let \{{x}_{n}\} be a sequence generated in the following manner:
where {W}_{n}=sup\{{\parallel p{x}_{n}\parallel}^{2}:p\in F(T)\cap S(f)\}, \{{\alpha}_{n}\} is a real number 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 number sequence such that {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{r}_{n}>0. Then the sequence \{{x}_{n}\} converges strongly to {P}_{F(T)\cap S(f)}{x}_{1}, where {P}_{F(T)\cap S(f)} is the metric projection from E onto F(T)\cap S(f).
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The authors are grateful to the editor and the anonymous reviewers’ suggestions which improved the contents of the article.
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JS design the algorithm and give the main convergence analysis. MC participated in the design of the study. Both authors read and approved the final manuscript.
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Song, J., Chen, M. On generalized asymptotically quasiϕnonexpansive mappings and a Ky Fan inequality. Fixed Point Theory Appl 2013, 237 (2013). https://doi.org/10.1186/168718122013237
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DOI: https://doi.org/10.1186/168718122013237