Abstract
In this paper, we prove strong and weak convergence theorems for a mapping defined on a bounded, closed and convex subset of a uniformly convex Banach space, satisfying the RCSC condition. This condition was introduced by Karapınar (Dynamical Systems and Methods, 2012). We first establish the demiclosed principle for the mapping satisfying the RCSC condition. Then, using this principle, we establish the weak and strong convergence theorems. Results in the paper extend and improve a number of important results in this literature such as Khan and Suzuki (Nonlinear Anal. 80:211-215, 2013) and Reich (J. Math. Anal. Appl. 67:274-276, 1979).
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1 Introduction
Let C be a nonempty closed convex subset of a Banach space X. A mapping \(T \colon C\to C\) is said to be nonexpansive if \(\Vert Tx-Ty\Vert \leq \Vert x-y\Vert \) for all \(x, y \in C\). It is called quasi-nonexpansive [1] if \(F(T)\neq\emptyset\) and \(\Vert Tx-p\Vert \leq \Vert x-p\Vert \) for all \(x \in C\) and for all \(p \in F(T)\), where \(F(T)\) is the set of fixed points of T, i.e., \(F(T) = \{x \in C : Tx = x\}\). Every nonexpansive mapping with \(F(T)\neq \emptyset\) is a quasi-nonexpansive mapping.
In 2008, Suzuki [2] introduced a mapping satisfying condition (C). More accurately, a mapping \(T \colon C\to C\) is said to satisfy condition (C) if
for all \(x,y \in C\). Every nonexpansive mapping satisfies condition (C); also if a mapping satisfies condition (C) and has a fixed point, then it is a quasi-nonexpansive mapping [2].
Fixed point theorems for a mapping satisfying condition (C) were studied by Dhompongsa et al. [3] and Phuengrattana [4]. Khan and Suzuki [5] proved a weak convergence theorem for a mapping satisfying condition (C) in uniformly convex Banach spaces whose dual has the Kadec-Klee property.
In 2013, Karapınar [6] suggested a new modification of mappings satisfying condition (C) to a mapping satisfying (RCSC)-condition.
Definition 1.1
Let T be a mapping on a subset C of a Banach space X. Then T is said to satisfy Reich-Chatterjea-Suzuki-(C) condition ((RCSC)-condition) if
for all \(x,y \in C\).
Motivated by the above mentioned works, in this paper, we prove some weak and strong convergence theorems for generalized nonexpansive ((RCSC)-condition) mappings in a uniformly convex Banach space, which has the Kadec-Klee property. Our results generalize the results of Khan and Suzuki [5], Reich [7] to the case of a mapping satisfying (RCSC)-condition. For other works in this direction, please see Mogbademu [8], Saluja [9], Thakur [10] and Zheng [11].
2 Preliminaries
Throughout this paper, we denote by \(\mathbb{N}\) the set of positive integers and by \(\mathbb{R}\) the set of real numbers.
We now recall some definitions and results useful for our main results.
A Banach space X is called uniformly convex [12] if for each \(\varepsilon\in(0,2]\) there is \(\delta> 0\) such that for \(x, y \in X\),
Lemma 2.1
([12])
Let X be a uniformly convex Banach space. Let \(\{x_{n}\}\) and \(\{y_{n}\}\) be sequences in X satisfying \(\lim_{n\to\infty} \Vert x_{n}\Vert = 1\), \(\lim_{n\to\infty} \Vert y_{n}\Vert = 1\) and \(\lim_{n\to \infty} \Vert x_{n} + y_{n}\Vert = 2\). Then \(\lim_{n\to\infty} \Vert x_{n} - y_{n}\Vert = 0\).
Lemma 2.2
([5])
Let X be a uniformly convex Banach space and let \(\{u_{n}\}\), \(\{v_{n}\}\) and \(\{w_{n}\}\) be sequences in X. Let d and t be real numbers with \(d\in(0, \infty)\) and \(t \in(0, 1)\). Assume that \(\lim_{n\to\infty} \Vert u_{n} - v_{n}\Vert = d\), \(\limsup_{n\to \infty} \Vert u_{n}- w_{n}\Vert \leq(1 - t)d\) and \(\limsup_{n\to\infty} \Vert v_{n} - w_{n}\Vert \leq td\). Then \(\lim_{n\to\infty} \Vert tu_{n} + (1 - t) v_{n} - w_{n}\Vert = 0\).
Proposition 2.1
Let C be a nonempty subset of a Banach space X and \(T : C\to C\) be a mapping satisfying (RCSC)-condition. Then T has the following properties:
-
(i)
If T has a fixed point, then it is a quasi-nonexpansive mapping [6], Proposition 6.
-
(ii)
If C is closed, then \(F(T)\) is closed; further if X is strictly convex and C is convex, then \(F(T)\) is also convex [6], Proposition 10.
A Banach space X is said to have the Kadec-Klee property if, for every sequence \(\{x_{n}\}\) in X which converges weakly to a point \(x \in X\) with \(\Vert x_{n}\Vert \) converging to \(\Vert x\Vert \), \(\{x_{n}\}\) converges strongly to x. Every uniformly convex Banach space has the Kadec-Klee property [13].
Lemma 2.3
Let X be a reflexive Banach space whose dual has the Kadec-Klee property. Let \(\{x_{n}\}\) be a bounded sequence in X and let \(y, z\in X\) be weak subsequential limits of \(\{x_{n}\}\). Assume that for every \(t \in[0, 1]\), \(\lim_{n\to\infty} \Vert tx_{n} + (1 - t) y- z\Vert \) exists. Then \(y = z\).
Proposition 2.2
Let C be a nonempty subset of a Banach space X and \(T \colon C\to C\) be a mapping satisfying (RCSC)-condition. Then
-
(1)
\(\Vert x-Ty\Vert \leq9\Vert Tx-x\Vert + \Vert x-y\Vert \),
-
(2)
\(\Vert y-Ty\Vert \leq9\Vert Tx-x\Vert + 2\Vert x-y\Vert \)
hold for all \(x, y \in C\).
Proof
(1) follows from [6], Corollary 16.
For (2), it follows from (1) that
Thus we have (2). □
3 Main results
In this section, we prove weak and strong convergence theorems. First, we establish some auxiliary results.
The following lemma is an extension of Lemma 8 of [5] to the case of mappings satisfying (RCSC)-condition.
Lemma 3.1
Let C be a nonempty bounded convex subset of a uniformly convex Banach space X, and let \(T \colon C \to C\) be a mapping satisfying (RCSC)-condition. Suppose that for any \(\varepsilon> 0\), there exists \(\xi(\varepsilon) > 0\) such that \(\Vert Tu- u\Vert < \xi (\varepsilon)\), \(\Vert Tv - v\Vert <\xi(\varepsilon)\) for some \(u,v \in C\). Then, for any \(t\in[0, 1]\),
Proof
Assume to the contrary that there exist sequences \(\{u_{n}\}, \{v_{n}\} \in C\), \(\{t_{n}\} \in[0, 1]\) and \(\varepsilon> 0\) such that
and
Setting \(x_{n} = t_{n}u_{n} + (1 - t_{n})v_{n}\) and \(w_{n} = Tx_{n}\), from Proposition 2.2(ii), we get
Similarly, we can show that
and hence
Since C is bounded and
we get \(0 < \liminf_{n\to\infty} t_{n}\).
Similarly, we can show that \(\limsup_{n\to\infty} t_{n} <1\).
So, without loss of generality, we may assume that \(\Vert u_{n}- v_{n}\Vert \) converges to \(d\in(0, \infty)\) and \(t_{n}\) converges to \(t\in(0, 1)\) as \(n\to\infty\).
Since \(\lim_{n\to\infty} \Vert Tu_{n}-u_{n}\Vert = 0\) and \(0 < \liminf_{n\to\infty} \Vert u_{n}-x_{n}\Vert \), we obtain
for sufficiently large \(n\in\mathbb{N}\).
Since T satisfies (RCSC)-condition, for sufficiently large \(n\in \mathbb{N}\), we have
By similar arguments, we have
for sufficiently large \(n \in\mathbb{N}\).
Now, using the triangular inequality and Proposition 2.2(i), we have
and
It then follows from Lemma 2.2 that
which is a contradiction, and this completes the proof. □
We now establish the demiclosed principle for the mapping satisfying (RCSC)-condition.
Proposition 3.1
Let T be a mapping on a bounded and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition. Then \(I-T\) is demiclosed at zero. That is, if \(\{x_{n}\} \in C\) converges weakly to \(x_{0}\in C\) and \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert = 0\), then \(Tx_{0} = x_{0}\).
Proof
Let \(\xi\colon(0, \infty) \to(0, \infty)\) be a function satisfying the conclusion of Lemma 3.1. Let \(\{x_{n}\}\) be a sequence converging weakly to \(x_{0}\in C\) and \(\lim_{n\to \infty} \Vert Tx_{n}-x_{n}\Vert = 0\). For arbitrarily chosen \(\varepsilon> 0\), define a strictly decreasing sequence \(\{\varepsilon_{n}\}\) in \((0, \infty)\) by
It is obvious that \(\varepsilon_{n+1} < \xi(\varepsilon_{n})\). Choose a subsequence \(\{x_{f(n)}\}\) of \(\{x_{n}\}\) such that \(\Vert x_{f(n)}-Tx_{f(n)}\Vert < \xi(\varepsilon_{n})\). Since \(x_{0}\) belongs to the closed convex hull of \(\{x_{f(n)} : n \in\mathbb{N}\}\), it is a weak limit of \(\{x_{f(n)}\}\). Hence, there exist \(y\in C\) and \(v\in\mathbb{N}\) such that \(\Vert y-x_{0}\Vert < \varepsilon\) and y belongs to the convex hull of \(\{x_{f(n)} : n = 1, 2,\ldots, v\}\). Using Lemma 3.1, we have \(\Vert Ty- y\Vert < \varepsilon\). Using Proposition 2.2(ii), we obtain
Since \(\varepsilon> 0\) is arbitrary, we obtain \(Tx_{0} = x_{0}\). □
Lemma 3.2
Let T be a mapping on a bounded and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition. For arbitrary \(x_{1} \in C\) and a real number \(\alpha\in[1/2, 1)\), construct a sequence \(\{x_{n}\}\) in C by
If \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0\), then \(\lim_{n\to \infty }\Vert tx_{n} + (1-t) p-q\Vert \) exists, where \(p, q \in F(T)\) and \(t \in[0, 1]\).
Proof
Since T satisfies (RCSC)-condition, by Proposition 2.1, it is quasi-nonexpansive. Let \(S = \alpha T + (1-\alpha)I\), then S is a self-mapping on C, and \(F(S) = F(T)\) also S is quasi-nonexpansive, and
Thus, for any \(q \in F(S)\), we have
hence the sequence \(\{\Vert x_{n}-q\Vert \}\) is nonincreasing and bounded below. Therefore, it converges.
Since the sequence \(\{\Vert p-q\Vert \}\) obviously converges, we see that \(\lim_{n\to\infty} \Vert tx_{n}+(1-t)p-q\Vert \) exists for \(t=1\) and \(t=0\). Thus it remains to consider \(t\in(0, 1)\).
Let \(\lim_{n\to\infty} \Vert x_{n}-p\Vert = d\). If \(d = 0\), there is nothing to prove. Take \(d > 0\). We have
for all \(l\in\mathbb{N}\cup\{0\}\), where \(S^{0}\) is the identity mapping on C. Then there exists \(\nu\in\mathbb{N}\) such that
for all \(l\geq0\) and \(m, n\geq\nu\). Since T satisfies (RCSC)-condition and Proposition 2.2(i), we obtain
and hence
for all \(l\geq0\) and \(m, n \geq\nu\).
Let \(h \colon\mathbb{N} \to[0, \infty)\) be a function defined by
Take two subsequences \(\{f(n)\}\) and \(\{g(n)\}\) of \(\{n\}\) such that \(\nu< f(1)\), \(f(n) < g(n)\) for each \(n\in\mathbb{N}\) and
Set \(u_{n} = x_{g(n)}\), \(v_{n} = p\) and \(w_{n} = S^{g(n)-f(n)}(tx_{f(n)} + (1 - t) p)\). Then we have
and
By (3.2), (3.3), (3.4) and Lemma 2.2, we have
Substituting the value of \(u_{n}\), \(v_{n}\) and \(w_{n}\), we have
Using the quasi-nonexpansiveness of S, we get
Thus \(\lim_{n\to\infty} h(n) = \lim_{n\to\infty} \Vert tx_{n} + (1 - t) p - q\Vert \) exists. □
Now, we prove a weak convergence theorem.
Theorem 3.1
Let X be a uniformly convex Banach space whose dual has the Kadec-Klee property. Let T be a mapping on a bounded, closed and convex subset C of X. Assume that T satisfies (RCSC)-condition and define a sequence \(\{x_{n}\}\) in C by (3.1). If \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0\), then \(\{x_{n}\}\) converges weakly to a fixed point of T.
Proof
Let W be the set of all weak subsequential limits of \(\{x_{n}\}\). Since \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert \) is equal to 0, by Proposition 3.1 we have \(W \subset F(T)\). Using Lemma 2.3 and Lemma 3.2, W is singleton. But X is a uniformly convex Banach space, hence reflexive. So every sequence \(\{x_{n}\}\) has a subsequence converging weakly to the unique element of W. Since W is singleton, therefore \(\{x_{n}\}\) itself converges weakly to the unique element of W. □
Remark 1
Theorem 3.1 is a generalization of Theorem 11 of [5].
Since the dual of a reflexive Banach space with Fréchet differentiable norm has the Kadec-Klee property [16], as a direct consequence of Theorem 3.1, we get the following result.
Corollary 3.1
Let X be a uniformly convex Banach space whose norm is Fréchet differentiable. Let T be a mapping on a bounded, closed and convex subset C of X. Assume that T satisfies (RCSC)-condition and define a sequence \(\{x_{n}\}\) in C by (3.1). If \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0\), then \(\{x_{n}\}\) converges weakly to a fixed point of T.
Recall that a mapping \(T \colon C \to C\) is said to satisfy condition (I) [17] if there exists a nondecreasing function \(f :[0,\infty) \to[0,\infty)\) with \(f(0) = 0\), \(f(r) > 0\) for all \(r \in(0, \infty)\) such that \(d(x, Tx) \geq f(d(x, F(T)))\) for all \(x \in C\), where \(d(x, F(T)) = \inf_{p\in F(T)} d(x, p)\).
We now establish a strong convergence theorem.
Theorem 3.2
Let T be a mapping on a bounded, closed and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition and define a sequence \(\{x_{n}\}\) in C by (3.1). If \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0\) and T satisfies condition (I), then \(\{x_{n}\}\) converges strongly to a fixed point of T.
Proof
By Lemma 3.2, we know that \(\lim_{n\to\infty} \Vert x_{n}-p\Vert \) exists for all \(p\in F(T)\), and hence \(\lim_{n\to\infty} d(x_{n}, F(T))\) exists. Assume that \(\lim_{n\to\infty} \Vert x_{n}-p\Vert = r\) for some \(r \geq0\).
If \(r = 0\), then \(\{x_{n}\}\) converges strongly to p and the result follows.
Suppose \(r > 0\). From the hypothesis and condition (I), we have \(\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0\) and \(f(d(x_{n}, F(T))) \leq \Vert Tx_{n}-x_{n}\Vert \). This gives \(\lim_{n \to\infty}f(d(x_{n}, F(T))) = 0\). Since f is a nondecreasing function, we have \(\lim_{n\to\infty} d(x_{n}, F(T)) =0\). Thus, there exist a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) and a sequence \(\{y_{k}\} \subset F(T)\) such that
Again, we see that
Hence,
This shows that \(\{y_{k}\}\) is a Cauchy sequence in a complete space, and hence it converges to a point \(p \in X\). Since \(F(T)\) is closed, therefore \(p \in F(T)\) and then \(\{x_{n_{k}}\}\) converges strongly to p. Since \(\lim_{n\to\infty} \Vert x_{n}- p\Vert \) exists, \(x_{n} \to p \in F(T)\). This completes the proof. □
We now give an example of mapping T which satisfies (RCSC)-condition but fails to satisfy condition (C).
Example 1
Let \(X=\mathbb{R}\) with usual metric and \(C=[0,1]\subset X\). Define a mapping \(T\colon C\to C\) by the rule
Set \(x=\frac{9}{10}\) and \(y=\frac{3}{5}\), we see that
and
i.e.,
hence, T fails to satisfy condition (C).
To verify that T satisfies condition (RCSC), consider the following cases.
Case-I: Let \(x, y \in [0 , \frac{4}{5} )\), then we have
\(x, y \in [0 , \frac{4}{5} )\).
Case-II: Let \(x, y \in [\frac{4}{5} , 1 ]\), then
Since
and
which implies that
for all \(x, y \in [\frac{4}{5} , 1 ]\).
Case-III: Let \(x \in [0 , \frac{4}{5} )\) and \(y \in [\frac{4}{5} , 1 ]\) or \(x \in [\frac {4}{5} , 1 ]\) and \(y \in [0 , \frac{4}{5} )\). Then
Also,
We now have two subcases as follows.
Case-III(A): \(x \geq\frac{y}{2}\), then \(\vert x-\frac{y}{2}\vert = x-\frac{y}{2}\), and by (3.5) we have
Case-III(B): \(x < \frac{y}{2}\), then \(\vert x-\frac{y}{2}\vert = \frac{y}{2}-x\), and by (3.5) we have
Hence \(\vert Tx-Ty\vert \leq\frac{1}{3} [ \vert x-y\vert +\vert Tx-y\vert +\vert x-Ty\vert ]\) for all \(x \in [0 , \frac{4}{5} )\) and \(y \in [\frac{4}{5} , 1 ]\).
Case-IV: Let \(x \in [\frac{4}{5} , 1 ]\) and \(y \in [0 , \frac{4}{5} )\). By interchanging the role of x and y in Case-III, we can see that
for all \(x \in [\frac{4}{5} , 1 ]\) and \(y \in [0 , \frac{4}{5} )\).
In view of Case-I to Case-IV, we can say that T satisfies condition (RCSC) for all \(x,y\in C\).
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Acknowledgements
The first author would like to thank the Rajiv Gandhi National Fellowship, University Grants Commission, Government of India under the grant (F1-17.1/2011-12/RGNF-ST-CHH-6632). The second author is supported by the Chhattisgarh Council of Science and Technology, India (MRP-2015).
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Thakur, D., Thakur, B.S. & Postolache, M. Convergence theorems for generalized nonexpansive mappings in uniformly convex Banach spaces. Fixed Point Theory Appl 2015, 144 (2015). https://doi.org/10.1186/s13663-015-0397-z
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DOI: https://doi.org/10.1186/s13663-015-0397-z