Open Access

Viscosity approximations methods for \((\psi,\varphi)\)-weakly contractive mappings

Fixed Point Theory and Applications20162016:100

https://doi.org/10.1186/s13663-016-0585-5

Received: 5 April 2016

Accepted: 16 September 2016

Published: 21 November 2016

Abstract

In this paper, we study viscosity approximations with \((\psi,\varphi)\)-weakly contractive mappings. We show that Moudafi’s viscosity approximations follow from Browder and Halpern type convergence theorems. Our results generalize a number of convergence theorems including a strong convergence theorem of Song and Liu (Fixed Point Theory Appl. 2009:824374, 2009).

Keywords

\((\psi,\varphi)\)-weakly contractive mappings Browder type convergence Halpern type convergence Moudafi’s viscosity approximations

MSC

54H25 47H10

1 Introduction and preliminaries

Let \((M, d)\) be a metric space and \(f:M\to M\) a self-mapping. A point \(z\in M\) is said to be a fixed point of f if \(f(z)=z\). Throughout this paper, \(F(f)\) denotes the set of fixed points of f, \(\mathbb{N}\) the set of natural numbers and M a metric space \((M,d)\).

A mapping \(f:M \to M\) is a contraction if there exists \(r\in[0, 1)\) such that for all \(x,y \in M\),
$$ d\bigl(f(x),f(y)\bigr)\leq rd(x,y). $$
(1.1)
The classical Banach contraction principle (BCP) states that ‘Every contraction of a complete metric space has a unique fixed point.’ In 1969, Boyd and Wong [2] obtained the following interesting generalization of the BCP.

Theorem 1.1

Let \(f:M \to M\) a self-mapping of a complete metric space M such that for all \(x, y \in M\),
$$ d\bigl(f(x),f(y)\bigr)\leq\alpha\bigl(d(x,y)\bigr), $$
(1.2)
where \(\alpha:[0,\infty)\rightarrow [0,\infty)\) is upper semicontinuous from the right and \(\alpha(t)< t\) for all \(t>0\). Then f has a unique fixed point in M.

The mapping \(f:M\rightarrow M\) satisfying (1.2) is called a nonlinear contraction [2].

The mapping \(f:M\rightarrow M\) is called weakly contractive, if
$$ d\bigl(f(x),f(y)\bigr)\leq d(x,y)-\varphi\bigl(d(x,y)\bigr) $$
(1.3)
for all \(x, y \in M\), where \(\varphi:[0,\infty)\rightarrow [0,\infty)\) is a continuous and nondecreasing function such that \(\varphi(t)=0\) if and only if \(t=0\).

We note that (1.3) follows from Tasković [3, 4]. For an earlier work in this direction, we refer to Krasnosel’skiĭ et al. [5] and Dugundji and Granas [6]. Also, these mappings have been studied by Aĺber and Guerre-Delabriere [7] and Rhoades [8] as mentioned by Jachymski [9] (see also [10]).

In this paper, we use the following class of mappings satisfying the so-called \((\psi,\varphi)\)-condition (see for details [1119]).

A mapping \(f:M\rightarrow M\) is called \((\psi,\varphi)\)-weakly contractive if
$$ \psi\bigl(d\bigl(f(x),f(y)\bigr)\bigr)\leq\psi\bigl(d(x,y)\bigr)- \varphi\bigl(d(x,y)\bigr) $$
(1.4)
for all \(x,y \in M\), where \(\psi,\varphi:[0,\infty)\rightarrow [0,\infty)\) are both continuous and monotone nondecreasing functions with \(\psi(t)=0=\varphi(t)\) if and only if \(t=0\).

Remark 1.2

We remark that if \(\varphi(t)=(1-r)t\) with \(r\in(0, 1)\), then (1.3) reduces to (1.1). If \(\psi(t)=t\), then (1.4) recovers (1.3). In fact, weakly contractive mappings are also related closely to nonlinear contractions. If α is continuous and \(\varphi(t)=t-\alpha(t)\) then (1.3) turns into (1.2). We have the following irreversible implications (see [13], Example 2.2).
$$ (1.1) \Rightarrow(1.2)\Rightarrow (1.3) \Rightarrow(1.4). $$
Thus \((\psi,\varphi)\)-weakly contractive mappings are more general than its predecessors as listed above.

Theorem 1.3

([13], Theorem 2.1)

Every \((\psi, \varphi)\)-weakly contractive mapping of a complete metric space has a unique fixed point.

It was observed by Đorić [20] that the continuity of φ can be relaxed to lower semi-continuity in Theorem 1.3.

Definition 1.4

Let Y be a nonempty subset of a Banach space X. A mapping \(f:Y\rightarrow Y\) is said to be nonexpansive if for all \(x,y \in Y\),
$$ \bigl\| f(x)-f(y)\bigr\| \leq\|x-y\|. $$
Let X be a real Banach space with its dual space \(X^{\ast}\) and Y be a nonempty closed convex subset of X. Let \(\langle x,x^{\ast}\rangle\) be the dual pairing between \(x \in X\) and \(x^{\ast} \in X^{\ast}\), and \(J:X\rightarrow2^{X^{\ast}}\) be the normalized duality mapping on X defined by
$$ J(x)=\bigl\{ x^{\ast}\in X^{\ast}:\bigl\langle x,x^{\ast} \bigr\rangle =\|x\|^{2}=\bigl\| x^{\ast}\bigr\| ^{2}\bigr\} $$
for all \(x \in X\). Then X is said to be smooth or to have a Gâteaux differentiable norm if \(\lim_{t\rightarrow 0}\frac{\|x+ty\|-\|x\|}{t}\) exists for each \(x,y \in X\) with \(\|x\|=\|y\|=1\). A Banach space X is said to be uniformly smooth whenever given ε> 0 there exists \(\delta> 0\) such that for all \(x,y \in X\) with \(\|x\| = 1\) and \(\|y\| \leq \delta\), then \(\|x + y\| + \|x - y\| < 2 + \varepsilon\|y\|\).

Definition 1.5

[21]

Let Y be a nonempty closed convex subset of a Banach space X and Z a nonempty subset of Y. A retraction from Y to Z is a continuous mapping \(P : Y \rightarrow Z\) such that \(P(x) = x\) for \(x \in Z\). A retraction P from Y to Z is sunny if P satisfies the property: \(P(P(x) + t(x - P(x))) =P(x)\) for all \(x \in Y\) and \(t > 0\), whenever \(P(x)+t(x-P(x)) \in Y\). A retraction P from Y to Z is sunny nonexpansive if P is both sunny and nonexpansive [2224].

A well-known way to find a fixed point of a nonexpansive mapping is to use a contraction to approximate it (Browder [25, 26]). More precisely, fix \(z\in Y\) and define a mapping \(f_{t}:Y\rightarrow Y\) by \(f_{t} (x)=tz+(1-t)S(x)\) for all \(x\in Y\) and given \(t\in(0,1)\). It is easy to see that \(f_{t}\) is a contraction on Y and the BCP ensures that \(f_{t}\) has a unique fixed point \(u_{t}\in Y\), that is,
$$ u_{t}=tz+(1-t)S(u_{t}). $$
(1.5)
In 1967, Halpern [27] introduced the following iteration for an arbitrary \(z \in Y\) and a sequence \(\{\alpha_{n}\}\subset(0,1)\):
$$ u_{0} \in Y,\qquad u_{n+1}=\alpha_{n}z+(1- \alpha_{n})S(u_{n}) $$
(1.6)
for \(n \in\mathbb{N}\), where \(S:Y\to Y\) is a nonexpansive mapping.

In the case \(F(S)\neq\emptyset\), Browder [25] (respectively, Halpern [27]) showed that \(\{u_{t}\}\) (respectively, \(\{u_{n}\}\)) converges strongly to the fixed point of S that is nearest to z in a Hilbert space. A number of extensions and generalizations of their results have appeared in [1, 2834] and elsewhere.

Theorem 1.6

[28]

Let Y be a bounded closed convex subset of a uniformly smooth Banach space X and \(S:Y\to Y\) a nonexpansive mapping. Define a net \(\{x_{\alpha}\}\) in Y by
$$x_{\alpha}=\alpha z+ (1-\alpha)S(x_{\alpha}) $$
for \(\alpha\in(0,1)\), where \(z \in Y\) is fixed. Then \(\{x_{\alpha}\}\) converges strongly to \(P(z)\) as \(\alpha\to0^{+}\), where P is the unique sunny nonexpansive retraction from Y onto \(F(S)\).

Theorem 1.7

[31, 32]

Let X, Y, S, P and z be as in Theorem  1.6. Define a sequence \(\{u_{n}\}\) in Y by
$$u_{1} \in Y,\qquad u_{n+1}=\alpha_{n} z+ (1- \alpha_{n})S(u_{n}) $$
for \(n \in\mathbb{N}\), where \(\{\alpha_{n}\}\) is a real sequence in \((0,1)\) satisfying
$$(\mathrm{C}1)\quad \lim_{n\rightarrow\infty} \alpha_{n}=0,\qquad (\mathrm{C}2)\quad \sum_{n=1}^{\infty} \alpha_{n}=\infty, \qquad (\mathrm{C}3)\quad \lim_{n\rightarrow\infty} \frac{\alpha_{n+1}}{\alpha_{n}}=1. $$
Then \(\{u_{n}\}\) converges strongly to \(P(z)\).
In 2000, Moudafi [35] generalized Browder’s and Halpern’s theorems and proved that in a real Hilbert space H, for a given \(u_{0} \in Y\subseteq H\), the sequence \(\{u_{n}\}\) generated by the algorithm
$$ u_{n+1}=\alpha_{n}f(u_{n})+(1- \alpha_{n})S(u_{n}) $$
(1.7)
for \(n \in\mathbb{N}\cup\{0\}\), where \(f:Y\rightarrow Y\) is a contraction, \(S:Y\rightarrow Y\) a nonexpansive mapping and \(\{\alpha_{n}\}\subseteq(0,1)\), satisfying certain conditions, converges strongly to a fixed point of S in Y, which is the unique solution to the following variational inequality:
$$\bigl\langle (I-f)x^{\ast}, x^{\ast}-x\bigr\rangle \geq0, \quad \forall x \in F(S). $$
Moudafi’s generalizations are called viscosity approximations. These methods can be applied to convex optimization, linear programming, monotone inclusions, and elliptic differential equations [30]. In 2004, Xu [36] extended Moudafi’s results from Hilbert spaces to more general Banach spaces. Suzuki [30] used Meir-Keeler type contractions f in (1.7) to find fixed points of S in Banach spaces. Recently, Song and Liu [1] considered the following viscosity approximations:
$$\begin{aligned}& v_{n}=\alpha_{n} f(v_{n})+(1- \alpha_{n})S_{n}(v_{n}); \\& u_{n+1}=\alpha_{n} f(u_{n})+(1- \alpha_{n})S_{n}(u_{n}) \end{aligned}$$
for \(n \in\mathbb{N}\), where \(S_{n}:Y\to Y\) is a sequence of nonexpansive mappings and \(f:Y\to Y\) is a weakly contractive mapping.

In this paper, motivated by Moudafi [35], Kopecká and Reich [37], Suzuki [30] and Song and Liu [1], we study viscosity approximations with a more general class of weakly contractive mappings. We show that Moudafi’s viscosity approximations can be obtained from Browder and Halpern type convergence results.

2 Convergence results

Throughout this section, \(\psi,\varphi:[0,\infty)\rightarrow[0,\infty)\) are continuous and strictly increasing functions such that
$$\psi(t)=0=\varphi(t)\quad \mbox{if and only if}\quad t=0. $$
Our main results are prefaced by the following lemmas and propositions.

Lemma 2.1

[24, 33]

Let Y be a nonempty convex subset of a smooth Banach space X and Z a nonempty subset of Y. Let J be the duality mapping from X into \(X^{\ast}\), and \(P:Y\rightarrow Z\) a retraction. Then P is both sunny and nonexpansive if and only if
$$ \bigl\langle x-P(x),J\bigl(y-P(x)\bigr)\bigr\rangle \leq0 $$
for all \(x \in Y\) and \(y\in Z\).

Lemma 2.2

[38]

Let \(\{\alpha_{n}\}\) be a sequence of positive reals and \(\{\beta_{n}\}\) a sequence of nonnegative reals such that
$$\lim_{n\to\infty}\alpha_{n}=0,\qquad \sum _{n=1}^{\infty}\alpha_{n}=\infty \quad\textit{and} \quad\lim_{n\to\infty}\frac{\beta_{n}}{\alpha_{n}}=0. $$
Further, consider a sequence of nonnegative reals \(\{\ell_{n}\}\) and the recursive inequality
$$\ell_{n+1}\leq\ell_{n}-\alpha_{n}\xi( \ell_{n})+\beta_{n} $$
for \(n\in\mathbb{N}\cup\{0\}\), where \(\xi(\ell)\) is continuous strictly increasing for \(\ell\geq0\) and \(\xi(0)=0\). Then
  1. (1)

    \(\lim_{n\to\infty}\ell_{n}=0\);

     
  2. (2)
    there exists a subsequence \(\{\ell_{n_{k}}\}\) of \(\{\ell_{n}\}\) such that
    $$\begin{aligned}& \ell_{n_{k}}\leq \xi^{-1} \biggl(\frac{1}{\sum_{m=0}^{n_{k}}\alpha_{m}}+ \frac{\beta _{n_{k}}}{\alpha_{n_{k}}} \biggr), \\ & \ell_{n_{k}+1}\leq \xi^{-1} \biggl(\frac{1}{\sum_{m=0}^{n_{k}}\alpha_{m}}+ \frac{\beta _{n_{k}}}{\alpha_{n_{k}}} \biggr)+\beta_{n_{k}}, \\ & \ell_{n}\leq \ell_{n_{k}+1}-\sum_{m=n_{k+1}}^{n-1} \frac{\alpha_{m}}{\theta_{m}},\quad n_{k}+1< n< n_{k+1},\theta_{m}=\sum_{i=0}^{m} \alpha_{i}, \\ & \ell_{n+1}\leq \ell_{0}-\sum_{m=1}^{n} \frac{\alpha_{m}}{\theta_{m}}\leq\ell_{0},\quad 1\leq n< n_{k}-1, \\ & 1\leq n_{k}\leq s_{\max}=\max \Biggl\{ s, \sum _{m=0}^{s}\frac{\alpha_{m}}{\theta_{m}}\leq \ell_{0} \Biggr\} . \end{aligned}$$
     

Definition 2.3

[30]

Let \(\{S_{n}\}\) be sequence of nonexpansive mappings on a closed convex subset Y of a Banach space X and \(F=\bigcap_{n=0}^{\infty }F(S_{n})\neq\emptyset\).
  1. (A)
    Let \(\{\alpha_{n}\}\) be a sequence in \((0,1]\) with \({\lim}_{n\rightarrow\infty}\alpha_{n}=0\). Then \((X,Y,\{S_{n}\},\{ \alpha_{n}\})\) is said to satisfy Browder property if for each \(z \in Y\), a sequence \(\{v_{n}\}\) defined by
    $$ v_{n}=\alpha_{n}z+(1-\alpha_{n})S_{n}(v_{n}), $$
    (2.1)
    for \(n \in\mathbb{N}\), converges strongly.
     
  2. (B)
    Let \(\{\alpha_{n}\}\) be a sequence in \([0,1]\) with \(\lim_{n\rightarrow\infty}\alpha_{n}=0\) and \(\sum_{n = 1}^{\infty}\alpha_{n}=\infty\). Then \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) is said to satisfy Halpern’s property if for each \(z \in Y\), a sequence \(\{v_{n}\}\) defined by
    $$ v_{1}\in Y, \qquad v_{n+1}= \alpha_{n}z+(1-\alpha_{n})S_{n}(v_{n}), $$
    (2.2)
    for \(n \in\mathbb{N}\), converges strongly.
     

It is well known that if X is a Hilbert space, Y is bounded and \(\{ S_{n}\}\) is a constant sequence S, then \((X,Y,\{S_{n}\},\{1/n\})\) has both the Browder and the Halpern properties (cf. [24, 31, 32, 34]).

Example 2.4

Let \(X=[0,\infty)\) equipped with the norm \(\|\cdot\|\) defined by \(\|x\| =|x|\) and \(Y=[0,1]\) a closed convex subset of X. Define a sequence of nonexpansive mappings \(S_{n}:Y\to Y\) by \(S_{n}(x)=\frac{x}{n}\) for all \(x\in Y\) and \(n\in\mathbb{N}\). Let \(\{\alpha_{n}\}\) be a sequence in \((0,1]\) defined by \(\alpha_{n}=\frac{1}{n^{2}+1}\).

Then
  1. (i)
    It is easy to see that the quadruple \((X,Y,\{S_{n}\},\{\alpha _{n}\})\) satisfies the Browder property and the sequence \(\{v_{n}\}\) defined by
    $$ v_{n}=\alpha_{n}z+(1-\alpha_{n})S_{n}(v_{n}), $$
    for each \(z \in Y\) and \(n \in\mathbb{N}\), converges strongly to 0.
     
  2. (ii)

    However, the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) does not satisfy the Halpern property, as the series \(\sum_{n = 1}^{\infty}\alpha_{n}\) is a convergent series.

     
  3. (iii)

    If we take \(\alpha_{n}=\frac{1}{n}\), then the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies both the Browder and the Halpern properties. We note that \(\{S_{n}\}\) is not a constant sequence here.

     

Proposition 2.5

([30], Proposition 4)

Let the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies Browder’s property and \(\{v_{n}\}\) is a sequence in Y, defined by (2.1). If \(P(z)=\lim_{n\to\infty}v_{n}\) for each \(z \in Y\) then P is a nonexpansive mapping on Y.

Proposition 2.6

([30], Proposition 5)

Let the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies Halpern’s property and \(\{v_{n}\}\) is a sequence in Y, defined by (2.2). If \(P(z)=\lim_{n\to\infty}v_{n}\) for each \(z \in Y\) then
  • \(P:Y\to Y\) is a nonexpansive mapping;

  • \(P(z)\) does not depend on the initial point \(v_{1}\).

Proposition 2.7

Let Y be a closed convex subset of a smooth Banach space X and Z a nonempty subset of Y. Let \(S:Y\rightarrow Y\) be a nonexpansive mapping, \(P:Y\to Z\) a unique sunny nonexpansive retraction, and \(T:Y\rightarrow Y\) a \((\psi, \varphi)\)-weakly contractive mapping with ψ convex. Then
  1. (a)

    the composite mapping \(S\circ T\) is \((\psi, \varphi )\)-weakly contractive on Y;

     
  2. (b)
    the mapping \(S_{t}=tT+(1-t)S\) for \(t\in(0,1)\) is \((\psi, \phi)\)-weakly contractive on Y and \(u_{t}\) is the unique solution of the fixed point equation
    $$ u_{t}=tT(u_{t})+(1-t)S(u_{t}), $$
    where \(\phi(s)=t \varphi(s)\) for each fixed \(t\in(0,1)\);
     
  3. (c)
    \(P(T(z))=z\) if and only if \(z \in Y\) is the unique solution of the variational inequality
    $$ \bigl\langle T(z)-z,J(y-z)\bigr\rangle \leq0 $$
    (2.3)
    for all \(y\in Z\).
     

Proof

  1. (a)
    For any \(x,y \in Y\) we have
    $$\bigl\| S\bigl(T(x)\bigr)-S\bigl(T(y)\bigr) \bigr\| \leq \bigl\| T(x)-T(y) \bigr\| . $$
    Since ψ is nondecreasing and T is a \((\psi, \varphi)\)-weakly contractive, the above inequality reduces to
    $$\begin{aligned} \psi\bigl( \bigl\| S\bigl(T(x)\bigr)-S\bigl(T(y)\bigr) \bigr\| \bigr) \leq& \psi\bigl( \bigl\| T(x)-T(y) \bigr\| \bigr) \\ \leq&\psi\bigl( \|x-y \| \bigr)-\varphi\bigl( \|x-y \| \bigr). \end{aligned}$$
    Therefore the mapping \(S\circ T\) is a \((\psi, \varphi)\)-weakly contractive.
     
  2. (b)
    Let \(x,y \in Y\). Then for each fixed \(t\in(0,1)\), we have
    $$\begin{aligned} \bigl\| S_{t}(x)-S_{t}(y) \bigr\| =& \bigl\| \bigl(tT(x)+(1-t)S(x)\bigr)- \bigl(tT(y)+(1-t)S(y)\bigr)\bigr\| \\ \leq&(1-t)\bigl\| S(x)-S(y)\bigr\| + t \bigl\| T(x)-T(y)\bigr\| \\ \leq&(1-t) \|x-y \| + t \bigl\| T(x)-T(y)\bigr\| . \end{aligned}$$
    Since ψ is nondecreasing, the above inequality reduces to
    $$\begin{aligned} \psi\bigl( \bigl\| S_{t}(x)-S_{t}(y)\bigr\| \bigr) \leq&\psi \bigl((1-t) \|x-y \|+t \bigl\| T(x)-T(y)\bigr\| \bigr). \end{aligned}$$
    Convexity of ψ implies
    $$\begin{aligned} \psi\bigl( \bigl\| S_{t}(x)-S_{t}(y)\bigr\| \bigr) \leq&(1-t)\psi\bigl( \|x-y \| \bigr)+t\psi\bigl(\bigl\| T(x)-T(y)\bigr\| \bigr). \end{aligned}$$
    Since T is \((\psi, \varphi)\)-weakly contractive, we have
    $$\begin{aligned} \psi\bigl(\bigl\| S_{t}(x)-S_{t}(y)\bigr\| \bigr) \leq&(1-t)\psi\bigl(\|x-y \|\bigr)+t \bigl[\psi\bigl(\|x-y\| \bigr)-\varphi\bigl(\|x-y \|\bigr) \bigr] \\ =&\psi\bigl(\|x-y \|\bigr)-t\varphi\bigl(\|x-y \|\bigr). \end{aligned}$$
    Let \(\phi(s)=t \varphi(s)\). Then
    $$\begin{aligned} \psi\bigl(\bigl\| S_{t}(x)-S_{t}(y)\bigr\| \bigr) \leq& \psi\bigl(\|x-y \|\bigr)-\phi\bigl(\|x-y \|\bigr). \end{aligned}$$
    Thus, the mappings \(S_{t}\) is \((\psi,\phi)\)-weakly contractive and by Theorem 1.3, \(S_{t}\) has a unique fixed point \(u_{t}\) in Y.
     
  3. (c)
    By (a) and Proposition 2.5, the mapping \(P\circ T\) is \((\psi,\phi)\)-weakly contractive. By Theorem 1.3, \(P\circ T\) has a unique fixed point \(P(T(z))=z\in Z\). By Lemma 2.1, such a \(z\in Z\) satisfies (2.3). Next, we show that the variational inequality (2.3) has a unique solution. Let \(w\in Y\) be another solution of (2.3). Then
    $$ \bigl\langle T(w)-w, J(z-w) \bigr\rangle \leq0 $$
    (2.4)
    and
    $$ \bigl\langle T(z)-z, J(w-z) \bigr\rangle \leq0. $$
    (2.5)
    Adding (2.4) and (2.5)
    $$\begin{aligned} 0 \geq& \bigl\langle w-z-\bigl(T(w)-T(z)\bigr), J(w-z) \bigr\rangle \\ =& \bigl\| w-z-\bigl(T(w)-T(z)\bigr) \bigr\| \|w-z \| \\ \geq& \|w-z \|^{2}- \bigl\| T(w)-T(z)\bigr\| \|w-z \| \\ =& \|w-z \| \bigl\{ \|w-z \|- \bigl\| T(w)-T(z)\bigr\| \bigr\} , \end{aligned}$$
    which implies that
    $$\|w-z \|- \bigl\| T(w)-T(z)\bigr\| \leq0 \quad\text{or}\quad \|w-z \|\leq\bigl\| T(w)-T(z)\bigr\| . $$
    Since ψ is nondecreasing and T is a \((\psi, \varphi)\)-weakly contractive, we have
    $$\begin{aligned} \psi\bigl( \|w-z \| \bigr) \leq& \psi\bigl( \bigl\| T(w)-T(z)\bigr\| \bigr) \\ \leq&\psi\bigl( \|w-z \| \bigr)-\varphi\bigl( \|w-z \| \bigr). \end{aligned}$$
     
Therefore \(\varphi( \|w-z \| )\leq0\), and \(w=z\). □

Now we present our first convergence result.

Theorem 2.8

Suppose the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies Browder’s property and \(T:Y\rightarrow Y\) is a \((\psi,\varphi)\)-weakly contractive mapping with ψ convex. For each \(z \in Y\), put \(P(z)=\lim_{n\to\infty}v_{n}\), where \(\{v_{n}\}\) is a sequence in Y defined by (2.1). Then the sequence \(\{u_{n} \}\subset Y\) defined by
$$ u_{n}=\alpha_{n}T(u_{n})+(1- \alpha_{n})S_{n}(u_{n}), $$
for \(n\in\mathbb{N}\), converges strongly to \(P(T(z))=z\in Y\).

Proof

Proposition 2.7(b) ensures the existence and uniqueness of \(\{u_{n}\}\). It follows from Proposition 2.7(a) and Proposition 2.5 that \(P\circ T\) is a \((\psi, \varphi)\)-weakly contractive mapping on Y. Therefore, by Theorem 1.3 there exists a unique element \(z \in Y\) such that \(P(T(z)) = z\). Define a sequence \(\{v_{n} \}\) in Y by
$$ v_{n}=\alpha_{n}T(z)+(1-\alpha_{n})S_{n}(v_{n}) $$
for \(n\in\mathbb{N}\). Then it is easy to see that \(\{v_{n}\}\) converges strongly to \(P(T(z))\).
Now, for \(n \in\mathbb{N}\),
$$\begin{aligned} \|u_{n}-v_{n} \| \leq&(1-\alpha_{n}) \bigl\| S_{n}(u_{n})-S_{n}(v_{n}) \bigr\| + \alpha_{n} \bigl\| T(u_{n})-T(z)\bigr\| \\ \leq& (1-\alpha_{n}) \|u_{n}-v_{n} \|+ \alpha_{n} \bigl\| T(u_{n})-T(z)\bigr\| \end{aligned}$$
or
$$\|u_{n}-v_{n} \|\leq\bigl\| T(u_{n})-T(z)\bigr\| . $$
Since ψ is nondecreasing, we have
$$\begin{aligned} \psi\bigl(\|u_{n}-v_{n} \|\bigr) \leq& \psi\bigl( \bigl\| T(u_{n})-T(z)\bigr\| \bigr). \end{aligned}$$
Further, \((\psi,\varphi)\)-weak contractivity of T implies
$$\begin{aligned} \psi\bigl(\|u_{n}-v_{n} \|\bigr) \leq& \psi\bigl(\|u_{n}-z \|\bigr)- \varphi\bigl(\|u_{n}-z\|\bigr) \\ \leq& \psi\bigl( \|u_{n}-v_{n} \|+\|v_{n}-z\|\bigr)- \varphi\bigl(\|u_{n}-z\|\bigr). \end{aligned}$$
Since \(v_{n}\to z\) as \(n\to\infty\), we get
$$\begin{aligned} \mathop{ \overline{\lim}} _{n \to\infty} \psi\bigl(\|u_{n}-v_{n} \|\bigr) \leq& \mathop{ \overline{\lim}} _{n \to\infty}\psi\bigl( \|u_{n}-v_{n} \|+\|v_{n}-z \|\bigr)- \mathop{\underline{\lim} } _{n \to\infty}\varphi\bigl( \|u_{n}-z\|\bigr) \\ =& \mathop{ \overline{\lim}} _{n \to\infty }\psi\bigl(\|u_{n}-v_{n} \|\bigr) -\mathop{\underline{\lim} } _{n \to \infty}\varphi\bigl(\|u_{n}-z\|\bigr), \end{aligned}$$
or
$$\begin{aligned} \mathop{\underline{\lim} } _{n \to\infty }\varphi\bigl(\|u_{n}-z\|\bigr) \leq& 0. \end{aligned}$$
The continuity of φ and \(\varphi(0)=0\) imply that
$$\underset{n\rightarrow\infty}{\lim}\|u_{n}-z \|=0. $$
Therefore \(\{u_{n}\}\) converges strongly to z. □

Corollary 2.9

Let X, Y, \(\{S_{n}\}\), \(\{v_{n}\}\), P, z and \(\{\alpha_{n}\}\) be as in Theorem  2.8 and \(T:Y\rightarrow Y\) a weakly contractive mapping. Then the sequence \(\{u_{n} \}\subset Y\) defined by
$$ u_{n}=\alpha_{n}T(u_{n})+(1- \alpha_{n})S_{n}(u_{n}), $$
for \(n\in\mathbb{N}\), converges strongly to \(P(T(z))=z\in Y\).

Proof

This follows from Theorem 2.8 when \(\psi(t)=t\). □

Example 2.10

Let \((X, \|\cdot\|)\) and Y be as in Example 2.4. Define the mappings \(S_{n},T:Y\to Y\) by
$$S_{n}(x)=x\quad\text{for all $x\in Y$ and $n\in\mathbb{N}$}\quad \text{and}\quad T(x) =1-\frac{x}{2}\quad\text{for all $x\in Y$.} $$
Let \(\psi,\varphi:[0,\infty)\rightarrow[0,\infty)\) be the functions defined by
$$\psi(t)=t \quad\text{and}\quad \varphi(t)=\frac{t}{2}. $$
Then the mapping \(S_{n}\) is nonexpansive for each \(n\in\mathbb{N}\) and T is \((\psi,\varphi)\)-weakly contractive.
Let \(\{\alpha_{n}\}\) be a sequence in \((0,1]\) defined by \(\alpha_{n}=\frac {1}{n+1}\). Then the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies the Browder property and for each \(z\in Y\), we have
$$\begin{aligned} v_{n} =& z \alpha_{n} +(1-\alpha_{n})S_{n}(v_{n}) \\ =& z \frac{1}{n+1} + \biggl(1-\frac{1}{1+n} \biggr)v_{n}, \end{aligned}$$
or
$$v_{n}=z. $$
By Theorem 2.8, put \(P(z)=\lim_{n\to\infty} v_{n}=z\). Then P is an identity mapping. Now
$$\begin{aligned} u_{n} =& \alpha_{n} T(u_{n})+(1- \alpha_{n})S_{n}(u_{n}) \\ =&\frac{1}{n+1} \biggl[1-\frac{u_{n}}{2} \biggr] + \biggl(1- \frac {1}{1+n} \biggr)u_{n}, \end{aligned}$$
or
$$u_{n}= \frac{2}{3}. $$
Now \(\lim_{n\to\infty} u_{n}=\frac{2}{3}=P(T(\frac{2}{3}))\). Thus the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies all the conditions of Theorem 2.8 and the sequence \(\{u_{n}\}\) strongly converges to \(\frac{2}{3}\).

The following theorem is our second convergence result.

Theorem 2.11

Suppose the quadruple \((X,Y,\{S_{n}\},\{\alpha_{n}\})\) satisfies Halpern’s property and \(T:Y\rightarrow Y\) is a \((\psi,\varphi)\)-weakly contractive mapping with ψ is convex. Put \(P(z)=\lim_{n\to\infty}v_{n}\) for each \(z \in Y\), where \(\{v_{n}\}\) is defined by (2.2). Then the sequence \(\{u_{n} \}\subset Y\) defined by
$$ x_{1}\in Y\quad \textit{and}\quad u_{n+1}= \alpha_{n}T(u_{n})+(1-\alpha_{n})S_{n}(u_{n}) $$
for \(n\in\mathbb{N}\), converges strongly to a unique point \(P(T(z))=z\in Y\).

Proof

By Propositions 2.6 and 2.7(a), the mapping \(P\circ T\) is \((\psi,\varphi)\)-weakly contractive on Y. From Theorem 1.3, there exists a unique \(z\in Y\) such that \(z=P(T(z))\). For \(n\in\mathbb{N}\), we define a sequence \(\{v_{n} \}\) in Y by
$$ v_{n+1}=\alpha_{n}T(z)+(1-\alpha_{n})S_{n}(v_{n}). $$
Then by the assumption \(\{v_{n}\}\) converges strongly to \(P(T(z))\).
Now for \(n\in\mathbb{N}\), we have
$$\|u_{n+1}-v_{n+1} \| = \bigl\| \alpha_{n} \bigl(T(u_{n})-T(z)\bigr)+(1-\alpha _{n}) \bigl(S_{n}(u_{n})-S_{n}(v_{n})\bigr) \bigr\| . $$
Since \(S_{n}\) is nonexpansive and ψ is nondecreasing and convex, we have
$$\begin{aligned} \psi\bigl(\|u_{n+1}-v_{n+1}\|\bigr) =& \psi\bigl(\alpha_{n} \bigl\| T(u_{n})-T(z)\bigr\| +(1-\alpha_{n})\bigl\| S_{n}(u_{n})-S_{n}(v_{n}) \bigr\| \bigr) \\ \leq& \alpha_{n}\psi\bigl(\bigl\| T(u_{n})-T(z)\bigr\| \bigr)+(1- \alpha_{n}) \psi\bigl(\bigl\| S_{n}(u_{n})-S_{n}(v_{n}) \bigr\| \bigr) \\ \leq& \alpha_{n}\psi\bigl(\bigl\| T(u_{n})-T(z)\bigr\| \bigr)+(1- \alpha_{n})\psi\bigl(\|u_{n}-v_{n}\|\bigr). \end{aligned}$$
By \((\psi, \varphi)\)-weak contractivity of T, we get
$$\begin{aligned} \psi\bigl(\|u_{n+1}-v_{n+1}\|\bigr) \leq&\alpha_{n}\bigl[ \psi\bigl(\|u_{n}-z\|\bigr)-\varphi\bigl(\| u_{n}-z\|\bigr)\bigr] +(1- \alpha_{n})\psi\bigl(\| u_{n}-v_{n}\|\bigr) \\ \leq&\alpha_{n}\bigl[\psi\bigl(\|u_{n}-v_{n}\|+ \|v_{n}-z\|\bigr)-\varphi \bigl(\|u_{n}-v_{n}\|+\| v_{n}-z\|\bigr)\bigr] \\ &{}+ (1-\alpha_{n})\psi \bigl(\|u_{n}-v_{n}\|\bigr). \end{aligned}$$
The continuity of \(\psi, \varphi\) and \(v_{n}\to z\) as \(n\to\infty\) imply that
$$\begin{aligned}& \lim_{n\rightarrow\infty}\psi \bigl( \|u_{n+1}-v_{n+1} \| \bigr) \\& \quad \leq\lim_{n\rightarrow\infty} \alpha_{n}\bigl[\psi\bigl( \|u_{n}-v_{n}\|\bigr)-\varphi\bigl(\|u_{n}-v_{n} \|\bigr)+ (1-\alpha_{n})\psi \bigl(\| u_{n}-v_{n}\|\bigr)\bigr] \\& \quad=\lim_{n\rightarrow\infty} \bigl[\psi\bigl(\|u_{n}-v_{n} \|\bigr)-\alpha_{n}\varphi\bigl(\|u_{n}-v_{n}\|\bigr)\bigr] \\& \quad\leq\lim_{n\rightarrow\infty} \bigl[\psi\bigl(\|u_{n}-v_{n} \|\bigr)-\alpha_{n}\varphi\bigl(\|u_{n}-v_{n}\|\bigr)+ \alpha_{n}\|v_{n}-z\|\bigr]. \end{aligned}$$
Thus, for some (sufficiently large) \(N_{0}\leq n\), we have
$$\begin{aligned} \psi\bigl( \|u_{n+1}-v_{n+1} \| \bigr) \leq& \psi \bigl( \|u_{n}-v_{n} \|\bigr)-\alpha_{n}\varphi \bigl(\|u_{n}-v_{n}\|\bigr)+ \alpha_{n}\|v_{n}-z\|. \end{aligned}$$
For \(\ell_{n}=\psi(\|u_{n}-v_{n}\|)\), we get the following recursive inequality:
$$\ell_{n+1}\leq\ell_{n}-\alpha_{n}\varphi( \ell_{n})+\beta_{n}, $$
where \(\varepsilon_{n}=\|v_{n}-z\|\) and \(\beta_{n}=\alpha_{n}\varepsilon_{n}\). Now, by Lemma 2.2,
$$\lim_{n\to\infty}\psi\bigl(\|u_{n}-v_{n}\|\bigr)=0. $$
The continuity of ψ and the fact that \(\psi(0)=0\) imply that
$$\lim_{n\to\infty}\|u_{n}-v_{n}\|=0. $$
By the triangle inequality, we have
$$ \lim_{n\rightarrow\infty} \|u_{n}-z \|\leq \lim _{n\rightarrow\infty} \|u_{n}-v_{n} \|+ \lim _{n\rightarrow\infty} \|v_{n}-z \|=0. $$
Therefore \(\{u_{n}\}\) strong convergence of to \(z=P(T(z))\). □

Corollary 2.12

[1]

Let \(X,Y,\{S_{n}\}, \{v_{n}\}, P, z\) and \(\{\alpha_{n}\}\) be as in Theorem  2.11 and \(T:Y\rightarrow Y\) a weakly contractive mapping. Then the sequence \(\{u_{n} \}\subset Y\) defined by
$$ x_{1}\in Y\quad \textit{and}\quad u_{n+1}= \alpha_{n}T(u_{n})+(1-\alpha_{n})S_{n}(u_{n}), $$
for \(n\in\mathbb{N}\), converges strongly to a unique point \(P(T(z))=z\in Y\).

Proof

This follows from Theorem 2.11 when \(\psi(t)=t\). □

Declarations

Acknowledgements

The authors are indebted to the referees and the Editor for their constructive comments and suggestions, which have been useful for the improvement of the paper.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematical Sciences and Computing, Walter Sisulu University
(2)
Department of Mathematics, Visvesvaraya National Institute of Technology

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