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

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).

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 [11–19]).

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 [22–24].

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, 28–34] 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) $\lim_{n\to\infty}\ell_{n}=0$;

• (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 $F=\bigcap_{n=0}^{\infty }F(S_{n})\neq\emptyset$.

• (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.

• (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

• (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.

• (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.

• (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

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

• (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 $t\in(0,1)$;

• (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

• (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.

• (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.

• (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 $\psi(t)=t$. □