Mann–Dotson’s algorithm for a countable family of non-self strict pseudo-contractive mappings

Abstract

The aim of this paper to present some weak and strong convergence results for countable family of non-self mappings. More precisely, we employ the Mann–Dotson’s algorithm to approximate common fixed points of a countable family of non-self k-strict Pseudocontractive mappings in q-uniformly smooth Banach spaces.

1 Introduction

Strict pseudocontractive mappings were introduced by Browder and Petryshyn [1] in 1967. They proved the first convergence result for k-strict pseudocontractive self-mappings in real Hilbert spaces. They proved weak and strong convergence theorems by using the Krasnosel’skiĭ-Mann algorithm with a constant control sequence. This class of mappings properly includes the class of nonexpansive mappings. Iterative algorithms for nonexpansive mappings have received a lot of attention from researchers, on the other hand, Iterative algorithms for strict pseudocontractive mappings are far less developed. Since pseudocontractive mappings have more applications in solving inverse problems than nonexpansive mappings, developing new iterative algorithms for strict pseudocontractive mappings is interesting [2]. Rhoades [3] generalized results presented in [1] and proved fixed point results using Krasnosel’skiĭ-Mann algorithm with a variable control sequence under some conditions. Recently, Marino and Xu [4] proved that the sequence generated by Krasnosel’skiĭ-Mann algorithm converges weakly to a fixed point of k-strict pseudocontractive self mapping under some assumptions.

Finding fixed points of nonexpansive mappings and k-strict pseudocontractive by Krasnosel’skiĭ-Mann algorithm [5,6,7,8,9] have been extensively studied in the last few decades for a self-mapping ( [10, 11] see also). If the k-strict pseudocontractive mapping is non-self then most of the Krasnoselskii-Mann type algorithms are based on the nearest point projection technique. But in many applications, calculating the nearest point projection is not easy and it may require approximation algorithm by itself, even in the case of Hilbert spaces [12, 13]. To overcome this problem Colao and Marino [12] used inwardness condition on non-self mapping and introduced a new line of research and proved some fixed point results for non-self nonexpansive mapping using Mann–Dotson algorithm. Recently, Colao et al. [14] proved the following convergence results for non-self k-strict pseudocontractive mapping which satisfies inward condition in Hilbert spaces.

Theorem 1.1

Suppose $\mathcal {E}$ be a closed, strictly convex, and nonempty subset of a given Hilbert space $\mathcal {M}$, $T: \mathcal {E} \rightarrow \mathcal {M}$ a k-strict pseudocontractive mapping which satisfies the inward condition with $\emptyset \ne F(T)$. Choose $\varepsilon >0$ such that $\tilde{k}=k+\varepsilon <1$. Then sequence $\{x_n\}$ given by

$$\begin{aligned} {\left\{ \begin{array}{ll} x_0 \in \mathcal {E},\\ \alpha _0= \max \{ \tilde{k}, v(x_0)\},\\ x_{n+1}=\alpha _n x_n+(1-\alpha _n)T(x_n),\\ \alpha _{n+1}=\max \{ \alpha _n, v(x_{n+1})\}, \end{array}\right. } \end{aligned}$$

(1.1)

converges weakly to a fixed point of the mapping T. If $\sum _n (1-\alpha _n)< \infty$, then the convergence is strong.

Marino and Muglia [15] generalized Colao et al. [14] results from Hilbert space to q-uniformly smooth Banach space and proved the following theorem

Theorem 1.2

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-uniformly smooth Banach space $\mathcal {X}$, $T: \mathcal {E} \rightarrow \mathcal {X}$ a k-strict pseudocontractive mapping which satisfies the inward condition with $\emptyset \ne F(T)$. Then sequence $\{x_n\}$ given by

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha _0= \min \{k, h_{\mathcal {E},T}(x_0)\},\\ x_{n+1}=(1-\alpha _n) x_n+ \alpha _n T(x_n),\\ \alpha _{n+1}=\min \{ \alpha _n, h_{\mathcal {E},T}(x_{n+1})\}, \end{array}\right. } \end{aligned}$$

(1.2)

(a) converges strongly to a fixed point of the mapping T if $\sum \nolimits _{n=1}^{\infty } \alpha _n< \infty$ and $\mathcal {E}$ is strictly convex (b) converges weakly to a fixed point of the mapping T if $\sum \nolimits _{n=1}^{\infty } \alpha _n = \infty$, $\mathcal {X}$ is uniformly convex and $\mathcal {E}$ is a nonexpansive retract of $\mathcal {X}$ and (c) if $\sum \nolimits _{n=1}^{\infty } \alpha _n = \infty$ and $\mathcal {X}$ satisfies Opial’s condition then the sequence converges weakly to a fixed point of the mapping T.

In recent years, there are many researchers who investigated fixed point results for strict pseudocontractive mappings with different settings and conditions [1, 4, 16,17,18,19,20,21].

In this paper, using inwardness condition we prove some weak and strong convergence results using Mann–Dotson’s algorithm for a countable family of non-self k-strict pseudocontractive mappings. We ensure that Mann–Dotson’s algorithm converges (strongly and weakly) to common fixed points under different conditions on countable family of nonself mappings.

2 Preliminaries

Let $(\mathcal {X},\Vert \cdot \Vert )$ be a real Banach space, $\mathcal {E}$ a subset of $\mathcal {X}.$ A non-self mapping $T:\mathcal {E} \rightarrow \mathcal {X}$ is said to be Lipschitz if for each $x,y \in \mathcal {E}$ there exists a real constant $L \ge 0$, such that

$$\begin{aligned} \Vert T(x)-T(y)\Vert \le L \Vert x-y\Vert . \end{aligned}$$

If Lipschitz constant $L=1,$ then the mapping T is said to be a non-self nonexpansive mapping. A point $p \in \mathcal {E}$ is said to be a fixed point of the mapping T if $T(p)=p.$ The set of all fixed points for mapping T is denoted by F(T).

Definition 2.1

Suppose J be a normalized duality mapping on a real Banach space $\mathcal {X}$, $J:\mathcal {X} \rightarrow 2^{\mathcal {X}^*}$ defined as $J(x)= \{h \in \mathcal {X}^*: \langle x, h \rangle =\Vert x\Vert \Vert h\Vert , \Vert h\Vert = \Vert x\Vert \}$, for all $x \in \mathcal {X}$, where $\mathcal {X}^*$ is dual space of $\mathcal {X}$ and the pair $\langle \cdot , \cdot \rangle$ denotes the generalized duality pairing. If the dual space $\mathcal {X}^*$ is strictly convex, then the normalized duality J is single valued, and the single valued duality mapping is denoted by j.

• 1. A mapping $T: D(T) \subset \mathcal {X} \rightarrow \mathcal {X}$ is said to be pseudocontractive, if for all $x, y \in D(T)$, there is $j(x-y) \in J(x-y)$ such that

  $$\begin{aligned} \langle T(x)- T(y), j(x-y) \rangle \le \Vert x-y\Vert ^2. \end{aligned}$$

  (2.1)

• 2. A mapping $T: D(T) \subset \mathcal {X} \rightarrow \mathcal {X}$ is said to be k-strict pseudocontractive in the Browder-Petryshyn sense, if there exists $k \in (0,1)$ such that for each $x, y \in D(T)$, there exists $j(x-y) \in J(x-y)$ such that

  $$\begin{aligned} \langle T(x)- T(y), j(x-y) \rangle \le \Vert x-y\Vert ^2-k\Vert x-y-(T(x)-T(y))\Vert ^2. \end{aligned}$$

  (2.2)

Remark 2.2

It is also easy to get that every k-strict pseudocontractive mapping is L-Lipschitz mapping for $L= \frac{k+1}{k}$ [22].

Now we will recollect some basic concepts, definitions, and facts from the literature.

• We use $\rightharpoonup$ to denote weak convergence, $\rightarrow$ to denote strong convergence;

• $\omega _{w}(x_n)$ represents cluster points ($\omega$-limit) set of a sequence $\{x_n\},$ that is, $\omega _{w}(x_n):=\{x: \exists ~ x_{n_k} \rightharpoonup x\}.$

Lemma 2.3

[23]. Let us assume that $\{p_n\},$ $\{q_n\}$ and $\{r_n\}$ be any sequences of positive real numbers in such a way that

$$\begin{aligned} p_{n+1} \le (1+q_n)p_n+r_n \text { for all } n \in \mathbb {N}. \end{aligned}$$

If $\sum \nolimits _{n=1}^{\infty }q_n < \infty$ and $\sum \nolimits _{n=1}^{\infty }r_n < \infty .$ Then the limit $\lim \limits _{n \rightarrow \infty } p_n$ exists [24].

Definition 2.4

[10]. A subset $\mathcal {E}$ of $\mathcal {X}$ is said to be strictly convex if it is convex as well as it satisfies the property

$$\begin{aligned} \delta x+(1-\delta ) y \in \text {interior}(\mathcal {E}) \end{aligned}$$

for all $x,y \in \partial \mathcal {E}$ and $\delta \in (0,1)$. Otherwise, there are no segments lies within the boundary $\partial \mathcal {E}$.

Definition 2.5

[10]. A Banach space $\mathcal {X}$ is said to be uniformly convex if for every $0 < \varepsilon \le 2$ there exists a $\delta > 0$ in such a way that, for each $x,y \in \mathcal {X}$ satisfying $\Vert x\Vert = \Vert y\Vert = 1,$ the condition $\Vert x -y\Vert \ge \varepsilon$ implies $\left\| \frac{x+y}{2}\right\| \le 1-\delta .$

Definition 2.6

[25]. A Banach space $\mathcal {X}$ is said to satisfy the Opial condition if, for any given sequence $\{x_n\}$ which converges weakly and have the weak limit $x \in \mathcal {X}$ satisfies:

$$\begin{aligned} \liminf \limits _{n \rightarrow \infty } \Vert x_n-x\Vert <\liminf \limits _{n \rightarrow \infty } \Vert x_n-y\Vert \end{aligned}$$

for all $y \in \mathcal {X}$ with $x \ne y.$

All the $\ell ^p$ $(1\le p < \infty )$ spaces, all Hilbert spaces and all finite dimensional Banach spaces satisfy Opial condition. A Banach space having a weakly sequentially continuous duality mapping also satisfies the Opial condition. But $L_p$ $(0<p< \infty ,~ p \ne 2)$ spaces do not satisfy Opial condition [10].

Definition 2.7

[13]. The norm defined on a given Banach space $\mathcal {X}$ is said to be Fréchet differentiable at any given unit vector x if for each y with $\Vert y\Vert =1,$ the limit

$$\begin{aligned} \lim \limits _{\lambda \rightarrow 0} \frac{\Vert x+\lambda y\Vert -\Vert x\Vert }{\lambda } \end{aligned}$$

(2.3)

exists uniformly. The norm defined on a Banach space $\mathcal {X}$ is said to be Fréchet differentiable if it is Fréchet differentiable at the each unit vector $x \in \mathcal {X}.$

In this case, for all bounded $u,v \in \mathcal {X}$ the following inequality holds (see [23]):

$$\begin{aligned} \frac{1}{2} \Vert u\Vert ^2+ \langle v, j(u) \rangle \le \frac{1}{2} \Vert u+v\Vert ^2 \le \frac{1}{2} \Vert u\Vert ^2 + \langle v, j(u) \rangle +g(\Vert v\Vert ) \end{aligned}$$

(2.4)

where $j(\cdot ): \mathcal {X} \rightarrow \mathcal {X}^*$ is the normalized duality mapping of $\mathcal {X}$ and $g:[0,\infty ) \rightarrow \mathbb {R}$ is a mapping in such a way that $\lim \limits _{t \rightarrow 0} \frac{g(t)}{t}=0$ (or $g(t)=o(t)$ as $t \rightarrow 0$).

Definition 2.8

[13] A Banach space $\mathcal {X}$ is said to be uniformly smooth if the limit defined in equation (2.3) is exists uniformly in x too.

Remark 2.9

[26] If $\mathcal {X}$ is a q-uniformly smooth Banach space, there exists a constant $C_q >0$ then following holds

$$\begin{aligned} \Vert x+y\Vert ^q \le \Vert x\Vert ^q + q \langle y, j_q(x) \rangle + C_q\Vert y\Vert ^q \end{aligned}$$

for each $x,y \in \mathcal {X}$, where

$$\begin{aligned} j_q (x)= \{ x^* \in \mathcal {X}^*: \langle x, x^* \rangle = \Vert x\Vert ^q, \Vert x^*\Vert = \Vert x\Vert ^{q-1}\}. \end{aligned}$$

Definition 2.10

[13] Suppose the space $\mathcal {X}$ is q-uniformly smooth with $(q \in (1,2])$. The mapping $T: D(T) \subset \mathcal {X} \rightarrow \mathcal {X}$ is said to be k-strict pseudocontractive mapping with $k \in (0,1)$ if for each $x,y \in D(T)$

$$\begin{aligned} \langle T(x) -T(y), j_q(x-y) \rangle \le \Vert x-y\Vert ^q -k \Vert (x-y)-(T(x)-T(y))\Vert ^q. \end{aligned}$$

Lemma 2.11

[16] Suppose $\mathcal {X}$ be a real q-uniformly smooth Banach space and $\mathcal {E}$ a convex and nonempty subset of $\mathcal {X}$. Suppose $T: \mathcal {E} \rightarrow \mathcal {X}$ be a k-strict pseudocontractive mapping. Let $\mu \in \left( 0, \min \left\{ \frac{qk^{\frac{1}{k-1}}}{C_q},1 \right\} \right)$. Then the mapping $S = \mu I + (1-\mu )T$ is nonexpansive. If $F(T) \ne \emptyset$ then $F(S)= F(T)$.

Lemma 2.12

[27]. Suppose $\mathcal {X}$ be uniformly convex Banach space, $\{x_n\}$, $\{y_n\}$ $\in \mathcal {X}$ are two sequences in such a way that $\limsup \limits _{n \rightarrow \infty } \Vert x_n\Vert \le \delta ,$ $\limsup \limits _{n \rightarrow \infty } \Vert y_n\Vert \le \delta ,$ and $\lim \limits _{n \rightarrow \infty } \Vert \alpha _nx_n+(1-\alpha _n)y_n\Vert = \delta ,$ where $\{\alpha _n\} \subseteq [a,b] \subset [0,1]$ and $\delta \ge 0.$ Then $\lim \limits _{n \rightarrow \infty } \Vert x_n-y_n\Vert =0$ (see also [28]).

Lemma 2.13

[26]. Suppose $\mathcal {X}$ be a given uniformly convex Banach space, $\mathcal {B}_s:=\{ x \in \mathcal {X}: \Vert x\Vert \le s\},$ for any $s>0.$ Then there is a strictly increasing convex continuous function $\psi : [0, \infty ) \rightarrow [0, \infty )$ with $\psi (0)=0$ such that

$$\begin{aligned} \Vert \mu x+(1-\mu )y\Vert ^2 \le \mu \Vert x\Vert ^2 +(1-\mu ) \Vert y\Vert ^2 -\mu (1-\mu ) \psi (\Vert x-y\Vert ) \end{aligned}$$

(2.5)

for each $x,y \in \mathcal {B}_s$ and $\mu \in [0,1].$

Lemma 2.14

[29]. Suppose $\mathcal {E}$ be a convex, closed, bounded and nonempty subset of a given uniformly rotund Banach space $\mathcal {X}.$ Then there is a strictly increasing continuous convex function $\tau : \mathbb {R}^{+} \rightarrow \mathbb {R}^{+}$ with $\tau (0)=0$ in such a way that for every contraction mapping $T: \mathcal {E} \rightarrow \mathcal {X},$ for each $x,y \in \mathcal {E}$ and $\beta \in [0,1],$ following holds:

$$\begin{aligned} \tau \left( \Vert \beta T(x)+(1-\beta )T(y)-T\{\beta x+(1-\beta )y\} \Vert \right) \le \Vert x-y\Vert -\Vert T(x)-T(y)\Vert . \end{aligned}$$

Lemma 2.15

(Demiclosedness principle) [30]. Suppose $\mathcal {X}$ be a given uniformly convex Banach space, $\mathcal {E}$ a convex closed and nonempty subset of $\mathcal {X}$, $T:\mathcal {E} \rightarrow \mathcal {X}$ a nonexpansive mapping. Suppose $\{x_n\}$ is a given sequence in $\mathcal {X}$ in such a way that $\{x_n\}$ converges weakly to x, $\lim \limits _{n \rightarrow \infty } \Vert x_n-T(x_n)\Vert =0.$ Then $T(x)=x.$ That is, $I- T$ is demiclosed at zero.

Definition 2.16

[31]. Let $\mathcal {E} \ne \emptyset$ be a subset of a Hilbert space $\mathcal {H}$ and $\{T_n\}$ a family of mappings from $\mathcal {E}$ into $\mathcal {H}$ with $\mathcal {F}=\mathop \cap \nolimits _{n=1}^{\infty } F(T_n) \ne \emptyset .$ The family of mappings $\{T_n\}$ is said to be uniformly weakly closed if for any convergent sequence $\{x_n\} \subset \mathcal {E}$ such that

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\Vert x_n-T_n(x_n)\Vert =0 \text { implies } \omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_n). \end{aligned}$$

The following definitions are useful in dealing with countable family of mappings. Suppose $\{T_n\}$, $\mathfrak {T}$ are two families of non-self mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne F(\mathfrak {T})=\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ where $F(T_n)$ denotes the set of all fixed points of mappings $T_n$, $F(\mathfrak {T})$ is the set of all common fixed points of all mappings in $\mathfrak {T}.$

• (i) The family of mappings $\{T_n\}$ satisfies AKTT-condition (I) if for every bounded subset $\mathcal {B}$ of $\mathcal {E},$ $\sum \nolimits _{n=1}^{\infty } \sup \{ \Vert T_{n+1}(x)-T_n(x)\Vert :x \in \mathcal {B} \} < \infty$ [32].

• (ii) The family of mappings $\{T_n\}$ satisfies AKTT-condition (II) if for every bounded subset $\mathcal {B}$ of $\mathcal {E},$ and every increasing sequence $\{n_i\}$ of natural numbers $\exists$ a mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ along with $I-T$ is demiclosed at 0 and a subsequence $\{n_{i_j}\}$ of $\{n_i\}$ in such a way that [33]

  $$\begin{aligned} \lim \limits _{j \rightarrow \infty } \sup \{ \Vert T_{n_{i_j}}(x)-T(x)\Vert :x \in \mathcal {B} \} =0, \mathop \cap \nolimits _{n=1}^{\infty } F(T_n) = F(T). \end{aligned}$$

• (iii) The family of mappings $\{T_n\}$ satisfies NST-condition if for every bounded sequence $\{x_n\}$ in $\mathcal {E},$ [34]

  $$\begin{aligned} \lim \limits _{n \rightarrow \infty }\Vert x_n-T_n(x_n)\Vert =0 \implies \omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_n). \end{aligned}$$

• (iv) The family of mappings $\{T_n\}$ satisfies NST-condition (I) along with $\mathfrak {T}$ if for every given bounded sequence $\{x_n\}$ in $\mathcal {E}$

  $$\begin{aligned} \lim \limits _{n \rightarrow \infty }\Vert x_n-T_n(x_n)\Vert =0 \implies \lim \limits _{n \rightarrow \infty }\Vert x_n-T(x_n)\Vert =0 \end{aligned}$$

  for each $T \in \mathfrak {T}$ [35].

• (v) The family of mappings $\{T_n\}$ is said to satisfy NST-condition (II) if for every given bounded sequence $\{x_n\}$ in $\mathcal {E}$

  $$\begin{aligned} \lim \limits _{n \rightarrow \infty }\Vert x_{n+1}-T_n(x_n)\Vert =0 \text { implies } \lim \limits _{n \rightarrow \infty }\Vert x_n-T_m(x_n)\Vert =0 \end{aligned}$$

  for each $m \in \mathbb {N}$ [35].

Motivated by the above conditions we consider a new type of following condition:

Definition 2.17

The family of mappings $\{T_n\}$ is said to satisfy AKTT$^*$-condition with $\mathfrak {T}$ if the family of mappings satisfies AKTT-condition (I) and NST-condition (I) both.

Remark 2.18

It can be seen that if $\{T_n\}$ is weakly closed then the family of mappings $\{T_n\}$ satisfies NST-condition.

Lemma 2.19

[32]. Suppose $\mathcal {X}$ be a given Banach space, $\mathcal {E}$ a closed, nonempty subset of $\mathcal {X}$. Suppose the family of mappings $\{T_n\}: \mathcal {E}\rightarrow \mathcal {X}$ satisfies AKTT-condition (I). Then, $\forall x \in \mathcal {E},$ $\{T_n(x)\}$ converges strongly to some point of $\mathcal {X}.$ Further, suppose $T:\mathcal {E} \rightarrow \mathcal {X}$ be a mapping defined for $x \in \mathcal {E}$

$$\begin{aligned} T(x)=\lim \limits _{n \rightarrow \infty } T_n(x). \end{aligned}$$

(2.6)

Then, $\lim \limits _{n \rightarrow \infty } \sup \{\Vert T(x)-T_n(x)\Vert :x \in \mathcal {B}\}=0$ for any bounded subset $\mathcal {B}$ of $\mathcal {E}.$ In particular, if $I-T$ is demiclosed at 0, $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n)= F(T) \ne \emptyset ,$ then the given family of mappings $\{T_n\}$ satisfies AKTT-condition (II).

Definition 2.20

[36]. The mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ is called a inward mapping if $\forall x \in \mathcal {E},$ we have

$$\begin{aligned} T(x) \in I_{\mathcal {E}}(x):= \{x+c(y-x): c \ge 1 \text { and } y \in \mathcal {E} \}. \end{aligned}$$

Definition 2.21

[36]. The mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ is called weakly inward if $\forall x \in \mathcal {E},$ we have

$$\begin{aligned} T(x) \in \overline{I_{\mathcal {E}}(x)}, \text { where } \overline{I_{\mathcal {E}}(x)} \text { is the closure of } I_{\mathcal {E}}(x). \end{aligned}$$

(2.7)

For more details and properties of weakly inward mappings, one may refer to [37, 38].

Definition 2.22

[13]. Suppose $\mathcal {X}$ be a Banach space and $\mathcal {E}$ a convex, closed and nonempty subset of $\mathcal {X}$ and $T: \mathcal {E} \rightarrow \mathcal {X}$ a mapping, define a function $h_{\mathcal {E},T}:\mathcal {E} \rightarrow \mathbb {R}$ as

$$\begin{aligned} h_{\mathcal {E},T}(x):= \inf \{ \Gamma \ge 0: \Gamma x+(1-\Gamma ) T(x) \in \mathcal {E}\}. \end{aligned}$$

(2.8)

Lemma 2.23

[13]. Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given Banach space $\mathcal {X}.$ Suppose $T: \mathcal {E} \rightarrow \mathcal {X}$ be a mapping and $h_{\mathcal {E},T}:\mathcal {E} \rightarrow \mathbb {R}$ a function defined in (2.8). Then the following properties hold:

• (Z1) $\forall$ $x \in \mathcal {E}$ and $\forall$ $\alpha \in [h_{\mathcal {E},T}(x),1],$ $\alpha x+(1-\alpha ) T(x) \in \mathcal {E};$

• (Z2) $\forall$ $x \in \mathcal {E}$ and $\forall$ $\beta \in [0,h_{\mathcal {E},T}(x)),$ $\beta x+(1-\beta ) T(x) \notin \mathcal {E};$

• (Z3) $\forall$ $x \in \mathcal {E},$ $T(x) \in \mathcal {E}$ if and only if $h_{\mathcal {E},T}(x)=0;$

• (Z4) If $T(x) \notin \mathcal {E},$ then $h_{\mathcal {E},T}(x)x+(1-h_{\mathcal {E},T}(x))T(x) \in \partial \mathcal {E}.$

Lemma 2.24

[39]. Suppose $\mathcal {E}$ be a convex, closed and nonempty subset of a given Banach space $\mathcal {X}$ and $T: \mathcal {E} \rightarrow \mathcal {X}$ a mapping which is weakly inward. Then $h_{\mathcal {E},T}(x) <1$ for all $x \in \mathcal {E}.$

Definition 2.25

[40]. The mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ is said to be semicompact if for each bounded sequence $\{x_n\} \in \mathcal {X}$ such that $x_n-T(x_n) \rightarrow y$ for some $y \in \mathcal {X},$ there exists a convergent subsequence.

Definition 2.26

[41]. The mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ is said to be closed if a sequence $\{x_n\} \in \mathcal {E}$ satisfying $x_n \rightarrow x$ and $T(x_n) \rightarrow y$ then $x \in \mathcal {E}$ and $T(x)=y.$

Definition 2.27

[42] The mapping $T: \mathcal {E} \rightarrow \mathcal {X}$ with $F(T) \ne \emptyset$ satisfies condition (I) if $\exists$ a nondecreasing function $f: [0, \infty ) \rightarrow [0, \infty )$ along with $f(r)>0, f(0)=0$ $\forall$ $r \in (0, \infty )$ in such a way that $f(d(x, F(T))) \le \Vert x-T(x)\Vert$ $\forall x \in \mathcal {E},$ where $d(x, F(T)) = \inf \{\Vert x-y\Vert :y \in F(T)\}.$

Definition 2.28

[41]. Any finite family of mappings $\mathfrak {T}: \mathcal {E} \rightarrow \mathcal {X}$ along with $\emptyset \ne F(\mathfrak {T})$ is said to satisfy condition (II) if $\exists$ a nondecreasing function $f: [0, \infty ) \rightarrow [0, \infty )$ along with $f(r)>0, f(0)=0$ $\forall r \in (0, \infty ),$ in such a way that

$$\begin{aligned} \max \{\Vert x-T(x)\Vert : T \in \mathfrak {T}\}\ge f(d(x, F(\mathfrak {T}))) \ \ \forall x \in \mathcal {E}. \end{aligned}$$

Lemma 2.29

[16] Suppose $\mathcal {E}$ be a convex and nonempty subset of a real q-uniformly smooth Banach space (in short q-USBS) $\mathcal {X}.$ Suppose $C_q>0$ and T be a k-strict pseudocontractive mapping. Suppose $\mu \in \left( 0, \min \left\{ \left( \frac{kq}{C_q} \right) ^{\frac{1}{q-1}},1 \right\} \right)$ then

$$\begin{aligned} T_{\mu }= (1-\mu )I+\mu T \end{aligned}$$

is a nonexpansive mapping. If $F(T) \ne \emptyset$ then $F(T)= F(T_{\mu })$.

Proposition 2.30

[15] Suppose $\mathcal {E}$ be nonempty subset of a space $\mathcal {X}$ with $F(T) \ne \emptyset$. If the mapping T is inward, then for all $\mu \in (0,1)$ the average mapping $T_{\mu }= (1-\mu )I+ \mu T$ is inward too.

3 Weak convergence results

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-uniformly smooth Banach space $\mathcal {X}$, $\{T_n\}$ a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}.$ Assume that $x_1 \in \mathcal {E},$ we can generate a sequence $\{x_n\}$ in $\mathcal {E}$ as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha _1= \max \{ \frac{1}{2}, h_{\mathcal {E},T_{\mu _1}}(x_1)\}\\ x_{n+1}=\alpha _n x_n+(1-\alpha _n)T_{\mu _n}(x_n)\\ \alpha _{n+1}=\max \{ \alpha _n, h_{\mathcal {E},T_{\mu _{n+1}}}(x_{n+1})\}. \end{array}\right. } \end{aligned}$$

(3.1)

From Lemma 2.24, for any weakly inward mapping $T_{\mu }: \mathcal {E} \rightarrow \mathcal {X},$ we have $h_{\mathcal {E},T_{\mu }}(x) <1$ $\forall x \in \mathcal {E}.$ Now it can be seen that for each $n \in \mathbb {N},$ $\alpha _{n+1} \in [h_{\mathcal {E},T_{\mu _{n+1}}}(x_{n+1}),1].$ Thus from Lemma 2.23 (Z1), we get

$$\begin{aligned} x_{n+1}=\alpha _n x_n+(1-\alpha _n)T_{\mu _n}(x_n) \in \mathcal {E}. \end{aligned}$$

Hence the algorithm (3.1) is well defined.

Now, we present some important lemmas which can be utilized to prove the main convergence results.

Lemma 3.1

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a q-uniformly smooth Banach space (in short q-USBS) $\mathcal {X}.$ Suppose $\{T_n\}$ be any given family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. Suppose $\{x_n\}$ be a sequence given by (3.1). Then for each $p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ $\lim \limits _{n \rightarrow \infty } \Vert x_n -p\Vert$ exists.

Proof

Suppose $p \in \mathop \cap \nolimits _{n=1}^{\infty }F(T_n).$ From (3.1) $\forall n \in \mathbb {N},$ we have

$$\begin{aligned} \Vert x_{n+1}-p \Vert= & {} \Vert \alpha _n x_n+(1-\alpha _n)T_{\mu _n}(x_n)- p \Vert \nonumber \\\le & {} \alpha _n \Vert x_n- p \Vert +(1-\alpha _n) \Vert T_{\mu _n}(x_n)- p \Vert \nonumber \\= & {} \alpha _n \Vert x_n- p \Vert +(1-\alpha _n) \Vert T_{\mu _n}(x_n)- T_{\mu _n}(p) \Vert \nonumber \\\le & {} \alpha _n \Vert x_n- p \Vert +(1-\alpha _n) \Vert x_n- p \Vert \nonumber \\\le & {} \Vert x_n- p \Vert . \end{aligned}$$

(3.2)

Hence we get the required conclusion.

Lemma 3.2

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. Suppose $\{x_n\}$ be a sequence given by (3.1). Then there exists a strictly increasing convex continuous function $\psi : [0, \infty ) \rightarrow [0, \infty )$ along with $\psi (0)=0$ such that $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n)\psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ) < \infty .$

Proof

By Lemma 3.1, both the sequences $\{x_n-p\}$ and $\{T_{\mu _n} (x_n)-p\}$ are bounded, so these are contained in $\mathcal {B}_s:=\{ x \in \mathcal {X}: \Vert x\Vert \le s\}$ for sufficiently large $s>0.$ In view of Lemma 2.13, there is a strictly increasing convex continuous function $\psi : [0, \infty ) \rightarrow [0, \infty )$ along with $\psi (0)=0$ such that (2.5) holds. Thus, we have

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^2= & {} \Vert \alpha _n x_n+(1-\alpha _n)T_{\mu _n}(x_n)- p \Vert ^2\\= & {} \Vert \alpha _n (x_n-p)+(1-\alpha _n)(T_{\mu _n}(x_n)- p) \Vert ^2\\\le & {} \alpha _n \Vert x_n-p \Vert ^2 +(1-\alpha _n)\Vert T_{\mu _n}(x_n)- p \Vert ^2-\alpha _n (1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert )\\\le & {} \alpha _n \Vert x_n-p \Vert ^2 +(1-\alpha _n)\Vert x_n- p \Vert ^2-\alpha _n (1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert )\\= & {} \Vert x_n-p \Vert ^2 -\alpha _n (1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ). \end{aligned}$$

So,

$$\begin{aligned} \alpha _n (1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ) \le \Vert x_n-p \Vert ^2 -\Vert x_{n+1}-p\Vert ^2. \end{aligned}$$

(3.3)

Since $\alpha _n \in \left[ \frac{1}{2},1\right] ,$

$$\begin{aligned} \frac{1}{2}(1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ) \le \alpha _n (1-\alpha _n) \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ). \end{aligned}$$

Therefore, from (3.3) and using the hypothesis of the Theorem, we get the desired conclusion.

Lemma 3.3

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\mathop \cap \nolimits _{n=1}^{\infty }F(T_n) \ne \emptyset$. Suppose $\mathcal {E}$ is a nonexpansive retract of $\mathcal {X}$ and $\{x_n\}$ be a sequence given by (3.1). Then for any $\beta \in [0,1]$, $p, q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ the limit $\lim \limits _{n \rightarrow \infty } \Vert \beta x_n +(1-\beta ) p-q\Vert$ exists.

Proof

In view of Lemma 3.1, the given sequence $\{x_n\}$ is bounded. Let $p, q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$ and set

$$\begin{aligned} \zeta _n(\beta ):= \Vert \beta x_n +(1-\beta ) p-q\Vert . \end{aligned}$$

Then $\lim \limits _{n \rightarrow \infty } \zeta _n(0)=\Vert p-q\Vert .$ By Lemma 3.1, $\lim \limits _{n \rightarrow \infty } \zeta _n(1)=\Vert x_n-q\Vert$ exists. Now it only remains to check for the case $\beta \in (0,1).$ Since, we compute the sequence $\{\alpha _n \}$ by (3.1), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha _1= \max \{ \frac{1}{2}, h_{\mathcal {E},T_{\mu _1}}(x_1)\}\\ \alpha _{n+1}=\max \{ \alpha _n, h_{\mathcal {E},T_{\mu _{n+1}} }(x_{n+1})\}, \end{array}\right. } \end{aligned}$$

which depends on the initial point $x_1.$ Now, we define a mapping $S_{n,x_1}: \mathcal {E} \rightarrow \mathcal {X}$ by

$$\begin{aligned} S_{n,x_1}:= \alpha _n I+(1-\alpha _n) T_n \text { for all }n \in \mathbb {N}. \end{aligned}$$

Then using Lemma 2.11 we can say $S_{n,x_1}$ is nonexpansive. Moreover, $x_{n+1}=S_{n,x_1}(x_n)$ and $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n) \subseteq \mathop \cap \nolimits _{n=1}^{\infty } F(S_{n,x_1}).$ Let $V_{n,m}: \mathcal {E} \rightarrow \mathcal {X}$ be the mapping defined as

$$\begin{aligned} V_{n,m}=S_{n+m-1,x_1}R_{\mathcal {E}}S_{n+m-2,x_1}R_{\mathcal {E}}...S_{n+1,x_1}R_{\mathcal {E}}S_{n,x_1}R_{\mathcal {E}}. \end{aligned}$$

Where $R_{\mathcal {E}}:\mathcal {X} \rightarrow \mathcal {E}$ is the nonexpansive retraction. Since a retraction does not move point into $\mathcal {E},$ it can be seen that $x_{n+m}=V_{n,m}(x_n)$ and $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n) \subseteq \mathop \cap \nolimits _{n=1}^{\infty } F(V_{n,m}).$ Moreover,

$$\begin{aligned} \Vert V_{n,m}(x)-V_{n,m}(y)\Vert \le \Vert x-y\Vert \text { for all }x,y \in \mathcal {E}. \end{aligned}$$

Set

$$\begin{aligned} \xi _{n,m}(\beta ):=\Vert \beta V_{n,m}(x_n)+(1-\beta )p-V_{n,m}\{\beta x_n +(1-\beta )p \}\Vert . \end{aligned}$$

Now, by Lemma 2.14, $\exists$ a strictly increasing convex continuous function $\tau : \mathbb {R}^{+} \rightarrow \mathbb {R}^{+}$ satisfying $\tau (0)=0$ such that

$$\begin{aligned} \tau \left( \xi _{n,m}(\beta ) \right)= & {} \tau \left( \Vert \beta V_{n,m}(x_n)+(1-\beta )p-V_{n,m}\{\beta x_n +(1-\beta ) p\}\Vert \right) \\\le & {} \Vert x_n-p\Vert - \Vert V_{n,m}(x_n)-V_{n,m}(p)\Vert \\= & {} \Vert x_n-p\Vert - \Vert x_{n+m}-p\Vert . \end{aligned}$$

Since $\lim \limits _{n \rightarrow \infty } \Vert x_n-p\Vert$ exists and hence last difference is zero. Therefore $\lim \limits _{n,m \rightarrow \infty } \tau \left( \xi _{n,m}(\beta ) \right) =0$ and $\lim \limits _{n,m \rightarrow \infty } \xi _{n,m}(\beta )=0.$ Now, we have

$$\begin{aligned} \zeta _{n+m}(\beta )= & {} \Vert \beta x_{n+m} +(1-\beta ) p-q\Vert \\= & {} \Vert \beta V_{n,m}(x_{n}) +(1-\beta ) p-q\Vert \\\le & {} \xi _{n,m}(\beta )+\Vert V_{n,m}\{\beta x_n +(1-\beta )p \}-q\Vert \\\le & {} \xi _{n,m}(\beta )+\Vert \beta x_n +(1-\beta )p-q\Vert \\\le & {} \xi _{n,m}(\beta )+\zeta _{n}(\beta ). \end{aligned}$$

Hence

$$\begin{aligned} \limsup \limits _{n \rightarrow \infty } \zeta _{n}(\beta )\le & {} \lim \limits _{n \rightarrow \infty } \xi _{n,m}(\beta )+ \liminf \limits _{n \rightarrow \infty } \zeta _{n}(\beta )\\\le & {} \liminf \limits _{n \rightarrow \infty } \zeta _{n}(\beta ). \end{aligned}$$

That is, there exists $\lim \limits _{n \rightarrow \infty } \Vert \beta x_n +(1-\beta ) p-q\Vert$ $\forall \beta \in (0,1).$ Hence the proof is complete.

Lemma 3.4

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty }F(T_n)$. Assume that $\{x_n\}$ is a sequence given by (3.1) then $\omega _{w}(x_n)$ is a singleton set.

Proof

Since $\{x_n\}$ is a bounded sequence in $\mathcal {X}$, $\lim \limits _{n \rightarrow \infty } \Vert \beta x_n +(1-\beta ) p-q\Vert$ exists $\forall \beta \in [0,1]$, $p,q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$ and the norm defined on $\mathcal {X}$ is Fréchet differentiable, by (2.4), we have

$$\begin{aligned} \frac{1}{2} \Vert p-q\Vert ^2+ \beta \langle x_n-p, j(p-q) \rangle\le & {} \frac{1}{2} \Vert \beta x_n +(1-\beta ) p-q\Vert ^2 \\\le & {} \frac{1}{2} \Vert p-q\Vert ^2 + \beta \langle x_n-p, j(p-q) \rangle \\{} & {} +g(\beta \Vert x_n-p\Vert ). \end{aligned}$$

Since the middle term admits limit, $\lim \limits _{n \rightarrow \infty } \Vert x_n-p\Vert$ exists. Thus

$$\begin{aligned} \frac{1}{2} \Vert p-q\Vert ^2 \!+\! \beta \limsup \limits _{n \rightarrow \infty } \langle x_n-p, j(p-q) \rangle \!\le & {} \! \frac{1}{2} \!\lim \limits _{n \rightarrow \infty }\! \Vert \beta x_n +(1-\beta ) p-q\Vert ^2 \\&\!\le&\! \frac{1}{2} \Vert p-q\Vert ^2\! + \! \beta \! \liminf \limits _{n \rightarrow \infty }\! \langle x_n-p, j(p-q) \rangle \\ {}{} & {} +o(\beta ). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \limsup \limits _{n \rightarrow \infty } \langle x_n-p, j(p-q) \rangle \le \liminf \limits _{n \rightarrow \infty } \langle x_n-p, j(p-q) \rangle +\frac{o(\beta )}{\beta }, \end{aligned}$$

if $\beta \rightarrow 0,$ $\lim \limits _{n \rightarrow \infty } \langle x_n-p, j(p-q) \rangle$ exists $\forall p,q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ Now, we prove that the set $\omega _{w}(x_n)$ is singleton. For this, assume $\tilde{p},\tilde{q} \in \omega _{w}(x_n)$ and there exist two subsequences $\{x_{n_i}\},$ $\{x_{n_j}\}$ of $\{x_{n}\}$ such that $x_{n_i} \rightharpoonup \tilde{p}$ and $x_{n_j} \rightharpoonup \tilde{q},$ thus

$$\begin{aligned} \langle \tilde{p}-p, j(p-q) \rangle= & {} \lim \limits _{i \rightarrow \infty } \langle x_{n_i}-p, j(p-q) \rangle \\= & {} \lim \limits _{j \rightarrow \infty } \langle x_{n_j}-p, j(p-q) \rangle \\= & {} \langle \tilde{q}-p, j(p-q) \rangle . \end{aligned}$$

That is,

$$\begin{aligned} \langle \tilde{p}-\tilde{q}, j(p-q)\rangle =0 \end{aligned}$$

for all $p,q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ Hence $\tilde{p}=\tilde{q}$ and $\omega _{w}(x_n)$ is a singleton set.

Theorem 3.5

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}$. Suppose $\mathcal {E}$ is a nonexpansive retract of $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. If $\{T_{\mu _n}\}$ satisfies NST-condition, then the sequence $\{x_n\}$ defined by (3.1) converges weakly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [a,b] \subset (0,1).$

Proof

In view of Lemma 3.3, the limit $\lim \limits _{n \rightarrow \infty } \Vert \beta x_n +(1-\beta ) p-q\Vert$ exists $\forall \beta \in [0,1]$, $p, q \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ Using Lemma 3.1 sequence $\{x_n\}$ is bounded, so $\exists$ a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\{x_{n_k}\}$ converges weakly to some $x^{*} \in \omega _w\{x_n\} \subset \mathcal {E}.$ By Lemma 3.4, the set $\omega _w\{x_n\}$ is singleton. Thus $\{x_n\}$ converges weakly to $x^{*} \in \omega _w\{x_n\}.$ Since $\{\alpha _n\} \subseteq [a,b] \subset (0,1),$ by Lemma 3.2, we get $\sum \nolimits _{n=1}^{\infty } \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ) < \infty .$ This implies that $\lim \limits _{n \rightarrow \infty } \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert )=0$, $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0.$ By the assumption that $\{T_{\mu _n}\}$ satisfies NST-condition, hence we have $\omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})$ and hence $\omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$.

Theorem 3.6

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}$. Suppose $\mathcal {E}$ is a nonexpansive retract of $\mathcal {X}$, $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. If $\{T_{\mu _n}\}$ satisfies AKTT-condition (I), mapping $T_{\mu }$ is defined by (2.6) and $F(T_{\mu })=\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n}),$ then sequence $\{x_n\}$ given by (3.1) converges weakly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [0,1]$ with $\sum \limits _{n=1}^{\infty } (1-\alpha _n) = \infty .$

Proof

First, we prove that $\lim \limits _{n \rightarrow \infty } \Vert x_n-T(x_n)\Vert$ exists. Since $\{T_{\mu _n}\}$ satisfies AKTT-condition (I) and sequence $\{x_n\}$ is bounded, we have

$$\begin{aligned} \sum \nolimits _{n=1}^{\infty } \sup \{ \Vert T_{\mu _{n+1}}(x)-T_{\mu _n}(x)\Vert :x \in \{x_n\} \} < \infty . \end{aligned}$$

(3.4)

Moreover, by (3.1) and the triangle inequality, we get

$$\begin{aligned} \Vert x_{n+1}-T_{\mu _{n+1}}(x_{n+1})\Vert= & {} \Vert \alpha _n x_n+(1-\alpha _n)T_{\mu _n}(x_n)-T_{\mu _{n+1}}(x_{n+1})\Vert \\\le & {} \alpha _n \Vert x_n -T_{\mu _n}(x_n) \Vert +\Vert T_{\mu _n}(x_n)-T_{\mu _{n+1}}(x_{n+1})\Vert \\\le & {} \alpha _n \Vert x_n -T_{\mu _n}(x_n) \Vert +\Vert T_{\mu _n}(x_n)-T_{\mu _n}(x_{n+1})\Vert \\{} & {} +\Vert T_{\mu _n}(x_{n+1})-T_{\mu _{n+1}}(x_{n+1})\Vert \\\le & {} \alpha _n \Vert x_n -T_{\mu _n}(x_n) \Vert +\Vert x_n-x_{n+1}\Vert +\Vert T_{\mu _n}(x_{n+1})-T_{\mu _{n+1}}(x_{n+1})\Vert \\= & {} \alpha _n \Vert x_n -T_{\mu _n}(x_n) \Vert +(1-\alpha _n) \Vert x_n -T_{\mu _n}(x_n) \Vert \\ {}{} & {} +\Vert T_{\mu _n}(x_{n+1})-T_{\mu _{n+1}}(x_{n+1})\Vert \\= & {} \Vert x_n-T_{\mu _n}(x_n)\Vert +\Vert T_{\mu _n}(x_{n+1})-T_{\mu _{n+1}}(x_{n+1})\Vert \\\le & {} \Vert x_n-T_{\mu _n}(x_n)\Vert + \sup \{\Vert T_{\mu _n}(x)-T_{\mu _{n+1}}(x)\Vert : x \in \{x_n\} \}. \end{aligned}$$

Now, using (3.4) and the Lemma (), $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert$ exists. Since $\sum \limits _{n=1}^{\infty } (1-\alpha _n) = \infty$ and using Lemma 3.2, we have $\liminf \limits _{n \rightarrow \infty } \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert )=0.$ This implies that $\liminf \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0.$ Hence $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0.$ Now, we prove $\omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ Let $p \in \omega _w\{x_n\},$ then there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ in such a way that $x_{n_k} \rightharpoonup p.$ The mapping $T_{\mu }$ is nonexpansive and $I-T_{\mu }$ is demiclosed at zero. From Lemma 2.19, we have

$$\begin{aligned} \lim \limits _{k \rightarrow \infty } \sup \{\left\| T_{\mu }(x)-T_{\mu _{n_k}}(x)\right\| :x \in \{x_n\}\}=0. \end{aligned}$$

(3.5)

From (3.5) and the triangle inequality, we get

$$\begin{aligned} \Vert x_{n_k}-T_{\mu }(x_{n_k}) \Vert\le & {} \Vert x_{n_k}-T_{\mu _{n_k}}(x_{n_k}) \Vert +\left\| T_{\mu _{n_k}}(x_{n_k}) -T_{\mu }(x_{n_k}) \right\| \\\le & {} \Vert x_{n_k}-T_{\mu _{n_k}}(x_{n_k}) \Vert + \sup \{\left\| T_{\mu _{n_k}}(x) -T_{\mu }(x)\right\| : x \in \{x_n\}\} \end{aligned}$$

making $k \rightarrow \infty ,$ $\Vert x_{n_k}-T_{\mu }(x_{n_k}) \Vert \rightarrow 0.$ Using the demiclosedness of mapping $I-T_{\mu },$ $p \in F(T_{\mu })$ and $\omega _w\{x_n\} \subset F(T_{\mu })= \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n}).$ Hence $\{T_{\mu _n}\}$ satisfies NST-condition. Now the proof directly follows from Theorem 3.5.

Theorem 3.7

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}$ which has Opial condition. Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. If $\{T_{\mu _n}\}$ satisfies NST-condition, then sequence $\{x_n\}$ given by (3.1) converges weakly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [a,b] \subset (0,1)$ with $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n) = \infty .$

Proof

Since $\{\alpha _n\} \subseteq [a,b] \subset (0,1),$ using Lemma 3.2, we get $\sum \nolimits _{n=1}^{\infty } \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert ) < \infty .$ This implies that $\lim \limits _{n \rightarrow \infty } \psi (\Vert x_n-T_{\mu _n}(x_n)\Vert )=0$ and $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0.$ By the assumption that $\{T_{\mu _n}\}$ satisfies the NST-condition, we have $\omega _w\{x_n\} \subset \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n}).$ In view of Lemma 3.1, sequence $\{x_n\}$ is bounded. Then there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\{x_{n_k}\}$ converges weakly to some $x^{*} \in \omega _w\{x_n\} \subset \mathcal {E}.$ Thus, $x^{*} \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=\mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ To prove weak convergence of the sequence $\{x_n\}$ to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{n}),$ it is suffice to prove that $\omega _w\{x_n\}$ is singleton. Arguing by contradiction, let $\tilde{p}, \tilde{q} \in \omega _w\{x_n\},$ $\{x_{n_k}\}$ and $\{x_{n_j}\}$ be two subsequences of $\{x_n\}$ such that $x_{n_k} \rightharpoonup \tilde{p}$ and $x_{n_j} \rightharpoonup \tilde{q},$ respectively with $\tilde{p} \ne \tilde{q}.$ Since $\lim \limits _{n \rightarrow \infty } \Vert x_n -p\Vert$ exists for each $p \in \mathop \cap \nolimits ^{\infty }_{n=1} F(T_{n}),$ from the Opial condition, we have

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \Vert x_n-\tilde{p}\Vert= & {} \lim \limits _{k \rightarrow \infty } \Vert x_{n_k}-\tilde{p}\Vert \\< & {} \lim \limits _{k \rightarrow \infty } \Vert x_{n_k}-\tilde{q}\Vert =\lim \limits _{j \rightarrow \infty } \Vert x_{n_j}-\tilde{q}\Vert \\< & {} \lim \limits _{j \rightarrow \infty } \Vert x_{n_j}-\tilde{p}\Vert =\lim \limits _{n \rightarrow \infty } \Vert x_n-\tilde{p}\Vert , \end{aligned}$$

a contradiction. This completes the proof.

Theorem 3.8

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}$ which has Opial condition. Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. If $\{T_{\mu _n}\}$ satisfies AKTT-condition (I), mapping $T_{\mu }$ is defined by (2.6) and $F(T_{\mu })=\mathop \cap \nolimits _{n=1}^{\infty } F(T_{{\mu }_n}),$ then the sequence $\{x_n\}$ given by (3.1) converges weakly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [0,1]$ with $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n) = \infty .$

Proof

Since the sequence of mapping $\{T_{\mu _n}\}$ satisfies AKTT-condition (I) along with $F(T_{\mu })=\mathop \cap \nolimits _{n=1}^{\infty } F(T_{{\mu }_n}),$ following the proof of Theorem 3.6, it can be shown that $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{{\mu }_n}(x_n)\Vert =0$ and $\{T_{{\mu }_n}\}$ satisfies NST-condition. Following largely the similar argument for application of Opial condition as in proof of Theorem 3.7, we can conclude that the sequence $\{x_n\}$ converges weakly to $p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$

4 Strong convergence results

Theorem 4.1

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$. If $\{T_{\mu _n}\}$ satisfies AKTT-condition (I), mapping $T_\mu$ is defined by (2.6) and $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})= F(T_\mu ),$ then sequence $\{x_n\}$ given by (3.1) converges strongly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n) < \infty .$

Proof

By Lemma 3.1, $\forall p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ $\lim \limits _{n \rightarrow \infty } \Vert x_n -p\Vert$ exists. Since

$$\begin{aligned} \Vert x_n-x_{n+1}\Vert =(1-\alpha _n)\Vert x_n-T_{\mu _n}(x_n)\Vert , \end{aligned}$$

using boundedness of the sequences $\{x_n\}$, $\{T_{\mu _n}(x)\},$

$$\begin{aligned} \sum \nolimits _{n=1}^{\infty } \Vert x_n-x_{n+1}\Vert < \infty . \end{aligned}$$

That is, $\{x_n\}$ is a strongly Cauchy sequence. Therefore, $\exists$ $x^{*} \in \mathcal {E}$ such that

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } x_n =x^{*}. \end{aligned}$$

Now, it suffices to show that $x^{*} \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$

Since the family $\{T_{\mu _n}\}$ is weakly inward $\forall n \in \mathbb {N},$ $T_{\mu _n}(x) \in \overline{I_{\mathcal {E}}(x)}$ $\forall n \in \mathbb {N}.$ Hence

$$\begin{aligned} T_{\mu }(x)=\lim \limits _{n \rightarrow \infty } T_{\mu _n}(x) \in \overline{I_{\mathcal {E}}(x)}, \end{aligned}$$

and $T_\mu$ is weakly inward mapping. Using Lemma 2.24, we know that $h_{\mathcal {E},T_\mu }(x) <1$ $\forall x \in \mathcal {E}.$ Lemma 2.23 (Z1) implies that for all $\delta \in (h_{\mathcal {E},T_\mu }(x^*),1)$

$$\begin{aligned} \delta x^{*}+(1-\delta )T_\mu (x^{*}) \in \mathcal {E}. \end{aligned}$$

(4.1)

On the other hand, $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n) < \infty$ ensures that $\lim \limits _{n \rightarrow \infty } \alpha _n=1$ where $\alpha _{n}=\max \{ \alpha _{n-1}, h_{n}(x_{n})\}.$ So we can choose a subsequence $\{x_{n_j}\}$ with the property that $\left\{ h_{\mathcal {E},T_{\mu _{n_j}}}(x_{n_j})\right\}$ is nondecreasing and $h_{\mathcal {E},T_{\mu _{n_j}}}(x_{n_j}) \rightarrow 1.$ In particular, Lemma 2.23 (Z2) ensures that for any fixed $\delta <1,$

$$\begin{aligned} \delta x_{n_j} +(1-\delta ) T_{\mu _{n_j}}(x_{n_j}) \notin \mathcal {E} \text { for sufficiently large }j. \end{aligned}$$

(4.2)

Take two positive real numbers $\delta _1, \delta _2 \in (h_{\mathcal {E},T_\mu }(x^*),1)$ with $\delta _1 \ne \delta _2$ and set $\rho _1=\delta _1 x^{*} +(1-\delta _1) T_\mu (x^*)$ and $\rho _2=\delta _2 x^{*} +(1-\delta _2) T_\mu (x^*).$ Now, for any $\delta \in [\delta _1, \delta _2]$ by (4.1), we get

$$\begin{aligned} \rho =\delta x^{*}+(1-\delta )T_\mu (x^{*}) \in \mathcal {E}. \end{aligned}$$

(4.3)

Now, we prove that $T_{\mu _n}(x_n) \rightarrow T_\mu (x^*)$ as $n \rightarrow \infty .$ For this, take $\mathcal {C}=\mathcal {B}_r(x^*) \cap \mathcal {E},$ $\forall r>0,$ then $\mathcal {C}$ is a bounded subset of $\mathcal {E}.$ Using triangle inequality, we have

$$\begin{aligned} \Vert T_{\mu _n}(x_n)-T_\mu (x^*)\Vert\le & {} \Vert T_{\mu _n}(x_n)-T_{\mu _n}(x^*)\Vert +\Vert T_{\mu _n}(x^*)-T_\mu (x^*)\Vert \\\le & {} \Vert x_n-x^*\Vert +\sup \{\Vert T_{\mu _n}(x)-T_\mu (x)\Vert : x \in \mathcal {C} \}. \end{aligned}$$

Since $x_n \rightarrow x^*$ as $n \rightarrow \infty$ and from Lemma 2.19, we have $T_{\mu _n}(x_n) \rightarrow T_\mu (x^*)$ as $n \rightarrow \infty .$ By (4.2), $\delta x_{n_j} +(1-\delta ) T_{\mu _{n_j}}(x_{n_j}) \notin \mathcal {E}.$ Since $x_n \rightarrow x^*$ and $T_{\mu _n}(x_n) \rightarrow T_\mu (x^*)$ as $n \rightarrow \infty ,$ we get

$$\begin{aligned} \lim \limits _{j \rightarrow \infty } \delta x_{n_j} +(1-\delta ) T_{\mu _{n_j}}(x_{n_j})=\rho , \text { and }\rho \in \partial \mathcal {E}. \end{aligned}$$

Since $\rho$ is any arbitrary point in segment $[\rho _1, \rho _2],$ the entire segment $[\rho _1, \rho _2] \subset \partial \mathcal {E}.$ The strict convexity of $\mathcal {E}$ implies that $\rho _1 =\rho _2,$ that is,

$$\begin{aligned} \delta _1 x^{*} +(1-\delta _1) T_\mu (x^*)=\delta _2 x^{*} +(1-\delta _2) T_\mu (x^*), \end{aligned}$$

hence $T_\mu (x^*)=x^*,$ so $x^* \in F(T_\mu ).$ Since $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n)= \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})= F(T),$ therefore, $x^* \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$

Corollary 4.2

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}$, $\{T_n\}$ a family of weakly inward nonexpansive mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ If $\{T_{\mu _n}\}$ satisfies AKTT-condition (I) and mapping $T_\mu$ defined by (2.6), $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=F(T_\mu ),$ then sequence $\{x_n\}$ given by (3.1) converges strongly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\sum \nolimits _{n=1}^{\infty } (1-\alpha _n) < \infty .$

Theorem 4.3

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ If $\{T_{\mu _n}\}$ satisfies AKTT$^*$-condition with $\mathfrak {T}_\mu ,$ and $\mathfrak {T}_\mu$ is a family of closed mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=F(\mathfrak {T}_\mu ).$ If there exists a mapping $\bar{T}_\mu \in \mathfrak {T}_\mu$ such that $\bar{T}_\mu$ is semicompact, then sequence $\{x_n\}$ given by (3.1) converges strongly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [0,1]$ with $\sum \nolimits _{n=1}^{\infty }(1-\alpha _n)= \infty$.

Proof

From Theorem 3.6, we have $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0$, and since family of mappings $\{T_{\mu _n}\}$ satisfies AKTT$^*$-condition, we have

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \Vert x_n-T_\mu (x_n)\Vert =0 \ \ \forall \ T_\mu \in \mathfrak {T}_\mu . \end{aligned}$$

(4.4)

In particular,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \Vert x_n-\bar{T}_\mu (x_n)\Vert =0. \end{aligned}$$

Since mapping $\bar{T}_\mu \in \mathfrak {T}_\mu$ is semicompact, we can find a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $x_{n_j} \rightarrow p \in \mathcal {E}$ as $j \rightarrow \infty .$ From (4.4)

$$\begin{aligned} \lim \limits _{j \rightarrow \infty } \Vert x_{n_j}-T_\mu (x_{n_j})\Vert =0 \ \ \forall \ T_\mu \in \mathfrak {T}_\mu . \end{aligned}$$

Using triangle inequality,

$$\begin{aligned} \Vert T_\mu (x_{n_j})-p \Vert \le \Vert T_\mu (x_{n_j})-x_{n_j}\Vert +\Vert x_{n_j}-p \Vert . \end{aligned}$$

Thus $\Vert T_\mu (x_{n_j})-p \Vert \rightarrow 0$ as $j \rightarrow \infty .$ Since each mapping $T_\mu \in \mathfrak {T}_\mu$ is closed, it confirms that $T_\mu (p)=p$ $\forall$ $T_\mu \in \mathfrak {T}_\mu ,$ hence $p \in F(\mathfrak {T}_\mu ).$ Using Lemma 3.1 and $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=F(\mathfrak {T}_\mu ),$ it can be easily seen that the sequence $\{x_n\}$ converges strongly to $p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})$ and hence $p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n)$.

Theorem 4.4

Suppose $\mathcal {E}$ be a convex closed and nonempty subset of a given q-USBS $\mathcal {X}.$ Suppose $\{T_n\}$ be a family of $k_n$-strict pseudocontractive weakly inward mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\emptyset \ne \mathop \cap \nolimits _{n=1}^{\infty } F(T_n).$ If the family of mappings $\{T_{\mu _n}\}$ satisfies AKTT$^*$-condition with $\mathfrak {T}_\mu ,$ and $\mathfrak {T}_\mu$ is a family of closed mappings from $\mathcal {E}$ into $\mathcal {X}$ with $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=F(\mathfrak {T}_\mu ).$ If $\mathfrak {T}_\mu$ is finite and satisfies condition (II), then sequence $\{x_n\}$ given by (3.1) converges strongly to a point in $\mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ provided $\{\alpha _n\} \subseteq [0,1]$ with $\sum \nolimits _{n=1}^{\infty }(1-\alpha _n)= \infty$.

Proof

Lemma 3.1 confirms that for each $p \in \mathop \cap \nolimits _{n=1}^{\infty } F(T_n),$ $\lim \limits _{n \rightarrow \infty } \Vert x_n -p\Vert$ exists. Now, we prove that the sequence $\{x_n\}$ is a Cauchy sequence. From (3.2) we get

$$\begin{aligned} \Vert x_{n+1}-p\Vert \le \Vert x_n-p\Vert . \end{aligned}$$

Following this way, $\forall$ $n,m \in \mathbb {N},$ we can write

$$\begin{aligned} \Vert x_{n+m}-p\Vert \le \Vert x_n-p\Vert . \end{aligned}$$

(4.5)

From Theorem 3.6, we have $\lim \limits _{n \rightarrow \infty } \Vert x_n-T_{\mu _n}(x_n)\Vert =0$, and since family of mappings $\{T_{\mu _n}\}$ satisfies AKTT$^*$-condition, we have

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \Vert x_n-T_\mu (x_n)\Vert =0 \ \ \forall \ T_\mu \in \mathfrak {T}_\mu . \end{aligned}$$

(4.6)

Since $\mathfrak {T}_\mu$ satisfies condition (II), there exists a function f such that

$$\begin{aligned} f(d(x_n, F(\mathfrak {T}_\mu ))) \le \max \{\Vert x_n-T_\mu (x_n)\Vert : T_\mu \in \mathfrak {T}_\mu \}. \end{aligned}$$

From (4.6), we get

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } d(x_n, F(\mathfrak {T}_\mu ))=0. \end{aligned}$$

(4.7)

Thus for any given $\varepsilon >0,$ there exists a natural number $n_0$ such that

$$\begin{aligned} d(x_{n_0}, F(\mathfrak {T}_\mu ))< \frac{\varepsilon }{2}. \end{aligned}$$

Since $F(\mathfrak {T}_\mu )$ is closed, from (4.7), there exists a point $x^* \in F(\mathfrak {T}_\mu ))$ such that $\Vert x_{n_0}-x^*\Vert < \frac{\varepsilon }{2}.$ From (4.5), we have for all $n \ge n_0$ and $m\ge 1,$

$$\begin{aligned} \Vert x_{n+m}-x_n\Vert\le & {} \Vert x_{n+m}-p\Vert +\Vert x_n-p\Vert \\\le & {} 2\Vert x_n-p\Vert \\\le & {} 2\Vert x_{n_0}-p\Vert < 2 \frac{\varepsilon }{2} =\varepsilon . \end{aligned}$$

Thus the sequence $\{x_n\}$ is a Cauchy sequence in $\mathcal {E}.$ Since $\mathcal {X}$ is complete and $\mathcal {E}$ is a closed subset of $\mathcal {X},$ $\mathcal {E}$ is also complete. Therefore the sequence $\{x_n\}$ converges to some $\tilde{p} \in \mathcal {E}.$ Since the mapping $T_\mu$ is closed, from (4.6), it can be easily follows that $T_\mu (\tilde{p} )=\tilde{p}$ $\forall$ $T_\mu \in \mathfrak {T}_\mu$ and $\tilde{p} \in F(\mathfrak {T}_\mu ).$ Since $\mathop \cap \nolimits _{n=1}^{\infty } F(T_{\mu _n})=\mathop \cap \nolimits _{n=1}^{\infty } F(T_n) = F(\mathfrak {T}_\mu ),$ sequence $\{x_n\}$ converges strongly to $\tilde{p} \in \mathop \cap \nolimits _{n=1}^{\infty }F(T_n).$