Strong convergence of some algorithms for $\lambda$-strict pseudo-contractions in Banach spaces

Abstract

In this paper, we introduce a new iterative algorithm for finding a common element of the set of common fixed points of an infinite family of strictly pseudo-contractive mappings in a real 2-uniformly smooth Banach space. Then we proved a strong convergence theorem under some suitable conditions. Our results generalize and improve several recent results.

1 Introduction

Throughout the paper unless otherwise stated, let E be a real Banach space and $E^{*}$ the dual space of E. Let $\{x_{n}\}$ be any sequence in E, then $x_{n}\rightarrow x$(respectively, $x_{n}\rightharpoonup x$, $x_{n}\rightharpoondown x$) will denote strong (respectively, weak, $weak^{*}$) convergence of the sequence $\{x_{n}\}$. Let C be a nonempty, closed and convex subset of E and T be a self-mapping of C. We use F(T) to denote the fixed points of T. The normalized duality mapping $J:E\rightarrow 2^{E^{*}}$ is defined by

$$\begin{aligned} J(x)=\{f\in E^{*}: \langle x,f\rangle =\Vert x\Vert ^{2} \quad {and} \quad \Vert f\Vert =\Vert x\Vert \},\quad \forall x \in E, \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ denotes the generalized duality pairing. In the sequel we shall donate single-valued duality mappings by j.

We recall that the modulus of smoothness of E is the function $\rho _{E} :[0,\infty ) \rightarrow [0,\infty )$ defined by

$$\begin{aligned} \rho _{E}(t):=\sup \left\{ \frac{1}{2}(\Vert x+y\Vert +\Vert x-y\Vert )-1: \Vert x\Vert \le 1,\Vert y\Vert \le t\right\} . \end{aligned}$$

E is said to be uniformly smooth if $\lim _{t\rightarrow 0}\frac{\rho _{E}(t)}{t}=0$.

Let $q>1$. E is said to be q-uniformly smooth if there exists a constant $c>0$ such that $\rho _{E}(t)\le ct^{q}$. It is well-known that E is uniformly smooth if and only if the norm of E is uniformly Fréchet differentiable. If E is q-uniformly smooth, then $q \le 2$ and E is uniformly smooth, and hence the norm of E is uniformly Fréchet differentiable. If E is uniformly smooth, then the normalized duality map j is single-valued and norm to norm uniformly continuous.

If a Banach space E admits a sequentially continuous duality mapping J from weak topology to weak star topology, from Lemma 1 of [1], it follows that the duality mapping J is single-valued, and also E is smooth. In this case, duality mapping J is said to be weakly sequentially continuous, i.e., for each $\{x_{n}\} \subset E$ with $x_{n}\rightharpoonup x$; then $J(x_{n})\rightharpoondown J(x)$ (see [1]).

A Banach space E is said to satisfy Opial’s condition if for any sequence $\{x_{n}\}$ in E, $x_{n} \rightharpoonup x (n\rightarrow \infty )$ implies

$$\begin{aligned} \limsup _{n\rightarrow \infty }\Vert x_{n}-x\Vert <\limsup _{n\rightarrow \infty }\Vert x_{n}-y\Vert ,\quad \forall y\in E \quad with \quad x\ne y. \end{aligned}$$

By Theorem 1 of [1], we know that if E admits a weakly sequentially continuous duality mapping, then E satisfies Opial’s condition, and E is smooth; for the details, see [1].

Let C be a subset of a real Hilbert space H. Recall that a mapping $T : C \rightarrow C$ is said to be strictly pseudo-contractive if there exists a constant $0< \lambda < 1$ such that

$$\begin{aligned} \Vert Tx-Ty\Vert ^{2}\le \Vert x-y\Vert ^{2}+\lambda \Vert (I-T)x-(I-T)y\Vert ^{2},\quad x,y\in C. \end{aligned}$$

(1.1)

Let C be a subset of a real Banach space E. Recall that a mapping $T : C \rightarrow C$ is said to be strictly pseudo-contractive if there exists a constant $0< \lambda < 1$ such that

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

(1.2)

for every $x,y\in C$ and for some $j(x-y)\in J(x-y)$.

From (1.2) we can prove that if T is $\lambda$-strict pseudo-contractive, then T is Lipschitz continuous with the Lipschitz constant $L = \frac{1+\lambda }{\lambda }$.

It is clear that the class of strictly pseudo-contractive mappings strictly includes the class of nonexpansive mappings, which are mappings T on C such that

$$\begin{aligned} \Vert Tx-Ty\Vert \le \Vert x-y\Vert ,\quad \forall x,y\in C. \end{aligned}$$

(1.3)

Let C be a subset of E. Then $P_{C} : E \rightarrow C$ is called a retraction from E onto C if $P_{C}(x)=x$ for all $x\in C$. A retraction $P_{C} : E \rightarrow C$ is said to be sunny if $P_{C}(x+t(x-P_{C}(x)))=P_{C}(x)$ for all $x\in E$ and $t\ge 0$. A subset C of E is said to be a sunny nonexpansive retract of E if there exists a sunny nonexpansive retraction of E onto C.

Proposition 1.1

(See, e.g., Bruck [2], Reich [3], Goebel and Reich [4]) Let E be a smooth Banach space and let C be a nonempty subset of E. Let $P_{C} : E \rightarrow C$ be a retraction and let J be the normalized duality mapping on E. Then the following are equivalent:

• (a) $P_{C}$ is sunny and nonexpansive.

• (b) $\Vert P_{C}x-P_{C}y\Vert ^{2} \le \langle x-y, J (P_{C}x-P_{C}y)\rangle , \forall x, y \in E$.

• (c) $\langle x -P_{C}x, J(y-P_{C}x)\rangle \le 0, \forall x \in E, y \in C$.

In 2011, Yao et al. [5] in a real Hilbert space, introduced the following iterative algorithm: for $x_{0} =x\in C$,

$$\begin{aligned} x_{n+1} = P_{C}((1 - k - \alpha _{n})x_{n} + kT x_{n});\quad n \ge 0, \end{aligned}$$

(1.4)

where $\{\alpha _{n}\}$ is a real sequence in (0; 1). He obtained that the sequence $\{x_{n}\}$ generated by (1.4) converges strongly to the minimum-norm fixed point of T.

In this paper, motivated and inspired by Yao et al. [5], we will introduce a new iterative scheme in a real 2-uniformly smooth Banach space which admits a weakly sequentially continuous duality mapping defined as:given $x_{1} =x\in C$,

$$\begin{aligned} x_{n+1} = P_{C}\left( (1 - k - \alpha _{n})x_{n} + k\sum _{i=1}^{\infty } \eta _{i}^{(n)}T_{i}x_{n}\right) ; \quad n \ge 1, \end{aligned}$$

(1.5)

where $\{\alpha _{n}\}$ is a real sequence in (0; 1), $k \in (0; \frac{2\lambda }{C_{2}})$ and $\{\eta _{i}\}^{\infty }_{i=1}$ is a positive sequence such that $\sum _{i=1}^{\infty } \eta _{i}= 1$. We will prove that if the parameters satisfy appropriate conditions, then the sequence $\{x_{n}\}$ generated by (1.5) converges strongly to a common element of the fixed points of an infinite family of $\lambda _{i}$-strictly pseudo-contractive mappings.

2 Preliminaries

In order to prove our main results, we need the following lemmas.

Lemma 2.1

[6] Let E be a 2-uniformly smooth Banach space, then exists a constant $C_{2}>0$ such that

$$\begin{aligned} \Vert x+y\Vert ^{2}\le \Vert x\Vert ^{2}+2\langle y,j(x)\rangle +C_{2}\Vert y\Vert ^{2},\quad \forall x,y \in E. \end{aligned}$$

Lemma 2.2

[7, 8] Let $\{s_{n}\}$ be a sequence of non-negative real numbers satisfying

$$\begin{aligned} s_{n+1}\le (1-\lambda _{n})s_{n}+\lambda _{n}\delta _{n}+\gamma _{n},\quad n\ge 0, \end{aligned}$$

where $\{\lambda _{n}\}$, $\{\delta _{n}\}$ and $\{\gamma _{n}\}$ satisfy the following conditions: (i) $\{\lambda _{n}\}\subset [0,1]$ and $\sum _{n=0}^{\infty }\lambda _{n}=\infty$, (ii) $\limsup _{n\rightarrow \infty }\delta _{n}\le 0$ or $\sum _{n=0}^{\infty }\lambda _{n}\delta _{n}<\infty$, (iii) $\gamma _{n}\ge 0(n\ge 0),\sum _{n=0}^{\infty }\gamma _{n}<\infty$. Then $\lim _{n\rightarrow \infty }s_{n}=0$.

Lemma 2.3

(See [9, Lemma 1.3]) Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E. Suppose that the normalized duality mapping $J : E \rightarrow E^{*}$ is weakly sequentially continuous at zero. Let $T : C \rightarrow E$ be a $\lambda$-strict pseudo-contraction with $0 < \lambda < 1$. Then for any $\{x_{n}\}\subset C$, if $x_{n} \rightharpoonup x$, and $x_{n} -T x_{n} \rightarrow y \in E$, then $x - T x = y$.

Lemma 2.4

(See [10, Lemma 2.11]) Let E be a 2-uniformly smooth Banach space which admits a weakly sequentially continuous duality mapping J from E to $E^{*}$ and C be a nonempty convex subset of E. Assume that $T_{i} : C \rightarrow E$ is a countable family of $\lambda _{i}$-strict pseudocontraction for some $0 < \lambda _{i} < 1$ and $\inf \{\lambda _{i} : i \in \mathbb {N}\} > 0$ such that $F =\bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset$. Assume that $\{\eta _{i}\}^{\infty }_{i=1}$ is a positive sequence such that $\sum _{i=1}^{\infty } \eta _{i}= 1$. Then $\sum _{i=1}^{\infty } \eta _{i}T_{i} : C \rightarrow E$ is a $\lambda$-strict pseudocontraction with $\lambda = \inf \{\lambda _{i} : i \in \mathbb {N}\}$ and $F\left( \sum _{i=1}^{\infty } \eta _{i}T_{i}\right) = F$.

3 Main result

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E which admits a weakly sequentially continuous duality mapping J. Let C be also a sunny nonexpansive retraction of E and $T : C \rightarrow C$ be a $\lambda$-strict pseudo-contraction. Let $k \in (0; \frac{2\lambda }{C_{2}})$ be a constant. For each $t \in (0; 1)$, we consider the mapping $T_{t}$ given by

$$\begin{aligned} T_{t}x=P_{C}((1-k-t)x+kTx),\quad \forall x\in C. \end{aligned}$$

It is easy to check that $T_{t} : C \rightarrow C$ is a contraction for a small enough t. As a matter of fact, from Lemma 2.2 and (1.2),

$$\begin{aligned} \Vert T_{t}x-T_{t}y\Vert ^{2}&=\Vert P_{C}((1-k-t)x+kTx)-P_{C}((1-k-t)y+kTy)\Vert ^{2} \nonumber \\&\le \Vert (1-t)(x-y)-k((x-Tx)-(y-Ty))\Vert ^{2} \nonumber \\&\le (1-t)^{2}\Vert x-y\Vert ^{2}-2k(1-t)\langle (I-T)x-(I-T)y,J(x-y)\rangle \nonumber \\&\quad +C_{2}\Vert k[(I-T)x-(I-T)y]\Vert ^{2} \nonumber \\&\le (1-t)^{2}\Vert x-y\Vert ^{2}-2k(1-t)\lambda \Vert (I-T)x-(I-T)y]\Vert ^{2} \nonumber \\&\quad +C_{2}k^{2}\Vert (I-T)x-(I-T)y]\Vert ^{2} \nonumber \\&=(1-t)^{2}\Vert x-y\Vert ^{2}-k[2(1-t)\lambda -C_{2}k]\Vert (I-T)x-(I-T)y]\Vert ^{2}. \end{aligned}$$

(3.1)

We can choose a small enough t such that $2\lambda -C_{2}k>0$. Then, from (3.1),

$$\begin{aligned} \Vert T_{t}x-T_{t}y\Vert \le (1-t)\Vert x-y\Vert ,\quad \forall x,y \in C, \end{aligned}$$

(3.2)

which implies that $T_{t}$ is a contraction. Using the Banach contraction principle, there exists a unique fixed point $x_{t}$ of $T_{t}$ in C, that is,

$$\begin{aligned} x_{t}=P_{C}((1-k-t)x_{t}+kTx_{t}). \end{aligned}$$

(3.3)

Theorem 3.1

Suppose that $F(T)\ne \emptyset$. Then, as $t\rightarrow 0$, the net $\{x_{t}\}$ generated by (3.3) converges strongly to the minimum-norm fixed point of T.

Proof

First, we prove that $\{x_{t}\}$ is bounded. Take $p\in F(T)$. From (3.3) and (3.2),

$$\begin{aligned} \Vert x_{t}-p\Vert&=\Vert P_{C}((1-k-t)x_{t}+kTx_{t})-P_{C}p\Vert \\&\le \Vert (1-k-t)(x_{t}-p)+k(Tx_{t}-p)-tp\Vert \\&\le (1-t)\Vert x_{t}-p\Vert +t\Vert p\Vert , \end{aligned}$$

that is, $\Vert x_{t}-p\Vert \le \Vert p\Vert$, which implies that $\{x_{t}\}$ is bounded and so is $\{Tx_{t}\}$.

From (3.3),

$$\begin{aligned} \Vert x_{t}-Tx_{t}\Vert&=\Vert P_{C}((1-k-t)x_{t}+kTx_{t})-P_{C}Tx_{t}\Vert \\&\le \Vert (1-k)(x_{t}-Tx_{t})-tx_{t}\Vert \\&\le (1-k)\Vert x_{t}-Tx_{t}\Vert +t\Vert x_{t}\Vert . \end{aligned}$$

It follows that

$$\begin{aligned} \Vert x_{t}-Tx_{t}\Vert \le \frac{t}{k}\Vert x_{t}\Vert \rightarrow 0. \end{aligned}$$

(3.4)

Next we show that $\{x_{t}\}$ is relatively norm compact as $t\rightarrow 0$. Let $\{t_{n}\}\subset (0; 1)$ be a sequence such that $t_{n}\rightarrow 0$ as $n\rightarrow \infty$. Put $x_{n} := x_{t_{n}}$. It follows from (3.4) that

$$\begin{aligned} \Vert x_{n}-Tx_{n}\Vert \rightarrow 0\quad as\quad n\rightarrow \infty . \end{aligned}$$

(3.5)

Setting $y_{t} = (1-k-t)x_{t} + kT x_{t}$, we then have $x_{t} = P_{C}y_{t}$, and, for any $p \in F(T)$,

$$\begin{aligned} x_{t}-p&=x_{t}-y_{t}+y_{t}-p \nonumber \\&=x_{t}-y_{t}+(1-k-t)(x_{t}-p)+ k(T x_{t}-p)-tp. \end{aligned}$$

(3.6)

By Proposition 1.1, we have

$$\begin{aligned} \langle x_{t}-y_{t},J(x_{t}-p)\rangle \le 0. \end{aligned}$$

(3.7)

Combining (3.6) and (3.7),

$$\begin{aligned} \Vert x_{t}-p\Vert ^{2}&=\langle x_{t}-y_{t},J(x_{t}-p)\rangle +(1-k-t)\langle x_{t}-p,J(x_{t}-p)\rangle \nonumber \\&\quad + k\langle T x_{t}-p,J(x_{t}-p)\rangle -t\langle p,J(x_{t}-p)\rangle \\&\le \Vert (1-k-t)(x_{t}-p)+ k(T x_{t}-p)\Vert \Vert x_{t}-p\Vert -t\langle p,J(x_{t}-p)\rangle \\&\le (1-t)\Vert x_{t}-p\Vert ^{2}-t\langle p,J(x_{t}-p)\rangle , \end{aligned}$$

which implies that $\Vert x_{t}-p\Vert ^{2}\le \langle p,J(p-x_{t})\rangle$. In particular,

$$\begin{aligned} \Vert x_{n}-p\Vert ^{2}\le \langle p,J(p-x_{n})\rangle ,\quad \forall p\in F(T). \end{aligned}$$

(3.8)

Since $\{x_{n}\}$ is bounded we may assume, without loss of generality, that $\{x_{n}\}$ converges weakly to a point $x^{*}\in C$. From (3.5) and Lemma 2.4, we have that $x^{*}\in F(T)$. Hence it follows from (3.8) that

$$\begin{aligned} \Vert x_{n}-x^{*}\Vert ^{2}\le \langle x^{*},J(x^{*}-x_{n})\rangle . \end{aligned}$$

Since J is weak sequentially continuous and $x_{n} \rightharpoonup x^{*}$, we have that $x_{n} \rightarrow x^{*}$. So we prove that the relative norm compactness of the net $\{x_{t}\}$ as $t\rightarrow 0$.

To show that the entire net $\{x_{t}\}$ converges to $x^{*}$, assume $x_{t_{m}}\rightarrow \bar{x} \in F(T)$, where $t_{m} \rightarrow 0$. Put $x_{m} = x_{t_{m}}$. Similarly, we obtain

$$\begin{aligned} \Vert x_{m}-x^{*}\Vert ^{2}\le \langle x^{*},J(x^{*}-x_{m})\rangle \end{aligned}$$

and hence

$$\begin{aligned} \Vert \bar{x}-x^{*}\Vert ^{2}\le \langle x^{*},J(x^{*}-\bar{x})\rangle . \end{aligned}$$

(3.9)

Interchanging $x^{*}$ and $\bar{x}$, we have

$$\begin{aligned} \Vert x^{*}-\bar{x}\Vert ^{2}\le \langle \bar{x},J(\bar{x}-x^{*})\rangle . \end{aligned}$$

(3.10)

Adding (3.9) and (3.10), we obtain

$$\begin{aligned} 2\Vert x^{*}-\bar{x}\Vert ^{2}\le \Vert x^{*}-\bar{x}\Vert ^{2}, \end{aligned}$$

which implies that $\bar{x} = x^{*}$.

Finally, we return to (3.8) and take the limit as $n\rightarrow \infty$ to get

$$\begin{aligned} \Vert x^{*}-p\Vert ^{2}\le \langle p,J(p-x^{*})\rangle ,\quad \forall p\in F(T). \end{aligned}$$

Equivalently,

$$\begin{aligned} \Vert x^{*}\Vert ^{2}\le \langle x^{*},J(p)\rangle ,\quad \forall p\in F(T). \end{aligned}$$

This clearly implies that

$$\begin{aligned} \Vert x^{*}\Vert \le \Vert p\Vert \quad \forall p\in F(T). \end{aligned}$$

Therefore, $x^{*}$ is a minimum-norm fixed point of T. This completes the proof. $\square$

Corollary 3.2

Suppose that $F(T)\ne \emptyset$ and the origin 0 belongs to C. Then, as $t\rightarrow 0_{+}$, the net $\{x_{t}\}$ generated by the algorithm

$$\begin{aligned} x_{t}=(1-k-t)x_{t}+kTx_{t} \end{aligned}$$

converges strongly to the minimum-norm fixed point of T.

Now we propose the following iterative algorithm which is the discretisation of the implicit method (3.3).

Theorem 3.3

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E which admits a weakly sequentially continuous duality mapping J from E to $E^{*}$. Let C be also a sunny nonexpansive retraction of E, $T_{i} : C \rightarrow C$ be $\lambda _{i}$-strictly pseudo-contractive mapping such that $F = \bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset$ and $\lambda = \inf \{\lambda _{i} : i \in \mathbb {N}\}>0$. Assume for each n, $\{\eta _{i}^{n}\}^{\infty }_{i=1}$ be an infinity sequence of positive number such that $\sum _{i=1}^{\infty } \eta _{i}^{n}= 1$ and for all n , $\eta _{i}^{n} > 0$. For given $x_{1} \in C$ arbitrarily, let the sequence $\{x_{n}\}$ be generated iteratively by

$$\begin{aligned} x_{n+1}=P_{C}((1-k-\alpha _{n})x_{n}+k\sum _{i=1}^{\infty } \eta _{i}^{n}T_{i}x_{n}),\quad \forall n\ge 1, \end{aligned}$$

(3.11)

where $\{\alpha _{n}\}$ is a real sequence in (0; 1) and $k\in (0,\frac{2\lambda }{C_{2}})$. The following control conditions are satisfied:

• (A1) $\lim _{n\rightarrow \infty }\alpha _{n}=0$, $\sum _{n=1}^{\infty }\alpha _{n}=\infty$, $\sum _{n=1}^{\infty }|\alpha _{n+1}-\alpha _{n}|<\infty$;

• (A2) $\sum _{n=1}^{\infty }\sum _{i=1}^{\infty }|\eta _{i}^{n+1}-\eta _{i}^{n}|<\infty$, $\eta _{i}=\lim _{n\rightarrow \infty }\eta _{i}^{n}>0$.

Then the sequence $\{x_{n}\}$ generated by (3.11) strongly converges to the minimum-norm fixed point $x^{*} \in F$.

Proof

For each $n\ge 1$, put $B_{n}=\sum _{i=1}^{\infty } \eta _{i}^{n}T_{i}$. By Lemma 2.4, each $B_{n}$ is a $\lambda$-strict pseudocontraction on C and $F(B_{n})= F$ for all n.

First, we show that the sequence $\{x_{n}\}$ is bounded. Take $p \in F$, it follows from (3.11) that

$$\begin{aligned} \Vert x_{n+1}-p\Vert&=\Vert P_{C}((1-k-\alpha _{n})x_{n}+kB_{n}x_{n})-p\Vert \nonumber \\&\le \Vert (1-k-\alpha _{n})(x_{n}-p)+k(B_{n}x_{n}-p)\Vert +\alpha _{n}\Vert p\Vert . \end{aligned}$$

(3.12)

From (3.2), we note that

$$\begin{aligned} \Vert (1-k-\alpha _{n})(x_{n}-p)+k(B_{n}x_{n}-p)\Vert \le (1-\alpha _{n})\Vert x_{n}-p\Vert . \end{aligned}$$

(3.13)

It follows from (3.12) and (3.13) that

$$\begin{aligned} \Vert x_{n+1}-p\Vert&\le (1-\alpha _{n})\Vert x_{n}-p\Vert +\alpha _{n}\Vert p\Vert \\&\le \max \{\Vert x_{n}-p\Vert ,\Vert p\Vert \}\\&\le \max \{\Vert x_{1}-p\Vert ,\Vert p\Vert \}. \end{aligned}$$

Hence, $\{x_{n}\}$ is bounded and so is $\{B_{n}x_{n}\}$.

We now estimate $\Vert x_{n+1}- x_{n}\Vert$. From (3.11),

$$\begin{aligned} \Vert x_{n+1}-x_{n}\Vert&=\Vert P_{C}((1-k-\alpha _{n})x_{n}+kB_{n}x_{n})\!-\!P_{C}((1-k-\alpha _{n-1})x_{n-1}+kB_{n-1}x_{n-1})\Vert \\&\le \Vert (1-k-\alpha _{n})(x_{n}-x_{n-1})+k(B_{n}x_{n}-B_{n}x_{n-1})\!+\!k(B_{n}x_{n-1}-B_{n-1}x_{n-1}) \nonumber \\&\quad +(\alpha _{n-1}-\alpha _{n})x_{n-1}\Vert \\&\le \Vert (1\!-\!k\!-\!\alpha _{n})(x_{n}\!-\!x_{n-1})+k(B_{n}x_{n}\!-\!B_{n}x_{n-1})\Vert +k\Vert B_{n}x_{n-1}-B_{n-1}x_{n-1}\Vert \nonumber \\&\quad +|\alpha _{n-1}-\alpha _{n}|\Vert x_{n-1}\Vert \\&\le (1-\alpha _{n})\Vert x_{n}-x_{n-1}\Vert +k\Vert B_{n}x_{n-1}-B_{n-1}x_{n-1}\Vert +|\alpha _{n-1}-\alpha _{n}|\Vert x_{n-1}\Vert \\&\le (1-\alpha _{n})\Vert x_{n}-x_{n-1}\Vert +k\sum _{i=1}^{\infty } |\eta _{i}^{n}-\eta _{i}^{n-1}|\Vert T_{i}x_{n-1}\Vert +|\alpha _{n-1}-\alpha _{n}|\Vert x_{n-1}\Vert \\&\le (1-\alpha _{n})\Vert x_{n}-x_{n-1}\Vert +M\left[ \sum _{i=1}^{\infty } |\eta _{i}^{n}-\eta _{i}^{n-1}|+|\alpha _{n-1}-\alpha _{n}|\right] , \end{aligned}$$

where $M=\max \{\sup _{i\ge 1}\sup _{n\ge 1}\Vert T_{i}x_{n-1}\Vert ,\sup _{n\ge 1}\Vert x_{n-1}\Vert \}$. By Lemma 2.3, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n+1}-x_{n}\Vert =0. \end{aligned}$$

(3.14)

On the other hand, we note that

$$\begin{aligned} \Vert x_{n}-B_{n}x_{n}\Vert&\le \Vert x_{n}-x_{n+1}\Vert +\Vert x_{n+1}-B_{n}x_{n}\Vert \\&\le \Vert x_{n}-x_{n+1}\Vert +(1-k)\Vert x_{n}-B_{n}x_{n}\Vert +\alpha _{n}\Vert x_{n}\Vert , \end{aligned}$$

which implies

$$\begin{aligned} \Vert x_{n}-B_{n}x_{n}\Vert \le \frac{1}{k}(\Vert x_{n}-x_{n+1}\Vert +\alpha _{n}\Vert x_{n}\Vert ). \end{aligned}$$

Noticing conditions (A1) and (3.14), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n}-B_{n}x_{n}\Vert =0. \end{aligned}$$

(3.15)

Define $B=\sum _{i=1}^{\infty }\eta _{i}T_{i}$, then $B : C \rightarrow C$ is a $\lambda$-strict pseudocontraction such that $F(B) =\bigcap _{i=1}^{\infty }F(T_{i})=F$ by Lemma 2.4, furthermore $B_{n}x \rightarrow Bx$ as $n \rightarrow \infty$ for all $x \in C$. We observe that

$$\begin{aligned} \Vert x_{n}-Bx_{n}\Vert \le \Vert x_{n}-B_{n}x_{n}\Vert +\Vert B_{n}x_{n}-Bx_{n}\Vert . \end{aligned}$$

By (3.15) and (A2), we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n}-Bx_{n}\Vert =0. \end{aligned}$$

Let the net $\{x_{t}\}$ be defined by (3.3). By Theorem 3.1, $x_{t} \rightarrow x^{*}$ as $t\rightarrow 0$. Next we prove that $\limsup _{n\rightarrow \infty }\langle x^{*},J(x^{*}-x_{n})\rangle \le 0.$

Set $y_{t} = (1-k-t)x_{t} + kBx_{t}$. It follows that

$$\begin{aligned} \Vert x_{t}-x_{n}\Vert ^{2}&=\langle x_{t}-y_{t},J(x_{t}-x_{n})\rangle +\langle y_{t}-x_{n},J(x_{t}-x_{n})\rangle \\&\le \langle y_{t}-x_{n},J(x_{t}-x_{n})\rangle \\&=\langle (1-k-t)(x_{t}-x_{n})+k(B x_{t}-Bx_{n}),J(x_{t}-x_{n})\rangle \nonumber \\&\quad +k\langle Bx_{n}-x_{n},J(x_{t}-x_{n})\rangle -t\langle x_{n},J(x_{t}-x_{n})\rangle \\&\le (1-t)\Vert x_{t}-x_{n}\Vert ^{2}+k\Vert Bx_{n}-x_{n}\Vert \Vert x_{t}-x_{n}\Vert \nonumber \\&\quad -t\langle x_{n}-x_{t},J(x_{t}-x_{n})\rangle -t\langle x_{t},J(x_{t}-x_{n})\rangle \\&=\Vert x_{t}-x_{n}\Vert ^{2}+k\Vert Bx_{n}-x_{n}\Vert \Vert x_{t}-x_{n}\Vert -t\langle x_{t},J(x_{t}-x_{n})\rangle , \end{aligned}$$

and hence that

$$\begin{aligned} \langle x_{t},J(x_{t}-x_{n})\rangle \le \frac{k}{t}\Vert Bx_{n}-x_{n}\Vert \Vert x_{t}-x_{n}\Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \limsup _{t\rightarrow 0}\limsup _{n\rightarrow \infty }\langle x_{t},J(x_{t}-x_{n})\rangle \le 0. \end{aligned}$$

(3.16)

From $x_{t} \rightarrow x^{*}$ as $t\rightarrow 0$, we have $x_{t}-x_{n} \rightarrow x^{*}-x_{n}$ as $t \rightarrow 0$. Noticing that J is single valued and norm to norm uniformly continuous on bounded sets of a uniformly smooth Banach space E, we obtain

$$\begin{aligned}&|\langle x^{*},J(x^{*}-x_{n})\rangle -\langle x_{t},J(x_{t}-x_{n})\rangle |\\&\quad =|\langle x^{*},J(x^{*}-x_{n})-J(x_{t}-x_{n})\rangle +\langle x^{*}-x_{t},J(x_{t}-x_{n})\rangle |\\&\quad \le |\langle x^{*},J(x^{*}-x_{n})-J(x_{t}-x_{n})\rangle |+\Vert x^{*}-x_{t}\Vert \Vert x_{t}-x_{n}\Vert \rightarrow 0 \quad as\quad t\rightarrow 0. \end{aligned}$$

Hence, $\forall \epsilon > 0$; $\exists ~\delta > 0$ such that $\forall t \in (0;\delta )$, for all $n \ge 1$, we have

$$\begin{aligned} \langle x^{*},J(x^{*}-x_{n})\rangle \le \langle x_{t},J(x_{t}-x_{n})\rangle +\epsilon . \end{aligned}$$

By (3.16), we obtain

$$\begin{aligned} \limsup _{n\rightarrow \infty }\langle x^{*},J(x^{*}-x_{n})\rangle&=\limsup _{t\rightarrow 0}\limsup _{n\rightarrow \infty }\langle x^{*},J(x^{*}-x_{n})\rangle \\&\le \limsup _{t\rightarrow 0}\limsup _{n\rightarrow \infty }\langle x_{t},J(x_{t}-x_{n})\rangle +\epsilon \\&\le \epsilon . \end{aligned}$$

Since $\epsilon$ is arbitrary, we have

$$\begin{aligned} \limsup _{n\rightarrow \infty }\langle x^{*},J(x^{*}-x_{n})\rangle \le 0. \end{aligned}$$

(3.17)

Finally, we show that $x_{n} \rightarrow x^{*}$. Set $y_{n} = (1-k-\alpha _{n})x_{n} + kB_{n} x_{n}$ for all $n\ge 1$. From (3.11), we observe

$$\begin{aligned} \Vert x_{n+1}-x^{*}\Vert ^{2}&=\langle x_{n+1}-y_{n},J(x_{n+1}-x^{*})\rangle +\langle y_{n}-x^{*},J(x_{n+1}-x^{*})\rangle \\&\le \langle y_{n}-x^{*},J(x_{n+1}-x^{*})\rangle \\&=\langle (1-k-\alpha _{n})(x_{n}-x^{*})+k(B_{n} x_{n}-x^{*}),J(x_{n+1}-x^{*})\rangle \nonumber \\&\quad +\alpha _{n}\langle x^{*},J(x^{*}-x_{n+1})\rangle \\&\le \Vert (1-k-\alpha _{n})(x_{n}-x^{*})+k(B_{n} x_{n}-x^{*})\Vert \Vert x_{n+1}-x^{*}\Vert \nonumber \\&\quad +\alpha _{n}\langle x^{*},J(x^{*}-x_{n+1})\rangle \\&\le (1-\alpha _{n})\Vert x_{n}-x^{*}\Vert \Vert x_{n+1}-x^{*}\Vert +\alpha _{n}\langle x^{*},J(x^{*}-x_{n+1})\rangle \\&\le \frac{1-\alpha _{n}}{2}(\Vert x_{n}-x^{*}\Vert ^{2}+\Vert x_{n+1}-x^{*}\Vert ^{2})+\alpha _{n}\langle x^{*},J(x^{*}-x_{n+1})\rangle , \end{aligned}$$

which implies

$$\begin{aligned} \Vert x_{n+1}-x^{*}\Vert ^{2}\le (1-\alpha _{n})\Vert x_{n}-x^{*}\Vert ^{2}+\frac{2\alpha _{n}}{1+\alpha _{n}}\langle x^{*},J(x^{*}-x_{n+1})\rangle . \end{aligned}$$

(3.18)

Apply Lemma 2.3 to (3.18), we obtain $x_{n} \rightarrow x^{*}$ as $n\rightarrow \infty$. This completes the proof. $\square$

Corollary 3.4

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E which admits a weakly sequentially continuous duality mapping J from E to $E^{*}$. Let C be also a sunny nonexpansive retraction of E. Let $T_{i}$, $\lambda$, k and $\eta _{i}^{n}$ be as in Theorem 3.3. Suppose that $F = \bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset$ and the origin 0 belongs to C. Assume that the conditions (A1) and (A2) are satisfied. Then the sequence $\{x_{n}\}$ generated by the algorithm

$$\begin{aligned} x_{n+1}=(1-k-\alpha _{n})x_{n}+k\sum _{i=1}^{\infty } \eta _{i}^{n}T_{i}x_{n},\quad \forall n\ge 1, \end{aligned}$$

converges strongly to the minimum-norm fixed point $x^{*}$ of F.

Corollary 3.5

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E which admits a weakly sequentially continuous duality mapping J from E to $E^{*}$. Let C be also a sunny nonexpansive retraction of E. Let $T : C \rightarrow C$ be a $\lambda$-strict pseudo-contraction. Suppose that $F = F(T)\ne \emptyset$. Assume that the condition (A1) is satisfied. Then the sequence $\{x_{n}\}$ generated by the algorithm

$$\begin{aligned} x_{n+1}=(1-k-\alpha _{n})x_{n}+kTx_{n},\quad \forall n\ge 1, \end{aligned}$$

converges strongly to the minimum-norm fixed point $x^{*}$ of F.

Corollary 3.6

Let C be a nonempty closed convex subset of a Hilbert space H, $T_{i} : C \rightarrow C$ be $\lambda _{i}$-strictly pseudo-contractive mapping such that $F = \bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset$ and $\lambda = \inf \{\lambda _{i} : i \in \mathbb {N}\}>0$. Assume for each n, $\{\eta _{i}^{n}\}^{\infty }_{i=1}$ be an infinity sequence of positive number such that $\sum _{i=1}^{\infty } \eta _{i}^{n}= 1$ and for all n, $\eta _{i}^{n} > 0$. For given $x_{1} \in C$ arbitrarily, let the sequence $\{x_{n}\}$ be generated iteratively by

$$\begin{aligned} x_{n+1}=P_{C}\left( (1-k-\alpha _{n})x_{n}+k\sum _{i=1}^{\infty } \eta _{i}^{n}T_{i}x_{n}\right) ,\quad \forall n\ge 1, \end{aligned}$$

(3.19)

where $\{\alpha _{n}\}$ is a real sequence in (0; 1) and $k \in (0; 1 - \lambda )$. The following control conditions are satisfied:

• (A1) $\lim _{n\rightarrow \infty }\alpha _{n}=0$, $\sum _{n=1}^{\infty }\alpha _{n}=\infty$, $\sum _{n=1}^{\infty }|\alpha _{n+1}-\alpha _{n}|<\infty$;

• (A2) $\sum _{n=1}^{\infty }\sum _{i=1}^{\infty }|\eta _{i}^{n+1}-\eta _{i}^{n}|<\infty$, $\eta _{i}=\lim _{n\rightarrow \infty }\eta _{i}^{n}>0$.

Then the sequence $\{x_{n}\}$ generated by (3.19) strongly converges to the minimum-norm fixed point $x^{*} \in F$.

Remark 3.7

Our results improve and extend the results of Yao et al. [5] in the following aspects:

• (i) Hilbert space is replaced by a 2-uniformly smooth Banach space E which admits a weakly sequentially continuous duality mapping;

• (ii) Theorem 3.3 extends Theorem 3.3 of Yao et al. [5] from one strictly pseudo-contractive mapping to an infinite family of strictly pseudo-contractive mappings.