1 Introduction

Since 1994, the split feasibility problem (SFP) [1–3] has received much attention, owing to its applications in many optimization problems, signal processing and medical image reconstruction with special progress in intensity-modulated radiation therapy [4–6]. Let us recall the SFP: to find a point \(q\in B_{1} \) such that

$$ \begin{aligned} q\in C \quad \text{such that } Aq\in Q , \end{aligned} $$
(1.1)

where \(A: B_{1}\longrightarrow B_{2} \) is a bounded linear operator, C and Q are nonempty closed convex subsets of two real Hilbert spaces \(B_{1} \) and \(B_{2} \), respectively.

It is easy to see that problem (1.1) is equivalent to the following fixed point equation:

$$ \begin{aligned} u=P_{C}\bigl(I-\delta A^{*}(I-P_{Q})A\bigr)u ,\quad u\in C , \end{aligned} $$
(1.2)

where \(A^{*} \) is the corresponding adjoint operator of A, the stepsize δ is a properly chosen real number, and \(P_{C} \) and \(P_{Q} \) are the metric projections from \(B_{1} \) and \(B_{2} \) onto C and Q, respectively. If \(\delta \in (0,\frac{2}{\|A\|^{2}}) \), then the CQ algorithm converges to a solution of (1.1), whenever the solution set is nonempty. However, in order to actualize the CQ algorithm, computing the operator norm of A is a very complicated work in practice.

As a prolongation of problem (1.1), the split common fixed point problem (SCFPP) has been extensively researched in recent years. The SCFPP is an inverse problem, which aims to find an element in a fixed point set so that the image under a bounded linear operator belongs to another fixed point set. More specifically, the SCFPP is looking for a \(q\in B_{1} \) such that

$$ \begin{aligned} q\in F(U) \quad \text{such that } Aq\in F(T), \end{aligned} $$
(1.3)

where \(A:B_{1}\longrightarrow B_{2} \) is the bounded linear operator, and \(U:B_{1}\longrightarrow B_{1}\), \(T:B_{2}\longrightarrow B_{2} \) are the two nonlinear operators. We denote by \(F(U) \) and \(F(T) \) the sets of fixed points of U and T, respectively. Δ denotes the set of solutions of SCFPP, that is,

$$ \begin{aligned} \Delta = \bigl\lbrace q\in F(U) : Aq\in F(T) \bigr\rbrace . \end{aligned} $$

In particularly, if T and U are both the identity operator, then the SCFPP is clearly changed to the SFP.

A typical method for solving the SCFPP is to use the following iterative algorithm:

$$ \begin{aligned} u_{n+1}=U\bigl(I-\delta A^{*}(I-T)A\bigr)u_{n},\quad n\geq 0 . \end{aligned} $$
(1.4)

It is shown in [7] that, if the stepsize \(\delta \in (0,\frac{2}{\|A\|^{2}}) \) and the operators in (1.3) are directed, then the sequence generated by algorithm (1.4) converges weakly to a solution of the SCFPP whenever such a solution exists.

Moudafi [8] introduced an iteration scheme for demicontractive mappings and obtained a weak convergence theorem for the SCFPP in Hilbert spaces. Since then, many authors have studied the SCFPP of other mappings in the frameworks of two Hilbert spaces (see, for instance, [9–13])

In 2015, Tang et al. [14] obtained a weak convergence theorem of the SCFPP for the asymptotically nonexpansive mapping S in Banach spaces of the following algorithm:

$$ \textstyle\begin{cases} z_{n}=u_{n}+\delta J_{1}^{-1}A^{*}J_{2}(T-I)Au_{n}, \\ u_{n+1}=(1-\alpha _{n})z_{n}+\alpha _{n} S^{n}z_{n}. \end{cases} $$
(1.5)

They showed that the sequence \(\{u_{n}\} \) generated by (1.5) converges weakly to a \(q\in \Delta \).

Recently, Tang et al. [15] studied and proved a strong convergence theorem for the SCFPP (1.3) in infinite dimensional real Hilbert spaces based on the viscosity approximation, a single-step regularized method working as follows:

$$ \begin{aligned} u_{n+1}=\alpha _{n} u_{n}+\beta _{n} h(u_{n})+\gamma _{n} S \bigl(I-\xi _{n} A^{*}(I-T)A\bigr)u_{n}, \quad n\geq 0 , \end{aligned} $$
(1.6)

where S and T are firmly nonexpansive mappings for which both \(I-S \) and \(I-T \) are demiclosed at zero, and \(h:H\longrightarrow H \) is an α-contraction mapping with \(\alpha \in (0, 1) \).

In this article, inspired by the above results, we consider and study the SCFPP for asymptotically nonexpansive mappings in the frameworks of two real Banach spaces. That is, we present an iterative algorithm to approximate a solution of the SCFPP and show some strong convergence theorems under appropriate conditions, which also solve some variational inequalities. Therefore, we extend the main results of Tang et al. [14] and Hong et al. [16] from Hilbert spaces to Banach spaces and from firmly nonexpensive mappings to asymptotically nonexpansive mappings. In some cases, some other results are also improved (see [8, 15, 17, 18]).

2 Preliminaries

The following are some definitions and lemmas that will be used in the proof of the main results in the next section.

Throughout this paper, let B be a real Banach space and \(B ^{*}\) be the dual space of B. The normalized duality mapping \(J: B \rightarrow 2^{B^{*}} \) is defined by

$$ J(x)=\bigl\{ f\in B^{*}:\langle x,f\rangle = \Vert x \Vert \Vert f \Vert , \Vert x \Vert = \Vert f \Vert \bigr\} ,\quad \forall x\in B, $$

where \(\langle \centerdot , \centerdot \rangle \) denotes the duality pairing. As is well known (see e.g. [6]), the operator J is well defined and J is multiple-valued and nonlinear in general. And J is an identity mapping if and only if B is a Hilbert space.

A Banach space B is said to be strictly convex if \(\frac{\|u+v\|}{2}<1 \) for \(\|u\|=\|v\|=1 \) and \(u\neq v\). The modulus of convexity of B is defined by

$$ \delta _{B}(\epsilon )=\inf \biggl\lbrace 1- \biggl\Vert \frac{1}{2}(u+v) \biggr\Vert : \Vert u \Vert , \Vert v \Vert \leq 1, \Vert u-v \Vert \geq \epsilon \biggr\rbrace , $$

for all \(0\leq \epsilon \leq 2 \). B is called uniformly convex, if for all \(0<\epsilon \leq 2 \) such that \(\delta _{B}(0)=0 \) and \(\delta _{B}(\epsilon )>0 \).

Let \(\rho _{B}:[0,+\infty )\longrightarrow [0,+\infty )\) be the modulus of smoothness of B which is defined by

$$ \rho _{B}(s)=\sup \biggl\lbrace \frac{1}{2}\bigl( \Vert u+v \Vert + \Vert u-v \Vert \bigr)-1: \Vert u \Vert =1, \Vert v \Vert \leq s \biggr\rbrace . $$

A Banach space B is called uniformly smooth if \(\frac{\rho _{B}(s)}{s}\rightarrow 0 \) as \(s\rightarrow 0 \). Then a Banach space B is called q-uniformly smooth, if, for all \(s>0 \), there exists a constant \(c>0 \) such that \(\rho _{B}(s)\geq cs^{q} \). It is known that every q-uniformly smooth Banach space is uniformly smooth.

Definition 2.1

Let C be a nonempty closed convex subset of a Banach space B and \(T:C\longrightarrow C\) be a mapping, then

  • T is called a contraction if there exists a constant \(k\in (0,1)\) satisfying

    $$ \bigl\Vert T(u)-T(v) \bigr\Vert \leq k \Vert u-v \Vert ,\quad \forall u, v\in C. $$

  • T is called a nonexpansive mapping if the above inequality is also true for \(k=1\).

  • T is called a firmly nonexpansive mapping if

    $$ \Vert Tu-Tv \Vert ^{2}\leq \langle Tu-Tv,u-v\rangle . $$

  • T is called an asymptotically nonexpansive mapping, if, for all \(u, v \in C\), there exists a sequence \(\{k_{n}\}\) with \(\lim_{n\rightarrow \infty } k_{n} = 1\) such that

    $$ \bigl\Vert T^{n} u-T^{n} v \bigr\Vert \leq k_{n} \Vert u-v \Vert . $$

It is easy to see that every nonexpansive mapping is an asymptotically nonexpansive mapping.

Definition 2.2

Let C be a nonempty closed convex subset of a real Banach space B, mapping \(U:C\longrightarrow B\) is said to be uniformly regular if

$$ \begin{aligned} \lim_{n\rightarrow \infty }\sup_{u\in C} \bigl\Vert U^{n+1}u-U^{n}u \bigr\Vert =0. \end{aligned} $$

Lemma 2.1

([19])

If B is a 2-uniformly smooth Banach space with the best smoothness constant \(t>0 \), we have the relation

$$ \Vert u+v \Vert ^{2}\leq \Vert u \Vert ^{2}+2\langle v,Ju\rangle +2 \Vert tv \Vert ^{2}. $$

Lemma 2.2

([20])

Let \(\{w_{n}\}\) and \(\{z_{n}\}\) be bounded sequences in a Banach space B and \(\{\alpha _{n}\}\) be a sequence in \([0,1]\) with \(0<\liminf_{n \rightarrow \infty }\alpha _{n}\leq \limsup_{n\rightarrow \infty } \alpha _{n}<1\). Suppose that \(w_{n+1}=(1-\alpha _{n})w_{n}+\alpha _{n}z_{n}\) for all \(n\geq 0\) and \(\limsup_{n\rightarrow \infty }(\|z_{n+1}-z_{n}\|- \|w_{n+1}-w_{n}\|)\leq 0\). Then \(\lim_{n\rightarrow \infty }\|z_{n}-w_{n} \|=0\).

Lemma 2.3

([21])

Let C be a nonempty bounded and closed convex subset of a reflexive smooth Banach space B and J be a weakly sequential continuous normal duality mapping \(T: C \rightarrow C\) be an asymptotical nonexpansive mapping. Then \(I-T\) is demiclosed at zero, i.e., if \(u_{n}\rightharpoonup u\) weakly and \(u_{n}-Tu_{n} \rightarrow 0\) strongly, then \(u\in F(T)\).

Lemma 2.4

([22])

Assume \(\{p_{n}\}\) is a sequence of nonnegative real numbers such that

$$ p_{n+1}\leq (1-\sigma _{n})p_{n}+\xi _{n}, \quad n\geq 0, $$

where \(\{\sigma _{n}\}\) is a sequence in \((0,1)\) and \(\{\xi _{n}\}\) is a real sequence such that

$$ \begin{aligned} &(1)\quad \lim_{n\rightarrow \infty }\sigma _{n}=0 \quad \textit{and}\quad \sum ^{\infty }_{n=0}\sigma _{n}=\infty ; \\ &(2)\quad \limsup_{n\rightarrow \infty } \frac{\xi _{n}}{\sigma _{n}}\leq 0\quad \textit{or} \quad \sum ^{\infty }_{n=1} \vert \xi _{n} \vert < \infty . \end{aligned} $$

Then \(\lim_{n\rightarrow \infty }p_{n}=0\).

3 Main results

Theorem 3.1

Let \(B_{1} \) be a real strictly convex and 2-uniformly smooth Banach space with the best smoothness constant t satisfying \(0< t<\frac{1}{\sqrt{2}} \) and a weakly sequential continuous normal duality mapping J, \(B_{2} \) be a real smooth Banach space. Suppose that \(h:B_{1}\longrightarrow B_{1} \) is a contraction mapping with contractive coefficient \(k\in (0,1) \) and \(A:B_{1}\longrightarrow B_{2} \) is a bounded linear operator and \(A^{*} \) is the adjoint of A. Let \(T:B_{2}\longrightarrow B_{2} \) be a nonexpansive mapping and \(U:B_{1}\longrightarrow B_{1} \) be an asymptotically nonexpansive mapping with asymptotical coefficient sequence \(\{k_{n}\} \) and \(F(U)\neq \emptyset \). Assume that the SCFPP (1.3) has a nonempty solution set Δ and U is uniformly regular in Δ. Let \(\{u_{n}\} \) be a sequence generated by

$$ \textstyle\begin{cases} u_{1}\in B_{1}, \\ v_{n}=u_{n}-\delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(I-T)Au_{n}, \\ u_{n+1}=\alpha _{n} u_{n}+\gamma _{n} h(u_{n})+\zeta _{n} U^{n}v_{n}, \end{cases} $$
(3.1)

where \(\{\alpha _{n}\},\{\gamma _{n}\},\{\zeta _{n}\}\subset (0,1) \), satisfying the following conditions:

$$ \begin{aligned} &(i) \quad \alpha _{n}+\gamma _{n}+ \zeta _{n}=1,\qquad 0< \liminf_{n \rightarrow \infty }\alpha _{n}\leq \limsup_{n\rightarrow \infty } \alpha _{n}< 1; \\ &(ii) \quad \lim_{n\rightarrow \infty }\gamma _{n}=0,\qquad \sum ^{\infty }_{n=0} \gamma _{n}=\infty ,\qquad \eta \gamma _{n}=k_{n}-1,\qquad 0< \eta < 1-k; \\ &(iii) \quad \delta \in \biggl(0,\frac{1-2t^{2}}{ \Vert A \Vert ^{2}}\biggr). \end{aligned} $$

Then the sequence \(\{u_{n}\} \) generated by (3.1) converges strongly to a point \(q=P_{\Delta }h(q)\in \Delta \), which also solves the variational inequality:

$$ \bigl\langle (I-h)q,j(q-w)\bigr\rangle \leq 0,\quad \forall w\in \Delta . $$

Proof

Since \(P_{\Delta }h \) is a contraction on \(B_{1} \), there exists an unique element \(q\in B_{1} \) such that \(q=P_{\Delta }h(q) \) by the Banach contraction principle. So there is a \(q\in \Delta \). Now, we split the proof into five steps.

Step 1 First we show that the sequence \(\{u_{n}\} \) is bounded. For any given \(q\in \Delta \), it follows from (3.1), condition (iii) and Lemma 2.1 that

$$ \begin{aligned} \Vert v_{n}-q \Vert ^{2}&= \bigl\Vert (u_{n}-q)+\delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(T-I)Au_{n} \bigr\Vert ^{2} \\ &\leq \delta ^{2} \bigl\Vert J_{B_{1}}^{-1}A^{*}J_{B_{2}}(T-I)Au_{n} \bigr\Vert ^{2}+2 \delta \bigl\langle u_{n}-q,A^{*}J_{B_{2}}(T-I)Au_{n} \bigr\rangle \\ &\quad {}+2t^{2} \Vert u_{n}-q \Vert ^{2} \\ &\leq \delta ^{2} \Vert A \Vert ^{2} \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2}+2t^{2} \Vert u_{n}-q \Vert ^{2} \\ &\quad {}+2\delta \bigl\langle Au_{n}-Aq,J_{B_{2}}(T-I)Au_{n} \bigr\rangle \\ &=\delta ^{2} \Vert A \Vert ^{2} \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2}+2t^{2} \Vert u_{n}-q \Vert ^{2}-2 \delta \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2} \\ &\quad {}+2\delta \bigl\langle TAu_{n}-Aq,J_{B_{2}}(T-I)Au_{n} \bigr\rangle \\ &\leq \bigl(\delta ^{2} \Vert A \Vert ^{2}-2\delta \bigr) \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2}+2t^{2} \Vert u_{n}-q \Vert ^{2} \\ &\quad {}+\delta \bigl( \Vert TAu_{n}-Aq \Vert ^{2}+ \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2}\bigr) \\ &\leq 2t^{2} \Vert u_{n}-q \Vert ^{2}+\bigl( \delta ^{2} \Vert A \Vert ^{2}-\delta \bigr) \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2}+\delta \Vert Au_{n}-Aq \Vert ^{2} \\ &\leq \bigl(2t^{2}+\delta \Vert A \Vert ^{2}\bigr) \Vert u_{n}-q \Vert ^{2}-\delta \bigl(1-\delta \Vert A \Vert ^{2}\bigr) \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2} \\ &\leq \Vert u_{n}-q \Vert ^{2}. \end{aligned} $$
(3.2)

Because \(0\leq k<1 \), by (3.1), (3.2) and condition (ii), we have

$$\begin{aligned} \Vert u_{n+1}-q \Vert &= \bigl\Vert \alpha _{n} u_{n}+\gamma _{n} h(u_{n})+ \zeta _{n} U^{n}v_{n}-q \bigr\Vert \\ &=\|\alpha _{n}(u_{n}-q)+\gamma _{n}(h(u_{n})-h(q)+ \gamma _{n}\bigl(h(q)-q\bigr)+ \zeta _{n}\bigl(U^{n}v_{n}-q \bigr)\| \\ &\leq \alpha _{n} \Vert u_{n}-q \Vert +k\gamma _{n} \Vert u_{n}-q \Vert +\gamma _{n} \bigl\Vert h(q)-q \bigr\Vert +k_{n}\zeta _{n} \Vert v_{n}-q \Vert \\ &\leq (\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}) \Vert u_{n}-q \Vert +\gamma _{n} \bigl\Vert h(q)-q \bigr\Vert \\ &= \bigl(1- \bigl(\gamma _{n}(1-k)-\zeta _{n}(k_{n}-1) \bigr) \bigr) \Vert u_{n}-q \Vert +\gamma _{n} \bigl\Vert h(q)-q \bigr\Vert \\ &\leq \bigl(1-\bigl(\gamma _{n}(1-k)-\eta \gamma _{n}\bigr) \bigr) \Vert u_{n}-q \Vert + \gamma _{n} \bigl\Vert h(q)-q \bigr\Vert \\ &= \bigl(1-\gamma _{n}(1-k-\eta ) \bigr) \Vert u_{n}-q \Vert +\gamma _{n}(1-k-\eta ) \frac{ \Vert h(q)-q \Vert }{1-k-\eta } \\ &\leq \max \biggl\lbrace \Vert u_{n}-q \Vert ,\frac{ \Vert h(q)-q \Vert }{1-k-\eta } \biggr\rbrace . \end{aligned}$$
(3.3)

By induction, we readily obtain

$$ \Vert u_{n+1}-q \Vert \leq \max \biggl\lbrace \Vert u_{0}-q \Vert , \frac{ \Vert h(q)-q \Vert }{1-k-\eta } \biggr\rbrace . $$

This implies that \(\{u_{n}\} \) is bounded, and so are \(\{v_{n}\} \), \(\{h(u_{n})\}\), \(\{U^{n}v_{n}\}\).

Step 2 We show that \(\lim_{n\rightarrow \infty }\|u_{n+1}-u_{n}\|=0 \) and \(\lim_{n\rightarrow \infty }\|v_{n+1}-v_{n}\|=0 \). To see this, we set \(z_{n}=\frac{u_{n+1}-\alpha _{n} u_{n}}{1-\alpha _{n}} \), \(\forall n \geq 0 \). We have

$$ \begin{aligned} z_{n+1}-z_{n}&= \frac{u_{n+2}-\alpha _{n+1} u_{n+1}}{1-\alpha _{n+1}}- \frac{u_{n+1}-\alpha _{n} u_{n}}{1-\alpha _{n}} \\ &= \frac{\gamma _{n+1}h(u_{n+1})+\zeta _{n+1}U^{n+1}v_{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}h(u_{n})+\zeta _{n}U^{n}v_{n}}{1-\alpha _{n}} \\ &= \frac{\gamma _{n+1}h(u_{n+1})+(1-\alpha _{n+1}-\gamma _{n+1})U^{n+1}v_{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}h(u_{n})+(1-\alpha _{n}-\gamma _{n})U^{n}v_{n}}{1-\alpha _{n}} \\ &=\frac{\gamma _{n+1}}{1-\alpha _{n+1}} \bigl[h(u_{n+1})-h(u_{n}) \bigr]+\biggl( \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}}\biggr)h(u_{n}) \\ &\quad {}-\biggl(\frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}}\biggr)U^{n}v_{n}+ \biggl(1- \frac{\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \bigl(U^{n+1}v_{n+1}-U^{n}v_{n} \bigr) \\ &=\frac{\gamma _{n+1}}{1-\alpha _{n+1}} \bigl[h(u_{n+1})-h(u_{n}) \bigr]+\biggl( \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}}\biggr) \bigl(h(u_{n})-U^{n}v_{n} \bigr) \\ &\quad {}+\biggl(1-\frac{\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \bigl(U^{n+1}v_{n+1}-U^{n+1}v_{n} \bigr)+\biggl(1- \frac{\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \bigl(U^{n+1}v_{n}-U^{n}v_{n} \bigr). \end{aligned} $$

This implies that

$$ \begin{aligned} \Vert z_{n+1}-z_{n} \Vert &\leq \frac{k\gamma _{n+1}}{1-\alpha _{n+1}} \Vert u_{n+1}-u_{n} \Vert + \biggl\vert \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}} \biggr\vert \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert \\ &\quad {}+\sup_{v\in \Delta } \bigl\Vert U^{n+1}v-U^{n}v \bigr\Vert +k_{n+1}\biggl(1- \frac{\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \Vert v_{n+1}-v_{n} \Vert . \end{aligned} $$
(3.4)

By Lemma 2.1 and condition \((iii) \), we can get

$$\begin{aligned} \Vert v_{n+1}-v_{n} \Vert ^{2}&= \bigl\Vert (u_{n+1}-u_{n})- \bigl( \delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(I-T)Au_{n+1}- \delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(I-T)Au_{n} \bigr) \bigr\Vert ^{2} \\ &\leq \delta ^{2} \Vert A \Vert ^{2} \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2}+2\delta ^{2} \Vert u_{n+1}-u_{n} \Vert ^{2} \\ &\quad {}-2\delta \bigl\langle Au_{n+1}-Au_{n},J_{B_{2}}(I-T)Au_{n+1}-J_{B_{2}}(I-T)Au_{n} \bigr\rangle \\ &= \delta ^{2} \Vert A \Vert ^{2} \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2}+2\delta ^{2} \Vert u_{n+1}-u_{n} \Vert ^{2} \\ &\quad {}-2\delta \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2} \\ &\quad {}+2\delta \bigl\langle TAu_{n+1}-TAu_{n},J_{B_{2}}(I-T)Au_{n+1}-J_{B_{2}}(I-T)Au_{n} \bigr\rangle \\ &\leq \bigl(\delta ^{2} \Vert A \Vert ^{2}-2\delta \bigr) \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2}+2 \delta ^{2} \Vert u_{n+1}-u_{n} \Vert ^{2} \\ &\quad {}+\delta \bigl( \Vert TAu_{n+1}-TAu_{n} \Vert ^{2}+ \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2}\bigr) \\ &\leq \bigl(\delta ^{2} \Vert A \Vert ^{2}-\delta \bigr) \bigl\Vert (I-T)Au_{n+1}-(I-T)Au_{n} \bigr\Vert ^{2}+2 \delta ^{2} \Vert u_{n+1}-u_{n} \Vert ^{2} \\ &\quad {}+\delta \Vert A \Vert ^{2} \Vert u_{n+1}-u_{n} \Vert ^{2} \\ &=\bigl(2\delta ^{2}+\delta \Vert A \Vert ^{2}\bigr) \Vert u_{n+1}-u_{n} \Vert ^{2}-\delta \bigl(1- \delta \Vert A \Vert ^{2}\bigr) \bigl\Vert (I-T) (Au_{n+1}-Au_{n}) \bigr\Vert ^{2} \\ &\leq \Vert u_{n+1}-u_{n} \Vert ^{2}. \end{aligned}$$
(3.5)

Thus, we have \(\|y_{n+1}-y_{n}\|\leq \|x_{n+1}-x_{n}\| \). Then, from (3.4), (3.5) and condition \((ii) \), it follows that

$$ \begin{aligned} \Vert z_{n+1}-z_{n} \Vert &\leq \biggl( \frac{k\gamma _{n+1}}{1-\alpha _{n+1}}+k_{n+1}\biggl(1- \frac{\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \biggr) \Vert u_{n+1}-u_{n} \Vert \\ &\quad {}+ \biggl\vert \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}} \biggr\vert \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert +\sup _{v \in \Delta } \bigl\Vert U^{n+1}v-U^{n}v \bigr\Vert \\ &=\frac{k\gamma _{n+1}+k_{n+1}\zeta _{n+1}}{1-\alpha _{n+1}} \Vert u_{n+1}-u_{n} \Vert \\ &\quad {}+ \biggl\vert \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}} \biggr\vert \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert +\sup _{v \in \Delta } \bigl\Vert U^{n+1}v-U^{n}v \bigr\Vert \\ &\leq \biggl(1-\frac{(1-k-\eta )\gamma _{n+1}}{1-\alpha _{n+1}}\biggr) \Vert u_{n+1}-u_{n} \Vert \\ &\quad {}+ \biggl\vert \frac{\gamma _{n+1}}{1-\alpha _{n+1}}- \frac{\gamma _{n}}{1-\alpha _{n}} \biggr\vert \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert +\sup _{v \in \Delta } \bigl\Vert U^{n+1}v-U^{n}v \bigr\Vert . \end{aligned} $$

Therefore, by condition \((ii) \), we have

$$ \begin{aligned} \limsup_{n\rightarrow \infty }\bigl( \Vert z_{n+1}-z_{n} \Vert - \Vert u_{n+1}-u_{n} \Vert \bigr)\leq 0. \end{aligned} $$

It follows form Lemma 2.2 and condition \((i) \) that

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert z_{n}-u_{n} \Vert = 0. \end{aligned} $$

Note that \(z_{n}=\frac{u_{n+1}-\alpha _{n} u_{n}}{1-\alpha _{n}} \), it is easy to see that

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert u_{n+1}-u_{n} \Vert =0. \end{aligned} $$
(3.6)

Clearly, from (3.5) we obtain

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert v_{n+1}-v_{n} \Vert = 0. \end{aligned} $$

Step 3 We prove that \(\|u_{n}-Uu_{n}\|\rightarrow 0 \), as \(n\rightarrow \infty \). We have

$$ \begin{aligned} \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert &= \bigl\Vert U^{n}v_{n}u_{n}+\gamma _{n} h(u_{n})- \alpha _{n}U^{n}v_{n}- \gamma _{n}U^{n}v_{n} \bigr\Vert \\ &= \bigl\Vert \alpha _{n}\bigl(u_{n}-U^{n}v_{n} \bigr)+\gamma _{n}\bigl(h(u_{n})-U^{n}v_{n} \bigr) \bigr\Vert \\ &\leq \alpha _{n} \Vert u_{n}-u_{n+1} \Vert + \alpha _{n} \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert + \gamma _{n} \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert . \end{aligned} $$

We can get

$$ \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert \leq \frac{\alpha _{n}}{1-\alpha _{n}} \Vert u_{n+1}-u_{n} \Vert + \frac{\gamma _{n}}{1-\alpha _{n}} \bigl\Vert h(u_{n})-U^{n}v_{n} \bigr\Vert . $$

By (3.6) and condition \((ii) \), we have

$$ \begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert =0. \end{aligned} $$
(3.7)

By (3.2) and condition \((iii) \), we obtain

$$ \begin{aligned} \Vert u_{n+1}-q \Vert ^{2}&= \bigl\Vert \alpha _{n} u_{n}+\gamma _{n} h(u_{n})+ \zeta _{n} U^{n}v_{n}-q \bigr\Vert ^{2} \\ &\leq \alpha _{n} \Vert u_{n}-q \Vert ^{2}+ \gamma _{n} \bigl\Vert h(u_{n})-q \bigr\Vert ^{2}+\zeta _{n} \bigl\Vert U^{n}v_{n}-q \bigr\Vert ^{2} \\ &\leq \alpha _{n} \Vert u_{n}-q \Vert ^{2}+ \gamma _{n} \bigl\Vert h(u_{n})-q \bigr\Vert ^{2}+\zeta _{n} k_{n}^{2} \Vert v_{n}-q \Vert ^{2} \\ &\leq \alpha _{n} \Vert u_{n}-q \Vert ^{2}+ \gamma _{n} \bigl\Vert h(u_{n})-q \bigr\Vert ^{2} \\ &\quad {}+\zeta _{n} k_{n}^{2} \bigl[ \bigl(2t^{2}+\delta \Vert A \Vert ^{2}\bigr) \Vert u_{n}-q \Vert ^{2}-\delta \bigl(1-\delta \Vert A \Vert ^{2}\bigr) \bigl\Vert (T-I)Au_{n} \bigr\Vert ^{2} \bigr] \\ &\leq \bigl(\alpha _{n}+\zeta _{n} k_{n}^{2} \bigr) \Vert u_{n}-q \Vert ^{2} +\gamma _{n} \bigl\Vert h(u_{n})-q \bigr\Vert ^{2} \\ &\quad {}-\zeta _{n} k_{n}^{2}\delta \bigl(1- \delta \Vert A \Vert ^{2}\bigr) \Vert TAu_{n}-Au_{n} \Vert ^{2}. \end{aligned} $$

By the last inequality and condition (i) we can get

$$ \begin{aligned} \Vert TAu_{n}-Au_{n} \Vert ^{2}&\leq \frac{ (\alpha _{n}+\zeta _{n} k_{n}^{2}-1 ) \Vert u_{n}-q \Vert ^{2}+\gamma _{n} \Vert h(u_{n})-q \Vert ^{2}}{\zeta _{n} k_{n}^{2}\delta (1-\delta \Vert A \Vert ^{2})} \\ &\quad {}+ \frac{( \Vert u_{n}-q \Vert + \Vert u_{n+1}-q \Vert ) \Vert u_{n+1}-u_{n} \Vert }{\zeta _{n} k_{n}^{2}\delta (1-\delta \Vert A \Vert ^{2})} \\ &= \frac{ (\zeta _{n} (k_{n}^{2}-1)-\gamma _{n} ) \Vert u_{n}-q \Vert ^{2}+\gamma _{n} \Vert h(u_{n})-q \Vert ^{2}}{\zeta _{n} k_{n}^{2}\delta (1-\delta \Vert A \Vert ^{2})} \\ &\quad {}+ \frac{( \Vert u_{n}-q \Vert + \Vert u_{n+1}-q \Vert ) \Vert u_{n+1}-u_{n} \Vert }{\zeta _{n} k_{n}^{2}\delta (1-\delta \Vert A \Vert ^{2})}. \end{aligned} $$

By condition \((ii) \) and applying Step 2, we have

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert TAu_{n}-Au_{n} \Vert =0. \end{aligned} $$
(3.8)

Considering the bounded sequence \(\{u_{n}\}\), it must have a convergent subsequence \(\{u_{n_{k}}\} \). There exists a subsequence \(\{u_{n_{k_{j}}}\} \) of \(\{u_{n_{k}}\} \) such that \(u_{n_{k_{j}}}\rightharpoonup w\in B_{1} \). Without loss of generality, we assume that \(u_{n_{k}}\rightharpoonup w \) as \(k\rightarrow \infty \). Therefore, \(Au_{n_{k}}\rightharpoonup Aw\) as \(k\rightarrow \infty \) and

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert TAu_{n_{k}}-Au_{n_{k}} \Vert =0. \end{aligned} $$

Since \(v_{n}=u_{n}-\delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(I-T)Au_{n} \), \(\|u_{n}-v_{n}\|=\|\delta J_{B_{1}}^{-1}A^{*}J_{B_{2}}(I-T)Au_{n}\| \), we can get

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert u_{n}-v_{n} \Vert =0. \end{aligned} $$
(3.9)

Moreover, we have

$$ \begin{aligned} \bigl\Vert u_{n}-U^{n}u_{n} \bigr\Vert &\leq \Vert u_{n}-u_{n+1} \Vert + \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert + \bigl\Vert U^{n}v_{n}-U^{n}u_{n} \bigr\Vert \\ &\leq \Vert u_{n+1}-u_{n} \Vert + \bigl\Vert u_{n+1}-U^{n}v_{n} \bigr\Vert +k_{n} \Vert v_{n}-u_{n} \Vert . \end{aligned} $$

In view of Step 2 and (3.7), (3.9), we obtain

$$ \begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert u_{n}-U^{n}u_{n} \bigr\Vert = 0 . \end{aligned} $$
(3.10)

Furthermore, we have

$$ \begin{aligned} \Vert u_{n}-Uu_{n} \Vert &= \bigl\Vert u_{n}-u_{n+1}+u_{n+1}-U^{n+1}u_{n+1}+U^{n+1}u_{n+1}-U^{n+1}u_{n}U^{n+1}u_{n}-Uu_{n} \bigr\Vert \\ &\leq \Vert u_{n+1}-u_{n} \Vert + \bigl\Vert u_{n+1}-U^{n+1}u_{n+1} \bigr\Vert + \bigl\Vert U^{n+1}u_{n+1}-U^{n+1}u_{n} \bigr\Vert \\ &\quad {}+ \bigl\Vert U^{n+1}u_{n}-Uu_{n} \bigr\Vert \\ &\leq \Vert u_{n+1}-u_{n} \Vert + \bigl\Vert u_{n+1}-U^{n+1}u_{n+1} \bigr\Vert +k_{n+1} \Vert u_{n+1}-u_{n} \Vert +k_{1} \bigl\Vert U^{n} u_{n}-u_{n} \bigr\Vert \\ &=(1+k_{n+1}) \Vert u_{n+1}-u_{n} \Vert + \bigl\Vert u_{n+1}-U^{n+1}u_{n+1} \bigr\Vert +k_{1} \bigl\Vert U^{n} u_{n}-u_{n} \bigr\Vert . \end{aligned} $$

By (3.6) and (3.10), we can get

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert u_{n}-Uu_{n} \Vert =0. \end{aligned} $$
(3.11)

Step 4 Since \(B_{1} \) is a reflexive Banach space and \(\{u_{n}\} \) is bounded, there exists a subsequence \(u_{n_{k}}\rightharpoonup w\in B_{1} \) as \(n\rightarrow \infty \). And

$$ \begin{aligned} \lim_{n\rightarrow \infty }\bigl\langle (I-h)q,j(q-u_{n_{k}}) \bigr\rangle =\limsup_{n\rightarrow \infty }\bigl\langle (I-h)q,j(q-u_{n}) \bigr\rangle . \end{aligned} $$

By Step 3, we know \(Au_{n}\rightharpoonup Aw \). Because \(B_{1} \) and \(B_{2} \) are reflexive smooth Banach spaces, it follows from Step 3 and Lemma 2.3, that \(Aw\in F(T) \). That is, \(w\in \Delta \).

On the other hand, since \(q\in \Delta \) satisfies

$$ \bigl\langle (I-h)q, j(q-w)\bigr\rangle \leq 0,\quad \forall w\in \Delta , $$

and because J is a weakly sequential continuous duality mapping, we obtain

$$ \begin{aligned} \limsup_{n\rightarrow \infty }\bigl\langle (I-h)q, j(q-u_{n}) \bigr\rangle &=\lim_{k\rightarrow \infty } \bigl\langle (I-h)q, j(q-u_{n_{k}}) \bigr\rangle \\ &=\bigl\langle (I-h)q, j(q-w)\bigr\rangle \leq 0. \end{aligned} $$
(3.12)

Step 5 Finally, we prove that \(\{u_{n}\} \) converges strongly to \(q\in \Delta \). We have

$$ \begin{aligned} \Vert u_{n+1}-q \Vert ^{2}&=\bigl\langle \alpha _{n} u_{n}+\gamma _{n} h(u_{n})+ \zeta _{n} U^{n}v_{n}-q,j(u_{n+1}-q) \bigr\rangle \\ &=\alpha _{n}\bigl\langle u_{n}-q,j(u_{n+1}-q)\bigr\rangle +\gamma _{n}\bigl\langle h(u_{n})-q,j(u_{n+1}-q) \bigr\rangle \\ &\quad {}+\zeta _{n}\bigl\langle U^{n}v_{n}-q,j(u_{n+1}-q) \bigr\rangle \\ &\leq \alpha _{n}\bigl\langle u_{n}-q,j(u_{n+1}-q) \bigr\rangle +\gamma _{n} \bigl\langle h(u_{n})-h(q),j(u_{n+1}-q) \bigr\rangle \\ &\quad {}+\gamma _{n}\bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle +\zeta _{n} \bigl\langle U^{n}v_{n}-q,j(u_{n+1}-q) \bigr\rangle \\ &\leq \alpha _{n} \Vert u_{n}-q \Vert \Vert u_{n+1}-q \Vert +k\gamma _{n} \Vert u_{n}-q \Vert \Vert u_{n+1}-q \Vert \\ &\quad {}+\gamma _{n}\bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle +k_{n}\zeta _{n} \Vert v_{n}-q \Vert \Vert u_{n+1}-q \Vert \\ &\leq (\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}) \Vert u_{n}-q \Vert \Vert u_{n+1}-q \Vert +\gamma _{n}\bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle \\ &\leq \frac{\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}}{2} \Vert u_{n}-q \Vert ^{2}+ \frac{\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}}{2} \Vert u_{n+1}-q \Vert ^{2} \\ &\quad {}+\gamma _{n}\bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle . \end{aligned} $$

This implies that

$$ \begin{aligned} \biggl(1-\frac{\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}}{2}\biggr) \Vert u_{n+1}-q \Vert ^{2}&\leq \frac{\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}}{2} \Vert u_{n}-q \Vert ^{2} \\ &\quad {}+\gamma _{n}\bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle . \end{aligned} $$

That is,

$$ \begin{aligned} \Vert u_{n+1}-q \Vert ^{2}&\leq \frac{\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n}}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} \Vert u_{n}-q \Vert ^{2} \\ &\quad {}+ \frac{2\gamma _{n}}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} \bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle \\ &\leq \biggl(1- \frac{2\gamma _{n}(1-k-\eta )}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})}\biggr) \Vert u_{n}-q \Vert ^{2} \\ &\quad {}+ \frac{2\gamma _{n}}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} \bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle . \end{aligned} $$
(3.13)

Let

$$ \sigma _{n}= \frac{2\gamma _{n}(1-k-\eta )}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} ,\qquad \xi _{n}= \frac{2\gamma _{n}}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} \bigl\langle h(q)-q,j(u_{n+1}-q)\bigr\rangle . $$

By conditions \((i)\) and \((ii) \), we know that

$$ \lim_{n\rightarrow \infty }\sigma _{n}=0 \quad \text{and}\quad \sigma _{n}= \frac{2\gamma _{n}(1-k-\eta )}{2-(\alpha _{n}+k\gamma _{n}+k_{n}\zeta _{n})} \geq \gamma _{n}(1-k-\eta ). $$

Because \(\sum ^{\infty }_{n=0}\gamma _{n}=\infty \), \(\sum ^{\infty }_{n=0}\sigma _{n}=\infty \). In addition, by (3.12) we have

$$ \limsup_{n\rightarrow \infty } \frac{\xi _{n}}{\sigma _{n}}=\limsup_{n \rightarrow \infty } \frac{\langle h(q)-q,j(u_{n+1}-q)\rangle }{1-k-\eta }\leq 0 . $$

Thus, applying Lemma 2.4 and (3.13), we conclude that

$$ \begin{aligned} \lim_{n\rightarrow \infty } \Vert u_{n}-q \Vert =0. \end{aligned} $$

This completes the proof. □

Remark 3.1

We know that each firmly nonexpansive mapping is a nonexpansive mapping and every nonexpansive mapping is an asymptotically nonexpansive mapping. In this paper, we research the SCFPP for asymptotically nonexpansive mappings in 2-uniformly Banach space. So there are five features to explain in detail:

  1. 1

    If \(\gamma _{n}=0\) in Theorem 3.1, then \(\{u_{n}\}\) converges strongly to a fixed point of U. It is the main result of Tang et al. [14].

  2. 2

    Since Hilbert space, \(L^{p}(1< p\leq 2) \) space, etc. are 2-uniformly convex spaces, if U is a firmly nonexpansive mapping and \(B_{1}\), \(B_{2}\) are Hilbert spaces in Theorem 3.1, then we obtain the main results of Tang et al. [15].

  3. 3

    If \(g(u_{n})=c \), U is a firmly nonexpansive mapping and \(B_{1}\), \(B_{2}\) are Hilbert spaces in the iterative sequence \(\{u_{n}\} \) in Theorem 3.1, then we obtain the main results of Hong et al. [16].

  4. 4

    If U is a nonexpansive mapping and \(B_{1} \) is a Hilbert space in Theorem 3.1, then we obtain the main results of Tang et al. [17].

  5. 5

    It is well known that a firmly nonexpansive mapping includes resolvents and projection operators. Let C, D be nonempty closed convex subsets of \(B_{1}\), \(B_{2} \), respectively. When \(U^{n}=P_{C}\), \(T=P_{Q} \), then (3.1) can also solve the split feasibility problem. That is, our Theorem 3.1 generalizes and improves the main results of Deepho J and Kuman P [18].