Hybrid simultaneous algorithms for the split equality problem with applications

Abstract

The split equality problem has board applications in many areas of applied mathematics. Many researchers studied this problem and proposed various algorithms to solve it. From the literature we know that most algorithms for the split equality problems came from the idea of the projected Landweber algorithm proposed by Byrne and Moudafi (Working paper UAG, 2013), and few algorithms came from the idea of the alternating CQ-algorithm given by Moudafi (Nonlinear Anal. 79:117-121, 2013). Hence, it is important and necessary to give new algorithms from the idea of the alternating CQ-algorithm. In this paper, we first present a hybrid projected Landweber algorithm to study the split equality problem. Next, we propose a hybrid alternating CQ-algorithm to study the split equality problem. As applications, we consider the split feasibility problem and linear inverse problem. Finally, we give numerical results for the split feasibility problem to demonstrate the efficiency of the proposed algorithms.

1 Introduction

Let H be a real Hilbert space with inner product $\langle\cdot ,\cdot\rangle$ and norm $\|\cdot\|$. We denote the strong convergence and weak convergence of $\{x_{n}\}_{n\in\mathbb{N}}$ to $x\in H$ by $x_{n}\to x$ and $x_{n}\rightharpoonup x$, respectively. The symbols $\mathbb{N}$ and $\mathbb{R}$ are used to denote the sets of positive integers and real numbers, respectively. For each $x\in H$, there is a unique element $\bar{x}\in C$ such that $\|x-\bar {x}\|=\min_{y\in C}\|x-y\|$. In this study, we set $P_{C}x=\bar{x}$, and $P_{C}$ is called the metric projection from H onto C.

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces. Let $A:H_{1}\rightarrow H_{2}$ and $A^{*}:H_{2}\rightarrow H_{1}$ be two linear and bounded operators. Then $A^{*}$ is called the adjoint of A if $\langle Az,w\rangle=\langle z,A^{*}w\rangle$ for all $z\in H_{1}$ and $w\in H_{2}$. It is known that the adjoint operator of a linear and bounded operator on a Hilbert space always exists and is linear, bounded, and unique. Further, we know that $\|A\|=\|A^{*}\|$.

Let $H_{1}$, $H_{2}$, and $H_{3}$ be real Hilbert spaces. Let C and Q be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{3}$ and $B:H_{2}\rightarrow H_{3}$ be linear and bounded operators with adjoint operators $A^{* }$ and $B^{* }$, respectively. The following problem is the split equality problem, which was studied by Moudafi [2, 3]:

$$ (\mathrm{SEP})\quad \mbox{Find }\bar{x}\in C\mbox{ and }\bar{y}\in Q\mbox{ such that }A\bar {x}=B\bar{y}. $$

Let $\Omega:=\{(x,y)\in C\times Q: Ax=By\}$ be the solution set of problem (SEP). Further, we observed that $(x,y)$ is a solution of the split equality problem if and only if

$$ \left \{ \textstyle\begin{array}{l} x=P_{C}(x-\rho_{1}A^{* }(Ax-By)), \\ y=P_{Q}(y+\rho_{2}B^{* }(Ax-By)), \end{array}\displaystyle \right . $$

for all $\rho_{1}>0$ and $\rho_{2}>0$ (for details, see [4]).

As mentioned by Moudafi [2], the interest of the split equality problem covers many situations, for instance, in decomposition methods for PDEs, game theory, and intensity modulated radiation therapy (IMRT). For details, see [2, 5, 6]. To solve problem (SEP), Moudafi [3] proposed the alternating CQ-algorithm:

$$ (\mathrm{ACQA})\quad \left \{ \textstyle\begin{array}{l} x_{n+1}:=P_{C}(x_{n}-\rho_{n}A^{* }(Ax_{n}-By_{n})), \\ y_{n+1}:=P_{Q}(y_{n}+\rho_{n}B^{* }(Ax_{n+1}-By_{n})),\quad n\in\mathbb{N}, \end{array}\displaystyle \right . $$

where $H_{1}=\mathbb{R}^{N}$, $H_{2}=\mathbb{R}^{M}$, $P_{C}$ is the metric projection mapping from $H_{1}$ onto C, and $P_{Q}$ is the metric projection mapping from $H_{2}$ onto Q, $\varepsilon>0$, A is a $J\times N$ matrix, B is a $J\times M$ matrix, $\lambda_{A}$ and $\lambda_{B}$ are the spectral radii of $A^{* }A$ and $B^{* }B$, respectively, and $\{\rho_{n}\}$ is a sequence in $( \varepsilon,\min \{ \frac{1}{\lambda_{A}},\frac {1}{\lambda_{B}} \} -\varepsilon )$.

In 2013, Byrne and Moudafi [1] presented a simultaneous algorithm, which was called the projected Landweber algorithm, to study the split equality problem

$$ (\mathrm{PLA})\quad \left \{ \textstyle\begin{array}{l} x_{n+1}:=P_{C}(x_{n}-\rho_{n}A^{* }(Ax_{n}-By_{n})), \\ y_{n+1}:=P_{Q}(y_{n}+\rho_{n}B^{* }(Ax_{n}-By_{n})),\quad n\in\mathbb{N}, \end{array}\displaystyle \right . $$

where $H_{1}=\mathbb{R}^{N}$, $H_{2}=\mathbb{R}^{M}$, $P_{C}$ is the metric projection mapping from $H_{1}$ onto C, and $P_{Q}$ is the metric projection mapping from $H_{2}$ onto Q, $\varepsilon>0$, A is a $J\times N$ matrix, B is a $J\times M$ matrix, $\lambda_{A}$ and $\lambda_{B}$ are the spectral radii of $A^{* }A$ and $B^{* }B$, respectively, and $\{\rho_{n}\}$ is a sequence in $( \varepsilon, \frac{2}{\lambda_{A}+\lambda_{B}} )$.

Besides, we also observed that Chen et al. [7] gave the following modification of (ACQA) by using the Tikhonov regularization method and proved a convergence theorem under suitable conditions:

$$ (\mathrm{TRA}) \quad \left \{ \textstyle\begin{array}{l} x_{n+1}:=P_{C}((1-\varepsilon_{n}\rho_{n})x_{n}-\rho _{n}A^{*}(Ax_{n}-By_{n})), \\ y_{n+1}:=P_{Q}((1-\varepsilon_{n}\rho_{n})y_{n}+\rho _{n}B^{*}(Ax_{n+1}-By_{n})),\quad n\in\mathbb{N}, \end{array}\displaystyle \right . $$

where $\{\varepsilon_{n}\}_{n\in\mathbb{N}}$ is a sequence in $(0,\infty)$. Besides, many researchers studied problem (SEP) and gave various algorithms. For more details about the algorithms for the split equality problem, we refer to [8, 9] and related references.

Besides, from the literature we know that most algorithms in the literature come from the idea of the projected Landweber algorithm, and few algorithms come from the idea of the alternating CQ-algorithm. Hence, it is important and necessary to give new algorithms from the idea of the alternating CQ-algorithm. In this paper, motivated by the works mentioned on the split equality problem, we present a hybrid projected Landweber algorithm and a hybrid alternating CQ-algorithm to study the split equality problem and give convergence theorems for the proposed algorithms. As applications, we consider the split feasibility problem and linear inverse problem in real Hilbert spaces. Finally, we give numerical results for the split feasibility problem to demonstrate the efficiency of the proposed algorithms.

2 Main results

In the sequel, we need the following lemma, which is a crucial tool for our results.

Lemma 2.1

[10]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $P_{C}$ be the metric projection from H onto C. Then:

• (i) $\langle x-P_{C}x,P_{C}x-y\rangle\geq0$ for all $x\in H$ and $y\in C$;

• (ii) $\|x-P_{C}x\|^{2}+\|P_{C}x-y\|^{2}\leq\|x-y\|^{2}$ for all $x\in H$ and $y\in C$;

• (iii) $\|P_{C}x-P_{C}y\|^{2}\leq\langle x-y,P_{C}x-P_{C}y\rangle$ for all $x,y\in H$.

2.1 Hybrid projected Landweber algorithm

Let $H_{1}$, $H_{2}$, and $H_{3}$ be real Hilbert spaces with inner product $\langle\cdot,\cdot\rangle_{H_{i}}$ and norm $\|\cdot \|_{H_{i}}$, $i=1,2,3$. For simplicity, we write $\langle\cdot,\cdot\rangle$ and $\|\cdot\|$. Let C and Q be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{3}$ and $B:H_{2}\rightarrow H_{3}$ be linear and bounded operators with adjoint operators $A^{* }$ and $B^{* }$, respectively. Choose $\delta\in(0,1)$. Let Ω be the solution set of the split equality problem and suppose that $\Omega\neq\emptyset$. Let $\{\rho_{n}\}_{n\in\mathbb{N}}$ be a sequence in $(0,\infty)$.

Now we present a hybrid projected Landweber algorithm to study the split equality problem.

Algorithm 2.1

For given $x_{n}\in H_{1}$ and $y_{n}\in H_{2}$, find the approximate solution by the following iterative process.

Step 1.:

Compute the next iterate $(u_{n},v_{n})$ as follows:

$$ \left \{ \textstyle\begin{array}{l} u_{n}=P_{C}[x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n})], \\ v_{n}=P_{Q}[y_{n}+\rho_{n}B^{*}(Ax_{n}-By_{n})],\end{array}\displaystyle \right . $$

where $\rho_{n}>0$ satisfies

$$\begin{aligned}& \rho _{n}^{2}\bigl(\bigl\Vert A^{*}(Ax_{n}-By_{n})-A^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}+\bigl\Vert B^{*}(Ax_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}\bigr) \\& \quad \leq \delta \Vert x_{n}-u_{n}\Vert ^{2}+\delta \Vert y_{n}-v_{n}\Vert ^{2}, \quad 0< \delta< 1. \end{aligned}$$

(2.1)

Step 2.:

If $x_{n}=u_{n}$ and $y_{n}=v_{n}$, then $(x_{n},y_{n})$ is a solution of problem (SEP) and stop. Otherwise, go to Step 3.

Step 3.:

Compute the next iterate $(x_{n+1},y_{n+1})$ as follows:

$$ \left \{ \textstyle\begin{array}{l} D_{(n,1)}:=x_{n}-u_{n}+\rho_{n}[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n})], \\ D_{(n,2)}:=y_{n}-v_{n}-\rho_{n}[B^{*}(Au_{n}-Bv_{n})-B^{*}(Ax_{n}-By_{n})], \\ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{(n,1)}\rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle}{\|D_{(n,1)}\|^{2}+\|D_{(n,2)}\|^{2}}, \\ x_{n+1}=P_{C}[x_{n}-\alpha_{n} D_{(n,1)}], \\ y_{n+1}=P_{Q}[y_{n}-\alpha_{n} D_{(n,2)}]. \end{array}\displaystyle \right . $$

Next, update $n:=n+1$ and go to Step 1.

Remark 2.1

If $0<\rho_{n}\leq\frac{\sqrt{\delta}}{\sqrt {2}(\|A\|^{2}+\|B\|^{2})}$, then (2.1) holds.

Proof

Without loss of generality, we may assume that $x_{n}\neq u_{n}$ and $y_{n}\neq v_{n}$. We know that

$$\begin{aligned}& \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}(Ax_{n}-By_{n})-A^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}+\bigl\Vert B^{*}(Ax_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}\bigr) \\& \quad \leq \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}\bigr\Vert ^{2}+\bigl\Vert B^{*}\bigr\Vert ^{2}\bigr)\cdot\bigl\Vert (Ax_{n}-By_{n})-(Au_{n}-Bv_{n}) \bigr\Vert ^{2} \\& \quad \leq \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}\bigr\Vert ^{2}+\bigl\Vert B^{*}\bigr\Vert ^{2}\bigr)\cdot \bigl(\Vert A\Vert \cdot \Vert x_{n}-u_{n}\Vert + \Vert B\Vert \cdot \Vert y_{n}-v_{n}\Vert \bigr)^{2} \\& \quad \leq 2\rho_{n}^{2}\cdot\bigl(\Vert A\Vert ^{2}+\Vert B\Vert ^{2}\bigr)\cdot\bigl(\Vert A\Vert ^{2}\cdot \Vert x_{n}-u_{n}\Vert ^{2}+\Vert B\Vert ^{2}\cdot \Vert y_{n}-v_{n} \Vert ^{2}\bigr) \\& \quad \leq 2\rho_{n}^{2}\cdot\bigl(\Vert A\Vert ^{2}+\Vert B\Vert ^{2}\bigr)^{2}\cdot \bigl( \Vert x_{n}-u_{n}\Vert ^{2}+\Vert y_{n}-v_{n}\Vert ^{2}\bigr) \\& \quad \leq 2\cdot\frac{\delta}{2(\Vert A\Vert ^{2}+\Vert B\Vert ^{2})^{2}}\cdot \bigl(\Vert A\Vert ^{2}+ \Vert B\Vert ^{2}\bigr)^{2}\cdot\bigl(\Vert x_{n}-u_{n}\Vert ^{2}+\Vert y_{n}-v_{n}\Vert ^{2}\bigr) \\& \quad = \delta\cdot\bigl(\Vert x_{n}-u_{n}\Vert ^{2}+\Vert y_{n}-v_{n}\Vert ^{2} \bigr). \end{aligned}$$

Therefore, the proof is completed. □

Theorem 2.1

Let $\{\rho_{n}\}_{n\in\mathbb{N}}$ be a sequence in $(0,2/(\|A\|^{2}+\|B\|^{2}))$ such that (2.1) holds and assume that $\liminf_{n\rightarrow\infty}\rho_{n}(2-\rho_{n}(\|A\|^{2}+\|B\|^{2}))>0$. Then, for the sequence $\{(x_{n},y_{n})\}_{n\in\mathbb{N}}$ in Algorithm 2.1, there exists $(\bar{x},\bar{y})\in\Omega$ such that $x_{n}\rightharpoonup\bar {x}$ and $y_{n}\rightharpoonup\bar{y}$ as $n\rightarrow\infty$.

Proof

Take any $n\in\mathbb{N}$ and let n be fixed. Take any $(\bar{u},\bar{v})\in\Omega$ and let $(\bar{u},\bar{v})$ be fixed. Then $\bar{u}\in C$, $\bar{v}\in Q$, and $A\bar{u}=B\bar{v}$. First, we set

$$ \left \{ \textstyle\begin{array}{l} \varepsilon_{n,1}:=\rho_{n}[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n})], \\ \varepsilon_{n,2}:=\rho_{n}[B^{*}(Ax_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n})]. \end{array}\displaystyle \right . $$

Then

$$\begin{aligned}& \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle \\& \quad = \langle x_{n}-u_{n},x_{n}-u_{n}+ \varepsilon_{n,1}\rangle+\langle y_{n}-v_{n},y_{n}-v_{n}+ \varepsilon_{n,2}\rangle \\& \quad = \Vert x_{n}-u_{n}\Vert ^{2}+\langle x_{n}-u_{n},\varepsilon_{n,1}\rangle+\Vert y_{n}-v_{n}\Vert ^{2}+\langle y_{n}-v_{n},\varepsilon_{n,2}\rangle \\& \quad = \frac{1}{2}\Vert x_{n}-u_{n}\Vert ^{2}+\langle x_{n}-u_{n},\varepsilon _{n,1}\rangle+\frac{1}{2}\Vert x_{n}-u_{n} \Vert ^{2} \\& \qquad {} +\frac{1}{2}\Vert y_{n}-v_{n}\Vert ^{2}+\langle y_{n}-v_{n},\varepsilon_{n,2} \rangle +\frac{1}{2}\Vert y_{n}-v_{n}\Vert ^{2} \\& \quad \geq \frac{1}{2}\Vert x_{n}-u_{n}\Vert ^{2}+\langle x_{n}-u_{n},\varepsilon _{n,1}\rangle+ \frac{1}{2}\Vert \varepsilon_{n,1} \Vert ^{2} \\& \qquad {} +\frac{1}{2}\Vert y_{n}-v_{n}\Vert ^{2}+\langle y_{n}-v_{n},\varepsilon_{n,2} \rangle +\frac{1}{2}\Vert \varepsilon_{n,2}\Vert ^{2} \\& \quad = \frac{1}{2}\Vert x_{n}-u_{n}+ \varepsilon_{n,1}\Vert ^{2}+\frac{1}{2}\Vert y_{n}-v_{n}+\varepsilon_{n,2}\Vert ^{2} \\& \quad = \frac{1}{2}\Vert D_{(n,1)}\Vert ^{2}+ \frac{1}{2}\Vert D_{(n,2)}\Vert ^{2}. \end{aligned}$$

(2.2)

By (2.2) we know that

$$ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{(n,1)}\rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle}{\|D_{(n,1)}\|^{2}+\|D_{(n,2)}\|^{2}}\geq\frac{1}{2}. $$

(2.3)

Next, by Lemma 2.1 we know that

$$ \Vert x_{n}-\alpha_{n} D_{(n,1)}-x_{n+1}\Vert ^{2}+\Vert x_{n+1}- \bar{u}\Vert ^{2}\leq \Vert x_{n}-\alpha_{n} D_{(n,1)}-\bar{u}\Vert ^{2} $$

(2.4)

and

$$ \Vert y_{n}-\alpha_{n} D_{(n,2)}-y_{n+1}\Vert ^{2}+\Vert y_{n+1}- \bar{v}\Vert ^{2}\leq \Vert y_{n}-\alpha_{n} D_{(n,2)}-\bar{v}\Vert ^{2}. $$

(2.5)

Hence, by (2.4),

$$\begin{aligned}& \Vert x_{n}-\bar{u}\Vert ^{2}-\Vert x_{n+1}-\bar{u}\Vert ^{2} \\& \quad \geq \Vert x_{n}-\bar{u}\Vert ^{2}-\Vert x_{n}-\alpha_{n} D_{(n,1)}-\bar {u}\Vert ^{2}+\Vert x_{n+1}-x_{n}+\alpha_{n} D_{(n,1)}\Vert ^{2} \\& \quad \geq \Vert x_{n}-\bar{u}\Vert ^{2}-\Vert x_{n}-\alpha_{n} D_{(n,1)}-\bar{u}\Vert ^{2} \\& \quad = \Vert x_{n}-\bar{u}\Vert ^{2}-\Vert x_{n}-\bar{u}\Vert ^{2}-\alpha_{n}^{2} \Vert D_{(n,1)}\Vert ^{2}+ 2\alpha_{n}\langle x_{n}-\bar{u}, D_{(n,1)}\rangle \\& \quad = 2\alpha_{n}\langle x_{n}-\bar{u}, D_{(n,1)} \rangle-\alpha_{n}^{2}\Vert D_{(n,1)}\Vert ^{2}. \end{aligned}$$

(2.6)

Similarly, we have

$$ \|y_{n}-\bar{v}\|^{2}-\|y_{n+1}- \bar{v}\|^{2}\geq2\alpha_{n}\langle y_{n}-\bar{v} , D_{(n,2)}\rangle-\alpha_{n}^{2}\| D_{(n,2)} \|^{2}. $$

(2.7)

By (2.6) and (2.7) we get

$$\begin{aligned}& \Vert x_{n+1}-\bar{u}\Vert ^{2}+\Vert y_{n+1}-\bar{v}\Vert ^{2} \\& \quad \leq \Vert x_{n}-\bar{u}\Vert ^{2}+\Vert y_{n}-\bar{v}\Vert ^{2}-2\alpha_{n}\langle x_{n}-\bar{u},D_{(n,1)}\rangle-2\alpha_{n}\langle y_{n}-\bar{v}, D_{(n,2)}\rangle \\& \qquad {}+\alpha_{n}^{2}\bigl(\Vert D_{(n,1)} \Vert ^{2}+\Vert D_{(n,2)}\Vert ^{2}\bigr). \end{aligned}$$

(2.8)

Next, we know that

$$\begin{aligned}& \langle u_{n}-\bar{u},D_{(n,1)}\rangle+\langle v_{n}-\bar{v}, D_{(n,2)}\rangle \\& \quad = \bigl\langle u_{n}-\bar{u},x_{n}-u_{n}+ \rho _{n}\bigl[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n}) \bigr]\bigr\rangle \\& \qquad {} +\bigl\langle v_{n}-\bar{v},y_{n}-v_{n}- \rho _{n}\bigl[B^{*}(Au_{n}-Bv_{n})-B^{*}(Ax_{n}-By_{n}) \bigr]\bigr\rangle \\& \quad = \bigl\langle u_{n}-\bar{u},x_{n}-u_{n}- \rho_{n}A^{*}(Ax_{n}-By_{n})\bigr\rangle +\bigl\langle u_{n}-\bar{u},\rho_{n}A^{*}(Au_{n}-Bv_{n}) \bigr\rangle \\& \qquad {} +\bigl\langle v_{n}-\bar{v},y_{n}-v_{n}+ \rho_{n}B^{*}(Ax_{n}-By_{n})\bigr\rangle -\bigl\langle v_{n}-\bar{v},\rho_{n}B^{*}(Au_{n}-Bv_{n}) \bigr\rangle . \end{aligned}$$

(2.9)

By Lemma 2.1,

$$ \bigl\langle u_{n}-\bar{u},x_{n}- \rho_{n}A^{*}(Ax_{n}-By_{n})-u_{n}\bigr\rangle \geq0 $$

(2.10)

and

$$ \bigl\langle v_{n}-\bar{v},y_{n}+ \rho_{n}B^{*}(Ax_{n}-By_{n})-v_{n}\bigr\rangle \geq0. $$

(2.11)

Besides, we also have

$$\begin{aligned}& \bigl\langle u_{n}-\bar{u},A^{*}(Au_{n}-Bv_{n}) \bigr\rangle -\bigl\langle v_{n}-\bar {v},B^{*}(Au_{n}-Bv_{n}) \bigr\rangle \\& \quad = \langle Au_{n}-A\bar{u},Au_{n}-Bv_{n} \rangle-\langle Bv_{n}-B\bar {v},Au_{n}-Bv_{n} \rangle \\& \quad = \langle Au_{n}-Bv_{n}-A\bar{u}+B \bar{v},Au_{n}-Bv_{n}\rangle \\& \quad = \langle Au_{n}-Bv_{n},Au_{n}-Bv_{n} \rangle \\& \quad = \|Au_{n}-Bv_{n}\|^{2}\geq0. \end{aligned}$$

(2.12)

So, by (2.9), (2.10), (2.11), and (2.12) we determine that

$$ \langle u_{n}-\bar{u},D_{(n,1)}\rangle+\langle v_{n}-\bar{v}, D_{(n,2)}\rangle\geq0, $$

(2.13)

which implies that

$$ \langle x_{n}-\bar{u},D_{(n,1)}\rangle+\langle y_{n}-\bar{v}, D_{(n,2)}\rangle\geq\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle. $$

(2.14)

By (2.2), (2.8), and (2.14),

$$\begin{aligned}& \|x_{n+1}-\bar{u}\|^{2}+\|y_{n+1}- \bar{v}\|^{2} \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-2\alpha_{n}\langle x_{n}-\bar{u} ,D_{(n,1)}\rangle-2\alpha_{n}\langle y_{n}-\bar{v}, D_{(n,2)}\rangle \\& \qquad {} +\alpha_{n}^{2}\bigl(\|D_{(n,1)} \|^{2}+\|D_{(n,2)}\|^{2}\bigr) \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-2\alpha_{n}\bigl(\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle\bigr) \\& \qquad {}+\alpha_{n}^{2}\bigl(\|D_{(n,1)} \|^{2}+\|D_{(n,2)}\|^{2}\bigr) \\& \quad = \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-\alpha_{n}\bigl(\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle\bigr) \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}. \end{aligned}$$

(2.15)

So, $\{\|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v}\|^{2}\}$ is a decreasing sequence, and $\lim_{n\rightarrow\infty}\|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v}\|^{2}$ exists. Further, $\{x_{n}\}_{n\in\mathbb{N}}$ and $\{y_{n}\}_{n\in \mathbb{N}}$ are bounded sequences, and

$$ \lim_{n\rightarrow\infty}\langle x_{n}-u_{n},D_{(n,1)} \rangle +\langle y_{n}-v_{n}, D_{(n,2)}\rangle=0. $$

(2.16)

Besides, we know that

$$\begin{aligned}& \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle \\& \quad = \langle x_{n}-u_{n},x_{n}-u_{n}+ \varepsilon_{n,1}\rangle+\langle y_{n}-v_{n},y_{n}-v_{n}+ \varepsilon_{n,2}\rangle \\& \quad = \|x_{n}-u_{n}\|^{2}+\langle x_{n}-u_{n},\varepsilon_{n,1}\rangle+ \|y_{n}-v_{n}\|^{2}+\langle y_{n}-v_{n}, \varepsilon_{n,2}\rangle, \end{aligned}$$

(2.17)

which implies that

$$\begin{aligned}& \|x_{n}-u_{n}\|^{2}+ \|y_{n}-v_{n}\|^{2} \\& \quad = \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle -\langle x_{n}-u_{n},\varepsilon_{n,1}\rangle-\langle y_{n}-v_{n},\varepsilon _{n,2}\rangle \\& \quad \leq \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle+ \|x_{n}-u_{n}\|\cdot \|\varepsilon_{n,1}\|+ \|y_{n}-v_{n}\|\cdot\|\varepsilon_{n,2}\| \\& \quad \leq \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle+\frac{1}{2}\bigl(\|x_{n}-u_{n}\|^{2}+ \| \varepsilon_{n,1}\|^{2}+\|y_{n}-v_{n} \|^{2}+\|\varepsilon_{n,2}\|^{2}\bigr) \\& \quad \leq \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle+\frac{1+\delta}{2}\cdot\bigl(\|x_{n}-u_{n}\|^{2} + \|y_{n}-v_{n}\|^{2}\bigr). \end{aligned}$$

(2.18)

Hence, by (2.18) we derive that

$$ (1-\delta) \bigl(\|x_{n}-u_{n} \|^{2}+\|y_{n}-v_{n}\|^{2}\bigr)\leq2 \langle x_{n}-u_{n},D_{(n,1)}\rangle+2\langle y_{n}-v_{n},D_{(n,2)}\rangle. $$

(2.19)

By (2.16) and (2.19) we know that

$$ \lim_{n\rightarrow\infty}\|x_{n}-u_{n}\|= \lim_{n\rightarrow \infty}\|y_{n}-v_{n}\|=0. $$

(2.20)

By Lemma 2.1 again,

$$\begin{aligned} \Vert u_{n}-\bar{u}\Vert ^{2} = &\bigl\Vert P_{C}\bigl[x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n}) \bigr]-P_{C}[\bar{u}]\bigr\Vert ^{2} \\ \leq &\bigl\Vert x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n})- \bar{u}\bigr\Vert ^{2} \\ \leq &\Vert x_{n}-\bar{u}\Vert ^{2}+ \rho_{n}^{2}\Vert A\Vert ^{2}\cdot \Vert Ax_{n}-By_{n}\Vert ^{2}-2\rho _{n} \langle Ax_{n}-By_{n},Ax_{n}-A\bar{u}\rangle. \end{aligned}$$

(2.21)

Similarly,

$$ \|v_{n}-\bar{v}\|^{2}\leq\|y_{n}- \bar{v}\|^{2}+\rho_{n}^{2}\|B\|^{2}\cdot\|Ax_{n}-By_{n}\|^{2}+2\rho_{n} \langle Ax_{n}-By_{n},By_{n}-B\bar{v}\rangle. $$

(2.22)

By (2.21) and (2.22),

$$\begin{aligned}& \|u_{n}-\bar{u}\|^{2}+\|v_{n}- \bar{v}\|^{2} \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}+\rho_{n}^{2}\bigl(\|A\|^{2}+\|B \|^{2}\bigr) \cdot\|Ax_{n}-By_{n}\|^{2} \\& \qquad {}-2\rho_{n}\langle Ax_{n}-By_{n},Ax_{n}-A \bar{u}\rangle+2\rho_{n}\langle Ax_{n}-By_{n},By_{n}-B \bar{v}\rangle \\& \quad = \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-\rho_{n}\bigl(2-\rho_{n}\bigl(\|A \|^{2}+\|B\|^{2}\bigr)\bigr) \cdot\|Ax_{n}-By_{n} \|^{2}. \end{aligned}$$

(2.23)

We also have

$$\begin{aligned} \|u_{n}-\bar{u}\|^{2}+\|v_{n}- \bar{v}\|^{2} = &\|u_{n}-x_{n}\|^{2}+2 \langle u_{n}-x_{n},x_{n}-\bar{u}\rangle+ \|x_{n}-\bar{u}\|^{2} \\ &{}+\|v_{n}-y_{n}\|^{2}+2\langle v_{n}-y_{n},y_{n}-\bar{v}\rangle+ \|y_{n}-\bar{v}\|^{2}. \end{aligned}$$

(2.24)

By (2.20), (2.23), and (2.24) we get

$$ \lim_{n\rightarrow\infty}\|Ax_{n}-By_{n} \|=0. $$

(2.25)

Since $\{x_{n}\}_{n\in\mathbb{N}}$ and $\{y_{n}\}_{n\in\mathbb{N}}$ are bounded sequences, there exist subsequences $\{x_{n_{k}}\}_{k\in\mathbb{N}}$ and $\{y_{n_{k}}\}_{k\in\mathbb{N}}$ of $\{x_{n}\}_{n\in\mathbb{N}}$ and $\{y_{n}\}_{n\in\mathbb{N}}$, respectively, such that $x_{n_{k}}\rightharpoonup\bar{x}$ and $y_{n_{k}}\rightharpoonup\bar{y}$ for some $\bar{x}\in H_{1}$ and $\bar{y}\in H_{2}$. Since $\{x_{n}\} _{n=2}^{\infty}$ is a sequence in C, we know that $\bar{x}\in C$. Also, $\bar{y}\in Q$. Since $x_{n_{k}}\rightharpoonup\bar{x}$ and $y_{n_{k}}\rightharpoonup\bar{y}$, it is easy to see that $Ax_{n_{k}}\rightharpoonup A\bar{x}$ and $By_{n_{k}}\rightharpoonup B\bar {y}$ by using the properties of A and B. Further, $Ax_{n_{k}}-By_{n_{k}}\rightharpoonup A\bar{x}-B\bar{y}$, and the lower semicontinuity of the squared norm implies

$$ \|A\bar{x}-B\bar{y}\|^{2}\leq\liminf _{k\rightarrow \infty}\|Ax_{n_{k}}-By_{n_{k}}\|^{2}=\lim _{n\rightarrow\infty}\|Ax_{n}-By_{n}\|^{2}=0. $$

(2.26)

Then $A\bar{x}=B\bar{y}$ and $(\bar{x},\bar{y})\in\Omega$.

Next, let $\{x^{\prime}_{n_{k}}\}$ and $\{y^{\prime}_{n_{k}}\}$ be other subsequences of $\{x_{n}\}_{n\in\mathbb{N}}$ and $\{y_{n}\}_{n\in \mathbb{N}}$ such that $x^{\prime}_{n_{k}}\rightharpoonup\hat{x}$ and $y^{\prime}_{n_{k}}\rightharpoonup\hat{y}$, respectively. Following the same argument as before, we get that $(\hat{x},\hat{y})\in\Omega$. Besides, we have

$$\begin{aligned}& \|x_{n}-\bar{x}\|^{2}+\|y_{n}- \bar{y}\|^{2} \\& \quad = \|x_{n}-\hat{x}\|^{2}+\|\hat{x}-\bar{x} \|^{2}+2\langle x_{n}-\hat {x},\hat{x}-\bar{x}\rangle \\& \qquad {}+ \|y_{n}-\hat{y}\|^{2}+\|\hat{y}-\bar {y}\|^{2}+2 \langle y_{n}-\hat{y},\hat{y}-\bar{y}\rangle \end{aligned}$$

(2.27)

and

$$\begin{aligned}& \|x_{n}-\hat{x}\|^{2}+\|y_{n}- \hat{y}\|^{2} \\& \quad = \|x_{n}-\bar{x}\|^{2}+\|\hat{x}-\bar{x} \|^{2}+2\langle x_{n}-\bar {x},\bar{x}-\hat{x}\rangle \\& \qquad {}+ \|y_{n}-\bar{y}\|^{2}+\|\hat{y}-\bar {y}\|^{2}+2 \langle y_{n}-\bar{y},\bar{y}-\hat{y}\rangle. \end{aligned}$$

(2.28)

Clearly, $\lim_{n\rightarrow\infty}\|x_{n}-\bar {x}\|^{2}+\|y_{n}-\bar{y}\|^{2}$ exists, and $\lim_{n\rightarrow\infty}\|x_{n}-\hat {x}\|^{2}+\|y_{n}-\hat{y}\|^{2}$ exists. Hence, by (2.27) we get

$$\begin{aligned}& \lim_{n\rightarrow\infty}\bigl(\Vert x_{n}-\bar{x} \Vert ^{2}+\Vert y_{n}-\bar{y}\Vert ^{2} \bigr) \\& \quad = \lim_{k\rightarrow\infty}\bigl(\bigl\Vert x^{\prime}_{n_{k}}- \bar {x}\bigr\Vert ^{2}+\bigl\Vert y^{\prime }_{n_{k}}- \bar{y}\bigr\Vert ^{2}\bigr) \\& \quad = \lim_{k\rightarrow\infty}\bigl(\bigl\Vert x^{\prime}_{n_{k}}- \hat {x}\bigr\Vert ^{2}+\bigl\Vert y^{\prime }_{n_{k}}- \hat{y}\bigr\Vert ^{2}\bigr)+\Vert \hat{x}-\bar{x}\Vert ^{2}+\Vert \hat{y}-\bar{y}\Vert ^{2} \\& \qquad {} +\lim_{k\rightarrow\infty}2\bigl\langle x^{\prime}_{n_{k}}- \hat {x},\hat{x}-\bar{x}\bigr\rangle +2\lim_{k\rightarrow\infty}\bigl\langle y^{\prime }_{n_{k}}-\hat{y},\hat{y}-\bar{y}\bigr\rangle \\& \quad = \lim_{k\rightarrow\infty}\bigl(\bigl\Vert x^{\prime}_{n_{k}}- \hat {x}\bigr\Vert ^{2}+\bigl\Vert y^{\prime }_{n_{k}}- \hat{y}\bigr\Vert ^{2}\bigr)+\Vert \hat{x}-\bar{x}\Vert ^{2} +\Vert \hat{y}-\bar{y}\Vert ^{2} \\& \quad = \lim_{n\rightarrow\infty}\bigl(\Vert x_{n}-\hat{x} \Vert ^{2}+\Vert y_{n}-\hat {y}\Vert ^{2} \bigr)+\Vert \hat{x}-\bar{x}\Vert ^{2} +\Vert \hat{y}- \bar{y}\Vert ^{2}. \end{aligned}$$

(2.29)

Similarly, by (2.28) we have

$$ \lim_{n\rightarrow\infty}\bigl(\|x_{n}-\hat{x} \|^{2}+\|y_{n}-\hat{y}\|^{2}\bigr) = \lim _{n\rightarrow\infty}\bigl(\|x_{n}-\bar{x}\|^{2}+ \|y_{n}-\bar {y}\|^{2}\bigr)+\|\hat{x}-\bar{x} \|^{2} +\|\hat{y}-\bar{y}\|^{2}. $$

(2.30)

By (2.29) and (2.30) we know that $\bar{x}=\hat{x}$ and $\bar{y}=\hat{y}$. Therefore, $x_{n}\rightharpoonup\bar{x}$ and $y_{n}\rightharpoonup\bar{y}$, and the proof is completed. □

Remark 2.2

In Theorem 2.1, if we choose $\{\rho_{n}\} _{n\in\mathbb{N}}$ from $( 0,\frac{\delta}{\sqrt {2}(\|A\|^{2}+\|B\|^{2})} ]$, then we only need to assume that $\liminf_{n\rightarrow\infty}\rho _{n}>0$.

Proof

Since $\rho_{n}\in ( 0,\frac{\delta }{\sqrt{2}(\|A\|^{2}+\|B\|^{2})} ]$, we have

$$ \rho_{n}\bigl(\|A\|^{2}+\|B\|^{2}\bigr)\leq \sqrt{2}\cdot\rho_{n}\cdot \bigl(\|A\|^{2}+\|B \|^{2}\bigr)\leq\delta, \quad \forall n\in\mathbb{N}, $$

(2.31)

which implies that

$$ \bigl(2-\rho_{n}\bigl(\|A\|^{2}+\|B\|^{2}\bigr) \bigr)\geq2-\delta>1,\quad \forall n\in \mathbb{N}. $$

(2.32)

Since $\liminf_{n\rightarrow\infty}\rho_{n}>0$, we may assume that there is κ such that $\rho_{n}\geq\kappa>0$ for all $n\in \mathbb{N}$. Hence, we determine

$$ \rho_{n}\bigl(2-\rho_{n}\bigl(\|A\|^{2}+\|B \|^{2}\bigr)\bigr)\geq\kappa\cdot(2-\delta )>\kappa, \quad \forall n\in \mathbb{N}. $$

(2.33)

By (2.33) we get the conclusion of Remark 2.2. □

2.2 Hybrid alternating CQ-algorithm

In this subsection, we present a hybrid alternating CQ-algorithm to study the split equality problem.

Algorithm 2.2

For given $x_{n}\in H_{1}$ and $y_{n}\in H_{2}$, find the approximate solution by the following iterative process.

Step 1.:

Compute the next iterate $(u_{n},v_{n})$ as follows:

$$ \left \{ \textstyle\begin{array}{l} u_{n}=P_{C}[x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n})], \\ v_{n}=P_{Q}[y_{n}+\rho_{n}B^{*}(Au_{n}-By_{n})],\end{array}\displaystyle \right . $$

where $\rho_{n}>0$ satisfies

$$\begin{aligned}& \rho _{n}^{2}\bigl(\bigl\Vert A^{*}(Ax_{n}-By_{n})-A^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}+\bigl\Vert B^{*}(Au_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}\bigr) \\& \quad \leq \delta \Vert x_{n}-u_{n}\Vert ^{2}+\delta \Vert y_{n}-v_{n}\Vert ^{2}, \quad 0< \delta< 1. \end{aligned}$$

(2.34)

Step 2.:

If $x_{n}=u_{n}$ and $y_{n}=v_{n}$, then $(x_{n},y_{n})$ is a solution of problem (SEP) and stop. Otherwise, go to Step 3.

Step 3.:

Compute the next iterate $(x_{n+1},y_{n+1})$ as follows:

$$ \left \{ \textstyle\begin{array}{l} D_{(n,1)}:=x_{n}-u_{n}+\rho_{n}[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n})], \\ D_{(n,2)}:=y_{n}-v_{n}-\rho_{n}[B^{*}(Au_{n}-Bv_{n})-B^{*}(Au_{n}-By_{n})], \\ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{(n,1)}\rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle}{\|D_{(n,1)}\|^{2}+\|D_{(n,2)}\|^{2}}, \\ x_{n+1}=P_{C}[x_{n}-\alpha_{n} D_{(n,1)}], \\ y_{n+1}=P_{Q}[y_{n}-\alpha_{n} D_{(n,2)}]. \end{array}\displaystyle \right . $$

Next, update $n:=n+1$ and go to Step 1.

Remark 2.3

If $0<\rho_{n}\leq{\frac{\sqrt{\delta}}{\max\{\sqrt{2}\cdot \|A\|^{2},\sqrt{2\cdot\|A\|^{2}\cdot|B\|^{2}+\|B\|^{3}}\}}}$, then (2.34) holds.

Proof

Without loss of generality, we may assume that $x_{n}\neq u_{n}$ and $y_{n}\neq v_{n}$. We have

$$\begin{aligned}& \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}(Ax_{n}-By_{n})-A^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}+\bigl\Vert B^{*}(Au_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n}) \bigr\Vert ^{2}\bigr) \\& \quad \leq \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}\bigr\Vert ^{2}\cdot \bigl\Vert (Ax_{n}-By_{n})-(Au_{n}-Bv_{n}) \bigr\Vert ^{2}+\Vert B\Vert ^{3}\cdot \Vert y_{n}-v_{n}\Vert ^{2}\bigr) \\& \quad \leq \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}\bigr\Vert ^{2}\cdot \bigl(\Vert Ax_{n}-Au_{n} \Vert +\Vert By_{n}-Bv_{n}\Vert \bigr)^{2}+ \Vert B\Vert ^{3}\cdot \Vert y_{n}-v_{n} \Vert ^{2}\bigr) \\& \quad \leq \rho_{n}^{2}\cdot\bigl(\bigl\Vert A^{*}\bigr\Vert ^{2}\cdot \bigl(2\Vert Ax_{n}-Au_{n} \Vert ^{2}+2\Vert By_{n}-Bv_{n}\Vert ^{2}\bigr)+\Vert B\Vert ^{3}\cdot \Vert y_{n}-v_{n}\Vert ^{2}\bigr) \\& \quad \leq \rho_{n}^{2}\cdot\bigl(2\bigl\Vert A^{*}\bigr\Vert ^{4}\cdot \Vert x_{n}-u_{n}\Vert ^{2}+\bigl(2\Vert A\Vert ^{2}\cdot \Vert B\Vert ^{2}+\Vert B\Vert ^{3}\bigr)\cdot \Vert y_{n}-v_{n}\Vert ^{2}\bigr) \\& \quad \leq \delta \Vert x_{n}-u_{n}\Vert ^{2}+\delta \Vert y_{n}-v_{n}\Vert ^{2}. \end{aligned}$$

Therefore, the proof is completed. □

Theorem 2.2

Let $\{\rho_{n}\}_{n\in\mathbb{N}}$ be a sequence in $(0,1/\max \{\|A\|^{2},\|B\|^{2}\})$ such that (2.34) holds and assume that ${\liminf_{n\rightarrow\infty}}\rho_{n}(1-\rho _{n}\|A\|^{2})>0$ or ${\liminf_{n\rightarrow\infty}}\rho_{n}(1-\rho _{n}\|B\|^{2})>0$. Then, for the sequence $\{(x_{n},y_{n})\}_{n\in\mathbb{N}}$ in Algorithm 2.2, there exists $(\bar{x},\bar{y})\in\Omega$ such that $x_{n}\rightharpoonup\bar{x}$ and $y_{n}\rightharpoonup\bar{y}$ as $n\rightarrow\infty$.

Proof

Take any $n\in\mathbb{N}$ and let n be fixed. Take any $(\bar{u},\bar{v})\in\Omega$ and let $(\bar{u},\bar{v})$ be fixed. Then $\bar{u}\in C$, $\bar{v}\in Q$, and $A\bar{u}=B\bar{v}$. First, we set

$$ \left \{ \textstyle\begin{array}{l} \varepsilon_{n,1}:=\rho_{n}[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n})], \\ \varepsilon_{n,2}:=\rho_{n}[B^{*}(Au_{n}-By_{n})-B^{*}(Au_{n}-Bv_{n})].\end{array}\displaystyle \right . $$

Then

$$ \langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle\geq \frac{1}{2}\|D_{(n,1)}\|^{2}+\frac{1}{2} \|D_{(n,2)}\|^{2}. $$

(2.35)

By (2.35) we have that

$$ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{(n,1)}\rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle}{\|D_{(n,1)}\|^{2}+\|D_{(n,2)}\|^{2}}\geq\frac{1}{2}. $$

(2.36)

Next, by Lemma 2.1 we have

$$ \|x_{n}-\alpha_{n} D_{(n,1)}-x_{n+1} \|^{2}+\|x_{n+1}-\bar{u}\|^{2}\leq \|x_{n}-\alpha_{n} D_{(n,1)}-\bar{u}\|^{2} $$

(2.37)

and

$$ \|y_{n}-\alpha_{n} D_{(n,2)}-y_{n+1} \|^{2}+\|y_{n+1}-\bar{v}\|^{2}\leq \|y_{n}-\alpha_{n} D_{(n,2)}-\bar{v}\|^{2}. $$

(2.38)

Hence, by (2.37),

$$ \|x_{n}-\bar{u}\|^{2}-\|x_{n+1}- \bar{u}\|^{2} \geq2\alpha_{n}\langle x_{n}- \bar{u}, D_{(n,1)}\rangle-\alpha_{n}^{2}\| D_{(n,1)}\|^{2}. $$

(2.39)

Also, by (2.38),

$$ \|y_{n}-\bar{v}\|^{2}-\|y_{n+1}- \bar{v}\|^{2}\geq2\alpha_{n}\langle y_{n}- \bar{v}, D_{(n,2)}\rangle-\alpha_{n}^{2}\| D_{(n,2)}\|^{2}.$$

(2.40)

By (2.39) and (2.40) we get

$$\begin{aligned}& \|x_{n+1}-\bar{u}\|^{2}+\|y_{n+1}- \bar{v}\|^{2} \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-2\alpha_{n}\langle x_{n}- \bar{u},D_{(n,1)}\rangle-2\alpha_{n}\langle y_{n}-\bar{v}, D_{(n,2)}\rangle \\& \qquad {}+\alpha_{n}^{2}\bigl(\|D_{(n,1)} \|^{2}+\|D_{(n,2)}\|^{2}\bigr). \end{aligned}$$

(2.41)

Next, we have

$$\begin{aligned}& \langle u_{n}-\bar{u},D_{(n,1)}\rangle+\langle v_{n}-\bar{v}, D_{(n,2)}\rangle \\& \quad = \bigl\langle u_{n}-\bar{u},x_{n}-u_{n}+ \rho _{n}\bigl[A^{*}(Au_{n}-Bv_{n})-A^{*}(Ax_{n}-By_{n}) \bigr]\bigr\rangle \\& \qquad {} +\bigl\langle v_{n}-\bar{v},y_{n}-v_{n}- \rho _{n}\bigl[B^{*}(Au_{n}-Bv_{n})-B^{*}(Au_{n}-By_{n}) \bigr]\bigr\rangle \\& \quad = \bigl\langle u_{n}-\bar{u},x_{n}-u_{n}- \rho_{n}A^{*}(Ax_{n}-By_{n})\bigr\rangle +\bigl\langle u_{n}-\bar{u},\rho_{n}A^{*}(Au_{n}-Bv_{n}) \bigr\rangle \\& \qquad {} +\bigl\langle v_{n}-\bar{v},y_{n}-v_{n}+ \rho_{n}B^{*}(Au_{n}-By_{n})\bigr\rangle -\bigl\langle v_{n}-\bar{v},\rho_{n}B^{*}(Au_{n}-Bv_{n}) \bigr\rangle . \end{aligned}$$

(2.42)

By Lemma 2.1,

$$ \bigl\langle u_{n}-\bar{u},x_{n}- \rho_{n}A^{*}(Ax_{n}-By_{n})-u_{n}\bigr\rangle \geq0 $$

(2.43)

and

$$ \bigl\langle v_{n}-\bar{v},y_{n}+ \rho_{n}B^{*}(Au_{n}-By_{n})-v_{n}\bigr\rangle \geq0. $$

(2.44)

Besides, we also have

$$ \bigl\langle u_{n}-\bar{u},A^{*}(Au_{n}-Bv_{n}) \bigr\rangle -\bigl\langle v_{n}-\bar{v},B^{*}(Au_{n}-Bv_{n}) \bigr\rangle =\|Au_{n}-Bv_{n}\|^{2}\geq0. $$

(2.45)

So, by (2.42), (2.43), (2.44), and (2.45) we determine that

$$ \langle u_{n}-\bar{u},D_{(n,1)}\rangle+\langle v_{n}-\bar{v}, D_{(n,2)}\rangle\geq0, $$

(2.46)

which implies that

$$ \langle x_{n}-\bar{u},D_{(n,1)}\rangle+\langle y_{n}-\bar {v},D_{(n,2)}\rangle\geq\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle. $$

(2.47)

By (2.35), (2.41), and (2.47),

$$\begin{aligned}& \|x_{n+1}-\bar{u}\|^{2}+\|y_{n+1}- \bar{v}\|^{2} \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-2\alpha_{n}\langle x_{n}- \bar{u},D_{(n,1)}\rangle-2\alpha_{n}\langle y_{n}-\bar{v}, D_{(n,2)}\rangle \\& \qquad {} +\alpha_{n}^{2}\bigl(\|D_{(n,1)} \|^{2}+\|D_{(n,2)}\|^{2}\bigr) \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-2\alpha_{n}\bigl(\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle\bigr) \\& \qquad {} +\alpha_{n}^{2}\bigl(\|D_{(n,1)} \|^{2}+\|D_{(n,2)}\|^{2}\bigr) \\& \quad = \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}-\alpha_{n}\bigl(\langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n}, D_{(n,2)}\rangle\bigr) \\& \quad \leq \|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v} \|^{2}. \end{aligned}$$

(2.48)

So, $\{\|x_{n}-\bar{u}\|^{2}+\|y_{n}-\bar{v}\|^{2}\}$ is a decreasing sequence, $\lim_{n\rightarrow\infty}\|x_{n}-\bar {u}\|^{2}+\|y_{n}-\bar{v}\|^{2}$ exists, $\{x_{n}\}_{n\in\mathbb{N}}$ and $\{y_{n}\}_{n\in\mathbb{N}}$ are bounded sequences, and

$$ \lim_{n\rightarrow\infty}\langle x_{n}-u_{n},D_{(n,1)} \rangle +\langle y_{n}-v_{n}, D_{(n,2)}\rangle=0. $$

(2.49)

Besides, we have

$$\begin{aligned}& \langle x_{n}-u_{n},D_{(n,1)} \rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle \\& \quad = \|x_{n}-u_{n}\|^{2}+\langle x_{n}-u_{n},\varepsilon_{n,1}\rangle+ \|y_{n}-v_{n}\|^{2}+\langle y_{n}-v_{n}, \varepsilon_{n,2}\rangle, \end{aligned}$$

(2.50)

which implies that

$$\begin{aligned}& \|x_{n}-u_{n}\|^{2}+ \|y_{n}-v_{n}\|^{2} \\& \quad \leq \langle x_{n}-u_{n},D_{(n,1)}\rangle+ \langle y_{n}-v_{n},D_{(n,2)}\rangle+\frac{1+\delta}{2}\cdot\bigl(\|x_{n}-u_{n}\|^{2} + \|y_{n}-v_{n}\|^{2}\bigr). \end{aligned}$$

(2.51)

Hence, by (2.51) we derive that

$$ (1-\delta) \bigl(\|x_{n}-u_{n} \|^{2}+\|y_{n}-v_{n}\|^{2}\bigr)\leq2 \langle x_{n}-u_{n},D_{(n,1)}\rangle+2\langle y_{n}-v_{n},D_{(n,2)}\rangle. $$

(2.52)

By (2.49) and (2.52) we get that

$$ \lim_{n\rightarrow\infty}\|x_{n}-u_{n} \|=\lim_{n\rightarrow \infty}\|y_{n}-v_{n}\|=0. $$

(2.53)

By Lemma 2.1 again,

$$\begin{aligned} \Vert u_{n}-\bar{u}\Vert ^{2} = &\bigl\Vert P_{C}\bigl[x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n}) \bigr]-P_{C}[\bar{u}]\bigr\Vert ^{2} \\ \leq &\bigl\Vert x_{n}-\rho_{n}A^{*}(Ax_{n}-By_{n})- \bar{u}\bigr\Vert ^{2} \\ \leq &\Vert x_{n}-\bar{u}\Vert ^{2}+ \rho_{n}^{2}\Vert A\Vert ^{2}\cdot \Vert Ax_{n}-By_{n}\Vert ^{2} \\ &{}-2\rho _{n} \langle Ax_{n}-By_{n},Ax_{n}-A\bar{u}\rangle \\ = &\Vert x_{n}-\bar{u}\Vert ^{2}-\rho_{n} \cdot\bigl(2-\rho_{n}\Vert A\Vert ^{2}\bigr)\cdot \Vert Ax_{n}-By_{n}\Vert ^{2} \\ &{}-2\rho_{n}\langle Ax_{n}-By_{n},By_{n}-A \bar{u}\rangle. \end{aligned}$$

(2.54)

Similarly,

$$\begin{aligned} \Vert v_{n}-\bar{v}\Vert ^{2} = &\bigl\Vert P_{Q}\bigl[y_{n}+\rho_{n}B^{*}(Au_{n}-By_{n}) \bigr]-P_{Q}[\bar{v}]\bigr\Vert ^{2} \\ \leq &\bigl\Vert y_{n}+\rho_{n}B^{*}(Au_{n}-By_{n})- \bar{v}\bigr\Vert ^{2} \\ \leq &\Vert y_{n}-\bar{v}\Vert ^{2}+ \rho_{n}^{2}\Vert B\Vert ^{2}\cdot \Vert Au_{n}-By_{n}\Vert ^{2} \\ &{}+2\rho _{n} \langle Au_{n}-By_{n}, By_{n}-B\bar{v}\rangle \\ = &\Vert y_{n}-\bar{v}\Vert ^{2}-\rho_{n} \bigl(2-\rho_{n}\Vert B\Vert ^{2}\bigr)\cdot \Vert Au_{n}-By_{n}\Vert ^{2} \\ &{}+2\rho_{n} \langle Au_{n}-By_{n}, Au_{n}-B\bar{v}\rangle. \end{aligned}$$

(2.55)

We also have

$$ 2\langle Ax_{n}-By_{n},By_{n}-A \bar{u}\rangle=\|Ax_{n}-A\bar{u}\|^{2}- \|Ax_{n}-By_{n}\|^{2}-\|By_{n}-A\bar{u} \|^{2} $$

(2.56)

and

$$ 2\langle Au_{n}-By_{n}, Au_{n}-B \bar{v}\rangle=\|Au_{n}-B\bar{v}\|^{2}+ \|Au_{n}-By_{n}\|^{2}-\|By_{n}-B\bar{v} \|^{2}. $$

(2.57)

By (2.54), (2.55), (2.56), and (2.57),

$$\begin{aligned}& \Vert u_{n}-\bar{u}\Vert ^{2}+\Vert v_{n}-\bar{v}\Vert ^{2} \\& \quad \leq \Vert x_{n}-\bar{u}\Vert ^{2}+\Vert y_{n}-\bar{v}\Vert ^{2}-\rho_{n}\bigl(1-\rho _{n}\Vert A\Vert ^{2}\bigr)\cdot \Vert Ax_{n}-By_{n}\Vert ^{2} \\& \qquad {} -\rho_{n}\bigl(1-\rho_{n}\Vert B\Vert ^{2}\bigr)\cdot \Vert Au_{n}-By_{n}\Vert ^{2}+\rho_{n}\bigl(\Vert Au_{n}-A\bar {u}\Vert ^{2}-\Vert Ax_{n}-A\bar{u}\Vert ^{2}\bigr) \\& \quad \leq \Vert x_{n}-\bar{u}\Vert ^{2}+\Vert y_{n}-\bar{v}\Vert ^{2}-\rho_{n}\bigl(1-\rho _{n}\Vert A\Vert ^{2}\bigr)\cdot \Vert Ax_{n}-By_{n}\Vert ^{2} \\& \qquad {} -\rho_{n}\bigl(1-\rho_{n}\Vert B\Vert ^{2}\bigr)\cdot \Vert Au_{n}-By_{n}\Vert ^{2} \\& \qquad {} +\rho_{n}\cdot \Vert A\Vert \cdot \Vert u_{n}-x_{n}\Vert \cdot\bigl(\Vert Au_{n}-A \bar {u}\Vert +\Vert Ax_{n}-A\bar{u}\Vert \bigr). \end{aligned}$$

(2.58)

We also have

$$\begin{aligned} \|u_{n}-\bar{u}\|^{2}+\|v_{n}- \bar{v}\|^{2} = &\|u_{n}-x_{n}\|^{2}+2 \langle u_{n}-x_{n},x_{n}-\bar{u}\rangle+ \|x_{n}-\bar{u}\|^{2} \\ &{}+\|v_{n}-y_{n}\|^{2}+2\langle v_{n}-y_{n},y_{n}-\bar{v}\rangle+ \|y_{n}-\bar{v}\|^{2}. \end{aligned}$$

(2.59)

Case 1: $\liminf_{n\rightarrow\infty}\rho_{n}(1-\rho_{n}\|A\|^{2})>0$.

By (2.53), (2.58), and (2.59) we get

$$ \lim_{n\rightarrow\infty}\|Ax_{n}-By_{n} \|=0. $$

(2.60)

Case 2: Suppose that $\liminf_{n\rightarrow\infty}\rho_{n}(1-\rho_{n}\|B\|^{2})>0$.

By (2.53), (2.58), and (2.59) we get

$$ \lim_{n\rightarrow\infty}\|Au_{n}-By_{n} \|=0. $$

(2.61)

By (2.53) and (2.61) we determine

$$ \lim_{n\rightarrow\infty}\|Ax_{n}-By_{n} \|=0. $$

(2.62)

Next, following the same argument as the final proof of Theorem 2.1, we get the conclusion of Theorem 2.2. □

Remark 2.4

Suppose that $\{\rho_{n}\}_{n\in\mathbb{N}}$ satisfy the following inequality:

$$ 0< \kappa\leq\rho_{n}\leq\frac{\delta}{\max\{\sqrt{2}\cdot \|A\|^{2},\sqrt{2\cdot\|A\|^{2}\cdot\|B\|^{2}+\|B\|^{3}},\|B\|^{2}\}}. $$

Then $\{\rho_{n}\}_{n\in\mathbb{N}}$ satisfy the conditions in Remark 2.3 and Theorem 2.2.

3 Applications of the split equality problem

3.1 The split feasibility problem

Let $H_{1}$ and $H_{2}$ be real Hilbert spaces. Let C and Q be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a linear and bounded operator with adjoint operator $A^{* }$. The following problem is the split feasibility problem in Hilbert spaces, which was first introduced by Censor and Elfving [11]:

$$ (\mathrm{SFP}) \quad \text{Find } \bar{x}\in H_{1}\text{ such that } \bar{x}\in C \text{ and }A\bar{x}\in Q. $$

Here, let $\Omega_{1}:=\{x\in C: Ax\in Q\}$ be the solution set of problem (SFP). It is worth noting that this problem is a particular case of the split equality problem when $H_{3}=H_{2}$ and B is the identity mapping on $H_{2}$. For additional details, one can refer to [6, 11–24] and related literature.

By Algorithm 2.2, we get the following algorithm to study problem (SFP).

Algorithm 3.1

For given $x_{n}\in H_{1}$ and $y_{n}\in H_{2}$, find the approximate solution by the following iterative process.

Step 1.:

For $n\in\mathbb{N}$, let $u_{n}$ and $v_{n}$ be defined by

$$ \left \{ \textstyle\begin{array}{l} u_{n}=P_{C}[x_{n}-\rho_{n}A^{*}(Ax_{n}-y_{n})], \\ v_{n}=P_{Q}[y_{n}+\rho_{n}(Au_{n}-y_{n})],\end{array}\displaystyle \right . $$

where $\rho_{n}>0$ satisfies

$$\begin{aligned}& \rho_{n}^{2}\bigl(\bigl\Vert A^{*}(Ax_{n}-y_{n})-A^{*}(Au_{n}-v_{n}) \bigr\Vert ^{2}+\bigl\Vert (Au_{n}-y_{n})-B^{*}(Au_{n}-v_{n}) \bigr\Vert ^{2}\bigr) \\& \quad \leq \delta \Vert x_{n}-u_{n}\Vert ^{2}+\delta \Vert y_{n}-v_{n}\Vert ^{2},\quad 0< \delta< 1. \end{aligned}$$

(3.1)

Step 2.:

If $x_{n}=u_{n}$ and $y_{n}=v_{n}$, then $(x_{n},y_{n})$ is a solution of problem (SFP) and stop. Otherwise, go to Step 3.

Step 3.:

Compute the next iterate $(x_{n+1},y_{n+1})$ as follows:

$$ \left \{ \textstyle\begin{array}{l} D_{(n,1)}:=x_{n}-u_{n}+\rho_{n}[A^{*}(Au_{n}-v_{n})-A^{*}(Ax_{n}-y_{n})], \\ D_{(n,2)}:=y_{n}-v_{n}-\rho_{n}[(Au_{n}-v_{n})-(Au_{n}-y_{n})], \\ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{(n,1)}\rangle+\langle y_{n}-v_{n},D_{(n,2)}\rangle}{\|D_{(n,1)}\|^{2}+\|D_{(n,2)}\|^{2}}, \\ x_{n+1}=P_{C}[x_{n}-\alpha_{n} D_{(n,1)}], \\ y_{n+1}=P_{Q}[y_{n}-\alpha_{n} D_{(n,2)}].\end{array}\displaystyle \right . $$

Next, update $n:=n+1$ and go to Step 1.

We get the following convergence theorem for the split feasibility problem by using Theorem 2.2.

Theorem 3.1

Let $H_{1}$ and $H_{2}$ be real Hilbert spaces. Let C and Q be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a linear and bounded operator with adjoint operator $A^{* }$. Choose $\delta\in(0,1)$. Let $\Omega_{1}$ be the solution set of the split feasibility problem and suppose that $\Omega_{1}\neq\emptyset$. Let $\{\rho _{n}\}_{n\in\mathbb{N}}$ be a sequence in $(0,1/\max\{\|A\|^{2},1\} )$ such that (3.1) hold and assume that ${\liminf_{n\rightarrow\infty}}\rho_{n}(1-\rho _{n}\|A\|^{2})>0$ or ${\liminf_{n\rightarrow\infty}}\rho_{n}(1-\rho_{n})>0$ s. Then, for the sequence $\{(x_{n},y_{n})\}_{n\in\mathbb{N}}$ in Algorithm 3.1, there exists $\bar{x}\in\Omega_{1}$ such that $x_{n}\rightharpoonup\bar {x}$ as $n\rightarrow\infty$.

3.2 Linear inverse problem

In this subsection, we study an inverse problem by our algorithms and convergence theorems. Let $H_{1}$ and $H_{2}$ be real Hilbert spaces. Let C be a nonempty closed convex subset of $H_{1}$, and $A:H_{1}\rightarrow H_{2}$ be a linear and bounded operator with adjoint operator $A^{*}$. Given $b\in H_{2}$. Then we consider the following inverse problem in this section:

$$ (\mathrm{IV}) \quad \mbox{Find }\bar{x}\in C\mbox{ such that }A\bar{x}=b. $$

This is a particular case of the split equality problem if $H_{2}=H_{3}$, $Q=\{b\}$, and $B(x)=x$ for all $x\in H_{2}$. Next, take any $(x_{1},y_{1})\in H_{1}\times H_{2}$ with $y_{1}=b$. Then, by Algorithm 2.2 we get the following algorithm to study problem (IV).

Algorithm 3.2

For given$x_{n}\in H_{1}$, find the approximate solution by the following iterative process.

Step 1.:

Compute the next iterate $u_{n}$ as follows:

$$ u_{n}=P_{C}\bigl[x_{n}-\rho_{n}A^{*}(Ax_{n}-b) \bigr], $$

where $\rho_{n}>0$ satisfies

$$ \rho_{n}^{2}\cdot\bigl\Vert A^{*}(Ax_{n})-A^{*}(Au_{n})\bigr\Vert ^{2}\leq \delta \Vert x_{n}-u_{n}\Vert ^{2},\quad 0< \delta< 1. $$

(3.2)

Step 2.:

If $x_{n}=u_{n}$, then $x_{n}$ is a solution of problem (IV) and stop. Otherwise, go to Step 3.

Step 3.:

Compute the next iterate $x_{n+1}$ as follows:

$$ \left \{ \textstyle\begin{array}{l} D_{n}:=x_{n}-u_{n}+\rho_{n}[A^{*}(Au_{n})-A^{*}(Ax_{n})], \\ \alpha_{n}:=\frac{\langle x_{n}-u_{n},D_{n}\rangle}{\|D_{n}\|^{2}}, \\ x_{n+1}=P_{C}[x_{n}-\alpha_{n} D_{n}].\end{array}\displaystyle \right . $$

Next, update $n:=n+1$ and go to Step 1.

We get the following convergence theorem for the linear inverse problem by using Theorem 2.2.

Theorem 3.2

Let $H_{1}$ and $H_{2}$ be real Hilbert spaces. Let C be a nonempty closed convex subset of $H_{1}$, and $A:H_{1}\rightarrow H_{2}$ be a linear and bounded operator with adjoint operator  $A^{* }$. Given $b\in H_{2}$ and $\delta \in(0,1)$. Let $\Omega_{2}$ be the solution set of (IV) and suppose that $\Omega _{2}\neq\emptyset$. Let $\{\rho_{n}\}_{n\in\mathbb{N}}$ be a sequence in $(0,1/\max\{\|A\|^{2},1\})$ such that (3.2) holds and assume that ${\liminf_{n\rightarrow\infty}}\rho _{n}(1-\rho_{n}\|A\|^{2})>0$ or ${\liminf_{n\rightarrow\infty}}\rho _{n}(1-\rho_{n})>0$. Then, for the sequence $\{x_{n}\}_{n\in\mathbb {N}}$ in Algorithm 3.2, there exists $\bar{x}\in\Omega_{2}$ such that $x_{n}\rightharpoonup \bar{x}$ as $n\rightarrow \infty$.

Remark 3.1

By Algorithm 2.1 and Theorem 2.1, we can get the related algorithms and convergence theorems for the split feasibility problem and the inverse problems.

4 Numerical results

All codes were written in R language (version 3.2.4 (2016-03-10), the R Foundation for Statistical Computing Platform: x86-64-w64-mingw32/x64).

Example 4.1

Let $H_{1}=H_{2}=H_{3}=\mathbb{R}^{2}$, $C:=\{x\in\mathbb{R}^{2}: \|x\|\leq1\}$, $Q:=\{x=(u,v)\in\mathbb{R}^{2}: (u-6)^{2}+(v-8)^{2}\leq25\}$,

$$A:= \left[ \textstyle\begin{array}{@{}c@{\quad}c@{}} 5 & 0\\ 0 & 5 \end{array}\displaystyle \right],\qquad B:= \left[ \textstyle\begin{array}{@{}c@{\quad}c@{}} 1 & 0\\ 0 & 1 \end{array}\displaystyle \right]. $$

Then problem (SEP) has a unique solution $(\bar{x},\bar{y})\in \mathbb{R}^{2}\times\mathbb{R}^{2}$, where $\bar{x}:=(\bar{x}_{1},\bar {x}_{2})$, $\bar{y}:=(\bar{y}_{1},\bar{y}_{2})$. Indeed, $\bar{x}_{1}=0.6$, $\bar{x}_{2}=0.8$, $\bar{y}_{1}=3$, $\bar{y}_{2}=4$. Let $\varepsilon>0$ and the algorithm stop if $\|x_{n}-\bar{x}\|+\|y_{n}-\bar {y}\|<\varepsilon$.

In Table 1, setting $\varepsilon=10^{-1}$, $x_{1}=(10,10)^{T}$, $y_{1}=(1,1)^{T}$, and $\rho_{n}=0.01$ for all $n\in\mathbb{N}$, we get the numerical results.

Table 1 $\pmb{\varepsilon=10^{-1}}$ , $\pmb{x_{1}=(10,10)^{T}}$ , $\pmb{y_{1}=(1,1)^{T}}$ , $\pmb{\rho_{n}=0.01}$

In Table 2, setting $\varepsilon=10^{-1}$, $x_{1}=(5,5)^{T}$, $y_{1}=(1,1)^{T}$, and $\rho_{n}=0.01$ for all $n\in\mathbb{N}$, we get the numerical results.

Table 2 $\pmb{\varepsilon=10^{-1}}$ , $\pmb{x_{1}=(5,5)^{T}}$ , $\pmb{y_{1}=(1,1)^{T}}$ , $\pmb{\rho_{n}=0.01}$

In Table 3, setting $\varepsilon=4\times10^{-2}$, $x_{1}=(-12,-50)^{T}$, $y_{1}=(-40,20)^{T}$, and $\rho_{n}=0.01$ for all $n\in\mathbb{N}$, we get the numerical results.

Table 3 $\pmb{\varepsilon=4\times10^{-2}}$ , $\pmb{x_{1}=(12,-50)^{T}}$ , $\pmb{y_{1}=(-40,20)^{T}}$ , $\pmb{\rho_{n}=0.01}$