Alternating iterative algorithms for the split equality problem without prior knowledge of operator norms

Abstract

In this paper, we study the alternating CQ algorithm for solving the split equality problem in Hilbert spaces. It is, however, not easy to implement since its selection of the stepsize requires prior information on the norms of bounded linear operators. To avoid this difficulty, we propose several modified algorithms in which the selection of the stepsize is independent of the norms. In particular, we consider the case whenever the convex sets involved are level sets of given convex functions.

1 Introduction

The split feasibility problem (SFP) requires finding a point $x\in H_{1}$ satisfying the property

$$\begin{aligned} x\in C \quad \text{and} \quad Ax\in Q, \end{aligned}$$

(1)

where $A: H_{1}\rightarrow H_{2}$ is a bounded linear operator, and C and Q are two nonempty closed and convex subsets of Hilbert spaces $H_{1}$ and $H_{2}$, respectively. The SFP was first introduced by Censor and Elfving [5] and has a very broad range of applications in many disciplines including signal processing, image reconstruction problem, and radiation therapy; see [2, 3, 11]. Various iterative algorithms have been constructed to solve SFP (1); see [10, 21–24, 26]. An iterative algorithm for solving the SFP, called the CQ algorithm, has the following iterative step:

$$\begin{aligned} x_{n+1}=P_{C}\bigl(x_{n}-\gamma A^{*}(I-P_{Q})Ax_{n}\bigr), \end{aligned}$$

(2)

where $\gamma \in (0,\frac{2}{\|A\|^{2}})$, I denotes the identity operator, $A^{*}$ denotes the adjoint of A, and $P_{C}$ and $P_{Q}$ are projections onto C and Q, respectively. The SFP can be also solved by a different method [19, 29]:

$$\begin{aligned} x_{n+1}=x_{n}-\gamma _{n} \bigl[(I-P_{C})x_{n}+A^{*}(I-P_{Q})Ax_{n} \bigr], \end{aligned}$$

(3)

where

$$ \gamma _{n}= \frac{ \Vert (I-P_{C})x_{n} \Vert ^{2}+ \Vert (I-P_{Q})Ax_{n} \Vert ^{2}}{ \Vert (I-P_{C})x_{n}+A^{*}(I-P_{Q})Ax_{n} \Vert ^{2}}. $$

It is known that both (2) and (3) converge weakly to a solution of the SFP if it is consistent, that is, its solution set is nonempty.

Recently, Moudafi [13] introduced the split equality problem (SEP):

$$\begin{aligned} \text{Find} \quad x\in C, y\in Q \quad \text{such that} \quad Ax=By. \end{aligned}$$

(4)

Here, $C\subseteq H_{1}$, $Q\subseteq H_{2}$ are two nonempty closed and convex subsets, and $A: H_{1}\rightarrow H_{3}$, $B: H_{2}\rightarrow H_{3}$ are two bounded linear operators, where $H_{1}$, $H_{2}$, $H_{3}$ are Hilbert spaces. In what follows, we always assume that the SEP is consistent, namely

$$ S=\bigl\{ (x, y)\in C \times Q \mid Ax=By\bigr\} \neq \emptyset . $$

Many algorithms for solving the SEP have been proposed; see [8, 9, 12, 13, 15, 20]. In particular, Byrne and Moudafi [4, 14] introduced the simultaneous CQ algorithm:

$$ \textstyle\begin{cases} x_{n+1}=P_{C}(x_{n}-\gamma _{n} A^{*}(Ax_{n}-By_{n})), \\ y_{n+1}=P_{Q}(y_{n}+\gamma _{n} B^{*}(Ax_{n}-By_{n})), \end{cases} $$

(5)

where $\gamma _{n}\in (\epsilon , 2/(\|A\|^{2}+\|B\|^{2})-\epsilon )$ for small enough $\epsilon >0$. To determine stepsize $\gamma _{n}$ in (5), one needs first to calculate or estimate the norms $\|A\|$ and $\|B\|$, which is in general difficult or even impossible. To overcome this drawback, many authors have conducted worthwhile works[6–8, 19]. Among these works, Wang [19] suggested a novel variable-step:

$$ \tau _{n}= \frac{\rho _{n}}{\max ( \Vert A^{*}(Ax_{n}-By_{n}) \Vert , \Vert B^{*}(Ax_{n}-By_{n}) \Vert )}, $$

(6)

where $\{\rho _{n}\}$ is a sequence of positive real numbers such that

$$ \sum_{n=0}^{\infty }\rho _{n}=\infty , \qquad \sum_{n=0}^{\infty } \rho _{n}^{2}< \infty . $$

(7)

For solving the SEP, Moudafi [13] also introduced the alternating CQ-algorithm:

$$ \textstyle\begin{cases} x_{n+1}=P_{C}(x_{n}-\gamma _{n}A^{*}(Ax_{n}-By_{n})), \\ y_{n+1}=P_{Q}(y_{n}+\gamma _{n}B^{*}(Ax_{n+1}-By_{n})), \end{cases} $$

(8)

where $\{\gamma _{n}\}$ is a nondecreasing sequence such that $\gamma _{n}\in (\epsilon , \min (\frac{1}{\|A\|^{2}}, \frac{1}{\|B\|^{2}})-\epsilon )$ for small enough $\epsilon >0$. It is worth noting that in algorithm (8) the choice of the stepsize still depends on the norms $\|A\|$ and $\|B\|$. Thus, a similar question arises: Does there exist a way to select the stepsize in algorithm (8) that does not depend on the operator norms? It is the purpose of this paper to answer this question affirmatively. Motivated by the choice of stepsize (6), we propose three alternating iterative algorithms for the SEP, in which the choice of the stepsize is independent of the norms $\|A\|$ and $\|B\|$. To the best of our knowledge, this is the first work to study alternating iterative algorithms for the SEP without prior knowledge of operator norms.

2 Preliminaries

Throughout this paper, we always assume that H is a real Hilbert space with the inner product $\langle \cdot , \cdot \rangle$ and the norm $\|\cdot \|$, and C is a nonempty closed convex subset in H. We denote by I the identity operator on H, and by $\operatorname{Fix}(T)$ the set of the fixed points of an operator T. Given a sequence $\{x_{n}\}$ in H, $\omega _{w}(x_{n})$ stands for the set of cluster points in the weak topology. The notation → stands for strong convergence and ⇀ stands for weak convergence.

Definition 2.1

Let $T:H\rightarrow H$ be an operator. Then T is

• (i) nonexpansive if

  $$ \Vert Tx-Ty \Vert \leq \Vert x-y \Vert , \quad \forall x, y\in H; $$

• (ii) firmly nonexpansive if

  $$ \Vert Tx-Ty \Vert ^{2}\leq \Vert x-y \Vert ^{2}- \bigl\Vert (I-T)x-(I-T)y \bigr\Vert ^{2}, \quad \forall x,y \in H. $$

Definition 2.2

For any $x\in H$, the metric projection onto C is defined as

$$ P_{C} x= \operatorname{argmin}\bigl\{ \Vert y-x \Vert \mid y\in C \bigr\} . $$

The projection $P_{C}$ has the following well-known properties.

Lemma 2.3

([1, 17])

Let $x,y\in H$and $z\in C$. Then

• (i) $\langle x-P_{C} x, z-P_{C} x\rangle \leq 0$;

• (ii) $P_{C}$is firmly nonexpansive;

• (iii) $\langle P_{C}x-P_{C}y, x-y\rangle \geq \|P_{C}x-P_{C}y\|^{2}$.

Definition 2.4

Let $T: H\rightarrow H$ be an operator with $\operatorname{Fix}(T)\neq \emptyset$. Then $I-T$ is said to be demiclosed at zero if, for any $\{x_{n}\}$ in H, the following implication holds:

$$ x_{n}\rightharpoonup x\quad \text{and}\quad (I-T)x_{n}\rightarrow 0 \quad \Longrightarrow\quad x\in \operatorname{Fix}(T). $$

It is well known that $I-T$ is demiclosed at zero if T is nonexpansive [1, 17]. Since $P_{C}$ is clearly nonexpansive, then $I-P_{C}$ is demiclosed at zero.

Definition 2.5

Let $\lambda \in (0,1)$ and $f: H\rightarrow (-\infty , +\infty ]$ be a proper function.

• (i) The function f is convex if

  $$ f\bigl(\lambda x+(1-\lambda )y\bigr)\leq \lambda f(x)+(1-\lambda )f(y),\quad \forall x,y\in H. $$

• (ii) A vector $u\in H$ is a subgradient of f at a point x if

  $$ f(y)\geq f(x)+\langle u, y-x\rangle , \quad \forall y\in H. $$

• (iii) The set of all subgradients of f at x, denoted by $\partial f(x)$, is called the subdifferential of f.

Next, we state the following lemmas which will be used in the sequel.

Lemma 2.6

([1])

Let $f:H\rightarrow (-\infty , +\infty ]$be a proper convex function. Then f is semicontinuous if and only if it is weakly semicontinuous.

Lemma 2.7

([18])

Suppose that $\{a_{n}\}$and $\{b_{n}\}$are two sequences of nonnegative numbers such that

$$ a_{n+1}\leq a_{n} +b_{n}, \quad n\geq 0. $$

If $\sum_{n=0}^{\infty } b_{n}$converges, then $\lim_{n\to \infty } a_{n}$exists.

Lemma 2.8

([25])

Let $\{x_{n}\}$be a sequence in H satisfying the properties:

• (i) $\lim_{n\rightarrow \infty }\|x_{n}-x\|$exists for each $x\in C$;

• (ii) $\omega _{w}(x_{n})\subseteq C$.

Then $\{x_{n}\}$is weakly convergent to a point in C.

Lemma 2.9

Assume $\{a_{n}\}$, $\{\gamma _{n}\}$, $\{b_{n}\}$, and $\{c_{n}\}$are sequences of nonnegative real numbers such that

$$\begin{aligned} a_{n+1}\leq (1+\gamma _{n}) (a_{n}-b_{n}+c_{n}), \quad n\geq 0, \end{aligned}$$

(9)

where $\sum_{n=0}^{\infty }\gamma _{n}<\infty$and $\sum_{n=0}^{\infty }c_{n}<\infty$. Then

• (i) $\lim_{n\to \infty } a_{n}$exists,

• (ii) $\sum_{n=0}^{\infty }b_{n}<\infty$.

Proof

(i) Since $\sum_{n=0}^{\infty }\gamma _{n}$ is convergent, we have $\prod_{n=0}^{\infty }(1+\gamma _{n})$ is convergent. Let $\prod_{n=0}^{\infty }(1+\gamma _{n})=\gamma$. It follows from (9) that

$$\begin{aligned} a_{n}&\leq (1+\gamma _{n-1})a_{n-1}+ (1+\gamma _{n-1})c_{n-1} \\ &\leq (1+\gamma _{n-1}) (1+\gamma _{n-2})a_{n-2}+(1+ \gamma _{n-1}) (1+ \gamma _{n-2})c_{n-2}+ (1+\gamma _{n-1})c_{n-1} \\ &\leq \cdots \\ &\leq \prod_{k=0}^{n-1}(1+\gamma _{k})a_{0}+\sum_{j=0}^{n-1} \Biggl( \prod_{k=j}^{n-1}(1+\gamma _{k})c_{j} \Biggr) \\ &\leq \gamma a_{0}+\gamma \sum_{j=0}^{n-1}c_{j} \\ &\leq \gamma a_{0}+\gamma c, \end{aligned}$$

where $c=\sum_{n=0}^{\infty }c_{n}$. This implies that

$$\begin{aligned} a_{n+1}&\leq (1+\gamma _{n})a_{n} +(1+\gamma _{n})c_{n} \\ &\leq a_{n}+\gamma _{n}(\gamma a_{0}+\gamma c)+ \gamma c_{n}. \end{aligned}$$

Since $\sum_{n=0}^{\infty }\gamma _{n}< \infty$ and $\sum_{n=0}^{\infty }c_{n}< \infty$, by Lemma 2.7, part (i) holds.

(ii) It follows from (9) that

$$\begin{aligned} a_{n+1}&\leq (1+\gamma _{n})a_{n}-(1+\gamma _{n})b_{n}+(1+\gamma _{n})c_{n} \\ &\leq (1+\gamma _{n}) (1+\gamma _{n-1})a_{n-1}-(1+ \gamma _{n}) (1+ \gamma _{n-1})b_{n-1}- (1+\gamma _{n})b_{n} \\ &\quad {}+(1+\gamma _{n}) (1+\gamma _{n-1})c_{n-1}+ (1+ \gamma _{n})c_{n} \\ &\leq \cdots \\ &\leq \prod_{k=0}^{n}(1+\gamma _{k}) a_{0}-\sum_{j=0}^{n} \Biggl( \prod_{k=j}^{n}(1+\gamma _{k})b_{j} \Biggr)+\sum_{j=0}^{n} \Biggl(\prod_{k=j}^{n}(1+ \gamma _{k})c_{j} \Biggr) \\ &\leq \gamma a_{0}+\gamma c-\sum_{j=0}^{n}b_{j}, \end{aligned}$$

where the last inequality holds since $\prod_{k=j}^{n}(1+\gamma _{k})\geq 1$. So we obtain

$$ \sum_{j=0}^{n}b_{j}\leq \gamma a_{0}+\gamma c, $$

which means that $\sum_{n=0}^{\infty }b_{n}<\infty$. □

Lemma 2.10

Let $\{u_{n}\}$, $\{\rho _{n}\}$be sequences of nonnegative real numbers such that

• (i) $\sum_{k=0}^{\infty }\rho _{n} u_{n}<\infty$;

• (ii) there exists some $M>0$such that, for all $k\geq 0$, $|u_{n+1}-u_{n}|\leq M\rho _{n}$;

• (iii) $\{\rho _{n}\}$satisfies condition (7).

Then $\lim_{n\to \infty }u_{n}=0$.

Proof

It follows from (i) and the assumption $\sum_{k=0}^{\infty }\rho _{n}=\infty$ that

$$ \liminf_{n\to \infty }u_{n}=0. $$

On the other hand, observe that

$$\begin{aligned} u_{n+1}^{2}&=u_{n}^{2}+2u_{n}(u_{n+1}-u_{n})+(u_{n+1}-u_{n})^{2} \\ &\leq u_{n}^{2}+2M\rho _{n} u_{n}+M^{2} \rho _{n}^{2}. \end{aligned}$$

(10)

It is clear that $\sum_{n=0}^{\infty }(2M\rho _{n} u_{n}+M^{2}\rho _{n}^{2})< \infty$ due to (i) and (7). Applying Lemma 2.7 to (10), we obtain that $\lim_{n\to \infty }u_{n}$ exists. Hence $\lim_{n\to \infty }u_{n}=0$, since we have shown that $\liminf_{n\to \infty }u_{n}=0$. □

3 Alternating iterative algorithm I

As shown in the introduction, one needs first to calculate (or at least estimate) the norms $\|A\|$ and $\|B\|$ when algorithm (8) is implemented. But this is difficult or even impossible. To overcome this difficulty, we aim to introduce the following alternating iterative algorithm which does not depend on the norms. In this and next sections, we mainly consider the case whenever the projections $P_{C}$ and $P_{Q}$ have closed-form expressions; for example, half spaces and closed balls.

Algorithm 3.1

Let $(x_{0}, y_{0})\in H_{1}\times H_{2}$ be arbitrary, $\delta >0$ be a constant, and $\{\rho _{n}\}$ be a sequence of positive real numbers. Given $(x_{n}, y_{n})$, construct $(x_{n+1}, y_{n+1})$ via the formula

$$ \textstyle\begin{cases} x_{n+1}=P_{C}(x_{n}-\tau _{n}A^{*}(Ax_{n}-By_{n})), \\ y_{n+1}=P_{Q}(y_{n}+\tau _{n}B^{*}(Ax_{n+1}-By_{n})), \end{cases} $$

(11)

where $\tau _{n}=\rho _{n}(\max \{\|A^{*}(Ax_{n}-By_{n})\|,\|B^{*}(Ax_{n}-By_{n}) \|, \delta \})^{-1}$.

Theorem 3.2

Let $\{(x_{n}, y_{n})\}$be the sequence generated by Algorithm 3.1. If $\{\rho _{n}\}$satisfies condition (7), then $\{(x_{n}, y_{n})\}$converges weakly to a solution of SEP (4).

Proof

Let $(x^{*}, y^{*})\in S$ be arbitrarily chosen. Then $x^{*}\in C$, $y^{*}\in Q$ and $Ax^{*}=By^{*}$. Let

$$ z_{n}=P_{Q}\bigl(y_{n}+\tau _{n}B^{*}(Ax_{n}-By_{n}) \bigr). $$

From Lemma 2.3 and the obvious fact that $x_{n}\in C$, it follows that

$$\begin{aligned} \Vert x_{n+1}-x_{n} \Vert &= \bigl\Vert P_{C}\bigl(x_{n}-\tau _{n}A^{*}(Ax_{n}-By_{n}) \bigr)-P_{C} x_{n} \bigr\Vert \\ &\leq \tau _{n} \bigl\Vert A^{*}(Ax_{n}-By_{n}) \bigr\Vert \leq \rho _{n}, \end{aligned}$$

(12)

and

$$\begin{aligned} \Vert y_{n+1}-z_{n} \Vert &= \bigl\Vert P_{Q}\bigl(y_{n}+\tau _{n}B^{*}(Ax_{n+1}-By_{n}) \bigr)-P_{Q}\bigl(y_{n}+ \tau _{n}B^{*}(Ax_{n}-By_{n}) \bigr) \bigr\Vert \\ &\leq \bigl\Vert \tau _{n}B^{*}(Ax_{n+1}-Ax_{n}) \bigr\Vert \\ &\leq \tau _{n} \Vert B \Vert \Vert A \Vert \Vert x_{n+1}-x_{n} \Vert \\ &\leq \rho _{n}^{2}\frac{1}{\delta } \Vert B \Vert \Vert A \Vert . \end{aligned}$$

(13)

On the other hand, in view of (11) and Lemma 2.3, we have

$$\begin{aligned} \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}&= \bigl\Vert P_{C}\bigl(x_{n}-\tau _{n}A^{*}(Ax_{n}-By_{n}) \bigr)-x^{*} \bigr\Vert ^{2} \\ &\leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}-2 \tau _{n}\bigl\langle A^{*}(Ax_{n}-By_{n}), x_{n}-x^{*} \bigr\rangle +\tau _{n}^{2} \bigl\Vert A^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}-2 \tau _{n}\bigl\langle Ax_{n}-By_{n}, Ax_{n}-Ax^{*} \bigr\rangle +\rho _{n}^{2}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}&= \bigl\Vert P_{Q}\bigl(y_{n}+\tau _{n}B^{*}(Ax_{n}-By_{n}) \bigr)-y^{*} \bigr\Vert ^{2} \\ &\leq \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}+2 \tau _{n}\bigl\langle B^{*}(Ax_{n}-By_{n}), y_{n}-y^{*} \bigr\rangle +\tau _{n}^{2} \bigl\Vert B^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}+2 \tau _{n}\bigl\langle Ax_{n}-By_{n}, By_{n}-By^{*} \bigr\rangle +\rho _{n}^{2}. \end{aligned}$$

By adding the last two inequalities and using the fact that $Ax^{*}=By^{*}$, we finally obtain

$$\begin{aligned} & \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2} \\ &\quad \leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2\tau _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}+2 \rho _{n}^{2}. \end{aligned}$$

(14)

It then follows from Young’s inequality, (13), and (14) that

$$\begin{aligned} & \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n+1}-y^{*} \bigr\Vert ^{2} \\ &\quad = \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n+1}-z_{n}+z_{n}-y^{*} \bigr\Vert ^{2} \\ &\quad \leq \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl(1+\rho _{n}^{2}\bigr) \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}+\biggl(1+ \frac{1}{\rho _{n}^{2}}\biggr) \Vert y_{n+1}-z_{n} \Vert ^{2} \\ &\quad \leq \bigl(1+\rho _{n}^{2}\bigr) \biggl( \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}+ \rho _{n}^{2} \Vert A \Vert ^{2} \Vert B \Vert ^{2}\frac{1}{\delta ^{2}} \biggr) \\ &\quad \leq \bigl(1+\rho _{n}^{2}\bigr) \biggl( \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2 \tau _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}+\rho _{n}^{2} \biggl(2+ \Vert A \Vert ^{2} \Vert B \Vert ^{2} \frac{1}{\delta ^{2}} \biggr) \biggr). \end{aligned}$$

Now, by setting $a_{n}=\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*}\|^{2}$, $\gamma _{n}=\rho _{n}^{2}$, $b_{n}=2\tau _{n}\|Ax_{n}-By_{n}\|^{2}$, and $c_{n}=\rho _{n}^{2} (2+\|A\|^{2}\|B\|^{2}\frac{1}{\delta ^{2}} )$, we obtain

$$\begin{aligned} a_{n+1}\leq (1+\gamma _{n}) (a_{n}-b_{n}+c_{n}). \end{aligned}$$

(15)

Applying Lemma 2.9 to (15), we conclude that there exists $a^{*}\geq 0$ such that $\lim_{n\rightarrow \infty }a_{n}=a^{*}$ and

$$ \sum_{n=0}^{\infty }\tau _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}< \infty . $$

Hence $\{a_{n}\}$ is bounded, and so are $\{x_{n}\}$ and $\{y_{n}\}$. Since A and B are bounded linear operators, there exists $M_{1}>0$ such that $\|A^{*}(Ax_{n}-By_{n})\|\leq M_{1}$, $\|B^{*}(Ax_{n}-By_{n})\|\leq M_{1}$, and $\|Ax_{n+1}-By_{n+1}\|+\|Ax_{n}-By_{n}\|\leq M_{1}$. So we have

$$ \sum_{n=0}^{\infty }\frac{\rho _{n}}{\max (M_{1},\delta )} \Vert Ax_{n}-By_{n} \Vert ^{2}\leq \sum _{n=0}^{\infty }\tau _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}< \infty . $$

This implies that

$$ \sum_{n=0}^{\infty }\rho _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}< \infty . $$

We next show $\lim_{n\to \infty }\|Ax_{n}-By_{n}\|^{2}=0$. From (13) and the obvious fact that $y_{n}\in Q$, it follows that

$$\begin{aligned} \Vert y_{n+1}-y_{n} \Vert &\leq \Vert y_{n+1}-z_{n} \Vert + \Vert z_{n}-y_{n} \Vert \\ &= \Vert y_{n+1}-z_{n} \Vert + \bigl\Vert P_{Q}\bigl(y_{n}+\tau _{n}B^{*}(Ax_{n}-By_{n}) \bigr)-P_{Q} y_{n} \bigr\Vert \\ &\leq \rho _{n}^{2}\frac{1}{\delta } \Vert B \Vert \Vert A \Vert +\tau _{n} \bigl\Vert B^{*}(Ax_{n}-By_{n}) \bigr\Vert \\ &\leq \rho _{n}^{2}\frac{1}{\delta } \Vert B \Vert \Vert A \Vert +\rho _{n}. \end{aligned}$$

Let $u_{n}=\|Ax_{n}-By_{n}\|^{2}$. By the last inequality and (12), we have

$$\begin{aligned} \vert u_{n+1}-u_{n} \vert &= \bigl\vert \Vert Ax_{n+1}-By_{n+1} \Vert - \Vert Ax_{n}-By_{n} \Vert \bigr\vert \bigl( \Vert Ax_{n+1}-By_{n+1} \Vert + \Vert Ax_{n}-By_{n} \Vert \bigr) \\ &\leq \bigl\Vert (Ax_{n+1}-By_{n+1})-(Ax_{n}-By_{n}) \bigr\Vert M_{1} \\ &\leq \bigl( \Vert A \Vert \Vert x_{n+1}-x_{n} \Vert + \Vert B \Vert \Vert y_{n+1}-y_{n} \Vert \bigr)M_{1} \\ &\leq \rho _{n}\biggl( \Vert A \Vert + \Vert B \Vert +\rho _{n}\frac{1}{\delta } \Vert B \Vert ^{2} \Vert A \Vert \biggr)M_{1} \\ &\leq \rho _{n}M_{2}, \end{aligned}$$

where $M_{2}$ is a positive number such that $(\|A\|+\|B\|+\rho _{n}\frac{1}{\delta }\|B\|^{2}\|A\|)M_{1}\leq M_{2}$ for all $n\geq 0$. Therefore, by Lemma 2.10, we have $\lim_{n\to \infty }\|Ax_{n}-By_{n}\|^{2}=0$.

Now we turn to prove that $\omega _{w}(x_{n}, y_{n})\subseteq S$. Let x̄ and ȳ be weak cluster points of the sequences $\{x_{n}\}$ and $\{y_{n}\}$, respectively. We assume that $x_{n_{k}}\rightharpoonup \bar{x}$ and $y_{n_{k}}\rightharpoonup \bar{y}$, where $\{x_{n_{k}}\}$ and $\{y_{n_{k}}\}$ are subsequences of $\{x_{n}\}$ and $\{y_{n}\}$, respectively. Since $\{x_{n_{k}}\}\subseteq C$, $\{y_{n_{k}}\}\subseteq Q$ and C and Q are closed and convex, we have $\bar{x}\in C$ and $\bar{y}\in Q$. Furthermore, the weak convergence of $\{Ax_{n_{k}}-By_{n_{k}}\}$ to $A\bar{x}-B\bar{y}$ and the weakly lower semicontinuity of the squared norm imply

$$ \Vert A\bar{x}-B\bar{y} \Vert ^{2}\leq \liminf _{k\to \infty } \Vert Ax_{n_{k}}-By_{n_{k}} \Vert ^{2}=0. $$

Hence $(\bar{x}, \bar{y})\in S$.

It is readily seen that $\lim_{n\rightarrow \infty }\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*} \|^{2}$ exists for each $(x^{*}, y^{*})\in S$ and $\omega _{w}(x_{n}, y_{n})\subseteq S$. Therefore, it follows from Lemma 2.8 that $\{(x_{n}, y_{n})\}$ converges weakly to a solution of SEP (4). □

Remark 3.3

It is clear that our choice of the stepsize in Algorithm 3.1 does not need any information on the values of $\|A\|$ and $\|B\|$.

4 Alternating iterative algorithm II

In this section, we propose another alternating iterative algorithm for problem (4), in which the choice of the stepsize does not need any prior information of operator norms.

Algorithm 4.1

Let $(x_{0}, y_{0})\in H_{1}\times H_{2}$ be arbitrary, $\delta >0$ be a constant, and $\{\rho _{n}\}$ be a sequence of positive real numbers. Given $(x_{n}, y_{n})$, construct $(x_{n+1}, y_{n+1})$ via the formula

$$ \textstyle\begin{cases} x_{n+1}=x_{n}-\tau _{n}[(I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n})], \\ y_{n+1}=y_{n}-\tau _{n}[(I-P_{Q})y_{n}-B^{*}(Ax_{n+1}-By_{n})], \end{cases} $$

(16)

where $\tau _{n}=\rho _{n}(\max \{\|(I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n})\|, \|(I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n})\|, \delta \})^{-1}$.

Theorem 4.2

Let $\{(x_{n}, y_{n})\}$be the sequence generated by Algorithm 4.1. If $\{\rho _{n}\}$satisfies condition (7), then $\{(x_{n}, y_{n})\}$converges weakly to a solution of SEP (4).

Proof

Let $(x^{*}, y^{*})\in S$ be arbitrarily chosen. Then $x^{*}\in C$, $y^{*}\in Q$, and $Ax^{*}=By^{*}$. Let

$$ z_{n}=y_{n}-\tau _{n}\bigl[(I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n}) \bigr]. $$

Then we have

$$\begin{aligned} \Vert y_{n+1}-z_{n} \Vert ^{2}&= \bigl\Vert \tau _{n}B^{*}(Ax_{n+1}-Ax_{n}) \bigr\Vert ^{2} \\ &\leq \tau _{n}^{2} \Vert B \Vert ^{2} \Vert A \Vert ^{2} \Vert x_{n+1}-x_{n} \Vert ^{2} \\ &\leq \tau _{n}^{2} \Vert B \Vert ^{2} \Vert A \Vert ^{2}\rho _{n}^{2} \\ &\leq \rho _{n}^{4}\frac{1}{\delta ^{2}} \Vert B \Vert ^{2} \Vert A \Vert ^{2}. \end{aligned}$$

(17)

It follows from (16) that

$$\begin{aligned} \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}&= \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}-2\tau _{n}\bigl\langle (I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n}), x_{n}-x^{*}\bigr\rangle \\ &\quad {}+\tau _{n}^{2} \bigl\Vert (I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}-2 \tau _{n}\bigl\langle (I-P_{C})x_{n}, x_{n}-x^{*} \bigr\rangle -2\tau _{n}\bigl\langle Ax_{n}-By_{n}, Ax_{n}-Ax^{*}\bigr\rangle \\ &\quad {}+\tau _{n}^{2} \bigl\Vert (I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}-2 \tau _{n} \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}-2\tau _{n} \bigl\langle Ax_{n}-By_{n}, Ax_{n}-Ax^{*}\bigr\rangle +\rho _{n}^{2}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}&= \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2\tau _{n}\bigl\langle (I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n}), y_{n}-y^{*}\bigr\rangle \\ &\quad {}+\tau _{n}^{2} \bigl\Vert (I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2 \tau _{n}\bigl\langle (I-P_{Q})y_{n}, y_{n}-y^{*} \bigr\rangle +2\tau _{n}\bigl\langle Ax_{n}-By_{n}, By_{n}-By^{*}\bigr\rangle \\ &\quad {}+\tau _{n}^{2} \bigl\Vert (I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n}) \bigr\Vert ^{2} \\ &\leq \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2 \tau _{n} \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+2\tau _{n} \bigl\langle Ax_{n}-By_{n}, By_{n}-By^{*}\bigr\rangle +\rho _{n}^{2}. \end{aligned}$$

By adding the last two inequalities and using the fact $Ax^{*}=By^{*}$, we finally obtain

$$\begin{aligned} \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}&\leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2\tau _{n} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2} \\ &\quad {}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr)+2\rho _{n}^{2}. \end{aligned}$$

(18)

It then follows from Young’s inequality, (17), and (18) that

$$\begin{aligned} & \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n+1}-y^{*} \bigr\Vert ^{2} \\ &\quad = \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n+1}-z_{n}+z_{n}-y^{*} \bigr\Vert ^{2} \\ &\quad \leq \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl(1+\rho _{n}^{2}\bigr) \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}+\biggl(1+ \frac{1}{\rho _{n}^{2}}\biggr) \Vert y_{n+1}-z_{n} \Vert ^{2} \\ &\quad \leq \bigl(1+\rho _{n}^{2}\bigr) \biggl( \bigl\Vert x_{n+1}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert z_{n}-y^{*} \bigr\Vert ^{2}+ \rho _{n}^{2}\frac{1}{\delta ^{2}} \Vert B \Vert ^{2} \Vert A \Vert ^{2} \biggr) \\ &\quad \leq \bigl(1+\rho _{n}^{2}\bigr) \biggl[ \bigl\Vert x_{n}-x^{*} \bigr\Vert ^{2}+ \bigl\Vert y_{n}-y^{*} \bigr\Vert ^{2}-2 \tau _{n} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2} \\ &\qquad {}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr)+\rho _{n}^{2} \biggl(2+\frac{1}{\delta ^{2}} \Vert B \Vert ^{2} \Vert A \Vert ^{2} \biggr) \biggr]. \end{aligned}$$

Now, by setting $a_{n}=\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*}\|^{2}$, $\gamma _{n}=\rho _{n}^{2}$, $c_{n}=\rho _{n}^{2} (2+\frac{1}{\delta ^{2}}\|B\|^{2}\|A\|^{2} )$ and

$$\begin{aligned} b_{n}=2\tau _{n} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr), \end{aligned}$$

we can immediately obtain

$$\begin{aligned} a_{n+1}\leq (1+\gamma _{n}) (a_{n}-b_{n}+c_{n}). \end{aligned}$$

(19)

Applying Lemma 2.9 to (19), we conclude that $\lim_{n\rightarrow \infty }a_{n}$ exists and

$$ \sum_{n=0}^{\infty }2\tau _{n} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr)< \infty . $$

Therefore $\{a_{n}\}$ is bounded, and so are $\{x_{n}\}$ and $\{y_{n}\}$. Observe that

$$ \bigl\Vert P_{C}x_{n}-x^{*} \bigr\Vert \leq \bigl\Vert x_{n}-x^{*} \bigr\Vert ,\qquad \bigl\Vert P_{Q}y_{n}-y^{*} \bigr\Vert \leq \bigl\Vert y_{n}-y^{*} \bigr\Vert . $$

This implies that $\{(I-P_{C})x_{n}\}$ and $\{(I-P_{Q})y_{n}\}$ are also bounded. Since A and B are bounded linear operators, there exists $M_{3}>0$ such that, for all $n\ge 0$, $\|(I-P_{C})x_{n}+A^{*}(Ax_{n}-By_{n})\|\leq M_{3}$, $\|(I-P_{Q})y_{n}-B^{*}(Ax_{n}-By_{n})\|\leq M_{3}$, $\|(I-P_{C})x_{n+1}\|+\|(I-P_{C})x_{n}\|< M_{3}$, and $\|(I-P_{Q})y_{n+1}\|+\|(I-P_{Q})y_{n}\|< M_{3}$. Thus, we have

$$\begin{aligned} &\sum_{n=0}^{\infty }\frac{\rho _{n}}{\max (M_{3},\delta )} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr) \\ &\quad \leq \sum_{n=0}^{\infty }\tau _{n} \bigl( \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}+ \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}+ \Vert Ax_{n}-By_{n} \Vert ^{2} \bigr)< \infty . \end{aligned}$$

This implies that

$$\begin{aligned} \sum_{n=0}^{\infty }\rho _{n} \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2}< \infty ,\qquad \sum_{n=0}^{ \infty }\rho _{n} \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2}< \infty , \end{aligned}$$

(20)

and

$$\begin{aligned} \sum_{n=0}^{\infty }\rho _{n} \Vert Ax_{n}-By_{n} \Vert ^{2}< \infty . \end{aligned}$$

We next prove that $\lim_{n\to \infty }\|(I-P_{C})x_{n}\|^{2}=0$, $\lim_{n\to \infty }\|(I-P_{Q})y_{n}\|^{2}=0$, and $\lim_{n\to \infty }\|Ax_{n}-By_{n}\|^{2}=0$. From (17), we get

$$ \Vert y_{n+1}-y_{n} \Vert \leq \Vert y_{n+1}-z_{n} \Vert + \Vert z_{n}-y_{n} \Vert \leq \rho _{n}^{2} \frac{1}{\delta } \Vert B \Vert \Vert A \Vert +\rho _{n}. $$

Observe that

$$\begin{aligned} & \bigl\vert \bigl\Vert (I-P_{C})x_{n+1} \bigr\Vert ^{2}- \bigl\Vert (I-P_{C})x_{n} \bigr\Vert ^{2} \bigr\vert \\ &\quad = \bigl\vert \bigl\Vert (I-P_{C})x_{n+1} \bigr\Vert - \bigl\Vert (I-P_{C})x_{n} \bigr\Vert \bigr\vert \bigl( \bigl\Vert (I-P_{C})x_{n+1} \bigr\Vert + \bigl\Vert (I-P_{C})x_{n} \bigr\Vert \bigr) \\ &\quad \leq \bigl\Vert (I-P_{C})x_{n+1}-(I-P_{C})x_{n} \bigr\Vert M_{3} \\ &\quad \leq \Vert x_{n+1}-x_{n} \Vert M_{3} \\ &\quad \leq M_{3}\rho _{n}. \end{aligned}$$

This along with (20) and Lemma 2.10 implies that $\lim_{n\to \infty }\|(I-P_{C})x_{n}\|^{2}=0$. Similarly, we have

$$\begin{aligned} & \bigl\vert \bigl\Vert (I-P_{Q})y_{n+1} \bigr\Vert ^{2}- \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert ^{2} \bigr\vert \\ &\quad = \bigl\vert \bigl\Vert (I-P_{Q})y_{n+1} \bigr\Vert - \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert \bigr\vert \bigl( \bigl\Vert (I-P_{Q})y_{n+1} \bigr\Vert + \bigl\Vert (I-P_{Q})y_{n} \bigr\Vert \bigr) \\ &\quad \leq \bigl\Vert (I-P_{Q})y_{n+1}-(I-P_{Q})y_{n} \bigr\Vert M_{3} \\ &\quad \leq \Vert y_{n+1}-y_{n} \Vert M_{3} \\ &\quad \leq \rho _{n}\biggl(\rho _{n}\frac{1}{\delta } \Vert B \Vert \Vert A \Vert +1\biggr)M_{3} \\ &\quad \leq M_{4}\rho _{n}, \end{aligned}$$

where $M_{4}$ is a positive number such that $(\rho _{n}\frac{1}{\delta }\|B\|\|A\|+1)M_{3}\leq M_{4}$ for all $n\geq 0$. Hence, by Lemma 2.10, we conclude that $\lim_{n\to \infty }\|(I-P_{Q})y_{n}\|^{2}=0$. Using a similar method in Theorem 3.2, we obtain $\lim_{n\to \infty }\|Ax_{n}-By_{n}\|^{2}=0$.

Now, we show that $\omega _{w}(x_{n}, y_{n})\subseteq S$. Let x̄ and ȳ be weak cluster points of the sequences $\{x_{n}\}$ and $\{y_{n}\}$, respectively. We assume that $\{x_{n_{k}}\}$ and $\{y_{n_{k}}\}$ are subsequences of $\{x_{n}\}$ and $\{y_{n}\}$ such that $x_{n_{k}}\rightharpoonup \bar{x}$ and $y_{n_{k}}\rightharpoonup \bar{y}$, respectively. Since $I-P_{C}$ and $I-P_{Q}$ are demiclosed at zero, it follows from $(I-P_{C})x_{n_{k}}\rightarrow 0$ and $(I-P_{Q})y_{n_{k}}\rightarrow 0$ that $\bar{x}\in C$ and $\bar{y}\in Q$. Furthermore, the weak convergence of $\{Ax_{n_{k}}-By_{n_{k}}\}$ to $A\bar{x}-B\bar{y}$ and the weakly lower semicontinuity of the squared norm imply

$$ \Vert A\bar{x}-B\bar{y} \Vert ^{2}\leq \liminf _{k\to \infty } \Vert Ax_{n_{k}}-By_{n_{k}} \Vert ^{2}=0. $$

Hence $(\bar{x}, \bar{y})\in S$. In summary, we have proved that $\lim_{n\rightarrow \infty }\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*} \|^{2}$ exists for each $(x^{*}, y^{*})\in S$ and $\omega _{w}(x_{n}, y_{n})\subseteq S$. Thus, we conclude from Lemma 2.8 that $\{(x_{n}, y_{n})\}$ converges weakly to a solution of SEP (4). □

5 A relaxed alternating iterative algorithm

In this section, we consider the case whenever $P_{C}$ or $P_{Q}$ fails to have a closed-form expression. Indeed, Moudafi [12] considered one of such cases when C and Q are level sets:

$$\begin{aligned} C=\bigl\{ x\in H_{1} \mid c(x)\leq 0\bigr\} \end{aligned}$$

(21)

and

$$\begin{aligned} Q=\bigl\{ y\in H_{2} \mid q(y)\leq 0\bigr\} , \end{aligned}$$

(22)

where $c:H_{1}\rightarrow \mathbb{R}$ and $q:H_{2}\rightarrow \mathbb{R}$ are two convex lower semicontinuous and subdifferential functions on $H_{1}$ and $H_{2}$, respectively. Here the subdifferential operators ∂c and ∂q of c and q are assumed to be bounded, i.e., bounded on bounded sets. In this case, it is known that the associated projections are very hard to calculate. To overcome this difficulty, Moudafi [12] presented the relaxed alternating CQ-algorithm (RACQA):

$$ \textstyle\begin{cases} x_{n+1}=P_{C_{n}}(x_{n}-\gamma A^{*}(Ax_{n}-By_{n})), \\ y_{n+1}=P_{Q_{n}}(y_{n}+\gamma B^{*}(Ax_{n+1}-By_{n})), \end{cases} $$

(23)

where $\{C_{n}\}$ and $\{Q_{n}\}$ are two sequences of closed convex sets defined by

$$\begin{aligned} C_{n}=\bigl\{ x\in H_{1} \mid c(x_{n})+\langle \xi _{n}, x-x_{n}\rangle \leq 0 \bigr\} , \quad \xi _{n}\in \partial c(x_{n}) \end{aligned}$$

(24)

and

$$\begin{aligned} Q_{n}=\bigl\{ y\in H_{2} \mid q(y_{n})+\langle \eta _{n}, y-y_{n}\rangle \leq 0 \bigr\} , \quad \eta _{n}\in \partial q(y_{n}). \end{aligned}$$

(25)

Since $C_{n}$ and $Q_{n}$ are clearly half-spaces, the associated projections thus have closed form expressions. This indicates that the implementation of RACQA is very easy. Under suitable conditions, Moudafi [12] proved that the RACQA converges weakly to a solution of (4).

Following the RACQA and our proposed algorithms, we now present a relaxed alternating iterative algorithm in which we just need projections onto half-spaces [16, 27, 28].

In what follows, we will treat SEP (4) under the following assumptions.

(A1):

The sets C and Q are given in (21) and (22), respectively.

(A2):

For any $x\in H_{1}$ and $y\in H_{2}$, at least one subgradient $\xi \in \partial c(x)$ and $\eta \in \partial q(y)$ can be calculated.

Remark 5.1

It follows from Lemma 2.6 that both c and q are weakly lower semicontinuous by condition (A2), since $c:H_{1}\rightarrow \mathbb{R}$ and $q:H_{2}\rightarrow \mathbb{R}$ are convex.

We are now in a position to present a relaxed alternative iterative algorithm that does not depend on operator norms for solving SEP (4).

Algorithm 5.2

Let $(x_{0}, y_{0})$ be arbitrary, $\delta >0$ be a constant, and $\{\rho _{n}\}$ be a sequence of positive real numbers. Given $(x_{n}, y_{n})$, construct $(x_{n+1}, y_{n+1})$ via the formula

$$ \textstyle\begin{cases} x_{n+1}=x_{n}-\tau _{n} [(I-P_{C_{n}})x_{n}+A^{*}(Ax_{n}-By_{n}) ], \\ y_{n+1}=y_{n}-\tau _{n} [(I-P_{Q_{n}})y_{n}-B^{*}(Ax_{n+1}-By_{n}) ], \end{cases} $$

(26)

where $C_{n}$ and $Q_{n}$ are given as in (24) and (25), respectively, and

$$ \tau _{n}=\rho _{n}\bigl(\max \bigl\{ \bigl\Vert (I-P_{C_{n}})x_{n}+A^{*}(Ax_{n}-By_{n}) \bigr\Vert , \bigl\Vert (I-P_{Q_{n}})y_{n}-B^{*}(Ax_{n}-By_{n}) \bigr\Vert , \delta \bigr\} \bigr)^{-1}. $$

Remark 5.3

By the definition of the subgradient, it is obvious that $C\subseteq C_{n}$ and $Q\subseteq Q_{n}$ for all $n\geq 0$. Since $C_{n}$ and $Q_{n}$ are both half-spaces, the projections onto $C_{n}$ and $Q_{n}$ can be directly calculated. Thus Algorithm 5.2 is easily implementable.

Theorem 5.4

Let $\{(x_{n}, y_{n})\}$be the sequence generated by Algorithm 5.2. If $\{\rho _{n}\}$satisfies condition (7), then $\{(x_{n}, y_{n})\}$converges weakly to a solution of SEP (4).

Proof

Take $(x^{*}, y^{*})\in S$, i.e., $x^{*}\in C$ (and thus $x^{*}\in C_{n}$); $y^{*}\in Q$ (and thus $y^{*}\in Q_{n}$), $Ax^{*}=By^{*}$. Similarly as in the proof of Theorem 4.2, we can conclude that $\lim_{n\to \infty } (\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*}\|^{2} )$ exists, $\lim_{n\to \infty }\|(I-P_{C_{n}})x_{n}\|^{2}=0$, $\lim_{n\to \infty }\|(I-P_{Q_{n}})y_{n}\|^{2}=0$, and $\lim_{n\to \infty }\|Ax_{n}-By_{n}\|^{2}=0$.

We next show that $\omega _{w}(x_{n}, y_{n})\subseteq S$. Since $\lim_{n\to \infty } (\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*}\|^{2} )$ exists, the sequences $\{x_{n}\}$ and $\{y_{n}\}$ are bounded. Let x̄ and ȳ be weak cluster points of the sequences $\{x_{n}\}$ and $\{y_{n}\}$, respectively. Without loss of generality, we assume that $x_{n}\rightharpoonup \bar{x}$ and $y_{n}\rightharpoonup \bar{y}$. Since ∂c is bounded on bounded sets, there is a constant $\sigma _{1}>0$ such that $\|\xi _{n}\|\leq \sigma _{1}$ for all $n\geq 0$. From (24) and the fact that $P_{C_{n}}(x_{n})\in C_{n}$, it follows that

$$\begin{aligned} c(x_{n})&\leq \langle \xi _{n}, x_{n}-P_{C_{n}}x_{n} \rangle \leq \sigma _{1} \bigl\Vert (I-P_{C_{n}})x_{n} \bigr\Vert . \end{aligned}$$

The weakly lower semicontinuity of c leads to

$$\begin{aligned} c(\bar{x})\leq \liminf_{n\rightarrow +\infty }c(x_{n})\leq \sigma _{1} \liminf_{n\to +\infty } \bigl\Vert (I-P_{C_{n}})x_{n} \bigr\Vert =0, \end{aligned}$$

and therefore $\bar{x}\in C$. Likewise, since ∂q is bounded on bounded sets, there is a constant $\sigma _{2} >0$ such that $\|\eta _{n}\|\leq \sigma _{2}$ for all $n\geq 0$. From (25) and the fact that $P_{Q_{n}}(y_{n})\in Q_{n}$, it follows that

$$\begin{aligned} q(y_{n})&\leq \langle \eta _{n}, y_{n}-P_{Q_{n}}y_{n} \rangle \leq \sigma _{2} \bigl\Vert (I-P_{Q_{n}})y_{n} \bigr\Vert . \end{aligned}$$

Again, the weakly lower semicontinuity of q leads to

$$\begin{aligned} q(\bar{y})\leq \liminf_{n\rightarrow +\infty }q(y_{n})\leq \sigma _{2} \liminf_{n\to +\infty } \bigl\Vert (I-P_{Q_{n}})y_{n} \bigr\Vert =0, \end{aligned}$$

and therefore $\bar{y}\in Q$. Furthermore, the weak convergence of $\{Ax_{n}-By_{n}\}$ to $A\bar{x}-B\bar{y}$ and the weakly lower semicontinuity of the squared norm imply

$$\begin{aligned} \Vert A\bar{x}-B\bar{y} \Vert ^{2}\leq \liminf _{n\rightarrow +\infty } \Vert Ax_{n}-By_{n} \Vert ^{2}=0. \end{aligned}$$

Hence $(\bar{x}, \bar{y})\in S$.

Finally, we deduce from Lemma 2.8 that $\{(x_{n}, y_{n})\}$ converges weakly to a solution of SEP (4), since $\lim_{n\rightarrow \infty }\|x_{n}-x^{*}\|^{2}+\|y_{n}-y^{*} \|^{2}$ exists for each $(x^{*}, y^{*})\in S$ and $\omega _{w}(x_{n}, y_{n})\subseteq S$. □