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An iterative method for split hierarchical monotone variational inclusions
Fixed Point Theory and Applicationsvolume 2015, Article number: 121 (2015)
Abstract
In this paper, we introduce a split hierarchical monotone variational inclusion problem (SHMVIP) which includes split variational inequality problems, split monotone variational inclusion problems, split hierarchical variational inequality problems, etc., as special cases. An iterative algorithm is proposed to compute the approximate solutions of an SHMVIP. The weak convergence of the sequence generated by the proposed algorithm is studied. We present an example to illustrate our algorithm and convergence result.
Introduction
Let $H_{1}$ and $H_{2}$ be real Hilbert spaces, $C \subseteq H_{1}$ and $Q \subseteq H_{2}$ be nonempty, closed, and convex sets, $A : H_{1}\rightarrow H_{2}$ be a bounded linear operator, and $f : H_{1} \rightarrow H_{1}$ and $g : H_{2}\rightarrow H_{2}$ be two given operators. Recently, Censor et al. [1] introduced the following split variational inequality problem (SVIP):
and such that
Let Λ denote the solution set of the SVIP, that is,
If f and g are convex and differentiable, then the SVIP is equivalent to the following split minimization problem:
and such that
For further details on the equivalence between a variational inequality and an optimization problem, we refer to [2]. The SVIP also contains the split feasibility problem (SFP) [3] as a special case. For further details of the SFP, we refer to [4, 5] and the references therein.
If the sets C and Q are the set of fixed points of the operators $T : H_{1} \to H_{1}$ and $S : H_{2} \to H_{2}$, respectively, then the SVIP is called a split hierarchical variational inequality problem (SHVIP). It is introduced and studied by Ansari et al. [6]. Several special cases of a SHVIP, namely, the split convex minimization problem, the split variational inequality problem defined over the solution set of monotone variational inclusion problem, the split variational inequality problem defined over the solution set of equilibrium problem, are also considered in [6].
Let $B_{1} : H_{1} \rightrightarrows H_{1}$ and $B_{2} : H_{2} \rightrightarrows H_{2}$ be setvalued mappings with nonempty values, and let $f : H_{1} \to H_{1}$ and $g : H_{2} \to H_{2}$ be mappings. Then, inspired by the work in [1], Moudafi [7] introduced the following split monotone variational inclusion problem (SMVIP):
and such that
Let Γ denote the solution set of SMVIP, that is,
To solve the SMVIP, Moudafi [7] proposed the following iterative method: Let $\lambda> 0$ and $x_{0}$ be the initial guess. Compute
where $\gamma\in(0, 1/L)$ with L being the spectral radius of the operator $A^{*}A$, $U = J_{\lambda}^{B_{1}}(I\lambda f)$, $T = J_{\lambda}^{B_{2}}(I\lambda g)$, and $J_{\lambda}^{B_{1}}$ and $J_{\lambda}^{B_{2}}$ are the resolvents of $B_{1}$ and $B_{2}$, respectively, with parameter λ (see [8]). He obtained the following weak convergence result for iterative method (1.7).
Theorem 1.1
([7], Theorem 2.1)
Given a bounded linear operator $A : H_{1} \rightarrow H_{2}$. Let $f : H_{1}\rightarrow H_{1}$ and $g : H_{2}\rightarrow H_{2}$ be $\alpha_{1}$ and $\alpha_{2}$ inverse strongly monotone operators on $H_{1}$ and $H_{2}$, respectively, and $B_{1}$, $B_{2}$ be two maximal monotone operators, and set $\alpha:= \min\{\alpha_{1},\alpha_{2}\}$. Consider the operator $U:= J_{\lambda}^{B_{1}}(I\lambda f)$, $T:= J_{\lambda}^{B_{2}}(I\lambda g)$ with $\lambda\in(0,2\alpha)$. Then the sequence $\{x_{n}\}$ generated by (1.7) converges weakly to an element $x^{*}\in\Gamma$, provided that $\Gamma\neq\emptyset$ and $\gamma \in(0, 1/L)$.
Let $T : H_{1}\rightarrow H_{1}$ and $S : H_{2}\rightarrow H_{2}$ be operators such that $\operatorname{Fix}(T)\neq\emptyset$ and $\operatorname{Fix}(S)\neq\emptyset$, where $\operatorname{Fix}(T)$ and $\operatorname{Fix}(S)$ denote the set of fixed points of T and S, respectively. Inspired by the work in [6] and [7], in this paper, we introduce the following split hierarchical monotone variational inclusion problem (SHMVIP):
and such that
We denote by Ψ the set of solutions of the SHMVIP, that is,
We propose an iterative algorithm to compute the approximate solutions of the SHMVIP. The weak convergence of the sequence generated by the proposed algorithm is studied. An example is presented to illustrate the proposed algorithm and result.
Preliminaries
Let H be a real Hilbert space whose inner product and norm are denoted by $\langle\cdot,\cdot\rangle$ and $\\cdot\$, respectively. We denote by $x_{n} \rightarrow x$ (respectively, $x_{n}\rightharpoonup x$) the strong (respectively, weak) convergence of the sequence $\{x_{n}\}$ to x. Let $T : H \rightarrow H$ be an operator whose range is denoted by $R(T)$. The set of all fixed points of T is denoted by $\operatorname{Fix}(T)$, that is, $\operatorname{Fix}(T) = \{ x \in H : x = Tx \}$.
Definition 2.1
An operator $T : H \rightarrow H$ is said to be:

(a)
nonexpansive if $\TxTy\ \leq\xy\$, for all $x, y \in H$;

(b)
strongly nonexpansive [6, 9, 10] if T is nonexpansive and
$$\lim_{n\rightarrow\infty}\bigl\ (x_{n}y_{n})(Tx_{n}Ty_{n}) \bigr\ =0, $$whenever $\{x_{n}\}$ and $\{y_{n}\}$ are bounded sequences in H and $\lim_{n\rightarrow\infty}(\x_{n}y_{n}\\Tx_{n}Ty_{n}\)=0$;

(c)
averaged nonexpansive [11] if it can be written as
$$T = (1\alpha)I +\alpha S, $$where $\alpha\in(0,1)$, I is the identity operator on H, and $S:H\rightarrow H$ is a nonexpansive mapping;

(d)
firmly nonexpansive if $\TxTy\^{2}\leq\langle xy, TxTy\rangle$, for all $x,y \in H$;

(e)
αinverse strongly monotone (αism) if there exists a constant $\alpha> 0$ such that
$$\langle TxTy, xy\rangle\geq\alpha\TxTy\^{2}, \quad \mbox{for all } x ,y \in H. $$
The following example shows that every nonexpansive operator is not necessarily strongly nonexpansive.
Example 2.1
Let $T : [1,1] \to \mathbb {R}$ be defined by $Tx = x$, for all $x \in[1,1]$. Then T is nonexpansive but not strongly nonexpansive.
Indeed, let $x_{n} = 1$ and $y_{n}= 0$, for all n. Then $\{x_{n}\}$ and $\{ y_{n} \}$ are bounded sequences. Also,
Thus, T is not strongly nonexpansive.
The following result will be used in the sequel.
Proposition 2.1
[11]
Let $T : H \rightarrow H$ be an operator.

(i)
If T is νism, then for $\gamma> 0$, γT is $\frac{\nu}{\gamma}$ism.

(ii)
T is averaged if and only if the complement $IT$ is νism for some $\nu> \frac{1}{2}$. Indeed, for $\alpha\in(0,1)$, T is αaveraged if and only if $IT$ is $\frac{1}{2\alpha}$ism.

(iii)
The composite of finitely many averaged mappings is averaged.
Let $\varphi: H \rightarrow H$ be a given singlevalued αinverse strongly monotone operator and $\lambda\in(0, 2\alpha)$. Then $(I\lambda\varphi)$ is averaged. Indeed, since φ is αinverse strongly monotone, λφ is $\frac{\alpha}{\lambda}$ism. Thus, $I\lambda\varphi$ is averaged as $\frac{\alpha}{\lambda} > \frac{1}{2}$.
Recall that a Banach space X is said to satisfy Opial’s condition [12] if whenever $\{x_{n}\}$ is a sequence in X which converges weakly to x as $n\to\infty$, then
It is well known that every Hilbert space satisfies Opial’s condition.
Lemma 2.1
(Demiclosedness principle) [12], Lemma 2
Let C be a nonempty, closed, and convex subset of a real Hilbert space H and $T : C \rightarrow C $ be a nonexpansive operator with $\operatorname{Fix}(T) \neq\emptyset$. If the sequence $\{x_{n}\} \subseteq C$ converges weakly to x and the sequence $\{(IT)x_{n}\}$ converges strongly to y, then $(IT)x = y$. In particular, if $y = 0$, then $x \in\operatorname{Fix}(T)$.
Let $B : H \rightrightarrows H$ be a setvalued mapping. The domain, range, and inverse of B are denoted by
respectively.
Definition 2.2
The setvalued mapping $B : H \rightrightarrows H$ with nonempty values is said to be:

(a)
monotone if
$$\langle uv, xy\rangle\geq0, \quad\mbox{for all } u \in Bx, v \in By; $$ 
(b)
maximal monotone if it is monotone and the graph
$$G(B) =\bigl\{ (x,u) \in H \times H : u \in Bx \bigr\} $$of B is not properly contained in the graph of any other monotone operator.
Algorithm and convergence result
It is well known that when the setvalued mapping $B : H \rightrightarrows H$ is maximal monotone, then for each $x \in H$ and $\lambda> 0$, there is a unique $z \in H $ such that $x \in(I+\lambda B)z$ [13, 14]. In this case, the operator $J_{\lambda}^{B} := (I+\lambda B)^{1}$ is called resolvent of B with parameter λ. It is well known that $J_{\lambda}^{B}$ is a singlevalued and firmly nonexpansive mapping.
Indeed, for any given $u \in H$, let $x, y \in J_{\lambda}^{B}(u)$. Then $x, y \in(I+\lambda B)^{1}(u)$, and thus $ux \in\lambda Bx$ and $uy \in\lambda B y$. The monotonicity of λB implies that
This implies that $\xy\ \leq0$, and thus $x=y$. Hence, $J_{\lambda }^{B}$ is singlevalued.
Next we show that $J_{\lambda}^{B}$ is firmly nonexpansive mapping. For any $x, y \in H $, let $J_{\lambda}^{B}(x)=(I+\lambda B)^{1}(x)$ and $J_{\lambda}^{B}(y)=(I+\lambda B)^{1}(y)$. This implies that $x \in(I+\lambda B)(J_{\lambda}^{B}(x))$ and $y \in(I+\lambda B)(J_{\lambda}^{B}(y))$. It follows that $\frac{1}{\lambda} ( x(J_{\lambda}^{B}(x)) ) \in B(J_{\lambda}^{B}(x))$ and $\frac{1}{\lambda} ( y(J_{\lambda}^{B}(y)) ) \in B(J_{\lambda}^{B}(y))$. The monotonicity of B implies that
that is,
Thus, $J_{\lambda}^{B}$ is firmly nonexpansive.
Let $\phi: H \rightarrow H$ be a given singlevalued operator. Then
Indeed, let $x^{*} \in\operatorname{Fix}(J_{\lambda}^{B}(I\lambda\phi )(x^{*}))$. Then $x^{*} = J_{\lambda}^{B}(I\lambda\phi)(x^{*})$. It follows that
Since $J_{\lambda}^{B}$ is firmly nonexpansive, and therefore, averaged. It is well known that the composition of averaged mapping is averaged, thus $J_{\lambda}^{B}(I\lambda\varphi)$ is averaged. Since every averaged mapping is strongly nonexpansive [10], it follows that $J_{\lambda }^{B}(I\lambda\varphi)$ is also strongly nonexpansive.
Let $B_{1} : H_{1} \rightrightarrows H_{1}$ and $B_{2} : H_{2} \rightrightarrows H_{2}$ be setvalued mappings with nonempty values, and let $f : H_{1} \to H_{1}$ and $g : H_{2} \to H_{2}$ be mappings. Let $T : H_{1}\rightarrow H_{1}$ and $S : H_{2}\rightarrow H_{2}$ be operators such that $\operatorname{Fix}(T)\neq\emptyset$ and $\operatorname{Fix}(S)\neq\emptyset$. Let $U := J_{\lambda}^{B_{1}}(I\lambda f)$ and $V := J_{\lambda}^{B_{2}}(I\lambda g)$. With the help of (3.1), (1.8), and (1.9) can be rewritten as
and such that
Now we propose the following algorithm to compute the approximate solutions of the SHMVIP.
Algorithm 3.1
Initialization: Let $\lambda> 0$ and take arbitrary $x_{0} \in H_{1}$.
Iterative step: For a given current $x_{n}\in H_{1}$, compute
where $\gamma\in (0, \frac{1}{\A\^{2}} )$.
Last step: Update $n := n+1$.
Next we prove the weak convergence of the sequence generated by Algorithm 3.1.
Theorem 3.1
Let $A : H_{1}\rightarrow H_{2}$ be a bounded linear operator, $f : H_{1}\rightarrow H_{1}$ be an $\alpha_{1}$inverse strongly monotone operator, $T : H_{1}\rightarrow H_{1}$ be a strongly nonexpansive operator such that $\operatorname{Fix}(T) \neq\emptyset$, $g : H_{2}\rightarrow H_{2}$ be an $\alpha_{2}$inverse strongly monotone operator, $S : H_{2}\rightarrow H_{2}$ be a nonexpansive operator such that $\operatorname{Fix}(S)\neq\emptyset$, and $\alpha:= \min\{\alpha_{1},\alpha_{2}\}$. Consider the operator $U := J_{\lambda}^{B_{1}}(I\lambda f)$ and $V := J_{\lambda}^{B_{2}}(I\lambda g)$ with $\lambda\in(0,2\alpha)$, and $B_{1} : H_{1} \rightrightarrows H_{1}$ and $B_{2} : H_{2} \rightrightarrows H_{2}$ are two maximal monotone setvalued mappings with nonempty values. Then the sequence $\{x_{n}\}$ generated by Algorithm 3.1 converges weakly to an element $x^{*}\in\Psi$, provided $\Psi\neq\emptyset$.
Proof
Let $p\in\Psi$. Then $Tp=p$, $Up=p$, $SAp=Ap$, and $VAp=Ap$. Let $y_{n}:= x_{n}+\gamma A^{*}(SVI)Ax_{n}$ and consider
Consider the third term of inequality (3.5), we have
Combining (3.5) and (3.6), we obtain
Since $\gamma\in(0,\frac{1}{\A\^{2}})$, we have $\gamma(1\gamma\ A\^{2}) >0$, and thus,
From the above inequality (3.7), we have
This shows that $\x_{n+1}p\\leq\x_{n}p\$ and this implies that $\{\x_{n}p\\}_{n=1}^{\infty}$ is a monotonic decreasing sequence, also $\{x_{n}\}$ is a bounded sequence, see [15], Theorem 4.5.10 and, hence, $\lim_{n\rightarrow\infty}\x_{n}p\$ exists. Taking the limit at both sides in (3.9), and noticing that $\gamma(1\gamma\A\^{2}) >0$, we have
and since $y_{n} := x_{n}+\gamma A^{*}(SVI)Ax_{n}$, we have $\ y_{n}x_{n}\= \gamma\A\\(SVI)Ax_{n}\$ thus in view of (3.10) we have
Since $\{x_{n}\}$ is a bounded sequence and it has a weakly convergent subsequence say, $x_{n_{i}}\rightharpoonup x^{*}$, [10] or with the help of Opial’s condition [12], we can see that $x_{n}\rightharpoonup x^{*}$. Thus, we have $Ax_{n}\rightharpoonup Ax^{*}$. Since SV is nonexpansive, from (3.10) and the closedness of $SVI$ at 0 we obtain $SV Ax^{*}=Ax^{*}$. Next we show that $VAx^{*}=Ax^{*}$. We have
Taking the limit at both sides and by using (3.10), we obtain
Since $S Ap = Ap$ and $V Ap = Ap$, by the nonexpansiveness of S and V, we have
and therefore
Thus, we have
From (3.12) and (3.13), we obtain
Since V is averaged nonexpansive and every averaged nonexpansive map is strongly nonexpansive, we see that V is strongly nonexpansive. Since $\{Ap\}$ and $\{Ax_{n}\}$ are bounded sequences, by the definition of strong nonexpansiveness of V, we have
Since V is nonexpansive, by the demiclosedness principle, we have
Now, we show that $Tx^{*}=x^{*}$ and $Ux^{*}=x^{*}$. By using the nonexpansiveness of T and U, in view of (3.4) and (3.8), we have
This implies that
From (3.8), we see that $\{y_{n}\}$ is a bounded sequence and thus $\{Uy_{n}\}$ is bounded and since $\{p\}$ is a constant sequence thus $\{p\}$ is also bounded and since T is strongly nonexpansive, we have
In view of (3.4) and (3.8), by using the nonexpansiveness of TU, we have
It follows that
By using the nonexpansiveness of T and U, we have
and therefore
Thus, from (3.17), we have
From (3.8) we see that $\{y_{n}\}$ is a bounded sequence and $\{p\}$ being a constant sequence, also bounded, by the strong nonexpansiveness of U, we have
Next consider, for all $f \in H$,
Since $x_{n}\rightharpoonup x^{*}$ and from (3.11), we have $\lim_{n\rightarrow\infty}\f(y_{n})f(x^{*})\=0$, thus $y_{n}\rightharpoonup x^{*}$. Thus, in view of (3.19) and by applying the demiclosedness principle, we have $Ux^{*}=x^{*}$. Again, since $y_{n}\rightharpoonup x^{*}$, in view of (3.19), we have $U y_{n}\rightharpoonup x^{*}$. Thus, again in view of (3.16) and by applying the demiclosedness principle, we have $Tx^{*}=x^{*}$. □
Now, we illustrate Algorithm 3.1 and Theorem 3.1 by the following example.
Example 3.1
Let $H_{1} = H_{2} = H = \mathbb{R}$ and $B : H \rightrightarrows H$ be defined by
Then, as shown in [16], B is a setvalued maximal monotone mapping. We define the mappings $A, f, h, T, S : H \rightarrow H$ by
and
respectively. It is easy to show that A is a bounded linear operator, f and h are $\frac{1}{3}$ism, T is firmly nonexpansive, and thus T is strongly nonexpansive [10] and S is nonexpansive. Let $B_{1}(x)= B_{2}(x)=Bx$. Then $B_{1}$ and $B_{2}$ are maximal monotone setvalued mappings. Let $J_{\lambda}^{B_{1}}(x) = J_{\lambda}^{B_{2}}(x)= \frac{x}{2}$ be the resolvent operator. The values of $\{x_{n}\}$ with different values of n are reported in the Table 1. All codes were written in Matlab R2010.
Table 1 shows that the sequence $\{x_{n}\}$ converges to 0, which is the required solution.
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MSC
 49J40
 47J20
 49J52
 49M05
Keywords
 split hierarchical monotone variational inclusions
 fixed point problems
 iterative method
 maximal monotone setvalued mappings
 resolvent operators
 convergence analysis