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# Hierarchical convergence of an implicit double-net algorithm for nonexpansive semigroups and variational inequality problems

Fixed Point Theory and Applications20112011:101

https://doi.org/10.1186/1687-1812-2011-101

• Accepted: 20 December 2011
• Published:

## Abstract

In this paper, we show the hierarchical convergence of the following implicit double-net algorithm:

${x}_{s,t}=s\left[tf\left({x}_{s,t}\right)+\left(1-t\right)\left({x}_{s,t}-\mu A{x}_{s,t}\right)\right]+\left(1-s\right)\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(v\right){x}_{s,t}d\nu ,\phantom{\rule{1em}{0ex}}\forall s,t\in \left(0,1\right),$

where f is a ρ-contraction on a real Hilbert space H, A : H → H is an α-inverse strongly monotone mapping and S = {T(s)}s ≥ 0: HH is a nonexpansive semi-group with the common fixed points set Fix(S) ≠ , where Fix(S) denotes the set of fixed points of the mapping S, and, for each fixed t (0, 1), the net {x s, t } converges in norm as s → 0 to a common fixed point x t Fix(S) of {T(s)}s ≥ 0and, as t → 0, the net {x t } converges in norm to the solution x* of the following variational inequality:

$\left\{\begin{array}{c}{x}^{*}\in Fix\left(S\right);\hfill \\ \hfill \begin{array}{cc}\hfill 〈A{x}^{*},\phantom{\rule{2.77695pt}{0ex}}x-{x}^{*}〉\ge 0,\hfill & \hfill \forall x\in Fix\left(S\right).\hfill \end{array}\hfill \end{array}\right\$

MSC(2000): 49J40; 47J20; 47H09; 65J15.

## Keywords

• fixed point
• variational inequality
• double-net algorithm
• hierarchical convergence
• Hilbert space

## 1 Introduction

In nonlinear analysis, a common approach to solving a problem with multiple solutions is to replace it by a family of perturbed problems admitting a unique solution and to obtain a particular solution as the limit of these perturbed solutions when the perturbation vanishes.

In this paper, we introduce a more general approach which consists in finding a particular part of the solution set of a given fixed point problem, i.e., fixed points which solve a variational inequality. More precisely, the goal of this paper is to present a method for finding hierarchically a fixed point of a nonexpansive semigroup S = {T(s)}s ≥ 0with respect to another monotone operator A, namely,

Find x* Fix(S) such that
$〈A{x}^{*},\phantom{\rule{2.77695pt}{0ex}}x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall x\in Fix\left(S\right).$
(1.1)

This is an interesting topic due to the fact that it is closely related to convex programming problems. For the related works, refer to .

This paper is devoted to solve the problem (1.1). For this purpose, we propose a double-net algorithm which generates a net {x s ,t} and prove that the net {x s ,t} hierarchically converges to the solution of the problem (1.1), that is, for each fixed t (0, 1), the net {x s ,t} converges in norm as s → 0 to a common fixed point x t Fix(S) of the nonexpansive semigroup {T(s)}s ≥ 0and, as t → 0, the net {x t } converges in norm to the unique solution x* of the problem (1.1).

## 2 Preliminaries

Let H be a real Hilbert space with inner product 〈·, ·〉 and norm ||·||, respectively. Recall a mapping f : H → H is called a contraction if there exists ρ [0, 1) such that
$||f\left(x\right)-f\left(y\right)||\phantom{\rule{2.77695pt}{0ex}}\le \rho ||x-y||,\phantom{\rule{1em}{0ex}}\forall x,\phantom{\rule{2.77695pt}{0ex}}y\in H.$
A mapping T : C → C is said to be nonexpansive if
$||Tx-Ty||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||x-y||,\phantom{\rule{1em}{0ex}}\forall x,\phantom{\rule{2.77695pt}{0ex}}y\in H.$

Denote the set of fixed points of the mapping T by Fix(T).

Recall also that a family S : = {T(s)}s ≥ 0of mappings of H into itself is called a nonexpansive semigroup if it satisfies the following conditions:

(S1) T(0)x = x for all x H;

(S2) T(s + t) = T(s)T(t) for all s, t ≥ 0;

(S3) ||T(s)x - T(s)y|| ≤ ||x - y|| for all x, y H and s ≥ 0;

(S4) for all x H, s → T(s)x is continuous.

We denote by Fix(T(s)) the set of fixed points of T(s) and by Fix(S) the set of all common fixed points of S, i.e., Fix(S) = s ≥ 0Fix(T(s)). It is known that Fix(S) is closed and convex (, Lemma 1).

A mapping A of H into itself is said to be monotone if
$〈Au\phantom{\rule{2.77695pt}{0ex}}-Av,\phantom{\rule{2.77695pt}{0ex}}u-v〉\phantom{\rule{2.77695pt}{0ex}}\ge 0,\phantom{\rule{1em}{0ex}}\forall u,\phantom{\rule{2.77695pt}{0ex}}v\phantom{\rule{2.77695pt}{0ex}}\in H,$
and A : C → H is said to be α-inverse strongly monotone if there exists a positive real number α such that
$〈Au-Av,\phantom{\rule{2.77695pt}{0ex}}u-v〉\ge \alpha {||Au-Av||}^{2},\phantom{\rule{1em}{0ex}}\forall u,\phantom{\rule{2.77695pt}{0ex}}v\in H.$

It is obvious that any α-inverse strongly monotone mapping A is monotone and $\frac{1}{\alpha }$-Lipschitz continuous.

Now, we introduce some lemmas for our main results in this paper.

Lemma 2.1.  Let H be a real Hilbert space. Let the mapping A : H → H be α-inverse strongly monotone and μ > 0 be a constant. Then, we have
${||\left(I-\mu A\right)x-\left(I-\mu A\right)y||}^{2}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}{||x-y||}^{2}+\phantom{\rule{2.77695pt}{0ex}}\mu \left(\mu -2\alpha \right){||Ax-Ay||}^{2},\phantom{\rule{1em}{0ex}}\forall x,\phantom{\rule{2.77695pt}{0ex}}y\in H.$

In particular, if 0 ≤ μ ≤ 2α, then I - μA is nonexpansive.

Lemma 2.2.  Let C be a nonempty bounded closed convex subset of a Hilbert space H and {T(s)}s ≥ 0be a nonexpansive semigroup on C. Then, for all h ≥ 0,
$\underset{t\to \infty }{lim}\underset{x\in C}{sup}∥\frac{1}{t}{\int }_{0}^{t}T\left(s\right)xds-T\left(h\right)\frac{1}{t}{\int }_{0}^{t}T\left(s\right)xds∥=0.$

Lemma 2.3.  (Demiclosedness Principle for Nonexpansive Mappings) Let C be a nonempty closed convex subset of a real Hilbert space H and T : C → C be a nonexpansive mapping with Fix(T) ≠ . If {x n } is a sequence in C converging weakly to a point x C and {(I - T)x n } converges strongly to a point y C, then (I - T)x = y. In particular, if y = 0, then x Fix(T).

Lemma 2.4. Let H be a real Hilbert space. Let f : H → H be a ρ-contraction with coefficient ρ [0, 1) and A : H → H be an α-inverse strongly monotone mapping. Let μ (0, 2α) and t (0, 1). Then, the variational inequality
$\left\{\begin{array}{c}{x}^{*}\in Fix\left(S\right);\hfill \\ \hfill 〈tf\left(z\right)\phantom{\rule{2.77695pt}{0ex}}+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-z〉\phantom{\rule{2.77695pt}{0ex}}\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right),\hfill \end{array}\right\$
(2.1)
is equivalent to its dual variational inequality
$\left\{\begin{array}{c}{x}^{*}\in Fix\left(S\right);\hfill \\ \hfill 〈tf\left({x}^{*}\right)+\left(1-t\right)\left(I-\mu A\right){x}^{*}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-z〉\phantom{\rule{2.77695pt}{0ex}}\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).\hfill \end{array}\right\$
(2.2)
Proof. Assume that x* Fix(S) solves the problem (2.1). For all y Fix(S), set
$x={x}^{*}+s\left(y-{x}^{*}\right)\in Fix\left(S\right),\phantom{\rule{1em}{0ex}}\forall s\in \left(0,1\right).$
We note that
$〈tf\left(x\right)+\left(1-t\right)\left(I-\mu A\right)x-x,\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-x〉\ge 0.$
Hence, we have
$〈tf\left({x}^{*}+s\left(y-{x}^{*}\right)\right)+\left(1-t\right)\left(I-\mu A\right)\left({x}^{*}+s\left(y-{x}^{*}\right)\right)-{x}^{*}-s\left(y-{x}^{*}\right),\phantom{\rule{2.77695pt}{0ex}}s\left({x}^{*}-y\right)〉\ge 0,$
which implies that
$〈tf\left({x}^{*}+s\left(y-{x}^{*}\right)\right)+\left(1-t\right)\left(I-\mu A\right)\left({x}^{*}+s\left(y-{x}^{*}\right)\right)-{x}^{*}-s\left(y-{x}^{*}\right),\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-y〉\ge 0.$
Letting s → 0, we have
$〈tf\left({x}^{*}\right)+\left(1-t\right)\left(I-\mu A\right)\left({x}^{*}\right)-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-y〉\ge 0,$

which implies the point x* Fix(S) is a solution of the problem (2.2).

Conversely, assume that the point x* Fix(S) solves the problem (2.2). Then, we have
$〈tf\left({x}^{*}\right)+\left(1-t\right)\left(I-\mu A\right){x}^{*}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-z〉\phantom{\rule{2.77695pt}{0ex}}\ge 0.$
Noting that I - f and A are monotone, we have
$〈\left(I-f\right)z-\left(I-f\right){x}^{*},\phantom{\rule{2.77695pt}{0ex}}z\phantom{\rule{2.77695pt}{0ex}}-{x}^{*}〉\ge 0$
and
$〈Az-A{x}^{*},\phantom{\rule{2.77695pt}{0ex}}z-{x}^{*}〉\ge 0.$
Thus, it follows that
$t〈\left(I-f\right)z-\left(I-f\right){x}^{*},\phantom{\rule{2.77695pt}{0ex}}z-{x}^{*}〉+\left(1-t\right)\mu 〈Az-A{x}^{*},\phantom{\rule{2.77695pt}{0ex}}z-{x}^{*}〉\ge 0,$
which implies that
$\begin{array}{c}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-z〉\\ \phantom{\rule{7.1em}{0ex}}\ge 〈tf\left({x}^{*}\right)+\left(1-t\right)\left(I-\mu A\right){x}^{*}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}{x}^{*}-z〉\\ \phantom{\rule{7.1em}{0ex}}\ge 0.\hfill \end{array}$

This implies that the point x* Fix(S) solves the problem (2.1). This completes the proof. □

## 3 Main results

In this section, we first introduce our double-net algorithm and then prove a strong convergence theorem for this algorithm.

Throughout, we assume that

(C1) H is a real Hilbert space;

(C2) f : H → H is a ρ-contraction with coefficient ρ [0, 1), A : H → H is an α-inverse strongly monotone mapping, and S = {T(s)}s ≥ 0: H → H is a nonexpansive semigroup with Fix(S) ≠ ;

(C3) the solution set Ω of the problem (1.1) is nonempty;

(C4) μ (0, 2α) is a constant, and {λ s }0 < s < 1is a continuous net of positive real numbers satisfying lims→ 0λ s = +∞.

For any s, t (0, 1), we define the following mapping
$x↦{W}_{s,t}x:=s\left[tf\left(x\right)\phantom{\rule{2.77695pt}{0ex}}+\left(1\phantom{\rule{2.77695pt}{0ex}}-t\right)\left(x\phantom{\rule{2.77695pt}{0ex}}-\mu Ax\right)\right]+\left(1\phantom{\rule{2.77695pt}{0ex}}-s\right)\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right)xd\nu .$
We note that the mapping W s, t is a contraction. In fact, we have
which implies that W s, t is a contraction. Hence, by Banach's Contraction Principle, W s, t has a unique fixed point, which is denoted x s, t H, that is, x s, t is the unique solution in H of the fixed point equation
$\begin{array}{c}{x}_{s,t}=s\left[tf\left({x}_{s,t}\right)+\left(1-t\right)\left({x}_{s,t}-\mu A{x}_{s,t}\right)\right]\\ \phantom{\rule{2.5em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\left(1\phantom{\rule{2.77695pt}{0ex}}-s\right)\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu ,\phantom{\rule{1em}{0ex}}\forall s,\phantom{\rule{2.77695pt}{0ex}}t\phantom{\rule{2.77695pt}{0ex}}\in \left(0,1\right).\end{array}$
(3.1)

Now, we give some lemmas for our main result.

Lemma 3.1. For each fixed t (0, 1), the net {x s, t } defined by (3.1) is bounded.

Proof. Taking any z Fix(S), since I - μA is nonexpansive (by Lemma 2.1), it follows from (3.1) that
This implies that
$\begin{array}{c}∥{x}_{s,t}-z∥\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}\frac{1}{\left(1-\rho \right)t}\left(t||f\left(z\right)-z||\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\left(1-t\right)\mu ||Az||\right)\\ \phantom{\rule{4.8em}{0ex}}\le \phantom{\rule{1em}{0ex}}\frac{1}{\left(1-\rho \right)t}max\left\{||f\left(z\right)-z||,\phantom{\rule{2.77695pt}{0ex}}\mu ||Az||\right\}.\end{array}$

Thus, it follows that, for each fixed t (0, 1), {x s, t } is bounded and so are the nets {f(x s, t )} and {(I - μA)x s, t }. This completes the proof. □

Lemma 3.2. x s, t → x t Fix(S) as s → 0.

Proof. For each fixed t (0, 1), we set ${R}_{t}:=\frac{1}{\left(1-\rho \right)t}max\left\{||f\left(z\right)-z||,\phantom{\rule{2.77695pt}{0ex}}\mu ||Az||\right\}$. It is clear that, for each fixed t (0, 1), {x s, t } B(p, R t ), where B(p, R t ) denotes a closed ball with the center p and radius R t . Notice that
$∥\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu -p∥\le \phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-p||\phantom{\rule{2.77695pt}{0ex}}\le {R}_{t}.$
Moreover, we observe that if x B(p, R t ), then
$||T\left(s\right)x-p||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||T\left(s\right)x-T\left(s\right)p||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||x-p||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}{R}_{t},$
that is, B(p, R t ) is T(s)-invariant for all s (0, 1). From (3.1), it follows that
$\begin{array}{c}∥T\left(\tau \right){x}_{s,t}-{x}_{s,t}∥\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}∥T\left(\tau \right){x}_{s,t}-T\left(\tau \right)\frac{1}{{\lambda }_{s}}\underset{0}{\overset{{\lambda }_{s}}{\int }}T\left(\nu \right){x}_{s,t}d\nu ∥\\ \phantom{\rule{9.5em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+∥T\left(\tau \right)\frac{1}{{\lambda }_{s}}\underset{0}{\overset{{\lambda }_{s}}{\int }}T\left(\nu \right){x}_{s,t}d\nu -\frac{1}{{\lambda }_{s}}\underset{0}{\overset{{\lambda }_{s}}{\int }}T\left(\nu \right){x}_{s,t}d\nu ∥\\ \phantom{\rule{6.9em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+∥\frac{1}{{\lambda }_{s}}\underset{0}{\overset{{\lambda }_{s}}{\int }}T\left(\nu \right){X}_{s,t}\phantom{\rule{2.77695pt}{0ex}}d\nu \phantom{\rule{2.77695pt}{0ex}}-{x}_{s,t}∥\\ \phantom{\rule{7.8em}{0ex}}\le \phantom{\rule{1em}{0ex}}∥T\phantom{\rule{2.77695pt}{0ex}}\left(\tau \right)\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu -\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu ∥\\ \phantom{\rule{7.6em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+2∥{x}_{s,t}-\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){X}_{s,t}\phantom{\rule{2.77695pt}{0ex}}d\nu ∥\\ \phantom{\rule{7.5em}{0ex}}\le \phantom{\rule{1em}{0ex}}{2}_{S}∥tf\left({x}_{s,t}\right)+\left(1\phantom{\rule{2.77695pt}{0ex}}-t\right)\phantom{\rule{2.77695pt}{0ex}}\left({x}_{s,t}-\mu A{x}_{s,t}\right)-\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu ∥\\ \phantom{\rule{9.6em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}+∥T\left(\tau \right)\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu -\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){x}_{s,t}d\nu ∥.\end{array}$
By Lemma 2.2, for all 0 ≤ τ < ∞ and fixed t (0, 1), we deduce
$\underset{s\to 0}{lim}||T\left(\tau \right){x}_{s,t}-{x}_{s,t}||=0.$
(3.2)
Next, we show that, for each fixed t (0, 1), the net {x s, t } is relatively norm-compact as s → 0. In fact, from Lemma 2.1, it follows that
$||{x}_{s,t}-\mu A{x}_{s,t}-\left(z-\mu Az\right)|{|}^{2}\le ||{x}_{s,t}-z|{|}^{2}+\mu \left(\mu -2\alpha \right)||A{x}_{s,t}-Az|{|}^{2}.$
(3.3)
By (3.1), we have
$\begin{array}{c}\phantom{\rule{1.6em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}||{x}_{s,t}-z|{|}^{2}\hfill \\ =\phantom{\rule{1em}{0ex}}st〈f\left({x}_{s,t}\right)-f\left(z\right),\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉+st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+s\left(1-t\right)〈\left(I-\mu A\right){x}_{s,t}-\left(I-\mu A\right)z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+s\left(1-t\right)〈\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+\left(1-s\right)〈\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){X}_{s,t}\phantom{\rule{2.77695pt}{0ex}}d\nu \phantom{\rule{2.77695pt}{0ex}}-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \le \phantom{\rule{1em}{0ex}}st||f\left({x}_{s,t}\right)-f\left(z\right)||\phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-z||+\phantom{\rule{2.77695pt}{0ex}}st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+s\left(1-t\right)||\left(I-\mu A\right){x}_{s,t}-\left(I-\mu A\right)z||\phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-z||-\phantom{\rule{2.77695pt}{0ex}}s\left(1-t\right)\mu 〈Az,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+\left(1-s\right)∥\frac{1}{{\lambda }_{s}}{\int }_{0}^{{\lambda }_{s}}T\left(\nu \right){X}_{s,t}\phantom{\rule{2.77695pt}{0ex}}d\nu \phantom{\rule{2.77695pt}{0ex}}-z||\phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-z∥\hfill \\ \le \phantom{\rule{1em}{0ex}}st\rho ||{x}_{s,t}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+s\left(1-t\right)||\left(I-\mu A\right){x}_{s,t}-\left(I-\mu A\right)z||\phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-z||+\phantom{\rule{2.77695pt}{0ex}}\left(1-s\right)||{x}_{s,t}-z|{|}^{2}\hfill \\ \le \phantom{\rule{1em}{0ex}}st\rho ||{x}_{s,t}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+\frac{s\left(1-t\right)}{2}\left(||\left(I-\mu A\right){x}_{s,t}-\left(I-\mu A\right)z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}||{x}_{s,t}-z|{|}^{2}\right)+\left(1-s\right)||{x}_{s,t}-z|{|}^{2}.\hfill \end{array}$
This together with (3.3) imply that
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}||{x}_{s,t}-z|{|}^{2}\hfill \\ \le \phantom{\rule{1em}{0ex}}st\rho ||{x}_{s,t}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}+\frac{s\left(1-t\right)}{2}\left(||{x}_{s,t}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\mu \left(\mu -2\alpha \right)||A{x}_{s,t}-Az|{|}^{2}+||{x}_{s,t}-z|{|}^{2}\right)+\left(1-s\right)||{x}_{s,t}-z|{|}^{2}\hfill \\ \le \phantom{\rule{1em}{0ex}}\left[1-\left(1-\rho \right)st\right]||{x}_{s,t}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}st〈f\left(z\right)-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.6em}{0ex}}-s\left(1-t\right)\mu 〈Az,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉,\hfill \end{array}$
which implies that
$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}||{x}_{s,t}-z|{|}^{2}\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{s,t}-z〉,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).\hfill \end{array}$
(3.4)
Assume that {s n } (0, 1) is such that s n 0 as n → ∞. By (3.4), we obtain immediately that
$\begin{array}{c}\phantom{\rule{1.5em}{0ex}}||{x}_{{s}_{n},t}-z|{|}^{2}\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{{s}_{n},t}-z〉,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).\hfill \end{array}$
(3.5)

Since $\left\{{x}_{{s}_{n},t}\right\}$ is bounded, without loss of generality, we may assume that, as s n 0, $\left\{{x}_{{s}_{n},t}\right\}$ converges weakly to a point x t . From (3.2) and Lemma 2.3, we get x t Fix(S).

Further, if we substitute x t for z in (3.5), then it follows that
$||{x}_{{s}_{n},t}-{x}_{t}|{|}^{2}\le \frac{1}{\left(1-\rho \right)t}〈tf\left({x}_{t}\right)+\left(1-t\right)\left(I-\mu A\right){x}_{t}-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{{s}_{n},t}-{x}_{t}〉.$

Therefore, the weak convergence of $\left\{{x}_{{s}_{n},t}\right\}$ to x t actually implies that ${x}_{{s}_{n},t}\to {x}_{t}$ strongly. This has proved the relative norm-compactness of the net {x s, t } as s → 0.

Now, if we take the limit as n → ∞ in (3.5), we have
$\begin{array}{c}\phantom{\rule{1.3em}{0ex}}{∥{x}_{t}-z∥}^{2}\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-z〉,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).\hfill \end{array}$
In particular, x t solves the following variational inequality:
$\left\{\begin{array}{c}{x}_{t}\in Fix\left(S\right);\hfill \\ 〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-z〉\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right),\hfill \end{array}\right\$
or the equivalent dual variational inequality (see Lemma 2.4):
$\left\{\begin{array}{c}{x}_{t}\in Fix\left(S\right);\hfill \\ 〈tf\left({x}_{t}\right)+\left(1-t\right)\left(I-\mu A\right){x}_{t}-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-z〉\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).\hfill \end{array}\right\$
(3.6)

Notice that (3.6) is equivalent to the fact that x t = PFix(S)(tf + (1-t)(I - μA))x t , that is, x t is the unique element in Fix(S) of the contraction PFix(S)(tf +(1-t)(I -μA)). Clearly, it is sufficient to conclude that the entire net {x s, t } converges in norm to x t Fix(S) as s → 0. This completes the proof. □

Lemma 3.3. The net {x t } is bounded.

Proof. In (3.6), if we take any y Ω, then we have
$〈tf\left({x}_{t}\right)+\left(1-t\right)\left(I-\mu A\right){x}_{t}-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\ge 0.$
(3.7)
By virtue of the monotonicity of A and the fact that y Ω, we have
$〈\left(I-\mu A\right){x}_{t}-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\le 〈\left(I-\mu A\right)y-y,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\le 0.$
(3.8)
Thus, it follows from (3.7) and (3.8) that
$〈f\left({x}_{t}\right)-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in \Omega$
(3.9)
and hence
$\begin{array}{c}\parallel {x}_{t}-y{\parallel }^{2}\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}〈{x}_{t}-y,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉+〈f\left({x}_{t}\right)-{x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\hfill \\ \phantom{\rule{5.4em}{0ex}}=\phantom{\rule{1em}{0ex}}〈f\left({x}_{t}\right)-f\left(y\right),\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉+〈f\left(y\right)-y,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉\hfill \\ \phantom{\rule{5.4em}{0ex}}\le \phantom{\rule{1em}{0ex}}\rho \parallel {x}_{t}-y{\parallel }^{2}+\phantom{\rule{2.77695pt}{0ex}}〈f\left(y\right)-y,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉.\hfill \end{array}$
Therefore, we have
$||{x}_{t}-y|{|}^{2}\le \frac{1}{1-\rho }〈f\left(y\right)-y,\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-y〉,\phantom{\rule{1em}{0ex}}\forall y\in \Omega .$
(3.10)
In particular,
$||{x}_{t}-y||\le \frac{1}{1-\rho }||\phantom{\rule{2.77695pt}{0ex}}f\left(y\right)-y||,\phantom{\rule{1em}{0ex}}\forall t\in \left(0,1\right),$

which implies that {x t } is bounded. This completes the proof. □

Lemma 3.4. If the net {x t } converges in norm to a point x* Ω, then the point solves the variational inequality
$〈\left(I-f\right){x}^{*},\phantom{\rule{2.77695pt}{0ex}}x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall x\in \Omega .$
(3.11)

Proof. First, we note that the solution of the variational inequality (3.11) is unique.

Next, we prove that ω w (x t ) Ω, that is, if (t n ) is a null sequence in (0, 1) such that ${x}_{{t}_{n}}\to {x}^{\prime }$ weakly as n → ∞, then x' Ω. To see this, we use (3.6) to get
$〈\mu A{x}_{t},\phantom{\rule{2.77695pt}{0ex}}z-{x}_{t}〉\ge \frac{t}{1-t}〈\left(I-f\right){x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-z〉,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).$
However, since A is monotone, we have
$〈Az,\phantom{\rule{2.77695pt}{0ex}}z-{x}_{t}〉\ge 〈A{x}_{t},\phantom{\rule{2.77695pt}{0ex}}z-{x}_{t}〉.$
Combining the last two relations yields that
$〈\mu Az,\phantom{\rule{2.77695pt}{0ex}}z-{x}_{t}〉\ge \frac{t}{1-t}〈\left(I-f\right){x}_{t},\phantom{\rule{2.77695pt}{0ex}}{x}_{t}-z〉,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).$
(3.12)
Letting t = t n 0 as n → ∞ in (3.12), we get
$〈Az,\phantom{\rule{2.77695pt}{0ex}}z-{x}^{\prime }〉\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right),$
which is equivalent to its dual variational inequality
$〈A{x}^{\prime },\phantom{\rule{2.77695pt}{0ex}}z-{x}^{\prime }〉\ge 0,\phantom{\rule{1em}{0ex}}\forall z\in Fix\left(S\right).$

That is, x' is a solution of the problem (1.1) and hence x' Ω.

Finally, we prove that x' = x*, the unique solution of the variational inequality (3.11). In fact, by (3.10), we have
$||{x}_{{t}_{n}}-{x}^{\prime }|{|}^{2}\le \frac{1}{1-\rho }〈f\left({x}^{\prime }\right)-{x}^{\prime },\phantom{\rule{2.77695pt}{0ex}}{x}_{{t}_{n}}-{x}^{\prime }〉,\phantom{\rule{1em}{0ex}}\forall {x}^{\prime }\in \Omega .$
Therefore, the weak convergence to x' of $\left\{{x}_{{t}_{n}}\right\}$ implies that ${x}_{{t}_{n}}\to {x}^{\prime }$ in norm. Thus, if we let t = t n 0 in (3.10), then we have
$〈f\left({x}^{\prime }\right)-{x}^{\prime },\phantom{\rule{2.77695pt}{0ex}}y-{x}^{\prime }〉\le 0,\phantom{\rule{1em}{0ex}}\forall y\in \Omega ,$

which implies that x' Ω solves the problem (3.11). By the uniqueness of the solution, we have x' = x* and it is sufficient to guarantee that x t → x* in norm as t → 0. This completes the proof. □

Thus, by the above lemmas, we can obtain immediately the following theorem.

Theorem 3.5. For each (s, t) (0, 1) × (0, 1), let {x s, t } be a double-net algorithm defined implicitly by (3.1). Then, the net {x s, t } hierarchically converges to the unique solution x* of the hierarchical fixed point problem and the variational inequality problem (1.1), that is, for each fixed t (0, 1), the net {x s, t } converges in norm as s → 0 to a common fixed point x t Fix(S) of the nonexpansive semigroup {T(s)}s ≥ 0. Moreover, as t → 0, the net {x t } converges in norm to the unique solution x* Ω and the point x* also solves the following variational inequality.
$\left\{\begin{array}{c}{x}^{*}\in \Omega ;\hfill \\ 〈\left(I-f\right){x}^{*},\phantom{\rule{2.77695pt}{0ex}}x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall x\in \Omega .\hfill \end{array}\right\$

## Declarations

### Acknowledgements

Yonghong Yao was supported in part by Colleges and Universities Science and Technology Development Foundation (20091003) of Tianjin and NSFC 11071279. Yeol Je Cho was supported by the Korea Research Foundation Grant funded by the Korean Government (KRF-2008-313-C00050). Yeong-Cheng Liou was supported in part by NSC 99-2221-E-230-006.

## Authors’ Affiliations

(1)
Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300160, People's Republic of China
(2)
Department of Mathematics Education and the RINS, Gyeongsang National University, Chinju, 660-701, Republic of Korea
(3)
Department of Information Management, Cheng Shiu University, Kaohsiung, 833, Taiwan

## References 