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

Abstract

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

$$x_{s,t}=s\left[tf\left(x_{s,t}\right)+\left(1-t\right)\left(x_{s,t}-\mu A{x}_{s,t}\right)\right]+\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(v\right)x_{s,t}d\nu ,\forall s,t\in \left(0,1\right),$$

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}: H → H 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 ≥ 0}and, as t → 0, the net {x _t } converges in norm to the solution x* of the following variational inequality:

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

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

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 ≥ 0}with respect to another monotone operator A, namely,

Find x* ∈ Fix(S) such that

$$〈A{x}^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(S\right).$$

(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 [1–19].

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 ≥ 0}and, 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

$$\left|\right|f\left(x\right)-f\left(y\right)\left|\right|\le \rho \left|\right|x-y\left|\right|,\forall x,y\in H.$$

A mapping T : C → C is said to be nonexpansive if

$$\left|\right|Tx-Ty\left|\right|\le \left|\right|x-y\left|\right|,\forall x,y\in H.$$

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

Recall also that a family S : = {T(s)}_{s ≥ 0}of 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 ≥ 0}Fix(T(s)). It is known that Fix(S) is closed and convex ([20], Lemma 1).

A mapping A of H into itself is said to be monotone if

$$〈Au-Av,u-v〉\ge 0,\forall u,v\in H,$$

and A : C → H is said to be α-inverse strongly monotone if there exists a positive real number α such that

$$〈Au-Av,u-v〉\ge \alpha {\left|\right|Au-Av\left|\right|}^2,\forall u,v\in H.$$

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. [21] 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

$${\left|\right|\left(I-\mu A\right)x-\left(I-\mu A\right)y\left|\right|}^2\le {\left|\right|x-y\left|\right|}^2+\mu \left(\mu -2\alpha \right){\left|\right|Ax-Ay\left|\right|}^2,\forall x,y\in H.$$

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

Lemma 2.2. [22] Let C be a nonempty bounded closed convex subset of a Hilbert space H and {T(s)}_{s ≥ 0}be 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. [23] (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

$$\left\{\begin{array}{c}{x}^{*}\in Fix\left(S\right);\hfill \\ \hfill 〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x^{*}-z〉\ge 0,\forall z\in Fix\left(S\right),\hfill \end{array}\right.$$

(2.1)

is equivalent to its dual variational inequality

$$\left\{\begin{array}{c}{x}^{*}\in Fix\left(S\right);\hfill \\ \hfill 〈tf\left(x^{*}\right)+\left(1-t\right)\left(I-\mu A\right)x^{*}-x^{*},x^{*}-z〉\ge 0,\forall z\in Fix\left(S\right).\hfill \end{array}\right.$$

(2.2)

Proof. Assume that x* ∈ Fix(S) solves the problem (2.1). For all y ∈ Fix(S), set

$$x=x^{*}+s\left(y-x^{*}\right)\in Fix\left(S\right),\forall s\in \left(0,1\right).$$

We note that

$$〈tf\left(x\right)+\left(1-t\right)\left(I-\mu A\right)x-x,x^{*}-x〉\ge 0.$$

Hence, we have

$$〈tf\left(x^{*}+s\left(y-x^{*}\right)\right)+\left(1-t\right)\left(I-\mu A\right)\left(x^{*}+s\left(y-x^{*}\right)\right)-x^{*}-s\left(y-x^{*}\right),s\left(x^{*}-y\right)〉\ge 0,$$

which implies that

$$〈tf\left(x^{*}+s\left(y-x^{*}\right)\right)+\left(1-t\right)\left(I-\mu A\right)\left(x^{*}+s\left(y-x^{*}\right)\right)-x^{*}-s\left(y-x^{*}\right),x^{*}-y〉\ge 0.$$

Letting s → 0, we have

$$〈tf\left(x^{*}\right)+\left(1-t\right)\left(I-\mu A\right)\left(x^{*}\right)-x^{*},x^{*}-y〉\ge 0,$$

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

$$〈tf\left(x^{*}\right)+\left(1-t\right)\left(I-\mu A\right)x^{*}-x^{*},x^{*}-z〉\ge 0.$$

Noting that I - f and A are monotone, we have

$$〈\left(I-f\right)z-\left(I-f\right)x^{*},z-x^{*}〉\ge 0$$

and

$$〈Az-A{x}^{*},z-x^{*}〉\ge 0.$$

Thus, it follows that

$$t〈\left(I-f\right)z-\left(I-f\right)x^{*},z-x^{*}〉+\left(1-t\right)\mu 〈Az-A{x}^{*},z-x^{*}〉\ge 0,$$

which implies that

$$\begin{array}{c}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x^{*}-z〉\\ \ge 〈tf\left(x^{*}\right)+\left(1-t\right)\left(I-\mu A\right)x^{*}-x^{*},x^{*}-z〉\\ \ge 0.\hfill \end{array}$$

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 < 1}is a continuous net of positive real numbers satisfying lim_{s→ 0}λ _s = +∞.

For any s, t ∈ (0, 1), we define the following mapping

$$x\mapsto W_{s,t}x:=s\left[tf\left(x\right)+\left(1-t\right)\left(x-\mu Ax\right)\right]+\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)xd\nu .$$

We note that the mapping W _{s, t} is a contraction. In fact, we have

$$\begin{array}{c}∥W_{s,t}x-W_{s,t}y∥=∥s\left[tf\left(x\right)+\left(1-t\right)\left(x-\mu Ax\right)\right]+\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)xd\nu \\ -s\left[tf\left(y\right)+\left(1-t\right)\left(y-\mu Ay\right)\right]-\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)yd\nu ∥\\ \hfill \le st∥f\left(x\right)-f\left(y\right)\left|\right|+s\left(1-t\right)\left|\right|\left(x-\mu Ax\right)-\left(y-\mu Ay\right)\left|\right|\hfill \\ +\left(1-s\right)\left|\right|\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}\left(T\left(\nu \right)x-T\left(\nu \right)y\right)d\nu ∥\\ \le st\rho \left|\right|x-y\left|\right|+s\left(1-t\right)\left|\right|x-y\left|\right|+\left(1-s\right)\left|\right|x-y\left|\right|\\ =\left[1-\left(1-\rho \right)st\right]\left|\right|x-y\left|\right|,\hfill \end{array}$$

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

$$\begin{array}{c}{x}_{s,t}=s\left[tf\left(x_{s,t}\right)+\left(1-t\right)\left(x_{s,t}-\mu A{x}_{s,t}\right)\right]\\ +\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu ,\forall s,t\in \left(0,1\right).\end{array}$$

(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

$$\begin{array}{c}\left|\right|x_{s,t}-z\left|\right|\hfill \\ =∥s\left[tf\left(x_{s,t}\right)+\left(1-t\right)\left(I-\mu A\right)x_{s,t}\right]+\left(1-s\right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu -z∥\\ \le s\left|\right|tf\left(x_{s,t}\right)+\left(1-t\right)\left(I-\mu A\right)x_{s,t}-z\left|\right|+\left(1-s\right)∥\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu -z∥\\ \le s\left[t\left|\right|f\left(x_{s,t}\right)-f\left(z\right)\left|\right|+t\left|\right|f\left(z\right)-z\left|\right|+\left(1-t\right)\left|\right|\left(I-\mu A\right)x_{s,t}-\left(I-\mu A\right)z\left|\right|\right.\\ \left.+\left(1-t\right)\left|\right|\left(I-\mu A\right)z-z\left|\right|\right]+\left(1-s\right)\left|\right|x_{s,t}-z\left|\right|\hfill \\ \le s\left[t\rho \left|\right|x_{s,t}-z\left|\right|+t\left|\right|f\left(z\right)-z\left|\right|+\left(1-t\right)\left|\right|x_{s,t}-z\left|\right|+\left(1-t\right)\mu \left|\right|Az\left|\right|\right]\\ +\left(1-s\right)\left|\right|x_{s,t}-z\left|\right|\hfill \\ =\left[1-\left(1-\rho \right)st\right]\left|\right|x_{s,t}-z\left|\right|+st\left|\right|f\left(z\right)-z\left|\right|+s\left(1-t\right)\mu \left|\right|Az\left|\right|.\hfill \end{array}$$

This implies that

$$\begin{array}{c}∥x_{s,t}-z∥\le \frac{1}{\left(1-\rho \right)t}\left(t\left|\right|f\left(z\right)-z\left|\right|+\left(1-t\right)\mu \left|\right|Az\left|\right|\right)\\ \le \frac{1}{\left(1-\rho \right)t}max\left\{\left|\right|f\left(z\right)-z\left|\right|,\mu \left|\right|Az\left|\right|\right\}.\end{array}$$

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 $R_t:=\frac{1}{\left(1-\rho \right)t}max\left\{\left|\right|f\left(z\right)-z\left|\right|,\mu \left|\right|Az\left|\right|\right\}$. 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

$$∥\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu -p∥\le \left|\right|x_{s,t}-p\left|\right|\le R_t.$$

Moreover, we observe that if x ∈ B(p, R _t ), then

$$\left|\right|T\left(s\right)x-p\left|\right|\le \left|\right|T\left(s\right)x-T\left(s\right)p\left|\right|\le \left|\right|x-p\left|\right|\le R_t,$$

that is, B(p, R _t ) is T(s)-invariant for all s ∈ (0, 1). From (3.1), it follows that

$$\begin{array}{c}∥T\left(\tau \right)x_{s,t}-x_{s,t}∥\le ∥T\left(\tau \right)x_{s,t}-T\left(\tau \right)\frac{1}{{\lambda }_s}\underset{0}{\overset{{\lambda }_s}{\int }}T\left(\nu \right)x_{s,t}d\nu ∥\\ +∥T\left(\tau \right)\frac{1}{{\lambda }_s}\underset{0}{\overset{{\lambda }_s}{\int }}T\left(\nu \right)x_{s,t}d\nu -\frac{1}{{\lambda }_s}\underset{0}{\overset{{\lambda }_s}{\int }}T\left(\nu \right)x_{s,t}d\nu ∥\\ +∥\frac{1}{{\lambda }_s}\underset{0}{\overset{{\lambda }_s}{\int }}T\left(\nu \right)X_{s,t}d\nu -x_{s,t}∥\\ \le ∥T\left(\tau \right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu -\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu ∥\\ +2∥x_{s,t}-\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)X_{s,t}d\nu ∥\\ \le 2_S∥tf\left(x_{s,t}\right)+\left(1-t\right)\left(x_{s,t}-\mu A{x}_{s,t}\right)-\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu ∥\\ +∥T\left(\tau \right)\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu -\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)x_{s,t}d\nu ∥.\end{array}$$

By Lemma 2.2, for all 0 ≤ τ < ∞ and fixed t ∈ (0, 1), we deduce

$$\underset{s\to 0}{lim}\left|\right|T\left(\tau \right)x_{s,t}-x_{s,t}\left|\right|=0.$$

(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

$$\left|\right|x_{s,t}-\mu A{x}_{s,t}-\left(z-\mu Az\right)|{|}^2\le ||x_{s,t}-z|{|}^2+\mu \left(\mu -2\alpha \right)\left|\right|A{x}_{s,t}-Az|{|}^2.$$

(3.3)

By (3.1), we have

$$\begin{array}{c}\left|\right|x_{s,t}-z|{|}^2\hfill \\ =st〈f\left(x_{s,t}\right)-f\left(z\right),x_{s,t}-z〉+st〈f\left(z\right)-z,x_{s,t}-z〉\hfill \\ +s\left(1-t\right)〈\left(I-\mu A\right)x_{s,t}-\left(I-\mu A\right)z,x_{s,t}-z〉\hfill \\ +s\left(1-t\right)〈\left(I-\mu A\right)z-z,x_{s,t}-z〉\hfill \\ +\left(1-s\right)〈\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)X_{s,t}d\nu -z,x_{s,t}-z〉\hfill \\ \le st\left|\right|f\left(x_{s,t}\right)-f\left(z\right)\left|\right|\left|\right|x_{s,t}-z\left|\right|+st〈f\left(z\right)-z,x_{s,t}-z〉\hfill \\ +s\left(1-t\right)\left|\right|\left(I-\mu A\right)x_{s,t}-\left(I-\mu A\right)z\left|\right|\left|\right|x_{s,t}-z\left|\right|-s\left(1-t\right)\mu 〈Az,x_{s,t}-z〉\hfill \\ +\left(1-s\right)∥\frac{1}{{\lambda }_s}{\int }_0^{{\lambda }_s}T\left(\nu \right)X_{s,t}d\nu -z\left|\right|\left|\right|x_{s,t}-z∥\hfill \\ \le st\rho \left|\right|x_{s,t}-z|{|}^2+st〈f\left(z\right)-z,x_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,x_{s,t}-z〉\hfill \\ +s\left(1-t\right)\left|\right|\left(I-\mu A\right)x_{s,t}-\left(I-\mu A\right)z\left|\right|\left|\right|x_{s,t}-z\left|\right|+\left(1-s\right)\left|\right|x_{s,t}-z|{|}^2\hfill \\ \le st\rho \left|\right|x_{s,t}-z|{|}^2+st〈f\left(z\right)-z,x_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,x_{s,t}-z〉\hfill \\ +\frac{s\left(1-t\right)}{2}\left(\left|\right|\left(I-\mu A\right)x_{s,t}-\left(I-\mu A\right)z|{|}^2+||x_{s,t}-z|{|}^2\right)+\left(1-s\right)\left|\right|x_{s,t}-z|{|}^2.\hfill \end{array}$$

This together with (3.3) imply that

$$\begin{array}{c}\left|\right|x_{s,t}-z|{|}^2\hfill \\ \le st\rho \left|\right|x_{s,t}-z|{|}^2+st〈f\left(z\right)-z,x_{s,t}-z〉-s\left(1-t\right)\mu 〈Az,x_{s,t}-z〉\hfill \\ +\frac{s\left(1-t\right)}{2}\left(\left|\right|x_{s,t}-z|{|}^2+\mu \left(\mu -2\alpha \right)||A{x}_{s,t}-Az|{|}^2+\left|\right|x_{s,t}-z|{|}^2\right)+\left(1-s\right)\left|\right|x_{s,t}-z|{|}^2\hfill \\ \le \left[1-\left(1-\rho \right)st\right]\left|\right|x_{s,t}-z|{|}^2+st〈f\left(z\right)-z,x_{s,t}-z〉\hfill \\ -s\left(1-t\right)\mu 〈Az,x_{s,t}-z〉,\hfill \end{array}$$

which implies that

$$\begin{array}{c}\left|\right|x_{s,t}-z|{|}^2\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x_{s,t}-z〉,\forall z\in Fix\left(S\right).\hfill \end{array}$$

(3.4)

Assume that {s _n } ⊂ (0, 1) is such that s _n → 0 as n → ∞. By (3.4), we obtain immediately that

$$\begin{array}{c}\left|\right|x_{s_n,t}-z|{|}^2\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x_{s_n,t}-z〉,\forall z\in Fix\left(S\right).\hfill \end{array}$$

(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

$$\left|\right|x_{s_n,t}-x_t|{|}^2\le \frac{1}{\left(1-\rho \right)t}〈tf\left(x_t\right)+\left(1-t\right)\left(I-\mu A\right)x_t-x_t,x_{s_n,t}-x_t〉.$$

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

$$\begin{array}{c}{∥x_t-z∥}^2\hfill \\ \le \frac{1}{\left(1-\rho \right)t}〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x_t-z〉,\forall z\in Fix\left(S\right).\hfill \end{array}$$

In particular, x _t solves the following variational inequality:

$$\left\{\begin{array}{c}{x}_t\in Fix\left(S\right);\hfill \\ 〈tf\left(z\right)+\left(1-t\right)\left(I-\mu A\right)z-z,x_t-z〉\ge 0,\forall z\in Fix\left(S\right),\hfill \end{array}\right.$$

or the equivalent dual variational inequality (see Lemma 2.4):

$$\left\{\begin{array}{c}{x}_t\in Fix\left(S\right);\hfill \\ 〈tf\left(x_t\right)+\left(1-t\right)\left(I-\mu A\right)x_t-x_t,x_t-z〉\ge 0,\forall z\in Fix\left(S\right).\hfill \end{array}\right.$$

(3.6)

Notice that (3.6) is equivalent to the fact that x _t = P_Fix(S)(tf + (1-t)(I - μA))x _t , that is, x _t is the unique element in Fix(S) of the contraction P_Fix(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

$$〈tf\left(x_t\right)+\left(1-t\right)\left(I-\mu A\right)x_t-x_t,x_t-y〉\ge 0.$$

(3.7)

By virtue of the monotonicity of A and the fact that y ∈ Ω, we have

$$〈\left(I-\mu A\right)x_t-x_t,x_t-y〉\le 〈\left(I-\mu A\right)y-y,x_t-y〉\le 0.$$

(3.8)

Thus, it follows from (3.7) and (3.8) that

$$〈f\left(x_t\right)-x_t,x_t-y〉\ge 0,\forall y\in \Omega$$

(3.9)

and hence

$$\begin{array}{c}\parallel x_t-y{\parallel }^2\le 〈x_t-y,x_t-y〉+〈f\left(x_t\right)-x_t,x_t-y〉\hfill \\ =〈f\left(x_t\right)-f\left(y\right),x_t-y〉+〈f\left(y\right)-y,x_t-y〉\hfill \\ \le \rho \parallel x_t-y{\parallel }^2+〈f\left(y\right)-y,x_t-y〉.\hfill \end{array}$$

Therefore, we have

$$\left|\right|x_t-y|{|}^2\le \frac{1}{1-\rho }〈f\left(y\right)-y,x_t-y〉,\forall y\in \Omega .$$

(3.10)

In particular,

$$\left|\right|x_t-y\left|\right|\le \frac{1}{1-\rho }\left|\right|f\left(y\right)-y\left|\right|,\forall t\in \left(0,1\right),$$

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

$$〈\left(I-f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in \Omega .$$

(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

$$〈\mu A{x}_t,z-x_t〉\ge \frac{t}{1-t}〈\left(I-f\right)x_t,x_t-z〉,\forall z\in Fix\left(S\right).$$

However, since A is monotone, we have

$$〈Az,z-x_t〉\ge 〈A{x}_t,z-x_t〉.$$

Combining the last two relations yields that

$$〈\mu Az,z-x_t〉\ge \frac{t}{1-t}〈\left(I-f\right)x_t,x_t-z〉,\forall z\in Fix\left(S\right).$$

(3.12)

Letting t = t _n → 0 as n → ∞ in (3.12), we get

$$〈Az,z-x^{\prime }〉\ge 0,\forall z\in Fix\left(S\right),$$

which is equivalent to its dual variational inequality

$$〈A{x}^{\prime },z-x^{\prime }〉\ge 0,\forall z\in Fix\left(S\right).$$

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

$$\left|\right|x_{t_n}-x^{\prime }|{|}^2\le \frac{1}{1-\rho }〈f\left(x^{\prime }\right)-x^{\prime },x_{t_n}-x^{\prime }〉,\forall x^{\prime }\in \Omega .$$

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

$$〈f\left(x^{\prime }\right)-x^{\prime },y-x^{\prime }〉\le 0,\forall y\in \Omega ,$$

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.

$$\left\{\begin{array}{c}{x}^{*}\in \Omega ;\hfill \\ 〈\left(I-f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in \Omega .\hfill \end{array}\right.$$