New Methods with Perturbations for Non-Expansive Mappings in Hilbert Spaces

Abstract

In this paper, we suggest and analyze two iterative algorithms with perturbations for non-expansive mappings in Hilbert spaces. We prove that the proposed iterative algorithms converge strongly to a fixed point of some non-expansive mapping.

2000 Mathematics Subject Classification 47H09, 47H10.

1. Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that a mapping T: C → C is said to be non-expansive if

$$\left|\right|Tx-Ty\left|\right|\le \left|\right|x-y\left|\right|\mathsf{\text{for}}\mathsf{\text{all}}x,y\in C.$$

Denote by Fix(T) the set of fixed points of T; that is, Fix(T) = {x ∈ C : Tx = x}.

Recently, iterative methods for finding fixed points of non-expansive mappings have received vast investigations due to its extensive applications in a variety of applied areas of inverse problem, partial differential equations, image recovery, and signal processing; see [1–34] and the references therein. There are perturbations always occurring in the iterative processes because the manipulations are inaccurate. It is no doubt that researching the convergent problems of iterative methods with perturbation members is a significant job.

It is our purpose in this paper that we suggest and analyze two iterative algorithms with errors for non-expansive mappings in Hilbert spaces. We prove that the proposed iterative algorithms converge strongly to a fixed point of some non-expansive mapping.

2. Preliminaries

Let H be a real Hilbert space with inner product 〈·,·〉 and norm || · ||, respectively. Recall that the nearest point (or metric) projection from H onto a nonempty closed convex subset C of H is defined as follows: for each point x ∈ H, P _C (x)] is the unique point in C with the property:

$$\left|\right|x-P_C\left(x\right)\left|\right|\le \left|\right|x-y\left|\right|,\forall y\in C.$$

A characterization for P _C is described below. Given x ∈ H and z ∈ C. Then z = P _C (x) if and only if there holds the inequality

$$⟨x-z,y-z⟩\le 0,\forall y\in C.$$

(2.1)

It is known that P _C is non-expansive. The following well-known lemmas play an important role in our argument in the next sections.

Lemma 2.1. (Demiclosedness principle) Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a non-expansive mapping with$Fix\left(T\right)\ne \varnothing$. Then, T is demiclosed on C, i.e., if x _n → x ∈ C weakly and x _n - Tx _n → y strongly, then (I - T)x = y.

Lemma 2.2. (Suzuki's lemma) Let {x _n } and {y _n } be bounded sequences in a Banach space X and {β _n } be a sequence in [0, 1] with 0 < lim inf _n→∞ β _n ≤ lim sup_n→∞, β _n < 1. Suppose that x_n+1= (1 - β _n )y _n + β _n x _n for all n ≥ 0 and lim sup_n→∞(||y_n+1- y _n ||- ||x_n+1- x _n ||) ≤ 0. Then, lim_n→∞||y _n - x _n || = 0.

Lemma 2.3. (Liu's lemma) Assume {a _n } is a sequence of nonnegative real numbers such that

$$a_{n+1}\le \left(1-{\gamma }_n\right)a_n+{\gamma }_n{\delta }_n+{\sigma }_n,n\ge 0,$$

where {γ _n } is a sequence in (0,1), and {δ _n } and {σ _n } are two sequences in R such that

• (i) ${\sum }_{n=0}^{\infty }{\gamma }_n=\infty$ ;

• (ii) $lim{sup}_{n\to \infty }{\delta }_n\le 0or\sum _{n=0}^{\infty }\left|{\delta }_n{\gamma }_n\right|<\infty$ ;

• (iii) ${\sum }_{n=0}^{\infty }\left|{\sigma }_n\right|<\infty$.

Then lim_n→∞a _n = 0.

3. Main results

In this section, we introduce our algorithms with perturbations and state our main results.

Algorithm 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a non-expansive mapping. For given x₀ ∈ C, define a sequence {x _m } by the following manner:

$$x_m=P_C\left({\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{x}_m\right),m\ge 0,$$

(3.1)

where {α _m } is a sequence in [0, 1], and the sequence {u _m } ⊂ H is a small perturbation for the m-step iteration satisfying ||u _m || → 0 as m → ∞.

Remark 3.2. In this point, we want to point out that we permit the perturbation {u _m } in the whole space H. If {u _m } ⊂ C, then (3.1) reduces to

$$x_m={\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{x}_m,m\ge 0.$$

(3.2)

Theorem 3.3. Suppose$Fix\left(T\right)\ne \varnothing$. Then, as α _m → 0, the sequence {x _m } generated by the implicit method (3.1) converges to$\tilde{x}\in Fix\left(T\right)$.

Proof. We first show that {x _m } is bounded. Indeed, take an x* ∈ Fix(T) to derive that

$$\begin{array}{ll}\hfill \left|\right|x_m-x^{*}\left|\right|& =\left|\right|P_C\left({\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{x}_m\right)-x^{*}\left|\right|\\ \le {\alpha }_m\left|\right|x^{*}\left|\right|+\left(1-{\alpha }_m\right)\left|\right|T{x}_m-x^{*}\left|\right|+{\alpha }_m\left|\right|u_m\left|\right|\\ \le \left(1-{\alpha }_m\right)\left|\right|x_m-x^{*}\left|\right|+{\alpha }_m\left|\right|x^{*}\left|\right|+{\alpha }_m\left|\right|u_m\left|\right|.\end{array}$$

This implies that

$$\left|\right|x_m-x^{*}\left|\right|\le \left|\right|x^{*}\left|\right|+\left|\right|u_m\left|\right|.$$

Since ||u _m || → 0, there exists a constant M > 0 such that sup _m {||u _m ||} ≤ M. Hence, ||x _m - x*|| ≤ ||x*|| + M for all n ≥ 0. It follows that {x _m } is bounded, so is the sequence {Tx _m }.

Since x _m ∈ C and also Tx _m ∈ C, we get

$$\left|\right|x_m-T{x}_m\left|\right|=\left|\right|P_C\left({\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{x}_m\right)-P_C\left(T{x}_m\right)\left|\right|\le {\alpha }_m\left|\right|u_m-T{x}_m\left|\right|\to 0.$$

(3.3)

Setting y _m = α _m u _m + (1 - α _m )Tx _m for all n ≥ 0, we then have x _m = P _C (y _m ), and for any x* ∈ Fix(T),

$$\begin{array}{ll}\hfill x_m-x^{*}& =P_C\left(y_m\right)-y_m+y_m-x^{*}\\ =P_C\left(y_m\right)-y_m+{\alpha }_m{u}_m+\left(1-{\alpha }_m\right)\left(T{x}_m-x^{*}\right)-{\alpha }_m{x}^{*}.\end{array}$$

Noting that the fact by (2.1) that

$$⟨P_C\left(y_m\right)-y_m,P_C\left(y_m\right)-x^{*}⟩\le 0.$$

Hence, we have

$$\begin{array}{c}\left|\right|x_m-x^{*}|{|}^2=⟨P_C\left(y_m\right)-y_m,x_m-x^{*}⟩+{\alpha }_m⟨u_m,x_m-x^{*}⟩\\ +\left(1-{\alpha }_m\right)⟨T{x}_m-x^{*},x_m-x^{*}⟩-{\alpha }_m⟨x^{*},x_m-x^{*}⟩\\ \le \left(1-{\alpha }_m\right)\left|\right|T{x}_m-x^{*}\left|\right|\left|\right|x_m-x^{*}\left|\right|+{\alpha }_m\left|\right|u_m\left|\right|\left|\right|x_m-x^{*}\left|\right|-{\alpha }_m⟨x^{*},x_m-x^{*}⟩\\ \le \left(1-{\alpha }_m\right)\left|\right|x_m-x^{*}|{|}^2+{\alpha }_m|\left|u_m\right|\left|\right||x_m-x^{*}||-{\alpha }_m⟨x^{*},x_m-x^{*}⟩.\end{array}$$

It turns out that

$$\left|\right|x_m-x^{*}|{|}^2\le ⟨x^{*},x^{*}-x_m⟩+|\left|u_m\right||\left(\left|\right|x^{*}\left|\right|+M\right),x^{*}\in Fix\left(T\right).$$

(3.4)

Since {x _m } is bounded, without loss of generality, we may assume that {x _m } converges weakly to a point $\tilde{x}\in C$. Noticing (3.3), we can use Lemma 2.1 to get $\tilde{x}\in Fix\left(T\right)$. Therefore, we can substitute $\tilde{x}$ for x* in (3.4) to get

$$\left|\right|x_m-\tilde{x}|{|}^2\le ⟨\tilde{x},\tilde{x}-x_m⟩+|\left|u_m\right||\left(\left|\right|x^{*}\left|\right|+M\right).$$

Consequently, the weak convergence of {x _m } to $\tilde{x}$ actually implies that $x_m\to \tilde{x}$ strongly. Finally, in order to complete the proof, we have to prove that the weak cluster points set ω _w (x _m ) is singleton. As a matter of fact, if $x_{m_i}\rightharpoonup \widehat{x}\in Fix\left(T\right)$ and $x_{m_k}\rightharpoonup \bar{x}\in Fix\left(T\right)$, then we have $x_{m_i}\to \widehat{x}\in Fix\left(T\right)$ and $x_{m_k}\to \bar{x}\in Fix\left(T\right)$. From (3.4), we have

$$\left|\right|x_{m_i}-\bar{x}|{|}^2\le ⟨\bar{x},\bar{x}-x_{m_i}⟩+|\left|u_{m_i}\right||\left(\left|\right|\bar{x}\left|\right|+M\right),$$

and

$$\left|\right|x_{m_k}-\widehat{x}|{|}^2\le ⟨\widehat{x},\widehat{x}-x_{m_k}⟩+|\left|u_{m_k}\right||\left(\left|\right|\widehat{x}\left|\right|+M\right).$$

Hence, we have $\left|\right|\widehat{x}-\bar{x}|{|}^2\le ⟨\bar{x},\bar{x}-\widehat{x}⟩$ and $\left|\right|\bar{x}-\widehat{x}|{|}^2\le ⟨\widehat{x},\widehat{x}-\bar{x}⟩$. Therefore, we obtain

$$2\left|\right|\widehat{x}-\bar{x}|{|}^2\le ⟨\bar{x}-\widehat{x},\bar{x}-\widehat{x}⟩=||\widehat{x}-\bar{x}|{|}^2.$$

We have immediately $\widehat{x}=\bar{x}$. This completes the proof.

From Theorem 3.3, we have the following corollary.

Corollary 3.4. Suppose$Fix\left(T\right)\ne \varnothing$. Then, as α _m → 0, the sequence {x _m } generated by the implicit method (3.2) converges to$\tilde{x}\in Fix\left(T\right)$.

Next, we introduce an explicit algorithm.

Algorithm 3.5. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a non-expansive mapping. For given x₀ ∈ C, define a sequence {x _n } by the following manner:

$$x_{n+1}=\left(1-{\beta }_n\right)x_n+{\beta }_n{P}_C\left({\alpha }_n{u}_n+\left(1-{\alpha }_n\right)T{x}_n\right),n\ge 0,$$

(3.5)

where {α _n } and {β _n } are two sequences in (0,1), and the sequence {u _n } ⊂ H is a perturbation for the n-step iteration.

Remark 3.6. If {u _n } ⊂ C, then (3.5) reduces to

$$x_{n+1}=\left(1-{\beta }_n\right)x_n+{\beta }_n\left[{\alpha }_n{u}_n+\left(1-{\alpha }_n\right)T{x}_n\right],n\ge 0,$$

(3.6)

Theorem 3.7. Suppose$Fix\left(T\right)\ne \varnothing$. Assume the following conditions are satisfied:

• (i) lim_n→∞α _n = 0 and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$;

• (ii) 0 < lim inf_n→∞ β _n ≤ lim sup_n→∞ β _n < 1;

• (iii) ${\sum }_{n=0}^{\infty }{\alpha }_n\left|\right|u_n\left|\right|<\infty$.

Then, the sequence {x _n } generated by the explicit iterative method (3.5) converges to$\tilde{x}\in Fix\left(T\right)$.

Proof. First, we show that {x _n } is bounded. Take an x* ∈ Fix(T) to derive that

$$\begin{array}{ll}\hfill \left|\right|x_{n+1}-x^{*}\left|\right|& =\left|\right|\left(1-{\beta }_n\right)\left(x_n-x^{*}\right)+{\beta }_n\left(P_C\left({\alpha }_n{u}_n+\left(1-{\alpha }_n\right)T{x}_n\right)-x^{*}\right)\left|\right|\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-x^{*}\left|\right|+{\beta }_n\left|\right|{\alpha }_n{u}_n+\left(1-{\alpha }_n\right)T{x}_n-x^{*}\left|\right|\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-x^{*}\left|\right|+{\beta }_n\left[{\alpha }_n\left|\right|u_n\left|\right|+\left(1-{\alpha }_n\right)\left|\right|T{x}_n-x^{*}\left|\right|+{\alpha }_n\left|\right|x^{*}\left|\right|\right]\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-x^{*}\left|\right|+{\beta }_n\left[{\alpha }_n\left|\right|u_n\left|\right|+\left(1-{\alpha }_n\right)\left|\right|x_n-x^{*}\left|\right|+{\alpha }_n\left|\right|x^{*}\left|\right|\right]\\ =\left(1-{\beta }_n{\alpha }_n\right)\left|\right|x_n-x^{*}\left|\right|+{\beta }_n{\alpha }_n\left|\right|x^{*}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|\\ \le max\left\{\left|\right|x_n-x^{*}\left|\right|,\left|\right|x^{*}\left|\right|\right\}+{\alpha }_n\left|\right|u_n\left|\right|.\end{array}$$

By induction, we get

$$\left|\right|x_n-x^{*}\left|\right|\le max\left\{\left|\right|x_0-x^{*}\left|\right|,\left|\right|x^{*}\left|\right|\right\}+\sum _{n=0}^{n-1}{\alpha }_n\left|\right|u_n\left|\right|.$$

Thus, {x _n } is bounded, so is the sequence {Tx _n }. Next, we show that

$$\left|\right|x_{n+1}-x_n\left|\right|\to 0.$$

(3.7)

Indeed, we write x_n+1= (1 - β _n )x _n + β _n y _n , n ≥ 0. It is clear that y _n = P _C (α _n u _n + (1 - α _n )Tx _n ) for all n ≥ 0. Then, we have

$$\begin{array}{ll}\hfill \left|\right|y_{n+1}-y_n\left|\right|& \le \left|\right|{\alpha }_{n+1}{u}_{n+1}+\left(1-{\alpha }_{n+1}\right)T{x}_{n+1}-{\alpha }_n{u}_n-\left(1-{\alpha }_n\right)T{x}_n\left|\right|\\ =\left|\right|{\alpha }_{n+1}{u}_{n+1}-{\alpha }_n{u}_n+\left(1-{\alpha }_{n+1}\right)\left(T{x}_{n+1}-T{x}_n\right)+\left({\alpha }_n-{\alpha }_{n+1}\right)T{x}_n\left|\right|\\ \le \left(1-{\alpha }_{n+1}\right)\left|\right|T{x}_{n+1}-T{x}_n\left|\right|+\left({\alpha }_n+{\alpha }_{n+1}\right)\left|\right|T{x}_n\left|\right|+{\alpha }_{n+1}\left|\right|u_{n+1}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|\\ \le \left(1-{\alpha }_{n+1}\right)\left|\right|x_{n+1}-x_n\left|\right|+\left({\alpha }_n+{\alpha }_{n+1}\right)\left|\right|T{x}_n\left|\right|+{\alpha }_{n+1}\left|\right|u_{n+1}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|.\end{array}$$

It follows that

$$\left|\right|y_{n+1}-y_n\left|\right|-\left|\right|x_{n+1}-x_n\left|\right|\le {\alpha }_{n+1}\left|\right|x_{n+1}-x_n\left|\right|+\left({\alpha }_n+{\alpha }_{n+1}\right)\left|\right|T{x}_n\left|\right|+{\alpha }_{n+1}\left|\right|u_{n+1}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|.$$

This together with (i) and (iii) implies that

$$\underset{n\to \infty }{limsup}\left(\left|\right|y_{n+1}-y_n\left|\right|-\left|\right|x_{n+1}-x_n\left|\right|\right)\le 0.$$

Hence, by Lemma 2.2, we get

$$\underset{n\to \infty }{lim}\left|\right|y_n-x_n\left|\right|=0$$

(3.8)

Consequently, lim _n→∞||x_n+1- x _n || = lim_n→∞β _n ||y _n - x _n || = 0. We now show that

$$\left|\right|x_n-T{x}_n\left|\right|\to 0.$$

Notice that

$$\begin{array}{ll}\hfill \left|\right|x_n-T{x}_n\left|\right|& \le \left|\right|x_n-x_{n+1}\left|\right|+\left|\right|x_{n+1}-T{x}_n\left|\right|\\ \le \left|\right|x_n-x_{n+1}\left|\right|+\left(1-{\beta }_n\right)\left|\right|x_n-T{x}_n\left|\right|+{\beta }_n\left|\right|y_n-P_C\left(T{x}_n\right)\left|\right|\\ \le \left|\right|x_n-x_{n+1}\left|\right|+\left(1-{\beta }_n\right)\left|\right|x_n-T{x}_n\left|\right|+{\beta }_n\left[{\alpha }_n\left|\right|u_n\left|\right|+{\alpha }_n\left|\right|T{x}_n\left|\right|\right].\end{array}$$

Hence,

$$\left|\right|x_n-T{x}_n\left|\right|\le \frac{1}{{\beta }_n}\left|\right|x_n-x_{n+1}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|+{\alpha }_n\left|\right|T{x}_n\left|\right|.$$

Therefore,

$$\underset{n\to \infty }{lim}\left|\right|x_n-T{x}_n\left|\right|=0.$$

(3.9)

We next show that

$$\underset{n\to \infty }{limsup}⟨\tilde{x},\tilde{x}-x_n⟩\le 0,$$

where $\tilde{x}=\underset{m\to \infty }{lim}{y}_m$ and {y _m } be the sequence defined by the implicit method (3.1). Since x _n ∈ C and 〈y _m - [α _m u _m + (1 - α _m )Ty _m ], y _m - x _n 〉 ≤ 0, we have

$$\begin{array}{ll}\hfill \left|\right|y_m-x_n|{|}^2& =⟨y_m-x_n,y_m-x_n⟩\\ =⟨y_m-\left[{\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{y}_m\right],y_m-x_n⟩+⟨{\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{y}_m-x_n,y_m-x_n⟩\\ \le ⟨{\alpha }_m{u}_m+\left(1-{\alpha }_m\right)T{y}_m-x_n,y_m-x_n⟩\\ ={\alpha }_m⟨u_m,y_m-x_n⟩+\left(1-{\alpha }_m\right)⟨T{y}_m-x_n,y_m-x_n⟩-{\alpha }_m⟨x_n,y_m-x_n⟩\\ ={\alpha }_m⟨u_m,y_m-x_n⟩+\left(1-{\alpha }_m\right)⟨T{y}_m-T{x}_n,y_m-x_n⟩\\ +\left(1-{\alpha }_m\right)⟨T{x}_n-x_n,y_m-x_n⟩-{\alpha }_m⟨x_n-y_m,y_m-x_n⟩-{\alpha }_m⟨y_m,y_m-x_n⟩\\ \le {\alpha }_m\left|\right|u_m\left|\right|\left|\right|y_m-x_n\left|\right|+\left(1-{\alpha }_m\right)\left|\right|T{y}_m-T{x}_n\left|\right|\left|\right|y_m-x_n\left|\right|\\ +\left(1-{\alpha }_m\right)\left|\right|T{x}_n-x_n\left|\right|\left|\right|y_m-x_n\left|\right|+{\alpha }_m\left|\right|y_m-x_n|{|}^2-{\alpha }_m⟨y_m,y_m-x_n⟩\\ \le {\alpha }_m\left|\right|u_m\left|\right|M_1+\left|\right|y_m-x_n|{|}^2+||T{x}_n-x_n||M_1-{\alpha }_m⟨y_m,y_m-x_n⟩,\end{array}$$

where M₁ > 0 such that sup{||y _m - x _n ||, m, n ≥ 0} ≤ M₁. It follows that

$$⟨y_m,y_m-x_n⟩\le \left|\right|u_m\left|\right|M_1+\frac{\left|\right|T{x}_n-x_n\left|\right|M_1}{{\alpha }_m}.$$

Therefore,

$$\underset{m\to \infty }{limsup}\underset{n\to \infty }{limsup}⟨y_m,y_m-x_n⟩\le 0.$$

(3.10)

We note that

$$\begin{array}{ll}\hfill ⟨\tilde{x},\tilde{x}-x_n⟩& =⟨\tilde{x},\tilde{x}-y_m⟩+⟨\tilde{x}-y_m,y_m-x_n⟩+⟨y_m,y_m-x_n⟩\\ \le ⟨\tilde{x},\tilde{x}-y_m⟩+\left|\right|\tilde{x}-y_m\left|\right|\left|\right|y_m-x_n\left|\right|+⟨y_m,y_m-x_n⟩\\ \le ⟨\tilde{x},\tilde{x}-y_m⟩+\left|\right|\tilde{x}-y_m\left|\right|M+⟨y_m,y_m-x_n⟩.\end{array}$$

This together with $y_m\to \tilde{x}$ and (3.10) implies that

$$\underset{n\to \infty }{limsup}⟨\tilde{x},\tilde{x}-x_n⟩=\underset{m\to \infty }{limsup}\underset{n\to \infty }{limsup}⟨\tilde{x},\tilde{x}-x_n⟩\le \underset{m\to \infty }{limsup}\underset{n\to \infty }{limsup}⟨y_m,y_m-x_n⟩\le 0$$

(3.11)

From (3.8), (3.9) and (3.11), we have

$$\underset{n\to \infty }{limsup}⟨\tilde{x},\tilde{x}-y_n⟩\le 0\mathsf{\text{and}}\underset{n\to \infty }{limsup}⟨\tilde{x},\tilde{x}-T{x}_n⟩\le 0.$$

(3.12)

Finally, we show that $x_n\to \tilde{x}$. Set z _n = α _n u _n + (1 - α _n )Tx _n ,n ≥ 0. Since $\tilde{x}\in C$ and y _n = P _C (z _n ). Hence $⟨y_n-z_n,y_n-\tilde{x}⟩\le 0$. From (3.5), we have

$$\begin{array}{ll}\hfill \left|\right|x_{n+1}-\tilde{x}|{|}^2& =\left|\right|\left(1-{\beta }_n\right)\left(x_n-\tilde{x}\right)+{\beta }_n\left(y_n-\tilde{x}\right)|{|}^2\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n||y_n-\tilde{x}|{|}^2\\ =\left(1-{\beta }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n⟨y_n-z_n,y_n-\tilde{x}⟩+{\beta }_n⟨z_n-\tilde{x},y_n-\tilde{x}⟩\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n⟨z_n-\tilde{x},y_n-\tilde{x}⟩\\ =\left(1-{\beta }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n\left(1-{\alpha }_n\right)⟨T{x}_n-\tilde{x},y_n-\tilde{x}⟩\\ +{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\beta }_n{\alpha }_n⟨u_n,y_n-\tilde{x}⟩.\end{array}$$

Note that

$$⟨T{x}_n-\tilde{x},y_n-\tilde{x}⟩\le \left|\right|T{x}_n-\tilde{x}\left|\right|\left|\right|y_n-\tilde{x}\left|\right|\le \frac{1}{2}\left(\left|\right|x_n-\tilde{x}|{|}^2+||y_n-\tilde{x}|{|}^2\right)\le \frac{1}{2}\left(\left|\right|x_n-\tilde{x}|{|}^2+||z_n-\tilde{x}|{|}^2\right),$$

and

$$\begin{array}{ll}\hfill \left|\right|z_n-\tilde{x}|{|}^2& =\left|\right|\left(1-{\alpha }_n\right){\left(T{x}_n-\tilde{x}\right)}^2-{\alpha }_n\tilde{x}+{\alpha }_n{u}_n|{|}^2\\ \le \left|\right|\left(1-{\alpha }_n\right){\left(T{x}_n-\tilde{x}\right)}^2-{\alpha }_n\tilde{x}|{|}^2+{\alpha }_n|\left|u_n\right|\left|\right||\left(1-{\alpha }_n\right){\left(T{x}_n-\tilde{x}\right)}^2-{\alpha }_n\tilde{x}||+{\alpha }_n^2|\left|u_n\right|{|}^2\\ \le \left(1-{\alpha }_n\right)\left|\right|T{x}_n-\tilde{x}|{|}^2-2{\alpha }_n\left(1-{\alpha }_n\right)⟨\tilde{x},T{x}_n-\tilde{x}⟩+{\alpha }_n^2|\left|\tilde{x}\right|{|}^2\\ +{\alpha }_n\left|\right|u_n\left|\right|\left|\right|\left(1-{\alpha }_n\right){\left(T{x}_n-\tilde{x}\right)}^2-{\alpha }_n\tilde{x}\left|\right|+{\alpha }_n^2\left|\right|u_n|{|}^2\\ \le \left(1-{\alpha }_n\right)\left|\right|x_n-\tilde{x}|{|}^2-2{\alpha }_n\left(1-{\alpha }_n\right)⟨\tilde{x},T{x}_n-\tilde{x}⟩+{\alpha }_n^2|\left|\tilde{x}\right|{|}^2+{\alpha }_n\left|\right|u_n\left|\right|M_1,\end{array}$$

where M₂ is a constant such that ${sup}_n\left\{\left|\right|\left(1-{\alpha }_n\right){\left(T{x}_n-\tilde{x}\right)}^2-{\alpha }_n\tilde{x}\left|\right|+{\alpha }_n\left|\right|u_n\left|\right|\right\}\le M_2$. Hence, we have

$$\begin{array}{c}\left|\right|x_{n+1}-\tilde{x}|{|}^2\le \left(1-{\beta }_n\right)||x_n-\tilde{x}|{|}^2+\frac{{\beta }_n\left(1-{\alpha }_n\right)}{2}\left(\left|\right|x_n-\tilde{x}|{|}^2+||z_n-\tilde{x}|{|}^2\right)\\ +{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\beta }_n{\alpha }_n{M}_2\left|\right|u_n\left|\right|\\ \le \left(1-{\beta }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+\frac{{\beta }_n\left(1-{\alpha }_n\right)}{2}||x_n-\tilde{x}|{|}^2\\ +\frac{{\beta }_n\left(1-{\alpha }_n\right)}{2}\left(\left(1-{\alpha }_n\right)\left|\right|x_n-\tilde{x}|{|}^2-2{\alpha }_n\left(1-{\alpha }_n\right)⟨\tilde{x},T{x}_n-\tilde{x}⟩\right.\\ \left.+{\alpha }_n^2\left|\right|\tilde{x}|{|}^2+{\alpha }_n|\left|u_n\right||M_2\right)+{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\beta }_n{\alpha }_n{M}_2\left|\right|u_n\left|\right|\\ \le \left(1-{\beta }_n{\alpha }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n{\alpha }_n{\left(1-{\alpha }_n\right)}^2⟨\tilde{x},\tilde{x}-T{x}_n⟩\\ +{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\beta }_n{\alpha }_n^2\left|\right|\tilde{x}|{|}^2+2M_2{\alpha }_n|\left|u_n\right||\\ =\left(1-{\beta }_n{\alpha }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\beta }_n{\alpha }_n\left\{{\left(1-{\alpha }_n\right)}^2⟨\tilde{x},\tilde{x}-T{x}_n⟩\right.\\ \left.+{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\alpha }_n\left|\right|\tilde{x}|{|}^2\right\}+2M_2{\alpha }_n\left|\right|u_n\left|\right|\\ =\left(1-{\gamma }_n\right)\left|\right|x_n-\tilde{x}|{|}^2+{\gamma }_n{\delta }_n+{\sigma }_n,\end{array}$$

where ${\gamma }_n={\beta }_n{\alpha }_n,{\delta }_n={\left(1-{\alpha }_n\right)}^2⟨\tilde{x},\tilde{x}-T{x}_n⟩+{\beta }_n{\alpha }_n⟨\tilde{x},\tilde{x}-y_n⟩+{\alpha }_n\left|\right|\tilde{x}|{|}^2$ and σ _n = 2M₂α _n ||u _n ||. Now, applying Lemma 2.3 to the last inequality, we conclude that $x_n\to \tilde{x}$. This completes the proof.

Corollary 3.8. Suppose$Fix\left(T\right)\ne \varnothing$. Assume the following conditions are satisfied:

• (i) lim_n→∞ α _n = 0 and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$;

• (ii) 0 < lim inf_n→∞ β _n ≤ lim sup_n→∞, β _n < 1;

• (iii) ${\sum }_{n=0}^{\infty }{\alpha }_n\left|\right|u_n\left|\right|<\infty$.

Then, the sequence {x _n } generated by the explicit iterative method (3.6) converges to$\tilde{x}\in Fix\left(T\right)$.

Remark 3.9. We would like to point out that our algorithms (3.1) and (3.5) converge strongly to the minimum-norm fixed point $\tilde{x}$ of T. As a matter of fact, from (3.4), as m → ∞, we deduce

$$\left|\right|\tilde{x}-x^{*}|{|}^2\le ⟨x^{*},x^{*}-\tilde{x}⟩,\forall x^{*}\in Fix\left(T\right),$$

which is equivalent to

$$\left|\right|\tilde{x}|{|}^2\le ⟨x^{*},\tilde{x}⟩\le |\left|x^{*}\right|\left|\right|\left|\tilde{x}\right||,\forall x^{*}\in Fix\left(T\right).$$

Therefore,

$$\left|\right|\tilde{x}\left|\right|\le \left|\right|x^{*}\left|\right|,\forall x^{*}\in Fix\left(T\right).$$

That is, $\tilde{x}$ is the minimum-norm fixed point of T.

Minimum-norm solutions are important in applied problems, e.g., defining the pseudoinverse of a bounded linear operator, and many other problems in signal processing. Therefore, using iterative methods to find the minimum-norm solution of a given nonlinear problem is of significant value. Finding the minimum-norm solution of a nonlinear problem has recently been received a lot of attention, and for some related works, please see [35–37]. Our paper provides such iterative methods (an implicit and an explicit) for finding minimum-norm solutions of nonlinear operator equations governed by non-expansive mappings.