The contraction-proximal point algorithm with square-summable errors

Abstract

In this paper, we study the contraction-proximal point algorithm for approximating a zero of a maximal monotone mapping. The norm convergence of such an algorithm has been established under two new conditions. This extends a recent result obtained by Ceng, Wu and Yao to a more general case.

MSC:47J20, 49J40, 65J15, 90C25.

1 Introduction

We consider the problem of finding $\hat{x}\in \mathcal{H}$ so that

$$0\in A\hat{x},$$

where ℋ is a Hilbert space and $A:\mathcal{H}\rightrightarrows \mathcal{H}$ is a given maximal monotone mapping. This problem is essential due to its various application in some concrete disciplines, including convex programming and variational inequalities. A classical way to solve such a problem is the proximal point algorithm (PPA) [1]. For any initial guess $x_0\in \mathcal{H}$, the PPA generates an iterative sequence as

$$x_{n+1}=J_{c_n}(x_n+e_n),$$

(1)

where $J_{c_n}$ stands for the resolvent of A and $(e_n)$ is the error sequence. In general, the following accuracy criterion on the error sequence:

$$\parallel e_n\parallel \le {ϵ}_n\text{with }\sum _{n=0}^{\mathrm{\infty }}{ϵ}_n<\mathrm{\infty },$$

(I)

is needed to ensure the convergence of PPA. In [1], Rockefeller also presented another accuracy criterion on the error sequence:

$$\parallel e_n\parallel \le {\eta }_n\parallel {\tilde{x}}_n-x_n\parallel \text{with }\sum _{n=0}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty },$$

where

$${\tilde{x}}_n=J_{c_n}(x_n+e_n).$$

This criterion was then improved by Han and He [2] as

$$\parallel e_n\parallel \le {\eta }_n\parallel {\tilde{x}}_n-x_n\parallel \text{with }\sum _{n=0}^{\mathrm{\infty }}{\eta }_n^2<\mathrm{\infty }.$$

(II)

It is well known that the PPA does not necessarily converge strongly [3]. Then how to modify the PPA so that the strong convergence is guaranteed attracts serious attention of many researchers (see, e.g., [4–8]). In particular, one method for doing this has the following scheme:

$$x_{n+1}={\lambda }_{n}u+(1-{\lambda }_n)J_{c_n}(x_n+e_n),$$

(2)

where $u\in \mathcal{H}$ is fixed and $({\lambda }_n)$ is a real sequence. This algorithm, introduced independently by Xu [8] and Kamimura-Takahashi [5], is known as the contraction-proximal point algorithm (CPPA) [9], which is indeed a combination of Halpern’s iteration and the PPA. There are various conditions that ensure the norm convergence of the CPPA with criterion (I) (cf. [7, 10–12]) and the weakest one so far may be the following [13]:

• (i) $c_n\ge c>0$;

• (ii) ${lim}_n{\lambda }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$;

• (iii) $\parallel e_n\parallel \le {\eta }_n$, ${\sum }_{n=0}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty }$.

Let us now turn our attention to the CPPA under criterion (II). In this situation, Ceng, Wu and Yao [14] obtained the norm convergence under the following conditions:

• (i) ${lim}_n{c}_n=\mathrm{\infty }$;

• (ii) ${lim}_n{\lambda }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$;

• (iii) $\parallel e_n\parallel \le {\eta }_n\parallel {\tilde{x}}_n-x_n\parallel$ with ${\sum }_{n=0}^{\mathrm{\infty }}{\eta }_n^2<\mathrm{\infty }$.

In the hypothesis mentioned above, the sequence $(c_n)$ is assumed to tend to infinity, so it is natural to ask whether the norm convergence is still guaranteed for bounded $(c_n)$, especially for constant sequence. In the present paper, we shall answer this question affirmatively and relax condition ${lim}_n{c}_n=\mathrm{\infty }$ to a more general case:

$$c_n\ge c>0,$$

that is, we only need to assume the sequence $(c_n)$ is bounded below away from zero. The paper is organized as follows. In Section 2, we prove two useful lemmas that are very useful for proving the boundedness of the iteration. In Section 3, we establish norm convergence of the CPPA under two different conditions. As a result, we extend the corresponding result obtained in [14].

2 Some lemmas

We denote by ‘→’ strong convergence, and ‘⇀’ weak convergence. An operator $A:\mathcal{H}\rightrightarrows \mathcal{H}$ is called monotone if

$$〈u-v,x-y〉\ge 0,$$

for any $u\in Ax$, $v\in Ay$; maximal monotone if its graph

$$\mathcal{G}(A)=\{(x,y):x\in \mathcal{D}(A),y\in Ax\}$$

is not properly contained in the graph of any other monotone operator.

Let C be a nonempty, closed and convex subset of ℋ. We use $P_C$ to denote the projection from ℋ onto C; namely, for $x\in \mathcal{H}$, $P_{C}x$ is the unique point in C with the property:

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{min}\parallel x-y\parallel .$$

It is well known that $P_{C}x$ is characterized by

$$〈x-P_{C}x,z-P_{C}x〉\le 0,\mathrm{\forall }z\in C.$$

(3)

A mapping $T:\mathcal{H}\to \mathcal{H}$ is called firmly nonexpansive if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel (I-T)x-(I-T)y\parallel }^2$$

for all $x,y\in \mathcal{H}$. Here and hereafter, we denote by

$$J_c={(I+cA)}^{-1}$$

the resolvent of A, where $c>0$ and I is the identity operator. The zero set of A is denoted by $S:=\{x\in \mathcal{D}(A):0\in Ax\}$. The resolvent operator has the following properties (see [15]).

Lemma 1 Let A be a maximal monotone operator. Then

• (i) $\mathcal{D}(J_c)=\mathcal{H}$;

• (ii) $J_c$ is single-valued and firmly nonexpansive;

• (iii) $Fix(J_c)=S$, where $Fix(J_c)$ denotes the fixed point set of $J_c$;

• (iv) its graph $\mathcal{G}(A)$ is weak-to-strong closed in $\mathcal{H}\times \mathcal{H}$.

Since $J_c$ is firmly nonexpansive, this implies that

$${\parallel J_{c}x-z\parallel }^2\le {\parallel x-z\parallel }^2-{\parallel (I-J_c)x\parallel }^2$$

(4)

for all $x\in \mathcal{H}$ and all $z\in S$. In what follows, we present two lemmas that are very useful for proving the boundedness of the iterative sequence.

Lemma 2 Given $\beta >0$, let $(s_n)$ be a nonnegative real sequence satisfying

$$s_{n+1}\le (1-{\lambda }_n)(1+{ϵ}_n)s_n+{\lambda }_n\beta ,$$

where $({\lambda }_n)\subset (0,1)$ and $({ϵ}_n)\in {\ell }_1$ are real sequences. Then $(s_n)$ is bounded; more precisely,

$$s_n\le max\{\beta ,s_0\}exp\left(\sum _{n=0}^{\mathrm{\infty }}{ϵ}_n\right)<\mathrm{\infty }.$$

Proof

We first show the following estimates:

$$s_{n+1}\le max\{\beta ,s_0\}\prod _{k=0}^n(1+{ϵ}_k),\mathrm{\forall }n\ge 0.$$

(5)

For $n=0$, we have

$$\begin{array}{rcl}{s}_1& \le & (1-{\lambda }_0)(1+{ϵ}_0)s_0+{\lambda }_0\beta \\ \le & (1+{ϵ}_0)[(1-{\lambda }_0)s_0+{\lambda }_0\beta ]\\ \le & max\{\beta ,s_0\}(1+{ϵ}_0).\end{array}$$

Assume $s_n\le max\{\beta ,s_0\}{\prod }_{k=0}^{n-1}(1+{ϵ}_k)$. Since $max\{\beta ,s_0\}{\prod }_{k=0}^{n-1}(1+{ϵ}_k)\ge \beta$, we have

$$\begin{array}{rcl}{s}_{n+1}& \le & (1+{ϵ}_n)(1-{\lambda }_n)s_n+{\lambda }_n\beta \\ \le & (1+{ϵ}_n)[(1-{\lambda }_n)s_n+{\lambda }_n\beta ]\\ \le & (1+{ϵ}_n)max\{s_0,\beta \}\prod _{k=0}^{n-1}(1+{ϵ}_k)\\ =& max\{\beta ,s_0\}\prod _{k=0}^n(1+{ϵ}_k).\end{array}$$

We thus verify inequality (5) by induction. Hence,

$$\begin{array}{rcl}{s}_{n+1}& \le & max\{\beta ,s_0\}\prod _{k=0}^n(1+{ϵ}_k)\\ =& max\{\beta ,s_0\}exp(\sum _{k=0}^{n}ln(1+{ϵ}_k))\\ \le & max\{\beta ,s_0\}exp\left(\sum _{k=0}^{\mathrm{\infty }}{ϵ}_k\right)<\mathrm{\infty },\end{array}$$

where the last inequality follows from the basic inequality: $ln(1+x)<x$ for all $x>0$. □

Lemma 3 Given $\beta >0$, let $(s_n)$ be a nonnegative real sequence satisfying

$$s_{n+1}\le (1-{\lambda }_n)(1+{ϵ}_n)s_n+{\lambda }_n\beta ,$$

where $({\lambda }_n)\subset (0,1)$ and $({ϵ}_n)\subseteq [0,\mathrm{\infty })$ are real sequences. If $2{ϵ}_n(1-{\lambda }_n)\le {\lambda }_n$, then $(s_n)$ is bounded; more precisely, $s_n\le max\{2\beta ,s_0\}<\mathrm{\infty }$.

Proof Let ${\tau }_n={\lambda }_n-{ϵ}_n(1-{\lambda }_n)$. Then ${\tau }_n\in (0,1)$. It follows that

$$\begin{array}{rcl}{s}_{n+1}& \le & (1+{ϵ}_n)(1-{\lambda }_n)s_n+{\lambda }_n\beta \\ =& (1-{\tau }_n)s_n+{\tau }_n({\lambda }_n\beta /{\tau }_n)\\ \le & max\{{\lambda }_n\beta /{\tau }_n,s_n\}.\end{array}$$

Since $2{ϵ}_n(1-{\lambda }_n)\le {\lambda }_n$, we have

$$\frac{{\lambda }_n}{{\tau }_n}=\frac{{\lambda }_n}{{\lambda }_n-{ϵ}_n(1-{\lambda }_n)}\le 2,$$

which implies that

$$s_{n+1}\le max\{2\beta ,s_n\}.$$

By induction, we can show the result as desired. □

We end this section by two useful lemmas. The first one is due to Maingé [16] and the second one is due to Xu [8].

Lemma 4 Let $(s_n)$ be a real sequence that does not decrease at infinity, in the sense that there exists a subsequence $(s_{n_k})$ so that

$$s_{n_k}\le s_{n_k+1}\mathit{\text{for all}}k\ge 0.$$

For every $n>n_0$ define an integer sequence $(\tau (n))$ as

$$\tau (n)=max\{n_0\le k\le n:s_k<s_{k+1}\}.$$

Then $\tau (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and for all $n>n_0$

$$max(s_{\tau (n)},s_n)\le s_{\tau (n)+1}.$$

(6)

Lemma 5 Let $\{s_n\}$, $\{c_n\}\subset {\mathbb{R}}^{+}$, $\{{\lambda }_n\}\subset (0,1)$ and $\{b_n\}\subset \mathbb{R}$ be sequences such that

$$s_{n+1}\le (1-{\lambda }_n)s_n+b_n+c_n\mathit{\text{for all}}n\ge 0.$$

If ${\lambda }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$, ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$ and ${\overline{lim}}_n{b}_n/{\lambda }_n\le 0$, then ${lim}_n{s}_n=0$.

3 Convergence analysis

In what follows, we assume that A is a maximal monotone mapping and its zero set S is nonempty. To establish the convergence, we need the following lemma, which is indeed proved in [2]. We present here a different proof that is mainly based on property of firmly nonexpansive mappings.

Lemma 6 Let $\eta \in (0,1/2)$, $x,e\in \mathcal{H}$ and $\tilde{x}:=J_c(x+e)$. If $\parallel e\parallel \le \eta \parallel x-\tilde{x}\parallel$, then

$${\parallel \tilde{x}-z\parallel }^2\le (1+{(2\eta )}^2){\parallel x-z\parallel }^2-\frac{1}{2}{\parallel \tilde{x}-x\parallel }^2,\mathrm{\forall }z\in S.$$

(7)

Proof Since $z\in Fix(J_c)$, it follows from (4) that

$$\begin{array}{rcl}{\parallel \tilde{x}-z\parallel }^2& \le & {\parallel x+e-z\parallel }^2-{\parallel x+e-J_c(x+e)\parallel }^2\\ =& {\parallel (x-z)+e\parallel }^2-{\parallel (x-\tilde{x})+e\parallel }^2\\ =& {\parallel x-z\parallel }^2+2〈\tilde{x}-z,e〉-{\parallel \tilde{x}-x\parallel }^2.\end{array}$$

(8)

By using inequality $2〈a,b〉\le 2{\eta }^2{\parallel a\parallel }^2+{\parallel b\parallel }^2/2{\eta }^2$, we have

$$2〈\tilde{x}-z,e〉\le 2{\eta }^2{\parallel \tilde{x}-z\parallel }^2+\frac{1}{2{\eta }^2}{\parallel e\parallel }^2.$$

Subsisting this into (8) and noting $\parallel e\parallel \le \eta \parallel x-\tilde{x}\parallel$, we see that

$${\parallel \tilde{x}-z\parallel }^2\le {\parallel x-z\parallel }^2+2{\eta }^2{\parallel \tilde{x}-z\parallel }^2-\frac{1}{2}{\parallel \tilde{x}-x\parallel }^2,$$

from which it follows that

$${\parallel \tilde{x}-z\parallel }^2\le (1+\frac{2{\eta }^2}{1-2{\eta }^2}){\parallel x-z\parallel }^2-\frac{1}{2(1-2{\eta }^2)}{\parallel \tilde{x}-x\parallel }^2.$$

Consequently, the desired inequality (7) follows from the fact $\eta \in (0,1/2)$. □

We now are ready to prove our main results.

Theorem 1 For any $x_0\in \mathcal{H}$, the sequence $(x_n)$ generated by

$$[\begin{array}{l}{\tilde{x}}_n=J_{c_n}(x_n+e_n),\\ x_{n+1}={\lambda }_{n}u+(1-{\lambda }_n){\tilde{x}}_n,\end{array}$$

(9)

converges strongly to $P_S(u)$, provided that

• (i) $c_n\ge c>0$;

• (ii) ${lim}_n{\lambda }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$;

• (iii) $\parallel e_n\parallel \le {\eta }_n\parallel {\tilde{x}}_n-x_n\parallel$, ${\sum }_{n=0}^{\mathrm{\infty }}{\eta }_n^2<\mathrm{\infty }$.

Proof Let $z=P_S(u)$. By our hypothesis, we may assume without loss of generality that ${\eta }_n\in (0,1/2)$. Then by Lemma 6, we have

$${\parallel {\tilde{x}}_n-z\parallel }^2\le (1+{ϵ}_n){\parallel x_n-z\parallel }^2-\frac{1}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2,$$

(10)

where ${ϵ}_n:={(2{\eta }_n)}^2$ satisfying ${\sum }_{n=0}^{\mathrm{\infty }}{ϵ}_n<\mathrm{\infty }$. It then follows from (9) that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel (1-{\lambda }_n){({\tilde{x}}_n-z)}^2+{\lambda }_n(u-z)\parallel }^2\\ \le & (1-{\lambda }_n){\parallel {\tilde{x}}_n-z\parallel }^2+{\lambda }_n{\parallel u-z\parallel }^2,\end{array}$$

which together with (10) yields

$${\parallel x_{n+1}-z\parallel }^2\le (1-{\lambda }_n)(1+{ϵ}_n){\parallel x_n-z\parallel }^2+{\lambda }_n{\parallel u-z\parallel }^2.$$

Applying Lemma 2 to the last inequality, we conclude that $(x_n)$ is bounded.

It follows from the subdifferential inequality that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel (1-{\lambda }_n)({\tilde{x}}_n-z)+{\lambda }_n(u-z)\parallel }^2\\ \le & (1-{\lambda }_n){\parallel {\tilde{x}}_n-z\parallel }^2+2{\lambda }_n〈u-z,x_{n+1}-z〉.\end{array}$$

Combining this with (10) yields

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & (1-{\lambda }_n){\parallel x_n-z\parallel }^2-\frac{1-{\lambda }_n}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2\\ +2{\lambda }_n〈u-z,x_{n+1}-z〉+M{ϵ}_n,\end{array}$$

(11)

where $M>0$ is a sufficiently large number. Since $({ϵ}_n)\in {\ell }_1$, we assume that

$$s:=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n{ϵ}_k<\mathrm{\infty }$$

and define $t_n:=s-{\sum }_{k=0}^{n-1}{ϵ}_k$. Setting $s_n={\parallel x_n-z\parallel }^2+M{t}_n$, we rewrite (11) as

$$s_{n+1}-s_n+{\lambda }_n{\parallel x_n-z\parallel }^2+\frac{1-{\lambda }_n}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2\le 2{\lambda }_n〈u-z,x_{n+1}-z〉.$$

(12)

It is obvious that $s_n\to 0\iff \parallel x_n-z\parallel \to 0$.

We next consider two possible cases on the sequence $(s_n)$.

Case 1. $(s_n)$ is eventually decreasing (i.e., there exists $N\ge 0$ such that $(s_n)$ is decreasing for $n\ge N$). In this case, $(s_n)$ must be convergent, and from (12) it follows

$$\frac{1-{\lambda }_n}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2\le (s_n-s_{n+1})+M{\lambda }_n\to 0,$$

from which we have $\parallel {\tilde{x}}_n-x_n\parallel \to 0$. Extract a subsequence $(x_{n_k})$ from $(x_n)$ so that $(x_{n_k})$ converges weakly to $\hat{x}$ and

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{n+1}-z〉=\underset{k\to \mathrm{\infty }}{lim}〈u-z,x_{n_k}-z〉.$$

By noting the fact that ${\tilde{x}}_n=J_n(x_n+e_n)$, this implies

$$0\leftarrow \frac{x_{n_k}+e_{n_k}-{\tilde{x}}_{n_k}}{c_{n_k}}\in A({\tilde{x}}_{n_k})$$

and ${\tilde{x}}_{n_k}=x_{n_k}+({\tilde{x}}_{n_k}-x_{n_k})\rightharpoonup \hat{x}$. Hence, the weak-to-strong closedness of $\mathcal{G}(A)$ implies $0\in A(\hat{x})$, i.e., $\hat{x}\in S$. Consequently, we have

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{n+1}-z〉=〈u-z,\hat{x}-z〉\le 0,$$

where the inequality follows from (3). Again it follows from (12) that

$${\parallel x_{n+1}-z\parallel }^2\le (1-{\lambda }_n){\parallel x_n-z\parallel }^2+2{\lambda }_n〈u-z,x_{n+1}-z〉+M{ϵ}_n.$$

By using Lemma 5, we conclude that $\parallel x_n-z\parallel \to 0$.

Case 2. $(s_n)$ is not eventually decreasing. Hence, we can find a subsequence $(s_{n_k})$ so that $s_{n_k}\le s_{n_k+1}$ for all $k\ge 0$. In this case, we may define an integer sequence $(\tau (n))$ as in Lemma 4. Since $s_{\tau (n)}\le s_{\tau (n)+1}$ for all $n>n_0$, it follows again from (12) that

$$\frac{1-{\lambda }_{\tau (n)}}{2}{\parallel {\tilde{x}}_{\tau (n)}-x_{\tau (n)}\parallel }^2\le M{\lambda }_{\tau (n)}\to 0,$$

so that $\parallel {\tilde{x}}_{\tau (n)}-x_{\tau (n)}\parallel \to 0$ as $n\to \mathrm{\infty }$. Analogously,

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{\tau (n)}-z〉\le 0.$$

(13)

On the other hand, we deduce from (9) that

$$\parallel x_{\tau (n)}-x_{\tau (n)+1}\parallel \le {\lambda }_{\tau (n)}\parallel u-x_{\tau (n)}\parallel +\parallel {\tilde{x}}_{\tau (n)}-x_{\tau (n)}\parallel \to 0,$$

which together with (13) gets

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{\tau (n)+1}-z〉\le 0.$$

(14)

Noting $s_{\tau (n)+1}-s_{\tau (n)}\ge 0$ and dividing by ${\lambda }_{\tau (n)}$ in (12), we arrive at

$${\parallel x_{\tau (n)}-z\parallel }^2\le 2〈u-z,x_{\tau (n)+1}-z〉$$

for all $n>n_0$, which together with (14) yields

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}\parallel x_{\tau (n)}-z\parallel \le 0.$$

In view of (6), we have

$${\parallel x_n-z\parallel }^2+M{t}_n\le {\parallel x_{\tau (n)+1}-z\parallel }^2+M{t}_{\tau (n)+1}.$$

Since $\parallel x_{\tau (n)}-z\parallel \to 0$ and $\parallel x_{\tau (n)+1}-x_{\tau (n)}\parallel \to 0$ implies $\parallel x_{\tau (n)+1}-z\parallel \to 0$, this together with the fact $t_n\to 0$ immediately yields $x_n\to z$. □

For criterion (I), Boikanyo and Morosanu [10] introduced a new condition:

$$\parallel e_n\parallel \le {\eta }_n\text{with }\underset{n\to \mathrm{\infty }}{lim}\frac{{\eta }_n}{{\lambda }_n}=0$$

to ensure the convergence of the CPPA. In the following theorem, we shall present a similar condition under the accuracy criterion (II).

Theorem 2 For any $x_0\in \mathcal{H}$, the sequence $(x_n)$ generated by

$$[\begin{array}{l}{\tilde{x}}_n=J_{c_n}(x_n+e_n),\\ x_{n+1}={\lambda }_{n}u+(1-{\lambda }_n){\tilde{x}}_n,\end{array}$$

(15)

converges strongly to $P_S(u)$, provided that

• (i) $c_n\ge c>0$;

• (ii) ${lim}_n{\lambda }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$;

• (iii) $\parallel e_n\parallel \le {\eta }_n\parallel {\tilde{x}}_n-x_n\parallel$, ${lim}_n{\eta }_n^2/{\lambda }_n=0$.

Proof Let $z=P_S(u)$. Similarly, we have

$${\parallel x_{n+1}-z\parallel }^2\le (1-{\lambda }_n)(1+{ϵ}_n){\parallel x_n-z\parallel }^2+{\lambda }_n{\parallel u-z\parallel }^2,$$

(16)

where ${ϵ}_n:={(2{\eta }_n)}^2$ satisfying ${ϵ}_n/{\lambda }_n\to 0$, so we assume without loss of generality that $2{ϵ}_n(1-{\lambda }_n)\le {\lambda }_n$. Applying Lemma 3, we conclude that $(x_n)$ is bounded.

From inequality (11), we also obtain

$$s_{n+1}-s_n+{\lambda }_n{s}_n+\frac{1-{\lambda }_n}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2\le 2{\lambda }_n〈u-z,x_{n+1}-z〉+s_n{ϵ}_n,$$

(17)

where we define $s_n:={\parallel x_n-z\parallel }^2$.

To show $s_n\to 0$, we consider two possible cases for $(s_n)$.

Case 1. $(s_n)$ is eventually decreasing (i.e., there exists $N\ge 0$ such that $(s_n)$ is decreasing for $n\ge N$). In this case, $(s_n)$ must be convergent, and from (12) it follows

$$\frac{1-{\lambda }_n}{2}{\parallel {\tilde{x}}_n-x_n\parallel }^2\le (s_n-s_{n+1})+M({ϵ}_n+{\lambda }_n)\to 0,$$

where $M>0$ a sufficiently large number. Analogous to the previous theorem,

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{n+1}-z〉\le 0.$$

Rearranging terms in (17) yields that

$$s_{n+1}\le (1-{\lambda }_n)s_n+{\lambda }_{n}M(2〈u-z,x_{n+1}-z〉+{ϵ}_n/{\lambda }_n).$$

We note that by our hypothesis ${ϵ}_n/{\lambda }_n$ goes to zero, and thus apply Lemma 5 to the previous inequality to conclude that $s_n\to 0$.

Case 2. $(s_n)$ is not eventually decreasing. In this case, we may define an integer sequence $(\tau (n))$ as in Lemma 4. Since $s_{\tau (n)}\le s_{\tau (n)+1}$ for all $n>n_0$, it follows again from (17) that

$$\frac{1-{\lambda }_{\tau (n)}}{2}{\parallel {\tilde{x}}_{\tau (n)}-x_{\tau (n)}\parallel }^2\le M({\lambda }_{\tau (n)}+{ϵ}_{\tau (n)})\to 0,$$

so that $\parallel {\tilde{x}}_{\tau (n)}-x_{\tau (n)}\parallel \to 0$, and furthermore $\parallel x_{\tau (n)}-x_{\tau (n)+1}\parallel \to 0$. Analogously,

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}〈u-z,x_{\tau (n)+1}-z〉\le 0.$$

It follows from (17) that for all $n>n_0$

$$s_{\tau (n)}\le 2〈u-z,x_{\tau (n)+1}-z〉+\frac{M{ϵ}_{\tau (n)}}{{\lambda }_{\tau (n)}}.$$

By combining the last two inequalities, we have

$$\underset{n\to \mathrm{\infty }}{\overline{lim}}{s}_{\tau (n)}\le 0,$$

from which we arrive at

$$\begin{array}{rcl}\sqrt{s_{\tau (n)+1}}& =& \parallel (x_{\tau (n)}-z)-(x_{\tau (n)}-x_{\tau (n)+1})\parallel \\ \le & \sqrt{s_{\tau (n)}}+\parallel x_{\tau (n)}-x_{\tau (n)+1}\parallel \to 0.\end{array}$$

Consequently, $s_n\to 0$ follows from (6) immediately. □