Approximation of fixed points for nonexpansive semigroups in Hilbert spaces

Abstract

In this paper, we propose two new algorithms for finding a common fixed point of a nonexpansive semigroup in Hilbert spaces and prove some strong convergence theorems for nonexpansive semigroups. Our results improve and generalize the corresponding results given by Shimizu and Takahashi (J. Math. Anal. Appl. 211:71-83, 1997), Shioji and Takahashi (Nonlinear Anal. TMA 34:87-99, 1998), Lau et al. (Nonlinear Anal. TMA 67:1211-1225, 2007) and many others.

MSC:47H05, 47H10, 47H17.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. A mapping $T:C\to C$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

Recall that a family $S:={\{T(s)\}}_{s\ge 0}$ of mappings of C into itself is called a nonexpansive semigroup if it satisfies the following conditions:

• (S1) $T(0)x=x$ for all $x\in C$;

• (S2) $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$;

• (S3) $\parallel T(s)x-T(s)y\parallel \le \parallel x-y\parallel$ for all $x,y\in C$ and $s\ge 0$;

• (S4) for each $x\in H$, $s\to T(s)x$ 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., $Fix(S)={\bigcap }_{s\ge 0}Fix(T(s))$. It is known that $Fix(S)$ is closed and convex [[1], Lemma 1].

Approximation of fixed points of nonexpansive mappings by a sequence of finite means has been considered by many authors; see, for instance, [1–27]. This work was originated with the beautiful work of Baillon [5] in 1975 (see also [6] and [7] for a generalization): If C is a closed convex subset of a Hilbert space and T is a nonexpansive mapping from C into itself such that the set $Fix(T)$ of fixed points of T is nonempty, then for each $x\in C$, the Cesàro mean

$$\frac{1}{n}\sum _{k=1}^n{T}^{k}x$$

converges weakly to $x^{\ast }\in Fix(T)$. In this case, if we put $x^{\ast }=P_{Fix(T)}x$ for each $x\in C$, then $P_{Fix(T)}$ is a nonexpansive retraction from C onto $F(T)$. In [18], Takahashi proved the existence of such a retraction for an amenable semigroup of nonexpansive mappings on a Hilbert space. In [19], Rodé also found a sequence of means on a semigroup generalizing the Cesàro means and extended Baillon’s theorem. In [28], Lau, Shioji and Takahashi extended Takahashi’s result and Rode’s result to a closed convex subset of a uniformly convex Banach space.

In the literature, a nonlinear ergodic theorem for nonexpansive semigroups has been considered by many authors (see [29–46]). Especially, Shioji and Takahashi [17] introduced an implicit iteration $\{x_n\}$ in a Hilbert space defined by

$$x_n={\alpha }_{n}x+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds,\mathrm{\forall }n\ge 0,$$

(1.1)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ and $\{{\lambda }_n\}$ is a sequence of positive real numbers divergent to ∞. Under certain restrictions on the sequence $\{{\alpha }_n\}$, Shioji and Takahashi [17] proved strong convergence of $\{x_n\}$ generated by (1.1) to a member of $Fix(T(s))$. In [16], Shimizu and Takahashi studied the strong convergence of the iterative sequence $\{x_n\}$ defined by

$$x_{n+1}={\alpha }_{n}x+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds,\mathrm{\forall }n\ge 0.$$

(1.2)

The corresponding viscosity approximations of (1.1) and (1.2) have been extended in [29]. Lau et al. [37] studied the iterative schemes of Browder and Halpern types for a nonexpansive semigroup ${\{T(s)\}}_{s\ge 0}$ on a compact convex subset C of a smooth (and strictly convex) Banach space with respect to a sequence $\{{\mu }_n\}$ of strongly asymptotically invariant means defined on an appropriate invariant subspace of $l^{\mathrm{\infty }}(S)$, the space of bounded real-valued functions on a semigroup S.

Motivated and inspired by the works in the literature, in this paper, we introduce two new algorithms for finding a common fixed point of a nonexpansive semigroup ${\{T(s)\}}_{s\ge 0}$ in Hilbert spaces and prove that both approaches converge strongly to a common fixed point of ${\{T(s)\}}_{s\ge 0}$.

2 Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. The metric (or nearest point) projection from H onto C is the mapping $P_C:H\to C$ which assigns to each point $x\in C$ the unique point $P_{C}x\in C$ satisfying the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{inf}\parallel x-y\parallel =:d(x,C).$$

It is well known that $P_C$ is a nonexpansive mapping and satisfies

$$〈x-y,P_{C}x-P_{C}y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2,\mathrm{\forall }x,y\in H.$$

Moreover, $P_C$ is characterized by the following properties:

$$〈x-P_{C}x,y-P_{C}x〉\le 0$$

(2.1)

and

$${\parallel x-y\parallel }^2\ge {\parallel x-P_{C}x\parallel }^2+{\parallel y-P_{C}x\parallel }^2,\mathrm{\forall }x\in H,y\in C.$$

We need the following lemmas for proving our main results.

Lemma 2.1 [16]

Let C be a nonempty bounded closed convex subset of a Hilbert space H and ${\{T(s)\}}_{s\ge 0}$ be a nonexpansive semigroup on C. Then, for any $h\ge 0$,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in C}{sup}\parallel \frac{1}{t}{\int }_0^{t}T(s)xds-T(h)\frac{1}{t}{\int }_0^{t}T(s)xds\parallel =0.$$

Lemma 2.2 [8]

Let C be a closed convex subset of a real Hilbert space H and $S:C\to C$ be a nonexpansive mapping. Then the mapping $I-S$ is demiclosed. That is, if $\{x_n\}$ is a sequence in C such that $x_n\to x^{\ast }$ weakly and $(I-S)x_n\to y$ strongly, then $(I-S)x^{\ast }=y$.

Lemma 2.3 [13]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X and $\{{\gamma }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that

$$x_{n+1}=(1-{\gamma }_n)x_n+{\gamma }_n{y}_n,\mathrm{\forall }n\ge 0,$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_n-y_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel )\le 0.$$

Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.4 [12]

Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n{\gamma }_n,\mathrm{\forall }n\ge 1,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

• (a) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (b) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n{\gamma }_n|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

3 Main results

In this section, we show our main results.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\{T(s)\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix(S)\ne \mathrm{\varnothing }$. Let ${\{{\gamma }_t\}}_{0<t<1}$ and ${\{{\lambda }_t\}}_{0<t<1}$ be two continuous nets of positive real numbers such that ${\gamma }_t\in (0,1)$, ${lim}_{t\to 0}{\gamma }_t=1$ and ${lim}_{t\to 0}{\lambda }_t=+\mathrm{\infty }$. Let $\{x_t\}$ be the net defined in the following implicit manner:

$$x_t=P_C[t({\gamma }_t{x}_t)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds],\mathrm{\forall }t\in (0,1).$$

(3.1)

Then, as $t\to 0+$, the net $\{x_t\}$ strongly converges to $x^{\ast }\in Fix(S)$.

Proof First, we note that the net $\{x_t\}$ defined by (3.1) is well defined. We define the mapping

$$Wx:=P_C[t({\gamma }_{t}x)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)xds],\mathrm{\forall }t\in (0,1).$$

It follows that

$$\begin{array}{rcl}\parallel Wx-Wy\parallel & \le & \parallel t{\gamma }_t(x-y)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}(T(s)x-T(s)y)ds\parallel \\ \le & t{\gamma }_t\parallel x-y\parallel +(1-t)\parallel \frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}(T(s)x-T(s)y)ds\parallel \\ \le & t{\gamma }_t\parallel x-y\parallel +(1-t)\parallel x-y\parallel \\ =& [1-(1-{\gamma }_t)t]\parallel x-y\parallel .\end{array}$$

This implies that the mapping W is a contraction and so it has a unique fixed point. Therefore, the net $\{x_t\}$ defined by (3.1) is well defined.

Take $p\in Fix(S)$. By (3.1), we have

$$\begin{array}{rcl}\parallel x_t-p\parallel & =& \parallel P_C[t({\gamma }_t{x}_t)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds]-p\parallel \\ \le & \parallel t{\gamma }_t(x_t-p)-t(1-{\gamma }_t)p+(1-t)(\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds-p)\parallel \\ \le & t{\gamma }_t\parallel x_t-p\parallel +t(1-{\gamma }_t)\parallel p\parallel +(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}\parallel T(s)x_t-T(s)p\parallel ds\\ \le & t{\gamma }_t\parallel x_t-p\parallel +t(1-{\gamma }_t)\parallel p\parallel +(1-t)\parallel x_t-p\parallel \\ =& [1-(1-{\gamma }_t)t]\parallel x_t-p\parallel +t(1-{\gamma }_t)\parallel p\parallel .\end{array}$$

It follows that

$$\parallel x_t-p\parallel \le \parallel p\parallel ,$$

which implies that the net $\{x_t\}$ is bounded. Set $R:=\parallel p\parallel$. It is clear that $\{x_t\}\subset B(p,R)$. Notice that

$$\parallel \frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds-p\parallel \le \parallel x_t-p\parallel \le R.$$

Moreover, we observe that if $x\in B(p,R)$, then

$$\parallel T(s)x-p\parallel \le \parallel T(s)x-T(s)p\parallel \le \parallel x-p\parallel \le R,$$

i.e., $B(p,R)$ is $T(s)$-invariant for all s. Set $y_t=t({\gamma }_t{x}_t)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds$. Then $x_t=P_C[y_t]$. It follows that

By Lemma 2.1, we deduce that for all $0\le \tau <\mathrm{\infty }$,

$$\underset{t\to 0}{lim}\parallel T(\tau )x_t-x_t\parallel =0.$$

(3.2)

Note that $x_t=P_C[y_t]$. By using the property of the metric projection (2.1), we have

$$\begin{array}{rcl}{\parallel x_t-p\parallel }^2& =& 〈x_t-y_t,x_t-p〉+〈y_t-p,x_t-p〉\\ \le & 〈y_t-p,x_t-p〉\\ =& t{\gamma }_t〈x_t-p,x_t-p〉-t(1-{\gamma }_t)〈p,x_t-p〉\\ +(1-t)〈\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds-p,x_t-p〉\\ \le & [1-(1-{\gamma }_t)t]{\parallel x_t-p\parallel }^2-t(1-{\gamma }_t)〈p,x_t-p〉.\end{array}$$

Therefore, we have

$${\parallel x_t-p\parallel }^2\le 〈p,p-x_t〉,\mathrm{\forall }p\in Fix(S).$$

(3.3)

From this inequality, immediately it follows that ${\omega }_w(x_t)={\omega }_s(x_t)$, where ${\omega }_w(x_t)$ and ${\omega }_s(x_t)$ denote the sets of weak and strong cluster points of $\{x_t\}$, respectively.

Let $\{t_n\}\subset (0,1)$ be a sequence such that $t_n\to 0$ as $n\to \mathrm{\infty }$. Put $x_n:=x_{t_n}$, $y_n:=y_{t_n}$ and ${\lambda }_n:={\lambda }_{t_n}$. Since $\{x_n\}$ is bounded, without loss of generality, we may assume that the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in C$. Also, $y_n\to x^{\ast }$ weakly. Noticing (3.2), we can use Lemma 2.2 to get $x^{\ast }\in Fix(S)$. From (3.3), we have

$${\parallel x_n-p\parallel }^2\le 〈p,p-x_n〉,\mathrm{\forall }p\in Fix(S).$$

(3.4)

In particular, if we substitute $x^{\ast }$ for p in (3.4), then we have

$${\parallel x_n-x^{\ast }\parallel }^2\le 〈x^{\ast },x^{\ast }-x_n〉.$$

(3.5)

However, $x_n\rightharpoonup x^{\ast }$. This together with (3.5) guarantees that $x_n\to x^{\ast }$ and so the net $\{x_t\}$ is relatively compact, as $t\to 0^{+}$, in the norm topology.

Now, in (3.4), taking $n\to \mathrm{\infty }$, we get

$${\parallel x^{\ast }-p\parallel }^2\le 〈p,p-x^{\ast }〉,\mathrm{\forall }p\in Fix(S).$$

This is equivalent to the following:

$$0\le 〈x^{\ast },p-x^{\ast }〉,\mathrm{\forall }p\in Fix(S).$$

Therefore, $x^{\ast }=P_{Fix(T)}(0)$, which is obviously unique. This is sufficient to conclude that the entire net $\{x_t\}$ converges in norm to $x^{\ast }$. This completes the proof. □

Remark 3.2

It is known that the algorithm

$$x_t=P_C[t{x}_t+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds],\mathrm{\forall }t\in (0,1),$$

has only weak convergence. However, our similar algorithm (3.1) (with ${\gamma }_t\to 1$) has strong convergence.

Next, we introduce an explicit algorithm for the nonexpansive semigroup $S={\{T(s)\}}_{s\ge 0}:C\to C$ and prove the strong convergence theorems of this algorithm.

Theorem 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\{T(s)\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix(S)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence generated iteratively by the following explicit algorithm:

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{P}_C[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds],\mathrm{\forall }n\ge 0,$$

(3.6)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are sequences of real numbers in $[0,1]$ and $\{{\lambda }_n\}$ is a sequence of positive real numbers. Suppose that the following conditions are satisfied:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (iii) ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n-1}}{{\lambda }_n}=1$.

Then the sequence $\{x_n\}$ generated by (3.6) strongly converges to a point $x^{\ast }\in Fix(S)$.

Proof Take $p\in Fix(S)$. From (3.6), we have

It follows that, by induction,

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\parallel p\parallel \}.$$

Set $y_n=P_C[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)z_n]$ for all $n\ge 0$, where $z_n=\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds$. We have

and

Therefore, we have

where $M>0$ is a constant such that

$$\underset{n\ge 1}{sup}\{\parallel x_{n-1}\parallel ,\parallel z_{n-1}\parallel ,2\parallel x_{n-1}-p\parallel \}\le M.$$

Hence we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_n-y_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel )\le 0.$$

This together with Lemma 2.3 implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-x_n\parallel =0.$$

Therefore, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}{\beta }_n\parallel y_n-x_n\parallel =0.$$

Note that

$$\begin{array}{rcl}\parallel T(\tau )x_n-x_n\parallel & \le & \parallel T(\tau )x_n-T(\tau )\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds\parallel \\ +\parallel T(\tau )\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds-\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds\parallel \\ +\parallel \frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds-x_n\parallel \\ \le & \parallel T(\tau )\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds-\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds\parallel \\ +2\parallel x_n-\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds\parallel .\end{array}$$

(3.7)

From (3.6), we have

It follows that

(3.8)

From (3.7), (3.8) and Lemma 2.1, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T(\tau )x_n-x_n\parallel =0,\mathrm{\forall }\tau \ge 0.$$

(3.9)

Notice that $\{x_n\}$ is a bounded sequence and $\tilde{x}$ is a weak limit of $\{x_n\}$. Putting $x^{\ast }=P_{Fix(S)}(0)$. Then there exists a positive number R such that $B(x^{\ast },R)$ contains $\{x_n\}$. Moreover, $B(x^{\ast },R)$ is $T(s)$-invariant for all $s\ge 0$ and so, without loss of generality, we can assume that ${\{T(s)\}}_{s\ge 0}$ is a nonexpansive semigroup on $B(x^{\ast },R)$. By the demiclosedness principle (Lemma 2.2) and (3.9), we have $\tilde{x}\in Fix(S)$ and hence

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast },x_{n+1}-x^{\ast }〉=\underset{n\to \mathrm{\infty }}{lim}〈x^{\ast },\tilde{x}-x^{\ast }〉\le 0.$$

Finally, we prove that $x_n\to x^{\ast }$. Set $u_n={\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds$. It follows that $y_n=P_C[u_n]$ for all $n\ge 0$. By using the property of the metric projection (2.1), we have

$$〈y_n-u_n,y_n-x^{\ast }〉\le 0$$

and so

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& =& 〈y_n-x^{\ast },y_n-x^{\ast }〉\\ =& 〈y_n-u_n,y_n-x^{\ast }〉+〈u_n-x^{\ast },y_n-x^{\ast }〉\\ \le & 〈u_n-x^{\ast },y_n-x^{\ast }〉\\ =& {\alpha }_n{\gamma }_n〈x_n-x^{\ast },y_n-x^{\ast }〉-{\alpha }_n(1-{\gamma }_n)〈x^{\ast },y_n-x^{\ast }〉\\ +(1-{\alpha }_n)〈z_n-x^{\ast },y_n-x^{\ast }〉\\ \le & {\alpha }_n{\gamma }_n\parallel x_n-x^{\ast }\parallel \parallel y_n-x^{\ast }\parallel -{\alpha }_n(1-{\gamma }_n)〈x^{\ast },y_n-x^{\ast }〉\\ +(1-{\alpha }_n)\parallel z_n-x^{\ast }\parallel \parallel y_n-x^{\ast }\parallel \\ \le & [1-(1-{\gamma }_n){\alpha }_n]\parallel x_n-x^{\ast }\parallel \parallel y_n-x^{\ast }\parallel -{\alpha }_n(1-{\gamma }_n)〈x^{\ast },y_n-x^{\ast }〉\\ \le & \frac{1-(1-{\gamma }_n){\alpha }_n}{2}{\parallel x_n-x^{\ast }\parallel }^2+\frac{1}{2}\parallel y_n-x^{\ast }\parallel -{\alpha }_n(1-{\gamma }_n)〈x^{\ast },y_n-x^{\ast }〉,\end{array}$$

that is,

$${\parallel y_n-x^{\ast }\parallel }^2\le [1-(1-{\gamma }_n){\alpha }_n]{\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n(1-{\gamma }_n)〈x^{\ast },y_n-x^{\ast }〉.$$

By the convexity of the norm, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & (1-{\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n{\parallel y_n-x^{\ast }\parallel }^2\\ \le & [1-(1-{\gamma }_n){\alpha }_n{\beta }_n]{\parallel x_n-x^{\ast }\parallel }^2-2(1-{\gamma }_n){\alpha }_n{\beta }_n〈x^{\ast },y_n-x^{\ast }〉.\end{array}$$

Hence all the conditions of Lemma 2.4 are satisfied. Therefore, we immediately deduce that $x_n\to x^{\ast }$. This completes the proof. □

In Theorem 3.3, if we put ${\beta }_n=1$ for each $n\ge 1$, we have the following corollary.

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\{T(s)\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix(S)\ne \mathrm{\varnothing }$. Let the sequence $\{x_n\}$ be generated iteratively by the following explicit algorithm:

$$x_{n+1}=P_C[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds],\mathrm{\forall }n\ge 0,$$

(3.10)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are sequences of real numbers in $[0,1]$ and $\{{\lambda }_n\}$ is a sequence of positive real numbers. Suppose that the following conditions are satisfied:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n-1}}{{\lambda }_n}=1$.

Then the sequence $\{x_n\}$ generated by (3.10) strongly converges to a point $x^{\ast }\in Fix(S)$.

Remark 3.5

It is known that the algorithm

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{P}_C[{\alpha }_n{x}_n+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds],\mathrm{\forall }n\ge 0,$$

has only weak convergence. However, our similar algorithm (3.6) (with ${\gamma }_n\to 1$) has strong convergence.