Viscosity Approximation Methods for Zeros of Accretive Operators and Fixed Point Problems in Banach Spaces

Abstract

In this paper, we introduce iterative algorithms for finding a zero of set-valued accretive operators by the viscosity approximation method based on Meir–Keeler-type contractions in a reflexive Banach space which admits a weakly continuous duality mapping. We obtain some strong convergence theorems under suitable conditions. As applications, we apply our results for finding common fixed point of nonexpansive semigroups and for solving equilibrium problem, optimization problem, and variational inequalities.

1 Introduction

Throughout this paper, we denote by $X$ and $X^*$ a real Banach space and the dual space of $X$, respectively. The duality mapping $J : X\longrightarrow 2^{X^*}$ is defined by

$$\begin{aligned} J(x)=\{f\in X^*:\langle x,f\rangle =\Vert x\Vert ^2,\Vert f\Vert =\Vert x\Vert \}, \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ denotes the duality pairing between $X$ and $X^*$. If $X:=H$ is a real Hilbert space, then $J=I$ where $I$ is the identity mapping. It is well known that if $X$ is smooth, then $J$ is single-valued, which is denoted by $j$ (see [30]). Let $C$ be a nonempty, closed and convex subset of $X$ and $T$ be a self-mapping on $C$. We denote the fixed points set of the mapping $S$ by $Fix(T)=\{x\in C: Tx=x\}$. A mapping $T:C\longrightarrow C$ is said to be $L$ -Lipschitzian if there exists a constant $L > 0$ such that

$$\begin{aligned} \Vert Tx-Ty\Vert \le L\Vert x-y\Vert ,~~\forall x,y\in C. \end{aligned}$$

If $0 < L < 1$, then $T$ is a contraction and if $L = 1$, then $T$ is a nonexpansive mapping.

Let $A:X\longrightarrow 2^{X}$ be a set-valued mapping. We denote $D(A)$ by domain of $A$, that is, $D(A)=\{x\in X:Ax\ne \emptyset \}$. A set-valued mapping $A:D(A)\subset X\longrightarrow 2^X$ is said to be accretive if for all $x, y\in D(A)$ there exist $j(x-y)\in J(x-y)$ such that

$$\begin{aligned} \langle u-v,j(x-y)\rangle \ge 0~~\text {for}~~u\in Ax~~\text {and}~~v\in Ay. \end{aligned}$$

In a Hilbert space, an accretive operator is also called monotone. Let $A:D(A)\subset X\longrightarrow 2^X$ be an accretive mapping, we can define a single-valued mapping $J_{r}^{A}:X\longrightarrow D(A)$ by $J_{r}^{A}=(I+r A)^{-1}$, which is called the resolvent operator associated with $A$, where $r>0$ and also denote $A^{-1}0$ by the set of zeros of $A$, that is, $A^{-1}0=\{x\in D(A):0\in Ax\}$. It is well known that $J_{r}^{A}$ is nonexpansive and $Fix(J_{r}^{A})=\{x\in X:J_{r}^{A}x=x\}$ (see [30]). An operator $A$ is called $m$ -accretive if it is accretive and $R(I + rA)$, range of $I + rA$, is $X$ for all $r > 0$; and $A$ is said to satisfy the range condition if $\overline{D(A)}=R(I + rA),~~\forall r> 0$, where $\overline{D(A)}$ denotes the closure of the domain of $A$.

Interest in accretive mappings stems mainly from their firm connection with equations of evolution. It is known (see, e.g., [1]) that many physically significant problems can be modeled by initial-value problems of the form

$$\begin{aligned} x'(t)+Ax(t)=0,~~x(0)=x_0, \end{aligned}$$

(1.1)

where $A$ is an accretive operator in an appropriate Banach space. Typical examples where such evolution equations occur can be found in the heat, wave, or Schrödinger equations. One of the fundamental results in the theory of accretive operators, due to Browder [2], states that if $A$ is locally Lipschitzian and accretive then $A$ is $m$-accretive. This result was subsequently generalized by Martin [3] to the continuous accretive operators. If in (1.1) $x(t)$ is independent of $t$, then (1.1) reduces to $Au = 0$ whose solutions correspond to the equilibrium points of system (1.1). Consequently, considerable research effects have been devoted, especially within the past 20 years or so, to iterative methods for approximating these equilibrium points.

In recent years, many authors have constructed the several iterative methods which is related fixed points problems in several settings, see, e.g., ([4–13, 16–21]).

In 1974, Bruck [22] introduced an iteration process and proved, in Hilbert space setting, the convergence of the process to a zero of a maximal monotone operator. In [23], Reich extended this result to uniformly smooth Banach spaces provided that the operator is $m$-accretive. By the inspiration of the regularization method for Rockafellars proximal point algorithm [40] and the iterative methods of Halpern [24], in 2003, Benavides et al. [25] studied the Halpern type iteration process (1.2) to find a zero of an $m$-accretive operator $A$ in a uniformly smooth Banach space with a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$ in virtue of the resolvent $J_r=(I+rA)^{-1}$ of $A$ for all $r>0$:

$$\begin{aligned} x_{n+1}=\alpha _nu+(1-\alpha _n)J_{r_n}x_n,~~\forall n\ge 1. \end{aligned}$$

(1.2)

On the other hand, Takahashi [14] introduced the following proximal point algorithm in a reflexive Banach space with a uniformly Gâteaux differentiable norm by the viscosity approximation method:

$$\begin{aligned} x_{n+1}=\alpha _nf(x_n)+(1-\alpha _n)J_{r_n}x_n,~~\forall n\ge 1, \end{aligned}$$

(1.3)

where $f:C\longrightarrow C$ is a contraction and $J_{r}=(I+rA)^{-1}$ is a resolvent of $A$ for all $r>0$. Under some mild conditions on the parameters $\{\alpha _n\}$ and $\{r_n\}$, he proved that the sequence $\{x_n\}$ defined by (1.3) converges strongly to a point in $A^{-1}0$.

Later, Petruşel and Yao [15] also studied strong convergence theorem of proximal point algorithm (1.3) by the viscosity approximation method with a generalized contraction mapping $f$.

Recently, Song et al. [17] studied the following strong convergence of the proximal point algorithm in a reflexive Banach space which admits a weakly sequentially continuous duality mapping:

$$\begin{aligned} x_{n+1}=J_{r_n}(\alpha _nu+(1-\alpha _n)x_n),~~\forall n\ge 1, \end{aligned}$$

(1.4)

where $\{\alpha _n\}\subset (0,1), \{r_n\}\subset (0,\infty )$ and $J_{r}=(I+rA)^{-1}$ is a resolvent of $A$ for all $r>0$.

In this paper, motivated by Petruşel and Yao [15] and Song et al. [17], we introduce iterative algorithms for finding a zero of set-valued accretive operators by the viscosity approximation method based on Meir–Keeler-type contractions in a reflexive Banach space which admits a weakly continuous duality mapping. We obtain some strong convergence theorems under suitable conditions. As applications, we apply our results for finding common fixed point of nonexpansive semigroups and for solving equilibrium problem, optimization problem, and variational inequalities.

2 Preliminaries

A Banach space $X$ is said to be strictly convex if $\frac{\Vert x+y\Vert }{2}<1$ for all $x, y\in X$ with $\Vert x\Vert =\Vert y\Vert =1$ and $x\ne y$. A Banach space $X$ is said to be uniformly convex if for each $\epsilon >0$, there exists $\delta >0$ such that for $x,y\in X$ with $\Vert x\Vert , \Vert y\Vert \le 1$ and $\Vert x-y\Vert \ge \epsilon , \frac{\Vert x+y\Vert }{2}\le 1-\delta$ holds. It is well known that a uniformly convex Banach space is reflexive and strictly convex (see [30]). Let $S(X)=\{x\in X:\Vert x\Vert =1\}$ denote the unit sphere of a Banach space $X$. The norm of $X$ is said to be Gâteaux differentiable (or $X$ is said to be smooth) if the limit

$$\begin{aligned} \lim _{t\longrightarrow 0}\frac{\Vert x+ty\Vert -\Vert x\Vert }{t} \end{aligned}$$

(2.1)

exists for each $x,y\in S(X)$. The norm of $X$ is said to be uniformly Gâteaux differentiable, if for each $y\in S(X)$, the limit (2.1) is attained uniformly for $x\in S(X)$.

Let $\varphi :[0,\infty ):=\mathbb {R}^{+}\longrightarrow \mathbb {R}^{+}$ be a continuous strictly increasing function such that $\varphi (0)=0$ and $\varphi (t)\longrightarrow \infty$ as $t\longrightarrow \infty$. This function $\varphi$ is called a gauge function. The duality mapping $J_\varphi :X\longrightarrow 2^{X^*}$ associated with a gauge function $\varphi$ is defined by

$$\begin{aligned} J_\varphi (x)=\{f^*\in X^*:\langle x,f^*\rangle =\Vert x\Vert \varphi (\Vert x\Vert ),~\Vert f^*\Vert =\varphi (\Vert x\Vert ),~~\forall x\in X\}, \end{aligned}$$

where $\langle \cdot ,\cdot \rangle$ denotes the generalized duality paring. In particular, the duality mapping with the gauge function $\varphi (t)=t$, denoted by $J$ is referred to as the normalized duality mapping. Clearly, there holds the relation $J_\varphi (x)=\frac{\varphi (\Vert x\Vert )}{\Vert x\Vert }J(x)$ for each $x\ne 0$ (see [31]).

Browder [31] initiated the study of certain classes of nonlinear operators by means of the duality mapping $J_\varphi$. Following Browder [31], we say Banach space $X$ has a weakly continuous duality mapping if there exits a gauge function $\varphi$ for which the duality mapping $J_\varphi (x)$ is single-valued and continuous from the weak topology to the $\text {weak}^{*}$ topology, that is, for each $\{x_n\}$ with $x_n\rightharpoonup x$, the sequence $\{J(x_n)\}$ converges $\text {weakly}^{*}$ to $J_\varphi (x)$.

A Banach space $X$ is said to satisfy Opials condition if, for any sequence $\{x_n\}$ in $X, x_n\rightharpoonup x$ implies

$$\begin{aligned} \limsup _{n\longrightarrow \infty }\Vert x_n-x\Vert <\limsup _{n\longrightarrow \infty }\Vert x_n-y\Vert ,~~\forall y\in X~~\text {with}~~ x\ne y. \end{aligned}$$

By Theorem 3.2.8 of [29], we know that, if $X$ admits the weakly continuous duality mapping $J_{\varphi }$ with gauge function $\varphi$, then $X$ satisfy Opial’s condition.

Lemma 2.1

(Cioranescu [32]) Assume that a Banach space $X$ has a weakly continuous duality mapping $J_\varphi$ with gauge $\varphi$. For all $x, y \in X$, the following inequality holds

$$\begin{aligned} \Phi (\Vert x+y\Vert )\le \Phi (\Vert x\Vert )+\langle y,J_\varphi (x+y)\rangle . \end{aligned}$$

A mapping $\psi :\mathbb {R^+}\longrightarrow \mathbb {R^+}$ is said to be an L-function if $\psi (0)=0,~\psi (t)>0$ for each $t>0$ and for every $s>0$, there exists $u>s$ such that $\psi (t)\le s$ for each $t\in [s,u]$. As a consequence, every $L$-function $\psi$ satisfies $\psi (t)<t$, for each $t>0.$

Definition 2.2

Let $(X,d)$ be a metric space. A mapping $f:X\longrightarrow X$ is said to be

• (1) a $(\psi ,L)$ -contraction if $\psi : \mathbb {R^+}\longrightarrow \mathbb {R^+}$ is an $L$-function and $d(f(x),f(y))< \psi (d(x,y)),~\forall x,y\in X$, with $x\ne y$,

• (2) a Meir–Keeler-type mapping if for each $\epsilon >0$ there exists $\delta =\delta (\epsilon )>0$ such that for each $x,y\in X$, with $\epsilon \le d(x,y)<\epsilon +\delta$, we have $d(f(x),f(y))<\epsilon$.

Lemma 2.3

(Lim [27]) Let $(X,d)$ be a metric space and $f:X \longrightarrow X$ be a mapping. The following assertions are equivalent:

• (i) $f$ is a Meir–Keeler-type mapping;

• (ii) there exists an $L$-function $\psi :\mathbb {R}^{+}\longrightarrow \mathbb {R}^{+}$ such that $f$ is a $(\psi ,L)$-contraction.

Lemma 2.4

(Petrusel and Yao [28]) Let $C$ be a convex subset of a Banach space $X$. Let $T : C \longrightarrow C$ be a nonexpansive mapping and $f$ be a $(\psi ,L)$-contraction. Then the following assertions hold

• (i) $T\circ f$ is a $(\psi ,L)$-contraction on $C$ and has a unique fixed point in $C$;

• (ii) for each $\alpha \in (0,1)$, the mapping $x \longrightarrow \alpha f(x)+(1-\alpha )Tx$ is Meir–Keeler-type and it has a unique fixed point in $C$.

Lemma 2.5

(Suzuki [26]) Let $C$ be a convex subset of a Banach space $X$. Let $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. Then for each $\epsilon >0$ there exists $k\in (0,1)$ such that

$$\begin{aligned} \text {for}~\text {each}~x,y\in C~\text {with}~\Vert x-y\Vert \ge \epsilon ~\text {we}~\text {have}~\Vert f(x)-f(y)\Vert \le k\Vert x-y\Vert . \end{aligned}$$

From now on, by a generalized contraction mapping, we mean a Meir–Keeler-type mapping or a $(\psi ,L)$-contraction. In the rest of paper, we suppose that the $L$-function from the definition of $(\psi ,L)$-contraction is continuous, strictly increasing and $\lim _{t \longrightarrow \infty } \eta (t) = \infty ,$ where $\eta (t):= t - \psi (t)~,\forall t \in \mathbb {R^+}.$ As a consequence, we have that $\eta$ is a bijection on $\mathbb {R^+}.$

Lemma 2.6

(Xu [33]) Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$\begin{aligned} a_{n+1}\le (1-\sigma _n)a_n+\delta _n, \end{aligned}$$

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

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

• (ii) $\limsup _{n\longrightarrow \infty }\frac{\delta _n}{\sigma _n}\le 0$ or $\sum _{n=0}^{\infty }|\delta _n|<\infty$.

Then, $\lim _{n\longrightarrow \infty }a_n=0$.

3 Main Results

Theorem 3.1

Let $C$ be a nonempty, closed and convex subset of a reflexive Banach space $X$ which admits a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$. Let $T:C\longrightarrow C$ be a nonexpansive mapping such that $Fix(T)\ne \emptyset$ and $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. Then the net $\{x_t\}$ defined by

$$\begin{aligned} x_t=tf(x_t)+(1-t)Tx_t. \end{aligned}$$

(3.1)

converges strongly to an element $x^*\in Fix(T)$, where $x^*$ is the unique solution of the variational inequality

$$\begin{aligned} \langle f(x^*)-x^*,J_{\varphi }(z-x^*)\rangle \le 0,~~\forall z\in Fix(T). \end{aligned}$$

(3.2)

Proof

Note that from Lemma 2.4(i), we have $\{x_t\}$ defined by (3.1) is well defined. First, we show the uniqueness of a solution of the variational inequality (3.2). Suppose that $\tilde{x},x^*\in Fix(T)$ are solutions of (3.2). Since $f$ is a Meir–Keeler-type contraction, then for each $\epsilon >0$ such that $\Vert \tilde{x}-x^*\Vert \ge \epsilon$. By Lemma 2.5, there exists $k_{\epsilon }\in (0,1)$ such that $\Vert f(\tilde{x})-f(x^*)\Vert \le k_{\epsilon }\Vert \tilde{x}-x^*\Vert$. Then we have

$$\begin{aligned} \langle f(x^*)-x^*,J_\varphi (\tilde{x}-x^*)\rangle \le 0. \end{aligned}$$

(3.3)

Interchange $x^*$ and $\tilde{x}$ to obtain

$$\begin{aligned} \langle f(\tilde{x})-\tilde{x},J_\varphi (x^*-\tilde{x})\rangle \le 0. \end{aligned}$$

(3.4)

Adding up (3.3) and (3.4), we have

$$\begin{aligned} 0\ge & {} \langle ( f(x^*)-x^*)-(f(\tilde{x})-\tilde{x}),J_\varphi (\tilde{x}-x^*)\rangle \\= & {} \langle \tilde{x}-x^*,J_\varphi (\tilde{x}-x^*)\rangle -\langle f(\tilde{x})-f(x^*),J_\varphi (\tilde{x}-x^*)\rangle \\\ge & {} \Vert \tilde{x}-x^*\Vert \varphi (\Vert \tilde{x}-x^*\Vert )-\Vert f(\tilde{x})-f(x^*)\Vert \Vert J_\varphi (\tilde{x}-x^*)\Vert \\\ge & {} \Vert \tilde{x}-x^*\Vert \varphi (\Vert \tilde{x}-x^*\Vert )-k\Vert \tilde{x}-x^*\Vert \varphi (\Vert \tilde{x}-x^*\Vert )\\= & {} (1-k)\Phi (\Vert \tilde{x}-x^*\Vert ), \end{aligned}$$

which is a contradiction, we must have $\tilde{x}=x^*$ and the uniqueness is proved. Below, we use $x^*$ to denote the unique solution of the variational inequality (3.2).

Next, we show that $\{x_t\}$ is bounded. Take $p\in Fix(T)$, fixed $\epsilon _0$, for each $t\in (0,1)$.

Case 1 :

$\Vert x_t-p\Vert <\epsilon _0$. In this case, it easy see that $\{x_t\}$ is bounded.

Case 2 :

$\Vert x_t-p\Vert \ge \epsilon _0$. By Lemma 2.5, there exists $k_{\epsilon _0}\in (0,1)$ such that $\Vert f(x_t)-f(p)\Vert \le k_{\epsilon _0}\Vert x_t-p\Vert$. Then, we have

$$\begin{aligned} \Vert x_t-p\Vert= & {} \Vert t(f(x_t)-p)+(1-t)(Tx_t-p)\Vert \\\le & {} t\Vert f(x_t)-p\Vert +(1-t)\Vert Tx_t-p\Vert \\\le & {} t\Vert f(x_t)-f(p)\Vert +t\Vert f(p)-p\Vert +(1-t)\Vert Tx_t-p\Vert \\\le & {} \big (1-(1-k_{\epsilon _0})t\big )\Vert x_t-p\Vert +t\Vert f(p)-p\Vert . \end{aligned}$$

It follows that

$$\begin{aligned} \Vert x_t-p\Vert \le \frac{1}{1-k_{\epsilon _0}}\Vert f(p)-p\Vert . \end{aligned}$$

Thus, $\{x_t\}$ is bounded, so are $\{f(x_t)\}$ and $\{Tx_t\}$.

By definition of $\{x_t\}$, we have

$$\begin{aligned} \Vert x_t-Tx_t\Vert= & {} \Vert t f(x_t)+(1-t)Tx_t-Tx_t\Vert \nonumber \\= & {} t\Vert f(x_t)-Tx_t\Vert \longrightarrow 0~~\text {as}~~t\longrightarrow 0. \end{aligned}$$

(3.5)

Assume that $\{t_n\}\subset (0,1)$ is a sequence such that $t_n\longrightarrow 0$ as $n\longrightarrow \infty$. Put $x_n:=x_{t_n}$. By reflexivity of a Banach space $X$ and boundedness of $\{x_n\}$, there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup \tilde{x}\in C$ as $i\longrightarrow \infty$. Let us to show $\tilde{x}\in Fix(T)$. Suppose that $\tilde{x}\not \in Fix(T)$, i.e., $\tilde{x}\ne T\tilde{x}$. By the Opial’s condition and (3.5), we have

$$\begin{aligned} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-\tilde{x}\Vert< & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-T\tilde{x}\Vert \\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\big (\Vert x_{n_i}-Tx_{n_i}\Vert +\Vert Tx_{n_i}-T\tilde{x}\Vert \big )\\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-\tilde{x}\Vert , \end{aligned}$$

which is a contradiction. Thus, we obtain $\tilde{x}\in Fix(T)$.

Next, we show that $\{x_{n}\}$ is relatively sequentially compact. For each $z\in Fix(T)$, suppose the contrary, there exists $\epsilon >0$ and a subsequence $\{x_{m_i}\}$ of $\{x_{n_i}\}$ such that $\Vert x_{m_i}-z\Vert \ge \epsilon$. By Lemma 2.5, there exists $k_{\epsilon }\in (0,1)$ such that $\Vert f(x_{m_i})-f(z)\Vert \le k_{\epsilon }\Vert x_{m_i}-z\Vert$. Then we have

$$\begin{aligned} \Phi (\Vert x_{m_i}-z\Vert )= & {} \Phi (\Vert (1-t_{m_i})(Tx_{m_i}-z)+t_{m_i}(f(x_{m_i})-z)\Vert )\\= & {} \Phi (\Vert (1-t_{m_i})(Tx_{m_i}-z)+t_{m_i}(f(x_{m_i})-f(z))+t_{m_i}(f(z)-z\Vert )\\\le & {} \Phi (\Vert (1-t_{m_i})(Tx_{m_i}-z)+t_{m_i}(f(x_{m_i})-f(z))\Vert )+t_{m_i}\langle f(z)\\&-z,J_{\varphi }(x_{m_i}-z)\rangle \\\le & {} \big (1-(1-k_{\epsilon })t_{m_i}\big )\Phi (\Vert x_{m_i}-z\Vert )+t_{m_i}\langle f(z)-z,J_{\varphi }(x_{m_i}-z)\rangle . \end{aligned}$$

It follows that

$$\begin{aligned} \Phi (\Vert x_{m_i}-z\Vert )\le \frac{1}{1-k_{\epsilon }}\langle f(z)-z,J_{\varphi }(x_{m_i}-z)\rangle . \end{aligned}$$

(3.6)

Since $J_\varphi$ is single-valued and weakly continuous duality mapping, it follows that from (3.6) that $\Phi (\Vert x_{m_i}-\tilde{x}\Vert )\longrightarrow 0$. By the properties $\Phi$ implies that $x_{m_i}\longrightarrow \tilde{x}$, that is $\Vert x_{m_i}-\tilde{x}\Vert <\epsilon$, which is a contradiction. Thus, we obtain $x_{n_i}\longrightarrow \tilde{x}$ as $i\longrightarrow \infty$.

Next, we show that $\tilde{x}$ solves the variational inequality (3.2). Since

$$\begin{aligned} x_t=tf(x_t)+(1-t)Tx_t, \end{aligned}$$

we can derive that

$$\begin{aligned} (I-f)x_t=-\frac{1}{t}(I-T)x_t+(I-T)x_t. \end{aligned}$$

(3.7)

Since $T$ is nonexpansive, we have that $I - T$ is accretive. Note that for all $z\in Fix(T)$, it follows from (3.7) that

$$\begin{aligned} \langle (I-f)x_t,J_{\varphi }(x_t-z)\rangle= & {} -\frac{1}{t}\langle (I-T)x_t-(I-T)z,J_{\varphi }(x_t-z)\rangle \nonumber \\&+\,\langle (I-T)x_t,J_{\varphi }(x_t-z)\rangle \nonumber \\\le & {} \langle (I-T)x_t,J_{\varphi }(x_t-z)\rangle \nonumber \\\le & {} \Vert x_t-Tx_t\Vert M, \end{aligned}$$

(3.8)

where $M>0$ is a constant such that $M=\sup _{t\in (0,1)}\{\Vert J_{\varphi }(x_t-z)\Vert \}$. Now, replacing $t$ in (3.8) with $t_n$ and taking the limit as $n\longrightarrow \infty$, we noticing that $x_{t_n}-Tx_{t_n}\longrightarrow \tilde{x}-T\tilde{x}=0$ for $\tilde{x}\in Fix(T)$, we obtain $\langle (f-I)\tilde{x},J_{\varphi }(z-\tilde{x})\rangle \le 0$. Hence $\tilde{x}\in Fix(T)$ is the solution of the variational inequality (3.2). Consequently, $x^*=\tilde{x}$ by uniqueness. Therefore $x_t\longrightarrow x^*$ as $t\longrightarrow 0$. This completes the proof. $\square$

Using Theorem 3.1, we get the following strong convergence theorems for proximal point algorithm by the viscosity approximation method based on Meir–Keeler-type contractions for zeros of accretive operators in Banach spaces. The proofs are similarly related to Theorems 3.2 and 3.3 in [17]; see also, [14, 21] and [15].

Theorem 3.2

Let $C$ be a nonempty, closed and convex subset of a reflexive Banach space $X$ which admits a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$. Let $A:D(A)\subset X\longrightarrow 2^{X}$ be an accretive operator such that $A^{-1}0\ne \emptyset$ which satisfies the condition $\overline{D(A)}\subset C\subset \bigcap _{r>0}R(I+rA)$. Let $J_r$ be the resolvent of $A$ for all $r>0$ and $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} x_{n+1}=\alpha _n f(x_n)+(1-\alpha _n)J_{r_n}x_n,~~\forall n\ge 1, \end{aligned}$$

(3.9)

where $\{\alpha _n\}\subset (0,1)$ and $\{r_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim _{n\longrightarrow \infty }r_n=\infty$.

Then the sequence $\{x_n\}$ defined by (3.9) converges strongly to an element $x^*\in A^{-1}0$, where $x^*$ is the unique solution of the variational inequality

$$\begin{aligned} \langle f(x^*)-x^*,J_{\varphi }(z-x^*)\rangle \le 0,~~\forall z\in A^{-1}0. \end{aligned}$$

(3.10)

Proof

First, we show that $\{x_n\}$ is bounded. Take $p\in A^{-1}0$, fixed $\epsilon _0$, for all $n\ge 1$.

Case 1 :

$\Vert x_n-p\Vert <\epsilon _0$. In this case, it easy see that $\{x_n\}$ is bounded.

Case 2 :

$\Vert x_n-p\Vert \ge \epsilon _0$. By Lemma 2.5, there exists $k_{\epsilon _0}\in (0,1)$ such that $\Vert f(x_n)-f(p)\Vert \le k_{\epsilon _0}\Vert x_n-p\Vert$.

Then, we have

$$\begin{aligned} \Vert x_{n+1}-p\Vert= & {} \Vert \alpha _n(f(x_n)-p)+(1-\alpha _n)(J_{r_n}x_n-p)\Vert \\\le & {} \alpha _n\Vert f(x_n)-p\Vert +(1-\alpha _n)\Vert J_{r_n}x_n-p)\Vert \\\le & {} \alpha _n\Vert f(x_n)-f(p)\Vert +\alpha _n\Vert f(p)-p\Vert +(1-\alpha _n)\Vert x_n-p\Vert \\\le & {} \big (1-(1-k_{\epsilon _0})\alpha _n\big )\Vert x_n-p\Vert +\alpha _n\Vert f(p)-p\Vert \\= & {} \big (1-(1-k_{\epsilon _0})\alpha _n\big )\Vert x_n-p\Vert +(1-k_{\epsilon _0})\alpha _n\frac{\Vert f(p)-p\Vert }{1-k_{\epsilon _0}}\\\le & {} \max \bigg \{\Vert x_n-p\Vert ,\frac{\Vert f(p)-p\Vert }{1-k_{\epsilon _0}}\bigg \}. \end{aligned}$$

By induction, we have

$$\begin{aligned} \Vert x_n-p\Vert \le \max \bigg \{\Vert x_1-p\Vert ,\frac{\Vert f(p)-p\Vert }{1-k_{\epsilon _0}}\bigg \},~~\forall n\ge 1. \end{aligned}$$

Thus $\{x_n\}$ is bounded, so are $\{f(x_n)\}$ and $\{J_{r_n}x_n\}$.

Next, we show that $\lim _{n\longrightarrow \infty }\Vert x_n-J_{r}x_n\Vert =0$. By the condition $(C1)$, we have

$$\begin{aligned} \Vert x_{n+1}-J_{r_n}x_n\Vert =\alpha _n\Vert f(x_n)-J_{r_n}x_n\Vert \longrightarrow 0~~\text {as}~~n\longrightarrow \infty . \end{aligned}$$

(3.11)

For $r>0$, we note that

$$\begin{aligned} \Vert J_{r}J_{r_n}x_n-J_{r_n}x_n\Vert= & {} \Vert (I-J_r)J_{r_n}x_n\Vert =r\Vert A_rJ_{r_n}x_n\Vert \le r|AJ_{r_n}x_n|\le r\Vert A_{r_n}x_n\Vert \\= & {} r\frac{\Vert x_n-J_{r_n}x_n\Vert }{r_n}. \end{aligned}$$

It follows from the condition $(C2)$ that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert J_{r}J_{r_n}x_n-J_{r_n}x_n\Vert =0. \end{aligned}$$

(3.12)

We observe that

$$\begin{aligned} \Vert x_{n+1}-J_rx_{n+1}\Vert\le & {} \Vert x_{n+1}-J_{r_n}x_n\Vert +\Vert J_{r_n}x_n-J_rJ_{r_n}x_n\Vert +\Vert J_rJ_{r_n}x_n-J_rx_{n+1}\Vert \\\le & {} 2\Vert x_{n+1}-J_{r_n}x_n\Vert +\Vert J_{r_n}x_n-J_rJ_{r_n}x_n\Vert . \end{aligned}$$

It follows from (3.11) and (3.12) that $\lim _{n\longrightarrow \infty }\Vert x_{n+1}-J_rx_{n+1}\Vert =0$, and hence

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert x_n-J_{r}x_n\Vert =0,~~\forall r>0. \end{aligned}$$

(3.13)

Next, we show that

$$\begin{aligned} \limsup \nolimits _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(x_n-x^*)\rangle \le 0, \end{aligned}$$

where $x^*$ is the same as in Theorem 3.1. To show this, we take a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\begin{aligned} \limsup \nolimits _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(x_n-x^*)\rangle =\lim \nolimits _{i\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(x_{n_i}-x^*)\rangle . \end{aligned}$$

By reflexivity of a Banach space $X$ and boundedness of $\{x_n\}$, there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup z$ as $i\longrightarrow \infty$. Let us to show $z\in A^{-1}0$. Suppose that $z\not \in A^{-1}0$, i.e., $z\ne J_rz$. By the Opial’s condition and (3.13), we have

$$\begin{aligned} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-z\Vert< & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-J_rz\Vert \\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\big (\Vert x_{n_i}-J_rx_{n_i}\Vert +\Vert J_rx_{n_i}-J_rz\Vert \big )\\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert x_{n_i}-z\Vert , \end{aligned}$$

which is a contradiction. Thus, we obtain $z\in A^{-1}0$. Since $J_\varphi$ is single-valued and weakly continuous duality mapping, we obtain that

$$\begin{aligned} \limsup _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(x_n-x^*)\rangle= & {} \lim _{i\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(x_{n_i}-x^*)\rangle \nonumber \\= & {} \langle f(x^*)-x^*,J_{\varphi }(z-x^*)\rangle \le 0. \end{aligned}$$

(3.14)

Finally, we show that $x_n\longrightarrow x^*$ as $n\longrightarrow \infty$. Suppose the contrary, $\{x_n\}$ does not converge strongly to $x^*\in A^{-1}0$. Then there exist $\epsilon >0$ and a subsequence $\{x_{n_j}\}$ of $\{x_{n}\}$ such that $\Vert x_{n_j}-x^*\Vert \ge \epsilon$ for all $j\in \mathbb {N}$. By Lemma 2.5, there exists $k_{\epsilon }\in (0,1)$ such that

$$\begin{aligned} \Vert f(x_{n_j})-f(x^*)\Vert \le k_{\epsilon }\Vert x_{n_j}-x^*\Vert ~~\quad \text {for}~\text {all}~~j\in \mathbb {N}. \end{aligned}$$

By Lemma 2.1, we have

$$\begin{aligned} \Phi (\Vert x_{{n_j}+1}-x^*\Vert )= & {} \Phi (\Vert (1-\alpha _{n_j})(J_{r_{n_j}}x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-x^*)\Vert )\nonumber \\= & {} \Phi (\Vert (1-\alpha _{n_j})(J_{r_{n_j}}x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-f(x^*))\nonumber \\&+\, \alpha _{n_j}(f(x^*)-x^*)\Vert )\nonumber \\\le & {} \Phi (\Vert (1-\alpha _{n_j})(J_{r_{n_j}}x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-f(x^*))\Vert )\nonumber \\&+\, \alpha _{n_j}\langle f(x^*)-x^*,J_{\varphi }(x_{{n_j}+1}-x^*)\rangle \nonumber \\\le & {} \big (1-(1-k_{\epsilon })\alpha _{n_j}\big )\Phi (\Vert x_{n_j}-x^*\Vert )+\alpha _{n_j}\langle f(x^*)\nonumber \\&-\,x^*,J_{\varphi }(x_{{n_j}+1}-x^*)\rangle . \end{aligned}$$

(3.15)

Put $\sigma _{n_j}:=(1-k_{\epsilon })\alpha _{n_j}$ and $\delta _{n_j}:=\frac{1}{1-k_\epsilon }\langle f(x^*)-x^*,J_{\varphi }(x_{{n_j}+1}-x^*)\rangle$. Then (3.15) reduces to formula

$$\begin{aligned} \Phi (\Vert x_{{n_j}+1}-x^*\Vert )\le (1-\sigma _{n_j})\Phi (\Vert x_{n_j}-x^*\Vert )+\sigma _{n_j}\delta _{n_j}. \end{aligned}$$

It is easily seen that $\sum _{j=1}^{\infty }\sigma _{n_j}=\infty$ and (using (3.15))

$$\begin{aligned} \limsup \nolimits _{j\longrightarrow \infty }\delta _{n_j}=\limsup \nolimits _{j\longrightarrow \infty }\frac{1}{1-k_{\epsilon }}\langle f(x^*)-x^*,J_{\varphi }(x_{{n_j}+1}-x^*)\rangle \le 0. \end{aligned}$$

Then by Lemma 2.6, we have $\Phi (\Vert x_{n_j}-x^*\Vert )\longrightarrow 0$. This implies that $x_{n_j}\longrightarrow x^*$, that is $\Vert x_{n_j}-x^*\Vert <\epsilon _0$, which is contradiction. Therefore, we conclude that $x_{n}\longrightarrow x^*$. This completes the proof. $\square$

Theorem 3.3

Let $C$ be a nonempty, closed and convex subset of a reflexive Banach space $X$ which admits a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$. Let $A:D(A)\subset X\longrightarrow 2^{X}$ be an accretive operator such that $A^{-1}0\ne \emptyset$ which satisfies the condition $\overline{D(A)}\subset C\subset \bigcap _{r>0}R(I+rA)$. Let $J_r$ be the resolvent of $A$ for all $r>0$ and $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} x_{n+1}=J_{r_n}(\alpha _n f(x_n)+(1-\alpha _n)x_n),~~\forall n\ge 1, \end{aligned}$$

(3.16)

where $\{\alpha _n\}\subset (0,1)$ and $\{r_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim _{n\longrightarrow \infty }r_n=\infty$.

Then the sequence $\{x_n\}$ defined by (3.16) converges strongly to an element $x^*\in A^{-1}0$, where $x^*$ is the unique solution of the variational inequality (3.2).

Proof

By the similar method to the proof technique of Theorem 3.2, we show that the sequence $\{x_n\}$ is bounded firstly. Let $y_n=\alpha _n f(x_n)+(1-\alpha _n)x_n$. Take $p\in A^{-1}0$, fixed $\epsilon _0$, for all $n\ge 1$.

Case 1 :

$\Vert x_n-p\Vert <\epsilon _0$. In this case, it easy see that $\{x_n\}$ is bounded.

Case 2 :

$\Vert x_n-p\Vert \ge \epsilon _0$. By Lemma 2.5, there exists $k_{\epsilon _0}\in (0,1)$ such that $\Vert f(x_n)-f(p)\Vert \le k_{\epsilon _0}\Vert x_n-p\Vert$. Then, we have

$$\begin{aligned} \Vert x_{n+1}-p\Vert= & {} \Vert J_{r_n}y_n-p\Vert \le \Vert y_n-p\Vert \\= & {} \Vert \alpha _n(f(x_n)-p)+(1-\alpha _n)(x_n-p)\Vert \\\le & {} \alpha _n\Vert f(x_n)-f(p)\Vert +\alpha _n\Vert f(p)-p\Vert +(1-\alpha _n)\Vert x_n-p\Vert \\\le & {} \big (1-(1-k_{\epsilon _0})\big )\Vert x_n-p\Vert +\alpha _n\Vert f(p)-p\Vert . \end{aligned}$$

By induction, we have

$$\begin{aligned} \Vert x_n-p\Vert \le \max \bigg \{\Vert x_1-p\Vert ,\frac{\Vert f(p)-p\Vert }{1-k_{\epsilon _0}}\bigg \},~~\forall n\ge 1. \end{aligned}$$

Thus $\{x_n\}$ is bounded, so are $\{f(x_n)\}$ and $\{J_{r_n}y_n\}$.

Next, we show that $\lim _{n\longrightarrow \infty }\Vert y_n-J_ry_n\Vert =0$. By the condition $(C1)$, we have

$$\begin{aligned} \Vert y_n-x_n\Vert =\alpha _n\Vert f(x_n)-x_n\Vert \longrightarrow 0~~\text {as}~~n\longrightarrow \infty . \end{aligned}$$

(3.17)

For $r>0$, we note that

$$\begin{aligned} \Vert x_{n+1}-J_rx_{n+1}\Vert= & {} \Vert J_{r_n}y_n-J_rJ_{r_n}y_n\Vert =\Vert (I-J_r)J_{r_n}y_n\Vert \\= & {} r\Vert A_rJ_{r_n}y_n\Vert \le r|AJ_{r_n}y_n|\le r\Vert A_{r_n}y_n\Vert = r\frac{\Vert y_n-J_{r_n}y_n\Vert }{r_n}. \end{aligned}$$

It follows from the condition $(C2)$ that $\lim _{n\longrightarrow \infty }\Vert x_{n+1}-J_rx_{n+1}\Vert =0$, and hence

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert x_n-J_rx_n\Vert =0,~~\forall r>0. \end{aligned}$$

(3.18)

Observe that

$$\begin{aligned} \Vert y_n-J_ry_n\Vert\le & {} \Vert y_n-x_n\Vert +\Vert x_n-J_rx_n\Vert +\Vert J_rx_n-J_ry_n\Vert \\\le & {} 2\Vert x_n-y_n\Vert +\Vert x_n-J_rx_n\Vert . \end{aligned}$$

It follows from (3.17) and (3.18) that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert y_n-J_ry_n\Vert =0. \end{aligned}$$

(3.19)

Next, we show that

$$\begin{aligned} \limsup \nolimits _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(y_n-x^*)\rangle \le 0, \end{aligned}$$

where $x^*$ is the same as in Theorem 3.1. To show this, we take a subsequence $\{y_{n_i}\}$ of $\{y_n\}$ such that

$$\begin{aligned} \limsup \nolimits _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(y_n-x^*)\rangle =\lim \nolimits _{i\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(y_{n_i}-x^*)\rangle . \end{aligned}$$

By reflexivity of a Banach space $X$ and boundedness of $\{y_n\}$, there exists a subsequence $\{y_{n_i}\}$ of $\{y_n\}$ such that $y_{n_i}\rightharpoonup z$ as $i\longrightarrow \infty$. Let us to show $z\in A^{-1}0$. Suppose that $z\not \in A^{-1}0$, i.e., $z\ne J_rz$. By the Opial’s condition and (3.19), we have

$$\begin{aligned} \liminf \nolimits _{n\longrightarrow \infty }\Vert y_{n_i}-z\Vert< & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert y_{n_i}-J_rz\Vert \\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\big (\Vert y_{n_i}-J_ry_{n_i}\Vert +\Vert J_ry_{n_i}-J_rz\Vert \big )\\\le & {} \liminf \nolimits _{n\longrightarrow \infty }\Vert y_{n_i}-z\Vert , \end{aligned}$$

which is a contradiction. Thus, we obtain $z\in A^{-1}0$. Since $J_\varphi$ is single-valued and weakly continuous duality mapping, we obtain that

$$\begin{aligned} \limsup _{n\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(y_n-x^*)\rangle= & {} \lim _{i\longrightarrow \infty }\langle f(x^*)-x^*,J_{\varphi }(y_{n_i}-x^*)\rangle \nonumber \\= & {} \langle f(x^*)-x^*,J_{\varphi }(z-x^*)\rangle \le 0. \end{aligned}$$

(3.20)

Finally, we show that $x_n\longrightarrow x^*$ as $n\longrightarrow \infty$. Suppose the contrary, $\{x_n\}$ does not converge strongly to $x^*\in A^{-1}0$. Then there exist $\epsilon >0$ and a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $\Vert x_{n_j}-x^*\Vert \ge \epsilon$ for all $j\in \mathbb {N}$. By Lemma 2.5, there exists $k_{\epsilon }\in (0,1)$ such that

$$\begin{aligned} \Vert f(x_{n_j})-f(x^*)\Vert \le k_{\epsilon }\Vert x_{n_j}-x^*\Vert ~~\quad \text {for}~\text {all}~~j\in \mathbb {N}. \end{aligned}$$

By Lemma 2.1, we have

$$\begin{aligned} \Phi (\Vert x_{{n_j}+1}-x^*\Vert )= & {} \Phi (\Vert J_{r_{n_j}}y_{n_j}-x^*\Vert )\le \Phi (\Vert y_{n_j}-x^*\Vert )\nonumber \\= & {} \Phi (\Vert (1-\alpha _{n_j})(x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-x^*)\Vert )\nonumber \\= & {} \Phi (\Vert (1-\alpha _{n_j})(x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-f(x^*))\nonumber \\&+\,\alpha _{n_j}(f(x^*)-x^*)\Vert )\nonumber \\\le & {} \Phi (\Vert (1-\alpha _{n_j})(x_{n_j}-x^*)+\alpha _{n_j}(f(x_{n_j})-f(x^*))\Vert )\nonumber \\&+\,\alpha _{n_j}\langle f(x^*)-x^*,J_{\varphi }(y_{n_j}-x^*)\rangle \nonumber \\\le & {} \big (1-(1-k_{\epsilon })\alpha _{n_j}\big )\Phi (\Vert x_{n_j}-x^*\Vert )+\alpha _{n_j}\langle f(x^*)-x^*,\nonumber \\&J_{\varphi }(y_{n_j}-x^*)\rangle . \end{aligned}$$

(3.21)

Put $\sigma _{n_j}:=(1-k_{\epsilon })\alpha _{n_j}$ and $\delta _{n_j}:=\frac{1}{1-k_\epsilon }\langle f(x^*)-x^*,J_{\varphi }(y_{n_j}-x^*)\rangle$. Then (3.21) reduces to formula

$$\begin{aligned} \Phi (\Vert x_{{n_j}+1}-x^*\Vert )\le (1-\sigma _{n_j})\Phi (\Vert x_{n_j}-x^*\Vert )+\sigma _{n_j}\delta _{n_j}. \end{aligned}$$

It is easily seen that $\sum _{j=1}^{\infty }\sigma _{n_j}=\infty$ and (using (3.20))

$$\begin{aligned} \limsup \nolimits _{j\longrightarrow \infty }\delta _{n_j}=\limsup \nolimits _{j\longrightarrow \infty }\frac{1}{1-k_{\epsilon }}\langle f(x^*)-x^*,J_{\varphi }(y_{n_j}-x^*)\rangle \le 0. \end{aligned}$$

Then by Lemma 2.6, we have $\Phi (\Vert x_{n_j}-x^*\Vert )\longrightarrow 0$. This implies that $x_{n_j}\longrightarrow x^*$, that is $\Vert x_{n_j}-x^*\Vert <\epsilon _0$, which is contradiction. Therefore, we conclude that $x_{n}\longrightarrow x^*$. This completes the proof. $\square$

4 Application to Nonexpansive Semigroup

Definition 4.1

Let $C$ be a nonempty, closed and convex subset of a real Banach space $X$. A one-parameter family $\mathcal {S} = \{T(t):t>0\}$ from $C$ into itself is said to be a nonexpansive semigroup on $C$ if it satisfies the following conditions:

• (i) $T(0)x=x ~~\text {for all} ~~x\in C$;

• (ii) $T(s+t)x=T(s) T(t)x$ for all $x\in C$ and $s,t>0$;

• (iii) for each $x\in C$ the mapping $t\mapsto T(t)x$ is continuous;

• (iv) $\Vert T(t)x-T(t)y\Vert \le \Vert x-y\Vert$ for all $x,y\in C$ and $t>0$.

Remark 4.2

We denote by $Fix(\mathcal {S})$ the set of all common fixed points of $\mathcal {S},$ that is $Fix(\mathcal {S}):=\bigcap _{t>0}Fix(T(t))=\{{x\in C:T(t)x=x}\}$. We know that $Fix(\mathcal {S})$ is nonempty if $C$ is bounded (see [44]).

Now, we present the concept of a uniformly asymptotically regular semigroup (see [34–36]).

Definition 4.3

Let $C$ be a nonempty, closed and convex subset of a real Banach space $X$ and $\mathcal {S}=\{T(t):t>0\}$ be a semigroup of nonexpansive operators. Then $\mathcal {S}$ is called uniformly asymptotically regular (in short, u.a.r.) on $C$ if for all $h\ge 0$ and for any bounded subset $B$ of $C$ such that

$$\begin{aligned} \lim _{t\longrightarrow \infty }\sup _{x\in B}\Vert T(h)T(t)x-T(t)x\Vert =0. \end{aligned}$$

The nonexpansive semigroup $\{\sigma _t:t>0\}$ defined by the following lemma is an example of u.a.r. operator semigroup. Other examples of u.a.r. operator semigroup can be found in [34]. The following lemma found in [37].

Lemma 4.4

(see [37]) Let $C$ be a nonempty, closed and convex subset of a smooth Banach space $X$ and let $\mathcal {S}=\{T(h):h>0\}$ be a u.a.r. nonexpansive semigroup on $C$ such that $Fix(\mathcal {S})=\bigcap _{h>0}Fix(T(h))\ne \emptyset$ and at least there exists a $T(h)$ which is demicompact. Then, for each $x\in C$, there exists a sequence $\{T(t_k):t_k>0,~k\in \mathbb {N}\}\subset \{T(h):h>0\}$ such that $\{T(t_k)x\}$ converges strongly to some point in $Fix(\mathcal {S})$, where $\lim _{k\longrightarrow \infty }t_k=\infty$.

Using Lemma 4.4 and Theorems 3.2 and 3.3, we have the following results.

Theorem 4.5

Let $C$ be a nonempty, closed and convex subset of a uniformly smooth Banach space $X$ which admits a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$. Let $\mathcal {S}=\{T(t):t>0\}$ be a u.a.r. nonexpansive semigroup from $C$ into itself such that $Fix(\mathcal {S}):=\bigcap _{h>0}Fix(T(h))\ne \emptyset$ and least there exists a $T(h)$ which is demicompact and $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} x_{n+1}=\alpha _n f(x_n)+(1-\alpha _n)T(t_n)x_n,~~\forall n\ge 1, \end{aligned}$$

(4.1)

where $\{\alpha _n\}\subset (0,1)$ and $\{t_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim _{n\longrightarrow \infty }t_n=\infty$.

Then the sequence $\{x_n\}$ defined by (4.1) converges strongly to an element $x^*\in Fix(\mathcal {S})$.

Proof

By using the same arguments and techniques as those of Theorem 3.2, we only need show that

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert x_n-Tx_n\Vert =0. \end{aligned}$$

By the condition $(C1)$, we have

$$\begin{aligned} \Vert x_{n+1}-T(t_n)x_n\Vert =\alpha _n\Vert f(x_n)-T(t_n)x_n\Vert \longrightarrow 0~~\text {as}~~n\longrightarrow \infty . \end{aligned}$$

(4.2)

Since $\{T(t):t>0\}$ is a u.a.r. nonexpansive semigroup and $\lim _{n\longrightarrow \infty }t_n=\infty$, then for all $h>0$, and for any bounded subset $B$ of $C$ containing $\{x_n\}$, we obtain that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert T(h)T(t_n)x_n-T(t_n)x_n\Vert \le \lim _{n\longrightarrow \infty }\sup _{\omega \in B}\Vert T(h)T(t_n)\omega -T(t_n)\omega \Vert =0. \end{aligned}$$

(4.3)

For all $h>0$, observe that

$$\begin{aligned} \Vert x_{n+1}-T(h)x_{n+1}\Vert\le & {} \Vert x_{n+1}-T(t_n)x_n\Vert +\Vert T(t_n)x_n-T(h)T(t_n)x_n\Vert \\&+\,\Vert T(h)T(t_n)-T(h)x_{n+1}\Vert \\\le & {} 2\Vert x_{n+1}-T(t_n)x_n\Vert +\Vert T(t_n)x_n-T(h)T(t_n)x_n\Vert . \end{aligned}$$

It follows from (4.2) and (4.3) that $\lim _{n\longrightarrow \infty }\Vert x_{n+1}-T(h)x_{n+1}\Vert =0$, and hence

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert x_n-T(h)x_n\Vert =0~~\text {for}~\text {all}~h\ge 0. \end{aligned}$$

(4.4)

Since $\{T(h):h>0\}$ is a u.a.r. nonexpansive semigroup, by Lemma 4.4, for each $x\in C$, there exists a sequence $\{T(t_k):t_k>0,k\in \mathbb {N}\}\subset \{T(h):h>0\}$ such that $\{T(t_k)x\}$ converges strongly to some point in $Fix(\mathcal {S})$, where $t_k\longrightarrow \infty$ as $k\longrightarrow \infty$. Define a mapping $T:C\longrightarrow C$ by

$$\begin{aligned} Tx=\lim \nolimits _{k\longrightarrow \infty }T(t_k)x,~~\forall x\in C. \end{aligned}$$

By [37], Remark 3.4], we see that the mapping $T$ is nonexpansive such that $Fix(T)=Fix(\mathcal {S})$. From (4.4), we obtain that

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert x_n-Tx_n\Vert= & {} \lim \nolimits _{n\longrightarrow \infty }\lim \nolimits _{k\longrightarrow \infty }\Vert x_n-T(t_k)x_n\Vert \\= & {} \lim \nolimits _{k\longrightarrow \infty }\lim \nolimits _{n\longrightarrow \infty }\Vert x_n-T(t_k)x_n\Vert =0. \end{aligned}$$

This completes the proof. $\square$

Theorem 4.6

Let $C$ be a nonempty, closed and convex subset of a uniformly smooth Banach space $X$ which admits a weakly continuous duality mapping $J_\varphi$ with gauge function $\varphi$. Let $\mathcal {S}=\{T(t):t>0\}$ be a u.a.r. nonexpansive semigroup from $C$ into itself such that $Fix(\mathcal {S}):=\bigcap _{h>0}Fix(T(h))\ne \emptyset$ and least there exists a $T(h)$ which is demicompact and $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} x_{n+1}=T(t_n)[\alpha _n f(x_n)+(1-\alpha _n)x_n],~~\forall n\ge 1, \end{aligned}$$

(4.5)

where $\{\alpha _n\}\subset (0,1)$ and $\{t_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim \nolimits _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim \nolimits _{n\longrightarrow \infty }t_n=\infty$.

Then the sequence $\{x_n\}$ defined by (4.5) converges strongly to an element $x^*\in Fix(\mathcal {S})$.

5 Application to Equilibrium Problem

Let $C$ be a nonempty, closed convex subset of a real Hilbert space $H$ and $\Theta :C\times C\longrightarrow \mathbb {R}$ be a bifunction, where $\mathbb {R}$ is the set of all real numbers. The equilibrium problem is to find $x\in C$ such that

$$\begin{aligned} \Theta (x,y)\ge 0,~~\forall y\in C. \end{aligned}$$

(5.1)

The set of solutions of the equilibrium problem (5.1) is denoted by $EP(\Theta )$. Given a mapping $T:C\longrightarrow H$, let $\Theta (x,y)=\langle Tx,y-x\rangle$ for all $x,y\in C$. Then $z\in EP(\Theta )$ if and only if $\langle Tx,y-x\rangle \ge 0$ for all $y\in C$, i.e., $z$ is a solution of the variational inequality.

Numerous problems in physics, optimization and economics reduce to find a solution of the equilibrium problem (5.1). Some methods have been proposed to solve the equilibrium problems (see, for instance, BlumOettli [38] and Combettes and Hirstoaga [39]).

For solving the equilibrium problem, let us assume that a bifunction $\Theta :C\times C\longrightarrow \mathbb {R}$ satisfies the following conditions:

• (A1) $\Theta (x,x)=0$ for all $x\in C$;

• (A2) $\Theta$ is monotone, i.e., $\Theta (x,y)+\Theta (y,x)\le 0$ for each $x,y\in C$;

• (A3) $\Theta$ is upper-semicontinuous, i.e., for each $x,y,z\in C$,

  $$\begin{aligned} \limsup \nolimits _{t\longrightarrow 0^{+}}\Theta (tz+(1-t)x,y)\le \Theta (x,y); \end{aligned}$$

• (A4) $\Theta (x,\cdot )$ is convex and weakly lower semicontinuous for each $x \in C$.

The following lemmas were also given in [38] and [39], respectively.

Lemma 5.1

(see [38], corollary 1]) Let $C$ be a nonempty, closed and convex subset of $H$ and let $\Theta :C\times C\longrightarrow \mathbb {R}$ satisfying the conditions (A1)–(A4). Let $r>0$ and $x\in H$. Then there exists $z\in C$ such that

$$\begin{aligned} \Theta (z,y)+\frac{1}{r}\langle y-z,z-x\rangle \ge 0,~~\forall y\in C. \end{aligned}$$

Lemma 5.2

(see [39], lemma 2.12]) Assume that $\Theta :C\times C\longrightarrow \mathbb {R}$ satisfies the conditions (A1)–(A4). For $r>0$ and $x\in H$, define a mapping $T_r:H\longrightarrow C$ as follows:

$$\begin{aligned} T_r(x)=\big \{z\in C:\Theta (z,y)+\frac{1}{r}\langle y-z,z-x\rangle \ge 0,~~\forall y\in C\big \},~~\forall x\in H. \end{aligned}$$

Then the following hold

• (1) $T_r$ is single-valued.

• (2) $T_r$ is firmly nonexpansive, i.e., for each $x,y\in H$,

  $$\begin{aligned} \Vert T_rx-T_ry\Vert ^2\le \langle T_rx-T_ry,x-y\rangle . \end{aligned}$$

• (3) $Fix(T_r)=EP(\Theta )$.

• (4) $EP(\Theta )$ is closed and convex.

Theorem 5.3

Let $C$ be a nonempty, closed and convex subset of a real Hilbert space $H$. Let $\Theta :C\times C\longrightarrow \mathbb {R}$ be a bifunction satisfies the conditions (A1)–(A4) with $EP(\Theta )\ne \emptyset$. Let $f:C\longrightarrow C$ be a Meir–Keeler contraction-type. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{l} \Theta (u_n,y)+\frac{1}{r_n}\langle y-u_n,u_n-x_n\rangle \ge 0,\\ x_{n+1}=\alpha _n f(x_n)+(1-\alpha _n)u_n,~~\forall n\ge 1, \end{array} \end{array}\right. } \end{aligned}$$

(5.2)

where $\{\alpha _n\}\subset (0,1)$ and $\{r_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim \nolimits _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim \nolimits _{n\longrightarrow \infty }r_n=\infty$.

Then the sequence $\{x_n\}$ defined by (5.2) converges strongly to an element $x^*\in EP(\Theta )$.

Proof

By using the same arguments and techniques as those of Theorem 3.2, we only need show that there exists a number $r>0$ such that

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert x_n-T_rx_n\Vert =0. \end{aligned}$$

From definition of $T_r$, we have

$$\begin{aligned} \Theta (T_{r_n}x_n,y)+\frac{1}{r_n}\langle y-T_{r_n}x_n,T_{r_n}x_n-x_n\rangle \ge 0,~~\forall y\in C, \end{aligned}$$

(5.3)

and

$$\begin{aligned} \Theta (T_rT_{r_n}x_n,y)+\frac{1}{r_n}\langle y-T_rT_{r_n}x_n,T_rT_{r_n}x_n-x_n\rangle \ge 0,~~\forall y\in C. \end{aligned}$$

(5.4)

Substituting $y=T_rT_{r_n}x_n$ in (5.3) and $y=T_{r_n}x_n$ in (5.4). Then, add these two inequalities and from the condition $(A2)$, we obtain

$$\begin{aligned} \bigg \langle T_{r_n}x_n-T_rT_{r_n}x_n,\frac{T_rT_{r_n}x_n-x_n}{r}-\frac{T_{r_n}x_n-x_n}{r_n}\bigg \rangle \ge 0, \end{aligned}$$

and hence, for each $r>0$,

$$\begin{aligned} \frac{\Vert T_rT_{r_n}x_n-T_{r_n}x_n\Vert ^2}{r}\le & {} \bigg \langle T_{r_n}x_n-T_rT_{r_n}x_n,\frac{1}{r_n}(x_n-T_{r_n}x_n)\bigg \rangle \\\le & {} \Vert T_rT_{r_n}x_n-T_{r_n}x_n\Vert \frac{1}{r_n}\Vert T_{r_n}x_n-x_n\Vert , \end{aligned}$$

which implies that

$$\begin{aligned} \begin{array}{lcl} \Vert T_rT_{r_n}x_n-T_{r_n}x_n\Vert \le \frac{r\Vert T_{r_n}x_n-x_n\Vert }{r_n}. \end{array} \end{aligned}$$

It follows from the condition (C2) that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\Vert T_rT_{r_n}x_n-T_{r_n}x_n\Vert =0. \end{aligned}$$

(5.5)

Noticing $u_n=T_{r_n}x_n$, we note that

$$\begin{aligned} \Vert x_{n+1}-T_rx_{n+1}\Vert\le & {} \Vert x_{n+1}-T_{r_n}x_n\Vert +\Vert T_{r_n}x_n-T_rT_{r_n}x_n\Vert +\Vert T_rT_{r_n}x_n-T_rx_{n+1}\Vert \\\le & {} 2\Vert x_{n+1}-u_n\Vert +\Vert T_{r_n}x_n-T_rT_{r_n}x_n\Vert \end{aligned}$$

It follows from (5.5) that $\lim \nolimits _{n\longrightarrow \infty }\Vert x_{n+1}-T_rx_{n+1}\Vert =0$, and hence

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert x_n-T_rx_n\Vert =0,~~\forall r>0. \end{aligned}$$

This completes the proof. $\square$

6 Application to Optimization Problem

Let $H$ be a real Hilbert space and $\phi : H \longrightarrow (-\infty ,+\infty ]$ a proper convex lower semicontinuous function. Then the subdifferential $\partial \phi$ of $\phi$ is defined as follows:

$$\begin{aligned} \partial \phi =\big \{y\in H:\phi (z)\ge \phi (x)+\langle z-x,y\rangle ,~~z\in H\big \}~~\forall x\in H. \end{aligned}$$

From [40], we know that $\partial \phi$ is maximal monotone. It is easy to verify that $0\in \partial \phi$ if and only if $\phi (x)=\min _{y\in H}\phi (y)$ see also [40–42].

Consider a kind of optimization problem with a nonempty set of solutions

$$\begin{aligned} \min \nolimits _{x\in C} h(x), \end{aligned}$$

(6.1)

where $h(x)$ is a convex and lower semicontinuous functional defined on a closed convex subset $C$ of a real Hilbert space $H$. We denote by $arg\min (h)$ the set of solutions of (6.1). Define a bifunction $\Theta :C\times C\longrightarrow \mathbb {R}$ by $\Theta (x,y):=h(y)-h(x)$. It is obvious that $EP(\Theta )=arg\min (h)$. In addition, it is easy to see that $\Theta (x,y)$ satisfies the conditions (A1)–(A4) in the Sect. 5.

Using Theorem 5.3, we have the following result.

Theorem 6.1

Let $C$ be a nonempty, closed and convex subset of a real Hilbert space $H$. Let $h$ be a convex and lower semicontinuous functional on $C$ such that $arg\min (h)\ne \emptyset$. Let $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{l} u_n=arg\min \nolimits _{y\in C}\bigg \{h(y)+\frac{1}{2r_n}\Vert y-x_n\Vert ^2\bigg \},\\ x_{n+1}=\alpha _n f(x_n)+(1-\alpha _n)u_n,~~\forall n\ge 1, \end{array} \end{array}\right. } \end{aligned}$$

(6.2)

where $\{\alpha _n\}\subset (0,1)$ and $\{r_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim \nolimits _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim \nolimits _{n\longrightarrow \infty }r_n=\infty$.

Then the sequence $\{x_n\}$ defined by (6.2) converges strongly to an element $x^*\in arg\min (h)$.

7 Application to Variational Inequalities

Let $C$ be a nonempty, closed and convex subset of a real Hilbert space $H$ and $A:C\longrightarrow H$ be a mapping. The classical variational inequality is to find $x\in C$ such that

$$\begin{aligned} \langle Ax,y-x\rangle \ge 0,~~\forall y\in C. \end{aligned}$$

(7.1)

The set of solutions of the classical variational inequality (7.1) is denoted by $VI(C,A)$. We recall that a mapping $A:C\longrightarrow H$ is monotone if

$$\begin{aligned} \langle Ax-Ay,x-y\rangle \ge 0,~~\forall x,y\in C. \end{aligned}$$

The following lemmas found in [43].

Lemma 7.1

(Bose [43]) Let $C$ be a nonempty, closed and convex subset of a real Hilbert space $H$. Let $A$ be a continuous monotone mapping of $C$ into $H$. Define a bifunction $\Theta :C\times C\longrightarrow \mathbb {R}$ as follows:

$$\begin{aligned} \Theta (x,y):=\langle Ax,y-x\rangle ,~~\forall x,y\in C. \end{aligned}$$

Then the following hold

• (1) $\Theta$ is satisfies the conditions (A1)–(A4) in Sect. 5 and $VI(C,A)=EP(\Theta )$.

• (2) for each $x\in H, z\in C$ and $r>0$,

  $$\begin{aligned} \Theta (z,y)+\frac{1}{r}\langle y-z,z-x\rangle \ge 0~\forall y\in C,\Longleftrightarrow z=P_C(x-rAx), \end{aligned}$$

  where $P_C$ is projection operator from $H$ into $C$.

Using Lemma 7.1 and Theorem 5.3, we have the following result.

Theorem 7.2

Let $C$ be a nonempty, closed and convex subset of a real Hilbert space $H$. Let $A$ be a continuous monotone mapping of $C$ such that $VI(C,A)\ne \emptyset$. Let $f:C\longrightarrow C$ be a Meir–Keeler-type contraction. For given $x_1\in C$, let $\{x_n\}$ be a sequence defined by

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{ll} u_n=P_C(x_n-r_nAx_n),\\ x_{n+1}=\alpha _n f(x_n)+(1-\alpha _n)u_n,~~\forall n\ge 1, \end{array} \end{array}\right. } \end{aligned}$$

(7.2)

where $\{\alpha _n\}\subset (0,1)$ and $\{r_n\}\subset (0,\infty )$ are sequences which satisfy the following conditions:

• (C1) $\lim \nolimits _{n\longrightarrow \infty }\alpha _n=0$ and $\sum _{n=1}^{\infty }\alpha _n=\infty$;

• (C2) $\lim \nolimits _{n\longrightarrow \infty }r_n=\infty$.

Then the sequence $\{x_n\}$ defined by (7.2) converges strongly to an element $x^*\in VI(C,A)$.