A new iterative algorithm for the sum of two different types of finitely many accretive operators in Banach space and its connection with capillarity equation

Abstract

In this paper, we present a new iterative algorithm with errors to solve the problems of finding zeros of the sum of finitely many m-accretive operators and finitely many α-inversely strongly accretive operators in a real smooth and uniformly convex Banach space. Strong convergence theorems are established, which extend the corresponding works given by some authors. Moreover, the relationship among the zero of the sum of m-accretive operator and α-inversely strongly accretive operator, the solution of one kind variational inequality, and the solution of the capillarity equation is investigated.

1 Introduction and preliminaries

Let E be a real Banach space with norm $\|\cdot\|$ and let $E^{*}$ denote the dual space of E. We use ‘→’ and ‘⇀’ to denote strong and weak convergence either in E or in $E^{*}$, respectively. We denote the value of $f \in E^{*}$ at $x \in E$ by $\langle x,f \rangle$.

A Banach space E is said to be uniformly convex if, for each $\varepsilon\in(0,2]$, there exists $\delta> 0$ such that

$$\|x\| = \|y\| = 1, \qquad\|x - y\| \geq \varepsilon\quad \Rightarrow\quad \biggl\Vert \frac{x+y}{2}\biggr\Vert \leq 1-\delta. $$

A Banach space E is said to be smooth if

$$\lim_{t \rightarrow0}\frac{\|x+ty\|-\|x\|}{t} $$

exists for each $x , y \in\{z \in E : \|z\| = 1\}$.

In addition, we define a function $\rho_{E}: [0,+\infty) \rightarrow [0,+\infty)$ called the modulus of smoothness of E as follows:

$$\rho_{E}(t) = \sup\biggl\{ \frac{1}{2}\bigl(\Vert x+y\Vert + \|x-y\|\bigr)-1 : x,y \in E, \|x\|=1, \|y\| \leq t\biggr\} . $$

It is well known that E is uniformly smooth if and only if $\frac{\rho_{E}(t)}{t}\rightarrow0$, as $t \rightarrow0$. Let $q > 1$ be a real number. A Banach space E is said to be q-uniformly smooth if there exists a positive constant C such that $\rho_{E}(t)\leq Ct^{q}$. It is obvious that q-uniformly smooth Banach space must be uniformly smooth.

The normalized duality mapping $J: E \rightarrow2^{E^{*}}$ is defined by

$$Jx : = \bigl\{ f \in E^{*}: \langle x, f\rangle= \|x\|^{2} = \|f \|^{2}\bigr\} , \quad x \in E. $$

It is well known that J is single-valued and norm-to-norm uniformly continuous on each bounded subsets of E if E is a real smooth and uniformly convex Banach space; see [1]. Moreover, $J(cx) = cJx$, for all $x \in E$ and $c \in R^{1}$. In what follows, we still denote by J the single-valued normalized duality mapping. If, E is reduced to the Hilbert space H, then $J \equiv I$ is the identity mapping. The normalized duality mapping J is said to be weakly sequentially continuous if $\{x_{n}\}$ is a sequence in E which converges weakly to x it follows that $\{Jx_{n}\}$ converges in weak^∗ to Jx. J is said to be weakly sequentially continuous at zero if $\{x_{n}\}$ is a sequence in E which converges weakly to 0 it follows that $\{Jx_{n}\}$ converges in weak^∗ to 0.

Let C be a nonempty, closed, and convex subset of E and let Q be a mapping of E onto C. Then Q is said to be sunny [2] if $Q(Q(x)+t(x-Q(x))) = Q(x)$, for all $x \in E$ and $t \geq 0$.

A mapping Q of E into E is said to be a retraction [2] if $Q^{2} = Q$. If a mapping Q is a retraction, then $Q(z) = z$ for every $z \in R(Q)$, where $R(Q)$ is the range of Q.

For a mapping $U: C \rightarrow C$, we use $\operatorname{Fix}(U)$ to denote the fixed point set of it; that is, $\operatorname{Fix}(U) : = \{x\in C: Ux = x\}$.

For an operator $A: D(A) \subset E \rightarrow2^{E}$, we use $A^{-1} 0$ to denote the set of zeros of it; that is, $A^{-1} 0 : = \{x \in D(A) : Ax = 0\}$.

Let $T : C \rightarrow E$ be a mapping. Then T is said to be

• (1) nonexpansive if

  $$\|Tx - Ty\| \leq\|x-y\|, \quad \mbox{for }\forall x,y \in C; $$

• (2) k-Lipschitz if there exists $k > 0$ such that

  $$\|Tx - Ty\| \leq k \|x - y\|,\quad \mbox{for }\forall x,y \in C. $$

  In particular, if $0 < k < 1$, then T is called a contraction and if $k = 1$, then T reduces to a nonexpansive mapping;

• (3) accretive if for all $x, y \in C$, there exists $j(x-y) \in J(x-y)$ such that

  $$\bigl\langle Tx - Ty, j(x-y)\bigr\rangle \geq0, $$

  where J is the normalized duality mapping;

• (4) α-inversely strongly accretive if for all $x, y \in C$, there exists $j(x-y) \in J(x-y)$ such that

  $$\bigl\langle Tx - Ty, j(x-y)\bigr\rangle \geq\alpha\| Tx - Ty\|^{2}, $$

  for some $\alpha> 0$;

• (5) m-accretive if T is accretive and $R(I+\lambda T) = E$, for $\forall\lambda> 0$;

• (6) strongly positive (see [3]) if E is a real smooth Banach space and there exists $\overline{\gamma} > 0$ such that

  $$\langle Tx,Jx\rangle\geq\overline{\gamma} \|x\|^{2}, \quad \mbox{for } \forall x \in C. $$

  In this case,

  $$\|aI-bT\| = \sup_{\|x\| \leq1}\bigl\vert \bigl\langle (aI - bT)x, J(x)\bigr\rangle \bigr\vert , $$

  where I is the identity mapping and $a \in [0,1]$, $b \in[-1,1]$.

We denote by $J_{r}^{A}$ (for $r > 0$) the resolvent of the accretive operator A; that is, $J^{A}_{r} : = (I + rA)^{-1}$. It is well known that $J^{A}_{r}$ is nonexpansive and $\operatorname{Fix}(J_{r}^{A}) = A^{-1}0$.

A subset C of E is said to be a sunny nonexpansive retract of E if there exists a sunny nonexpansive retraction of E onto C.

Many practical problems can be reduced to finding zeros of the sum of two accretive operators; that is, $0 \in(A+B)x$. Forward-backward splitting algorithms, which have recently received much attention to many mathematicians, were proposed by Lions and Mercier [4], by Passty [5], and, in a dual form for convex programming, by Han and Lou [6].

The classical forward-backward splitting algorithm is given in the following way:

$$ x_{n+1} = (I+r_{n} B)^{-1}(I-r_{n} A)x_{n},\quad n \geq0. $$

(1)

Based on iterative algorithm (1), much work has been done for finding $x \in H$ such that $x \in(A+B)^{-1}0$, where A and B are α-inversely strongly accretive operator and m-accretive operator defined in the Hilbert space H, respectively. However, most of the existing work are undertaken in the frame of Hilbert spaces; see [4–10], etc.

Recently, Qin et al., presented the following iterative algorithm in the frame of q-uniformly smooth Banach spaces E in [11]:

$$ x_{0} \in C,\quad x_{n+1}= \alpha_{n} f(x_{n}) + \beta_{n} (I+r_{n} B)^{-1}\bigl[(I-r_{n} A)x_{n} +e_{n}\bigr] + \gamma_{n} f_{n},\quad n \geq0, $$

(2)

where $\{e_{n}\}$ is the error sequence, f is a contraction, A and B are α-inversely strongly accretive operator and m-accretive operator, respectively. If $(A+B)^{-1}0 \neq\emptyset$, they proved that $\{x_{n}\}$ converges strongly to $x = Q_{(A+B)^{-1}0}f(x)$, where $Q_{(A+B)^{-1}0}$ is the unique sunny nonexpansive retraction of E onto $(A+B)^{-1}0$, under some conditions.

On the other hand, there are some excellent work done on approximating fixed points of nonexpansive mappings. For example, in 2009, Yao et al. presented the following iterative algorithm in the frame of Hilbert space in [12]:

$$ \begin{cases} x_{0} \in C, \\ y_{n} = P_{C}[(1-\alpha_{n})x_{n}], \\ x_{n+1}= (1- \beta_{n})x_{n} + \beta_{n} Ty_{n},\quad n \geq0, \end{cases} $$

(3)

where $P_{C}$ is the metric projection from H onto C and $T : C \rightarrow C$ is a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$. They proved that $\{x_{n}\}$ constructed by (3) converges strongly to a fixed point of T.

In 2006, Marino and Xu, presented the following iterative algorithm in the frame of Hilbert spaces in [13]:

$$ x_{0} \in C,\quad x_{n+1}= \alpha_{n} \gamma f(x_{n}) + (I-\alpha_{n} A)Tx_{n},\quad n \geq0, $$

(4)

where f is a contraction, A is a strongly positive linear bounded operator, and T is nonexpansive. If $\operatorname{Fix}(T)\neq \emptyset$, they proved that $\{x_{n}\}$ converges strongly to $p \in \operatorname{Fix}(T)$, which solves the variational inequality $\langle (\gamma f - A)p, z - p\rangle\leq0$, for $\forall z \in \operatorname{Fix}(T)$, under some conditions.

Our paper is organized in the following way: in Section 2, inspired by the work in [11–13], we shall present the following iterative algorithm with errors in a real smooth and uniformly convex Banach space:

$$ \begin{cases} x_{0} \in C, \\ y_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})], \\ z_{n}= (1- \beta_{n})x_{n} + \beta_{n} [a_{0}y_{n} + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}(y_{n}-r_{n,i}B_{i}y_{n})], \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I-\gamma_{n}T)z_{n},\quad n \geq0, \end{cases} $$

(A)

where C is a nonempty, closed, and convex sunny nonexpansive retract of E, $Q_{C}$ is the sunny nonexpansive retraction of E onto C, $\{e_{n}\} \subset E$ is the error sequence, $\{A_{i}\}_{i = 1}^{N}$ is a finite family of m-accretive operators and $\{B_{i}\}_{i = 1}^{N}$ is a finite family of α-inversely strongly accretive operators. $T : E \rightarrow E$ is a strongly positive linear bounded operator with coefficient $\overline{\gamma}$ and $f : E \rightarrow E$ is a contraction with coefficient $k \in(0,1)$. $J_{r_{n,i}}^{A_{i}} = (I + r_{n,i} A_{i} )^{-1}$, for $i = 1,2,\ldots,N$, $\sum_{m = 0}^{N}a_{m} = 1$, $0 < a_{m} < 1$, for $m = 0 , 1, 2, \ldots, N$. More detail of iterative algorithm (A) will be presented in Section 2. Then $\{x_{n}\}$ is proved to converge strongly to $p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$, which is also a solution of one kind variational inequality.

Our main contributions in Section 2 are:

• (i) the discussion is undertaken in the frame of real smooth and uniformly convex Banach space, which is more general than that in Hilbert space or in q-uniformly smooth Banach space;

• (ii) the assumption that ‘the normalized duality mapping J is weakly sequentially continuous’ in most of the existing related work is weaken to ‘J is weakly sequentially continuous at zero’;

• (iii) a new path convergence theorem (Lemma 8) is obtained which is a direct extension of the corresponding result in [13] from Hilbert space to real smooth and uniformly convex Banach space;

• (iv) the connection between zeros of the sum of m-accretive operators and α-inversely strongly accretive operators and the solution of one kind variational inequalities is being set up.

In Section 3, one kind capillarity equation is discussed, from which we can see the connection among the unique solution of this equation, the unique solution of one kind variational inequality and the iterative algorithm presented in Section 2.

Next, we list some results we need in sequel:

Lemma 1

(see [1])

Let E be a Banach space and $f: E \rightarrow E$ be a contraction. Then f has a unique fixed point $u \in E$.

Lemma 2

(see [14])

Let E be a real uniformly convex Banach space, C be a nonempty, closed, and convex subset of E and $T: C \rightarrow E$ be a nonexpansive mapping such that $\operatorname{Fix}(T) \neq \emptyset$, then $I-T$ is demiclosed at zero.

Lemma 3

(see [15])

In a real Banach space E, the following inequality holds:

$$\|x+y\|^{2} \leq\|x\|^{2} + 2\bigl\langle y, j(x+y)\bigr\rangle , \quad \forall x,y \in E, $$

where $j(x+y) \in J(x+y)$.

Lemma 4

(see [16])

Let $\{a_{n}\}$ and $\{c_{n}\}$ be two sequences of nonnegative real numbers satisfying

$$a_{n+1}\leq(1-t_{n})a_{n} + b_{n}+c_{n}, \quad \forall n \geq0, $$

where $\{t_{n}\}\subset(0,1)$ and $\{b_{n}\}$ is a number sequence. Assume that $\sum_{n=0}^{\infty}t_{n} = +\infty$, $\limsup_{n \rightarrow\infty} \frac{b_{n}}{t_{n}} \leq0$, and $\sum_{n=0}^{\infty}c_{n} < +\infty$. Then $\lim_{n \rightarrow\infty }a_{n} = 0$.

Lemma 5

(see [17])

Let E be a Banach space and let A be an m-accretive operator. For $\lambda>0$, $\mu>0$, and $x \in E$, one has

$$J_{\lambda}x = J_{\mu}\biggl(\frac{\mu}{\lambda}x+\biggl(1- \frac{\mu}{\lambda}\biggr)J_{\lambda}x\biggr), $$

where $J_{\lambda}= (I+\lambda A)^{-1}$ and $J_{\mu}= (I+\mu A)^{-1}$.

Lemma 6

(see [18])

Let E be a real Banach space and let C be a nonempty, closed, and convex subset of E. Suppose $A: C \rightarrow E$ is a single-valued operator and $B: E \rightarrow 2^{E}$ is m-accretive. Then

$$\operatorname{Fix}\bigl((I+rB)^{-1}(I-rA)\bigr) = (A+B)^{-1}0, \quad \textit{for }\forall r > 0. $$

Lemma 7

(see [19])

Assume T is a strongly positive bounded operator with coefficient $\overline{\gamma} > 0$ on a real smooth Banach space E and $0 < \rho\leq \|T\|^{-1}$. Then $\|I-\rho T \| \leq1- \rho\overline{\gamma}$.

2 Strong convergence theorems

Lemma 8

Let E be a real smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny nonexpansive retract of E, and let $Q_{C}$ be the sunny nonexpansive retraction of E onto C. Let $f : E \rightarrow E$ be a fixed contractive mapping with coefficient $k \in(0,1)$, $T: E \rightarrow E$ be a strongly positive linear bounded operator with coefficient $\overline{\gamma}$ and $U : C \rightarrow C$ be a nonexpansive mapping. Suppose that the duality mapping $J : E \rightarrow E^{*}$ is weakly sequentially continuous at zero, $0 < \eta< \frac{\overline{\gamma}}{2k}$ and $\operatorname{Fix}(U) \neq\emptyset$. If for each $t \in(0,1)$, define $T_{t} : E \rightarrow E$ by

$$ T_{t} x : = t \eta f(x) + (I-t T)UQ_{C}x, $$

(5)

then $T_{t}$ has a fixed point $x_{t}$, for each $0 < t \leq \|T\|^{-1}$, which is convergent strongly to the fixed point of U, as $t \rightarrow0$. That is, $\lim_{t\rightarrow0}x_{t} = p_{0} \in \operatorname{Fix}(U)$. Moreover, $p_{0}$ satisfies the following variational inequality: for $\forall z \in \operatorname{Fix}(U)$,

$$ \bigl\langle (T-\eta f)p_{0}, J(p_{0}-z)\bigr\rangle \leq0. $$

(6)

Proof

Step 1. $T_{t}$ is a contraction, for $0 < t < \|T\|^{-1}$.

In fact, noticing Lemma 7, we have

$$\begin{aligned} \Vert T_{t}x - T_{t}y\Vert &\leq t\eta\bigl\Vert f(x) - f(y)\bigr\Vert + \bigl\Vert (I-tT) (UQ_{C}x-UQ_{C}y) \bigr\Vert \\ & \leq kt\eta \Vert x - y\Vert + (1-t\overline{\gamma})\Vert x-y\Vert \\ & = \bigl[1-t(\overline{\gamma}-k\eta)\bigr]\Vert x - y\Vert , \end{aligned}$$

which implies that $T_{t}$ is a contraction since $0 < \eta< \frac{\overline{\gamma}}{2k}$.

Then Lemma 1 implies that $T_{t}$ has a unique fixed point, denoted by $x_{t}$, which uniquely solves the fixed point equation $x_{t} = t\eta f(x_{t}) + (I-tT)UQ_{C}x_{t}$.

Step 2. $\{x_{t}\}$ is bounded, for $t \in(0, \|T\|^{-1})$.

For $p \in \operatorname{Fix}(U)\subset C$, we have $p = UQ_{C}p$, then

$$\begin{aligned} \Vert x_{t} - p\Vert &= \bigl\Vert (I-tT) (UQ_{C}x_{t}-p) + t\bigl(\eta f(x_{t})-Tp\bigr)\bigr\Vert \\ & \leq(1-t\overline{\gamma})\Vert x_{t}-p\Vert + t\bigl\Vert \eta f(x_{t})-Tp\bigr\Vert \\ & = (1-t\overline{\gamma})\Vert x_{t}-p\Vert + t\bigl\Vert \eta \bigl(f(x_{t})-f(p)\bigr)+\bigl(\eta f(p)-Tp\bigr)\bigr\Vert \\ & \leq(1-t\overline{\gamma})\Vert x_{t}-p\Vert + t\bigl(k\eta \Vert x_{t}-p\Vert +\bigl\Vert \eta f(p)-Tp\bigr\Vert \bigr) \\ & = \bigl[1-t(\overline{\gamma}-k\eta)\bigr]\Vert x_{t}-p\Vert +t \bigl\Vert \eta f(p)-Tp\bigr\Vert . \end{aligned}$$

This ensures that

$$\|x_{t} - p\|\leq\frac{\|\eta f(p)-Tp\|}{\overline{\gamma}-k\eta}. $$

Thus $\{x_{t}\}$ is bounded, which implies that both $\{f(x_{t})\}$ and $\{TUQ_{C}x_{t}\}$ are bounded.

Step 3. $x_{t} - UQ_{C}x_{t} \rightarrow0$, as $t \rightarrow0$.

Noticing the result of Step 2, we have $\|x_{t} - UQ_{C}x_{t}\| = t \|\eta f(x_{t})- TUQ_{C}x_{t}\| \rightarrow0$, as $t \rightarrow 0$.

Step 4. $\langle(T- \eta f)x - (T-\eta f)y, J(x-y)\rangle\geq (\overline{\gamma}- k \eta)\|x-y\|^{2}$, for $\forall x,y \in E$.

In fact,

$$\begin{aligned}& \bigl\langle (T- \eta f)x - (T-\eta f)y, J(x-y)\bigr\rangle \\& \quad = \bigl\langle Tx - Ty, J(x-y)\bigr\rangle - \eta\bigl\langle f(x) - f(y), J(x-y)\bigr\rangle \\& \quad \geq\overline{\gamma}\|x-y\|^{2} - k\eta\|x-y\|^{2} = (\overline{\gamma }- k \eta)\|x-y\|^{2}. \end{aligned}$$

Step 5. If the variational inequality (6) has a solution, then the solution must be unique.

Suppose both $u_{0}\in \operatorname{Fix}(U)$ and $v_{0} \in \operatorname{Fix}(U)$ are the solutions of the variational inequality (6). Then we have

$$ \bigl\langle (T- \eta f)v_{0}, J(v_{0}-u_{0}) \bigr\rangle \leq0 $$

(7)

and

$$ \bigl\langle (T- \eta f)u_{0}, J(u_{0}-v_{0}) \bigr\rangle \leq0. $$

(8)

Adding up (7) and (8), we obtain

$$\bigl\langle (T- \eta f)u_{0} - (T-\eta f)v_{0}, J(u_{0}-v_{0})\bigr\rangle \leq0. $$

In view of the result of Step 4, we have $u_{0} = v_{0}$.

Step 6. $x_{t} \rightarrow p_{0}\in \operatorname{Fix}(U)$, as $t \rightarrow0$, which satisfies the variational inequality (6).

For $\forall z \in \operatorname{Fix}(U)$, $x_{t} - z = t(\eta f(x_{t})-Tz)+(I-tT)(UQ_{C}x_{t}-z)$. Thus Lemma 3 implies that

$$\begin{aligned} \|x_{t} - z\|^{2} &\leq \|I-tT\|^{2} \|UQ_{C}x_{t}-UQ_{C}z\|^{2}+2t \bigl\langle \eta f(x_{t})-Tz, J(x_{t}-z)\bigr\rangle \\ & \leq(1-t\overline{\gamma})\|x_{t}-z\|^{2}+ 2t \bigl\langle \eta f(x_{t})-Tz, J(x_{t}-z)\bigr\rangle . \end{aligned}$$

Then

$$\begin{aligned} \|x_{t} - z\|^{2} &\leq\frac{2}{\overline{\gamma}} \bigl\langle \eta f(x_{t})-Tz, J(x_{t}-z)\bigr\rangle \\ & = \frac{2}{\overline{\gamma}} \bigl[\eta\bigl\langle f(x_{t})-f(z), J(x_{t}-z)\bigr\rangle +\bigl\langle \eta f(z)-T(z), J(x_{t}-z)\bigr\rangle \bigr] \\ & \leq\frac{2}{\overline{\gamma}} \bigl[\eta k \|x_{t}-z\|^{2}+\bigl\langle \eta f(z)-Tz, J(x_{t}-z)\bigr\rangle \bigr]. \end{aligned}$$

Therefore, for $\forall z \in \operatorname{Fix}(U)$, we have

$$ \|x_{t} - z\|^{2}\leq \frac{2}{\overline{\gamma}-2k\eta} \bigl\langle \eta f(z)-Tz, J(x_{t}-z)\bigr\rangle . $$

(9)

Since $\{x_{t}\}$ is bounded as $t \rightarrow0^{+}$, we can choose $\{t_{n}\}\subset(0,1)$ such that $t_{n} \rightarrow0^{+}$ and $x_{t_{n}}\rightharpoonup p_{0}$. From Lemma 2 and the result of Step 3, we see that $p_{0} = UQ_{C} p_{0} = Up_{0}$. Thus $p_{0} \in \operatorname{Fix}(U)$. Substituting z by $p_{0}$ in (9), then we can deduce that $x_{t_{n}}\rightarrow p_{0}$ since J is weakly sequentially continuous at zero. Next, we shall prove that $p_{0}$ solves the variational inequality (6).

Since $x_{t} = t\eta f(x_{t}) + (I-tT)UQ_{C}x_{t}$,

$$(T-\eta f)x_{t} = - \frac{1}{t}(I-tT) (I-UQ_{C})x_{t}. $$

For $\forall z \in \operatorname{Fix}(U)$, since U is nonexpansive,

$$\begin{aligned} \begin{aligned}[b] &\bigl\langle (T-\eta f)x_{t}, J(x_{t} - z) \bigr\rangle \\ &\quad = - \frac{1}{t} \bigl\langle (I-tT) (I-UQ_{C})x_{t}, J(x_{t}-z)\bigr\rangle \\ &\quad = -\frac{1}{t} \bigl\langle (I-UQ_{C})x_{t}-(I-UQ_{C})z, J(x_{t}-z)\bigr\rangle +\bigl\langle T(I-UQ_{C})x_{t}, J(x_{t}-z)\bigr\rangle \\ &\quad = -\frac{1}{t} \bigl[\|x_{t} - z\|^{2}- \bigl\langle UQ_{C}x_{t}-UQ_{C}z, J(x_{t}-z) \bigr\rangle \bigr]+\bigl\langle T(I-UQ_{C})x_{t}, J(x_{t}-z)\bigr\rangle \\ &\quad \leq\bigl\langle T(I-UQ_{C})x_{t}, J(x_{t}-z)\bigr\rangle . \end{aligned} \end{aligned}$$

(10)

Since $x_{t_{n}} \rightarrow p_{0}$, we have $(I-UQ_{C})x_{t_{n}} \rightarrow (I-UQ_{C})p_{0} = 0$, as $n \rightarrow\infty$. Since $\{x_{t_{n}}\}$ is bounded, $(T - \eta f)x_{t_{n}} \rightarrow(T - \eta f)p_{0}$ and J is uniformly continuous on each bounded subset of E, taking the limits on both sides of (10) we have $\langle(T-\eta f)p_{0}, J(p_{0} - z)\rangle\leq0$, for $z \in \operatorname{Fix}(U)$. Thus $p_{0}$ satisfies (6).

In a summary, we infer that each cluster point of $\{x_{t}\}$ is equal to $p_{0}$, which is the unique solution of the variational inequality (6).

This completes the proof. □

Remark 9

Lemma 8 is a direct extension of Theorem 3.2 in [13] from Hilbert space to real smooth and uniformly convex Banach space.

Theorem 10

Let E be a real smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny nonexpansive retract of E, and let $Q_{C}$ be the sunny nonexpansive retraction of E onto C. Let $f : E \rightarrow E$ be a fixed contractive mapping with coefficient $k \in(0,1)$, $T: E \rightarrow E$ be a strongly positive linear bounded operator with coefficient $\overline{\gamma}$. Suppose that the duality mapping $J : E \rightarrow E^{*}$ is weakly sequentially continuous at zero, and $0 < \eta< \frac{\overline{\gamma}}{2k}$. Let $A_{i} : C \rightarrow2^{E}$ be m-accretive operator and $B_{i} : C \rightarrow E$ be α-inversely strongly accretive operator, where $i = 1,2,\ldots, N$. Suppose that, for $\forall r > 0$ and $i = 1,2,\ldots, N$,

$$\bigl\langle B_{i}x - B_{i}y, J\bigl[(I-rB_{i})x - (I-rB_{i})y\bigr]\bigr\rangle \geq0. $$

Let $\{x_{n}\}$ be generated by the iterative algorithm (A), $0 < a_{m} < 1$, for $m = 0, 1,2,\ldots,N$, $\sum_{m = 0}^{N} a_{m} = 1$. Suppose $\{e_{n}\}_{n=0}^{\infty} \subset E$, $\{\alpha_{n}\}$, $\{\beta_{n}\}$, and $\{\gamma_{n}\}$ are three sequences in $(0,1)$ and $\{r_{n,i}\}\subset(0,+\infty)$ satisfying the following conditions:

• (i) $\sum_{n=0}^{\infty}\gamma_{n} = \infty$, $\gamma_{n} \rightarrow0$, $\frac{\gamma_{n-1}}{\gamma_{n}} \rightarrow1$, $\beta_{n} \rightarrow1$, $\alpha_{n} \rightarrow0$, as $n \rightarrow\infty$;

• (ii) $\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty$, $\sum_{n=0}^{\infty}|\beta_{n+1} - \beta_{n}| < +\infty$, $\sum_{n=0}^{\infty}(1-\gamma_{n} \overline{\gamma})\alpha_{n}\beta_{n} < +\infty$;

• (iii) $\sum_{n=0}^{\infty}|r_{n+1,i} - r_{n,i}| < +\infty$ and $r_{n,i}\geq\varepsilon> 0$, for $n \geq0$ and $i = 1,2,\ldots, N$;

• (iv) $\sum_{n=0}^{\infty}\|e_{n}\| < +\infty$.

If $\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \neq\emptyset$, then $\{x_{n}\}$ converges strongly to a point $p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$, which is the unique solution of the following variational inequality: for $\forall z \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$,

Proof

Let $u_{n,i} = (I - r_{n,i}B_{i})y_{n}$, $v_{n} = a_{0} y_{n} + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}u_{n,i}$, for $n \geq0$, where $i = 1,2, \ldots, N$.

We shall split the proof into five steps:

Step 1. $\{x_{n}\}$, $\{u_{n,i}\}$ ($i = 1,2, \ldots, N$), $\{y_{n}\}$, $\{v_{n}\}$, and $\{z_{n}\}$ are all bounded.

From the assumptions on $B_{i}$, in view of Lemma 3, we have, for $\forall x,y \in C$,

$$\begin{aligned} \bigl\Vert (I-rB_{i})x - (I-rB_{i})y\bigr\Vert ^{2} &\leq\|x - y\|^{2} - 2r \bigl\langle B_{i}x - B_{i}y, J\bigl[(I-rB_{i})x - (I-rB_{i})y\bigr] \bigr\rangle \\ &\leq\|x - y\|^{2}, \end{aligned}$$

which implies that $(I - rB_{i})$ is nonexpansive, for $r >0$.

Then noticing the facts that both $(I-r_{n,i}B_{i})$ and $J_{r_{n,i}}^{A_{i}}$ are nonexpansive, for $i = 1,2,\ldots, N$, we have, for $\forall p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \subset C$,

$$ \|y_{n} - p\| \leq(1-\alpha_{n}) \|x_{n} - p\|+\|e_{n}\|+\alpha_{n} \|e_{n} + p\|. $$

(11)

Using Lemma 6, we have, for $p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$,

$$\begin{aligned} \|z_{n} - p\|&\leq (1-\beta_{n}) \|x_{n} - p\| + \beta_{n} \Biggl(a_{0} \|y_{n} - p\| + \sum_{i = 1}^{N} a_{i} \bigl\Vert J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) (y_{n} - p)\bigr\Vert \Biggr) \\ & \leq (1-\beta_{n})\|x_{n} - p\| + \beta_{n} a_{0}\|y_{n} - p\| + \beta_{n}\sum _{i = 1}^{N} a_{i} \|y_{n} - p\| \\ & = (1-\beta_{n})\|x_{n} - p\| + \beta_{n} \|y_{n} - p\|. \end{aligned}$$

(12)

Then Lemma 7 implies that for $n \geq0$,

$$\begin{aligned} \Vert x_{n+1} - p\Vert &\leq\gamma_{n} \bigl\Vert \eta f(x_{n}) - Tp\bigr\Vert + \bigl\Vert (I- \gamma_{n}T) (z_{n} - p)\bigr\Vert \\ & \leq\gamma_{n}\eta k \Vert x_{n}- p\Vert + \gamma_{n} \bigl\Vert \eta f(p) - Tp\bigr\Vert + (1- \gamma_{n}\overline{\gamma})\Vert z_{n} - p\Vert . \end{aligned}$$

(13)

Noticing (11)-(13), we have, for $n \geq0$,

$$\begin{aligned}& \Vert x_{n+1} - p\Vert \\& \quad \leq \gamma_{n}\eta k \Vert x_{n}- p\Vert + \gamma_{n}\bigl\Vert \eta f(p) - Tp\bigr\Vert \\& \qquad {}+ (1- \gamma_{n}\overline{\gamma})\bigl[(1-\beta_{n})\Vert x_{n} - p\Vert + \beta_{n} \Vert y_{n} - p \Vert \bigr] \\& \quad \leq\bigl[\gamma_{n}\eta k+(1-\gamma_{n}\overline{ \gamma}) (1-\beta_{n}) +(1-\gamma_{n}\overline{\gamma}) (1- \alpha_{n})\beta_{n}\bigr]\Vert x_{n}- p\Vert + \gamma_{n}\bigl\Vert \eta f(p) - Tp\bigr\Vert \\& \qquad {}+(1-\gamma_{n}\overline{\gamma})\beta_{n}\Vert e_{n}\Vert +(1-\gamma_{n}\overline {\gamma}) \beta_{n}\alpha_{n}\Vert e_{n} + p\Vert \\& \quad = \bigl[1-\alpha_{n}\beta_{n}(1- \gamma_{n} \overline{\gamma})-\gamma_{n} (\overline{\gamma}- k \eta)\bigr]\Vert x_{n} -p\Vert + \gamma_{n} \bigl\Vert \eta f(p) - Tp\bigr\Vert \\& \qquad {}+ (1-\gamma_{n}\overline{\gamma})\beta_{n} \Vert e_{n}\Vert + (1-\gamma_{n}\overline{\gamma}) \alpha_{n} \beta_{n} \Vert e_{n} + p\Vert \\& \quad \leq\bigl[1-\gamma_{n} (\overline{\gamma}- k \eta)\bigr]\Vert x_{n} -p\Vert + \gamma_{n} \bigl\Vert \eta f(p) - Tp \bigr\Vert + 2\Vert e_{n}\Vert + (1-\gamma_{n} \overline{\gamma})\alpha_{n} \beta_{n} \Vert p\Vert \\& \quad \leq \max\biggl\{ \Vert x_{n} - p\Vert , \frac{\Vert \eta f(p) -Tp\Vert }{\overline{\gamma}-k \eta} \biggr\} +2 \Vert e_{n}\Vert +(1-\gamma_{n}\overline{ \gamma})\alpha_{n} \beta_{n} \Vert p\Vert . \end{aligned}$$

(14)

By using the inductive method, we can easily get the following result from (14):

$$\|x_{n+1}-p\| \leq \max\biggl\{ \|x_{0} - p\|, \frac{\|\eta f(p) -Tp\|}{\overline{\gamma}-k \eta}\biggr\} + 2\sum_{k=0}^{n} \|e_{k}\|+\|p\|\sum_{k=0}^{n}(1- \gamma_{k}\overline{\gamma })\alpha_{k} \beta_{k}. $$

Therefore, from assumptions (ii) and (iv), we know that $\{x_{n}\}$ is bounded.

For $\forall p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$, since $\|y_{n} - p\| \leq\|(1-\alpha_{n})(x_{n}+e_{n})-p\|\leq \|x_{n}\|+\|e_{n}\|+\|p\|$, $\{y_{n}\}$ is bounded, which implies that $\{u_{n,i}\}$ is bounded in view of the fact that $I-r_{n,i}B_{i}$ is nonexpansive, for each $i = 1,2, \ldots, N$.

Moreover, $\{J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}\}$ is bounded since $J_{r_{n,i}}^{A_{i}}$ is nonexpansive, for $i =1,2,\ldots,N$. Thus $\{v_{n}\}$ is bounded, which ensures that $\{z_{n}\}$ is bounded. Since $r_{n,i}\geq\varepsilon> 0$, $B_{i}y_{n} = \frac{y_{n}-u_{n,i}}{r_{n,i}}$ is bounded, for $n \geq0$ and $i = 1,2, \ldots, N$.

Set $M_{1} = \sup\{\|u_{n,i}\|, \|J^{A_{i}}_{r_{n,i}}u_{n,i}\|, \|B_{i}y_{n}\|, \|x_{n}\|, \|v_{n}\|, \|z_{n}\|, \|Tz_{n}\|, \eta\|f(x_{n})\|: n \geq0, i = 1,2, \ldots, N \}$.

Step 2. $\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0$.

First, we shall discuss $\|J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1, i}\|$, for $n \geq1$.

If $r_{n-1,i}\leq r_{n,i}$, then by using Lemma 5, we have

$$\begin{aligned}& \bigl\Vert J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1,i} \bigr\Vert \\& \quad = \biggl\Vert J_{r_{n-1,i}}^{A_{i}}\biggl( \frac{r_{n-1,i}}{r_{n,i}}u_{n,i} +\biggl(1-\frac{r_{n-1,i}}{r_{n,i}} \biggr)J_{r_{n,i}}^{A_{i}}u_{n,i}\biggr) - J_{r_{n-1,i}}^{A_{i}}u_{n-1,i}\biggr\Vert \\& \quad \leq \biggl\Vert \frac{r_{n-1,i}}{r_{n,i}}u_{n,i}+\biggl(1- \frac{r_{n-1,i}}{r_{n,i}}\biggr)J_{r_{n,i}} ^{A_{i}}u_{n,i} - u_{n-1,i}\biggr\Vert \\& \quad \leq \frac{r_{n-1,i}}{r_{n,i}}\Vert u_{n,i}-u_{n-1,i}\Vert + \biggl(1-\frac {r_{n-1,i}}{r_{n,i}}\biggr)\bigl\Vert J_{r_{n,i}} ^{A_{i}}u_{n,i} - u_{n-1,i}\bigr\Vert \\& \quad \leq \Vert u_{n,i}-u_{n-1,i}\Vert +\frac{r_{n,i}-r_{n-1,i}}{\varepsilon} \bigl\Vert J_{r_{n,i}} ^{A_{i}}u_{n,i} - u_{n-1,i} \bigr\Vert . \end{aligned}$$

(15)

If $r_{n,i}\leq r_{n-1,i}$, then imitating the proof of (15), we have

$$ \bigl\Vert J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1,i} \bigr\Vert \leq \Vert u_{n,i}-u_{n-1,i}\Vert + \frac{r_{n-1,i}-r_{n,i}}{\varepsilon}\bigl\Vert J_{r_{n,i}} ^{A_{i}}u_{n,i} - u_{n-1,i}\bigr\Vert . $$

(16)

Combining (15) and (16), we have, for $n \geq1$,

$$\begin{aligned}& \bigl\Vert J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1,i} \bigr\Vert \\& \quad \leq \Vert u_{n,i}-u_{n-1,i}\Vert +\frac{|r_{n-1,i}-r_{n,i}|}{\varepsilon} \bigl\Vert J_{r_{n,i}} ^{A_{i}}u_{n,i} - u_{n-1,i} \bigr\Vert \\& \quad \leq \Vert u_{n,i}-u_{n-1,i}\Vert + \frac{2|r_{n-1,i}-r_{n,i}|}{\varepsilon}M_{1} \\& \quad \leq\bigl\Vert (I-r_{n,i}B_{i}) (y_{n} - y_{n-1})\bigr\Vert +|r_{n,i}-r_{n-1,i}|\Vert B_{i}y_{n-1}\Vert +\frac {2|r_{n-1,i}-r_{n,i}|}{\varepsilon}M_{1} \\& \quad \leq \Vert y_{n} - y_{n-1}\Vert +|r_{n,i}-r_{n-1,i}|\Vert B_{i}y_{n-1} \Vert +\frac {2|r_{n-1,i}-r_{n,i}|}{\varepsilon}M_{1}. \end{aligned}$$

(17)

Let $M_{2} = (\frac{2}{\varepsilon}+1)M_{1}$, and using (17), we have, for $n \geq1$,

$$\begin{aligned} \|v_{n} - v_{n-1}\| &\leq a_{0} \|y_{n}-y_{n-1}\|+ \sum_{i = 1}^{N} a_{i} \bigl\Vert J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1,i} \bigr\Vert \\ & \leq\|y_{n} - y_{n-1}\|+M_{2}\sum _{i = 1}^{N} a_{i}|r_{n,i}-r_{n-1,i}|. \end{aligned}$$

(18)

Using (18), we have, for $n \geq1$,

$$\begin{aligned} \Vert z_{n} - z_{n-1}\Vert \leq&(1- \beta_{n})\Vert x_{n}-x_{n-1}\Vert + | \beta_{n}-\beta_{n-1}| \Vert x_{n-1}\Vert \\ &{}+ \beta_{n}\Vert v_{n}-v_{n-1}\Vert +| \beta_{n}-\beta_{n-1}|\Vert v_{n-1}\Vert \\ \leq&(1-\beta_{n})\Vert x_{n}-x_{n-1}\Vert + |\beta_{n}-\beta_{n-1}| \Vert x_{n-1}\Vert + \beta_{n} \Vert y_{n}-y_{n-1}\Vert \\ &{} + \beta_{n} M_{2}\sum_{i = 1}^{N} a_{i}|r_{n,i}-r_{n-1,i}|+|\beta_{n}- \beta_{n-1}|\Vert v_{n-1}\Vert . \end{aligned}$$

(19)

Noticing that for $n \geq1$,

$$\begin{aligned} \begin{aligned}[b] \|y_{n} - y_{n-1}\| \leq{}& (1- \alpha_{n})\|x_{n}-x_{n-1}\|+ | \alpha_{n}-\alpha_{n-1}| \|x_{n-1}\| \\ &{}+(1-\alpha_{n})\|e_{n}-e_{n-1}\|+| \alpha_{n}-\alpha _{n-1}|\|e_{n-1}\|. \end{aligned} \end{aligned}$$

(20)

Using (19) and (20), we have, for $n \geq1$,

$$\begin{aligned}& \Vert x_{n+1} - x_{n}\Vert \\& \quad \leq\gamma_{n} \eta\bigl\Vert f(x_{n})-f(x_{n-1}) \bigr\Vert + \eta|\gamma_{n}-\gamma_{n-1}|\bigl\Vert f(x_{n-1})\bigr\Vert + \Vert I-\gamma_{n}T\Vert \Vert z_{n} - z_{n-1}\Vert \\& \qquad {}+|\gamma_{n}-\gamma_{n-1}|\Vert Tz_{n-1} \Vert \\& \quad \leq\gamma_{n} \eta k\Vert x_{n}-x_{n-1} \Vert + \eta|\gamma_{n}-\gamma_{n-1}|\bigl\Vert f(x_{n-1})\bigr\Vert + (1-\gamma_{n}\overline{\gamma}) \Vert z_{n} - z_{n-1}\Vert \\& \qquad {}+|\gamma_{n}-\gamma_{n-1}|\Vert Tz_{n-1} \Vert \\& \quad \leq\bigl[(1-\gamma_{n}\overline{\gamma}) (1- \alpha_{n}\beta_{n})+\gamma_{n} \eta k\bigr] \Vert x_{n}-x_{n-1}\Vert +2M_{1}| \gamma_{n}-\gamma_{n-1}| \\& \qquad {}+2M_{1}(1-\gamma_{n}\overline{\gamma})| \beta_{n}-\beta_{n-1}|+(1-\gamma _{n}\overline{ \gamma})M_{2}\beta_{n}\sum_{i = 1}^{N} a_{i}|r_{n,i}-r_{n-1,i}| \\& \qquad {}+(1-\gamma_{n}\overline{\gamma})\beta_{n}| \alpha_{n} - \alpha_{n-1}|\bigl(M_{1}+\Vert e_{n-1}\Vert \bigr)+(1-\gamma_{n}\overline{\gamma})\beta _{n}(1-\alpha_{n})\Vert e_{n}-e_{n-1} \Vert \\& \quad \leq\bigl[1-\gamma_{n}(\overline{\gamma}- \eta k)\bigr]\Vert x_{n}-x_{n-1}\Vert +2M_{1}|\gamma_{n}- \gamma_{n-1}| \\& \qquad {}+2M_{1}(1-\gamma_{n}\overline{\gamma})| \beta_{n}-\beta_{n-1}|+(1-\gamma _{n}\overline{ \gamma})M_{2}\beta_{n}\sum_{i = 1}^{N} a_{i}|r_{n,i}-r_{n-1,i}| \\& \qquad {}+(1-\gamma_{n}\overline{\gamma})\beta_{n}| \alpha_{n} - \alpha_{n-1}|\bigl(M_{1}+\Vert e_{n-1}\Vert \bigr)+(1-\gamma_{n}\overline{\gamma})\beta _{n}(1-\alpha_{n})\Vert e_{n}-e_{n-1} \Vert . \end{aligned}$$

(21)

From the assumptions on $\{\alpha_{n}\}$, $\{\beta_{n}\}$, $\{\gamma_{n}\}$, $\{r_{n,i}\}$, and $\{e_{n}\}$, in view of (21), and Lemma 4, we have $\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0$.

Step 3. Set $W_{n} = [a_{0} I + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})]$, then $W_{n} : C\rightarrow C$ is nonexpansive and $\operatorname{Fix}(W_{n}) = \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$.

It is obvious that $W_{n}$ is nonexpansive and $\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\subset \operatorname{Fix}(W_{n})$. So we are left to show that $\operatorname{Fix}(W_{n}) \subset\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$.

In fact, if $p \in \operatorname{Fix}(W_{n})$, then $W_{n}p = p$. For $\forall q \in \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\subset \operatorname{Fix}(W_{n})$, we have

$$\begin{aligned} \Vert p-q\Vert &\leq a_{0} \Vert p-q\Vert +a_{1} \bigl\Vert J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p - q\bigr\Vert + \cdots+ a_{N}\bigl\Vert J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p - q\bigr\Vert \\ & \leq(a_{0} +a_{1} + \cdots+a_{N-1})\Vert p-q \Vert + a_{N}\bigl\Vert J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p - q\bigr\Vert \\ & = (1-a_{N})\Vert p-q\Vert +a_{N}\bigl\Vert J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p - q\bigr\Vert \\ & \leq \Vert p-q\Vert . \end{aligned}$$

Therefore, $\|p-q\|= (1-a_{N}) \|p-q\|+a_{N} \|J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q\|$, which implies that $\|p-q\|= \|J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q\|$. Similarly, $\|p-q\|= \|J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q\|=\cdots= \|J_{r_{n,N-1}}^{A_{N-1}}(I-r_{n,N-1}B_{N-1})p-q\|$.

Then $\|p-q\|= \|\frac{a_{1}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q)+ \frac{a_{2}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,2}}^{A_{2}}(I-r_{n,2}B_{2})p-q)+ \cdots +\frac{a_{N}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q)\|$, which implies from the strictly convexity of E that $p-q = J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q = J_{r_{n,2}}^{A_{2}}(I-r_{n,2}B_{2})p-q = \cdots= J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p - q$.

Therefore, $J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})p = p$, for $i = 1,2,\ldots,N$. Then $p \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$. Thus $\operatorname{Fix}(W_{n})\subset\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$.

Step 4. $W_{n} y_{n} - y_{n} \rightarrow0$, as $n \rightarrow\infty$, where $W_{n}$ is the same as that in Step 3.

In fact, since both $\{x_{n}\}$ and $\{W_{n}y_{n}\}$ are bounded and $\beta_{n} \rightarrow1$, as $n \rightarrow+\infty$,

$$z_{n} - W_{n} y_{n} = (1-\beta_{n}) (x_{n} - W_{n}y_{n}) \rightarrow0,\quad \mbox{as }n \rightarrow +\infty. $$

Since both $\{f(x_{n})\}$ and $\{Tz_{n}\}$ are bounded and $\gamma_{n} \rightarrow0$, as $n \rightarrow+\infty$,

$$x_{n+1} - z_{n}= \gamma_{n}\bigl[\eta f(x_{n}) - Tz_{n}\bigr] \rightarrow0, \quad \mbox{as }n \rightarrow +\infty. $$

Therefore

$$x_{n} - W_{n}y_{n} = (x_{n}-x_{n+1})+(x_{n+1}-z_{n})+(z_{n}-W_{n}y_{n}) \rightarrow0, $$

as $n \rightarrow+\infty$, in view of the fact of Step 2.

Since $\sum_{n=0}^{\infty}e_{n} <+\infty$ and $\alpha_{n} \rightarrow 0$, as $n \rightarrow\infty$,

$$\begin{aligned} \|W_{n}y_{n}-y_{n}\|&= \bigl\Vert Q_{C}W_{n}y_{n}-Q_{C}\bigl[(1-\alpha _{n}) (x_{n}+e_{n})\bigr]\bigr\Vert \\ & \leq\|W_{n}y_{n}-x_{n}\|+ \alpha_{n}\|x_{n}\| +(1-\alpha_{n}) \|e_{n}\|\rightarrow0, \quad \mbox{as }n \rightarrow\infty. \end{aligned}$$

Moreover, $x_{n+1}-y_{n} \rightarrow0$, as $n \rightarrow\infty$.

Step 5. $\limsup_{n\rightarrow+\infty}\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle\leq0$, where $p_{0} \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$, which is the unique solution of the variational inequality (∗).

Noticing the result of Step 3 and using Lemma 8, we know that there exists $z_{t}$ such that $z_{t} = t\eta f(z_{t})+(I-tT)W_{n}Q_{C}z_{t}$ for $t \in(0,1)$. Moreover, $z_{t} \rightarrow p_{0} \in \operatorname{Fix}(W_{n}) = \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0$, as $t \rightarrow0$. And, $p_{0}$ is the unique solution of the variational inequality (∗).

Since $\|z_{t}\|\leq\|z_{t} - p_{0} \|+\|p_{0}\|$, then $\{z_{t}\}$ is bounded, as $t \rightarrow0$. Using Lemma 3, we have

$$\begin{aligned} \Vert z_{t} - y_{n}\Vert ^{2} =& \Vert z_{t} - W_{n}y_{n}+W_{n}y_{n} - y_{n}\Vert ^{2} \\ \leq&\Vert z_{t} - W_{n}y_{n}\Vert ^{2} + 2\bigl\langle W_{n}y_{n} - y_{n}, J(z_{t} - y_{n}) \bigr\rangle \\ =& \bigl\Vert t\eta f(z_{t}) + (I-tT)W_{n}Q_{C}z_{t} - W_{n}y_{n}\bigr\Vert ^{2}+ 2\bigl\langle W_{n}y_{n} - y_{n}, J(z_{t} - y_{n}) \bigr\rangle \\ \leq&\Vert W_{n}Q_{C}z_{t} - W_{n}y_{n}\Vert ^{2} + 2t\bigl\langle \eta f(z_{t}) - TW_{n}Q_{C}z_{t}, J(z_{t} - W_{n}y_{n}) \bigr\rangle \\ &{}+ 2\bigl\langle W_{n}y_{n} - y_{n}, J(z_{t} - y_{n}) \bigr\rangle \\ \leq&\Vert z_{t} - y_{n}\Vert ^{2} + 2t \bigl\langle \eta f(z_{t}) - TW_{n}Q_{C}z_{t}, J(z_{t} - W_{n}y_{n}) \bigr\rangle \\ &{}+ 2 \Vert W_{n}y_{n} - y_{n}\Vert \Vert z_{t} - y_{n}\Vert , \end{aligned}$$

which implies that

$$ t\bigl\langle TW_{n}Q_{C}z_{t}-\eta f(z_{t}), J(z_{t} - W_{n}y_{n}) \bigr\rangle \leq\| W_{n}y_{n} - y_{n}\| \|z_{t}-y_{n}\|. $$

So, $\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\langle TW_{n}Q_{C}z_{t}-\eta f(z_{t}), J(z_{t} - W_{n}y_{n}) \rangle\leq0$ in view of Step 4.

Since $z_{t} \rightarrow p_{0}$, $W_{n} Q_{C}z_{t} \rightarrow W_{n} Q_{C}p_{0} = p_{0}$, as $t \rightarrow0$ in view of Step 3. Noticing the fact that

$$\begin{aligned}& \bigl\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\bigr\rangle \\& \quad = \bigl\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n}) - J(z_{t} -W_{n}y_{n})\bigr\rangle + \bigl\langle Tp_{0}- \eta f(p_{0}), J(z_{t} -W_{n}y_{n}) \bigr\rangle \\& \quad = \bigl\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n}) - J(z_{t} -W_{n}y_{n})\bigr\rangle \\& \qquad {} + \bigl\langle Tp_{0}-\eta f(p_{0})-TW_{n}Q_{C}z_{t}+ \eta f(z_{t}), J(z_{t} -W_{n}y_{n}) \bigr\rangle \\& \qquad {} + \bigl\langle TW_{n}Q_{C}z_{t}-\eta f(z_{t}), J(z_{t} -W_{n}y_{n})\bigr\rangle , \end{aligned}$$

we have $\limsup_{n\rightarrow+\infty}\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\rangle\leq0$.

Since $\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})\rangle= \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})-J(p_{0} - W_{n}y_{n})\rangle + \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\rangle$ and $x_{n+1}-W_{n}y_{n} \rightarrow0$, then $\limsup_{n \rightarrow \infty} \langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0}) \rangle\leq0$.

Step 6. $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$, where $p_{0} \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0 \subset C$ is the same as that in Step 5.

Let $M_{3} = \sup\{\|(1-\alpha_{n})(x_{n}+e_{n})-p_{0}\|: n \geq0\}$. By using Lemma 3 again, we have

$$ \|y_{n} - p_{0}\|^{2} \leq(1- \alpha_{n})^{2} \|x_{n} - p_{0} \|^{2} +2 \bigl\langle (1-\alpha_{n})e_{n}- \alpha_{n}p_{0}, J\bigl[(1-\alpha_{n}) (x_{n}+e_{n})-p_{0}\bigr]\bigr\rangle . $$

(22)

Using (22) and the result of Step 3, we have

$$\begin{aligned} \|z_{n} - p_{0}\|^{2} \leq&(1- \beta_{n})\|x_{n}-p_{0}\|^{2}+ \beta_{n} \|W_{n}y_{n}-W_{n}p_{0} \|^{2} \\ \leq&(1-\beta_{n})\|x_{n}-p_{0} \|^{2}+\beta_{n} \|y_{n}-p_{0} \|^{2} \\ \leq&(1-\alpha_{n}\beta_{n})\|x_{n}-p_{0} \|^{2} \\ &{}+2\beta_{n} \bigl\langle (1-\alpha_{n})e_{n}- \alpha_{n}p_{0}, J\bigl[(1-\alpha_{n}) (x_{n}+e_{n})-p_{0}\bigr]\bigr\rangle . \end{aligned}$$

(23)

Using (23) and Lemma 3, we have, for $n \geq0$,

$$\begin{aligned}& \Vert x_{n+1} - p_{0}\Vert ^{2} \\& \quad = \bigl\Vert \gamma_{n}\bigl(\eta f(x_{n})-Tp_{0} \bigr)+(I- \gamma_{n} T) (z_{n} - p_{0})\bigr\Vert ^{2} \\& \quad \leq(1-\gamma_{n} \overline{\gamma})^{2}\Vert z_{n}-p_{0}\Vert ^{2}+2\gamma_{n} \bigl\langle \eta f(x_{n})-Tp_{0}, J(x_{n+1}-p_{0}) \bigr\rangle \\& \quad \leq(1-\gamma_{n} \overline{\gamma})^{2}(1- \alpha_{n}\beta_{n})\Vert x_{n}-p_{0} \Vert ^{2}+2\gamma_{n}\eta \bigl\langle f(x_{n})-f(p_{0}), J(x_{n+1}-p_{0})-J(x_{n}-p_{0}) \bigr\rangle \\& \qquad {}+2\gamma_{n} \eta\bigl\langle f(x_{n})- f(p_{0}), J(x_{n}-p_{0})\bigr\rangle +2 \gamma_{n}\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle \\& \qquad {}+ 2(1-\gamma_{n} \overline{\gamma})^{2} \beta_{n} (1-\alpha_{n})\bigl\langle e_{n}, J \bigl[(1-\alpha_{n}) (x_{n}+e_{n})-p_{0} \bigr]\bigr\rangle \\& \qquad {}- 2\alpha_{n}\beta_{n} (1-\gamma_{n} \overline{\gamma})^{2} \bigl\langle p_{0}, J\bigl[(1- \alpha_{n}) (x_{n}+e_{n})-p_{0}\bigr] \bigr\rangle \\& \quad \leq\bigl[1-\gamma_{n} (\overline{\gamma}-2\eta k )\bigr] \Vert x_{n}-p_{0}\Vert ^{2}+2M_{3} \bigl[\Vert e_{n}\Vert + \alpha_{n} \beta_{n}(1-\gamma_{n} \overline{\gamma})\Vert p_{0}\Vert \bigr] \\& \qquad {}+\gamma_{n} \bigl[2\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle +2\eta \Vert x_{n}-p_{0}\Vert \Vert x_{n+1}-x_{n} \Vert \bigr]. \end{aligned}$$

(24)

Let $\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)$, $\delta_{n}^{(2)} = \gamma_{n}[2\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle+2\eta\|x_{n}-p_{0}\|\|x_{n+1}-x_{n}\|]$, $\delta_{n}^{(3)} = 2M_{3}[\|e_{n}\|+ \alpha_{n} \beta_{n}(1-\gamma_{n} \overline{\gamma})\|p_{0}\|]$. Then (24) can be simplified as $\|x_{n+1}-p_{0}\|^{2} \leq(1-\delta_{n}^{(1)})\|x_{n}-p_{0}\|^{2} + \delta_{n}^{(2)}+\delta_{n}^{(3)}$.

Using the assumptions (ii) and (iv), the results of Steps 1, 2, and 5 and by using Lemma 4, we know that $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$.

This completes the proof. □

Remark 11

The assumption that ‘the α-inversely strongly accretive operator $B_{i} : C \rightarrow E$ satisfies for $\forall r > 0$ and $i = 1,2,\ldots, N$, $\langle B_{i}x - B_{i}y, J[(I-rB_{i})x - (I-rB_{i})y]\rangle\geq0$’ is valid, and we can find an example in Section 3 (Remark 26).

Lemma 12

(see [11])

Let E be a real q-uniformly smooth Banach space with constant $K_{q}$ and C be a nonempty, closed, and convex subset of E. Let $A: C \rightarrow E$ be an α-inversely strongly accretive operator. Then for $\forall r \leq(\frac{q \alpha}{K_{q}})^{\frac{1}{q-1}}$, $(I-rA)$ is nonexpansive.

Corollary 13

Let E be a real q-uniformly smooth Banach space with constant $K_{q}$ and also be a uniformly convex Banach space. Let C, $Q_{C}$, f, k, η, T, J, $A_{i}$, $a_{m}$ ($m = 0, 1,2,\ldots,N$), $\overline{\gamma}$, $\{e_{n}\}$, $\{\alpha_{n}\}$, $\{\beta_{n}\}$, $\{\gamma_{n}\}$, and $\{r_{n,i}\}$ satisfy the same conditions as those in Theorem  10. Let $B_{i} : C \rightarrow E$ be α-inversely strongly accretive operator, where $i = 1,2,\ldots, N$. Let $\{x_{n}\}$ be generated by the iterative algorithm (A). Suppose further that

• (v) $r_{n,i}\leq(\frac{q \alpha}{K_{q}})^{\frac{1}{q-1}}$, for $n \geq0$ and $i = 1,2,\ldots, N$.

If $\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \neq\emptyset$, then $\{x_{n}\}$ converges strongly to a point $p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0$, which is the unique solution of the variational inequality (∗).

Proof

Lemma 12 ensures that $(I-r_{n,i}B_{i})$ is nonexpansive, for $n \geq0$ and $i = 1,2,\ldots, N$. Then copy the proof of Theorem 10, the result follows.

This completes the proof. □

Corollary 14

If $i \equiv1$, then iterative algorithm (A) becomes the following one:

$$ \begin{cases} x_{0} \in E, \\ y_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})], \quad n\geq0, \\ z_{n}= (1- \beta_{n})x_{n} + \beta_{n} [a_{0}y_{n} + (1-a_{0}) J_{r_{n}}^{A}(y_{n}-r_{n}By_{n})], \quad n\geq0, \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I- \gamma_{n}T)z_{n}, \quad n \geq0. \end{cases} $$

(B)

Let E, C, $Q_{C}$, f, k, η, T, J, $\overline{\gamma}$, $\{e_{n}\}$, $\{\alpha_{n}\}$, $\{\beta_{n}\}$, and $\{\gamma_{n}\}$ satisfy the same conditions as those in Theorem  10. Let $A: C \rightarrow E$ be m-accretive operator and $B: C \rightarrow E$ be α-inversely strongly accretive operator satisfying that

$$\bigl\langle Bx - By, J\bigl[(I-r B)x -(I-rB)y\bigr]\bigr\rangle \geq0, \quad \textit{for }\forall r >0, \forall x,y \in C. $$

Suppose that $0 < a_{0} <1$, $\{r_{n}\} \subset(0,+\infty)$ such that $\sum_{n = 0}^{\infty} |r_{n+1}-r_{n}| <+\infty$ and $r_{n} \geq\varepsilon> 0$ for $n \geq0$.

If $(A+B)^{-1}0 \neq\emptyset$, then $\{x_{n}\}$ generated by the iterative algorithm (B) converges strongly to $p_{0} \in(A+B)^{-1}0$, which is the unique solution of the following variational inequality: for $\forall z \in(A+B)^{-1}0$,

Corollary 15

If $B_{i} \equiv0$, then iterative algorithm (A) becomes the following one for approximating common zeros of finitely many m-accretive operators:

$$ \begin{cases} x_{0} \in E, \\ y_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})], \quad n\geq0, \\ z_{n}= (1- \beta_{n})x_{n} + \beta_{n} (a_{0}y_{n} + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}y_{n}),\quad n\geq0, \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I- \gamma_{n}T)z_{n},\quad n \geq0. \end{cases} $$

(C)

Let E, C, $Q_{C}$, f, k, η, T, J, $\overline{\gamma}$, $a_{m}$ ($m = 1,2, \ldots,N$), $\{e_{n}\}$, $\{\alpha_{n}\}$, $\{\beta_{n}\}$, and $\{\gamma_{n}\}$, $\{r_{n,i}\}$ satisfy the same conditions as those in Theorem  10. Let $A_{i}: C \rightarrow E$ be m-accretive operator, $i = 1,2, \ldots,N$.

If $\bigcap_{i = 1}^{N} A_{i}^{-1}0 \neq\emptyset$, then $\{x_{n}\}$ generated by (C) converges strongly to a point $p_{0} \in\bigcap_{i = 1}^{N} A_{i}^{-1}0$, which is the unique solution of the following variational inequality: for $\forall z \in\bigcap_{i = 1}^{N} A_{i}^{-1}0$,

Corollary 16

If $A_{i} \equiv0$, then iterative algorithm (A) becomes to the following one for approximating common zeros of finitely many α-inversely strongly accretive operators:

$$ \begin{cases} x_{0} \in E, \\ y_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})],\quad n\geq0, \\ z_{n}= (1- \beta_{n})x_{n} + \beta_{n} [a_{0}y_{n} + \sum_{i = 1}^{N} a_{i} (y_{n}-r_{n,i}B_{i} y_{n})],\quad n\geq0, \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I- \gamma_{n}T)z_{n},\quad n \geq0. \end{cases} $$

(D)

Let E, C, $Q_{C}$, f, k, η, T, J, $\overline{\gamma}$, $a_{m}$ ($m = 1,2, \ldots,N$), $\{e_{n}\}$, $\{\alpha_{n}\}$, $\{\beta_{n}\}$, and $\{\gamma_{n}\}$, $\{r_{n,i}\}$ satisfy the same conditions as those in Theorem  10. Let $B_{i}: C \rightarrow E$ be an α-inversely strongly accretive operator satisfying for $\forall r > 0$ and $i = 1,2,\ldots, N$,

$$\bigl\langle B_{i}x - B_{i}y, J\bigl[(I-rB_{i})x - (I-rB_{i})y\bigr]\bigr\rangle \geq0. $$

If $\bigcap_{i = 1}^{N} B_{i}^{-1}0 \neq\emptyset$, then $\{x_{n}\}$ generated by (D) converges strongly to a point $p_{0} \in\bigcap_{i = 1}^{N} B_{i}^{-1}0$, which is the unique solution of the following variational inequality: for $\forall z \in\bigcap_{i = 1}^{N} B_{i}^{-1}0$,

3 Connection with nonlinear capillarity equation

Remark 17

In the next of this paper, we have four purposes: (1) give a new example to show that the assumption that ‘the set of zeros of the sum of an m-accretive operator and an α-inversely strongly monotone operator is nonempty’ is valid; that is, $(A+B)^{-1}0 \neq\emptyset$ is meaningful; (2) set up the relation between the solution of the capillarity equation and the zero of the sum of an m-accretive operator and an α-inversely strongly accretive operator; (3) apply the iterative algorithm studied in Section 2 to approximate the solution of the capillarity equation; (4) set up the relationship between the solution of the capillarity equation and the solution of one kind variational inequality.

Remark 18

In the following, assume $\frac{2N}{N+1} < p < +\infty$, $1 \leq q ,r < +\infty$ if $p \geq N$, and $1 \leq q , r \leq\frac{Np}{N-p}$ if $p < N$, where $N \geq1$. $\|\cdot\|_{p}$ denotes the norm in $L^{p}(\Omega)$. Let $\frac{1}{p} + \frac{1}{p'} = 1$.

We shall examine the following capillarity equation, which is a special case in [20]:

$$ \begin{cases} -\operatorname{div}[(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u]+\lambda(|u|^{q-2}u + |u|^{r-2}u)+ u(x) = 0,\quad \text{a.e. in } \Omega, \\ -\langle\vartheta,(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u\rangle = 0,\quad \text{a.e. on } \Gamma, \end{cases} $$

(E)

where Ω is a bounded conical domain of a Euclidean space $R^{N}$ with its boundary $\Gamma\in C^{1}$ (cf. [21]). $|\cdot|$ denotes the Euclidean norm in $R^{N}$, $\langle \cdot,\cdot\rangle$ the Euclidean inner-product and ϑ the exterior normal derivative of Γ. λ is a nonnegative constant.

Theorem 19

(see [20])

The capillarity equation (E) has a unique solution $u(x) \in L^{p}(\Omega)$.

Lemma 20

(see [20])

Define the mapping $B_{p,q,r}:W^{1,p}(\Omega)\rightarrow(W^{1,p}(\Omega))^{*}$ by

$$\begin{aligned} \langle v,B_{p,q,r}u \rangle =& \int_{\Omega}\biggl\langle \biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}}\biggr)|\nabla u|^{p-2}\nabla u , \nabla v \biggr\rangle \, dx \\ &{}+ \lambda\int_{\Omega}\bigl\vert u(x)\bigr\vert ^{q-2}u(x) v(x)\, dx \\ &{}+ \lambda \int_{\Omega}\bigl\vert u(x)\bigr\vert ^{r-2}u(x) v(x)\, dx, \end{aligned}$$

for any $u,v \in W^{1,p}(\Omega)$. Then $B_{p,q,r}$ is everywhere defined, strictly monotone, hemi-continuous and coercive.

Lemma 21

(see [20])

Define a mapping $A:L^{p}(\Omega)\rightarrow2^{L^{p}(\Omega)}$ as follows:

$$D(A) = \bigl\{ u\in L^{p}(\Omega)|\textit{there exists an } f \in L^{p}(\Omega),\textit{such that } f \in B_{p,q,r}u\bigr\} . $$

For $u \in D(A)$, let $Au = \{ f \in L^{p}(\Omega)|f \in B_{p,q,r}u\}$.

Then $A:L^{p}(\Omega)\rightarrow2^{L^{p}(\Omega)}$ is m-accretive.

Lemma 22

Define a mapping $C :L^{p}(\Omega)\rightarrow L^{p}(\Omega)$ by $C u = u(x)$, for $\forall u(x) \in L^{p}(\Omega)$.

Then C is 1-inversely strongly accretive.

Proof

Let $J_{p}:L^{p}(\Omega)\rightarrow L^{p'}(\Omega)$ denote the normalized duality mapping. Then it is easy to check that $J_{p}u = |u|^{p-1} \operatorname{sgn} u \|u\|^{2-p}_{p}$, $\forall u\in L^{p}(\Omega)$.

Thus for $\forall u(x), v(x)\in L^{p}(\Omega)$, $\langle Cu - Cv, J_{p}(u-v)\rangle= \int_{\Omega}|u - v|^{p} \|u-v\|^{2-p}_{p}\, dx = \|u-v\|^{2}_{p}$, which implies that C is 1-inversely strongly accretive.

This completes the proof. □

Theorem 23

$u(x) \in L^{p}(\Omega)$ is the unique solution of (E) if and only if $u(x) \in(A+C)^{-1}0$.

Proof

If $u(x)$ is the solution of (E), then

$$-\operatorname{div}\biggl[\biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}}\biggr)|\nabla u|^{p-2}\nabla u\biggr]+\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x) = 0, \quad \mbox{a.e. in }\Omega. $$

Thus for $\forall\varphi\in C_{0}^{\infty}(\Omega)$, by using the property of generalized functions, we have

$$\begin{aligned} 0 =& \biggl\langle \varphi, -\operatorname{div}\biggl[\biggl(1+\frac{|\nabla u|^{p}}{\sqrt {1+|\nabla u|^{2p}}} \biggr)|\nabla u|^{p-2}\nabla u\biggr]+\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x)\biggr\rangle \\ =& \int_{\Omega}-\operatorname{div}\biggl[\biggl(1+ \frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}}\biggr)|\nabla u|^{p-2}\nabla u\biggr] \varphi \,dx \\ &{}+ \int_{\Omega}\bigl[\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x)\bigr]\varphi \,dx \\ =& \int_{\Omega}\biggl\langle \biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}} \biggr)|\nabla u|^{p-2}\nabla u, \nabla\varphi\biggr\rangle \,dx \\ &{}+ \int _{\Omega}\bigl[\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x)\bigr]\varphi \,dx \\ =& \langle\varphi,B_{p,q,r}u + Cu\rangle= \langle\varphi, Au + Cu \rangle. \end{aligned}$$

Then $u(x) \in(A + C)^{-1}0$.

On the other hand, if $u(x) \in(A+C)^{-1}0$, then for $\forall \varphi\in C_{0}^{\infty}(\Omega)$,

$$\begin{aligned} 0 = &\langle\varphi, Au+ Cu\rangle= \langle\varphi, B_{p,q,r}u + Cu \rangle \\ =& \int_{\Omega}\biggl\langle \biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}} \biggr)|\nabla u|^{p-2}\nabla u , \nabla\varphi \biggr\rangle \,dx \\ &{}+ \lambda\int_{\Omega}\bigl(|u|^{q-2}u+|u|^{r-2}u \bigr) \varphi \,dx + \int_{\Omega} u(x) \varphi \,dx \\ =& \biggl\langle \varphi, -\operatorname{div}\biggl[\biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}} \biggr)|\nabla u|^{p-2}\nabla u\biggr] \\ &{}+\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x)\biggr\rangle . \end{aligned}$$

Then $-\operatorname{div}[(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u]+\lambda(|u|^{q-2}u + |u|^{r-2}u)+ u(x) = 0$, a.e. $x\in\Omega$. By using the Green’s formula, we know that for any $v \in W^{1,p}(\Omega)$,

$$\begin{aligned}& \int_{\Gamma}\biggl\langle \vartheta,\biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}} \biggr)|\nabla u|^{p-2}\nabla u \biggr\rangle v|_{\Gamma}\, d \Gamma(x) \\& \quad = \int_{\Omega}\operatorname{div}\biggl[\biggl(1+ \frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}}\biggr)|\nabla u|^{p-2}\nabla u \biggr]v \,dx \\& \qquad {}+ \int _{\Omega} \biggl\langle \biggl(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}}\biggr)|\nabla u|^{p-2}\nabla u ,\nabla v\biggr\rangle \,dx \\& \quad = \int_{\Omega}\bigl[\lambda\bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)+ u(x)\bigr] \,dx + \langle v, B_{p,q,r}u\rangle \\& \qquad {}- \int_{\Omega}\lambda \bigl(|u|^{q-2}u + |u|^{r-2}u\bigr)\,dx \\& \quad = \langle v, Au+Cu\rangle= 0. \end{aligned}$$

Thus $- \langle \vartheta,(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u \rangle = 0$, a.e. on Γ.

Then $u(x) \in(A+C)^{-1}0$ implies that $u(x)$ is the solution of (E).

This completes the proof. □

Theorem 24

Suppose A and C are the same as those in Lemmas 21 and 22, respectively. Let $T : L^{p}(\Omega)\rightarrow L^{p}(\Omega)$ be any strongly positive linear bounded operator with coefficient $\overline{\gamma}$ and $f: L^{p}(\Omega)\rightarrow L^{p}(\Omega)$ be a contraction with coefficient k. Suppose the following conditions are satisfied:

• (i) $0 < \eta< \frac{\overline{\gamma}}{2k}$, and $0 < a < 1$;

• (ii) $\{e_{n}\}_{n=0}^{\infty} \subset L^{p}(\Omega)$, $\sum_{n=0}^{\infty}\|e_{n}\| < +\infty$;

• (iii) $\{\alpha_{n}\}$, $\{\beta_{n}\}$, and $\{\gamma_{n}\}$ are three sequences in $(0,1)$. $\gamma_{n} \rightarrow0$, $\frac{\gamma_{n-1}}{\gamma_{n}} \rightarrow1$, $\beta_{n} \rightarrow1$, $\alpha_{n} \rightarrow0$, as $n \rightarrow \infty$. $\sum_{n=0}^{\infty}\gamma_{n} = \infty$, $\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty$, $\sum_{n=0}^{\infty}|\beta_{n+1} - \beta_{n}| < +\infty$, $\sum_{n=0}^{\infty}(1-\gamma_{n} \overline{\gamma})\alpha_{n}\beta_{n} < +\infty$;

• (iv) $\{r_{n}\}\subset(0,1)$ such that $\sum_{n=0}^{\infty}|r_{n+1} - r_{n}| < +\infty$, and $1 \geq r_{n} \geq\varepsilon> 0$ for $n \geq0$.

If we construct the following iterative algorithm:

$$ \begin{cases} u_{0}(x) \in L^{p}(\Omega), \\ v_{n}(x) = (1-\alpha_{n})(u_{n}(x)+e_{n}(x)), \\ w_{n}(x)= (1- \beta_{n})u_{n}(x) + \beta_{n} [av_{n}(x) + (1-a) J_{r_{n}}^{A}(v_{n}(x)-r_{n}Cv_{n}(x))], \\ u_{n+1}(x)=\gamma_{n} \eta f(u_{n})+(I-\gamma_{n}T)w_{n}(x), \quad n \geq0, \end{cases} $$

(F)

then $u_{n}(x)$ converges strongly to $u(x)\in(A+C)^{-1}0$, which is the unique solution of the capillarity equation (E) and satisfies the following variational inequality: for $\forall z(x)\in (A+C)^{-1}0$,

$$\bigl\langle (T-\eta f)u(x), J_{p}\bigl(u(x)-z\bigr)\bigr\rangle \leq0. $$

Remark 25

From Theorem 24 we can easily see the relationship among the solution of the capillarity equation, the solution of a variational inequality, and the zero of sum of an m-accretive operator and an α-inversely strongly accretive operator.

Remark 26

Let C be the 1-inversely strong accretive operator defined in Lemma 22, then it is obvious that C satisfies

$$\bigl\langle C x - Cy, J_{p}\bigl[(I-rC)x - (I-rC)y\bigr] \bigr\rangle \geq 0,\quad \mbox{for }1 \geq r >0, x,y \in L^{p}(\Omega). $$

Thus the assumption imposed on $B_{i}$ in Theorem 10 is valid.