# Splitting-midpoint method for zeros of the sum of accretive operator and μ-inversely strongly accretive operator in a q-uniformly smooth Banach space and its applications

## Abstract

Combining the implicit midpoint method and the splitting method, we present a new iterative algorithm with errors to solve the problems of finding zeros of the sum of m-accretive operators and μ-inversely strongly accretive operators in a real q-uniformly smooth and uniformly convex Banach space. We obtain some strong convergence theorems, which demonstrate the relationship between the zero of the sum of m-accretive operator and μ-inversely strongly accretive operator and the solution of one kind variational inequality. Moreover, the applications of the main results on the nonlinear problems with Neumann boundaries and Signorini boundaries are demonstrated.

## 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

$$\lim_{t \rightarrow 0}\frac{\|x+ty\|-\|x\|}{t}$$

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 $$\frac{\rho_{E}(t)}{t}\rightarrow 0$$, as $$t \rightarrow 0$$. 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 a q-uniformly smooth Banach space must be uniformly smooth.

The generalized duality mapping $$J_{q}: E \rightarrow 2^{E^{*}}$$ is defined by

$$J_{q}x : = \bigl\{ f \in E^{*}: \langle x, f\rangle = \|x \|^{q}, \|f\| = \|x\|^{q-1}\bigr\} ,\quad x \in E.$$

In particular, $$J \equiv J_{2}$$ is called the normalized duality mapping and $$J_{q}(x) = \|x\|^{q-2}J(x)$$ for $$x \neq 0$$. If E is reduced to the Hilbert space H, then $$J_{q} \equiv I$$ is the identity mapping. It is well known that J is single-valued and norm-to-norm uniformly continuous on each bounded subset 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. 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.

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

For an operator $$A: D(A) \subset E \rightarrow 2^{E}$$, 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 $$T : E \rightarrow E$$ be a mapping. Then T is said to be

1. (1)

nonexpansive if

$$\|Tx - Ty\| \leq \|x-y\|\quad \mbox{for } \forall x,y \in E;$$
2. (2)

k-Lipschitz if there exists $$k > 0$$ such that

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

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

3. (3)

accretive if for all $$x, y \in E$$, there exists $$j_{q}(x-y) \in J_{q}(x-y)$$ such that

$$\bigl\langle Tx - Ty, j_{q}(x-y)\bigr\rangle \geq 0;$$
4. (4)

μ-inversely strongly accretive if for all $$x, y \in E$$, there exists $$j_{q}(x-y) \in J_{q}(x-y)$$ such that

$$\bigl\langle Tx - Ty, j_{q}(x-y)\bigr\rangle \geq \mu \| Tx - Ty \|^{q}$$

for some $$\mu > 0$$;

5. (5)

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

6. (6)

strongly positive (see [2]) if E is a real smooth Banach space and there exists $$\overline{\gamma} > 0$$ such that

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

in this case,

$$\|aI-bT\| = \sup_{\|x\| \leq 1}\bigl\vert \bigl\langle (aI - bT)x, J(x)\bigr\rangle \bigr\vert ,$$

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 $$\operatorname{Fix}(J_{r}^{A}) = A^{-1}0$$.

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 from many mathematicians, were proposed by Lions and Mercier [3], by Passty [4], and, in a dual form for convex programming, by Han and Lou [5].

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

$$x_{n+1} = (I+r_{n} B)^{-1}(I-r_{n} A)x_{n}, \quad n \geq 0.$$
(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 μ-inversely strongly accretive operator and m-accretive operator defined in the Hilbert space H, respectively. However, most of the existing work is undertaken in the frame of Hilbert spaces, see [39], etc.

Recently, Qin et al., presented the following iterative algorithm in the frame of q-uniformly smooth Banach spaces E in [10]:

$$x_{0} \in E, \quad x_{n+1}= \alpha_{n} f(x_{n}) + \beta_{n} (I+r_{n} B)^{-1}\bigl[(I-r_{n} A)x_{n} +e_{n} \bigr] + \gamma_{n} f_{n}, \quad n \geq 0,$$
(2)

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

On the other hand, there is some excellent work done on approximating fixed points of nonexpansive mappings. For example, in 2006, Marino and Xu presented the following iterative algorithm in the frame of Hilbert spaces in [11], which sets up the relationship between fixed point of a nonexpansive mapping and the solution of one kind variational inequality

$$x_{0} \in C,\quad x_{n+1}= \alpha_{n} \gamma f(x_{n}) + (I-\alpha_{n} A)Tx_{n}, \quad n \geq 0,$$
(3)

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 $$\langle (\gamma f - A)p, z - p\rangle \leq 0$$ for $$\forall z \in \operatorname{Fix}(T)$$ under some conditions.

The implicit midpoint rule (IMR) is one of the powerful numerical methods for solving ordinary differential equations, which is extensively studied recently by Alghamdi et al. They presented the following implicit midpoint rule for approximating fixed point of nonexpansive mapping in a Hilbert space in [12]:

$$x_{0} \in H,\quad x_{n+1}= (1- \alpha_{n})x_{n} + \alpha_{n} T\biggl( \frac{x_{n}+x_{n+1}}{2}\biggr), \quad n \geq 0,$$
(4)

where T is nonexpansive from H to H. If $$\operatorname{Fix}(T) \neq \emptyset$$, then $$\{x_{n}\}$$ converges weakly to $$p_{0} \in \operatorname{Fix}(T)$$, under some conditions.

Inspired by the work in [1012], we shall present the following iterative algorithm with errors in a real q-uniformly smooth and uniformly convex Banach space E:

$$\textstyle\begin{cases} x_{0} \in E, \\ y_{n}= \alpha_{n} x_{n} + (1-\alpha_{n})J_{r_{n}}^{A}[\frac{x_{n}+y_{n}}{2}-r_{n}B(\frac{x_{n}+y_{n}}{2})], \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I-\gamma_{n}T)y_{n}+ e_{n}, \quad n \geq 0, \end{cases}$$
(A)

where $$\{e_{n}\}$$ is the error sequence, $$A: E \rightarrow E$$ is an m-accretive operator and $$B: E \rightarrow E$$ is a μ-inversely strongly accretive operator. $$T : E \rightarrow E$$ is a strongly positive linear bounded operator with coefficient $$\overline{\gamma}$$ and $$f : E \rightarrow E$$ is a contraction with coefficient $$k \in (0,1)$$. $$J_{r_{n}}^{A} = (I + r_{n} A)^{-1}$$. More details of iterative algorithm (A) will be presented in Section 2. Then $$\{x_{n}\}$$ is proved to converge strongly to $$p_{0} \in (A+B)^{-1}0$$, which is also a solution of the following variational inequality: $$\forall z \in (A+B)^{-1}0$$, $$\langle (T- \eta f )p_{0}, J(p_{0}-z) \rangle \leq 0$$. In Section 3, we shall present two examples, one of which is the generalized p-Laplacian problems with Neumann boundaries and the other is Laplacian problems with Signorini boundaries, to demonstrate the applications of the main results in Section 2.

Our main contributions are:

1. (i)

the iterative algorithm is new in the sense that it combines the idea of iterative algorithms (1)-(4);

2. (ii)

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

3. (iii)

the assumption that ‘the normalized duality mapping J is weakly sequentially continuous’ in most of the existing related work is weakened to ‘J is weakly sequentially continuous at zero’;

4. (iv)

a new path convergence theorem for nonexpansive mapping is proved, which extends the corresponding result in [11] from a Hilbert space to a real smooth and uniformly convex Banach space;

5. (v)

compared to the work done in [12], strong convergence theorems are obtained instead of weak convergence theorems;

6. (vi)

compared to the work done in [10], the connection between zeros of the sum of m-accretive operators and μ-inversely strongly accretive operators and the solution of one kind variational inequalities is being set up;

7. (vii)

the applications of the main results on the nonlinear problems with Neumann boundaries and Signorini boundaries are demonstrated, from which we can see the connections among variational inequalities, nonlinear boundary value problems and iterative algorithms.

Next, we list some results we need in the 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 [13])

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 [14])

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 [15])

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

$$a_{n+1}\leq (1-t_{n})a_{n} + b_{n}+c_{n}, \quad \forall n \geq 0,$$

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

### Lemma 5

(see [16])

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

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

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

### Lemma 6

(see [17])

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 [18])

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 $$\|I-\rho T \| \leq 1- \rho \overline{\gamma}$$.

## 2 Strong convergence theorems

### Lemma 8

Let E be a real smooth and uniformly convex Banach space. 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 $$U : E \rightarrow E$$ 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

$$T_{t} x : = t \eta f(x) + (I-t T)Ux,$$
(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 $$t \rightarrow 0$$. That is, $$\lim_{t\rightarrow 0}x_{t} = p_{0} \in \operatorname{Fix}(U)$$. Moreover, $$p_{0}$$ satisfies the following variational inequality: for $$\forall z \in \operatorname{Fix}(U)$$,

$$\bigl\langle (T-\eta f)p_{0}, J(p_{0}-z)\bigr\rangle \leq 0.$$
(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) (Ux-Uy)\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 $$x_{t} = t\eta f(x_{t}) + (I-tT)Ux_{t}$$.

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

For $$p \in \operatorname{Fix}(U)$$, then

\begin{aligned} \Vert x_{t} - p\Vert &= \bigl\Vert (I-tT) (Ux_{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 $$\{TUx_{t}\}$$ are bounded.

Step 3. $$x_{t} - Ux_{t} \rightarrow 0$$, as $$t \rightarrow 0$$.

Noticing the result of Step 2, we have $$\|x_{t} - Ux_{t}\| = t \|\eta f(x_{t})- TUx_{t}\| \rightarrow 0$$, 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 solutions, 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

$$\bigl\langle (T- \eta f)v_{0}, J(v_{0}-u_{0}) \bigr\rangle \leq 0$$
(7)

and

$$\bigl\langle (T- \eta f)u_{0}, J(u_{0}-v_{0}) \bigr\rangle \leq 0.$$
(8)

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

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

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 $$t \rightarrow 0$$, which satisfies the variational inequality (6).

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

\begin{aligned} \|x_{t} - z\|^{2} &\leq \|I-tT\|^{2} \|Ux_{t}-Uz\|^{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 $$t \rightarrow 0^{+}$$, then we can choose $$\{t_{n}\}\subset (0,1)$$ such that $$t_{n} \rightarrow 0^{+}$$ and $$x_{t_{n}}\rightharpoonup p_{0}$$. From Lemma 2 and the result of Step 3, we see that $$p_{0} = Up_{0}$$. 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 $$x_{t} = t\eta f(x_{t}) + (I-tT)Ux_{t}$$, then

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

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

\begin{aligned}& \bigl\langle (T-\eta f)x_{t}, J(x_{t} - z) \bigr\rangle \\& \quad = - \frac{1}{t} \bigl\langle (I-tT) (I-U)x_{t}, J(x_{t}-z)\bigr\rangle \\& \quad = -\frac{1}{t} \bigl\langle (I-U)x_{t}-(I-U)z, J(x_{t}-z)\bigr\rangle +\bigl\langle T(I-U)x_{t}, J(x_{t}-z)\bigr\rangle \\& \quad = -\frac{1}{t} \bigl[\|x_{t} - z\|^{2}- \bigl\langle Ux_{t}-Uz, J(x_{t}-z)\bigr\rangle \bigr]+\bigl\langle T(I-U)x_{t}, J(x_{t}-z)\bigr\rangle \\& \quad \leq \bigl\langle T(I-U)x_{t}, J(x_{t}-z)\bigr\rangle . \end{aligned}
(10)

Since $$x_{t_{n}} \rightarrow p_{0}$$, then $$(I-U)x_{t_{n}} \rightarrow (I-U)p_{0} = 0$$, 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, then taking limits on both sides of (10) we have $$\langle(T-\eta f)p_{0}, J(p_{0} - z)\rangle \leq 0$$ 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. □

### Lemma 9

(see [10])

Let E be a real q-uniformly smooth Banach space with constant $$K_{q}$$. Let $$A: E\rightarrow E$$ be a μ-inversely strongly accretive operator. Then for $$\forall r \leq (\frac{q \mu}{K_{q}})^{\frac{1}{q-1}}$$, $$(I-rA)$$ is nonexpansive.

### Theorem 10

Let E be a real q-uniformly smooth Banach space with constant $$K_{q}$$ and also be a uniformly convex Banach space. 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 $$A: E \rightarrow 2^{E}$$ be an m-accretive operator and $$B: E \rightarrow E$$ be a μ-inversely strongly accretive operator. Let $$\{x_{n}\}$$ be generated by the iterative algorithm (A). Suppose $$\{e_{n}\}\subset E$$, $$\{\alpha_{n}\}$$ and $$\{\gamma_{n}\}$$ are two sequences in $$(0,1)$$ and $$\{r_{n}\}\subset (0,+\infty)$$ satisfying the following conditions:

1. (i)

$$\sum_{n=0}^{\infty}\gamma_{n} = \infty$$, $$\gamma_{n} \rightarrow 0$$, $$\alpha_{n} \rightarrow 0$$, as $$n \rightarrow \infty$$;

2. (ii)

$$\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty$$, $$\sum_{n=0}^{\infty}|\gamma_{n+1} - \gamma_{n}| < +\infty$$;

3. (iii)

$$\sum_{n=0}^{\infty}|r_{n+1} - r_{n}| < +\infty$$, $$0 < \varepsilon \leq r_{n}\leq(\frac{q \mu }{K_{q}})^{\frac{1}{q-1}}$$ for $$n \geq 0$$;

4. (iv)

$$\sum_{n=0}^{\infty} \|e_{n}\| < +\infty$$.

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

$$\bigl\langle (T-\eta f)p_{0}, J(p_{0}-z)\bigr\rangle \leq 0.$$
(11)

### Proof

Let $$u_{n} = (I - r_{n}B)(\frac{x_{n}+y_{n}}{2})$$ for $$n \geq 0$$.

We shall split the proof into six steps.

Step 1. $$\{y_{n}\}$$ is well defined.

Define $$W_{t}: E \rightarrow E$$ by $$W_{t} x: = t u + (1-t)W(\frac{u+x}{2})$$, where $$W : E \rightarrow E$$ is nonexpansive for $$x , u \in E$$, then $$W_{t}$$ is a contraction for $$0 \leq t < 1$$.

In fact,

\begin{aligned} \|W_{t} x - W_{t} y\|&\leq (1-t)\biggl\Vert \frac{u+x}{2}-\frac{u+y}{2}\biggr\Vert \\ &\leq \frac{1-t}{2} \|x-y \|, \end{aligned}

which implies that $$W_{t}$$ is a contraction. Thus there exists $$x_{t}$$ such that $$W_{t} x_{t} = x_{t}$$. That is, $$x_{t} = t u + (1-t)W(\frac{u+x}{2})$$ for $$0 \leq t < 1$$.

From Lemma 9 we know that $$J_{r_{n}}^{A}(I-r_{n} B)$$ is nonexpansive, therefore $$\{y_{n}\}$$ is well defined.

Step 2. $$\{x_{n}\}$$, $$\{u_{n}\}$$, $$\{y_{n}\}$$ are all bounded.

In view of Lemmas 6 and 9, we have, for $$\forall p \in (A+B)^{-1}0$$,

\begin{aligned} \begin{aligned}[b] \|y_{n} - p\|&\leq \alpha_{n} \|x_{n} - p\| + (1-\alpha_{n})\biggl\Vert \frac{x_{n}+y_{n}}{2} - p\biggr\Vert \\ & \leq \alpha_{n}\|x_{n} - p\| + \frac{1-\alpha_{n}}{2} \|x_{n} - p\|+ \frac{1-\alpha_{n}}{2}\|y_{n} - p\|, \end{aligned} \end{aligned}
(12)

which implies that $$\|y_{n} - p\|\leq \|x_{n} -p\|$$.

Using Lemma 7 and (12), we have, for $$p \in (A+B)^{-1}0$$ and $$n \geq 0$$,

\begin{aligned} \|x_{n+1} - p\|&\leq \gamma_{n}\bigl\Vert \eta f(x_{n}) - Tp\bigr\Vert + \bigl\Vert (I-\gamma_{n}T) (y_{n} - p)\bigr\Vert + \|e_{n}\| \\ & \leq \gamma_{n}\eta k \|x_{n}- p\| + \gamma_{n} \bigl\Vert \eta f(p) - Tp\bigr\Vert + (1- \gamma_{n}\overline{\gamma})\|y_{n} - p\|+ \|e_{n} \| \\ & \leq \bigl[1-\gamma_{n} (\overline{\gamma}- k \eta)\bigr] \|x_{n} -p\|+\gamma_{n}(\overline{\gamma}- k \eta) \frac{\|\eta f(p)-Tp\|}{\overline{\gamma}- k \eta}+\|e_{n}\| \\ & \leq \max\biggl\{ \|x_{n} -p\|, \frac{\|\eta f(p)-Tp\|}{\overline{\gamma}- k \eta}\biggr\} + \|e_{n}\|. \end{aligned}
(13)

By using the inductive method, we can easily get the following result from (13):

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

which implies that $$\{x_{n}\}$$ is bounded. Then (12) implies that $$\{y_{n}\}$$ is bounded.

Since $$J_{r_{n}}^{A}$$ and $$(I-r_{n}B)$$ are nonexpansive, f is a contraction and T is bounded, then $$\{u_{n}\}$$, $$\{f(x_{n})\}$$, $$\{J_{r_{n}}^{A}u_{n}\}$$, $$\{B(\frac{x_{n}+y_{n}}{2})\}$$ and $$\{Ty_{n}\}$$ are all bounded.

Set $$M_{1} = \sup\{\|u_{n}\|, \|J^{A}_{r_{n}}u_{n}\|, \|B(\frac{x_{n}+y_{n}}{2})\|, \|x_{n}\|, \eta\|f(x_{n})\|, \|Ty_{n}\|: n \geq 0\}$$.

Step 3. $$\lim_{n \rightarrow \infty} \|x_{n+1} - x_{n} \| = 0$$.

First, we shall discuss $$\|J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A}u_{n-1}\|$$ for $$n \geq 1$$.

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

\begin{aligned}& \bigl\Vert J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A}u_{n-1} \bigr\Vert \\& \quad = \biggl\Vert J_{r_{n-1}}^{A}\biggl( \frac{r_{n-1}}{r_{n}}u_{n} +\biggl(1-\frac{r_{n-1}}{r_{n}} \biggr)J_{r_{n}}^{A}u_{n}\biggr) - J_{r_{n-1}}^{A}u_{n-1}\biggr\Vert \\& \quad \leq \biggl\Vert \frac{r_{n-1}}{r_{n}}u_{n}+\biggl(1- \frac{r_{n-1}}{r_{n}}\biggr)J_{r_{n}} ^{A}u_{n} - u_{n-1}\biggr\Vert \\& \quad \leq \frac{r_{n-1}}{r_{n}}\Vert u_{n}-u_{n-1}\Vert + \biggl(1-\frac{r_{n-1}}{r_{n}}\biggr)\bigl\Vert J_{r_{n}} ^{A}u_{n} - u_{n-1}\bigr\Vert \\& \quad \leq \Vert u_{n}-u_{n-1}\Vert +\frac{r_{n}-r_{n-1}}{\varepsilon} \bigl\Vert J_{r_{n}} ^{A}u_{n} - u_{n-1} \bigr\Vert . \end{aligned}
(14)

If $$r_{n}\leq r_{n-1}$$, then imitating the proof of (14), we have

$$\bigl\Vert J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A}u_{n-1} \bigr\Vert \leq \|u_{n}-u_{n-1}\|+\frac{r_{n-1}-r_{n}}{\varepsilon}\bigl\Vert J_{r_{n}} ^{A}u_{n} - u_{n-1}\bigr\Vert .$$
(15)

Combining (14) and (15), we have, for $$n \geq 1$$,

$$\bigl\Vert J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A}u_{n-1} \bigr\Vert \leq \|u_{n}-u_{n-1}\|+\frac{|r_{n-1}-r_{n}|}{\varepsilon}\bigl\Vert J_{r_{n}} ^{A}u_{n} - u_{n-1}\bigr\Vert .$$
(16)

Note that

\begin{aligned} \|u_{n}-u_{n-1}\|&\leq \biggl\Vert (I-r_{n}B) \biggl(\frac{x_{n}+y_{n}}{2}-\frac{x_{n-1}+y_{n-1}}{2}\biggr)\biggr\Vert + |r_{n-1}-r_{n}|\biggl\Vert B\biggl( \frac{x_{n-1}+y_{n-1}}{2}\biggr)\biggr\Vert \\ & \leq \frac{\|x_{n} - x_{n-1}\|}{2}+ \frac{\|y_{n}-y_{n-1}\|}{2}+ |r_{n-1}-r_{n}| \biggl\Vert B\biggl(\frac{x_{n-1}+y_{n-1}}{2}\biggr)\biggr\Vert . \end{aligned}
(17)

Using (16) and (17), for $$n \geq 1$$, we have

\begin{aligned} \|y_{n} - y_{n-1}\| \leq& \alpha_{n} \|x_{n}-x_{n-1}\|+ |\alpha_{n} - \alpha_{n-1}|\|x_{n-1}\| \\ &{} +(1-\alpha_{n})\bigl\Vert J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A} u_{n-1}\bigr\Vert +|\alpha_{n}-\alpha_{n-1}|\bigl\Vert J_{r_{n-1}}^{A} u_{n-1}\bigr\Vert \\ \leq& \alpha_{n}\|x_{n}-x_{n-1}\|+ | \alpha_{n} - \alpha_{n-1}|\|x_{n-1}\|+(1-\alpha_{n})\|u_{n}-u_{n-1}\| \\ &{}+ \frac{|r_{n}-r_{n-1}|}{\varepsilon}\bigl\Vert J_{r_{n}}^{A}u_{n}-u_{n-1} \bigr\Vert +|\alpha_{n}-\alpha_{n-1}|\bigl\Vert J_{r_{n-1}}^{A} u_{n-1}\bigr\Vert \\ \leq& \frac{1+\alpha_{n}}{2}\|x_{n}-x_{n-1}\|+ \frac{1-\alpha_{n}}{2}\|y_{n}-y_{n-1}\| \\ &{}+2M_{1}| \alpha_{n}-\alpha_{n-1}|+\biggl(1+\frac{2}{\varepsilon} \biggr)M_{1}|r_{n}-r_{n-1}|. \end{aligned}
(18)

From (18), we know that for $$n \geq 1$$,

$$\|y_{n} - y_{n-1}\|\leq \|x_{n}-x_{n-1} \|+ 4M_{1}|\alpha_{n}-\alpha_{n-1}|+2M_{1} \biggl(1+\frac{2}{\varepsilon}\biggr)|r_{n}-r_{n-1}|.$$
(19)

Using (19), we have for $$n \geq 1$$,

\begin{aligned} \|x_{n+1} - x_{n}\| \leq& \gamma_{n} \eta k\|x_{n}-x_{n-1}\|+ \eta|\gamma_{n}-\gamma_{n-1}|\bigl\Vert f(x_{n-1}) \bigr\Vert + (1-\gamma_{n}\overline{\gamma})\|y_{n} - y_{n-1}\| \\ &{}+|\gamma_{n}-\gamma_{n-1}|\|Ty_{n-1}\|+ \|e_{n+1}-e_{n}\| \\ \leq& \bigl[1-\gamma_{n}(\overline{\gamma}-\eta k)\bigr] \|x_{n}-x_{n-1}\| \\ &{}+2M_{1}\biggl[|\gamma_{n}- \gamma_{n-1}|+2|\alpha_{n} -\alpha_{n-1}|+\biggl(1+ \frac{2}{\varepsilon}\biggr)|r_{n}-r_{n-1}|\biggr] \\ &{}+ \|e_{n+1}-e_{n}\|. \end{aligned}
(20)

From the assumptions on $$\{e_{n}\}$$, $$\{\alpha_{n}\}$$, $$\{\gamma_{n}\}$$ and $$\{r_{n}\}$$, in view of (20) and Lemma 4, we have $$\lim_{n \rightarrow \infty} \|x_{n+1} - x_{n} \| = 0$$.

Step 4. Set $$W_{n} = J_{r_{n}}^{A}(I-r_{n}B)$$, then $$W_{n} y_{n} - y_{n} \rightarrow 0$$, as $$n \rightarrow \infty$$.

It is obvious that $$W_{n}$$ is nonexpansive and $$(A+B)^{-1}0 = \operatorname{Fix}(W_{n})$$.

Since both $$\{x_{n}\}$$ and $$\{W_{n}(\frac{x_{n}+y_{n}}{2})\}$$ are bounded and $$\alpha_{n} \rightarrow 0$$, as $$n \rightarrow +\infty$$, then

$$y_{n} - W_{n}\biggl(\frac{x_{n}+y_{n}}{2}\biggr) = \alpha_{n}\biggl[x_{n} - W_{n}\biggl( \frac{x_{n}+y_{n}}{2}\biggr)\biggr] \rightarrow 0, \quad \mbox{as } n \rightarrow + \infty.$$

Since both $$\{f(x_{n})\}$$ and $$\{Ty_{n}\}$$ are bounded and $$\gamma_{n} \rightarrow 0$$, as $$n \rightarrow +\infty$$, then

$$x_{n+1} - y_{n} = \gamma_{n}\bigl[\eta f(x_{n}) - Ty_{n}\bigr] + e_{n} \rightarrow 0, \quad \mbox{as } n \rightarrow +\infty.$$

Therefore, in view of the result of Step 3,

\begin{aligned} \|W_{n}y_{n}-y_{n}\|&\leq \biggl\Vert W_{n}y_{n}-W_{n}\biggl(\frac{x_{n}+y_{n}}{2}\biggr) \biggr\Vert +\biggl\Vert W_{n}\biggl(\frac{x_{n}+y_{n}}{2} \biggr)-y_{n}\biggr\Vert \\ &\leq \frac{\|x_{n}-y_{n}\|}{2}+\alpha_{n}\biggl\Vert x_{n}-W_{n} \biggl(\frac{x_{n}+y_{n}}{2}\biggr)\biggr\Vert \rightarrow 0, \quad \mbox{as } n \rightarrow \infty. \end{aligned}

Step 5. $$\limsup_{n\rightarrow +\infty}\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle \leq 0$$, where $$p_{0} \in (A+B)^{-1}0$$, which is the unique solution of the variational inequality (11).

Since $$W_{n}$$ is nonexpansive, then Lemma 8 implies that there exists $$z_{t}$$ such that $$z_{t} = t\eta f(z_{t})+(I-tT)W_{n}z_{t}$$ for $$t \in (0,1)$$. Moreover, $$z_{t} \rightarrow p_{0} \in \operatorname{Fix}(W_{n}) = (A+B)^{-1}0$$, as $$t \rightarrow 0$$; and $$p_{0}$$ is the unique solution of the variational inequality (11).

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

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

which implies that

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

So, $$\lim_{t \rightarrow 0}\limsup_{n\rightarrow +\infty}\langle TW_{n}z_{t}-\eta f(z_{t}), J(z_{t} - W_{n}y_{n}) \rangle \leq 0$$ in view of Step 4.

Since $$z_{t} \rightarrow p_{0}$$, then $$W_{n}z_{t} \rightarrow W_{n}p_{0} = p_{0}$$, as $$t \rightarrow 0$$. Noticing the following 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}z_{t}+ \eta f(z_{t}), J(z_{t} -W_{n}y_{n})\bigr\rangle \\& \qquad {} + \bigl\langle TW_{n}z_{t}-\eta f(z_{t}), J(z_{t} -W_{n}y_{n})\bigr\rangle , \end{aligned}

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

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

Step 6. $$x_{n} \rightarrow p_{0}$$, as $$n \rightarrow +\infty$$, where $$p_{0} \in (A+B)^{-1}0$$ is the same as that in Step 5.

Since

$$\|y_{n} - p_{0}\| \leq \alpha_{n} \|x_{n} - p_{0}\| + (1-\alpha_{n})\biggl\Vert \frac{x_{n} + y_{n}}{2}-p_{0}\biggr\Vert \leq \frac{1+\alpha_{n}}{2} \|x_{n} - p_{0}\| + \frac{1-\alpha_{n}}{2}\|y_{n}-p_{0} \|,$$

then $$\|y_{n} - p_{0}\| \leq \|x_{n} -p_{0}\|$$.

Using Lemma 3 and letting $$M_{2} = \max\{M_{1}, \|p_{0}\|\}$$, we have for $$n \geq 0$$,

\begin{aligned}& \|x_{n+1} - p_{0}\|^{2} \\& \quad = \bigl\Vert \gamma_{n}\bigl(\eta f(x_{n})-Tp_{0} \bigr)+(I- \gamma_{n} T) (y_{n} - p_{0})+e_{n} \bigr\Vert ^{2} \\& \quad \leq (1-\gamma_{n} \overline{\gamma})^{2} \|y_{n}-p_{0}\|^{2}+2\gamma_{n} \bigl\langle \eta f(x_{n})-Tp_{0}, J(x_{n+1}-p_{0}) \bigr\rangle \\& \qquad {}+2\bigl\langle e_{n}, J(x_{n+1}-p_{0}) \bigr\rangle \\& \quad \leq (1-\gamma_{n} \overline{\gamma})\|x_{n}-p_{0} \|^{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 \\& \qquad {}+2 \gamma_{n}\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle +2\|e_{n}\| \|x_{n+1}-p_{0}\| \\& \quad \leq \bigl[1-\gamma_{n} (\overline{\gamma}-2\eta k )\bigr] \|x_{n}-p_{0}\|^{2} \\& \qquad {} +\gamma_{n} \bigl[2\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle +2\eta \|x_{n}-p_{0} \|\|x_{n+1}-x_{n}\|\bigr]+4M_{2}\|e_{n}\|. \end{aligned}
(21)

Let $$\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)$$, $$\delta_{n}^{(2)} = \gamma_{n}[2\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle +2\eta \|x_{n}-p_{0}\|\|x_{n+1}-x_{n}\|]$$ and $$\delta_{n}^{(3)}= 4M_{2} \|e_{n}\|$$. Then (21) 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, the results of Steps 2, 3 and 5 and by using Lemma 4, we know that $$x_{n} \rightarrow p_{0}$$, as $$n \rightarrow +\infty$$.

This completes the proof. □

### Theorem 11

If $$e_{n} \equiv 0$$, then iterative algorithm (A) becomes the following accurate iterative algorithm:

$$\textstyle\begin{cases} x_{0} \in E, \\ y_{n} = \alpha_{n} x_{n} + \beta_{n} J_{r_{n}}^{A}[\frac{x_{n}+y_{n}}{2}-r_{n}B(\frac{x_{n}+y_{n}}{2})], \quad n\geq 0, \\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I- \gamma_{n}T)y_{n}, \quad n \geq 0. \end{cases}$$
(B)

If $$(A+B)^{-1}0 \neq \emptyset$$, then under the assumptions except on $$\{e_{n}\}$$ of Theorem  10, $$\{x_{n}\}$$ generated by the iterative algorithm (B) converges strongly to $$p_{0} \in (A+B)^{-1}0$$, which is the unique solution of the variational inequality (11).

## 3 Applications

In this section, we shall demonstrate the applications of Theorem 10 to the nonlinear problems with Neumann boundaries and Signorini boundaries, respectively.

### Example 1

Now, we shall present an example of nonlinear Neumann boundary value problem involving the generalized p-Laplacian, which comes from [19]:

$$\textstyle\begin{cases} -\operatorname{div}[(C(x)+|\nabla u|^{2})^{\frac{p-2}{2}}\nabla u] + \varepsilon |u|^{q-2}u+g(x,u(x),\nabla u(x))= h(x), \quad \mbox{a.e. }x \in\Omega, \\ - \langle\vartheta,(C(x)+|\nabla u|^{2})^{\frac{p-2}{2}}\nabla u \rangle \in \beta_{x}(u(x)), \quad \mbox{a.e. }x\in\Gamma. \end{cases}$$
(C)

In (C), Ω is a bounded conical domain of a Euclidean space $$R^{N}$$ with its boundary $$\Gamma\in C^{1}$$ (see [20]). $$h(x)\in L^{2}(\Omega)$$ is a given function. ε is a nonnegative constant and ϑ denotes the exterior normal derivative of Γ, $$0 \leq C(x) \in L^{p}(\Omega)$$.

Let $$\varphi:\Gamma\times R\rightarrow R$$ be a given function such that, for each $$x\in\Gamma$$, $$\varphi_{x}= \varphi(x,\cdot):R\rightarrow R$$ is a proper, convex and lower-semi-continuous function with $$\varphi_{x}(0)=0$$. Let $$\beta_{x}$$ be the subdifferential of $$\varphi_{x}$$, i.e., $$\beta_{x}\equiv\partial\varphi_{x}$$. Suppose that $$0\in \beta_{x}(0)$$ and for each $$t\in R$$, the function $$x\in\Gamma \rightarrow (I+\lambda \beta_{x})^{-1}(t)\in R$$ is measurable for $$\lambda >0$$.

Suppose that $$g:\Omega \times R^{N+1} \rightarrow R$$ is a given function satisfying the following conditions:

1. (a)

Carathéodory’s conditions:

\begin{aligned}& x \rightarrow g(x,r)\quad \mbox{is measurable on }\Omega \mbox{ for all } r \in R^{N+1}; \\& r \rightarrow g(x,r)\quad \mbox{is continuous on }R^{N+1}\mbox{ for almost all } x\in \Omega. \end{aligned}
2. (b)

Nonexpansive with respect to $$r_{1}$$, i.e.,

$$\bigl\vert g(x,r_{1},\ldots,r_{N+1})-g(x,t_{1}, \ldots,t_{N+1})\bigr\vert \leq |r_{1}-t_{1}|,$$

where $$(r_{1}, r_{2}, \ldots, r_{N+1}), (t_{1},\ldots,t_{N+1})\in R^{N+1}$$.

3. (c)

Monotone with respect to $$r_{1}$$, i.e.,

$$\bigl(g(x,r_{1},\ldots,r_{N+1})-g(x,t_{1}, \ldots,t_{N+1})\bigr) (r_{1} - t_{1}) \geq 0$$

for all $$x \in \Omega$$ and $$(r_{1},\ldots,r_{N+1}),(t_{1},\ldots,t_{N+1})\in R^{N+1}$$.

Assume $$\frac{2N}{N+1} < p <+\infty$$, $$\frac{2N}{N+1}< q<+\infty$$, where $$N \geq 1$$. Let $$\frac{1}{p}+\frac{1}{p'} = 1$$. We use $$\| \cdot \|_{2}$$ to denote the norm of $$L^{2}(\Omega)$$ and $$\langle\cdot,\cdot\rangle$$ to denote the inner product in $$R^{N}$$, respectively.

### Lemma 12

([19])

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

$$\langle v,B_{p,q}u\rangle = \int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u , \nabla v \bigr\rangle \, dx + \varepsilon \int_{\Omega}\bigl\vert u(x)\bigr\vert ^{q-2}u(x) v(x) \, dx$$

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

### Lemma 13

([19])

The mapping $$\Phi_{p}:W^{1,p}(\Omega)\rightarrow R$$ defined by $$\Phi_{p}(u)= \int_{\Gamma}\varphi_{x} (u|_{\Gamma}(x))\, d\Gamma(x)$$ for any $$u\in W^{1,p}(\Omega)$$ is proper convex and lower-semi-continuous on $$W^{1,p}(\Omega)$$.

### Lemma 14

([19])

The mapping $$A : L^{2}(\Omega) \rightarrow 2^{L^{2}(\Omega)}$$ defined by

$$A u = \bigl\{ f \in L^{2}(\Omega): f \in B_{p,q} u + \partial \Phi_{p}(u)\bigr\} \quad \textit{for } \forall u\in D(A),$$

where

$$D(A) = \bigl\{ u \in L^{2}(\Omega): \textit{there exists } f \in L^{2}(\Omega) \textit{ such that } f \in B_{p,q} u + \partial \Phi_{p}(u)\bigr\}$$

is m-accretive.

### Lemma 15

Define $$S: D(A) \subset L^{2}(\Omega)\rightarrow L^{2}(\Omega)$$ by

$$Su(x) = g\bigl(x,u(x), \nabla u(x)\bigr)- h(x)$$

for $$u(x) \in D(S)$$, where $$h(x)$$ is the same as that in (C). Then S is inversely strongly accretive.

### Proof

From assumptions (c) and (b) on g, we know that if $$r_{1} \leq t_{1}$$, then

\begin{aligned} g(x,t_{1},\ldots, t_{N+1}) - g(x,r_{1}, \ldots, r_{N+1})&= \bigl\vert g(x,r_{1},\ldots, r_{N+1}) - g(x,t_{1}, \ldots, t_{N+1})\bigr\vert \\ &\leq |r_{1} - t_{1}| = t_{1} - r_{1}; \end{aligned}

if $$r_{1} \geq t_{1}$$, then

\begin{aligned} g(x,r_{1},\ldots, r_{N+1}) - g(x,t_{1}, \ldots, t_{N+1})&= \bigl\vert g(x,r_{1},\ldots, r_{N+1}) - g(x,t_{1}, \ldots, t_{N+1})\bigr\vert \\ &\leq |r_{1} - t_{1}| = r_{1} - t_{1}. \end{aligned}

Thus

$$\bigl[g(x,r_{1},\ldots, r_{N+1}) - g(x,t_{1}, \ldots, t_{N+1})\bigr]^{2}\leq \bigl[g(x,r_{1}, \ldots, r_{N+1}) - g(x,t_{1}, \ldots, t_{N+1}) \bigr](r_{1}-t_{1}),$$

which implies that

\begin{aligned} \langle Su - Sv, u - v\rangle &= \int_{\Omega}\bigl[g \bigl(x,u(x), \nabla u(x)\bigr)-g\bigl(x,v(x), \nabla v(x)\bigr)\bigr](u - v)\, dx \\ &\geq \int_{\Omega}\bigl[g\bigl(x,u(x), \nabla u(x)\bigr)-g \bigl(x,v(x), \nabla v(x)\bigr)\bigr]^{2} \, dx = \|Su - Sv \|_{2}^{2}. \end{aligned}

Then S is inversely strongly monotone.

This completes the proof. □

### Lemma 16

([19])

For $$h(x) \in L^{2}(\Omega)$$, nonlinear boundary value problem (C) has a unique solution in $$L^{2}(\Omega)$$.

### Lemma 17

([20])

For $$\forall \varphi \in C_{0}^{\infty}(\Omega)$$, $$\langle \varphi, \partial \Phi_{p}(u)\rangle = 0$$, $$u \in W^{1,p}(\Omega)$$.

### Lemma 18

$$u(x)\in L^{2}(\Omega)$$ is the solution of (C) if and only if $$u(x) \in (A+S)^{-1}0$$.

### Proof

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

$$-\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{p-2}{2}}\nabla u\bigr] + \varepsilon |u|^{q-2}u+g \bigl(x,u(x),\nabla u(x)\bigr)= h(x),\quad \mbox{a.e. }x \in \Omega.$$

Thus, for $$\forall \varphi \in C_{0}^{\infty}(\Omega)$$, by using the property of generalized function and Lemma 17, we have

\begin{aligned} 0 =& \int_{\Omega}-\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u\bigr] \varphi\, dx + \varepsilon \int_{\Omega}|u|^{q-2}u \varphi \, dx \\ &{}+ \int _{\Omega}g\bigl(x,u(x),\nabla u(x)\bigr)\varphi \, dx - \int _{\Omega}h \varphi \, dx \\ = &\int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u, \nabla \varphi\bigr\rangle \, dx+ \int_{\Omega}\varepsilon |u|^{q-2}u \varphi\, dx \\ &{}+ \int _{\Omega}g\bigl(x,u(x),\nabla u(x)\bigr)\varphi \, dx - \int _{\Omega}h \varphi\, dx \\ =& \langle \varphi,B_{p,q}u + Su\rangle = \bigl\langle \varphi, B_{p,q}u +\partial \Phi_{p}(u)+ Su \bigr\rangle = \langle \varphi, Au + Su\rangle. \end{aligned}

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

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

\begin{aligned} 0 =& \langle \varphi, Au+ Su\rangle = \bigl\langle \varphi, B_{p,q}u + \partial \Phi_{p}(u)+ Su\bigr\rangle \\ =& \int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{p-2}{2}}\nabla u, \nabla \varphi \bigr\rangle \,dx+\varepsilon \int _{\Omega}|u|^{q-2}u \varphi \,dx \\ &{}+ \int _{\Omega}g\bigl(x,u(x),\nabla u(x)\bigr)\varphi \,dx - \int _{\Omega}h \varphi \,dx \\ = &\int_{\Omega}\bigl\{ -\operatorname{div}\bigl[\bigl(C(x)+| \nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u\bigr]+ \varepsilon |u|^{q-2}u + g\bigl(x,u(x),\nabla u(x)\bigr) - h \bigr\} \varphi \,dx. \end{aligned}

Therefore, $$-\operatorname{div}[(C(x)+|\nabla u|^{2})^{\frac{p-2}{2}}\nabla u]+\varepsilon |u|^{q-2}u + g(x,u(x),\nabla u(x)) = h(x)$$, a.e. $$x\in\Omega$$. By using Green’s formula, we know that for any $$v \in W^{1,p}(\Omega)$$,

\begin{aligned}& \int_{\Gamma}\bigl\langle \vartheta,\bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u\bigr\rangle v|_{\Gamma}\, d \Gamma(x) \\& \quad = \int_{\Omega}\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{p-2}{2}}\nabla u\bigr]v \,dx + \int _{\Omega} \bigl\langle \bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{p-2}{2}}\nabla u,\nabla v\bigr\rangle \,dx \\& \quad = \int_{\Omega}\bigl[\varepsilon |u|^{q-2}u + g \bigl(x,u(x),\nabla u(x)\bigr) - h(x)\bigr] v(x)\,dx + \langle v, B_{p,q}u\rangle - \int_{\Omega}\varepsilon |u|^{q-2}u v\, dx \\& \quad = \bigl\langle v, Au+Su-\partial \Phi_{p}(u)\bigr\rangle = \bigl\langle v, -\partial \Phi_{p}(u)\bigr\rangle \\& \quad = - \int_{\Gamma}\beta_{x}(u)v|_{\Gamma} \, d\Gamma(x). \end{aligned}

Thus $$- \langle \vartheta,(C(x)+|\nabla u|^{2})^{\frac{p-2}{2}}\nabla u\rangle \in \beta_{x}(u)$$, a.e. on Γ.

This completes the proof. □

### Theorem 19

Let A and S be the same as those in Lemma  14 and Lemma  15, respectively. Let $$f : L^{2}(\Omega)\rightarrow L^{2}(\Omega)$$ be a fixed contractive mapping with coefficient $$k \in (0,1)$$ and $$T: L^{2}(\Omega) \rightarrow L^{2}(\Omega)$$ be a strongly positive linear bounded operator with coefficient $$\overline{\gamma}$$. Suppose that $$0 < \eta < \frac{\overline{\gamma}}{2k}$$. Let $$\{u_{n}\}$$ be generated by the iterative algorithm (D)

$$\textstyle\begin{cases} u_{0} \in L^{2}(\Omega), \\ y_{n}= \alpha_{n} u_{n} + (1-\alpha_{n})(I+r_{n}A)^{-1}[\frac{u_{n}+y_{n}}{2}-r_{n}S(\frac{u_{n}+y_{n}}{2})], \\ u_{n+1}=\gamma_{n} \eta f(u_{n})+(I-\gamma_{n}T)y_{n}+ e_{n}, \quad n \geq 0. \end{cases}$$
(D)

Suppose $$\{e_{n}\}\subset L^{2}(\Omega)$$, $$\{\alpha_{n}\}$$ and $$\{\gamma_{n}\}$$ are two sequences in $$(0,1)$$ and $$\{r_{n}\}\subset (0,+\infty)$$ satisfies the conditions presented in Theorem  10. Then $$\{u_{n}(x)\}$$ converges strongly to a point $$p_{0}(x) \in (A+S)^{-1}0$$, which is the common solution of nonlinear boundary value problem (C) and the following variational inequality: for $$\forall z \in (A+S)^{-1}0$$,

$$\bigl\langle (T-\eta f)p_{0}, p_{0}-z \bigr\rangle \leq 0.$$
(22)

### Example 2

Next, we shall consider the following Laplacian equation with Signorini boundary value conditions, which can be found in [21]:

$$\textstyle\begin{cases} -\Delta u + u = h(x),\quad x \in\Omega, \\ u \geq Mu, \frac{\partial u}{\partial \vartheta} \geq 0, \quad (u - Mu)\frac{\partial u}{\partial \vartheta} = 0,\quad \mbox{a.e. }x\in\Gamma, \end{cases}$$
(E)

where Ω is a bounded domain of $$R^{N}$$ ($$N \geq 1$$) with its boundary Γ sufficiently smooth, ϑ is the exterior normal derivative of Γ. $$h(x) \in L^{2}(\Omega)$$, $$\overline{h}\in H^{\frac{1}{2}}(\Gamma)$$, $$\phi \in H^{\frac{1}{2}}(\Gamma)$$ are given functions. $$Mu = \overline{h}(x) - \int_{\Gamma}\phi\frac{\partial u}{\partial \vartheta}d\Gamma(x)$$ for $$x \in \Gamma$$ and ϕ is a nonnegative function defined on Γ.

From [21], we know that (E) can be expressed in the form of the following quasi-variational inequality:

$$\textstyle\begin{cases} u \in H_{L}^{1}(\Omega), \qquad u \in Q(u), \\ \langle Lu, u -w\rangle \leq \langle h(x),u - w\rangle, \quad \forall w \in Q(u), \end{cases}$$
(F)

where $$H_{L}^{1}(\Omega):= \{u \in H^{1}(\Omega): Lu \in L^{2}(\Omega)\}$$ with norm $$\|\cdot\|_{H_{L}^{1}(\Omega)}= (\|\cdot\|^{2}_{H^{1}(\Omega)}+\|\cdot\|^{2}_{L^{2}(\Omega)})^{\frac{1}{2}}$$ for $$u \in H_{L}^{1}(\Omega)$$, $$L = L_{1}+L_{2}$$ and $$L_{1} u = - \Delta u$$, $$L_{2} u = u$$ for $$u \in H_{L}^{1}(\Omega)$$. $$Q(u) : = \{v \in H^{1}(\Omega): v \geq \overline{h}(x) - \langle \phi, \frac{\partial u}{\partial \vartheta}\rangle _{\Gamma}, \mbox{a.e. } x \in \Gamma\}$$, here $$\langle\cdot, \cdot\rangle$$ denotes the generalized duality pairing between $$H^{\frac{1}{2}}(\Gamma)$$ and $$H^{-\frac{1}{2}}(\Gamma)$$.

### Lemma 20

([21])

Quasi-variational inequality (F) has a solution, which implies that (E) has a solution $$u(x) \in H_{L}^{1}(\Omega)$$.

### Lemma 21

If $$u(x) \in (L_{1}+L_{3})^{-1}0$$, where $$L_{3} u = L_{2}u - h(x)$$, then $$u(x) \in H_{L}^{1}(\Omega)$$ is the solution of (E).

### Proof

It is easy to check that if $$u(x) \in (L_{1}+L_{3})^{-1}0$$, then $$u(x)$$ satisfies (F), which implies that the result is true in view of Lemma 20.

This completes the proof. □

### Theorem 22

Let $$L_{1}$$ and $$L_{3}$$ be the same as those above. Let $$f : H_{L}^{1}(\Omega) \rightarrow H_{L}^{1}(\Omega)$$ be a fixed contractive mapping with coefficient $$k \in (0,1)$$, $$T: H_{L}^{1}(\Omega) \rightarrow H_{L}^{1}(\Omega)$$ be a strongly positive linear bounded operator with coefficient $$\overline{\gamma}$$. Suppose that $$0 < \eta < \frac{\overline{\gamma}}{2k}$$. Let $$\{u_{n}(x)\}$$ be generated by the iterative algorithm (G):

$$\textstyle\begin{cases} u_{0} \in H_{L}^{1}(\Omega), \\ y_{n}= \alpha_{n} u_{n} + (1-\alpha_{n})(I+r_{n}L_{1})^{-1}[\frac{u_{n}+y_{n}}{2}-r_{n}L_{3}(\frac{u_{n}+y_{n}}{2})], \\ u_{n+1}=\gamma_{n} \eta f(u_{n})+(I-\gamma_{n}T)y_{n}+ e_{n}, \quad n \geq 0. \end{cases}$$
(G)

Suppose $$\{e_{n}\}\subset H_{L}^{1}(\Omega)$$, $$\{\alpha_{n}\}$$ and $$\{\gamma_{n}\}$$ are two sequences in $$(0,1)$$ and $$\{r_{n}\}\subset (0,+\infty)$$ satisfies the conditions presented in Theorem  10. Then $$\{u_{n}(x)\}$$ converges strongly to a point $$p_{0}(x) \in (L_{1}+L_{3})^{-1}0$$, which is the common solution of the Laplacian equation with Signorini boundary value condition (E) and the following variational inequality: for $$\forall z \in (L_{1}+L_{3})^{-1}0$$,

$$\bigl\langle (T-\eta f)p_{0}, p_{0}-z \bigr\rangle \leq 0.$$
(23)

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## Acknowledgements

This paper is supported by the National Natural Science Foundation of China (No. 11071053), Natural Science Foundation of Hebei Province (No. A2014207010), Key Project of Science and Research of Hebei Educational Department (ZH2012080) and Key Project of Science and Research of Hebei University of Economics and Business (2013KYZ01).

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Wei, L., Shi, A. Splitting-midpoint method for zeros of the sum of accretive operator and μ-inversely strongly accretive operator in a q-uniformly smooth Banach space and its applications. J Inequal Appl 2015, 183 (2015). https://doi.org/10.1186/s13660-015-0704-6