A new iterative algorithm for the sum of infinite m-accretive mappings and infinite $\mu_{i}$-inversely strongly accretive mappings and its applications to integro-differential systems

Abstract

A new three-step iterative algorithm for approximating the zero point of the sum of an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings in a real q-uniformly smooth and uniformly convex Banach space is presented. The computational error in each step is being considered. A strong convergence theorem is proved by means of some new techniques, which extend the corresponding work by some authors. The relationship between the zero point of the sum of an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings and the solution of one kind variational inequalities is investigated. As an application, an integro-differential system is exemplified, from which we construct an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings. Moreover, the iterative sequence of the solution of the integro-differential systems is obtained.

1 Introduction and preliminaries

Throughout this paper, we assume that E is a real Banach space with norm $\Vert \cdot \Vert$ and $E^{*}$ is the dual space of E. We use ‘→’ and ‘⇀’ (or ‘$w-\lim$’) 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

$$\Vert x\Vert = \Vert y\Vert = 1,\qquad \Vert x - y\Vert \geq \varepsilon\quad \Rightarrow\quad \biggl\Vert \frac{x+y}{2}\biggr\Vert \leq1-\delta. $$

A Banach space E is said to be smooth if

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

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

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

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

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.

An operator $B: E\rightarrow E^{*}$ is said to be monotone [1] if $\langle u - v, Bu - Bv\rangle\geq0$, for all $u,v \in D(B)$. The monotone operator B is said to be maximal monotone if the graph of B, $G(B)$, is not contained properly in any other monotone subset of $E \times E^{*}$. An operator $B: E \rightarrow E^{*}$ is said to be coercive if $\lim_{n \rightarrow \infty}\frac{\langle x_{n}, Bx_{n}\rangle}{ \Vert x_{n}\Vert }= +\infty$ for $\{x_{n}\} \subset D(B)$ such that $\lim_{n \rightarrow \infty} \Vert x_{n}\Vert = +\infty$.

A single-valued mapping $F:D(F) = E \rightarrow E^{*}$ is said to be hemi-continuous [1] if $w-\lim_{t\rightarrow0}F(x+ty) = Fx$, for any $x,y\in E$. A single-valued mapping $F:D(F) = E \rightarrow E^{*}$ is said to be demi-continuous [1] if $w-\lim_{n\rightarrow\infty}Fx_{n} = Fx$, for any sequence $\{x_{n}\}$ strongly convergent to x in E.

Following from [1] or [2], the function h is said to be a proper convex function on E if h is defined from E onto $(-\infty, +\infty]$, h is not identically +∞ such that $h((1-\lambda)x+\lambda y)\leq(1-\lambda)h(x)+\lambda h(y)$, whenever $x,y \in E$ and $0 \leq\lambda\leq1$. h is said to be strictly convex if $h((1-\lambda)x+\lambda y)< (1-\lambda)h(x)+\lambda h(y)$, for all $0 < \lambda< 1$ and $x,y \in E$ with $x \neq y$, $h(x) <+\infty$ and $h(y) < +\infty$. The function $h: E \rightarrow(-\infty, +\infty]$ is said to be lower-semi-continuous on E if $\liminf_{y \rightarrow x}h(y) \geq h(x)$, for any $x \in E$. Given a proper convex function h on E and a point $x \in E$, we denote by $\partial h(x)$ the set of $x^{*} \in E^{*}$ such that $h(x) \leq h(y)+\langle x - y, x^{*}\rangle$ for any $y \in E$. Such elements $x^{*}$ are called subgradients of h at x, and $\partial h(x)$ is called the subdifferential of h at x.

For $q > 1$, the generalized duality mapping $J_{q}: E \rightarrow 2^{E^{*}}$ is defined by

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

In particular, $J: = J_{2}$ is called the normalized duality mapping and $J_{q}(x) = \Vert x\Vert ^{q-2}J(x)$ for $x \neq0$. It is well known that if E is smooth, then $J_{q}$ is single-valued. If E is reduced to the Hilbert space H, then $J_{q} \equiv I$ is the identity mapping. It can be seen from [2] that the normalized duality mapping J has the following properties:

• (i) if E is uniformly smooth, then J is norm-to-norm uniformly continuous on each bounded subset in E;

• (ii) the reflexivity of E and strict convexity of $E^{*}$ imply that J is single-valued, monotone, and demi-continuous.

In the following, we still denote by J and $J_{q}$ the single-valued normalized duality mapping and the single-valued generalized duality mapping.

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

Let $T : D(T) \sqsubseteq E \rightarrow E$ be a mapping. Then T is said to be

• (1) non-expansive if

  $$\Vert Tx - Ty\Vert \leq \Vert x-y\Vert \quad \mbox{for } \forall x,y \in D(T); $$

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

  $$\Vert Tx - Ty\Vert \leq k \Vert x - y\Vert \quad \mbox{for } \forall x,y \in D(T); $$

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

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

  $$\bigl\langle Tx - Ty, j_{q}(x-y)\bigr\rangle \geq0; $$

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

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

  for some $\mu> 0$;

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

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

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

  in this case,

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

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

• (7) demiclosed at p if whenever $\{x_{n}\}$ is a sequence in $D(T)$ such that $x_{n} \rightharpoonup x \in D(T)$ and $Tx_{n} \rightarrow p$ then $Tx =p$;

• (8) strongly accretive if, for all $x, y \in D(T)$, there exists $j(x-y) \in J(x-y)$ such that

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

  for some $\epsilon> 0$.

For the accretive mapping A, we use $N(A)$ to denote the set of zero points of it; that is, $N(A): = \{x \in D(A) : Ax = 0\}$. If A is accretive, then we can define, for each $r>0$, a single-valued mapping $J_{r}^{A} : R(I+rA)\rightarrow D(A)$ by $J_{r}^{A} : = (I+rA)^{-1}$, which is called the resolvent of A [1]. It is well known that $J^{A}_{r}$ is non-expansive and $N(A) = \operatorname {Fix}(J_{r}^{A})$.

Let C be a nonempty, closed and convex subset of E and Q be a mapping of E onto C. Then Q is said to be sunny [4] 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 [4] 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.

A subset C of E is said to be a sunny non-expansive retract of E [4] if there exists a sunny non-expansive retraction of E onto C and it is called a non-expansive retract of E if there exists a non-expansive 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 [5], by Passty [6], and, in a dual form for convex programming, by Han and Lou [7].

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 $x \in N(A+B)$, where A and B are μ-inversely strongly accretive mapping and m-accretive mapping defined in the Hilbert space H, respectively. In 2015, Wei et al. extended the related work from the Hilbert space to the real smooth and uniformly convex Banach space and presented the following iterative algorithm with errors [8]:

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

(2)

where C is a nonempty, closed, and convex sunny non-expansive retract of E, $Q_{C}$ is the sunny non-expansive 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 mappings and $\{B_{i}\}_{i = 1}^{N}$ is a finite family of μ-inversely strongly accretive mappings. $T : E \rightarrow E$ is a strongly positive linear bounded operator with coefficient γ̅ and $f : E \rightarrow E$ is a contraction with coefficient $k \in(0,1)$. $J^{A_{i}}_{r_{n,i}}= (I+r_{n,i}A_{i})^{-1}$, for $i = 1,2, \ldots, N$, $\sum_{m = 0}^{N}a_{m} = 1$, $0 < a_{m} < 1$, for $m = 0 , 1, 2, \ldots, N$. Then $\{x_{n}\}$ is proved to converge strongly to $p_{0} \in\bigcap_{i = 1}^{N} N(A_{i}+B_{i})$, which solves the variational inequality

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

for $\forall z\in\bigcap_{i = 1}^{N} N(A_{i}+B_{i})$ 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 the fixed point of a non-expansive mapping in a Hilbert space in [9]:

$$ x_{0} \in H,\quad x_{n+1}= (1-\alpha_{n})x_{n} + \alpha_{n} T\biggl(\frac {x_{n}+x_{n+1}}{2}\biggr),\quad n \geq0, $$

(3)

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

Inspired by the work in [8] and [9], we shall present the following iterative algorithm with errors in a real q-uniformly smooth and uniformly convex Banach space:

$$ \textstyle\begin{cases} x_{0}\in C \quad \text{chosen arbitrarily},\\ y_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e'_{n})],\\ z_{n}= \delta_{n}y_{n} + \beta_{n}\sum_{i = 1}^{\infty}a_{i} J_{r_{n,i}}^{A_{i}}[\frac{y_{n}+z_{n}}{2}-r_{n,i}B_{i}(\frac{y_{n}+z_{n}}{2})] +\zeta_{n} e''_{n},\\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I-\gamma_{n}T)z_{n}+e'''_{n}, \quad n \geq0, \end{cases} $$

(A)

where C is a nonempty, closed, and convex sunny non-expansive retract of E, $Q_{C}$ is the sunny non-expansive retraction of E onto C, $\{e'_{n}\}$, $\{e''_{n}\}$, and $\{e'''_{n}\}$ are three error sequences, $A_{i}: C \rightarrow E$ is m-accretive and $B_{i}: C \rightarrow E$ is $\mu_{i}$-inversely strongly accretive, where $i \in \mathbb{N^{+}}$. $T : E \rightarrow E$ is a strongly positive linear bounded operator with coefficient γ̅ 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 $i \in\mathbb {N^{+}}$. $\sum_{m = 1}^{\infty}a_{m} = 1$, $0 < a_{m} < 1$, for $m \in \mathbb{N^{+}}$. $\delta_{n} + \beta_{n} +\zeta_{n} \equiv1$, for $n \geq 0$. More details of iterative algorithm (A) will be presented in Section 2. Then $\{x_{n}\}$ is proved to converge strongly to $p_{0} \in\bigcap_{i = 1}^{\infty} N(A_{i}+B_{i})$, which is also a solution of one kind of variational inequality.

Our main contributions are:

• (i) A new three-step iterative algorithm is designed by combining the ideas of famous iterative algorithms such as proximal methods, Halpern methods, convex combination methods, viscosity methods, and implicit midpoint methods.

• (ii) The assumption that ‘the duality mapping J is weakly sequentially continuous’ or ‘J is weakly sequentially continuous at zero’ in most of the existing related work is deleted.

• (iii) ‘$B_{i}$ is μ-inversely strongly accretive’ in most of the related work is replaced by ‘$B_{i}$ is $\mu _{i}$-inversely strongly accretive’, which makes the discussion more general. Moreover, the design of the iterative algorithm is extended from finite case of the sum of m-accretive mappings and μ-inversely strongly accretive mappings to the infinite case.

• (iv) 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.

• (v) In each step of the three-step iterative algorithm, computational error is being considered - that is, we consider three error sequences $\{e'_{n}\}$, $\{e''_{n}\}$, and $\{e'''_{n}\}$.

• (vi) All of the three sequences $\{y_{n}\}$, $\{z_{n}\}$, and $\{x_{n}\}$ constructed in the new iterative algorithm are proved to be strongly convergent to the zero point of the sum of an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings.

• (vii) The connection between the zero point of the sum of m-accretive mappings and $\mu_{i}$-inversely strongly accretive mappings and the solution of one kind variational inequalities is being set up.

• (viii) In Section 3, the application of the main result in Section 2 to one kind integro-differential systems is demonstrated, from which we can see the connections between variational inequalities, integro-differential equations, and iterative algorithms.

Next, we list some results we need in the sequel.

Lemma 1

(see [2])

Let E be a Banach space and C be a nonempty closed and convex subset of E. Let $f: C \rightarrow C$ be a contraction. Then f has a unique fixed point $u \in C$.

Lemma 2

(see [10])

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 non-expansive mapping such that $\operatorname {Fix}(T) \neq \emptyset$, then $I-T$ is demiclosed at zero.

Lemma 3

(see [11])

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

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

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

Lemma 4

(see [12])

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 [1])

Let E be a Banach space and let A be an m-accretive mapping. 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 [13])

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 mapping and $B: E \rightarrow 2^{E}$ is m-accretive. Then

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

Lemma 7

(see [14])

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

Lemma 8

(see [15])

Let E be a real strictly convex Banach space and let C be a nonempty closed and convex subset of E. Let $T_{m}: C \rightarrow C$ be a non-expansive mapping for each $m \geq1$. Let $\{a_{m}\}$ be a real number sequence in (0,1) such that $\sum_{m = 1}^{\infty}a_{m} = 1$. Suppose that $\bigcap_{m=1}^{\infty} \operatorname {Fix}(T_{m}) \neq\emptyset$. Then the mapping $\sum_{m = 1}^{\infty}a_{m} T_{m}$ is non-expansive with $\operatorname {Fix}(\sum_{m = 1}^{\infty}a_{m}T_{m}) = \bigcap_{m = 1}^{\infty} \operatorname {Fix}(T_{m})$.

Lemma 9

(See [16])

Let C be a nonempty, closed, and convex subset of a real q-uniformly smooth Banach space E with constant $K_{q}$. Let the mapping $A: C\rightarrow E$ be a μ-inversely strongly accretive mapping. Then the following inequality holds:

$$\bigl\Vert (I-rA)x-(I-rA)y\bigr\Vert ^{q} \leq \Vert x-y\Vert ^{q} - r\bigl(q \mu- K_{q} r^{q-1}\bigr)\Vert Ax-Ay\Vert ^{q}. $$

In particular, if $0 < r \leq(\frac{q \mu}{K_{q}})^{\frac{1}{q-1}}$, then $(I-rA)$ is non-expansive.

2 Strong convergence theorems

Lemma 10

Let E be a real uniformly smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny non-expansive retract of E, and let $Q_{C}$ be the sunny non-expansive 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 γ̅ and $U : C \rightarrow C$ be a non-expansive mapping. Suppose that $0 < \eta< \frac{\overline{\gamma}}{2k}$ and $\operatorname {Fix}(U) \neq\emptyset$. If for each $t \in(0,1)$, define $T_{t} : E \rightarrow E$ by

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

(4)

then $T_{t}$ has a fixed point $x_{t}$, for each $0 < t \leq \Vert T\Vert ^{-1}$, 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 $\forall z \in \operatorname {Fix}(U)$,

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

(5)

Proof

Copying Steps 1 to 5 of Lemma 8 in [8], we have the following results:

• (a) $T_{t}$ is a contraction, for $0 < t < \Vert T\Vert ^{-1}$.

• (b) $T_{t}$ has a unique fixed point $x_{t}$.

• (c) $\{x_{t}\}$ is bounded, for $0 < t < \Vert T\Vert ^{-1}$.

• (d) $x_{t} - UQ_{C}x_{t} \rightarrow0$, as $t \rightarrow0$.

• (e) If the inequality (5) has a solution, then the solution must be unique.

Finally, we are to show that $x_{t} \rightarrow p_{0} \in \operatorname {Fix}(U)$, as $t \rightarrow0$, which satisfies the variational inequality (5).

Assume $t_{n} \rightarrow0$. Set $x_{n} : = x_{t_{n}}$ and define $\mu: E \rightarrow\mathbb{R}$ by

$$\mu(x) = \operatorname {LIM}\Vert x_{n} - x\Vert ^{2},\quad x \in E, $$

where LIM is the Banach limit on $l^{\infty}$. Let

$$K = \Bigl\{ x \in E : \mu(x) = \min_{x \in E}\operatorname {LIM}\Vert x_{n} - x\Vert ^{2}\Bigr\} . $$

It is easily seen that K is a nonempty, closed, convex, and bounded subset of E. Since $x_{n} - UQ_{C}x_{n} \rightarrow0$, for $x \in K$,

$$\mu(UQ_{C}x) = \operatorname {LIM}\Vert x_{n} - UQ_{C}x \Vert ^{2}\leq \operatorname {LIM}\Vert x_{n} - x\Vert ^{2} = \mu(x), $$

it follows that $UQ_{C}(K) \subset K$; that is, K is invariant under $UQ_{C}$. Since a uniformly smooth Banach space has the fixed point property for non-expansive mappings, $UQ_{C}$ has a fixed point, say $p_{0}$, in K. That is, $UQ_{C} p_{0} = p_{0} \in C$, which ensures that $p_{0} = Up_{0}$ from the definition of U and then $p_{0} \in \operatorname {Fix}(U)$. Since $p_{0}$ is also a minimizer of μ over E, it follows that, for $t \in(0,1)$

$$\begin{aligned} 0 \leq{}&\frac{\mu(p_{0}+\eta t f(p_{0})-tTp_{0})-\mu(p_{0})}{t} \\ ={}& \operatorname {LIM}\frac{\Vert x_{n}-p_{0}-\eta t f(p_{0})+tTp_{0}\Vert ^{2}-\Vert x_{n} - p_{0}\Vert ^{2}}{t} \\ ={}& \operatorname {LIM}\frac{ \langle x_{n}-p_{0}-\eta t f(p_{0}) +t Tp_{0}, J(x_{n}-p_{0}-\eta t f(p_{0}) +t Tp_{0})\rangle -\Vert x_{n} - p_{0}\Vert ^{2}}{t} \\ ={}& \operatorname {LIM}\bigl( \bigl\langle x_{n}-p_{0}, J \bigl(x_{n}-p_{0}-\eta t f(p_{0}) +t Tp_{0}\bigr)\bigr\rangle \\ &{}+ t\bigl\langle Tp_{0}-\eta f(p_{0}), J \bigl(x_{n}-p_{0}-\eta t f(p_{0}) +t Tp_{0}\bigr)\bigr\rangle -\Vert x_{n} - p_{0} \Vert ^{2}\bigr)/{t}. \end{aligned}$$

Since E is uniformly smooth, then by letting $t \rightarrow0$, we find the two limits above can be interchanged and obtain

$$ \operatorname {LIM}\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n} - p_{0})\bigr\rangle \leq0. $$

(6)

Since $x_{t} - p_{0} = t(\eta f(x_{t})-Tp_{0})+(I-tT)(UQ_{C}x_{t} - p_{0})$, then

$$\begin{aligned} \Vert x_{t} - p_{0}\Vert ^{2} ={}& t \bigl\langle \eta f(x_{t}) - Tp_{0}, J(x_{t}-p_{0}) \bigr\rangle + \bigl\langle (I-tT) (UQ_{C}x_{t}-p_{0}), J(x_{t}-p_{0})\bigr\rangle \\ \leq{}& t \eta\bigl\langle f(x_{t}) - f(p_{0}), J(x_{t}-p_{0})\bigr\rangle +t \bigl\langle \eta f(p_{0}) - Tp_{0}, J(x_{t}-p_{0}) \bigr\rangle \\ &{}+\Vert I -tT\Vert \Vert x_{t} -p_{0}\Vert ^{2} \\ \leq{}&\bigl[1-t(\overline{\gamma}-\eta k)\bigr]\Vert x_{t} -p_{0}\Vert ^{2}+t \bigl\langle \eta f(p_{0}) - Tp_{0}, J(x_{t}-p_{0})\bigr\rangle . \end{aligned}$$

Therefore,

$$\Vert x_{t} - p_{0}\Vert ^{2} \leq \frac{1}{\overline{\gamma}-\eta k}\bigl\langle \eta f(p_{0}) - Tp_{0}, J(x_{t}-p_{0})\bigr\rangle . $$

Hence by (6)

$$\operatorname {LIM}\Vert x_{n} - p_{0}\Vert ^{2} \leq \frac{1}{\overline{\gamma}-\eta k}\operatorname {LIM}\bigl\langle \eta f(p_{0}) - Tp_{0}, J(x_{n}-p_{0})\bigr\rangle \leq0, $$

which implies that $\operatorname {LIM}\Vert x_{n} - p_{0}\Vert ^{2}=0$, and then there exists a subsequence which is still denoted by $\{x_{n}\}$ such that $x_{n} \rightarrow p_{0}$.

Next, we shall show that $p_{0}$ solves the variational inequality (5).

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 $\forall z \in \operatorname {Fix}(U)$,

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

Taking the limits on both sides of the above inequality, $\langle (T-\eta f)p_{0}, J(p_{0}-z)\rangle\leq0$ since $x_{n} \rightarrow p_{0}$ and J is uniformly continuous on each bounded subsets of E.

Thus $p_{0}$ satisfies the variational inequality (5).

Now assume there exists another subsequence $\{x_{m}\}$ of $\{x_{t}\}$ satisfying $x_{m} \rightarrow q_{0}$. Then result (d) implies that $UQ_{C}x_{m} \rightarrow q_{0}$. From Lemma 2, we know that $I-UQ_{C}$ is demiclosed at zero, then $q_{0} = UQ_{C}q_{0}$, which ensures that $q_{0}\in \operatorname {Fix}(U)$. Repeating the above proof, we can also know that $q_{0}$ solves the variational inequality (5). Thus $p_{0} = q_{0}$ by using the result of (e).

Hence $x_{t} \rightarrow p_{0}$, as $t \rightarrow0$, which is the unique solution of the variational inequality (5).

This completes the proof. □

Theorem 11

Let E be a real q-uniformly smooth Banach space with constant $K_{q}$ and also be a uniformly convex Banach space. Let C be a nonempty, closed, and convex sunny non-expansive retract of E, and $Q_{C}$ be the sunny non-expansive retraction of E onto C. Let $f : E \rightarrow E$ be a contraction with coefficient $k \in(0,1)$, $T: E \rightarrow E$ be a strongly positive linear bounded operator with coefficient γ̅. Let $A_{i}: C \rightarrow E$ be m-accretive mappings, $B_{i}: C \rightarrow E$ be $\mu_{i}$-inversely strongly accretive mappings, for $i \in\mathbb{N^{+}}$. Let $D : = \bigcap_{i = 1}^{\infty}N(A_{i}+B_{i})\neq\emptyset$. Suppose $0 < \eta< \frac{\overline{\gamma}}{2k}$. Suppose $\{\alpha_{n}\}$, $\{\delta_{n}\}$, $\{\beta_{n}\}$, $\{\zeta_{n}\}$, $\{\gamma_{n}\} \subset(0,1)$, and $\{r_{n,i}\}\subset(0,+\infty)$ for $i \in\mathbb{N^{+}}$. Suppose $\{a_{i}\}_{i = 1}^{\infty}\subset(0,1)$ with $\sum_{i = 1}^{\infty}a_{i} = 1$, $\{e''_{n}\}\subset C$, and $\{e'_{n}\}, \{e'''_{n}\} \subset E$ are three error sequences. Let $\{x_{n}\}$ be generated by the iterative algorithm (A). Further suppose that the following conditions are satisfied:

• (i) $\sum_{n=0}^{\infty} \alpha_{n} <+\infty$;

• (ii) $\sum_{n=0}^{\infty} \gamma_{n} = \infty$, $\gamma_{n} \rightarrow0$, as $n \rightarrow\infty$ and $\sum_{n=1}^{\infty} \vert \gamma_{n} -\gamma_{n-1}\vert <+\infty$;

• (iii) $\sum_{n=0}^{\infty} \vert r_{n+1,i} - r_{n,i}\vert < +\infty$, $0 < \varepsilon\leq r_{n,i}\leq (\frac{q\mu_{i}}{K_{q}})^{\frac{1}{q-1}}$, for $n \geq0$ and $i \in\mathbb{N^{+}}$;

• (iv) $\delta_{n} + \beta_{n} + \zeta_{n} \equiv1$, for $n \geq0$, $\sum_{n=1}^{\infty} \vert \delta_{n} -\delta_{n-1}\vert <+\infty$, $\sum_{n=1}^{\infty} \vert \beta_{n}-\beta_{n-1}\vert <+\infty$, $\sum_{n=0}^{\infty}\frac{\zeta_{n}}{\beta_{n}}<+\infty$, and $\beta_{n} \rightarrow1$, as $n \rightarrow\infty$;

• (v) $\sum_{n=0}^{\infty} \Vert e'_{n}\Vert < +\infty$, $\sum_{n=0}^{\infty} \Vert e''_{n}\Vert < +\infty$, $\sum_{n=0}^{\infty} \Vert e'''_{n}\Vert < +\infty$.

Then three sequences $\{x_{n}\}$, $\{y_{n}\}$, and $\{z_{n}\}$ converge strongly to the unique element $p_{0} \in D$, which satisfies the following variational inequality: for $\forall y \in D$,

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

(7)

Proof

We shall split the proof into seven steps.

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

In fact, it suffices to show that $\{z_{n}\}$ is well defined.

For $t,s,r \in(0,1)$ and $t+s+r \equiv1$, define $U_{t,s,r}: C \rightarrow C$ by $U_{t,s,r} x: = tu + sU(\frac{u+x}{2})+rv$, where $U: C \rightarrow C$ is non-expansive for $x,u, v \in C$. Then

$$\Vert U_{t,s,r} x - U_{t,s,r}y\Vert \leq s\biggl\Vert \frac{u+x}{2}-\frac{u+y}{2}\biggr\Vert \leq\frac{s}{2} \Vert x-y\Vert . $$

Thus $U_{t,s,r}$ is a contraction, which ensures from Lemma 1 that there exists $x_{t,s,r}\in C$ such that $U_{t,s,r} x_{t,s,r} = x_{t,s,r}$. That is, $x_{t,s,r} = tu + sU(\frac{u+x_{t,s,r}}{2})+rv$.

Since $J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i})$ is non-expansive in view of Lemma 9 and $\sum_{i = 1}^{\infty}a_{i} = 1$, $\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i})$ is non-expansive, which implies that $\{z_{n}\}$ is well defined, and then $\{x_{n}\}$ is well defined.

Step 2. $D:= \bigcap_{i = 1}^{\infty}N(A_{i}+B_{i}) = \operatorname {Fix}(\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i}))$.

Lemma 6 implies that $N(A_{i}+B_{i}) = \operatorname {Fix}(J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i}))$, where $i\in\mathbb{N^{+}}$. Then Lemma 8 ensures that $\bigcap_{i = 1}^{\infty}N(A_{i}+B_{i}) = \bigcap_{i = 1}^{\infty} \operatorname {Fix}(J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i})) = \operatorname {Fix}(\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}(I-r_{n,i} B_{i}))$.

Step 3. $\{x_{n}\}$, $\{y_{n}\}$, and $\{z_{n}\}$ are all bounded.

$\forall p \in D$, we see that, for $n \geq0$,

$$ \Vert y_{n} - p\Vert \leq(1-\alpha_{n})\Vert x_{n}-p\Vert +(1-\alpha_{n})\bigl\Vert e'_{n}\bigr\Vert +\alpha_{n} \Vert p \Vert . $$

(8)

Therefore, for $p \in D$ and $n \geq0$, we have

$$\begin{aligned} \Vert z_{n} - p\Vert &\leq\delta_{n}\Vert y_{n}-p\Vert +\beta_{n}\Biggl\Vert \sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}} \biggl[\frac{y_{n}+z_{n}}{2}-r_{n,i}B_{i}\biggl( \frac {y_{n}+z_{n}}{2}\biggr)\biggr]-p\Biggr\Vert +\zeta_{n}\bigl\Vert e''_{n}-p\bigr\Vert \\ &\leq \biggl(\delta_{n}+\frac{\beta_{n}}{2}\biggr)\Vert y_{n}-p\Vert +\frac{\beta_{n}}{2}\Vert z_{n}-p\Vert + \zeta _{n}\bigl\Vert e''_{n}-p \bigr\Vert \\ &\leq \biggl(1-\frac{\beta_{n}}{2}\biggr)\Vert y_{n}-p\Vert + \frac{\beta_{n}}{2}\Vert z_{n}-p\Vert +\zeta_{n}\bigl\Vert e''_{n}-p\bigr\Vert , \end{aligned}$$

which implies that

$$ \Vert z_{n} - p\Vert \leq \Vert y_{n}-p\Vert + \frac{2\zeta_{n}}{2-\beta_{n}}\bigl\Vert e''_{n}-p\bigr\Vert \leq \Vert y_{n} - p\Vert +2\bigl\Vert e''_{n}\bigr\Vert +\frac{2\zeta_{n}}{2-\beta_{n}} \Vert p\Vert . $$

(9)

Noticing (8) and (9), using Lemma 7, we have, for $n \geq0$,

$$ \begin{aligned}[b] \Vert x_{n+1} - p\Vert \leq{}& \gamma_{n}\eta k \Vert x_{n}- p\Vert + \gamma _{n}\bigl\Vert \eta f(p) - Tp\bigr\Vert + (1-\gamma_{n} \overline{\gamma})\Vert z_{n} - p\Vert + \bigl\Vert e'''_{n}\bigr\Vert \\ \leq{}&\bigl[1-\gamma_{n} (\overline{\gamma}- k \eta)\bigr]\Vert x_{n} -p\Vert + \gamma_{n} \bigl\Vert \eta f(p) - Tp \bigr\Vert \\ &{}+ \bigl\Vert e'_{n}\bigr\Vert + 2\bigl\Vert e''_{n}\bigr\Vert +\bigl\Vert e'''_{n}\bigr\Vert + \alpha_{n} \Vert p\Vert +\frac{2\zeta_{n}}{2-\beta_{n}}\Vert p\Vert . \end{aligned} $$

(10)

By using the inductive method, we can easily get the following result from (10):

$$\begin{aligned} \Vert x_{n+1}-p\Vert \leq{}& \max\biggl\{ \Vert x_{0} - p\Vert , \frac{\Vert \eta f(p) -Tp\Vert }{\overline{\gamma}-k \eta} \biggr\} + \sum_{k=0}^{n}\bigl\Vert e'_{k}\bigr\Vert +2 \sum_{k=0}^{n} \bigl\Vert e''_{k}\bigr\Vert \\ & {}+ \sum_{k=0}^{n}\bigl\Vert e'''_{k}\bigr\Vert +\Vert p \Vert \Biggl(\sum_{k=0}^{n} \alpha_{k}+\sum_{k=0}^{n} \frac{2\zeta_{k}}{2-\beta_{k}}\Biggr). \end{aligned}$$

Therefore, from assumptions (i), (iv), and (v), we know that $\{x_{n}\}$ is bounded. Then $\{y_{n}\}$ and $\{z_{n}\}$ are bounded in view of (8) and (9), respectively.

Let $u_{n,i} = (I - r_{n,i}B_{i})(\frac{y_{n}+z_{n}}{2})$, then $\{u_{n,i}\}$ is bounded in view of Lemma 9, for $n \geq0$ and $i \in\mathbb{N^{+}}$.

Since $\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i}\Vert \leq \Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i} - \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})p\Vert +\Vert p\Vert \leq \Vert \frac{y_{n}+z_{n}}{2}-p\Vert +\Vert p\Vert$ in view of Step 2, then $\{\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i}\}$ is bounded. Moreover, we can easily know that $\{f(x_{n})\}$, $\{Tz_{n}\}$, $\{B_{i}(\frac{y_{n}+z_{n}}{2})\}$, and $\{J_{r_{n,i}}^{A_{i}}u_{n,i}\}$ are all bounded, for $n \geq0$ and $i\in\mathbb{N^{+}}$.

Set $M' = \sup\{ \Vert u_{n,i}\Vert , \Vert x_{n}\Vert , \Vert Tz_{n}\Vert , \Vert y_{n}\Vert , \Vert f(x_{n})\Vert , \Vert J_{r_{n,i}}^{A_{i}}u_{n,i}\Vert , \Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i}\Vert , \Vert B_{i}(\frac{y_{n}+z_{n}}{2})\Vert : n \geq0, i \in\mathbb{N^{+}}\}$. Then $M'$ is a positive constant.

Step 4. $\lim_{n \rightarrow\infty} \Vert x_{n+1} - x_{n} \Vert = 0$.

In fact, if $r_{n,i}\leq r_{n+1,i}$, then, using Lemma 5,

$$ \begin{aligned}[b] &\bigl\Vert J_{r_{n+1,i}}^{A_{i}}u_{n+1,i}-J_{r_{n,i}}^{A_{i}}u_{n,i} \bigr\Vert \\ &\quad \leq \Vert u_{n+1,i}-u_{n,i}\Vert +\frac{r_{n+1,i}-r_{n,i}}{\varepsilon} \bigl\Vert J_{r_{n+1,i}}^{A_{i}}u_{n+1,i}-u_{n,i}\bigr\Vert \\ & \quad \leq \Vert u_{n+1,i}-u_{n,i}\Vert +2 \frac{r_{n+1,i}-r_{n,i}}{\varepsilon}M'. \end{aligned} $$

(11)

If $r_{n+1,i}\leq r_{n,i}$, then imitating the proof of (11), we have

$$ \bigl\Vert J_{r_{n+1,i}}^{A_{i}}u_{n+1,i}-J_{r_{n,i}}^{A_{i}}u_{n,i} \bigr\Vert \leq \Vert u_{n+1,i}-u_{n,i}\Vert +2 \frac{r_{n,i}-r_{n+1,i}}{\varepsilon}M'. $$

(12)

Combining (11) and (12), we have, for $n \geq0$ and $i \in\mathbb{N^{+}}$,

$$ \bigl\Vert J_{r_{n+1,i}}^{A_{i}}u_{n+1,i}-J_{r_{n,i}}^{A_{i}}u_{n,i} \bigr\Vert \leq \Vert u_{n+1,i}-u_{n,i}\Vert +2 \frac{\vert r_{n,i}-r_{n+1,i}\vert }{\varepsilon}M'. $$

(13)

Then in view of Lemma 9

$$ \begin{aligned}[b] \Vert u_{n+1,i}-u_{n,i}\Vert ={}& \biggl\Vert (I-r_{n+1,i}B_{i}) \biggl(\frac {y_{n+1}+z_{n+1}}{2}- \frac{y_{n}+z_{n}}{2}\biggr)\biggr\Vert \\ &{} + \vert r_{n,i}-r_{n+1,i}\vert \biggl\Vert B_{i}\biggl(\frac{y_{n}+z_{n}}{2}\biggr)\biggr\Vert \\ \leq{}& \biggl\Vert \frac{y_{n+1}-y_{n}}{2}\biggr\Vert +\biggl\Vert \frac{z_{n+1}-z_{n}}{2}\biggr\Vert +\vert r_{n,i}-r_{n+1,i}\vert M'. \end{aligned} $$

(14)

In view of (13) and (14), we have

$$\begin{aligned} &\Vert z_{n+1}-z_{n}\Vert \\ &\quad \leq \delta_{n+1}\Vert y_{n+1}-y_{n} \Vert +\vert \delta_{n+1}-\delta_{n}\vert \Vert y_{n}\Vert +\vert \beta _{n+1}-\beta_{n} \vert \Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i} \Biggr\Vert \\ &\qquad {}+\beta_{n+1} \Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}}u_{n+1,i} - \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n,i} \Biggr\Vert + \bigl\Vert \zeta _{n+1}e''_{n+1}- \zeta_{n} e''_{n}\bigr\Vert \\ &\quad \leq \biggl(\delta_{n+1}+\frac{\beta_{n+1}}{2}\biggr)\Vert y_{n+1}-y_{n}\Vert +\vert \beta_{n+1}-\beta _{n}\vert M'+\vert \delta_{n+1}- \delta_{n}\vert M'+\frac{\beta_{n+1}}{2} \Vert z_{n+1} - z_{n}\Vert \\ &\qquad {}+\biggl(1+\frac{2}{\varepsilon}\biggr)\beta_{n+1} \vert r_{n,i}-r_{n+1,i}\vert M' +\bigl\Vert \zeta_{n+1}e''_{n+1}- \zeta_{n} e''_{n}\bigr\Vert , \end{aligned} $$

which implies that

$$\begin{aligned} \Vert z_{n+1}-z_{n}\Vert \leq{}& \frac{\delta_{n+1}+\frac{\beta_{n+1}}{2}}{1-\frac{\beta_{n+1}}{2}} \Vert y_{n+1}-y_{n}\Vert + \frac{2\vert \beta_{n+1}-\beta_{n}\vert M'}{2-\beta_{n+1}} \\ &{}+\frac{2\vert \delta _{n+1}-\delta_{n}\vert M'}{2-\beta_{n+1}} +\frac{2(1+\frac{2}{\varepsilon})\beta_{n+1} \vert r_{n,i}-r_{n+1,i}\vert M'}{2-\beta_{n+1}} +\frac{2\Vert \zeta_{n+1}e''_{n+1}-\zeta_{n} e''_{n}\Vert }{2-\beta_{n+1}} \\ \leq{}&\Vert y_{n+1}-y_{n}\Vert + 2\vert \beta_{n+1}-\beta_{n}\vert M'+2\vert \delta_{n+1}-\delta_{n}\vert M' \\ &{}+2\biggl(1+\frac{2}{\varepsilon}\biggr)\beta_{n+1} \vert r_{n,i}-r_{n+1,i}\vert M' +2\bigl\Vert \zeta_{n+1}e''_{n+1}- \zeta_{n} e''_{n}\bigr\Vert . \end{aligned}$$

(15)

On the other hand,

$$ \begin{aligned}[b]\Vert y_{n+1}-y_{n}\Vert \leq{}& (1- \alpha_{n+1})\Vert x_{n+1}-x_{n}\Vert +\vert \alpha_{n+1}-\alpha_{n}\vert \Vert x_{n}\Vert \\ &{}+(1-\alpha_{n+1})\bigl\Vert e'_{n+1}-e'_{n} \bigr\Vert +\vert \alpha_{n+1}-\alpha_{n}\vert \bigl\Vert e'_{n}\bigr\Vert . \end{aligned} $$

(16)

Thus in view of (15) and (16), we have, for $n \geq1$,

$$ \begin{aligned}[b] &\Vert x_{n+1}-x_{n}\Vert \\ &\quad \leq\gamma_{n} \eta\bigl\Vert f(x_{n})-f(x_{n-1}) \bigr\Vert + \eta \vert \gamma_{n}-\gamma_{n-1}\vert \bigl\Vert f(x_{n-1})\bigr\Vert \\ &\qquad {} + \Vert I-\gamma_{n}T\Vert \Vert z_{n} - z_{n-1}\Vert +\vert \gamma_{n}-\gamma_{n-1} \vert \Vert Tz_{n-1}\Vert +\bigl\Vert e'''_{n}-e'''_{n-1} \bigr\Vert \\ & \quad \leq\gamma_{n} \eta k\Vert x_{n}-x_{n-1} \Vert + \eta \vert \gamma_{n}-\gamma_{n-1}\vert \bigl\Vert f(x_{n-1})\bigr\Vert + (1-\gamma_{n}\overline{ \gamma})\Vert z_{n} - z_{n-1}\Vert \\ &\qquad {} +\vert \gamma_{n}-\gamma_{n-1}\vert \Vert Tz_{n-1}\Vert +\bigl\Vert e'''_{n}-e'''_{n-1} \bigr\Vert \\ & \quad \leq\bigl[1-\gamma_{n}(\overline{\gamma}- \eta k)\bigr] \Vert x_{n}-x_{n-1}\Vert +(1+\eta) M'\vert \gamma_{n}-\gamma_{n-1}\vert +\bigl\Vert e'''_{n}-e'''_{n-1} \bigr\Vert \\ &\qquad {}+(1-\gamma_{n}\overline {\gamma})\biggl[M' \vert \alpha_{n}-\alpha_{n-1}\vert + 2M' \vert \beta_{n}-\beta_{n-1}\vert +2M'\vert \delta_{n}-\delta_{n-1}\vert \\ &\qquad {}+2M'\biggl(1+\frac{2}{\varepsilon }\biggr)\vert r_{n,i}-r_{n-1,i}\vert +\bigl\Vert e'_{n} \bigr\Vert + 2\bigl\Vert e'_{n-1}\bigr\Vert +2\bigl\Vert e''_{n}\bigr\Vert +2\bigl\Vert e''_{n-1}\bigr\Vert \biggr]. \end{aligned} $$

(17)

Using Lemma 4, we have from (17) $\lim_{n \rightarrow\infty} \Vert x_{n+1} - x_{n} \Vert = 0$.

Step 5. $\lim_{n \rightarrow\infty} \Vert y_{n}-z_{n}\Vert = 0$, $\lim_{n \rightarrow\infty} \Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})(\frac{y_{n}+z_{n}}{2})-z_{n}\Vert = 0$ and $\lim_{n \rightarrow\infty} \Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}-y_{n}\Vert = 0$.

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

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

In view of Step 4, $x_{n}- z_{n} \rightarrow0$, as $n \rightarrow+\infty$. Since $\alpha_{n} \rightarrow0$, $\Vert y_{n}- Q_{C}x_{n}\Vert \leq\alpha_{n}\Vert x_{n}\Vert +(1-\alpha_{n})\Vert e'_{n}\Vert \rightarrow0$, as $n \rightarrow+\infty$. Therefore

$$y_{n}- z_{n} = y_{n} - Q_{C} z_{n} = y_{n} -Q_{C}x_{n} + Q_{C}x_{n} - Q_{C} z_{n} \rightarrow0,\quad \mbox{as } n \rightarrow+\infty. $$

Since $\delta_{n} + \beta_{n} + \zeta_{n} \equiv1$, $\beta_{n} \rightarrow1$, as $n \rightarrow\infty$, and $\{\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})(\frac{y_{n}+z_{n}}{2})\}$ is bounded,

$$\begin{aligned} & \Biggl\Vert z_{n} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) \biggl(\frac{y_{n}+z_{n}}{2}\biggr)\Biggr\Vert \\ & \quad \leq\delta_{n}\Biggl\Vert y_{n} - \sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) \biggl(\frac{y_{n}+z_{n}}{2}\biggr)\Biggr\Vert \\ &\qquad {}+\zeta_{n}\Biggl\Vert e''_{n}- \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) \biggl( \frac{y_{n}+z_{n}}{2}\biggr)\Biggr\Vert \rightarrow 0, \end{aligned}$$

as $n \rightarrow+\infty$. Using the above facts, we have

$$\begin{aligned} & \Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}-y_{n} \Biggr\Vert \\ &\quad \leq\Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}- \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) \biggl( \frac{y_{n}+z_{n}}{2}\biggr)\Biggr\Vert \\ &\qquad {} +\Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) \biggl(\frac{y_{n}+z_{n}}{2}\biggr)-z_{n}\Biggr\Vert +\Vert z_{n}-y_{n}\Vert \rightarrow 0,\quad \mbox{as } n \rightarrow\infty. \end{aligned} $$

Step 6. $\limsup_{n\rightarrow+\infty}\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle\leq0$, where $p_{0} \in D$, which is the unique solution of the variational inequality (7).

Since $\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}): C \rightarrow C$ is non-expansive, using Lemma 10, we know that there exists $z_{t}$ such that $z_{t} = t\eta f(z_{t})+(I-tT)\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}$ for $t \in (0,\Vert T\Vert ^{-1})$. Moreover, $z_{t} \rightarrow p_{0} \in D$, as $t \rightarrow 0$, which is the unique solution of the variational inequality (7).

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

$$\begin{aligned} &\Vert z_{t} - y_{n}\Vert ^{2} \\ &\quad = \Biggl\Vert z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}+ \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}\Biggr\Vert ^{2} \\ &\quad \leq\Biggl\Vert z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr\Vert ^{2} + 2\Biggl\langle \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}, J(z_{t} - y_{n}) \Biggr\rangle \\ & \quad = \Biggl\Vert t\eta f(z_{t}) + (I-tT)\sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr\Vert ^{2} \\ &\qquad{} + 2\Biggl\langle \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}, J(z_{t} - y_{n}) \Biggr\rangle \\ &\quad \leq\Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr\Vert ^{2} \\ &\qquad {}+ 2t\Biggl\langle \eta f(z_{t}) - T\sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}, J\Biggl(z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr) \Biggr\rangle \\ &\qquad {}+ 2\Biggl\langle \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}, J(z_{t} - y_{n}) \Biggr\rangle \\ &\quad \leq \Vert z_{t} - y_{n}\Vert ^{2} \\ &\qquad {}+ 2t\Biggl\langle \eta f(z_{t}) - T\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}, J\Biggl(z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr) \Biggr\rangle \\ &\qquad {} + 2 \Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}\Biggr\Vert \Vert z_{t} - y_{n} \Vert , \end{aligned}$$

which implies that

$$\begin{aligned} & t\Biggl\langle T\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}- \eta f(z_{t}), J\Biggl(z_{t} - \sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr) \Biggr\rangle \\ &\quad \leq\Biggl\Vert \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} - y_{n}\Biggr\Vert \Vert z_{t}-y_{n}\Vert . \end{aligned} $$

So, $\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\langle T\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}-\eta f(z_{t}), J(z_{t} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I- r_{n,i}B_{i})y_{n}) \rangle\leq 0$ in view of Step 5.

Since $z_{t} \rightarrow p_{0}$, $\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t} \rightarrow\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i}) Q_{C}p_{0} = p_{0}$, as $t \rightarrow0$. Noticing the following fact:

$$\begin{aligned} & \Biggl\langle Tp_{0}-\eta f(p_{0}), J \Biggl(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle \\ &\quad = \Biggl\langle Tp_{0}-\eta f(p_{0}), J \Biggl(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr) - J\Biggl(z_{t} -\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle \\ &\qquad {}+ \Biggl\langle Tp_{0}-\eta f(p_{0}), J \Biggl(z_{t} -\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle \\ &\quad = \Biggl\langle Tp_{0}-\eta f(p_{0}), J \Biggl(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr) - J\Biggl(z_{t} -\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle \\ &\qquad {} + \Biggl\langle Tp_{0}-\eta f(p_{0})-T\sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}+ \eta f(z_{t}), \\ &\qquad {} J\Biggl(z_{t} -\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle \\ & \qquad {}+ \Biggl\langle T\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})Q_{C}z_{t}- \eta f(z_{t}), J\Biggl(z_{t} -\sum _{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n} \Biggr)\Biggr\rangle , \end{aligned}$$

we have $\limsup_{n\rightarrow+\infty}\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n})\rangle\leq0$.

Since $\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})\rangle= \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})-J(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n})\rangle+ \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - \sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n})\rangle$ and $x_{n+1}-\sum_{i = 1}^{\infty} a_{i} J_{r_{n,i}}^{A_{i}}(I- r_{n,i}B_{i})y_{n} \rightarrow 0$ in view of Step 5, then $\limsup_{n \rightarrow\infty} \langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0}) \rangle\leq0$.

Step 7. $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$, where $p_{0} \in D$ is the same as that in Step 6.

Let $M'' = \sup \{\Vert (1-\alpha_{n})(x_{n}+e_{n})-p_{0}\Vert , \Vert x_{n}-p_{0}\Vert , M'\Vert p_{0}\Vert , \Vert e''_{n}-p_{0}\Vert ^{2}: n \geq0\}$. By using Lemma 3 again, we have

$$ \begin{aligned}[b] &\Vert y_{n} - p_{0}\Vert ^{2} \\ &\quad \leq(1-\alpha_{n})^{2} \Vert x_{n} - p_{0}\Vert ^{2} +2 \bigl\langle (1-\alpha_{n})e'_{n}- \alpha_{n}p_{0}, J\bigl[(1-\alpha_{n}) \bigl(x_{n}+e'_{n}\bigr)-p_{0}\bigr] \bigr\rangle . \end{aligned} $$

(18)

Since

$$\begin{aligned} \Vert z_{n} - p_{0}\Vert ^{2} &\leq\delta_{n} \Vert y_{n}-p_{0} \Vert ^{2}+\beta_{n} \biggl\Vert \frac{y_{n}+z_{n}}{2}-p_{0} \biggr\Vert ^{2}+\zeta _{n}\bigl\Vert e''_{n}-p_{0}\bigr\Vert ^{2} \\ & \leq \biggl(\delta_{n}+\frac{\beta_{n}}{2}\biggr)\Vert y_{n}-p_{0}\Vert ^{2}+\frac{\beta_{n}}{2} \Vert z_{n}-p_{0}\Vert ^{2}+ \zeta_{n}\bigl\Vert e''_{n}-p_{0} \bigr\Vert ^{2}, \end{aligned} $$

combining (18), we have

$$ \begin{aligned}[b] &\Vert z_{n} - p_{0}\Vert ^{2} \\ &\quad \leq \Vert y_{n}-p_{0}\Vert ^{2}+2 \zeta_{n}\bigl\Vert e''_{n}-p_{0} \bigr\Vert ^{2} \\ &\quad \leq(1-\alpha_{n})^{2} \Vert x_{n} - p_{0}\Vert ^{2} +2 \bigl\langle (1-\alpha_{n})e'_{n}- \alpha_{n}p_{0}, J\bigl[(1-\alpha_{n}) \bigl(x_{n}+e'_{n}\bigr)-p_{0}\bigr] \bigr\rangle \\ &\qquad {}+2\zeta_{n}\bigl\Vert e''_{n}-p_{0} \bigr\Vert ^{2}. \end{aligned} $$

(19)

Using (19) and Lemma 3, we have, for $n \geq0$,

$$\begin{aligned} &\Vert x_{n+1} - p_{0}\Vert ^{2} \\ &\quad = \bigl\Vert \gamma_{n}\bigl(\eta f(x_{n})-Tp_{0} \bigr)+(I- \gamma_{n} T) (z_{n} - p_{0})+e'''_{n} \bigr\Vert ^{2} \\ &\quad \leq(1-\gamma_{n} \overline{\gamma})^{2}\Vert z_{n}-p_{0}\Vert ^{2}+2\gamma_{n} \bigl\langle \eta f(x_{n})-Tp_{0}, J(x_{n+1}-p_{0}) \bigr\rangle \\ &\qquad {}+ 2\bigl\langle e'''_{n}, J(x_{n+1}-p_{0})\bigr\rangle \\ &\quad \leq(1-\gamma_{n} \overline{\gamma})^{2}(1- \alpha_{n})^{2}\Vert x_{n}-p_{0} \Vert ^{2}+ 2\bigl\langle e'''_{n}, J(x_{n+1}-p_{0})\bigr\rangle \\ &\qquad {}+2\gamma_{n}\eta\bigl\langle f(x_{n})-f(p_{0}), J(x_{n+1}-p_{0})-J(x_{n}-p_{0})\bigr\rangle \\ &\qquad {} +2\gamma_{n} \eta\bigl\langle f(x_{n})- f(p_{0}), J(x_{n}-p_{0})\bigr\rangle +2 \gamma_{n}\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle \\ & \qquad {}+ 2(1-\gamma_{n} \overline{\gamma})^{2}(1- \alpha_{n})\bigl\langle e'_{n}, J\bigl[(1- \alpha_{n}) \bigl(x_{n}+e'_{n} \bigr)-p_{0}\bigr]\bigr\rangle \\ &\qquad {} - 2\alpha_{n} (1-\gamma_{n} \overline{ \gamma})^{2} \bigl\langle p_{0}, J\bigl[(1-\alpha _{n}) \bigl(x_{n}+e'_{n} \bigr)-p_{0}\bigr]\bigr\rangle + 2(1-\gamma_{n} \overline{ \gamma})^{2} \zeta_{n} \bigl\Vert e''_{n}-p_{0} \bigr\Vert ^{2} \\ & \quad \leq\bigl[1-\gamma_{n} (\overline{\gamma}-2\eta k )\bigr] \Vert x_{n}-p_{0}\Vert ^{2}+2M'' \bigl[\bigl\Vert e'_{n}\bigr\Vert + \bigl\Vert e'''_{n}\bigr\Vert +(1- \gamma_{n} \overline{\gamma})^{2}(\alpha_{n} + \zeta_{n})\bigr] \\ & \qquad {}+2\gamma_{n} \bigl[\bigl\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\bigr\rangle +\eta \Vert x_{n}-p_{0}\Vert \Vert x_{n+1}-x_{n} \Vert \bigr]. \end{aligned}$$

(20)

Let $\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)$, $\delta_{n}^{(2)} = 2\gamma_{n}[\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle+\eta \Vert x_{n}-p_{0}\Vert \Vert x_{n+1}-x_{n}\Vert ]$, $\delta_{n}^{(3)} = 2M''[\Vert e'_{n}\Vert + \Vert e'''_{n}\Vert +(1-\gamma_{n} \overline{\gamma})^{2}(\alpha_{n} +\zeta_{n})]$. Then (20) can be simplified as $\Vert x_{n+1}-p_{0}\Vert ^{2} \leq (1-\delta_{n}^{(1)})\Vert x_{n}-p_{0}\Vert ^{2} + \delta_{n}^{(2)}+\delta_{n}^{(3)}$.

From the assumptions (i), (ii), (iv), and (v), the results of Steps 1, 4, and 6 and Lemma 4, we know that $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$.

Combine the result of Step 5, $y_{n} \rightarrow p_{0}$ and $z_{n} \rightarrow p_{0}$, as $n \rightarrow\infty$.

This completes the proof. □

Remark 12

The assumptions imposed on the real number sequences in Theorem 11 are reasonable if we take $\alpha_{n} = \frac{1}{n^{2}}$, $\gamma_{n} = \frac{1}{n}$, $\delta_{n} = 1-\frac{1}{n^{2}}-\frac{n}{n+1}$, $\beta_{n} = \frac{n}{n+1}$, and $\zeta_{n} = \frac{1}{n^{2}}$ for $n \geq0$.

Remark 13

Three sequences $\{x_{n}\}$, $\{y_{n}\}$, and $\{z_{n}\}$ are proved to be strongly convergent to the zero point of the sum of an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings. The strongly convergent point $p_{0}$ is the unique solution of a variational inequality.

Remark 14

Compared to the previous work, the computational error is considered in each step and the work on finding zero point of the sum of a finite family of m-accretive mappings and an finite family of μ-inversely strongly accretive mapping is extended to the infinite case. Compared to the work in [8], the construction of $z_{n}$ in the iterative algorithm (A) is implicit and a different $B_{i}$ corresponds to a different $\mu_{i}$, which makes the iterative algorithm (A) more general. Moreover, the assumption that ‘the normalized duality mapping J is weakly sequentially continuous at zero’ is deleted.

Remark 15

If $e'_{n} = e''_{n} = e'''_{n} \equiv0$, then iterative algorithm (A) becomes an accurate one.

Remark 16

If $C \equiv E$, then the iterative algorithm (A) becomes the following one:

$$\textstyle\begin{cases} x_{0}\in E,\\ y_{n} = (1-\alpha_{n})(x_{n}+e'_{n}),\\ z_{n}= \delta_{n}y_{n} + \beta_{n} \sum_{i = 1}^{\infty}a_{i} J_{r_{n,i}}^{A_{i}}[\frac {y_{n}+z_{n}}{2}-r_{n,i}B_{i}(\frac{y_{n}+z_{n}}{2})] +\zeta_{n} e''_{n},\\ x_{n+1}=\gamma_{n} \eta f(x_{n})+(I-\gamma_{n}T)z_{n}+e'''_{n}, \quad n \geq0. \end{cases} $$

3 Integro-differential systems and iterative algorithms

In this section, we have five purposes: (1) based on one kind nonlinear integro-differential system, construct an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings; (2) prove that under some conditions, the nonlinear integro-differential systems discussed exist solutions; (3) show the connections between the solution of the integro-differential systems and the zero point of the sum of an infinite family of m-accretive mappings and an infinite family of $\mu_{i}$-inversely strongly accretive mappings; (4) construct the iterative approximate sequence of the solution of the integro-differential systems; (5) demonstrate the relationship between the solution of the nonlinear integro-differential systems and the solution of one kind variational inequalities.

3.1 Discussion of integro-differential systems

We shall study the following nonlinear integro-differential systems involving the generalized $p_{i}$-Laplacian:

$$ \textstyle\begin{cases} \frac{\partial u^{(i)}(x,t)}{\partial t} -\operatorname {div}[(C(x,t)+\vert Du^{(i)}\vert ^{2})^{\frac{p_{i}-2}{2}}Du^{(i)}] + \varepsilon \vert u^{(i)}\vert ^{r_{i}-2}u^{(i)} \\ \quad {}+ g(x,u^{(i)}, Du^{(i)})+ a\frac{\partial}{\partial t} \int_{\Omega}u^{(i)}\,dx = f(x,t),\quad (x,t) \in\Omega\times(0,T), \\ -\langle\vartheta,(C(x,t)+\vert Du^{(i)}\vert ^{2})^{\frac{p_{i}-2}{2}}Du^{(i)} \rangle\in\beta_{x}(u^{(i)}),\quad (x,t) \in\Gamma\times(0,T),\\ u^{(i)}(x,0) = u^{(i)}(x, T),\quad x \in\Omega, i \in\mathbb{N^{+}}, \end{cases} $$

(21)

where Ω is a bounded conical domain of a Euclidean space $R^{N}$ ($N\geq1$), Γ is the boundary of Ω with $\Gamma\in C^{1}$ [17] and ϑ denotes the exterior normal derivative to Γ. $\langle\cdot,\cdot\rangle$ and $\vert \cdot \vert$ denote the Euclidean inner-product and Euclidean norm in $R^{N}$, respectively. T is a positive constant. $Du^{(i)} = (\frac{\partial u^{(i)}}{\partial x_{1}}, \frac{\partial u^{(i)}}{\partial x_{2}}, \ldots, \frac{\partial u^{(i)}}{\partial x_{N}})$ and $x = (x_{1}, x_{2}, \ldots, x_{N}) \in\Omega$. $\beta_{x}$ is the subdifferential of $\varphi_{x}$, where $\varphi_{x}= \varphi(x,\cdot):R\rightarrow R$ for $x\in\Gamma$. a and ε are non-expansive constants, $0 \leq C(x,t) \in \bigcap_{i = 1}^{\infty}V_{i} : = \bigcap_{i = 1}^{\infty} L^{p_{i}}(0 , T; W^{1,p_{i}}(\Omega))$, $f(x,t)\in\bigcap_{i = 1}^{\infty} W_{i} : = \bigcap_{i = 1}^{\infty} L^{\max\{p_{i},p_{i}'\}}(0,T; L^{\max\{p_{i},p_{i}'\}}(\Omega))$ and $g:\Omega\times R^{N+1} \rightarrow R$ are given functions.

Our discussion of (21) is based on the following assumptions, some of which can be found in [18–20].

Assumption 1

$\{p_{i}\}_{i=1}^{\infty}$ is a real number sequence with $\frac{2N}{N+1} < p_{i} < +\infty$, $\{\mu_{i}\}_{i=1}^{\infty}$ is any real number sequence in $(0,1]$ and $\{r_{i}\}_{i=1}^{\infty}$ is a real number sequence satisfying $\frac{2N}{N+1} < r_{i} \leq \min\{p_{i},p_{i}'\} < +\infty$. $\frac{1}{p_{i}}+\frac{1}{p'_{i}} = 1$ and $\frac{1}{r_{i}}+\frac{1}{r'_{i}} = 1$ for $i \in\mathbb{N^{+}}$.

Assumption 2

Green’s formula is available.

Assumption 3

For each $x\in\Gamma, \varphi_{x}= \varphi(x,\cdot):R\rightarrow R$ is a proper, convex and lower- semi-continuous function and $\varphi_{x}(0)=0$.

Assumption 4

$0 \in\beta_{x}(0)$ and for each $t \in R$, the function $x \in\Gamma\rightarrow(I+\lambda\beta_{x})^{-1}(t)\in R$ is measurable for $\lambda > 0$.

Assumption 5

Suppose that $g:\Omega\times R^{N+1} \rightarrow R$ satisfies the following conditions:

• (a) Carathéodory’s conditions;

• (b) growth condition:

  $$\bigl\vert g(x,r_{1},\ldots,r_{N+1})\bigr\vert ^{\max\{p_{i},p_{i}'\}}\leq\bigl\vert h_{i}(x,t)\bigr\vert ^{p_{i}}+ b_{i} \vert r_{1}\vert ^{p_{i}}, $$

  where $(r_{1}, r_{2}, \ldots, r_{N+1})\in R^{N+1}$, $h_{i}(x,t)\in W_{i}$, and $b_{i}$ is a positive constant, for $i \in\mathbb{N^{+}}$;

• (c) monotone condition: g is monotone in the following sense:

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

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

Assumption 6

For $i \in\mathbb{N^{+}}$, let $V^{*}_{i}$ denote the dual space of $V_{i}$. The norm in $V_{i}$, $\Vert \cdot \Vert _{V_{i}}$, is defined by

$$\bigl\Vert u(x,t)\bigr\Vert _{V_{i}} = \biggl( \int_{0}^{T} \bigl\Vert u(x,t)\bigr\Vert _{W^{1,p_{i}}(\Omega)} ^{p_{i}}\,dt\biggr)^{\frac{1}{p_{i}}},\quad u(x,t) \in V_{i}. $$

Lemma 17

(see [19])

For $i \in\mathbb{N^{+}}$, define the operator $B_{i}: V_{i} \rightarrow V^{*}_{i}$ by

$$\langle w,B_{i}u \rangle= \int_{0}^{T} \int_{\Omega}\bigl\langle \bigl(C(x,t)+\vert Du\vert ^{2}\bigr)^{\frac {p_{i}-2}{2}}Du, Dw\bigr\rangle \,dx\,dt +\varepsilon \int_{0}^{T} \int_{\Omega} \vert u\vert ^{r_{i}-2}uw \,dx\,dt, $$

for $u,w \in V_{i}$. Then $B_{i}$ is maximal monotone and coercive, where $i \in\mathbb{N^{+}}$.

Lemma 18

(see [19])

For $i \in\mathbb{N^{+}}$, define the function $\Phi_{i}: V_{i} \rightarrow R$ by

$$\Phi_{i}(u)= \int_{0}^{T} \int_{\Gamma}\varphi_{x}\bigl(u|_{\Gamma}(x,t) \bigr)\,d\Gamma(x)\,dt, $$

for $u(x,t) \in V_{i}$. Then $\Phi_{i}$ is a proper, convex and lower-semi-continuous mapping on $V_{i}$. Therefore, the subdifferential $\partial\Phi_{i}: V_{i}\rightarrow V^{*}_{i}$ is maximal monotone.

Lemma 19

(see [19])

For $i \in\mathbb{N^{+}}$, define $S_{i}: D(S_{i}) = \{u(x,t) \in V_{i}: \frac{\partial u }{\partial t} \in V^{*}_{i}, u(x,0) = u(x,T)\} \rightarrow V^{*}_{i}$ by

$$S_{i}u = \frac{\partial u }{\partial t}+ a\frac{\partial}{\partial t} \int_{\Omega}u \,dx. $$

Then $S_{i}$ is linear maximal monotone operator possessing a dense domain in $V_{i}$, where $i \in\mathbb{N^{+}}$.

Definition 20

For $i \in\mathbb{N^{+}}$, define a mapping $A_{i}: W_{i} \rightarrow2^{W_{i}}$ as follows:

$$D(A_{i})=\bigl\{ u \in W_{i} |\mbox{ there exists an }f \in W_{i}\mbox{ such that }f \in B_{i}u + \partial \Phi_{i}(u) + S_{i}u\bigr\} . $$

For $u \in D(A_{i})$, we set $A_{i}u =\{f\in W_{i} | f \in B_{i}u + \partial \Phi_{i}(u) + S_{i}u\}$.

Proposition 21

The mapping $A_{i}: W_{i} \rightarrow2^{W_{i}}$ is m-accretive, where $i \in\mathbb{N^{+}}$.

Proof

Similar to the proof of Lemmas 3.5 and 3.7 in [18] or the proof of Proposition 2.5 in [19], we have $R(I+\lambda A_{i})= W_{i}$, for $\forall\lambda>0$.

Let $J_{i}: W_{i} \rightarrow W_{i}^{*}$ denote the generalized duality mapping. Then, for $u(x,t) \in W_{i}$,

$$J_{i} u = \textstyle\begin{cases} \vert u\vert ^{p_{i}-1}\operatorname {sgn}u, & p_{i} \geq2,\\ \vert u\vert ^{p'_{i}-1}\operatorname {sgn}u,& 1< p_{i} < 2. \end{cases} $$

In fact, if $p_{i} \geq2$, then $\langle u, J_{i} u\rangle= \int^{T}_{0}\int_{\Omega} \vert u\vert ^{p_{i}}\,dx\,dt = \Vert u\Vert _{W_{i}}^{p_{i}}$ and $\Vert J_{i}u\Vert _{W_{i}^{*}}= (\int^{T}_{0}\int_{\Omega} \vert u\vert ^{(p_{i}-1)p'_{i}}\,dx\,dt)^{\frac{1}{p'_{i}}} = \Vert u\Vert _{W_{i}}^{\frac{p_{i}}{p'_{i}}}= \Vert u\Vert _{W_{i}}^{p_{i}-1}$. Thus $J_{i} u = \vert u\vert ^{p_{i}-1}\operatorname {sgn}u$, if $p_{i} \geq2$. Similarly, $J_{i} u = \vert u\vert ^{p'_{i}-1}\operatorname {sgn}u$, if $1< p_{i}<2$.

By using a similar method as that of Proposition 2.4 in [19], we can prove that for any $u,v\in D(A_{i})$, $\langle A_{i}u - A_{i}v, J_{i}(u-v)\rangle\geq0$. Thus $A_{i}$ is accretive. The result follows. This completes the proof. □

Remark 22

Noticing Proposition 21, an infinite family of m-accretive mappings $\{A_{i}\}_{i = 1}^{\infty}$ is constructed.

Definition 23

Define $C_{i} : D(C_{i}) = L^{\max\{p_{i},p'_{i}\}}(0,T;W^{1,\max\{p_{i},p'_{i}\}}(\Omega))\subset W_{i} \rightarrow W_{i}$ by

$$(C_{i}u) (x,t) = g(x,u,Du)-f(x,t) $$

for $\forall u(x,t) \in D(C_{i})$ and $f(x,t)$ is the same as that in (21), where $i \in\mathbb{N^{+}}$.

Lemma 24

The mapping $C_{i}: D(C_{i})\subset W_{i} \rightarrow W_{i}$ is continuous and strongly accretive. If, further assume that $g(x,r_{1},\ldots,r_{N+1}) \equiv r_{1}$, then $C_{i}$ is $\mu_{i}$-inversely strongly accretive, where $i \in\mathbb{N^{+}}$.

Proof

Similar to Proposition 2.6 in [19], we know that for $u \in D(C_{i})$, $x \rightarrow g(x, u, Du)$ is measurable on Ω, and then $C_{i}$ is everywhere defined and continuous.

Our next discussion is divided into two cases.

Case 1. $p_{i} \geq2$. From assumption 5, we know that

$$\begin{aligned}[b] & \bigl\langle C_{i}u - C_{i}v, \widetilde{J_{i}}(u-v)\bigr\rangle \\ &\quad = \int_{0}^{T} \int_{\Omega} \bigl(g(x,u,Du) - g(x, v, Dv)\bigr)\Vert u-v \Vert ^{2-p_{i}}_{W_{i}}\vert u-v\vert ^{p_{i}-1} \operatorname {sgn}(u-v) \,dx\,dt \\ &\quad \geq \Vert u-v\Vert ^{2-p_{i}}_{W_{i}} \int_{0}^{T} \int_{\Omega} \vert u-v\vert ^{p_{i}}\,dx\,dt = \Vert u-v\Vert ^{2}_{W_{i}}, \end{aligned} $$

where $\widetilde{J_{i}}: W_{i} \rightarrow W_{i}^{*}$ is the normalized duality mapping, which implies that $C_{i}$ is strongly accretive.

If, furthermore, $g(x,r_{1},\ldots,r_{N+1}) \equiv r_{1}$, since $\{\mu_{i}\} \subset(0,1]$, then we have

$$\bigl\langle C_{i}u - C_{i}v, J_{i}(u-v)\bigr\rangle = \int_{0}^{T} \int_{\Omega} \vert u-v\vert ^{p_{i}}\,dx\,dt = \Vert C_{i}u-C_{i}v\Vert ^{p_{i}}_{W_{i}}\geq \mu_{i}\Vert C_{i}u-C_{i}v\Vert _{W_{i}}^{p_{i}}, $$

where $J_{i}: W_{i} \rightarrow W_{i}^{*}$ is the generalized duality mapping in Proposition 21, which implies that $C_{i}$ is $\mu_{i}$-inversely strongly accretive.

Case 2. $1< p_{i} < 2$. Similar to Case 1, the result follows.

This completes the proof. □

Remark 25

Noticing Lemma 24, we have constructed an infinite family of $\mu_{i}$-inversely strongly accretive mappings.

Lemma 26

([18, 19])

(1) If $w(x,t) \in\partial\Phi_{i}(u)$, then $w(x,t) \in\partial\beta_{x}(u)$, a.e. on $\Gamma\times (0,T)$. (2) $\langle\varphi, \partial\Phi_{i}(u)\rangle\equiv0$, $\forall\varphi\in C_{0}^{\infty}(0, T; \Omega)$.

Lemma 27

([21])

Let E be a smooth Banach space, let $A: D(A) \subset E \rightarrow2^{E}$ be an m-accretive mapping, and $S: D(S) \subset E \rightarrow E$ be a continuous and strongly accretive mapping with $\overline{D(A)}\subset D(S)$. Then, for any $z \in E$, the equation $z \in Sx+\lambda Ax$ has a unique solution $x_{\lambda}$, $\lambda> 0$.

Theorem 28

For $f(x,t)\in\bigcap_{i = 1}^{\infty}W_{i}$, there exists unique $u^{(i)} \in W_{i}$ satisfying the following:

• (a) $\frac{\partial u^{(i)}(x,t)}{\partial t} -\operatorname {div}[(C(x,t)+\vert D u^{(i)}\vert ^{2})^{\frac{p_{i}-2}{2}}Du^{(i)}] + \varepsilon \vert u^{(i)}\vert ^{r_{i}-2}u^{(i)} + g(x,u^{(i)},Du^{(i)})+ a \frac{\partial}{\partial t} \int_{\Omega}u^{(i)}\,dx = f(x,t)$, $(x,t) \in\Omega\times(0,T)$;

• (b) $-\langle\vartheta,(C(x,t)+\vert Du^{(i)}\vert ^{2})^{\frac{p_{i}-2}{2}}Du^{(i)} \rangle\in\beta_{x}(u^{(i)}(x,t))$, $(x,t) \in\Gamma\times (0,T)$;

• (c) $u^{(i)}(x,0) = u^{(i)}(x,T)$, $x \in\Omega$, where $i \in\mathbb{N^{+}}$.

Proof

Using Proposition 21, Lemmas 24 and 27, we know that for $\theta\in W_{i}$, there exists unique $u^{(i)}(x,t)\in D(A_{i})\subset W_{i}$ such that

$$ \theta= C_{i}u^{(i)} + A_{i}u^{(i)}. $$

(22)

Then, for $\varphi\in C^{\infty}_{0}(0,T;\Omega)$, we have

$$\langle\varphi, \theta\rangle= \bigl\langle \varphi, C_{i}u^{(i)} \bigr\rangle + \bigl\langle \varphi, B_{i}u^{(i)}\bigr\rangle + \bigl\langle \varphi, \partial\Phi_{i}\bigl(u^{(i)}\bigr)\bigr\rangle + \bigl\langle \varphi, S_{i}u^{(i)}\bigr\rangle , $$

which implies that

$$\begin{aligned} & \int_{0}^{T} \int_{\Omega} f \varphi \,dx\,dt \\ &\quad = \int_{0}^{T} \int_{\Omega}\frac{\partial u^{(i)}}{\partial t }\varphi \,dx\,dt + a \int_{0}^{T} \int_{\Omega} \biggl(\frac{\partial}{\partial t} \int_{\Omega}u^{(i)}\,dx\biggr) \varphi \,dx\,dt \\ &\qquad {}+ \int_{0}^{T} \int_{\Omega}\bigl\langle \bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac{p_{i}-2}{2}}Du^{(i)}, D\varphi\bigr\rangle \,dx\,dt + \varepsilon \int_{0}^{T} \int_{\Omega}\bigl\vert u^{(i)}\bigr\vert ^{r_{i}-2} u^{(i)}\varphi \,dx\,dt \\ &\qquad {}+ \int_{0}^{T} \int_{\Omega} g\bigl(x,u^{(i)},Du^{(i)}\bigr) \varphi \,dx\,dt \\ &\quad = \int_{0}^{T} \int_{\Omega}\frac{\partial u^{(i)}}{\partial t }\varphi \,dx\,dt + a \int_{0}^{T} \int_{\Omega} \biggl(\frac{\partial}{\partial t} \int_{\Omega}u^{(i)}\,dx\biggr) \varphi \,dx\,dt \\ &\qquad {}+ \int_{0}^{T} \int_{\Omega}-\operatorname {div}\bigl[\bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac {p_{i}-2}{2}}Du^{(i)} \bigr]\varphi \,dx\,dt + \varepsilon \int_{0}^{T} \int_{\Omega}\bigl\vert u^{(i)}\bigr\vert ^{r_{i}-2} u^{(i)} \varphi \,dx\,dt \\ &\qquad {}+ \int_{0}^{T} \int_{\Omega} g\bigl(x,u^{(i)}, Du^{(i)}\bigr) \varphi \,dx\,dt. \end{aligned}$$

Therefore, from the property of the generalized function, we know that (a) is true.

From the definition of $S_{i}$, we know that (c) is trivial.

By using the results of (a), the Green’s formula and (22), we have, for $w \in W_{i}$,

$$\begin{aligned} & \int^{T}_{0} \int_{\Gamma}\bigl\langle \vartheta, \bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac{p_{i}-2}{2}}Du^{(i)} \bigr\rangle w d\Gamma(x)\,dt \\ &\quad = \int_{0}^{T} \int_{\Omega} \operatorname {div}\bigl[\bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac{p_{i}-2}{2}}Du^{(i)} \bigr] w \,dx\,dt \\ &\qquad {}+ \int_{0}^{T} \int_{\Omega}\bigl\langle \bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac{p_{i}-2}{2}}Du^{(i)}, Dw\bigr\rangle \,dx\,dt \\ &\quad = \int_{0}^{T} \int_{\Omega}g\bigl(x,u^{(i)},Du^{(i)} \bigr)w\,dx\,dt+ \int_{0}^{T} \int_{\Omega}\frac{\partial u^{(i)}}{\partial t} w \,dx\,dt\\ &\qquad {}+ \int_{0}^{T} \int_{\Omega} \biggl(a \frac{\partial}{\partial t} \int_{\Omega}u^{(i)}\,dx\biggr) w \,dx\,dt + \varepsilon \int_{0}^{T} \int_{\Omega}\bigl\vert u^{(i)}\bigr\vert ^{r_{i}-2} u^{(i)}w \,dx\,dt\\ &\qquad {}+ \int_{0}^{T} \int_{\Omega}\bigl\langle \bigl(C(x,t)+\bigl\vert Du^{(i)}\bigr\vert ^{2}\bigr)^{\frac{p_{i}-2}{2}}Du^{(i)}, Dw\bigr\rangle \,dx\,dt - \int_{0}^{T} \int_{\Omega}f(x,t) w \,dx\,dt \\ &\quad = \int_{0}^{T} \int _{\Omega}-\partial\Phi_{i}\bigl(u^{(i)} \bigr) w \,dx\,dt. \end{aligned}$$

Thus $- \langle\vartheta, (C(x,t)+\vert Du^{(i)}\vert ^{2})^{\frac{p_{i}-2}{2}}Du^{(i)}\rangle \in \partial\Phi_{i}(u^{(i)})$. In view of Lemma 26, (b) follows.

This completes the proof. □

3.2 Applications of iterative algorithms to integro-differential systems

Theorem 29

If $\varepsilon\equiv0$, $g(x,r_{1},\ldots,r_{N+1}) \equiv r_{1}$ and $f(x,t) \equiv k$, were k is a constant, then $u(x,t)\equiv k$ is the unique solution of the integro-differential system (21). Moreover, $\{u(x,t)\in\bigcap_{i = 1}^{\infty}W_{i}| u(x,t)\equiv k\textit{ satisfying }\mbox{(21)}\} = \bigcap_{i = 1}^{\infty}N(A_{i}+C_{i})$.

Proof

From Theorem 28, we know that (21) has a unique solution for this special case. It is easy to check that $u(x,t)\equiv k$ satisfies (21), which implies that $u(x,t)\equiv k$ is the unique solution of (21) for this special case.

Next, we show that $\bigcap_{i = 1}^{\infty}N(A_{i}+C_{i})$ is a singleton in this special case.

In fact, if $A_{i} u +C_{i} u \equiv0$ and $A_{i} v+C_{i} v \equiv0$, then $A_{i} u + u \equiv A_{i}v + v$, which implies that $0 \leq \langle A_{i} u -A_{i} v, J_{i}(u-v)\rangle= \langle v - u, J_{i}(u-v)\rangle\leq0$, and then $u(x,t) \equiv v(x,t)$. That is, $\bigcap_{i = 1}^{\infty}N(A_{i}+C_{i})$ is a singleton.

The result $u(x,t)\equiv k \in\bigcap_{i = 1}^{\infty}N(A_{i}+C_{i})$ follows from the definitions of $A_{i}$ and $C_{i}$, which implies that $\{u(x,t)\in\bigcap_{i = 1}^{\infty}W_{i}| u(x,t)\equiv k\mbox{ satisfying (21)}\} = \bigcap_{i = 1}^{\infty}N(A_{i}+C_{i})$.

This completes the proof. □

Remark 30

Combining the results of Proposition 21, Lemma 24, and Theorem 29, we set up the relationship between the solution of one kind integro-differential systems and the zero point of the sum of infinite m-accretive mappings and infinite $\mu_{i}$-inversely strongly accretive mappings.

Remark 31

Set $p:= \inf_{i \in N^{+}}(\min\{p_{i},p_{i}'\})$ and $q:= \sup_{i \in N^{+}}(\max\{p_{i},p_{i}'\})$.

Let $E:= L^{\min\{p,p'\}}(0,T; L^{\min\{p,p'\}}(\Omega))$, where $\frac{1}{p}+\frac{1}{p'} = 1$.

Let $X:= L^{\max\{q,q'\}}(0,T; W^{1,\max\{q,q'\}}(\Omega))$, where $\frac{1}{q}+\frac{1}{q'} = 1$.

Then $E = L^{p}(0,T;L^{p}(\Omega))$, $X = L^{q}(0,T;W^{1,q}(\Omega))$ and $X \subset W_{i} \subset E$, $\forall i \in\mathbb{N^{+}}$. Our next discussion of Theorem 32 will be based on X and E.

Theorem 32

Suppose $A_{i}$ and $C_{i}$ are the same as those in Proposition  21 and Lemma  24, respectively. Let $f : E\rightarrow E$ be a fixed contractive mapping with coefficient $k \in(0,1)$ and $T: E \rightarrow E$ be any strongly positive linear bounded operator with coefficient γ̅. Suppose that $0 < \eta< \frac{\overline{\gamma}}{2k}$, $\{\alpha_{n}\}$, $\{\delta_{n}\}$, $\{\beta_{n}\}$, $\{\zeta_{n}\}$, $\{\gamma_{n}\} \subset(0,1)$ and $\{r_{n,i}\}\subset(0,+\infty)$ for $i\in\mathbb {N^{+}}$. Suppose $\{a_{i}\}_{i = 1}^{\infty}\subset(0,1)$ with $\sum_{i = 1}^{\infty}a_{i} = 1$, $\{e''_{n}\}\subset X$, and $\{e'_{n}\}, \{e'''_{n}\} \subset E$. Furthermore, suppose that the following conditions are satisfied:

• (i) $\sum_{n=0}^{\infty} \alpha_{n} <+\infty$;

• (ii) $\sum_{n=0}^{\infty} \gamma_{n} = \infty$, $\gamma_{n} \rightarrow0$, as $n \rightarrow\infty$, and $\sum_{n=1}^{\infty} \vert \gamma_{n} -\gamma_{n-1}\vert <+\infty$;

• (iii) $\sum_{n=0}^{\infty} \vert r_{n+1,i} - r_{n,i}\vert < +\infty$ and $0 < \varepsilon\leq r_{n,i}\leq (\frac{p\mu_{i}}{K_{p}})^{\frac{1}{p-1}}$, for $n \geq0$ and $i\in\mathbb{N^{+}}$;

• (iv) $\delta_{n} + \beta_{n} + \zeta_{n} \equiv1$, for $n \geq0$, $\sum_{n=1}^{\infty} \vert \delta_{n} -\delta_{n-1}\vert <+\infty$, $\sum_{n=1}^{\infty} \vert \beta_{n}-\beta_{n-1}\vert <+\infty$, $\sum_{n=0}^{\infty}\frac{\zeta_{n}}{\beta_{n}}<+\infty$, and $\beta_{n} \rightarrow1$, as $n \rightarrow\infty$;

• (v) $\sum_{n=0}^{\infty} \Vert e'_{n}\Vert < +\infty$, $\sum_{n=0}^{\infty} \Vert e''_{n}\Vert < +\infty$, $\sum_{n=0}^{\infty} \Vert e'''_{n}\Vert < +\infty$.

Let $\{u_{n}\}$ be generated by the iterative algorithm (C)

$$ \textstyle\begin{cases} u_{0}(x,t) \in X,\quad \text{chosen arbitrarily},\\ v_{n}(x,t)= Q_{X}[(1-\alpha_{n})(u_{n}(x,t)+e_{n}')],\\ w_{n}(x,t) = \delta_{n} v_{n}(x,t) + \beta_{n} \sum_{i=1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} [\frac{w_{n}+v_{n}}{2}-r_{n,i}C_{i}(\frac{w_{n}+v_{n}}{2})]+\zeta_{n}e''_{n},\\ u_{n+1}(x,t)=\gamma_{n} \eta f(u_{n})+(I-\gamma_{n}T)w_{n}(x,t)+ e_{n}''', \quad n \geq 0. \end{cases} $$

(C)

If, in the integro-differential systems (21), $\varepsilon\equiv0$, $g(x,r_{1},\ldots,r_{N+1})\equiv r_{1}$, and $f(x,t)\equiv k$, then three sequences $\{u_{n}(x,t)\}$, $\{v_{n}(x,t)\}$, and $\{w_{n}(x,t)\}$ converge strongly to the unique solution $u(x,t)$ of (21), which is also the unique element in $\bigcap_{i=1}^{\infty }N(A_{i}+C_{i})$ and satisfies the following variational inequality: for $\forall y \in \bigcap_{i=1}^{\infty}N(A_{i}+C_{i})$,

$$\bigl\langle (T - \eta f)u(x,t), J\bigl(u(x,t) - y\bigr)\bigr\rangle \leq0. $$

Remark 33

From the work done in this section, we can find the connection between integro-differential systems, variational inequalities, and iterative algorithms. This may emphasize the significance of the work in this paper.