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