1 Introduction
Throughout this paper, let X be a real Banach space with norm \(\|\cdot \|\) and \(X^{*}\) be the dual space of X. Let K be a non-empty closed and convex subset of X. Let \(\langle x, f\rangle\) be the value of \(f \in X^{*}\) at \(x \in X\). We write \(x_{n} \rightarrow x\) to denote that \(\{ x_{n}\}\) converges strongly to x and \(x_{n} \rightharpoonup x\) to denote that \(\{x_{n}\}\) converges weakly to x.
Suppose that
A is a multi-valued operator from
X into
\(X^{*}\).
A is said to be monotone [
1] if for
\(\forall v_{i} \in Au_{i}\),
\(i = 1,2\), one has
\(\langle u_{1} - u_{2}, v_{1} - v_{2}\rangle\geq0\). The monotone operator
A is called maximal monotone if
\(R(J+k A) = X^{*}\), for
\(k > 0\), where
\(J: X \rightarrow2^{X^{*}}\) is the normalized duality mapping defined by
$$J(x) = \bigl\{ f \in X^{*}: \langle x, f\rangle= \|x\|^{2} = \|f \|^{2} \bigr\} ,\quad \forall x \in X. $$
A point
\(x \in D(A)\) is called a zero of
A if
\(Ax = 0\). The set of zeros of
A is denoted by
\(A^{-1}0\).
Suppose that the Lyapunov functional
\(\phi: X \times X \rightarrow (0,+\infty)\) is defined as follows:
$$\phi(x,y) = \|x\|^{2}-2 \bigl\langle x,j(y) \bigr\rangle + \|y \|^{2},\quad \forall x , y \in X, j(y) \in J(y). $$
Let
T be a single-valued mapping of
K into itself.
(1)
If \(Tp = p\), then p is called a fixed point of T. And \(\operatorname{Fix}(T)\) denotes the set of fixed points of T;
(2)
If there exists a sequence
\(\{x_{n}\}\subset K\) which converges weakly to
\(p\in K\) such that
\(x_{n} - Tx_{n} \rightarrow0\), as
\(n \rightarrow\infty\), then
p is called an asymptotic fixed point of
T [
2]. And
\(\widehat{\operatorname{Fix}}(T)\) denotes the set of asymptotic fixed points of
T;
(3)
If there exists a sequence
\(\{x_{n}\}\subset K\) which converges strongly to
\(p\in K\) such that
\(x_{n} - Tx_{n} \rightarrow0\), as
\(n \rightarrow\infty\), then
p is called a strong asymptotic fixed point of
T [
2]. And
\(\widetilde{\operatorname{Fix}}(T)\) denotes the set of strong asymptotic fixed points of
T;
(4)
T is called strongly relatively non-expansive [
2] if
\(\widehat{\operatorname{Fix}}(T) = \operatorname{Fix}(T)\neq\emptyset\) and
\(\phi(p, Tx)\leq\phi (p,x)\) for
\(x \in K\) and
\(p \in \operatorname{Fix}(T)\);
(5)
T is called weakly relatively non-expansive [
2] if
\(\widetilde{\operatorname{Fix}}(T) = \operatorname{Fix}(T)\neq\emptyset\) and
\(\phi(p, Tx)\leq\phi (p,x)\) for
\(x \in K\) and
\(p \in \operatorname{Fix}(T)\).
If
X is a real reflexive and strictly convex Banach space and
K is a non-empty closed and convex subset of
X, then for each
\(x \in X\) there exists a unique point
\(x_{0} \in K\) such that
\(\|x - x_{0}\| = \inf \{ \|x - y\|: y \in K\}\). In this case, we can define the metric projection mapping
\(P_{K}: X \rightarrow K\) by
\(P_{K}x = x_{0}\) for
\(\forall x \in X\) [
3].
If
X is a real reflexive, strictly convex, and smooth Banach space and
K is a non-empty closed and convex subset of
X, then for
\(\forall x \in X\), there exists a unique point
\(x_{0} \in K\) such that
\(\phi(x_{0}, x) = \inf \{\phi(y,x) : y \in K\}\). In this case, we can define the generalized projection mapping
\(\Pi_{K}: X \rightarrow K\) by
\(\Pi_{K} x = x_{0}\) for
\(\forall x \in X\) [
3].
Note that if X is a Hilbert space H, then \(P_{K}\) and \(\Pi_{K}\) are coincidental.
Since maximal monotone operators and weakly (or strongly) relatively non-expansive mappings have close connection with practical problems, one has a good reason to study them. During past years, much work has been done in designing iterative schemes to approximate a common element of the set of zeros of maximal monotone operators and the set of fixed points of weakly (or strongly) relatively non-expansive mappings. Among them, a projection iterative scheme is considered as one of the effective iterative schemes which almost always generates strongly convergent iterative sequences (see [
4‐
8] and the references therein). Next, we list some recent closely related work.
Klin-eam et al. [
5] presented the following projection iterative scheme for maximal monotone operator
A and two strongly relatively non-expansive mappings
B and
C in a real uniformly convex and uniformly smooth Banach space
X.
$$ \textstyle\begin{cases} v_{n} = J^{-1}[\beta_{n} Jx_{n}+(1-\beta_{n})JC(J+r_{n}A)^{-1}Jx_{n}], \\ y_{n} = J^{-1}[\alpha_{n} Jx_{n}+(1-\alpha_{n})JBv_{n}], \\ H_{n}= \{p \in K: \phi(p,y_{n})\leq\phi(p,x_{n})\}, \\ V_{n} = \{p \in K: \langle p - x_{n}, Jx_{1} - Jx_{n}\rangle\leq0\}, \\ x_{n+1}= \Pi_{H_{n} \cap V_{n}}(x_{1}),\quad n \in N. \end{cases} $$
(1.1)
Then
\(\{x_{n}\}\) generated by (
1.1) converges strongly to
\(\Pi_{A^{-1}0 \cap \operatorname{Fix}(B)\cap \operatorname{Fix}(C)}(x_{1})\).
Compared to (
1.1), the following so-called monotone projection iterative scheme for maximal monotone operator
A and strongly relatively non-expansive mapping
B in a real uniformly convex and uniformly smooth Banach space
X is presented [
4].
$$ \textstyle\begin{cases} x_{1} \in X,\qquad r_{1} > 0, \\ y_{n} = (J+r_{n}A)^{-1}J(x_{n}+e_{n}), \\ z_{n} = J^{-1}[\alpha_{n} Jx_{n}+(1-\alpha_{n})Jy_{n}], \\ u_{n} = J^{-1}[\beta_{n} Jx_{n}+(1-\beta_{n})JBz_{n}], \\ H_{1}= \{p \in X: \phi(p,z_{1})\leq\alpha_{1}\phi(p,x_{1})+(1-\alpha_{1})\phi (p,x_{1}+e_{1})\}, \\ V_{1} = \{p \in X: \phi(p,u_{1})\leq\beta_{1}\phi(p,x_{1})+(1-\beta_{1})\phi (p,z_{1})\}, \\ W_{1} = X, \\ H_{n}= \{p \in H_{n-1}\cap V_{n-1}\cap W_{n-1}: \phi(p,z_{n})\leq \alpha_{n}\phi(p,x_{n})+(1-\alpha_{n})\phi(p,x_{n}+e_{n})\}, \\ V_{n} = \{p \in H_{n-1}\cap V_{n-1}\cap W_{n-1}: \phi(p,u_{n})\leq \beta_{n}\phi(p,x_{n})+(1-\beta_{n})\phi(p,z_{n})\}, \\ W_{n} = \{p \in H_{n-1}\cap V_{n-1}\cap W_{n-1}: \langle p - x_{n}, Jx_{1} - Jx_{n}\rangle\leq0\}, \\ x_{n+1}= \Pi_{H_{n} \cap V_{n} \cap W_{n}}(x_{1}), \quad n \in N. \end{cases} $$
(1.2)
Then
\(\{x_{n}\}\) generated by (
1.2) converges strongly to
\(\Pi_{A^{-1}0 \cap \operatorname{Fix}(B)}(x_{1})\).
In recent work, Wei et al. [
8] extended the corresponding topic to the case for infinite maximal monotone operators
\(A_{i}\) and infinite weakly relatively non-expansive mappings
\(B_{i}\).
$$ \textstyle\begin{cases} x_{1} \in X,\qquad r_{1,i}\in(0,+\infty),\quad i \in N, \\ y_{n,i} = (J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),\quad i \in N, \\ z_{n,i} = J^{-1}[\alpha_{n}Jx_{n}+(1-\alpha_{n})JB_{i}y_{n,i}], \quad i \in N, \\ V_{1} = X = W_{1}, \\ V_{n+1,i} = \{p \in X: \langle y_{n,i}-p, J(x_{n}+e_{n}) - Jy_{n,i}\rangle \geq0\}, \quad i \in N, \\ V_{n+1}= (\bigcap_{i = 1}^{\infty} V_{n+1,i})\cap V_{n}, \\ W_{n+1,i} = \{p \in V_{n+1,i}: \phi(p,z_{n,i}) \leq\alpha_{n} \phi (p,x_{n}) + (1-\alpha_{n})\phi(p,y_{n,i})\},\quad i \in N, \\ W_{n+1}= (\bigcap_{i = 1}^{\infty} W_{n+1,i})\cap W_{n}, \\ U_{n+1} = \{p \in W_{n+1}: \|x_{1} - p\|^{2} \leq\|P_{W_{n+1}}(x_{1})- x_{1}\| ^{2}+\lambda_{n+1}\}, \\ x_{n+1} \in U_{n+1},\quad n \in N. \end{cases} $$
(1.3)
Then
\(\{x_{n}\}\) generated by (
1.3) converges strongly to
$$P_{\bigcap _{n=1}^{\infty}W_{n}}(x_{1})\in\Biggl(\bigcap _{i = 1}^{\infty }A_{i}^{-1}0\Biggr)\cap \Biggl(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}) \Biggr). $$
Compared to traditional (monotone) projection iterative schemes (e.g., (
1.1) and (
1.2)), some different ideas in (
1.3) can be seen. (1) Metric projection mapping
\(P_{W_{n+1}}\) instead of generalized projection mapping Π is involved in (
1.3). (2) The iterative item
\(x_{n+1}\) can be chosen arbitrarily in the set
\(U_{n+1}\), while
\(x_{n+1}\) in both (
1.1) and (
1.2) and some others are needed to be the unique value of generalized projection mapping Π. (3)
\(\{x_{n}\}\) in (
1.3) converges strongly to the unique value of metric projection mapping
P, while
\(\{ x_{n}\}\) in both (
1.1) and (
1.2) converges strongly to the unique value of the generalized projection mapping Π.
A special case of (
1.3) is presented as Corollary 2.13 in [
8]. Now, we rewrite it as follows:
$$ \textstyle\begin{cases} x_{1} \in H, \qquad e_{1} \in H, \\ y_{n} = (I+r_{n}A)^{-1}(x_{n}+e_{n}), \\ z_{n} = \alpha_{n}x_{n}+(1-\alpha_{n})By_{n}, \\ U_{1} = H = V_{1}, \\ U_{n+1} = \{p \in U_{n}: (y_{n}-p)(x_{n}+e_{n} - y_{n}) \geq0, \\ \|p-z_{n}\|^{2} \leq\alpha_{n}\|p-x_{n}\|^{2} + (1-\alpha_{n})\|p-y_{n}\|^{2}\} , \\ V_{n+1} = \{p \in U_{n+1}: \|x_{1} - p\|^{2} \leq\|P_{U_{n+1}}(x_{1})- x_{1}\| ^{2}+\lambda_{n+1}\}, \\ x_{n+1} \in V_{n+1},\quad n \in N. \end{cases} $$
(1.4)
Based on iterative scheme (
1.4), an iterative sequence is defined as follows after taking
\(H = (-\infty,+\infty)\),
\(Ax = 2x\),
\(Bx = x\) for
\(x \in(-\infty,+\infty)\),
\(e_{n} = \alpha_{n} = \lambda_{n} = \frac{1}{n}\), and
\(r_{n} = 2^{n-1}\):
$$ \textstyle\begin{cases} x_{1} = 1, \\ y_{n} = \frac{x_{n}+e_{n}}{1+2r_{n}},\quad n \in N, \\ x_{n+1}= \frac{x_{1}+y_{n}-\sqrt{(x_{1}-y_{n})^{2}+\lambda_{n+1}}}{2},\quad n \in N. \end{cases} $$
(1.5)
A computational experiment based on (
1.5) is conducted in [
8], from which we can see the effectiveness of iterative scheme (
1.4).
Inspired by the work of [
8], three questions come to our mind. (1) In iterative scheme (
1.3), in each iterative step
n, countable sets
\(V_{n+1,i}\) and
\(W_{n+1,i}\) are needed to be evaluated. It is formidable. Can we avoid it? (2)
\(x_{n+1}\) in either (
1.3) or (
1.4) can be chosen arbitrarily in a set, can a different choice of
\(x_{n+1}\) in
\(V_{n+1}\) lead to a different rate of convergence? (3) Which one is better, our new one or those in [
8]? In this paper, we shall answer the questions, construct new simple projection sets in theoretical sense, and do computational experiments for some special cases.
2 Preliminaries
In this section, we list some definitions and results we need later. The modulus of convexity of
X,
\(\delta_{X}: [0,2] \rightarrow[0,1]\), is defined as follows [
9]:
$$\delta_{X}(\epsilon) = \sup \biggl\{ 1-\frac{\|x+y\|}{2}: x,y \in X, \|x\|= \| y\| = 1, \|x - y\| \geq \epsilon \biggr\} $$
for
\(\forall\epsilon\in[0,2]\). A Banach space
X is called uniformly convex [
9] if
\(\delta_{X}(\epsilon)> 0\) for
\(\forall \epsilon\in[0,2]\). A Banach space
X is called uniformly smooth [
9] if the limit
\(\lim_{t \rightarrow0}\frac{\|x+ty\|-\|x\|}{t}\) is attained uniformly for
\((x,y)\in X\times X\) with
\(\|x\|= \|y\| = 1\).
X is said to have Property (H): if for every sequence \(\{x_{n}\} \subset X\) converging weakly to \(x \in X\) and \(\|x_{n}\| \rightarrow\|x\| \), one has \(x_{n} \rightarrow x\), as \(n \rightarrow\infty\). The uniformly convex and uniformly smooth Banach space X has Property (H).
It is well known that if
X is a real uniformly convex and uniformly smooth Banach space, then the normalized duality mapping
J is single-valued, surjective and
\(J(kx) = kJ(x)\) for
\(x \in X\) and
\(k \in (-\infty,+\infty)\). Moreover,
\(J^{-1}\) is also the normalized duality mapping from
\(X^{*}\) into
X, and both
J and
\(J^{-1}\) are uniformly continuous on each bounded subset of
X or
\(X^{*}\), respectively [
9].
Let
\(\{K_{n}\}\) be a sequence of non-empty closed and convex subsets of
X. Then the strong lower limit of
\(\{K_{n}\}\),
\(s\mbox{-}\liminf K_{n}\), is defined as the set of all
\(x \in X\) such that there exists
\(x_{n} \in K_{n}\) for almost all
n and it tends to
x as
\(n \rightarrow\infty\) in the norm; the weak upper limit of
\(\{K_{n}\}\),
\(w\mbox{-}\limsup K_{n}\), is defined as the set of all
\(x \in X\) such that there exists a subsequence
\(\{K_{n_{m}}\}\) of
\(\{K_{n}\}\) and
\(x_{n_{m}} \in K_{n_{m}}\) for every
\(n_{m}\) and it tends to
x as
\(n_{m} \rightarrow\infty\) in the weak topology; the limit of
\(\{K_{n}\}\),
\(\lim K_{n}\), is the common value when
\(s\mbox{-}\liminf K_{n} = w\mbox{-}\limsup K_{n}\) [
12].
3 Main results
In this section, our discussion is based on the following conditions:
(\(I_{1}\))
X is a real uniformly convex and uniformly smooth Banach space and \(J: X \rightarrow X^{*}\) is the normalized duality mapping;
(\(I_{2}\))
\(A_{i}: X \rightarrow X^{*}\) is maximal monotone and \(B_{i} : X \rightarrow X\) is weakly relatively non-expansive for each \(i\in N\). And \((\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i})) \neq\emptyset\);
(\(I_{3}\))
\(\{e_{n}\} \subset X\) is the error sequence such that \(e_{n} \rightarrow0\), as \(n \rightarrow\infty\);
(\(I_{4}\))
\(\{r_{n,i}\}\) and \(\{\lambda_{n}\}\) are two real number sequences in \((0,+\infty)\) with \(\inf_{n}r_{n,i} > 0\) for \(i \in N\) and \(\lambda _{n} \rightarrow0\), as \(n \rightarrow\infty\);
(\(I_{5}\))
\(\{a_{n,i}\}\) and \(\{b_{i}\}\) are two real number sequences in \((0,1)\) and \(\sum_{i = 1}^{\infty} a_{n,i} = 1 = \sum_{i = 1}^{\infty} b_{i}\) for \(n \in N\);
(\(I_{6}\))
\(\{\alpha_{n}\}\) and \(\{\beta_{n}\}\) are two real number sequences in \([0,1)\).
Proof
We split the proof into seven steps.
Step 1. \(U_{n}\) is a non-empty closed and convex subset of X for each \(n \in N\).
Noticing the definition of Lyapunov functional, we have
$$\begin{aligned}& \phi(v, y_{n}) \leq\alpha_{n} \phi(v, x_{n})+ (1-\alpha_{n}) \phi(v, x_{n}+e_{n}) \\& \quad \Longleftrightarrow\quad 2\alpha_{n} \langle v, Jx_{n}\rangle+2(1-\alpha _{n}) \bigl\langle v, J(x_{n}+e_{n}) \bigr\rangle - 2 \langle v, Jy_{n} \rangle \\& \hphantom{\quad \Longleftrightarrow\quad}\quad \leq \alpha_{n}\|x_{n} \|^{2}+ (1-\alpha_{n}) \|x_{n}+e_{n} \|^{2}- \|y_{n}\|^{2} \end{aligned}$$
and
$$\begin{aligned}& \phi(v, z_{n}) \leq\beta_{n} \phi(v, x_{n})+ (1-\beta_{n}) \phi(v, y_{n}) \\& \quad \Longleftrightarrow\quad 2\beta_{n} \langle v, Jx_{n}\rangle+2(1-\beta _{n})\langle v, Jy_{n} \rangle- 2 \langle v, Jz_{n}\rangle \\& \hphantom{\quad \Longleftrightarrow\quad}\quad \leq \beta_{n}\|x_{n} \|^{2}+ (1-\beta_{n}) \|y_{n}\|^{2}- \|z_{n}\|^{2}. \end{aligned}$$
Thus \(U_{n}\) is closed and convex for each \(n \in N\).
Next, we shall prove that \((\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap (\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\subset U_{n}\), which implies that \(U_{n} \neq\emptyset\).
For this, we shall use inductive method. Now, \(\forall q \in(\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty }\operatorname{Fix}(B_{i}))\).
If
\(n=1\), then
\(q \in U_{1} = X\) is obviously true. In view of the convexity of
\(\|\cdot\|^{2}\) and Lemma
2.5, we have
$$\begin{aligned} \phi(q, y_{1}) =& \Vert q \Vert ^{2} - 2 \Biggl\langle q, \alpha _{1}Jx_{1}+(1-\alpha_{1})\sum _{i = 1}^{\infty }a_{1,i}J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \Biggr\rangle \\ &{}+ \Biggl\Vert \alpha_{1}Jx_{1}+(1- \alpha_{1})\sum_{i = 1}^{\infty }a_{1,i}J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \Biggr\Vert ^{2} \\ \leq& \Vert q \Vert ^{2} - 2 \alpha_{1}\langle q, Jx_{1}\rangle-2(1-\alpha_{1})\sum _{i = 1}^{\infty}a_{1,i} \bigl\langle q, J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr\rangle \\ &{}+ \alpha_{1} \Vert x_{1} \Vert ^{2}+(1- \alpha_{1})\sum_{i = 1}^{\infty}a_{1,i} \bigl\Vert (J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr\Vert ^{2} \\ =& \alpha_{1} \phi(q, x_{1}) + (1-\alpha_{1}) \sum_{i = 1}^{\infty}a_{1,i}\phi \bigl(q, (J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr) \\ \leq&\alpha_{1} \phi(q, x_{1}) + (1-\alpha_{1}) \phi(q, x_{1}+e_{1}). \end{aligned}$$
Moreover, from the definition of weakly relatively non-expansive mapping, we have
$$\begin{aligned} \phi(q, z_{1}) \leq& \|q\|^{2} - 2 \beta_{1} \langle q, Jx_{1}\rangle-2(1-\beta_{1})\sum _{i = 1}^{\infty}b_{i}\langle q, JB_{i}y_{1}\rangle \\ &{}+ \beta_{1}\|x_{1}\|^{2}+(1- \beta_{1})\sum_{i = 1}^{\infty}b_{i} \|B_{i}y_{1}\|^{2} \\ =& \beta_{1} \phi(q, x_{1}) + (1-\beta_{1}) \sum_{i = 1}^{\infty}b_{i}\phi (q,B_{i}y_{1})\leq\beta_{1} \phi(q, x_{1}) + (1-\beta_{1})\phi(q, y_{1}). \end{aligned}$$
Thus \(q \in U_{2}\).
Suppose the result is true for
\(n = k+1\). Then, if
\(n = k+2\), we have
$$\begin{aligned} \phi(q, y_{k+1}) =& \Vert q \Vert ^{2} - 2 \Biggl\langle q, \alpha _{k+1}Jx_{k+1}+(1-\alpha_{k+1})\sum _{i = 1}^{\infty }a_{k+1,i}J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \Biggr\rangle \\ &{}+ \Biggl\Vert \alpha_{k+1}Jx_{k+1}+(1- \alpha_{k+1})\sum_{i = 1}^{\infty }a_{k+1,i}J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \Biggr\Vert ^{2} \\ \leq& \Vert q \Vert ^{2}- 2 \alpha_{k+1}\langle q, Jx_{k+1}\rangle \\ &{}-2(1-\alpha _{k+1})\sum _{i = 1}^{\infty}a_{k+1,i} \bigl\langle q, J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr\rangle \\ &{}+ \alpha_{k+1} \Vert x_{k+1} \Vert ^{2}+(1- \alpha_{k+1})\sum_{i = 1}^{\infty }a_{k+1,i} \bigl\Vert (J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr\Vert ^{2} \\ =& \alpha_{k+1} \phi(q, x_{k+1}) + (1-\alpha_{k+1}) \sum_{i = 1}^{\infty }a_{k+1,i}\phi \bigl(q, (J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr) \\ \leq&\alpha_{k+1} \phi(q, x_{k+1}) + (1-\alpha_{k+1}) \phi(q, x_{k+1}+e_{k+1}). \end{aligned}$$
Moreover,
$$\begin{aligned} \phi(q, z_{k+1}) \leq& \Vert q \Vert ^{2} - 2 \beta_{k+1}\langle q, Jx_{k+1}\rangle-2(1-\beta_{k+1}) \sum_{i = 1}^{\infty}b_{i}\langle q, JB_{i}y_{k+1}\rangle \\ &{}+ \beta_{k+1} \Vert x_{k+1} \Vert ^{2}+(1- \beta_{k+1})\sum_{i = 1}^{\infty}b_{i} \Vert B_{i}y_{k+1} \Vert ^{2} \\ =& \beta_{k+1} \phi(q, x_{k+1}) + (1-\beta_{k+1}) \sum_{i = 1}^{\infty }b_{i} \phi(q,B_{i}y_{k+1}) \\ \leq&\beta_{k+1} \phi(q, x_{k+1}) + (1-\beta_{k+1}) \phi(q, y_{k+1}). \end{aligned}$$
Then \(q \in U_{k+2}\). Therefore, by induction, \((\bigcap_{i = 1}^{\infty }A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\subset U_{n}\) for \(n \in N\).
Step 2. \(P_{U_{n+1}}(x_{1}) \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\), as \(n \rightarrow\infty\).
It follows from Lemma
2.6 that
\(\lim U_{n}\) exists and
\(\lim U_{n} = \bigcap_{n = 1}^{\infty} U_{n} \neq\emptyset\). Since
X has Property (H), then Lemma
2.7 implies that
\(P_{U_{n+1}}(x_{1}) \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow\infty\).
Step 3. \(V_{n} \neq\emptyset\), for \(N \cup\{0\}\), which ensures that \(\{x_{n}\}\) is well defined.
Since \(\|P_{U_{n+1}}(x_{1}) - x_{1}\| = \inf_{y \in U_{n+1}}\|y - x_{1}\|\), then for \(\lambda_{n+1}\), there exists \(\delta_{n+1} \in U_{n+1}\) such that \(\|x_{1} - \delta_{n+1}\|^{2} \leq(\inf_{y \in U_{n+1}} \|x_{1} - y\|)^{2} + \lambda_{n+1} = \|P_{U_{n+1}}(x_{1})-x_{1}\|^{2}+ \lambda_{n+1}\). This ensures that \(V_{n+1}\neq\emptyset\) for \(n \in N \cup\{0\}\).
Step 4. Both \(\{x_{n}\}\) and \(\{P_{U_{n+1}}(x_{1})\}\) are bounded.
Since \(\lambda_{n} \rightarrow0\), then there exists \(M_{1} > 0\) such that \(\lambda_{n} < M_{1}\) for \(n \in N\). Step 2 implies that \(\{ P_{U_{n+1}}(x_{1})\}\) is bounded, and then there exists \(M_{2} > 0\) such that \(\|P_{U_{n+1}}(x_{1})\| \leq M_{2}\) for \(n \in N\). Set \(M = (M_{2}+\| x_{1}\|)^{2}+M_{1}\). Since \(x_{n+1} \in V_{n+1}\), then \(\|x_{1} - x_{n+1}\|^{2} \leq\|P_{U_{n+1}}(x_{1})-x_{1}\|^{2}+ \lambda _{n+1}\leq M\), \(\forall n \in N\). Thus \(\{x_{n}\}\) is bounded.
Step 5. \(x_{n+1} - P_{U_{n+1}}(x_{1})\rightarrow0\), as \(n \rightarrow \infty\).
Since
\(x_{n+1} \in V_{n+1}\subset U_{n+1}\) and
\(U_{n}\) is a convex subset of
X, then for
\(\forall k \in(0,1)\),
\(kP_{U_{n+1}}(x_{1})+(1-k)x_{n+1}\in U_{n+1}\). Thus
$$ \bigl\Vert P_{U_{n+1}}(x_{1}) - x_{1} \bigr\Vert \leq \bigl\Vert k P_{U_{n+1}}(x_{1})+(1-k)x_{n+1}-x_{1} \bigr\Vert . $$
(3.2)
Since
\(\{x_{n}\}\) is bounded, it follows from (
3.2) and Lemma
2.8 that
$$\begin{aligned} \bigl\Vert P_{U_{n+1}}(x_{1}) - x_{1} \bigr\Vert ^{2} \leq& \bigl\Vert k P_{U_{n+1}}(x_{1})+(1-k)x_{n+1}-x_{1} \bigr\Vert ^{2} \\ \leq& k \bigl\Vert P_{U_{n+1}}(x_{1})- x_{1} \bigr\Vert ^{2} + (1-k) \Vert x_{n+1}-x_{1} \Vert ^{2} \\ &{}-k(1-k)\eta \bigl( \bigl\Vert P_{U_{n+1}}(x_{1})-x_{n+1} \bigr\Vert \bigr). \end{aligned}$$
Therefore, \(k\eta(\|P_{U_{n+1}}(x_{1})-x_{n+1}\|)\leq\|x_{n+1}-x_{1}\|^{2}-\| P_{U_{n+1}}(x_{1})-x_{1}\|^{2} \leq\lambda_{n+1}\). Letting \(k \rightarrow1\) first and then \(n \rightarrow\infty\), we know that \(P_{U_{n+1}}(x_{1})-x_{n+1} \rightarrow0\), as \(n \rightarrow\infty\).
Step 6. \(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), \(y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\) and \(z_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as \(n \rightarrow \infty\).
From Step 2 and Step 5, we know that
\(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow\infty\). And then
\(x_{n+1} - x_{n} \rightarrow0\), as
\(n \rightarrow\infty\). Since
\(x_{n+1} \in V_{n+1}\subset U_{n+1}\) and
\(e_{n} \rightarrow0\), then
$$\begin{aligned} 0 \leq&\phi(x_{n+1}, y_{n})\leq\alpha_{n} \phi(x_{n+1}, x_{n})+ (1-\alpha_{n}) \phi(x_{n+1}, x_{n}+e_{n}) \\ =& \alpha_{n} \Vert x_{n+1} \Vert ^{2}+ \alpha_{n} \Vert x_{n} \Vert ^{2} - 2 \alpha_{n} \langle x_{n+1}, Jx_{n}\rangle \\ &{}+(1-\alpha_{n}) \Vert x_{n+1} \Vert ^{2} + (1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert ^{2} -2 (1-\alpha _{n}) \bigl\langle x_{n+1}, J(x_{n}+e_{n}) \bigr\rangle \\ =& \Vert x_{n+1} \Vert ^{2} - \alpha_{n} \Vert x_{n} \Vert ^{2} - (1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert ^{2} \\ &{}+2\alpha_{n}\langle x_{n}-x_{n+1}, Jx_{n}\rangle +2 (1-\alpha_{n}) \bigl\langle x_{n}+e_{n}-x_{n+1}, J(x_{n}+e_{n}) \bigr\rangle \\ \leq& \bigl( \Vert x_{n+1} \Vert ^{2}- \Vert x_{n}+e_{n} \Vert ^{2} \bigr)+ \alpha_{n} \bigl( \Vert x_{n}+e_{n} \Vert ^{2} - \Vert x_{n} \Vert ^{2} \bigr)+2 \alpha_{n} \Vert x_{n} \Vert \Vert x_{n+1}-x_{n} \Vert \\ &{}+2(1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert \Vert x_{n}+e_{n}-x_{n+1} \Vert \rightarrow0. \end{aligned}$$
Then Lemma
2.4 implies that
\(x_{n+1} - y_{n}\rightarrow0\) and then
\(y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow \infty\).
Since
\(x_{n+1} \in V_{n+1}\subset U_{n+1}\) and
J is uniformly continuous on each bounded subset of
X, then
$$\begin{aligned} 0 \leq&\phi(x_{n+1}, z_{n})\leq\beta_{n} \phi(x_{n+1}, x_{n})+ (1-\beta_{n}) \phi(x_{n+1}, y_{n}) \\ =& \beta_{n} \bigl(\langle x_{n+1}, Jx_{n+1}- Jx_{n}\rangle+\langle x_{n} - x_{n+1}, Jx_{n}\rangle \bigr)+(1-\beta_{n})\phi(x_{n+1},y_{n}) \\ \leq&\beta_{n}\|x_{n+1}\|\|Jx_{n+1}- Jx_{n}\|+\beta_{n}\|x_{n}\|\|x_{n+1}-x_{n} \| +(1-\beta_{n})\phi(x_{n+1},y_{n})\rightarrow0. \end{aligned}$$
Using Lemma
2.4 again, we have
\(x_{n+1} - z_{n}\rightarrow0\) and then
\(z_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow\infty\).
Step 7. \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in(\bigcap_{i = 1}^{\infty}A_{i}^{-1}0) \cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\).
First, we shall show that \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in \bigcap_{i = 1}^{\infty}A_{i}^{-1}0\).
From (
3.1) and Lemma
2.5, for
\(\forall q \in(\bigcap_{i = 1}^{\infty }A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\), we have
$$\begin{aligned} \phi(q, y_{n}) \leq&\alpha_{n} \phi(q, x_{n})+ (1-\alpha _{n})\sum_{i = 1}^{\infty} a_{n,i}\phi \bigl(q, (J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}) \bigr) \\ \leq&\alpha_{n} \phi(q, x_{n}) \\ &{}+ (1-\alpha_{n}) \sum_{i = 1}^{\infty} a_{n,i} \bigl[ \phi(q, x_{n}+e_{n})-\phi \bigl((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n} \bigr) \bigr]. \end{aligned}$$
Then
$$\begin{aligned}& (1-\alpha_{n})\sum_{i = 1}^{\infty} a_{n,i}\phi \bigl((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n} \bigr) \\& \quad \leq\alpha_{n} \phi(q, x_{n})-\phi(q, y_{n})+ (1-\alpha_{n})\phi(q, x_{n}+e_{n}) \\& \quad = \alpha_{n} \bigl[\phi(q, x_{n})- \phi(q, x_{n}+e_{n}) \bigr]+ \bigl[\phi(q, x_{n}+e_{n})- \phi(q, y_{n}) \bigr] \\& \quad \leq \Vert x_{n} \Vert ^{2} - \Vert x_{n}+e_{n} \Vert ^{2}+2 \Vert q \Vert \bigl\Vert J(x_{n}+e_{n})-Jx_{n} \bigr\Vert \\& \qquad {}+ \Vert x_{n}+e_{n} \Vert ^{2}- \Vert y_{n} \Vert ^{2}+2 \Vert q \Vert \bigl\Vert Jy_{n}-J(x_{n}+e_{n}) \bigr\Vert . \end{aligned}$$
Since
\(0 \leq \sup_{n}\alpha_{n} < 1\), then
\(\sum_{i = 1}^{\infty} a_{n,i}\phi((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n}) \rightarrow0\), which implies from Lemma
2.4 that
\((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n})- (x_{n}+e_{n}) \rightarrow0\), as
\(n \rightarrow\infty\). Thus from Step 6,
\((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}) \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\), as
\(n \rightarrow\infty\).
Denote
\(u_{n,i} = (J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n})\), then
\(Ju_{n,i} + r_{n,i}A_{i}u_{n,i} = J(x_{n}+e_{n})\). Since
\(u_{n,i} \rightarrow P_{\bigcap _{m = 1}^{\infty}U_{m}}(x_{1})\),
\(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\),
\(e_{n} \rightarrow0\),
\(\inf_{n}r_{n,i} > 0\) and
J is uniformly continuous on each bounded subset of
X, then
\(A_{i}u_{n,i} \rightarrow0\) for
\(i \in N\), as
\(n \rightarrow\infty\). Using Lemma
2.2,
\(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in\bigcap_{i = 1}^{\infty}A_{i}^{-1}0\).
Next, we shall show that \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\in \bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i})\).
Since \(z_{n} = J^{-1}[ \beta_{n}Jx_{n} + (1-\beta_{n})\sum_{i = 1}^{\infty }b_{i}JB_{i}y_{n}]\), then \(Jz_{n} - Jx_{n} = (1-\beta_{n})(\sum_{i = 1}^{\infty }b_{i}JB_{i}y_{n} - Jx_{n})\). Since both J and \(J^{-1}\) are uniformly continuous on each bounded subset of X, \(z_{n} \rightarrow P_{\bigcap _{m = 1}^{\infty}U_{m}}(x_{1})\), \(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\) and \(0 \leq \sup_{n} \beta_{n} < 1\), then \(\sum_{i = 1}^{\infty }b_{i}JB_{i}y_{n} - Jx_{n} \rightarrow0\), which implies that \(J^{-1}(\sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n}) \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\), as \(n \rightarrow\infty\).
Employing Lemma
2.9, for
\(\forall q \in(\bigcap_{i = 1}^{\infty }A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\), we have
$$\begin{aligned}& \phi \Biggl(q, J^{-1} \Biggl(\sum_{i = 1}^{\infty }b_{i}JB_{i}y_{n} \Biggr) \Biggr) \\& \quad = \Vert q \Vert ^{2} - 2 \Biggl\langle q, \sum _{i = 1}^{\infty}b_{i}JB_{i}y_{n} \Biggr\rangle + \Biggl\Vert \sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} \Biggr\Vert ^{2} \\& \quad \leq \Vert q \Vert ^{2} - 2 \sum _{i = 1}^{\infty}b_{i}\langle q, JB_{i}y_{n}\rangle +\sum_{i = 1}^{\infty}b_{i} \Vert B_{i}y_{n} \Vert ^{2}-b_{1}b_{k} \eta \bigl( \Vert JB_{1}y_{n} - JB_{k} y_{n} \Vert \bigr) \\& \quad = \sum_{i = 1}^{\infty}b_{i} \phi(q,B_{i}y_{n}) - b_{1}b_{k}\eta \bigl( \Vert JB_{1}y_{n} - JB_{k} y_{n} \Vert \bigr). \end{aligned}$$
(3.3)
Since
\(Jy_{n} \rightarrow JP_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\) and
\(\sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} \rightarrow JP_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\), then from the definition of weakly relatively non-expansive mapping and (
3.3), we have
$$\begin{aligned}& b_{1}b_{k}\eta \bigl( \Vert JB_{1}y_{n} - JB_{k} y_{n} \Vert \bigr) \\& \quad \leq\sum_{i = 1}^{\infty}b_{i} \phi(q,B_{i}y_{n}) - \phi \Biggl(q,J^{-1} \Biggl( \sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} \Biggr) \Biggr) \\& \quad \leq\sum_{i = 1}^{\infty}b_{i} \phi(q,y_{n}) - \phi \Biggl(q,J^{-1} \Biggl(\sum _{i = 1}^{\infty}b_{i}JB_{i}y_{n} \Biggr) \Biggr) \\& \quad = \Vert y_{n} \Vert ^{2} - 2 \langle q, Jy_{n}\rangle+ 2\sum_{i = 1}^{\infty }b_{i} \langle q, JB_{i}y_{n}\rangle- \Biggl\Vert \sum _{i = 1}^{\infty }b_{i}JB_{i}y_{n} \Biggr\Vert ^{2}\rightarrow0, \end{aligned}$$
as
\(n \rightarrow\infty\). This ensures that
\(JB_{1}y_{n} - JB_{k}y_{n} \rightarrow0\) for
\(k \neq1\), as
\(n \rightarrow\infty\).
Since \(y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), then \(\{ y_{n}\}\) is bounded. Since \((\|q\| - \|B_{i}y_{n}\|)^{2} \leq\phi(q, B_{i}y_{n}) \leq\phi(q, y_{n})\leq(\|q\|+\|y_{n}\|)^{2}\), then \(\|B_{i}y_{n}\| \leq\|q\|\) or \(\|B_{i}y_{n}\| \leq2\|q\|+\|y_{n}\|\), \(i \in N\). Set \(K = \sup\{\|y_{n}\|: n \in N\}+2\|q\|\), then \(K < +\infty\).
Since \(\sum_{i = 1}^{\infty}b_{i} = 1\), then for \(\forall\varepsilon> 0\), there exists \(m_{0} \in N\) such that \(\sum_{i = m_{0}+1}^{\infty}b_{i} < \frac{\varepsilon}{4K}\).
Since
\(JB_{1}y_{n} - JB_{k}y_{n} \rightarrow0\), as
\(n \rightarrow\infty\), for
\(\forall k \in\{1,2,\ldots, m_{0}\}\), then we can choose
\(n_{0} \in N\) such that
\(\|JB_{1}y_{n} - JB_{k}y_{n}\|< \frac{\varepsilon}{2}\) for all
\(n \geq n_{0}\) and
\(k \in\{2,\ldots, m_{0}\}\). Then, if
\(n \geq n_{0}\),
$$\begin{aligned} \Biggl\Vert JB_{1}y_{n} - \sum _{i = 1}^{\infty}b_{i}JB_{i}y_{n} \Biggr\Vert \leq&\sum_{i = 2}^{m_{0}}b_{i} \Vert JB_{1}y_{n} - JB_{i}y_{n} \Vert + \sum_{i = m_{0}+1}^{\infty}b_{i} \Vert JB_{1}y_{n} - JB_{i}y_{n} \Vert \\ < & \Biggl(\sum_{i = 2}^{m_{0}}b_{i} \Biggr)\frac{\varepsilon}{2}+ \Biggl(\sum_{i = m_{0}+1}^{\infty}b_{i} \Biggr)2K < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \end{aligned}$$
This implies that
\(JB_{1}y_{n} - \sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} \rightarrow0\), and then
\(JB_{1}y_{n} \rightarrow JP_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow\infty\). Thus
\(B_{1}y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as
\(n \rightarrow \infty\). Lemma
2.1 implies that
\(P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\in \operatorname{Fix}(B_{1})\).
Repeating the above process for showing \(P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\in \operatorname{Fix}(B_{1})\), we can also prove that \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\in \operatorname{Fix}(B_{k})\), \(\forall k \in N\). Therefore, \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\in\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i})\).
This completes the proof. □
Proof
Copy Steps 2, 3, 4, and 5 of Theorem
3.1, and do small changes in Steps 1, 6, and 7 in the following way.
Step 1. \(U_{n}\) is a non-empty closed and convex subset of X.
We notice that
$$\begin{aligned}& \phi(v, y_{n}) \leq\alpha_{n} \phi(v, x_{1})+ (1-\alpha_{n}) \phi(v, x_{n}+e_{n}) \\& \quad \Longleftrightarrow\quad 2\alpha_{n} \langle v, Jx_{1}\rangle+2(1-\alpha _{n}) \bigl\langle v, J(x_{n}+e_{n}) \bigr\rangle - 2 \langle v, Jy_{n} \rangle \\& \hphantom{\quad \Longleftrightarrow\quad }\quad \leq \alpha_{n}\|x_{1} \|^{2}+ (1-\alpha_{n}) \|x_{n}+e_{n} \|^{2}- \|y_{n}\|^{2} \end{aligned}$$
and
$$\begin{aligned}& \phi(v, z_{n}) \leq\beta_{n} \phi(v, x_{1})+ (1-\beta_{n}) \phi(v, y_{n}) \\& \quad \Longleftrightarrow\quad 2\beta_{n} \langle v, Jx_{1}\rangle+2(1-\beta _{n})\langle v, Jy_{n} \rangle- 2 \langle v, Jz_{n}\rangle \\& \hphantom{\quad \Longleftrightarrow\quad }\quad \leq \beta_{n}\|x_{1} \|^{2}+ (1-\beta_{n}) \|y_{n}\|^{2}- \|z_{n}\|^{2}. \end{aligned}$$
Thus \(U_{n}\) is closed and convex for \(n \in N\).
Next, we shall prove that \((\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap (\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\subset U_{n}\), which ensures that \(U_{n} \neq\emptyset\).
For this, we shall use inductive method. Now, \(\forall q \in(\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty }\operatorname{Fix}(B_{i}))\).
If
\(n=1\),
\(q \in U_{1} = X \) is obviously true. In view of the convexity of
\(\|\cdot\|^{2}\) and Lemma
2.5, we have
$$\begin{aligned} \phi(q, y_{1}) =& \Vert q \Vert ^{2} - 2 \Biggl\langle q, \alpha _{1}Jx_{1}+(1-\alpha_{1})\sum _{i = 1}^{\infty }a_{1,i}J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \Biggr\rangle \\ &{}+ \Biggl\Vert \alpha_{1}Jx_{1}+(1- \alpha_{1})\sum_{i = 1}^{\infty }a_{1,i}J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \Biggr\Vert ^{2} \\ \leq& \Vert q \Vert ^{2} - 2 \alpha_{1}\langle q, Jx_{1}\rangle-2(1-\alpha_{1})\sum _{i = 1}^{\infty}a_{1,i} \bigl\langle q, J(J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr\rangle \\ &{}+ \alpha_{1} \Vert x_{1} \Vert ^{2}+(1- \alpha_{1})\sum_{i = 1}^{\infty}a_{1,i} \bigl\Vert (J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr\Vert ^{2} \\ =& \alpha_{1} \phi(q, x_{1}) + (1-\alpha_{1}) \sum_{i = 1}^{\infty}a_{1,i}\phi \bigl(q, (J+r_{1,i}A_{i})^{-1}J(x_{1}+e_{1}) \bigr) \\ \leq&\alpha_{1} \phi(q, x_{1}) + (1-\alpha_{1}) \phi(q, x_{1}+e_{1}). \end{aligned}$$
Moreover, from the definition of weakly relatively non-expansive mapping, we have
$$\begin{aligned} \phi(q, z_{1}) \leq& \Vert q \Vert ^{2} - 2 \beta_{1}\langle q, Jx_{1}\rangle-2(1-\beta_{1}) \sum_{i = 1}^{\infty}b_{i}\langle q, JB_{i}y_{1}\rangle \\ &{}+ \beta_{1} \Vert x_{1} \Vert ^{2}+(1-\beta_{1})\sum _{i = 1}^{\infty}b_{i} \Vert B_{i}y_{1} \Vert ^{2} \\ =& \beta_{1} \phi(q, x_{1}) + (1-\beta_{1}) \sum_{i = 1}^{\infty}b_{i}\phi (q,B_{i}y_{1})\leq\beta_{1} \phi(q, x_{1}) + (1-\beta_{1})\phi(q, y_{1}). \end{aligned}$$
Thus \(q \in U_{2}\).
Suppose the result is true for
\(n = k+1\). Then, if
\(n = k+2\), we have
$$\begin{aligned} \phi(q, y_{k+1}) =& \Vert q \Vert ^{2} - 2 \Biggl\langle q, \alpha _{k+1}Jx_{1}+(1-\alpha_{k+1})\sum _{i = 1}^{\infty }a_{k+1,i}J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \Biggr\rangle \\ &{}+ \Biggl\Vert \alpha_{k+1}Jx_{1}+(1- \alpha_{k+1})\sum_{i = 1}^{\infty }a_{k+1,i}J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \Biggr\Vert ^{2} \\ \leq& \Vert q \Vert ^{2} - 2 \alpha_{k+1}\langle q, Jx_{1}\rangle \\ &{}-2(1-\alpha _{k+1})\sum _{i = 1}^{\infty}a_{k+1,i} \bigl\langle q, J(J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr\rangle \\ &{}+ \alpha_{k+1} \Vert x_{1} \Vert ^{2}+(1- \alpha_{k+1})\sum_{i = 1}^{\infty }a_{k+1,i} \bigl\Vert (J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr\Vert ^{2} \\ =& \alpha_{k+1} \phi(q, x_{1}) + (1-\alpha_{k+1}) \sum_{i = 1}^{\infty }a_{k+1,i}\phi \bigl(q, (J+r_{k+1,i}A_{i})^{-1}J(x_{k+1}+e_{k+1}) \bigr) \\ \leq&\alpha_{k+1} \phi(q, x_{1}) + (1-\alpha_{k+1}) \phi(q, x_{k+1}+e_{k+1}). \end{aligned}$$
Moreover,
$$\begin{aligned} \phi(q, z_{k+1}) \leq& \Vert q \Vert ^{2} - 2 \beta_{k+1}\langle q, Jx_{1}\rangle-2(1-\beta_{k+1}) \sum_{i = 1}^{\infty}b_{i}\langle q, JB_{i}y_{k+1}\rangle \\ &{}+ \beta_{k+1} \Vert x_{1} \Vert ^{2}+(1- \beta_{k+1})\sum_{i = 1}^{\infty}b_{i} \Vert B_{i}y_{k+1} \Vert ^{2} \\ =& \beta_{k+1} \phi(q, x_{1}) + (1-\beta_{k+1}) \sum_{i = 1}^{\infty }b_{i} \phi(q,B_{i}y_{k+1}) \\ \leq&\beta_{k+1} \phi(q, x_{1}) + (1-\beta_{k+1}) \phi(q, y_{k+1}). \end{aligned}$$
Then \(q \in U_{k+2}\). Therefore, by induction, \(\emptyset\neq(\bigcap_{i = 1}^{\infty}A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty }\operatorname{Fix}(B_{i}))\subset U_{n}\), for \(n \in N\).
Step 6. \(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), \(y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), and \(z_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as \(n \rightarrow \infty\).
Following from the results of Step 2 and Step 5, \(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as \(n \rightarrow\infty\). And then \(x_{n+1} - x_{n} \rightarrow0\), as \(n \rightarrow\infty\).
Since
\(x_{n+1} \in V_{n+1}\subset U_{n+1}\),
\(\alpha_{n} \rightarrow0\), and
\(e_{n} \rightarrow0\), then
$$\begin{aligned} 0 \leq&\phi(x_{n+1}, y_{n})\leq\alpha_{n} \phi(x_{n+1}, x_{1})+ (1-\alpha_{n}) \phi(x_{n+1}, x_{n}+e_{n}) \\ =& \alpha_{n} \Vert x_{n+1} \Vert ^{2}+ \alpha_{n} \Vert x_{1} \Vert ^{2} - 2 \alpha_{n} \langle x_{n+1}, Jx_{1}\rangle \\ &{}+(1-\alpha_{n}) \Vert x_{n+1} \Vert ^{2} + (1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert ^{2} -2 (1-\alpha _{n}) \bigl\langle x_{n+1}, J(x_{n}+e_{n}) \bigr\rangle \\ =& \Vert x_{n+1} \Vert ^{2} - \alpha_{n} \Vert x_{1} \Vert ^{2} - (1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert ^{2} \\ &{}+2\alpha_{n}\langle x_{1}-x_{n+1}, Jx_{1}\rangle + 2 (1-\alpha_{n}) \bigl\langle x_{n}+e_{n}-x_{n+1}, J(x_{n}+e_{n}) \bigr\rangle \\ \leq& \bigl( \Vert x_{n+1} \Vert ^{2}- \Vert x_{n}+e_{n} \Vert ^{2} \bigr)+ \alpha_{n} \bigl( \Vert x_{n}+e_{n} \Vert ^{2} - \Vert x_{1} \Vert ^{2} \bigr)+2 \alpha_{n} \Vert x_{1} \Vert \Vert x_{n+1}-x_{1} \Vert \\ &{}+2(1-\alpha_{n}) \Vert x_{n}+e_{n} \Vert \Vert x_{n}+e_{n}-x_{n+1} \Vert \rightarrow0, \end{aligned}$$
as
\(n \rightarrow\infty\). Lemma
2.4 implies that
\(x_{n+1} - y_{n}\rightarrow0\) and then
\(y_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\) as
\(n \rightarrow\infty\).
Since
\(x_{n+1} \in V_{n+1}\subset U_{n+1}\) and
\(\beta_{n} \rightarrow0\), then
$$0 \leq\phi(x_{n+1}, z_{n})\leq\beta_{n} \phi(x_{n+1}, x_{1})+ (1-\beta_{n}) \phi(x_{n+1}, y_{n})\rightarrow0. $$
Lemma
2.4 implies that
\(x_{n+1} - z_{n}\rightarrow0\) and then
\(z_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\) as
\(n \rightarrow \infty\).
Step 7. \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in(\bigcap_{i = 1}^{\infty}A_{i}^{-1}0) \cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\).
First, we shall show that \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in \bigcap_{i = 1}^{\infty}A_{i}^{-1}0\).
From (
3.4) and Lemma
2.5, for
\(\forall q \in(\bigcap_{i = 1}^{\infty }A_{i}^{-1}0)\cap(\bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i}))\), we have
$$\begin{aligned} \phi(q, y_{n}) \leq&\alpha_{n} \phi(q, x_{1})+ (1-\alpha _{n})\sum_{i = 1}^{\infty} a_{n,i}\phi \bigl(q, (J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}) \bigr) \\ \leq&\alpha_{n} \phi(q, x_{1}) \\ &{}+ (1-\alpha_{n}) \sum_{i = 1}^{\infty} a_{n,i} \bigl[ \phi(q, x_{n}+e_{n})-\phi \bigl((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n} \bigr) \bigr]. \end{aligned}$$
Thus
$$\begin{aligned}& (1-\alpha_{n})\sum_{i = 1}^{\infty} a_{n,i}\phi \bigl((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n} \bigr) \\& \quad \leq\alpha_{n} \phi(q, x_{1})-\phi(q, y_{n})+ (1-\alpha_{n})\phi(q, x_{n}+e_{n}) \\& \quad = \alpha_{n} \bigl[\phi(q, x_{1})- \phi(q, x_{n}+e_{n}) \bigr]+ \bigl[\phi(q, x_{n}+e_{n})- \phi(q, y_{n}) \bigr] \\& \quad \leq\alpha_{n} \bigl[\phi(q, x_{1})- \phi(q, x_{n}+e_{n}) \bigr]+ \bigl( \Vert x_{n}+e_{n} \Vert ^{2}- \Vert y_{n} \Vert ^{2} \bigr)+2 \Vert q \Vert \bigl\Vert J(x_{n}+e_{n})-Jy_{n} \bigr\Vert . \end{aligned}$$
Since
\(\alpha_{n} \rightarrow0\), then
\(\sum_{i = 1}^{\infty} a_{n,i}\phi((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}),x_{n}+e_{n}) \rightarrow0\), which implies from Lemma
2.4 that
\((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n})- (x_{n}+e_{n}) \rightarrow0\), as
\(n \rightarrow\infty\). Thus
\((J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n}) \rightarrow P_{\bigcap_{m = 1}^{\infty }U_{m}}(x_{1})\), as
\(n \rightarrow\infty\).
Let
\(u_{n,i} = (J+r_{n,i}A_{i})^{-1}J(x_{n}+e_{n})\), then
\(Ju_{n,i} + r_{n,i}A_{i}u_{n,i} = J(x_{n}+e_{n})\). Since
\(u_{n,i} \rightarrow P_{\bigcap _{m = 1}^{\infty}U_{m}}(x_{1})\),
\(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\),
\(e_{n} \rightarrow0\), and
\(\inf_{n}r_{n,i} > 0\), then
\(A_{i}u_{n,i} \rightarrow0\) for
\(i \in N\), as
\(n \rightarrow\infty \). Using Lemma
2.2,
\(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1}) \in\bigcap_{i = 1}^{\infty}A_{i}^{-1}0\).
Next, we shall show that \(P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\in \bigcap_{i = 1}^{\infty}\operatorname{Fix}(B_{i})\).
Since \(z_{n} = J^{-1}[ \beta_{n}Jx_{1} + (1-\beta_{n})\sum_{i = 1}^{\infty }b_{i}JB_{i}y_{n}]\), then \(Jz_{n} - Jx_{n} = \beta_{n}(Jx_{1}-Jx_{n})+(1-\beta _{n})(\sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} - Jx_{n})\). Since both J and \(J^{-1}\) are uniformly continuous on each bounded subset of X, \(z_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), \(x_{n} \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), and \(\beta_{n} \rightarrow0\), then \(\sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n} - Jx_{n} \rightarrow0\), which implies that \(J^{-1}(\sum_{i = 1}^{\infty}b_{i}JB_{i}y_{n}) \rightarrow P_{\bigcap_{m = 1}^{\infty}U_{m}}(x_{1})\), as \(n \rightarrow\infty\).
The following proof is the same as the corresponding part in Step 7 of Theorem
3.1.
This completes the proof. □
Proof
We can easily see from iterative scheme (
3.9) that
$$ y_{n} = \frac{x_{n}+e_{n}}{1+2r_{n}} \quad \mbox{for }n \in N, $$
(3.13)
and
$$ z_{n} = \beta_{n}x_{n}+(1-\beta_{n})y_{n}\quad \mbox{for }n \in N. $$
(3.14)
From (
3.14), we can see that
\((v - z_{n})^{2} \leq\beta_{n} (v - x_{n})^{2} + (1-\beta_{n})(v-y_{n})^{2}\) is always true for
\(v \in(-\infty, +\infty)\). Then we can simplify
\(U_{n+1}\) and
\(V_{n+1}\) as follows:
$$ U_{n+1} = U_{n} \cap \bigl\{ v \in(-\infty, +\infty): 2(x_{n}+e_{n}-y_{n})v \leq (x_{n}+e_{n})^{2}-y_{n}^{2} \bigr\} \quad \mbox{for }n \in N, $$
(3.15)
and
$$\begin{aligned} \begin{aligned}[b] &V_{n+1} = U_{n+1} \cap \bigl[x_{1} - \sqrt{ \bigl(x_{1}-P_{U_{n+1}}(x_{1}) \bigr)^{2}+ \lambda _{n+1}}, x_{1} + \sqrt{ \bigl(x_{1}-P_{U_{n+1}}(x_{1}) \bigr)^{2}+\lambda_{n+1}} \bigr] \\ &\quad \mbox{for }n \in N. \end{aligned} \end{aligned}$$
(3.16)
Next, we split the proof into three parts.
Part 1. We shall show that both
\(\{x_{n}\}\) and
\(\{y_{n}\}\) generated by (
3.10) converge strongly to
\(0 \in A^{-1}0\cap \operatorname{Fix}(B)\), as
\(n\rightarrow\infty\).
By using inductive method, we first show that the following is true:
$$ \textstyle\begin{cases} x_{1} = 1,\qquad x_{2} = 1 - \frac{\sqrt{2}}{2}, \\ 0 < (1+r_{n+1})y_{n+1} < 1,\quad n \in N, \\ U_{1} = (-\infty, +\infty) = V_{1}, \\ U_{2} = (-\infty,\frac{4}{3}],\qquad V_{2} = [1-\frac{\sqrt{2}}{2}, \frac {4}{3}], \\ U_{n+1} = (-\infty, w_{n}],\quad n \in N\setminus\{1\}, \\ V_{n+1} = [x_{1} - \sqrt{(x_{1}-w_{n})^{2}+\lambda_{n+1}},w_{n}],\quad n \in N\setminus\{1\}, \\ \mbox{we may choose }x_{n+1} = x_{1} - \sqrt{(x_{1}-w_{n})^{2}+\lambda_{n+1}}, \quad n \in N \setminus\{1\}. \end{cases} $$
(3.17)
In fact, if
\(n = 1\),
\(y_{1} = \frac{x_{1}+e_{1}}{1+2r_{1}} = \frac{2}{3}\). Since
\((x_{1}+e_{1})-y_{1} = 2r_{1}y_{1} = 2y_{1} = \frac{4}{3} > 0\), then from (
3.15),
\(U_{2} = (-\infty, +\infty)\cap(-\infty,(1+r_{1})y_{1}]= (-\infty ,\frac{4}{3}]\). Thus
\(P_{U_{2}}(x_{1}) = x_{1}\). From (
3.16),
\(V_{2} = U_{2} \cap[1-\frac {\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}] = [1-\frac{\sqrt{2}}{2}, \frac{4}{3}]\). So, we may choose
\(x_{2} = 1 - \frac{\sqrt{2}}{2}\).
If
\(n = 2\),
\(y_{2} = \frac{x_{2}+e_{2}}{1+2r_{2}} = \frac{3}{10}-\frac{\sqrt {2}}{10}\) and
\(w_{2} = \min\{(1+r_{1})y_{1}, (1+r_{2})y_{2}\} = \frac{9-3\sqrt {2}}{10} = (1+r_{2})y_{2}\). It is easy to see that
\(0 < (1+r_{2})y_{2} < 1\), and then
\(x_{2} + e_{2} - y_{2} = 2r_{2}y_{2} > 0\). From (
3.15),
\(U_{3} = U_{2} \cap(-\infty, 3y_{2}] = [-\infty, \frac{4}{3}] \cap(-\infty, \frac {9-3\sqrt{2}}{10}] = (-\infty, w_{2}]\), and then
\(P_{U_{3}}(x_{1}) = w_{2} = \frac{9-3\sqrt{2}}{10}\). From (
3.16),
\(V_{3} = U_{3} \cap[x_{1}-\sqrt {(x_{1}-w_{2})^{2} + \lambda_{3}}, x_{1}+\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}}] = [1-\sqrt{(\frac{1+3\sqrt{2}}{10})^{2}+\frac{1}{3}},\frac{9-3\sqrt {2}}{10}] = [x_{1}-\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}},w_{2}]\). Then we may choose
\(x_{3} = x_{1}-\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}}\).
Suppose that (
3.17) is true for
\(n = k\). We now begin the discussion for
\(n = k+1\).
Since
\(0 < (1+r_{k+1})y_{k+1} < 1\), then
\(x_{k+1}+e_{k+1}-y_{k+1} = 2r_{k+1}y_{k+1} > 0\). From (
3.15) and (
3.13),
\(U_{k+2} = U_{k+1} \cap(-\infty, (1+r_{k+1})y_{k+1}] = (-\infty, w_{k+1}]\), and then
\(P_{U_{k+2}}(x_{1}) = w_{k+1}\).
Note that
\(w_{k+1} < 1 = x_{1} < x_{1} + \sqrt{(x_{1}-w_{k+1})^{2}+\lambda _{k+2}}\) and
\(\sqrt{(x_{1}-w_{k+1})^{2}+\lambda_{k+2}}> x_{1}-w_{k+1}> 0\), then from (
3.16) we know that
$$V_{k+2} = \bigl[x_{1} - \sqrt{(x_{1}-w_{k+1})^{2}+ \lambda_{k+2}},w_{k+1} \bigr]. $$
Then we may choose
$$x_{k+2} = x_{1} - \sqrt{(x_{1}-w_{k+1})^{2}+ \lambda_{k+2}}. $$
Since
\(y_{k+2} = \frac{x_{k+2}+e_{k+2}}{1+2r_{k+2}} = \frac {x_{k+2}}{1+2^{k+2}}+\frac{1}{(k+2)(1+2^{k+2})}\), then
\((1+r_{k+2})y_{k+2} = \frac{1+r_{k+2}}{1+2r_{k+2}}(x_{k+2}+e_{k+2})\). Note that
$$\begin{aligned} (1+r_{k+2})y_{k+2} > 0&\quad \Longleftrightarrow\quad x_{k+2}+e_{k+2} > 0 \\ &\quad\Longleftrightarrow\quad 1 + \frac{1}{k+2} > \sqrt{(1-w_{k+1})^{2}+ \lambda _{k+2}} \\ &\quad\Longleftrightarrow\quad 1+\frac{2}{k+2}+\frac{1}{(k+2)^{2}} > (1-w_{k+1})^{2}+\frac{1}{k+2} \\ &\quad\Longleftrightarrow\quad 1+\frac{1}{k+2}+\frac{1}{(k+2)^{2}} > (1-w_{k+1})^{2}. \end{aligned}$$
This is obviously true. Then
\((1+r_{k+2})y_{k+2} > 0\). Since
$$\begin{aligned} x_{k+2}+\frac{1}{k+2} =& 1-\sqrt{(1-w_{k+1})^{2}+ \frac {1}{k+2}}+\frac{1}{k+2} < w_{k+1}+ \frac{1}{k+2} \\ < & 1+ \frac{1}{k+2} < \frac{1+2^{k+2}}{1+2^{k+1}}= \frac{1+2r_{k+2}}{1+r_{k+2}}, \end{aligned}$$
then
\((1+r_{k+2})y_{k+2}= \frac{1+r_{k+2}}{1+2r_{k+2}} (x_{k+2}+e_{k+2}) < 1\).
By now, we have proved that (
3.17) is true.
In this part, it is left to prove that \(x_{n} \rightarrow0\), \(y_{n} \rightarrow0\), as \(n \rightarrow\infty\).
From (
3.17),
\(\{(1+r_{n})y_{n}\}\) is bounded, which implies that
\(\{w_{n}\}\) is bounded. Thus
\(\{x_{n}\}\) is bounded. Let
\(\{x_{n_{i}}\}\) be any subsequence of
\(\{x_{n}\}\) such that
\(\lim_{i \rightarrow\infty}x_{n_{i}} = a\). Then
\(w_{n_{i}} \rightarrow a\) and
\(y_{n_{i}}\rightarrow0\) as
\(i \rightarrow\infty\). Since
\(0 < w_{n_{i}} \leq(1+r_{n_{i}})y_{n_{i}} < 1\), then
\(0 \leq a \leq \lim_{i \rightarrow\infty}(1+r_{n_{i}})y_{n_{i}}\leq 1\). That is,
\(0 \leq a \leq \lim_{i \rightarrow\infty }r_{n_{i}}y_{n_{i}}\leq1\). From the fact that
\(2r_{n}y_{n} = x_{n} + e_{n} - y_{n}\), we have
\(\lim_{i \rightarrow\infty}(1+r_{n_{i}})y_{n_{i}} = \frac{a}{2}\). By now, we know that
\(0 \leq a \leq\frac{a}{2} \leq1\), then
\(a = 0\). This means that each strongly convergent subsequence of
\(\{x_{n}\}\) converges strongly to 0. Thus
\(x_{n} \rightarrow0 \in A^{-1}0 \cap \operatorname{Fix}(B)\), as
\(n \rightarrow\infty\). And then
\(y_{n} \rightarrow0\),
\(w_{n} \rightarrow0\), as
\(n \rightarrow\infty\).
Part 2. We shall show that both
\(\{x_{n}\}\) and
\(\{y_{n}\}\) generated by (
3.11) converge strongly to
\(0 \in A^{-1}0\cap \operatorname{Fix}(B)\), as
\(n\rightarrow\infty\).
First, we shall use inductive method to show that the following is true:
$$ \textstyle\begin{cases} x_{1} = 1, \\ 0 < (1+r_{n+1})y_{n+1} < (1+ r_{n})y_{n}, \quad n \in N, \\ \frac{1+2^{n+1}}{(n+2)2^{n+1}}< (1+r_{n+1})y_{n+1}< 1,\quad n \in N\setminus \{1\}, \\ U_{1} = (-\infty, +\infty) = V_{1},\qquad V_{2} = [1-\frac{\sqrt{2}}{2}, \frac {4}{3}], \qquad V_{3} = [1-\frac{\sqrt{3}}{3}, \frac{11}{10}], \\ U_{n+1} = (-\infty,(1+r_{n})y_{n}],\quad n \in N\setminus\{1\}, \\ V_{n+1} = [x_{1} - \sqrt{[x_{1}-(1+r_{n})y_{n}]^{2}+\lambda_{n+1}}, (1+r_{n})y_{n}],\quad n \in N\setminus\{1, 2\}, \\ \mbox{we may choose }x_{n+1} = (1+r_{n})y_{n},\quad n \in N. \end{cases} $$
(3.18)
In fact, if
\(n = 1\),
\(y_{1} = \frac{x_{1}+e_{1}}{1+2r_{1}} = \frac{2}{3}\). Since
\((x_{1}+e_{1})-y_{1} = 2r_{1}y_{1} = 2y_{1} = \frac{4}{3} > 0\), then from (
3.15),
\(U_{2} = (-\infty, +\infty)\cap(-\infty,(1+r_{1})y_{1}]= (-\infty ,\frac{4}{3}]\). Thus
\(P_{U_{2}}(x_{1}) = x_{1}\). From (
3.16),
\(V_{2} = U_{2} \cap[1-\frac {\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}] = [1-\frac{\sqrt{2}}{2}, \frac{4}{3}]\). Then we may choose
\(x_{2} = (1+r_{1})y_{1} = \frac{4}{3}\).
If
\(n = 2\),
\(y_{2} = \frac{x_{2}+e_{2}}{1+2r_{2}} = \frac{11}{30}\). It is easy to see that
\(0 < (1+r_{2})y_{2} = \frac{11}{10} < (1+r_{1})y_{1} = \frac {4}{3}\). From (
3.15),
\(U_{3} = U_{2} \cap(-\infty, 3y_{2}] = (-\infty, \frac{11}{10}] = (-\infty, (1+r_{2})y_{2}]\), and then
\(P_{U_{3}}(x_{1}) = x_{1}\). From (
3.16),
\(V_{3} = U_{3} \cap[1-\frac{\sqrt{3}}{3}, 1+\frac {\sqrt{3}}{3}]= [1-\frac{\sqrt{3}}{3}, \frac{11}{10}]\). Then we may choose
\(x_{3} = (1+r_{2})y_{2} = \frac{11}{10}\). Thus
\(y_{3} = \frac {x_{3}+e_{3}}{1+2r_{3}} = \frac{43}{270}\). It is easy to check that
\(0 < (1+r_{3})y_{3} = \frac{43}{54}< \frac{11}{10} = (1+r_{2})y_{2}\) and
\(\frac {1+2^{3}}{(2+2)2^{3}} < (1+r_{3})y_{3} < 1\).
Suppose that (
3.18) is true for
\(n = k\). Next, we show the result is true for
\(n = k+1\).
Since
\(0 < (1+r_{k+1})y_{k+1} < (1+ r_{k})y_{k} < 1\), then (
3.15) implies that
\(U_{k+2} = U_{k+1} \cap(-\infty, (1+r_{k+1})y_{k+1}] = (-\infty,(1+r_{k+1})y_{k+1}]\) and
\(P_{U_{k+2}}(x_{1}) = (1+r_{k+1})y_{k+1}\).
Note that
\((1+r_{k+1})y_{k+1} < 1 = x_{1} < x_{1} + \sqrt {[x_{1}-(1+r_{k+1})y_{k+1}]^{2}+\lambda_{k+2}}\) and
\(x_{1} - \sqrt{[x_{1}-(1+r_{k+1})y_{k+1}]^{2}+\lambda _{k+2}}<(1+r_{k+1})y_{k+1}\). Then, from (
3.16), we know that
$$V_{k+2} = \bigl[x_{1} - \sqrt{ \bigl[x_{1}-{(1+r_{k+1})y_{k+1} \bigr]^{2}+\lambda_{k+2}}}, (1+r_{k+1})y_{k+1} \bigr]. $$
Thus we may choose
$$x_{k+2} = (1+r_{k+1})y_{k+1}. $$
And then,
\(y_{k+2} = \frac{x_{k+2}+e_{k+2}}{1+2r_{k+2}} = \frac {x_{k+2}}{1+2^{k+2}}+\frac{1}{(k+2)(1+2^{k+2})}\). So
\((1+r_{k+2})y_{k+2} = \frac{1+r_{k+2}}{1+2r_{k+2}}(x_{k+2}+e_{k+2})\). Note that
$$(1+r_{k+2})y_{k+2} > 0\quad \Longleftrightarrow\quad x_{k+2}+e_{k+2} > 0 \quad \Longleftrightarrow\quad (1+r_{k+1})y_{k+1} + \frac {1}{k+2} > 0, $$
which is obviously true from the assumption. Thus
\((1+r_{k+2})y_{k+2} > 0\).
Since
\((1+r_{k+1})y_{k+1}< 1\), then
\(\frac {1+2^{k+1}}{1+2^{k+2}}[(1+r_{k+1})y_{k+1}+\frac{1}{k+2}]<\frac {1+2^{k+1}}{1+2^{k+2}} \frac{k+3}{k+2}< 1\). Thus
$$(1+r_{k+2})y_{k+2} = (1+r_{k+2}) \frac {x_{k+2}+e_{k+2}}{1+2r_{k+2}} = \frac{1+2^{k+1}}{1+2^{k+2}} \biggl[(1+r_{k+1})y_{k+1}+\frac{1}{k+2} \biggr]< 1. $$
Note that
$$\begin{aligned} &(1+r_{k+2})y_{k+2}< (1+r_{k+1})y_{k+1} \\ &\quad \Longleftrightarrow\quad \frac {1+r_{k+2}}{1+2r_{k+2}}(x_{k+2}+e_{k+2}) < (1+r_{k+1})y_{k+1} \\ &\quad\Longleftrightarrow \quad \frac {1+2^{k+1}}{1+2^{k+2}} \biggl[(1+r_{k+1})y_{k+1}+ \frac{1}{k+2} \biggr] < (1+r_{k+1})y_{k+1} \\ &\quad\Longleftrightarrow\quad \frac {2^{k+2}-2^{k+1}}{1+2^{k+2}}(1+r_{k+1})y_{k+1}> \frac {1+2^{k+1}}{(k+2)(1+2^{k+2})} \\ &\quad\Longleftrightarrow\quad (1+r_{k+1})y_{k+1}> \frac{1+2^{k+1}}{(k+2)2^{k+1}}, \end{aligned}$$
which is true from the assumption.
Compute the following:
$$\begin{aligned} (1+r_{k+2})y_{k+2} =& \frac {1+r_{k+2}}{1+2r_{k+2}}(x_{k+2}+e_{k+2}) \\ =& \frac {1+2^{k+1}}{1+2^{k+2}} \biggl[(1+r_{k+1})y_{k+1}+ \frac{1}{k+2} \biggr] \\ >& \frac{1+2^{k+1}}{1+2^{k+2}} \biggl[\frac{1+2^{k+1}}{(k+2)2^{k+1}}+\frac {1}{k+2} \biggr] \\ =& \frac{1+2^{k+1}}{(k+2)2^{k+1}} > \frac{1+2^{k+2}}{(k+3)2^{k+2}}. \end{aligned}$$
By now, we have proved that (
3.18) is true.
In this part, it is left to prove that \(x_{n} \rightarrow0\), \(y_{n} \rightarrow0\), as \(n \rightarrow\infty\).
Since
\(\{(1+r_{n})y_{n}\}\) is decreasing and bounded in
\((0,1)\), then
\(\lim_{n \rightarrow\infty}(1+r_{n})y_{n} = \lim_{n \rightarrow\infty}x_{n} = a\). Coming back to (
3.13), we know that
\(r_{n}y_{n} \rightarrow0\), as
\(n \rightarrow\infty\). Then
\(y_{n} \rightarrow0\), and then
\(x_{n} \rightarrow0\), as
\(n \rightarrow\infty\).
Part 3. We shall show that both
\(\{x_{n}\}\) and
\(\{y_{n}\}\) generated by (
3.12) converge strongly to
\(0 \in A^{-1}0\cap \operatorname{Fix}(B)\), as
\(n\rightarrow\infty\).
First, we shall use inductive method to show that the following is true:
$$ \textstyle\begin{cases} x_{1} = 1, \qquad x_{2} = \frac{7}{6} - \frac{\sqrt{2}}{4}, \\ 0 < (1+r_{n+1})y_{n+1} < 1, \quad n \in N, \\ U_{1} = (-\infty, +\infty) = V_{1}, \\ U_{2} = (-\infty,\frac{4}{3}],\qquad V_{2} = [1-\frac{\sqrt{2}}{2}, \frac {4}{3}], \\ U_{n+1} = (-\infty, w_{n}],\quad n \in N\setminus\{1\}, \\ V_{n+1} = [x_{1} - \sqrt{(x_{1}-w_{n})^{2}+\lambda_{n+1}},w_{n}],\quad n \in N\setminus\{1\}, \\ \mbox{we may choose }x_{n+1} = \frac{x_{1} - \sqrt{(x_{1}-w_{n})^{2}+\lambda _{n+1}}+w_{n}}{2},\quad n \in N \setminus\{1\}. \end{cases} $$
(3.19)
In fact, if
\(n = 1\),
\(y_{1} = \frac{x_{1}+e_{1}}{1+2r_{1}} = \frac{2}{3}\). Since
\((x_{1}+e_{1})-y_{1} = 2r_{1}y_{1} = 2y_{1} = \frac{4}{3} > 0\), then from (
3.15),
\(U_{2} = (-\infty, +\infty)\cap(-\infty,(1+r_{1})y_{1}]= (-\infty ,\frac{4}{3}]\). Then
\(P_{U_{2}}(x_{1}) = x_{1}\). From (
3.16),
\(V_{2} = U_{2} \cap[1-\frac {\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}] = [1-\frac{\sqrt{2}}{2}, \frac{4}{3}]\). Thus we may choose
\(x_{2} = \frac{1-\frac{\sqrt{2}}{2}+\frac{4}{3}}{2} = \frac{7}{6} - \frac{\sqrt{2}}{4}\).
If
\(n = 2\),
\(y_{2} = \frac{x_{2}+e_{2}}{1+2r_{2}} = \frac{1}{3}-\frac{\sqrt {2}}{20}\) and
\(w_{2} = \min\{(1+r_{1})y_{1}, (1+r_{2})y_{2}\} = 1-\frac {3\sqrt{2}}{20}\). It is easy to see that
\(0 < (1+r_{2})y_{2} = 1 -\frac {3\sqrt{2}}{20}< 1\). Thus from (
3.15),
\(U_{3} = U_{2} \cap(-\infty, 3y_{2}] = (-\infty, \frac{4}{3}] \cap(-\infty, 1-\frac{3\sqrt {2}}{20}] = (-\infty, w_{2}]\), and then
\(P_{U_{3}}(x_{1}) = 1-\frac{3\sqrt {2}}{20} = w_{2}\).
From (
3.16),
\(V_{3} = U_{3} \cap[x_{1}-\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}}, x_{1}+\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}}] = [1-\sqrt{\frac{18}{400}+\frac {1}{3}}, 1- \frac{3\sqrt{2}}{20}] = [x_{1}-\sqrt{(x_{1}-w_{2})^{2} + \lambda _{3}},w_{2}]\). Then we may choose
\(x_{3} = \frac{x_{1}-\sqrt{(x_{1}-w_{2})^{2} + \lambda_{3}}+w_{2}}{2}= 1- \frac{3\sqrt{2}}{40} - \frac{\sqrt{1362}}{120}\). We can easily check that
\(0 < (1+r_{3})y_{3} = 5y_{3} = \frac{20}{27}-\frac {9\sqrt{2}+\sqrt{1362}}{216} < 1\).
Suppose that (
3.19) is true for
\(n = k\). Next, we shall show that (
3.19) is true for
\(n = k+1\).
Since
\(0 <(1+r_{k+1})y_{k+1}< 1\), then
\(x_{k+1}+e_{k+1}-y_{k+1} = 2r_{k+1}y_{k+1} > 0\). From (
3.15),
\(U_{k+2} = U_{k+1} \cap(-\infty, (1+r_{k+1})y_{k+1}] = (-\infty, w_{k+1}]\), and
\(P_{U_{k+2}}(x_{1}) = w_{k+1}\). From (
3.16),
\(V_{k+2} = U_{k+2} \cap[x_{1} - \sqrt {(x_{1}-w_{k+1})^{2}+\lambda_{k+2}},x_{1} + \sqrt{(x_{1}-w_{k+1})^{2}+\lambda_{k+2}}]\).
Note that
\(w_{k+1} < 1 = x_{1} < x_{1} + \sqrt{(x_{1}-w_{k+1})^{2}+\lambda _{k+2}}\) and
\(\sqrt{(x_{1}-w_{k+1})^{2}+\lambda_{k+2}}> x_{1}- w_{k+1}> 0\). Then
\(V_{k+2} = [x_{1} - \sqrt{(x_{1}-w_{k+1})^{2}+\lambda_{k+2}}, w_{k+1}]\). Thus we may choose
$$x_{k+2} = \frac{x_{1} - \sqrt{(x_{1}-w_{k+1})^{2}+\lambda_{k+2}}+w_{k+1}}{2}. $$
Note that
$$\begin{aligned} (1+r_{k+2})y_{k+2} > 0&\quad \Longleftrightarrow\quad x_{k+2}+e_{k+2} > 0 \\ &\quad\Longleftrightarrow\quad \frac{1-\sqrt{(1-w_{k+1})^{2}+\frac {1}{k+2}}+w_{k+1}}{2} + \frac{1}{k+2} > 0 \\ &\quad\Longleftrightarrow\quad \frac{1+w_{k+1}}{2} + \frac{1}{k+2} > \frac{\sqrt {(1-w_{k+1})^{2}+\frac{1}{k+2}}}{2} \\ &\quad\Longleftrightarrow\quad \biggl(\frac{k+4}{k+2} \biggr)^{2}+ \frac{2(k+4)}{k+2}w_{k+1} > 1-2w_{k+1}+\frac{1}{k+2} \\ &\quad\Longleftrightarrow\quad \biggl(\frac{k+4}{k+2} \biggr)^{2}+ \frac{12+4k}{k+2}w_{k+1} > \frac {k+3}{k+2}, \end{aligned}$$
which is obviously true since
\((\frac{k+4}{k+2})^{2} > 1+\frac{1}{k+2}\). Then
\((1+r_{k+2})y_{k+2} > 0\).
Moreover,
$$\begin{aligned} &(1+r_{k+2})y_{k+2}< 1 \\ &\quad\Longleftrightarrow\quad \frac {1+r_{k+2}}{1+2r_{k+2}} \biggl(x_{k+2}+\frac{1}{k+2} \biggr) < 1 \\ &\quad\Longleftrightarrow\quad \frac{1+w_{k+1}-\sqrt{(1-w_{k+1})^{2}+\frac {1}{k+2}}}{2}< \frac{1+2r_{k+2}}{1+r_{k+2}}- \frac{1}{k+2} \\ &\quad\Longleftrightarrow\quad 1+w_{k+1}< \sqrt{(1-w_{k+1})^{2}+ \frac{1}{k+2}}+\frac {2(1+2^{k+2})}{1+2^{k+1}}-\frac{2}{k+2} \\ &\quad\Longleftrightarrow\quad w_{k+1}< \sqrt{(1-w_{k+1})^{2}+ \frac{1}{k+2}}+\frac {1+3\cdot2^{k+1}}{1+2^{k+1}}-\frac{2}{k+2} \end{aligned}$$
which is true since
\(\frac{1+3\cdot2^{k+1}}{1+2^{k+1}}-\frac{2}{k+2} > 1\). Then
\((1+r_{k+2})y_{k+2}= \frac{1+r_{k+2}}{1+2r_{k+2}} (x_{k+2}+e_{k+2}) < 1\).
By now, we have proved that (
3.19) is true.
In this part, it is left to prove that \(x_{n} \rightarrow0\), \(y_{n} \rightarrow0\), as \(n \rightarrow\infty\).
From (
3.19),
\(\{(1+r_{n})y_{n}\}\) is bounded, which implies that
\(\{w_{n}\}\) is bounded. Then we can easily check that
\(\{x_{n}\}\) is bounded. Let
\(\{ x_{n_{i}}\}\) be any subsequence of
\(\{x_{n}\}\) such that
\(\lim_{i \rightarrow\infty}x_{n_{i}} = a\). Then
\(w_{n_{i}} \rightarrow a\) and
\(y_{n_{i}}\rightarrow0\) as
\(i \rightarrow\infty\). Since
\(2r_{n}y_{n} = x_{n} + e_{n} - y_{n}\), then
\(\lim_{i \rightarrow\infty}(1+r_{n_{i}})y_{n_{i}} = \frac {a}{2}\). Since
\(0 < w_{n_{i}} \leq(1+r_{n_{i}})y_{n_{i}} < 1\), then
\(0 \leq a \leq\frac{a}{2}\leq1\). Thus
\(a = 0\). This means that each strongly convergent subsequence of
\(\{x_{n}\}\) converges strongly to 0. Thus
\(x_{n} \rightarrow0\), as
\(n \rightarrow\infty\). And then
\(y_{n} \rightarrow 0\),
\(w_{n} \rightarrow0\), as
\(n \rightarrow\infty\).
This completes the proof. □
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.