Abstract
In this paper, we present a new iterative algorithm with errors to solve the problems of finding zeros of the sum of finitely many m-accretive operators and finitely many α-inversely strongly accretive operators in a real smooth and uniformly convex Banach space. Strong convergence theorems are established, which extend the corresponding works given by some authors. Moreover, the relationship among the zero of the sum of m-accretive operator and α-inversely strongly accretive operator, the solution of one kind variational inequality, and the solution of the capillarity equation is investigated.
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1 Introduction and preliminaries
Let E be a real Banach space with norm \(\|\cdot\|\) and let \(E^{*}\) denote the dual space of E. We use ‘→’ and ‘⇀’ to denote strong and weak convergence either in E or in \(E^{*}\), respectively. We denote the value of \(f \in E^{*}\) at \(x \in E\) by \(\langle x,f \rangle\).
A Banach space E is said to be uniformly convex if, for each \(\varepsilon\in(0,2]\), there exists \(\delta> 0\) such that
A Banach space E is said to be smooth if
exists for each \(x , y \in\{z \in E : \|z\| = 1\}\).
In addition, we define a function \(\rho_{E}: [0,+\infty) \rightarrow [0,+\infty)\) called the modulus of smoothness of E as follows:
It is well known that E is uniformly smooth if and only if \(\frac{\rho_{E}(t)}{t}\rightarrow0\), as \(t \rightarrow0\). Let \(q > 1\) be a real number. A Banach space E is said to be q-uniformly smooth if there exists a positive constant C such that \(\rho_{E}(t)\leq Ct^{q}\). It is obvious that q-uniformly smooth Banach space must be uniformly smooth.
The normalized duality mapping \(J: E \rightarrow2^{E^{*}}\) is defined by
It is well known that J is single-valued and norm-to-norm uniformly continuous on each bounded subsets of E if E is a real smooth and uniformly convex Banach space; see [1]. Moreover, \(J(cx) = cJx\), for all \(x \in E\) and \(c \in R^{1}\). In what follows, we still denote by J the single-valued normalized duality mapping. If, E is reduced to the Hilbert space H, then \(J \equiv I\) is the identity mapping. The normalized duality mapping J is said to be weakly sequentially continuous if \(\{x_{n}\}\) is a sequence in E which converges weakly to x it follows that \(\{Jx_{n}\}\) converges in weak∗ to Jx. J is said to be weakly sequentially continuous at zero if \(\{x_{n}\}\) is a sequence in E which converges weakly to 0 it follows that \(\{Jx_{n}\}\) converges in weak∗ to 0.
Let C be a nonempty, closed, and convex subset of E and let Q be a mapping of E onto C. Then Q is said to be sunny [2] if \(Q(Q(x)+t(x-Q(x))) = Q(x)\), for all \(x \in E\) and \(t \geq 0\).
A mapping Q of E into E is said to be a retraction [2] if \(Q^{2} = Q\). If a mapping Q is a retraction, then \(Q(z) = z\) for every \(z \in R(Q)\), where \(R(Q)\) is the range of Q.
For a mapping \(U: C \rightarrow C\), we use \(\operatorname{Fix}(U)\) to denote the fixed point set of it; that is, \(\operatorname{Fix}(U) : = \{x\in C: Ux = x\}\).
For an operator \(A: D(A) \subset E \rightarrow2^{E}\), we use \(A^{-1} 0\) to denote the set of zeros of it; that is, \(A^{-1} 0 : = \{x \in D(A) : Ax = 0\}\).
Let \(T : C \rightarrow E\) be a mapping. Then T is said to be
-
(1)
nonexpansive if
$$\|Tx - Ty\| \leq\|x-y\|, \quad \mbox{for }\forall x,y \in C; $$ -
(2)
k-Lipschitz if there exists \(k > 0\) such that
$$\|Tx - Ty\| \leq k \|x - y\|,\quad \mbox{for }\forall x,y \in C. $$In particular, if \(0 < k < 1\), then T is called a contraction and if \(k = 1\), then T reduces to a nonexpansive mapping;
-
(3)
accretive if for all \(x, y \in C\), there exists \(j(x-y) \in J(x-y)\) such that
$$\bigl\langle Tx - Ty, j(x-y)\bigr\rangle \geq0, $$where J is the normalized duality mapping;
-
(4)
α-inversely strongly accretive if for all \(x, y \in C\), there exists \(j(x-y) \in J(x-y)\) such that
$$\bigl\langle Tx - Ty, j(x-y)\bigr\rangle \geq\alpha\| Tx - Ty\|^{2}, $$for some \(\alpha> 0\);
-
(5)
m-accretive if T is accretive and \(R(I+\lambda T) = E\), for \(\forall\lambda> 0\);
-
(6)
strongly positive (see [3]) if E is a real smooth Banach space and there exists \(\overline{\gamma} > 0\) such that
$$\langle Tx,Jx\rangle\geq\overline{\gamma} \|x\|^{2}, \quad \mbox{for } \forall x \in C. $$In this case,
$$\|aI-bT\| = \sup_{\|x\| \leq1}\bigl\vert \bigl\langle (aI - bT)x, J(x)\bigr\rangle \bigr\vert , $$where I is the identity mapping and \(a \in [0,1]\), \(b \in[-1,1]\).
We denote by \(J_{r}^{A}\) (for \(r > 0 \)) the resolvent of the accretive operator A; that is, \(J^{A}_{r} : = (I + rA)^{-1}\). It is well known that \(J^{A}_{r}\) is nonexpansive and \(\operatorname{Fix}(J_{r}^{A}) = A^{-1}0\).
A subset C of E is said to be a sunny nonexpansive retract of E if there exists a sunny nonexpansive retraction of E onto C.
Many practical problems can be reduced to finding zeros of the sum of two accretive operators; that is, \(0 \in(A+B)x\). Forward-backward splitting algorithms, which have recently received much attention to many mathematicians, were proposed by Lions and Mercier [4], by Passty [5], and, in a dual form for convex programming, by Han and Lou [6].
The classical forward-backward splitting algorithm is given in the following way:
Based on iterative algorithm (1), much work has been done for finding \(x \in H\) such that \(x \in(A+B)^{-1}0\), where A and B are α-inversely strongly accretive operator and m-accretive operator defined in the Hilbert space H, respectively. However, most of the existing work are undertaken in the frame of Hilbert spaces; see [4–10], etc.
Recently, Qin et al., presented the following iterative algorithm in the frame of q-uniformly smooth Banach spaces E in [11]:
where \(\{e_{n}\}\) is the error sequence, f is a contraction, A and B are α-inversely strongly accretive operator and m-accretive operator, respectively. If \((A+B)^{-1}0 \neq\emptyset\), they proved that \(\{x_{n}\}\) converges strongly to \(x = Q_{(A+B)^{-1}0}f(x)\), where \(Q_{(A+B)^{-1}0}\) is the unique sunny nonexpansive retraction of E onto \((A+B)^{-1}0\), under some conditions.
On the other hand, there are some excellent work done on approximating fixed points of nonexpansive mappings. For example, in 2009, Yao et al. presented the following iterative algorithm in the frame of Hilbert space in [12]:
where \(P_{C}\) is the metric projection from H onto C and \(T : C \rightarrow C\) is a nonexpansive mapping with \(\operatorname{Fix}(T)\neq\emptyset\). They proved that \(\{x_{n}\}\) constructed by (3) converges strongly to a fixed point of T.
In 2006, Marino and Xu, presented the following iterative algorithm in the frame of Hilbert spaces in [13]:
where f is a contraction, A is a strongly positive linear bounded operator, and T is nonexpansive. If \(\operatorname{Fix}(T)\neq \emptyset\), they proved that \(\{x_{n}\}\) converges strongly to \(p \in \operatorname{Fix}(T)\), which solves the variational inequality \(\langle (\gamma f - A)p, z - p\rangle\leq0\), for \(\forall z \in \operatorname{Fix}(T)\), under some conditions.
Our paper is organized in the following way: in Section 2, inspired by the work in [11–13], we shall present the following iterative algorithm with errors in a real smooth and uniformly convex Banach space:
where C is a nonempty, closed, and convex sunny nonexpansive retract of E, \(Q_{C}\) is the sunny nonexpansive retraction of E onto C, \(\{e_{n}\} \subset E \) is the error sequence, \(\{A_{i}\}_{i = 1}^{N}\) is a finite family of m-accretive operators and \(\{B_{i}\}_{i = 1}^{N}\) is a finite family of α-inversely strongly accretive operators. \(T : E \rightarrow E\) is a strongly positive linear bounded operator with coefficient \(\overline{\gamma}\) and \(f : E \rightarrow E \) is a contraction with coefficient \(k \in(0,1)\). \(J_{r_{n,i}}^{A_{i}} = (I + r_{n,i} A_{i} )^{-1}\), for \(i = 1,2,\ldots,N\), \(\sum_{m = 0}^{N}a_{m} = 1\), \(0 < a_{m} < 1\), for \(m = 0 , 1, 2, \ldots, N\). More detail of iterative algorithm (A) will be presented in Section 2. Then \(\{x_{n}\}\) is proved to converge strongly to \(p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\), which is also a solution of one kind variational inequality.
Our main contributions in Section 2 are:
-
(i)
the discussion is undertaken in the frame of real smooth and uniformly convex Banach space, which is more general than that in Hilbert space or in q-uniformly smooth Banach space;
-
(ii)
the assumption that ‘the normalized duality mapping J is weakly sequentially continuous’ in most of the existing related work is weaken to ‘J is weakly sequentially continuous at zero’;
-
(iii)
a new path convergence theorem (Lemma 8) is obtained which is a direct extension of the corresponding result in [13] from Hilbert space to real smooth and uniformly convex Banach space;
-
(iv)
the connection between zeros of the sum of m-accretive operators and α-inversely strongly accretive operators and the solution of one kind variational inequalities is being set up.
In Section 3, one kind capillarity equation is discussed, from which we can see the connection among the unique solution of this equation, the unique solution of one kind variational inequality and the iterative algorithm presented in Section 2.
Next, we list some results we need in sequel:
Lemma 1
(see [1])
Let E be a Banach space and \(f: E \rightarrow E\) be a contraction. Then f has a unique fixed point \(u \in E\).
Lemma 2
(see [14])
Let E be a real uniformly convex Banach space, C be a nonempty, closed, and convex subset of E and \(T: C \rightarrow E\) be a nonexpansive mapping such that \(\operatorname{Fix}(T) \neq \emptyset\), then \(I-T\) is demiclosed at zero.
Lemma 3
(see [15])
In a real Banach space E, the following inequality holds:
where \(j(x+y) \in J(x+y)\).
Lemma 4
(see [16])
Let \(\{a_{n}\}\) and \(\{c_{n}\}\) be two sequences of nonnegative real numbers satisfying
where \(\{t_{n}\}\subset(0,1)\) and \(\{b_{n}\}\) is a number sequence. Assume that \(\sum_{n=0}^{\infty}t_{n} = +\infty\), \(\limsup_{n \rightarrow\infty} \frac{b_{n}}{t_{n}} \leq0\), and \(\sum_{n=0}^{\infty}c_{n} < +\infty\). Then \(\lim_{n \rightarrow\infty }a_{n} = 0\).
Lemma 5
(see [17])
Let E be a Banach space and let A be an m-accretive operator. For \(\lambda>0\), \(\mu>0\), and \(x \in E\), one has
where \(J_{\lambda}= (I+\lambda A)^{-1}\) and \(J_{\mu}= (I+\mu A)^{-1}\).
Lemma 6
(see [18])
Let E be a real Banach space and let C be a nonempty, closed, and convex subset of E. Suppose \(A: C \rightarrow E\) is a single-valued operator and \(B: E \rightarrow 2^{E}\) is m-accretive. Then
Lemma 7
(see [19])
Assume T is a strongly positive bounded operator with coefficient \(\overline{\gamma} > 0\) on a real smooth Banach space E and \(0 < \rho\leq \|T\|^{-1}\). Then \(\|I-\rho T \| \leq1- \rho\overline{\gamma}\).
2 Strong convergence theorems
Lemma 8
Let E be a real smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny nonexpansive retract of E, and let \(Q_{C}\) be the sunny nonexpansive retraction of E onto C. Let \(f : E \rightarrow E\) be a fixed contractive mapping with coefficient \(k \in(0,1)\), \(T: E \rightarrow E\) be a strongly positive linear bounded operator with coefficient \(\overline{\gamma}\) and \(U : C \rightarrow C\) be a nonexpansive mapping. Suppose that the duality mapping \(J : E \rightarrow E^{*}\) is weakly sequentially continuous at zero, \(0 < \eta< \frac{\overline{\gamma}}{2k}\) and \(\operatorname{Fix}(U) \neq\emptyset\). If for each \(t \in(0,1)\), define \(T_{t} : E \rightarrow E\) by
then \(T_{t}\) has a fixed point \(x_{t}\), for each \(0 < t \leq \|T\|^{-1}\), which is convergent strongly to the fixed point of U, as \(t \rightarrow0\). That is, \(\lim_{t\rightarrow0}x_{t} = p_{0} \in \operatorname{Fix}(U)\). Moreover, \(p_{0}\) satisfies the following variational inequality: for \(\forall z \in \operatorname{Fix}(U)\),
Proof
Step 1. \(T_{t}\) is a contraction, for \(0 < t < \|T\|^{-1}\).
In fact, noticing Lemma 7, we have
which implies that \(T_{t}\) is a contraction since \(0 < \eta< \frac{\overline{\gamma}}{2k}\).
Then Lemma 1 implies that \(T_{t}\) has a unique fixed point, denoted by \(x_{t}\), which uniquely solves the fixed point equation \(x_{t} = t\eta f(x_{t}) + (I-tT)UQ_{C}x_{t}\).
Step 2. \(\{x_{t}\}\) is bounded, for \(t \in(0, \|T\|^{-1})\).
For \(p \in \operatorname{Fix}(U)\subset C\), we have \(p = UQ_{C}p\), then
This ensures that
Thus \(\{x_{t}\}\) is bounded, which implies that both \(\{f(x_{t})\}\) and \(\{TUQ_{C}x_{t}\}\) are bounded.
Step 3. \(x_{t} - UQ_{C}x_{t} \rightarrow0\), as \(t \rightarrow0\).
Noticing the result of Step 2, we have \(\|x_{t} - UQ_{C}x_{t}\| = t \|\eta f(x_{t})- TUQ_{C}x_{t}\| \rightarrow0\), as \(t \rightarrow 0\).
Step 4. \(\langle(T- \eta f)x - (T-\eta f)y, J(x-y)\rangle\geq (\overline{\gamma}- k \eta)\|x-y\|^{2}\), for \(\forall x,y \in E\).
In fact,
Step 5. If the variational inequality (6) has a solution, then the solution must be unique.
Suppose both \(u_{0}\in \operatorname{Fix}(U)\) and \(v_{0} \in \operatorname{Fix}(U)\) are the solutions of the variational inequality (6). Then we have
and
Adding up (7) and (8), we obtain
In view of the result of Step 4, we have \(u_{0} = v_{0}\).
Step 6. \(x_{t} \rightarrow p_{0}\in \operatorname{Fix}(U)\), as \(t \rightarrow0\), which satisfies the variational inequality (6).
For \(\forall z \in \operatorname{Fix}(U)\), \(x_{t} - z = t(\eta f(x_{t})-Tz)+(I-tT)(UQ_{C}x_{t}-z)\). Thus Lemma 3 implies that
Then
Therefore, for \(\forall z \in \operatorname{Fix}(U)\), we have
Since \(\{x_{t}\}\) is bounded as \(t \rightarrow0^{+}\), we can choose \(\{t_{n}\}\subset(0,1)\) such that \(t_{n} \rightarrow0^{+}\) and \(x_{t_{n}}\rightharpoonup p_{0}\). From Lemma 2 and the result of Step 3, we see that \(p_{0} = UQ_{C} p_{0} = Up_{0}\). Thus \(p_{0} \in \operatorname{Fix}(U)\). Substituting z by \(p_{0}\) in (9), then we can deduce that \(x_{t_{n}}\rightarrow p_{0}\) since J is weakly sequentially continuous at zero. Next, we shall prove that \(p_{0}\) solves the variational inequality (6).
Since \(x_{t} = t\eta f(x_{t}) + (I-tT)UQ_{C}x_{t}\),
For \(\forall z \in \operatorname{Fix}(U)\), since U is nonexpansive,
Since \(x_{t_{n}} \rightarrow p_{0}\), we have \((I-UQ_{C})x_{t_{n}} \rightarrow (I-UQ_{C})p_{0} = 0\), as \(n \rightarrow\infty\). Since \(\{x_{t_{n}}\}\) is bounded, \((T - \eta f)x_{t_{n}} \rightarrow(T - \eta f)p_{0}\) and J is uniformly continuous on each bounded subset of E, taking the limits on both sides of (10) we have \(\langle(T-\eta f)p_{0}, J(p_{0} - z)\rangle\leq0\), for \(z \in \operatorname{Fix}(U)\). Thus \(p_{0}\) satisfies (6).
In a summary, we infer that each cluster point of \(\{x_{t}\}\) is equal to \(p_{0}\), which is the unique solution of the variational inequality (6).
This completes the proof. □
Remark 9
Lemma 8 is a direct extension of Theorem 3.2 in [13] from Hilbert space to real smooth and uniformly convex Banach space.
Theorem 10
Let E be a real smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny nonexpansive retract of E, and let \(Q_{C}\) be the sunny nonexpansive retraction of E onto C. Let \(f : E \rightarrow E\) be a fixed contractive mapping with coefficient \(k \in(0,1)\), \(T: E \rightarrow E\) be a strongly positive linear bounded operator with coefficient \(\overline{\gamma}\). Suppose that the duality mapping \(J : E \rightarrow E^{*}\) is weakly sequentially continuous at zero, and \(0 < \eta< \frac{\overline{\gamma}}{2k}\). Let \(A_{i} : C \rightarrow2^{E}\) be m-accretive operator and \(B_{i} : C \rightarrow E\) be α-inversely strongly accretive operator, where \(i = 1,2,\ldots, N\). Suppose that, for \(\forall r > 0\) and \(i = 1,2,\ldots, N\),
Let \(\{x_{n}\}\) be generated by the iterative algorithm (A), \(0 < a_{m} < 1\), for \(m = 0, 1,2,\ldots,N\), \(\sum_{m = 0}^{N} a_{m} = 1\). Suppose \(\{e_{n}\}_{n=0}^{\infty} \subset E\), \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), and \(\{\gamma_{n}\}\) are three sequences in \((0,1)\) and \(\{r_{n,i}\}\subset(0,+\infty)\) satisfying the following conditions:
-
(i)
\(\sum_{n=0}^{\infty}\gamma_{n} = \infty\), \(\gamma_{n} \rightarrow0\), \(\frac{\gamma_{n-1}}{\gamma_{n}} \rightarrow1\), \(\beta_{n} \rightarrow1\), \(\alpha_{n} \rightarrow0\), as \(n \rightarrow\infty\);
-
(ii)
\(\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty\), \(\sum_{n=0}^{\infty}|\beta_{n+1} - \beta_{n}| < +\infty\), \(\sum_{n=0}^{\infty}(1-\gamma_{n} \overline{\gamma})\alpha_{n}\beta_{n} < +\infty\);
-
(iii)
\(\sum_{n=0}^{\infty}|r_{n+1,i} - r_{n,i}| < +\infty\) and \(r_{n,i}\geq\varepsilon> 0\), for \(n \geq0\) and \(i = 1,2,\ldots, N\);
-
(iv)
\(\sum_{n=0}^{\infty}\|e_{n}\| < +\infty\).
If \(\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \neq\emptyset\), then \(\{x_{n}\}\) converges strongly to a point \(p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\), which is the unique solution of the following variational inequality: for \(\forall z \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\),
Proof
Let \(u_{n,i} = (I - r_{n,i}B_{i})y_{n}\), \(v_{n} = a_{0} y_{n} + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}u_{n,i}\), for \(n \geq0\), where \(i = 1,2, \ldots, N\).
We shall split the proof into five steps:
Step 1. \(\{x_{n}\}\), \(\{u_{n,i}\}\) (\(i = 1,2, \ldots, N\)), \(\{y_{n}\}\), \(\{v_{n}\}\), and \(\{z_{n}\}\) are all bounded.
From the assumptions on \(B_{i}\), in view of Lemma 3, we have, for \(\forall x,y \in C\),
which implies that \((I - rB_{i})\) is nonexpansive, for \(r >0\).
Then noticing the facts that both \((I-r_{n,i}B_{i})\) and \(J_{r_{n,i}}^{A_{i}}\) are nonexpansive, for \(i = 1,2,\ldots, N\), we have, for \(\forall p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \subset C\),
Using Lemma 6, we have, for \(p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\),
Then Lemma 7 implies that for \(n \geq0\),
Noticing (11)-(13), we have, for \(n \geq0\),
By using the inductive method, we can easily get the following result from (14):
Therefore, from assumptions (ii) and (iv), we know that \(\{x_{n}\}\) is bounded.
For \(\forall p \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\), since \(\|y_{n} - p\| \leq\|(1-\alpha_{n})(x_{n}+e_{n})-p\|\leq \|x_{n}\|+\|e_{n}\|+\|p\|\), \(\{y_{n}\}\) is bounded, which implies that \(\{u_{n,i}\}\) is bounded in view of the fact that \(I-r_{n,i}B_{i}\) is nonexpansive, for each \(i = 1,2, \ldots, N\).
Moreover, \(\{J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})y_{n}\}\) is bounded since \(J_{r_{n,i}}^{A_{i}}\) is nonexpansive, for \(i =1,2,\ldots,N\). Thus \(\{v_{n}\}\) is bounded, which ensures that \(\{z_{n}\}\) is bounded. Since \(r_{n,i}\geq\varepsilon> 0\), \(B_{i}y_{n} = \frac{y_{n}-u_{n,i}}{r_{n,i}}\) is bounded, for \(n \geq0\) and \(i = 1,2, \ldots, N\).
Set \(M_{1} = \sup\{\|u_{n,i}\|, \|J^{A_{i}}_{r_{n,i}}u_{n,i}\|, \|B_{i}y_{n}\|, \|x_{n}\|, \|v_{n}\|, \|z_{n}\|, \|Tz_{n}\|, \eta\|f(x_{n})\|: n \geq0, i = 1,2, \ldots, N \}\).
Step 2. \(\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0\).
First, we shall discuss \(\|J_{r_{n,i}}^{A_{i}}u_{n,i}-J_{r_{n-1,i}}^{A_{i}}u_{n-1, i}\|\), for \(n \geq1\).
If \(r_{n-1,i}\leq r_{n,i}\), then by using Lemma 5, we have
If \(r_{n,i}\leq r_{n-1,i}\), then imitating the proof of (15), we have
Combining (15) and (16), we have, for \(n \geq1\),
Let \(M_{2} = (\frac{2}{\varepsilon}+1)M_{1}\), and using (17), we have, for \(n \geq1\),
Using (18), we have, for \(n \geq1\),
Noticing that for \(n \geq1\),
Using (19) and (20), we have, for \(n \geq1\),
From the assumptions on \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), \(\{\gamma_{n}\}\), \(\{r_{n,i}\}\), and \(\{e_{n}\}\), in view of (21), and Lemma 4, we have \(\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0\).
Step 3. Set \(W_{n} = [a_{0} I + \sum_{i = 1}^{N} a_{i} J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})]\), then \(W_{n} : C\rightarrow C\) is nonexpansive and \(\operatorname{Fix}(W_{n}) = \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\).
It is obvious that \(W_{n}\) is nonexpansive and \(\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\subset \operatorname{Fix}(W_{n})\). So we are left to show that \(\operatorname{Fix}(W_{n}) \subset\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\).
In fact, if \(p \in \operatorname{Fix}(W_{n})\), then \(W_{n}p = p\). For \(\forall q \in \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\subset \operatorname{Fix}(W_{n})\), we have
Therefore, \(\|p-q\|= (1-a_{N}) \|p-q\|+a_{N} \|J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q\|\), which implies that \(\|p-q\|= \|J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q\|\). Similarly, \(\|p-q\|= \|J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q\|=\cdots= \|J_{r_{n,N-1}}^{A_{N-1}}(I-r_{n,N-1}B_{N-1})p-q\|\).
Then \(\|p-q\|= \|\frac{a_{1}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q)+ \frac{a_{2}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,2}}^{A_{2}}(I-r_{n,2}B_{2})p-q)+ \cdots +\frac{a_{N}}{\sum_{i=1}^{N} a_{i}}(J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p-q)\|\), which implies from the strictly convexity of E that \(p-q = J_{r_{n,1}}^{A_{1}}(I-r_{n,1}B_{1})p-q = J_{r_{n,2}}^{A_{2}}(I-r_{n,2}B_{2})p-q = \cdots= J_{r_{n,N}}^{A_{N}}(I-r_{n,N}B_{N})p - q\).
Therefore, \(J_{r_{n,i}}^{A_{i}}(I-r_{n,i}B_{i})p = p\), for \(i = 1,2,\ldots,N\). Then \(p \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\). Thus \(\operatorname{Fix}(W_{n})\subset\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\).
Step 4. \(W_{n} y_{n} - y_{n} \rightarrow0\), as \(n \rightarrow\infty\), where \(W_{n}\) is the same as that in Step 3.
In fact, since both \(\{x_{n}\}\) and \(\{W_{n}y_{n}\}\) are bounded and \(\beta_{n} \rightarrow1\), as \(n \rightarrow+\infty\),
Since both \(\{f(x_{n})\}\) and \(\{Tz_{n}\}\) are bounded and \(\gamma_{n} \rightarrow0\), as \(n \rightarrow+\infty\),
Therefore
as \(n \rightarrow+\infty\), in view of the fact of Step 2.
Since \(\sum_{n=0}^{\infty}e_{n} <+\infty\) and \(\alpha_{n} \rightarrow 0\), as \(n \rightarrow\infty\),
Moreover, \(x_{n+1}-y_{n} \rightarrow0\), as \(n \rightarrow\infty\).
Step 5. \(\limsup_{n\rightarrow+\infty}\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle\leq0\), where \(p_{0} \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\), which is the unique solution of the variational inequality (∗).
Noticing the result of Step 3 and using Lemma 8, we know that there exists \(z_{t}\) such that \(z_{t} = t\eta f(z_{t})+(I-tT)W_{n}Q_{C}z_{t}\) for \(t \in(0,1)\). Moreover, \(z_{t} \rightarrow p_{0} \in \operatorname{Fix}(W_{n}) = \bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0\), as \(t \rightarrow0\). And, \(p_{0}\) is the unique solution of the variational inequality (∗).
Since \(\|z_{t}\|\leq\|z_{t} - p_{0} \|+\|p_{0}\|\), then \(\{z_{t}\}\) is bounded, as \(t \rightarrow0\). Using Lemma 3, we have
which implies that
So, \(\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\langle TW_{n}Q_{C}z_{t}-\eta f(z_{t}), J(z_{t} - W_{n}y_{n}) \rangle\leq0\) in view of Step 4.
Since \(z_{t} \rightarrow p_{0}\), \(W_{n} Q_{C}z_{t} \rightarrow W_{n} Q_{C}p_{0} = p_{0}\), as \(t \rightarrow0\) in view of Step 3. Noticing the fact that
we have \(\limsup_{n\rightarrow+\infty}\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\rangle\leq0\).
Since \(\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})\rangle= \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})-J(p_{0} - W_{n}y_{n})\rangle + \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\rangle\) and \(x_{n+1}-W_{n}y_{n} \rightarrow0\), then \(\limsup_{n \rightarrow \infty} \langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0}) \rangle\leq0\).
Step 6. \(x_{n} \rightarrow p_{0}\), as \(n \rightarrow+\infty\), where \(p_{0} \in\bigcap_{i = 1}^{N}(A_{i}+B_{i})^{-1}0 \subset C\) is the same as that in Step 5.
Let \(M_{3} = \sup\{\|(1-\alpha_{n})(x_{n}+e_{n})-p_{0}\|: n \geq0\}\). By using Lemma 3 again, we have
Using (22) and the result of Step 3, we have
Using (23) and Lemma 3, we have, for \(n \geq0\),
Let \(\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)\), \(\delta_{n}^{(2)} = \gamma_{n}[2\langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle+2\eta\|x_{n}-p_{0}\|\|x_{n+1}-x_{n}\|]\), \(\delta_{n}^{(3)} = 2M_{3}[\|e_{n}\|+ \alpha_{n} \beta_{n}(1-\gamma_{n} \overline{\gamma})\|p_{0}\|]\). Then (24) can be simplified as \(\|x_{n+1}-p_{0}\|^{2} \leq(1-\delta_{n}^{(1)})\|x_{n}-p_{0}\|^{2} + \delta_{n}^{(2)}+\delta_{n}^{(3)}\).
Using the assumptions (ii) and (iv), the results of Steps 1, 2, and 5 and by using Lemma 4, we know that \(x_{n} \rightarrow p_{0}\), as \(n \rightarrow+\infty\).
This completes the proof. □
Remark 11
The assumption that ‘the α-inversely strongly accretive operator \(B_{i} : C \rightarrow E\) satisfies for \(\forall r > 0\) and \(i = 1,2,\ldots, N\), \(\langle B_{i}x - B_{i}y, J[(I-rB_{i})x - (I-rB_{i})y]\rangle\geq0\)’ is valid, and we can find an example in Section 3 (Remark 26).
Lemma 12
(see [11])
Let E be a real q-uniformly smooth Banach space with constant \(K_{q}\) and C be a nonempty, closed, and convex subset of E. Let \(A: C \rightarrow E\) be an α-inversely strongly accretive operator. Then for \(\forall r \leq(\frac{q \alpha}{K_{q}})^{\frac{1}{q-1}}\), \((I-rA)\) is nonexpansive.
Corollary 13
Let E be a real q-uniformly smooth Banach space with constant \(K_{q}\) and also be a uniformly convex Banach space. Let C, \(Q_{C}\), f, k, η, T, J, \(A_{i}\), \(a_{m}\) (\(m = 0, 1,2,\ldots,N\)), \(\overline{\gamma}\), \(\{e_{n}\}\), \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), \(\{\gamma_{n}\}\), and \(\{r_{n,i}\}\) satisfy the same conditions as those in Theorem 10. Let \(B_{i} : C \rightarrow E\) be α-inversely strongly accretive operator, where \(i = 1,2,\ldots, N\). Let \(\{x_{n}\}\) be generated by the iterative algorithm (A). Suppose further that
-
(v)
\(r_{n,i}\leq(\frac{q \alpha}{K_{q}})^{\frac{1}{q-1}}\), for \(n \geq0\) and \(i = 1,2,\ldots, N\).
If \(\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0 \neq\emptyset\), then \(\{x_{n}\}\) converges strongly to a point \(p_{0} \in\bigcap_{i = 1}^{N} (A_{i}+B_{i})^{-1}0\), which is the unique solution of the variational inequality (∗).
Proof
Lemma 12 ensures that \((I-r_{n,i}B_{i})\) is nonexpansive, for \(n \geq0\) and \(i = 1,2,\ldots, N\). Then copy the proof of Theorem 10, the result follows.
This completes the proof. □
Corollary 14
If \(i \equiv1\), then iterative algorithm (A) becomes the following one:
Let E, C, \(Q_{C}\), f, k, η, T, J, \(\overline{\gamma}\), \(\{e_{n}\}\), \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), and \(\{\gamma_{n}\}\) satisfy the same conditions as those in Theorem 10. Let \(A: C \rightarrow E\) be m-accretive operator and \(B: C \rightarrow E\) be α-inversely strongly accretive operator satisfying that
Suppose that \(0 < a_{0} <1\), \(\{r_{n}\} \subset(0,+\infty)\) such that \(\sum_{n = 0}^{\infty} |r_{n+1}-r_{n}| <+\infty\) and \(r_{n} \geq\varepsilon> 0\) for \(n \geq0\).
If \((A+B)^{-1}0 \neq\emptyset\), then \(\{x_{n}\}\) generated by the iterative algorithm (B) converges strongly to \(p_{0} \in(A+B)^{-1}0 \), which is the unique solution of the following variational inequality: for \(\forall z \in(A+B)^{-1}0\),
Corollary 15
If \(B_{i} \equiv0\), then iterative algorithm (A) becomes the following one for approximating common zeros of finitely many m-accretive operators:
Let E, C, \(Q_{C}\), f, k, η, T, J, \(\overline{\gamma}\), \(a_{m}\) (\(m = 1,2, \ldots,N\)), \(\{e_{n}\}\), \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), and \(\{\gamma_{n}\}\), \(\{r_{n,i}\}\) satisfy the same conditions as those in Theorem 10. Let \(A_{i}: C \rightarrow E\) be m-accretive operator, \(i = 1,2, \ldots,N\).
If \(\bigcap_{i = 1}^{N} A_{i}^{-1}0 \neq\emptyset\), then \(\{x_{n}\}\) generated by (C) converges strongly to a point \(p_{0} \in\bigcap_{i = 1}^{N} A_{i}^{-1}0\), which is the unique solution of the following variational inequality: for \(\forall z \in\bigcap_{i = 1}^{N} A_{i}^{-1}0\),
Corollary 16
If \(A_{i} \equiv0\), then iterative algorithm (A) becomes to the following one for approximating common zeros of finitely many α-inversely strongly accretive operators:
Let E, C, \(Q_{C}\), f, k, η, T, J, \(\overline{\gamma}\), \(a_{m}\) (\(m = 1,2, \ldots,N\)), \(\{e_{n}\}\), \(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), and \(\{\gamma_{n}\}\), \(\{r_{n,i}\}\) satisfy the same conditions as those in Theorem 10. Let \(B_{i}: C \rightarrow E\) be an α-inversely strongly accretive operator satisfying for \(\forall r > 0\) and \(i = 1,2,\ldots, N\),
If \(\bigcap_{i = 1}^{N} B_{i}^{-1}0 \neq\emptyset\), then \(\{x_{n}\}\) generated by (D) converges strongly to a point \(p_{0} \in\bigcap_{i = 1}^{N} B_{i}^{-1}0\), which is the unique solution of the following variational inequality: for \(\forall z \in\bigcap_{i = 1}^{N} B_{i}^{-1}0\),
3 Connection with nonlinear capillarity equation
Remark 17
In the next of this paper, we have four purposes: (1) give a new example to show that the assumption that ‘the set of zeros of the sum of an m-accretive operator and an α-inversely strongly monotone operator is nonempty’ is valid; that is, \((A+B)^{-1}0 \neq\emptyset\) is meaningful; (2) set up the relation between the solution of the capillarity equation and the zero of the sum of an m-accretive operator and an α-inversely strongly accretive operator; (3) apply the iterative algorithm studied in Section 2 to approximate the solution of the capillarity equation; (4) set up the relationship between the solution of the capillarity equation and the solution of one kind variational inequality.
Remark 18
In the following, assume \(\frac{2N}{N+1} < p < +\infty\), \(1 \leq q ,r < +\infty\) if \(p \geq N\), and \(1 \leq q , r \leq\frac{Np}{N-p}\) if \(p < N\), where \(N \geq1\). \(\|\cdot\|_{p}\) denotes the norm in \(L^{p}(\Omega)\). Let \(\frac{1}{p} + \frac{1}{p'} = 1\).
We shall examine the following capillarity equation, which is a special case in [20]:
where Ω is a bounded conical domain of a Euclidean space \(R^{N}\) with its boundary \(\Gamma\in C^{1}\) (cf. [21]). \(|\cdot|\) denotes the Euclidean norm in \(R^{N}\), \(\langle \cdot,\cdot\rangle \) the Euclidean inner-product and ϑ the exterior normal derivative of Γ. λ is a nonnegative constant.
Theorem 19
(see [20])
The capillarity equation (E) has a unique solution \(u(x) \in L^{p}(\Omega)\).
Lemma 20
(see [20])
Define the mapping \(B_{p,q,r}:W^{1,p}(\Omega)\rightarrow(W^{1,p}(\Omega))^{*} \) by
for any \(u,v \in W^{1,p}(\Omega)\). Then \(B_{p,q,r}\) is everywhere defined, strictly monotone, hemi-continuous and coercive.
Lemma 21
(see [20])
Define a mapping \(A:L^{p}(\Omega)\rightarrow2^{L^{p}(\Omega)}\) as follows:
For \(u \in D(A) \), let \(Au = \{ f \in L^{p}(\Omega)|f \in B_{p,q,r}u\}\).
Then \(A:L^{p}(\Omega)\rightarrow2^{L^{p}(\Omega)}\) is m-accretive.
Lemma 22
Define a mapping \(C :L^{p}(\Omega)\rightarrow L^{p}(\Omega)\) by \(C u = u(x)\), for \(\forall u(x) \in L^{p}(\Omega)\).
Then C is 1-inversely strongly accretive.
Proof
Let \(J_{p}:L^{p}(\Omega)\rightarrow L^{p'}(\Omega)\) denote the normalized duality mapping. Then it is easy to check that \(J_{p}u = |u|^{p-1} \operatorname{sgn} u \|u\|^{2-p}_{p}\), \(\forall u\in L^{p}(\Omega)\).
Thus for \(\forall u(x), v(x)\in L^{p}(\Omega)\), \(\langle Cu - Cv, J_{p}(u-v)\rangle= \int_{\Omega}|u - v|^{p} \|u-v\|^{2-p}_{p}\, dx = \|u-v\|^{2}_{p}\), which implies that C is 1-inversely strongly accretive.
This completes the proof. □
Theorem 23
\(u(x) \in L^{p}(\Omega)\) is the unique solution of (E) if and only if \(u(x) \in(A+C)^{-1}0\).
Proof
If \(u(x)\) is the solution of (E), then
Thus for \(\forall\varphi\in C_{0}^{\infty}(\Omega)\), by using the property of generalized functions, we have
Then \(u(x) \in(A + C)^{-1}0\).
On the other hand, if \(u(x) \in(A+C)^{-1}0\), then for \(\forall \varphi\in C_{0}^{\infty}(\Omega)\),
Then \(-\operatorname{div}[(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u]+\lambda(|u|^{q-2}u + |u|^{r-2}u)+ u(x) = 0\), a.e. \(x\in\Omega\). By using the Green’s formula, we know that for any \(v \in W^{1,p}(\Omega)\),
Thus \(- \langle \vartheta,(1+\frac{|\nabla u|^{p}}{\sqrt{1+|\nabla u|^{2p}}})|\nabla u|^{p-2}\nabla u \rangle = 0\), a.e. on Γ.
Then \(u(x) \in(A+C)^{-1}0\) implies that \(u(x)\) is the solution of (E).
This completes the proof. □
Theorem 24
Suppose A and C are the same as those in Lemmas 21 and 22, respectively. Let \(T : L^{p}(\Omega)\rightarrow L^{p}(\Omega)\) be any strongly positive linear bounded operator with coefficient \(\overline{\gamma}\) and \(f: L^{p}(\Omega)\rightarrow L^{p}(\Omega)\) be a contraction with coefficient k. Suppose the following conditions are satisfied:
-
(i)
\(0 < \eta< \frac{\overline{\gamma}}{2k}\), and \(0 < a < 1\);
-
(ii)
\(\{e_{n}\}_{n=0}^{\infty} \subset L^{p}(\Omega)\), \(\sum_{n=0}^{\infty}\|e_{n}\| < +\infty\);
-
(iii)
\(\{\alpha_{n}\}\), \(\{\beta_{n}\}\), and \(\{\gamma_{n}\}\) are three sequences in \((0,1)\). \(\gamma_{n} \rightarrow0\), \(\frac{\gamma_{n-1}}{\gamma_{n}} \rightarrow1\), \(\beta_{n} \rightarrow1\), \(\alpha_{n} \rightarrow0\), as \(n \rightarrow \infty\). \(\sum_{n=0}^{\infty}\gamma_{n} = \infty\), \(\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty\), \(\sum_{n=0}^{\infty}|\beta_{n+1} - \beta_{n}| < +\infty\), \(\sum_{n=0}^{\infty}(1-\gamma_{n} \overline{\gamma})\alpha_{n}\beta_{n} < +\infty\);
-
(iv)
\(\{r_{n}\}\subset(0,1)\) such that \(\sum_{n=0}^{\infty}|r_{n+1} - r_{n}| < +\infty\), and \(1 \geq r_{n} \geq\varepsilon> 0\) for \(n \geq0\).
If we construct the following iterative algorithm:
then \(u_{n}(x)\) converges strongly to \(u(x)\in(A+C)^{-1}0\), which is the unique solution of the capillarity equation (E) and satisfies the following variational inequality: for \(\forall z(x)\in (A+C)^{-1}0\),
Remark 25
From Theorem 24 we can easily see the relationship among the solution of the capillarity equation, the solution of a variational inequality, and the zero of sum of an m-accretive operator and an α-inversely strongly accretive operator.
Remark 26
Let C be the 1-inversely strong accretive operator defined in Lemma 22, then it is obvious that C satisfies
Thus the assumption imposed on \(B_{i}\) in Theorem 10 is valid.
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Acknowledgements
This paper is supported by the National Natural Science Foundation of China (No. 11071053), Natural Science Foundation of Hebei Province (No. A2014207010), Key Project of Science and Research of Hebei Educational Department (ZH2012080) and Key Project of Science and Research of Hebei University of Economics and Business (2013KYZ01).
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Wei, L., Duan, L. A new iterative algorithm for the sum of two different types of finitely many accretive operators in Banach space and its connection with capillarity equation. Fixed Point Theory Appl 2015, 25 (2015). https://doi.org/10.1186/s13663-015-0269-6
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DOI: https://doi.org/10.1186/s13663-015-0269-6