Splitting-midpoint method for zeros of the sum of accretive operator and μ-inversely strongly accretive operator in a q-uniformly smooth Banach space and its applications
- Li Wei^{1}Email author and
- Aifen Shi^{1}
https://doi.org/10.1186/s13660-015-0704-6
© Wei and Shi 2015
Received: 28 October 2014
Accepted: 19 May 2015
Published: 9 June 2015
Abstract
Combining the implicit midpoint method and the splitting method, we present a new iterative algorithm with errors to solve the problems of finding zeros of the sum of m-accretive operators and μ-inversely strongly accretive operators in a real q-uniformly smooth and uniformly convex Banach space. We obtain some strong convergence theorems, which demonstrate the relationship between the zero of the sum of m-accretive operator and μ-inversely strongly accretive operator and the solution of one kind variational inequality. Moreover, the applications of the main results on the nonlinear problems with Neumann boundaries and Signorini boundaries are demonstrated.
Keywords
μ-inversely strongly accretive operator sum zero implicit midpoint method splitting method strong convergence variational inequalityMSC
47H05 47H09 47H101 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 \(\langle x,f \rangle\).
It is well known that E is uniformly smooth if and only if \(\frac{\rho_{E}(t)}{t}\rightarrow 0\), as \(t \rightarrow 0\). Let \(q > 1\) be a real number. A Banach space E is said to be q-uniformly smooth if there exists a positive constant C such that \(\rho_{E}(t)\leq Ct^{q}\). It is obvious that a q-uniformly smooth Banach space must be uniformly smooth.
For a mapping \(T: E \rightarrow E\), we use \(\operatorname{Fix}(T)\) to denote the fixed point set of it; that is, \(\operatorname{Fix}(T) : = \{x\in E: Tx = x\}\).
For an operator \(A: D(A) \subset E \rightarrow 2^{E}\), we use \(A^{-1} 0\) to denote the set of zeros of it; that is, \(A^{-1} 0 : = \{x \in D(A) : Ax = 0\}\).
- (1)nonexpansive if$$\|Tx - Ty\| \leq \|x-y\|\quad \mbox{for } \forall x,y \in E; $$
- (2)k-Lipschitz if there exists \(k > 0\) such thatin particular, if \(0 < k < 1\), then T is called a contraction and if \(k = 1\), then T reduces to a nonexpansive mapping;$$\|Tx - Ty\| \leq k \|x - y\| \quad \mbox{for } \forall x,y \in E; $$
- (3)accretive if for all \(x, y \in E\), there exists \(j_{q}(x-y) \in J_{q}(x-y)\) such that$$\bigl\langle Tx - Ty, j_{q}(x-y)\bigr\rangle \geq 0; $$
- (4)μ-inversely strongly accretive if for all \(x, y \in E\), there exists \(j_{q}(x-y) \in J_{q}(x-y)\) such thatfor some \(\mu > 0\);$$\bigl\langle Tx - Ty, j_{q}(x-y)\bigr\rangle \geq \mu \| Tx - Ty \|^{q} $$
- (5)
m-accretive if T is accretive and \(R(I+\lambda T) = E\) for \(\forall \lambda > 0\);
- (6)strongly positive (see [2]) if E is a real smooth Banach space and there exists \(\overline{\gamma} > 0\) such thatin this case,$$\langle Tx,Jx\rangle \geq \overline{\gamma} \|x\|^{2}\quad \mbox{for } \forall x \in E; $$where I is the identity mapping and \(a \in [0,1]\), \(b \in [-1,1]\).$$\|aI-bT\| = \sup_{\|x\| \leq 1}\bigl\vert \bigl\langle (aI - bT)x, J(x)\bigr\rangle \bigr\vert , $$
We denote by \(J_{r}^{A}\) (for \(r > 0 \)) the resolvent of the accretive operator A; that is, \(J^{A}_{r} : = (I + rA)^{-1}\). It is well known that \(J^{A}_{r}\) is nonexpansive and \(\operatorname{Fix}(J_{r}^{A}) = A^{-1}0\).
Many practical problems can be reduced to finding zeros of the sum of two accretive operators; that is, \(0 \in (A+B)x\). Forward-backward splitting algorithms, which have recently received much attention from many mathematicians, were proposed by Lions and Mercier [3], by Passty [4], and, in a dual form for convex programming, by Han and Lou [5].
Based on iterative algorithm (1), much work has been done for finding \(x \in H\) such that \(x \in (A+B)^{-1}0\), where A and B are μ-inversely strongly accretive operator and m-accretive operator defined in the Hilbert space H, respectively. However, most of the existing work is undertaken in the frame of Hilbert spaces, see [3–9], etc.
- (i)
the iterative algorithm is new in the sense that it combines the idea of iterative algorithms (1)-(4);
- (ii)
the discussion is undertaken in the frame of a real q-uniformly smooth and uniformly convex Banach space, which is more general than that in a Hilbert space;
- (iii)
the assumption that ‘the normalized duality mapping J is weakly sequentially continuous’ in most of the existing related work is weakened to ‘J is weakly sequentially continuous at zero’;
- (iv)
a new path convergence theorem for nonexpansive mapping is proved, which extends the corresponding result in [11] from a Hilbert space to a real smooth and uniformly convex Banach space;
- (v)
compared to the work done in [12], strong convergence theorems are obtained instead of weak convergence theorems;
- (vi)
compared to the work done in [10], the connection between zeros of the sum of m-accretive operators and μ-inversely strongly accretive operators and the solution of one kind variational inequalities is being set up;
- (vii)
the applications of the main results on the nonlinear problems with Neumann boundaries and Signorini boundaries are demonstrated, from which we can see the connections among variational inequalities, nonlinear boundary value problems and iterative algorithms.
Next, we list some results we need in the sequel.
Lemma 1
(see [1])
Let E be a Banach space and \(f: E \rightarrow E\) be a contraction. Then f has a unique fixed point \(u \in E\).
Lemma 2
(see [13])
Let E be a real uniformly convex Banach space, C be a nonempty, closed, and convex subset of E and \(T: C \rightarrow E\) be a nonexpansive mapping such that \(\operatorname{Fix}(T) \neq \emptyset\), then \(I-T\) is demiclosed at zero.
Lemma 3
(see [14])
Lemma 4
(see [15])
Lemma 5
(see [16])
Lemma 6
(see [17])
Lemma 7
(see [18])
Assume T is a strongly positive bounded operator with coefficient \(\overline{\gamma} > 0\) on a real smooth Banach space E and \(0 < \rho \leq \|T\|^{-1}\). Then \(\|I-\rho T \| \leq 1- \rho \overline{\gamma}\).
2 Strong convergence theorems
Lemma 8
Proof
Step 1. \(T_{t}\) is a contraction for \(0 < t < \|T\|^{-1}\).
Then Lemma 1 implies that \(T_{t}\) has a unique fixed point, denoted by \(x_{t}\), which uniquely solves the fixed point equation \(x_{t} = t\eta f(x_{t}) + (I-tT)Ux_{t}\).
Step 2. \(\{x_{t}\}\) is bounded for \(t \in (0, \|T\|^{-1})\).
Step 3. \(x_{t} - Ux_{t} \rightarrow 0\), as \(t \rightarrow 0\).
Noticing the result of Step 2, we have \(\|x_{t} - Ux_{t}\| = t \|\eta f(x_{t})- TUx_{t}\| \rightarrow 0\), as \(t \rightarrow 0\).
Step 4. \(\langle(T- \eta f)x - (T-\eta f)y, J(x-y)\rangle \geq (\overline{\gamma}- k \eta)\|x-y\|^{2}\) for \(\forall x,y \in E\).
Step 5. If the variational inequality (6) has solutions, then the solution must be unique.
In view of the result of Step 4, we have \(u_{0} = v_{0}\).
Step 6. \(x_{t} \rightarrow p_{0}\in \operatorname{Fix}(U)\), as \(t \rightarrow 0\), which satisfies the variational inequality (6).
Since \(\{x_{t}\}\) is bounded as \(t \rightarrow 0^{+}\), then we can choose \(\{t_{n}\}\subset (0,1)\) such that \(t_{n} \rightarrow 0^{+}\) and \(x_{t_{n}}\rightharpoonup p_{0}\). From Lemma 2 and the result of Step 3, we see that \(p_{0} = Up_{0}\). Thus \(p_{0} \in \operatorname{Fix}(U)\). Substituting z by \(p_{0}\) in (9), then we can deduce that \(x_{t_{n}}\rightarrow p_{0}\) since J is weakly sequentially continuous at zero. Next, we shall prove that \(p_{0}\) solves the variational inequality (6).
Since \(x_{t_{n}} \rightarrow p_{0}\), then \((I-U)x_{t_{n}} \rightarrow (I-U)p_{0} = 0\), as \(n \rightarrow \infty\). Since \(\{x_{t_{n}}\}\) is bounded, \((T - \eta f)x_{t_{n}} \rightarrow (T - \eta f)p_{0}\) and J is uniformly continuous on each bounded subset of E, then taking limits on both sides of (10) we have \(\langle(T-\eta f)p_{0}, J(p_{0} - z)\rangle \leq 0\) for \(z \in \operatorname{Fix}(U)\). Thus \(p_{0}\) satisfies (6).
In a summary, we infer that each cluster point of \(\{x_{t}\}\) is equal to \(p_{0}\), which is the unique solution of the variational inequality (6).
This completes the proof. □
Lemma 9
(see [10])
Let E be a real q-uniformly smooth Banach space with constant \(K_{q}\). Let \(A: E\rightarrow E\) be a μ-inversely strongly accretive operator. Then for \(\forall r \leq (\frac{q \mu}{K_{q}})^{\frac{1}{q-1}}\), \((I-rA)\) is nonexpansive.
Theorem 10
- (i)
\(\sum_{n=0}^{\infty}\gamma_{n} = \infty\), \(\gamma_{n} \rightarrow 0\), \(\alpha_{n} \rightarrow 0\), as \(n \rightarrow \infty\);
- (ii)
\(\sum_{n=0}^{\infty}|\alpha_{n+1} - \alpha_{n}| < +\infty\), \(\sum_{n=0}^{\infty}|\gamma_{n+1} - \gamma_{n}| < +\infty\);
- (iii)
\(\sum_{n=0}^{\infty}|r_{n+1} - r_{n}| < +\infty\), \(0 < \varepsilon \leq r_{n}\leq(\frac{q \mu }{K_{q}})^{\frac{1}{q-1}}\) for \(n \geq 0\);
- (iv)
\(\sum_{n=0}^{\infty} \|e_{n}\| < +\infty\).
Proof
Let \(u_{n} = (I - r_{n}B)(\frac{x_{n}+y_{n}}{2})\) for \(n \geq 0\).
We shall split the proof into six steps.
Step 1. \(\{y_{n}\}\) is well defined.
Define \(W_{t}: E \rightarrow E\) by \(W_{t} x: = t u + (1-t)W(\frac{u+x}{2})\), where \(W : E \rightarrow E\) is nonexpansive for \(x , u \in E\), then \(W_{t}\) is a contraction for \(0 \leq t < 1\).
From Lemma 9 we know that \(J_{r_{n}}^{A}(I-r_{n} B)\) is nonexpansive, therefore \(\{y_{n}\}\) is well defined.
Step 2. \(\{x_{n}\}\), \(\{u_{n}\}\), \(\{y_{n}\}\) are all bounded.
Since \(J_{r_{n}}^{A}\) and \((I-r_{n}B)\) are nonexpansive, f is a contraction and T is bounded, then \(\{u_{n}\}\), \(\{f(x_{n})\}\), \(\{J_{r_{n}}^{A}u_{n}\}\), \(\{B(\frac{x_{n}+y_{n}}{2})\}\) and \(\{Ty_{n}\}\) are all bounded.
Set \(M_{1} = \sup\{\|u_{n}\|, \|J^{A}_{r_{n}}u_{n}\|, \|B(\frac{x_{n}+y_{n}}{2})\|, \|x_{n}\|, \eta\|f(x_{n})\|, \|Ty_{n}\|: n \geq 0\}\).
Step 3. \(\lim_{n \rightarrow \infty} \|x_{n+1} - x_{n} \| = 0\).
First, we shall discuss \(\|J_{r_{n}}^{A}u_{n}-J_{r_{n-1}}^{A}u_{n-1}\|\) for \(n \geq 1\).
From the assumptions on \(\{e_{n}\}\), \(\{\alpha_{n}\}\), \(\{\gamma_{n}\}\) and \(\{r_{n}\}\), in view of (20) and Lemma 4, we have \(\lim_{n \rightarrow \infty} \|x_{n+1} - x_{n} \| = 0\).
Step 4. Set \(W_{n} = J_{r_{n}}^{A}(I-r_{n}B)\), then \(W_{n} y_{n} - y_{n} \rightarrow 0\), as \(n \rightarrow \infty\).
It is obvious that \(W_{n}\) is nonexpansive and \((A+B)^{-1}0 = \operatorname{Fix}(W_{n})\).
Step 5. \(\limsup_{n\rightarrow +\infty}\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle \leq 0\), where \(p_{0} \in (A+B)^{-1}0\), which is the unique solution of the variational inequality (11).
Since \(W_{n}\) is nonexpansive, then Lemma 8 implies that there exists \(z_{t}\) such that \(z_{t} = t\eta f(z_{t})+(I-tT)W_{n}z_{t}\) for \(t \in (0,1)\). Moreover, \(z_{t} \rightarrow p_{0} \in \operatorname{Fix}(W_{n}) = (A+B)^{-1}0\), as \(t \rightarrow 0\); and \(p_{0}\) is the unique solution of the variational inequality (11).
So, \(\lim_{t \rightarrow 0}\limsup_{n\rightarrow +\infty}\langle TW_{n}z_{t}-\eta f(z_{t}), J(z_{t} - W_{n}y_{n}) \rangle \leq 0\) in view of Step 4.
Since \(\langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})\rangle = \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - x_{n+1})-J(p_{0} - W_{n}y_{n})\rangle + \langle Tp_{0}-\eta f(p_{0}), J(p_{0} - W_{n}y_{n})\rangle \) and \(x_{n+1}-W_{n}y_{n} \rightarrow 0\), then \(\limsup_{n \rightarrow \infty} \langle\eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0}) \rangle \leq 0\).
Step 6. \(x_{n} \rightarrow p_{0}\), as \(n \rightarrow +\infty\), where \(p_{0} \in (A+B)^{-1}0\) is the same as that in Step 5.
Let \(\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)\), \(\delta_{n}^{(2)} = \gamma_{n}[2\langle \eta f(p_{0})-Tp_{0}, J(x_{n+1}-p_{0})\rangle +2\eta \|x_{n}-p_{0}\|\|x_{n+1}-x_{n}\|]\) and \(\delta_{n}^{(3)}= 4M_{2} \|e_{n}\|\). Then (21) can be simplified as \(\|x_{n+1}-p_{0}\|^{2} \leq (1-\delta_{n}^{(1)})\|x_{n}-p_{0}\|^{2} + \delta_{n}^{(2)}+\delta_{n}^{(3)}\).
Using the assumptions, the results of Steps 2, 3 and 5 and by using Lemma 4, we know that \(x_{n} \rightarrow p_{0}\), as \(n \rightarrow +\infty\).
This completes the proof. □
Theorem 11
If \((A+B)^{-1}0 \neq \emptyset\), then under the assumptions except on \(\{e_{n}\}\) of Theorem 10, \(\{x_{n}\}\) generated by the iterative algorithm (B) converges strongly to \(p_{0} \in (A+B)^{-1}0 \), which is the unique solution of the variational inequality (11).
3 Applications
In this section, we shall demonstrate the applications of Theorem 10 to the nonlinear problems with Neumann boundaries and Signorini boundaries, respectively.
Example 1
In (C), Ω is a bounded conical domain of a Euclidean space \(R^{N}\) with its boundary \(\Gamma\in C^{1}\) (see [20]). \(h(x)\in L^{2}(\Omega)\) is a given function. ε is a nonnegative constant and ϑ denotes the exterior normal derivative of Γ, \(0 \leq C(x) \in L^{p}(\Omega)\).
Let \(\varphi:\Gamma\times R\rightarrow R\) be a given function such that, for each \(x\in\Gamma\), \(\varphi_{x}= \varphi(x,\cdot):R\rightarrow R\) is a proper, convex and lower-semi-continuous function with \(\varphi_{x}(0)=0\). Let \(\beta_{x}\) be the subdifferential of \(\varphi_{x}\), i.e., \(\beta_{x}\equiv\partial\varphi_{x}\). Suppose that \(0\in \beta_{x}(0)\) and for each \(t\in R\), the function \(x\in\Gamma \rightarrow (I+\lambda \beta_{x})^{-1}(t)\in R\) is measurable for \(\lambda >0\).
- (a)Carathéodory’s conditions:$$\begin{aligned}& x \rightarrow g(x,r)\quad \mbox{is measurable on }\Omega \mbox{ for all } r \in R^{N+1}; \\& r \rightarrow g(x,r)\quad \mbox{is continuous on }R^{N+1}\mbox{ for almost all } x\in \Omega. \end{aligned}$$
- (b)Nonexpansive with respect to \(r_{1}\), i.e.,where \((r_{1}, r_{2}, \ldots, r_{N+1}), (t_{1},\ldots,t_{N+1})\in R^{N+1}\).$$\bigl\vert g(x,r_{1},\ldots,r_{N+1})-g(x,t_{1}, \ldots,t_{N+1})\bigr\vert \leq |r_{1}-t_{1}|, $$
- (c)Monotone with respect to \(r_{1}\), i.e.,for all \(x \in \Omega\) and \((r_{1},\ldots,r_{N+1}),(t_{1},\ldots,t_{N+1})\in R^{N+1}\).$$\bigl(g(x,r_{1},\ldots,r_{N+1})-g(x,t_{1}, \ldots,t_{N+1})\bigr) (r_{1} - t_{1}) \geq 0 $$
Assume \(\frac{2N}{N+1} < p <+\infty\), \(\frac{2N}{N+1}< q<+\infty\), where \(N \geq 1\). Let \(\frac{1}{p}+\frac{1}{p'} = 1\). We use \(\| \cdot \|_{2}\) to denote the norm of \(L^{2}(\Omega)\) and \(\langle\cdot,\cdot\rangle\) to denote the inner product in \(R^{N}\), respectively.
Lemma 12
([19])
Lemma 13
([19])
The mapping \(\Phi_{p}:W^{1,p}(\Omega)\rightarrow R\) defined by \(\Phi_{p}(u)= \int_{\Gamma}\varphi_{x} (u|_{\Gamma}(x))\, d\Gamma(x)\) for any \(u\in W^{1,p}(\Omega)\) is proper convex and lower-semi-continuous on \(W^{1,p}(\Omega)\).
Lemma 14
([19])
Lemma 15
Proof
Then S is inversely strongly monotone.
This completes the proof. □
Lemma 16
([19])
For \(h(x) \in L^{2}(\Omega)\), nonlinear boundary value problem (C) has a unique solution in \(L^{2}(\Omega)\).
Lemma 17
([20])
For \(\forall \varphi \in C_{0}^{\infty}(\Omega)\), \(\langle \varphi, \partial \Phi_{p}(u)\rangle = 0\), \(u \in W^{1,p}(\Omega)\).
Lemma 18
\(u(x)\in L^{2}(\Omega)\) is the solution of (C) if and only if \(u(x) \in (A+S)^{-1}0\).
Proof
Then \(u(x) \in (A + S)^{-1}0\).
This completes the proof. □
Theorem 19
Example 2
Lemma 20
([21])
Quasi-variational inequality (F) has a solution, which implies that (E) has a solution \(u(x) \in H_{L}^{1}(\Omega)\).
Lemma 21
If \(u(x) \in (L_{1}+L_{3})^{-1}0\), where \(L_{3} u = L_{2}u - h(x)\), then \(u(x) \in H_{L}^{1}(\Omega)\) is the solution of (E).
Proof
It is easy to check that if \(u(x) \in (L_{1}+L_{3})^{-1}0\), then \(u(x)\) satisfies (F), which implies that the result is true in view of Lemma 20.
This completes the proof. □
Theorem 22
Declarations
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).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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