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Forward-Backward Splitting Methods for Accretive Operators in Banach Spaces
Splitting methods have recently received much attention due to the fact that many nonlinear problems arising in applied areas such as image recovery, signal processing, and machine learning are mathematically modeled as a nonlinear operator equation and this operator is decomposed as the sum of two (possibly simpler) nonlinear operators. Most of the investigation on splitting methods is however carried out in the framework of Hilbert spaces. In this paper, we consider these methods in the setting of Banach spaces. We shall introduce two iterative forward-backward splitting methods with relaxations and errors to find zeros of the sum of two accretive operators in the Banach spaces. We shall prove the weak and strong convergence of these methods under mild conditions. We also discuss applications of these methods to variational inequalities, the split feasibility problem, and a constrained convex minimization problem.
Splitting methods have recently received much attention due to the fact that many nonlinear problems arising in applied areas such as image recovery, signal processing, and machine learning are mathematically modeled as a nonlinear operator equation and this operator is decomposed as the sum of two (possibly simpler) nonlinear operators. Splitting methods for linear equations were introduced by Peaceman and Rachford  and Douglas and Rachford . Extensions to nonlinear equations in Hilbert spaces were carried out by Kellogg  and Lions and Mercier  (see also [5–7]). The central problem is to iteratively find a zero of the sum of two monotone operators and in a Hilbert space , namely, a solution to the inclusion problem
Many problems can be formulated as a problem of form (1.1). For instance, a stationary solution to the initial value problem of the evolution equation can be recast as (1.1) when the governing maximal monotone is of the form . In optimization, it often needs  to solve a minimization problem of the form where are proper lower semicontinuous convex functions from to the extended real line , and is a bounded linear operator on . As a matter of fact, (1.3) is equivalent to (1.1) (assuming that and have a common point of continuity) with and . Here is the adjoint of and is the subdifferential operator of in the sense of convex analysis. It is known [8, 9] that the minimization problem (1.3) is widely used in image recovery, signal processing, and machine learning.
A splitting method for (1.1) means an iterative method for which each iteration involves only with the individual operators and , but not the sum . To solve (1.1), Lions and Mercier  introduced the nonlinear Peaceman-Rachford and Douglas-Rachford splitting iterative algorithms which generate a sequence by the recursion and respectively, a sequence by the recursion Here we use to denote the resolvent of a monotone operator ; that is, .
The nonlinear Peaceman-Rachford algorithm (1.4) fails, in general, to converge (even in the weak topology in the infinite-dimensional setting). This is due to the fact that the generating operator for the algorithm (1.4) is merely nonexpansive. However, the mean averages of can be weakly convergent . The nonlinear Douglashere-Rachford algorithm (1.5) always converges in the weak topology to a point and is a solution to (1.1), since the generating operator for this algorithm is firmly nonexpansive, namely, the operator is of the form , where is nonexpansive.
There is, however, little work in the existing literature on splitting methods for nonlinear operator equations in the setting of Banach spaces (though there was some work on finding a common zero of a finite family of accretive operators [10–12]).
The main difficulties are due to the fact that the inner product structure of a Hilbert space fails to be true in a Banach space. We shall in this paper use the technique of duality maps to carry out certain initiative investigations on splitting methods for accretive operators in Banach spaces. Namely, we will study splitting iterative methods for solving the inclusion problem (1.1), where and are accretive operators in a Banach space .
We will consider the case where is single-valued accretive and is possibly multivalued -accretive operators in a Banach space and assume that the inclusion (1.1) has a solution. We introduce the following two iterative methods which we call Mann-type and respectively, Halpern-type forward-backward methods with errors and which generate a sequence by the recursions where is the resolvent of the operator of order (i.e., ), and is a sequence in . We will prove weak convergence of (1.6) and strong convergence of (1.7) to a solution to (1.1) in some class of Banach spaces which will be made clear in Section 3.
The paper is organized as follows. In the next section we introduce the class of Banach spaces in which we shall study our splitting methods for solving (1.1). We also introduce the concept of accretive and -accretive operators in a Banach space. In Section 3, we discuss the splitting algorithms (1.6) and (1.7) and prove their weak and strong convergence, respectively. In Section 4, we discuss applications of both algorithms (1.6) and (1.7) to variational inequalities, fixed points of pseudocontractions, convexly constrained minimization problems, the split feasibility problem, and linear inverse problems.
Throughout the paper, is a real Banach space with norm , distance , and dual space . The symbol denotes the pairing between and , that is, , the value of at . will denote a nonempty closed convex subset of , unless otherwise stated, and the closed ball with center zero and radius . The expressions and denote the strong and weak convergence of the sequence , respectively, and stands for the set of weak limit points of the sequence .
The modulus of convexity of is the function defined by Recall that is said to be uniformly convex if for any . Let . We say that is -uniformly convex if there exists a constant so that for any .
The modulus of smoothness of is the function defined by Recall that is called uniformly smooth if . Let . We say that is -uniformly smooth if there is a so that for . It is known that is -uniformly convex if and only if is -uniformly smooth, where (. For instance, spaces are -uniformly convex and -uniformly smooth if , whereas -uniformly convex and -uniformly smooth if .
The norm of is said to be the Fréchet differentiable if, for each , exists and is attained uniformly for all such that . It can be proved that is uniformly smooth if the limit (2.3) exists and is attained uniformly for all such that . So it is trivial that a uniformly smooth Banach space has a Fréchet differentiable norm.
The subdifferential of a proper convex function is the set-valued operator defined as If is proper, convex, and lower semicontinuous, the subdifferential for any , the interior of the domain of . The generalized duality mapping is defined by If , the corresponding duality mapping is called the normalized duality mapping and denoted by . It can be proved that, for any , Thus we have the following subdifferential inequality, for any : In particular, we have, for , Some properties of the duality mappings are collected as follows.
Proposition 2.1 (see Cioranescu ). Let .(i)The Banach space is smooth if and only if the duality mapping is single valued.(ii)The Banach space is uniformly smooth if and only if the duality mapping is single-valued and norm-to-norm uniformly continuous on bounded sets of .
Among the estimates satisfied by -uniformly convex and -uniformly smooth spaces, the following ones will come in handy.
Lemma 2.2 (see Xu ). Let be given.(i)If is uniformly convex, then there exists a continuous, strictly increasing and convex function with such that
where .(ii)If is -uniformly smooth, then there exists a constant such that
The best constant satisfying (2.10) will be called the -uniform smoothness coefficient of . For instance , for , is 2-uniformly smooth with , and for , is -uniformly smooth with , where is the unique solution to the equation
In a Banach space with the Fréchet differentiable norm, there exists a function such that and for all
Recall that is a nonexpansive mapping if , for all . From now on, denotes the fixed point set of . The following lemma claims that the demiclosedness principle for nonexpansive mappings holds in uniformly convex Banach spaces.
Lemma 2.3 (see Browder ). Let be a nonempty closed convex subset of a uniformly convex space and a nonexpansive mapping with . If is a sequence in such that and , then . In particular, if , then .
A set-valued operator , with domain and range , is said to be accretive if, for all and every , It follows from Lemma 1.1 of Kato  that is accretive if and only if, for each , there exists such that An accretive operator is said to be -accretive if the range for some . It can be shown that an accretive operator is -accretive if and only if for all .
Given and , we say that an accretive operator is -inverse strongly accretive (-isa) of order if, for each , there exists such that When , we simply say -isa, instead of -isa of order 2; that is, is -isa if, for each , there exists such that
Given a subset of and a mapping , recall that is a retraction of onto if for all . We say that is sunny if, for each and , we have whenever .
The first result regarding the existence of sunny nonexpansive retractions onto the fixed point set of a nonexpansive mapping is due to Bruck.
Theorem 2.4 (see Bruck ). If is strictly convex and uniformly smooth and if is a nonexpansive mapping having a nonempty fixed point set , then there exists a sunny nonexpansive retraction of onto .
3. Splitting Methods for Accretive Operators
In this section we assume that is a real Banach space and is a nonempty closed subset of . We also assume that is a single-valued and -isa operator for some and is an -accretive operator in , with and . Moreover, we always use to denote the resolvent of of order ; that is,
It is known that the -accretiveness of implies that is single valued, defined on the entire , and firmly nonexpansive; that is, Below we fix the following notation:
Lemma 3.1. For , .
Proof. From the definition of , it follows that
This lemma alludes to the fact that in order to solve the inclusion problem (1.1), it suffices to find a fixed point of . Since is already “split,” an iterative algorithm for corresponds to a splitting algorithm for (1.1). However, to guarantee convergence (weak or strong) of an iterative algorithm for , we need good metric properties of such as nonexpansivity. To this end, we need geometric conditions on the underlying space (see Lemma 3.3).
Lemma 3.2. Given and , there holds the relation
Proof. Note that (. By the accretivity of , we have such that It turns out that This along with the triangle inequality yields that
We notice that though the resolvent of an accretive operator is always firmly nonexpansive in a general Banach space, firm nonexpansiveness is however insufficient to estimate useful bounds which are required to prove convergence of iterative algorithms for solving nonlinear equations governed by accretive operations. To overcome this difficulty, we need to impose additional properties on the underlying Banach space . Lemma 3.3 below establishes a sharper estimate than nonexpansiveness of the mapping , which is useful for us to prove the weak and strong convergence of algorithms (1.6) and (1.7).
Lemma 3.3. Let be a uniformly convex and -uniformly smooth Banach space for some . Assume that is a single-valued -isa of order in . Then, given , there exists a continuous, strictly increasing and convex function with such that, for all , where is the -uniform smoothness coefficient of (see Lemma 2.2).
Proof. Put and . Since , it follows from the accretiveness of that
Since , by the accretivity of it is easy to show that there exists such that ; hence, for is nonexpansive. Now since is uniformly convex, we can use Lemma 2.2 to find a continuous, strictly increasing and convex function , with , satisfying where the last inequality follows from the nonexpansivity of the resolvent . Letting and combining (3.10) and (3.11) yield
On the other hand, since is also -uniformly smooth and is -isa of order , we derive that Finally the required inequality (3.9) follows from (3.12) and (3.13).
Remark 3.4. Note that from Lemma 3.3 one deduces that, under the same conditions, if , then the mapping is nonexpansive.
3.1. Weak Convergence
Mann's iterative method  is a widely used method for finding a fixed point of nonexpansive mappings . We have proved that a splitting method for solving (1.1) can, under certain conditions, be reduced to a method for finding a fixed point of a nonexpansive mapping. It is therefore the purpose of this subsection to introduce and prove its weak convergence of a Mann-type forward-backward method with errors in a uniformly convex and -uniformly smooth Banach space. (See  for a similar treatment of the proximal point algorithm [23, 24] for finding zeros of monotone operators in the Hilbert space setting.) To this end we need a lemma about the uniqueness of weak cluster points of a sequence, whose proof, included here, follows the idea presented in [21, 25].
Lemma 3.5. Let be a closed convex subset of a uniformly convex Banach space with a Fréchet differentiable norm, and let be a sequence of nonexpansive self-mappings on with a nonempty common fixed point set . If and , where , then for all and all weak limit points of .
Proof. We first claim that the sequence is bounded. As a matter of fact, for each fixed and any ,
As , we can apply Lemma 2.5 to find that exists. In particular, is bounded.
Let us next prove that, for every and , the limit exists. To see this, we set which is nonexpansive. It is to see that we can rewrite in the manner where
By nonexpansivity, we have that and the summability of implies that Set
Let be a closed bounded convex subset of containing and . A result of Bruck  assures the existence of a strictly increasing continuous function with such that
for all nonexpansive, and . Applying (3.21) to each , we obtain
Now since exists, (3.19) and (3.22) together imply that Furthermore, we have
After taking first and then in (3.24) and using (3.19) and (3.23), we get Hence the limit (3.15) exists.
If we replace now and in (2.12) with and , respectively, we arrive at Since the exists, we deduce that where . Consequently, we deduce that Setting tend to , we conclude that exists. Therefore, for any two weak limit points and of , ; that is, .
Theorem 3.6. Let be a uniformly convex and -uniformly smooth Banach space. Let be an -isa of order and an -accretive operator. Assume that . We define a sequence by the perturbed iterative scheme
where , , and . Assume that(i) and ;(ii);(iii).
Then converges weakly to some .
Proof. Write . Notice that we can write
where . Then the iterative formula (3.29) turns into the form
Thus, by nonexpansivity of ,
Therefore, condition (i) implies
Take to deduce that, as and is nonexpansive,
Due to (3.33), Lemma 2.5 is applicable and we get that exists; in particular, is bounded. Let be such that , for all , and let . By (2.7) and Lemma 3.3, we have
From (3.35), assumptions (ii) and (iii), and (3.33), it turns out that Consequently, Since , there exists such that for all . Then, by Lemma 3.2,
By Lemmas 3.3 and 3.1, is nonexpansive and . We can therefore make use of Lemma 2.3 to assure that Finally we set and rewrite scheme (3.31) as where the sequence satisfies . Since is a sequence of nonexpansive mappings with as its nonempty common fixed point set, and since the space is uniformly convex with a Fréchet differentiable norm, we can apply Lemma 3.5 together with (3.39) to assert that the sequence has exactly one weak limit point; it is therefore weakly convergent.
3.2. Strong Convergence
Halpern's method  is another iterative method for finding a fixed point of nonexpansive mappings. This method has been extensively studied in the literature [28–30] (see also the recent survey ). In this section we aim to introduce and prove the strong convergence of a Halpern-type forward-backward method with errors in uniformly convex and -uniformly smooth Banach spaces. This result turns out to be new even in the setting of Hilbert spaces.
Theorem 3.7. Let be a uniformly convex and -uniformly smooth Banach space. Let be an -isa of order and an -accretive operator. Assume that . We define a sequence by the iterative scheme
where , . Assume the following conditions are satisfied:(i); (ii); (iii).
Then converges in norm to , where is the sunny nonexpansive retraction of onto .
Proof. Let , where is the sunny nonexpansive retraction of onto whose existence is ensured by Theorem 2.4. Let be a sequence generated by where we abbreviate . Hence to show the desired result, it suffices to prove that . Indeed, since and are both nonexpansive, it follows that where . According to condition (i), we can apply Lemma 2.5(ii) to conclude that as .
We next show . Indeed, since and is nonexpansive, we have Hence, we can apply Lemma 2.5(i) to claim that is bounded.
Using the inequality (2.7) with , we derive that
Let us define for all . Depending on the asymptotic behavior of the sequence we distinguish two cases.
Case 1. Suppose that there exists such that the sequence is nonincreasing; thus, exists. Since and , it follows immediately from (3.47) that Consequently,
By condition (iii), there exists such that for all . Then, by Lemma 3.2, we get
The demiclosedness principle (i.e., Lemma 2.3) implies that
Note that from inequality (3.47) we deduce that
Next we prove that
Equivalently (should ), we need to prove that
To this end, let satisfy . By Reich's theorem , we get as . Using subdifferential inequality, we deduce that where is a constant such that
Then it follows from (3.55) that
Then, letting and noting the fact that the duality map is norm-to-norm uniformly continuous on bounded sets, we get (3.54) as desired. Due to (3.53), we can apply Lemma 2.5(ii) to (3.52) to conclude that ; that is, .
Case 2. Suppose that there exists such that . Let us define
Obviously since for any . Set
Note that the sequence is nonincreasing and . Moreover, and
for any (see Lemma 3.1 of Maingé  for more details). From inequality (3.47) we get
It turns out that
Now repeating the argument of the proof of (3.53) in Case 1, we can get