Convergence analysis of a Halpern type algorithm for accretive operators

Nonlinear Analysis 75 (2012) 5027–5031

Contents lists available at SciVerse ScienceDirect

Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Convergence analysis of a Halpern type algorithm for accretive operators Youli Yu ∗ School of Mathematics and Information Engineering, Taizhou University, Linhai 317000, China

article

info

Article history: Received 6 November 2011 Accepted 9 April 2012 Communicated by Enzo Mitidieri

abstract In this paper, we study the convergence of a Halpern type proximal point algorithm for accretive operators in Banach spaces. Our results fill the gap in the work of Zhang and Song (2012) [1] and, consequently, all the results there can be corrected accordingly. © 2012 Elsevier Ltd. All rights reserved.

MSC: 47H06 47J05 47J25 47H10 47H17 Keywords: Proximal point algorithm Accretive operator Uniformly Gâteaux differentiable Uniformly convex Weakly continuous

1. Introduction Very recently, in [1], Zhang and Song considered the convergence analysis of the following Halpern type proximal point algorithms: xn+1 = αn u + βn xn + (1 − αn − βn )JrAn xn ,

(1.1)

xn+1 = βn xn + (1 − βn )JrAn (αn u + (1 − αn )xn ),

(1.2)

and where A is an accretive operator, {αn } and {βn } are two sequences in (0, 1) and {rn } ⊂ (0, +∞). They proved that under some conditions, if 0 ∈ R(A), then as n → ∞, the sequence {xn } defined by (1.1) (or (1.2)) converges strongly to a zero p of A, which is the unique solution of the variational inequality

⟨u − p, Jϕ (x − p)⟩ ≤ 0,

∀x ∈ A−1 (0).

While carefully reading their work, we discovered that there is a gap in the proof of [1, Theorem 3.1]. In [1, Theorem 3.1], in order to prove limn→∞ g (∥xn − JrAn xn ∥) = 0 (see page 5), they discussed two cases: Case 1. 41 (1 − αn − βn )g (∥xn − JrAn xn ∥) ≤ αn ∥u − p∥2 for all n ≥ 0. Case 2. 41 (1 − αn − βn )g (∥xn − JrAn xn ∥) > αn ∥u − p∥2 for all n ≥ 0.

∗

Tel.: +86 576 85137063; fax: +86 576 85137063. E-mail addresses: [email protected], [email protected].

0362-546X/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2012.04.017

(1.3)

5028

Y. Yu / Nonlinear Analysis 75 (2012) 5027–5031

It is clear that the authors ignored another possible case: that there exist two subsequences {xni } and {xnj } of {xn } satisfying 1 4

(1 − αni − βni )g (∥xni − JrAn xni ∥) ≤ αni ∥u − p∥2 for all i ≥ 0 i

and 1 4

(1 − αnj − βnj )g (∥xnj − JrAn xnj ∥) > αnj ∥u − p∥2 for all j ≥ 0. j

Hence, the proof of [1, Theorem 3.1] is incomplete. Consequently, all theorems in [1, Theorems 3.1, 3.2, 3.7 and 3.8] are questionable. The main purpose of the present paper is to correct these gaps in [1]. Consequently, all the results in [1] can be corrected. Some related works on Halpern’s method and the proximal point algorithm for accretive operators have been studied extensively. Please see [2–5] and the references therein. 2. Preliminaries Let E be a Banach space with the norm ∥ · ∥ and the dual space E ∗ . The normalized duality mapping J from E into 2E is defined by J (x) = {x∗ ∈ E ∗ : ⟨x, x∗ ⟩ = ∥x∥ ∥x∗ ∥, ∥x∥ = ∥x∗ ∥}. It is well known that E is smooth if and only if J is single-valued. In the sequel, we shall denote the single-valued normalized duality map by j. Let S := {x ∈ E : ∥x∥ = 1} denote the unit sphere of a Banach space E. The space E is said to have a Gâteaux differentiable norm (or E is said to be smooth) if the limit lim

∥x + ty∥ − ∥x∥ t

t →0

exists for each x, y ∈ S, and E is said to have a uniformly Gâteaux differentiable norm if for each y ∈ S the above limit is attained uniformly for any x ∈ S. A Banach space E is said to be uniformly convex if, for any ϵ ∈ (0, 2], there exists δ > 0 such that, for any x, y ∈ S,

  x + y  ≤ 1 − δ.  ∥x − y∥ ≥ ϵ implies  2  It is known that a uniformly convex Banach space is reflexive and strictly convex. An operator A : D(A) ⊂ E → 2E is called accretive if for all x, y ∈ D(A) there exists j(x − y) ∈ J (x − y) such that

⟨u − v, j(x − y)⟩ ≥ 0,

for u ∈ Ax and v ∈ Ay.

Let A : D(A) ⊂ E → 2 be an accretive operator and A−1 (0) = {x ∈ D(A); 0 ∈ Ax}. An operator A is called m-accretive if it is accretive and R(I + rA), the range of (I + rA), is the whole space E for all r > 0; and A is said to satisfy the range condition if D(A) ⊂ R(I + rA) for all r > 0, where I is the identity operator of E and D(A) denotes the closure of the domain E

I −J A

of A. We define JrA = (I + rA)−1 and Ar = r r , r > 0, which are known as the resolvent and Yosida’s approximation of A, respectively. A Banach space E is said to have a weakly continuous duality mapping [6] if there exists a gauge function ϕ such that the duality mapping Jϕ is single-valued and weak–weak∗ sequentially continuous, where the gauge function ϕ : [0, ∞) → [0, ∞) is a continuous strictly increasing function with ϕ(0) = 0 and limt →∞ ϕ(t ) = ∞ and the duality mapping Jϕ is defined by Jϕ (x) = {x∗ ∈ E ∗ ; ⟨x, x∗ ⟩ = ∥x∥ ∥x∗ ∥, ∥x∗ ∥ = ϕ(∥x∥)}. Let

Φ (t ) =

t



ϕ(τ )dτ ,

t ≥ 0.

0

Then, Φ (kt ) ≤ kΦ (t ) for all 0 < k < 1. Lemma 2.1 ([7]). Let E be a Banach space with a weakly continuous duality mapping Jϕ . Then: (i) Φ (∥x + y∥) ≤ Φ (∥x∥) + ⟨y, Jϕ (x + y)⟩ for all x, y ∈ E; (ii) lim supn→∞ Φ (∥xn − y∥) = lim supn→∞ Φ (∥xn − x∥) + Φ (∥y − x∥) for all y ∈ E provided that the sequence {xn } converges weakly to x. Lemma 2.2 ([8]). Let E be a reflexive Banach space having a weakly continuous duality mapping Jϕ . Suppose that K is a nonempty, closed and convex subset of E, and T : K → K is a non-expansive mapping with F (T ) ̸= ∅. Then for each u ∈ K , there exists an

Y. Yu / Nonlinear Analysis 75 (2012) 5027–5031

5029

x ∈ F (T ), for all y ∈ F (T ), such that

⟨u − x, Jϕ (y − x)⟩ ≤ 0. Lemma 2.3 ([9]). Let q > 1 and r > 0 be two fixed real numbers. Then a Banach space E is uniformly convex if and only if there exists a continuous strictly increasing convex function g : [0, +∞) → [0, +∞) with g (0) = 0 such that

∥λx + (1 − λ)y∥q ≤ λ∥x∥2 + (1 − λ)∥y∥2 − ωq (λ)g (∥x − y∥), for all x, y ∈ Br (0) := {x ∈ E : ∥x∥ ≤ r } and λ ∈ [0, 1], where ωq (λ) = λq (1 − λ) + λ(1 − λ)q . Lemma 2.4 ([10]). Assume that {an } is a sequence of nonnegative real numbers such that an+1 ≤ (1 − γn )an + δn where {γn } is a sequence in (0, 1) and {δn } is a sequence such that: (1) n=1 γn = ∞; ∞ (2) lim supn→∞ δn /γn ≤ 0 or n=1 |δn | < ∞. Then limn→∞ an = 0.

∞

Lemma 2.5 ([11]). Let {sn } be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence {sni } of {sn } such that sni ≤ sni +1 for all i ≥ 0. For every n ≥ n0 , define an integer sequence {τ (n)} as

τ (n) = max{k ≤ n : sni < sni +1 }. Then τ (n) → ∞ as n → ∞ and for all n ≥ n0 , max{sτ (n) , sn } ≤ sτ (n)+1 . 3. The main results Theorem 3.1. Let E be a uniformly convex Banach space which has a weakly continuous duality mapping Jϕ with a gauge function ϕ . Let A : D(A) ⊂ E → 2E be  an accretive operator satisfying the range condition and K be a nonempty, closed and convex subset of E such that D(A) ⊂ K ⊂ r >0 R(I + rA). Let {αn } and {βn } be two sequences in (0, 1) and {rn } ⊂ (0, +∞) satisfying the conditions: (C1) limn→∞ αn = 0 and n=0 αn = ∞; (C2) lim supn→∞ βn < 1 and lim infn→∞ rn > 0. Then the sequence {xn } defined by (1.1) converges strongly to p = PA−1 (0) (u), which is the unique solution of the variational inequality (1.3).

∞

Proof. Define p = PA−1 (0) (u). First, by using arguments similar to those of [1], we can prove that the sequence {xn } is bounded and from (1.1) and Lemma 2.3, we also have 1

∥xn+1 − p∥2 ≤ αn ∥u − p∥2 + ∥xn − p∥2 − (1 − αn − βn )g (∥xn − JrAn xn ∥). 4

It follows that 1

∥xn+1 − p∥2 − ∥xn − p∥2 + (1 − αn − βn )g (∥xn − JrAn xn ∥) ≤ αn ∥u − p∥2 . 4

Since limn→∞ αn = 0 and lim supn→∞ βn < 1, without loss of generality, we may assume that 41 (1 − αn − βn ) > k > 0 for all n. Thus,

∥xn+1 − p∥2 − ∥xn − p∥2 + kg (∥xn − JrAn xn ∥) ≤ αn ∥u − p∥2 .

(3.1)

In order to prove that xn → p as n → ∞, we consider two possible cases. Case 1. Assume that {∥xn − p∥} is a monotone sequence. In other words, for n0 large enough, {∥xn − p∥}n≥n0 is either nondecreasing or non-increasing; hence {∥xn − p∥} is convergent (since {∥xn − p∥} is bounded). Therefore, we derive from (3.1) that lim g (∥xn − JrAn xn ∥) = 0.

n→∞

Hence lim ∥xn − JrAn xn ∥ = 0.

n→∞

By arguments similar to that of [1], we can prove that xn → p as n → ∞.

5030

Y. Yu / Nonlinear Analysis 75 (2012) 5027–5031

Case 2. Assume that {∥xn − p∥} is not a monotonic sequence. Then, we can define an integer sequence {τ (n)} for all n ≥ n0 (for some n0 large enough) by

τ (n) := max{k ∈ N; k ≤ n, ∥xk − p∥ ≤ ∥xk+1 − p∥}. Clearly, τ is a non-decreasing sequence such that τ (n) → +∞ as n → ∞ and

∥xτ (n) − p∥ ≤ ∥xτ (n)+1 − p∥ for all n ≥ n0 . In this case, we derive from (3.1) that lim g (∥xτ (n) − JrAτ (n) xτ (n) ∥) = 0.

n→∞

Hence lim ∥xτ (n) − JrAτ (n) xτ (n) ∥ = 0.

n→∞

Note that ∥JλA x − JµA x∥ ≤ (|λ − µ|/λ)∥x − JλA x∥ holds for all λ, µ > 0. So, 0 ≤ ∥xτ (n) − JrA xτ (n) ∥ ≤ ∥xτ (n) − JrAτ (n) xτ (n) ∥ + ∥JrAτ (n) xτ (n) − JrA xτ (n) ∥

≤ ∥xτ (n) − JrAτ (n) xτ (n) ∥ +

|rτ (n) − r | rτ (n)

∥xτ (n) − JrAτ (n) xτ (n) ∥

→ 0. Following an argument similar to that in the proof in [1], we can prove that lim sup⟨u − p, Jϕ (xτ (n)+1 − p)⟩ ≤ 0.

(3.2)

n→∞

Consequently, for any n ≥ n0 , from (1.1) and Lemma 2.1, we derive

Φ (∥xτ (n)+1 − p∥) = Φ (∥αn (u − p) + βn (xn − p) + (1 − αn − βn )(JrAn xn − p)∥) ≤ (1 − αn )Φ (∥xτ (n) − p∥) + αn ⟨u − p, Jϕ (xτ (n)+1 − p)⟩. This, together with (3.2) and Lemma 2.4, implies that lim Φ (∥xτ (n) − p∥) = 0

n→∞

and

lim Φ (∥xτ (n)+1 − p∥) = 0.

n→∞

Hence lim ∥xτ (n) − p∥ = 0

n→∞

and

lim ∥xτ (n)+1 − p∥ = 0.

n→∞

From Lemma 2.5, we have max{∥xτ (n) − p∥, ∥xn − p∥} ≤ ∥xτ (n)+1 − p∥. So, 0 ≤ ∥xn − p∥ ≤ max{∥xτ (n) − p∥, ∥xn − p∥} ≤ ∥xτ (n)+1 − p∥. Therefore, limn→∞ ∥xn − p∥ = 0. This completes the proof.



By using an argument similar to that in the proof of Theorem 3.1 and [1], we can obtain the following theorems. Since the proofs are similar, we omit them. Theorem 3.2. Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Let A : D(A) ⊂ E → 2E be an accretive operator satisfying the range condition and K be a nonempty, closed and convex subset of E such that D(A) ⊂ K ⊂ r >0 R(I + rA). Let {αn } and {βn } be two sequences in (0, 1) and {rn } ⊂ (0, +∞) satisfying the conditions: (C1) limn→∞ αn = 0 and n=0 αn = ∞; (C2) lim supn→∞ βn < 1 and lim infn→∞ rn > 0.

∞

Then the sequence {xn } defined by (1.1) converges strongly to p = PA−1 (0) (u), which is the unique solution of the variational inequality (1.3). Theorem 3.3. Let E be a uniformly convex Banach space which has a weakly continuous duality mapping Jϕ with a gauge function ϕ . Let A : D(A) ⊂ E → 2E be  an accretive operator satisfying the range condition and K be a nonempty, closed and convex subset of E such that D(A) ⊂ K ⊂ r >0 R(I + rA). Let {αn } and {βn } be two sequences in (0, 1) and {rn } ⊂ (0, +∞) satisfying the conditions:

Y. Yu / Nonlinear Analysis 75 (2012) 5027–5031

5031

(C1) limn→∞ αn = 0 and n=0 αn = ∞; (C2) lim supn→∞ βn < 1 and lim infn→∞ rn > 0.

∞

Then the sequence {xn } defined by (1.2) converges strongly to p = PA−1 (0) (u), which is the unique solution of the variational inequality (1.3). Theorem 3.4. Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Let A : D(A) ⊂ E → 2E be an accretive operator satisfying the range condition and K be a nonempty, closed and convex subset of E such that D(A) ⊂ K ⊂ r >0 R(I + rA). Let {αn } and {βn } be two sequences in (0, 1) and {rn } ⊂ (0, +∞) satisfying the conditions: (C1) limn→∞ αn = 0 and n=0 αn = ∞; (C2) lim supn→∞ βn < 1 and lim infn→∞ rn > 0.

∞

Then the sequence {xn } defined by (1.2) converges strongly to p = PA−1 (0) (u), which is the unique solution of the variational inequality (1.3). Acknowledgments The author is extremely grateful to the referees for their useful comments and suggestions which helped to improve this paper. This research was partially supported by the Youth Foundation of Taizhou University (2011QN11). References [1] Q. Zhang, Y. Song, Halpern type proximal point algorithm of accretive operators, Nonlinear Anal. 75 (2012) 1859–1868. [2] R.E. Bruck, S. Reich, Nonexpansive projections and resolvents of accretive operators in Banach spaces, Houston J. Math. 3 (1997) 459–470. [3] O. Nevanlinna, S. Reich, Strong convergence of contraction semigroups and of iterative methods for accretive operators in Banach spaces, Israel J. Math. 32 (1997) 44–58. [4] S. Reich, Strong convergence theorems for resolvents of accretive operators in Banach spaces, J. Math. Anal. Appl. 75 (1980) 287–292. [5] G. Lopez, V. Martin-Marquez, H.K. Xu, Halpern’s iteration for nonexpansive mappings, Contemp. Math. 513 (2010) 211–231. [6] F.E. Browder, Convergence theorems for sequences of nonlinear operators in Banach spaces, Math. Z. 100 (1967) 201–225. [7] J.P. Gossez, E.L. Dozo, Some geometric properties related to the fixed point theory for nonexpansive mappings, Pacific J. Math. 40 (1972) 565–573. [8] S. Reich, Asymptotic behavior of contractions in Banach spaces, J. Math. Anal. Appl. 44 (1973) 57–70. [9] H.K. Xu, Inequality in Banach spaces with applications, Nonlinear Anal. 16 (1991) 1127–1138. [10] H.K. Xu, Iterative algorithms for nonlinear operators, J. Lond. Math. Soc. 66 (2002) 240–256. [11] P.E. Mainge, Strong convergence of projected subgradient methods for nonsmooth and nonstrictly convex minimization, Set-Valued Anal. 16 (2008) 899–912.