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Weak and strong convergence theorems for common zeros of accretive operators

Abstract

In this paper, we propose two proximal point algorithms for investigating common zeros of a family of accretive operators. Weak and strong convergence of the two algorithms are obtained in a Banach space.

MSC:47H06, 47H09, 47J25.

1 Introduction and preliminaries

Let E be a real Banach space, and let E be the dual space of E. Let R + be a positive real number set. Let φ:[0,]:= R + R + be a continuous strictly increasing function such that φ(0)=0 and φ(t) as t. This function φ is called a gauge function. The duality mapping J φ :E E associated with a gauge function φ is defined by

J φ (x)= { f E : x , f = x φ ( x ) , f = φ ( x ) } ,xE,

where , denotes the generalized duality pairing. In the case that φ(t)=t, we write J for J φ and call J the normalized duality mapping.

Following Browder [1], we say that a Banach space E has a weakly continuous duality mapping if there exists a gauge φ for which the duality mapping J φ (x) is single-valued and weak-to-weak sequentially continuous (i.e., if { x n } is a sequence in E weakly convergent to a point x, then the sequence J φ ( x n ) converges weakly to J φ ). It is known that l p has a weakly continuous duality mapping with a gauge function φ(t)= t p 1 for all 1<p<.

Let U E ={xE:x=1}. E is said to be smooth or is said to be have a Gâteaux differentiable norm if the limit lim t 0 x + t y x t exists for each x,y U E . E is said to have a uniformly Gâteaux differentiable norm if for each y U E , the limit is attained uniformly for all x U E . E is said to be uniformly smooth or is said to have a uniformly Fréchet differentiable norm if the limit is attained uniformly for x,y U E .

It is well known that Fréchet differentiability of the norm of E implies Gâteaux differentiability of the norm of E. It is known that if the norm of E is uniformly Gâteaux differentiable, then the duality mapping J is single-valued and uniformly norm-to-weak continuous on each bounded subset of E.

Let D be a nonempty subset of a set C. Let Proj D :CD. Q is said to be

  1. (1)

    a contraction if Proj D 2 = Proj D ;

  2. (2)

    sunny if for each xC and t(0,1), we have Proj D (tx+(1t) Proj D x)= Proj D x;

  3. (3)

    a sunny nonexpansive retraction if Proj D is sunny, nonexpansive and a contraction.

D is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C onto D. The following result, which was established in [24], describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Let E be a smooth Banach space, and let C be a nonempty subset of E. Let Proj C :EC be a retraction and J φ be the duality mapping on E. Then the following are equivalent:

  1. (1)

    Proj C is sunny and nonexpansive;

  2. (2)

    Proj C x Proj C y 2 xy, J φ ( Proj C x Proj C y), x,yE;

  3. (3)

    x Proj C x, J φ (y Proj C x)0, xE, yC.

It is well known that if E is a Hilbert space, then a sunny nonexpansive retraction Proj C is coincident with the metric projection from E onto C. Let C be a nonempty closed convex subset of a smooth Banach space E, let xE and let x 0 C. Then we have from the above that x 0 = Proj C x if and only if x x 0 , J φ (y x 0 )0 for all yC, where Proj C is a sunny nonexpansive retraction from E onto C.

A Banach space E is said to be strictly convex if and only if

x=y= ( 1 λ ) x + λ y

for x,yE and 0<λ<1 implies that x=y.

E is said to be uniformly convex if for any ϵ(0,2] there exists δ>0 such that for any x,y U E ,

xyϵimplies x + y 2 1δ.

It is known that a uniformly convex Banach space is reflexive and strictly convex.

Let C be a nonempty closed convex subset of E. Let S:CC be a mapping. In this paper, we use F(S) to denote the set of fixed points of S. Recall that S is said to be nonexpansive iff SxSyxy, x,yC. For the existence of fixed points of a nonexpansive mapping, we refer the readers to [5]. Let x be a fixed element in C, and let S be a nonexpansive mapping with a nonempty fixed point set. For each t(0,1), let x t be the unique solution of the equation y=tx+(1t)Sy. In the framework of uniformly smooth Banach spaces, Reich [6] proved that { x t } converges strongly to a fixed point Proj F ( S ) x, where Proj F ( S ) is the unique sunny nonexpansive retraction from C onto F(S), of S as t0. Xu [7] further extended the results to the framework of reflexive Banach spaces; for more details, see [7] and [8] and the references therein.

Let I denote the identity operator on E. An operator AE×E with domain D(A)={zE:Az} and range R(A)={Az:zD(A)} is said to be accretive if for each x i D(A) and y i A x i , i=1,2, there exists j( x 1 x 2 )J( x 1 x 2 ) such that y 1 y 2 ,j( x 1 x 2 )0. An accretive operator A is said to be M-accretive if R(I+rA)=E for all r>0. In this paper, we use A 1 (0) to denote the set of zero points of A. For an accretive operator A, we can define a nonexpansive single-valued mapping J r :R(I+rA)D(A) by J r = ( I + r A ) 1 for each r>0, which is called the resolvent of A. Set

Φ(t)= 0 t φ(τ)dτ,t0,

then

J φ (x)=Φ ( x ) ,xE,

where denotes the sub-differential in the sense of convex analysis.

Zero problems of accretive operators recently have been extensively studied (see [621] and the references therein) because of their important applications in real world. Proximal point algorithm, which was proposed by Martinet [22, 23] and generalized by Rockafellar [24, 25], is a classical method for investigating zeros of monotone operators. In this paper, we propose two proximal point algorithms for investigating common zeros of a family of m-accretive operators. Weak and strong convergence theorems are established in Banach spaces.

The first part of the next lemma is an immediate consequence of the subdifferential inequality and the proof of the second part can be found in [26].

Lemma 1.1 Assume that a Banach space E has a weakly continuous duality mapping J φ with gauge φ.

  1. (i)

    For all x,yE, the following inequality holds:

    Φ ( x + y ) Φ ( x ) + y , J φ ( x + y ) .

In particular, for all x,yE,

x + y 2 x 2 +2 y , J ( x + y ) .
  1. (ii)

    Assume that a sequence { x n } in E converges weakly to a point xE.

Then the following identity holds:

lim sup n Φ ( x n y ) = lim sup n Φ ( x n x ) +Φ ( y x ) ,x,yE.

Lemma 1.2 [7]

Let E be a reflexive Banach space and have a weakly continuous duality map J φ (x) with gauge φ. Let C be a closed convex subset of E, and let S:CC be a nonexpansive mapping. Fix xC and t(0,1). Let x t C be the unique fixed point of the mapping tx+(1t)S. Then S has a fixed point if and only if { x t } remains bounded as t 0 + , and in this case, { x t } converges as t 0 + strongly to a fixed point of S. Define a mapping Proj F ( S ) :CF(S) by Proj F ( S ) x:= lim t 0 x t . Then Proj F ( S ) is the sunny nonexpansive retraction from C onto F(S).

Lemma 1.3 [27]

Let C be a closed convex subset of a strictly convex Banach space E. Let N1 be some positive integer, and let S m :CC be a nonexpansive mapping. Suppose that m = 1 N F( S m ) is nonempty. Then the mapping m = 1 N β m S m , where { β m } is a real number sequence in (0,1) such that m = 1 N β m =1, is nonexpansive with F( m = 1 N β m S m )= m = 1 N F( S m ).

Lemma 1.4 [28]

Let { a n }, { b n } and { c n } be three nonnegative real sequences satisfying b n + 1 (1 a n ) b n + a n c n , n n 0 , where n 0 is some positive integer, { a n } is a number sequence in (0,1) such that n = n 0 a n =, { c n } is a number sequence such that lim sup n c n 0. Then lim n a n =0.

Lemma 1.5 [29]

Let E be a uniformly convex Banach space, s>0 be a positive number, and B s (0) be a closed ball of E. There exits a continuous, strictly increasing and convex function g:[0,)[0,) with g(0)=0 such that

m = 1 N β m x m 2 m = 1 N β m x m 2 β 1 β 2 g ( x 1 x 2 )

for all x 1 , x 2 ,, x N B s (0)={xE:x<s} and β 1 , β 2 ,, β N (0,1) such that m = 1 N β m =1.

Lemma 1.6 [30]

Let E be a uniformly convex Banach space. Let C be a nonempty closed convex subset of E, and let S:CC be a nonexpansive mapping. Then IS is demiclosed at zero.

2 Main results

Theorem 2.1 Let E be a uniformly convex Banach space which has the Opial condition. Let N1 be some positive integer, and let A m be an M-accretive operator in E for each 1mN. Assume that m = 1 N D ( A i ) ¯ is convex. Let { α n } and { β n , m } be real number sequences in (0,1), and let { r m } be a positive real number sequence. Assume that m = 1 N A m 1 (0) is not empty. Let { x n } be a sequence generated in the following manner: x 1 m = 1 N D ( A i ) ¯ and

x n + 1 = α n x n +(1 α n ) m = 1 N β n , m J r m x n ,n1,

where J r m = ( I + r m A m ) 1 . Assume that the following restrictions are satisfied:

  1. (a)

    0<a α n b<1;

  2. (b)

    m = 1 N β n , m =1 and 0<c β n , m <1,

where a, b and c are real numbers. Then the sequence { x n } converges weakly to x m = 1 N A m 1 (0).

Proof We start the proof with the boundedness of the sequence { x n }. Fixing p m = 1 N A m 1 (0), we find that

x n + 1 p α n x n p + ( 1 α n ) m = 1 N β n , m J r m x n p α n x n p + ( 1 α n ) m = 1 N β n , m J r m x n p x n p .

This shows that the limit lim n x n p exists. This implies that { x n } is bounded. Using Lemma 1.5, we find that

x n + 1 p 2 α n x n p 2 + ( 1 α n ) m = 1 N β n , m J r m x n p 2 α n ( 1 α n ) g ( x n m = 1 N β n , m J r m x n ) x n p 2 α n ( 1 α n ) g ( m = 1 N β n , m x n J r m x n ) .

This implies that

α n (1 α n )g ( m = 1 N β n , m x n J r m x n ) x n p 2 x n + 1 p 2 .

In view of restriction (a), we find that lim n g( m = 1 N β n , m x n J r m x n )=0. It follows that lim n m = 1 N β n , m x n J r m x n =0. Using restriction (b), we arrive at lim n x n J r m x n =0 for each m{1,2,,N}. Since { x n } is bounded, we see that there exists a subsequence { x n i } of { x n } converging weakly to x . Using Lemma 1.6, we obtain that x F( J r m ). This proves that x m = 1 N A m 1 (0).

Next we show that { x n } converges weakly to x . Supposing the contrary, we see that there exists some subsequence { x n j } of { x n } such that { x n j } converges weakly to x ˆ C, where x ˆ x . Similarly, we can show x ˆ m = 1 N A m 1 (0). Note that we have proved that lim n x n p exists for every p m = 1 N A m 1 (0). Assume that lim n x n x =d, where d is a nonnegative number. Since the space has the Opial condition [31], we see that

d= lim inf i x n i x < lim inf i x n i x ˆ = lim inf j x n j x ˆ < lim inf j x n j x =d.

This is a contradiction. Hence x = x ˆ . This completes the proof. □

Theorem 2.2 Let E be a strictly convex and reflexive Banach space which has a weakly continuous duality map J φ . Let N1 be some positive integer, and let A m be an M-accretive operator in E for each 1mN. Assume that m = 1 N D ( A i ) ¯ is convex. Let { α n } and { β n , m } be real number sequences in (0,1), and let { r m } be a positive real number sequence for each 1mN. Assume that m = 1 N A m 1 (0) is not empty. Let { x n } be a sequence generated in the following manner: x 1 m = 1 N D ( A i ) ¯ and

x n + 1 = α n x+(1 α n ) m = 1 N β n , m J r m x n ,n1,

where x is a fixed element in m = 1 N D ( A m ) ¯ and J r m = ( I + r m A m ) 1 . Assume that the following restrictions are satisfied:

  1. (a)

    lim n α n =0, n = 1 α n = and n = 1 | α n + 1 α n |<;

  2. (b)

    m = 1 N β n , m =1, lim n β n , m = β m and n = 1 | β n + 1 , m β n , m |<.

Then the sequence { x n } converges strongly to x = Proj m = 1 N A m 1 ( 0 ) x, where Proj m = 1 N A m 1 ( 0 ) is the unique sunny nonexpansive retract from m = 1 N D ( A m ) ¯ onto m = 1 N A m 1 (0).

Proof We start the proof with the boundedness of the sequence { x n }. Fixing p m = 1 N A m 1 (0), we find that

x n + 1 p α n x p + ( 1 α n ) m = 1 N β n , m J r m x n p α n x p + ( 1 α n ) m = 1 N β n , m J r m x n p α n x p + ( 1 α n ) x n p .

This implies that x n + 1 pmax{xp, x 1 p}. This shows that { x n } is bounded. Put y n = m = 1 N β n , m J r m x n . It follows that

y n y n 1 m = 1 N β n , m J r m x n m = 1 N β n , m J r m x n 1 + m = 1 N β n , m J r m x n 1 m = 1 N β n 1 , m J r m x n 1 x n x n 1 + m = 1 N | β n , m β n 1 , m | J r m x n 1 .

This implies that

x n + 1 x n ( 1 α n ) y n y n 1 + | α n α n 1 | x y n 1 ( 1 α n ) x n x n 1 + m = 1 N | β n , m β n 1 , m | J r m x n 1 + | α n α n 1 | x y n 1 .

In light of restrictions (a) and (b), we find that

lim n x n + 1 x n =0.
(2.1)

Set S= m = 1 N β m J r m . It follows from Lemma 1.3 that S is nonexpansive with F(S)= m = 1 N F( J r m )= m = 1 N A m 1 (0). Note that

S x n x n x n x n + 1 + x n + 1 S x n x n x n + 1 + α n x S x n + β n m = 1 N β n , m J r m x n S x n x n x n + 1 + α n x S x n + m = 1 N | β n , m β m | J r m x n .

In view of (2.1), we find from the restrictions (a) and (b) that

lim n S x n x n =0.
(2.2)

Now, we are in a position to prove

lim sup n x Proj m = 1 N A m 1 ( 0 ) x , J φ ( x n Proj m = 1 N A m 1 ( 0 ) x ) 0.
(2.3)

By Lemma 1.2, we have the sunny nonexpansive retraction Proj m = 1 N A m 1 ( 0 ) : m = 1 N D ( A m ) ¯ m = 1 N A m 1 (0). Take a subsequence { x n k } of { x n } such that

lim sup n x Proj m = 1 N A m 1 ( 0 ) x , J φ ( x n Proj m = 1 N A m 1 ( 0 ) ) x = lim k x Proj m = 1 N A m 1 ( 0 ) x , J φ ( x n k Proj m = 1 N A m 1 ( 0 ) ) x .
(2.4)

Since E is reflexive, we may further assume that x n k x ¯ for some x ¯ m = 1 N D ( A m ) ¯ . Since j φ is weakly continuous, we have from Lemma 1.1

lim sup k Φ ( x n k y ) = lim sup k Φ ( x n k x ¯ ) +Φ ( y x ¯ ) ,yE.

Put f(y)= lim sup k Φ( x n k y), yE. It follows that

f(y)=f( x ¯ )+Φ ( y x ¯ ) ,yE.
(2.5)

From (2.2), we have

f ( S x ¯ ) = lim sup k Φ ( x n k S x ¯ ) = lim sup k Φ ( S x n k S x ¯ ) lim sup k Φ ( x n k x ¯ ) = f ( x ¯ ) .
(2.6)

Using (2.5), we have

f(S x ¯ )=f( x ¯ )+Φ ( S x ¯ x ¯ ) .
(2.7)

Combining (2.6) with (2.7), we obtain that

Φ ( S x ¯ x ¯ ) 0.

Hence S x ¯ = x ¯ ; that is, x ¯ F(S)= m = 1 N A m 1 (0). It follows from (2.4) that

lim sup n x Proj m = 1 N A m 1 ( 0 ) x , j φ ( x n Proj m = 1 N A m 1 ( 0 ) x ) 0.

Finally, we prove that x n Proj m = 1 N A m 1 ( 0 ) x as n. Using Lemma 1.1, we find that

Φ ( x n + 1 Proj m = 1 N A m 1 ( 0 ) x ) = Φ ( α n ( x Proj m = 1 N A m 1 ( 0 ) x ) + ( 1 α n ) ( m = 1 N β n , m J r m x n Proj m = 1 N A m 1 ( 0 ) x ) ) ( 1 α n ) Φ ( x n Proj m = 1 N A m 1 ( 0 ) x ) + α n x Proj m = 1 N A m 1 ( 0 ) x , J φ ( x n + 1 Proj m = 1 N A m 1 ( 0 ) x ) .

Using Lemma 1.4, we see that Φ( x n Proj m = 1 N A m 1 ( 0 ) x)0. This implies that

lim n x n Proj m = 1 N A m 1 ( 0 ) x=0.

This completes the proof. □

3 Applications

In this section, we give an application of Theorem 2.1 in the framework of Hilbert spaces.

Let C be a nonempty closed and convex subset of a Hilbert space H. Let F be a bifunction of C×C into , where denotes the set of real numbers. Recall the following equilibrium problem:

Find xC such that F(x,y)0,yC.
(3.1)

To study the equilibrium problem (3.1), we may assume that F satisfies the following restrictions:

(A1) F(x,x)=0 for all xC;

(A2) F is monotone, i.e., F(x,y)+F(y,x)0 for all x,yC;

(A3) for each x,y,zC, lim t 0 F(tz+(1t)x,y)F(x,y);

(A4) for each xC, yF(x,y) is convex and lower semi-continuous.

Lemma 3.1 [32]

Let F be a bifunction from C×C to which satisfies (A1)-(A4), and let A F be a multivalued mapping of H into itself defined by

A F x={ { z H : F ( x , y ) y x , z , y C } , x C , , x C .

Then A F is a maximal monotone operator with the domain D( A F )C, EP(F)= A F 1 (0), where EP(F) stands for the solution set of (3.1), and

T r x= ( I + r A F ) 1 x,xH,r>0,

where T s is defined by

T r x= { z C : F ( z , y ) + 1 r y z , z x 0 , y C } ,xH.

Corollary 3.2 Let C be a nonempty closed and convex subset of a Hilbert space H. Let N1 be some positive integer, and let F m :C×CR be a bifunction satisfying (A1)-(A4). Let { α n } and { β n , m } be real number sequences in (0,1), and let { r m } be a positive real number sequence. Assume that m = 1 N EP( F m ) is not empty. Let { x n } be a sequence generated in the following manner: x 1 C and

x n + 1 = α n x n +(1 α n ) m = 1 N β n , m T r m x n ,n1,

where T r m = ( I + r m A F m ) 1 . Assume that the following restrictions are satisfied:

  1. (a)

    0<a α n b<1;

  2. (b)

    m = 1 N β n , m =1 and 0<c β n , m <1,

where a, b and c are real numbers. Then the sequence { x n } converges weakly to x m = 1 N EP( F m ).

Corollary 3.3 Let C be a nonempty closed and convex subset of a Hilbert space H. Let N1 be some positive integer, and let F m :C×CR be a bifunction satisfying (A1)-(A4). Let { α n } and { β n , m } be real number sequences in (0,1), and let { r m } be a positive real number sequence. Assume that m = 1 N EP( F m ) is not empty. Let { x n } be a sequence generated in the following manner: x 1 C and

x n + 1 = α n x+(1 α n ) m = 1 N β n , m T r m x n ,n1,

where x is a fixed element in C and T r m = ( I + r m A F m ) 1 . Assume that the following restrictions are satisfied:

  1. (a)

    lim n α n =0, n = 1 α n = and n = 1 | α n + 1 α n |<;

  2. (b)

    m = 1 N β n , m =1, lim n β n , m = β m and n = 1 | β n + 1 , m β n , m |<.

Then the sequence { x n } converges strongly to x = Proj m = 1 N EP ( F m ) x, where Proj m = 1 N EP ( F m ) is the metric projection from C onto m = 1 N EP( F m ).

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Acknowledgements

This first author thanks the Fundamental Research Funds for the Central Universities (2014ZD44). The authors are grateful to the reviewers’ suggestions which improved the contents of the article.

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Correspondence to Shenghua Wang.

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The authors declare that they have no competing interests.

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Both authors designed the algorithms and established weak and strong convergence analysis. Both authors read and approved the final manuscript.

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Wang, S., Li, T. Weak and strong convergence theorems for common zeros of accretive operators. J Inequal Appl 2014, 282 (2014). https://doi.org/10.1186/1029-242X-2014-282

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  • DOI: https://doi.org/10.1186/1029-242X-2014-282

Keywords

  • accretive operator
  • convergence
  • resolvent
  • proximal point algorithm
  • zero point