Open Access

Another type of Mann iterative scheme for two mappings in a complete geodesic space

Journal of Inequalities and Applications20142014:72

https://doi.org/10.1186/1029-242X-2014-72

Received: 15 October 2013

Accepted: 23 January 2014

Published: 13 February 2014

Abstract

In this paper, we show a Δ-convergence theorem for a Mann iteration procedure in a complete geodesic space with two quasinonexpansive and Δ-demiclosed mappings. The proposed method is different from known procedures with respect to the order of taking the convex combination.

1 Introduction

The fixed point approximation has been studied in a variety of ways and its results are useful for the other studies. In 1953, Mann [1] introduced an iteration procedure for approximating fixed points of a nonexpansive mapping T in a Hilbert space. Later, Reich [2] discussed this iteration procedure in a uniformly convex Banach space whose norm is Fréchet differentiable. In 1998, Takahashi and Tamura [3] considered an iteration procedure with two nonexpansive mappings and obtained weak convergence theorems for this procedure in a uniformly convex Banach space which satisfies Opial’s condition or whose norm is Fréchet differentiable. On the other hand, in 2008, Dhompongsa and Panyanak [4] proved the following theorem.

Theorem 1.1 Let C be a bounded closed convex subset of a complete CAT ( 0 ) space and T : C C a nonexpansive mapping. For any initial point x 0 in C, define the Mann iterative sequence { x n } by
x n + 1 = ( 1 t n ) x n t n T x n , n = 0 , 1 , 2 , ,

where { t n } is a sequence in [ 0 , 1 ] , with the restrictions that n = 0 t n diverges and lim sup n t n < 1 . Then { x n } Δ-converges to a fixed point of T.

Further, in a CAT ( 1 ) space, Kimura et al. [5] proved the Δ-convergence theorem for a family of nonexpansive mappings including the following scheme:
x n + 1 = ( 1 α n ) x n α n ( ( 1 β n ) S x n β n T x n ) .
In a Hilbert space H, the following equality holds for any x , y , z H :
α x + ( 1 α ) ( β y + ( 1 β ) z ) = γ ( δ x + ( 1 δ ) y ) + ( 1 γ ) z ,
where α , β , γ , δ ] 0 , 1 [ such that α = γ δ and β = γ ( 1 δ ) / ( 1 γ δ ) . However, in CAT ( κ ) spaces with κ > 0 , it does not hold in general, that is, the value of the convex combination taken twice depends on their order. Thus, the following formulas are different in general:
x n + 1 = ( 1 α n ) x n α n ( ( 1 β n ) S x n β n T x n ) , x n + 1 = ( 1 α n ) ( β n x n ( 1 β n ) S x n ) α n ( ( 1 β n ) x n β n T x n ) .
(1)

In this paper, we show an analogous result to Theorem 1.1 using the iterative scheme (1) in a complete CAT ( 1 ) space with two quasinonexpansive and Δ-demiclosed mappings. We also deal with the image recovery problem for two closed convex sets.

2 Preliminaries

Let X be a metric space. For x , y X , a mapping c : [ 0 , l ] X is said to be a geodesic if c satisfies c ( 0 ) = x , c ( l ) = y and d ( c ( s ) , c ( t ) ) = | s t | for all s , t [ 0 , l ] . An image [ x , y ] of c is called a geodesic segment joining x and y. For r > 0 , X is said to be an r-geodesic metric space if, for any x , y X with d ( x , y ) < r , there exists a geodesic segment [ x , y ] . In particular, if a segment [ x , y ] is unique for any x , y X with d ( x , y ) < r , then X is said to be a uniquely r-geodesic metric space. In what follows, we always assume d ( x , y ) < π / 2 for any x , y X . Thus, we say X is a geodesic metric space instead of a π / 2 -geodesic metric space. For the more general case, see [6].

Let X be a uniquely geodesic metric space. A geodesic triangle is defined by ( x , y , z ) = [ x , y ] [ y , z ] [ z , x ] . Let M be the two-dimensional unit sphere in R 3 . For x ¯ , y ¯ , z ¯ M , a triangle ( x ¯ , y ¯ , z ¯ ) M is called a comparison triangle of ( x , y , z ) if d ( x , y ) = d M ( x ¯ , y ¯ ) , d ( y , z ) = d M ( y ¯ , z ¯ ) , d ( z , x ) = d M ( z ¯ , x ¯ ) . Further, for any x , y X and t ] 0 , 1 [ , if z [ x , y ] satisfies d ( x , z ) = ( 1 t ) d ( x , y ) and d ( z , y ) = t d ( x , y ) , then z is denoted by z = t x ( 1 t ) y . A point z ¯ [ x ¯ , y ¯ ] is called a comparison point of z [ x , y ] if d ( x , z ) = d M ( x ¯ , z ¯ ) . X is said to be a CAT ( 1 ) space if, for any p , q ( x , y , z ) X and its comparison points p ¯ , q ¯ ( x ¯ , y ¯ , z ¯ ) M , the inequality d ( p , q ) d M ( p ¯ , q ¯ ) holds.

Let X be a geodesic metric space and { x n } a bounded sequence of X. For x X , we put r ( x , { x n } ) = lim sup n d ( x , x n ) . The asymptotic radius of { x n } is defined by r ( { x n } ) = inf x X r ( x , { x n } ) . Further, the asymptotic center of { x n } is defined by A C ( { x n } ) = { x X : r ( x , { x n } ) = r ( { x n } ) } . If, for any subsequences { x n k } of { x n } , A C ( { x n k } ) = { x 0 } , i.e., their asymptotic center consists of the unique element x 0 , then we say { x n } Δ-converges to x 0 and we denote it by x n Δ x 0 .

Let X be a metric space. A mapping T : X X is said to be a nonexpansive if T satisfies d ( T x , T y ) d ( x , y ) for any x , y X . The set of fixed points of T is denoted by F ( T ) = { z X : T z = z } . Further, a mapping T : X X with F ( T ) is said to be a quasinonexpansive if T satisfies d ( T x , z ) d ( x , z ) for any x X and z F ( T ) . Moreover, T is said to be Δ-demiclosed if, for any bounded sequence { x n } X and x 0 X satisfying d ( x n , T x n ) 0 and x n Δ x 0 , we have x 0 F ( T ) .

3 Tools for the main results

In this section, we introduce some tools for using the main theorem.

Theorem 3.1 (Kimura and Satô [7])

Let ( x , y , z ) be a geodesic triangle in a CAT ( 1 ) space such that d ( x , y ) + d ( y , z ) + d ( z , x ) < 2 π . Let u = t x ( 1 t ) y for some t [ 0 , 1 ] . Then
cos d ( u , z ) sin d ( x , y ) cos d ( x , z ) sin t d ( x , y ) + cos d ( y , z ) sin ( 1 t ) d ( x , y ) .

Corollary 3.2 (Kimura and Satô [8])

Let ( x , y , z ) be a geodesic triangle in a CAT ( 1 ) space such that d ( x , y ) + d ( y , z ) + d ( z , x ) < 2 π . Let u = t x ( 1 t ) y for some t [ 0 , 1 ] . Then
cos d ( u , z ) t cos d ( x , z ) + ( 1 t ) cos d ( y , z ) .

Theorem 3.3 (Espínola and Fernández-León [9])

Let X be a complete CAT ( 1 ) space and { x n } a sequence in X. If r ( { x n } ) < π / 2 , then the following hold.
  1. (i)

    A C ( { x n } ) consists of exactly one point;

     
  2. (ii)

    { x n } has a Δ-convergent subsequence.

     

Theorem 3.4 (Kimura and Satô [8])

Let X be a metric space and T a mapping from X into itself. If T is a nonexpansive with F ( T ) , then T is quasinonexpansive and Δ-demiclosed.

The following lemmas are important properties of real numbers and they are easy to show. Thus, we omit the proofs.

Lemma 3.5 Let δ be a real number such that 1 < δ < 0 and { b n } , { c n } real sequences satisfying δ b n 1 , δ c n 1 and lim inf n b n c n 1 . Then lim n b n = lim n c n = 1 .

Lemma 3.6 Let s ] 0 , [ and { b n } , { c n } bounded real sequences satisfying b n 0 , s < c n and lim n b n / c n = 0 . Then lim n b n = 0 .

Lemma 3.7 Let { b n } and { c n } be bounded real sequences satisfying lim n ( b n c n ) = 0 . Then lim inf n b n = lim inf n c n .

4 The main result

In this section, we show the main result.

Theorem 4.1 Let X be a complete CAT ( 1 ) space such that for any u , v X , d ( u , v ) < π / 2 . Let S and T be quasinonexpansive and Δ-demiclosed mappings from X into itself with F ( S ) F ( T ) . Let { α n } , { β n } and { γ n } be sequences of [ a , b ] ] 0 , 1 [ . Define a sequence { x n } X by the following recurrence formula: x 1 X and
{ u n = ( 1 β n ) x n β n S x n , v n = ( 1 γ n ) x n γ n T x n , x n + 1 = ( 1 α n ) u n α n v n

for n N . Then { x n } Δ-converges to a common fixed point of S and T.

Proof Let z F ( S ) F ( T ) . By Corollary 3.2, we have
cos d ( u n , z ) ( 1 β n ) cos d ( x n , z ) + β n cos d ( S x n , z ) ( 1 β n ) cos d ( x n , z ) + β n cos d ( x n , z ) = cos d ( x n , z ) , cos d ( v n , z ) ( 1 γ n ) cos d ( x n , z ) + γ n cos d ( T x n , z ) ( 1 γ n ) cos d ( x n , z ) + γ n cos d ( x n , z ) = cos d ( x n , z ) .
Then, by Corollary 3.2 again, we have
cos d ( x n + 1 , z ) ( 1 α n ) cos d ( u n , z ) + α n cos d ( v n , z ) ( 1 α n ) cos d ( x n , z ) + α n cos d ( x n , z ) cos d ( x n , z ) .

So, we get d ( x n + 1 , z ) d ( x n , z ) for all n N and there exists d 0 = lim n d ( x n , z ) d ( x 1 , z ) < π / 2 .

Furthermore, by Theorem 3.1, we have
cos d ( u n , z ) sin d ( x n , S x n ) cos d ( x n , z ) sin ( 1 β n ) d ( x n , S x n ) + cos d ( S x n , z ) sin β n d ( x n , S x n ) 2 cos d ( x n , z ) sin d ( x n , S x n ) 2 cos ( 1 2 β n ) d ( x n , S x n ) 2
(2)
and
cos d ( v n , z ) sin d ( x n , T x n ) cos d ( x n , z ) sin ( 1 γ n ) d ( x n , T x n ) + cos d ( T x n , z ) sin γ n d ( x n , T x n ) 2 cos d ( x n , z ) sin d ( x n , T x n ) 2 cos ( 1 2 γ n ) d ( x n , T x n ) 2 .
(3)
Let d n = d ( x n , z ) , s n = d ( x n , S x n ) / 2 and t n = d ( x n , T x n ) / 2 for n N . If there exists n 0 N such that s n 0 = t n 0 = 0 , then we have x n 0 F ( S ) F ( T ) and since
x n 0 + 1 = ( 1 α n 0 ) ( ( 1 β n 0 ) x n 0 β n 0 S x n 0 ) α n 0 ( ( 1 γ n 0 ) x n 0 γ n 0 T x n 0 ) = ( 1 α n 0 ) x n 0 α n 0 x n 0 = x n 0 ,

and the proof is finished. So, we may assume s n 0 or t n 0 for all n N .

If s n = 0 and t n 0 , then we have u n = x n . From (2), (3), and Corollary 3.2, we get
2 cos d n + 1 sin t n cos t n = cos d n + 1 sin 2 t n ( 1 α n ) cos d ( u n , z ) sin 2 t n + α n cos d ( v n , z ) sin 2 t n 2 ( 1 α n ) cos d n sin t n cos t n + 2 α n cos d n sin t n cos ( 1 2 γ n ) t n .
Dividing by 2 sin t n > 0 , we get
cos d n + 1 cos t n ( 1 α n ) cos d n cos t n + α n cos d n cos ( 1 2 γ n ) t n .
(4)
If t n = 0 and s n 0 , then we have v n = x n . In a similar way as above, we get
cos d n + 1 cos s n ( 1 α n ) cos d n cos ( 1 2 β n ) s n + α n cos d n cos s n .
(5)
If s n 0 and t n 0 , then from (2), (3), and Corollary 3.2, we get
cos d n + 1 sin 2 s n sin 2 t n ( 1 α n ) cos d ( u n , z ) sin 2 s n sin 2 t n + α n cos d ( v n , z ) sin 2 s n sin 2 t n 4 cos d n sin s n sin t n ( ( 1 α n ) cos t n cos ( 1 2 β n ) s n + α n cos s n cos ( 1 2 γ n ) t n ) .
Dividing by 4 sin s n sin t n > 0 , we get
cos d n + 1 cos s n cos t n ( 1 α n ) cos d n cos t n cos ( 1 2 β n ) s n + α n cos d n cos s n cos ( 1 2 γ n ) t n .
(6)
Therefore, (4) and (5) can be reduced to the inequality (6) and it is equivalent to
( ε n cos s n α n cos ( 1 2 β n ) s n 1 α n α n ) ( ε n cos t n ( 1 α n ) cos ( 1 2 γ n ) t n α n 1 α n ) 1 ,
where ε n = cos d n + 1 / cos d n for n N . It follows that lim n ε n = cos d 0 / cos d 0 = 1 . Since { α n } [ a , b ] ] 0 , 1 [ for n N , we get
lim inf n ( cos s n α n cos ( 1 2 β n ) s n 1 α n α n ) ( cos t n ( 1 α n ) cos ( 1 2 γ n ) t n α n 1 α n ) 1 .
(7)
Then we show that there exists n 0 N such that for all n n 0 , the following hold:
1 2 cos s n α n cos ( 1 2 β n ) s n 1 α n α n 1
(8)
and
1 2 cos t n ( 1 α n ) cos ( 1 2 γ n ) t n α n 1 α n 1 .
(9)
First, we show the right inequality of (8). Since { β n } [ a , b ] ] 0 , 1 [ for n N , we get cos s n cos | 1 2 β n | s n = cos ( 1 2 β n ) s n . Hence we get
cos s n α n cos ( 1 2 β n ) s n 1 α n α n 1 α n 1 α n α n = 1 .
By the same method as above, the right inequality of (9) also holds. Next, let us show the left inequality of (8). If it does not hold, then letting
σ n = cos s n α n cos ( 1 2 β n ) s n 1 α n α n and τ n = cos t n ( 1 α n ) cos ( 1 2 γ n ) t n α n 1 α n ,
we can find a subsequence { σ n i } { σ n } such that σ n i < 1 / 2 for i N and lim i σ n i = σ 1 / 2 . Since { α n } , { γ n } [ a , b ] ] 0 , 1 [ and { t n } [ 0 , π / 4 [ [ 0 , π / 2 [ , we have { τ n } is bounded. Therefore, by taking a subsequence again if necessary, we may assume that { τ n i } converges to τ R . Then, by (7), we get σ τ = lim i σ n i τ n i lim inf n σ n τ n 1 . Hence we may assume that τ n i < 0 for all i N . Since 2 / 2 < cos s n , 2 / 2 < cos t n , 0 < cos ( 1 2 β n ) s n 1 , 0 < cos ( 1 2 γ n ) t n 1 and { α n } [ a , b ] ] 0 , 1 [ , we also have
0 < 2 2 b cos s n α n cos ( 1 2 β n ) s n and 0 < 2 2 ( 1 a ) cos t n ( 1 α n ) cos ( 1 2 γ n ) t n .
(10)
Let ρ be a real number such that
0 < ρ < min { 2 2 b , 2 2 ( 1 a ) , 1 b b + a 1 a } .
(11)
Then, by (10), we get
ρ 1 α n i α n i σ n i < 0 and ρ α n i 1 α n i τ n i < 0 .
(12)
Then, by (11) and (12), we have
σ n i τ n i ( ρ 1 α n i α n i ) ( ρ α n i 1 α n i ) = ρ 2 ( 1 α n i α n i + α n i 1 α n i ) ρ + 1 ρ 2 ( 1 b b + a 1 a ) ρ + 1 = ρ ( ρ ( 1 b b + a 1 a ) ) + 1 .
Thus, as i , we have
1 σ τ ρ ( ρ ( 1 b b + a 1 a ) ) + 1 < 1 .
(13)
This is a contradiction. We also obtain the left inequality of (9) in a similar way. Hence we get
lim n ( cos s n α n cos ( 1 2 β n ) s n 1 α n α n ) = lim n ( cos t n ( 1 α n ) cos ( 1 2 γ n ) t n α n 1 α n ) = 1
(14)
by Lemma 3.5, (8), and (9). Furthermore, from (14), we get
lim n cos s n cos ( 1 2 β n ) s n α n cos ( 1 2 β n ) s n = 0 .
(15)
By Lemma 3.6 and (15), we get
lim n ( cos s n cos ( 1 2 β n ) s n ) = 0 .
(16)
Moreover, by Lemma 3.7 and (16), we get
lim inf n cos s n = lim inf n cos ( 1 2 β n ) s n = lim inf n cos | 1 2 β n | s n .
Hence we get
lim sup n s n = lim sup n ( | 1 2 β n | s n ) lim sup n | 1 2 β n | lim sup n s n ,
and we have
0 ( 1 lim sup n | 1 2 β n | ) lim sup n s n = lim inf n ( 1 | 1 2 β n | ) lim sup n s n .

Since { β n } [ a , b ] ] 0 , 1 [ for n N , we have lim inf n ( 1 | 1 2 β n | ) > 0 and thus, lim sup n s n 0 . Therefore, we get lim sup n s n = 0 and we also get lim sup n t n = 0 . It implies d ( x n , S x n ) 0 and d ( x n , T x n ) 0 .

Next, let { y k } be a subsequence of { x n } . Since r ( { x n } ) d 0 < π / 2 , by Theorem 3.3(i), there exists a unique asymptotic center x 0 of { y k } . Moreover, since r ( { y k } ) < π / 2 , by Theorem 3.3(ii), there exists a subsequence { z l } of { y k } such that z l Δ z 0 X . Further, since d ( z l , S z l ) 0 , d ( z l , T z l ) 0 and S, T are Δ-demiclosed, we have z 0 F ( S ) F ( T ) . Then we can show that x 0 = z 0 , i.e., x 0 F ( S ) F ( T ) . If not, from the uniqueness of the asymptotic centers x 0 , z 0 of { y k } , { z l } , respectively, due to Theorem 3.3(i), we have
lim sup k d ( y k , x 0 ) < lim sup k d ( y k , z 0 ) = lim n d ( x n , z 0 ) = lim sup l d ( z l , z 0 ) < lim sup l d ( z l , x 0 ) lim sup k d ( y k , x 0 ) .
This is a contradiction. Hence we get x 0 F ( S ) F ( T ) . Next, we show that for any subsequences of { x n } , their asymptotic center consists of the unique element. Let { u k } , { v k } be subsequences of { x n } , x 0 A C ( { u k } ) and x 0 A C ( { v k } ) . We show x 0 = x 0 by using contradiction. Assume x 0 x 0 . Then x 0 A C ( { u k } ) and x 0 A C ( { v k } ) by Theorem 3.3(i). It follows that
lim sup k d ( u k , x 0 ) < lim sup k d ( u k , x 0 ) = lim n d ( x n , x 0 ) = lim sup k d ( v k , x 0 ) < lim sup k d ( v k , x 0 ) = lim n d ( x n , x 0 ) = lim sup k d ( u k , x 0 ) .

This is a contradiction. Hence we get x 0 = x 0 . Therefore, we have { x n } Δ-converges to a common fixed point of S and T. □

By Theorem 3.4, we know that a nonexpansive mapping having a fixed point satisfies the assumptions in Theorem 4.1. Thus, we get the following result.

Corollary 4.2 Let X be a complete CAT ( 1 ) space such that for any u , v X , d ( u , v ) < π / 2 . Let S and T be nonexpansive mappings of X into itself such that F ( S ) F ( T ) . Let { α n } , { β n } and { γ n } be sequences in [ a , b ] ] 0 , 1 [ . Define a sequence { x n } X as the following recurrence formula: x 1 X and
{ u n = ( 1 β n ) x n β n S x n , v n = ( 1 γ n ) x n γ n T x n , x n + 1 = ( 1 α n ) u n α n v n

for n N . Then { x n } Δ-converges to a common fixed point of S and T.

5 An application to the image recovery

The image recovery problem is formulated as to find the nearest point in the intersection of family of closed convex subsets from a given point by using corresponding metric projection of each subset. In this section, we consider this problem for two subsets of a complete CAT ( 1 ) space.

Theorem 5.1 Let X be a complete CAT ( 1 ) space such that for any u , v X , d ( u , v ) < π / 2 . Let C 1 and C 2 be nonempty closed convex subsets of X such that C 1 C 2 . Let P 1 and P 2 be metric projections onto C 1 and C 2 , respectively. Let { α n } , { β n } and { γ n } be real sequences in [ a , b ] ] 0 , 1 [ . Define a sequence { x n } X by the following recurrence formula: x 1 X and
{ u n = ( 1 β n ) x n β n P 1 x n , v n = ( 1 γ n ) x n γ n P 2 x n , x n + 1 = ( 1 α n ) u n α n v n

for n N . Then { x n } Δ-converges to a fixed point of the intersection of C 1 and C 2 .

Proof We see that P 1 and P 2 are quasinonexpansive [9] and Δ-demiclosed [8]. Further, we also get F ( P 1 ) = C 1 and F ( P 2 ) = C 2 . Thus, letting S = P 1 and T = P 2 in Theorem 4.1, we obtain the desired result. □

Declarations

Acknowledgements

The authors thank the anonymous referees for their valuable comments and suggestions. The first author is supported by Grant-in-Aid for Scientific Research No. 22540175 from the Japan Society for the Promotion of Science.

Authors’ Affiliations

(1)
Department of Information Science, Toho University, Miyama

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© Kimura and Nakagawa; licensee Springer. 2014

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