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

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:CC 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,zH:

αx+(1α) ( β y + ( 1 β ) z ) =γ ( δ x + ( 1 δ ) y ) +(1γ)z,

where α,β,γ,δ]0,1[ such that α=γδ and β=γ(1δ)/(1γδ). 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,yX, 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))=|st| 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,yX with d(x,y)<r, there exists a geodesic segment [x,y]. In particular, if a segment [x,y] is unique for any x,yX 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,yX. 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,yX and t]0,1[, if z[x,y] satisfies d(x,z)=(1t)d(x,y) and d(z,y)=td(x,y), then z is denoted by z=tx(1t)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 xX, 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 AC({ x n })={xX:r(x,{ x n })=r({ x n })}. If, for any subsequences { x n k } of { x n }, AC({ 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:XX is said to be a nonexpansive if T satisfies d(Tx,Ty)d(x,y) for any x,yX. The set of fixed points of T is denoted by F(T)={zX:Tz=z}. Further, a mapping T:XX with F(T) is said to be a quasinonexpansive if T satisfies d(Tx,z)d(x,z) for any xX and zF(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)<2π. Let u=tx(1t)y for some t[0,1]. Then

cosd(u,z)sind(x,y)cosd(x,z)sintd(x,y)+cosd(y,z)sin(1t)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)<2π. Let u=tx(1t)y for some t[0,1]. Then

cosd(u,z)tcosd(x,z)+(1t)cosd(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)

    AC({ 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,vX, 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 nN. Then { x n } Δ-converges to a common fixed point of S and T.

Proof Let zF(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 nN 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 nN. 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 nN.

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 2sin 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(12 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(12 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 4sin 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 nN. It follows that lim n ε n =cos d 0 /cos d 0 =1. Since { α n }[a,b]]0,1[ for nN, 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 nN, we get cos s n cos|12 β n | s n =cos(12 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 iN and lim i σ n i =σ1/2. 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 τR. Then, by (7), we get στ= 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 iN. Since 2 /2<cos s n , 2 /2<cos t n , 0<cos(12 β n ) s n 1, 0<cos(12 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 and0< 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 <0andρ 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

1στρ ( 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(12 β n ) s n = lim inf n cos|12 β n | s n .

Hence we get

lim sup n s n = lim sup n ( | 1 2 β n | s n ) lim sup n |12 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 nN, we have lim inf n (1|12 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 AC({ u k }) and x 0 AC({ v k }). We show x 0 = x 0 by using contradiction. Assume x 0 x 0 . Then x 0 AC({ u k }) and x 0 AC({ 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,vX, 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 nN. 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,vX, 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 nN. 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. □

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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.

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Kimura, Y., Nakagawa, K. Another type of Mann iterative scheme for two mappings in a complete geodesic space. J Inequal Appl 2014, 72 (2014). https://doi.org/10.1186/1029-242X-2014-72

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Keywords

  • Nonexpansive Mapping
  • Common Fixed Point
  • Nonempty Closed Convex Subset
  • Iteration Procedure
  • Real Sequence