R E S E A R C H
Open Access
Another type of Mann iterative scheme for
two mappings in a complete geodesic space
Yasunori Kimura and Koichi Nakagawa
**Correspondence:
[email protected] Department of Information Science, Toho University, Miyama, Funabashi, Chiba 274-8510, Japan
Abstract
In this paper, we show a
-convergence theorem for a Mann iteration procedure in a
complete geodesic space with two quasinonexpansive and-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 , Mann [] introduced an iteration procedure for ap-proximating fixed points of a nonexpansive mappingT in a Hilbert space. Later, Reich [] discussed this iteration procedure in a uniformly convex Banach space whose norm is Fréchet differentiable. In , Takahashi and Tamura [] considered an iteration proce-dure 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 , Dhompongsa and Panyanak [] proved the following theorem.
Theorem . Let C be a bounded closed convex subset of a completeCAT()space and T:C→C a nonexpansive mapping.For any initial point xin C,define the Mann iterative sequence{xn}by
xn+= ( –tn)xn⊕tnTxn, n= , , , . . . ,
where {tn} is a sequence in [, ], with the restrictions that
∞
n=tn diverges and
lim supn→∞tn< .Then{xn}-converges to a fixed point of T.
Further, in aCAT() space, Kimuraet al.[] proved the-convergence theorem for a family of nonexpansive mappings including the following scheme:
xn+= ( –αn)xn⊕αn
( –βn)Sxn⊕βnTxn
.
In a Hilbert spaceH, the following equality holds for anyx,y,z∈H:
αx+ ( –α)βy+ ( –β)z=γδx+ ( –δ)y+ ( –γ)z,
whereα,β,γ,δ∈], [ such thatα=γ δandβ=γ( –δ)/( –γ δ). However, inCAT(κ) spaces withκ> , 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 gen-eral:
xn+= ( –αn)xn⊕αn
( –βn)Sxn⊕βnTxn
,
xn+= ( –αn)
βnxn⊕( –βn)Sxn
⊕αn
( –βn)xn⊕βnTxn
.
()
In this paper, we show an analogous result to Theorem . using the iterative scheme () in a completeCAT() space with two quasinonexpansive and-demiclosed mappings. We also deal with the image recovery problem for two closed convex sets.
2 Preliminaries
LetXbe a metric space. Forx,y∈X, a mappingc: [,l]→Xis said to be a geodesic ifc
satisfiesc() =x,c(l) =yandd(c(s),c(t)) =|s–t|for alls,t∈[,l]. An image [x,y] ofcis called a geodesic segment joiningxandy. Forr> ,Xis said to be anr-geodesic metric space if, for anyx,y∈Xwithd(x,y) <r, there exists a geodesic segment [x,y]. In particular, if a segment [x,y] is unique for anyx,y∈Xwithd(x,y) <r, thenXis said to be a uniquely
r-geodesic metric space. In what follows, we always assumed(x,y) <π/ for anyx,y∈X. Thus, we sayXis a geodesic metric space instead of aπ/-geodesic metric space. For the more general case, see [].
LetXbe a uniquely geodesic metric space. A geodesic triangle is defined by(x,y,z) = [x,y]∪[y,z]∪[z,x]. LetMbe the two-dimensional unit sphere in R. For x¯,y¯,z¯∈M,
a triangle (x¯,y¯,z¯)⊂Mis called a comparison triangle of(x,y,z) ifd(x,y) =dM(x¯,y¯),
d(y,z) =dM(y¯,¯z),d(z,x) =dM(z¯,x¯). Further, for anyx,y∈Xandt∈], [, ifz∈[x,y] satis-fiesd(x,z) = ( –t)d(x,y) andd(z,y) =td(x,y), thenzis denoted byz=tx⊕( –t)y. A point
¯
z∈[x¯,y¯] is called a comparison point ofz∈[x,y] ifd(x,z) =dM(x¯,z¯).Xis said to be aCAT() space if, for anyp,q∈ (x,y,z)⊂Xand its comparison pointsp¯,q¯∈ (x¯,y¯,z¯)⊂M, the in-equalityd(p,q)≤dM(p¯,q¯) holds.
Let Xbe a geodesic metric space and{xn} a bounded sequence ofX. Forx∈X, we putr(x,{xn}) =lim supn→∞d(x,xn). The asymptotic radius of{xn}is defined byr({xn}) =
infx∈Xr(x,{xn}). Further, the asymptotic center of{xn} is defined byAC({xn}) ={x∈X:
r(x,{xn}) =r({xn})}. If, for any subsequences {xnk} of {xn}, AC({xnk}) ={x}, i.e., their
asymptotic center consists of the unique elementx, then we say{xn}-converges tox
and we denote it byxn–
x.
Let X be a metric space. A mappingT :X→X is said to be a nonexpansive if T
satisfies d(Tx,Ty)≤d(x,y) for any x,y∈X. The set of fixed points ofT is denoted by
F(T) ={z∈X:Tz=z}. Further, a mappingT:X→XwithF(T)=∅is said to be a quasi-nonexpansive ifT satisfiesd(Tx,z)≤d(x,z) for anyx∈Xandz∈F(T). Moreover,T is said to be-demiclosed if, for any bounded sequence {xn} ⊂X andx∈X satisfying d(xn,Txn)→ andxn–
x, we havex∈F(T).
3 Tools for the main results
Theorem .(Kimura and Satô []) Let(x,y,z)be a geodesic triangle in aCAT()space such that d(x,y) +d(y,z) +d(z,x) < π.Let u=tx⊕( –t)y for some t∈[, ].Then
cosd(u,z)sind(x,y)≥cosd(x,z)sintd(x,y) +cosd(y,z)sin( –t)d(x,y).
Corollary .(Kimura and Satô []) Let(x,y,z)be a geodesic triangle in aCAT()space such that d(x,y) +d(y,z) +d(z,x) < π.Let u=tx⊕( –t)y for some t∈[, ].Then
cosd(u,z)≥tcosd(x,z) + ( –t)cosd(y,z).
Theorem .(Espínola and Fernández-León []) Let X be a completeCAT()space and
{xn}a sequence in X.If r({xn}) <π/,then the following hold. (i) AC({xn})consists of exactly one point;
(ii) {xn}has a-convergent subsequence.
Theorem .(Kimura and Satô []) 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-demiclosed.
The following lemmas are important properties of real numbers and they are easy to show. Thus, we omit the proofs.
Lemma . Letδ be a real number such that – <δ< and{bn}, {cn} real sequences
satisfyingδ≤bn≤,δ≤cn≤andlim infn→∞bncn≥.Thenlimn→∞bn=limn→∞cn= .
Lemma . Let s∈],∞[and{bn},{cn}bounded real sequences satisfying bn≤,s<cn
andlimn→∞bn/cn= .Thenlimn→∞bn= .
Lemma . Let{bn}and{cn}be bounded real sequences satisfyinglimn→∞(bn–cn) = .
Thenlim infn→∞bn=lim infn→∞cn.
4 The main result
In this section, we show the main result.
Theorem . Let X be a completeCAT()space such that for any u,v∈X,d(u,v) <π/.
Let S and T be quasinonexpansive and -demiclosed mappings from X into itself with F(S)∩F(T)=∅.Let{αn},{βn}and{γn}be sequences of[a,b]⊂], [.Define a sequence
{xn} ⊂X by the following recurrence formula:x∈X and ⎧
⎪ ⎨ ⎪ ⎩
un= ( –βn)xn⊕βnSxn,
vn= ( –γn)xn⊕γnTxn,
xn+= ( –αn)un⊕αnvn
Proof Letz∈F(S)∩F(T). By Corollary ., we have
cosd(un,z)≥( –βn)cosd(xn,z) +βncosd(Sxn,z)
≥( –βn)cosd(xn,z) +βncosd(xn,z) =cosd(xn,z),
cosd(vn,z)≥( –γn)cosd(xn,z) +γncosd(Txn,z)
≥( –γn)cosd(xn,z) +γncosd(xn,z) =cosd(xn,z).
Then, by Corollary . again, we have
cosd(xn+,z)≥( –αn)cosd(un,z) +αncosd(vn,z)
≥( –αn)cosd(xn,z) +αncosd(xn,z)
≥cosd(xn,z).
So, we getd(xn+,z)≤d(xn,z) for alln∈Nand there existsd=limn→∞d(xn,z)≤d(x,z) <
π/.
Furthermore, by Theorem ., we have
cosd(un,z)sind(xn,Sxn)
≥cosd(xn,z)sin( –βn)d(xn,Sxn) +cosd(Sxn,z)sinβnd(xn,Sxn)
≥cosd(xn,z)sin
d(xn,Sxn)
cos
( – βn)d(xn,Sxn)
()
and
cosd(vn,z)sind(xn,Txn)
≥cosd(xn,z)sin( –γn)d(xn,Txn) +cosd(Txn,z)sinγnd(xn,Txn)
≥cosd(xn,z)sin
d(xn,Txn)
cos
( – γn)d(xn,Txn)
. ()
Letdn=d(xn,z),sn=d(xn,Sxn)/ andtn=d(xn,Txn)/ forn∈N. If there existsn∈N
such thatsn=tn= , then we havexn∈F(S)∩F(T) and since
xn+= ( –αn)
( –βn)xn⊕βnSxn
⊕αn
( –γn)xn⊕γnTxn
= ( –αn)xn⊕αnxn =xn,
and the proof is finished. So, we may assumesn= ortn= for alln∈N.
Ifsn= andtn= , then we haveun=xn. From (), (), and Corollary ., we get cosdn+sintncostn
≥( –αn)cosd(un,z)sintn+αncosd(vn,z)sintn
≥( –αn)cosdnsintncostn+ αncosdnsintncos( – γn)tn.
Dividing by sintn> , we get
cosdn+costn≥( –αn)cosdncostn+αncosdncos( – γn)tn. () Iftn= andsn= , then we havevn=xn. In a similar way as above, we get
cosdn+cossn≥( –αn)cosdncos( – βn)sn+αncosdncossn. () Ifsn= andtn= , then from (), (), and Corollary ., we get
cosdn+sinsnsintn
≥( –αn)cosd(un,z)sinsnsintn+αncosd(vn,z)sinsnsintn
≥cosdnsinsnsintn
( –αn)costncos( – βn)sn+αncossncos( – γn)tn
.
Dividing by sinsnsintn> , we get
cosdn+cossncostn
≥( –αn)cosdncostncos( – βn)sn+αncosdncossncos( – γn)tn. ()
Therefore, () and () can be reduced to the inequality () and it is equivalent to
εncossn αncos( – βn)sn
– –αn αn
εncostn ( –αn)cos( – γn)tn
– αn –αn
≥,
whereεn=cosdn+/cosdnforn∈N. It follows thatlimn→∞εn=cosd/cosd= . Since
{αn} ⊂[a,b]⊂], [ forn∈N, we get
lim inf
n→∞
cossn αncos( – βn)sn
– –αn αn
costn ( –αn)cos( – γn)tn
– αn –αn
≥. ()
Then we show that there existsn∈Nsuch that for alln≥n, the following hold:
– ≤
cossn αncos( – βn)sn
– –αn αn ≤
()
and
– ≤
costn ( –αn)cos( – γn)tn
– αn –αn≤
. ()
First, we show the right inequality of (). Since{βn} ⊂[a,b]⊂], [ forn∈N, we get
cossn≤cos| – βn|sn=cos( – βn)sn. Hence we get
cossn αncos( – βn)sn
– –αn αn
≤
αn
– –αn αn
By the same method as above, the right inequality of () also holds. Next, let us show the left inequality of (). If it does not hold, then letting
σn=
cossn αncos( – βn)sn
– –αn αn
and τn=
costn ( –αn)cos( – γn)tn
– αn –αn
,
we can find a subsequence{σni} ⊂ {σn}such thatσni< –/ fori∈Nandlimi→∞σni=
σ ≤–/. Since{αn},{γn} ⊂[a,b]⊂], [ and{tn} ⊂[,π/[⊂[,π/[, we have{τn}is bounded. Therefore, by taking a subsequence again if necessary, we may assume that{τni}
converges toτ ∈R. Then, by (), we getσ τ=limi→∞σniτni≥lim infn→∞σnτn≥. Hence
we may assume thatτni< for alli∈N. Since
√
/ <cossn,
√
/ <costn, <cos( – βn)sn≤, <cos( – γn)tn≤ and{αn} ⊂[a,b]⊂], [, we also have
<
√
b ≤
cossn αncos( – βn)sn
and <
√
( –a)≤
costn ( –αn)cos( – γn)tn
. ()
Letρbe a real number such that
<ρ<min
√
b,
√
( –a),
–b
b +
a
–a
. ()
Then, by (), we get
ρ– –αni
αni
≤σni< and ρ–
αni
–αni
≤τni< . ()
Then, by () and (), we have
σniτni≤ ρ–
–αni
αni
ρ– αni
–αni
=ρ– –αni
αni
+ αni
–αni
ρ+
≤ρ– –b
b +
a
–a
ρ+
=ρ ρ– –b
b +
a
–a
+ .
Thus, asi→ ∞, we have
≤σ τ≤ρ ρ– –b
b +
a
–a
+ < . ()
This is a contradiction. We also obtain the left inequality of () in a similar way. Hence we get
lim
n→∞
cossn αncos( – βn)sn
– –αn αn
= lim
n→∞
costn ( –αn)cos( – γn)tn
– αn –αn
= ()
by Lemma ., (), and (). Furthermore, from (), we get
lim
n→∞
cossn–cos( – βn)sn αncos( – βn)sn
By Lemma . and (), we get
lim
n→∞
cossn–cos( – βn)sn
= . ()
Moreover, by Lemma . and (), we get
lim inf
n→∞ cossn=lim infn→∞ cos( – βn)sn=lim infn→∞ cos| – βn|sn. Hence we get
lim sup
n→∞ sn=lim supn→∞
| – βn|sn
≤lim sup
n→∞ | – βn|lim supn→∞ sn,
and we have
≥
–lim sup
n→∞ | – βn|
lim sup
n→∞
sn=lim inf n→∞
–| – βn|
lim sup
n→∞
sn.
Since {βn} ⊂[a,b]⊂], [ forn∈N, we have lim infn→∞( –| – βn|) > and thus,
lim supn→∞sn≤. Therefore, we getlim supn→∞sn= and we also getlim supn→∞tn= . It impliesd(xn,Sxn)→ andd(xn,Txn)→.
Next, let{yk} be a subsequence of{xn}. Sincer({xn})≤d<π/, by Theorem .(i),
there exists a unique asymptotic centerxof{yk}. Moreover, sincer({yk}) <π/, by The-orem .(ii), there exists a subsequence{zl}of{yk}such thatzl–
z∈X. Further, since d(zl,Szl)→,d(zl,Tzl)→ andS,T are-demiclosed, we havez∈F(S)∩F(T). Then
we can show thatx=z,i.e.,x∈F(S)∩F(T). If not, from the uniqueness of the
asymp-totic centersx,zof{yk},{zl}, respectively, due to Theorem .(i), we have
lim sup
k→∞
d(yk,x) <lim sup
k→∞
d(yk,z)
= lim
n→∞d(xn,z)
=lim sup
l→∞
d(zl,z)
<lim sup
l→∞
d(zl,x)
≤lim sup
k→∞
d(yk,x).
This is a contradiction. Hence we getx∈F(S)∩F(T). Next, we show that for any
sub-sequences of{xn}, their asymptotic center consists of the unique element. Let{uk},{vk} be subsequences of{xn},x∈AC({uk}) andx∈AC({vk}). We showx=xby using
con-tradiction. Assumex=x. Thenx∈/ AC({uk}) andx∈/AC({vk}) by Theorem .(i). It follows that
lim sup
k→∞
d(uk,x) <lim sup
k→∞
duk,x
= lim
n→∞d
xn,x
=lim sup
k→∞ d
<lim sup
k→∞
d(vk,x)
= lim
n→∞d(xn,x)
=lim sup
k→∞
d(uk,x).
This is a contradiction. Hence we getx=x. Therefore, we have{xn}-converges to a
common fixed point ofSandT.
By Theorem ., we know that a nonexpansive mapping having a fixed point satisfies the assumptions in Theorem .. Thus, we get the following result.
Corollary . Let X be a completeCAT()space such that for any u,v∈X,d(u,v) <π/.
Let S and T be nonexpansive mappings of X into itself such that F(S)∩F(T)=∅.Let{αn},
{βn} and{γn}be sequences in[a,b]⊂], [.Define a sequence{xn} ⊂X as the following
recurrence formula:x∈X and ⎧
⎪ ⎨ ⎪ ⎩
un= ( –βn)xn⊕βnSxn,
vn= ( –γn)xn⊕γnTxn,
xn+= ( –αn)un⊕αnvn
for n∈N.Then{xn}-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 intersec-tion 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 completeCAT() space.
Theorem . Let X be a completeCAT()space such that for any u,v∈X,d(u,v) <π/.
Let Cand Cbe nonempty closed convex subsets of X such that C∩C=∅.Let Pand P be metric projections onto Cand C,respectively.Let{αn},{βn}and{γn}be real sequences
in[a,b]⊂], [.Define a sequence{xn} ⊂X by the following recurrence formula:x∈X and
⎧ ⎪ ⎨ ⎪ ⎩
un= ( –βn)xn⊕βnPxn,
vn= ( –γn)xn⊕γnPxn,
xn+= ( –αn)un⊕αnvn
for n∈N.Then{xn}-converges to a fixed point of the intersection of Cand C.
Proof We see thatP andP are quasinonexpansive [] and-demiclosed []. Further,
we also getF(P) =CandF(P) =C. Thus, lettingS=PandT=Pin Theorem ., we
obtain the desired result.
Competing interests
Authors’ contributions
The authors have contributed in this work on an equal basis. All authors read and approved the final manuscript.
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.
Received: 15 October 2013 Accepted: 23 January 2014 Published:13 Feb 2014 References
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