This paper introduces a new variational inequality problem in $CAT(0)$ spaces involving nonself multivalued nonexpansive mappings, expanding the theoretical framework of variational inequalities in metric spaces.
Contribution
It presents a novel variational inequality problem tailored for $CAT(0)$ spaces with nonself multivalued nonexpansive mappings, extending existing VIP theories.
Findings
01
Defines a new VIP in $CAT(0)$ spaces
02
Establishes foundational properties of the new VIP
03
Lays groundwork for future solution methods
Abstract
In this paper, we introduce a new variational inequality problem(VIP) associated with nonself multivalued nonexpansive mappings in CAT(0) spaces.
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Taxonomy
TopicsOptimization and Variational Analysis · Contact Mechanics and Variational Inequalities · Point processes and geometric inequalities
Full text
Variational Inequalities on Geodesic Spaces
Emirhan Hacioglu
Department of Mathematics, Yildiz Technical
University, Davutpasa Campus, Esenler, 34220 Istanbul, Turkey
Let (X,d) be a metric space then the family of nonempty, closed
and convex subsets of X, the family of nonempty compact and convex subsets
of X, the family of nonempty compact subsets of X, the family of
nonempty closed and bounded convex subsets of X will be denoted by C(X),KC(X),K(X),CB(X), respectively. Let H be a Haussdorf Metric on
CB(X), defined by
[TABLE]
where d(x,B)=inf{d(x,y);y∈B}. A multivalued mappings T:X→2X is called nonexpansive if for all x,y∈X
[TABLE]
is satisfied. A point is called fixed point of T if x∈Tx and the
set of all fixed points of T is denoted by F(T).Many iterative processes
to find a fixed point of multivalued mappings have been introduced in metric
spaces and Banach spaces. One of them is defined by Nadler[1] as
generalization of Picard as follows;
[TABLE]
A multivalued version of Mann and Ishikawa fixed point procedures goes as
follow;
[TABLE]
and
[TABLE]
where {αn} and {βn} are sequences in[0,1].
Gursoy and Karakaya [17] (see also [18]) introduced Picard-S
iteration as follows;
[TABLE]
where {αn} and {βn} are sequences in[0,1].
They have showed that it converges to fixed point of contraction mappings
faster than Ishikawa, Noor, SP, CR, S and some other iterations. Also they
use it to solve differential equations. Now, we define multivalued version
of Picad-S iteration in CAT(0) spaces as follows; Let K⊂CAT(0) be
a nonempty, closed and convex subset, T:K→C(K) is a mapping, x0∈K. then for any n≥0, the proximal multivalued Picard-S
iteration is defined by
[TABLE]
where PK is a metric projection, {αn} and {βn}
are sequences in[0,1] with liminfn(1−βn)βn>0, un∈Tyn,vn∈Tzn and wn∈Txn.
Before the results we give some definitions and lemmas about CAT(0)
and Δ−convergences.
Let (X,d) be a metric space, x,y∈X and C⊆X nonempty
subset. A geodesic path (or shortly a geodesic) joining x and y is a map
c:[0,t]⊆R→X such that c(0)=x, c(t)=y and d(c(r),c(s))=∣r−s∣ for
all r,s∈[0,t]. In particular c is an isometry and d(c(0),c(t))=t. The image of c,c([0,t]) is called geodesic segment
from x to y and it is unique (it not necessarily be unique) then it is
denoted by [x,y].z∈[x,y] if and only if for an λ∈[0,1] such that d(z,x)=(1−λ)d(x,y) and d(z,y)=λd(x,y). The point z is denoted by z=(1−λ)x⊕λy. If
for every x,y∈X there is a geodesic path then (X,d) called geodesic
space and uniquely geodesic space if that geodesic path is unique for any
pair x,y. A subset C⊆X is called convex if it contains all
geodesic segment joining any pair of points in it.
In geodesic metric space (X,d), a geodesic triangle Δ(x,y,z)
consist of three point x,y,z as vertices and three geodesic segments of
any pair of these points, that is, q∈Δ(x,y,z) means that q∈[x,y]∪[x,z]∪[y,z]. The triangle Δ(x,y,z) in (R2,d2) is called comparison triangle for the triangle Δ(x,y,z)
such that d(x,y)=d2(x,y),d(x,z)=d2(x,z) and d(y,z)=d2(y,z) A point point
z∈[x,y] called comparison point
for z∈[x,y] if d(x,z)=d2(x,z). A
geodesic triangle Δ(x,y,z) in X is satisfied CAT(0) inequality ifd(p,q)≤d2(p,q) for all p,q∈Δ(x,y,z) where p,q∈Δ(x,y,z) are the comparison points of p,q respectively.
A geodesic space is called CAT(0) space if for all geodesic triangles
satisfies CAT(0) inequality or alternatively: A geodesic space is called CAT(0) space if and only if the inequality
[TABLE]
satisfied for every x,y,z∈X, λ∈[0,1].
Proposition 1.1**.**
[10*]*Let (X,d) be a CAT(0) space Then, for any x,y,z∈X and λ∈[0,1], we have
[TABLE]
Let {xn} be a bounded sequence on X and x∈X. Then, with
setting
[TABLE]
the asymptotic radius of {xn} is defined by
[TABLE]
the asymptotic radius of {xn} with respect to K⊆X is
defined by
[TABLE]
and the asymptotic center of {xn} is defined by
[TABLE]
and let ωw(xn):=∪A({xn}) where union is taken on all
subsequences of {xn}.
Definition 1.2**.**
[12]A sequence {xn}⊂X is said to be Δ− convergent to x∈X if x is the unique asymptotic center of
all subsequence {un} of {xn}, i.e.ωw(xn):=∪A({xn})={x} . In this case we write Δ−limnxn=x.
Every bounded sequence in a complete CAT(0) space has a Δ-convergent subsequence
2. ii)
If K is a closed convex subset of a complete
CAT(0) and if {xn} is a bounded sequence in K, then the asymptotic center of {xn} is in K
Lemma 1.4**.**
[10]** If {xn} is a bounded sequence in X
with A({xn})={x} and {un} is a subsequence of {xn}
with A({un})=u and the sequence {d(xn,u)} converges, then x=u
Theorem 1.5**.**
[11*]*Let X be a bounded, complete and uniformly convex
metric space. If T is a multivalued nonexpansive mapping which assigns to
each point of X a nonempty compact subset of X, then T has a fixed
point in X.
In a complete CAT(0) space, the metric projection PK(x) of x onto a
nonempty, closed and convex subset K is singleton and nonexpansive.
The concept of inner-product has been generalized from Hilbert space to a CAT(0) space X by Berg and Nikolaev [16]. as follows: For any a,b∈X, with denoting ab as a vector in X,
quasi-linearization mapping defined as
[TABLE]
for all a,b,c,d∈X and satisfies following properties
[TABLE]
for all a,b,c,d,e∈X The last properties is known as a Cauchy-Schwarz
inequality and it is a characterization of CAT(0) space: A geodesic metric
space is a CAT(0) if and only if it satisfies Cauchy-Schwarz inequality.
Lemma 1.6**.**
[16*]*Let X be a CAT(0) and K be a nonempty and
convex subset of X, x∈X and u∈K. Then u=PK(x) if and only
if
[TABLE]
Let X be a real Hilbert space and K⊂X be nonempty closed and
convex. A operator A:K→2X is called monotone if and only if
[TABLE]
for all x,y∈X,x∗∈Ax,y∗∈Ay. If A is a monotone
operator then the variational inequality associated with A is finding (u,x)u∈Ax such that
[TABLE]
The VIPs associated with monotone operators have applications in applied
mathematics. For interested readers can find more informations about VIPs
and their applications in the book by Kinderlehrer and Stampacchia (see
[2, 3]
).
Now let X be a complete CAT(0) space, K⊂X be nonempty, closed
and convex and T:K→X be a nonexpansive mapping. In 2015,
Khatibzadeh, & Ranjbar [15] defined the variational inequality
associated with the nonexpansive mapping T as follows
[TABLE]
They prove some existence and convergence results for this problem.
In this paper, we define variational inequality associated with the a
non-self multivalued nonexpansive mapping T:K→KC(X) as follows
[TABLE]
and we prove some existence and convergence theorems for this problem.
2. Existence of A Solution
In this section, it is assumed that X is a complete CAT(0) and K is a
nonempty, closed and convex subset of X.
Definition 2.1**.**
If K is also bounded subset of X and T:K→C(X). Then the
projection PKT of multivalued mapping T onto K is defined by
[TABLE]
where PK is metric projection and D(x′,K)=infv′∈Kd(x′,v′).
Lemma 2.2**.**
PK∗T(x)* is multivalued nonexpansive mapping from K to 2K*
Proof.
Since K is closed, convex and bounded, PK∗(Tx)⊂K. We
also have
[TABLE]
by the nonexpansiveness of PK. ∎
Lemma 2.3**.**
If T is compact valued then PK∗ is compact valued.
Proof.
Let (vn)⊂PK∗T(x) be a sequence then there is a
sequence (xn′)⊂Tx such that for all n∈N, there vn=PK(xn′).Since T have compact values then (xn′) have convergent subsequence (xnk′) with limk→∞xnk′=z∈Tx and since for all k∈N,
[TABLE]
we get that the sequence (vn)=(PK(xn′)) have convergent
subsequence (vnk)=(PK(xnk′)) with limk→∞(PK(xnk′))=PK(z)∈PK∗T(x) therefore PK∗Tx is compact.
∎
Theorem 2.4**.**
If T:K→KC(X). Then there exists a solution (u,x)u∈Tx of the variational inequality (1.2)
Proof.
Since X is uniformly convex and T is compact valued, PKT have fixed
point p∈PKT(p)⊂K by Theorem 1.5. There exist p′∈Tp such that p=PK(p′) by definition of PKT. we have ⟨p′p,yp⟩≤0 for all y∈K
by Lemma 1.6. Hence we have
[TABLE]
where p′∈Tp, that is (p′,p)p′∈Tp
is a solution of the problem (1.2).
∎
Theorem 2.5**.**
If x∈int(K) and (u,x)u∈Tx is a solution of problem (1.2) then x∈F(T), i.e., u=x.
Proof.
There exists ϵ>0 such that B(x,ϵ)⊂K. Let take t∈(0,1) such that tx⊕(1−t)u∈B(x,ϵ), that is, d(x,tx⊕(1−t)u)=(1−t)d(x,u)<ϵ . Since B(x,ϵ)⊂K
then tx⊕(1−t)u∈K and d(u,tx⊕(1−t)u)=td(u,x) so we have
[TABLE]
and which implies
[TABLE]
since t∈(0,1) then d(x,u)=0. Hence u=x∈Tx
∎
If K is not bounded, the problem (1.2) does not always have a
solution. However if o∈X be arbitrary and setting Kr=K∩B(o,r)
then if Kr=∅ By Theorem 2.4 there is xr∈Kr such that (ur,xr)ur∈Tx is a solution of problem
[TABLE]
Theorem 2.6**.**
The problem (1.2) have a solution if and only if there is
a r>0 such that the solution of the problem (2.1) (ur,xr)ur∈Txr,xr∈Kr satisfies d(o,xr)<r.
Proof.
If the problem 1.2 have a solution (u,x)x∈Tx then (u,x)x∈Tx is a solution of the problem (2.1) and d(o,x)<r is satisfied.
Now, let there is a r>0 such that the solution of the problem (2.1) (ur,xr)ur∈Txr, xr∈Kr satisfies d(o,xr)<r
and y∈K be arbitrary. Then we can chose t∈(0,1) such that (1−t)xr⊕ty∈B(o,r), that is, (1−t)xr⊕ty⊂Kr
and d(xr,(1−t)xr⊕ty)=td(xr,y). Then
[TABLE]
Hence
[TABLE]
that is (ur,xr)ur∈Txr is a solution of the problem (1.2) .
∎
Theorem 2.7**.**
Let T:K→KC(X) and o∈X be fixed. If there exist x0∈K
and u0∈Tx0 such that
[TABLE]
where u∈Tx such that d(x,u)=d(x,Tx) then the problem (1.2)
have a solution.
Proof.
Let R,M∈R such that d(u0,u0)<M, d(x0,o)<r and
[TABLE]
for all x∈K,d(x,o)≥r. Then
[TABLE]
for r=d(x,o). If (ur,xr)ur∈Txr is a solution of the
problem (2.1) then since
[TABLE]
holds so we have d(xr,o)<r. Hence by Theorem 2.6 the problem (1.2) has a solution.
∎
3. Convergence Results to The Solutions
In this section, it is assumed that X is a complete CAT(0) and K is a
nonempty, closed and convex subset of X.
Theorem 3.1**.**
If T:K→KC(X) is a nonexpansive mapping and {xn} is a bounded sequence in K with Δ−limn→∞xn=z and limn→∞d(xn,Txn)=0 then z∈K and z∈T(z).
Proof.
By Lemma 1.3, z∈K.We can find a sequence {yn} such
that yn∈Txn,d(xn,yn)=d(xn,Txn), so we have limn→∞d(xn,yn)=0 and we can find a sequence {zn} in Tz such that d(yn,zn)=d(yn,Tz). Then Since Tz
is compact, there is a convergent subsequence {zni} of {zn},
say limi→∞zni=u∈Tz.
[TABLE]
implies that limsupi→∞d(xni,u)≤limsupi→∞H(Txni,Tz) and Δ−limi→∞xni=z Because of T is multivalued
nonexpansive mapping,
[TABLE]
which implies that
[TABLE]
which implies that z=u∈Tz.
∎
Lemma 3.2**.**
If T:K→KC(X) is a nonexpansive mapping and {xn} is a bounded sequence in K with limn→∞d(xn,Txn)=0 and {d(xn,p)} converges for all p∈F(T) then ωw(xn)⊆F(T) and ωw(xn) include exactly
one point.
Proof.
Let take u∈ωw(xn) then there exist subsequence {un}
of {xn} with A({un})={u}.Then by Lemma 1.3
there exist subsequence {vn} of {un} with Δ−limn→∞vn=v∈K . Then by Theorem 3.1
we have v∈F(T) and by Lemma 1.4 we conclude that u=v,
hence we get ωw(xn)⊆F(T). Let take subsequence {un} of {xn}with A({un})={u} and A({xn})={x}.
Because of v∈ωw(xn)⊆F(T), {d(xn,u)}
converges, so by Lemma 1.4 we have x=u, this means that ωw(xn) include exactly one point.
∎
Theorem 3.3**.**
If T:K→C(X) is a nonexpansive mapping with F(T)=∅ and Tp={p} for all p∈F(T) and {xn} is a
sequence in K defined by (1.1) with liminfn→∞βn(1−βn)>0 then {xn} is bounded, limn→∞d(xn,Txn)=0 and {d(xn,p)} converges
for all p∈F(T).
Proof.
Let p∈F(T) then for any x∈K,we have that
[TABLE]
since metric projection PK is nonexpansive and PK(p)={x∈K:d(p,x)=d(p,K)}={p} we have
[TABLE]
and
[TABLE]
Here we have d2(xn+1,p)≤d2(xn,p) implies that limn→∞d(xn,p) exists, it is bounded,and d(xn+1,p)≤d(yn,p)≤d(xn,p) implies limn→∞[d(xn,p)−d(yn,p)]=0. Since βn(1−βn)d2(Txn,xn))≤d2(xn,p)−d2(yn,p),by assumption
we have that limn→∞d2(Txn,xn)=0,so limn→∞d(Txn,xn)=0
∎
Theorem 3.4**.**
If T:K→KC(X) is a nonexpansive mapping with F(T)=∅ and Tp={p} for all p∈F(T) and {xn} is a
sequence in K defined by (1.1) with liminfn→∞βn(1−βn)>0 then {xn} is Δ−convergent
to p∈F(T) where (p,p) is a solution of the problem (1.2)
Proof.
Since we have limn→∞d(xn,Txn)=0, {d(xn,p)} converges for all p∈F(T) and {xn} is bounded by
Theorem 3.3 then it follows from Lemma 3.2 that ωw(xn)⊆F(T) and ωw(xn) include exactly
one point p∈F(T) where (p,p) is a solution of the problem 1.2
∎
Theorem 3.5**.**
Let K be also compact and T:K→C(X) be a nonexpansive mapping
with F(T)=∅ and Tp={p} for all p∈F(T). If {xn}
is a sequence in K defined by (1.1) with liminfn→∞βn(1−βn)>0 then {xn} strongly converges to q∈F(T).where (q,q) is a solution of the problem (1.2)
Proof.
By Theorem 3.3, we have that limn→∞d(Txn,xn)=0 and limn→∞d(xn,p) exists for
all p∈F(T) Since K is compact there is a convergent subsequence {xni} of {xn}, say limi→∞xni=q. Then we have
[TABLE]
and taking limit on i,continuity of T implies that q∈Tq.
∎
4. Common Solution of System of Variational Inequalities
Let X be a CAT(0) space and Ki⊂X be a nonempty, closed and
convex subsets with i=1⋂NKi=∅. If Ti:Ki→C(X) are mappings for i=1...N. then the system of
variational inequalities problem is
[TABLE]
It is obvious that for N=1 the problem is reduced the problem (1.2).
The importance of studying the problem (4.1) is underlying on fact
that it is unification most of the problems; for example taking if we take Ti=0 for all i=1,...,N the reduce the problem (4.1) to convex
feasibility problem,
[TABLE]
or if every Ti is self operator and K=i=1⋂NF(Ti)
then it turn to common fixed point problem. We will show that the algorithm
defined by (4.2) is convergent to common fixed point of family of
non-self multivalued nonexpansive mappings {Ti}i=1N which is
also a solution of system of variational inequalities problem (4.1)
Let K=i=1⋂NKi=∅ and x1∈K. then
for any n≥0, the modified proximal multivalued Picard-S iteration is
defined by
[TABLE]
where un,i∈Tiyn,wn,i∈Tixn,vn,i∈Tizn,{λn,i},{αn,i},{βn,i} and {γn,i} are the sequences satisfies i=1∑Nλn,i=1,i=1∑N(αn,i+βn,i)=1,i=0∑Nγn,i=1 in [b,c] for some b,c∈(0,1)
Lemma 4.1**.**
[19]** Let (X,d,W) be a uniformly convex hyperbolic space with modulus
of uniform convexity δ. For any r>0,ϵ∈(0,2),λ∈[0,1] and a,x,y∈X, if d(x,a)≤r, d(y,a)≤r and d(x,y)≥∈r then d((1−λ)x⊕λy,z)≤(1−2λ(1−λ)δ(r,∈))r.
Proposition 4.2**.**
[20]** Assume that X is a CAT(0) space. Then X is uniformly convex
and
[TABLE]
is a modulus of uniform convexity.
Lemma 4.3**.**
[9]** Let (X,d) be a complete CAT(0) space, {x1,x2,..xn}⊂X and {λ1,λ2,..,λn}⊂[0,1] with i=1∑nλi=1.Then d(i=1⨁nλixi,z)≤i=1∑nλid(xi,z) for every z∈X.
Lemma 4.4**.**
[20]** Let X be a complete CAT(0) space with modulus
of convexity δ(r,∈) and let x∈E. Suppose that δ(r,∈) increases with r (for a fixed ∈ ) and suppose {tn} is a
sequence in [b,c] for some b,c∈(0,1), {xn} and {yn} are
the sequences in X such that limsupn→∞d(xn,x)≤r,limsupn→∞d(yn,x)≤r
and limn→∞d((1−tn)xn⊕tnyn,x)=r for some r≥0. Then limn→∞d(xn,yn)=0.
The following Lemma is very important to our results.
Lemma 4.5**.**
Let X be a complete CAT(0) space with modulus of convexity
δ(r,∈) and let x∈X. Suppose that δ(r,∈) increases
with r (for a fixed ∈ ) and suppose {tn,i} with i=1∑Ntn,i=1 is a sequence in [b,c] for some b,c∈(0,1), {xn,i}n=1∞ are the sequences for i∈{1,2,..,N} in X such that limsupn→∞d(xn,i,x)≤r and limn→∞d(i=1⨁Ntn,ixn,i,x)=r for some r≥0. Then limn→∞d(xn,k,xn,l)=0 for k,l∈{1,2,..,N}.
Proof.
If r=0 then it is obvious let r>0.Since limsupn→∞d(xn,i,x)≤r for each i=1,2,..N, then, by Lemma 4.3, for every m=1,2,..,N,
[TABLE]
Let assume that d(xn,k,xn,l)↛0 for fixed k,l∈{1,2,..,N} with k=l then there is subsequence denoted by (without
loss of generality) {xn,k} and {xn,l} such that infnd(xn,k,xn,l)>0.Since
[TABLE]
then
[TABLE]
and since tn,k,tn,l∈[b,c] and by positivity of d, d(i=1,i=k⨁N1−tn,ktn,ixn,i,xn,k)↛0. therefore there is subsequence again
denoted by {xn,k} for some k=1,2,..N such that d(i=1⨁N1−tn,ktn,ixn,i,xn,k)>0 so d(xn,k,x)≤r,d(i=1,i=k⨁N1−tn,ktn,ixn,k,x)≤r and limn→∞d(i=1⨁Ntn,ixn,i,x)=limn→∞d((1−tn,m)[i=1,i=k⨁N1−tn,ktn,ixn,i]⊕tn,mxn,m,x)=r hence we can apply Lemma 4.4.
∎
From this point, it is assumed that X is a complete CAT(0) and K=i=1⋂NKi is a nonempty, closed and convex subset of X where Ki⊂X be a nonempty, closed and convex subsets with K=i=1⋂NKi=∅ for all i=1,2,...,N.
Lemma 4.6**.**
Let {Ti}i=1N be multivalued nonexpansive
mappings from K to CC(X) with F=i=1⋂NF(Ti)=∅,Tip={p} for all p∈F. If {xn} is the
sequence defined by (4.2) then {xn} is bounded and limn→∞d(xn,p) exist for all p∈F.
Proof.
Let p∈F.Then from definition of {xn},
[TABLE]
and
[TABLE]
and
[TABLE]
Hence d(yn,p)≤d(xn,p),d(zn,p)≤d(xn,p) and d(xn+1,p)≤d(xn,p) and so limn→∞d(xn,p)
exists and {xn} is bounded sequence.
∎
Lemma 4.7**.**
Let {Ti}i=1N be multivalued nonexpansive
mappings from K to C(X) with F=i=1⋂NF(Ti)=∅,Tip={p} for all p∈F. If {xn} is the
sequence defined by (4.2) then limn→∞d(xn,Tixn) exist for all i=1,2..N.
Proof.
Let p∈F.From the Lemma 4.6limn→∞d(xn,p) exist and {xn} is bounded sequence. so let limn→∞d(xn,p)=c. Since d(yn,p)≤d(xn,p) and d(un,i,p)≤d(yn,p),limsupn→∞d(yn,p)≤c and limsupn→∞d(un,i,p)≤c
and again from Lemma 4.6 similarly limsupn→∞d(zn,p)≤c and limsupn→∞d(vn,i,p)≤c
and limsupn→∞d(xn,p)≤c and limsupn→∞d(wn,i,p)≤c.Moreover we have
[TABLE]
implies limn→∞d(⨁i=1Nλn,iun,i,p)=c. We find that limn→∞d(un,i,un,j)=0 for all i,j=1,2,..,N by Lemma 4.5. Then
[TABLE]
and then liminfn→∞d(un,m,p)≥c for all m=1,2,..,N. Since limsupn→∞d(un,i,,p)≤c and d(un,i,p)≤d(yn,p) thus we have limn→∞d(un,i,p)=c and limn→∞d(yn,p)=c.
[TABLE]
which implies that limn→∞d(⨁i=1Nαn,iwn,i⊕⨁i=1Nβn,ivn,i,p)=c. Also we have limn→∞d(vn,i,vn,j)=0, limn→∞d(vn,i,wn,j)=0 and limn→∞d(wn,i,wn,j)=0
for all i,j=1,...,N by Lemma 4.7. Then
[TABLE]
and since limn→∞d(vn,i,wn,j)=0 and limn→∞d(wn,i,wn,j)=0 for all i,j=1,...,N then liminfn→∞d(vn,m,p)≥c for all m=1,2,..,N and
since limsupn→∞d(vn,i,,p)≤c and d(vn,i,p)≤d(zn,p) thus we have limn→∞d(vn,i,p)=c.and limn→∞d(zn,p)=c. Finally
[TABLE]
which implies that limn→∞[d(γn,0xn⊕⨁i=1Nγn,iwn,i,p)=c and since limsupn→∞d(xn,p)≤c and limsupn→∞d(wn,i,p)≤c we find that limn→∞d(xn,wn,i)=0 and limn→∞d(wn,i,wn,j)=0
for all i,j by Lemma 4.5..Hence d(xn,Tixn)≤d(xn,wn,i) for all i=1,2..N and limn→∞d(xn,Tixn)=0
∎
Theorem 4.8**.**
Let {Ti}i=1N be multivalued nonexpansive mappings from K to KC(X) with F=⋂i=1NF(Ti)=∅,Tip={p} for all p∈F . Then a sequence {xn} defined by (4.2) Δ-converges to p∈F where (p,p) is a common
solution of the problem (4.1).
Proof.
It follows from Lemma 4.6 and Lemma 4.7 that limn→∞d(xn,Tixn)=0 for all i∈{1,2..N},
limn→∞d(xn,p) exists for all p∈F. Let ωw(xn):=∪A({un}) where union take on all subsequence {un} of {xn}. To show that Δ-convergence of {xn}
it is enough to show that ωw(xn)⊆F and ωw(xn) contains single point. First of all ωw(xn)⊂K by Lemma 1.3. Let takeu∈ωw(xn), then
there exist subsequence {un} of {xn} such that A{un}={u}.By Lemma 1.3 and Lemma 1.4
there exist a subsequence (vn) of {un} which Δ−convergent to v. Let fix i∈{1,2...N},Since Tiv is compact,
then for each n≥1 we can pick up zn,i∈Tiv satisfies d(vn,zn,i)=d(vn,Tiv) and compactness of Tiv implies there
exist a convergent subsequence {znk,i} of {zn,i}. Let znk,i→wi∈Tiv. Since Ti is nonexpansive map
we have;
[TABLE]
Hence we have
[TABLE]
which implies
[TABLE]
Hence by uniqueness of asymptotic centers, we have wi=v∈Tiv.
Since i was arbitrary we have v∈F=i=1⋂NF(Ti)
so limn→∞d(xn,v) exist by Lemma 4.6 which implies u=v∈F by Lemma 1.4.Thus we have ωw(xn)⊆F. If we take subsequence {un} of {xn} with A{un}={u} and A{xn}={x} then, since u∈ωw(xn)⊆F and limn→∞d(xn,v) exist, we have u=x by Lemma 1.4.
∎
Theorem 4.9**.**
If K is also compact {Ti}i=1N are multivalued nonexpansive
mappings from K to C(X) with F=⋂i=1NF(Ti)=∅,Tip={p} for all p∈F then the sequence {xn}
defined by (4.2) strongly strongly converges to p∈F where (p,p)
is a common solution of the problem (4.1)
Proof.
By Lemma 4.6 and Lemma 4.7,we have that limn→∞d(xn,Tixn)=0 for all i∈{1,2..N},
limn→∞d(xn,p) exists for all p∈F Since K is
compact there is a convergent subsequence {xnk} of {xn},
say limi→∞xnk=q. Then for all i∈{1,2..N}, we have
[TABLE]
and taking limit on k,implies that q∈Tiq for all i∈{1,2..N}. Hence p∈F
∎
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