Some Fixed Point Theorems for Multi-Valued Weakly Uniform Increasing Operators.
ISHAKALTUN(*) - DURANTURKOGLU(**)
ABSTRACT- In the present work, we introduce the concept of weakly uniform in- creasing operators onanordered spaces. We also give some commonfixed point theorems for a pair of multivalued weakly uniform increasing operators in partially ordered metric space and in partially ordered Banach space.
1.Introduction.
The study on the existence of fixed points for single valued increasing operators is bounteous and successful, the results obtained are widely used to investigate the existence of solutions to the ordinary and partial dif- ferential equations (see [6], [7] and reference therein).
It is natural to extend this study to multi-valued case. Recently in [4], [7], [8], the authors have verified some fixed point theorems for multi-va- lued increasing operators.
Motivated by [3], [4], [9], [10] and [11] we prove some fixed point the- orems for multi-valued operators satisfying some weakly uniform in- creasing properties in metric space and Banach space, respectively.
2.Preliminaries
In this section some notations and preliminaries are given.
LetXbe a topological space andbe a partial order endowed onX.
(*) Indirizzo dell'A.: Department of Mathematics, Faculty of Science and Arts, Kirikkale University, 71450 Yahsihan, Kirikkale, Turkey.
E-mail: ialtun@kku.edu.tr, ishakaltun@yahoo.com
(**) Indirizzo dell'A.: Department of Mathematics, Faculty of Science and Arts, Gazi University, 06500 Teknikokullar, Ankara, Turkey.
E-mail: dturkoglu@gazi.edu.tr
In order to define the multi-valued weakly increasing and multi-valued weakly uniform increasing operators, we give the following relations be- tweentwo subsets ofX.
DEFINITION1 ([4]). Let A;B be two nonempty subsets of X, the rela- tions between A and B are defined as follows:
(1) If for every a2A, there exists b2B such that ab; then A1B.
(2) If for every b2B, there exists a2A such that ab; then A2B.
(3) If A1B and A2B, then AB.
REMARK1. 1 and2 are different relations between A and B. For example, let XR, A[12;1], B[0;1], be usual order on X, then A1B but A62B;if A[0;1], B[0;12], then A2B while A61B.
REMARK 2. 1, 2 and are reflexive and transitive, but are not antisymmetric. For instance, let XR, A[0;3], B[0;1][ 2;3],be usual order on X, then AB and BA, but A6B. Hence, they are not partial orders on2X.
We canshow that is a partial order on CC(R), which denotes the family of closed and convex subsets ofR.
Let us recall some basic definitions and facts from multivalued analysis (see details [2]).
Let 2X denote the family of all nonempty subset of X. A multivalued mapT:X!2X is called upper semi continuous (u.s.c.) on X if for each x02X, the setTx0is a nonempty, closed subset ofX, and if for each open setVofXcontainingTx0, there exists anopenneighborhoodUofx0such thatT(U)V.
Tis said to be completely continuous ifT(B) is relatively compact for every bounded subsetBX.
If the multivalued map T is completely continuous with nonempty compact values, then T is u.s.c. if and only if T has a closed graph (i.e.
un!u0;vn!v0;vn2Tunimplyv02Tu0).
DEFINITION 2 ([4]). If fxng X satisfies x1x2 xn or x1x2 xn , thenfxngis called a monotone sequence.
DEFINITION 3 ([4]). A multivalued operator T:X!2Xn; is called order closed if for monotone sequencesfung;fvng X;un !u0;vn!v0
and vn2Tunimply v02Tu0.
DEFINITION4 ([4]). A function f :X!Ris called order upper (lower) semi continuous, if, for monotone sequencefung X;and u02X
un!u0)lim sup
n!1 f(un)f(u0); (f(u0)lim inff(un)):
REMARK 3. An operator with closed graph must be an order closed operator and an upper (lower) semi continuous function is an order upper (lower) semi continuous. There exist some examples in[4],showing that the converse is not true.
Let X be a real Banach space with a normk k X:A non-empty closed subsetPofXis said to be anorder cone inXif
(i) PPP;
(ii) lPP, forl0 an d
(iii) P\ P fug, whereudenotes the zero element ofX.
Thenthe relationxy if and only if y x2P defines the partial ordering in X: P is called regular, if every nondecreasing and bounded above inorder sequence inX has a limit. i.e. if fxng X satisfies x1x2 xn y( y2X ), thenthere exists x2X such that
xn x
k kX!0 asn! 1. We do not require any property of the cones in the present discussion, however, the details of order cones and their properties may be found in Guo and Lakshmikantham [5].
Now we introduce the following definition.
DEFINITION 5. Let X be a topological space, be a partial order endowed on X and S;T:X!2X be two maps. The pair(S;T)is said to be weakly uniform increasing with respect to1if for any x2X we have Sx1Sy;Sx1Sz for all y2Sx;z2Tx and Tx1Ty;Tx1Tz for all y2Tx;z2Sx. Similarly the pair (S;T) is said to be weakly uniform increasing with respect to2if for any x2X we have Sy2Sx;Sz2Sx for all y2Sx;z2Tx and Ty2Tx;Tz2Tx for all y2Tx;z2Sx.
Now we give some examples.
EXAMPLE1. LetRdenote the real line with the usual order relation and let X0;1 R. Consider two mappings S;T:X!2X defined
by
Sx (0;1]; if x0 f1g; if x60 8<
: and Tx
n1 3;1
2
o; if x0 n2
3
o; if x60 8>
><
>>
:
for x20;1. Then the pair of mappings (S;T) is weakly uniform in- creasing with respect to1but not weakly uniform increasing with respect to2on0;1:Indeed, since
Sx (0;1]; if x0 f1g; if x60 (
1f1g
Sy; for all y2Sx
;
Sx (0;1]; if x0 f1g; if x60 (
1f1g
Sz; for all z2Tx;
and
Tx 1
3;1 2
; if x0
2 3
; if x60
8>
>>
<
>>
>: 1
2 3
Ty; for all y2Tx
;
Tx 1
3;1 2
; if x0
2 3
; if x60
8>
>>
<
>>
>: 1
2 3
Tz; for all z2Sx so(S;T)is weakly uniform increasing with respect to1. Note that the pair (S;T)is not weakly uniform increasing with respect to2since S12 f1g 62(0;1]S0for122S0.
EXAMPLE2. Let X[1;1)andbe usual order on X. Consider two mappings S;T:X!2X defined by Sx[1;x2] and Tx[1;2x] for all x2X. Then the pair of mappings (S;T) is weakly uniform increasing with respect to2but not1. Indeed, since
Sy[1;y2]2[1;x2]Sx for all y2Sx;
Sz[1;z2]2[1;x2]Sx for all z2Tx;
Ty[1;2y]2[1;2x]Tx for all y2Tx;
Tz[1;2z]2 [1;2x]Tx for all z2Sx;
then the pair(S;T)is weakly uniform increasing with respect to2. But S2[1;4]61f1g S1for12S2;so the pair(S;T)is not weakly uniform increasing with respect to1:
3.Fixed point theorems.
Inthis section, we prove some fixed point theorems for multivalued weakly uniform increasing operators with respect to1and2 inpartial order metric space and partial order Banach space.
We will use the following lemma in our theorems.
LEMMA1 ([1]). Let(X;d)be a metric space,W:X!R. Define the re- lationon X as follows:
xy()d(x;y)W(x) W(y):
Thenis a partial order on X and X is called a partial ordered metric space.
Now we give a fixed point theorem.
THEOREM 1. Let (X;d) be a complete metric space, W:X!Rbe a bounded below function, be a partial order induced by W and S;T:X!2Xn;be two order closed with respect toand weakly uniform increasing operator with respect to1. Then there exist z1;z22X such that z12Sz2and z22Tz1. Further if Sz21fz2gand Tz11fz1g, then S and T have a common fixed point (z1z2) in X.
PROOF. Let x0 2X be arbitrary point. SinceSx06 ;, we canchoose x12Sx0. Now since (S;T) is weakly uniform increasing with respect to1; we havex12Sx01Syfor ally2Sx0and sox12Sx01Sx1. Again, since (S;T) is weakly uniform increasing with respect to1we haveSx11Sz for all z2Tx1. Now we choose x22Tx1, then Sx11Sx2 and so x12Sx01Sx11Sx2. Now using the definition of 1, we canchoose x32Sx2 such that x1x3. Similarly, since (S;T) is weakly uniform in- creasing with respect to1, we havex22Tx11Tyfor ally2Tx1and so x22Tx11Tx2. Againsince (S;T) is weakly uniform increasing with re- spect to1, we haveTx21Tzfor allz2Sx2. Now sincex32Sx2, we have Tx21Tx3 and so x22Tx11Tx21Tx3. Now using the definition of
1we canchoosex42Tx3such thatx2x4. Continue this process, we will get a sequencefxng1n1which its subsequencesfx2n1g1n0andfx2n2g1n0 are increasing and satisfiesx2n12Sx2n;x2n22Tx2n1;n0;1;2; :::.
By the definitionwe have
W(x1)W(x3) W(x2n1) and
W(x2)W(x4) W(x2n2) :
SinceWis bounded from below, thenfW(x2n1)g1n0 andfW(x2n2)g1n0 are convergent. Thus fore>0, there existsN, ifn>m>Nthen
d(x2m1;x2n1)W(x2m1) W(x2n1)<e and
d(x2m2;x2n2)W(x2m2) W(x2n2)<e
which indicatesfx2n1g1n0andfx2n2g1n0are Cauchy sequence inX. Due to completeness of X, we know there exist z1;z22X, x2n1!z1 and x2n2!z2. Since S and T are order closed, fx2n1g1n0 and fx2n2g1n0 monotone and x2n12Sx2n;x2n22Tx2n1, we deduce that z12Sz2 and z22Tz1. Now ifSz21fz2g, thenz1z2 and ifTz11fz1g, thenz2z1 and soz1z2is a commonfixed point ofSandT. p The proof of the following theorem carries over in the same manner.
THEOREM 2. Let (X;d) be a complete metric space, W:X!R be a bounded above function, be a partial order induced by W and S;T:X!2Xn;be two order closed with respect toand weakly uniform increasing operator with respect to2. Then there exist z1;z2 2X such that z12Sz2and z22Tz1. Further iffz2g 2Sz2andfz1g 2Tz1, then S and T have a common fixed point (z1z2) in X.
PROOF. Letx02X be arbitrary point. SinceSx06 ;, we canchoose x12Sx0. Now since (S;T) is weakly uniform increasing with respect to2; we haveSx11Sx0 forx12Sx0. Again, since (S;T) is weakly uniform in- creasing with respect to 2 we have Sz2Sx1 for all z2Tx1. Now we choosex2 2Tx1, thenSx2 2Sx1and soSx22Sx12Sx0. Now using the definition of2, we canchoosex32Sx2such thatx3x1. Similarly, since (S;T) is weakly uniform increasing with respect to2, we haveTx21Tx1
x22Tx1. Againsince (S;T) is weakly uniform increasing with respect to2,
we haveTz2Tx2for allz2Sx2. Now sincex32Sx2, we haveTx32Tx2
and so Tx32Tx22Tx1. Now using the definition of2we canchoose x42Tx3 such thatx4x2. Continue this process, we will get a sequence fxng1n1which its subsequencesfx2n1g1n0andfx2n2g1n0are decreasing and satisfiesx2n12Sx2n;x2n22Tx2n1;n0;1;2; :::.
By the definitionwe have
W(x1)W(x3) W(x2n1) and
W(x2)W(x4) W(x2n2) :
SinceWis bounded from above, thenfW(x2n1)g1n0 andfW(x2n2)g1n0 are convergent. Thus fore>0, there existsN, ifn>m>N then
d(x2m1;x2n1)W(x2n1) W(x2m1)<e and
d(x2m2;x2n2)W(x2n2) W(x2m2)<e
which indicatesfx2n1g1n0andfx2n2g1n0are Cauchy sequence inX. Due to completeness of X, we know there exist z1;z22X, x2n1!z1 and x2n2!z2. Since S and T are order closed, fx2n1g1n0 and fx2n2g1n0 monotone and x2n12Sx2n;x2n22Tx2n1, we deduce that z12Sz2 and z22Tz1. Now iffz2g 2Sz2, then z2z1 and iffz1g 2Tz1, thenz1z2
and soz1z2is a commonfixed point ofSandT. p Now we give some fixed point theorem in ordered Banach space.
THEOREM3. Let X be a Banach space, PX be a regular cone,be a partial order induced by P and S;T:X!2Xn;be two order closed with respect toand weakly uniform increasing operator with respect to1. If the following condition hold:
(i) there exists some u2X such that Sx1fugand Tx1fugfor all x2X.
Then there exist z1;z22X such that z12Sz2 and z22Tz1. Further if Sz21fz2g and Tz11fz1g, then S and T have a common fixed point (z1z2) in X.
PROOF. We can construct a sequencefxng1n1 inX as inthe proof of Theorem 1. By (i), and the definition of 1, we have xnu;n1;2; :::.
SincePis regular,fx2n1g1n0andfx2n2g1n0are nondecreasing and boun- ded above inorder, there exist z1;z22X such that x2n1!z1 and x2n2!z2. Now the proof cancomplete as inthe proof of Theorem 1.u p
The following theorem can be similarly verified.
THEOREM4. Let X be a Banach space, PX be a regular cone,be a partial order induced by P and S;T:X!2Xn;be two order closed with respect toand weakly uniform increasing operator with respect to2. If the following condition hold:
(ii) there exists some u2X such thatfug 2Sx andfug 2Tx for all x2X.
Then there exist z1;z22X such that z12Sz2and z22Tz1. Further if fz2g 2Sz2 andfz1g 2Tz1, then S and T have a common fixed point (z1z2) in X.
REMARK4. By Theorems1, 2, 3 and4,we give some improved ver- sions of Theorems3.2, 3.3, 4.2and4.3of [4],respectively.
Now we introduce the following condition.
CONDITION1. Let X be a Banach space. Two maps S;T:X!2X are said to satisfy Condition1if for any countable subset A of X and for any fixed a2X the condition
A fag [S(A)[T(A)implies A is compact;
here T(A) [x2ATx.
Using the above condition, we can give the following theorems.
THEOREM 5. Let X be a Banach space, PX be a cone (not ne- cessary regular),be a partial order induced by P and S;T:X!2Xn;
be two order closed with respect to and weakly uniform increasing operator with respect to 1. Suppose that S and T are satisfy the Condition 1,then there exist z1;z22X such that z12Sz2 and z22Tz1. Further if Sz21fz2g and Tz11fz1g, then S and T have a common fixed point (z1z2) in X.
PROOF. We can construct a sequencefxng1n1 inXas inthe proof of Theorem 1.
Now letA fx0;x1; :::g. SinceAis countable, A fx0g [ fx1;x3; :::g [ fx2;x4; :::g
fx0g [S(A)[T(A);
andSandTare satisfy the Condition 1, thenAis compact. Thusfx2n1g1n0 and fx2n2g1n0 have convergent subsequences which converges to say z1;z22X, respectively. However fx2n1g1n0 andfx2n2g1n0 are increas- ing, sox2n1!z1andx2n2!z2forn! 1. Now the proof cancomplete
as inthe proof of Theorem 1. p
The following theorem can be similarly verified.
THEOREM6. Let X be a Banach space, PX be a cone (not necessary regular),be a partial order induced by P and S;T:X!2Xn;be two order closed with respect to and weakly uniform increasing operator with respect to2. Suppose that S and T are satisfy the Condition1,then there exist z1;z22X such that z12Sz2and z22Tz1. Further iffz2g 2Sz2 andfz1g 2Tz1, then S and T have a common fixed point (z1z2) in X.
REMARK5. By Theorems5and6,we give some improved versions of Theorem3.1 of[3]and Theorem4of [12].
Acknowledgments. The authors are grateful to the referees for their valuable comments in modifying the first version of this paper.
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Manoscritto pervenuto in redazione il 21 maggio 2007.