doi:10.5899/2012/jnaa-00168 Research Article
Quasi Contraction and Fixed Points
Mehdi Roohi1∗, Mohsen Alimohammady2
(1)Department of Mathematics, Faculty of Sciences, Golestan University, P.O.Box. 155, Gorgan, Iran. (2)Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran.
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Copyright 2012 c⃝Mehdi Roohi and Mohsen Alimohammady. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and re-production in any medium, provided the original work is properly cited.
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Abstract
In this note, we establish and improve some results on fixed point theory in topological vector spaces. As a generalization of contraction maps, the concept of quasi contraction multivalued maps on a topological vector space will be defined. Further, it is shown that a quasi contraction and closed multivalued map on a topological vector space has a unique fixed point if it is bounded value.
Keywords: : Topological vector space, Quasi contraction, Fixed point, Multi-valued map.
1
Introduction
Fixed point theory has many applications in almost all branches of mathematics. The Banach contraction principles occupies a central position in fixed point theory. Banach’s original theorem is expressed in metric spaces and some authors have extended this re-sult in some other versions [1], [4], [6] and [7]. Kirk [5] and Edelstien [3] studied and achieved some basic results in fixed point theory. Here, we would improve their results for multivalued maps in topological vector spaces.
For two setsX andY and each element xof X we associate a nonempty subset F(x) ofY and this correspondencex7→F(x) is called amultivalued mapping or amultifunction fromXintoY; i.e.,F is a function fromXtoP∗(Y), whereP∗(Y) is the set of all nonempty
subsets of Y. Thelower inverse of a multivalued mappingF is the multi-valued mapping
Fl of Y intoX defined by
Fl(y) ={x∈X:y∈F(x)},
also for any nonempty subsetB of Y we have,
Fl(B) ={x∈X:F(x)∩B ̸=∅},
finally it is understood thatFl(∅) =∅. The set {x∈X:F(x) ⊆B} is the upper inverse
of B and is denoted byFu(B). F is lower semicontinuous (upper semicontinuous), if for
every open setU ⊆Y,Fl(U)(Fu(U)) is open in X. It is well known [2] that F is upper
semi-continuous atx0 if for each open setV containingF(x0) there exists a neighborhood
U of x0 such thatx∈U implies thatF(x)⊆V.
For two multivalued maps F : X → P∗(Y) and G : Y → P∗(Z) the composition
GoF :X → P∗(Z) is defined by
GoF(x) = ∪
y∈F(x)
G(y).
2
Main Results
The following definition of quasi contraction is an extended definition of contraction in metric spaces and we achieve some results which they extend some results in invariant metric spaces.
Definition 2.1. Suppose (X, τ) is a topological vector space. A multi-valued map F :
X→ P∗(X) is said to be :
(a) quasi contraction map if for all x, y∈X and any open neighborhoodU of 0 there is a constant 0≤c <1 such that x−y∈U implies that F(x)−F(y)⊆cU.
(b) closed map if xn→x, yn→y and yn∈F(xn) imply that y ∈F(x).
(c) bounded valued map if F(x) is a bounded set in X for all x∈X.
(d) upper semi-continuous at x0 if for each open set V containing F(x0) there exists
a neighborhoodU(x0) such that x∈U(x0) implies that F(x)⊆V.
It is well known that any contraction multivalued map between metric spaces is con-tinuous. Extending this fact is our next aim.
Theorem 2.1. Suppose (X, τ) is a topological vector space andF :X→ P∗(X) is quasi
contraction. If0∈F(0) then F is upper semicontinuous at 0.
Proof. SupposeV is any open neighborhood ofF(0). Consider open neighborhoodU
of 0 for whichcU ⊆V. ThenU =U− {0} ⊆U. Therefore, F(U)⊆F(U)−F(0)⊆cU ⊆
V.
Corollary 2.1. Suppose (X, τ) is a topological vector space andF :X→ P∗(X) is quasi
contraction. Then F is upper semicontinuous.
Proof. Consider x0 ∈ X, y0 ∈ F(x0) and V an open neighborhood of F(x0). Then
V −y0 =U is an open neighborhood of 0. DefineG(x) =F(x+x0)−y0. We claim that
g is contraction, too. To see this, ifW is an open neighborhood of 0 and x−y∈W, then
G(x)−G(y) = (F(x+x0)−y0)−(F(y+x0)−y0)⊆cW.
Also, G(0) = F(x0)−y0 contains 0, so from Theorem 2.1 for V −y0, there is open
Lemma 2.1. Suppose (X, τ) is a locally convex space and x ∈X. If F :X → P∗(X) is
a contraction bounded valued map, then any sequence {yn} is a Cauchy sequence, where
yn∈Fn(x) for alln∈N.
Proof. LetU be an open convex neighborhood of 0 which is also balanced. From the assumptionF(x) and soF(x)−xare bounded sets. SinceU is absorbent, there isα0>0
such that
F(x)−x∈αU, for all α with|α| ≥α0,
so F(x)−x∈ α0U. Then Fn+1(x)−Fn(x)⊆cn(α0)U. Since 0 ≤c <1, there is N ∈N
such that (cm+· · ·+cn)α
0 <1 for all m, n≥N. Therefore, ifm > n≥N we have
Fm+1(x)−Fn+1(x) ⊆ cmα0U+· · ·+cnα0U
= (cm+· · ·+cn)α0U
⊆ U.
This completes the proof.
The following result is other version of Banach contraction Theorem. First we need to the next lemma.
Theorem 2.2. Suppose (X, τ) is a sequentially complete locally convex space and F :
X → P∗(X) is a quasi contraction bounded valued map. If F is closed multi-valued map,
thenF has a unique fixed point.
Proof. SupposeU is any open neighborhood of 0 andx is any element inX. Make a sequence {yn} in Y by induction, wherey1 ∈F(x) and yn+1 ∈F(yn) for all n∈N.
Ap-plying Lemma 2.1,{yn} is a Cauchy sequence. SinceX is sequentially complete, so{yn}
converges to an elementy∈X. Thaty∈F(y) follows from closedness ofF,yn∈F(yn−1)
and yn−1 → y. For the uniqueness, suppose x, y ∈ X are two distinct fixed points of
F. Then there is a convex open neighborhood U of 0 such that x−y /∈ U. Since U is absorbent so there is α0 >0 such that x−y∈α0U. Therefore, F(x)−F(y)⊆cα0U and
so x−y∈cnα
0U for eachn∈N. Since 0≤c <1, we can assume that cnα0<1 for some
n∈N. On the other hand,U is convex socnα0U ⊆U. Consequently,x−y∈U which is a contradiction.
Remark 2.1. It should be noticed that closedness in Theorem 2.2 could be reduced to the following condition
yn−→y and yn∈F(yn−1) =⇒ y∈F(y).
As a special case of quasi contraction multi-valued maps, we introduce the quasi con-traction maps.
Corollary 2.2. Suppose(X, τ)is a sequentially complete locally convex space, also suppose
f :X →X is quasi contraction. Then f has a unique fixed point.
Proof. It is a direct result of Theorem 2.2.
Theorem 2.3. Let(X, τ)be a locally convex space andf :X→X be a quasi contraction map. If for some xo ∈ X there exists a convergence subsequence fni(x0) to an element
u∈X, then u is a fixed point for f.
Proof. F is quasi contractive, so (fn(x
0))n is a Cauchy sequence from the proof of
Lemma 2.1. Hence from the assumption fn(x
0) −→u. From Theorem 2.1, f is
continu-ous, so f(u) =f(limfn(x
0)) = limfn+1(x0) =u.
Definition 2.3. A family{Aj :j∈J}of sets in X has finite intersection property if each
finite subfamily of it, has nonempty intersection.
For a multi-valued map F : X → P∗(X), set O(Fn(x)) =
∪
m≥nFm(x), where it is
understood that F0(x) = {x}. The following result would improve a result of Ciric [4]. First we need to the following lemma.
Lemma 2.2. [1] SupposeF :X→ P∗(X)is a multi-valued map and there isx0 ∈X such
that O(x0) has finite intersection property. Then F has a fixed point if O(F2(x))⊆F(x)
for all x∈X.
Proof. It is easy to see that F(O(x0)) ⊆ O(x0). Set K = {A ⊆ O(x0) : A ̸=
∅, F(A)⊆A}. Then partially ordered K by inclusion. Since O(x0) has finite intersection
property, so from Zorns lemma K has minimal element, say C. Then F(C) ⊆ C, but
F(F(C))⊆F(C) implies thatF(C) =C. Now, ifu /∈F(u) for each u∈C, then from as-sumption u /∈O(F2(u)). Since u∈C, soF(u)⊆F(C) =C, therefore Fk(u)⊆C for any
nonnegative integer k. Now O(F2(u)) = C follows from minimality of C. Consequently,
u∈O(F2(u)) which is a contradiction.
Theorem 2.4. Suppose X is a topological vector space, O(x) has the finite intersection property for each x∈X and F :X → P∗(X) is a multi-valued map. Then F has a fixed
point if x /∈F(x) implies that x /∈O(Fm(x)) for allm≥2.
Proof. Assume x /∈ F(x). We claim that x /∈ O(F2(x)). On the contrary there are two cases :
(a) x= limiyni whereni ≥2, yni ∈Fni(x)
(b) x∈Fm(x) for somem≥2.
If (a) satisfies, then x ∈ (O(F m(x)))′ ⊆ O(Fm(x)) hence, from the assumption
x ∈ F(x), which is a contradiction. Assume (b) satisfies, then x ∈ O(Fm(x)) which
is impossible again. Therefore, x /∈ O(F2(x)) and so O(F2(x)) ⊆ F(x). That F has a
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