Volume 2010, Article ID 321594,12pages doi:10.1155/2010/321594
Research Article
A Continuation Method for Weakly Kannan Maps
David Ariza-Ruiz and Antonio Jim ´enez-Melado
Departamento de An´alisis Matem´atico, Facultad de Ciencias, Universidad de M´alaga, 29071 M´alaga, Spain
Correspondence should be addressed to Antonio Jim´enez-Melado,[email protected]
Received 25 September 2009; Revised 4 December 2009; Accepted 6 December 2009
Academic Editor: Mohamed A. Khamsi
Copyrightq2010 D. Ariza-Ruiz and A. Jim´enez-Melado. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The first continuation method for contractive maps in the setting of a metric space was given by Granas. Later, Frigon extended Granas theorem to the class of weakly contractive maps, and recently Agarwal and O’Regan have given the corresponding result for a certain type of quasicontractions which includes maps of Kannan type. In this note we introduce the concept of weakly Kannan maps and give a fixed point theorem, and then a continuation method, for this class of maps.
1. Introduction
Suppose thatX, dis a metric space and that f : D ⊂ X → X is a map. We say thatf
is contractive if there exists α ∈ 0,1such that dfx, fy ≤ αdx, yfor all x, y ∈ D. The well-known Banach fixed point theorem states that f has a fixed point if D X and X, dis complete. In 1962, Rakotch1obtained an extension of Banach theorem replacing the constant αby a function of dx, y,α αdx, y, provided that α is nonincreasing and 0 ≤ αt < 1 for all t > 0 for a recent refinement of this result see 2. A similar generalization of the contractive condition was considered by Dugundji and Granas3, who extended Banach theorem to the class of weakly contractive mappingsi.e.,ααx, y, with sup{αx, y:a≤dx, y≤b}<1 for all 0< a≤b.
Another focus of attention in Fixed Point Theory is to establish fixed point theorems for non-self mappings. In the setting of a Banach space, Gatica and Kirk4proved that if
f : U → X is contractive, withUan open neighborhood of the origin, then f has a fixed point if it satisfies the well-known Leray-Schauder condition:
fx/λx, forx∈∂U, λ >1. L-S
Granas6, which is known as continuation method for contractive maps. In fact, the jump from a Banach space setting to the metric space setting was given by Granas himself in6
for more information on this topic see, for instance,7–9. After Granas, Frigon8gave a similar result for weakly contractive maps.
A variant of the Banach contraction principle was given by Kannan10, who proved that a mapf:X → X, whereX, dis a complete metric space, has a unique fixed point iff
is what we call a Kannan map, that is, there existsα∈0,1such that, for allx, y∈X,
dfx, fy≤ α
2
dx, fxdy, fy. 1.1
In this note, following the pattern of Dugundji and Granas3, we extend Kannan theorem to the class of weakly Kannan mapsi.e.,ααx, y, with sup{αx, y:a≤dx, y≤ b}< 1 for all 0 < a≤b. This is done inSection 2. InSection 3we use a local version of the previous result to obtain a continuation method for weakly Kannan maps.
2. Weakly Kannan Maps
In this section we follow the pattern of Dugundji and Granas3to introduce the concept of weakly Kannan maps.
Definition 2.1. LetX, dbe a metric space,D ⊂X, andf :D → X. Thereforef is a weakly Kannan map if there existsα: D×D → 0,1, withθa, b: sup{αx, y : a≤ dx, y ≤ b}<1 for every 0< a≤bsuch that, for allx, y∈D,
dfx, fy≤ α
x, y
2
dx, fxdy, fy. 2.1
Remark 2.2. Clearly, any weakly Kannan mapfhas at most one fixed point: ifxfxand
yfy, then
dx, ydfx, fy≤ 1
2
dx, fxdy, fy0. 2.2
Remark 2.3. Notice that iff:D⊂X → Xis a weakly Kannan map and we defineαfx, yon D×Das
αf
x, y
⎧ ⎪ ⎨ ⎪ ⎩
2dfx, fy
dx, fxdy, fy ifd
x, fxdy, fy/0,
0 otherwise,
2.3
thenαf is well defined, takes values in0,1, satisfies sup{αfx, y : a≤ dx, y ≤ b} < 1
for all 0< a≤bforαf is smaller than anyαassociated tof, and also satisfies2.1, withα
replaced byαf, for allx, y∈D. Conversely, ifαf is defined as in2.3and satisfies the above
Remark 2.4. Although Kannan showed that the concept of Kannan map is independent of the concept of contractive map, Janos11observed that any contractive mapf :D ⊂ X → X
whose Lipschitz constant defined by
Lfsup
dfx, fy
dx, y :x, y∈X, x /y
2.4
is less than 1/3 is a Kannan map. Next, we exhibit an example of a weakly Kannan map
f, withLf 1/3, which is not a Kannan map, thus showing that the constant 1/3 in the aforementioned result by Janos is sharp.
Example 2.5. Consider the metric spaceX 0,∞with the usual metricdx, y |x−y|, and letf :X → Xbe the function defined asfx 1/3log1ex. Then,Lf 1/3 andfis
a weakly Kannan map, but not a Kannan map.
The equality Lf 1/3 follows from the fact that|fx| < 1/3 for all x ∈ 0,∞
together with
lim
x→ ∞
dfx, f0
dx,0 1
3. 2.5
We also have thatfis not a Kannan map because
lim
x→ ∞
2dfx, f0
dx, fxd0, f0 1. 2.6
To check thatf is a weakly Kannan map, consider the functionα: X×X → 0,∞
given by2.3. This function is well defined and also takes values in0,1sinceLf 1/3. Next, assume that 0< a≤band let us see thatθa, b sup{αx, y: a≤dx, y≤b}<1. To see this, observe that|u−fu| → ∞asu → ∞, so there isM >0 such that|u−fu|> bfor allu > M. Observe also thatfM, the restriction offto0, M, is a Kannan map with constant αM∈0,1, due to the fact thatLfM<1/3, forfMis continuously differentiable on0, M
and|fu| < 1/3 for allu ∈ 0, M. We will seeθa, b ≤ max{2/3, αM}. To do it, suppose
thatx, y ∈0,∞witha≤ |x−y| ≤band 0≤x < y. Then, ify > M, use|y−fy|> band thatLf≤1/3 to obtainαx, y≤2/3. Otherwise, we would have 0≤x < y≤Mand then
αx, y≤αM.
Although the way we have introduced the concept of weakly Kannan map has been by analogy with the work done by Dugundji and Granas in3, we would like to mention that this extension may be done in some different ways. For instance, Pathak et al. 12, Theorem 3.1have proved the following result.
Theorem A. LetX, dbe a complete metric space and suppose thatf :X → Xis a map such that
dfx, fy≤α1dx, fxdx, fxα2dy, fydy, fy, 2.7
Observe that relation2.7can be written in the following more general form:
dfx, fy≤A1xdx, fxA2ydy, fy, 2.8
for allx, y ∈X, whereAi :X → 0,1,i1,2, and notice that any map satisfying2.8also
satisfies the relation2.1withαx, y 2 max{A1x, A2y}. In fact, the arguments used by the authors in the proof of Theorem A are also valid for this class of maps. Next, we state this slightly more general result and include the proof for the sake of completeness. Then, we obtain, as a consequence, a fixed point theorem for weakly Kannan maps.
Theorem 2.6. LetX, dbe a complete metric space and assume thatA : X ×X → 0,∞is a bounded function satisfying the following condition: for any sequence{xn}inXandu∈X,
xn−→u⇒lim supAxn, u<1. ∗
Assume also thatf :X → Xis a map such that
dfx, fy≤Ax, ydx, fxdy, fy, 2.9
for allx, y ∈X. If there exists a sequence{xn}inXwithdxn, fxn → 0, thenf has a unique
fixed pointuinX, andxn → u.
Proof. SinceA is bounded, there exists M > 0 such that |Ax, y| ≤ M for all x, y ∈ X. Suppose that{xn}is a sequence inXwithdxn, fxn → 0 and use2.9to obtain that, for
alln, m∈N,
dfxn, fxm
≤Mdxn, fxn
dxm, fxm
. 2.10
This implies that{fxn}is a Cauchy sequence. SinceX, dis complete, the sequence {fxn}is convergent, say tou∈X. Thenxn → ubecausedxn, fxn → 0. Thus, by∗,
lim supAxn, u<1.
That u fu is a consequence of the following relation and the fact that lim supAxn, u<1, then
du, fulimdfxn, fu
≤lim supAxn, u
dxn, fxn
du, fu
du, fulim supAxn, u.
2.11
Finally,uis the unique fixed point offbecause ifzfz:
du, z dfu, fz≤Au, zdu, fudz, fz0. 2.12
Corollary 2.7. Let X, dbe a complete metric space and suppose that f : X → X is a weakly Kannan map. Then,fhas a unique fixed pointu∈ Xand, for anyx0 ∈ X, the sequence of iterates
Proof. Sincef is a weakly Kannan map, there exists a function α : X×X → 0,1with
θa, b:sup{αx, y:a≤dx, y≤b}<1 for all 0< a≤b, satisfying2.1for allx, y∈X. Hence, the functionA :X ×X → 0,1/2given asAx, y 1/2αx, yis bounded and satisfies the conditions∗and2.9.
Consider anyx0 ∈ X and define xn fxn−1, n 1,2, . . . . We may assume that dx0, x1>0 because otherwise we have finished. We will prove thatdxn, fxn → 0 and
hence, byTheorem 2.6,{xn}will converge to a pointuwhich is the unique fixed point off.
First of all, observe that the inequality
dxn1, xn≤αxn, xn−1dxn, xn−1 2.13
holds for alln≥1. In fact, it is a consequence of the following one, which is true by2.1:
dxn1, xn≤ αxn, xn−1
2 dxn1, xn dxn, xn−1. 2.14
From 2.13 we obtain that the sequence {dxn, xn−1} is nonincreasing, for 0 ≤ αxn, xn−1≤1, and then it is convergent to the real number
dinf{dxn, xn−1:n1,2, . . .}. 2.15
To prove thatd0, suppose thatd >0 and arrive to a contradiction as follows: use
0< d≤dxn, xn−1≤dx1, x0 2.16
and the definition of θ θd, dx1, x0to obtain αxn, xn−1 ≤ θ for all n 1,2, . . . .This,
together with2.13, gives that
d≤dxn1, xn≤θndx1, x0, 2.17
for alln1,2, . . ., which is impossible sinced >0 and 0≤θ <1.
Remark 2.8. We do not know whether Theorem A is, or not, a particular case ofTheorem 2.6, although that is the case if the functionsα1, α2satisfy the additional assumption sup{α1t α2t : t ≥ 0} < 2. To see this, suppose that the mapf : X → X is in the conditions of Theorem A, that is,fsatisfies relation2.7for some given functionsαi:R → 0,1,i1,2,
and suppose also that the functionsα1, α2 satisfy in addition sup{α1t α2t: t≥ 0} <2. DefineA:X×X → 0,∞asAx, y max{ax, ay}, wherea:X → 0,1is given by
az 1
2
Let us see that, with this functionA,f satisfies the hypotheses ofTheorem 2.6. Indeed,Ais clearly bounded and also satisfies∗; if{xn}is a sequence inX andu ∈ X, withxn → u,
then
sup{axn:n1,2, . . .} ≤sup
α1t α2t
2 :t≥0
<1. 2.19
Since we also have thatau<1, we obtain that sup{Axn, u:n1,2, . . .}<1.
Finally, to see thatf satisfies relation2.9, use relation2.7withx, y ∈ X, together with the same relation interchanging the roles ofxandy, and the fact thatdfx, fy dfy, fx, to obtain that
dfx, fy≤axdx, fxaydy, fy, 2.20
from which the result follows.
To prove the homotopy result of the next section, we will need the following local version ofCorollary 2.7.
Corollary 2.9. Assume thatX, dis a complete metric space,x0∈X, r >0, andf:Bx0, r → X
is a weakly Kannan map with associated functionαsatisfying2.1. Ifθis defined as usual, and
dx0, fx0< 1
3min r
2, r
1−θr
2, r
, 2.21
thenfhas a fixed point.
Proof. In view of Corollary 2.7, it suffices to show that the closed ballBx0, ris invariant underf. To prove it, consider anyx∈Bx0, rand obtain the relation
dx0, fx≤dx0, fx0dfx0, fx
≤dx0, fx0αx0, x
2
dx0, fx0dx, fx
≤dx0, fx0αx0, x
2
dx0, fx0dx, x0 dx0, fx,
2.22
from which, having in mind thatαx0, x≤1,
dx0, fx≤3dx0, fx0αx0, xdx0, x. 2.23
which
dx0, fx≤r1−θr
2, r
rθr
2, r
r. 2.24
3. A Homotopy Result
In 1974 ´Ciri´c13introduced the concept of quasicontractions and proved the following fixed point theorem: suppose that X, d is a complete metric space and that f : X → X is a quasicontraction, that is, there existsq∈0,1such that, for allx, y∈X,
dfx, fy≤qmaxdx, y, dx, fx, dy, fy, dx, fy, dy, fx. 3.1
Then,fhas a fixed point inX.
Observe that any contractive map, as well as any Kannan map, is a quasicontraction; thus, the theorem by ´Ciri´c generalizes the well known fixed point theorems by Banach and Kannan.
On the other hand, Agarwal and O’Regan 14considered a certain class of quasi-contractions: those mapsf :X → X, whereX, dis a metric space, for which there exists
q∈0,1such that, for allx, y∈X,
dfx, fy
≤qmax
dx, y, dx, fx, dy, fy,1
2
dx, fydy, fx,
Q
and gave the following homotopy result.
Theorem B. LetX, dbe a complete metric space,Uan open subset ofX, andH:U×0,1 → X
satisfying the following properties:
iHx, λ/xfor allx∈∂Uand allλ∈0,1,
iithere existsq∈0,1such that for allx, y∈Uandλ∈0,1we have
dHx, λ, Hy, λ
≤qmax
dx, y, dx, Hx, λ, dy, Hy, λ,1
2
dx, Hy, λ, dy, Hx, λ,
3.2
iiiHx, λis continuous inλ, uniformly forx∈U.
IfH·,0has a fixed point inU, thenH·, λalso has a fixed point inUfor allλ∈0,1.
Theorem 3.1. LetX, dbe a complete metric space,Uan open subset ofX, andH:U×0,1 → X
satisfying the following properties:
P1Hx, λ/xfor allx∈∂Uand allλ∈0,1,
P2there existsα:U×U → 0,1such that for allx, y∈Uandλ∈0,1one has
dHx, λ, Hy, λ≤ α
x, y
2
dx, Hx, λ dy, Hy, λ, 3.3
andθa, b sup{αx, y:a≤dx, y≤b}<1for all0< a≤b,
P3there exists a continuous functionφ : 0,1 → Rsuch that, for everyx ∈Uandt, s ∈
0,1,dHx, t, Hx, s≤ |φt−φs|.
IfH·,0has a fixed point inU, thenH·, λalso has a fixed point inUfor allλ∈0,1.
Proof. Consider the nonempty set
A{λ∈0,1:Hx, λ xfor somex∈U}. 3.4
We will prove thatA 0,1, and for this it suffices to show thatAis both closed and open in0,1.
We start showing that A is closed in 0,1: suppose that {λn} is a sequence in A
converging toλ∈0,1and let us show thatλ∈A. By definition ofA, there exists a sequence
{xn}inUwithxn Hxn, λn. We will prove that {xn} converges to a pointx0 ∈ Uwith Hx0, λ x0, thus showing thatλ∈A.
That{xn}is a Cauchy sequence is a consequence of the following relation, where we
have usedP2,P3, and the fact thatxmHxm, λm:
dxn, xm dHxn, λn, Hxm, λm
≤dHxn, λn, Hxn, λm dHxn, λm, Hxm, λm
≤φλn−φλmαxn, xm
2 dxn, Hxn, λm dxm, Hxm, λm
φλn−φλmαxn2, xmdHxn, λn, Hxn, λm
≤ 3
2φλn−φλm.
Writex0limxnand let us see thatx0∈Uand also thatx0Hx0, λ. Thatx0Hx0, λis
a consequence of the following relation:
dx0, Hx0, λ≤dx0, xn dxn, Hx0, λ
≤dx0, xn dHxn, λn, Hxn, λ dHxn, λ, Hx0, λ
≤dx0, xn φλn−φλ
1
2dxn, Hxn, λ dx0, Hx0, λ
≤dx0, xn
3
2φλn−φλ 1
2dx0, Hx0, λ,
3.6
and thatx0 ∈Uis straightforward fromP1.
Next we prove that A is open in 0,1: suppose that λ0 ∈ A and let us show that λ0−δ, λ0δ∩0,1⊂A, for someδ >0. Sinceλ0∈A, there existsx0∈Uwithx0Hx0, λ0. Considerr >0 withBx0, r⊂Uand use the continuity ofφto obtainδ >0 such that
φλ−φλ0<minr 2, r
1−θr
2, r
, 3.7
for allλ∈λ0−δ, λ0δ∩0,1.
To show now that anyλ ∈λ0−δ, λ0δ∩0,1is also inA, it suffices to prove that the mapH·, λ:Bx0, r → Xhas a fixed point. And this is true byCorollary 2.9, since
dx0, Hx0, λ dHx0, λ0, Hx0, λ
≤φλ0−φλ
<minr 2, r
1−θr
2, r
.
3.8
Remark 3.2. A careful reading of the proof shows that hypothesisP3inTheorem 3.1can be easily replaced by the weaker hypothesisiiiin Theorem B.
Remark 3.3. The counterpart to Theorem 3.1 for weakly contractive maps was proved by Frigon8. In that result, it was assumed, in place of our3.3, an equivalent formulation of the following conditionH’:
dHx, λ, Hy, λ≤αx, ydx, y. H’
Observe that conditionH’means that all the mapsH·, λ: U → X,λ ∈ 0,1are weakly contractive, and with the same functionα. Our condition3.3is no surprise then. It also means that all the mapsH·, λare of weakly Kannan type, and with the same function
α.
of Kannan type since any Kannan map satisfies Q. Moreover, f will not be of weakly contractive type.
Example 3.4. Consider the metric spaceX, d, whereX −1,1anddx, y |x−y|, and let
f:−1,1 → −1,1be the map given as
fx
⎧ ⎨ ⎩
−sinx, −1≤x <1,
0, x1.
3.9
First of all, we will see that the mapfdoes not satisfy conditionQ. Define, forx, y∈
−1,1,
βx, ymax
dx, y, dx, fx, dy, fy,1
2
dx, fydy, fx. 3.10
Then, forx∈0,1, we have thatβx,−x 2|x|, since|sinx| ≤ |x|. Hence,
lim
x→0
dfx, f−x βx,−x xlim→0
|sinx|
|x| 1, 3.11
showing that noq∈0,1can be found to satisfyQ.
Secondly, observe thatfis not weakly contractive, since any weakly contractive map is continuous.
Next, let us check thatf is a weakly Kannan map. Sincef has 0 as unique fixed point then, the functionα:−1,1×−1,1 → 0,∞given byαx, y 2dfx, fy/dx, fx dy, fyifx, y/ 0,0,α0,0 0, is well defined. We have to check thatαonly takes values in0,1 and thatθa, b sup{αx, y : a ≤ |x−y| ≤ b, x, y ∈ −1,1} < 1 for all 0< a≤b. In fact, all this will follow if we just show that, for 0< a≤2,
θa,2≤max
2 3,1−
a
8
1−cosa 4
. 3.12
Thus, take 0< a≤2 and assume thatx, y∈−1,1, witha≤ |x−y|. If any of the pointsx, y
equals 1, for exampley 1, then use|xsinx||x||sinx|and|sinx| ≤ |x|to obtain that
αx, y |x2|sinx|
sinx|1
2|sinx| |x||sinx|1 ≤
2
3. 3.13
Otherwise, we would have thatx, y ∈ −1,1. In this case, since|x−y| ≥ a, then we may assume additionally that|x| ≥a/2, and we claim that
αx, y≤1−a 8
1−cosa 4
To be convinced of this, check the following chain of inequalities having in mind that
|zsinz| |z||sinz|for allz ∈ −1,1, that|sinz| ≤ |z|, and also that|cosx/2| ≤ cosa/4:
αx, y 2sinx−sin
y |xsinx|ysiny
≤ 2|sinx|2sin
y |x||sinx|ysiny
1− |x| − |sinx|y−sin
y |x||sinx|ysiny
≤1− |x| − |sinx| 4
≤1− 1 4
|x| −2sinx 2
cosx 2
≤1− |x| 4
1−cosx 2
≤1− a 8
1−cosa 4
.
3.15
Next, defineH : −1,1×0,1 → −1,1byHx, λ λfxand let us see thatH
satisfiesP1,P2, andP3.
It is obvious thatHsatisfiesP1. To checkP2, observe that
x−λfx|x|λfx, 3.16
for allλ ∈ 0,1and allx∈ −1,1, and hence, ifαx, yis the function previously defined, we have that, for allλ∈0,1and allx, y∈−1,1,
dHx, λ, Hy, λλfx−fy
≤λα
x, y
2 x−fxy−f y λα x, y 2
|x|fxyfy
≤ α
x, y
2
|x|λfxyλfy
α
x, y
2 x−λfxy−λf y α x, y 2
dx, Hx, λ dy, Hy, λ.
3.17
Acknowledgments
This research was partially supported by the Spanish Grant no. MTM2007-60854 and regional AndalusianGrants no. FQM210 and no. FQM1504Governments.
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