Department of Nonlinear Analysis,Faculty of Mathematics and Computer Science, University of Łódź, Banacha 22, 90-238 Łódź, Poland Abstract four kinds of contractions of Kannan type and,
Trang 1Department of Nonlinear Analysis,
Faculty of Mathematics and
Computer Science, University of
Łódź, Banacha 22, 90-238 Łódź,
Poland
Abstract
four kinds of contractions of Kannan type and, by techniques based on thesegeneralized pseudodistances, we prove fixed point theorems for such contractions.The results are new in uniform and locally convex spaces and even in metric spaces.Examples are given
MSC: 47H10; 47H09; 54E15; 46A03; 54E35
Keywords: uniform space, metric space, J-family of generalized pseudodistances;contractions of Kannan type, fixed point; iterative approximation
1 Introduction
, i.e ∀m∈{0}∪ {v m = T [m] (v0)}.Recall that maps satisfying the conditions (B) and (K) that are presented in Theo-rems 1.1 and 1.2 below are called in literature Banach contractions and Kannan con-tractions, respectively, and first arose in works [1,2] and [3,4], respectively
(B)∃λ∈[0,1)∀x,y ∈X {d(T(x), T(y)) ≤ λd(x, y)},
[0,1/2)∀x, y ÎX{d(T(x), T(y))≤ h [d(T(x), x) + d(T(y), y)]},
A great number of applications and extensions of these results have appeared in theliterature and plays an important role in nonlinear analysis The different line ofresearch focuses on the study of the following interesting aspects of fixed point theory
in metric spaces and has intensified in the past few decades: (a) the existence anduniqueness of fixed points of various generalizations of Banach and Kannan contrac-tions; (b) the similarity between Banach and Kannan contractions; and (c) the interplay
© 2011 W łodarczyk and Plebaniak; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and
Trang 2between metric completeness and the existence of fixed points of Banach and Kannan
contractions These aspects have been successfully studied in various papers; see, for
example, [5-21] and references therein
It is interesting that Theorem 1.2 is independent of Theorem 1.1 that every Banachcontraction and every Kannan contraction on a complete metric space has a unique
fixed point and that in Theorem 1.3 the completeness of the metric space is omitted
Clearly, Banach contractions are always continuous but Kannan contractions are not
necessarily continuous Next, it is worth noticing that Theorem 1.2 is not an extension
of Theorem 1.1 In [5], it is constructed an example of noncomplete metric space X
Theorem 1.1 does not characterize metric completeness In [6], it is proved that a
fixed point which implies that Theorem 1.2 characterizes the metric completeness
Similarity between Banach and Kannan contractions may be seen in [7,8] In complete
appli-cations in fixed point theory and among others generalizations of Banach and Kannan
contractions are introduced, many interesting extensions of Theorems 1.1 and 1.2 to
The above are some of the reasons why in metric spaces the study of Kannan tractions and generalizations of Kannan contractions plays a particularly important
con-part in fixed point theory
we construct four kinds of contractions of Kannan type (see conditions (C1)-(C4)) and,
by techniques based on these generalized pseudodistances, we prove fixed point
theo-rems for such contractions (see Theotheo-rems 2.1-2.8) The definitions and the results are
new in uniform and locally convex spaces and even in metric spaces Examples (see
Section 12) and some conclusions (see Section 13) are given
2 Statement of main results
Let X be a Hausdorff uniform space with uniformity defined by a saturated family
D = {d α : X2→ [0, ∞), α ∈ A} of pseudometrics da, α ∈ A, uniformly continuous on
(J 1)∀ α∈A∀x,y,z ∈X {J α (x, z) ≤ J α (x, y) + J α (y, z)}; and
Trang 3the following holds:
∀α∈A{ lim
It is the purpose of the present paper to prove the following results
∀α∈ A∃η α∈[0,1/2)∀x,y ∈X {J α (T(x), T(y)) ≤ η α [J α (T(x), x) + J α (T(y), y)]}and, additionally,
(D1)∃v0 ,ω∈X∀α∈A{limm→∞J α (v m , w) = 0}
∀α∈A {J α (w, w) = 0}
fol-lowing three conditions:
(C2) ∀α∈ A∃η α∈[0,1/2)∀x,y ∈X {J α (T(x), T(y)) ≤ η α [J α (T(x), x) + J α (T(y)), y]},(C3) ∀α∈ A∃η α∈[0,1/2)∀x,y ∈X {J α (T(x), T(y)) ≤ η α [J α (x, T(x)) + J α (y, T(y))]},(c4) ∀α∈ A∃η α∈[0,1/2)∀x,y ∈X {J α (T(x), T(y)) ≤ η α [J α (x, T(x)) + J α (T(y), y)]},and, additionally,
(D2) ∃v0 ,ω∈X∀α∈A{limm→∞J α (v m , w) = lim m→∞J α (w, v m) = 0}
∀α∈A {J α (w, w) = 0}
It is worth noticing that conditions (C1)-(C4) are different and conditions (D1) and
Clearly, (D1) include (D2) The following theorem shows that with some additionalconditions the converse holds
con-ditions (C2)-(C4) and, additionally, condition (D1) and at least one of the following
conditions (D3)-(D6):
(D3) ∀v0,w ∈X{limm→∞v m = w⇒ ∃q∈N{T [q] (w) = w}},(D4) ∀v0,w ∈X{limm→∞v m = w⇒ ∃q∈N{T [q] is continuous at a point w}},(D5) ∀v0,w∈X{limm→∞v m = w⇒ ∃q∈N{limm→∞T [q] (v m ) = T [q] (w)}},(D6) ∀v0,w ∈X{limm→∞v m = w⇒ ∃q∈N∀α∈A{limm→∞J α (T [q] (v m ), T [q] (w)) = 0}}
∀α∈A {J α (w, w) = 0}
The following theorem shows that if we assume that the uniform space is tially complete, then the conditions (D1) and (D2) can be omitted
Trang 4at least one of the conditions (C1)-(C4) and, additionally, at least one of the conditions
(c) ∀α∈A {J α (w, w) = 0}
these maps will be used to extend Theorem 2.4 to the uniform spaces which are not
sequentially complete and without any conditions (D1)-(D6)
J = {J α : X2→ [0, ∞), α ∈ A) be a J-family on X We say that T : X ® X is
J = {J α : X2→ [0, ∞), α ∈ A}be the J-family on X Let the map T : X ® X be
∀α∈A {J α (w, w) = 0}
Also, the following uniqueness results hold
con-ditions (C1)-(C4) and, additionally, the following concon-ditions (D7) and (D8):
is continuous at a point w;
∀α∈A {J α (w, w) = 0}
(C5) ∀α∈A∃η α∈[0,1/2)∀x,y ∈X {d α (T(x), T(y)) ≤ η α [d α (T(x), x) + d α (y, T(y))]}
In this section, we present some propositions that will be used in Sections 4-11
J = {J α : X2→ [0, ∞), α ∈ A}be a J-family If x ≠ y, x, y Î X, then
Trang 5(2.2) hold Consequently, by (J 2), we get (2.3) which implies ∀α∈A {d α (x, y) = 0}.
W α(1)(x, y) = max {φ α (x), J α (x, y)}, W α(1)(x, y) = max {φ α (x), J α (x, y)}and
W α(2)(x, y) = max {φ α (y), J α (x, y)}, x, yÎ X, are J-families on X
V α(2)(x, y) = φ α (y) + J α (x, y) V α(1)(x, y) = φ α (x) + J α (x, y)and V α(2)(x, y) = φ α (y) + J α (x, y),
Remark 3.2 ∀i∈ {1,2}∀α∈A∀x,y ∈X {J α (x, y) ≤ W (i)
α (x, y) ∧ J α (x, y) ≤ V (i)
α (x, y)}
W(2)
α (x, y) = max {φ α (z), J α (x, z)} ≤ max {φ α (y)+ φ α (z), J α (x, y)+J α (y, z)} ≤ W(2)
α (x, y)+W α(2)(y, z)}and
W(2)
α (x, y) = max {φ α (z), J α (x, z) } ≤ max {φ α (y)+ φ α (z), J α (x, y)+J α (y, z) } ≤ W(2)
α (x, y)+W α(2)(y, z)} Therefore,
∀α∈A{limn→∞supm>n W α (i) (x n , x m) = limm→∞W α (i) (x m , y m) = 0} Then, by Remark 3 2,
∀α∈A{limm→∞d α (x m , y m) = 0} which gives that (J 2) for family W (i) holds
X, V α(1)(x, z) = φ α (x) + J α (x, z) ≤ φ α (x) + J α (x, y) + φ α (y) + J α (y, z) = V α(1)(x, y) + V α(1)(y, z),and
V α(2)(x, z) = φ α (z) + J α (x, z) ≤ φ α (y) + J α (x, y) + φ α (z) + J α (y, z) = V α(2)(x, y) + V α(2)(y, z) Thus, for
∀α∈A{limn→∞supm >n V α (i) (x n , x m) = limm→∞V α (i) (x m , y m) = 0} Then, by Remark 3.2, we
∀α∈A{limm→∞d α (x m , y m) = 0}, ∀α∈A{limm→∞d α (x m , y m) = 0}.This gives that (J 2) for
∀α∈ A∃λ α∈[0,1) ∀x ∈X {max J α (T(x), T[2](x)), J α (T[2](x), T(x)) } ≤ λ αmax{J α (x, T(x)), J α (T(x), x)}}
∀α∈ ∃λ∈[0,1)∀x ∈X {J α (T[2](x), T(x)) + J α (T(x), T[2](x)) ≤ λ α [J α (T(x), x) + J α (x, T(x))]}
Trang 6Proof (a)The proof will be broken into two steps.
STEP 1 If (C1) holds, then the assertion holds
that ∀α∈A∃η α∈[0,1/2) ∀x ∈X {max{J α (T[2](x), T(x)), J α (T(x), T[2](x))} ≤ η α [J α (x, T(x))+J α (T(x), T[2](x))]≤ 2η α J α (x, T(x)) ≤ 2η αmax{Jα (x, T(x)), J α (T(x), x)}} It is
clear that ∀α∈ A {λ α= 2η α < 1}
STEP 1 If (C2) holds, then the assertion holds
By (C2), ∀α∈A∃η α∈[0,1/2) ∀x ∈X {J α (T[2](x), T(x)) ≤ η α (J α (T[2](x), T(x))+J α (x, T(x))]∧J α T(x), T[2](x)) ≤ η α [J α (T(x), x)+J α (T(x), T[2](x))]}.Hence, ∀α∈A∃λ α∈[0,1) ∀x ∈X {J α (T[2](x), T(x)) ≤ λ α J α (x, T(x)) ∧J α (T(x), T[2](x)) ≤ λ α J α (T(x), x)};
∀α∈A∃λ α∈[0,1)∀x ∈X {J α (T[2](x), T(x)) + J α (T(x), T[2](x)) ≤ λ α [J α (T(x), x) + J α (x, T(x))]}
STEP 2 If (C4) holds, then the assertion holds
By (C4), ∀α∈A∃η α∈[0,1/2) ∀x ∈X {J α (T[2](x), T(x)) ≤ η α (J α (T(x), T[2](x))+J α (T(x), x)]∧J α T(x), T[2](x)) ≤ η α [J α (x, T(x))+J α (T[2](x), T(x))]}.This gives ∀α∈A∃η α∈[0,1/2) ∀x ∈X {J α (T[2](x), T(x)) ≤ η2
α) / (1−n 2
α ) [J α (x, T(x))+J α (T(x), x)] = η α/ (1−ηα ) [J α x, (T(x))+J α (T(x), x)]
∀α∈A∃λ α∈[0,1)∀x ∈X {J α (T[2](x), T(x)) + J α T(x), T[2](x)) ≤ λ α [J α (x, T(x)) + J α (T(x), x)] □
(C1)-(C4) Then:
(a)∀α∈A∀u0∈X{limn→∞supm >n J α (u n , u m) = limn→∞supm >n J α (u m , u n) = 0}.(b) ∀α∈A∀u0∈X {limm→∞J α (u m , u m+1) = limm→∞J α (u m+1 , u m) = 0}.(c)∀α∈A∀u0∈X{limn→∞supm >n d α (u n , u m) = 0}
(z), then Fix(T) = {z} and
∀α∈A {J α (z, z) = 0}
STEP 1 If (C1) or (C3) holds, then the assertion holds
Trang 7Indeed, by Propositions 3.2(a) and 3.3(a), if
∀α∈A∀u0∈X{limn→∞supm >n J α (u n , u m) = 0}
W α(2)(u m , u n) ≤m k=n−1W(2)α (u k+1 , u k) ≤m k=n−1λ k
α W α(2)(u1, u0)≤ W(2)
α (u1, u0)λ n
α/ (1−λ α andlimn→∞supm >n W α(2)(u m , u n) = 0 Hence, by Remark 3.2,
∀α∈A∀u0∈X{limn→∞supm >n J α (u m , u n) = 0}
STEP 2 If (C2) or (C4) holds, then the assertion holds
∀α∈A∀u0∈X{limn→∞supm >n J α (u n , u m) = 0}
∀α∈A∀u0∈X{limn→∞supm >n J α (u m , u n) = 0}
∀α∈A∀u0∈X∀n∈N{J α (u n , u n+1)≤ supm>n J α (u n , u m)∧Jα (u n+1 , u n)≤ supm>n J α (u m , u n)}
Trang 8(c) Let u0Î X be arbitrary and fixed By (a),
∀α∈A∀ε>0∃n3=n3 (α, ε)∈∀m >n3{d α (u m , u j0+m)< ε /2}. (3:11)Let now α0∈A andε0 > 0 be arbitrary and fixed, let n0 = max {n2(a0, ε0), n3(a0,
d α0(u k , u l ) = d α0(u i0+n0, u j0+n0)≤ d α0(u n0, u i0+n0)+d α0(u n0, u j0+n0)< ε02+ε02 =ε0 Hence, we
conclude that ∀α∈A∀ε>0∃n0=n0 (α,ε)∈∀k,l∈ , k >l>n0{d α (u k , u l)< ε}
STEP 1 If (C1) or (C3) holds, then the assertions hold
Trang 9Next, we see that
∀α∈A {W(1)
α0 (T(z), z) > 0} and, since z = T[q](z) = T[2q](z) and if q + 1
∀α∈A {J α (a, T(a)) = J α (T(a), a) = 0} and ∀α∈A {J α (b, T(b)) = J α (T(b), b) = 0} Hence, if
∀α∈A {J α (a, b) = J α (b, a) = 0} which, by Remark 3.1, implies a = b Contradiction
STEP 2 If (C2) or (C4) holds, then the assertions hold
defined in the proof of (a), then
∀α∈A {V(1)
α0 (z, T(z)) > 0} and, consequently, by (3.3), since z = T[q]
Trang 10∀α∈A {J α (a, T(a)) = J α (T(a), a) = 0} and ∀α∈A {J α (b, T(b)) = J α (T(b), b) = 0} Hence, if
∀α∈A {J α (a, b) = J α (b, a) = 0} and, by Remark 3.1, we get a = b, which is impossible
4 Proof of Theorem 2.1
The proof will be broken into 11 steps
∀α∈A{limm→∞J α (v m , v m+1) = limm→∞J α (v m+1 , v m) = 0} This follows from Proposition
3.4(b)
Proposi-tion 3.4(e)
T(v m )) + J α (T(v m ), w) ≤ η α [J α (T(w), w) + J α (v m+1 , v m )] + J α (v m+1 , w)} Hence, by Step
∀α∈A∃η α∈[0,1 / 2) {limm→∞J α (T(w), w)≤ limm→∞{η α [J α (T(w), w)+J α (v m+1 , v m )]+J α (v m+1 , w)} = η α J α (T(w), w)} Thus,
∀α∈A∃η α∈[0,1 / 2) {J α (T(w), w) ≤ η α J α (T(w), w)}, so, since ∀α∈A {η α∈ [0, 1/2)}, we get
∀α∈A {J α (T(w), w) = 0}
Trang 11Indeed, by (C1) and Step 3, we have that
∀α∈A∃η α∈[0,1/2){J α (v m , T(w)) ≤ η α [J α (v m , v m−1) + J α (T(w), w)] = η α J α (v m , v m−1)} Hence, by
∀α∈A{limn→∞supm >n J α (v n , v m) = 0} Next, by Step 5, ∀α∈A{limm→∞J α (v m , T(w)) = 0}
∀α∈A{limm→∞d α (v m , T(w)) = lim m→∞d α (x m , y m) = 0}, i.e the limit limm ®∞ vm = T
(w) holds
Since X is Hausdorff, thus T(w) = w is a consequence of Steps 2 and 6
∀α∈A∃η α∈[0,1/2){J α (w, w) = J α (T(w), T(w)) ≤ η α [J α (T(w), w) + J α (T(w), w)] = 0}, i.e
∀α∈A {J α (w, w) = 0} holds
STEP 8 If v0, wÎ X satisfy (D1), then∀α∈A∀u0∈X{limm→∞J α (u m , w) = 0}
∀α∈A∃η α∈[0,1/2) ∀u0∈X{limm→∞J α (u m , w)≤ limm→∞[J α (u m , v m )+J α (v m , w)] ≤ η αlimm→∞[J α (u m , u m−1)+J α (v m , v m−1 )]+limm→∞J α (v m , w) = 0}
∀α∈A∀u0∈X{limn→∞supm>n J α (u n , u m) = 0} and ∀α∈A∀u0∈X{limm→∞J α (u m , w) = 0}
(2.3) which implies ∀α∈A∀u0∈X{limm→∞d α (u m , w) = 0}
Indeed, by (C1), Step 7 and Proposition 3.4(d) (for q = 1), we get that Fix (T) = {w}
STEP 11 The assertions (a)-(c) are satisfied
This is a consequence of Steps 10, 9 and 7
5 Proof of Theorem 2.2
The proof will be broken into seven steps
Trang 12STEP 1 If v0, w Î X satisfy (D2), then
∀α∈A{limm→∞J α (v m , v m+1) = limm→∞J α (v m+1 , v m) = 0} This follows from Proposition
3.4(b)
Proposi-tion 3.4(e)
Indeed, we consider three cases:
∀α∈A∃η α∈[0,1/2) ∀m∈ {J α (w, T(w)) ≤ J α (w, T(v m ))+J α (T(v m ), T(w)) ≤ J α (w, v m+1)+η α [J α (v m+1 , v m )+J α (w, T(w))]}
∀α∈A∃η α∈[0,1/2) {limm→∞J α (w, T(w))≤ limm→∞[J α (w, v m+1)+η α J α (v m+1 , v m)+η α J α (w, T(w))] = η α J α (w, T(w))}, i.e
∀α∈A∃η α∈[0,1/2){J α (w, T(w)) ≤ η α J α (w, T(w))} Hence, since ∀α∈A {η α∈ [0, 1/2)}, we
get
∀α∈A∃η α∈[0,1/2) ∀m∈ {J α (T(w), w) ≤ J α (T(w), T(v m ))+J α (T(v m ), w) ≤ η α [J α (T(w), w)+J α (v m , v m+1 )]+J α (v m+1 , w)}
∀α∈A∃η α∈[0,1/2){J α (T(w), w) ≤ η α J α (T(w), w)+lim m→∞ [η α J α (v m , v m+1 )+J α (v m+1 , w)] = η α J α (T(w), w)}and, since
∀α∈A {η α∈ [0, 1/2)}, ∀α∈A∃η α∈[0,1/2){J α (T(w), w) ≤ η α J α (T(w), w)} implies
From (5.1), (5.2) and Remark 3.1, we conclude that T(w) = w
∀α∈A∃η α∈[0,1/2) ∀m∈ {J α (w, T(w)) ≤ J α (w, T(v m ))+J α (T(v m ), T(w)) ≤ J α (w, v m+1)+η α [J α (v m , v m+1 )+J α (w, T(w))]}
∀α∈A∃η α∈[0,1/2) {limm→∞J α (w, T(w))≤ limm→∞[J α (w, v m+1)+η α J α (v m , v m+1)+η α J α (w, T(w))] = η α J α (w, T(w))} Therefore,
∀α∈A∃η α∈[0,1/2){J α (w, T(w)) ≤ η α J α (w, T(w))}, so, since ∀α∈A {η α∈ [0, 1/2)}, we get
Similarly,
∀α∈A∃η α∈[0,1/2) ∀m∈ {J α (T(w), w) ≤ J α (T(w), T(v m ))+J α (T(v m ), w) ≤ η α [J α (w, T(w))+J α (v m , v m+1 )]+J α (v m+1 , w)}
∀α∈A∃η α∈[0,1/2){J α (T(w), w) ≤ η α J α (w, T(w))+lim m→∞ [ηα J α (v m , v m+1 )+J α (v m+1 , w)] = 0} Therefore,
From (5.3), (5.4) and Remark 3.1, we conclude that T(w) = w
Trang 13∀α∈A∃η α∈[0,1/2){J α (w, T(w)) ≤ η α J α (T(w), w) ≤ J α (T(w), w)} (5:6)Clearly, (5.5) and (5.6) give
0< J α0(w, T(w)) ≤ η α0J α0(T(w), w) < J α0(T(w), w) = J α0(w, T(w)), which is absurd
Consequently, we have that
From (5.8) and Remark 3.1, we conclude that T(w) = w
∀α∈A {J α (w, w)≤ limm→∞ J α (w, v m) + limm→∞ J α (v m , w) = 0}
STEP 4 If v0, wÎ X satisfy (D2), then ∀α∈A∀u0∈X{limm→∞ J α (u m , w) = 0}.Indeed, we consider three cases:
∀α∈A∃η α∈[0,1/2) ∀u0∈X{limm→∞J α (u m , w)≤ limm→∞[J α (u m , v m )+J α (v m , w)] ≤ η α[limm→∞(J α (u m , u m−1)+J α (v m−1, v m))]+limm→∞J α (v m , w) = 0}, so
∀α∈A∀u0∈X{limm→∞J α (u m , w) = 0} holds
conclude that
∀α∈A∃η α∈[0,1/2) ∀u0∈X{limm→∞J α (u m , w)≤ limm→∞[J α (u m , v m )+J α (v m , w)] ≤ η α[limm→∞(J α (u m−1, u m )+J α (v m−1, v m))]+limm→∞J α (v m , w) = 0}, so
∀α∈A∀u0∈X{limm→∞J α (u m , w) = 0} holds
∀α∈A∃η α∈[0,1/2) ∀u0∈X{limm→∞J α (u m , w)≤ limm→∞[J α (u m , v m )+J α (v m , w)] ≤ η α[limm→∞(J α (u m−1, u m )+J α (v m , v m−1))]+limm→∞J α (v m , w) = 0}, so
∀α∈A∀u0∈X{limm→∞J α (u m , w) = 0} holds
STEP 5 If v0, wÎ X satisfy (D2), then ∀α∈A∀u0∈X{limn→∞d α (u n , w) = 0}
3.4(a), Step 4 and (J 2), ∀α∈A∀u0∈X{limm→∞d α (u m , w) = 0}
Indeed, assume that at least one of the conditions (C2)-(C4) holds Then, by Step 3and Proposition 3.4(d), we have that Fix(T) = {w}
STEP 7 The assertions (a)-(c) are satisfied
6 Proof of Theorem 2.3
The proof will be broken into six steps
lim
m→∞v
This follows from Proposition 3.4(e)
STEP 2 If at least one of the conditions (C2)-(C4) holds and, additionally, the tions (D1) and (D3) hold, then (D2) is satisfied