Maias xed point theorems in a space [625570]

Maia's xed point theorems in a space
with distance
Paula Homorodan
Abstract. In this paper we establish new xed point theorems for Maia's
xed point theorem in the setting of a space with a distance, more pre-
cisely when one of the metrics is replaced with a distance.
We also present some exemples to illustrate the theoretical results.
Keywords. Space with distance, H-distance space, Maia xed point the-
orem.
1. Introduction
In 1968, M. G. Maia [12] generalized Banach's xed point theorem for a
set X endowed with two metrics. Were obtained a serie of similar results and
generalizations of this theorem in many papers [1] – [3], [8] – [15], [17] – [19].
In 1977, Ioan A. Rus [18] stated and proved an interesting xed point
theorem of Maia type by replacing one of the conditions in Maia's theorem.
Many papers deal with xed point theorems of Maia type and with
applications of these theorems in sets with two or three metrics [14] – [15],
[19].
Theorem 1.1. [3]Let X be a nonempty set, dandbe two metrics on Xand
T:X!Xbe an operator. We suppose that:
(i)d(x;y)(x;y),8x;y2X
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is continuous;
(iv)T: (X;)!(X;)is ana-contraction, with a2[0;1):
Then:
(a)FT=fxg;
(b)T: (X;d)!(X;d)is Picard operator;
(c)T: (X;)!(X;)is Picard operator;
(d)(x;x)1
1l(x;T(x));8x2X:
Remark 1.1. [3]From Maia's xed point theorem when d , we get Banach's
xed point theorem;

2 Paula Homorodan
Remark 1.2. [3]Assumption (i) in Theorem 1.1may be weakened to
(i') There exists c>0such thatd(x;y)c(x;y),8x;y2X, or to
(i")There exists c>0such thatd(Tx;Ty )c(x;y),8x;y2X, which
is particularly useful when dealing with integral equations;
Remark 1.3. [3]Condition (iv) in Theorem 1.1may be replaced by one of the
following conditions:
"T: (X;)!(X;)is a Kannan mapping", or
"T: (X;)!(X;)is a Zamrescu mapping", or
"T: (X;)!(X;)is a- contraction", etc.
Denition 1.1. [11]LetT:X!XwhereXis a metric space with respect to
the metricd. ThenTis called a Kannan contraction if the following condition
holds:
d(Tx;Ty )a[d(x;Tx ) +d(y;Ty ) ];8x;y2X
where 0<a<1
2.
Denition 1.2. [6]LetT:X!XwhereXis a metric space with respect
to the metric d. ThenTis called a Bianchini contraction if the following
condition holds:
d(Tx;Ty )hmaxfd(x;Tx );d(y;Ty )g;8x;y2X
where 0<h< 1.
We will try to expose other Maia's theorems when the second metric
is replaced with a distance, but rst we will explain more exactly what is a
distance space.
2. Preliminaries
The problem of xed points consists in nding conditions under which
for a given mapping T:X!Xthe set of xed point
Fix(T) =fx2X:T(x) =xg
ofTis non-empty.
LetXbe a non-empty set and d:XX!Rbe a mapping such that
for allx;y2Xwe have:
(im)d(x;y)0;
(iim)d(x;y) +d(y;x) = 0 if and only if x=y.
Then (X;d) is called a distance space and d is called a distance onX.
Let (X;d) be a distance space, fxn:n2(N) =f1;2;:::ggbe a sequence
inXandx2X.
We say that the sequence fxn:n2(N)g:
1) is convergent toxif and only if lim n!1d(x;xn) = 0.
We denote this by xn!xorx= lim n!1xn;
2) is convergent if it converge to some point in X.
3) is Cauchy orfundamental if lim n;m!1d(xn;xm) = 0.

Maia's xed point theorems in a space with distance 3
A distance space ( X;d) iscomplet if every Cauchy sequence in Xcon-
verges to some point in X.
LetXbe a non-empty set and d(x;y) be a distance on Xwith the
following property:
(N) for each point x2Xand any >0 there exists =(x;)>0
such that from d(x;y)andd(y;z)it followsd(x;z).
Then (X;d) is called an N-distance space anddis called an N-distance
onX.
Ifdis a symmetric, then we say that dis an N-symmetric .
Ifdsatises the condition:
(F) for any >0 there exists =()>0 such that from d(x;y)
andd(y;z)it followsd(x;z),
thendis called an F-distance or a Frechet distance and (X;d) is called an
F-distance space .
Ifdis a symmetric and an F-distance on a spaceX, then we say that d
is an F-symmetric .
Remark 2.1. [7]If(X;d)is an F-symmetric space, then any convergent se-
quence is a Cauchy sequence. For N-symmetric spaces and for quasimetric
spaces this assertion is not more true.
Spaces with H-distances
A distance space ( X;d) is called an H-distance space if for any two
distinct points x;y2Xthere exists =(x;y)>0 such that B(x;d; )\
B(y;d; ) =;.
Remark 2.2. [7]Let(X;d)be a distance space. Then (X;d)is anH-distance
space if and only if any convergent sequence has a unique limit point.
Lemma 2.1. [7]Let(X;d)be a distance space and the space (X;(d))is
Hausdor. Then d is an H-distance.
Proposition 2.1. [7]Let(X;d)be anH-distance space, T:X!Xbe a
continuous mapping.
Then:
1. The setFix(T)of xed points of Tis closed.
2. If for some point x2Xthe Picard iteration O(T;x)is convergent,
then the set of xed points Fix(T)of the mapping Tis non-empty.
In a recent paper [10], we present Kannan and Bianchini xed point
theorems in the setting of a space with a distance.
Theorem 2.1. [10]Let(X;d)be a complete H-distance space and let T:X!
Xbe a mapping for wich there exists 0<a<1
2such that:
d(Tx;Ty )a[d(x;Tx ) +d(y;Ty ) ];8x;y2X (2.1)
Then the Picard iteration of the point xis convergent.
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.

4 Paula Homorodan
Theorem 2.2. [10]Let(X;d)be a complete H-distance space and let T:X!
Xbe a mapping. Assume that there exists a number h,0<h< 1such that
d(Tx;Ty )hmaxfd(x;Tx );d(y;Ty )g;8x;y2X (2.2)
Then the Picard iteration of the point xis convergent.
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.
In the next section we try to expose other Maia's theorems when the
second metric is replaced with a distance. The main aim is to see why is so
important that one of the metric to be continuous and what happends when
is not continuous.
3. Main results
Theorem 3.1. Let X be a nonempty set, da metric on X,is aH-distance
onXandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is continuous;
(iv)T: (X;)!(X;)is a Kannan contraction in a H-distance space.
Then the Picard iteration of the point xis convergent in (X;d)and its
limit is x.
Proof. Letx02Xbe arbitrary and consider the Picard sequence fxng1
n=0,
xn+1=Txn; n0.
In [10] we prove that (2.1) reduces to
(xn;xn+1)a
1an

(x0;x1); n= 0;1;:::: (3.1)
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1andy:=xn1
in (2.1) and so we have
(Txn+p1;Txn1)a[(xn+p1;xn+p) +(xn1;xn) ]()
(xn+p;xn)a[(xn+p1;xn+p) +(xn1;xn) ]
and by using (3.1) we have
(xn+p;xn)a"a
1an+p1
(x0;x1) +a
1an
(x0;x1)#
=
=a"a
1an+p1
+a
1an#
(x0;x1)
We obtain
(xn+p;xn)a
a
1an

"a
1ap1
+ 1#
(x0;x1):

Maia's xed point theorems in a space with distance 5
Since 0a
1a1 it results that an!0 as (n!1 ) from which we conclude
thatfxngis a Cauchy sequence in ( X;).
By (i) it result that fxng1
n=0is a Cauchy sequence in ( X;d) as well and
by (ii) we deduce that fxng1
n=0converges in ( X;d).
On the other hand Tis continuous on ( X;d) so denoting
x= lim
n!1xn2Fix(T) we nd
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x);
i.e., xis a xed point of T;
This show that for any x02X, the Picard iteration converges in ( X;d)
and its limit is a xed point of T. Since Thas at most one xed point
(cardF T1), we deduce that, for every choice of x02X, the Picard iteration
converges to the same value  x, that is, the unique xed point of T.

Remark 3.1.
1.Assumption (i) in Theorem 3.1may be weakened to
(i') There exists c>0such thatd(x;y)c(x;y),8x;y2X, or
to
(i")There exists c>0such thatd(Tx;Ty )c(x;y),8x;y2X;
2.Ifis actually a quasimetric, then Theorem 3.1reduces to the well
known Maia xed point theorem in metric spaces with a Kannan condi-
tion[11].
The next Corollary is a generalization of Maia's xed point theorem in
metric spaces with a Kannan contraction [11].
Corollary 3.1. Let X be a nonempty set, da metric on X,is a symmetric
H-distance on XandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is continuous;
(iv)T: (X;)!(X;)is a Kannan contraction in a symmetric
H-distance space.
Then the Picard iteration of the point xis convergent in (X;d)and its
limit is x.
Proof. If (X;) is a symmetric H-distance space, then it is a H-distance and
conclusion follows by Theorem 3.1. 
Theorem 3.2. Let X be a nonempty set, da metric on X,is aH-distance
onXandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is not continuous;
(iv)T: (X;)!(X;)is a Kannan contraction in a H-distance space.

6 Paula Homorodan
Then the Picard iteration of the point xis convergent in (X;d).
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.
Proof. Letx02Xbe arbitrary and consider the Picard sequence fxng1
n=0,
xn+1=Txn; n0.
In [10] we prove that (2.1) reduces to
(xn;xn+1)a
1an

(x0;x1); n= 0;1;:::: (3.2)
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1andy:=xn1
in (2.1) and so we have
(Txn+p1;Txn1)a[(xn+p1;xn+p) +(xn1;xn) ]()
(xn+p;xn)a[(xn+p1;xn+p) +(xn1;xn) ]
and by using (3.2) we have
(xn+p;xn)a"a
1an+p1
(x0;x1) +a
1an
(x0;x1)#
=
=a"a
1an+p1
+a
1an#
(x0;x1)
We obtain
(xn+p;xn)a
a
1an

"a
1ap1
+ 1#
(x0;x1):
Since 0a
1a1 it results that an!0 as (n!1 ) from which we conclude
thatfxngis a Cauchy sequence in ( X;) .
By (i) it result that fxng1
n=0is a Cauchy sequence in ( X;d) as well and
by (ii) we deduce that fxng1
n=0converges in ( X;d), which proves the rst
part of the theorem.
Because (X;d) is aH-distance space, we know from Remark 2.2, that
any convergent sequence has a unique xed point.
We denote  x= lim
n!1xn2Fix(T), then the uniqueness is immediate.
Indeed, suppose that Twould have two xed points
x;y2Fix(T), x6= y. Then
(x;y) =(Tx;Ty)a[(x;Tx) +(y;Ty) ] = 0;
a contradiction. So  x6= y, and hence  xis the unique xed point of T.
Example 3.1. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
metricdand theF-symmetric , denedd;:X!Ras follows:
d(x;x) = 0 ,d(0;x) =d(x;0) = 1 ,
d(1;2n) =d(2n;1) = 1 ,n6= 0,
d(21;2n) =d(2n;21) = 21,n6= 0;1;,
d(2n;2n1) =d(2n1;2n) = 1 ,n6= 0;1and
d(2m;2n) =j2m2nj;m+ 16=n, for allm;n2N

Maia's xed point theorems in a space with distance 7
and
(x;x) = 0 ,(0;x) =(x;0) = 1 ,(1;21) =(21;1) = 4 ,
(1;2n) =(2n;1) = 5 ,n6= 0;1,
(21;2n) =(2n;21) = 21,n6= 0;1,
(2n;2n1) =(2n1;2n) = 9 ,n6= 0;1and
(2m;2n) =j2m2nj;m+ 16=n, for allm;n2N.
Note thatdis a metric but is not a metric, because the triangle in-
equality is not satised: 5 =(1;2n)>(1;0) +(0;2n) = 2 .
Now we consider the mapping T:X!X, whereT(0) = 0 ,T(1) = 21
andT(2n) = 2n1.
It is easy to see that d(x;y)(x;y),8x;y2X.
For theF-symmetric we chech the condition (2.1) :
Claim 1)x= 0;y= 1, when condition (2.1) witha=1
4, reduces to
(T(0);T(1))1
4
[(0;T(0)) +(1;T(1))]() 11:
Claim 2)x= 0;y= 2n, when condition (2.1) witha=1
4, reduces to
(T(0);T(2n))1
4
[(0;T(0)) +(2n;T(2n))]() 19
4:
Claim 3)x= 1;y= 2n, when condition (2.1) witha=1
4, reduces to
(T(1);T(2n))1
4
[(1;T(1)) +(2n;T(2n))]()1
213
4:
Claim 4)x= 2m;y= 2n, when condition (2.1) witha=1
4, reduces to
(T(2m);T(2n))1
4
[(2m;T(2m) +(2n;T(2n)]()
1
2<9
2, ifm6=n.
Ifm=n, thend(T(2m);T(2n)) = 0 and claim 4 it is also true.
Therefore,Tis a Kannan contraction, i.e.,
(T(x);T(y))1
4
[(x;T(x) +(y;T(y)];x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0g.
The example it is alsow true for
(T(x);T(y))1
3
[(x;T(x) +(y;T(y)];x;y2X:
Remark 3.2.
1.Assumption (i) in Theorem 3.2may be weakened to
(i') There exists c>0such thatd(x;y)c(x;y),8x;y2X, or
to
(i")There exists c>0such thatd(Tx;Ty )c(x;y),8x;y2X;
2.In general, the limit xof the Picard sequence is not a xed point of T;

8 Paula Homorodan
The next Corollary is a generalization of Maia's xed point theorem in
metric spaces with a Kannan contraction [11].
Corollary 3.2. Let X be a nonempty set, da metric on X,is a symmetric
H-distance on XandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is not continuous;
(iv)T: (X;)!(X;)is a Kannan contraction in a symmetric H-
distance space.
Then the Picard iteration of the point xis convergent in (X;d).
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.
Proof. If (X;) is a symmetric H-distance space, then it is a H-distance and
conclusion follows by Theorem 3.2. 
We try to extend the notions from the article [10] and change the nature
offrom Theorem 3.1 and Theorem 3.2 with a Bianchini contraction.
Theorem 3.3. Let X be a nonempty set, da metric on X,is aH-distance
onXandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is continuous;
(iv)T: (X;)!(X;)is a Bianchini contraction in a H-distance
space.
Then the Picard iteration of the point xis convergent in (X;d)and its
limit is x.
Proof. Letx02Xbe arbitrary and consider the Picard sequence fxng1
n=0,
xn+1=Txn; n0.
In [10] we prove that (2.2) reduces to
(xn;xn+1)hmaxf(xn1;xn);(xn;xn+1)g; n0: (3.3)
and we obtain
(xn;xn+1)hn
(x0;x1); n0: (3.4)
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1andy:=xn1
in (2.2) and so we have
(Txn+p1;Txn1)hmaxf(xn+p1;xn+p);(xn1;xn)g ()
(xn+p;xn)hmaxf(xn+p1;xn+p);(xn1;xn)g;
and by using (3.4) we have
(xn+p;xn)h
maxfhn+p1
(x0;x1);hn1
(x0;x1)g=

Maia's xed point theorems in a space with distance 9
=h
(x0;x1) maxfhn+p1;hn1g=
=h
hn1
(x0;x1) =hn
(x0;x1):
Since 0a
1a1 it results that an!0 as (n!1 ) from which we conclude
thatfxngis a Cauchy sequence in ( X;)
By (i) it result that fxng1
n=0is a Cauchy sequence in ( X;d) as well and
by (ii) we deduce that fxng1
n=0converges in ( X;d).
On the other hand Tis continuous on ( X;d) so denoting
x= lim
n!1xn2Fix(T) we nd
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x);
i.e., xis a xed point of T;
This show that for any x02X, the Picard iteration converges in ( X;d)
and its limit is a xed point of T. Since Thas at most one xed point
(cardF T1), we deduce that, for every choice of x02X, the Picard iteration
converges to the same value  x, that is, the unique xed point of T. 
Remark 3.3.
1.Assumption (i) in Theorem 3.3may be weakened to
(i') There exists c>0such thatd(x;y)c(x;y),8x;y2X, or
to
(i")There exists c>0such thatd(Tx;Ty )c(x;y),8x;y2X;
2.Ifis actually a quasimetric, then Theorem 3.3reduces to the well
known Maia xed point theorem in metric spaces with a Bianchini con-
dition [6].
The next Corollary is a generalization of Maia's xed point theorem in
metric spaces with a Bianchini condition [6].
Corollary 3.3. Let X be a nonempty set, da metric,is a symmetric
H-distance on XandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is continuous;
(iv)T: (X;)!(X;)is a Bianchini contraction in a symmetric
H-distance space.
Then the Picard iteration of the point xis convergent in (X;d)and its
limit is x.
Proof. If (X;) is a symmetric H-distance space, then it is a H-distance and
conclusion follows by Theorem 3.3. 
Theorem 3.4. Let X be a nonempty set, da metric on X,is aH-distance
onXandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is not continuous;
(iv)T: (X;)!(X;)is a Bianchini contraction in a H-distance
space.

10 Paula Homorodan
Then the Picard iteration of the point xis convergent in (X;d).
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.
Proof. Letx02Xbe arbitrary and consider the Picard sequence fxng1
n=0,
xn+1=Txn; n0.
In [10] we prove that (2.2) reduces to
(xn;xn+1)hmaxf(xn1;xn);(xn;xn+1)g; n0: (3.5)
and we obtain
(xn;xn+1)hn
(x0;x1); n0: (3.6)
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1andy:=xn1
in (2.2) and so we have
(Txn+p1;Txn1)hmaxf(xn+p1;xn+p);(xn1;xn)g ()
(xn+p;xn)hmaxf(xn+p1;xn+p);(xn1;xn)g;
and by using (3.6) we have
(xn+p;xn)h
maxfhn+p1
(x0;x1);hn1
(x0;x1)g=
=h
(x0;x1) maxfhn+p1;hn1g=
=h
hn1
(x0;x1) =hn
(x0;x1):
Since 0a
1a1 it results that an!0 as (n!1 ) from which we conclude
thatfxngis a Cauchy sequence in ( X;) .
By (i) it result that fxng1
n=0is a Cauchy sequence in ( X;d) as well and
by (ii) we deduce that fxng1
n=0converges in ( X;d), which proves the rst
part of the theorem.
Because (X;d) is aH-distance space, we know from Remark 2.2, that
any convergent sequence has a unique xed point.
We denote  x= lim
n!1xn2Fix(T), then the uniqueness is immediate.
Indeed, suppose that Twould have two xed points
x;y2Fix(T), x6= y. Then
(x;y) =(Tx;Ty)h
maxf(x;Tx);(y;Ty)g= 0;
a contradiction. So  x6= y, and hence  xis the unique xed point of T.
Example 3.2. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
metricdand theF-symmetric , denedd;:X!Ras follows:
d(x;x) = 0 ,d(0;x) =d(x;0) = 1 ,
d(1;2n) =d(2n;1) = 1 ,n6= 0,
d(21;2n) =d(2n;21) = 1 ,n6= 0;1;2,
d(22;2n) =d(2n;22) = 2 ,n6= 0;1,
d(2n;2n1) =d(2n1;2n) = 1 ,n6= 0;1and
d(2m;2n) =j2m2nj;m+ 16=n, for allm;n2N

Maia's xed point theorems in a space with distance 11
and
(x;x) = 0 ,(0;x) =(x;0) = 1 ,
(1;2n) =(2n;1) = 5 ,n6= 0;1,
(21;2n) =(2n;21) = 4 ,n6= 0;1;2,
(22;2n) =(2n;22) = 2 ,n6= 0;1;2,
(2n;2n1) =(2n1;2n) = 4 ,n6= 0;1and
(2m;2n) =j2m2nj;m+ 16=n, for allm;n2N.
Note thatdis a metric but is not a metric, because the triangle in-
equality is not satised: 5 =(1;2n)>(1;0) +(0;2n) = 2 .
Now we consider the mapping T:X!X, whereT(0) = 0 ,T(1) = 22
andT(2n) = 2n1.
It is easy to see that d(x;y)(x;y),8x;y2X.
For theF-symmetric we chech the condition (2.2) :
Claim 1)x= 0;y= 1, when condition (2.2) witha=1
2, reduces to
(T(0);T(1))1
2
maxf(0;T(0));(1;T(1))g () 15
2:
Claim 2)x= 0;y= 2n, when condition (2.2) witha=1
2, reduces to
(T(0);T(2n))1
2
maxf(0;T(0));(2n;T(2n))g () 12:
Claim 3)x= 1;y= 2n, when condition (2.2) witha=1
2, reduces to
(T(1);T(2n))1
2
maxf(1;T(1));(2n;T(2n))g () 22:
Claim 4)x= 2m;y= 2n, when condition (2.2) witha=1
2, reduces to
(T(2m);T(2n))1
2
maxf(2m;T(2m);(2n;T(2n)g ()
1
2<2, ifm6=n.
Ifm=n, thend(T(2m);T(2n)) = 0 and claim 4 it is also true.
Therefore,Tis a Bianchini contraction, i.e.,
(T(x);T(y))1
2
maxf(x;T(x);(y;T(y)g;x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0g.
Remark 3.4.
1.Assumption (i) in Theorem 3.4may be weakened to
(i') There exists c>0such thatd(x;y)c(x;y),8x;y2X, or
to
(i")There exists c>0such thatd(Tx;Ty )c(x;y),8x;y2X;
2.In general, the limit xof the Picard sequence is not a xed point of T;
The next Corollary is a generalization of Maia's xed point theorem in
metric spaces with a Bianchini condition [6].

12 Paula Homorodan
Corollary 3.4. Let X be a nonempty set, da metric,is a symmetric
H-distance on XandT:X!Xbe a mapping. We suppose that:
(i)d(x;y)(x;y),8x;y2X;
(ii)(X;d)is a complete metric space;
(iii)T: (X;d)!(X;d)is not continuous;
(iv)T: (X;)!(X;)is a Bianchini contraction in a complete sym-
metricH-distance space.
Then the Picard iteration of the point xis convergent in (X;d).
If, additionally the limit xof the Picard sequence is a xed point of T, then
xis the unique xed point of T.
Proof. If (X;) is a symmetric H-distance space, then it is a H-distance and
conclusion follows by Theorem 3.4. 
Conclusion
Working in the general setting of a H-distance space, we obtained signicant
generalizations of Maia's xed point theorems with a Kannan and Bianchini
contraction in usual metric spaces.
We note that while the inequality of the triangle is not always necessary to
prove the existence of a xed point, instead the continuity of the metric is
very useful.
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Paula Homorodan
Department of Mathematics and Computer Science
North University Center at Baia Mare
Technical University of Cluj-Napoca
Victoriei 76, 430122, Baia Mare, Romania
e-mail: paula homorodan@yahoo.com