Kannan and Bianchini xed point theorems [625569]

Kannan and Bianchini xed point theorems
in H-distance space endowed with a partially
ordered set
Paula Homorodan
Abstract. In this paper we establish new xed point theorems for Kan-
nan and Bianchini type mappings in the very general setting of a space
with a distance. We will work on a partially ordered set where we sup-
pose there is a H-distance and a non-decreasing mapping. The xed
point is unique under some additional conditions.
We also present some exemples to illustrate the theoretical results.
2010 MSC: 47H10; 47H09.
Keywords. Space with distance, H-distance space, partially ordered,
Kannan type mappings, Bianchini type mappings.
1. Introduction
In 2003, A. Ran and M. Reurings discuss an analogy of Banach's xed
point theorem in partially ordered sets. The key in this xed point theorem is
that the contractivity condition on the nonlinear map is only assumed to hold
on elements that are comparable in the partial order. The map is assumed
to be monotone. They show that under such conditions the conclusions of
Banach's xed point theorem still hold.
Theorem 1.1. [23]LetXbe a partially ordered set such that every pair x;y2
Xhas a lower bound and an upper bond. Furthermore, let dbe a metric on X
such that (X;d)is a complete metric space. If Fis a continuous, monotone
(i.e., either order-preserving or order reversing) map from XintoXsuch
that
(1)90<c< 1 :d(F(x);F(y))c
d(x;y);8xy;
(2) 9x02X:x0F(x0)or x 0F(x0);
thatXhas a unique xed point x. Moreover, for every x2X,
lim
n!1Fnx= x:

2 Paula Homorodan
Letf:X!Xbe an operator. Then f0:= 1 x; f1:=f; :::; fn:=
ffn;n2N, denote the iterate operators of f. ByI(f) we will denote the
set of all nonempty invariant subsets of f, i.e.,I(f) :=fYXjf(Y)Yg.
Also, byFf:=fx2Xjx=f(x)gwe will denote the xed point set of
the operator f, whileAf(x) :=fx2Xjfn(x)!x;asn!+1gdenotes
the attractor basin of fwith respect to x2X.
LetXbe a nonempty set. Denote s(X) :=f(xn)n2Njxn2X;n2Ng.
Letc(X)s(X) be a subset of s(X) and let Lim:c(X)!Xbe an
operator. By denition, the triple ( X;c(X);Lim ) is called an L-space (Fr echet
[12] ) if the following conditions are satised:
(i) Ifxn=x;8n2N, then (xn)n2N2c(X) andLim(xn)n2N=x.
(ii) If (xn)n2N2c(X) andLim(xn)n2N=x, then for all subsequences,( xni)i2N
of (xn)n2Nwe have that ( xni)i2N2c(X) andLim(xni)i2N=x.
By denition an element of c(X) is a convergent sequence, x:=Lim(xn)n2N
is the limit of this sequence and we also write xn!xasn!+1.
In what follows we denote an L-space by ( X;!).
In this setting, an operator f:X!Xis called orbitally continuous
ifx2Xandfn(i)(x)!a2X, asi!+1implyfn(i)+1(x)!f(a), as
i!+1.
Denition 1.1. [22]LetXbe a nonempty set. Then, by denition, (X;!;)
is an ordered L-space if and only if:
(i)(X;!)is an L-space,
(ii)(X;)is a partially ordered set,
(iii) (xn)n2N!x;(yn)n2N!yandxnyn, for eachn2N=)xy.
Denition 1.2. [1][11] Let(X;)be a partially ordered set. A mapping T:
X!Xis said to be non-decreasing if for all x1;x22X;x 1x2implies
Tx1Tx2.
Denition 1.3. [16]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.
Kannan [16] proved that if Xis complete, then a Kannan mapping has
a xed point. It is interesting that Kannan's xed point theorem is indepen-
dent of the Banach contraction principle [2]. Kannan's xed point theorem is
very important because Subrahmanyam [29] proved that Kannan's theorems
characterizes the metric completness. That is, a metric space Xis complete
if and only if every Kannan mapping on Xhas a xed point.
Denition 1.4. [8]LetT:X!XBianchini where Xis a metric space
with respect to the metric d. ThenTis called a Bianchini contraction if the
following condition holds:
d(Tx;Ty )h
maxfd(x;Tx );d(y;Ty )g;8x;y2X
where 0<h< 1.

Theorems in H-distance space endowed with a partially ordered set 3
Bianchini [8] obtained an extension of the Kannan contraction. It is easy
to see that every Kannan mapping is a Bianchini mapping but the converse
is not more true.
Geraghty[13] introduced an extension of the Banach contraction map-
ping principle in which the contractive constant was replaced by a function
having some specied properties.
Kikkawa ans Suzuki proved a genetralization of Kannan's theorem[17],[30].
Binayak Choudhury and Amaresh Kundu[11] introduced a generalization of
Kannan mapping in partially ordered metric space and show that these map-
pings have xed point if the space is complete.
Denition 1.5. [13]LetSis the class of function :R+![0;1)with
(i)R+=ft2R=t> 0g,
(ii)(tn)!1implies t n!0.
With the help of the above class of functions, Geraghty [13] had estab-
lished a generalisation of the Banach contraction principle.
A. Amini-Harandi and H. Emami [1] have shown that the result which
Geraghty proved is also valid in complete partially ordered metric spaces.
Theorem 1.2. [1]Let(X;)be a partially ordered set and suppose that there
exists a metric dinXsuch that (X;d)is complete metric space.
LetT:X!Xbe a nondecreasing mapping such that
d(Tx;Ty )(d(x;y))d(x;y), forx;y2Xwithxy,
where2S. Assume that either Tis continuous or Xsatises the following
condition:
iffxngis a non decreasing sequence in Xsuch thatxn!x, then
xnx;8n2N.
Besides, suppose that for each x;y2Xthere exists z2Xwhich is comparable
to x and y. If there exists x02Xwithx0Tx0, then T has a unique xed
point.
Theorem 1.3. [11] Let(X;)be a partially ordered set and suppose there is
a metricdonXsuch that (X;d)is a complete metric space.
Letf:X!Xbe a non-decreasing mapping such that
d(fx;fy )d(x;fx ) +d(y;fy )
2d(x;fx ) +d(y;fy )
2
;
8x;y2Xsuch that x and y are comparable. Also suppose that either:
(a)fis continuous or
(b)Xhas the property, if a non-decreasing sequence fxng!x, thenxnx
for alln0.
If, there exists x02Xwithx0f(x0)then f has a xed point in X.
In the next section we will try to understand spaces with distance and
see some results in H-distance space endowed with a partial order set.

4 Paula Homorodan
2. Preliminaries
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.
LetXbe a non-empty set and d(x;y) be a distance on Xwith the
following property:
(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 .
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 2.1. [10]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. [10]Let(X;d)be a distance space and the space (X;(d))is
Hausdor. Then d is an H-distance.
Using the concept of Hausdor metric, Nadler [20] proved the xed
point theorem for multi-valued contraction maps, which is a generalization
of the Banach contraction principle [2].
Distances in uniform space were given by V alyi [31]. More gen-
eral concepts of distances were given by Wlodarczyk and Plebaniak [32] and
Wlodarczyk [33].
Letbe a (partial) order on a distance space ( X;d). A sequence
fxngn2Nis called:
{ non-decreasing if xnxn+1for eachn2N;
{ non-increasing if xnxn+1for eachn2N;
{ monotone if it is either non-decreasing on non-increasing.
If (X;d;) is an ordered distance space and g:X!Xis a mapping,
then (X;d;) is said to have the sequential g-monotone propety [9] if it
veries:
i) iffxngn2Nis an non-decreasing sequence and lim
n!1d(x;xn) = 0,
theng(xn)g(x) for alln2N;
ii) iffxngn2Nis an non-increasing sequence and lim
n!1d(x;xn) = 0,
theng(xn)g(x) for alln2N;
An ordered distance space ( X;d;) is called monotonically complete if
any monotone Cauchy sequence in Xconverges to some point in X.
Fixm2Nand a subset Lf1;2;:::;mg. Introduce on Xthe ordering
L: (x1;:::;x m)L(y1;:::;y m) ixiyifori2Landyjxjforj =2L.

Theorems in H-distance space endowed with a partially ordered set 5
By construction, ( Xm;dm;;L) and (Xm;dm;;L) are ordered dis-
tance spaces.
IfLf1;2;:::;mgandM=f1;2;:::;mgthenxLif and only if
yM. HenceyMis the dual (inverse) order of the order yL.
Remark 2.2. [7]Let(X;d;)be a monotonically complete distance spaces.
Then (Xm;dm;L)and(Xm;dm;L)are ordered monotonically complete
distance spaces, too.
Proof. It is obvious. 
In 2007 J. Nieto, R.Pouso, and R. Rodriguez-Lopez, proved some xed
point theorems in topological space. They nd that every Hausdor topolog-
ical space is an L-space: if X is a topological space it suce to dene c(X) as
the family of all convergent sequences and Lim as the mapping that sends ev-
ery convergent sequence into its limit. Note that it is not only that Hausdor
topological spaces are L-spaces, but they can be naturally equipped with an
L-space structure such that convergent sequences and their respective limits
are the same from the topological and from the L-space points of view.
The next results are applicable, in particular, to not necessarily complete
metric spaces.
Theorem 2.1. [21]Let X be a Hausdor topological space with a partial order
and letf:X!Xbe an operator. Suppose that:
– f is order-preserving.
– f is orbitally monotone-continuous, that is, if x2Xandfn(i)(x)!a
with (fn(i)(x))i2Nmonotone, then fn(i)+1(x)!f(a).
– There exists x02Xwith (x0;f(x0))2X.
– f maps monotone sequences into convergent sequences. Then there ex-
ists at least one xed point of f in X.
More generally:
Theorem 2.2. [21]Let X be a Hausdor topological space with a partial order
and letf:X!Xbe an operator. Suppose that:
– One of the following conditions is satised:
either f is orbitally continuous, of f is orbitally -continuous and if
(xm)m2N!xis such that (xm;xm+1)2X,8m2N, then there exists
(xm)mk2Na subsequence of (xm)m2Nsuch that (xmk;x)2X,8k2N.
-X2I(ff).
– There exists x02Xwith (x0;f(x0))2X.
– If(xm)m2Nis a sequence in Xsuch that (xm;xm+1)2X,8m2N,
then (f(xm))m2Nis a convergent sequence. Then there exists at least one
xed point of f in X.
In a recent paper [14], we present Kannan xed point theorem in the
setting of a space with a distance and Bianchini xed point theorem in the
setting of a space with a distance.

6 Paula Homorodan
Theorem 2.3. [14]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.
Theorem 2.4. [14]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 )h
maxfd(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 ofT.
We propose to extend the Theorem 2.3 and Theorem 2.4 into a space
with a H-distance endowed with a partially ordered set.
3. Main results
Theorem 3.1. Let(X;)be a partially ordered set and suppose there is a
H-distancedonXsuch that (X;d)is a complete H-distance space.
LetT:X!Xbe a non-decreasing mapping for wich there exists 0<a<1
2
then:
d(Tx;Ty )a
[d(x;Tx ) +d(y;Ty ) ];8x;y2X (3.1)
such that x and y are comparable.
Then the Picard iteration at the any point x2Xis convergent. If, addition-
ally, the limit xof the Picard sequence is a xed point of Tand existsx02X
withx0Tx0then xis the unique xed point of T.
Proof. Letx02Xand consider the Picard sequence fxng,xn+1=Txn;
n0. SinceTis a non-decreasing function, we have
x0Tx0=x1T2x0=Tx1=x2T3x0 (3.2)



Tnx0=xnTn+1×0=xn+1



We prove that
d(xn;xn+1)a
1an

d(x0;x1); n= 0;1;::: (3.3)
First, we note that a2
0;1
2

=)a
1a2(0;1). Inequality (3.3) is
obviously true for n= 0.
By (3.2) we see that, for all n1,xnandxn+1are comparable elements.
Now, we take x:=x0; y:=x1in (3.1) and obtain
d(Tx0;Tx 1)a
[d(x0;Tx 0) +d(x1;Tx 1) ];

Theorems in H-distance space endowed with a partially ordered set 7
that is,
d(x1;x2)a
[d(x0;x1) +d(x1;x2) ]
which yields
d(x1;x2)a
1a1

d(x0;x1);
and so (3.3) holds for n= 1.
Next, we take x:=xn1; y:=xnin (3:1) and obtain
d(xn;xn+1)a
[d(xn1;xn) +d(xn;xn+1) ]
and we have
d(xn;xn+1)a
1a

d(xn1;xn): (3.4)
Suppose (3.3) holds for n=k, that is,
d(xk;xk+1)a
1ak

d(x0;x1)
and prove that (3.3) also holds for n=k+ 1. Indeed, in view of (3.4)
d(xk+1;xk+2)a
1a
d(xk;xk+1)a
1a
a
1ak

d(x0;x1)
i.e.,
d(xk+1;xk+2)a
1ak+1

d(x0;x1):
Then, by mathematical induction we have that (3.3) holds for all n0.
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1and
y:=xn1in (3.1). By (3.2) we see that x:=xn+p1andy:=xn1are
comparable and so we have
d(Txn+p1;Txn1)a
[d(xn+p1;xn+p) +d(xn1;xn) ]()
d(xn+p;xn)a
[d(xn+p1;xn+p) +d(xn1;xn) ]
and by using (3.2) we have
d(xn+p;xn)a
"a
1an+p1
d(x0;x1) +a
1an
d(x0;x1)#
=
=a
"a
1an+p1
+a
1an#
d(x0;x1)
We obtain
d(xn+p;xn)a
a
1an

"a
1ap1
+ 1#
d(x0;x1); (3.5)
by which we immediately conclude that fxngis a Cauchy sequence. As ( X;d)
is complete, it follows that fxngis convergent, which proves the rst part of
the theorem.

8 Paula Homorodan
Now, if we denote  x= lim
n!1xn2Fix(T), then we will prove the unique-
ness. Indeed, suppose that Twould have two xed points y;z2Fix(T).
Additionally assumed that for any y;z2Xthere exists w2Xwhich is
comparable to yandzand is such that wTw.
Then we have two cases:
Case 1: Let yandzbe comparable. Then by (3.2), for all n1,
Tny=yis comparable to Tnz=zand hence by (3.1),
d(z;y) =d(Tz;Ty)a
[d(Tn1z;Tnz) +d(Tn1y;Tny)] = 0
Hencey=z.
Case 2: Ifyandzare not comparable, then there exists w2Xwhich
is comparable to yandz. Since T is nondecreasing, Tnwis comparable to
bothTny=yandTnz=z. From (3.1), for all n1,
d(z;Tnw) =d(Tnz;Tnw)a
[d(Tn1z;Tnz) +d(Tn1w;Tnw)]
d(Tn1z;Tnz) +d(Tn1w;Tnw)
Also, from (3.1), for all n1,
d(y;Tnw) =d(Tny;Tnw)a
[d(Tn1y;Tny) +d(Tn1w;Tnw)]
d(Tn1y;Tny) +d(Tn1w;Tnw)
Then from the above two inequalities
d(y;z) =d(y;Tnw) +d(Tnw;z)
d(Tn1y;Tny) +d(Tn1w;Tnw) +d(Tn1w;Tnw) +d(Tn1z;Tnz)
We assumed that wTw. IfThave a xed point then Tnw!x. Hence
d(Tn1w;Tnw)!0 asn!1 . Takingn!1 in the above inequality, we
haved(y;z)0 that isy=z. So xis the unique xed point of T 
Remark 3.1.
1.Letxn!xasn!1 . IfTis continous then the limit xof the Picard
sequence is always a xed point of T. Indeed, we have:
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x);
i.e., xis a xed point of T;
2.By(3.2) andxn!xasn!1 , we have thatfxngis a non-decreasing
sequence in Xwithxn!x. Suppose that Xhas the property, if a non-
decreasing sequence fxng!x, thenxnxfor alln0. This implies
that, for all n0,xnx. In view of xnx, substituting x= xand
y=xnin(3.1) , we get
d(Tx;Tx n)a
[d(x;Tx) +d(xn;Txn)]
or
d(Tx;xn+1)a
[d(x;Tx) +d(xn;xn+1)]
[d(x;Tx) +d(xn;xn+1)]
Takingn!1 , using
lim
n!1d(xn+1;xn) = 0

Theorems in H-distance space endowed with a partially ordered set 9
andxn!xasn!1 , we obtain d(x;Tx)a
d(x;Tx), which implies
thatd(x;Tx) = 0 , that is x=Tx. We proved that Thas a exd point
inX
3.In general, the limit xof the Picard sequence is not a xed point of T;
4.Ifdis actually a quasimetric, then Theorem 3.1reduces to the well
known Kannan xed point theorem in metric spaces [16].
Example 3.1. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
F-symmetric d, dened as: d(x;x) = 0 ,d(0;x) =d(x;0) = 0 ;
d(1;2n) =d(2n;1) = 2 ,n>0.
Note thatdis not a metric, because the triangle inequality is not satis-
ed: 1 =d(1;21)>d(1;0) +d(0;21) = 0 .
Now we consider the mapping T:X!X, where
T(x) =(
0;0x<1
2
1;1
2x1
Now we check the next cases:
1)x= 0;y= 21, when condition (3.1) witha=1
4, reduces to
d(T(0);T(21))1
4
[d(0;T(0)) +d(21;T(21))]() 01
2:
2)x= 0;y= 2n;n> 1, when condition (3.1) witha=1
4, reduces to
d(T(0);T(2n))1
4
[d(0;T(0)) +d(2n;T(2n))]() 00:
3)x= 0;y= 1, when condition (3.1) witha=1
4, reduces to
d(T(0);T(1))1
4
[d(0;T(0)) +d(1;T(1))]() 00:
4)x= 1;y= 21;, when condition (3.1) witha=1
4, reduces to
d(T(1);T(21))1
4
[d(1;T(1)) +d(21;T(21))]() 01
2:
5)x= 1;y= 2n;n> 1, when condition (3.1) witha=1
4, reduces to
d(T(1);T(2n))1
4
[d(1;T(1)) +d(2n;T(2n))]() 00:
6)x= 21;y= 2n;n> 1, when condition (3.1) witha=1
4, reduces to
d(T(21);T(2n))1
4
[d(21;T(21)) +d(2n;T(2n))]() 01
2:
Therefore,Tis a Kannan contraction, i.e.,
d(T(x);T(y))1
4
[d(x;T(x) +d(y;T(y)];x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0;g.
The next Corollary is a generalization of Kannan xed point theorem
in partially ordered metric spaces [16].

10 Paula Homorodan
Corollary 3.1. Let(X;)be a partially ordered set and suppose there is a H-
distancedonXsuch that (X;d)is a complete symmetric H-distance space.
LetT:X!Xbe a non-decreasing mapping for wich there exists 0<a<1
2
then:
d(Tx;Ty )a
[d(x;Tx ) +d(y;Ty ) ];8x;y2X
such that x and y are comparable.
Then:
(i) The Picard iteration at the point xis convergent.
(ii) If, additionally, the limit xof the Picard sequence is a xed point
ofTand existsx02Xwithx0Tx0then xis the unique xed point of T.
Proof. If (X;d) is a complete symmetric H-distance space, then it is a com-
pleteH-distance and conclusion follows by Theorem 3.1. 
Theorem 3.2. Let(X;)be a partially ordered set and suppose there is a
H-distancedonXsuch that (X;d)is a complete H-distance space.
LetT:X!Xbe a non-decreasing mapping for wich there exists 0<h< 1
then:
d(Tx;Ty )h
maxfd(x;Tx );d(y;Ty )g;8x;y2X: (3.6)
such that x and y are comparable.
Then the Picard iteration at the any point x2Xis convergent. If, addition-
ally, the limit xof the Picard sequence is a xed point of Tand existsx02X
withx0Tx0then xis the unique xed point of T.
Proof. Letx02Xand consider the Picard sequence fxng,xn+1=Txn;
n0. SinceTis a non decreasing function, we have
x0Tx0=x1T2x0=Tx1=x2T3x0 (3.7)



Tnx0=xnTn+1×0=xn+1



We prove that
d(xn;xn+1)h
maxfd(xn1;xn);d(xn;xn+1)g; n0: (3.8)
By (3.7) we see that, for all n1,xnandxn+1are comparable elements.
We have two cases:
Case 1.
If maxfd(xn1;xn);d(xn;xn+1)g=d(xn;xn+1) then by (3.8) it follows
d(xn;xn+1)h
d(xn;xn+1)<d(xn;xn+1), a contradiction.
Case 2.
If maxfd(xn1;xn);d(xn;xn+1)g=d(xn1;xn) then we obtain
d(xn;xn+1)h
d(xn1;xn): (3.9)
By (3.9), we have d(xn;xn+1)hn
d(x0;x1); n0
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1and
y:=xn1in (3.6). By (3.7) we see that x:=xn+p1andy:=xn1are
comparable and so we have

Theorems in H-distance space endowed with a partially ordered set 11
d(Txn+p1;Txn1)h
maxfd(xn+p1;xn+p);d(xn1;xn)g ()
d(xn+p;xn)h
maxfd(xn+p1;xn+p);d(xn1;xn)g;
and by using (3.9) we have
d(xn+p;xn)h
maxfhn+p1
d(x0;x1);hn1
d(x0;x1)g=
=h
d(x0;x1)
maxfhn+p1;hn1g=
=h
hn1
d(x0;x1) =hn
d(x0;x1); (3.10)
by which we immediately conclude that fxngis a Cauchy sequence. As ( X;d)
is complete, it follows that fxngis convergent, which proves the rst part of
the theorem.
Now, if we denote  x= lim
n!1xn2Fix(T), then we will prove the unique-
ness. Indeed, suppose that Twould have two xed points y;z2Fix(T).
Additionally assumed that for any y;z2Xthere exists w2Xwhich is
comparable to yandzand is such that wTw.
Then we have two cases:
Case 1: Let yandzbe comparable. Then by (3.7), for all n1,
Tny=yis comparable to Tnz=zand hence by (3.6),
d(z;y) =d(Tz;Ty)h
maxfd(Tn1z;Tnz);d(Tn1y;Tny)g= 0
Hencey=z.
Case 2: Ifyandzare not comparable, then there exists w2Xwhich
is comparable to yandz. Since T is nondecreasing, Tnwis comparable to
bothTny=yandTnz=z. From (3.6), for all n1,
d(z;Tnw) =d(Tnz;Tnw)h
maxf(.Tn1z;Tnz) +d(Tn1w;Tnw)g 
d(Tn1z;Tnz) +d(Tn1w;Tnw)
Also, from (3.6), for all n1,
d(y;Tnw) =d(Tny;Tnw)h
maxf(.Tn1y;Tny) +d(Tn1w;Tnw)g 
d(Tn1y;Tny) +d(Tn1w;Tnw)
Then from the above two inequalities
d(y;z) =d(y;Tnw) +d(Tnw;z)
d(Tn1y;Tny) +d(Tn1w;Tnw) +d(Tn1w;Tnw) +d(Tn1z;Tnz)
We assumed that wTw. IfThave a xed point then Tnw!x. Hence
d(Tn1w;Tnw)!0 asn!1 . Takingn!1 in the above inequality, we
haved(y;z)0 that isy=z. So xis the unique xed point of T 
Remark 3.2.

12 Paula Homorodan
1.Letxn!xasn!1 . IfTis continous then the limit xof the Picard
sequence is always a xed point of T. Indeed, we have:
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x);
i.e., xis a xed point of T;
2.By(3.7) andxn!xasn!1 , we have thatfxngis a non-decreasing
sequence in Xwithxn!x. Suppose that Xhas the property, if a non-
decreasing sequence fxng!x, thenxnxfor alln0. This implies
that, for all n0,xnx. In view of xnx, substituting x= xand
y=xnin(3.6) , we get
d(Tx;Tx n)h
maxfd(x;Tx) +d(xn;Txn)g
or
d(Tx;xn+1)h
maxfd(x;Tx) +d(xn;xn+1)g 
fd(x;Tx) +d(xn;xn+1)g
Takingn!1 , using
lim
n!1d(xn+1;xn) = 0
andxn!xasn!1 , we obtain d(x;Tx)h
d(x;Tx), which implies
thatd(x;Tx) = 0 , that is x=Tx. We proved that Thas a exd point
inX
3.In general, the limit xof the Picard sequence is not a xed point of T;
4.Ifdis actually a quasimetric, then Theorem 3.2reduces to the well
Bianchini xed point theorem in metric spaces [8].
Example 3.2. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
F-symmetric d, dened as: d(x;x) = 0 ,d(0;x) =d(x;0) = 0 ;
d(1;2n) =d(2n;1) = 11 ,n>0.
Note thatdis not a metric, because the triangle inequality is not satis-
ed: 1 =d(1;21)>d(1;0) +d(0;21) = 0 .
Now we consider the mapping T:X!X, where
T(x) =(
0;0x<1
2
1;1
2x1
Now we check the next cases:
1)x= 0;y= 21, when condition (3.6) withh=1
2, reduces to
d(T(0);T(21))1
2
maxfd(0;T(0)) +d(21;T(21))g () 011
2:
2)x= 0;y= 2n;n> 1, when condition (3.6) withh=1
2, reduces to
d(T(0);T(2n))1
2
maxfd(0;T(0)) +d(2n;T(2n))g () 00:
3)x= 0;y= 1, when condition (3.6) withh=1
2, reduces to
d(T(0);T(1))1
2
maxfd(0;T(0)) +d(1;T(1))g () 00:

Theorems in H-distance space endowed with a partially ordered set 13
4)x= 1;y= 21;, when condition (3.6) withh=1
2, reduces to
d(T(1);T(21))1
2
maxfd(1;T(1)) +d(21;T(21))g () 011
2:
5)x= 1;y= 2n;n> 1, when condition (3.6) withh=1
2, reduces to
d(T(1);T(2n))1
2
maxfd(1;T(1)) +d(2n;T(2n))g () 00:
6)x= 21;y= 2n;n> 1, when condition (3.6) withh=1
2, reduces to
d(T(21);T(2n))1
2
maxfd(21;T(21))+d(2n;T(2n))g () 011
2:
Therefore,Tis a Bianchini contraction, i.e.,
d(T(x);T(y))1
2
maxfd(x;T(x); d(y;T(y)g;x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0;g.
The next Corollary is a generalization of Bianchini xed point theorem
in partially ordered metric spaces [8].
Corollary 3.2. Let(X;)be a partially ordered set and suppose there is a H-
distancedonXsuch that (X;d)is a complete symmetric H-distance space.
LetT:X!Xbe a non-decreasing mapping for wich there exists 0<h< 1
then:
d(Tx;Ty )h
maxfd(x;Tx );d(y;Ty )g 8x;y2X
such that x and y are comparable.
Then:
(i) The Picard iteration at the point xis convergent.
(ii) If, additionally, the limit xof the Picard sequence is a xed point
ofTand existsx02Xwithx0Tx0then xis the unique xed point of T.
Proof. If (X;d) is a complete symmetric H-distance space, then it is a com-
pleteH-distance and conclusion follows by Theorem 3.2. 
Conclusion
By working in the general setting of a complete H-distance space endowed
with a partially ordered set, we obtained signicant generalizations of Kannan
and Bianchini xed point theorems in usual partially ordered metric spaces.
We note that, so far, we were not able to obtain a Banach type xed point
theorem in distance spaces which are not quasimetric spaces.

14 Paula Homorodan
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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