Existence of Fixed Point Theorems in Linear
2-Normed Spaces
R. Krishnakumar1, T. Mani2, D. Dhamodharan3 1
PG & Research Department of Mathematics, Urumu Dhanalakshmi College, Tamilnadu, India.
2
Department of Mathematics, Trichy Engineering College, Tamilnadu, India.
3
PG & Research Department of Mathematics, Jamal Mohammed College, Tamilnadu, India.
Abstract–In this paper, we discuss the concept of some existence of fixed point theorems for a pair of mappings in a setting of
2–normed spaces.
Keywords–Normed space, 2–normed space, Fixed point, Common fixed point.
I. INTRODUCTION
In 1963, S.Gahler [5] was initiated by 2–normed space & 2–Banach space. In 1975, K.Iseki was discussed the concept of the fundamental results on fixed point theorems of 2–metric space. Next Albert White and Y.J.Cho [1] investigate the important properties of linear mappings on linear 2–normed space in the year of 1984. Here after, many authors establish the fixed point theorem in 2–normed spaces and 2–Banach spaces, See [4,6,16–20]. Recently, generalise the concept of 2–normed space as well as 2–Banach space into 2–cone Banach space [3,12,22].
We now state some definitions before presenting our main results.
II. PRELIMINARIES
1) Definition 1. [4] Let X be a real linear space with dimension of X is greater than 1 and
||
.,.
||:
X
X
[
0
,
)
be a function. Then(i)
||
x
,
y
||
0
if and only ifx
andy
are linearly dependent, (ii)||
x
,
y
||
||
y
,
x
||
,(iii)
||
x
,
y
||
|
|
||
x
,
y
||
,(iv)
||
x
y
,
z
||
||
x
,
z
||
||
y
,
z
||
, where for allx
,
y
,
z
X
and
R
.If
||
.,.
||
is called a 2–norm and the pair(
X
,
||
.,.
||)
is called a linear 2–normed space. So a 2–norm||
x
,
y
||
always satisfies [24]||
,
||
||
,
||
x
y
x
x
y
for allx
,
y
X
and all scalars
.2) Definition 2. [4] A sequence
{
x
n}
in X is convergent to an elementx
X
,
if for eacha
X
,
lim
||
,
||
0
x
nx
a
n . If
}
{
x
n converges tox
we writex
n
x
asn
.3) Definition 3. [4] A sequence
{
x
n}
in X is said to be a Cauchy sequence if for eacha
X
,
lim
||
x
n
x
m,
a
||
0
as
m
n
,
.4) Definition 4. [4] A complete 2–normed space is one in which every Cauchy sequence in X converges to an element of X. A complete 2–normed space X is called 2–Banach space.
5) Definition 5. [16] Let
(
X
,
||
.,.
||)
be a linear 2–normed space. Then the mappingT
:
X
X
is said to be a contraction if there existsk
[
0
,
1
)
such that||
Tx
Ty
,
z
||
k
||
x
y
,
z
||
for allx
,
y
,
z
X
.6) Definition 6. [11] A function
:
[
0
,
)
[
0
,
)
is called an altering distance function if the following properties are satisfied:7) Definition 7. [14] An ultra-altering distance function is a continuous, non-decreasing mapping
:
[
0
,
)
[
0
,
)
such that)
,
0
[
,
0
)
(
t
t
and
(
0
)
0
.We denote this set with
u .Delbosco [2] and Skof [23] have proved a fixed point theorem for self-maps of complete normed spaces by introducing a class
of functions
:
[
0
,
)
[
0
,
)
satisfying the following conditions:a)
:
[
0
,
)
[
0
,
)
is continuous in R and strictly increasing in R. b)
(
t
)
0
if and only ift
0
.c)
(
t
)
Mt
for everyt
0
,
0
are constants.III. MAIN RESULT
1) Theorem 1. Let T be a self-map of a complete 2–normed space
(
X
,
||
.,.
||)
and
satisfying (i) and (ii).Furthermore, let
f
,
g
,
h
be three decreasing functions from R into[
0
,
1
)
such thatf
(
t
)
2
g
(
t
)
h
(
t
)
1
for every0
t
. Suppose T satisfies the following condition
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
u
T
y
u
T
x
u
y
x
h
u
T
y
u
T
x
u
y
x
g
u
y
x
u
y
x
f
u
Ty
Tx
x y
y x
(1)
where
x
,
y
,
u
X
, each two ofx
,
y
andu
are distinct. Then T has an unique fixed point. Proof.Let us takex
0be arbitrary point in X.define
x
n1
Tx
n;
n
0
,
1
,
2
,...
and
n
||
x
n
x
n1,
u
||
forn
0
,
1
,
2
,...;
and
n
n
. Then we have
1
1
nn
||
x
n1
x
n2,
u
||
||
Tx
n
Tx
n1,
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 1
1
1 1
1 1
1
u
Tx
x
u
Tx
x
u
x
x
h
u
Tx
x
u
Tx
x
u
x
x
g
u
x
x
u
x
x
f
n n
n n n
n
n n
n n n
n n
n n
n
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 1 2
1
2 1
1 1
1 1
u
x
x
u
x
x
u
x
x
h
u
x
x
u
x
x
u
x
x
g
u
x
x
u
x
x
f
n n n
n n
n
n n n
n n
n n
n n
n
f
(
n)
(
n)
g
(
n)
n
n1
(2)
implies n
n n n
n
g
g
f
)
(
1
)
(
)
(
1
.
Since
f
(
t
)
2
g
(
t
)
h
(
t
)
1
,f
(
n)
2
g
(
n)
1
which implies1
)
(
1
)
(
)
(
n n n
g
g
f
.
If we set
)
(
1
)
(
)
(
n n n
g
g
f
r
then from (2) we get
n1
r
n wherer
1
.So
n
r
n
0, such that
n
0
asn
.Thus
n
. Then
n
n
, since
is continuous. So
0
and hence by (ii),
0
implies0
n
.Now show that
{
x
n}
is a Cauchy sequence.We prove it by contradiction. Then for every positive integer
and for every positive integerk
there exist two positive integers)
(
k
m
andn
(
k
)
such that)
(
)
(
k
m
k
n
k
and||
x
m(k)
x
n(k),
u
||
(3) For each integer
k
letm
(
k
)
be the least integer for whichm
(
k
)
n
(
k
)
k
,
,
||
||
x
n(k)x
m(k) 1u
and||
x
n(k)
x
m(k),
u
||
Then we have
||
,
||
x
n(k)
x
m(k)u
||
x
n(k)
x
m(k),
x
m(k)1||
||
x
n(k)
x
m(k)1,
u
||
||
x
m(k)1
x
m(k),
u
||
(4) Now by (1), we have
||
x
n(k)
x
m(k),
x
m(k)1||
||
Tx
n(k)1
Tx
m(k)1,
x
m(k)1||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) (
k m k n k m k m k m k n k m k m k n k m k m k m k m k n k n k m k m k n k m k m k n k m k m k nx
Tx
x
x
Tx
x
x
x
x
h
x
Tx
x
x
Tx
x
x
x
x
g
x
x
x
x
x
x
f
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) (
k m k n k m k m k m k n k m k m k n k m k m k m k m k n k n k m k m k n k m k m k n k m k m k nx
Tx
x
x
Tx
x
x
x
x
h
x
Tx
x
x
Tx
x
x
x
x
g
x
x
x
x
x
x
f
0
which implies by (ii)
||
x
n(k)
x
m(k),
x
m(k)1||
0
.(5) So by (4) and (5) we get,
||
x
n(k)
x
m(k),
u
||
0
m(k)1. Since{
n}
converges to 0,
,
||
||
x
n(k)x
m(k)u
ask
. Again||
,
||
||
,
||
||
,
||
||
,
||
x
n(k)1
x
m(k)u
x
n(k)1
x
m(k)x
n(k)
x
n(k)1
x
n(k)u
x
n(k)
x
m(k)u
n(k)
||
x
n(k)
x
m(k),
u
||
,since,
||
x
n(k)1
x
m(k),
x
n(k)||
can be made 0 as we have done in equation (5). So||
x
n(k)1
x
m(k),
u
||
n(k)
||
x
n(k)
x
m(k),
u
||
ask
. In the similar way,||
,
||
||
,
||
||
,
||
||
,
||
x
n(k)2
x
m(k)u
x
n(k)2
x
m(k)x
n(k)1
x
n(k)2
x
n(k)1u
x
n(k)1
x
m(k)u
n(k)1
||
x
n(k)1
x
m(k),
u
||
,since
||
x
n(k)2
x
m(k),
x
n(k)1||
can be made 0 as we have done in equation (5).So
||
x
n(k)2
x
m(k),
u
||
n(k)1
||
x
n(k)1
x
m(k),
u
||
ask
and in similar fashion we can show
,
||
||
x
n(k)2
x
m(k)1,
u
||
||
Tx
n(k)1
Tx
m(k),
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 ) ( ) ( ) ( 1 ) ( ) ( 1 ) ( ) ( ) ( 1 ) ( 1 ) ( ) ( 1 ) ( ) ( 1 ) ( ) ( 1 ) (u
Tx
x
u
Tx
x
u
x
x
h
u
Tx
x
u
Tx
x
u
x
x
g
u
x
x
u
x
x
f
k n k m k m k n k m k n k m k m k n k n k m k n k m k n k m k n
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
2 ) ( ) ( 1 ) ( 1 ) ( ) ( 1 ) ( 1 ) ( ) ( 2 ) ( 1 ) ( ) ( 1 ) ( ) ( 1 ) ( ) ( 1 ) (u
x
x
u
x
x
u
x
x
h
u
x
x
u
x
x
u
x
x
g
u
x
x
u
x
x
f
k n k m k m k n k m k n k m k m k n k n k m k n k m k n k m k n
Letting
k
, we get
(
)
a
(
)
(
)
c
(
)
(
)
a
(
)
c
(
)
(
)
(
)
which is a contradiction. So
{
x
n}
is a Cauchy sequence. Since X is complete2–normed space,x
nz
X
n
lim
.Claim: Show that
Tz
z
. Again using (1) we have,
||
x
n(k)1
Tz
,
u
||
||
Tx
n(k)
Tz
,
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (u
Tx
z
u
Tz
x
u
z
x
h
u
Tz
z
u
Tx
x
u
z
x
g
u
z
x
u
z
x
f
k n k n k n k n k n k n k n k n
implies
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 ) ( ) ( ) ( 1 ) ( ) ( ) ( 1 ) ( ) ( 1 ) (u
x
z
u
Tz
x
u
z
x
h
u
Tz
z
u
x
x
u
z
x
g
u
z
x
u
z
x
f
u
Tz
x
k n k n k n k n k n k n k n k n k n
passing limit as
n
on both sides of the inequality we get,
||
z
Tz
,
u
||
0
which gives by (ii),||
z
Tz
,
u
||
0
i.e.,
Tz
z
.Next let
w
be another fixed point of T. Then
||
z
w
,
u
||
||
Tz
Tw
,
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
u
Tz
w
u
Tw
z
u
w
z
h
u
Tw
w
u
Tz
z
u
w
z
g
u
w
z
u
w
z
f
f
(||
z
w
,
u
||)
h
(||
z
w
,
u
||)
(||
z
w
,
u
||)
||)
,
(||
z
w
u
, Sincef
(
t
)
h
(
t
)
1
which is a contradiction leads to the fact that
z
w
and thus completes the proof. Next we verify the Theorem (1) by a proper example.a) Example 1. Let X R R and d be a 2–normed which expresses
||
x
y
,
u
||
as the area of the Euclidean triangle with verticesx
(
x
1,
x
2),
y
(
y
1,
y
2)
andu
(
u
1,
u
2)
. Then(
X
,
||
.,.
||)
is a complete 2–normed space[1].Now take
x
(
1
,
0
),
y
(
2
,
0
)
andu
(
1
,
1
)
also letT
:
X
X
be a mapping such that)
0
,
2
(
Tx
wherex
(
1
,
0
)
X
and)
0
,
3
(
Ty
wherey
(
2
,
0
)
X
Now setting
6
1
)
(
,
5
1
)
(
,
5
2
)
(
t
g
t
h
t
f
and(
t
)
t
2;
t
R
We observe that all the conditions of Theorem (1) satisfied except the condition (1). Also it is very clear that T has no fixed point in X in this case.
Next we establish a common fixed point theorem in this line.
2) Theorem 2. Let S and T be self-mappings of a complete 2–normed space
(
X
,
||
.,.
||)
and
satisfying (i) and (ii). Furthermore, letf
,
g
,
h
be three decreasing functions from Rinto[
0
,
1
)
such thatf
(
t
)
2
g
(
t
)
h
(
t
)
1
for every0
t
. Suppose S and T satisfies the following condition(6)
where
x
,
y
,
u
X
, each two ofx
,
y
andu
are distinct. Then S and T have a unique common fixed point in X. Proof. Letx
0
X
be arbitrary.define
x
2n
Sx
2n1 andx
2n1
Tx
2n ;n
0
,
1
,
2
,...,
also let
n
||
x
n
x
n1,
u
||
forn
0
,
1
,
2
,...;
and
n
n
.We also assume that
n
0
for everyn
. Now for an even integern
, we have
n
n
||
x
n
x
n1,
u
||
||
Sx
n1
Tx
n,
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 1
1
1 1
1
1 1
u
Sx
x
u
Tx
x
u
x
x
h
u
Tx
x
u
Sx
x
u
x
x
g
u
x
x
u
x
x
f
n n n
n n
n
n n n
n n
n
n n n
n
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 1 1
1 1
1
1 1
u
x
x
u
x
x
u
x
x
h
u
x
x
u
x
x
u
x
x
g
u
x
x
u
x
x
f
n n n
n n
n
n n n
n n
n
n n n
n
f
(
n1)
(
n1)
g
(
n1)
n1
n
implies 1
1 1 1
)
(
1
)
(
)
(
nn n n
n
g
g
f
(7)
Since
f
(
t
)
2
g
(
t
)
h
(
t
)
1
,f
(
n1)
2
g
(
n1)
1
which implies1
)
(
1
)
(
)
(
1 1
1
n n n
g
g
f
If we set
)
(
1
)
(
)
(
1 1 1
n n n
g
g
f
r
then from (7) we get
n
r
n1wherer
1
. So
n
0n
r
, such that
n
0
asn
.Since
n
n1and
is strictly increasing,
n
n1,
n
1
,
2
,...
.Thus
n
. Then
n
n
, since
is continuous. So
0
and hence by (ii),
0
implies0
n
.Now show that
{
x
}
is a Cauchy sequence.
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
u
Sx
y
u
Ty
x
u
y
x
h
u
Ty
y
u
Sx
x
u
y
x
g
u
y
x
u
y
x
f
u
Ty
Sx
We prove it by contradiction. Then for every positive integer
and for every positive integerk
there exist two positive integers)
(
2
p
k
and2
q
(
k
)
such that)
(
2
)
(
2
q
k
p
k
k
and||
x
2p(k)
x
2q(k),
u
||
.(8) For each integer
k
let2
p
(
k
)
be the least integer for which2
p
(
k
)
2
q
(
k
)
k
,
,
||
||
x
2q(k)x
2p(k) 2u
and||
x
2q(k)
x
2p(k),
u
||
Then we have||
,
||
||
,
||
||
,
||
||
,
||
x
2q(k)
x
2p(k)u
x
2q(k)
x
2p(k)x
2p(k) 2
x
2q(k)
x
2p(k) 2u
x
2p(k) 2
x
2p(k)u
Since we can easily show that
||
x
2q(k)
x
2p(k),
x
2p(k)2||
0
as we have shown in (5) of Theorem (1).||
,
||
||
,
||
||
,
||
x
2q(k)
x
2p(k)u
x
2q(k)
x
2p(k) 2u
x
2p(k) 2
x
2p(k)u
||
x
2q(k)
x
2p(k)2,
u
||
||
x
2p(k)2
x
2p(k),
x
2p(k)1||
||
x
2p(k)2
x
2p(k)1,
u
||
||
x
2p(k)1
x
2p(k),
u
||
again we can show like (5) of Theorem (1),
0
||
,
||
x
2p(k)2
x
2p(k)x
2p(k)1
. Thus1 ) ( 2 2 ) ( 2 )
( 2 ) (
2
,
||
0
||
x
q kx
p ku
p k
p k
.(9) Since
{
n}
converges to 0,||
x
2q(k)
x
2p(k),
u
||
.Now
||
x
2q(k)
x
2p(k)1,
u
||
||
x
2q(k)
x
2p(k)1,
x
2p(k)||
||
x
2q(k)
x
2p(k),
u
||
||
x
2p(k)
x
2p(k)1,
u
||
) ( 2 )
( 2 ) (
2
,
||
||
x
q k
x
p ku
p k
since we can show that
||
x
2q(k)
x
2p(k)1,
x
2p(k)||
0
as we have done in (5) of Theorem (1).So,
||
x
2q(k)
x
2p(k)1,
u
||
ask
. (10) Again||
,
||
||
,
||
||
,
||
||
,
||
x
2q(k)
x
2p(k)2u
x
2q(k)
x
2p(k)2x
2p(k)1
x
2q(k)
x
2p(k)1u
x
2p(k)1
x
2p(k)2u
||
,
||
||
,
||
x
2q(k)
x
2p(k)1u
x
2p(k)1
x
2p(k)2u
,(since
||
x
2q(k)
x
2p(k)2,
x
2p(k)1||
0
for similar reason as of (5) of Theorem (1).)||
,
||
||
,
||
||
,
||
||
,
||
2 ) ( 2 1 ) ( 2
1 ) ( 2 ) ( 2 )
( 2 ) ( 2 )
( 2 1 ) ( 2 ) ( 2
u
x
x
u
x
x
u
x
x
x
x
x
k p k
p
k p k
p k
p k q k
p k
p k
q
1 ) ( 2 ) ( 2 )
( 2 ) (
2
,
||
||
0
x
q kx
p ku
p k
p kwhich gives
||
x
2q(k)
x
2p(k)2,
u
||
ask
. (11) Similarly,
,
||
||
x
2q(k) 1x
2p(k) 2u
ask
. (12) Now from (6) we have
||
x
2p(k)2
x
2q(k)1,
u
||
||
Sx
2p(k)1
Tx
2q(k),
u
||
||
,
||
min
||
,
||
,
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
||
,
||
1 ) ( 2 ) ( 2 )
( 2 1 ) ( 2 )
( 2 1 ) ( 2
) ( 2 ) ( 2 1
) ( 2 1 ) ( 2 )
( 2 1 ) ( 2
) ( 2 1 ) ( 2 )
( 2 1 ) ( 2
u
Sx
x
u
Tx
x
u
x
x
h
u
Tx
x
u
Sx
x
u
x
x
g
u
x
x
u
x
x
f
k p k
q k
q k
p k
q k
p
k q k
q k
p k
p k
q k
p
k q k
p k
q k
p
passing limit as
