Fixed Points Results via Iterates of Four Maps in TVS-valued Cone Metric Spaces

Fixed Points Results via Iterates of Four Maps in

TVS-valued Cone Metric Spaces

Muhammad Arshad and Akbar Azam

Abstract—: We obtain common fixed points of four mappings satisfying a contractive type condition by demonstrating the convergence of their sequence of iterates in TVS valued cone metric spaces. Our results generalize some well-known recent results in the literature.

Index Terms— contractive type mapping; non-normal cones; fixed point; cone metric space.

I. INTRODUCTIONANDPRELIMINARIES Fixed point theorems are very important tools for providing evidence of the existence and uniqueness of solutions to various mathematical models (i.e., differential, integral and partial differential equations) representing phenomena happening in different fields, such as steady state temperature distribution, chemical equations, economic theories, financial analysis and biomedical research. The literature of the last four decades flourishes with results which discover fixed points of self and nonself nonlinear operators in a metric space. For most of them, their reference result is the Banach contraction theorem, which states that if

X

is a complete metric space and

T

a single valued contractive self mapping on then

T

has a unique fixed point in This theorem looks simple but plays a fundamental role in fixed point theory and has become even more important because being based on iteration, it can be easily implemented on a computer. Common fixed point theorems deals with the guarantee that a family

{

of self mappings on a set

,

X

.

X

}

Ω

∈

i

T

i

:

X

has

one or more common fixed points, i.e., the system of functional equations has one or more simultaneous solutions. Recently Beg, Azam and Arshad [4], studied common fixed points of a pair of maps on topological vector space(TVS) valued cone metric spaces which is bigger than that of introduced by Huang and Zhang [5]. In this paper we obtain common fixed points of a pair of mappings satisfying Banach contractive condition without the assumption of normality in TVS-valued cone metric spaces. Our results improve and generalize some significant recent results(e.g., see [1,5,8,10] ).

)

(

∈

Ω

=

T

x

i

x

_i

Manuscript received March 2, 2010, accepted March 14, 2010. This work was supported by Higher Education Commission (HEC), Pakistan. M. Arshad is with the Department of Mathematics, International Islamic University, H-10, Islamabad, 44000, Pakistan (Phone: 0092-3335103984, Fax: 0092-51-226877, e-mail: [email protected]).

A. Azam is with the Department of Mathematics, COMSATS Institute of Information Technology, Chak Shahzad, Islamabad, 44000, Pakistan (e-mail: [email protected]).

Let

(

E

,

τ

)

be always a topological vector space (TVS) and

P

a subset of

E

. Then,

P

is called a cone whenever

(i)

P

is closed, non-empty and (ii)

},

0

{

≠

P

P

by

ax

+

∈

for all and non-negative real numbers ,

(iii)

P

y

x

,

∈

b

a

,

}.

0

{

)

(

−

=

∩

P

P

For a given cone

P

⊆

E

,

we can define a partial ordering

≤

with respect to

P

by

x

≤

y

if and only if

.

P

x

y

−

∈

x

<

y

will stand for

x

≤

y

and

x

≠

y

,

while

x

<<

y

will stand for where

denotes the interior of

,

int

P

x

y

−

∈

P

int

P

.

Definition 1: Let

X

be a non-empty set. Suppose the mapping

d

:

X

×

X

→

E

satisfies



d

1



0

≤

d

(

x

,

y

)

for all

x

,

y

∈

X

and

0

)

,

(

x

y

=

d

if and only if

x

=

y

,

)

,

(

)

,

(

)

(

d

₂

d

x

y

=

d

y

x

for all



,

,

y

X

x

∈

d

3



d

(

x

,

y

)

≤

d

(

x

,

z

)

+

d

(

z

,

y

)

for all

.

,

,

y

z

X

x

∈

Then

d

is called a TVS-valued cone metric on

X

and is called a TVS-valued cone metric space.

)

,

(

X

d

If

E

is a real Banach space then is called cone metric space [3].

)

,

(

X

d

Definition 2: Let be a TVS-valued cone metric

space,

)

,

(

X

d

X

x

∈

and a sequence in Then



converges to

1

}

{

x

_{n n≥}

X

.

i



_{

_x

_n

_}

_n≥1

_x

_{whenever for every}

E

c

∈

with

0

<<

c

there is a natural number

N

such that

d

(

x

_n

,

x

)

<<

c

for all

n

≥

N

.

We denote this by or



is a Cauchy sequence whenever for every

x

x

_n n→∞

=

lim

x

_n

→

x

.

ii



{

x

_n

}

_n≥1

E

c

∈

with

0

<<

c

there is a natural number

N

such that

d

(

x

n

,

x

m

)

<<

c

for all



is a complete cone metric space if every Cauchy sequence is convergent.

.

,

m

N

n

≥

A pair of self-mappings on a TVS-valued cone metric space is said to be compatible if, for any

sequence with and for

arbitrary

c

there exists a natural number

(

f

,

T

)

)

X

X

x

n

∈

fx

n

→

t

,

Tx

n

→

t

,

,

int

P

∈

n

0

such that

.

all

for

;

)

;

(

fTx

Tfx

c

n

n

₀

d

_{n n}

<<

>

The pair is said to be weakly compatible if they commute at their coincidence point (i.e.,

whenever ). A self-mapping on a

(

f

,

T

Tfx

fTx

=

Tx

fx

=

T

TVS-valued cone metric is called continuous at a point

X

∈

0

x

X

if, for every sequence implies

→

,

X

x

n

∈

0

x

x

_n

→

Tx

_n

Tx

₀

II. MAIN RESULTS

Theorem 3: Let be a complete TVS-valued cone

metric space and the mappings satisfy:

)

,

(

X

d

X

X

g

f

T

S

,

,

,

:

→

d



Sx

,

Ty



≤



d



fx

,

gy



for all

x

,

y

∈

X

where

0

≤

λ

<

1

.

If

SX

⊆

gX

,

TX

⊆

fX

,

is continuous,

(

is compatible and

is weakly compatible, then and have a unique common fixed point.

f

S

,

f

)

(

T

,

g

)

,

,

T

S

f

g

Proof: Let be an arbitrary point in Choose a point in

0

x

X

,

1

x

X

such that

y

₁

=

gx

₁

=

Sx

₀

.

This can be done since Similarly, choose a point in

.

gX

SX

⊆

x

₂

X

such that Continuing this process and having chosen in We obtain

.

1 2

2

fx

Tx

y

=

=

n

x

X

.

x

n1

in

X

such that

y

2k1



gx

2k1



Sx

2k

y

2k2



fx

2k2



Tx

2k1

,

k



0, 1, 2, . . . .

Then,

d



gx

2k1

,

fx

2k2





d



Sx

2k

,

Tx

2k1



≤



d



fx

2k

,

gx

2k1



.

Similarly,

).

,

(

)

,

(

)

,

(

)

,

(

1 2 2 2

1 2 2 2

1 2 2 2 3

2 2 2

+ + +

+ + +

+

≤

≤

=

k k

k k

k k k

k

gx

fx

d

gx

fx

Tx

Sx

d

gx

fx

d

λ

λ

Now by induction, we obtain for each

k

=

0

,

1

,

2

,...,

d



fx

2k2

,

gx

2k3



≤



2k2

d



fx

0

,

gx

1



.

For all

n

,

we have

d



y

n1

,

y

n2



≤



d



y

n

,

y

n1



≤

. . .

≤



n1

_d

_

_y

0

,

y

1



.

It follows that for

m

>

n

,

[

]

).

,

(

1

)

,

(

...

)

,

(

...

)

,

(

)

,

(

)

,

(

1 0

1 0 1 1

1

2 1 1

y

y

d

y

y

d

y

y

d

y

y

d

y

y

d

y

y

d

n

m n

n

m m

n n n

n n

m

⎥

⎦

⎤

⎢

⎣

⎡

−

≤

+

+

+

≤

+

+

+

≤

− +

−

+ + +

λ

λ

λ

λ

λ

Let

0

<<

c

be given, choose a symmetric neighborhood of such that

V

0

c

+

V

⊆

intP Also, choose a natural number

N

1 such that

[ ]

₁

d

(

y

₀

,

y

₁

)

V

,

n

∈

−λ λ

for all Then,

.

1

N

n

≥

₁λ₋n_λ

d

(

y

₁

,

y

₀

)

<<

c

,

for all Thus,

.

1

N

n

≥

,

)

,

(

1

)

,

(

y

y

d

y

₀

y

₁

c

d

n

n

m

⎥

<<

⎦

⎤

⎢

⎣

⎡

−

≤

λ

λ

for all Therefore, is a Cauchy sequence. Since

.

n

m

>

{

y

n

}

n≥1

X

is complete, there exists

z

∈

X

such that

y

n

→

z

.

For its subsequence we obtain,

.

and

,

,

2 2 2 2 1

1

2

z

fx

z

Sx

z

Tx

z

gx

k+

→

k+

→

k

→

k+

→

Since

f

is continuous therefore

.

and

,

,

1 2

2 2

2 1

2

fz

fTx

fz

fSx

fz

ffx

fz

fgx

k

k k

k

→

→

→

→

+

+ +

As

fSx

_2k

→

fz

and

(

S

,

f

)

is compatible, for arbitrary

c

∈

int

P

,

there exists a natural numbers and such that

1

n

2

n

(

2 2

)

for

all

1

2

,

fSx

c

k

n

Sfx

d

k k

<<

>

(

)

for

all

.

2

,

2

2

k

n

c

fz

fSx

d

k

<<

>

It implies that

(

) (

) (

)

{

,

}

,

max

all

for

2

2

,

,

,

2 1 2

2 2 2

n

n

k

c

c

c

fz

fSx

fSx

Sfx

fz

Sfx

_{k k k k}

>

=

+

<<

+

≤

that is Choose natural numbers and such that

.

2

fz

Sfx

_k

→

,

,

,

2 3

1

N

N

N

N

₄

(

) (

)

for

all

,

4

1

,

Sfx

₂

c

k

N

₁

fz

d

_k

<<

−

λ

≥

(

)

(

)

(

2 1

)

(

)

3

2 2

all

for

4

1

,

,

all

for

4

1

,

N

k

c

gx

z

d

N

k

c

fz

ffx

d

k k

≥

−

<<

≥

−

<<

+

λ

λ

and

(

) (

)

for

all

.

4

1

,

4

1

2

k

N

c

z

Tx

d

k

≥

−

<<

+

λ

For , by performing a

simple calculation we can have

{

1

,

2

,

3

,

4

max

N

N

N

N

k

≥

}

(

,

)

<<

,

for

all

m

≥

1

.

m

c

z

fz

d

So, _mc

−

d

(

fz

,

z

)

∈

P

,

for all

m

≥

1

.

Since

(

0

→

m c

as and is closed,

But Therefore,

. Hence To assert choose natural numbers such that

)

∞

→

m

P

.

)

,

(

gu

Su

P

d

∈

−

d

(

fz

,

z

)

∈

P

.

(

fz

z

)

=

0

d

,

fz

=

z

.

Sz

=

z

,

,

,

5

4

N

N

(

2 1

)

for

all

5

2

,

gx

c

k

N

z

d

_k+

<<

≥

λ

and

(

)

for

all

.

2

,

6

1

2

k

N

c

z

Tx

d

k+

<<

≥

Now

,

for

k

≥

max

{

N

₅

,

N

₆

}

we have

d



Sz

,

z



≤

d



Sz

,

Tx

2k1





d



Tx

2k1

,

z



≤



d



fz

,

gx

2k1





d



Tx

2k1

,

z



≤



d



z

,

gx

2k1





d



Tx

2k1

,

z



≤



c

2





c

2



c

.

By a similar argument (as in the proof of ), we

have Now we shall show As

there exists such that

z

fz

=

.

z

Sz

=

Tz

=

gz

.

,

gX

SX

⊆

u

∈

X

.

gu

Sz

z

=

=

Since,

(

) (

)

(

)

( )

,

0

,

,

,

,

=

≤

≤

≤

z

z

d

fz

gu

d

Sz

Tu

gu

Tu

d

λ

d

λ

therefore

z

=

Tu

=

gu

and weakly compatibility of and implies that

T

g

gz



gTu



Tgu



Tz

.

To see

Tz

=

z

,

consider,

d



Tz

,

z





d



Tz

,

Sz



≤



d



gz

,

fz







d



Tz

,

z

.

It yields

Tz

=

z

,

which further implies

gz

=

z

.

Hence,

z



gz



fz



Sz



Tz

.

Example 4: Let

X

=

[ ]

0

,

1

and

E

be the set of all real valued functions on

X

which also have continuous derivatives on Then is a vector space over under the following operations:

.

X

E

R



x



y



t





x



t





y



t



,





x



t







x



t



,

for all

x

,

y

∈

E

,

α

∈

R

.

Let

τ

be the strongest vector (locally convex) topology on

E

.

Then

( )

E

,

τ

is a topological vector space which is not normable and is not even metrizable. Define

d

:

X

×

X

→

E

as follows:



d



x

,

y



t





|

x

−

y

|

e

t

_,

( )

0

for

all

}.

:

{(

x

E

x

t

t

X

P

=

∈

≥

∈

Then

(

X

,

d

)

is a TVS-valued cone metric space. Let be such that

X

X

g

f

T

S

,

,

,

:

→

.

8

and

5

,

16

,

10

x

gx

x

fx

x

x

Tx

x

x

Sx

=

=

+

=

+

=

x

x



10

−

y

y



16

e

t



x



y



16



−

y



x



10





x



10



y



16



e

t

≤

16

x

−

10

y

160

e

t

≤

1

2

x

5

−

y

8

e

t

_,

that is

(

)( )

(

,

)( )

for

all

.

2

1

,

Ty

t

d

fx

gy

t

t

X

Sx

d

≤

∈

It suffices to assume

λ

=

₂1

,

_{in order to satisfy all}

assumptions of the above theorem.

If in Theorem 3, we choose

S

=

T

and , we obtain the following corollary.

g

f

=

Corollary 5: Let be a complete TVS-valued cone metric space and the mappings

satisfy:

)

,

(

X

d

X

X

f

T

:

→

d



Tx

,

Ty



≤



d



fx

,

fy



for all

x

,

y

∈

X

where

0

≤

λ

<

1

.

If

TX

⊆

fX

,

is continuous, is compatible then

T

and

have a unique common fixed point.

f

(

T

,

f

)

f

In the following result continuity of is not required whereas completeness of

f

X

is replaced with the completeness of

TX

or

fX

.

Theorem 6: Let be a TVS-valued cone metric space and the mappings satisfy:

)

,

(

X

d

X

X

f

T

:

→

d



Tx

,

Ty



≤



d



fx

,

fy



for all

x

,

y

∈

X

where

0

≤

λ

<

1

.

If

TX

⊆

fX

,

TX

or

fX

is complete,

(

T

,

f

)

is compatible then

T

and

f

have a unique common fixed point.

Proof: As in the proof of Theorem 3, construct a Cauchy

sequence

{

fx

_n

}

_n≥1 in

fX

such that

y

2k1



fx

2k1



Tx

2k

y

2k2



fx

2k2



Tx

2k1

,

k



0, 1, 2, . . . .

By completeness of

fX

,

there exists

u

,

v

∈

X

such that

y

_n

→

v

=

fu

(this holds also if

TX

is complete with

v

∈

TX

)

. Choose a natural number

N

such that

.

all

for

2

)

,

(

y

v

c

n

N

d

_n

<<

≥

Hence, for all

n

≥

N

.

2

2

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

1 2 2

2

1 2 2

2

1 2 2

2

2 2 2

2

c

c

c

v

y

d

y

v

d

fu

fx

d

y

v

d

Tu

Tx

d

y

v

d

Tu

y

d

y

fu

d

Tu

fu

d

n n

n n

n n

n n

=

+

<<

+

≤

+

≤

+

≤

+

≤

+ +

+ +

+ +

+ +

λ

Thus,

.

1

all

for

,

)

,

(

<<

m

≥

m

c

Tu

fu

d

So, _mc

−

d

(

fu

,

Tu

)

∈

P

,

for all

m

≥

1

.

Since

)

(

0

→

∞

→

as

m

m

c

and

P

is closed,

,

)

,

(

fu

Su

P

d

∈

−

but

P

∩

( )

−

P

=

{

0

}.

Therefore,

0

)

,

(

fu

Tu

=

d

. Hence

v



fu



Tu

.

By weakly compatibility of

(

T

,

f

)

,

we have

Tv



Tfu



fTu



fv

.

Then

d



Tv

,

v





d



Tv

,

Tu



≤



d



fv

,

fu







d



Tv

,

v

.

Thus

v

is a unique common fixed point of

T

and

f

.

Corollary 7: Let be a TVS-valued cone metric space and the mappings satisfy:

)

,

(

X

d

X

X

f

T

:

→

)

,

(

)

,

(

and

d

T

x

T

y

d

fx

fy

fX

TX

⊆

n n

≤

λ

for all

x

,

y

∈

X

where

0

≤

λ

<

1

.

Then

T

and have a unique common fixed point, if

Tf

and one

f

fT

=

of the following conditions is satisfied:

•

fX

is complete

•

TX

is complete.

Proof : By Corollary 5 and Theorem 6, we obtain such that

X

v

∈

fv



T

n

v



v

.

The result then follows from the fact that

).

,

(

)

,

(

)

,

(

)

,

(

)

,

(

)

,

(

v

Tv

d

fv

Tfv

d

fv

fTv

d

v

T

Tv

T

d

v

T

v

TT

d

v

Tv

d

n n n n

λ

λ

λ

=

=

≤

=

=

Example 8: Let

X

=

C

([

1

,

3

],

R

),

( )

E

,

τ

is the topological vector space of Example 4. Define

as follows:

E

X

X

d

:

×

→

d



x

,

y



t





s∈1,3

sup |

x



s



−

y



s

|

e

t

( )

?

0

for

all

}.

:

{(

x

E

x

t

t

X

P

=

∈

∈

Then is a TVS-valued cone metric space. Define

by

(

X

,

d

)

X

X

T

:

→

( )

( )

(

( )

)

( )

( ) ( )

.

,

4

2 1

1

s

x

s

x

f

du

e

u

u

x

s

x

T

u

s

=

+

+

=

−

∫

For

x

,

y

∈

X

( )

[ ]

( )

( )

[ ]

(

( ) ( )

)

( )

,

.

2

sup

sup

)

,

(

2

2

3 , 1 3

1 3 , 1

t

t

s

t

s

e

y

x

d

e

e

du

e

u

y

u

x

e

s

Ty

s

Tx

t

Ty

Tx

d

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

≤

⎟

⎠

⎞

⎜

⎝

⎛

₋

=

∈ ∈

∫

Similarly,

d



T

n

_x

_,

_T

n

_y

_

_e

t

_

_≤

_e

2n

2

n

n

!

d



x

,

y



e

t

_

_.

Note that

.

38

if

100

53

!

2

2

₌

₌

n

n

e

n n

Thus for

λ

=

₁₀₀53

,

_n

=

38

,

all conditions of Corollary 7

are satisfied and so

T

has a unique fixed point, which is

the unique solution of the integral equation:

( )

s

(

x

( )

u

u

)

e

du

x

u

s

1 2

1

4

+

+

−

=

∫

or the differential equation:

x

′



s







x



s

2



e

s−1

,

s

∈



1, 3



,

x

1



4.

REFERENCES

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