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1. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
A Unique Common Fixed Point Theorem under psi –
varphi Contractive Condition in Partial Metric Spaces Using
Rational Expressions
K.P.R.Rao
Department of Applied Mathematics, AcharyaNagarjunaUnivertsity-Dr.M.R.Appa Row Campus,
Nuzvid- 521 201, Krishna District, Andhra Pradesh, India
E-mail address: kprrao2004@yahoo.com
G.n.V. Kishore
Department of Mathematics, Swarnandhra Institute of Engineering and Technology,Seetharampuram,
Narspur- 534 280 , West Godavari District, Andhra Pradesh, India
E-mail address: kishore.apr2@gmail.com
Abstract
In this paper, we obtain a unique common fixed point theorem for two self maps satisfying psi - varphi
contractive condition in partial metric spaces by using rational expressions.
Keywords: partial metric, weakly compatible maps, complete space.
1. Introduction
The notion of partial metric space was introduced by S.G.Matthews [3] as a part of the study of denotational
semantics of data flow networks. In fact, it is widely recognized that partial metric spaces play an important
role in constructing models in the theory of computation ([2, 5, 7 -12], etc).
S.G.Matthews [3], Sandra Oltra and Oscar Valero[4] and Salvador Romaguera [6]and I.Altun, Ferhan Sola,
HakanSimsek [1] prove fixed point theorems in partial metric spaces for a single map.
In this paper, we obtain a unique common fixed point theorem for two self mappings satisfying a
generalized ψ -φ contractive condition in partial metric spaces by using rational expression.
First we recall some definitions and lemmas of partial metric spaces.
2. Basic Facts and Definitions
Definition 2.1. [3].A partial metric on a nonempty set X is a function p :X × X → R+ such that for all x, y, z
∈ X:
(p1) x = y ⇔ p(x, x) = p(x, y) = p(y, y),
(p2) p(x, x) ≤ p(x, y), p(y, y) ≤ p(x, y),
(p3) p(x, y) = p(y, x),
(p4) p(x, y) ≤ p(x, z) + p(z, y) − p(z, z).
(X, p) is called a partial metric space.
It is clear that |p(x, y) − p(y, z)| ≤ p(x, z) for all x, y, z ∈ X.
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2. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
Also clear that p(x, y) = 0 implies x = y from (p1) and (p2).
But if x = y, p(x, y) may not be zero. A basic example of a partial metric space is the pair (R+, p), where
p(x, y) = max{x, y} for all x, y ∈ R+.
Each partial metric p on X generates τ0 topology τp on X which has a base the family of open p - balls
{Bp(x, ε) / x ∈ X, ε > 0} for all x ∈ X and ε > 0, where Bp(x, ε)= {y ∈ X / p(x, y) < p(x, x) + ε}
for all x ∈ X and ε > 0.
If p is a partial metric on X, then the function ps : X × X → R+ given by
Ps(x, y) = 2p(x, y) − p(x, x) − p(y, y) (2.1)
is a metric on X.
Definition 2.2. [3]. Let (X, p) be a partial metric space.
(i) A sequence {xn} in (X, p) is said to converge to a point x ∈ X if and only if p(x, x) = lim p(x, xn).
n→∞
(ii) A sequence {xn} in (X, p) is said to be Cauchy sequence if lim p(xm, xn) exists and is finite .
m,n→∞
(iii) (X, p) is said to be complete if every Cauchy sequence {xn} in X converges, w.r.to τp, to a point
x ∈ X such that p(x, x) = lim
m,n→∞
p(xm, xn).
Lemma 2.3. [3].Let (X, p) be a partial metric space.
(a) {xn} is a Cauchy sequence in (X, p) if and only if it is a Cauchy sequence in the metric space (X, ps).
(b) (X, p) is complete if and only if the metric space (X, ps) is complete. Furthermore,
lim ps(xn, x) = 0 if and only if p(x, x) = lim p(x, xn) = lim p(xm, xn).
n→∞ n→∞ m,n→∞
3. Main Result
Theorem 3.1. Let (X, p) be a partial metric space and let T, f : X → X be mappings such that
(i) ψ(p (Tx, Ty)) ≤ψ (M (x, y)) − φ (M (x, y)) , ∀ x, y ∈ X,
 1 + p(fx, Tx) 
where M (x, y) = max p(fy, Ty) , p(fx, fy) and ψ: [0, ∞) → [0, ∞) is continuous,
 1 + p(fx, fy) 
non-decreasing with ψ (t) = 0 if and only if t = 0 and φ:[0, ∞) → [0, ∞) is lower semi continuous with
φ(t) = 0 if and only if t = 0,
(ii) T(X) ⊆ f(X) and f(X) is a complete subspace of X and
(iii) the pair (f, T) is weakly compatible.
Then T, f have a unique common fixed point of the form α in X.
Proof: Let x0 ∈ X. From (ii), there exist sequences {xn} and {yn} in X such that
yn = fxn+1 = Txn, n = 0, 1, 2,3, · · ·
Case(a): Suppose yn = yn+1 for some n.
Then fz = Sz , where z = xn+1. Denote fz = Tz = α.
ψ(p(Tα, α)) = ψ (p(Tα, Tz))
≤ ψ(M(α, z)) − φ(M(α, z)).
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3. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
 1 + p(fα , Tα ) 
M(α, z) = max p(fz, Tz) , p(fα , fz)
 1 + p(fα , fz) 
 1 + p(Tα , Tα ) 
= max p(α , α ) , p(Tα ,α )
 1 + p(Tα , α ) 
= p(Tα, α), from (p2).
Therefore
ψ(p(Tα, α)) ≤ ψ(p(Tα, α)) − φ(p(Tα, α))
< ψ(p(Tα, α)).
It is a contradiction.
Hence fα = Tα = α.
Hence α is a common fixed point of f and T.
Let β be another common fixed point of f and T such that α ≠ β.
ψ(p(α, β)) = ψ(p(Tα, Tβ))
 1 + p(α , α )   
 − ϕ  max p(β , β ) 1 + p(α , α ) , p(α , β )
   
≤ ψ max p( β , β )
 , p(α , β )  


  1 + p(α , β )     1 + p(α , β )  
= ψ(p(α, β)) – φ(p(α, β)), from (p2)
< ψ(p(α, β)).
It is a contradiction. Hence α = β.
Thus α is the unique common fixed point of T and f.
Case(b): Suppose yn≠ yn+1for all n.
ψ(p(yn, yn+1)) = ψ(p(Txn, Txn+1))
≤ ψ(M(xn, xn+1)) − φ(M(xn, xn+1))
= ψ(max{p(yn, yn+1), p(yn, yn-1)}) − φ(max{p(yn, yn+1), p(yn, yn-1)}).
If p(yn, yn+1)is maximum, then
ψ(p(yn, yn+1)) ≤ ψ(p(yn, yn+1)) - φ(p(yn, yn+1))
< ψ(p(yn, yn+1)).
It is a contradiction.
Hence p(yn, yn-1) is maximum.
Therefore
ψ(p(yn, yn+1)) ≤ ψ(p(yn, yn-1)) - φ(p(yn, yn-1)) (3.1)
≤ ψ(p(yn, yn-1)).
Since ψ is non – decreasing, we have p(yn, yn+1) ≤ p(yn, yn-1)
Similarly p(yn+2, yn+1) ≤ p(yn, yn+1).
Thus p(yn, yn+1) ≤ p(yn, yn-1), n = 1, 2, 3, …
Thus { p(yn, yn+1) } is a non - increasing sequence of non-negative real numbers and must converge to a real
number, say, L ≥ 0.
Letting n→ ∞ in (3.1), we get
ψ(L) ≤ ψ(L) - φ(L) so that φ(L) ≤ 0. Hence L = 0.
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4. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
Thus
lim p(yn, yn+1) = 0. (3.2)
n→∞
Hence from (p2),
lim p(yn, yn) = 0. (3.3)
n→∞
By definition of ps, we have ps(yn, yn+1) ≤ 2p(yn, yn+1).
From (3.2), we have
lim ps(yn, yn+1) = 0. (3.4)
n→∞
Now we prove that {yn} is Cauchy sequence in metric space (X, ps). On contrary suppose that {yn} is not
Cauchy. Then there exists an ε> 0 for which we can find two subsequences {ym(k)}, {yn(k)} of {yn} such that
n(k) is the smallest index for which n(k) > m(k) > k,
ps(ym(k), yn(k)) ≥ ε (3.5)
and
ps(ym(k), yn(k)-1) < ε. (3.6)
From (3.5) and (3.6),
ε ≤ ps(ym(k), yn(k))
≤ ps(ym(k), yn(k)-1) + ps(yn(k)-1, yn(k))
< ε + ps(yn(k)-1, yn(k))
Letting k → ∞ and using (3.2), we have
lim ps(ym(k), yn(k)) = ε. (3.7)
k →∞
From (p2), we get
lim p(ym(k), yn(k)) = ε . (3.8)
k →∞ 2
Letting k → ∞ and using (3.2) and (3.7) in the inequality
| ps(ym(k )-1, yn(k)) – ps(ym(k), yn(k)) | ≤ ps(ym(k), ym(k)-1)
we get
lim ps(ym(k )-1, yn(k)) = ε. (3.9)
k →∞
From (p2), we get
lim p(ym(k )-1, yn(k)) = ε . (3.10)
k →∞ 2
Letting k → ∞ and using (3.2) and (3.7) in the inequality
| ps(ym(k ), yn(k)+1) – ps(ym(k), yn(k)) | ≤ ps(yn(k), yn(k)+1)
we get
lim ps(ym(k ), yn(k)+1) = ε. (3.11)
k →∞
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5. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
From (p2), we get
lim p(ym(k ), yn(k)+1) = ε . (3.12)
k →∞ 2
ψ(p(ym(k ), yn(k)+1)) = ψ(p(Txm(k ), Txn(k)+1))
≤ ψ(M(xm(k ), xn(k)+1)) – φ(M(xm(k ), xn(k)+1))
 
 1 + p(y m(k) -1, y m(k) ) 
 
= ψ max p(y m(k) , y m(k) +1 ) , p(y m(k) -1 , y n(k) ) 
  1 + p(y m(k) -1 , y n(k) )  
   
 
 1 + p(ym(k)-1, y m(k) ) 
 
- ϕ  max p(ym(k) , ym(k) +1 ) , p(ym(k)-1 , yn(k) )  .
  1 + p(ym(k) -1, y n(k) )  
   
Letting k → ∞ and using (3.12), (3.2), (3.10) and (3.8), we get
ε ε ε ε
ψ  ≤ ψ  − ϕ   < ψ 
2 2 2 2
It is a contradiction.
Hence {yn} is Cauchy sequence in (X, ps).
Thus lim ps(yn, ym) = 0.
m,n→∞
By definition of ps and from (3.2), we get
lim ps(yn, ym) = 0. (3.13)
m,n→∞
Suppose f(X) is complete.
Since {yn} ⊆ f(X) is a Cauchy sequence in the complete metric space (f(X), ps), it follows that {yn}
converges in (f(X), ps).
Thus lim ps(yn, α) = 0 for some α
n →∞
∈ f(X).
There exists w ∈ X such that α = fw.
Since {yn+1} is Cauchy in X and {yn} → α, it follows that {yn+1} → α.
From Lemma 2.3(b) and (3.13), we have
p(α, α) = lim p(yn, α) = lim p(yn+1, α) = lim p(yn, ym) = 0. (3.14)
n→∞ n→∞ m,n→∞
p(Tw, α) ≤ p(Tw, Txn+1) + p(Txn+1, α) – p(Txn+1, Txn+1)
≤ p(Tw, Txn+1) + p(Txn+1, α).
Letting n→∞, we have
p(Tw, α) ≤ lim p(Tw, Txn+1).
n→∞
Since ψ is is continuous and non - decreasing, we have
ψ(p(Tw, α)) ≤ lim ψ(p(Tw, Txn+1))
n→∞
33

6. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
 
= lim ψ max p(yn , yn +1 ) 1 + p(α , Tw) , p(α , yn ) 
 
n→∞  1 + p(α , y n )
  
  1 + p(α , Tw) 
- lim ϕ  max p(yn , y n +1 ) , p(α , y n ) 
n→∞  1 + p(α , yn ) 
  
= ψ(0) – φ(0) = 0.
It follows that Tw = α. Thus Tw = α = fw.
Since the pair (f, T) is weakly compatible, we have fα = Tα.
As in Case(a), it follows that α is the unique common fixed point of T and f.
Example 3.2. Let X = [0, 1] and p(x, y) = max{x, y} for all x, y ∈X . Let T, f : X → X, f(x) = x/3
and T(x) = x2/6, ψ : [0,∞) → [0,∞) by ψ(t) = t and φ: [0,∞) → [0,∞) by φ(t) = t/2 . Then the conditions
(ii) and (iii) are satisfied and
ψ(p(Tx, Ty)) = p(Tx, Ty)
 x 2 y2 
= max  , 
6 6
x y
≤ max  , 
6 6
1 1  1 + p(fx, Tx) 
= p(fx, fy) ≤ max p(fy, Ty) , p(fx, fy)
2 2  1 + p(fx, fy) 
= ψ(M(x, y)) – φ(M(x, y)).
 1 + p(fx, Tx) 
Where M(x, y) = max p(fy, Ty) , p(fx, fy)
 1 + p(fx, fy) 
Thus (i) is holds.
Hence all conditions of Theorem 3.1 are satisfied and 0 is the unique common fixed point of T and f.
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7. Mathematical Theory and Modeling www.iiste.org
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Vol.2, No.4, 2012
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