SinghHematulinPantAGT.dvi


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Applied General Topology

c© Universidad Politécnica de Valencia

Volume 10, No. 1, 2009

pp. 121-130

New coincidence and common fixed point
theorems

S. L. Singh, Apichai Hematulin and Rajendra Pant

Abstract. In this paper, we obtain some extensions and a gen-
eralization of a remarkable fixed point theorem of Proinov. Indeed,

we obtain some coincidence and fixed point theorems for asymptoti-

cally regular non-self and self-maps without requiring continuity and

relaxing the completeness of the space. Some useful examples and dis-

cussions are also given.

2000 AMS Classification: 54H25; 47H10.

Keywords: Coincidence point; fixed point; Banach contraction; quasi-
contraction; asymptotic regularity.

1. Introduction

The well-known Banach fixed point theorem has been generalized and ex-
tended by many authors in various ways. Recently, Proinov [15] has obtained
two types of generalizations of Banach’s fixed point theorem. The first type
involves Meir- Keeler type conditions (see, for instance, Cho et al. [3], Jachym-
ski [6], Lim [10], Matkowski [11], Park and Rhoades [14]) and the second type
involves contractive gauge functions (see, for instance, Boyd and Wong [1] and
Kim et al. [9]). Proinov [15] obtained equivalence between these two types of
contractive conditions and also obtained a new fixed point theorem. Inspired
by Jungck [7], Naimpally et al. [13], Proinov [15] and Romaguera [19], we
obtain coincidence theorems on a very general setting and derive various fixed
point theorems. Some special cases are also discussed.

In all that follows Y is an arbitrary non-empty set, (X, d) a metric space and
N := {1, 2, 3, ..., }. For T, f : Y → X, let C(T, f ) denote the set of coincidence
points of T and f , that is C(T, f ) := {z ∈ Y : T z = f z}.

The following definition comes from Sastry et al. [20] and S. L. Singh et al.
[21].



122 S. L. Singh, A. Hematulin and R. Pant

Definition 1.1. Let S, T and f be maps on Y with values in a metric space
(X, d). The pair (S, T ) is asymptotically regular with respect to f at x0 ∈ Y if
there exists a sequence {xn} in Y such that

f x2n+1 = Sx2n, f x2n+2 = T x2n+1, n = 0, 1, 2, ..., and

lim
n→∞

d(f xn, f xn+1) = 0.

If Y = X and S = T then we get the definition of asymptotic regularity
of T with respect to f due to Rhoades et al. [18]. Further if Y = X, S = T
and f is the identity map on X, then we get the usual definition of asymptotic
regularity for a map T due to Browder and Peteryshyn [2].

Definition 1.2 ([16]). Let (X, d) be a metric space and T, f : X → X. Then
the self-maps T and f are R-weakly commuting if there exists a positive real
number R such that

d(T f x, f T x) ≤ Rd(T x, f x) for all x ∈ X.

Following Itoh and Takahashi [5] and Singh and Mishra [22], we have the
following definition for a pair of self-maps on a metric space X.

Definition 1.3. Let T, f : X → X. Then the pair (T, f ) is (IT)-commuting
at z ∈ X if T f z = f T z. They are (IT)-commuting on X (also called weakly
compatible, by Jungck and Rhoades [8]) if T f z = f T z for all z ∈ X such that
T z = f z.

Definition 1.4 ([15] Definition 2.1 (i)). Let φ denote the class of all functions
ϕ : R+ → R+ satisfying: for any ε > 0 there exists δ > ε such that ε < t < δ
implies ϕ(t) ≤ ε.

2. Main Results

Proinov [15] obtained the following result generalizing some fixed point the-
orems of Jachymski [6] and Matkowski [11].

Theorem 2.1 ([15, Th. 4.1]). Let T be a continuous and asymptotically regular
self-map on a complete metric space (X, d) satisfying the following conditions:

(P1): d(T x, T y) ≤ ϕ(D(x, y)), for all x, y ∈ X;
(P2): d(T x, T y) < D(x, y), for all distinct x, y ∈ X,

where D(x, y) = d(x, y) + γ[d(x, T x) + d(y, T y)], γ ≥ 0 and ϕ ∈ φ.

Then T has a unique fixed point.
Moreover if D(x, y) = d(x, y) + d(x, T x) + d(y, T y) and ϕ is continuous and

satisfies ϕ(t) < t for all t > 0, then continuity of T can be dropped.

For a self-map T : X → X the quasi-contraction due to Ćirić [4] is as follows
(C) d(T x, T y) ≤ qM (x, y),
where M (x, y) = max{d(x, y), d(x, T x), d(y, T y), d(x, T y), d(y, T x)}, 0≤ q < 1.

We remark that following the listing of conditions due to Rhoades [17] the
condition (C) is the condition (24). According to Rhoades [17] the condition
(25):



New coincidence and common fixed point theorems 123

d(T x, T y) < M (x, y),
is the most general condition among the contractive conditions.

The following example shows that (P1) is more general than condition (C).

Example 2.2. Let X = {1, 2, 3} with the usual metric d and T : X → X such
that

T 1 = 1, T 2 = 3, T 3 = 1. Then T satisfies (C) with q > 1.

Clearly, the condition (P1) is satisfied with ϕ(t) = t
2

for all t > 0 and ϕ(0) = 0
and γ ≥ 1.

Evidently T can not satisfy the conditions (24) and (25) listed by Rhoades
[17].

First we extend the scope of Theorem 2.1 by introducing a dummy map f
in Theorem 2.1. This idea comes essentially from Jungck [7].

We remark that the requirement “ϕ(t) < t for all t > 0” in Theorem 2.1 is
redundant as this is the consequence of Definition 1.4. We shall use this fact
in the proof of the following theorem.

Theorem 2.3. Let T and f be self-maps on a complete metric space (X, d)
such that

(A1): T (X) ⊆ f (X);
(A2): d(T x, T y) ≤ ϕ(g(x, y)) for all x, y ∈ X,

where g(x, y) = d(f x, f y) + γ[d(f x, T x) + d(f y, T y)], γ ≥ 0 and ϕ ∈ φ
is continuous;

(A3): d(T x, T y) < g(x, y) for all distinct x, y ∈ Y ;
(A4): (T, f ) is asymptotically regular at x0 ∈ X.

If T is continuous then T has a fixed point provided that T and f are R-weakly
commuting. Further if f is continuous and γ = 1 then T and f have a unique
common fixed point provided that T and f are R-weakly commuting.

Proof. Pick x0 ∈ X. Define a sequence {yn} by yn+1 = T xn = f xn+1, n =
0, 1, 2, ... This can be done since the range of f contains the range of T . Let us
fix ε > 0. Since ϕ ∈ φ, there exists δ > ε such that for any t ∈ (0, ∞),

(2.1) ε < t < δ ⇒ ϕ(t) ≤ ε.

Without loss of generality we may assume that δ ≤ 2ε. Since the pair (T, f )
is asymptotically regular, lim

n→∞
d(yn, yn+1) = 0. Hence, there exists an integer

N ≥ 1 such that

(2.2) d(yn, yn+1) <
δ − ε

1 + 2γ
for all n ≥ N.

By induction we shall show that

(2.3) d(yn, ym) <
δ + 2γε

1 + 2γ
for all m, n ∈ N with m ≥ n ≥ N .



124 S. L. Singh, A. Hematulin and R. Pant

Let n ≥ N be fixed. Obviously, (2.3) holds for m = n. Assuming (2.3) to hold
for an integer m ≥ n, we shall prove it for m + 1. By the triangle inequality,
we get

d(yn, ym+1) ≤ d(yn, yn+1) + d(yn+1, ym+1)

or

(2.4) d(yn, ym+1) ≤ d(yn, yn+1) + d(T xn, T xm).

We claim that

(2.5) d(T xn, T xm) ≤ ε.

To prove (2.5), we consider two cases.

Case 1.: Let g(xn, xm) ≤ ε. By (A2) and (A3),

d(T xn, T xm) ≤ g(xn, xm) ≤ ε, and (2.5) holds.

Case 2.: Let g(xn, xm) > ε. By (A2),

(2.6) d(T xn, T xm) ≤ ϕ(g(xn, xm)).

By the definition of g(x, y),

g(xn, xm) = d(yn, ym) + γ[d(yn, yn+1) + d(ym, ym+1)].

From (2.2) and (2.3),

g(xn, xm) <
δ + 2γε

1 + 2γ
+ 2γ

δ − ε

1 + 2γ
= δ.

Now by (2.1),

ε < g(xn, xm) < δ ⇒ ϕ(g(xn, xm)) ≤ ε.

So (2.6) implies (2.5). From (2.5), (2.4) and (2.2), it follows that

d(yn, ym+1) ≤
δ − ε

1 + 2γ
+ ε =

δ + 2γε

1 + 2γ
. This proves(2.3).

Since δ ≤ 2ε, (2.3) implies that d(yn, ym) < 2ε for all integers m and n with
m ≥ n ≥ N . So {yn} is a Cauchy sequence. Since the space X is complete the
sequence {yn} has a limit. Call it z.

Suppose T is continuous. Then T T xn → T z and T f xn → T z. Since T and
f are R-weakly commuting,

d(T f xn, f T xn) ≤ Rd(T xn, f xn).

Making n → ∞,

f T xn → T z. If z 6= T z, then by (A2),

d(T xn, T T xn) ≤ ϕ(g(xn, T xn)

= ϕ(d(f xn, f T xn) + γ[d(f xn, T xn) + d(f T xn, T T xn)]).

Making n → ∞,

d(z, T z) ≤ ϕ(d(z, T z) < d(z, T z), a contradiction. It follows that z = T z.



New coincidence and common fixed point theorems 125

If f continuous and γ = 1. Then f f xn → f z and f T xn → f z. Since T and
f are R-weakly commuting,

d(T f xn, f T xn) ≤ Rd(T xn, f xn).

Making n → ∞,

T f xn → f z. If z 6= f z, then by (A2),

d(T xn, T f xn) ≤ ϕ(g(xn, f xn)

= ϕ(d(f xn, f f xn) + γ[d(f xn, T xn) + d(f f xn, T f xn)]).

Making n → ∞,

d(z, f z) ≤ ϕ(d(z, f z) < d(z, f z), a contradiction. It follows that z = f z.

Now if z 6= T z, then by (A2),

d(T z, T f xn) ≤ ϕ(g(z, f xn)

= ϕ(d(f z, f f xn) + [d(f z, T z) + d(f f xn, T f xn)]).

Making n → ∞,

d(T z, f z) ≤ ϕ(d(T z, f z) < d(T z, f z), a contradiction.

It follows that T z = f z = z, and z is a common fixed point of f and T .
Uniqueness follows easily. �

We remark that Theorem 2.1 is obtained from Theorem 2.3 as a corollary.
Notice that conditions (P1) and (P2) come respectively from (A2) and (A3)
when f is the identity map on X. Further, the continuity of only one map
is needed. The following example shows the superiority of Theorem 2.3 over
Theorem 2.1.

Example 2.4. Let X = [0, ∞) with usual metric d. Let T : X → X such that

T x =

{

x if x is rational,
0 if x is irrational.

Theorem 2.1 is not applicable to this map T as it is not continuous. However,
if we take a (dummy) map f : X → X such that f x = 2x for all x ∈ X then T
and f satisfy all the hypotheses of Theorem 2.3. Notice that f is continuous
and T 0 = f 0 = 0.

Now we modify certain requirements of Theorem 2.3 a slightly to obtain a
new result.

Theorem 2.5. Let T and f be maps on an arbitrary non-empty set Y with
values in a metric space (X, d) such that

(B1): T (Y ) ⊆ f (Y );
(B2): d(T x, T y) ≤ ϕ(g(x, y)) for all x, y ∈ Y ,

where g(x, y) = d(f x, f y) + γ[d(f x, T x) + d(f y, T y)], 0 ≤ γ ≤ 1, and
ϕ : R+ → R+ continuous;



126 S. L. Singh, A. Hematulin and R. Pant

(B3): (T, f ) is asymptotically regular at x0 ∈ Y .
If T (Y ) or f (Y ) is a complete subspace of X then

(i): C(T, f ) is non-empty.

Further, if Y = X, then
(ii): T and f have a unique common fixed point provided that T and f

are (IT)-commuting at a point u ∈ C(T, f ).

Proof. Pick x0 ∈ Y . Define a sequence {yn} by yn+1 = T xn = f xn+1, n =
0, 1, 2..., this can be done since the range of f contains the range of T . Since
the pair (f, T ) is asymptotically regular, lim

n→∞
d(yn, yn+1) = 0.

First we shall show that {yn} is a Cauchy sequence. Suppose {yn} is not
Cauchy. Then there exists µ > 0 and increasing sequences {mk} and {nk} of
positive integers such that for all n ≤ mk < nk,

d(ymk , ynk ) ≥ µ and d(ymk , ynk−1) < µ.

By the triangle inequality,

d(ymk , ynk ) ≤ d(ymk , ynk−1) + d(ynk−1, ynk ).

Making k → ∞,

d(ymk , ynk ) < µ.

Thus, d(ymk , ynk ) → µ as k → ∞. Now by (B2),

d(ymk+1, ynk+1) = d(T xmk , T xnk )

≤ ϕ(g(xmk , xnk ))

= ϕ(d(f xmk , f xnk ) + γ[d(f xmk , T xmk ) + d(f xnk , T xnk )]).

Making k → ∞,

µ ≤ ϕ(µ) < µ,

a contradiction. Therefore {yn} is Cauchy. Suppose f (Y ) is complete. Then
{yn} being contained in f (Y ) has a limit in f (Y ). Call it z. Let u ∈ f

−1z.
Then f u = z. Using (B2),

d(T u, T xn) ≤ ϕ(d(f u, f xn) + γ[d(T u, f u) + d(T xn, f xn)]).

Making n → ∞,

d(T u, z) ≤ ϕ(γd(T u, z)) < d(T u, z),

a contradiction. Therefore T u = z = f u. This proves (i). Now if Y = X and
the pair(T, f ) is (IT)-commuting at u then T f u = f T u and T T u = T f u =
f T u = f f u. In view of (B2), it follows that

d(T u, T T u) < ϕ(g(u, T u))

= ϕ(d(f u, f T u) + γ[d(T u, f u) + d(T T u, f T u)]) < d(T u, T T u),

a contradiction. Therefore T T u = T u and f T u = T T u = T u = z. This proves
(ii).



New coincidence and common fixed point theorems 127

In the case T (Y ) is a complete subspace of X, the condition (B1) implies
that sequence {yn} converges in f (Y ), and the previous proof works. The
uniqueness of common fixed point follows easily. �

The following result generalizes an important result of Proinov [15, Cor. 4.3]

Corollary 2.6. Let T and f be maps on an arbitrary non-empty set Y with
values in metric space (X, d) such that

(C1): T (Y ) ⊆ f (Y );
(C2): d(T x, T y) ≤ ϕ(M (x, y)), for all x, y ∈ Y ,

where M (x, y) = max{d(f x, f y), d(f x, T x), d(f y, T y), 1
2
[d(f x, T y)+

d(f y, T x)]} and ϕ : R+ → R+ continuous.
If T (Y ) or f (Y ) is a complete subspace of X then conditions (i) and

(ii) of above Theorem 2.5 hold.

Now we obtain a new common fixed point theorem for three non self-maps.

Theorem 2.7. Let S, T and f be maps on an arbitrary non-empty set Y with
values in a metric space (X, d). Let (S, T ) be asymptotically regular with respect
to f at x0 ∈ Y and the following conditions are satisfied:

(D1): S(Y ) ∪ T (Y ) ⊆ f (Y );
(D2): d(Sx, T y) ≤ ϕ(h(x, y)), for all x, y ∈ X,

where h(x, y) = d(f x, f y) + γ[d(Sx, f x) + d(T y, f y)], 0 ≤ γ ≤ 1, and
ϕ : R+ → R+ continuous.

If S(Y ) or T (Y ) or f (Y ) is a complete subspace of X then
(I): C(S, f ) is non-empty;
(II): C(T, f ) is non-empty.

Further, if Y=X then
(III): S and f have a common fixed point provided that S and f are

(IT)-commuting at a point u ∈ C(S, f ).
(IV): T and f have a common fixed point provided that T and f are

(IT)-commuting at a point v ∈ C(T, f ).
(V): S, T and f have a unique common fixed point provided that (III)

and (IV) both are true.

Proof. Let x0 be an arbitrary point in Y . Since (S, T ) is asymptotically regular
with respect to f , then there exists a sequence {xn} in Y such that

f x2n+1 = Sx2n, f x2n+2 = T x2n+1, n = 0, 1, 2, ..., and

lim
n→∞

d(f xn, f xn+1) = 0.

Now we shall show that {f xn} is Cauchy sequence. Suppose {f xn} is not
Cauchy. Then there exists µ > 0 and increasing sequences {mk} and {nk} of
positive integers, such that for all n ≤ mk < nk,

d(f xmk , f xnk ) ≥ µ and d(f xmk , f xnk−1) < µ.

By the triangle inequality,

d(f xmk , f xnk ) ≤ d(f xmk , f xnk−1) + d(f xnk−1, f xnk ).



128 S. L. Singh, A. Hematulin and R. Pant

Making k → ∞, we get

d(f xmk , f xnk ) < µ.

Thus

d(f xmk , f xnk ) → µ as k → ∞.

By (D2) we have

d(f xmk+1, f xnk+1) = d(Sxmk , T xnk )

≤ ϕ(h(xmk , xnk ))

= ϕ(d(f xmk , f xnk ) + γ[d(Sxmk , f xmk ) + d(T xnk , f xnk )]).

Making k → ∞

µ ≤ ϕ(µ) < µ, a contradiction.

Thus {f xn} is Cauchy sequence. Suppose f (Y ) is a complete subspace of
X. Then {yn} being contained in f (Y ) has a limit in f (Y ). Call it z. Let
u = f −1z. Thus f u = z for some u ∈ Y . Note that the subsequences {f x2n+1}
and {f x2n+2} also converge to z. Now by (D2),

d(Su, T2n+1) ≤ ϕ(d(f u, f2n+1) + γ[d(Su, f u) + d(T2n+1, f2n+1)]).

Making n → ∞,

d(Su, f u) ≤ ϕ(γd(Su, f u)) < d(Su, f u) a contradiction.

Therefore Su = f u = z. This proves (I). Since S(Y )∪T (Y ) ⊆ f (Y ). Therefore
there exists v ∈ Y such that Su = f v. We claim that f v = T v. Using (D2),

d(f v, T v) = d(Su, T v)

≤ ϕ(d(f u, f v) + γ[d(Su, f u) + d(T v, f v)])

= ϕ(γd(f v, T v)) < d(f v, T v),

which is a contradiction. Therefore T v = f v = Su = f u. This proves (II).
Now if Y = X, (S, f ) and (T, f ) are (IT)-commuting then Sf u = f Su and
SSu = Sf u = f Su = f f u, T f v = f T v and T T v = T f v = f T v = f f v. In
view of (D2), it follows that

d(SSu, Su) = d(SSu, T v)

≤ ϕ(d(f Su, f v) + γ[d(SSu, f Su) + d(T v, f v)])

= ϕ(γd(SSu, Su)) < d(SSu, Su).

Therefore SSu = Su = f Su, Su is a common fixed point of S and f . Similarly,
T v is a common fixed point of T and f . Since Su = T v, we conclude that Su is
a common fixed point of S, T and f . The proof is similar when S(Y ) or T (Y )
are complete subspaces of X since, S(Y ) ∪ T (Y ) ⊆ f (Y ). Uniqueness of the
common fixed point follows easily. �

Acknowledgements. The authors are indebted to the referee and Prof.
Salvador Romaguera for their perspicacious comments and suggestions.



New coincidence and common fixed point theorems 129

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130 S. L. Singh, A. Hematulin and R. Pant

Received August 2008

Accepted January 2009

S. L. Singh (vedicmri@gmail.com)
21, Govind Nagar Rishikesh 249201, India

Apichai Hematulin

Department of Mathematics, Nakhonratchasima Rajabhat University, Nakho-
ratchasima, Thailand

Rajendra Pant (pant.rajendra@gmail.com)
SRM University Modinagar, Ghaziabad (U.P.) 201204, India