# Introduction

et (X; d) be a metric space. If f is a self-map on , and x0 X, we denote by fx the f -image of .

As a weaker form of the metric , Kada et al. [5] introduced the notion of weak distance (or simply -distance) on as follows:

Definition 1.1. Let (X; d) be a metric space and satisfy the following conditions:

( ) for all ( ) For any is lower semi continuous in the second variable, that is lim inf whenever as for some , and

( ) Given , there is a such that imply that .

Then p is known as a w-distance on X.

Obviously, every metric d on X satisfies the conditions ( w1 ) -( w3 ) , that is d is a w -distance on X.

Example 1. 1. Let with metric d ( x; y ) = x + y if x 6 = y and d(x; y) = 0 if x = y for all x; y 2 X. Note that (X; d) is a complete metric space. Define p(x; y) = y. Then p a w -distance on X.

For other examples one can refer to [5].

Recently Ume and Sucheol [4] have proved two common fixed point theorems, given below, for self-maps on a complete metric space with a w -distance on X, which generalize and improve the results of Fisher [1], Dien [3] and Liu et al. [6].

Theorem 1.1 ([4], Theorem 3.1). Let X be a complete metric space (X; d) with w-distance p on it. Suppose that f; g : X ! X and : X ! [0; 1) satisfy the conditions: (1.4)

Then and will have a unique common fixed point.

Theorem 1.2 ([4], Theorem 3.6). Let X be a complete metric space (X; d) with w-distance p and the mappings f; g : X ! X satisfy the conditions (a) and (d). Suppose that : X ! [0; 1) are such that Then f and g will have a unique common fixed point.
L X ? x 0 d w X p : X × X ? [0, ?) w 1 p(x, y) ? p(x, z) + p(z, y) x, y, z ? X w 2 x ? X, p(x, ?) : X ? R + p(x, y )? n p(x, y n ) y n ?y n? ? x ? X >0 ? >0 p(z, x) ? ? and p(z, y) ? ? d(x, y) < X= 1 m : m = 1, 2, 3, ... {0} ? g(X) f(X) ? p(t, gx) ? rp(t, f x) + ?(f x) ? ?(gx) f or all x, y ? X, 0 ? r < 1 (b) (c) x n ? n=1 lim n?? p(t, f x n ) = 0 = lim n?? p(t, gx n ), lim n?? max p(t, f x n ), p(t, gx n ), p(f gx n , gf x n ) = 0,
The purpose of this paper is to establish two fixed point theorems, which generalize those of Brian Fisher [1], Dien [3] and Liu et al. [6].

II.


# Preliminaries

First we state the following lemma, proved in [5]: Lemma 2.1. Let X be a metric space with w-distance p on it. Then (g) p(x; y) = 0 and p(x; z) = 0 imply that y = z:
Also is a Cauchy sequence in X, provided (h) p(xn; xm) n for all m > n 1 (i) p(x; xn)
n for all n 1 for each x 2 X:

We now introduce an orbit notion that is followed in the rest of the paper. Definition 2.1. Let f and g be self-maps on X. Given 2 X, if there exit points in X such that
(2.1)
the sequence is called a g-orbit relative to f at or simply a (g; f)-orbit at . We call a base sequence associated with the g-orbit (2.1). Note that when f is the identity map i on X, (2.1) and the base sequence coincide with the g-orbit gx 0 ; gx 1 ; ::: at x 0 . This notion was adopted in [8]. The notion of (g; f)-orbit is not unique. For instance, Nesic [7] defined a (g; f)-orbit at x0 by the iterations:

(2.2) which was employed by Fisher [1] though no name was mentioned.

Remark 2.1. If the self-maps f and g on X satisfy the inclusion (1.1), then by a routine induction, it can be easily shown that (g; f)-orbit at each exists with the choice (2.1). Given 2 X, there can be more than one base sequence as the following examples reveal:

Example 2.1. Let X = R with usual metric for all x; y 2 X. Define f; g : X ! X by f(x) = and for x 2 X. Then (1.1) is obvious and hence by Remark 2.1, orbits can be specified at each . Given 2 X, choose for n 1.

Since each xn has two choices, several base sequences can be specified to get the respective (g; f)-orbit.

We now prove Lemma 2.2. Suppose that (X; d) is a metric space with w-distance p on X. Let f; g : X ! X and : X ! [0; 1) satisfy the inclusion (1.1) and the condition (b) of Theorem 1:1. If X is complete metric space and 2 X, then x; y ? X i ? [0, 1), i = 1, 2, 3, 4, 5 a
(2.3) Proof. Given 2 X, Suppose thata 1 + a 4 + a 5 < 1 1 + a 2 + a 3 + 2a 4 < 1 a x n ? n=1 ? X ? ? ? ? ? ? y n = gx n?1 = f x n n ? 1, y n ? n=1 x n ? n=1 x 2n?1 = gx 2n?2 , x 2n = f x 2n?1 n ? 1 for for x n ? n=1 d(x, y) = |x? y| g(x) = x 2 4 x n = ± x 0 2 n ? x n ? n=1 ? lim n?? f x n = lim n?? gx n = ?X. z z for some x n ? n=1 ? p(t, f x n ) = p(t, gx n?1 ) ? r.p(t, f x n?1 ) + ?(f x n?1 ) ? ?(gx n?1 ) k ? 2 k n=1 p(t, f x n ) ? r. k n=1 p(t, f x n?1 ) + k n=1 ?(f x n?1 ) ? ?(f x n ) x 1 , x 2 , x 3 , ... x 0 x 0 x 0 x 0 x 0 x 0 x 0 x 2 x 0 x 0 x 0 which gives
Showing that converges so that n th term tends to 0 as n ! 1, that is Now, by ( i ) of Lemma 2.1, it follows that is a Cauchy sequence in the (g; f)-orbit X.

Since X is complete, there is a z 2 X such that fx n ! z as n ! 1.

Similar argument shows converges to in X. Proceeding the limit as n ! 1 in (2.1) and using these limits, it follows that , proving the lemma.

Remark 2.2. The converse of Lemma 2.2 is not true. Infact, the example given below shows that we can find a metric space (X; d) with a w-distance p on it satisfying condition (a) and (b) of Theorem 1.1 such that for any 2 X and for any base sequence at , both and converge to the same point in X; but X is not complete.

Example 2.2. Let X = [0; 1) with for all x; y 2 X. Clearly (X; d) is an incomplete metric space. Define f; g : X ! X by and for x 2 X . Then and so that . Let for which will be a w-distance on X. Also for any t 2 X; and from which it follows for any x 2 X where (x) = 1 for all Note that for any x0 2 X there is only one base sequence given by for all n 1 so that both and are constant sequences with each term equal to ; and hence they converge to Lemma 2.3. Suppose (X; d) is a metric space with w-distance p on it. Let f; g : X ! X and : X ! [0;1) be such that (a) of Theorem 1.1 and (f) of Theorem 1.2 hold. If (X; d) is a complete metric space, then for any x0 2 X and for any base sequence at ,

both the sequences and converge to the same point in .

Proof. Suppose that is a base sequence at some  
p(t, f x n ) ? r 1 ? r p(t, f x 1 ) + 1 1 ? r ?(f x 0 ) ? ?(f x k ) < r 1 ? r p(t, f x 0 ) + 1 1 ? r ?(f x 0 ) ? n=2 p(t, f x n ) lim n?? p(t, f x n ) = 0. f x n ? n=1 gx n ? n=1 z z = z x n ? n=1 x 0 f x n ? n=1 gx n ? n=1 d(x, y) =|x ? y| f x= 2x + 1 4 gx = 1 2 g(X) = 1 2 f (X) = 1 4 , 34g ( X ) ? f ( X ( p(x, y) = 1 4 max |2x ? 1| , |2x ? 4y + 1| , 2 |x ? y| x, y ? X, p (t, gx) = 1 4 |2t ? 1| p(t, f x) = 1 8 max |2(2t ? 1)| , |4(t ? x)| , |(4t ? 2x ? 1)| p(t, gx) ? 2 p(t, f x) + ?(f x) ? ?(gx ( ? x ?X. 1 x n ? n=1 x n = 1 2 ? f x n ? n=1 gx n ? n=1 1 2 . 1 2 ? X ?, ? x n ? n=1 x 0 f x n ? n=1 gx n ? n=1 X x n ? n=1 ? n = p(f x n , f x n+1 ) = p(gx n?1 ,

# Main Results

Theorem 3.1. let (x; d) be a metric space with w -distance p on it. Suppose that f; g : X ! X and : X ! [0;1) satisfy the inclusion (1.1) and the condition (1.4) of Theorem 1:1. Also suppose that (j) there is a base sequence at some point 
? n = p(gx n?1 , gx n ) ? a 1 p(f x n?1 , f x n ) + a 2 p(f x n?1 , gx n?1 ) + a 3 p(f x n , gx n ) + a 4 p(f x n?1 , gx n ) + a 5 p(gx n?1 , f x n )d(f x n , gx n?1 ) + [?(f x n?1 ) ? ?(gx n?1 )] + [ (f x n ) ? (gx n )] ? a 1 ? n?1 + a 2 ? n?1 + a 3 ? n + a 4 (? n?1 + ? n ) + [?(f x n?1 ) ? ?(f x n )] + [ (f x n ) ? (f x n+1 )] = (a 1 + a 2 + a 4 )? n?1 + (a 3 + a 4 )? n + [?(f x n?1 ) ? ?(f x n )] + [ (f x n ) ? (f x n+1 )] ? n ? ?? n?1 + ?{[?(f x n?1 ) ? ?(f x n )] + [ (f x n ) ? (f x n+1 )]}, n ? 2, ? = a 1 + a 2 + a 4 1 ? a 3 ? a 4 ? = 1 1 ? a 3 ? a 4 2 k ? k n=2 ? n ? ? k n=2 ? n?1 + ? ?(f x 1 ) ? ?(f x k )] + [ (f x 2 ) ? (f x 3 ) k n=2 ? n ? ? 1 ? 1 ? ? + ? ?(f x 1 ) + (f x 2 ) 1 ? ? ? n=2 ? n ? p(f x n , f x m )? ? n ? n = ? n +? n+1 +? ? ?+? m?1 ? f x n ? n=1 gx n ? n=1 z gx n?1 =f x n ? z = z x n ? n=1 f x n ? n=1 gx n ? n=1 ? x n ? n=1 f x n ? n=1 gx n ? n=1
p(z, gx)?rp(z, f x) +?(f x)??(gx)

x ? X  Proof. The proof is similar to the first main result and is omitted here. Remark 3.2. In view of Remark 2.6, Theorem 3.2 generilzes Theorem 1.2. Also since d is a w-distance on X, the fixed point theorem of Fisher [1] is a partcular case of Theorem 3.2 with p = d.
0 ? r < 1 u n ? n=1 ? X lim n?? p(z, f u n ) = lim2![with u 6 = fu or u 6 = gu Author ?: Bank Colony, Opposite Survey of India, Uppal, Hyderabad, Telangana State, India. e-mail: vangalasrp@yahoo.co.in Author ?: Applied Analysis Division, School of Advanced Sciences, VIT-University, Vellore (T. N.), India. e-mail: drtp.indra@gmail.com {p(u; fx) + p(u; gx) + p(fgx; gfx) : x 2 X}> 0:](image-2.png "u 2 X")
![f; g are such that (2.1) holds good. Now, by condition (b) with x = xn 1, we have so that for any J = lim n?? gx n = t, Common ixed Point Theorems for Self-Maps on Metric Spaces with Weak Distance F p(gx, gy) ? a 1 p(f x, f y) + a 2 p(f x, gx) + a 3 p(f y, gy) + a 4 p(f x, gy) + a 5 p(gx, f y)d(f y, gx) + [?(f x) ? ?(gx)] + [ (f y) ? (gy)] lim n?? max p(t, f x n ), p(t, gx n ), p(f gx n , gf x n ) = 0,](image-3.png "")
![n?? p(z, gu n ) = 0, lim n?? max{p(z, f u n ), p(z, gu n ), p(f gu n , gf u n )} = 0. p(z, f x n+1 ) = p(z, gx n ) ? rp(z, f x n ) + ?(f x n ) ? ?(gx n ).](image-4.png "")
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: X ! [0;1) satisfy the inclusion (1.1) and the condition (1.4) of Theorem 1:1 and the condition (f) of Theorem 1.2. If (j) and l) hold good, then z is the unique common fixed point of f and g.

The uniqueness of the common fixed point z follows as in the proof of Theorem 3.1 of [4]. Remark 3.1. In view of Remark 2.3, Theorem 3.1 generalizes Theorem 1.2. Also since d is a w-distance, the results proved by Dien [3] and Liuet.al [6] will be particular cases of Theorem 3.1.

Then as in Lemma 2.2, we can prove that converges hence Using this in (3.1), it follows that . Now if z is not a common fixed point of f and g, then either fz 6 = z or gz 6 = z; and therefore, by the condition 
			
			

				


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