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\title{Common Fixed Point Theorems for Self-Maps on Metric Spaces with Weak Distance}
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             \author[1]{T.  Phaneendra}

             \author[2]{V. Siva Rama  Prasad}

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\date{\small \em Received: 11 December 2013 Accepted: 2 January 2014 Published: 15 January 2014}

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\begin{abstract}
        


Fixed point theorems on complete metric spaces with a weak distance proved by Ume and Yi [4] have been improved under weaker conditions. The results of this paper also generalize those of Brian Fisher [1], Dien [3] and Liu et al. [6].

\end{abstract}


\keywords{self-map, w-distance on a metric space, (g; f)-orbit at a point, Common fixed point.}

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\let\tabcellsep& 	 	 		 
\section[{Introduction}]{Introduction}\par
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 .\par
As a weaker form of the metric , Kada et al. \hyperref[b4]{[5]} introduced the notion of weak distance (or simply -distance) on as follows:\par
Definition 1.1. Let (X; d) be a metric space and satisfy the following conditions:\par
( ) for all ( ) For any is lower semi continuous in the second variable, that is lim inf whenever as for some , and\par
( ) Given , there is a such that imply that .\par
Then p is known as a w-distance on X.\par
Obviously, every metric d on X satisfies the conditions ( w1 ) -( w3 ) , that is d is a w -distance on X.\par
Example 1.  {\ref 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.\par
For other examples one can refer to \hyperref[b4]{[5]}.\par
Recently Ume and Sucheol \hyperref[b3]{[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 \hyperref[b0]{[1]}, Dien \hyperref[b2]{[3]} and Liu et al. \hyperref[b5]{[6]}.\par
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)\par
Then and will have a unique common fixed point.\par
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,\par
The purpose of this paper is to establish two fixed point theorems, which generalize those of Brian Fisher \hyperref[b0]{[1]}, Dien \hyperref[b2]{[3]} and Liu et al. \hyperref[b5]{[6]}.\par
II. 
\section[{Preliminaries}]{Preliminaries}\par
First we state the following lemma, proved in \hyperref[b4]{[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)\par
n for all n 1 for each x 2 X:\par
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)\par
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 \hyperref[b7]{[8]}. The notion of (g; f)-orbit is not unique. For instance, Nesic \hyperref[b6]{[7]} defined a (g; f)-orbit at x0 by the iterations:\par
(2.2) which was employed by Fisher \hyperref[b0]{[1]} though no name was mentioned.\par
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:\par
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.\par
Since each xn has two choices, several base sequences can be specified to get the respective (g; f)-orbit.\par
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\par
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.\par
Since X is complete, there is a z 2 X such that fx n ! z as n ! 1.\par
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.\par
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.\par
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 ,\par
both the sequences and converge to the same point in .\par
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 , 3\textbf{4}g ( 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 , 
\section[{Main Results}]{Main Results}\par
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\par
p(z, gx)?rp(z, f x) +?(f x)??(gx)\par
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 \hyperref[b0]{[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 ) = lim\begin{figure}[htbp]
\noindent\textbf{2}\includegraphics[]{image-2.png}
\caption{\label{fig_0}u 2 X}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-3.png}
\caption{\label{fig_2}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-4.png}
\caption{\label{fig_3}}\end{figure}
 			\footnote{© 2014 Global Journals Inc. (US)} 		 		\backmatter  			 \par
: 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.\par
The uniqueness of the common fixed point z follows as in the proof of Theorem 3.1 of \hyperref[b3]{[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 \hyperref[b2]{[3]} and Liuet.al \hyperref[b5]{[6]} will be particular cases of Theorem 3.1.\par
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 			 			  				\begin{bibitemlist}{1}
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