Mathematical Theory and Modeling ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012
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[ d ( y2n+1 , y2n )] = [ d ( Ax2n , Bx2 n−1 )] ≤ k1 [ d ( Ax2 n , Sx2 n ) d ( Bx2 n−1 , Tx2 n −1 ) + d ( Bx2 n−1 , Sx2 n ) d ( Ax2 n , Tx2 n −1 ] + k2 [ d ( Ax2 n , Sx2 n ) d ( Ax2 n , Tx2 n−1 ) + d ( Bx2 n −1 , Tx2 n−1 ) d ( Bx2 n −1 Sx2 n ) ] 2
2
= k1 [ d ( y2 n +1 , y2 n ) d ( y2 n , y2 n−1 ) + 0 ]
+ k2 [ d ( y2 n +1 , y2 n ) d ( y2 n +1 , y2 n −1 ) + 0]
This implies
d ( y2 n +1 , y2 n ) ≤ k1 d ( y2 n , y2 n −1 ) + k2 [ d ( y2 n +1 , y2 n ) + d ( y2 n , y2 n −1 ) ] d ( y2 n +1 , y2 n ) ≤ h d ( y2 n , y2 n −1 ) where h =
k1 + k2 <1 1 − k2
For every int eger p > 0, we get d ( yn , yn + p ) ≤ d ( yn , yn +1 ) + d ( yn+1 , yn + 2 ) + ............ + d ( yn+ p −1 , yn + p ) ≤ h n d ( y0 , y1 ) + h n +1d ( y0 , y1 ) + ............. + h n + p −1d ( y0 , y1 )
≤ ( h n + h n +1 + ............. + h n + p −1 ) d ( y0 , y1 )
≤ h n (1 + h + h 2 + ............. + h p −1 ) d ( y0 , y1 ) Since h<1, , hn → 0 as n→ ∞,
so that d(yn,yn+)→ 0. This shows that the sequence {yn} is a Cauchy
sequence in X and since X is a complete metric space, it converges to a limit, say z∈X. The converse of the Lemma is not true, that is A,B,S and T are self maps of a metric space (X,d) satisfying (1.1.4) and (1.1.6), even if for x0∈X and for associated sequence of x0 converges, space (X,d) need not be complete. The following example establishes this.
1.1.11. Example:
Let
X=(-1,1)
1 1 8 if − 1 ≤ x < 10 Ax = Bx = 1 1 if ≤ x <1 10 10
with d(x,y)=
x− y
1 1 if − 1 < x < 8 10 Sx = Tx = 1 1 − x if ≤ x <1 5 10
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the metric