SOME FIXED POINT THEOREMS FOR ORDERED REICH TYPE ...

◭◭ ◭ ◮ ◮◮Go backFull ScreenCloseQuitAs β ≥ 0, 0 ≤ α < 1, it follows thatαn1−α d 0 → θ, βα n−1 d 0 → θ, so by (a) and (d) of Remark 1,for every c ∈ E with θ ≪ c, there exists n 0 ∈ N such that d(x n , x n+2m ) ≪ c, d(x n , x n+2m+1 ) ≪ cfor all n > n 0 . Thus, {x n } is a Cauchy sequence in X. Since X is complete, so there exists u ∈ Xsuch thatlim x n = lim fx n−1 = un→∞ n→∞We shall show that u is a fixed point of f. By (IV), we have x n ⊑ u, therefore, it follows from (I)thatd(fx n−1 , fu) ≼ λd(x n−1 , u) + µd(x n−1 , fx n−1 ) + δd(u, fu)d(x n , fu) ≼ λd(x n−1 , u) + µd(x n−1 , x n ) + δ[d(u, x n+1 )+ d(x n+1 , x n ) + d(x n , fu)](1 − δ)d(x n , fu) ≼ λd(x n−1 , u) + δd(x n+1 , u) + µd n−1 + δd n .In view of (4) and the fact that 1 − δ ≥ 0, by (d) of Remark 1, there exists n 1 ∈ N such that forevery c ∈ E with θ ≪ c, d n−1 ≪ (1−δ)4µ c and d n ≪ (1−δ)4δc for all n > n 1 . Also x n → u, so thereexists n 2 ∈ N such that for every c ∈ E with θ ≪ c, d(x n−1 , u) ≪ (1−δ)4λc and d(x n+1, u) ≪ (1−δ)4δcfor all n > n 2 .Thus, we can choose n 3 ∈ N such that(8)Again by rectangular inequality,d(x n , fu) ≪ c for all n > n 3 .d(fu, u) ≼ d(fu, x n ) + d(x n , x n+1 ) + d(x n+1 , u)= d(fu, x n ) + d n + d(x n+1 , u).In view of (8),(4) and the fact that x n → u by (c) of Remark 1, we conclude that d(fu, u) = θ,i.e., fu = u. Thus, u is fixed point of f.