A Contraction Fixed Point Theorem in Partially Ordered Metric Spaces and Application to Fractional Differential Equations

and Applied Analysis 3 Besides, if for any x, y ∈ M, there exists z ∈ M which is comparable to x and y, 2.4 then f has a unique fixed point. Proof. We first show that f has a fixed point. Since x0 f x0 and f is an increasing function, we obtain by induction that x0 f x0 f2 x0 f3 x0 · · · f x0 · · · . 2.5 Put xn 1 f x0 , n 1, 2, . . . For each integer n ≥ 1, from 2.5 , we have xn xn 1, then by 2.2 d xn 1, xn d ( f x0 , fn−1 x0 ) ≤ θβ θd xn, xn−1 d xn, xn−1 ≤ Kθd xn, xn−1 . 2.6 If there exists n0 ∈ N such that d xn0 , xn0−1 0, then xn0 fn0−1 x0 f xn0−1 xn0−1 and xn0−1 is a fixed point; in this case, the proof is finished. Otherwise, for any n ∈ N, d xn, xn−1 / 0. Then by 2.6 , we have d xn 1, xn ≤ Kθd xn, xn−1 ≤ · · · ≤ Kθ n d xn0 , xn0−1 −→ 0, as n −→ ∞, 2.7 that is, d xn 1, xn −→ 0, as n −→ ∞. 2.8 Now, we show that xn is a Cauchy sequence. By the triangle inequality and 2.2 , we have d xn, xm ≤ d xn, xn 1 d xn 1, xm 1 d xm 1, xm ≤ d xn, xn 1 θβ θd xn, xm d xn, xm d xm 1, xm ≤ d xn, xn 1 θKd xn, xm d xm 1, xm , 2.9