Applying the Gronwall’s Lemma [14], one can confirm that Hence, the inequality is valid. Since the number k$_{0}$ satisfies (4.4), we have e ( KL$_{f}$ − ω ) k 0 ≤ ωαϵ 2 K ( M$_{f}$ + M$_{g}$ ) so that Therefore, the inequality holds for each integer k with 1 ≤ k ≤ k$_{0}$. It is clear that ( 1 + K ω − KL$_{f}$ ) αϵ < 1 . Thus, For n ≥ n$_{0}$, it can be obtained by using (4.4) one more time that Hence, d ( φ$_{t}$$_{n}$, φ ) → 0 as n → ∞ . Next, we will verify the presence of a positive number ϵ$_{0}$ and a sequence $^{{}$˜ τ$_{n}$ } , ˜ τ$_{n}$ → ∞ as n → ∞ , such that d ( φ$_{t}$$_{n}$$_{+}$$_{˜}$ $_{τ$_{n}$}$, φ$_{˜}$ $_{τ$_{n}$}$) ≥ ϵ$_{0}$ for all n ∈ N . Let N be a natural number such that ∞ ∑ k = N +1 2 − k ≤ ϵ$_{0}$ 2 . One can confirm that 6 N ∑ k =1 2 − $^{k}$ρ$_{k}$ ( g$_{t}$$_{n}$$_{+}$$_{τ}$$_{n}$, g$_{τ}$$_{n}$ ) ≥ ϵ$_{0}$ 2 . ρ$_{k}$ ( φ$_{t}$$_{n}$, φ ) < ( 1 + K ω − KL$_{f}$ ) αϵ, 1 ≤ k ≤ k$_{0}$. sup s ∈ [ − k,k ] ‖ φ ( t$_{n}$ + s ) − φ ( s ) ‖ < ( 1 + K ω − KL$_{f}$ ) αϵ ‖ φ ( t$_{n}$ + s ) − φ ( s ) ‖ < ( 1 + K ω − KL$_{f}$ ) αϵ, s ∈ [ − k$_{0}$, k$_{0}$ ] . ‖ φ ( t$_{n}$ + s ) − φ ( s ) ‖ < Kαϵ ω − KL$_{f}$ + 2 K ( M$_{f}$ + M$_{g}$ ) ω e ( KL$_{f}$ − ω )( s +2 k 0 ) ψ$_{n}$ ( s ) ≤ Kαϵ ω − KL$_{f}$ e ωs ( 1 − e ( KL$_{f}$ − ω )( s +2 k 0 ) ) + 2 K ( M$_{f}$ + M$_{g}$ ) ω e KL$_{f}$$^{s}$e 2( KL$_{f}$ − ω ) k 0 . d ( φ$_{t}$$_{n}$, φ ) = ∞ ∑ k =1 2 − $^{k}$ρ$_{k}$ ( φ$_{t}$$_{n}$, φ ) < ( 1 + K ω − KL$_{f}$ ) αϵ k 0 ∑ k =1 2 − k + ∞ ∑ k = k 0 +1 2 − k < ( 2 + K ω − KL$_{f}$ ) αϵ ≤ ϵ.