finite. Therefore, from the inequality (28), we have which contradicts (22). Case 2. τ ≤ 1 . Since the proof is similar to proof of Case 1 , we shall give the sketch of the proof. Dividing inequality (18) by z ( t ) , and then integrating from t − τ to t, it is obtained that Using the similar arguments in Case 1 , we get that lim inf t →∞ z ( t − τ ) z ( t ) is finite. So, from inequality (29), we have which contradicts (21). Moreover, dividing inequality (18) by z ( t ) , and then integrating from n + 1 − τ to n + 1 , it is obtained that By using the similar arguments in Case 1 , we get that lim inf n →∞ z ( n + 1 − τ ) z ( t ) is finite. 15 z ( n + 1 − τ ) z ( n + 1) ≥ $_{e}$  n +1 ∫ n +1 − τ P ( s ) z ( s − τ ) z ( s ) $_{ds}$  . (30) lim t $_{→∞}$inf t ∫ t − τ   ∏ s − τ<t$_{j}$ ≤ s (1 − b$_{j}$ $_{)}$  b ( s $_{) exp}$  s ∫ s − τ a ( u ) $_{du}$  ds ≤ 1 e , z ( t − τ ) z ( t ) ≥ $_{e}$  t ∫ t − τ P ( s ) z ( s − τ ) z ( s ) $_{ds}$  . (29) lim n $_{→∞}$inf n +1 ∫ n   ∏ n − 1 <t$_{j}$ ≤ s (1 − b$_{j}$ $_{)}$  c ( s $_{) exp}$  s ∫ n − 1 a ( u ) $_{du}$  ds ≤ 1 e ,