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2 Positive Periodic Solutions of the Equation x (t) = a(t)x(t) − τ f (t, x(h(t))) 35 c2 MT , . (1 − λ)(T − λ1 ) λ c4 > max Now for x ∞ K c4 , we have t+T ≥Aτ x≥ = sup τ 0≤t≤T G(t, s) f (s, x(h(s))) ds t t+T ≤ sup τ 0≤t≤T G(t, s)(a(s)x(h(s)) λ1 + M) ds t t+T ≤ sup τ 0≤t≤T ⎡ G(t, s)(a(s)≥x≥ λ1 + M) ds t t+T t+T ≤ τ ⎣ λ1 c4 sup G(t, s) ds ⎦ a(s)G(t, s) ds + sup M 0≤t≤T 0≤t≤T t ⎤ t M T 1−λ ≤ τ λ1 c4 + < c4 . Hence, Aτ : K c4 → K c4 . Next, we define a nonnegative concave continuous functional ψ on K by ψ(x) = min x(t).

A(β, x(β)) dβ − 1 0 In view of the above, we define an operator Aτ by t+T Aτ x = τ G(t, s) f (s, x(s − γ1 (s, x(s))), . . 30) t for every x ∞ X and t ∞ R. Clearly Aτ x(t + T ) = Aτ x(t) and Aτ : X → X . The Green’s kernel G(t, s) satisfies the inequality T c(β) dβ exp 1 ψ= T exp c(β) dβ − 1 0 0 ≤ |G(t, s)| ≤ T exp b(β) dβ − 1 0 for every 0 ≤ t ≤ s ≤ t + T . Let k1 = exp ⎜ T exp 0 c(β) dβ . Then, ψ= =ω ⎜ T 0 b(β) dβ and k2 = k2 k2 (k2 − 1) 1 ω , ω= , k1 ≤ k2 , and λ = = > 1. k2 − 1 k1 − 1 ψ (k1 − 1) For any x ∞ X , we have 48 2 Positive Periodic Solutions of Nonlinear Functional Differential Equations τk2 ≥Aτ x≥ ≤ k1 − 1 T f (s, x(s − γ1 (s, x(s))), .

10 Let τ = 1, f˜0 < 1, and f˜∈ < 1. In addition, assume that there exists ρ > 0 such that f (t, x) > a(t)|x| for μρ < |x| < ρ, where μ = T ex p{− 0 a(s) ds}. 5) has at least two positive T -periodic solutions x1 and x2 such that 0 < ≥x1 ≥ < ρ < ≥x2 ≥. 9. 1 Let f˜0 < T and f˜∈ < T , and assume that there exists a constant c2 > 0 such that either (H1 ) or (H8 ) holds. 5) has at least three positive T -periodic solutions for 1 λ−1 <τ < . 10. 10 it is 1. 1. 11 Let f˜∈ < T . Assume that there are constants 0 < c1 < c2 such that (H1 ) holds and (H9 ) f (t, x) < x for 0 ≤ x ≤ c1 and 0 ≤ t ≤ T .