By Michael Clausen

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Clearly, f ∈ < ω12 implies that there exist φ ∞ (0, ω12 ) and β > 0 such that f (t, x) < φx for x ≥ β and 0 ≤ t ≤ T . 1. It is easy to show that Aτ : K c4 → K c4 , where c4 > max ω2θ , λc2 . 1. Then for x ∞ K (ψ, c2 , c3 ), from (H5 ) we have ≥Aτ x≥ > c2 . In addition, for x ∞ K c1 , (H6 ) implies T ≥Aτ x≥ ≤ ωτ f (s, x(h(s))) ds 0 T c1 ds ω2 < ωτ < c1 . 1. 1) has at least three positive T -periodic solutions. The theorem is now proved. 4 Let f ∈ < such that (H5 ) holds and (H7 ) f (t, x) < (λ−1)2 c1 λ3 (λ−1)2 λ3 and assume there exist constants 0 < c1 < c2 for x ∞ K , 0 ≤ x ≤ c1 and 0 ≤ t ≤ T .

Xm ) = f (t, x1 , . . , xm ), γi (t + T, x) = γi (t, x) for any x ∞ R+ , t ∞ R, i = 1, . . , m, and T > 0 is a constant. 5 Han and Wang [3] Assume that a1 (t) ≤ a(t, x) ≤ a2 (t) for any (t, x) ∞ R × R+ , where a1 and a2 are nonnegative T -periodic continuous functions T on R and 0 a1 (s) ds > 0. Let lim sup a2 (t) f (t, u 1 , u 2 , . . , u m ) < |u| θ lim sup a2 (t) f (t, u 1 , u 2 , . . 3 Positive Periodic Solutions of the Equation x (t) = a(t)x(t) − τb(t) f (t, x(h(t))) uniformly for t ∞ R, where θ = exp( exp( T 0 T 0 a2 (t)dt)−1 a1 (t)dt)−1 45 .

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))), . . , x(s − γm (s, x(s)))) ds 0 and τ (Aτ x)(t) ≥ k2 − 1 T f (s, x(s − γ1 (s, x(s))), . . , x(s − γm (s, x(s)))) ds 0 1 (k1 − 1) ≥Aτ x≥ = ≥Aτ x≥. ≥ k2 (k2 − 1) λ Thus, if we define a cone K on X by K = x ∞ X : x(t) ≥ 1 ≥x≥ , λ then Aτ : K → K .