By Richard Bellman (ed.)

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Proof. 10 in [N’Gu´er´ekata (79)], for proof. 7. If we equip AA(X), the space of almost automorphic functions with the sup norm f ∞ = sup f (t) t∈R then it turns out to be a Banach space. If we denote KAA(X), the space of compact almost automorphic X-valued functions, then we have AP (X) ⊂ KAA(X) ⊂ AA(X) ⊂ BC(R, X) . 21. 42). Proof. 48) t∈R which proves the theorem. ii) This statement is straight forward. 22. If f ∈ AA(X) and its derivative f exists and is uniformly continuous on R, then f ∈ AA(X).

34) April 22, 2008 10:13 World Scientific Book - 9in x 6in Preliminaries Proof. stability 23 By definition, for Reλ > 0, ∞ fˆ(λ) = = = = e−λt f (t)dt 0 ∞ nτ e−λt f (t)dt n=1 ∞ (n−1)τ n=1 ∞ 0 n=1 τ = τ e−λ(t+(n−1)τ ) f (t + (n − 1)τ )dt e−λt f (t)dt 0 e−λt f (t)dt 0 ∞ e−(n−1)λτ n=1 τ = τ e−(n−1)λτ e −λt f (t)dt 0 1 . 35) holds true as well. e, λ = 2πin/τ for n ∈ Z, then λ ∈ σ(f ) because at this point fˆ(λ) has a holomorphic extension. Moreover, at λn = 2πin/τ , τ fˆ(λ) has a holomorphic extension if and only if 0 e−2πnt/τ f (t)dt = 0.

From the identity e−λs (A − λ)T (s)x ds we have e−Re λt x = e−Re λt T (t)x t ≤ x + 0 e−Re λs T (s)(λ − A)x ds t = x + e−Re λs ds 0 −Re λt (λ − A)x −1 (λ − A)x . −Re λ This proves the lemma for λ < 0. For λ > 0 the inequality follows from the Laplace transform representation of the resolvent. (ii): This is proved in the same way, after first substituting R(λ, A)x for x in the first formula and passing to the holomorphic extension. 6. Let (T (t))t≥0 be C0 -semigroup of contractions on a Banach space X, with generator A.