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By Hino Y., et al.

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We now show that 1 ∈ ρ(P ). For every x ∈ X put f (t) = U (t, 0)g(t)x for t ∈ [0, 1], where g(t) is any continuous function of t such that g(0) = g(1) = 0, and 1 g(t)dt = 1 . 0 Thus f (t) can be continued to a 1-periodic function on the real line which we denote also by f (t) for short. Put Sx = [L−1 (−f )](0) . Obviously, S is a bounded operator. We have 1 [L−1 (−f )](1) = U (1, 0)[L−1 (−f )](0) + U (1, ξ)U (ξ, 0)g(ξ)xdξ 0 Sx = P Sx + P x. Thus (I − P )(Sx + x) = P x + x − P x = x. So, I − P is surjective.

Now we show the implication “ weak admissibility ⇒ mild admissibility”. As A D(DM − AM ) contains all classical solutions on R, the proof of this implication can be done as in that of the previous ones. The proof of the corollary is complete if we prove the implication “mild admissibility ⇒ admissibility”. In fact, this can be shown as in the proof of i). CHAPTER 2. 2. 24). Recall that by definition Λ(X) = {f ∈ BU C(R, X) : sp(f ) ⊂ Λ} , where Λ is a given closed subset of R. Obviously, Λ(X) is a translation invariant closed subspace of BU C(R, X).

Proof. Let us define t U1 (t) = t f (s)ds, t ∈ R, u(s)ds, F1 (t) = 0 0 t Uk (t) = t Fk−1 (s)ds, t ∈ R, k ∈ N. Uk−1 (s)ds, Fk (t) = 0 0 Then, by definition, we have u(t) = Pn (t) + A(Un (t)) + Fn (t), t ∈ R, where Pn is a polynomial of order of n − 1. From the closedness of A, we have u ∗ φ(t) = Pn ∗ φ(t) + A(Un ∗ φ(t)) + Fn ∗ φ(t), t ∈ R. Since the Fourier transform φ has compact support all convolutions above are infinitely differentiable. From the closedness of A we have that (Un ∗ φ)(k) (t) ∈ D(A), t ∈ R and CHAPTER 2.

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