By Alessandra Lunardi

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7. 4, and then argue by reiteration. 8. 13. 15. 14; for the second statement replace v(λ) by w(λ) = λ2s R(λ, A)s R(λ, B)s x. 15. 7) Prove that (0, +∞) is a ray of minimal growth for the following operators: (a) A : D(A) = Cb1 (R) → Cb (R) (resp. A : D(A) = W 1,p (R) → Lp (R), 1 ≤ p < ∞), Af = f (b) A : D(A) = Cb2 (R) → Cb (R) (resp. A : D(A) = W 2,p (R) → Lp (R), 1 ≤ p < ∞), Af = f (c) A : D(A) = {f ∈ C 2 ([0, π]) : f (0) = f (π) = 0} → C([0, π]) (resp. 1). Due to the Hille-Yosida Theorem, if in addition D(A) is dense in X and for every n ∈ N (λR(λ, A))n L(X) ≤ M , then A is the infinitesimal generator of a strongly continuous semigroup T (t), and the following representation formula holds.

Setting f (σ) = σR(σ, A)x for σ > 0, we have f (σ) = R(σ, A)x − σR(σ, A)2 x = R(σ, A)(I − σR(σ, A))x = −R(σ, A)2 Ax and f (+∞) = x, so that ∞ R(σ, A)2 Axdσ, λ > 0, x − λR(λ, A)x = − λ and if x ∈ D(A2 ), ∞ R(σ, A)2 A2 xdσ, λ > 0. Ax = λAR(λ, A)x − λ Therefore, Ax ≤ λ(M + 1) x + M2 2 A x , λ > 0. 2) so that x D(A) ≤C x 1/2 x 1/2 , D(A2 ) x ∈ D(A2 ), that is, D(A) ∈ J1/2 (X, D(A2 )). Let us prove that D(A) ∈ K1/2 (X, D(A2 )). For every x ∈ D(A) split x as x = −R(λ, A)Ax + λR(λ, A)x, λ > 0, where M x D(A) , λ = λR(λ, A)x + λAR(λ, A)Ax R(λ, A)Ax ≤ λR(λ, A)x D(A2 ) ≤ M x + λ(M + 1) Ax so that setting t = λ−2 K(t, x, X, D(A2 )) ≤ R(t−1/2 , A)Ax + t t−1/2 R(t−1/2 , A)x ≤ M t1/2 x D(A) + M t x + (M + 1)t1/2 Ax , t > 0 D(A2 ) 52 Chapter 3 which implies that t → K(t, x, X, D(A2 )) is bounded in (0, 1] by (2M + 1) x it is bounded by x in (1, ∞), then x ∈ (X, D(A2 ))1/2,∞ and x (X,D(A2 ))1/2,∞ ≤ (2M + 1) x D(A) .

For every t ∈ R we have |F (it)| ≤ T f (it) g(it) Lq0 (Λ) |F (1 + it)| ≤ ≤ Lq0 (Λ) ≤ T T f (1 + it) T L(Lp0 (Ω),Lq0 (Λ)) Lq1 (Λ) L(Lp1 (Ω),Lq1 (Λ)) g(1 + it) a pθ /p1 Lpθ (Ω) a pθ /p0 Lpθ (Ω) b qθ /q0 q θ (Λ) L , Lq1 (Λ) b qθ /q1 q θ (Λ) L . 3) we get T a b ν(dx) ≤ (sup |F (it)|)1−θ (sup |F (1 + it)|)θ |F (θ)| = t∈R Λ ≤ T 1−θ L(Lp0 (Ω),Lq0 (Λ)) T θ L(Lp1 (Ω),Lq1 (Λ)) Since T a Lqθ (Λ) is the supremum of |F (θ)|/ b simple functions on Λ, we get Ta Lqθ (Λ) ≤ T t∈R 1−θ L(Lp0 (Ω),Lq0 (Λ)) q θ (Λ) L T a Lpθ (Ω) b q θ (Λ) L .