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By Wu X.

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Clearly Q is bounded (because ξ and Q0 are bounded). Finally, Q is invertible, because Q−1 (a1 , a2 , a3 , . . ) = a1 φ0 + Q−1 0 (a2 , a3 , a4 , . . ). 40 Chapter 2. 6 Diagonalizable operators and semigroups In this section we introduce diagonalizable operators, which can be described entirely in terms of their eigenvalues and eigenvectors, thus having a very simple structure. If a semigroup generator is diagonalizable, then so is the semigroup. Many examples of semigroups discussed in the PDEs literature are diagonalizable.

H Thus, Rs (sI − A)z = z for z ∈ D(A), so that s ∈ ρ(A) and Rs = (sI − A)−1 . 2. From the last proposition we see that T is uniquely determined by its generator A. 4. 3. 4), then Mω Re s − ω (sI − A)−1 ∀ s ∈ Cω . 2) Proof. 1, by estimating the integral: ∞ (sI − A)−1 z e−(Re s)t Tt · z dt ∀ z ∈ X. 0 It is often needed to approximate elements of X by elements of D(A) in a natural way. 6, another one is given below. 4. Let D(A) be a dense subspace of X and let A : D(A) → X be such that there exist λ0 0 and m > 0 such that (λ0 , ∞) ⊂ ρ(A) and λ(λI − A)−1 Then we have m lim λ(λI − A)−1 z = z λ→∞ ∀ λ > λ0 .

2, A is closed. 5. If A : D(A) → X and β, s ∈ ρ(A), then simple algebraic manipulations show that the following identity holds: (sI − A)−1 − (βI − A)−1 = (β − s)(sI − A)−1 (βI − A)−1 . This formula is known as the resolvent identity. 6. If T ∈ L(X) is such that T < 1, then I − T is invertible and (I − T )−1 = I + T + T 2 + T 3 + · · · , (I − T )−1 The proof of this lemma is easy and it is left to the reader. 1 . 2. 7. Suppose that A : D(A) → X, D(A) ⊂ X and β ∈ ρ(A). Denote rβ = (βI − A)−1 . If s ∈ C is such that |s − β| < r1β , then s ∈ ρ(A) and (sI − A)−1 rβ .

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