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4, every diagonalizable operator is closed. 4. For unbounded operators we must be careful with the definition of the domain and the situation is described in the following two propositions. 2. Let (φk ) be a Riesz basis in X and let (φ˜k ) be the biorthogonal sequence to (φk ). Let (λk ) be a sequence in C which is not dense in C. 1) k∈N λk z, φ˜k φk Az = ∀ z ∈ D(A). 2) k∈N Then A is diagonalizable, we have σp (A) = {λk | k ∈ N}, σ(A) is the closure of σp (A) and for every s ∈ ρ(A) we have (sI − A)−1 z = k∈N 1 z, φ˜k φk s − λk ∀ z ∈ X.

8. Let τ > 0, take X = L2 [0, τ ] and for every t ∈ R and z ∈ X define (Tt z)(x) = z(x + t) if x + t 0 else. τ, Then T is a strongly continuous semigroup. Clearly Tτ = 0 (the semigroup is vanishing in finite time), so that ω0 (T) = −∞. It is not difficult to verify that the generator of T is A= d , dx D(A) = {z ∈ H1 (0, τ ) | z(τ ) = 0} and σ(A) = ∅ (this last fact is impossible for bounded operators A). 9. Take X = L2 (R) and for every t > 0 and z ∈ X define 1 (Tt z)(x) = √ 4πt ∞ −∞ e− (x−σ)2 4t z(σ)dσ ∀ x ∈ R.

8. Suppose that A : D(A) → X is the generator of a strongly continuous semigroup T on X, and −A is the generator of a strongly continuous semigroup S on X. Extend the family T to all of R by putting T−t = St , for all t > 0. Then T is a strongly continuous group on X. Proof. 5, d Tt St z = ATt St z + Tt (−A)St z = 0 . dt This implies that Tt St z = z for all t 0. By a similar argument, St Tt z = z for all t 0. Since D(A) is dense in X, we conclude that Tt is invertible and its inverse is St . 4, T can be extended to a strongly continuous group in the manner described in the proposition.

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