By Hahl H., Salzmann H.

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**Example text**

1 corresponding to the Riesz basis (φk ). We deﬁne the operator Q ∈ L(H1 , l2 ) by Qz = (ξz, (Q0 z)1 , (Q0 z)2 , (Q0 z)3 , . ). It is clear that Qφk = ek+1 for all k ∈ {0, 1, 2, . }. 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.

For k ∈ N, let φk ∈ D(A) be deﬁned by φk (x) = 2 sin (kx) π ∀ x ∈ (0, π). 44 Chapter 2. Operator Semigroups Then (φk ) is an orthonormal basis in X and we have Aφk = − k 2 φk ∀ k ∈ N. Simple considerations about the diﬀerential equation Az = f , with f ∈ L2 [0, π], show that 0 ∈ ρ(A). Thus we have shown that A is diagonalizable. 5, A is the generator of a strongly continuous semigroup T on X given by e−k Tt z = 2 t z, φk φk ∀t 0, z ∈ X . 12) k∈N It is now clear that this semigroup is exponentially stable.

Combining the last two propositions, we see that if A is diagonalizable, then σ(A) is the closure of σp (A). 6), we obtain that for every s ∈ ρ(A), 1 inf |s − λk | (sI − A)−1 k∈N 1 M · . 8. 5. 3, A is the generator of a strongly continuous semigroup T on X if and only if sup Re λk < ∞. 8) k∈N If this is the case, then k∈N and for every t 0, eλk t z, φ˜k φk Tt z = k∈N ∀ z ∈ X. 9) 42 Chapter 2. Operator Semigroups A semigroup as in the last proposition is called diagonalizable. Proof. 7) holds. 9) deﬁnes a bounded operator Tt on X.