• April 20, 2017
• Symmetry And Group By Jean-Paul Pier

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Note that for t ∈ [0, 1], Tt is described by (Tt z)(x) = z(x + t) eα z(x + t − 1) if t + x else. 15. 13. This example shows the importance of imposing the condition ρ(A) = ∅ in the deﬁnition of a diagonalizable operator. 3. Let X = L2 [0, π] and let the operator A be deﬁned by D(A) = H2 (0, π), Az = d2 z dx2 ∀ z ∈ D(A). 8. 8) we have Aφk = − k 2 φk ∀ k ∈ N. 1. 5) does not hold for A. Indeed, consider the constant function z(x) = 1 for all x ∈ (0, π). 5) would yield a non-zero series (which is not convergent in X).

3. Let X = L2 [0, π] and let the operator A be deﬁned by D(A) = H2 (0, π), Az = d2 z dx2 ∀ z ∈ D(A). 8. 8) we have Aφk = − k 2 φk ∀ k ∈ N. 1. 5) does not hold for A. Indeed, consider the constant function z(x) = 1 for all x ∈ (0, π). 5) would yield a non-zero series (which is not convergent in X). 8 (there, this operator was denoted by A). Then clearly A is an extension of A1 . More precisely, if we denote by V the space of aﬃne functions on (0, π), then dim V = 2 and D(A) = D(A1 ) + V . Hence, the graph G(A) is the sum of G(A1 ) and a two-dimensional space.

Proof. If (A, C) is observable, then (as already mentioned) Qτ > 0 (for every τ > 0). Since Q Qτ , it follows that Q > 0. To prove the converse statement, suppose that (A, C) is not observable and take x ∈ Ker Ψτ , x = 0. Then CetA x = 0 for all t 0, hence Qx = 0, which contradicts Q > 0. The proof for R > 0 is similar, using the dual system. Chapter 2 Operator Semigroups In this chapter and the following one, we introduce the basics about strongly continuous semigroups of operators on Hilbert spaces, which are also called operator semigroups for short.