Is A−1 an infinitesimal generator?∗ - Applied Mathematics

(diag (S(nt))) t≥0 . Since a C 0 -semigroup is uniformly bounded on every compact interval,and since for all t ≥ 0‖diag (S(nt))‖ = sup ‖S(nt)‖,we conclude thatsup ‖S(t)‖ = sup sup ‖S(nt)‖ = sup ‖diag (S(nt))‖ < ∞.t≥0t∈[0,1]Thus we have proved the theorem.nnt∈[0,1]Remark 2.3. Note that if the space H in the above theorem is a Banach space, then thespace H ext is a Banach space as well. Thus the same theorem holds on a Banach space.3 e A−1t on a Banach spaceIn this section we show that the answer to our problem is negative if we would relax ourassumption on the space. That is, if we would consider the problem on a Banach space.However, before we present these negative results, we show that the answer is positive if Agenerates a (sectorially) bounded analytical semigroup. From Theorem 2.5.2 of [9] we recallthat A generates a (sectorially) bounded analytical semigroup, (‖T (t)‖ ≤ M for all t ∈ Cwith | arg(t)| < θ, θ > 0) if and only if‖(sI − A) −1 ‖ ≤ M |s|for arg(s) < π/2 + θ. (7)Using this characterization it is easy to prove that A −1 generates a sectorially bounded analyticsemigroup if and only if A does.Theorem 3.1. Let A be the generator of a sectorially bounded analytic semigroup and letA −1 exist as a closed, densely defined operator. Then A −1 generates a sectorially boundedanalytic semigroup.Proof. Choose s ∈ C \ {0} with argument less than π 2 + θ.Taking the inverse, we get(sI − A −1 ) =(A − 1 )s I · s · A −1((sI − A −1 ) −1 = A A − 1 ) −1s I · 1s=(A − 1 s I + 1 ) (s I A − 1 ) −1s I · 1s= 1 s + 1 s 2 (A − 1 s I ) −1. (8)DRAFT4