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

Is A −1 an infinitesimal generator? ∗Hans ZwartDepartment of Applied Mathematics, University of TwenteP.O. Box 217, 7500 AE Enschede, The Netherlandsh.j.zwart@math.utwente.nlNovember 18, 2004AbstractIn this paper we study the question whether A −1 is the infinitesimal generator of abounded C 0 -semigroup if A generates a bounded C 0 -semigroup. If the semigroup generatedby A is analytic and sectorially bounded, then the same holds for the semigroupgenerated by A −1 . However, we construct a contraction semigroup with growth boundminus infinity for which A −1 does not generate a bounded semigroup. Using this examplewe construct an infinitesimal generator of a bounded semigroup for which its inverse doesnot generate a semigroup. Hence we show that the question posed by deLaubenfels in [8]must be answered negatively. All these example are on Banach spaces. On a Hilbert spacethe question whether the inverse of a generator of a bounded semigroup also generates abounded semigroup still remains open.1 IntroductionBefore we begin with giving background information on the problem, we first formulate theproblem. Note that this is the Hilbert space version of the question posed by deLaubenfelsat the end of [8].Problem 1.1. Let H be a Hilbert space and let A be the infinitesimal generator of a boundedC 0 -semigroup (T (t)) t≥0on H, i.e., ‖T (t)‖ ≤ M for all t ≥ 0. Furthermore, assume that A −1exists as a closed and densely defined operator.Are the above conditions sufficient for A −1 to be the infinitesimal generator of a boundedC 0 -semigroup?□In Section 2 we show that the answer to our question is positive if (under the conditionsof the problem) any A −1 is the infinitesimal generator of a C 0 -semigroup. Thus we get theboundedness for ”free”. Since the question is posed on a Hilbert space, one may wonder whatis the solution to this problem on a Banach space. In Section 3 we show that for a Banach spacethe answer to the problem is negative. More precisely, we construct a contraction semigroupwith growth bound minus infinity for which the inverse generator does not generate a boundedsemigroup. Hence the question posed by deLaubenfels in [8] should be answered negatively.However, on general Banach spaces, there is a positive result as well. Namely, the inverse of∗ file:artikle/inversegen.texDRAFT1

the generator of a sectorially bounded semigroup is again a generator of a sectorially boundedanalytic semigroup. This result originates from deLaubenfels, [8], but we present a new proof.In Section 4 we summarize some results on Problem 1.1. On Hilbert spaces it is not hard toshow that the answer to our central problem is positive for the class of contraction semigroups.In the remaining of this section we discuss the motivation for our problem. The first motivationcomes from systems theory. Within infinite-dimensional systems theory one studiesthe following set of equations:ẋ(t) = Ax(t) + Bu(t) x(0) = x 0y(t) = Cx(t) + Du(t),where A : D(A) ⊂ H → H is the infinitesimal generator of a C 0 -semigroup on the Hilbertspace H, B is a bounded linear operator from the Hilbert space U to the dual of the domainof A ∗ , i.e., B ∈ L(U, D(A ∗ ) ′ ), C ∈ L(D(A), Y ), where Y is a third Hilbert space, andD ∈ L(U, Y ). In [1], Ruth Curtain introduced the following related system:ẋ 1 (t) = A −1 x 1 (t) + A −1 Bu 1 (t)y 1 (t) = −CA −1 x 1 (t) + (D − CA −1 B)u 1 (t).This system has the nice property that A −1 B ∈ L(U, H), CA −1 ∈ L(H, Y ). Hence this systemis a bounded linear systems as studied in Curtain and Zwart [2]. Furthermore, the systems(1) and (2) share many properties. For instance, (1) is input-state stable if and only if (2)is. Here input-state stability means that for all inputs u ∈ L 2 ((0, ∞); U) the solution of (1)exists and is (uniformly) bounded on [0, ∞).The only stability property which is not known is the state-state stability, i.e., if thesemigroup generated by A is (strongly) stable if and only if the semigroup generated by A −1(strongly) stable. It is not hard to show that this property holds if the answer to our problemis positive, see also Grabowski and Callier [5].The second motivation for our problem comes from numerical analysis. Consider the(abstract) differential equationẋ(t) = Ax(t), x(0) = x 0 . (3)A standard method for solving this differential equation is by using Crank-Nicolson method.In this method the differential equation (3) is replaced by the difference equationx d (n + 1) = (I + ∆A/2)(I − ∆A/2) −1 x d (n), x d (0) = x(0), (4)where ∆ is the time step.If H is finite-dimensional, and thus A is a matrix, then it is easy to show that the solutionsof (3) are bounded if and only if the solutions of (4) are bounded. Or equivalently,if and only ifsup ‖e At ‖ =: M c < ∞t≥0DRAFTsup ‖A n d ‖ =: M d < ∞,n∈N2(1)(2)

where A d = (I + ∆A/2)(I − ∆A/2) −1 . However, the best estimates for M d are dependingon M c and the dimension of H, see [4]. In Guo and Zwart [6] the following estimate wasobtained,M d = sup ‖A n d ‖ ≤ 2 · e(M c 2 + M c,−1 2 ), (5)where M c,−1 = sup ‖e A−1t ‖.t≥0n∈NIt is not hard to show that if the answer to our problem is positive, then M c1 −1 is a functionof M c only, and thus by (5) M d becomes a function of M c only. We close this section withthe remark that (5) holds for any Hilbert space, and so if A generates a bounded semigroupand if the answer to our problem is positive, then the solutions of the difference equation (4)are bounded.2 Equivalent formulation of the problemIn this section we consider the following weaker formulation of Problem 1.1.Problem 2.1. Let H be a Hilbert space and let A be the infinitesimal generator of a boundedC 0 -semigroup (T (t)) t≥0on H, i.e., ‖T (t)‖ ≤ M for all t ≥ 0. Furthermore, assume that A −1exists as a closed and densely defined operator.Are the above conditions sufficient for A −1 to be the infinitesimal generator of a C 0 -semigroup?□Hence the difference between Problem 1.1 and Problem 2.1 is that we do not requirethat the semigroup generated by A −1 is bounded. It is clear that if Problem 1.1 holds, thenProblem 2.1 holds. In the following theorem we show that both problems are equivalent.Note that this does not mean that for any A for which A −1 generates a C 0 -semigroup, thissemigroup will be bounded. It merely means that if the answer to Problem 2.1 is positive forall A’s and all Hilbert spaces, then the answer to Problem 1.1 is positive as well.Theorem 2.2. If Problem 2.1 holds for A’s and all Hilbert spaces, then Problem 1.1 holdsas well.Proof. Assume that A generates a bounded semigroup (T (t)) t≥0, and that A −1 exists as aclosed and densely defined operator. Since Problem 2.1 holds, we conclude that A −1 generatesa C 0 -semigroup S(t). It remains to show that this semigroup is bounded. In order to do this,we introduce the Hilbert space H ext aswith the norm∥ ∥∥∥∥∥∥⎛⎜⎝x 1x 2.H ext = H ⊕ H ⊕ . . .⎞⎟∑⎠= √ ∞ ‖x n ‖ 2 H . (6)∥ n=1HextDRAFTOn H ext we consider the operator A ext = diag (n −1 A). It is easy to see that this generatesthe C 0 -semigroup ( diag ( T (n −1 t) )) t≥0 . Since (T (t)) t≥0is bounded this semigroup whichis bounded. The operator A ext has as inverse diag (nA −1 ). By the conclusion of Problem2.1 this generates a C 0 -semigroup. It is not hard to see that this semigroup is given by3

(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

Since arg(s −1 ) = − arg(s), we may use our estimate for the resolvent of A, and we obtain forall s ∈ C \ {0} with argument less than π 2 + θ∥‖(sI − A −1 ) −1 ‖ ≤ 1|s| + 1 ∥∥∥∥ (A|s| 2 − 1 ) ∥−1∥∥∥∥s I≤ 1|s| + 1|s| 2 · M| 1 s |= M + 1 .|s|Using theorem Theorem 2.5.2 of [9], we conclude that A −1 generates a sectorially boundedanalytic semigroup.In the rest of this section the following function will play an important role.h ac (t) = 1 √tJ 1 (2 √ t), (9)where J 1 (·) is the Bessel function of the first kind and of the first order.properties can be found in the literature, see e.g. Watson [10];The followinga. h ac (·) is continuous on (0, ∞), and continuous from the right at zero with limit one.b. h ac (·) is bounded by one on [0, ∞).c. For large t, we haveh ac (t) ≈ √ 1 t − 3 4 cos(2 √ t − 3 π 4 π).d. The function h ac (·) is the derivative of −J 0 (2 √ t), where J 0 is the Bessel function of thefirst kind and of the zeroth order.Furthermore, using property d. and [3], we have that∫ ∞Using this equation we can prove the following lemma.0e −st h ac (t)dt = 1 − e −1/s . (10)Lemma 3.2. Let A generate an exponentially stable C 0 -semigroup (T (t)) t≥0on the Banachspace X. Then the following equality holdse A−1τ x 0 = x 0 −∫ ∞0τh ac (tτ)T (t)x 0 dt. (11)Proof. Since T (t) is exponentially ( ) stable, A −1 exists as a bounded operator. Thus it generatesthe C 0 -semigroup e A−1 t. Furthermore, combining the exponential stability withproperty b., we see that the integral in (11) is well-defined.DRAFTt≥05

To show the equality in (11) we take the Laplace-transform on both sides. The Laplacetransform of e A−1τ x 0 equals (sI −A −1 ) −1 x 0 , whereas the Laplace transform of x 0 equals x 0 /s.Thus it remains to calculate the Laplace transform of the integral term in (11).∫ ∞[∫ ∞]e −sτ τh ac (tτ)T (t)x 0 dt dτ00∫ ∞[∫ ∞]=e −sτ τh ac (tτ)dτ T (t)x 0 dt, using Fubini0 0∫ ∞[∫ ∞=e − s t ˜τ 1 ]0 0 t 2 ˜τh ac(˜τ)d˜τ T (t)x 0 dt ˜τ = tτ∫ ∞[ ∫ d ∞= −e − s 1t ˜τ h ac (˜τ)d˜τ]0 ds 0t T (t)x 0dt∫ ∞[ ] d 1= −0 ds (1 − e−t/s )t T (t)x 0dt by (10)∫ ∞=[e −t/s 1 ]s 2 T (t)x 0 dt0= 1 s 2 ( 1s I − A ) −1x 0 .Using equation (8) we have that the Laplace transform of the right-hand side of (11) is equalto1s x 0 − 1 ( ) 1 −1s 2 s I − A x 0 = ( sI − A −1) −1x0which proves the lemma.Under the assumption that A generates an exponentially stable semigroup, the abovelemma gives the relation between the semigroups generated by A −1 and A. Since the semigroupgenerated by A is exponentially stable, A −1 is a bounded operator, and thus it isthe infinitesimal generator of a C 0 -semigroup. Using convergence results of semigroups, itis not hard to show the following. Let A be the infinitesimal generator of a bounded semigroup.∫Then A −1 is the infinitesimal generator of a C 0 -semigroup if and only if the limit of∞0τh ac (tτ)e −εt T (t)x 0 dt for ε ↓ 0 exists for all x ∈ X.( )Using the formula (11), we can show that in general e A−1 tis not (uniformly) bounded.However, on the domain A, we have that this semigroup converges to zero.Theorem 3.3. Let A be the infinitesimal generator of an exponentially stable semigroup(T (t)) t≥0on the Banach space X. Then the following holdsa. If ‖T (t)‖ ≤ Me −ωt , with M, ω > 0, then for x 0 ∈ D(A) we haveb. For x 0 ∈ D(A), we have thatt≥0‖e A−1t x 0 ‖ ≤ M ω ‖Ax 0‖. (12)DRAFTlimt→∞ eA−1t x 0 = 0. (13)6

Proof. We begin by studying the integral term in (11). Substituting ˜t = tτ is this term gives∫ ∞0∫ ∞( ) ˜tτh ac (tτ)T (t)x 0 dt = h ac (˜t)T x 0 d˜t0τ) ( ) ∣ ˜t ∣∣∣∞ ∫ ∞ ) ( ) 1 ˜t= −J 0(2√˜t T x 0 + J 0(2√˜tτ0 0τ T Ax 0 d˜tτ∫ ∞ ) ( ) 1 ˜t= x 0 + J 0(2√˜tτ T Ax 0 d˜t,τ0where we have used property d. and the fact that J 0 (0) = 1, J 0 (∞) = 0. So we have thate A−1τ x 0 = −∫ ∞0) ( ) 1 ˜tJ 0(2√˜tτ T Ax 0 d˜t (14)τSince the Bessel function J 0 is bounded by one on the positive real line, we can bound thesemigroup generated by A −1 as‖e A−1τ x 0 ‖ ≤∫ ∞01τ Me− ω ˜t τ ‖Ax 0 ‖d˜t = M ω ‖Ax 0‖,which proves (12).In order to prove part b., we use that fact that there exists a t 1 > 0 such that |J 0 (2 √˜t)| ≤1 ˜t π −1/4 for ˜t > t 1 , see Watson [10, p. 199]. Using (14) we have‖e A−1τ x 0 ‖≤∫ t101τ M‖Ax 0‖d˜t += Mt 1‖Ax 0 ‖ +τ= Mt 1‖Ax 0 ‖ +τ≤ Mt 1‖Ax 0 ‖ + M τπ∫ ∞1∫ ∞t 11t∫ 1∞t 1 /τπ ˜t −1/4 1 τ Me− ω τ ˜t ‖Ax 0 ‖d˜tπ ˜t −1/4 1 τ Me− ω ˜t τ ‖Ax 0 ‖d˜tMπ t−1/4 τ −1/4 e −ωt ‖Ax 0 ‖dt1τ 1/4 ‖Ax 0‖∫ ∞0t −1/4 e −ωt dt.From this it is clear that ‖e A−1τ x 0 ‖ converges to zero for τ going to infinity.So we see that the semigroup generated by A −1 is uniformly bounded on the domain of A.In the next theorem we estimate this semigroup on the Banach space, and find it is boundedby a constant times ( t 1/4 ) . In Example 3.5 we show that this bound is the best possible. Thusthe semigroup e A−1 tneeds not to be bounded, even when (T (t)) t≥0is exponentiallyt≥0stable.Theorem 3.4. Let A be the infinitesimal generator of a exponentially stable semigroup(T (t)) t≥0on the Banach space X. Then for t ≥ 0 there exists an M 0 > 0 such thatDRAFT‖e A−1t ‖ ≤ 1 + M 0 t 1 4 . (15)7

Proof. We again concentrate on the integral term in (11). We have that, using property c.,∫ ∞0‖τh ac (tτ)T (t)x 0 ‖dt≤≤∫ ∞0∫ ∞Mτ|h ac (tτ)|Me −ωt ‖x 0 ‖dt0M 2= MM 2 ‖x 0 ‖τ 1 4τ(tτ) 3/4 e−ωt ‖x 0 ‖dt∫ ∞= MM 2 ‖x 0 ‖τ 1 π √ 24ω 1 4 Γ( 3 4 ).Combining this inequality with equation (11), we see that (15) holds.01t 3/4 e−ωt dtExample 3.5. In our example we take X = C 0 (0, 1), i.e., the space of all continuous functionson [0, 1), which have limit zero at one. This is a Banach space under the supremum norm.On this state space we choose as semigroup{f(t + η) t + η ∈ [0, 1)(T (t)f) (η) =(16)0 t + η ≥ 1It is not hard to see that this is a strongly continuous semigroup on X with ‖T (t)‖ ≤ 1, andT (t) = 0 for t ≥ 1. So it is a contraction semigroup with growth bound minus infinity. Inparticular, the semigroup is exponentially stable.From Lemma 3.2 we have that( )∫ 1−ηe A−1τ f (η) = f(η) − τh ac (tτ)f(t + η)dt= f(η) −0∫ (1−η)τ0( ) ˜th ac (˜t)fτ + η d˜t. (17)By constructing a suitable sequence of functions f’s, we show that ‖e A−1τ ‖ behaves like 4√ τ.Therefore we fix τ.Let φ ε be a continuous function on R which is positive, bounded by one, φ ε (0) = 0, andφ ε (t) = 1 for |t| ≥ ε. Furthermore, define ψ τ be a continuous function on [0, ∞) which ispositive, bounded by one, ψ τ (t) = 0 for t ≥ τ, and ψ τ (t) = 1 for t < τ − 1.Let t k , k = 1, . . . , K be the zeros of h ac in [0, τ].[ K]∏g ε (t) = φ ε (t − t k ) sign(h ac (t))ψ τ (t), t ≥ 0.k=1It is not hard to see that g ε is continuous on [0, ∞) with g ε (t) = 0 for t ≥ τ and sup t≥0 |g ε (t)| =1. Hence f(t) := −g ε (t/τ) is an element of X with norm one.Next choose in equation (17) η = 0 and f(t) = −g ε (τt), then( )∫ τ( ) ˜te A−1τ f (0) = f(0) − h ac (˜t)f0 τ + 0 d˜t∫≥ f(0) +|h ac (˜t)|d˜t.DRAFT{˜t∈[0,τ−1]||˜t−t k |≥ɛ, for k=1,...K}8

Since f has norm one, we see that∫‖e A−1τ ‖ ≥ 1 + sup= 1 +ε>0 {˜t∈[0,τ−1]||˜t−t k |≥ɛ, for k=1,...K}∫ τ−10|h ac (˜t)|d˜t|h ac (˜t)|d˜t. (18)By property c. we find that this is of the order τ 1/4 . So for the exponentially stable semigroup(16) the semigroup generated by A −1 is not uniformly bounded.As one can see from the above proof, it would simplify a lot if we could choose f(t) =h ac (tτ). However, since the sign function is not continuous, we have to approximate it withcontinuous functions.Remark 3.6. The technique of Theorem 3.4 and Example 3.5 can be used to prove a moregeneral result. Let h : [0, ∞) → R be locally integrable. Then the transformation:H(A, τ)x 0 =∫ ∞0τh(tτ)T (t)x 0 dtgives for every exponentially stable C 0 -semigroup T (t) an uniformly bounded operator-valuedfunction if and only if h is absolutely integrable. Furthermore, we have that‖H(A, τ)‖ ≤∫ ∞For more information, see also Chapter XV of [7].0|h(t)|dt · sup ‖T (t)‖.Using Example 3.5 and Remark 2.3 it is easy to construct an example of a bounded C 0 -semigroup for which A −1 does not generate a C 0 -semigroup. This example shows that thequestion posed by deLaubenfels in [8] must be answered negatively.Example 3.7. Let X = C 0 (0, 1) be the Banach space of Example 3.5 and let (T (t)) t≥0bethe shift semigroup of the same example and let A be its infinitesimal generator. By (S(t)) t≥0we denote the semigroup generated by A −1 . The Banach space X ext is defined aswith the norm∥ ∥∥∥∥∥∥⎛⎜⎝x 1x 2.t≥0X ext = X ⊕ X ⊕ . . .⎞⎟∑⎠= √ ∞ ‖x n ‖ 2 X .∥ n=1XextOn X ext we consider the operator A ext = diag (n −1 A). It is easy to see that this generatesthe C 0 -semigroup (T ext (t)) t≥0:= ( diag ( T (n −1 t) )) t≥0 . Using the fact that (T (t)) t≥0 is acontraction semigroup, it is easy to show that (T ext (t)) t≥0 is a contraction semigroup on X ext .If A −1ext would be the generator of a C 0-semigroup, then this semigroup would be given by(diag (S(nt))) t≥0. Since a semigroup is bounded on every compact interval, we findsup ‖S(t)‖ = sup sup ‖S(nt)‖ = sup ‖diag (S(nt))‖ < ∞.t≥0DRAFTt∈[0,1]nt∈[0,1]However, we known from Example 3.5 that (S(t)) t≥0is not uniformly bounded. Hence A −1extis not the infinitesimal generator of a C 0 -semigroup.9

4 e A−1t on a Hilbert spaceOn Hilbert spaces there is, apart from the class of (sectorially) bounded analytic semigroups,a second class of generators for which our problem has been solved. This is the class ofinfinitesimal generators of contraction semigroups. It is well known that A is the infinitesimalgenerator of a contraction semigroup on the Hilbert space H if and only if Re〈x, Ax〉 ≤ 0for all x ∈ D(A) and (λI − A) is invertible for some λ > 0. Using this, the following lemmafollows almost directly.Lemma 4.1. Let A be the infinitesimal generator of a contraction semigroup on the Hilbertspace H and assume that A −1 exists as a closed, densely defined operator. Then A −1 is alsothe infinitesimal generator of a contraction semigroup.Proof. Since the domain of A −1 equals the range of A, we have to study the inner productof A −1 x with x for x in the range of A. Since these elements can be written as Ay, withy ∈ D(A), we find thatRe 〈x, A −1 x〉 = Re 〈Ay, y〉Since by assumption this last expression is positive, we see that A −1 satisfies one of thecondition. Furthermore, it is easy to show that (λI − A −1 ) is boundedly invertible for someλ > 0, see e.g. (8). Hence A −1 is the infinitesimal generator of a contraction semigroup onH.Since on a Hilbert space, many bounded semigroups are similar to a contraction semigroup,we see that on Hilbert spaces the answer to our problem is yes in many cases. One case ofparticular interest is the class of bounded groups, see [12]. However, it is unknown whetherthe problem holds for any semigroup on a Hilbert space.References[1] R.F. Curtain, Regular linear systems and their reciprocals: applications to Riccati equations,Systems and Control Letters, 49, pp. 81–89, 2003.[2] R.F. Curtain and H.J. Zwart, An introduction to Infinite-Dimensional Linear SystemsTheory, Springer-Verlag, New York, 1995.[3] G. Doetsch, Introduction to the Theory and Application of the Laplace transformation,Springer Verlag, Berlin, 1970.[4] J.L.M. van Dorsselaer, J.F.B.M. Kraaijevanger, and M.N. Spijker, Linear stability analysisin the numerical solution of initial value problems, Act Numerica, pp. 199–237, 1993.[5] P. Grabowski and F.M. Callier, On the Circle Criterion for Boundary Control Systemsin Factor Form: Lyapunov stability and Lur’e equations, Report 2002/05, Departmentof Mathematics, University of Namur (FUNDP), April 2002[6] B.Z. Guo and H. Zwart, On the relation between stability of continuous- and discretetimeevolution equations via the Cayley transform, IOET, (to appear).DRAFT[7] E. Hille and R.S. Phllips, Functional Analysis and Semigroups, AMS Providence R.I.,1957.10

[8] R. deLaubenfels, Inverses of generators, Proc. AMS., 104(2) (1988), 443-448.[9] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations,Springer Verlag 1983.[10] G.N. Watson, Theory of Bessel Functions, second edition, Cambridge, 1962.[11] H. Zwart, B. Jacob and O. Staffans, Weak admissibility does not imply admissibility foranalytic semigroups, Systems & Control Letters, 48, (2003), 341–350.[12] H. Zwart. On the invertibility and bounded extension of C 0 -semigroups. SemigroupForum, 63(2):153–160, 2001.DRAFT11