Semigroup Forum Vol. 63 (2001) 305–320

c 2001 Springer-Verlag New York Inc.
DOI: 10.1007/s002330010036

RESEARCH ARTICLE

C0 -groups with Polynomial Growth M. Malejki Communicated by J. A. Goldstein

Abstract We consider some conditions guaranteeing that an unbounded operator is a generator of some strongly continuous group satisfying a polynomial growth condition. The result is applied to a Friedrichs model given by T f (x) = xf (x)+ < f, φ > ψ(x) and considered as an operator in Lp (R). Some sufficient condition on φ and ψ are found such that iT generates an example of a group with polynomial growth.

Introduction In the class of all C0 -groups in complex Banach spaces, there exist groups satisfying a polynomial growth condition. More exactly, if {T (t)}t∈R is a C0 -group of operators in a Banach space X then we say that {T (t)}t∈R satisfies a polynomial growth condition when it is possible to find M > 0 and s ≥ 0 such that T (t) ≤ M (1 + |t|s ) for every t ∈ R. Let an operator A in X be given by 1 Ax = lim (T (t)x − x), t→0 t

x ∈ D(A),

where D(A) is the set of x ∈ X for that the limit exists in X. In this case A is called a generator for {T (t)}t∈R .
∞ n If A is a bounded operator in X then T (t) = exp(tA) = n=0 tn! An , where the series is absolutely convergent in X. When A is unbounded the situation is much more complicated, but I will denote any group generated by A in this way. It seems to be interesting to know some concrete operators A, for which {T (t)}t∈R satisfies a polynomial growth condition. The C0 -groups with a polynomial growth appear in a very natural way as we consider the Weyl functional calculus, given by the inverse Fourier transform, for operators that generate such groups (see for example [2]). In this case the domain of the Weyl functional calculus is quite large. For example every function

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Malejki

from the Schwartz-class belongs to it. Much of the theory of n-parameter C0 groups with polynomial growth condition we are able to find in the book [1] of W.O. Amrein, A. Boutet de Monvel and V. Georgescu. The case of bounded operators is very special. In the book of Colojara and Foias [4] we can find that if A is a generator of groups satisfying a polynomial growth condition then iA is exactly a generalized spectral operator with real spectrum. We can find many interesting examples of such bounded operators in the paper [3] of Barnes. Unbounded generators of C0 -groups with polynomial growth are given by Kantorovitz in [8]. The author considers perturbations of some selfadjoint operators S by a suitable bounded S -Voltera operator V and gets many situations in which iS + ξV generates some C0 -groups with polynomial growth. A main aim of this paper are some necessary or sufficient conditions when a closed densely defined operator with the real spectrum is a generator of some C0 -group satisfying a polynomial growth condition. This problem was also studied by Kisy´ nski in [9]. A main result in the paper is to prove that {exp(itT )}t∈R satysfies a polynomial growth condition if and only if there exist some constants M > 0 and d ≥ 0 such that    1 1 −1 2 1 + d u2 , (1) (k ± i − T ) u dk ≤ M   R    1 1 (k ± i − T ∗ )−1 v2 dk ≤ M (2) 1 + d v2 ,   R for all  > 0 and u ∈ X and v ∈ X ∗ . Next, these abstract results are applied to an operator given by a Friedrichs model. The Friedrichs model is an operator defined by the formula T f (x) = xf (x) + f, φ ψ(x),

(3)

for f ∈ D(T ) . Here ψ and φ are some fixed functions. I am mainly interested in such operator, when ψ , φ ∈ L2 (R). If ψ ∈ Lp (R) and φ ∈ Lq (R), 1/p + 1/q = 1, p, q ≥ 1, then T operates in Lp (R). In this case T is densely defined as an operator in Lp (R) and the domain D(T ) = D(Mx ), where Mx throughout this paper denotes the operator of multiplication by the identity on R. The  +∞ symbol f, g denotes usually the integral −∞ f (x)g(x)dx, so in the case when f, g ∈ L2 (R) it is the inner product of f and g . In the abstract case when X is a Banach space we assume that f ∈ X , g ∈ X ∗ and f, g = g(f ). Define for z ∈ C \ R 

+∞

φ(x)ψ(x) dx x−z −∞   = 1 + (Mx − z)−1 ψ, φ .

D(z) = 1 +

(4)

307

Malejki Then the resolvent of T is given by (z − T )−1 f (x) = (z − x)−1 f (x) +

1 (z − Mx )−1 f, φ (z − x)−1 ψ(x). D(z)

(5)

Since the resolvent is known, it is natural to formulate condition on it. Generally it is not true that an operator like iT generates a group with polynomial growth. In fact, such growth implies that the spectrum of T is contined in R. The main theorems in this paper (Theorem 5, Corollary 6) are modifications the result of S. Naboko included in [10] and devoted to conditions for similarity to a selfadjoint operators, what is equivalent to uniformly boundedness of the group (by the Stone theorem). 1. Remarks on the Hille-Yosida theorem This section is devoted to a lemma like the Hille-Yosida theorem. I am going to reformulate this theorem in the case of generators of C0 -groups with polynomial growth condition, because it is not popular in literature (but see [9]) and it will be very useful in the proof of the main theorem in section 3. Moreover, we will try to apply it to a very simple Friedrichs model T = Mx + ., φ ψ, where φψ = 0 everywhere on R. Lemma 1. Let X be a Banach space and s ≥ 0 be a fixed real number. Assume that T is a densely defined, closed operator in X . Then the following two conditions are equivalent: 1. There exists M1 ≥ 0 such that −n

(λ − iT )

1  ≤ M1 n |λ|



 1+

n |λ|

s 

for all λ ∈ R \ {0} and n = 1, 2, . . . . 2 . iT generates a C0 -group in X and there exists M2 ≥ 0 such that  exp(itT ) ≤ M2 (1 + |t|s ) for all t ∈ R . Proof. Denote U = iT and R(z, U ) = (z − U )−1 . 2. ⇒ 1. Both of the operators U and −U are generators of some C0 semigroups, so we can write for them the integral formulas (see [6] or [5])  ∞ −1 e−λt exp(±tU )udt, (λ ∓ U ) u = 0

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when λ > 0. The polynomial growth condition guarantee that the spectrum of T is contained in R and the above integral formula exists. Then (−1)n n!(R(z, U ))n+1 = R(n) (z, U ) implies (−1)n n!(λ ± U ))−(n+1) u =



∞

e−λt (−t)n exp(±tU )udt,

0

for n = 0, 1, . . . . Therefore for n = 0, 1, . . . and λ > 0 n!(λ ± U )−(n+1) u ≤



∞

e−λt tn M2 (1 + ts )udt   n! Γ(n + s + 1) = M2 u. + λn+1 λn+s+1 0

Here Γ means the Euler gamma function. This gives us −n

(λ − U )

1  ≤ M2 n |λ|



Γ(n + s) 1 1+ (n − 1)! |λ|s

 ,

for n = 1, 2, . . . and λ ∈ R \ {0}. s But if s > 0 is fixed, so Γ(n+s) (n−1)! = O(n ). Finally there exists M1 > 0 such that (λ − U )−n  ≤ M1

1 |λ|n

 1+

ns |λ|s

 .

1. ⇒ 2. Denote Uλ± = −λ + λ2 (λ − ±U )−1 for λ > 0. Then Uλ± is a bounded operator on X. As in the proof of the Hille-Yosida theorem (see, for example, [5]), it is possible to prove that Uλ± f −→ ±U f for f ∈ D(U ), when λ → +∞. Moreover ± T ± (t)f = lim etUλ f, t > 0 λ→+∞

is a well defined strongly continuous semigroup with ±U as a generator. Let  T (t) =

T + (t), when t ≥ 0 T − (−t), when t < 0.

Then we have that T (t) is C0 -group and U is its generator (see [5]). So, it is enough to prove only the estimation for T (t) . But ±

±

T ± (t)f  = lim etUλ f  ≤ lim sup etUλ f  λ→+∞

309

Malejki and ±

−1

etUλ  = e−tλ etλ (λ∓U )  ∞

1 −tλ 2 −1 n (tλ (λ ∓ U ) ) = e I + n! n=1  ∞

1 −tλ n 2n −n t λ (λ ∓ U )  1+ ≤ e n! n=1   ∞

n s 

1 n 2n 1 −tλ t λ 1+ 1+ ≤ M1 e n! λn λ n=1 ∞

t n λ n n s ≤ M1 + M1 e−tλ n! λ n=1 2

 t [s]s t[s]  = M1 + M1 e−tλ  s−1 + · · · + [s]! λs−[s] λ    Ft (λ)

+ ts



 ∞

 (n + [s] + 1)s (tλ)n+[s]+1−s  , (n + [s] + 1)!  n=0    G(tλ)

where [s] means the greatest integer less than or equal to s. Notice that lim e−tλ Ft (λ) = 0,

λ→∞

for all t > 0.

Let α = [s] + 1 − s, then α > 0 and G(tλ) =

∞

(n + α + s)s (tλ)n+α . (n + α + s)! n=0

Using the Stirling formula we get (n + α + s)s es ≤ . (n + α + s)! Γ(n + α + 1) Therefore G(tλ) ≤ es

∞

(tλ)n+α . Γ(n + α + 1) n=0

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Malejki


∞ xn+α If we denote f (x) = n=0 Γ(n+α+1) then it is not difficult to check that  x −t α−1 ex e t dt. This gives f (x) ≤ ex and G(tλ) ≤ es etλ . Therefore f (x) = Γ(α) 0 ±

lim sup etUλ  ≤ M1 + lim es M1 e−tλ ts etλ λ→∞

λ→∞

≤ M2 (1 + ts ), for all t > 0 and some M2 > 0. Finally, T ± (t) ≤ M2 (1 + ts ) for t > 0 , and this completes the proof. The lemma is not very convenient to apply to concrete operators, it is possible to use it in the case of the special Friedrichs model. Corollary 2. Let A be a densely defined, closed generator of an uniformly bounded C0 -group in X and let ψ ∈ X and φ ∈ X ∗ . Assume that (z − A)−1 ψ, φ = 0

for every z ∈ C \ R.

(∗)

Define T f = Af + f, φ ψ,

D(T ) = D(A).

Then  exp(itT ) ≤ C(1 + |t|) for all t ∈ R and some C > 0. Proof.

Direct computations allow us to check that (z − T )−1 f = (z − A)−1 f + (z − A)−1 f, φ (z − A)−1 ψ.

Next, using (∗) we can prove by induction (z − T )−n f = (z − A)−n f + (z − A)−1 f, φ (z − A)−n ψ + (z − A)−2 f, φ (z − A)−(n−1) ψ + · · · + (z − A)−n f, φ (z − A)−1 ψ. So we have the following inequalities (z − T )−n f  ≤ (z − A)−n f  +

n

|(z − A)−k f, ψ |(z − A)−(n+1−k) φ

k=0

≤ (z − A)−n f  n

+ ψφ (z − A)−k (z − A)−(n+1−k) f . k=0

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Malejki By the assumption (z − A)−n  ≤

M , |z|n

therefore

M M M + ψφ ∗ |z|n |z|k |z|n+1−k n

(z − T )−n  ≤

k=0

M n = + M 2 ψφ . |z|n |z|n+1 T is densely defined and closed like A. Now, using the previous theorem we complete the proof.

on

R.

Now, we have some remarks: 1) Let A = Mx and φ, ψ ∈ L2 (R) satisfy φψ = 0 almost everywhere

Then T = A+ < ∗, φ > ψ satisfies all conditions of Corollary and the group generated by iT is polynomially bounded at infinity. 2) Even for simple examples of functions φ, ψ ∈ L2 (R) such that ψφ = 0 we have the Friedrichs model T , which may not be similar to a selfadjoint operator. For instance let  1  1 x ∈ (−1, 0) x ∈ (0, 1) γ , |x|γ , φ(x) = ψ(x) = x 0, x ∈ R \ (−1, 0), 0, x ∈ R \ (0, 1) for some 0 < γ < 12 . Then φ, ψ ∈ L2 (R) and ψφ = 0, but T is not similar to a selfadjoint operator. If it were, it should satisfy (z − T )−1  ≤

M |z|

for all z ∈ C \ R. But (z − T )−1 f = (z − Mx )−1 f + (z − Mx )−1 f, φ (z − Mx )−1 ψ, so   (z − T )−1  ≥ (z − Mx )−1  − ∗, (z − Mx )−1 φ (z − Mx )−1 ψ    1  −1 −1  − (z − Mx ) φ(z − Mx ) ψ . =  |z|

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Malejki

Let z = ib, b > 0, then  1 |φ(x)|2 1 dx dx = 2 2 2γ 2 2 −1 x + b 0 x (x + b )  1/b 1 1 1 = dt ≥ C 2γ bb2γ 0 t2γ (t2 + 1) bb

(ib − Mx )−1 φ2 =



0

and similarly (ib − Mx )−1 ψ2 ≥ C so (ib − T )−1  ≥ C

1 , bb2γ

1 , bb2γ

when b is close to 0 . Therefore T could not be similar to a selfadjoint operator. 2. The necessary conditions First, let us consider necessary conditions. Unfortunately this theorem is true only for operators in Hilbert spaces and generally it is not true in all Banach spaces. Theorem 3. Let T be an operator in a Hilbert space H and let iT generates a C0 -group in H with polynomial growth i.e. assume that there exists a constant C > 0 such that  exp(itT ) ≤ C(1 + |t|s ) for all t ∈ R . Then conditions (1) and (2) are satisfied with some constant M = M (C, s) > 0 and d = 2s. Proof. solvent

By the polynomial growth we have the integral formula for the re ∞ −1 eizt exp(−itT )udt (z − T ) u = −i 0

for all z ∈ C \ R. Therefore,  +∞ (k + i − T )−1 u2 dk −∞   ∞  +∞  ∞ ei(k+i)t exp(−itT )udt, ei(k+i)t exp(−itT )udt dk = −∞ 0 0  ∞  +∞  1 √ eikt (e−t exp(−itT )u)dt, = 2π 2π 0 −∞   ∞ 1 ikt −t √ e (e exp(−itT )u)dt dk. 2π 0

313

Malejki Now, using the Plancherel theorem we get 

+∞

−∞

(k + i − T )−1 u2 dk  ∞   −t e exp(−itT )u, e−t exp(−itT )u dt = 2π 0 ∞  ∞ −2t 2 2 = 2π e  exp(−itT )u dt ≤ 4πC e−2t (1 + t2s )u2 dt 0 0     1 1 (2s)! 1 2 2 = 4πC u ≤ M (C, s) u2 . + 1 + 2 (2)2s+1  2s

In similar way we can prove 

−1

R

(k + i − T )

1 u dk ≤ M (C, s) 



2

1 1 + 2s 

 u2 .

In order to prove (2) , notice that T ∗ generates C0 -group with polynomial growth condition, because (exp(itT ))∗  =  exp(−itT ∗ ) =  exp(itT ). Therefore, we can use similar way to check that 

+∞

−∞

∗ −1

(k ± i − T )

1 u dk ≤ M (C, s)  2



1 1 + 2s 

 u2 .

As an example that this theorem could not be satisfied in general Banach spaces let us take the Banach space X = C0 (R) (the space of continuous functions vanishing in the infinity with the norm of supremum) and the operator T = Mx . Then the group generated by iMx ( {eitMx }t ) is bounded (so polynomially bounded too), in fact eitMx  = 1. Let us assume that there exist  +∞ M, s ≥ 0 such that for all  > 0 and u ∈ C0 (R) is −∞ (k+i−Mx )−1 u2 dk ≤ M 1 (1 + 1s )u2 . Let uN ∈ C0 (R) be such that uN (x) = 1 if x ∈ [−N, N ] and |uN | ≤ 1 ; therefore, 

+∞

−∞

−1

(k + i − Mx )

 uN  dk ≥

N

2

−N  N

= −N N

 =

−N

(k + i − Mx )−1 uN 2 dk sup x∈R

nN (x) dk (k − x)2 + 2

1 1 dk = 2N 2 uN 2 . 2 

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Malejki

Finally, we get that there exist M, s ≥ 0 such that for all  > 0 and N > 0   1 M  1 + s ≥ 2N.  But this give us a contradiction, because the last inequolity is not satisfied, in general, for all , N > 0. 3. The sufficient conditions Having the previous theorem we look for sufficient conditions for a closed operator to be a generator of a C0 − group with a polynomial growth. Let us start with the lemma. Lemma 4. Let T be an operator in a Banach space X . Assume that T is densely defined, closed and σ(T ) ⊂ R . If T satisfies condition (1) then there exists C > 0 such that   1 −1 1 (k ± i − T )  ≤ C 1+ d  2 for all k ∈ R and  > 0 Proof.

Fix  > 0 , u ∈ X , v ∈ X and define F : {z > 0}  λ −→ (λ + i − T )−1 u, v ∈ C.

Then F ∈ H 2 ({z > 0}) because of (1) . Let F˜ be a boundary value for F . Then it must be F˜ (x) = (x + i − T )−1 u, v . Using the theory of the Hardy spaces (see [7]) we can write  +∞ 1 1 F (w) = (x + i − T )−1 u, v dx, 2πi −∞ x−w where w > 0. Therefore −1

(w + i − T )

1 u, v = 2πi



+∞

−∞

(x + i − T )−1 u, v

1 dx, x−w

w > 0.

This implies |(w + i/2 − T )−1 u, v |  +∞ 1/2  +∞ 1/2 2  1 1 −1   ≤ dx (x + i/2 − T ) u, v dx 2 2π −∞ |x − w|  −∞   +∞ 1/2  d 1/2 2 1 1  2 = uv dx M 1+ 2 2 2π   −∞ (x − w) + (w)    d 1/2 √ 2 π 1  2 √ M 1+ ≤ uv. 2π   w

315

Malejki Let w = k + i/2 , k ∈ R ,  > 0, then |(k + i − T )

−1

u, v | ≤ C

1



 1+

1 d/2

 uv,

for some constant M > 0, which does not depend on u, v, k, . Similarly we can obtain   1 1 |(k − i − T )−1 u, v | ≤ C 1 + d/2 uv.   So the proof is completed. Theorem 5. Let T be an operator in the Banach space X . Assume that T is densely defined, closed and σ(T ) ⊂ R . If (1) and (2) are fulfiled then iT generates a C0 -group in X and there exists a constant C > 0 such that  exp(itT ) ≤ C(1 + |t|d ) for all t ∈ R, where d ≥ 0 is the constant as in (1) or (2) . Proof. tions:

In order to prove this theorem we need the following integral equa−(n+1)

(λ − iT )

in u=− 2π

and (λ + iT )−(n+1) u = −

in 2π



+∞−i

−∞−i



+∞−i

−∞−i

1 (z − T )−1 u dz (z + iλ)n+1

(∗)

1 (z − T )−1 u dz, (z − iλ)n+1

(∗∗)

where λ/2 ≥  > 0. To obtain for instance (∗) notice that (because of the Riesz integral for holomorphic functions and the resolvent properties)  1 n! (−1)n n!(−iλ − T )−(n+1) u = (z − T )−1 u dz, 2πi Γ (z + iλ)n+1 where Γ is a closed positively oriented Jordan curve that surrounds the point z0 = −iλ and is included in {z < 0}. The integral above does not depend on the choice of such Γ surrounding the point −iλ , so we can choose Γ as a union of two curves Γ1 and Γ2 , such that Γ1 is contained in the line k → k − i and Γ2 is a part of the circle with the center in −iλ and some enough large radius r. When  ≥ λ/2 is fixed the above integral does not depend on r , so using Lemma 4 we have     1 −1   ≤ Const() length(Γ2 )/rn+1 (z − T ) u dz   n+1 Γ2 (z + iλ) 1 = Const n −→ 0, as r → ∞, r

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Malejki

so −(n+1)

(−1) n!(−iλ − T ) n

n! u= 2πi



+∞−i

−∞−i

1 (z − T )−1 u dz, (z + iλ)n+1

for n ≥ 1. Finally a simple transformation gives us (∗) . In similar way we get (∗∗) . Now, notice that for n ≥ 2 

+∞

−∞

1 (k + i − T )−1 u dk (k − i( − λ))n  ∞+i/2 1 = (z + i/2 − T )−1 u dz 3 n (z − i( −∞+i/2 2  − λ))  = Γ

1 (z −

i( 32 

− λ))n

(z + i/2 − T )−1 u dz,

(+)

where Γ is a curve included in the upper half plane. But if λ/2 ≥  > 0 then the integrand is a holomorphic functions in the upper half plane, so finally the last integral equals to 0 . Similarly 

+∞ −∞

1 (k − i − T )−1 udk = 0. (k + i( − λ))n

(++)

Using (∗) and (+) and the resolvent equality we have (λ − iT )−n u  1 in−1 +∞ =− ((k − i − T )−1 u − (k + i − T )−1 u) dk 2π −∞ (k − i( − λ))n  2i in−1 +∞ ((k − i − T )−1 (k + i − T )−1 u) dk, =− 2π −∞ (k − i( − λ))n if only n ≥ 2 , 0 <  ≤ λ/2 and u ∈ X. So, if u ∈ X , v ∈ X ∗ and n ≥ 2 , then |(λ − iT )−n u, v |  +∞ ≤ 1/π

   (k + i − T )−1 u, (k + i − T ∗ )−1 v  dk n −∞ |(k − i( − λ)) |  +∞  (k + i − T )−1 u(k + i − T ∗ )−1 v dk ≤ 1/π | − λ|n −∞

317

Malejki  +∞ 1/2  −1 2 (k + i − T ) u dk | − λ|n −∞  +∞ 1/2 × (k + i − T ∗ )−1 v2 dk −∞   1 1 1 + uv ≤ M/π (λ − )n d ≤ 1/π

Let  = λ/n , n ≥ 2 , then 

n d  M 1 + uv λ πλn 1 − n1 )n 

n d  4M 1 + uv. ≤ πλn λ

|(λ − iT )−n u, v | ≤

Finally, (λ − iT )−n  ≤

4M πλn

 1+

n d  λ

and similarly we can obtain (λ + iT )

−n

4M ≤ πλn

 1+

n d  λ

still assuming n ≥ 2 and λ > 0. Using Lemma 4 we have   d/2 1 M −1 (λ ± iT ) ≤ 1+ πλ λ   d 1 M” ≤ . 1+ πλ λ By Lemma 1 iT is generate a C0 -group with polynomial growth condition and there exists C > 0 such that for all t ∈ R  exp(itT ) ≤ C(1 + |t|d ). Corollary 6. such that

Let φ, ψ ∈ L2 (R) be such that there exist M > 0 and k ≥ 0      1  ≤M 1+ 1  .  D(z)  |z|k

(6)

Assume that T is a Friedrichs model given by (3) , which acts in L2 (R) and is associated with φ and ψ. Then the group generated by iT satisfies a polynomial growth condition.

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Malejki

Proof. T is densely defined. The spectrum of T is contined in R because of (5) and (6). Then it is enough to check (1) and (2) and apply Theorem 5. Using the resolvent formula for the Friedrichs model we get  (k ± i − T )−1 u2 dk R      u(x) 2   dx dk ≤ const   R R k ± i − x  2       ψ(x) 2 1 1   uφ + const M 1 + k  k ± i − x  dx dk  2 R R   1 2 ≤ const |u(x)| dk dx 2 + ()2 (k − x) R R 2     1 1 1 2 uφ |u(x)| dk dx + const M 1 + k 2 2 2   R (k − x) + () R   2 1 1 1 1 uφ φ ≤ const u + const M 1 + k   2    1 1 ≤C 1 + 2(k+1) u2 ,   where  > 0 and u ∈ L2 (R) and C > 0 is a constant, which does not depend on  and u. To prove the condition (2) in Theorem 5 it is enough to notice that T ∗ is the Friedrichs model too. So the group is of polynomial growth with 2(k+1) as a degree. 4. Remark on the discrete case This section is only a short remark about the discrete case. It is easy to obtain a modification of the theorem of Naboko ([10]) for a discrete group given by {T n }n∈Z , where T is a bounded operator in a Hilbert space H. Theorem 7. Let T be a bounded operator in H and assume that σ(T ) ⊂ {z ∈ C | |z| = 1}. Then the following conditions are equvalent: 1. There exist M > 0 and d ≥ 0 such that T n  ≤ M (1 + |n|d ) for all n ∈ Z. 2 . There exist C > 0, s ≥ 0 such that 

2π

iθ −1

(T − re ) 0

 u dθ ≤ Cu 1 + 2



r2 r2 − 1

s 

1 , r2 − 1

Malejki

319

  2 s  1 r , (T − re ) u dθ ≤ Cu 1 + 2 2 r −1 r −1 0      2π s 1 1 1 (T − e−iθ )−1 u2 dθ ≤ Cu 1 + , −2 r 1−r 1 − r−2 0 s     2π 1 1 1 (T ∗ − e−iθ )−1 u2 dθ ≤ Cu 1 + , −2 r 1−r 1 − r−2 0 

2π

∗

iθ −1

2

for all r > 1 . Proof. The proof of Naboko included in [10] works in this case very well. Moreover, we can fix the relationships between constants d and s in the following way: (1) =⇒ (2) s = 2d (2) =⇒ (1) d = s. References [1] Amrein, W. O., A. Boutet de Monvel and V. Georgescu, “C0 -groups, Commutator Methods and Spectral Theory of N-Body Hamiltonians”, Birkhauser, Basel, 1996. [2] Anderson R.F.V., The Weyl functional calculus, J. Functional Anal. 4 (1969), 240–264. [3] Barnes B. A., Operators which satisfy polynomial growth conditions, Pacific J. Math. 138(2) (1989), 209–219. [4] Colojoara, I. and C. Foias, Theory of spectral operators, “Mathematics and Its Applications”, Vol. 9, Gordon and Breach, New York/London/Paris, 1968. [5] Davies E. B., “One-Parameter Semigroups”, Academic Press, London, 1980. [6] Hille, E. and R.S. Phillips, “ Functional Analysis and Semigroup,” rev. ed., Amer. Math. Soc. Colloq. Publ. 31, Providence, R.I., 1957. [7] Hoffman, K., “Banach Spaces of Analytic Functions,” Prentice-Hall, Englewood Cliffs, NJ, 1962. [8] Kantorovitz, S., “Spectral Theory of Banach Space Operators,” Lecture Notes in Math., 1012, Springer, Berlin/Heideberg/New York, 1983.

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[9] Kisy´ nski, J., “Around Widder’s characterization of the Laplace transform of an element of L∞ (R+ ) ,” Ann. Pol. Math. 74 (2000), 161–200. [10] Naboko, S. N,, Conditions for similarity to unitary and selfadjoint operators, Funktsional. Anal. i Prilozhen. 18(1) (1984), 16–27 (in Russian).

Institute of Mathematics Polish Academy of Sciences Ul. Sw. Tomasza 30 31-027 Krakow, Poland [email protected]

Received April 30, 1999 and in final form December 21, 1999 Online publication June 29, 2001