Conjugacy of Strongly Continuous Semigroups ... - Semantic Scholar

Conjugacy of Strongly Continuous Semigroups Generated by Normal Operators Ricardo Rosa1 Running head: Conjugacy of Semigroups Mailing address: Ricardo Rosa The Institute for Scientific Computing & Applied Mathematics 618 East Third Street Bloomington, IN 47405 Telephone: (812) 855-7847 1 Departamento de Matem´ atica Aplicada, IM-UFRJ, Caixa Postal 68530, C.E.P.: 21945, Rio de Janeiro, RJ, Brazil. Present address: The Institute for Scientific Computing & Applied Mathematics, 618 East Third Street, Bloomington, IN 47405

1

Abstract In this work, we show that a strongly continuous semigroup generated by a normal operator N is conjugate to the semigroup generated by the real part of N , provided zero is not an eigenvalue of the real part of N . We also show that in case N satisfies a certain sectorial property, the homeomorphism establishing the conjugacy, as well as its inverse, is locally H¨ older continuous. Moreover, in case N satisfies the sectorial property and the real part of N has a pure point spectrum with an at most countable number of eigenvalues, the homeomorphism and its inverse are Lipschitz continuous.

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1 . INTRODUCTION In the finite dimensional case, there is a complete description up to a nonlinear conjugacy of all groups of linear operators generated by hyperbolic operators (see [4], for instance). In the case of bounded hyperbolic generators in infinite dimensional Banach spaces, the classification is more difficult due to the complexity related to all possible pairwise decompositions of such Banach spaces, but we can still conjugate a group to a simpler one in the same way as in the finite dimensional case (see [1]). The problem becomes much more subtle when we consider unbounded operators. We recall here the definition of conjugacy. We say that two strongly continuous semigroups {etA }t≥0 and {etB }t≥0 in a Banach space X are conjugate if there exists a homeomorphism h : X → X such that h(etA x) = etB h(x), for every t ≥ 0 and x ∈ X. In this case, we also say that there is a conjugacy between the two semigroups or that {etA }t≥0 is conjugate to {etB }t≥0 through h. It is easy to see that if the semigroups above are conjugate and one of them is a group then the other one is also a group and h(etA x) = etB h(x) holds for every t ∈ IR. In this note, we address the case of unbounded operators in a Hilbert space, presenting a result stating that a strongly continuous semigroup generated by a (possibly unbounded) normal operator N is conjugate to the strongly continuous semigroup generated by the real part Re N = (N +N ∗ )/2 of N , provided zero is not an eigenvalue of Re N . Since the generator is normal we can decompose it on its real and imaginary parts, which are “nice” self-adjoint operators. In this case, we can decompose the space using the spectral representation for the real part and reduce the problem to the conjugacy of a countable family of strongly continuous groups. The assumption that zero is not an eigenvalue of Re N is to be compared to the hyperbolicity condition present in previous results on conjugacy. In the hyperbolic case, the spectrum must be bounded away from zero, while in our result, zero may lye on the continuous spectrum. It is quite remarkable that the hyperbolicity assumption can be relaxed in this case. The article is organized as follows. In the next section, we prove a Proposition stating the result in case of groups. This Proposition will be applied to the restrictions of the semigroup to the spectral spaces relative to the decomposition of the real part of the generator. In the subsequent section, we set up the notation for the unbounded case and prove our main Theorem, as well as some regularity results for the homeomorphism establishing the conjugacy.

2

The case of groups

In this section, we present a Proposition for (possibly unbounded) normal operators generating a strongly continuous group. This result will be applied in the next section to the general case.

3

Proposition 2.1 Let H be a complex Hilbert space with norm | · |. Let A and B be two self-adjoint operators in H with A bounded and such that AB = BA in the domain of B. Assume also that k etA k ≤ e−at , k etA k ≤ e−bt ,

for t ≥ 0, for t ≤ 0,

(1) (2)

where k · k is the operator norm in H, and a and b are real numbers such that 0 < a ≤ b. Then {et(A+iB) }t∈IR is conjugate to {etA }t∈IR , i.e., there exists a homeomorphism h : H → H such that h(et(A+iB) x) = etA h(x),

∀t ∈ IR, ∀x ∈ H.

Moreover, h and its inverse h−1 satisfy |h(x)| = |x| = |h−1 (x)|,

∀x ∈ H,

and |h(x) − h(y)|, |h−1 (x) − h−1 (y)| ≤ g(x, y) + |y − x|,

∀x, y ∈ H,

(3)

where g(x, y) = g(y, x), 0 ≤ g(x, y) ≤ 2 min{|x|, |y|}, and for every x ∈ H, g(x, y) → 0 as y → x. Furthermore, if B is bounded, we have, more precisely,   a kBk b b g(x, y) ≤ |y − x| + min{|x|, |y|}|y − x| , ∀x, y ∈ H. (4) a a Proof : Note by (1), that for t ≤ 0, |x| = |e−tA etA x| ≤ eat |etA x|, so that |etA x| ≥ e−at |x|; while for t ≥ 0,

|etA x| ≤ e−at |x|.

Thus, for x 6= 0, the function t → |etA x| is surjective from IR onto (0, +∞). Hence, for any x = 6 0, there exists a s(x) ∈ IR such that |es(x)A x| = 1. To see that s(x) is uniquely determined, suppose s0 , s00 ∈ IR are such that 0

|es A x| = 1 = |es 4

00

A

x|.

Without loss of generality, we can assume that s0 ≥ s00 , thus 0

0

1 = |es A x| = |e(s −s

00

)A s00 A

e

0

00

x| ≤ e−a(s −s ) |es

00

A

0

00

x| = e−a(s −s ) .

But a > 0, hence s0 = s00 . Now, define h(x) =



0, eis(x)B x,

if x = 0, if x = 6 0.

Let us compute s(et(A+iB) x) for x 6= 0, t ∈ IR. We look for s0 such that 0

1 = |es A et(A+iB) x|. Since A commutes with B, we must have 0

0

1 = |eitB e(t+s )A x| = |e(t+s )A x|; thus s(x) = t + s0 . Hence s(et(A+iB) x) = s(x) − t.

(5)

Therefore, using the commutativity between A and B, h(et(A+iB) x) = ei(s(x)−t)B et(A+iB) x = etA eis(x)B x = etA h(x). We also have that h is invertible, with its inverse given by  0, if x = 0, −1 h (x) = e−is(x)B x, if x 6= 0. So, h will give the required conjugacy provided we prove that h and h−1 are continuous. The continuity of h and h−1 follow from estimate (3), that we are now going to prove. Let us give estimates for s(x) first. In case 0 < |x| ≤ 1, it is easy to see that s(x) ≤ 0; hence |x| = |e−s(x)A xes(x)A x| ≤ eas(x) |es(x)A x| = eas(x) . Thus

1 log |x|, for 0 < |x| ≤ 1. a In case |x| ≥ 1, it is easy to see that s(x) ≥ 0, so that 0 ≥ s(x) ≥

(6)

1 = |es(x)A x| ≤ e−as(x) |x|; then 0 ≤ s(x) ≤

1 log |x|, a 5

for |x| ≥ 1.

(7)

Consider now x, y ∈ H, x 6= 0, y 6= 0. If 0 ≤ s(x) ≤ s(y), then 1

= ≤ ≤ ≤

|es(y)A y| = |e(s(y)−s(x))A es(x)A y| e−a(s(y)−s(x)) |es(x)A x + es(x)A (y − x)| e−a(s(y)−s(x)) (1 + |es(x)A (y − x)|) e−a(s(y)−s(x)) (1 + e−as(x) |y − x|).

Hence ea(s(y)−s(x)) ≤ 1 + e−as(x) |y − x|. Therefore a(s(y) − s(x)) ≤ log(1 + e−as(x) |y − x|) ≤ e−as(x) |y − x|. Since a > 0 and s(x) ≥ 0, we have e−as(x) ≤ 1, thus |s(y) − s(x)| ≤

1 |y − x|, a

for 0 ≤ s(x) ≤ s(y).

(8)

In case s(x) ≤ s(y) with s(x) ≤ 0, we have 1

= ≤ ≤ ≤

|es(y)A y| = |e(s(y)−s(x))A es(x)A y| e−a(s(y)−s(x)) |es(x)A x + es(x)A (y − x)| e−a(s(y)−s(x)) (1 + |es(x)A (y − x)|) e−a(s(y)−s(x)) (1 + e−bs(x) |y − x|).

Thus ea(s(y)−s(x)) ≤ 1 + e−bs(x) |y − x|; hence a(s(y) − s(x)) ≤ log(1 + e−bs(x) |y − x|). Therefore, using the inequality log(1 + r) ≤ rα /α, which is true for any r ≥ 0 and 0 < α ≤ 1, we obtain a(s(y) − s(x)) ≤ thus s(y) − s(x) ≤

a b −bs(x) (e |y − x|) b ; a

a b −as(x) e |y − x| b . a2

Using (6), we finally arrive at |s(y) − s(x)| ≤

a b 1 |y − x| b , 2 a |x|

for s(x) ≤ s(y), s(x) ≤ 0.

Now, we are in position to prove estimate (3). First note that |h(x)| = |eis(x)B x| = |x|, 6

(9)

which proves (3) in case y = 0. For x, y ∈ H, x 6= 0, y 6= 0, write h(y) − h(x) = eis(y)B y − eis(x)B x = eis(y)B x − eis(x)B x + eis(y)B (y − x). Similarly, h(x) − h(y) = eis(x)B y − eis(y)B y + eis(x)B (x − y). Hence |h(y) − h(x)| ≤ g(x, y) + |y − x|, where g(x, y) = min{|eis(y)B x − eis(x)B x|, |eis(x)B y − eis(y)B y|}. Clearly, |g(x, y)| ≤ 2 min{|x|, |y|} and g(x, y) = g(y, x). Moreover, by (8) and (9), we know that x → s(x) is continuous, hence, by the strong continuity of {e−itB }t∈IR , we find that that g(x, y) → 0 as y → x, for every x ∈ H. In case B is bounded, we have Z s(y) g(x, y) ≤ |eis(y)B x − eis(x)B x| = iBeisB x ds s(x) Z s(y) ≤ k B k |eisB x| ds =k B k |s(y) − s(x)||x|. s(x) Hence, we obtain from (8) that

g(x, y) ≤

kBk |y − x| |x|, a

provided 0 ≤ s(x) ≤ s(y); and from (9) that g(x, y) ≤

a kBkb |y − x| b , a a

provided s(x) ≤ s(y), s(x) ≤ 0. Hence, by symmetry in x and y, we finally obtain (4). The proof of the estimates for h−1 is exactly the same. ¶

Corollary 2.2 Proposition 2.1 above remains valid if we consider a and b such that b ≤ a < 0. Proof : Just reverse time.

¶

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Corollary 2.3 Every strongly continuous group generated by a normal hyperbolic operator N in a Hilbert space H is conjugate to a strongly continuous group {T (t)}t∈IR of the form T (t)x = e−t x1 + et x2 ,

t ∈ IR,

(10)

where x = x1 +x2 , x1 ∈ H1 , x2 ∈ H2 , accordingly to an orthogonal decomposition H = H1 ⊕ H2 , where H1 and H2 are Hilbert spaces; any of them may be null. Proof : Since N is normal and hyperbolic, we can write N = N1 ⊕ N2 , accordingly to an orthogonal decomposition of the space H = H1 ⊕ H2 , where Re σ(N1 ) ⊂ (−∞, 0) and Re σ(N2 ) ⊂ (0, +∞). Then {etN }t∈IR is conjugate to a strongly continuous group {U (t)}t∈IR of the form U (t)x = etN1 x1 + etN2 x2 ,

t ∈ IR,

where x = x1 + x2 , xi ∈ Hi , and {etNi }t∈IR is a strongly continuous group in Hi , i=1,2. By Proposition 2.1 and Corollary 2.2, we have that {etNi }t∈IR is conjugate to {etAi }t∈IR , where Ai is the real part of Ni , so that it is bounded and hyperbolic, i = 1, 2. We have then (see [1], for instance) since σ(A1 ) ⊂ (−∞, 0) and σ(A2 ) ⊂ (0, +∞), that {etA1 }t∈IR is conjugate to {e−t }t∈IR in H1 and {etA2 }t∈IR is conjugate to {et }t∈IR in H2 . Therefore, we finally obtain that {etN }t∈IR is conjugate to a strongly continuous group {T (t)}t∈IR of the form (10). ¶

3

The case of semigroups

Let H be a complex Hilbert space. We consider a normal linear operator N in H, i.e., N is closed with dense domain D(N ) in H and N N ∗ = N ∗ N in D(N N ∗ ) = D(N ∗ N ). We let σ(N ) denote the spectrum of N and set Re σ(N ) ≡ {Re ζ; ζ ∈ σ(N )}. We recall that the point spectrum is the set of eigenvalues and we assume that zero is not in the point spectrum of the real part of N . We also assume that N generates a strongly continuous semigroup in H, which in this case is equivalent to say that σ(N ) is bounded from the right, i.e., Re σ(N ) ⊂ (−∞, Λ), for some Λ ∈ IR. Consider the real and imaginary parts of N , defined respectively by A=

1 (N + N ∗ ), 2

B=

1 (N − N ∗ ), 2i

so that N = A + iB, and A and B are self-adjoint operators with same dense domain as N and such that AB = BA. Here, we use that N is normal and then D(N ) = D(N ∗ ) (see [3], for instance), but of course A and B may have extensions. 8

Let {etN }t≥0 denotes the semigroup generated by N . By the assumptions above on N , we also have that A generates a strongly continuous semigroup in H, denoted by {etA }t≥0 , and iB generates a strongly continuous unitary group in H, denoted by {eitB }t∈IR . Since A and B commute to each other, we also have etN = et(A+iB) = etA eitB = eitB etA , ∀t ≥ 0. Later, we will also assume that N has the following sectorial property: there exists a θ, 0 < θ < π/2, such that σ(N ) ⊂ {ζ ∈ C l ; | arg(ζ)| < θ} ∪ {ζ ∈ C l ; | arg(ζ − π)| < θ}.

(11)

In this case, N generates an analytic semigroup (see [2], for instance). We will also consider the case when A has a pure point spectrum, which is to say that the space H is spanned by the eigenvalues of A. This does not imply that the spectrum of A is made only of eigenvalues, but it does imply that the eigenvalues are dense in the spectrum of A (see [3], for instance). Now, we are in position to state our main Theorem as follows. Theorem 3.1 Let N be a normal operator in a complex Hilbert space H. Assume N generates a strongly continuous semigroup in H and zero is not an eigenvalue of the real part of N . Then the semigroup generated by N is conjugate to the semigroup generated by the real part of N . More precisely, there exists a homeomorphism h : H → H such that h(etN x) = etA h(x),

∀t ≥ 0, ∀x ∈ H,

where A = (N + N ∗ )/2 denotes the real part of N . Remark We can deduce from Theorem 3.1 a caracterization for semigroups generated by normal operators, namely that every strongly continuous semigroup generated by a normal operator N in a Hilbert space H such that zero is not an eigenvalue of the real part of N is conjugate to a strongly continuous semigroup {T (t)}t≥0 of the form T (t)x = etA1 x1 + etA2 x2 ,

t ≥ 0,

(12)

where x = x1 + x2 , x1 ∈ H1 , x2 ∈ H2 , accordingly to an orthogonal decomposition of the space H = H1 ⊕ H2 , where H1 and H2 are Hilbert spaces, any of them may be null, A1 is a negative symmetric operator generating a strongly continuous semigroup of contractions, and A2 is a positive symmetric bounded operator. In case we have more information about N , we can obtain further informations about the regularity of the homeomorphism that gives the conjugacy, like locally H¨ older continuity or locally Lipschitz continuity. More precisely, we have the following results. 9

Theorem 3.2 Let N be as in Theorem 3.1. Assume moreover that N satisfies the sectorial property (11). Then, for any ν, 0 < ν < 1, we can choose h in Theorem 3.1 to satisfy |h(y) − h(x)|, |h−1 (x) − h−1 (x)| ≤ C(R, θ)|y − x|ν ,

(13)

for any x, y ∈ H, |x|, |y| ≤ R and |x − y| ≤ 1, where R ≥ 1, θ is given by (11), and C(R, θ) depends on R and θ, but not on ν. Theorem 3.3 Let N be as in Theorem 3.2. Assume moreover that the real part of N has a pure point spectrum with an at most countable number of eigenvalues. Then, we can choose h in Theorem 3.1 to satisfy |h(y) − h(x)|, |h−1 (x) − h−1 (x)| ≤ C(R, θ)|y − x|, for any x, y ∈ H, |x|, |y| ≤ R and |x − y| ≤ 1, where R ≥ 1, θ is given by (11) and C(R, θ) depends only on R and θ.

3.1

Proof of the main result

The proof of Theorem 3.1 above will be carried out in a series of Lemmas. In all those Lemmas, H will be a complex Hilbert space normed with | · |, N will be a normal operator in H generating a strongly continuous semigroup and with zero not in the point spectrum of Re N , and A and B will be respectively the real and imaginary parts of N . Lemma 3.4 Assume moreover that Re σ(N ) ⊂ (−∞, 0]. Then, given any sequence {λn }n∈ZZ with 0 < λn < λn+1 , λn → 0 as n → −∞ and λn → +∞ as n → +∞, there exists a sequence {Pn }n∈ZZ of pairwise orthogonal projectors in H such that H = ⊕+∞ n=−∞ Pn H, Bn ≡ B|Pn H : Pn H → Pn H and An ≡ A|Pn H ∈ B(Pn H) with σ(An ) ⊂ [−λn+1 , −λn ], for n ∈ ZZ; An and Bn symmetric. Proof : This fact is well known; the proof is given for completeness. Since A is self-adjoint, there is a spectral family {E(λ)}λ∈IR associated to A. We recall that E(λ) is strongly continuous from the left and E(λ) → 0,

as λ → −∞,

E(λ) → I,

as λ → +∞.

Since Re σ(N ) ⊂ (−∞, 0], we also have that the real part A of N is such that σ(A) ⊂ (−∞, 0]; hence E(λ) = I for λ > 0. Since zero is not an eigenvalue of A, it follows that E(λ) is continuous at λ = 0 (see [3], for instance); thus E(λ) → I,

as λ → 0. 10

Now, given the sequence {λn }n∈ZZ , let Pn = E(−λn ) − E(−λn+1 ), Then

q X

n ∈ ZZ.

Pn = E(−λ−p ) − E(λq ) → I − 0 = I,

n=−p

as p, q → +∞. Hence I=

+∞ X

Pn .

n=−∞

Thus H = ⊕+∞ n=−∞ Pn H and {Pn }n∈IN is pairwise orthogonal. We let then An = A|Pn H : Pn H → Pn H, so that An is symmetric and σ(An ) ⊂ [−λn+1 , −λn ],

∀n ∈ ZZ.

Thus An ∈ B(Pn H). That Bn ≡ B|Pn H : Pn H → Pn H, it follows from the fact that B commutes with A; and since B is symmetric, so is Bn . ¶ Lemma 3.5 Assume also that Re σ(N ) ⊂ (−∞, 0], and consider an arbitrary sequence {λn }n∈ZZ with 0 < λn < λn+1 , λn → 0 as n → −∞, and λn → +∞ as n → +∞. Let Pn , An and Bn be as in Lemma 3.4 for this sequence {λn }n∈ZZ . Then {et(An +iBn ) }t∈IR is conjugate to {etAn }t∈IR through a homeomorphism hn : Pn H → Pn H satisfying |hn (x)| = |x| = |h−1 n (x)|,

∀x ∈ Pn H,

and −1 |hn (y) − hn (x)|, |h−1 n (y) − hn (x)| ≤ gn (x, y) + |y − x|,

∀x, y ∈ Pn H,

where gn (x, y) = gn (y, x), 0 ≤ gn (x, y) ≤ 2 min{|x|, |y|}, and gn (x, y) → 0 as y → x, for any x ∈ Pn H fixed. Proof : Just apply Proposition 2.1 for each n ∈ ZZ.

¶

Lemma 3.6 Assume also that Re σ(N ) ⊂ (−∞, 0], and consider an arbitrary sequence {λn }n∈ZZ with 0 < λn < λn+1 , λn → 0 as n → −∞, and λn → +∞ as n → +∞. Let Pn and hn be as in Lemma 3.5 for this sequence {λn }n∈IN . Then, the function h : H → H given by h(x) =

+∞ X

hn (Pn x),

n=−∞

is well-defined and is a homeomorphism in H. 11

x ∈ H,

Proof : First note that h is well defined since for any x ∈ H, we have |hn (Pn x)| = |Pn x|, which gives |h(x)|2 =

+∞ X

|hn (Pn x)|2 =

n=−∞

+∞ X

|Pn x|2 = |x|2 .

n=−∞

Then |h(x)| = |x|,

∀x ∈ H.

−1 Since hn is invertible and its inverse also satisfies Pn h−1 n (Pn x) = hn (Pn x) −1 and |hn (Pn x)| = |Pn x|, for any x ∈ H, it is easy to see that h is invertible with inverse given by

h

−1

+∞ X

(x) =

h−1 n (Pn x),

∀x ∈ H;

n=−∞

and with |h−1 (x)| = |x|,

∀x ∈ H.

In order to prove the continuity of h, take x ∈ H fixed, and any sequence {yk }k∈IN converging to x as k goes to infinity. Thanks to Lemma 3.5, we have |h(yk ) − h(x)|2

=

+∞ X

|hn (Pn yk ) − hn (Pn x)|2

+∞ X

[2gn (Pn yk , Pn x)2 + 2|Pn yk − Pn x|2 ]

n=−∞

≤

n=−∞

≤ 2|yk − x|2 + 2

+∞ X

gn (Pn yk , Pn x)2 .

n=−∞

Since yk tends to x, we have Pn yk → Pn x as k → +∞, hence gn (Pn yk , Pn x) → 0; this holding for any n ∈ ZZ. Moreover, |gn (Pn yk , Pn x)|2 ≤ 4 min{|Pn yk |, |Pn x|}2 ≤ 4|Pn x|2 , and

+∞ X

|Pn x|2 = |x|2 < +∞.

+∞ X

gn (Pn yk , Pn x)2 → 0,

n=−∞

Therefore,

n=−∞

as k → +∞. Thus h(yk ) → h(x),

as k → +∞, 12

which proves the continuity of h. The proof for h−1 is exactly the same.

¶

Lemma 3.7 Theorem 3.1 is true for N such that Re σ(N ) ⊂ (−∞, 0]. Proof : Thanks to Lemma 3.6, we have only to show that the function h constructed there satisfies h(etN x) = etA h(x), ∀t ≥ 0 and ∀x ∈ H. First note that for any x ∈ H and any t ≥ 0, we have +∞ X

etN x =

et(An +iBn ) Pn x,

n=−∞

and etA x =

+∞ X

etAn Pn x.

n=−∞ t(An +iBn )

Therefore, since hn conjugates {e h(etN x) =

+∞ X

}t∈IR to {etAn }t∈IR , we obtain

hn (et(An +iBn ) Pn x) =

n=−∞

for any x ∈ H and t ≥ 0.

+∞ X

etAn hn (Pn x) = etA h(x),

n=−∞ ¶

Now, we are able to finish the proof of our main result. Proof of Theorem 3.1 : Let N be a normal operator in a complex Hilbert space H, with N generating a strongly continuous semigroup in H and with zero not in the point spectrum of Re N . Since N is normal, we can write N = N1 ⊕N2 , where Ni is a normal operator defined on a Hilbert space Hi , i = 1, 2, with H = H1 ⊕ H2 and Re σ(N1 ) ⊂ (−∞, 0],

Re σ(N2 ) ⊂ [0, Λ].

Moreover, since zero is not an eigenvalue of Re N , it follows that zero is neither an eigenvalue of Re N1 nor of Re N2 . We can then apply Lemma 3.7 to N1 to obtain a homeomorphism h1 : H1 → H1 such that h1 (etN1 x) = etA1 h1 (x),

∀t ≥ 0, ∀x ∈ H1 ,

where A1 = (N1 + N1∗ )/2 is the real part of N1 . Also, note that N2 is bounded, so that it generates a strongly continuous group on H2 and we can apply Proposition 2.1 to −N2 to obtain a homeomorphism h2 : H2 → H2 , such that h2 (etN2 x) = etA2 h2 (x), 13

∀t ∈ IR, ∀x ∈ H2 ,

where A2 = (N1 + N2∗ )/2 is the real part of N2 . Therefore, since the real part A of N is given by A = A1 ⊕ A2 , we have from above that h(etN x) = etA h(x), ∀t ≥ 0, ∀x ∈ H, where h : H → H is a homeomorphism of H given by h(x) = h1 (P1 x) + h2 (P2 x),

∀x ∈ H,

where Pi is the orthogonal projection of H onto Hi , i = 1, 2. This finishes the proof. ¶

3.2

Proof of the regularity results

We assume N as in Theorem 3.2, so that in particular N satisfies the sectorial property (11). Lemma 3.8 In case Re σ(N ) ⊂ (−∞, 0], Lemma 3.5 is true with   λn λn+1 λn+1 λn+1 |y − x| + min{|x|, |y|}|y − x| , gn (x, y) ≤ arctan θ λn λn for every x, y ∈ Pn H and every n ∈ ZZ. Proof : It follows also from Proposition 2.1 by taking into account that in case (11) holds, we have k Bn kB(Pn H) ≤ λn+1 arctan θ, for n ∈ ZZ, where we used that Bn is symmetric, so that its norm is equal to its spectral radius. ¶ Lemma 3.9 In case Re σ(N ) ⊂ (−∞, 0], the functions h and h−1 defined in Lemma 3.6 are locally H¨ older continuous in the sense that given R ≥ 1, there exists C > 0, such that |h(y) − h(x)|, |h−1 (y) − h−1 (x)| ≤ C|y − x|β ,

(14)

for any x, y ∈ H with |x|, |y| ≤ R and |x − y| ≤ 1, where β = minn∈ZZ λn /λn+1 and {λn }n∈IN is an arbitrary sequence such that 0 < λn < λn+1 ,λn → 0 as n → −∞, λn → +∞ as n → +∞, and   λn+1 − λn log |n| < +∞. (15) lim sup λn+1 |n|→+∞ 14

Moreover, C can be chosen by the expression     2 R π2 1 β 2 C2 = 4 + (arctan θ) 2 + + C + 2, λ β2 β 3 where Cλ = sup |n|

1− λλn

n+1

n∈Z Z




.

Proof : Take R ≥ 1 and consider x, y ∈ H with |x|, |y| ≤ R and such that 0 < |x − y| ≤ 1. Set then δ = |y − x|, for notational simplicity. From (15), we have lim|n|→+∞ λn+1 /λn = 1; hence the quotient λn+1 /λn must be uniformly bounded. Let then C1 > 1 be given by C1 = max n∈Z Z

λn+1 . λn

(16)

Then, it follows from Lemma 3.8 that λn

gn (Pn y, Pn x) ≤ C2 [|Pn y − Pn x| λn+1 + |Pn y − Pn x|], for n ∈ ZZ, where

C2 = (C12 + C1 R) arctan θ.

(17) (18)

Therefore, using Lemma 3.5 and (17), |h(y) − h(x)|2

+∞ X

=

|hn (Pn y) − hn (Pn x)|2

n=−∞

≤ 2

+∞ X   gn (Pn y, Pn x)2 + |Pn y − Pn x|2

n=−∞

≤ (4C22 + 2)

+∞ X

|Pn y − Pn x|2 + 4C22

n=−∞

≤

(4C22

+∞ X

2λn

|Pn y − Pn x| λn+1

n=−∞ 2

+ 2)|y − x| +

4C22

X

2λn

|Pn y − Pn x| λn+1

Z Z∗

+4C22 |P0 y

− P0 x|

2λ0 λ1

.

where ZZ ∗ = ZZ \ {0}. Let then β = min

n∈Z Z

λn 1 = , λn+1 C1

(19)

and note that 0 < β < 1. Hence |h(y) − h(x)|2 ≤ (4C22 + 2)δ 2 + 4C22

X Z Z∗

15

2λn

|Pn y − Pn x| λn+1 + 4C22 δ 2β .

(20)

Set N1 =

(

N2 =

(

and

n ∈ ZZ ∗ ; |Pn y − Pn x|
0 is given by C 2 = 4C22 (2 +

2 π2 + C3β ) + 2. 3

(27)

This completes the proof that h is locally H¨older continuous. The proof for h−1 is exactly the same. The estimate for C comes from (16), (18), (19), (24) and (27). ¶ Proof of Theorem 3.2 : Let N be a normal operator in a complex Hilbert space H generating a strongly continuous semigroup, with zero not in the point 17

spectrum of Re N and with N satisfying the sectorial property (11). By decomposing N into a direct sum of two operators as done in the proof of Theorem 3.1, we can assume without loss of generality that Re σ(−N ) ⊂ (−∞, 0]. Then given ν, 0 < ν < 1, we simply have to choose an increasing sequence {λn }n∈ZZ with λn → 0 as n → −∞, λn → +∞ as n → +∞,   λn+1 − λn lim sup log |n| < +∞, λn |n|→+∞ and also ν ≤ min

n∈Z Z

λn < 1. λn+1

The proof then follows from Lemmas 3.7 and 3.9 applied to N with this particular sequence. It is not difficult to see that we can choose this sequence to be, for instance, ( n+l for n ∈ ZZ, n ≥ 0, l , λn = l for n ∈ ZZ, n < 0, l−n , where l ∈ IN is chosen large enough such that ν≤

l . l+1

(28)

Now, it is not hard to see that for the sequence above, we can bound the constant C in (14) independently of the choice of l, because the upper bound given for C in Lemma 3.9 decreases when l increases. This implies then that the constant C in (14) is independent of ν since the right hand side of (28) can be made as close as we want to 1, by increasing l. Therefore, (13) holds with C(R, θ) depending on R and θ, but independent of ν. ¶ Finally, we have: Proof of Theorem 3.3 : Let N be a normal operator in a complex Hilbert space H generating a strongly continuous semigroup, with zero not in the point spectrum of Re N and satisfying the sectorial property (11). Assume that the real part A of N has a pure point spectrum and a countable number of eigenvalues. Denote the eigenvalues by {−σn }n∈IN . In this case we consider the eigenprojectors Pn associated to each eigenvalue −σn , and we have etA Pn = e−σn t Pn ,

∀t ∈ IR, ∀n ∈ IN.

(29)

We can then apply Proposition 2.1 to each operator N Pn in Pn H to obtain a homeomorphism hn of Pn H conjugating {etN Pn }t∈IR to {etA Pn }t∈IR . Since

18

N satisfies the sectorial property (11) and (29) holds, we have from Proposition 2.1 that −1 |hn (y) − hn (x)|, |h−1 (30) n (y) − hn (x)| ≤ C(R, θ)|y − x|, for all x, y ∈ Pn H, |x|, |y| ≤ R, |x − y| ≤ 1, where C(R, θ) = 1 + (1 + R) arctan θ, for suitable θ given by (11). We set then h(x) =

+∞ X

hn (Pn x),

∀x ∈ H,

n=1

and we obtain as in Lemma 3.7 that h(etN x) = etA h(x),

∀t ≥ 0, ∀x ∈ H.

Also, from (30), we find |h(y) − h(x)|2

=

+∞ X

|hn (Pn y) − hn (Pn x)|2

n=1

≤ C(R, θ)2

+∞ X

|Pn y − Pn x|2 = C(R, θ)2 |y − x|2 ,

n=1

for all x, y ∈ H, |x|, |y| ≤ R, |x − y| ≤ 1, which proves that h is Lipschitz continuous. Similarly for h−1 . The case of a finite number of eigenvalues is totally analogous.This finishes the proof. ¶

ACKNOWLEDGMENTS This work was partially supported by the CNPq-Bras´ılia, Brazil, under the grant 200308/92-0, by the Office of Naval Research under the grant NAVY N00014-91J-1140, and by the National Science Foundation under the grant DMS-9024769. I would like to take this opportunity to thank also Professor Ciprian Foias for his support and many valuable discussions and suggestions that led me to this work.

References [1] C. Foias, “Ordinary Differential Equations,” lecture notes given at Indiana University, 1993, unpublished. [2] D. Henry, Geometric Theory of Semilinear Parabolic Equations, in “Lecture Notes in Mathematics,” Vol. 840, Springer-Verlag, New York, 1981. 19

[3] T. Kato, “Perturbation Theory for Linear Operators,” Springer-Verlag, New York, 1966. [4] J. Palis Jr. & W. de Melo, “Geometric Theory of Dynamical Systems,” Springer-Verlag, New York, 1982. [5] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, in “Appl. Math. Sci.,” Vol. 44, Springer-Verlag, New York, 1983. [6] F. Riesz & B. SZ-Nagy, “Functional Analysis,” Frederick Ungar Publ., New York, 1955.

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