T. Diagana INTEGER POWERS OF SOME UNBOUNDED LINEAR ...

210 T. Diagana
PROPOSITION 3. Let z, z ′ ∈ J(λ). If A is a diagonal operator with (λ i ) i∈N ⊂
K as a corresponding sequence, then
(i) A z .A z′ = A z+z′ ;
(ii) (A z ) z′ = A zz′ .
Proof. (i) If x = ∑ x i e i ∈ D(A z .A z′ ), then A z A z′ x = ∑ z+z
λ ′ i x i e i . Next, we use
i∈N
i∈N
the fact that z, z ′ ∈ J(λ) yield z + z ′ ∈ J(λ) (Remark 2(i)).
(ii) Similarly, if x = ∑ x i e i ∈ D((A z ) z′ ), then (A z ) z′ x = ∑ zz
λ ′ i x i e i . Now
i∈N
i∈N
since zz ′ ∈ J(λ) (Remark 2(i)) it is clear that (A z ) z′ = A zz′ .
PROPOSITION 4. If z ∈ J(λ), then the integer power A z of the diagonal operator
A is self-adjoint. Furthermore, ρ(A z ) = {λ ∈ K : λ ̸= λi z , ∀i ∈ N}, and
for each λ ∈ ρ(A z ).
‖(A z − λ) −1 ‖ ≤
1
inf i∈N |λ z i − λ|
Proof. First of all, note that A z x = ∑ i∈N
λ i z x i e i , ∀x = (x i ) i∈N ∈ D(A z ) with D(A z ) =
{x = (x i ) i∈N ⊂ K : lim |λ i| z |x i | ‖e i ‖ = 0}. Since z ∈ J(λ) it follows that A z is
i→∞
well-defined. Note that A z is a diagonal operator corresponding to γ i = λ z i with
lim |γ i| = ∞, since z ∈ J(λ). So to complete the proof one follows along the same
i→∞
line as in the proof of Proposition 2.
PROPOSITION 5. Let A, B be diagonal operators on E ω . If λ = (λ i ) i∈N , µ =
(µ i ) i∈N are respectively the corresponding sequences to the diagonal operators A and
B, and if |λ i | ̸= |µ i | for each i ∈ N, and if J(λ) ∩ J(µ) ∩ Z + ̸= ∅ (Z + being the set of
all natural numbers), then
for each z ∈ J(λ) ∩ J(µ) ∩ Z + .
D((A + B) z ) = D(A z ) ∩ D(B z ) = D((A + B) ∗z ),
Proof. Using the argument that |λ i | ̸= |µ i | for each i ∈ N it easily follows that
(10) |λ i + µ i | z = max(|λ| z ,|µ i | z )
for all i ∈ N, z ∈ Z + .