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

206 T. Diagana
To complete the proof one needs to compute ρ(A). For that, we have to solve
the equation
(6)
where x = ∑ i
(A − λI)x = y,
x i e i ∈ D(A) = D(A − λI) and y = ∑ i
y i e i ∈ E ω .
Considering Eq. (6) on (e i ) i∈N and using the fact A is self-adjoint it follows
that ∀i ∈ N, (λ i − λ).〈e i , x〉 = 〈e i , y〉. Equivalently, ∀i ∈ N,
(7)
(λ i − λ).ω i x i = ω i y i .
Now if ∀i ∈ N, λ i ̸= λ, Eq. (7) has a unique solution x, moreover,
(8)
x = (A − λ) −1 y = ∑ i
y i
λ i − λ e i.
Let us show that x = (A − λ) −1 y given above is well-defined. For that it is
|y i |
sufficient to prove that lim
i |λ i − λ| ‖e i‖ = 0. According to Eq. (5), the sequence
( )
1
|y i |
is bounded, hence lim
|λ i − λ| i∈N
i |λ i − λ| ‖e i‖ = 0.
It remains to find conditions on λ so that x defined above belongs to D(A). For
that, it is sufficient to find conditions so that:
Indeed, since lim
i
|y i | ‖e i ‖ = 0,
lim
i
|y i |
|λ i − λ| ‖Ae i‖ = lim
i
|λ i |
|λ i − λ| |y i| ‖e i ‖ = 0.
0 ≤ lim
i
|y i |
|λ i − λ| ‖Ae i‖
|λ i |
≤ lim
i | |λ i | − |λ| | . lim i
= 0.
|y i | ‖e i ‖
From Eq. (8) it follows that
‖(A − λ) −1 y‖ =
|y i |‖e i ‖
sup
i∈N |λ i − λ|
≤
1
‖y‖ . sup
i∈N |λ i − λ|
≤
1
‖y‖ .
inf i∈N |λ i − λ|
< ∞,