It will always be assumed in this section that A is sectorial for the sector S_{−a,ϕ} where
a > 0. To begin with, here is a useful lemma which will be used in the presentation of
these fractional powers.
This is fairly easy to evaluate using contour integrals. Consider the following contour
called Γ_{R} for large R. As R →∞, the integral over the little circle converges
to 0 and so does the integral over the big circle. There is one singularity at
i.
PICT
Thus
∫ e(ln|z|+iarg(z))(1− 2α)
lRim→∞ -----1-+-z2-----dz =
Γ R
Changing variables and using Fubini’s theorem which is justified because of the abolute
convergence of the iterated integrals, which follows from Corollary 53.2.7, this
becomes
1 ∫ ∞ ∫ ∞ α−1 β−1
Γ-(α)Γ (β) t (u− t) S (u )dudt
0 t
∫ ∫
= ----1---- ∞ u tα−1 (u − t)β−1S (u)dtdu
Γ (α )Γ (β) 0 0
1 ∫ ∞ ∫ 1 α−1 β−1
= Γ-(α-)Γ (β) S (u) (ux) (u− ux) udxdu
( 0∫ 1 0 )∫ ∞
= ----1---- xα−1(1 − x )β−1 dx S (u)uα+β−1du
Γ (α )Γ (β) 0 0
----1----( ∫ 1 α−1 β−1 ) −(α+β)
= Γ (α )Γ (β) 0 x (1 − x ) dx Γ (α+ β)(− A)
− (α+β)
= (− A )
This proves the first part of the theorem.
Consider 53.2.9. Since A is a closed operator, and approximating the integral with an
appropriate sequence of Riemann sums,
(− A)
can be taken inside the integral and
so
∫ ∫
--1- ∞ 1−1 ∞
(− A)Γ (1) 0 t S (t)dt = 0 (− A)S (t)dt
∫
= ∞ − d-(S(t))dt = S (0) = I.
0 dt
Next let x ∈ D
(− A )
. Then
-1--∫ ∞ 1− 1 ∫ ∞
Γ (1) 0 t S (t)dt(− A)x = − 0 S (t) Axdt
∫ ∞ ∫ ∞ d
= − AS (t)xdt = − --(S (t))dt = Ix
0 0 dt
This shows that the integral in which α = 1 deserves to be called A^{−1} so the definition is
not bad notation. Also, by assumption, A^{−1} is one to one. Thus
(− A )−1(− A )−1 x = 0
implies
(− A)−1x = 0
hence x = 0 so that
(− A)
^{−2} is also one to one. Similarly,
(− A )
^{−m} is one to one for all
positive integers m.
From what was just shown, if
(− A)
^{−α}x = 0 for α ∈
(0,1)
, then
−1 −(1− α) −α
(− A ) x = (− A) (− A) x = 0
and so x = 0. This shows
(− A)
^{−α} is one to one for all α ∈
[0,1]
if is defined as
(− A)
^{0}≡ I.
What about α > 1? For such α, it is of the form m + β where β ∈ [0,1) and m is a
positive integer. Therefore, if