for X a Hilbert space, A = A^{∗}. We want to define A^{α} for α ∈
(0,1)
in such
a way that things work as they should provided that
(Ax,x)
≥ 0.
If A ∈
(0,∞ )
we can get A^{−α} for α ∈
(0,1)
as
1 ∫ ∞
A−α ≡ ----- e−Atta−1dt
Γ (α) 0
Indeed, you change the variable as follows letting u = At,
∫ ∞ ∫ ∞ ( )a−1
e−Atta−1dt = e−u u- 1-du
0 ∫0∞ A A
= e−uuα−1A1−α 1du = A −αΓ (α)
0 A
Next we need to define e^{−At} for A ∈ℒ
(X,X )
.
Definition 24.8.1By definition, e^{−At}x_{0}will be x
(t)
where x
(t)
is the solution tothe initial value problem
x′ + Ax = 0, x(0) = x0
Such a solution exists and is unique by standard contraction mapping arguments as inTheorem 24.7.3. Equivalently, one could consider for Φ
(t)
≡ e^{−At}the solution in ℒ
(X,X )
of
′
Φ (t)+ A Φ(t) = 0, Φ (0) = I.
Now the case of interest here is that A = A^{∗} and
(Ax, x)
≥ δ
|x|
^{2}. We need an estimate
for
∥∥e− At∥∥
.
Lemma 24.8.2Suppose A = A^{∗}and
(Ax, x)
≥ ε
|x|
^{2}. Then
∥ ∥
∥e−At∥ ≤ e− εt
Proof: Let
ˆx
(t)
= x
(t)
e^{εt}. Then the equation for e^{−At}x_{0}≡ x
(t)
becomes
xˆ′(t)− εˆx(t)+ Aˆx (t) = 0,ˆx(0) = x0
Then multiplying by
ˆx
(t)
and integrating gives
∫ ∫
1 2 t 2 1 2 t
2 |ˆx(t)|− ε 0 |ˆx(s)|ds − 2 |x0| + 0 (A ˆx,ˆx)ds = 0
and so, from the estimate,
|ˆx (t)| ≤ |x0|
Hence,
|x (t)| ≤ |x0|e− εt
Since x_{0} was arbitrary, it follows that
|||| −At||||
e
≤ e^{−εt}. ■
With this estimate, we can define A^{−α} for α ∈
(0,1)
if A = A^{∗} and
(Ax,x)
≥ ε
|x |
^{2}.
Definition 24.8.3Let A ∈ℒ
(X,X )
,A = A^{∗}and
(Ax, x)
≥ ε
|x|
^{2}. Then forα ∈
(0,1)
,
1 ∫ ∞
A−α ≡ ----- e−Atta− 1dt
Γ (α) 0
The integral is well defined thanks to the estimate of the above lemma which gives
|| ||
||e−At||
≤ e^{−εt}. You can let the integral be a standard improper Riemann integral sinceeverything in sight is continuous. For such A, define
A α ≡ AA −(1−α)
Note that from Lemma 24.8.2it follows that each A^{α}is Hermitian and commutes with everyoperator C which commutes with A.
Note the meaning of the integral. It takes place in ℒ
(X, X)
and Riemann sums
approximating this integral also take place in ℒ
(X,X )
. Thus
∫ ∞ ∫ ∞
e− Atta−1dt(x0) = ta− 1e−At (x0)dt
0 0
by approximating the improper integral with Riemann sums and using the obvious fact that
such an identification holds for each sum in the approximation. This is stated as part of the
following lemma.
Lemma 24.8.4For all x_{0},∫_{0}^{∞}e^{−At}t^{a−1}dt
(x0)
= ∫_{0}^{∞}t^{a−1}e^{−At}
(x0)
dt. Also A^{−α}is Hermitian whenever A is. (This is the case considered here.) Also we have thesemigroup property e^{−A(t+s)
} = e^{−At}e^{−As}for t,s ≥ 0. In addition to this, if CA = ACthen e^{−At}C = Ce^{−At}. In words, e^{−At}commutes with every C ∈ ℒ
(X,X )
whichcommutes with A. Also, e^{−At}is Hermitian whenever A is and so is A^{−α}. (This is whatis being considered here.)
Proof: The semigroup property follows right away from uniqueness considerations for the
ordinary initial value problem. Indeed, letting s,t ≥ 0, fix s and consider the following for
Φ
(t)
≡ e^{−At}. Thus Φ
(0)
= I and Φ
(t)
x_{0} is the solution to x^{′} + Ax = 0,x
(0)
= x_{0}. Then
define
t → Φ (t+ s)− Φ (t)Φ (s) ≡ Y (t)
Then taking the time derivative, you get the following ordinary differential equation in
ℒ
(X,X )
Y ′(t) = Φ′(t+ s)− Φ′(t)Φ(s) = − AΦ (t+ s)+ AΦ (t)Φ (s)
= − A(Φ (t +s) − Φ (t)Φ(s)) = − AY (t)
also, letting t = 0,Y
(0)
= Φ
(s)
− Φ
(0)
Φ
(s)
= 0. Thus, by uniqueness of solutions to ordinary
differential equations, Y
(t)
= 0 for all t ≥ 0 which shows the semigroup property. See
Theorem 24.35. Actually, this identity holds in this case for all s,t ∈ ℝ but this
is not needed and the argument given here generalizes well to situations where
one can only consider t ∈ [0,∞). Note how this shows that it is also the case that
Φ
(t)
Φ
(s)
= Φ
(s)
Φ
(t)
.
Now consider the claim about commuting with operators which commute with A. Let
CA = AC for C ∈ℒ
(X, X )
and let y
(t)
be given by
( )
y (t) ≡ Ce −Atx0 − e−AtCx0
Then y
(0)
= Cx_{0}− Cx_{0} = 0 and
′ ( (− At )) ( −At )
y (t) = C − A( e x0) +(Ae Cx)0
= − AC e−Atx0 +A e−AtCx0
and so
′ ( −At −At )
y (t)+ A Ce x0 − e Cx0 = 0
y′ + Ay = 0
Therefore, by uniqueness to the ODE one obtains y
(t)
= 0 which shows that C commutes
with e^{−At}.
Finally consider the claim about e^{−At} being Hermitian. For Φ
(t)
≡ e^{−At}
Φ′(t) +A Φ (t) = 0,Φ (0) = I
Φ∗′(t)+ Φ ∗(t)A = 0, Φ∗(0) = I
Note that Lemma 24.8.2 shows that AA^{−α} = A^{−α}A. Also A^{−α}A^{−β} = A^{−β}A^{−α} and in
fact, A^{−α} commutes with every operator which commutes with A. Next is a technical lemma
which will prove useful.
Lemma 24.8.5For α,β > 0,Γ
(α)
Γ
(β)
= Γ
(α + β)
∫_{0}^{1}
(1 − v)
^{α−1}v^{β−1}dv
Proof:
∫ ∞ ∫ ∞ ∫ ∞ ∫ ∞
Γ (α)Γ (β) ≡ e−(t+s)tα− 1sβ− 1dtds = e−u (u − s)α− 1sβ−1duds
0 0 0 s
∫ ∞ −u ∫ u α−1 β−1
= 0 e 0 (u − s) s dsdu
∫ ∞ ∫ u
= e−u (u − s)α−1sβ−1dsdu
∫0∞ ∫0 1
= e−u (u − uv)α−1(uv)β− 1udvdu
0 0
∫ ∞ −u ∫ 1 α−1 β α−1 β−1
= 0 e 0 u u (1− v) v dvdu
∫ 1 ∫ ∞
= (1− v)α−1vβ− 1dv uα+β−1e−udu
0 ∫ 0
1 α−1 β− 1
= Γ (α+ β) 0 (1 − v) v dv■
Now consider whether A^{−α} acts like it should. In particular, is A^{−α}A^{−}
(1−α)
= A^{−1}?
Lemma 24.8.6For α ∈
(0,1)
,A^{−α}A^{−(1−α)
} = A^{−1}. More generally, if α + β <
1,A^{−α}A^{−β} = A^{−(α+β)
}.
Proof: The product is the following where β = 1 − α