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
∫
−α -1--- ∞ −At a−1
A ≡ Γ (α) 0 e t dt
Indeed, you change the variable as follows letting u = At,
∫ ∞ −At a− 1 ∫ ∞ −u (u-)a−1 1
0 e t dt = 0 e A A du
∫ ∞ 1
= e−uuα− 1A1 −αA-du = A −αΓ (α )
0
Next we need to define e^{−At} for A ∈ℒ
(X,X )
.
Definition 17.13.1By definition, e^{−At}x_{0}will be x
(t)
where x
(t)
is the solution to theinitial value problem
′
x + Ax = 0, x(0) = x0
Such a solution exists and is unique by standard contraction mapping arguments as inTheorem??. 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 17.13.2Suppose 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
ˆx′(t)− εˆx(t)+ Axˆ(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,
|xˆ(t)| ≤ |x0|
Hence,
|x(t)| ≤ |x0|e−εt
Since x_{0} was arbitrary, it follows that
||||e−At||||
≤ e^{−εt}. ■
With this estimate, we can define A^{−α} for α ∈
(0,1)
if A = A^{∗} and
(Ax,x)
≥ ε
|x|
^{2}.
Definition 17.13.3Let 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
|||| −At||||
e
≤ e^{−εt}. You can let the integral be a standard improper Riemann integral sinceeverything in sight is continuous. For such A, define
Aα ≡ AA− (1−α)
Note that from Lemma 17.13.2it follows that each A^{α}is Hermitian and commutes withevery operator 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 17.13.4For all x_{0},∫_{0}^{∞}e^{−At}t^{a−1}dt
(x0)
= ∫_{0}^{∞}t^{a−1}e^{−At}
(x0)
dt. AlsoA^{−α}is Hermitian whenever A is. (This is the case considered here.) Also we havethe semigroup property e^{−A(t+s)
} = e^{−At}e^{−As}for t,s ≥ 0. In addition to this, ifCA = AC then e^{−At}C = Ce^{−At}. In words, e^{−At}commutes with every C ∈ℒ
(X, X)
which commutes with A. Also, e^{−At}is Hermitian whenever A is and so is A^{−α}.(This is what is 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) = − A Φ(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 17.12.75. 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
y′(t) = C (− A (e−Atx0))+ (Ae−AtCx0)
( −At ) ( −At )
= − AC e x0 + A e Cx0
and so
y′(t)+ A (Ce−Atx0 − e−AtCx0 ) = 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)+ AΦ (t) = 0,Φ (0) = I
Φ∗′(t)+ Φ∗(t)A = 0, Φ∗(0) = I
and so, from what was just shown about commuting,
Φ∗′(t)+ Φ∗(t)A = 0, Φ∗(0) = I
Φ′(t) +Φ (t)A = 0,Φ (0) = I
Thus Φ
(t)
and Φ^{∗}
(t)
satisfy the same initial value problem and so they are the same
Thanks to Theorem ??.
Note that Lemma 17.13.2 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 17.13.5For 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∞ ∫01
= e− u (u− uv)α−1(uv)β−1udvdu
0 0
∫ ∞ − u∫ 1 α−1 β α− 1 β−1
= e u u (1 − v) v dvdu
∫01 0 ∫ ∞
= (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 17.13.6For 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