572 CHAPTER 19. HILBERT SPACES

Lemma 19.13.6 For 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

1Γ(α)

∫∞

0e−Atta−1dt

1Γ(β )

∫∞

0e−Assβ−1ds

Then this equals

1Γ(α)Γ(β )

∫∞

0

∫∞

0e−A(t+s)tα−1sβ−1dtds

=1

Γ(α)Γ(β )

∫∞

0

∫∞

se−Au (u− s)α−1 sβ−1du ds

=1

Γ(α)Γ(β )

∫∞

0e−Au

∫ u

0(u− s)α−1 sβ−1dsdu

=1

Γ(α)Γ(β )

∫∞

0uα+β−1e−Au

∫ 1

0(1− v)α−1 vβ−1dvdu

=∫ 1

0(1− v)α−1 vβ−1dv

1Γ(α)Γ(β )

∫∞

0uα+β−1e−Audu

From the above lemma, this equals

1Γ(α +β )

∫∞

0uα+β−1e−Audu≡ A−(α+β )

Note how this shows that these powers of A all commute with each other. If α +β = 1, thisbecomes ∫

∞

0e−Audu

Is this the usual inverse or is it something else called A−1 but not being the real inverse?We show it is the usual inverse. To see this, consider

A∫

∞

0e−Audu(x0) =

∫∞

0Ae−Aux0du =

∫∞

0−x′ (u)du

where x′ (t)+Ax(t) = 0,x(0) = x0 and |x(t)| ≤ e−εt |x0| . The manipulation is routine fromapproximating with Riemann sums. Then the right side equals x0. Thus A

∫∞

0 e−Audu = I.Similarly, ∫

∞

0e−Audu A(x0) =

∫∞

0e−AuAx0du =

∫∞

0Ae−Aux0du = x0

and so this shows the desired result.Now it follows that if Aα ≡ AA−(1−α) as defined above, then A1−α = AA−(1−(1−α)) =

AA−α and so,Aα A1−α ≡ AA−(1−α)AA−α = A2A−1 = A