19.13. FRACTIONAL POWERS OF OPERATORS 569

Definition 19.13.1 By definition, e−Atx0 will be x(t) where x(t) is the solution to the initialvalue problem

x′+Ax = 0, x(0) = x0

Such a solution exists and is unique by standard contraction mapping arguments as in The-orem 19.12.3. Equivalently, one could consider for Φ(t) ≡ e−At the solution in L (X ,X)of

Φ′ (t)+AΦ(t) = 0, Φ(0) = I.

Now the case of interest here is that A = A∗ and (Ax,x) ≥ δ |x|2. We need an estimatefor∥∥e−At

∥∥.

Lemma 19.13.2 Suppose A = A∗ and (Ax,x)≥ ε |x|2 . Then∥∥e−At∥∥≤ e−εt

Proof: Let x̂(t) = x(t)eεt . Then the equation for e−Atx0 ≡ x(t) becomes

x̂′ (t)− ε x̂(t)+Ax̂(t) = 0, x̂(0) = x0

Then multiplying by x̂(t) and integrating gives

12|x̂(t)|2− ε

∫ t

0|x̂(s)|2 ds− 1

2|x0|2 +

∫ t

0(Ax̂, x̂)ds = 0

and so, from the assumed estimate,

12|x̂(t)|2− ε

∫ t

0|x̂(s)|2 ds− 1

2|x0|2 +

∫ t

0ε |x̂(s)|2 ds≤ 0

and so |x̂(t)| ≤ |x0|. Hence, |x(t)|=∣∣e−Atx0

∣∣≤ |x0|e−εt . Since x0 was arbitrary, it followsthat

∥∥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 19.13.3 Let A ∈L (X ,X) ,A = A∗ and (Ax,x)≥ ε |x|2 . Then for α ∈ (0,1) ,

A−α ≡ 1Γ(α)

∫∞

0e−Atta−1dt

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 since everything insight is continuous. For such A, define

Aα ≡ AA−(1−α)

Note the meaning of the integral. It takes place in L (X ,X) and Riemann sums approx-imating this integral also take place in L (X ,X) . Thus∫

∞

0e−Atta−1dt (x0) =

∫∞

0ta−1e−At (x0)dt