54.3. SECTORIAL OPERATORS AND ANALYTIC SEMIGROUPS 1731

≤ η

4+4MM+

η

2+

η

4+4MM < η

Thus S (t)(B) has an η net for every η > 0 and so S (t) is compact.Next suppose S (t) is compact for all t > 0. Then

(−A)−α =1

Γ(α)

∫∞

0tα−1S (t)dt

and the integral is a limit in norm of Riemann sums of the formm

∑k=1

tα−1k S (tk)∆tk

and each of these operators is compact. Since (−A)−α is the limit in norm of compactoperators, it must also be compact. This proves the proposition.

Here are some observations which are listed in the book by Henry [63]. Like the aboveproposition, these are exercises in this book.

Observation 54.3.23 For each x ∈H, t→ tAS (t) is continuous and limt→0+ tAS (t)x = 0.

The reason for this is that if x ∈ D(A) , then

tAS (t)x = |tS (t)Ax| → 0

as t→ 0. Now suppose y ∈ H is arbitrary. Then letting x ∈ D(A) ,

|tAS (t)y| ≤ |tAS (t)(y− x)|+ |tAS (t)x|≤ ε + |tAS (t)x|

provided x is close enough to y. The last term converges to 0 and so

lim supt→0+

|tAS (t)y| ≤ ε

where ε > 0 is arbitrary. Thuslim

t→0+|tAS (t)y|= 0.

Why is t→ tAS (t)x continuous on [0,T ]? This is true if x ∈ D(A) because t→ tS (t)Ax iscontinuous. If y ∈ H is arbitrary, let xn converge to y in H where xn ∈ D(A) . Then

|tAS (t)y− tAS (t)xn| ≤C |y− xn|

and so the convergence is uniform. Thus t → tAS (t)y is continuous because it is the uni-form limit of a sequence of continuous functions.

Observation 54.3.24 If x ∈H and A is sectorial for S−a,φ ,−a < 0, then for any α ∈ [0,1] ,

limt→0+

tα∣∣∣∣(−A)α S (t)x

∣∣∣∣= 0.

This follows as above because you can verify this is true for x ∈ D(A) and then use thefact shown above that

tα∣∣∣∣(−A)α S (t)

∣∣∣∣≤C

to extend it to x arbitrary.