142
Lecture 11
Proposition 11.4.4. Let {T (t) : t
0} be a C0 -semigroup. The family of operators
?
{T (t) : t 0} is a C0 -semigroup whose generator is L? .
Proof. The semigroup law is immediately checked. Let us prove the strong continuity.
Recall that by (11.1.1) we have kT (t)k = kT (t)? k  M e!t , where we may assume ! > 0.
First, notice that T (t)? x ! x weakly for every x 2 X as t ! 0. Indeed, by the strong
continuity of T (t) we have hT (t)? x, yi = hx, T (t)yi ! hx, yi as t ! 0 for every y 2 X.
From the estimate
Z t
M !t
hT (s)? x, yi ds 
(e
1)kxk kyk
!
0
and the Riesz Theorem we get the existence of xt 2 X such that
Z
1 t
hT (s)? x, yi ds = hxt , yi
8 y 2 X.
t 0
Therefore, for t > 0 and 0 < h < t we infer
|hT (h)? xt
xt , yi| = |hxt , T (h)yi hxt , yi|
Z
Z
1 t
1 t
=
hT (s)? x, T (h)yi ds
hT (s)? x, yi ds
t 0
t 0
Z
Z
1 t
1 t
?
=
hT (s + h) x, yi ds
hT (s)? x, yi ds
t 0
t 0
Z
Z
1 t+h
1 h
?
=
hT (s) x, yi ds
hT (s)? x, yi ds
t t
t 0
⇤
1
M ⇥ !(t+h)
 kxk kyk
(e
e!t ) + (e!h 1) .
t
!
Taking the supremum on kyk = 1, we deduce
lim kT (h)? xt
h!0
xt k = 0.
(11.4.7)
Since by linearity and an "/3 argument the set where (11.4.7) holds is a weakly closed
subspace Y , containing xt for all t > 0. Since for any x 2 X, xt ! x weakly as t ! 0, we
conclude that Y = X and that T (t)? is strongly continuous. Denoting by A its generator,
for x 2 D(L) and y 2 D(A) we have
hLx, yi = limht
t!0
1
I)x, yi = limhx, t
(T (t)
t!0
1
(T (t)?
I)yi = hx, Ayi,
so that A ⇢ L? . Conversely, for y 2 D(L? ), x 2 D(L) we have
?
hx, T (t) y
Z
t
yi = hT (t)x x, yi =
hLT (s)x, yi ds
0
Z t
Z t
?
=
hT (s)x, L yi ds =
hx, T (s)? L? yi ds.
0
0
Semigroups of Operators
143
We deduce
?
T (t) y
y=
Z
t
T (s)? L? y ds,
0
whence, dividing by t and letting t ! 0 we get Ay = L? y for every y 2 D(L? ) and
consequently L? ⇢ A.