Much more on semigroups is available in Yosida [?]. This is just an introduction to the
subject.
Definition 17.14.1A strongly continuous semigroup defined on H,a Banachspace is a function S : [0,∞) → H which satisfies the following for all x_{0}∈ H.
S(t) ∈ ℒ (H,H ),S(t+ s) = S (t)S (s),
t → S (t)x is continuous, lim S (t)x = x
0 t→0+ 0 0
Sometimes such a semigroup is said to be C_{0}. It is said to have the linear operator A asits generator if
{ S(h)x-−-x }
D(A ) ≡ x : lhim→0 h exists
and for x ∈ D
(A )
, A is defined by
lim S(h)x-−-x≡ Ax
h→0 h
The assertion that t → S
(t)
x_{0} is continuous and that S
(t)
∈ℒ
(H, H )
is not
sufficient to say there is a bound on
||S (t)||
for all t. Also the assertion that for each
x_{0},
lt→i0m+ S (t)x0 = x0
is not the same as saying that S
(t)
→ I in ℒ
(H,H )
. It is a much weaker assertion. The
next theorem gives information on the growth of
||S(t)||
. It turns out it has exponential
growth.
Lemma 17.14.2Let M ≡ sup
{||S (t)|| : t ∈ [0,T ]}
. Then M < ∞.
Proof: If this is not true, then there exists t_{n}∈
[0,T]
such that
||S (t)||
n
≥ n. That is
the operators S
(t )
n
are not uniformly bounded. By the uniform boundedness principle,
Theorem 15.1.8, there exists x ∈ H such that
||S(t )x||
n
is not bounded. However, this is
impossible because it is given that t → S
(t)
x is continuous on
[0,T ]
and so
t →
||S (t)x||
must achieve its maximum on this compact set.
Now here is the main result for growth of
||S(t)||
.
Theorem 17.14.3For M described in Lemma 17.14.2, there exists α suchthat
||S(t)|| ≤ M eαt.
In fact, α can be chosen such that M^{1∕T} = e^{α}.
Proof:Let t be arbitrary. Then t = mT + r
(t)
where 0 ≤ r
(t)
< T. Then by the
semigroup property
||S(t)|| = ||S (mT + r(t))||
= ||S(r(t))S(T)m || ≤ M m+1
Now mT ≤ t ≤ mT + r
(t)
≤
(m + 1)
T and so
t-
m ≤ T ≤ m + 1
Therefore,
( )t
||S (t)|| ≤ M (t∕T)+1 = M M 1∕T .
Let M^{1∕T}≡ e^{α} and then
αt
||S(t)|| ≤ M e
This proves the theorem.
Definition 17.14.4Let S
(t)
be a continuous semigroup as described above. It is calleda contractionsemigroup if for all t ≥ 0
||S (t)|| ≤ 1.
It is called a bounded semigroup if there exists M such that for all t ≥ 0,
||S (t)|| ≤ M
Note that for S
(t)
an arbitrary continuous semigroup satisfying
||S (t)|| ≤ M eαt,
It follows that the semigroup,
T (t) = e− αtS (t)
is a bounded semigroup which satisfies
||T (t)|| ≤ M.
Proposition 17.14.5Given a continuous semigroup S
(t)
, its generator A exists and isa closed densely defined operator. Furthermore, for
αt
||S(t)|| ≤ M e
and λ > α, λI −A is onto and
(λI − A)
^{−1}maps H onto D
(A )
and is in ℒ
(H,H )
. Alsofor these values of λ,
∫ ∞
(λI − A )− 1x = e−λtS (t)xdt.
0
For λ > α, the following estimate holds.
|| || M
||||(λI − A )−1|||| ≤------
|λ − α|
Proof: First note D
(A )
≠∅. In fact 0 ∈ D
(A)
. It follows from Theorem 17.14.3 that
for all λ large enough, one can define a Laplace transform,
∫
∞ −λt
R(λ)x ≡ 0 e S (t) xdt ∈ H.
Here the integral is the ordinary improper Riemann integral. I claim each of these is in
D
(A)
.
∫∞ −λt ∫∞ −λt
S-(h-)0--e--S-(t)-xdt−--0-e--S-(t)-xdt
h
Using the semigroup property and changing the variables in the first of the above
integrals, this equals
( ∫ ∫ )
= 1- eλh ∞ e−λtS (t)xdt − ∞ e−λtS (t)xdt
h h 0
( )
1 ( λh )∫ ∞ −λt λh∫ h −λt
= h- e − 1 0 e S(t)xdt− e 0 e S(t)xdt
The limit as h → 0exists and equals
λR (λ)x− x
Thus R
(λ)
x ∈ D
(A)
as claimed and
AR (λ)x = λR (λ)x− x
Hence
x = (λI − A )R(λ)x. (17.14.77)
(17.14.77)
Since x is arbitrary, this shows that for λ large enough, λI − A is onto.
( ∫ t ) ∫ t ( S (h) x− x)
y∗ S(s)Axds = y∗ S(s) lim --------- ds
0 0 h→0+ h
The difference quotient is given to have a limit and so the difference quotients are
bounded. Therefore, one can use the dominated convergence theorem to take the limit
outside the integral and write the above equals
∫ t ( S(h)x − x)
hli→m0+ y∗ S (s) ----h---- ds
0( ( ∫ ∫ ) )
= lim y∗ 1- t+h S(s)xds− tS (s)xds
h→0+ h h 0
(∫ t+h ∫ h )
= lim y∗ S (s) xds− S (s)x
h→0+ t 0
= y∗(S(t)x− x) .
Thus since y^{∗} is arbitrary, for x ∈ D
(A)
∫ t
S(t)x = x+ S(s)Axds
0
Why is A closed? Suppose x_{n}→ x and x_{n}∈ D
(A )
while Ax_{n}→ z. From what was
just shown
∫
S (t)x = x + tS (s)Ax ds
n n 0 n
and so, passing to the limit this yields
∫ t
S(t)x = x + S(s)zds
0
which implies
∫
S(t)x−-x- 1 t
tli→m0+ h = hli→m0+ t 0 S (s)zds = z
and this also proves the last estimate. Also from
17.14.77, R
(λ)
maps H onto D
(A)
. This proves the proposition.
The linear mapping
(λI − A)
^{−1} is called the resolvent.
The above proof contains an argument which implies the following corollary.
Corollary 17.14.6Let S
(t)
be a continuous semigroup and let A be its generator. Thenfor 0 < a < b and x ∈ D
(A)
∫ b
S(b)x − S (a)x = S (t) Axdt
a
and also for t > 0 you can take the derivative from the left,
lim S-(t)x-− S-(t−-h)x-= S(t)Ax
h→0+ h
Proof:Letting y^{∗}∈ H^{′},
( ∫ b ) ∫ b ( )
y∗ S(t)Axdt = y∗ S(t) lim S-(h)x−-x dt
a a h→0 h
The difference quotients are bounded because they converge to Ax. Therefore, from the
dominated convergence theorem,
( )
∗ ∫ b ∫ b ∗( S (h)x − x)
y a S (t)Axdt = lih→m0 a y S(t)----h---- dt
( ∫ )
∗ b S-(h)-x−-x
= lih→m0 y a S(t) h dt
( ∫ b+h ∫ b )
= lim y∗ 1- S (t) xdt− 1- S (t) xdt
h→0 h a+h h a
( ∫ b+h ∫ a+h )
= lim y∗ 1- S (t) xdt− 1- S(t)xdt
h→0 h b h a
= y∗ (S (b)x − S(a)x)
Since y^{∗} is arbitrary, this proves the first part. Now from what was just shown, if t > 0
and h is small enough,
S(t)x− S (t− h) x 1∫ t
-------h--------= h- S (s)Axds
t−h
which converges to S
(t)
Ax as h → 0 + . This proves the corollary.
Given a closed densely defined operator, when is it the generator of a bounded
semigroup? This is answered in the following theorem which is called the Hille Yosida
theorem.
Theorem 17.14.7Suppose A is a densely defined linear operator which has the propertythat for all λ > 0,
(λI − A)−1 ∈ ℒ (H,H )
which means that λI − A : D
(A )
→ H is one to one and onto with continuous inverse.Suppose also that for all n ∈ ℕ,