An ideal of H∞ with a bounded functional
calculus
Joint work with Felix Schwenninger
Hans Zwart
University of Twente and Technical University of Eindhoven
April 27, 2017
Introduction
Let −A
be the infinitesimal generator of the C0 -semigroup
e−At t≥0 on the Banach space V . Assume that this semigroup is
bounded.
Let h be an absolutely integrable function from [0, ∞) to C. We
denote the Laplace transform of h by f (s).
It is easy to see that
1. f (s) is bounded and holomorphic on
∞
C+
0 := {s ∈ C | Re(s) > 0}. That is f ∈ H ;
2.
Z
f (A) :=
∞
h(t)e−At dt
0
defines a bounded linear operator, i.e, f (A) ∈ L(V ). This is
known as the Phillips calculus.
Introduction
If for all Banach spaces and all infinitesimal generators of bounded
semigroups, you want to define a functional calculus of bounded
operators, then h must be an absolutely integrable function from
[0, ∞) to C.
For Hilbert spaces the situation is better.
The question is/was if you can extend the functional calculus to all
f ∈ H∞ .
For me the question is (already for a long time):
Inverse generator problem
Let −A generate an exponentially stable semigroup on the Hilbert
space X. Does the following hold
sup ke−A
−1 t
k<∞
t≥0
So study the norm of
e
−t
s
(A).
Some results
We assume from now on that −A generates an exponentially
stable semigroup on the Hilbert space X.
For f ∈ H∞ there exists an (unbounded) f (A) : D(A) ⊂ X 7→ X.
Furthermore, for all t > 0, we have that
f (A)e−At ∈ L(X).
So the natural question if the above property holds for t = 0.
Already for a long time it is known that the answer to this question
is negative. Even the following holds.
A negative class, [Zwart; Batty, Gomilko, and Tomilov]
If f ∈ H∞ is such that the limit
lim
s∈R,s→∞
f (s)
does not exists, then there exists a generator, −A, of an
exponentially stable, holomorphic semigroup such that f (A) is not
bounded.
Result not if and only if (Partington).
A positive result, introduction
We have seen that for all t > 0, we have that f (A)e−At ∈ L(X).
Moreover the following estimate holds [Haase and Rozendaal,
2013]:
There exists an mA such that
kf (A)e−At k ≤ mA | log(t)|kf k∞ ,
where kf k∞ is the the norm on H∞ , i.e.,
kf k∞ = sup |f (s)|.
Re(s)>0
t ∈ (0, 12 ),
A positive result
We introduce the following ideal of H∞
S = f ∈ H∞ | ∃c > 0, δ > 1, ∀ε ∈ (0, 21 )
ke
−ε·
g(·) − g(·)k∞
c
≤
| log(ε)|δ
.
Theorem
For g ∈ S and −A the generator of an exponentially stable
semigroup on the Hilbert space X we have g(A) ∈ L(X).
A positive result, proof
The proof is based on the following equality for εn > εn+1
e−εn · g(·)− e−εn+1 · g(·) (A)
= e−εn+1 · e−(εn −εn+1 )· g(·) − g(·) (A)
= e−Aεn+1 e−(εn −εn+1 )· g(·) − g(·) (A),
where we used the property of the functional calculus.
A positive result, proof
Using Haase and Rozendaal, we find
−ε ·
e n g(·)− e−εn+1 · g(·) (A)
≤ mA | log(εn+1 )| e−(εn −εn+1 )· g(·) − g(·)
∞
c
≤ mA | log(εn+1 )|
,
| log(εn − εn+1 )|δ
where we used that g ∈ S.
A positive result, proof
Now using
p
εn = e−n ,
with p such that p(δ − 1) > 1, and a telescope argument we find
−1
−ε ·
NX
e 1 g(·) − e−εN · g(·) (A) ≤
n=1
M
np(δ−1)
.
Combining this with the facts
I
(e−ε1 · g(·)) (A) = e−ε1 A g(A) is bounded,
I
εN → 0, and
I
p(δ − 1) > 1,
we conclude that g(A) ∈ L(X).
Q.E.D.
Charatrisation of S
The ideal S can be characterised in a different way.
Theorem
g ∈ S if and only if there exists c1 > 0 and δ > 1 such that
|g(iω)| ≤
c1
,
(log(|ω| + e))δ
for a.e.
ω ∈ R.
Closing remarks
For any constant β ∈ C we have β(A) = βI.
Theorem
Let f ∈ H∞ be such that there exists f∞ ∈ C, c1 > 0 and δ > 1
such that
|f (iω) − f∞ | ≤
then f (A) ∈ L(X).
c1
,
(log(|ω| + e))δ
for a.e.
ω ∈ R,
Closing remarks
Other estimates of f (A)T (t) lead to other classes S.
If for instance, the following holds,
p
t ∈ (0, 12 ),
kf (A)e−At k ≤ mA | log(t)|kf k∞ ,
then g(A) bounded for all g ∈ H∞ for which the following
estimate holds for some c1 > 0 and δ > 21
|g(iω)| ≤
c1
,
(log(|ω| + e))δ
for a.e.
ω ∈ R.
Closing remarks
There also holds a spectral mapping theorem.
Theorem
Let g ∈ S and λ ∈ C be such that (λ − g(·))−1 ∈ H∞ , then
λI − g(A) is boundedly invertible.
Thanks