
Spectral properties of the Stokes operator 37
Finally to show that B is one-to-one, we observe that the kernel of B is equal
to V
◦
σ
∩H, which is exactly V
⊥
σ
. By the definition of H as the closure of V
σ
in L
2
,
this space reduces to {0} and the proposition follows.
As the closure in H of the range of B (denoted by R(B)) is equal to the
orthogonal space of ker B, we infer from Proposition 2.2 that R(B)isdense
in H. This allows us to define the inverse of B as an unbounded operator A with
a dense domain of definition.
Definition 2.4 Let us denote by A the following operator
A
R(B) →H
u →f/Bf= u.
This is exactly the operator A defined in (1.2.1) on page 19.
Lemma 2.1 The operator A is self-adjoint with domain R(B).
This lemma is a classical result of operator theory. For the reader’s
convenience, we give a proof of it.
Proof of Lemma 2.1 Of course A is symmetric, so the only point to check
is that the domain D(A
∗
)ofA
∗
is R(B). By definition,
D(A
∗
)={v ∈H|∃C>0, ∀u ∈ R(B), (Au|v)
H
≤ Cu
H
}.
Since A is symmetric, we have of course R(B) ⊂D(A
∗
). Now let us prove
that D(A
∗
) ⊂ R(B). Let us define the graph norm
v
2
A
∗
def
= v
2
H
+ A
∗
v
2
H
.
The fact that (D(A
∗
), ·
A
∗
) is a Hilbert space is left as an exercise to the
reader. The equality D(A
∗
)=R(B) will result from the fact that R(B) is closed
and dense for the ·
A
∗
norm. For any f ∈H, we have A
∗
Bf = ABf = f hence
Bf
2
A
∗
= Bf
2
H
+ f
2
H
which immediately gives that f
2
H
≤Bf
2
A
∗
, hence that the space R(B)is
closed in (D(A
∗
), ·
A
∗
). Now let v be in the orthogonal space of R(B)inthe
sense of the (·|·)
A
∗
scalar product. By definition, we have, for any f in H,
(Bf|v)
H
+(A
∗
Bf|A
∗
v)
H
=0.
As B is self-adjoint and A
∗
B = Id, we get for all f in H,
(f|Bv)
H
+(f|A
∗
v)
H
=0.