53.2.3 An Interesting Example
In this section related to this example, for V a Banach space, V ′ will denote the space of
continuous conjugate linear functions defined on V . Usually the symbol has meant the
space of continuous linear functions but here they will be conjugate linear. That is f ∈ V ′
and f is continuous.
Let Ω be a bounded open set in ℝn and define
where Γ is some measurable subset of the boundary of Ω and C∞
restrictions of functions in
By Corollary 13.5.11 V 0
is dense in L2
Now define the following for u,v ∈ V 0.
where a > 0 and a
0 is a C1
function. Also define the following inner product
denote the corresponding norm.
Of course V 0 is not a Banach space because it fails to be complete. u ∈ V will mean
that u ∈ L2
and there exists a sequence
⊆ V 0
For u ∈ V, define ∇u to be that element of L2
the space of
vector valued L2
functions taken with respect to the measure a
Denote this space by W for simplicity of notation.
Observation 53.2.11 V is a Hilbert space with inner product given by
Everything is obvious except completeness. Suppose then that
is a Cauchy
Then there exists a unique u ∈ L2
is also a Cauchy sequence in
is a Cauchy sequence in
Thus the thing to which ∇wn
converges in W
is the definition of
and u ∈ V.
and the last term converges to 0. Hence V
is complete as claimed.
Then it is clear V is a Hilbert space. The next observation is a simple one involving
the Riesz map.
Definition 53.2.12 Let V be a Hilbert space and let V ′ be the space of continuous
conjugate linear functions defined on V . Then define R : V → V ′ by
This is called the Riesz map.
Lemma 53.2.13 The Riesz map is one to one and onto and linear.
Proof: It is obvious it is one to one and linear. The only challenge is to show it is
onto. Let z∗∈ V ′. If z∗
then letting z
it follows Rz
. If z∗
is a closed subspace. It is closed because z∗ is continuous and it is just z∗−1
is not everything in
and w≠0. Then
and so z∗
x ∈ ker
Therefore, for any x ∈ V,
so let z = w∕
and so R
is onto. This proves the lemma.
Now for the V described above,
Also, as noted above V is dense in H ≡ L2
and so if
is identified with H′,
Let A : D
be given by
. Then the numerical range for
is contained in (−∞,−a
] and so A
by Proposition 53.2.9
is closed and densely defined.
Why is D
dense? It is because it contains
which is dense in
follows from integration by parts which shows that for u,v ∈ Cc∞
and since Cc∞
is dense in
Why is A closed? If un ∈ D
un → u
while Aun → ξ
then it follows
from the definition that Run →−ξ
so for any
v ∈ V,
which shows Ru = −ξ ∈ H and so u ∈ D
is closed. This
completes the example.
Obviously you could follow identical reasoning to include many other examples of
more complexity. What does it mean for u ∈ D
? It means that in a weak
Since A is sectorial for S−a,ϕ for any 0 < ϕ < π∕2, this has shown the existence of a weak
solution to the partial differential equation along with appropriate boundary
What are these appropriate boundary conditions? u = 0 on Γ is one. the other would be a
variational boundary condition which comes from integration by parts. Letting v ∈ V,
formally do the following using the divergence theorem.
and so the other boundary condition is
To what extent this weak solution is really a classical solution depends on more technical