1
1.
2
nd
order differential operators
Notations 1.1.
Let
Ω
be a precompact open subset of
R
d
,A
ij
=
A
ji
,A
i
,A
0
∈
BC
∞
(
Ω
)
for
i, j
=1
,...,d,
p
(
x, ξ
):=
−
d
X
i,j
=1
A
ij
ξ
i
ξ
j
+
d
X
i
=1
A
i
ξ
i
+
A
0
and
L
=
p
(
x, ∂
)=
−
d
X
i,j
=1
A
ij
∂
i
∂
j
+
d
X
i
=1
A
i
∂
i
+
A
0
.
We also let
L
†
=
−
d
X
i,j
=1
∂
i
∂
j
M
A
ij
−
d
X
i
=1
∂
i
M
A
i
+
A
0
.
Remark
1.2
.
The operators
L
and
L
†
have the following properties.
(1) The operator
L
†
is the formal adjoint of
L,
i.e.
(
Lu, v
)=(
u, L
†
v
)
for all
u, v
∈
D
(
Ω
)=
C
∞
c
(
Ω
)
.
(2) We may view
L
as an operator on
D
0
(
Ω
)
via the formula
u
∈
D
0
(
Ω
)
→
Lu
∈
D
0
(
Ω
)
where
h
Lu, φ
i
:=
h
u, L
†
φ
i
for all
φ
∈
C
∞
c
(
Ω
)
.
(3) The restriction of
L
to
H
k
+2
(
Ω
)
gives a bounded linear transformation
L
:
H
k
+2
(
Ω
)
→
H
k
(
Ω
)
for
k
∈
N
0
.
Indeed,
L
may be written as
L
=
−
d
X
i,j
=1
M
A
ij
∂
i
∂
j
+
d
X
i
=1
M
A
i
∂
i
+
M
A
0
.
Now
∂
i
:
H
k
(
Ω
)
→
H
k
+1
(
Ω
)
is bounded and
M
ψ
:
H
k
(
Ω
)
→
H
k
(
Ω
)
is
bounded where
ψ
∈
BC
∞
(
Ω
)
. Therefore, for
k
∈
N
0
,L
:
H
k
+2
(
Ω
)
→
H
k
(
Ω
)
is bounded.
De
f
nition 1.3.
For
u
∈
D
0
(
Ω
)
,
let
k
u
k
H
−
1
(
Ω
)
:=
sup
0
6
=
φ
∈
D
(
Ω
)
h
u, φ
i
k
φ
k
H
1
0
(
Ω
)
and
H
−
1
(
Ω
):=
©
u
∈
D
0
(
Ω
):
k
u
k
H
−
1
(
Ω
)
<
∞
ª
.
Example 1.4.
Let
Ω
=
R
d
and
S
⊂
Ω
be the unit sphere in
R
d
.
Then de
f
ne
σ
∈
D
0
(
Ω
)
by
h
σ, φ
i
:=
Z
S
φdσ.
Let us shows that
σ
∈
H
−
1
(
Ω
)
.
For this let
T
:
H
1
(
Ω
)
→
L
2
(
S, dσ
)
denote the
trace operator, i.e. the unique bounded linear operator such that
Tφ
=
φ

S
for all
φ
∈
C
∞
c
¡
R
d
¢
.
Since
T
is bounded,
h
σ, φ
i
≤
σ
(
S
)
1
/
2
k
Tφ
k
L
2
(
S
)
≤
σ
(
S
)
1
/
2
k
T
k
L
(
H
1
(
Ω
)
,L
2
(
S
))
k
φ
k
H
−
1
(
Ω
)
.
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This shows
σ
∈
H
−
1
(
Ω
)
and
k
σ
k
H
−
1
(
Ω
)
≤
σ
(
S
)
1
/
2
k
T
k
L
(
H
1
(
Ω
)
,L
2
(
S
))
.
Lemma 1.5.
Suppose
Ω
is an open subset of
R
d
such that
¯
Ω
is a manifold with
C
0
— boundary and
Ω
=
¯
Ω
o
,
then the map
u
∈
£
H
1
0
(
Ω
)
¤
∗
→
u

D
(
Ω
)
∈
H
−
1
(
Ω
)
is a
unitary map of Hilbert spaces.
Proof.
By de
f
nition
C
∞
c
(
Ω
)
is dense in
H
1
0
(
Ω
)
,
and hence it follows that the
map
u
∈
£
H
1
0
(
Ω
)
¤
∗
→
u

D
(
Ω
)
∈
H
−
1
(
Ω
)
is isometric. If
u
∈
H
−
1
(
Ω
)
,
it has a unique
extension to
H
1
0
(
Ω
)=
C
∞
c
(
Ω
)
H
1
(
Ω
)
and this provides the inverse map.
If we identify
L
2
(
Ω
)=
H
0
(
Ω
)
with elements of
D
0
(
Ω
)
via
u
→
(
u,
·
)
L
2
(
Ω
)
,
then
D
0
(
Ω
)
⊃
H
−
1
(
Ω
)
⊃
H
0
(
Ω
)=
L
2
(
Ω
)
⊃
H
1
(
Ω
)
⊃
H
2
(
Ω
)
⊃
...
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