1716 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS
and so z∗ (x)w− z∗ (w)x ∈ ker(z∗) . Therefore, for any x ∈V,
0 =(
w,z∗ (x)w− z∗ (w)x)
= z∗ (x)(w,w)− z∗ (w)(w,x)
and so
z∗ (x) =
(z∗ (w)
||w||2,x
)so let z = w/ ||w||2 . Then Rz = z∗ and so R is onto. This proves the lemma.
Now for the V described above,
Ru(v) =∫
Ω
(auv+a(x)∇u ·∇v
)dx
Also, as noted above V is dense in H ≡ L2 (Ω) and so if H is identified with H ′, it follows
V ⊆ H = H ′ ⊆V ′.
Let A : D(A)→ H be given by
D(A)≡ {u ∈V : Ru ∈ H}
andA≡−R
on D(A). Then the numerical range for A is contained in (−∞,−a] and so A is sectorial byProposition 54.3.10 provided A is closed and densely defined.
Why is D(A) dense? It is because it contains C∞c (Ω) which is dense in L2 (Ω) . This
follows from integration by parts which shows that for u,v ∈C∞c (Ω) ,
−∫
Ω
auvdx−∫
Ω
a(x)∇u ·∇vdx
=−∫
Ω
auvdx+∫
Ω
∇ · (a(x)∇u)vdx
and since C∞c (Ω) is dense in H,
Au =−au+∇ · (a(x)∇u) ∈ L2 (Ω) = H.
Why is A closed? If un ∈ D(A) and un→ u in H while Aun→ ξ in H, then it followsfrom the definition that Run→−ξ and {un} converges to u in V so for any v ∈V,
Ru(v) = limn→∞
Run (v) = limn→∞
(Run,v)H = (−ξ ,v)H
which shows Ru =−ξ ∈H and so u ∈D(A) and Au = ξ . Thus A is closed. This completesthe example.