Page 75 - Textos de Matemática Vol. 34

ON THE PROBLEM OF THE NEGATIVE SPECTRUM 65
Proof. Step 1. First, note that the norm induced by (6.1) is equivalent to the one in(3.2)fors=1,p=q=2andΨ≡1. Duetod>n−2andDefinition4.2,we have
tr W1(Rn) = tr B1 (Rn) = B(1−(n−d)/2,Ψ−1/2)(Γ).
Γ 2 Γ 2,2
By (6.2) and with 1/q := 1/2 − 1/(2r), we get
2,2
1−n−d−d1−1=1−n−d− d >0 2 2 q 2 2r
and hence, by Theorem 5.1, we have the compact embedding (6.5) B(1−(n−d)/2,Ψ−1/2)(Γ) → L (Γ).
2,2 q
In particular,
(6.6) ∥tr f |L (Γ)∥ ≤ c∥f |B1 (Rn)∥ ≤ c′ ∥f |W1(Rn)∥, f ∈ W1(Rn).
Γq 2,2 2 2
The inequality in (6.6) together with H¨older’s inequality enable us to verify that

b(γ) (trΓf)(γ) (trΓg)(γ) dμ(γ), f, g ∈ W21(Rn),
defines a bounded sesquilinear form in W21(Rn), which generates a bounded self-
adjoint operator B in W21(Rn) with
(6.8) a(f, g) = (Bf, g)W21(Rn), f, g ∈ W21(Rn).
Step 2. We verify that the operator B in (6.8) is indeed the compact operator (id − ∆)−1 ◦ trΓb . Let f ∈ W21(Rn) and g ∈ S(Rn). We have

(6.7) a(f, g) :=
⟨trΓb f, g⟩ =
b(γ) (trΓf)(γ) g(γ) dμ(γ) = a(f, g)
Γ
Γ
=(Bf,g)W21(Rn) =((id−∆)1/2Bf,(id−∆)1/2g)L2(Rn)
= ((id − ∆)Bf, g)L2(Rn) = ⟨(id − ∆)Bf, g⟩,
where ⟨·, ·⟩ denotes the usual bilinear form defined on S ′ (Rn ) × S (Rn ). Hence,
trΓb f = (id − ∆)Bf, f ∈ W21(Rn), which proves the desired equality.
Step 3. Let f ∈ W21(Rn) be such that trΓf = 0. Clearly by (6.7) and (6.8) we get Bf=0. Ontheotherhand,iff∈W21(Rn)issuchthatBf=0,thenby(6.7) and (6.8) we may conclude that b(γ)(trΓf)(γ) = 0 μ-a.e on Γ. And of course if we assume that b(γ) ̸= 0 μ-a.e. on Γ, it turns out that trΓf = 0, which proves the remaining assertion about N(B).
Step 4. To estimate the eigenvalues of B by their magnitude we consider first the factorisation
B = id2 ◦ (id − ∆)−1 ◦ idΓ ◦ b ◦ id1 ◦ trΓ,