Spectral Theory in Hilbert Space

76 11. STURM-LIOUVILLE EQUATIONS
Proof. If the supports are inside [a, c], direct calculation shows
that the function is
c
x
c
θ(x, λ) uϕ(·, λ) + ϕ(x, λ) uθ(·, λ) v(x) dx .
a
a
This is obviously an entire function of λ.
Lemma 11.10. Let σ be increasing and differentiable at 0. Then
1
−1 ds 1 dσ(t)
√
−1 t2 +s2 converges.
Proof. Integrating by parts we have, for s = 0,
1
−1
dσ(t)
√ t 2 + s 2
= σ(1) − σ(−1)
√ 1 + s 2
−
1
−1
x
σ(t) − σ(0)
(t
t
d
dt
1
√ t 2 + s 2
) dt .
The first factor in the last integral is bounded since σ ′ (0) exists, and the
second factor is negative since (t2 + s2 1
− ) 2 decreases with |t|. Furthermore,
the integral with respect to t of the second factor is integrable
with respect to s, by calculation (check this!). Thus the double integral
is absolutely convergent.
As usual we denote the spectral projectors belonging to T by Et.
Lemma 11.11. Let u ∈ L 2 (a, b) have compact support in [a, b) and
assume c < d to be points of differentiability for both 〈Etu, u〉 and ρ(t).
Then
(11.3) 〈Edu, u〉 − 〈Ecu, u〉 =
d
c
|û(t)| 2 dρ(t).
Proof. Let Γ be the positively oriented rectangle with corners in
c ± i, d ± i. According to Lemma 11.9
〈Rλu, u〉 dλ = û(λ)û(λ)m(λ) dλ
Γ
if either of these integrals exist. However, by Lemma 11.9,
û(λ)û(λ)m(λ) dλ =
∞
û(λ)û(λ) ( 1 t
−
t − λ t2 ) dρ(t) dλ.
+ 1
Γ
Γ
Γ
The double integral is absolutely convergent except perhaps where t =
λ. The difficulty is thus caused by
1
−1
ds
µ+1
µ−1
−∞
û(µ + is)û(µ − is) dρ(t)
t − µ − is