Spectral Theory in Hilbert Space

86 12. INVERSE SPECTRAL THEORY
easy estimates give
g(x) ≤ c(λ)
|k|
x
0
|q| + 1
|k|
x
0
|q|g,
where c(λ) = |u(0)| + |u ′ (0)|/|k|. Integrating after multiplying by the
integrating factor |q(x)| exp(− x
|q|/|k|) we obtain
0
g(x) ≤ c(λ)(e x
0 |q|/√ |λ| − 1).
The estimate for u follows immediately from this.
Proof of Theorem 12.2. As noted, we only need to prove the
theorem for α = 0, so assume this. Now let λ = rµ, where µ is in
some fixed, compact subset of C \ R, and r > 0 is large. We define
ϕr(x, µ) = √ rϕ(x/ √ r, rµ) and θr(x, µ) = θ(x/ √ r, rµ). Then ϕr and
θr satisfy the initial conditions (12.3) for α = 0 and satisfy the equation
−u ′′ + qru = µu on (0, b √ r), where qr(x) = q(x/ √ r)/r (check!!). From
Lemma 12.3 it immediately follows that, locally uniformly in x, µ, we
have ϕr(x, µ) → sinh(x√−µ) √ and θr(x, µ) → cosh(x −µ
√ −µ) as r → ∞.
Now let mr(µ) = m(rµ)
√ r and make a change of variable x = y/ √ r in
(11.2). This gives b √ r
|θr + mrϕr| 0
2 = Im(mr(µ))/ Im µ, so if c > 0 we
have
(12.6)
c
0
|θr(·, µ) + mr(µ)ϕr(·, µ)| 2 ≤
Im mr(µ)
Im µ
as soon as b √ r ≥ c. The inequality may be rewritten as
|mr(µ) − Cr| ≤ Rr,
where Cr and Rr are easily expressed in terms of c
0 θrϕr, c
0 |θr| 2
and
c
0 |ϕr| 2 . The inequality therefore confines mr to a disk Kr(c), and is
clear from Lemma 12.3 that as r → ∞ the coefficients converge, locally
uniformly for µ ∈ C \ R, to those in the corresponding disk K(c) for
the case q = 0. Therefore, given any neighborhood Ω of K(c), we must
have mr(µ) ∈ Ω for all sufficiently large r.
This is true for any c > 0, and it is obvious from (12.6) that K(c)
decreases as a function of c. We shall show presently that only the
point − √ −µ is common to all K(c), and then it follows that mr(µ) →
− √ −µ, locally uniformly for µ ∈ C \ R. But this means that m(λ) =
− √ −λ(1 + o(1)) as λ → ∞ in a closed, non-real sector with vertex at
the origin, and thus proves the theorem.
It remains to show that ∩c>0K(c) = − √ −µ. But any point ℓ
in the intersection corresponds to a solution u(x) = cosh(x √ −µ) +
ℓ sinh(x √ −µ)/ √ −µ with u ∈ L 2 (0, ∞), since we have c
0 |u|2 ≤
Im ℓ
Im µ for
all c > 0. Thus the only possible value is ℓ = − √ −µ. On the other