Spectral Theory in Hilbert Space

78 11. STURM-LIOUVILLE EQUATIONS
Note that we have proved that F is an isometry from L 2 (a, b) to
L 2 ρ.
Lemma 11.13. The integral
K ûϕ(x, ·) dρ is in L2 (a, b) if K is a
compact interval and û ∈ L 2 ρ, and as K → R the integral converges
in L 2 (a, b). The limit F −1 (û) is called the inverse transform of û.
If u ∈ L 2 (a, b) then F −1 (F(u)) = u. F −1 (û) = 0 if and only if û is
orthogonal in L 2 ρ to all generalized Fourier transforms.
Proof. If û ∈ L 2 ρ has compact support, then u(x) = 〈û, ϕ(x, ·)〉ρ
is continuous, so uc ∈ L 2 (a, b) for c ∈ (a, b), and has a transform ûc.
We have
uc 2 =
c
a
∞
−∞
ûϕ(x, ·) dρ u(x) dx.
Considered as a double integral this is absolutely convergent, so changing
the order of integration we obtain
uc 2 =
∞
−∞
c a
uϕ(·, t) û(t) dρ(t)
= 〈û, ûc〉ρ ≤ ûρûcρ = ûρuc,
according to Lemma 11.12. Hence uc ≤ ûρ, so u ∈ L 2 (a, b), and
u ≤ ûρ. If now û ∈ L 2 ρ is arbitrary, this inequality shows (like in the
proof of Lemma 11.12) that
K û(t)ϕ(x, t) dρ(t) converges in L2 (a, b) as
K → R through compact intervals; call the limit u1. If v ∈ L 2 (a, b),
ˆv is its generalized Fourier transform, K is a compact interval, and
c ∈ (a, b), we have
K
c a
v(x)ϕ(x, t) dx û(t) dρ(t) =
c
a
v(x)
K
û(t)ϕ(x, t) dρ(t) dx
by absolute convergence. Letting c → b and K → R we obtain 〈û, ˆv〉ρ =
〈u1, v〉. If û is the transform of u, then by Lemma 11.12 u1 − u is
orthogonal to L 2 (a, b), so u1 = u. Similarly, u1 = 0 precisely if û is
orthogonal to all transforms.
We have shown the inverse transform to be the adjoint of the transform
as an operator from L 2 (a, b) into L 2 ρ. The basic remaining difficulty
is to prove that the transform is surjective, i.e., according to
Lemma 11.13, that the inverse transform is injective. The following
lemma will enable us to prove this.
Lemma 11.14. The transform of Rλu is û(t)/(t − λ).