Spectral Theory in Hilbert Space

112 15. SINGULAR PROBLEMS
in the proof of Lemma 15.8) that
K F (x, t)dP (t) û(t) converges in L2W as K → R through compact intervals; call the limit u1. If v ∈ L2 W , ˆv
is its generalized Fourier transform, K is a compact interval, and L a
compact subinterval of I, we have
( F ∗ (x, t)W (x)v(x) dx) ∗ dP (t) û(t)
K
L
=
L
v ∗
(x)W (x)
K
F (x, t)dP (t) û(t) dx
by absolute convergence. Letting L → I and K → R we obtain
〈û, ˆv〉P = 〈u1, v〉W . If û is the transform of u, then by Lemma 15.8
u1 − u is orthogonal to HT , so u1 = PT 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 L2 W into L2P . The basic remaining difficulty is
to prove that the transform is surjective, i.e., according to Lemma 15.9,
that the inverse transform is injective. The following lemma will enable
us to prove this.
Lemma 15.10. The transform of Rλu is û(t)/(t − λ).
Proof. By Lemma 15.8, 〈Etu, v〉W = t
−∞ ˆv∗ dP û, so that
〈Rλu, v〉W =
∞
−∞
d〈Etu, v〉
t − λ =
∞
−∞
By properties of the resolvent
Rλu 2 =
ˆv ∗ (t) dP (t) û(t)
t − λ
1
2i Im λ 〈Rλu − R λ u, u〉W =
∞
−∞
d〈Etu, u〉W
|t − λ| 2
= 〈û(t)/(t − λ), ˆv(t)〉P .
= û(t)/(t − λ)2 P .
Setting v = Rλu and using Lemma 15.8, it therefore follows that
û(t)/(t − λ)2 P = 〈û(t)/(t − λ), F(Rλu)〉P = F(Rλu)2 P . It follows
that we have û(t)/(t−λ)−F(Rλu)P = 0, which was to be proved.
Lemma 15.11. The generalized Fourier transform is unitary from
and the inverse transform is the inverse of this map.
HT to L 2 P