Spectral Theory in Hilbert Space

15. SINGULAR PROBLEMS 111
Proof. If u has compact support Lemma 15.7 shows that (15.4)
holds for a dense set of values a, b since functions of bounded variation
are a.e. differentiable. Since both Et and P are left-continuous we
obtain, by letting b ↑ t, a → −∞ through such values,
t
〈Etu, v〉 = ˆv ∗ (t) dP (t) û(t)
−∞
when u, v have compact supports; first for u = v and then in general
by polarization. If PT is the projection of L 2 W onto HT we obtain as
t → ∞ also that 〈PT u, v〉W = 〈û, ˆv〉P when u and v have compact
supports.
For arbitrary u ∈ L 2 W
we set, for a compact subinterval K of I,
uK(x) =
u(x)
0
for x ∈ K
otherwise
and obtain a transform ûK. If L is another compact subinterval of I it
follows that ûK − ûLP = PT (uK − uL)W ≤ uK − uLW , and since
uK → u in L2 W as K → I, Cauchy’s convergence principle shows that
ûK converges to an element û ∈ L2 P as K → I. The lemma now follows
in full generality by continuity.
Note that we have proved that F is an isometry on HT , and a
partial isometry on L 2 W .
Lemma 15.9. The integral
compact interval and û ∈ L 2 P
K F (x, t) dP (t) û(t) is in HT if K is a
, and as K → R the integral converges in
HT . The limit F −1 (û) is called the inverse transform of û. If u ∈ L2 W
then F −1 (F(u)) = PT u. F −1 (û) = 0 if and only if û is orthogonal in
L2 P to all generalized Fourier transforms.
Proof. If û ∈ L2 P has compact support, then u(x) = 〈û, F ∗ (x, ·)〉P
is continuous, so uK ∈ L2 W for compact subintervals K of I, and has a
transform ûK. We have
uK 2
W = u ∗ ∞
(x)W (x) F (x, t)dP (t)û(t) dx .
K
−∞
Considered as a double integral this is absolutely convergent, so changing
the order of integration we obtain
uK 2 W =
∞
−∞
K
F ∗ ∗ (x, t)W (x)u(x) dx dP (t) û(t)
= 〈û, ûK〉P ≤ ûP ûKP ≤ ûP uKW ,
according to Lemma 15.8. Hence uKW ≤ ûP , so u ∈ L 2 W
uW ≤ ûP . If now û ∈ L 2 P
, and
is arbitrary, this inequality shows (like