Spectral Theory in Hilbert Space

15. SINGULAR PROBLEMS 113
Proof. According to Lemma 15.9 we need only show that if û ∈
L2 P has inverse transform 0, then û = 0. Now, according to Lemma 15.10,
F(v)(t)/(t − λ) is a transform for all v ∈ L2 W and non-real λ. Thus
we have 〈û(t)/(t − λ), F(v)(t)〉P = 0 for all non-real λ if û is orthogonal
to all transforms. But we can view this scalar product as the
Stieltjes-transform of the measure t
−∞ F(v)∗ dP û, so applying the inversion
formula Lemma 6.5 we have
K F(v)∗dP û = 0 for all compact
intervals K, and all v ∈ L2 W . Thus the cutoff of û, which equals û in
K and 0 outside, is also orthogonal to all transforms, i.e., has inverse
transform 0 according to Lemma 15.9. It follows that
F (x, t)dP (t) û(t)
K
is the zero-element of L2 W for any compact interval K. Now multiply
this from the left with F ∗ (x, s)W (x) and integrate with respect to x
over a large compact subinterval L ⊂ I. We obtain
B(s, t)dP (t) û(t) = 0 for every s,
K
where B(s, t) =
L F ∗ (x, s)W (x)F (x, t) dx. Thus B(s, t)dP (t) û(t) is
the zero measure for all s. By Assumption 13.7 the matrix B(s, t)
is invertible for s = t, so by continuity it is, given s, invertible for t
sufficiently close to s. Thus, varying s, it follows that dP (t) û(t) is the
zero measure in a neighborhood of every point. But this means that
û = 0 as an element of L2 P .
Lemma 15.12. If (u, v) ∈ T , then ˆv(t) = tû(t). Conversely, if û
and tû(t) are in L 2 P , then F −1 (û) ∈ D(T ).
Proof. We have (u, v) ∈ T if and only if u = Rλ(v − λu), which
holds if and only if û(t) = (ˆv(t) − λû(t))/(t − λ), i.e., ˆv(t) = tû(t),
according to Lemmas 15.10 and 15.11.
This completes the proof of Theorem 15.5. We also have the following
analogue of Corollary 14.5.
Theorem 15.13. Suppose u ∈ D(T ). Then the inverse transform
〈û, F ∗ (x, ·)〉P converges locally uniformly to u(x).
Proof. The proof is very similar to that of Corollary 14.5. Put
v = ( ˜ T − i)u so that v ∈ HT and u = Riv. Let K be a compact
interval, and put uK(x) =
K F (x, t) dP (t) û(t) = F −1 (χû)(x), where
χ is the characteristic function for K. Define vK similarly. Then by
Lemma 15.10
RivK = F −1 ( χ(t)ˆv(t)
) = F
t − i
−1 (χû) = uK .