Spectral Theory in Hilbert Space

15. SINGULAR PROBLEMS 107
the double integral being absolutely convergent. Similarly
〈u, Rλv〉W = v ∗ (x)W (x)G ∗ (y, x, λ)W (y)u(y) dxdy,
and since the integrals are equal by Theorem 5.2 (2) and G(x, y, λ) −
G ∗ (y, x, λ) = F (x, λ)(M(λ) − M ∗ (λ))F ∗ (y, λ) we obtain
〈F (·, λ), v〉W (M(λ) − M ∗ (λ))〈u, F ∗ (·, λ)〉W = 0.
By Assumption 13.7 this implies that M(λ) = M ∗ (λ) and thus (4).
Finally, to prove (5) we use the resolvent relation Theorem 5.2(3).
For u ∈ L2 W this gives
〈u, G ∗ (x, ·, λ) − G ∗ (x, ·, µ)〉W = Rλu(x) − Rµu(x)
Now
Thus
= (λ − µ)RλRµu(x) = (λ − µ)〈Rµu, G ∗ (x, ·, λ)〉W
= 〈u, (λ − µ)RµG ∗ (x, ·, λ)〉W .
RµG ∗ (x, ·, λ)(y) = 〈G ∗ (x, ·, λ), G ∗ (y, ·, µ)〉W .
G(x, y, λ) − G(x, y, µ) = (λ − µ)〈G ∗ (y, ·, µ), G ∗ (x, ·, λ)〉W
= (λ − µ)RλG ∗ (y, ·, µ)(x),
since both sides are clearly in D(T ). This proves (5).
Before we proceed, we note the following corollary, which completes
our results for the case of a discrete spectrum.
Corollary 15.3. Suppose for some non-real λ that all solutions
of Ju ′ + Qu = λW u and Ju ′ + Qu = λW u are in L2 W . Then for any
selfadjoint realization T the resolvent of T is compact.
In other words, if the deficiency indices are maximal, then the resolvent
is compact. Actually, the assumptions are here a bit stronger
than needed. In fact, it is not difficult to show (Exercise 15.1) that if
all solutions are in L2 W for some λ, real or not, then the same is true
for all λ.
Proof. One could use a version of Theorem 8.7 valid for L2 W
and show that Rλ is a Hilbert-Schmidt operator. Here is an alternative
proof. Suppose uj ⇀ 0 weakly in L2 W and let I = (a, b).
Then x
a F ∗ (y, λ)W (y)uj(y) dy and b
x F ∗ (y, λ)W (y)uj(y) dy are both
bounded uniformly with respect to x by Cauchy-Schwarz and since the
columns of F (·, λ) are in L2 W . The latter fact also shows that the integrals
tend pointwise to 0 as j → ∞. Since also the columns of F (·, λ)
are in L2 W it follows that Rλuj → 0 strongly in L2 W by dominated
convergence.