Spectral theory in Hilbert spaces

Zentrum Mathematik SS 2012/2013
Teschnische Universität München
Prof. Dr. Simone Warzel
Spectral theory in Hilbert spaces
Solutions
28. (i) By Theorem(P.121) (i) and (iv) we have for all bounded E-measurable function ϕ, ψ and
all f, g ∈ H that
(g, ϕ E f) = (1 E g, ϕ E f) (i)
= (ϕ ∗ Eg, 1 E f) = (ϕ ∗ Eg, f),
and hence
∫
(g, ϕ E ψ E ) = (ϕ ∗ Eg, ψ E f) =
ϕ(t)ψ(t)d(g, E(t)f) = (g, (ϕψ) E f).
Therefore, for bounded E-measurable functions ϕ, ψ, we have ϕ E ψ E = (ϕψ) E .
Let f ∈ D(u E v E ), i.e. let f ∈ D(v E ) and v E f ∈ D(u E ). As the function ϕ n u is bounded
for fixed n ∈ N, it follows that ϕ n uψ m v → ϕ n uv in L 2 (R, ρ f ) as m → ∞. Consequently,
u E v E f = lim (ϕ nu) E ( lim (ψ mv)f)
n→∞ m→∞
= lim lim (ϕ nu) E (ψ m v) E f
n→∞ m→∞
= lim lim (ϕ nuψ m v) E f
n→∞ m→∞
= lim (ϕ nuv) E f.
n→∞
The existence of the limit means that the sequence (ϕ n uv) is a Cauchy sequence in
L 2 (R, ρ f ), which together with the fact that
ϕ n (t)u(t)v(t) → u(t)v(t)
∀ t ∈ R
implies uv ∈ L 2 (R, ρ f ). Consequently, f ∈ D((uv) E ) and u E v E f = (uv) E f.
If f ∈ D(v E ) ∩ D((uv) E ), then
(uv) E f = lim
n→∞
lim (ϕ nuψ m v)f = lim (ϕ nu) E v E f.
m→∞ n→∞
The existence of this limit means that u ∈ L 2 (R, ρ vE f ). Consequently, v E f ∈ D(u E )
and hence f ∈ D(u E v E ).
(ii) Note that (viii) implies that
and (ix) implies that
(1 I ) 2 E = (1 2 I) E = (1 I ) E ,
(1 I ) ∗ E = (1 ∗ I) E = (1 I ) E ,
we have that (1 I ) E is an orghogonal projection.
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29. We obtain by a direct calculation that
− 1 ∮
1
(y,
2πi Γ A − z x)dz
P.121(vi)
= − 1
2πi
∫ ∮
= (
which completes the proof.
Cauchy integral
=
a,b/∈σ(A)
R
∫ b
a
Γ
∮
Γ
∫
R
1
d(y, E(t)x)dz
t − z
1
dz)d(y, E(t)x)
z − t
1d(y, E(t)x)
= (y, (E(b) − E(a))x),
30. Using Theorem(P.129), we see that the limit is nothing but 2πi(y, E(t)x). Since A is a selfadjoint
compact operator, we have
⎧ ∑
P ⎪⎨
j t < 0,
j:λ
E(t) = 2πi
j ≤t
⎪⎩
P 0 + ∑
P j t ≥ 0,
and hence (y, E(t)x) = 2πi ∑
j:λ j ≤t
(y, P j x).
j:λ j≤t
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