Spectral Theory in Hilbert Space

40 7. THE SPECTRAL THEOREM
function of λ for any f, so we have
∞
1 t
(7.1) 〈Rλf, f〉 = α + βλ + −
t − λ 1 + t2
dσf,f(t),
−∞
where σf,f is increasing and α, β may depend on f. Since Rλ ≤ 1
|Im λ| ,
we find that f 2 is an upper bound for ν〈Riνf, f〉 for ν ∈ R, the
imaginary part of which is βν 2 + ∞
−∞
ν2 dσf,f (t)
t2 +ν2 . Hence β = 0, and by
Fatou’s lemma we get, as ν → ∞, that ∞
−∞ dσf,f ≤ f 2 . A more
elementary argument is the following: For ν, ε > 0 we have
1
1 + ε2 νε
∞
ν
dσf,f ≤
2
t2 + ν2 dσf,f(t) ≤ f 2 ,
−νε
−∞
1 since 1+ε2 ≤ ν2
ν2 +t2 for |t| ≤ νε, so letting ν → ∞, and then ε → 0, we
obtain the same bound. We may now assume σf,f to be normalized so
as to be left-continuous and σf,f(−∞) = 0. Clearly ∞ t
−∞ 1+t2 dσf,f(t)
is absolutely convergent, so this part of the integral in (7.1) may be
incorporated in the constant α. So, with absolute convergence, we have
〈Rλf, f〉 = α ′ + ∞ dσf,f (t)
. However, for λ → ∞ along the imaginary
−∞ t−λ
axis, both the left hand side and the integral → 0 (Exercise 7.1), so we
must have α ′ = 0. The proof is now finished in the case f = g.
By the polarization identity (Exercise 7.2)
〈Rλf, g〉 = 1
3
i
4
k 〈Rλ(f + i k g), f + i k g〉 ,
k=0
so we obtain 〈Rλf, g〉 = ∞ dσf,g(t)
−∞ t−λ
σf,g = 1
4
3
k=0
by setting
i k σ f+i k g,f+i k g.
The function σf,g has the correct normalization, so only the bound on
the total variation remains to be proved. But if ∆ is an interval, then
∆ dσf,g is a semi-scalar product on H, so Cauchy-Schwarz’ inequality
∆ dσf,g
2 ≤
∆ dσf,f
∆ dσg,g
is valid. For ∆ = R this shows that
R dσf,g is bounded by fg. If {∆j} ∞ 1 is a partition of R into disjoint
intervals we obtain
dσf,g
≤
∆j
∆j
dσf,f
≤
∆j
∆j
dσg,g
dσf,f
1
2
1
2
∆j
dσg,g
1
2
≤ fg,