Spectral Theory in Hilbert Space

36 6. NEVANLINNA FUNCTIONS
However, if one assumes that Re G ≥ 0 the boundary values exist at
least in the sense of measure, and one has the following theorem.
Theorem 6.3 (Riesz-Herglotz). Let G be analytic in the unit circle
with positive real part. Then there exists an increasing function σ on
[0, 2π] such that
G(z) = i Im G(0) + 1
2π
π
−π
eiθ + z
eiθ dσ(θ) .
− z
With a suitable normalization the function σ will also be unique,
but we will not use this. To prove Theorem 6.3 we need some kind
of compactness result, so that we can obtain the theorem as a limiting
case of Lemma 6.2. What is needed is weak ∗ compactness in the
dual of the continuous functions on a compact interval, provided with
the maximum norm. This is the classical Helly theorem. Since we assume
minimal knowledge of functional analysis we will give the classical
proof.
Lemma 6.4 (Helly).
(1) Suppose {ρj} ∞ 1 is a uniformly bounded1 sequence of increasing
functions on an interval I. Then there is a subsequence
converging pointwise to an increasing function.
(2) Suppose {ρj} ∞ 1 is a uniformly bounded sequence of increasing
functions on a compact interval I, converging pointwise to ρ.
Then
(6.3)
f dρj → f dρ as j → ∞,
I
for any function f continuous on I.
I
Proof. Let r1, r2, . . . be a dense sequence in I, for example an enumeration
of the rational numbers in I. By Bolzano-Weierstrass’ theorem
we may choose a subsequence {ρ1j} ∞ 1 of {ρj} ∞ 1 so that ρ1j(r1) converges.
Similarly, we may choose a subsequence {ρ2j} ∞ 1 of {ρ1j} ∞ 1 such
that ρ2j(r2) converges; as a subsequence of ρ1j(r1) the sequence ρ2j(r1)
still converges. Continuing in this fashion, we obtain a sequence of sequences
{ρkj} ∞ j=1, k = 1, 2, . . . such that each sequence is a subsequence
of those coming before it, and such that ρ(rn) = limj→∞ ρkj(rn) exists
for n ≤ k. Thus ρjj(rn) → ρ(rn) as j → ∞ for every n, since ρjj(rn) is
a subsequence of ρnj(rn) from j = n on. Clearly ρ is increasing, so if
x ∈ I but = rn for all n, we may choose an increasing subsequence rjk ,
k = 1, 2, . . . , converging to x, and define ρ(x) = limk→∞ ρ(rjk ).
Suppose x is a point of continuity of ρ. If rk < x < rn we get
ρjj(rk) − ρ(rn) ≤ ρjj(x) − ρ(x) ≤ ρjj(rn) − ρ(rk). Given ε > 0 we may
1 i.e., all the functions are bounded by a fixed constant