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Zero-free inner functions are unimodular multiples of singular inner functions

Statement

Assume the Axiom of Choice. Let S be holomorphic and zero-free on D with ∣S∣≤1, and suppose that its boundary function satisfies ∣S∗∣=1 m-almost everywhere (equivalently, S is an inner function without zeros). Then there are a unique λ∈T and a unique finite positive measure μ⊥m with S=λ Sμ,Sμ(z)=exp⁡(−∫TK(z,ζ) dμ(ζ)). With the normalizations Sμ(0)=e−μ(T)>0 and λ=S(0)/Sμ(0), the pair (λ,μ) is unique. In particular Sμ is the singular inner function of μ, and every inner function is, up to a unimodular constant, the product of a Blaschke product and a singular inner function.

Facts & Assumptions

Given: The Axiom of Choice, hence countable choice (The Axiom of Choice, The Axiom of Countable Choice (ACω)); a zero-free holomorphic S on D with ∣S∣≤1 and ∣S∗∣=1 m-almost everywhere; and, where asserted, an inner function θ.

[L1]

S zero-free means log⁡∣S∣ is harmonic on D, so u:=−log⁡∣S∣=−Re⁡log⁡S is a nonnegative harmonic function with u(0)=−log⁡∣S(0)∣<+∞; the kernel K(z,ζ)=(ζ+z)/(ζ−z) is holomorphic in z with Re⁡K=P (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm, Holomorphic functions are real analytic and smooth in their two real coordinates, The C2 real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair, Star-shaped plane domains are homologically simply connected, Inner, singular inner and outer functions, Plane harmonic functions).

[L2]

Herglotz representation: every nonnegative harmonic u on D is u=P[μ] for a unique finite nonnegative regular Borel measure μ with μ(T)=u(0), and conversely P[μ] is nonnegative harmonic (Positive harmonic boundary measures and compact normalized families).

[L3]

The functions H(z):=∫TK(z,ζ) dμ(ζ) are holomorphic on D with Re⁡H=P[μ], and Sμ=e−H is holomorphic, zero-free, with ∣Sμ∣=e−P[μ] and Sμ(0)=e−μ(T); Sμ is a singular inner function exactly when μ⊥m (Properties of the singular functions Sμ, Inner, singular inner and outer functions).

[L4]

Maximum modulus: a holomorphic function on a domain with constant modulus is constant; a holomorphic function on a domain whose modulus has an interior maximum is constant (Local maximum modulus principle).

[L5]

For θ∈H∞ with ∣θ∗∣=1 a.e. (an inner function), one has ∣θ∣≤1 on D and ∥θ∥∞=1, by the boundary-norm identity of the Fatou theorem for analytic H∞; the zero sequence of θ satisfies the Blaschke condition, and θ/B for the Blaschke product of its zeros is holomorphic and zero-free with nontangential boundary modulus 1 a.e., while its modulus is bounded by 1 by the maximum principle on expanding discs ∣z∣<R with ∣BN(Rζ)∣→1 (Fatou's boundary theorem for analytic Hardy spaces, The zero set of a Hardy function satisfies the Blaschke condition, Boundary values and zeros of a Blaschke product, F. Riesz factorization of a Hardy-space function, Blaschke factors and Blaschke products, Local maximum modulus principle).

Proof

technique · direct
1.1givenL1L2

The measure of the modulus. By [L1] the function u=−log⁡∣S∣ is nonnegative harmonic with u(0)<+∞, so [L2] provides a unique finite nonnegative regular Borel measure μ with u=P[μ], μ(T)=u(0).

2.1step 1.1L3L4algebra

S is a unimodular constant times Sμ. Let H be the holomorphic function with Re⁡H=P[μ]=u from [L3]. Then ∣S(z)eH(z)∣=∣S(z)∣eRe⁡H(z)=e−u(z)eu(z)=1(z∈D), so the holomorphic function SeH has constant modulus 1; by [L4] it is a constant λ with ∣λ∣=1, that is, S=λe−H=λSμ.

3.1step 2.1L3algebra

Singularity of μ. Since ∣S∗∣=1 a.e. and ∣λ∣=1, the relation S=λSμ passes to the a.e. boundary values: ∣Sμ∗∣=1 a.e. By the equivalence of [L3], this forces μ⊥m.

4.1step 2.1step 3.1L2L3algebra

Uniqueness. If S=λSμ=λ′Sμ′ with unimodular λ,λ′ and finite positive μ,μ′⊥m, then taking moduli gives log⁡∣S∣=−P[μ]=−P[μ′], so P[μ]=P[μ′] and the uniqueness clause of the Herglotz representation [L2] gives μ=μ′; then λ=S(0)/Sμ(0)=λ′.

4.2step 2.1step 3.1L5algebra

Every inner function factors. Let θ be an inner function. The inner function cannot be identically zero, because its boundary modulus is 1 almost everywhere. Let (an) be its zero sequence with multiplicity and B its Blaschke product. By [L5] the sequence is a Blaschke sequence, and S:=θ/B is holomorphic, zero-free, satisfies ∣S∣≤1 and has ∣S∗∣=1 a.e. By steps 1.1, 2.1 and 3.1 there are λ∈T and μ⊥m with S=λSμ, so θ=B⋅S=λ B Sμ: up to the unimodular constant λ, θ is the product of the Blaschke product B and the singular inner function Sμ.

5.1step 2.1step 3.1step 4.1step 4.2∎

Assembly. Steps 1.1, 2.1 and 3.1 produce the representation S=λSμ with μ⊥m for a zero-free S, step 4.1 proves the asserted uniqueness, and step 4.2 gives the factorisation of a general inner function.

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