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Cauchy representation of an H1 function from its boundary values

Statement

Assume countable choice, as in the Hardy and circle conventions. Let f∈H1(D) and let μ be the unique finite complex Borel measure on T with f=P[μ]. Then μ≪m and μ=f∗m for the boundary function f∗ of the Fatou theorem, and for every z∈D f(z)=P[f∗](z)=∫TP(z,ζ)f∗(ζ) dm(ζ)=12πi∮Tf∗(ζ)ζ−z dζ. Moreover ∥f∥H1=∥f∗∥1=∣μ∣(T) and ∥fr−f∗∥1→0 as r↑1.

Facts & Assumptions

Given: Countable choice and f∈H1(D).

[F1]

The CC analytic Hardy boundary theorem gives f∗∈L1, f=P[f∗], ∥fr−f∗∥1→0 and ∥f∗∥1=∥f∥H1. Its analytic H1 representing measure exists and is unique under CC. (Fatou's boundary theorem for analytic Hardy spaces, Analytic Hardy spaces on the unit disc, The Axiom of Countable Choice (ACω))

[F2]

The density measure f∗m is countably additive by dominated convergence; its finite positive variation is supplied by the local CC circle-variation lemma, and the direct density formula gives total variation integral ∫∣f∗∣dm, and finite complex circle measures with equal Poisson integrals are equal under CC. (A complex L^1 density defines a complex measure whose total variation is |h| dmu, Complex circle measures have finite regular total variation under countable choice, Finite complex circle measures are determined by Fourier coefficients and Poisson integrals)

[F3]

Cauchy's integral formula applies on every radius-r circle with |z|<r<1. The torus parametrization is ζ=e2πit with dm=dt. (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy, The one-dimensional torus and its normalized Haar integral)

[F4]

The Poisson integral is the kernel integral against the boundary datum. L1 convergence and uniform convergence of bounded weights imply convergence of their integrals, by the estimate ∣∫(vrwr−vw)dm∣≤∥vr−v∥1∥wr∥∞+∥v∥1∥wr−w∥∞. (The Poisson integral of a finite complex boundary measure)

Proof

1.1F1F2givenconstructalgebra

Apply [F1] and set μ=f∗m. By [F2] it is a finite complex representing measure for f, and its total variation is ∥f∗∥1=∥f∥H1. Every other finite complex representing measure has the same Poisson integral, so [F2] proves it equals mu. Thus the unique measure in the Statement is exactly this absolutely continuous density measure, including mu zero when f zero.

2.1step 1.1F1F4algebra

By [F1] and [F4], f=P[f∗]=∫P(z,ζ)f∗(ζ)dm, and ∥fr−f∗∥1→0. Together with step 1.1 this gives the full Poisson, absolute-continuity, uniqueness and norm assertions under CC.

3.1step 2.1F3F4algebra

The Cauchy formula. Fix z∈D and r∈(∣z∣,1). The function f is holomorphic on a neighbourhood of the closed disc of radius r, so [F3] gives f(z)=12πi∮∣ζ∣=rf(ζ)ζ−z dζ. Writing the circle integral in the torus parametrization, this equals ∫Trζrζ−zf(rζ) dm(ζ) (the factor rζ is dζ2πi dm). As r↑1, the weights rζrζ−z converge uniformly on T to ζζ−z and are uniformly bounded for r≥(1+∣z∣)/2, because ∣rζ−z∣≥r−∣z∣≥(1−∣z∣)/2>0; together with ∥fr−f∗∥1→0 from step 2.1, [F4] gives f(z)=∫Tζζ−zf∗(ζ) dm(ζ)=12πi∮Tf∗(ζ)ζ−z dζ, the last equality being the same parametrization at r=1.

4.1step 1.1step 2.1step 3.1∎

Assembly. Step 1.1 gives μ=f∗m with the norm identity, step 2.1 gives f=P[f∗]=∫P(z,ζ)f∗dm and the L1 convergence of the radial functions, and step 3.1 gives the Cauchy representation of f by its boundary values.

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