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The mean-value L2 bound for holomorphic functions on a polydisc

Facts & Assumptions

[A1]

The Axiom of Countable Choice ACω is the principle defined by The Axiom of Countable Choice (ACω). It is the only choice assumption below; the polar-measure and product-Lebesgue suppliers used here state it explicitly.

[F1]

If g is holomorphic on D(b,R) and 0<s<R, then g(b)=(2π)−1∫02πg(b+seiθ) dθ (A holomorphic function equals its average on every circle inside a larger concentric holomorphy disc).

[F2]

The chart surface measure of S1 agrees with its polar surface measure; under the angular chart ω=eiθ its density is dθ, so σ(S1)=2π (Surface integration on compact C1 hypersurfaces, Agreement with the existing polar sphere measure).

[F3]

For every nonnegative Borel h on R2, polar coordinates give ∫h dλ2=∫0∞∫S1h(sω)s dσ(ω) ds (The polar surface set function on the unit sphere, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F4]

Lebesgue measure is translation invariant on measurable sets (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F5]

Every nonnegative measurable function is the increasing limit of nonnegative simple functions, and increasing limits pass through the Lebesgue integral; thus the setwise translation invariance in [F4] extends from indicators and simple functions to nonnegative Borel integrals (Every nonnegative measurable function admits an explicit increasing sequence of simple approximations, Monotone convergence for the integral).

[F6]

Borel sets in a finite Euclidean product are product-measurable; Tonelli permits iterated integration of nonnegative product-measurable functions, and the finite product of planar Lebesgue measures agrees with Euclidean Lebesgue measure under R2m≅(R2)m (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}).

[F7]

A holomorphic function is continuous, hence ∣f∣2 is Borel (A holomorphic function of several variables is continuous and separately holomorphic); the one-variable case is also supplied by Complex differentiability at a point implies continuity there.

[F8]

Restricting the complex linear derivative in the definition of holomorphy to a coordinate line shows that every coordinate slice of a holomorphic function is holomorphic (Holomorphic functions on an open subset of Cm).

[F9]

A polydisc is the product of its coordinate discs, and Cm is identified with R2m with the corresponding Lebesgue convention (Balls, polydiscs and the distinguished boundary in Cm, Complex m-space and its real coordinate dictionary).

Statement

Assume ACω (The Axiom of Countable Choice (ACω)). Let m≥1, let Ω⊆Cm be open, let a∈Ω, and let r=(r0,…,rm−1) be a polyradius with each rk>0 such that Δ‾r(a)⊆Ω. If f is holomorphic on Ω, then

∣f(a)∣2≤1λ2m(Δr(a))∫Δr(a)∣f∣2 dλ2m,λ2m(Δr(a))=πm∏k<mrk2,

where λ2m is Lebesgue measure under Cm≅R2m. In particular, for a common radius r the coefficient is (πr2)−m.

Proof

technique · direct

Given: ACω, m≥1, the open set Ω, the point a, the positive polyradius r, and the holomorphic function f from the statement.

1.1F1algebra

For a one-variable holomorphic g on D(b,ρ) and each 0<s<ρ, [F1] gives the circular mean of g as g(b). Expanding the nonnegative integral of ∣g(b+seiθ)−g(b)∣2 and using that mean identity yields ∣g(b)∣2≤(2π)−1∫02π∣g(b+seiθ)∣2 dθ.

2.1step 1.1F2F3F4F5F7A1algebra

By [F2], the angular measure in step 1.1 is the polar surface measure with total mass 2π. The function h(x)=∣g(b+x)∣2 for x∈D(0,ρ) and h(x)=0 otherwise is nonnegative Borel by [F7]. Apply [F3] to h, using translation invariance [F4]–[F5]. Integrating the circle inequality in step 1.1 against 2s ds/ρ2 for 0<s<ρ gives ∣g(b)∣2≤1πρ2∫D(b,ρ)∣g(z)∣2 dλ2(z); the same polar formula applied to the indicator of the disc gives λ2(D(b,ρ))=πρ2.

3.1step 2.1F6F9A1

Let Dk=D(ak,rk). By [F6] and [F9], the measure of their product is the product of their planar measures, so step 2.1 gives λ2m(Δr(a))=∏k<mπrk2.

4.1step 2.1step 3.1F6F7F8F9A1given∎

For 0≤k≤m, let Ik be the integral of ∣f(z0,…,zk−1,ak,…,am−1)∣2 over D0×⋯×Dk−1 with product planar measure, with I0=∣f(a)∣2. For each fixed tuple of the first k coordinates, [F8] shows that the resulting one-variable slice is holomorphic on D(ak,rk). The estimate of step 2.1 applies to that slice; integrating over the preceding discs and using [F6] gives Ik≤(πrk2)−1Ik+1. Iterating for k=0,…,m−1 and using [F6], [F7] and [F9] to identify Im with the integral in the statement gives ∣f(a)∣2≤(∏k<mπrk2)−1∫Δr(a)∣f∣2 dλ2m. Step 3.1 identifies the coefficient with 1/λ2m(Δr(a)), completing the proof.

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