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Separate holomorphy forces local boundedness on smaller polydiscs

Statement

Let m≥1, let R>0, and let

ΔRm:={z=(z1,…,zm)∈Cm:∣zj∣<R for every j}.

If f:ΔRm→C is separately holomorphic, then for every 0<r<R the function f is bounded on the closed polydisc Δ‾rm.

In particular, every separately holomorphic function on an open subset of Cm is locally bounded.

Facts & Assumptions

Given: A separately holomorphic function f on ΔRm and a radius 0<r<R.

[L1]

Separate holomorphy is the condition that each coordinate slice is one-variable holomorphic (Separately holomorphic functions).

[L3]

A separately holomorphic function that is locally bounded is jointly holomorphic (Locally bounded and separately holomorphic implies holomorphic).

[L4]

Jointly holomorphic functions are smooth, so their mixed derivatives are holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

[L5]

Cauchy estimates on a smaller polydisc bound Taylor coefficients by the supremum on that smaller distinguished boundary (Cauchy estimates for mixed derivatives on a polydisc).

[L6]

For a holomorphic one-variable function that is not identically zero on the connected component under consideration, the logarithm of the modulus is subharmonic (The logarithm of the modulus of a holomorphic function is subharmonic), and subharmonic means upper semicontinuous together with the disc submean inequality (Subharmonic functions on plane domains).

[L7]

Fatou's lemma controls the liminf of integrals of nonnegative measurable functions (Fatou's lemma), and monotone convergence controls increasing nonnegative boundary approximations (Monotone convergence for the integral).

[L9]

A holomorphic function on a connected open set is determined by its values on any nonempty open subset (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L10]

Subharmonic functions satisfy harmonic comparison on discs; continuous circle data have harmonic Poisson extensions; and upper-semicontinuous circle data are Borel and bounded above (Subharmonicity is equivalent to harmonic comparison on compactly contained discs, The Poisson integral on the unit disc, The Poisson integral gives the unique continuous harmonic extension on the closed unit disc, Upper semicontinuous functions are Borel and their circle averages are defined).

Proof

technique · induction
1.1giveninduction

We prove a stronger local claim by induction on m: every separately holomorphic function on ΔRm is holomorphic on a neighborhood of each point of ΔRm. Once that is known, the displayed boundedness follows, because the compact set Δ‾rm is covered by finitely many such holomorphic neighborhoods and each holomorphic function is bounded on a smaller closed polydisc inside its neighborhood.

1.2L1baseih

Base case m=1: separate holomorphy is ordinary one-variable holomorphy by [L1], so the local claim and the boundedness statement are immediate. Assume now that the local claim is known in dimension m−1, and prove it in dimension m≥2.

1.3givenconstruct

Fix p∈ΔRm. Choose ρ>0 with 0<ρ<R and Δ‾2ρm(p)⊆ΔRm. After translating and rescaling each coordinate disc, it is enough to prove that a separately holomorphic function on Δ2m is holomorphic in a neighborhood of the origin. Write z=(z′,w) with z′∈Cm−1 and w∈C.

1.4L2inductionconstruct

Box claim. If a closed real box K=I1×⋯×Id⊆Rd is covered by countably many closed sets Fn, then some Fn contains a smaller closed real box with nondegenerate sides. We prove this by induction on d. For d=1 it is [L2]. Assume the claim in dimension d−1. Write K=I×K′. Enumerate the closed subboxes of K′ with rational endpoints in the coordinates of K′ as Q1,Q2,…. For each pair (n,j) let En,j:={x∈I:{x}×Qj⊆Fn}. Each En,j is closed. Fix x∈I. The sections Fn(x):={y∈K′:(x,y)∈Fn} are closed and cover K′, so the induction hypothesis in dimension d−1 gives some n such that Fn(x) contains a smaller closed box; shrinking slightly if needed, that box contains a rational-endpoint subbox Qj. Hence x∈En,j. So the countable family En,j covers I, and [L2] gives one pair (n,j) for which En,j contains a nondegenerate closed subinterval J. Then J×Qj⊆Fn, proving the claim.

1.5construct

Whenever 0≤d<1<η, one can choose radii d<s<r1<1<r2<η.

2.1L1step 1.2ih

For each positive integer B, define ΩB:={z′∈Δ‾1m−1:∣f(z′,w)∣≤B for every ∣w∣≤1}. For fixed w with ∣w∣≤1, the induction hypothesis applied to z′↦f(z′,w) makes that function holomorphic, hence continuous, on Δ2m−1. Therefore each set {z′:∣f(z′,w)∣≤B} is closed, and so every ΩB is closed. Also ⋃B≥1ΩB=Δ‾1m−1, because for fixed z′ the slice w↦f(z′,w) is holomorphic on Δ2 and therefore bounded on ∣w∣≤1.

3.1step 2.1step 1.4construct

Apply the box claim to the real box [−1/2,1/2]2m−2⊆Δ‾1m−1 and the closed cover (ΩB)B≥1. We obtain some B0 and a nondegenerate closed real box contained in ΩB0. Inside its relative interior choose a closed complex polydisc Δ‾2εm−1(a′) for some a′ and some ε>0. Its centre satisfies d:=max⁡1≤j<m∣aj′∣<1, and ∣f(z′,w)∣≤B0for z′∈Δ‾2εm−1(a′), ∣w∣≤1. Hence f is bounded on the product polydisc E:=Δ2εm−1(a′)×Δ1.

4.1L3step 3.1construct

By [L3], the separately holomorphic and bounded function f is jointly holomorphic on E. Put ζ=z′−a′ and retain the notation f(ζ,w) after this translation. The original first-variable domain contains the centred polydisc Δηm−1, where η:=2−d>1, while the original target z′=0 now has coordinate ζ=−a′. Thus f is separately holomorphic on Δηm−1×Δ2 and jointly holomorphic on Δ2εm−1×Δ1. Choose radii d<s<r1<1<r2<η, which is possible because d<1<η.

5.1L4L5L6step 1.2step 3.1step 4.1ih

For each multi-index α∈Nm−1, define cα(w):=1α! ∂ζαf(0,w)(∣w∣<1), using the jointly holomorphic function from step 4.1. By [L4], every cα is holomorphic on Δ1. For fixed w with ∣w∣<1, the induction hypothesis makes ζ↦f(ζ,w) holomorphic on Δηm−1, so these cα(w) are exactly its Taylor coefficients at 0. Since f is bounded by B0 on the larger product Δ2εm−1×Δ1, [L5] applied at the strictly smaller radius ε gives ∣cα(w)∣≤B0 ε−∣α∣(∣w∣<1). For each nonzero multi-index α with cα≢0, define uα(w):=∣α∣−1log⁡∣cα(w)∣. By [L6], each such uα is subharmonic on Δ1. If cα≡0, then the term cα(w)ζα vanishes identically and is already harmless for the later power-series tail estimate. The displayed estimate gives a uniform upper bound for the whole family (uα)α≠0, cα≢0.

6.1L5step 1.5step 5.1

Fix w∈Δ1. The Cauchy estimates [L5] applied to the holomorphic function ζ↦f(ζ,w) on Δηm−1, with the strictly smaller radius r2, show that ∣cα(w)∣≤M(w) r2−∣α∣ for some finite constant M(w) and every nonzero multi-index α. Therefore lim sup⁡∣α∣→∞cα≢0uα(w)≤−log⁡r2.

7.1L6step 5.1step 6.1

Let S:={α∈Nm−1∖{0}:cα≢0}. If S is finite, the required tail estimate is immediate. Otherwise enumerate it as (α(j))j≥1 with nondecreasing degrees and put vj=uα(j). By steps 5.1 and 6.1, the subharmonic functions vj have a common upper bound A on Δ1 and satisfy lim sup⁡jvj(w)≤C:=−log⁡r2 pointwise.

8.1L7L9L10step 7.1assume-contradischarge-contradiction

We claim that for every compact K⋐Δ1 and every δ>0, vj≤C+δ on K for all sufficiently large j. Otherwise choose jk↑∞ and qk∈K with vjk(qk)>C+δ, and pass to a subsequence with qk→q∈K. Choose t>0 with D(q,t)‾⊆Δ1, discard finitely many terms so qk∈D(q,t), and put Pk(θ):=P((qk−q)/t,eiθ). By the definition of S and [L9], no selected coefficient can vanish on a nonempty open subset of Δ1. More specifically for the boundary argument, vjk cannot be identically −∞ on the circle: if it were, every constant harmonic function −n would majorize its boundary values, so [L10] would give vjk(qk)≤−n for every n, contradicting the finite strict lower bound just chosen. For ζ on the circle define ϕk,n(ζ):=sup⁡η∈∂D(q,t)(vjk(η)−n∣ζ−η∣). Compactness and the common upper bound make each ϕk,n finite and continuous, and upper semicontinuity gives ϕk,n↓vjk pointwise. Harmonic comparison, their Poisson extensions, and monotone convergence in [L7] therefore give vjk(qk)≤12π∫02πPk(θ)vjk(q+teiθ) dθ. The kernels Pk tend uniformly to 1. The functions Pk(θ)(A−vjk(q+teiθ)) are nonnegative and measurable, so Fatou's lemma [L7] and the pointwise limsup bound make the lower limit of their normalized integrals at least A−C. Since each Pk has normalized integral 1, the preceding inequality gives lim sup⁡kvjk(qk)≤C, a contradiction.

9.1step 8.1algebra

Fix 0<σ<1. Apply step 8.1 to K=Δ‾σ and δ=log⁡(r2/r1)>0. For all sufficiently large j, vj(w)≤−log⁡r1(∣w∣≤σ), or equivalently ∣cα(j)(w)∣r1∣α(j)∣≤1.

10.1L8step 1.5step 9.1

Fix such a σ. If S is finite, then ∑αcα(w)ζα is a finite sum in ζ. Otherwise step 9.1 dominates its tail on Δ‾sm−1×Δ‾σ by the convergent product-geometric majorant ∑α(s/r1)∣α∣; the finitely many low-degree terms are harmless and the coefficients outside S vanish. Thus the series converges locally uniformly. Every partial sum is holomorphic, so [L8] gives a jointly holomorphic limit G(ζ,w) on Δsm−1×Δσ.

11.1step 4.1step 10.1L9

On the open set Δmin⁡(ε,s)m−1×Δσ, the Taylor expansion of the jointly holomorphic function from step 4.1 is exactly the series defining G. Thus G=f on that nonempty open set. For fixed w∈Δσ, both ζ↦G(ζ,w) and ζ↦f(ζ,w) are holomorphic on Δsm−1 and agree on a nonempty open subset, so [L9] gives equality on all of Δsm−1. Hence f=G on Δsm−1×Δσ, and f is jointly holomorphic there.

12.1step 1.1step 1.2step 1.3step 11.1discharge-induction∎

Because max⁡j∣−aj′∣=d<s and 0<σ, the translated coordinates of the original target, (−a′,0), lie in the product from step 11.1. Thus that step proves the required local holomorphicity at the original origin in dimension m. By the reductions in steps 1.1 and 1.3, every point of ΔRm has a holomorphic neighborhood. Therefore f is locally bounded on ΔRm, and in particular bounded on every smaller closed polydisc Δ‾rm. This closes the induction.

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