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Smoothness of the Bergman kernel and positivity of its diagonal on bounded domains

Statement

Assume the Axiom of Countable Choice ACω (The Axiom of Countable Choice (ACω)). Let m≥1 and let Ω⊆Cm be a nonempty open set. Then KΩ∈C∞(Ω×Ω) and z↦KΩ(z,z) is C∞ on Ω. If Ω is bounded, its Lebesgue measure satisfies 0<λΩ(Ω)<∞, and for every z∈Ω,

KΩ(z,z)≥1λΩ(Ω)>0.

In particular, on a bounded domain z↦log⁡KΩ(z,z) is C∞.

Facts & Assumptions

[A1]

The only choice principle assumed is ACω (The Axiom of Countable Choice (ACω)). It is inherited through the Bergman Hilbert-space and Riesz setup, and is used by the Borel and Euclidean-ball measure suppliers below; no full Axiom of Choice is used.

[F1]

Under ACω, A2(Ω) is a closed complex Hilbert subspace of L2(Ω) with the first-variable-linear pairing. Point evaluation has Riesz section kw=KΩ(⋅,w) and reproduces evaluation. The definition also gives A2(Cm)={0} and KCm≡0, so no positivity is asserted for every unbounded open set (The Bergman space A2(Ω) and the Bergman kernel, The complex L2 pairing on equivalence classes, L2 with the integral pairing is a Hilbert space).

[F2]

The definition KΩ(z,w)=kw(z) makes z↦KΩ(z,w) holomorphic. Reproduction gives KΩ(z,w)=⟨kw,kz⟩, so conjugate symmetry of the first-variable-linear L2 pairing gives KΩ(w,z)=KΩ(z,w)‾; hence the kernel is antiholomorphic in w (The Bergman space A2(Ω) and the Bergman kernel, The complex L2 pairing on equivalence classes, L2 with the integral pairing is a Hilbert space).

[F3]

On every nonempty compact K⊆Ω, sup⁡K∣f∣≤CK∥f∥2. The diagonal extremal identity is KΩ(w,w)=∥kw∥22=sup⁡∥f∥2≤1∣f(w)∣2 (Sup-norm and first-derivative bounds by the L2 norm on compact subsets, Reproducing property, Bergman projection and the extremal characterization).

[F4]

The first-variable-linear Hilbert pairing satisfies ∣⟨f,g⟩∣≤∥f∥2∥g∥2 (Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs).

[F5]

A separately holomorphic, locally bounded function on an open subset of Cn is jointly holomorphic there (Locally bounded and separately holomorphic implies holomorphic).

[F6]

A holomorphic function on an open subset of complex Euclidean space is C∞ in the underlying real coordinates (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

[F7]

Complex conjugation is a real-linear coordinate map, and finite-order smooth maps are closed under composition (Real and imaginary parts, complex conjugation, and modulus, Ck Euclidean maps and diffeomorphisms, Ck Euclidean maps are closed under componentwise algebra and composition).

[F9]

For x>0, log⁡′(x)=1/x and log⁡ is continuous. For each integer n≥1, ddxx−n=−nx−n−1; products and quotients of continuous functions are continuous where their denominators are nonzero (The natural logarithm as the inverse of the exponential function, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, Integer powers am, For a natural n≥1 the function x↦xn is differentiable everywhere with derivative ι(n) x n−1; for n=0 it is the constant 1, with derivative 0; for a natural n≥1 the function x↦x−n is differentiable at every x≠0 with derivative −ι(n) x−n−1; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0). A real function is C∞ when all iterated coordinate partial derivatives of every finite order exist and are continuous (Ck maps and multi-index derivative notation in Euclidean space).

Proof

technique · direct, using the Riesz sections, compact evaluation bounds and the constant-function extremal witness

Given: ACω, m≥1, a nonempty open Ω⊆Cm, its Bergman space A2(Ω), and its Bergman kernel KΩ.

1.1A1F1F2given

For each w∈Ω, KΩ(z,w)=kw(z) is holomorphic in z by [F1, F2]. Reproducing evaluation at z on kw gives KΩ(z,w)=⟨kw,kz⟩. Conjugate symmetry then gives KΩ(w,z)=⟨kz,kw⟩=KΩ(z,w)‾. Thus KΩ is antiholomorphic in w, and F(z,η):=KΩ(z,η‾) is separately holomorphic on Ω×Ω∗, where Ω∗:={η:η‾∈Ω}.

1.2A1F1F8given

Suppose Ω is bounded. Choose a∈Ω; openness gives r>0 with B(a,r)⊆Ω. Boundedness supplies c∈Cm and R0>0 with Ω⊆B(c,R0). The Euclidean triangle inequality then gives Ω⊆B(0,∥c∥+R0+1). By [F8], Ω is measurable and 0<λ(B(a,r))≤λΩ(Ω)≤λ(B(0,∥c∥+R0+1))<∞. Hence 0<λΩ(Ω)<∞.

2.1A1F3F4F8step 1.1

Let K,L⊆Ω be nonempty compact sets. The evaluation estimate [F3] and the diagonal extremal identity [F3] give KΩ(z,z)≤CK2 for z∈K and KΩ(w,w)≤CL2 for w∈L. Since KΩ(z,w)=⟨kw,kz⟩, Cauchy–Schwarz [F4] gives ∣KΩ(z,w)∣≤CKCL on K×L. Around any z0,w0∈Ω choose closed Euclidean ball neighborhoods K,L contained in Ω; they are compact by [F8]. This proves that F is locally bounded on Ω×Ω∗.

2.2A1F1F3step 1.2

Suppose Ω is bounded. For z∈Ω let f0:=λΩ(Ω)−1/2 be the constant function. By step 1.2 it lies in A2(Ω) and has norm one. The extremal identity [F3] gives KΩ(z,z)≥∣f0(z)∣2=1/λΩ(Ω)>0.

3.1F5F7F8step 1.1step 2.1

The set Ω∗ is open because complex conjugation is a Euclidean isometry, so Ω×Ω∗ is an open subset of C2m. By [F5], the separately holomorphic, locally bounded function F is jointly holomorphic.

4.1F6F7step 3.1

By [F6], F is C∞ in real coordinates. The map (z,w)↦(z,w‾) and the diagonal map z↦(z,z) are real-linear coordinate maps, hence smooth; [F7] makes their compositions with F smooth. Therefore KΩ(z,w)=F(z,w‾) is C∞ on Ω×Ω, and z↦KΩ(z,z) is C∞ on Ω.

5.1F7F9step 4.1step 2.2

Let D(z):=KΩ(z,z). On a bounded Ω, steps 4.1 and 2.2 give D∈C∞(Ω) with D>0. Set ℓ(x):=log⁡x for x>0. By [F9], ℓ′(x)=x−1, and induction using the negative-power derivative in [F9] gives ℓ(n)(x)=(−1)n−1(n−1)!x−n for every n≥1. These derivatives are continuous on (0,∞) by [F9], and ℓ itself is continuous there; hence ℓ∈C∞((0,∞)). The composition theorem [F7] now gives log⁡KΩ(z,z)=ℓ(D(z))∈C∞(Ω).

6.1step 2.1step 4.1step 1.2step 2.2step 5.1∎

Steps 2.1 and 4.1 establish joint C∞ smoothness and the smooth diagonal for every nonempty open Ω; steps 1.2, 2.2 and 5.1 establish the positive diagonal bound and smooth logarithmic potential whenever Ω is bounded. These are the two claims.

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