LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The Cauchy kernel expands as an absolutely and uniformly convergent multi-indexed geometric series

Statement

Fix $m \geq 1$ , a point $a \in C^{m}$ , a polyradius $r$ and a real $θ$ with $0 \leq θ < 1$ . For $ζ \in Γ_{r} (a)$ and $z \in {\overline{Δ}}_{θ r} (a)$ , that is $∣ ζ_{k} - a_{k} ∣ = r_{k}$ and $∣ z_{k} - a_{k} ∣ \leq θ r_{k}$ for every $k < m$ ,

$\prod_{k < m} \frac{1}{ζ_{k} - z_{k}} = \sum_{α} (z - a)^{α} \prod_{k < m} (ζ_{k} - a_{k})^{- α_{k} - 1},$

the multi-indexed series converging absolutely (Multi-indexed power series in $C^{m}$ and their absolute convergence). Each term is dominated by

$M_{α} : = \prod_{k < m} \frac{θ^{α_{k}}}{r_{k}}, \sum_{α} M_{α} = \prod_{k < m} \frac{1}{r_{k} (1 - θ)},$

independently of $ζ$ and $z$ , so the convergence is absolute and uniform in the pair $(ζ, z)$ over $Γ_{r} (a) \times {\overline{Δ}}_{θ r} (a)$ .

Facts & Assumptions

Given: $m \geq 1$ , $a \in C^{m}$ , a polyradius $r$ , a real $θ$ with $0 \leq θ < 1$ , and points $ζ \in Γ_{r} (a)$ , $z \in {\overline{Δ}}_{θ r} (a)$ ; $C^{m}$ is read through Complex $m$ -space and its real coordinate dictionary.

[L1]

A multi-indexed series converges absolutely at $z$ when the series along one, equivalently every, bijection $σ : N \to N^{m}$ converges absolutely, its sum is then independent of $σ$ , and its box partial sums over $B_{N} = {α : α_{k} \leq N}$ converge to that sum (Multi-indexed power series in $C^{m}$ and their absolute convergence).

[L2]

$Δ_{r} (a)$ , ${\overline{Δ}}_{r} (a)$ and $Γ_{r} (a)$ are defined coordinatewise by $∣ z_{k} - a_{k} ∣ < r_{k}$ , $\leq r_{k}$ and $= r_{k}$ (Balls, polydiscs and the distinguished boundary in $C^{m}$ ).

[L3]

A complex series converges absolutely when the real series of moduli converges (Complex series, absolute convergence, complex power series, and radius of convergence); an absolutely convergent complex series converges and every rearrangement has the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum).

[L4]

If $∣ f_{n} (x) ∣ \leq M_{n}$ on a set $X$ with $\sum M_{n}$ convergent, then $\sum f_{n} (x)$ converges absolutely for every $x$ and its partial sums converge uniformly on $X$ (Weierstrass M-test for complex-valued function series, Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary).

[L5]

Negative integer powers are defined exactly for nonzero complex bases (Integer powers in the complex field).

[L6]

For real $r^{'}$ with $∣ r^{'} ∣ < 1$ , $\sum_{k} r^{' k}$ converges with sum $1 / (1 - r^{'})$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges), and $(r^{' k})$ is null (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L7]

A nonnegative series converges exactly when its partial sums are bounded above (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L8]

Sums over finite index sets in a commutative monoid are well posed; finite Cartesian-box sums factor by repeated finite Fubini, and distributivity in $C$ permits the finite sum/product factorisations used below (A finite sum in a commutative monoid indexed by an arbitrary finite set, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L9]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1givenL2L5L9

For $k < m$ put $u_{k} = (z_{k} - a_{k}) / (ζ_{k} - a_{k})$ , legitimate by [L2] and [L5] since $∣ ζ_{k} - a_{k} ∣ = r_{k} > 0$ ; then $∣ u_{k} ∣ \leq θ < 1$ by [L2] and [L9], and $ζ_{k} - z_{k} = (ζ_{k} - a_{k}) (1 - u_{k})$ with $∣ 1 - u_{k} ∣ \geq 1 - θ > 0$ by [L9].

1.2givenL2L8L9

Each term satisfies $∣ (z - a)^{α} \prod_{k < m} (ζ_{k} - a_{k})^{- α_{k} - 1} ∣ = \prod_{k < m} (∣ z_{k} - a_{k} ∣^{α_{k}} r_{k}^{- α_{k} - 1}) \leq \prod_{k < m} θ^{α_{k}} / r_{k} = M_{α}$ , by [L2], [L8] and [L9].

2.1step 1.2L6L7L8

For $N \in N$ the box sum of the majorants factors as $\sum_{α \in B_{N}} M_{α} = \prod_{k < m} (r_{k}^{- 1} \sum_{j \leq N} θ^{j})$ by [L8], which is at most $\prod_{k < m} (r_{k} (1 - θ))^{- 1}$ by [L6]. Every finite subset of $N^{m}$ lies in some $B_{N}$ , so along any bijection $σ$ the partial sums of $\sum_{n} M_{σ (n)}$ are bounded by that number, and [L7] makes the series convergent; letting $N \to \infty$ in the factored identity and using [L6] gives $\sum_{α} M_{α} = \prod_{k < m} (r_{k} (1 - θ))^{- 1}$ .

2.2step 1.1L5L8algebra

The box partial sum factors: by [L8], $\sum_{α \in B_{N}} (z - a)^{α} \prod_{k < m} (ζ_{k} - a_{k})^{- α_{k} - 1} = \prod_{k < m} (\frac{1}{ζ_{k} - a_{k}} \sum_{j \leq N} u_{k}^{j}) = \prod_{k < m} \frac{1 - u_{k}^{N + 1}}{(ζ_{k} - a_{k}) (1 - u_{k})}$ , the last equality by the finite geometric identity $(1 - u) \sum_{j \leq N} u^{j} = 1 - u^{N + 1}$ and step 1.1.

3.1step 1.2step 2.1L1L3L4

By step 1.2 and step 2.1 the hypotheses of [L4] hold with the constants $M_{α}$ on the set $Γ_{r} (a) \times {\overline{Δ}}_{θ r} (a)$ , so the series converges absolutely at every such pair and its partial sums along $σ$ converge uniformly there; by [L1] and [L3] the sum is independent of $σ$ and the box partial sums converge to it.

3.2step 1.1step 2.2L6L8L9

By step 1.1 the target value is $\prod_{k < m} ((ζ_{k} - a_{k}) (1 - u_{k}))^{- 1}$ , so the difference from the box sum of step 2.2 has modulus at most $\prod_{k < m} (r_{k} (1 - θ))^{- 1} \cdot ∣ 1 - \prod_{k < m} (1 - u_{k}^{N + 1}) ∣$ by [L9]; since $∣ u_{k}^{N + 1} ∣ \leq θ^{N + 1}$ , [L9] and [L8] bound the second factor by $(1 + θ^{N + 1})^{m} - 1$ , which tends to $0$ as $N \to \infty$ by [L6].

4.1step 3.1step 3.2∎

Hence the box partial sums converge to $\prod_{k < m} (ζ_{k} - z_{k})^{- 1}$ , and by step 3.1 they also converge to the sum of the series; the two limits agree, which is the displayed expansion, with the majorant and the uniformity already recorded in steps 1.2 and 3.1.

The Cauchy kernel expands as an absolutely and uniformly convergent multi-indexed geometric series

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources