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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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A continuous separately holomorphic function is the sum of an absolutely convergent power series with Cauchy-integral coefficients on every smaller polydisc

Statement

Fix m1, aCm and a polyradius ρ, let f:Δρ(a)C be continuous and separately holomorphic, let r be a polyradius with rk<ρk for every k<m, and let Ck(t)=ak+rkexp(it) on [0,2π]. For each multi-index α set

cα:=1(2πi)mC0 ⁣ ⁣Cm1f(ζ)k<m(ζkak)αk1dζm1dζ0,

an iterated integral as in the polydisc Cauchy formula. Then, with M=supΓr(a)f,

cαMk<mrkαk,

and for every real θ with 0<θ<1 the series αcα(za)α converges absolutely and uniformly on Δθr(a) with

f(z)=αcα(za)α(zΔθr(a)).

Since every zΔr(a) lies in Δθr(a) for some θ<1, the expansion holds throughout Δr(a).

The coefficients are asserted here only as those iterated integrals. That cα equals αf(a)/α! needs termwise differentiation and is not claimed by this statement.

Facts & Assumptions

Given: The data above, with Cm read through Complex m-space and its real coordinate dictionary and f continuous and separately holomorphic on Δρ(a) (Separately holomorphic functions).

[L1]

Under these hypotheses, f(z)=(2πi)mC0Cm1f(ζ)k<m(ζkzk)1dζm1dζ0 for every zΔr(a), as an iterated integral each of whose integrands is continuous on its circle (The iterated Cauchy integral formula on a polydisc).

[L2]

For ζΓr(a) and zΔθr(a) with 0θ<1, k<m(ζkzk)1=α(za)αk<m(ζkak)αk1, with each term dominated by k<mθαk/rk and the convergence absolute and uniform in the pair (The Cauchy kernel expands as an absolutely and uniformly convergent multi-indexed geometric series).

[L3]

A multi-indexed series converges absolutely at z when the series along one, equivalently every, enumeration of Nm converges absolutely; its sum is independent of the enumeration; and its box partial sums over BN={α:αkN} converge to that sum (Multi-indexed power series in Cm and their absolute convergence).

[L4]

If continuous functions on the trace of a fixed rectifiable contour converge uniformly to a continuous function, their integrals converge to its integral (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L5]

If gK on the trace of a rectifiable contour γ, with K0, then γgdzKL(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length); complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand) and exist for continuous integrands (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L6]

An absolutely convergent complex series converges and every rearrangement has the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum); a dominated series with summable bounds converges absolutely and uniformly (Weierstrass M-test for complex-valued function series); a nonnegative series converges exactly when its partial sums are bounded (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum); for r<1, krk=1/(1r) (For r<1, k0rk=1/(1r), and for r1 the series diverges).

[L7]

If a property holds at 0 and passes from q to q+1, it holds for every natural number (The principle of mathematical induction).

[L8]

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

[L9]

Δr(a), Δr(a) and Γr(a) are defined coordinatewise by zkak<rk, rk and =rk (Balls, polydiscs and the distinguished boundary in Cm).

[L10]

The once-traversed circle of radius r>0 has length 2πr (Every circle has circumference 2 pi r and circumference-to-diameter ratio pi).

[L12]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive); finite sums are additive, scale and are monotone in their terms (Laws of finite sums and finite products).

Proof

technique · direct
1.1

Γr(a) is closed and bounded in Cm, hence compact by [L11] and [L9], and it lies in Δρ(a) because rk<ρk; so f is continuous on it and M=supΓr(a)f is a real number by [L11].

givenL9L11
1.2

An induction on the number of remaining integrations ([L7]) using [L5] and [L10] gives the iterated bound: if hK at every point of Γr(a), then the modulus of the iterated integral C0Cm1hdζm1dζ0 is at most Kk<m(2πrk), each step contributing one factor L(Ck)=2πrk.

givenL5L7L10
2.1

Applying step 1.2 to the integrand of cα, whose modulus on Γr(a) is at most Mk<mrkαk1 by [L9] and [L12], gives cα(2π)mMk<mrkαk1k<m(2πrk)=Mk<mrkαk.

step 1.1step 1.2L8L9L12
2.2

Write SN(ζ,z) for the box partial sum of the expansion in [L2]. It is a finite sum, so multiplying by f(ζ) and integrating iteratedly, [L5] and [L12] give (2πi)mC0Cm1f(ζ)SN(ζ,z)dζm1dζ0=αBNcα(za)α.

step 1.1L2L5L12
3.1

Fix θ with 0<θ<1 and zΔθr(a). By step 2.1 and [L9], cα(za)αMk<mθαk, and the box sums of the right-hand side are Mk<mjNθjMk<m(1θ)1 by [L12] and [L6]; every finite subset of Nm lies in a box, so [L6] makes αMkθαk convergent and the M-test gives absolute and uniform convergence of αcα(za)α on Δθr(a).

step 2.1L6L9L12
3.2

By [L2] the difference SN(ζ,z)k<m(ζkzk)1 tends to 0 uniformly for ζΓr(a), so multiplying by f(ζ) and using step 1.1 the products differ by at most MεN with εN0; step 1.2 then bounds the difference of the two iterated integrals by MεNk<m(2πrk)/(2π)m, which tends to 0. Hence αBNcα(za)α converges to the iterated integral of [L1], which is f(z).

step 1.1step 1.2step 2.2L1L2L4L12
4.1

By step 3.1 and [L3] the box partial sums also converge to the sum αcα(za)α; comparing with step 3.2 gives f(z)=αcα(za)α for every zΔθr(a). Since a point of Δr(a) has zkak<rk for each k, it lies in Δθr(a) for any θ<1 exceeding every zkak/rk, so the expansion holds on all of Δr(a).

step 3.1step 3.2L3L9

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