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24 results · all verified · 12 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Holomorphic Functions of Several Complex Variables

1 · Prerequisites

2 · Summary

This page transfers the one-variable complex toolkit to Cm through the Euclidean dictionary, the real total derivative, and coordinate slices. The background already provides the one-variable Cauchy formula, Cauchy inequalities, the identity theorem, the maximum principle, Liouville's theorem, and the open mapping theorem, and those results are repeatedly applied to the holomorphic functions obtained by freezing all but one variable or by restricting to a complex line.

The page defines balls, polydiscs, separate holomorphy, holomorphic maps, and the complex Jacobian, then proves the iterated Cauchy formula on a polydisc and derives power-series expansions, Cauchy estimates, smoothness, and the Cauchy-Riemann characterization. It then shows that continuity plus separate holomorphy, and later local boundedness plus separate holomorphy, force full holomorphy, proves the correct several-variable identity theorem, and closes with the several-variable maximum-modulus principle, Liouville theorem, and scalar open mapping theorem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Complex m-space and its real coordinate dictionary

Remark

Fix a natural number m1. Complex m-space Cm is the set of functions mC, so a point z has coordinates zk for k<m, indexed from 0 exactly as Rn is in this library. With coordinatewise addition and multiplication by complex scalars it is a vector space over the field C (Vector space over a field, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)), with the standard basis e0,,em1 of The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0.

The coordinate identification. Writing zk=xk+iyk with xk,yk real (Real and imaginary parts, complex conjugation, and modulus), define

Φ:CmR2m,Φ(z)=(x0,y0,x1,y1,,xm1,ym1).

The interleaved ordering is the one used throughout this page; the ordering that groups all real parts before all imaginary parts is a different bijection, and nothing below is stated for it. Φ is a bijection and is R-linear.

Norms agree. Put z:=(k<mzk2)1/2. Since zk2=xk2+yk2, this is the Euclidean norm Φ(z)2 of The p-norms xp for rational p1, and x and The Euclidean inner product x,y=k<nxkyk on Rn, and it is a norm on the real vector space underlying Cm in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms. Consequently zw=Φ(z)Φ(w)2, so the metric of Cm, its balls (Open ball, closed ball and sphere in a metric space), its open sets, its convergent sequences, its Cauchy sequences and its continuous maps are verbatim those of R2m under Φ. In particular convergence and continuity are coordinatewise (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions, Vector-valued functions f:ARm, their limits and continuity, with the dictionary to the metric notions), Cm is complete (For n1 a sequence in Rn converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and Rn is complete in every norm), and a subset of Cm is compact exactly when it is closed and bounded (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

At m=1 this is the published plane dictionary. For m=1 the map Φ is the bijection of C=R[x]/(x2+1) as the Euclidean plane and as a normed real algebra: what the identification preserves and every clause above reduces to a clause recorded there. Openness, connectedness and real total differentiability on Cm are always read through Φ, exactly as that remark reads them through its own identification.

What Φ does not carry. Φ respects the additive and the real scalar structure but not multiplication by i in any way visible to a general R-linear map of R2m: an R-linear map of Cm need not be C-linear. That distinction is the whole content of the criterion the page proves next, and it is why "linear" is always qualified below.

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Balls, polydiscs and the distinguished boundary in Cm

Definition

Fix m1 and read Cm through Complex m-space and its real coordinate dictionary. A polyradius is a function r:mR with rk>0 for every k<m; a single positive real r abbreviates the constant polyradius with every rk=r.

For aCm and a polyradius r, the open polydisc, the closed polydisc and the distinguished boundary are

Δr(a):={z:zkak<rk for every k<m}, Δr(a):={z:zkakrk for every k<m}, Γr(a):={z:zkak=rk for every k<m}.

Thus Γr(a) is the set of points all of whose coordinates lie on their own circle: it is the product of the m circles {ζkak=rk}.

The open ball and closed ball of centre a and radius ρ>0 are those of the norm of the dictionary, that is the sets B(a,ρ)={z:za<ρ} and B(a,ρ)={z:zaρ} of Open ball, closed ball and sphere in a metric space and Euclidean spheres and closed balls as subspaces of Rn.

Remarks

The distinguished boundary is not the topological boundary when m2. The topological boundary of Δr(a) consists of the points where at least one coordinate satisfies zkak=rk, whereas Γr(a) requires every coordinate to do so. For m=1 the two coincide. For m2 the inclusion Γr(a)Δr(a) is proper: the point whose first coordinate is a0+r0 and whose remaining coordinates are ak lies in the topological boundary and not in Γr(a).

Polydiscs are open and convex. Openness is coordinatewise: if zkak<rk for every k, then B(z,ρ)Δr(a) for ρ=mink<m(rkzkak)>0, because wkzkwz by the dictionary. Convexity in the sense of A convex subset of Rm contains every line segment between two of its points is also coordinatewise: for z,w in Δr(a) and t[0,1], (1t)zk+twkak(1t)zkak+twkak<rk by Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive. Hence a polydisc is star-shaped with respect to each of its points (Star-shaped open subsets of Euclidean space). The same computation gives convexity of the closed polydisc.

Slices are discs. Fixing all coordinates but the kth at values aj with ajaj<rj, the set of ζ with the resulting point in Δr(a) is exactly the open disc {ζ:ζak<rk}; this is what makes the one-variable theory applicable one coordinate at a time. Moduli, real and imaginary parts are those of Real and imaginary parts, complex conjugation, and modulus.

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Holomorphic functions on an open subset of Cm

Definition

Fix m1, read Cm through Complex m-space and its real coordinate dictionary, and let UCm be open and aU.

A map L:CmC is C-linear when L(u+v)=L(u)+L(v) and L(λu)=λL(u) for all u,vCm and all λC, the vector-space operations being those of Vector space over a field over the field C (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)). Requiring the second clause only for real λ gives the strictly weaker notion of an R-linear map (A linear map L:RmRn in Euclidean coordinates) read through the dictionary.

A function f:UC is complex differentiable at a when there is a C-linear L:CmC with

f(a+h)=f(a)+L(h)+r(h),r(h)h0  as h0,

the quotient being considered for h0 with a+hU and the norm being that of the dictionary (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). The map f is holomorphic on U when it is complex differentiable at every point of U.

Such an L is unique, so the notation Df(a):=L is well posed. If L1 and L2 both satisfy the condition, then T=L1L2 is C-linear and T(h)/h0; fixing h0 and taking h replaced by th for real t(0,1) small enough that a+thU, C-linearity gives T(h)/h=T(th)/th, whose limit as t0 is 0; so T(h)=0 for every h.

Remarks

No continuity and no local boundedness are built in. The definition asks for the linear approximation and nothing else. That a holomorphic function is continuous is proved on this page rather than assumed, and the two theorems that recover holomorphy from separate holomorphy — under continuity, and under local boundedness — are theorems precisely because those properties are not part of the definition. Defining holomorphy by local power-series representability or by the C1 Cauchy–Riemann system, as some treatments do, would make one or other of them a tautology.

At m=1 this is the published one-variable notion. A C-linear L:CC satisfies L(h)=hL(1), so with c=L(1) the condition reads f(a+h)=f(a)+ch+r(h) with r(h)/h0, which is exactly complex differentiability at a with f(a)=c in the sense of Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions; conversely that condition produces the C-linear map hf(a)h.

Relation to the real total derivative. Reading f as a map R2mR2 through the dictionary, the displayed condition is the total-differentiability condition of The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(h2) remainder with the extra requirement that the approximating linear map be C-linear and not merely R-linear. So a complex differentiable f is real totally differentiable with Df(a) as its real total derivative, and The total derivative at a point is unique says the two uses of the notation cannot disagree. The standard basis vectors ek of The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0 are the ones used to read off coordinates of L.

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Separately holomorphic functions

Definition

Fix m1, let UCm be open and let f:UC. For aU and k<m write

Ua,k:={ζC:(a0,,ak1,ζ,ak+1,,am1)U}

and let fa,k:Ua,kC be the kth slice fa,k(ζ)=f(a0,,ak1,ζ,ak+1,,am1).

The function f is separately holomorphic on U when for every aU and every k<m the slice fa,k is holomorphic on Ua,k in the one-variable sense of Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions.

Each Ua,k is open: if ζUa,k, the corresponding point of U has a ball B(,ρ)U by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, and changing only the kth coordinate by less than ρ moves the point by less than ρ in the norm of Complex m-space and its real coordinate dictionary, so the disc {ξ:ξζ<ρ} lies in Ua,k (Open ball, closed ball and sphere in a metric space).

Remarks

No continuity in the remaining variables is asked. The condition constrains each slice separately and says nothing about how the slices fit together: a separately holomorphic function is not assumed continuous as a function on U, and on this page the two theorems that supply joint regularity — Osgood's lemma under continuity, and the locally bounded theorem — are what close that gap.

The slice through a point of a polydisc is a disc. If U=Δr(a) is a polydisc (Balls, polydiscs and the distinguished boundary in Cm) and aU, then Ua,k is the disc {ζ:ζak<rk}, which is what lets the one-variable theory be applied one coordinate at a time with the others held fixed.

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Wirtinger operators in Cm

Definition

Fix m1, let UCm be open and let f:UC. Read Cm as R2m through Complex m-space and its real coordinate dictionary, with real coordinates xk,yk for k<m given by zk=xk+iyk (Real and imaginary parts, complex conjugation, and modulus), and let xkf and ykf be the partial derivatives of Directional derivatives and partial derivatives of a map URmRn applied to the two real components of f and recombined.

At a point where all 2m of these partial derivatives exist, define the Wirtinger operators

zkf:=12(xkfiykf),zˉkf:=12(xkf+iykf)(k<m).

The differential identity. Suppose in addition that f is real totally differentiable at a point aU, so that Df(a) is the R-linear map with Df(a)h=k<m((xkf(a))ξk+(ykf(a))ηk) for hk=ξk+iηk, by A total derivative computes every directional derivative, and its matrix is the Jacobian read in the standard basis (The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0). Substituting ξk=12(hk+hk) and ηk=12i(hkhk) and collecting the coefficients of hk and hk using finite sums in the additive commutative monoid of C (A finite sum in a commutative monoid indexed by an arbitrary finite set) and distributivity in the complex field (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)) gives

Df(a)h=k<m((zkf(a))hk+(zˉkf(a))hk).

Indeed the coefficient of hk is 12xkf(a)+12iykf(a)=12(xkf(a)iykf(a)), and the coefficient of hk is 12xkf(a)12iykf(a)=12(xkf(a)+iykf(a)).

Remarks

At m=1 these are the published Wirtinger derivatives. The two displayed formulas are literally those of The Wirtinger derivatives zf and zˉf, and antiholomorphic functions with x0,y0 written x,y, and the differential identity reduces to the identity Df(h)=(zf)h+(zˉf)h recorded there.

These are operators on real-differentiable functions, not on holomorphic ones. Nothing above assumes any complex differentiability: the definition needs only the 2m real partial derivatives, and the differential identity needs only real total differentiability. Which functions have all zˉkf=0 is the question the next lemma and the Cauchy–Riemann characterisation answer.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes

Statement

Fix m1 and let T:CmC be R-linear, that is additive with T(λh)=λT(h) for every real λ. Then there are unique c0,,cm1 and d0,,dm1 in C with

T(h)=k<mckhk+k<mdkhk(hCm),

namely ck=12(T(ek)iT(iek)) and dk=12(T(ek)+iT(iek)). Moreover T is C-linear if and only if dk=0 for every k<m.

Facts & Assumptions

Given: An R-linear T:CmC, with Cm read through Complex m-space and its real coordinate dictionary.

[L1]

A map between Euclidean spaces is linear when it preserves real linear combinations (A linear map L:RmRn in Euclidean coordinates); C-linear additionally requires T(λh)=λT(h) for every complex λ (Holomorphic functions on an open subset of Cm).

[L3]

For z=a+bi with a,b real, Rez=a, Imz=b and z=abi (Real and imaginary parts, complex conjugation, and modulus).

[L5]

The Wirtinger operators of a real totally differentiable f satisfy Df(a)h=k<m(zkf(a))hk+k<m(zˉkf(a))hk (Wirtinger operators in Cm).

Proof

technique · direct
1.1

Write hk=ξk+iηk with ξk,ηk real, as in [L3]. By [L2] and R-linearity, h=k<m(ξkek+ηk(iek)) and hence T(h)=k<m(ξkT(ek)+ηkT(iek)).

givenL1L2L3algebra
1.2

Put ck=12(T(ek)iT(iek)) and dk=12(T(ek)+iT(iek)); then ck+dk=T(ek) and i(ckdk)=T(iek).

givenalgebra
2.1

Substituting ξk=12(hk+hk) and ηk=12i(hkhk) from [L3] into step 1.1 and collecting, the coefficient of hk is 12T(ek)+12iT(iek)=ck and the coefficient of hk is 12T(ek)12iT(iek)=dk, so T(h)=k<mckhk+k<mdkhk.

step 1.1step 1.2L3algebra
2.2

The coefficients are unique: if kckhk+kdkhk represents T as well, evaluating at h=ek gives ck+dk=T(ek) and at h=iek gives i(ckdk)=T(iek), a system whose only solution is the pair of step 1.2.

step 1.2L2L3algebra
3.1

If every dk=0 then T(h)=kckhk, which satisfies T(λh)=λT(h) for every complex λ, so T is C-linear in the sense of [L1].

step 2.1L1algebra
3.2

Conversely, suppose T is C-linear. Taking λ=i in [L1] and using step 2.1 gives kck(ihk)+kdkihk=ikckhk+ikdkhk; since ihk=ihk by [L3], the left side is ikckhkikdkhk, so 2ikdkhk=0 for every h. Evaluating at h=ek gives dk=0 for each k<m.

step 2.1L1L3algebra
4.1

Steps 2.1, 2.2, 3.1 and 3.2 prove the representation, its uniqueness, and the stated equivalence; the case m=1 and the zero functional, for which every ck and dk vanishes, are included with no separate argument. By [L5] the representation applied to T=Df(a) has ck=zkf(a) and dk=zˉkf(a), so the criterion reads: the real differential is C-linear exactly when every zˉkf(a) vanishes.

step 2.1step 2.2step 3.1step 3.2L5
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A holomorphic function of several variables is continuous and separately holomorphic

Statement

Let UCm be open and let f:UC be holomorphic (Holomorphic functions on an open subset of Cm). Then f is continuous on U and separately holomorphic on U (Separately holomorphic functions). Moreover, for aU and k<m the slice fa,k is complex differentiable at ak with derivative Df(a)ek, and

Df(a)ek=zkf(a),Df(a)h=k<m(zkf(a))hk.

Facts & Assumptions

Given: An open UCm and a holomorphic f:UC; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

f is complex differentiable at a when there is a C-linear L with f(a+h)=f(a)+L(h)+r(h) and r(h)/h0; that L is unique and written Df(a) (Holomorphic functions on an open subset of Cm).

[L2]

f is separately holomorphic when every slice fa,k is holomorphic on the open set Ua,k in the one-variable sense (Separately holomorphic functions, Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L3]

An R-linear T:CmC has a unique representation T(h)=k<mckhk+k<mdkhk, and T is C-linear exactly when every dk=0; for T=Df(a) at a point of real total differentiability, ck=zkf(a) and dk=zˉkf(a) (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in Cm).

[L4]

For every linear L:RmRn there is K0 with Lh2Kh2 for every h (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0).

[L5]

A complex differentiable function of one variable is continuous (Complex differentiability at a point implies continuity there).

[L7]

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).

[L8]

Continuity of a map into Rn from a subset of a metric space is the usual εδ condition with the Euclidean norm (Vector-valued functions f:ARm, their limits and continuity, with the dictionary to the metric notions).

Proof

technique · direct
1.1

Fix aU and write L=Df(a) as in [L1]. Since L is C-linear it is in particular R-linear, so [L4] read through the dictionary gives K0 with L(h)Kh for every h; alternatively [L3] and [L6] give L(h)=k<mckhk with ck=L(ek), and [L7] bounds L(h) by (k<mck)h because hkh.

givenL1L3L4L6L7
1.2

The remainder satisfies r(h)/h0 by [L1], so there is δ>0 with r(h)h whenever 0<h<δ and a+hU.

givenL1
1.3

Fix k<m, let aU and let ζUa,k. The point a obtained from a by replacing its kth coordinate by ζ lies in U and agrees with a off the kth coordinate, so Ua,k=Ua,k and fa,k=fa,k.

givenL2
2.1

Combining steps 1.1 and 1.2, f(a+h)f(a)L(h)+r(h)(K+1)h for such h, which tends to 0 with h; by [L8] this is continuity of f at a, and aU was arbitrary.

step 1.1step 1.2L7L8
2.2

With a as in step 1.3 and h=(ξζ)ek for ξ near ζ, [L1] and [L6] give fa,k(ξ)fa,k(ζ)=Df(a)ek(ξζ)+r(h), and h=ξζ by the dictionary, so r(h)/ξζ0. Hence fa,k is complex differentiable at ζ with derivative Df(a)ek; as ζUa,k was arbitrary, the slice is holomorphic on Ua,k and f is separately holomorphic by [L2].

step 1.3L1L2L6
3.1

By [L3] applied to the C-linear Df(a), every dk vanishes and Df(a)h=k<mckhk with ck=Df(a)ek=zkf(a); step 2.2 at ζ=ak identifies that number with the derivative of the slice, and [L5] confirms the slice is continuous, consistently with step 2.1.

step 2.1step 2.2L3L5L6
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Multi-indexed power series in Cm and their absolute convergence

Definition

Fix m1 and read Cm through Complex m-space and its real coordinate dictionary. A multi-index is αNm, with α=k<mαk and α!=k<mαk! as in Ck maps and multi-index derivative notation in Euclidean space, every index running over k<m from 0. For wCm the complex monomial is

wα:=k<mwkαk,

a finite product in the multiplicative commutative monoid of C (The product g0g1gn1 of a finite list in a monoid, by recursion, with the empty product (n=0) equal to the identity, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)) of natural powers in C (Integer powers in the complex field); for the zero multi-index w0=1.

Enumerating the index set. N is countable and Every finite power of an at most countable set is at most countable makes Nm at most countable; it is infinite, so there is a bijection σ:NNm (Injection, surjection, bijection, Finite, countably infinite, countable, uncountable).

Let c:NmC and a,zCm. The multi-indexed power series αcα(za)α converges absolutely at z when the complex series ncσ(n)(za)σ(n) converges absolutely (Complex series, absolute convergence, complex power series, and radius of convergence) for one bijection σ, equivalently for every one. The two conditions agree, and the sums agree, because for bijections σ,τ the series along τ is a rearrangement of the series along σ: applying Every absolutely convergent complex series converges, and rearrangements preserve its sum to the nonnegative series of moduli transfers convergence, and applying it again to the series itself transfers the sum. That common value is written αcα(za)α and no other notion of unordered sum is introduced.

Box partial sums. For NN put BN:={αNm:αkN for every k<m}, a finite set, and let SN(z):=αBNcα(za)α be the corresponding finite sum (A finite sum in a commutative monoid indexed by an arbitrary finite set). If the series converges absolutely at z with sum S, then SN(z)S. Given ε>0, absolute convergence supplies n0 with nn0cσ(n)(za)σ(n)<ε and n<n0cσ(n)(za)σ(n)Sε; taking N large enough that BN contains σ(0),,σ(n01), every index of BN outside that finite list is σ(n) for some nn0, so SN(z)S2ε. The same argument bounds SN(z) and the tail of the series by the corresponding tails of the series of moduli.

The series converges absolutely and uniformly on a set SCm when there are reals Mα0 with cα(za)αMα for every zS and every α, and with nMσ(n) convergent. By Weierstrass M-test for complex-valued function series the partial sums along σ then converge uniformly on S (Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary) and the series converges absolutely at every point of S.

Remarks

Why a bijection is fixed rather than an unordered sum defined. The library already has one theory of complex series and one rearrangement theorem, and the clause above uses exactly those. Introducing a separate notion of summation over Nm would create a second convergence notion that every later statement would have to be matched against; instead every multi-indexed sum below means the sum of the one-variable series along any enumeration, which the rearrangement theorem makes unambiguous.

Where the series live. The natural regions here are the polydiscs of Balls, polydiscs and the distinguished boundary in Cm rather than balls. If every zkak is positive, absolute convergence at z controls the series on the closed polydisc with that polyradius. If some coordinate is zero, the same coordinatewise domination holds on the corresponding degenerate product set, but that radius vector is not called a polyradius. This is exactly the shape the kernel expansion and the Cauchy estimates on this page produce.

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

Statement

Fix m1, a point aCm, a polyradius r and a real θ with 0θ<1. For ζΓr(a) and zΔθr(a), that is ζkak=rk and zkakθrk for every k<m,

k<m1ζkzk=α(za)αk<m(ζkak)αk1,

the multi-indexed series converging absolutely (Multi-indexed power series in Cm and their absolute convergence). Each term is dominated by

Mα:=k<mθαkrk,αMα=k<m1rk(1θ),

independently of ζ and z, so the convergence is absolute and uniform in the pair (ζ,z) over Γr(a)×Δθr(a).

Facts & Assumptions

Given: m1, aCm, a polyradius r, a real θ with 0θ<1, and points ζΓr(a), zΔθr(a); Cm 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:NNm converges absolutely, its sum is then independent of σ, and its box partial sums over BN={α:αkN} converge to that sum (Multi-indexed power series in Cm and their absolute convergence).

[L2]

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

[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 fn(x)Mn on a set X with Mn convergent, then fn(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).

[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).

[L9]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

For k<m put uk=(zkak)/(ζkak), legitimate by [L2] and [L5] since ζkak=rk>0; then ukθ<1 by [L2] and [L9], and ζkzk=(ζkak)(1uk) with 1uk1θ>0 by [L9].

givenL2L5L9
1.2

Each term satisfies (za)αk<m(ζkak)αk1=k<m(zkakαkrkαk1)k<mθαk/rk=Mα, by [L2], [L8] and [L9].

givenL2L8L9
2.1

For NN the box sum of the majorants factors as αBNMα=k<m(rk1jNθj) by [L8], which is at most k<m(rk(1θ))1 by [L6]. Every finite subset of Nm lies in some BN, so along any bijection σ the partial sums of nMσ(n) are bounded by that number, and [L7] makes the series convergent; letting N in the factored identity and using [L6] gives αMα=k<m(rk(1θ))1.

step 1.2L6L7L8
2.2

The box partial sum factors: by [L8], αBN(za)αk<m(ζkak)αk1=k<m(1ζkakjNukj)=k<m1ukN+1(ζkak)(1uk), the last equality by the finite geometric identity (1u)jNuj=1uN+1 and step 1.1.

step 1.1L5L8algebra
3.1

By step 1.2 and step 2.1 the hypotheses of [L4] hold with the constants Mα on the set Γr(a)×Δθr(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.

step 1.2step 2.1L1L3L4
3.2

By step 1.1 the target value is k<m((ζkak)(1uk))1, so the difference from the box sum of step 2.2 has modulus at most k<m(rk(1θ))11k<m(1ukN+1) by [L9]; since ukN+1θN+1, [L9] and [L8] bound the second factor by (1+θN+1)m1, which tends to 0 as N by [L6].

step 1.1step 2.2L6L8L9
4.1

Hence the box partial sums converge to k<m(ζkzk)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.

step 3.1step 3.2
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The iterated Cauchy integral formula on a polydisc

Statement

Fix m1, a point aCm and a polyradius ρ. Let f:Δρ(a)C be continuous and separately holomorphic (Separately holomorphic functions), let r be a polyradius with rk<ρk for every k<m, and let Ck(t)=ak+rkexp(it) on [0,2π]. Then for every zΔr(a)

f(z)=1(2πi)mC0 ⁣ ⁣Cm1f(ζ0,,ζm1)k<m(ζkzk)dζm1dζ0.

The right-hand side is an iterated integral: the innermost integral is taken over ζm1 with ζ0,,ζm2 held fixed, then over ζm2, and so on. Each successive integrand is continuous on the circle it is integrated over, so each of the m integrals exists. No integral over the distinguished boundary is formed and the order of integration is never interchanged.

Facts & Assumptions

Given: m1, aCm, polyradii ρ and r with rk<ρk, a continuous separately holomorphic f:Δρ(a)C, the circles Ck, and zΔr(a); Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

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

[L2]

f is separately holomorphic when for every bU and k<m the slice ζf(b0,,bk1,ζ,bk+1,,bm1) is holomorphic on the open set of ζ for which the point lies in U (Separately holomorphic functions).

[L3]

If f is holomorphic on D(a,R), 0<r<R, za<r and γ(t)=a+rexp(it) on [0,2π], then f(z)=(2πi)1γf(ζ)(ζz)1dζ (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

[L5]

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

[L6]

For aC, r>0 and kZ, the contour a+rexp(ikt) on [0,2π] is a closed complex contour whose trace for k0 is {a=r} (A circle traversed k times has winding number k inside and 0 outside).

[L7]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

[L8]

Nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives), such functions are continuous (Complex differentiability at a point implies continuity there), and composites of continuous maps are continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

Proof

technique · direct
1.1

By [L6] each Ck is a closed complex contour with trace the circle {ζak=rk}. If ζjaj=rj for j<p and zjaj<rj for jp, then ζjaj<ρj and zjaj<ρj by the hypothesis rj<ρj, so every such mixed point lies in Δρ(a) by [L1].

givenL1L6
1.2

For the fixed point z, define Hm(ζ0,,ζm1;z):=f(ζ0,,ζm1). Then, for p=m1,m2,,0, define Hp(ζ0,,ζp1;z) by Hp:=12πiCpHp+1(ζ0,,ζp1,ζp;z)ζpzpdζp, whenever the integrand is continuous on Cp. By construction H0(;z) is exactly the iterated integral in the statement, divided by (2πi)m.

given
2.1

Claim, proved by induction on q=mp using [L5]: for every p with 0pm the quantity Hp is defined and Hp(ζ0,,ζp1;z)=f(ζ0,,ζp1,zp,,zm1). For q=0, that is p=m, this is the definition of Hm.

step 1.1step 1.2L5
2.2

The slice ξf(ζ0,,ζp1,ξ,zp+1,,zm1) is holomorphic on the disc ξap<ρp: by step 1.1 the corresponding point lies in Δρ(a) for every such ξ, and by [L1] and [L2] that disc is exactly the slice domain, on which separate holomorphy makes the slice holomorphic.

step 1.1L1L2
3.1

Assume the claim for p+1. Fix ζ0,,ζp1 on their circles. By the assumption, Hp+1(ζ0,,ζp1,ζp;z)=f(ζ0,,ζp,zp+1,,zm1), which by step 1.1 is a continuous function of ζp on the circle ζpap=rp; dividing by ζpzp, which is nonzero there because zpap<rp by [L1] and [L7], leaves a continuous integrand by [L8], so the integral defining Hp exists by [L4].

step 1.1step 2.1L1L4L7L8
4.1

Applying [L3] to the slice of step 2.2, with R=ρp, r=rp and z=zp, gives f(ζ0,,ζp1,zp,zp+1,,zm1)=12πiCpf(ζ0,,ζp1,ζp,zp+1,)ζpzpdζp, which by step 3.1 is Hp(ζ0,,ζp1;z). This is the claim for p, so the induction of step 2.1 closes.

step 3.1step 2.2L3
5.1

Taking p=0 in step 2.1 gives H0(;z)=f(z), and step 1.2 identifies H0(;z) with the iterated integral divided by (2πi)m; every one of the m integrals exists by step 3.1. Since zΔr(a) was arbitrary, the formula holds throughout Δr(a).

step 1.2step 2.1step 3.1step 4.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

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
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An absolutely convergent multi-indexed power series is holomorphic and differentiates termwise

Statement

Fix m1, aCm, a polyradius r, a real M0 and coefficients c:NmC with

cαMk<mrkαkfor every multi-index α.

Then:

  1. for every θ with 0<θ<1 the series αcα(za)α converges absolutely and uniformly on Δθr(a), so its sum g is defined on Δr(a);
  2. g is holomorphic on Δr(a), with Dg(z)h=k<mbk(z)hk,bk(z)=ααkcα(za)αek, the kth series running over the multi-indices with αk1 and converging absolutely on Δr(a); equivalently zkg=bk;
  3. for every θ with 0<θ<1 the derived coefficients αkcα, re-indexed as a power series, obey a bound of the same shape on the polyradius θr, so the differentiation may be iterated; and every iterated complex partial derivative zβg:=z0β0zm1βm1g exists on Δr(a) with zβg(a)=β!cβ.

Facts & Assumptions

Given: The data above; Cm is read through Complex m-space and its real coordinate dictionary and polydiscs are those of Balls, polydiscs and the distinguished boundary in Cm.

[L1]

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

[L2]

f is complex differentiable at z when there is a C-linear L with f(z+h)=f(z)+L(h)+r(h) and r(h)/h0; L is unique and written Df(z) (Holomorphic functions on an open subset of Cm).

[L3]

An R-linear T with T(h)=k<mckhk is C-linear, and for a differentiable f the coefficients are zkf (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in Cm).

[L4]

An absolutely convergent complex series converges and every rearrangement has the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum).

[L5]

If fn(x)Mn on a set with Mn convergent, then fn converges absolutely and uniformly there (Weierstrass M-test for complex-valued function series).

[L6]

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); for real r with r<1, krk=1/(1r) (For r<1, k0rk=1/(1r), and for r1 the series diverges); if 0akbk eventually and bk converges then ak converges (If 0akbk eventually, convergence of bk gives convergence of ak, and divergence of ak gives divergence of bk).

[L7]

For p>0 and rational α>0 the sequence kα/(1+p)k tends to 0, the numerator being the corresponding power of the canonical natural (For every p>0 and every positive rational α, nα/(1+p)n0).

[L8]

If f is holomorphic on D(a,R), 0<r<R and fK on the circle ζa=r, then f(n)(a)n!K/rn for every natural n (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L9]

(z+w)n=kn(nk)zkwnk for complex z,w and natural n, the binomial coefficients read as complex numbers (The binomial theorem over the complex field).

[L10]

Linear combinations and products of functions complex differentiable at a point are complex differentiable there with the usual formulas; constants have derivative 0 and the identity derivative 1 (Linearity, product, reciprocal, and quotient rules for complex derivatives); such functions are continuous (Complex differentiability at a point implies continuity there).

[L11]

Multi-indices satisfy α=k<mαk and α!=k<mαk! (Ck maps and multi-index derivative notation in Euclidean space), with 0!=1 and (j+1)!=j!(j+1) (The factorial n! and the falling factorial nk, defined by recursion in N).

[L12]

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

[L13]

Natural powers satisfy w0=1 and wj+1=wjw; negative integer powers need a nonzero base (Integer powers in the complex field).

[L14]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive); finite sums and products satisfy the additivity, scaling and product laws (Laws of finite sums and finite products).

Proof

technique · direct
1.1

Fix θ with 0<θ<1. For zΔθr(a) the hypothesis and [L14] give cα(za)αMk<mθαk; the box sums of the right side are Mk<mjNθjM(1θ)m by [L14] and [L6], and every finite subset of Nm lies in a box, so [L6] makes the majorant series convergent and [L1] and [L5] give absolute and uniform convergence on Δθr(a). Since each zΔr(a) lies in some such closed polydisc, g is defined on Δr(a). This is claim 1.

givenL1L5L6L14
1.2

For 0<θ<θ1<1 and every k<m the series ααkθα1 converges: by [L7] the sequence j(θ/θ1)j1 is null, hence bounded by some Kθ, so jθj1Kθθ1j1 and [L6] with [L14] bounds the box sums of ααkθα1 by Kθ(1θ1)1(1θ)(m1); [L6] then gives convergence.

givenL6L7L14
1.3

Fix θ with 0<θ<1, a point z0Δθr(a) and θ with θ<θ<1; write w=z0a, so wkθrk. For hCm with h0 put η=maxk<mhk/rk and assume η(θθ)/2, which holds for all small h because hkh; then z0+hΔθr(a)Δr(a) by [L14].

givenL14
2.1

For each α let ϕα(τ)=k<m(wk+τhk)αk, a polynomial in τ of degree at most α by [L9] and [L13], hence entire, with ϕα(τ)=jαϕα,jτj and ϕα,j=ϕα(j)(0)/j! by [L10] and [L11]. In particular ϕα,0=wα and, by the product rule of [L10] and an induction on the number of factors ([L12]), ϕα,1=k<mαkwαekhk, terms with αk=0 being 0.

step 1.3L9L10L11L12L13
2.2

The series bk(z0)=ααkcαwαek converges absolutely: αkcαwαekMrk1αkθα1 by the hypothesis and [L14], and step 1.2 makes that majorant summable.

step 1.2step 1.3L1L14
3.1

Put T=(θθ)/η, so T2 by step 1.3. For τT and every k, wk+τhkθrk+Thkθrk by [L14], so ϕα(τ)k<m(θrk)αk there. Applying [L8] to ϕα on the disc of radius T gives ϕα,jk<m(θrk)αkTj.

step 1.3step 2.1L8L14
4.1

Hence Rα:=ϕα(1)ϕα,0ϕα,1=2jαϕα,j satisfies Rαk<m(θrk)αkj2Tj2T2k<m(θrk)αk, using T2 and [L6]. With T2=η2(θθ)2 this is 2η2(θθ)2k<m(θrk)αk.

step 3.1L6L14
5.1

Summing against the coefficients, the hypothesis on cα gives αcαRα2η2M(θθ)2αk<mθαk=2η2M(θθ)2(1θ)m by [L6] and [L14]; since ηh/mink<mrk, this is at most a constant times h2.

step 1.3step 4.1L6L14
6.1

By steps 2.1, 5.1 and 2.2, and by [L1] and [L4] which allow the absolutely convergent series to be split term by term, g(z0+h)g(z0)k<mbk(z0)hk=αcαRα, whose modulus is O(h2) and therefore o(h). The map hk<mbk(z0)hk is C-linear by [L3] and [L15], so [L2] makes g complex differentiable at z0 with that differential, and zkg(z0)=bk(z0) by [L3]. As θ<1 and z0 were arbitrary, this is claim 2.

step 2.1step 5.1step 2.2L1L2L3L4L15
7.1

For claim 3 fix k<m and θ with 0<θ<1, and re-index the derived series by β=αek, so its coefficient at β is (βk+1)cβ+ek, of modulus at most M(βk+1)θβ+1l<m(θrl)βlrk1 by the hypothesis and [L14]. By [L7] the numbers (βk+1)θβ+1 are bounded by a constant M, so the derived coefficients satisfy a bound of the same shape with polyradius θr and constant MM/rk.

step 6.1L7L14
8.1

Iterating step 7.1 and step 6.1, an induction on β ([L12]) shows that every zβg exists on Δr(a) and is the termwise β-fold derived series, whose coefficient at αβ is (k<mαk(αk1)(αkβk+1))cα and which vanishes unless αβ componentwise. Evaluating at z=a, [L13] kills every monomial (za)αβ except the one with α=β, whose coefficient is β!cβ by [L11]; so zβg(a)=β!cβ.

step 6.1step 7.1L11L12L13L14
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Osgood's lemma: continuous and separately holomorphic implies holomorphic

Statement

Let m1, let UCm be open and let f:UC be continuous and separately holomorphic (Separately holomorphic functions). Then f is holomorphic on U (Holomorphic functions on an open subset of Cm).

Consequently, for a continuous f on an open U the following three conditions are equivalent: f is holomorphic; f is separately holomorphic; every point of U has a polydisc neighbourhood on which f is the sum of an absolutely convergent multi-indexed power series.

Facts & Assumptions

Given: An open UCm and a continuous separately holomorphic f:UC; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

For f continuous and separately holomorphic on Δρ(a) and a polyradius r with rk<ρk, the iterated-integral coefficients satisfy cαMk<mrkαk with M=supΓr(a)f, and f(z)=αcα(za)α on Δr(a), absolutely and uniformly on every Δθr(a) with θ<1 (A continuous separately holomorphic function is the sum of an absolutely convergent power series with Cauchy-integral coefficients on every smaller polydisc).

[L2]

If cαMk<mrkαk for every α, then αcα(za)α converges absolutely on Δr(a) and its sum is holomorphic there (An absolutely convergent multi-indexed power series is holomorphic and differentiates termwise).

[L3]

A holomorphic function of several variables is continuous and separately holomorphic (A holomorphic function of several variables is continuous and separately holomorphic).

[L4]

Holomorphic on U means complex differentiable at every point of U (Holomorphic functions on an open subset of Cm), and separate holomorphy is a condition on the slices through each point (Separately holomorphic functions).

[L5]

Δr(a) is defined coordinatewise by zkak<rk (Balls, polydiscs and the distinguished boundary in Cm), and a multi-indexed power series and its absolute convergence are those of Multi-indexed power series in Cm and their absolute convergence.

[L6]

A set is open exactly when each of its points admits a ball inside it, and B(x,ε)={y:d(x,y)<ε} (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L7]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

Fix aU. By [L6] there is ε>0 with B(a,ε)U; put ρk=ε/(2m) for every k<m. If zΔρ(a) then za2=k<mzkak2<mρ02=ε2/4 by [L5] and the dictionary, so Δρ(a)B(a,ε)U.

givenL5L6L7
1.2

The restriction of f to Δρ(a) is continuous, and it is separately holomorphic there: for bΔρ(a) and k<m the slice domain inside Δρ(a) is an open subset of the slice domain inside U, on which the slice is holomorphic by hypothesis, and a restriction of a holomorphic function of one variable to an open subset is holomorphic.

givenL4L5
2.1

Put rk=ρk/2, so rk<ρk. By [L1] applied on Δρ(a) there are coefficients cα with cαMk<mrkαk and f(z)=αcα(za)α for every zΔr(a).

step 1.1step 1.2L1L5
3.1

By [L2] the sum of that series is holomorphic on Δr(a); by step 2.1 it is f there, so f is complex differentiable at every point of Δr(a), in particular at a. Since aU was arbitrary, [L4] makes f holomorphic on U.

step 2.1L2L4
4.1

For the equivalence, let f be continuous on the open U. If f is holomorphic then it is separately holomorphic by [L3]; if it is separately holomorphic then step 2.1 gives the local power-series representation and step 3.1 gives holomorphy; and if it is locally such a sum then [L2] makes it holomorphic on a polydisc about each point, hence on U by [L4]. So the three conditions are equivalent.

step 2.1step 3.1L2L3L4
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Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic

Statement

Let m1, let UCm be open and let f:UC be holomorphic. Then:

  1. every point aU has a polydisc Δr(a)U on which f(z)=αcα(za)α with the series absolutely convergent, and the coefficients are cα=zαf(a)α!,zα:=z0α0zm1αm1;
  2. every iterated complex partial derivative zαf exists and is holomorphic on U;
  3. xkf=zkf and ykf=izkf for every k<m, and f is of class Cn in the real coordinates for every natural n, hence smooth.

Facts & Assumptions

Given: An open UCm and a holomorphic f:UC; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

A continuous separately holomorphic function on an open set is holomorphic, and for a continuous function holomorphy, separate holomorphy and local power-series representability agree (Osgood's lemma: continuous and separately holomorphic implies holomorphic).

[L2]

For f continuous and separately holomorphic on Δρ(a) and rk<ρk, there are coefficients with cαMk<mrkαk and f=αcα(za)α on Δr(a) (A continuous separately holomorphic function is the sum of an absolutely convergent power series with Cauchy-integral coefficients on every smaller polydisc).

[L3]

Under such a coefficient bound the sum is holomorphic on Δr(a), differentiates termwise with zk, the derived series obeys a bound of the same shape on every smaller polyradius, every iterated zβ of the sum exists, and zβ(sum)(a)=β!cβ (An absolutely convergent multi-indexed power series is holomorphic and differentiates termwise).

[L4]

A holomorphic function of several variables is continuous and separately holomorphic, with Df(a)h=k<m(zkf(a))hk (A holomorphic function of several variables is continuous and separately holomorphic).

[L5]

zkf=12(xkfiykf) and zˉkf=12(xkf+iykf) (Wirtinger operators in Cm).

[L6]

An R-linear T has the unique representation T(h)=kckhk+kdkhk and is C-linear exactly when every dk=0; for a real totally differentiable f these coefficients are zkf and zˉkf (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes).

[L7]

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

[L8]

Complex differentiability at a gives real total differentiability at a with the same differential (Holomorphic functions on an open subset of Cm).

[L9]

f is of class Cn on an open subset of R2m when every iterated coordinate partial derivative of order at most n exists and is continuous (Ck maps and multi-index derivative notation in Euclidean space), and α!=k<mαk! with 0!=1 (The factorial n! and the falling factorial nk, defined by recursion in N).

[L10]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L11]

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

Proof

technique · direct
1.1

By [L4] the function f is continuous and separately holomorphic on U, so the construction inside the proof of [L1] gives, at each aU, a polydisc Δρ(a)U and then rk=ρk/2 for which [L2] supplies coefficients with cαMk<mrkαk and f=αcα(za)α on Δr(a).

givenL1L2L4L11
1.2

By [L6] and [L8] the C-linearity of Df(a) makes every zˉkf(a) vanish, so [L5] gives xkf=zkf+zˉkf=zkf and ykf=i(zkfzˉkf)=izkf at every point of U.

givenL4L5L6L8
2.1

By [L3] applied to that series, f differentiates termwise on Δr(a), every iterated zαf exists there, and zαf(a)=α!cα; dividing by α!0 ([L9]) gives claim 1.

step 1.1L3L9
3.1

By [L3] the derived series for zkf again obeys a bound of the same shape on a smaller polyradius, so its sum is holomorphic there; since that sum is zkf by step 2.1, each zkf is holomorphic on a polydisc about every point of U, hence holomorphic on U. An induction on α ([L7]) repeats this for every iterated derivative, giving claim 2.

step 1.1step 2.1L3L7
4.1

By step 1.2 each first-order real partial derivative of f is zkf or izkf, which step 3.1 makes holomorphic and [L10] makes continuous; applying step 1.2 to those functions in turn, an induction on the order ([L7]) shows every iterated real coordinate partial derivative of f exists and is continuous on U. Taking real and imaginary parts, which are continuous together with f, [L9] makes f of class Cn for every natural n, which is claim 3.

step 1.2step 3.1L7L9L10
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The coefficients of a convergent multi-indexed power series are its derivative coefficients, hence unique

Statement

Let m1, aCm and let r be a polyradius. Suppose c,c:NmC both satisfy a bound cαMk<mrkαk and cαMk<mrkαk, and suppose

αcα(za)α=αcα(za)αfor every zΔr(a).

Then cα=cα for every multi-index α. In particular a function has at most one such power-series representation about a given centre, and its coefficients are zαf(a)/α!.

Facts & Assumptions

Given: Coefficient families c,c with the stated bounds whose sums agree on Δr(a).

[L1]

Under a bound cαMk<mrkαk the series converges absolutely on Δr(a), its sum is holomorphic there, every iterated complex partial derivative of the sum exists, and zα(sum)(a)=α!cα (An absolutely convergent multi-indexed power series is holomorphic and differentiates termwise).

[L2]

A holomorphic function is locally the sum of an absolutely convergent power series whose coefficients are zαf(a)/α!, and every iterated complex partial derivative is holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

[L4]

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

Proof

technique · direct
1.1

Let f be the common sum on Δr(a). By [L1] applied to c, the function f is holomorphic on Δr(a), every zαf exists there, and zαf(a)=α!cα.

givenL1L3L4
1.2

By [L1] applied to c, the same function f satisfies zαf(a)=α!cα; the derivatives are those of the single function f and so do not depend on which series it is written as.

givenL1L3L4
2.1

Comparing steps 1.1 and 1.2 gives α!cα=α!cα, and α!0 by [L3], so cα=cα for every α.

step 1.1step 1.2L3
3.1

Consequently a holomorphic f has at most one power-series representation about a subject to such a bound, and by [L2] the one it has is the derivative series αzαf(a)(za)α/α!; this is what licenses the definite article in "the coefficients of f at a".

step 2.1L2
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Cauchy estimates for mixed derivatives on a polydisc

Statement

Let m1, let aCm, let ρ be a polyradius and let f:Δρ(a)C be holomorphic. Let r be a polyradius with rk<ρk for every k<m, and put M=supΓr(a)f, the supremum over the distinguished boundary only. Then for every multi-index α

zαf(a)  α!Mk<mrkαk.

The bound uses no value of f outside Γr(a), which for m2 is a proper subset of the topological boundary of the closed polydisc.

Facts & Assumptions

Given: A holomorphic f on Δρ(a) and a polyradius r with rk<ρk; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

For f continuous and separately holomorphic on Δρ(a) and rk<ρk, the iterated-integral coefficients satisfy cαMk<mrkαk with M=supΓr(a)f, and f=αcα(za)α on Δr(a) (A continuous separately holomorphic function is the sum of an absolutely convergent power series with Cauchy-integral coefficients on every smaller polydisc).

[L2]

Every iterated complex partial derivative of a holomorphic function exists and is holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic). For coefficient families satisfying the geometric polyradius bound, the power-series representation about a fixed centre is unique and its coefficients are cα=zαf(a)/α! (The coefficients of a convergent multi-indexed power series are its derivative coefficients, hence unique).

[L3]

A holomorphic function of several variables is continuous and separately holomorphic (A holomorphic function of several variables is continuous and separately holomorphic).

[L4]

If gK on the trace of a rectifiable contour γ, then γgdzKL(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length), and the once-traversed circle of radius r>0 has length 2πr (Every circle has circumference 2 pi r and circumference-to-diameter ratio pi).

[L5]

For f holomorphic on D(a,R), 0<r<R and fK on the circle ζa=r, one has f(n)(a)n!K/rn (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L6]

Γr(a) is the set of points with zkak=rk for every k<m, and for m2 it is a proper subset of the topological boundary of Δr(a) (Balls, polydiscs and the distinguished boundary in Cm).

Proof

technique · direct
1.1

By [L3] the function f is continuous and separately holomorphic on Δρ(a), so [L1] applies with the given r and produces coefficients cα with cαMk<mrkαk, the constant M being the supremum of f on Γr(a) alone. That bound is what the m-fold application of the ML estimate of [L4] on the m circles of radius rk produces, one factor 2πrk cancelling each factor (2π)1, and it specialises at m=1 to the published one-variable inequality of [L5].

givenL1L3L4L5L6
2.1

By [L2] those same coefficients are cα=zαf(a)/α!, so multiplying the bound of step 1.1 by α! gives zαf(a)α!Mk<mrkαk, as claimed; α!0 by [L7] makes the division legitimate.

step 1.1L2L7
3.1

No value of f off Γr(a) entered: the constant M of step 1.1 is a supremum over that set, and by [L6] it is for m2 a proper subset of the topological boundary of the closed polydisc.

step 1.1step 2.1L6
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A bounded separately holomorphic function on a polydisc is Lipschitz on every smaller polydisc

Statement

Let m1, let aCm, let ρ be a polyradius, let f:Δρ(a)C be separately holomorphic (Separately holomorphic functions) with fM throughout Δρ(a), and let 0<θ<1. Then for all z,wΔθρ(a)

f(w)f(z)  M1θk<mwkzkρk  M1θ(k<m1ρk)wz.

In particular f is Lipschitz, hence continuous, on Δθρ(a). No continuity of f in the remaining variables is assumed at any point of the argument.

Facts & Assumptions

Given: A separately holomorphic f on Δρ(a) with fM there, and 0<θ<1; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

f is separately holomorphic when for every point of the open set and every k<m the kth slice is holomorphic on the open set of ζ for which the point lies in the domain (Separately holomorphic functions).

[L2]

Δr(a), Δr(a) and Γr(a) are defined coordinatewise by zkak<rk, rk and =rk, and polydiscs are convex (Balls, polydiscs and the distinguished boundary in Cm, A convex subset of Rm contains every line segment between two of its points).

[L3]

For g holomorphic on D(a,R), 0<r<R and gK on the circle ζa=r, one has g(n)(a)n!K/rn (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L4]

For an open convex VC, a holomorphic g on V and z,wV, g(w)g(z)=(wz)01g(z+t(wz))dt (On a convex open set the difference quotient is an average of the derivative along the segment).

[L5]

For an integrable F:[a,b]Rn with ab, abF2abF2 (For ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 is integrable); vector-valued integrals are componentwise and real-linear (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); and abhabH when hH pointwise (If fg on [a,b] and both are integrable then abfabg; and m(ba)abfM(ba)).

[L6]

A holomorphic g=u+iv on an open subset of C has (u,v) of class Ck for every natural k, hence smooth (Holomorphic functions are real analytic and smooth in their two real coordinates), and every holomorphic function has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle); in particular g is holomorphic and therefore continuous (Complex differentiability at a point implies continuity there).

[L8]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive), and B(x,s)={y:d(x,y)<s} (Open ball, closed ball and sphere in a metric space).

[L9]

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

[L10]

A nonempty set of reals bounded below has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

Proof

technique · direct
1.1

Fix z,wΔθρ(a) and define points v(0),,v(m) by v(0)=z and, for k<m, letting v(k+1) agree with v(k) except that its kth coordinate is wk; this is a finite recursion and v(m)=w. Every coordinate of every v(k) is a coordinate of z or of w, so vj(k)ajθρj for every j<m and each v(k) lies in Δθρ(a)Δρ(a) by [L2].

givenL2L9
2.1

By [L7] the difference telescopes: f(w)f(z)=k<m(f(v(k+1))f(v(k))).

step 1.1L7
2.2

Fix k<m and let gk(ξ) be the value of f at the point agreeing with v(k) except in its kth coordinate, which is ξ. By step 1.1 and [L2] that point lies in Δρ(a) whenever ξak<ρk, so [L1] makes gk holomorphic on the disc D(ak,ρk), and gkM there.

step 1.1L1L2
3.1

Let ξakθρk and let 0<s<(1θ)ρk. For ζξ<(1θ)ρk one has ζak<ρk by [L8], so gk is holomorphic on D(ξ,(1θ)ρk) and bounded by M on the circle ζξ=s; [L3] with n=1 gives gk(ξ)M/s. The set of such s is nonempty and the bound holds for each, so taking the infimum over s(0,(1θ)ρk) by [L10] gives gk(ξ)M/((1θ)ρk).

step 2.2L3L8L10
4.1

The disc D(ak,ρk) is convex by [L2] and [L8], and zk,wk lie in the closed disc of radius θρk about ak by step 1.1, so the whole segment between them satisfies ξakθρk by [L8]. By [L4], [L6] and [L5], gk(wk)gk(zk)wkzk01gk(zk+t(wkzk))dtwkzkM(1θ)ρk, using step 3.1 on the segment.

step 1.1step 3.1L2L4L5L6L8
5.1

Since f(v(k+1))f(v(k))=gk(wk)gk(zk) by the definition of gk in step 2.2, summing step 4.1 over k<m and using step 2.1 and [L7] gives f(w)f(z)M1θk<mwkzk/ρk; each wkzkwz by the dictionary, which yields the second displayed bound and makes f Lipschitz on Δθρ(a). Only the slices of f and the uniform bound M were used, never continuity of f in the remaining variables.

step 2.1step 2.2step 4.1L7L8
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Locally bounded and separately holomorphic implies holomorphic

Statement

Let m1, let UCm be open and let f:UC be separately holomorphic and locally bounded: every point of U has a neighbourhood on which f is bounded. Then f is continuous on U and holomorphic on U.

The local-boundedness hypothesis is used, and it is not shown here to be removable: this page carries no theorem that separate holomorphy alone implies holomorphy, and none of its results is applied as if it did.

Facts & Assumptions

Given: An open UCm and a separately holomorphic, locally bounded f:UC; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

If f is separately holomorphic on Δρ(a) with fM there and 0<θ<1, then f(w)f(z)M1θ(k<mρk1)wz for z,wΔθρ(a), so f is Lipschitz there (A bounded separately holomorphic function on a polydisc is Lipschitz on every smaller polydisc).

[L2]

A continuous separately holomorphic function on an open subset of Cm is holomorphic (Osgood's lemma: continuous and separately holomorphic implies holomorphic).

[L3]

Separate holomorphy is a condition on the slices through each point of the domain (Separately holomorphic functions), and holomorphic means complex differentiable at every point (Holomorphic functions on an open subset of Cm).

[L4]

A function whose restrictions to the members of an open cover are continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L5]

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

Proof

technique · direct
1.1

Fix aU. Local boundedness and [L6] give ε>0 and M0 with B(a,ε)U and fM on B(a,ε), after intersecting the bounding neighbourhood with a ball inside U. Put ρk=ε/(2m); then Δρ(a)B(a,ε)U, since za2=k<mzkak2<mρ02=ε2/4 for zΔρ(a) by [L5] and the dictionary.

givenL5L6L7
2.1

The restriction of f to Δρ(a) is separately holomorphic by [L3], since a slice domain inside Δρ(a) is an open subset of the corresponding slice domain inside U and a restriction of a one-variable holomorphic function to an open subset is holomorphic; and fM there by step 1.1.

step 1.1L3L5
3.1

Applying [L1] with θ=12, the function f is Lipschitz, hence continuous, on Δρ/2(a), and in particular on the open set Δρ/2(a), which contains a and is open by [L5] and [L6].

step 1.1step 2.1L1L5L6
4.1

The sets Δρ/2(a) obtained in step 3.1 as a ranges over U form an open cover of U on each member of which f is continuous, so f is continuous on U by [L4].

step 3.1L4L6
5.1

By step 4.1 the function f is continuous on U and separately holomorphic by hypothesis, so [L2] makes it holomorphic on U. The bound M entered only through step 1.1 and [L1]; nothing above removes it, and this page proves no statement that would.

givenstep 4.1L1L2L3
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For C1 functions, holomorphy, complex linearity of the real derivative, and the Cauchy–Riemann system agree

Statement

Let m1, let UCm be open and let f:UC be of class C1 in the real coordinates (Ck maps and multi-index derivative notation in Euclidean space, applied to the real and imaginary parts of f on U read as an open subset of R2m). Let aU. The following are equivalent.

  1. f is complex differentiable at a.
  2. f is real totally differentiable at a and its real total derivative Df(a) is C-linear.
  3. f is real totally differentiable at a and zˉkf(a)=0 for every k<m — the several-variable Cauchy–Riemann system.

The C1 hypothesis is used only to pass from the Cauchy–Riemann system to real total differentiability; the implications from 1 to 2 and between 2 and 3 hold at any point with no regularity beyond what each condition states.

Facts & Assumptions

Given: An open UCm, a C1 function f:UC and a point aU; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

f is complex differentiable at a when there is a C-linear L with f(a+h)=f(a)+L(h)+r(h) and r(h)/h0; that condition is the total-differentiability condition with the extra requirement that L be C-linear (Holomorphic functions on an open subset of Cm).

[L2]

f is totally differentiable at a when there is an R-linear L with f(a+h)f(a)Lh/h0 (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(h2) remainder), and such an L is unique (The total derivative at a point is unique).

[L3]

At a point where all 2m real partial derivatives exist, zkf=12(xkfiykf) and zˉkf=12(xkf+iykf); at a point of real total differentiability, Df(a)h=k<m(zkf(a))hk+k<m(zˉkf(a))hk (Wirtinger operators in Cm, Directional derivatives and partial derivatives of a map URmRn).

[L4]

An R-linear T has a unique representation T(h)=kckhk+kdkhk, and T is C-linear exactly when every dk=0 (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes).

[L5]

If every partial derivative of f exists near a and is continuous at a, then f is totally differentiable at a (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L6]

f is of class C1 when its first-order coordinate partial derivatives exist and are continuous (Ck maps and multi-index derivative notation in Euclidean space).

[L7]

For one complex variable, complex differentiability at a, real total differentiability with Df(a) multiplication by a complex number, and real total differentiability with zˉf(a)=0 are equivalent, and then f(a)=zf(a) (Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with zˉf=0, or with the Cauchy–Riemann equations).

[L8]

A holomorphic function of several variables is continuous and separately holomorphic with Df(a)h=k<m(zkf(a))hk (A holomorphic function of several variables is continuous and separately holomorphic), and it is smooth in the real coordinates (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

Proof

technique · direct
1.1

Condition 1 implies condition 2. If f is complex differentiable at a with C-linear L, then L is in particular R-linear and the same remainder condition is the one in [L2] read through the dictionary, so f is real totally differentiable at a and, by the uniqueness in [L2], Df(a)=L is C-linear.

givenL1L2
1.2

Condition 2 implies condition 1. If f is real totally differentiable at a with C-linear Df(a), the remainder condition of [L2] is exactly that of [L1] for the C-linear map Df(a), so f is complex differentiable at a.

givenL1L2
1.3

Conditions 2 and 3 are equivalent. Real total differentiability is common to both, and given it, [L3] represents Df(a) in the form of [L4] with ck=zkf(a) and dk=zˉkf(a); by the uniqueness in [L4] the map Df(a) is C-linear exactly when every zˉkf(a) vanishes.

givenL3L4
1.4

The C1 hypothesis enters only here: it makes the first-order real partial derivatives exist near a and be continuous by [L6], so [L5] supplies the real total differentiability that conditions 2 and 3 name. Without it, the Cauchy–Riemann system alone constrains the partial derivatives and asserts nothing about the existence of Df(a).

givenL5L6
2.1

Steps 1.1, 1.2, 1.3 and 1.4 give the three-way equivalence for a C1 function. At m=1 the statement is [L7], with Df(a) being C-linear exactly when it is multiplication by a complex number, namely zf(a); and by [L8] a holomorphic function of several variables is automatically C1, so the C1 hypothesis restricts only the direction that starts from the Cauchy–Riemann system.

step 1.1step 1.2step 1.3step 1.4L7L8
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Sums, products and nonvanishing quotients of holomorphic functions are holomorphic

Statement

Let m1, let UCm be open and let f,g:UC be holomorphic. Then λf+μg is holomorphic on U for all λ,μC, fg is holomorphic on U, and f/g is holomorphic on the open set {zU:g(z)0}, with

D(λf+μg)(a)=λDf(a)+μDg(a),D(fg)(a)=f(a)Dg(a)+g(a)Df(a), D(f/g)(a)=g(a)Df(a)f(a)Dg(a)g(a)2(g(a)0),

and correspondingly zk(fg)=fzkg+gzkf and zk(f/g)=(gzkffzkg)/g2 for each k<m. In particular the holomorphic functions on U form a commutative ring under pointwise operations, containing the constants.

Facts & Assumptions

Given: An open UCm and holomorphic f,g:UC; Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

f is complex differentiable at a when there is a C-linear L with f(a+h)=f(a)+L(h)+r(h) and r(h)/h0; L is unique and written Df(a) (Holomorphic functions on an open subset of Cm).

[L2]

A holomorphic function of several variables is continuous, and Df(a)h=k<m(zkf(a))hk (A holomorphic function of several variables is continuous and separately holomorphic).

[L3]

A map T(h)=k<mckhk is C-linear, and for a differentiable f the coefficients are zkf (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in Cm).

[L4]

For every linear L:RmRn there is K0 with Lh2Kh2 for every h (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0).

[L5]

Linear combinations, products and nonvanishing quotients of complex numbers obey the field laws (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)), and the one-variable derivative rules take the displayed forms (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L7]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

Fix aU and write f(a+h)=f(a)+Df(a)h+rf(h) and g(a+h)=g(a)+Dg(a)h+rg(h) as in [L1], with rf(h),rg(h)=o(h); by [L4] read through the dictionary there is K0 with Df(a)hKh and Dg(a)hKh.

givenL1L4
2.1

For λ,μC the map hλDf(a)h+μDg(a)h is C-linear by [L3] and [L8], and the remainder of λf+μg at a is λrf(h)+μrg(h), which is o(h) by [L7]; so [L1] makes λf+μg complex differentiable at a with the stated differential.

step 1.1L1L3L7L8
2.2

Multiplying the two expansions of step 1.1 and collecting, f(a+h)g(a+h)=f(a)g(a)+(f(a)Dg(a)h+g(a)Df(a)h)+ϱ(h), where ϱ(h)=Df(a)hDg(a)h+(f(a)+Df(a)h)rg(h)+(g(a)+Dg(a)h+rg(h))rf(h). By step 1.1 and [L7] the first term is at most K2h2 and the others are bounded quantities times o(h), so ϱ(h)=o(h); the first-order part is C-linear by [L3], so [L1] gives the product rule.

step 1.1L1L3L5L7
2.3

Suppose g(a)0. By [L2] the function g is continuous, so [L6] gives a ball B about a inside U on which gg(a)/2>0; in particular {zU:g(z)0} is open by [L6]. On B write 1g(a+h)1g(a)=g(a)g(a+h)g(a)g(a+h)=Dg(a)hrg(h)g(a)g(a+h); using g(a+h)g(a) this equals Dg(a)hg(a)2+o(h) by step 1.1 and [L7]. So 1/g is complex differentiable at a with differential hDg(a)h/g(a)2, which is C-linear by [L3].

step 1.1L1L2L3L5L6L7
3.1

Combining steps 2.2 and 2.3 gives the quotient rule for f/g at every point where g does not vanish, and reading each differential at h=ek with [L2], [L3] and [L8] gives the displayed formulas for zk.

step 2.2step 2.3L2L3L5L8
4.1

Steps 2.1 and 2.2 make the holomorphic functions on U closed under pointwise addition and multiplication; those operations are commutative, associative and distributive because the values lie in the field C ([L5]), and every constant function is holomorphic with zero differential by [L1]. So the holomorphic functions on U form a commutative ring containing the constants.

step 2.1step 2.2L1L5
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Locally uniform limits of holomorphic functions are holomorphic, with locally uniform convergence of all derivatives

Statement

Let m1, let UCm be open, let each fn:UC be holomorphic, and suppose fnf locally uniformly on U: every point of U has a neighbourhood on which the convergence is uniform. Then f is holomorphic on U, and for every multi-index α

zαfnzαf

locally uniformly on U.

Facts & Assumptions

[L1]

A continuous separately holomorphic function on an open subset of Cm is holomorphic (Osgood's lemma: continuous and separately holomorphic implies holomorphic).

[L2]

For g holomorphic on Δρ(a) and a polyradius r with rk<ρk, zαg(a)α!supΓr(a)gk<mrkαk (Cauchy estimates for mixed derivatives on a polydisc).

[L3]

If holomorphic functions of one variable converge locally uniformly on an open subset of C, the limit is holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[L4]

A uniform limit of continuous complex-valued functions on a metric space is continuous (A uniform limit of continuous complex-valued functions is continuous).

[L5]

Separate holomorphy is holomorphy of each slice on its open slice domain (Separately holomorphic functions).

[L6]

A holomorphic function of several variables is continuous and separately holomorphic (A holomorphic function of several variables is continuous and separately holomorphic); every iterated complex partial derivative of a holomorphic function is holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic); differences of holomorphic functions are holomorphic (Sums, products and nonvanishing quotients of holomorphic functions are holomorphic).

[L7]

Δr(a), Δr(a) and Γr(a) are defined coordinatewise (Balls, polydiscs and the distinguished boundary in Cm); multi-index notation is that of Ck maps and multi-index derivative notation in Euclidean space.

Proof

technique · direct
1.1

Each fn is continuous by [L6], and locally uniform convergence makes f continuous at every point by [L4] applied on a neighbourhood where the convergence is uniform.

givenL4L6
1.2

Fix aU and k<m, and let V be the kth slice domain of U through a, an open subset of C by [L5]. The slices of the fn are holomorphic on V by [L6], and they converge to the slice of f locally uniformly on V, since a neighbourhood in U of a point of the slice meets the slice in a neighbourhood there. So [L3] makes the slice of f holomorphic on V, and f is separately holomorphic by [L5].

givenL3L5L6L8
2.1

By steps 1.1 and 1.2 the limit f is continuous and separately holomorphic on U, so [L1] makes it holomorphic.

step 1.1step 1.2L1
3.1

Fix aU and a multi-index α. By [L8] choose ε>0 with the ball B(a,ε)U and put ρk=ε/(2m), so Δρ(a)U by [L7] and [L9], and put rk=ρk/2. Shrinking ε if necessary, the convergence fnf is uniform on Δρ(a).

step 2.1L7L8L9
4.1

Fix bΔr(a) and put σk:=ρkbkak for each k<m. Then σk>rk because bkak<rk=ρk/2, and if ζkbk<σk then ζkakζkbk+bkak<ρk, so Δσ(b)Δρ(a)U by [L7] and [L9]. Also Γr(b)Δρ(a) because bkak<rk and rk+rk=ρk. The difference fnf is holomorphic on U by [L6] and step 2.1, so [L2] applied on Δσ(b) with inner polyradius r gives zαfn(b)zαf(b)α!(supΔρ(a)fnf)k<mrkαk.

step 3.1L2L6L7L9
5.1

The right-hand side of step 4.1 does not depend on b and tends to 0 by the uniform convergence of step 3.1, so zαfnzαf uniformly on the neighbourhood Δr(a) of a. Since aU and α were arbitrary, the convergence is locally uniform for every multi-index.

step 3.1step 4.1L6L7
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The modulus of a holomorphic function on a closed polydisc is bounded by its supremum on the distinguished boundary

Statement

Let m1, let aCm, let r be a polyradius, and let f be continuous on the closed polydisc Δr(a) and holomorphic on Δr(a). Then

supΔr(a)f=supΓr(a)f,

and both suprema are attained. For m2 the bounding set Γr(a) is a proper subset of the topological boundary of Δr(a), so this is stronger than the bound by the topological boundary.

Facts & Assumptions

Given: f continuous on Δr(a) and holomorphic on Δr(a); Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

If Ω is a bounded complex domain and g is continuous on Ω and holomorphic on Ω, then there is ζΩ with g(z)g(ζ) for every zΩ (Boundary maximum modulus principle on a bounded domain).

[L2]

Δr(a), Δr(a) and Γr(a) are defined coordinatewise by zkak<rk, rk and =rk; polydiscs are convex; and for m2 the distinguished boundary is a proper subset of the topological boundary (Balls, polydiscs and the distinguished boundary in Cm, A convex subset of Rm contains every line segment between two of its points).

[L3]

A holomorphic function of several variables is continuous and separately holomorphic (A holomorphic function of several variables is continuous and separately holomorphic), separate holomorphy being holomorphy of each slice on its open slice domain (Separately holomorphic functions).

[L4]

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

[L6]

A continuous map from a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L7]

A complex domain is a nonempty, connected, open subset of C (A complex domain is a nonempty connected open subset of C).

[L8]

A set is open exactly when each of its points admits a ball inside it, a set is closed when its complement is open, and B(x,s)={y:d(x,y)<s} (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L9]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

[L10]

Continuity of a map into Rn from a subset of a metric space is the usual εδ condition with the Euclidean norm (Vector-valued functions f:ARm, their limits and continuity, with the dictionary to the metric notions).

Proof

technique · direct
1.1

The sets Δr(a) and Γr(a) are closed and bounded by [L2] and [L9], hence compact and nonempty by [L5], so f attains a maximum on each by [L5] and both suprema are attained real numbers; write P=supΓr(a)f. By [L6] the function f is uniformly continuous on Δr(a).

givenL2L5L6L9
2.1

For a polyradius s with sk<rk for every k, put Ap:=sup{f(w):wΔs(a) and wkak=sk for every k<p} for 0pm; these are attained maxima by the argument of step 1.1 applied to the corresponding closed bounded sets, and Am=supΓs(a)f.

step 1.1L2L5
2.2

Let 0<t<1 and take s=tr. Every wΓtr(a) satisfies a+(wa)/tΓr(a) and (a+(wa)/t)w=(1/t1)wa by [L9], which tends to 0 as t1 uniformly in w because wa is bounded on Δr(a); so by the uniform continuity of step 1.1 and [L10], supΓtr(a)fP+ε once t is close enough to 1, for any prescribed ε>0.

step 1.1L9L10
3.1

Fix such an s and p<m, and let wΔs(a) have wkak=sk for k<p. Replacing the pth coordinate of w by any ξ with ξap<rp leaves the point in Δr(a), because the other coordinates satisfy wkaksk<rk; so by [L3] the slice g(ξ) is holomorphic on the disc D(ap,rp) and in particular continuous on the closed disc D(ap,sp), which is the closure of D(ap,sp) since every point of the circle is a limit of interior points along its radius and the closed disc is closed by [L8] and [L9].

step 2.1L2L3L8L9
4.1

The disc D(ap,sp) is a bounded complex domain by [L2], [L7] and [L8], being nonempty, open, convex hence connected, and bounded, and its topological boundary is the circle ξap=sp by step 3.1. So [L1] applied to the slice gives ξ on that circle with f(w)g(ξ), and the point obtained from w by putting ξ in the pth slot lies in Δs(a) with its first p+1 coordinates on their circles; hence f(w)Ap+1. Taking the supremum over such w gives ApAp+1.

step 2.1step 3.1L1L2L7L8
5.1

By [L4] the chain of step 4.1 gives A0Am, that is supΔs(a)fsupΓs(a)f for every polyradius s with sk<rk.

step 2.1step 4.1L4
6.1

Let zΔr(a) and ε>0. For t<1 the point a+t(za) lies in Δtr(a)Δtr(a) by [L2] and [L9], so steps 5.1 and 2.2 give f(a+t(za))supΓtr(a)fP+ε for t close enough to 1; letting t1 and using the continuity of f at z gives f(z)P+ε, hence f(z)P.

step 5.1step 2.2L2L9L10
7.1

Step 6.1 gives supΔr(a)fP, and the reverse inequality holds because Γr(a)Δr(a) by [L2]; so the two suprema are equal and attained by step 1.1. For m2 the set Γr(a) is by [L2] a proper subset of the topological boundary, so the bound is by a strictly smaller set than in the one-variable statement.

step 1.1step 6.1L2
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Holomorphic maps CmCn and the complex Jacobian matrix

Definition

Fix m,n1, read Cm and Cn through Complex m-space and its real coordinate dictionary, and let UCm be open with aU. A map L:CmCn is C-linear when L(u+v)=L(u)+L(v) and L(λu)=λL(u) for all u,vCm and all λC, the operations being those of the C-vector spaces Cm and Cn (Vector space over a field, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)); requiring the second clause only for real λ gives the weaker notion of an R-linear map (A linear map L:RmRn in Euclidean coordinates).

A map F:UCn is holomorphic at a when there is a C-linear L:CmCn with

F(a+h)=F(a)+L(h)+r(h),r(h)h0  as h0,

the norms being those of the dictionary (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) and the quotient considered for h0 with a+hU. It is holomorphic on U when it is holomorphic at every point.

Such an L is unique. If L1,L2 both work, then T=L1L2 is C-linear with T(h)/h0; fixing h0 and replacing h by th for small real t>0 gives T(h)/h=T(th)/th0, so T(h)=0. Write DF(a):=L.

The complex Jacobian JCF(a) is the matrix of DF(a) relative to the standard ordered bases of Cm and Cn (The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0, Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases), an n×m matrix over the field C (Finite rectangular matrices over a commutative ring, their entries, rows and columns); its (j,k) entry is the jth coordinate of DF(a)ek.

Remarks

n=1 recovers the scalar definition. For n=1 the norm on C1 is the modulus and the displayed condition is that of Holomorphic functions on an open subset of Cm; the Jacobian is then the single row of the coordinates of DF(a).

The entries are the Wirtinger derivatives of the components. A map into Cn is holomorphic exactly when each of its components is shows that F is holomorphic exactly when each component Fj is, and then (JCF(a))jk=zkFj(a) with the operators of Wirtinger operators in Cm. That identification is proved there and is not assumed here: this definition fixes the Jacobian as the matrix of the differential and nothing more.

The target dimension n=0. C0 has exactly one element, so every map into it is holomorphic with zero differential and empty Jacobian; nothing below needs that case and it is recorded only so that the convention is not left open.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A map into Cn is holomorphic exactly when each of its components is

Statement

Let m,n1, let UCm be open, let aU and let F:UCn with components Fj=πjF for j<n. Then F is holomorphic at a (Holomorphic maps CmCn and the complex Jacobian matrix) if and only if every Fj is holomorphic at a (Holomorphic functions on an open subset of Cm), and in that case

DF(a)h=(DF0(a)h,,DFn1(a)h),(JCF(a))jk=zkFj(a).

For n=1 this is the scalar definition read back.

Facts & Assumptions

Given: An open UCm, aU and F:UCn with components Fj; the spaces are read through Complex m-space and its real coordinate dictionary.

[L1]

F is holomorphic at a when there is a C-linear L:CmCn with F(a+h)=F(a)+L(h)+r(h) and r(h)/h0; L is unique and its matrix in the standard bases is JCF(a), whose (j,k) entry is the jth coordinate of L(ek) (Holomorphic maps CmCn and the complex Jacobian matrix, Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

[L2]

The scalar case is the same condition with r(h)/h0 and L a C-linear functional (Holomorphic functions on an open subset of Cm).

[L4]

Under the interleaved real-coordinate identification fixed in the Given, wj2=(Rewj)2+(Imwj)2, so the Euclidean norm is w=(j<nwj2)1/2 (The p-norms xp for rational p1, and x, The Euclidean inner product x,y=k<nxkyk on Rn).

[L6]

Finite sums in the additive commutative monoid of C may be regrouped termwise, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)).

[L7]

If a scalar function is complex differentiable at a, its complex-linear real differential has the form Dg(a)h=k<m(zkg(a))hk (A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in Cm).

[L8]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

By [L4] and [L6], every wCn satisfies wjw for each j<n and wj<nwj, the first because wj2 is one term of a sum of nonnegative terms and the second because the square of the right-hand side dominates that sum.

L4L6L8
2.1

Suppose F is holomorphic at a with L=DF(a) as in [L1]. For each j<n the map πjL is C-linear, being a coordinate of a C-linear map, and Fj(a+h)=Fj(a)+(πjL)(h)+rj(h) with rj(h)r(h) by step 1.1; so rj(h)/h0 and [L2] makes Fj holomorphic at a with DFj(a)=πjL.

givenstep 1.1L1L2
2.2

Conversely, suppose every Fj is holomorphic at a with Lj=DFj(a), and set L(h)=(L0(h),,Ln1(h)). Then L is C-linear because each coordinate is and the operations on Cn are coordinatewise, and the remainder of F has r(h)j<nrj(h) by step 1.1; each summand is o(h) and there are finitely many, so [L6] makes the sum o(h) and [L1] makes F holomorphic at a with DF(a)=L.

givenstep 1.1L1L2L6
3.1

In either direction DF(a)h=(DF0(a)h,,DFn1(a)h) by steps 2.1 and 2.2 and the uniqueness in [L1]. Evaluating at h=ek and reading the jth coordinate, [L1], [L5] and [L7] give (JCF(a))jk=DFj(a)ek=zkFj(a). For n=1 the two conditions of [L1] and [L2] coincide.

step 2.1step 2.2L1L2L5L7
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The composite of holomorphic maps is holomorphic and its complex Jacobian is the product

Statement

Let m,n,p1, let UCm and VCn be open, let F:UCn have F(U)V and be holomorphic at aU, and let G:VCp be holomorphic at F(a). Then GF is holomorphic at a with

D(GF)(a)=DG(F(a))DF(a),JC(GF)(a)=JCG(F(a))JCF(a).

Facts & Assumptions

Given: Open sets UCm and VCn, a map F:UV holomorphic at a, and G:VCp holomorphic at F(a); the spaces are read through Complex m-space and its real coordinate dictionary.

[L1]

F is holomorphic at a when there is a C-linear L with F(a+h)=F(a)+L(h)+r(h) and r(h)/h0; L is unique, written DF(a), and JCF(a) is its matrix in the standard bases (Holomorphic maps CmCn and the complex Jacobian matrix, Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases, Finite rectangular matrices over a commutative ring, their entries, rows and columns).

[L2]

A map into Cn is holomorphic exactly when each component is, with DF(a)h the tuple of the component differentials (A map into Cn is holomorphic exactly when each of its components is).

[L3]

A holomorphic function of several variables is continuous (A holomorphic function of several variables is continuous and separately holomorphic).

[L4]

[ST]BD=[S]CD[T]BC for linear T:UV and S:VW with ordered bases B,C,D ([ST]BD=[S]CD[T]BC).

[L5]

For every linear L:RmRn there is K0 with Lh2Kh2 (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0), the notion of linear map being that of A linear map L:RmRn in Euclidean coordinates.

[L6]

If f is totally differentiable at a and g at f(a), then gf is totally differentiable at a with D(gf)(a)=Dg(f(a))Df(a) (The chain rule for total derivatives: D(gf)(a)=Dg(f(a))Df(a)).

Proof

technique · direct
1.1

Write F(a+h)=F(a)+DF(a)h+rF(h) and, for κCn small, G(F(a)+κ)=G(F(a))+DG(F(a))κ+rG(κ) as in [L1], with rF(h)=o(h), rG(κ)=o(κ) and rG(0)=0. By [L5], read through the dictionary, there are K,K0 with DF(a)hKh and DG(F(a))κKκ.

givenL1L5
2.1

Put κ(h)=DF(a)h+rF(h), so F(a+h)=F(a)+κ(h); by step 1.1 and [L8] there is δ>0 with κ(h)(K+1)h whenever h<δ and a+hU, and F(a+h)V because F(U)V.

step 1.1L8L9
3.1

Substituting, G(F(a+h))=G(F(a))+DG(F(a))DF(a)h+ϱ(h) with ϱ(h)=DG(F(a))rF(h)+rG(κ(h)). By step 1.1 the first summand has norm at most KrF(h)=o(h); by step 2.1 the second has norm o(κ(h)) with κ(h)(K+1)h, hence o(h), the value at h with κ(h)=0 being 0. So ϱ(h)=o(h) by [L8].

step 1.1step 2.1L8
4.1

The composite DG(F(a))DF(a) is C-linear, being a composite of C-linear maps, so step 3.1 and [L1] make GF holomorphic at a with D(GF)(a)=DG(F(a))DF(a); this agrees with the real chain rule of [L6] read through the dictionary, by the uniqueness in [L1].

step 3.1L1L6
5.1

Taking matrices in the standard bases, [L4] turns step 4.1 into JC(GF)(a)=JCG(F(a))JCF(a), the entries being read off at the basis vectors by [L2], [L3] and [L7].

step 4.1L1L2L3L4L7
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The complex Jacobian determinant of a composite of equidimensional holomorphic maps is the product

Statement

Let m1, let U,VCm be open, let F:UV be holomorphic at aU and let G:VCm be holomorphic at F(a). Then JCF(a), JCG(F(a)) and JC(GF)(a) are m×m matrices over C and

detJC(GF)(a)=detJCG(F(a))detJCF(a).

Consequently, if F is holomorphic on U with a holomorphic two-sided inverse F1:VU, then detJCF(a)0 for every aU.

Facts & Assumptions

Given: Equidimensional holomorphic maps F and G as above.

[L1]

For holomorphic F at a and G at F(a), the composite is holomorphic at a and JC(GF)(a)=JCG(F(a))JCF(a) (The composite of holomorphic maps is holomorphic and its complex Jacobian is the product).

[L2]

JCF(a) is the matrix of the C-linear differential in the standard bases, an n×m matrix over C (Holomorphic maps CmCn and the complex Jacobian matrix, Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

[L5]

If AMn(R) is invertible over a commutative ring then det(A) is a unit, with inverse det(A1) (An invertible square matrix over a commutative ring has unit determinant).

Proof

technique · direct
1.1

Since the source and target dimensions are all m, [L2] makes each of the three Jacobians an m×m matrix over C, which is a commutative ring by [L4].

givenL2L4
2.1

By [L1] the composite Jacobian is the matrix product JCG(F(a))JCF(a), so [L3] applied over C gives detJC(GF)(a)=detJCG(F(a))detJCF(a).

step 1.1L1L3L4
3.1

If F has a holomorphic two-sided inverse F1, applying step 2.1 to G=F1 gives detJCF1(F(a))detJCF(a)=detJC(id)(a)=1, since the identity map is holomorphic with identity differential by [L2]; so detJCF(a) is a unit of the field C, in particular nonzero, as [L5] also records.

step 2.1L2L4L5
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically

Statement

Let m1, let UCm be a nonempty connected open set, and let f:UC be holomorphic (Holomorphic functions on an open subset of Cm). If there is a nonempty open set WU such that f(z)=0 for every zW, then f0 on U.

This is the several-variable identity theorem at the strength the page supports: the hypothesis is a nonempty open set of zeros. An accumulation point of the zero set is neither assumed nor sufficient in several variables; the companion page records that stronger one-variable statement as false here.

Facts & Assumptions

Given: A nonempty connected open set UCm, a holomorphic function f:UC, and a nonempty open set WU on which f=0.

[L1]

Holomorphic functions of several variables are smooth; every point aU has a polydisc Δr(a)U on which f(z)=αcα(za)α,cα=zαf(a)α!; and all mixed complex derivatives are holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

[L3]

The mixed complex derivative notation zα and the zero-order identity z0f=f are those in the power-series and smoothness statement [L1]; Ck maps and multi-index derivative notation in Euclidean space supplies the underlying multi-index arithmetic.

[L4]

Polydiscs in Cm are the coordinatewise discs of Balls, polydiscs and the distinguished boundary in Cm.

Proof

technique · direct
1.1

For every multi-index α, the derivative zαf is holomorphic and therefore continuous on U by [L1]; since f=0 on the open set W, every derivative of f is also 0 on W, so the set A:={zU:zαf(z)=0 for every α} contains W and is therefore nonempty.

givenL1L3
2.1

The set A is closed in U, because it is the intersection over all multi-indices α of the closed zero sets of the continuous functions zαf.

step 1.1
2.2

The set A is open in U: if aA, choose a smaller polydisc ΔU centred at a; then every coefficient αf(a)/α! in the power-series expansion of f on Δ is 0, so [L1] gives f=0 on Δ, and hence every derivative vanishes on Δ as well, which means ΔA.

step 1.1L1L4
3.1

The set A is a nonempty subset of U that is both open in U and closed in U, so connectedness and [L2] force A=U; in particular f=z0f vanishes at every point of U, hence f0 on U.

step 2.1step 2.2L2L3

Remarks

  • Why the hypothesis is open-set vanishing and not an accumulation point. In one complex variable, accumulation of zeros implies equality by local factorisation and isolated zeros. In several variables the zero set of a nonzero holomorphic function can contain whole complex hypersurfaces, so the open-set hypothesis is the honest form at this stage.

  • What the proof really uses. The proof needs only two page-level tools: holomorphic smoothness and the local power-series expansion. Once every derivative at one point vanishes, the power series on a smaller polydisc is identically zero, and connectedness propagates that local vanishing to the whole set.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The holomorphic functions on a domain in Cm have no zero divisors

Statement

Let m1 and let UCm be a nonempty connected open set. Under pointwise addition and multiplication, the holomorphic functions UC form an integral domain: if f,g:UC are holomorphic and fg0, then f0 or g0.

Facts & Assumptions

Given: A nonempty connected open set UCm and holomorphic functions f,g:UC.

[L1]

A holomorphic function vanishing on a nonempty open subset of a connected open set in Cm vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L2]

Sums, products and nonvanishing quotients of holomorphic functions are holomorphic (Sums, products and nonvanishing quotients of holomorphic functions are holomorphic).

[L3]

A holomorphic function of several variables is continuous (A holomorphic function of several variables is continuous and separately holomorphic).

[L4]

An integral domain is a commutative ring with 10 and no zero divisors (Zero divisor, and integral domain: a commutative ring with 10 and no zero divisors).

Proof

technique · direct
1.1

By [L2], the holomorphic functions on U are closed under pointwise addition and multiplication, and pointwise operations are commutative and associative because they are so in C; the constant functions 0 and 1 are holomorphic, and 10, so this is a nonzero commutative ring.

givenL2
1.2

Suppose fg0 and f≢0. Since f is continuous by [L3], the set V:={zU:f(z)0} is open in U; it is nonempty because f is not identically zero; and for every zV the equality f(z)g(z)=0 in C forces g(z)=0, so g vanishes on the nonempty open set V.

givenL3
2.1

Apply [L1] to g and the open set V: then g0 on U. So fg0 implies f0 or g0, and with step 1.1 this is exactly the zero-divisor clause of [L4]; therefore the ring of holomorphic functions on U is an integral domain.

step 1.1step 1.2L1L4

Remarks

  • Connectedness matters. On a disconnected open set, a function may vanish on one component and not on another, so the product of two nonzero holomorphic functions can be zero. The corollary is therefore genuinely about domains, not arbitrary open sets.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

An interior local maximum of the modulus forces a scalar holomorphic function to be constant

Statement

Let m1, let UCm be a nonempty connected open set, and let f:UC be holomorphic. Suppose there are aU and an open ball BU centred at a such that

f(z)f(a)(zB).

Then f is constant on U.

Facts & Assumptions

Given: A nonempty connected open set UCm, a holomorphic function f:UC, a point aU, and an open ball BU centred at a such that f(z)f(a) for every zB.

[L1]

The composite of holomorphic maps is holomorphic and its complex Jacobian is the product (The composite of holomorphic maps is holomorphic and its complex Jacobian is the product).

[L2]

If the modulus of a holomorphic function on a complex domain has an interior local maximum, then the function is constant (Local maximum modulus principle).

[L3]

A holomorphic function vanishing on a nonempty open subset of a connected open set in Cm vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L4]

Balls in Cm are the Euclidean balls of Balls, polydiscs and the distinguished boundary in Cm.

[L5]

Sums, products and nonvanishing quotients of holomorphic functions are holomorphic (Sums, products and nonvanishing quotients of holomorphic functions are holomorphic).

Proof

technique · direct
1.1

Because B is an open ball centred at a, there is r>0 such that {zCm:za<r}B in the Euclidean norm of [L4].

givenL4
2.1

Fix a vector vCm with v=1, and define gv(ξ):=f(a+ξv) on the disc D(0,r). The map ξa+ξv is holomorphic, so [L1] makes gv holomorphic on D(0,r); and for every ξD(0,r) one has a+ξvB, hence gv(ξ)=f(a+ξv)f(a)=gv(0). Therefore [L2] makes gv constant on all of D(0,r).

step 1.1L1L2
3.1

Let zB be arbitrary. If z=a there is nothing to prove. Otherwise put v:=(za)/za and ξ:=za; then v=1, ξ<r, and z=a+ξv, so step 2.1 gives f(z)=gv(ξ)=gv(0)=f(a). Thus f is constant on the nonempty open set B, and [L5] makes ff(a) holomorphic on U; applying [L3] to ff(a) and the open set B yields ff(a) on U.

step 2.1L3L5

Remarks

  • The argument gives constancy on the whole local ball. The slice theorem is applied on the entire disc cut out by the ball, not merely near the origin, so the proof first shows that f is constant on all of B and only then extends that constancy to U by the several-variable identity theorem.
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A bounded holomorphic function on all of Cm is constant

Statement

Let m1 and let f:CmC be holomorphic. If there is a real M0 such that f(z)M for every zCm, then f is constant.

Facts & Assumptions

Given: A holomorphic function f:CmC and a real M0 such that f(z)M for every zCm.

[L1]

The composite of holomorphic maps is holomorphic and its complex Jacobian is the product (The composite of holomorphic maps is holomorphic and its complex Jacobian is the product).

[L2]

Every bounded entire function of one complex variable is constant (Liouville's theorem: every bounded entire function is constant).

[L3]

Holomorphic functions on open subsets of Cm are those of Holomorphic functions on an open subset of Cm, and they are continuous (A holomorphic function of several variables is continuous and separately holomorphic).

Proof

technique · direct
1.1

Fix zCm. The map z:CCm defined by z(ξ)=ξz is holomorphic, so by [L1] the composite gz:=fz:CC is holomorphic; and for every ξC one has gz(ξ)=f(ξz)M, so [L2] makes gz constant on C.

givenL1L2L3
2.1

Evaluating the constant function gz at ξ=0 and ξ=1 gives f(z)=gz(1)=gz(0)=f(0). Since z was arbitrary, f is constant on Cm.

step 1.1

Remarks

  • No connectedness argument is needed. Every point is compared directly with the origin along the complex line it spans, so the conclusion is pointwise and not topological.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A nonconstant scalar holomorphic function on a domain in Cm is an open map

Statement

Let m1, let UCm be a nonempty connected open set, and let f:UC be holomorphic and nonconstant. Then f is an open map: for every open set OU, the image f(O) is open in C.

This theorem is about scalar-valued holomorphic functions. It asserts nothing for holomorphic maps into Cn with n2.

Facts & Assumptions

Given: A nonempty connected open set UCm, a nonconstant holomorphic function f:UC, and an open set OU.

[L1]

A holomorphic function vanishing on a nonempty open subset of a connected open set in Cm vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L2]

The composite of holomorphic maps is holomorphic and its complex Jacobian is the product (The composite of holomorphic maps is holomorphic and its complex Jacobian is the product).

[L3]

Every nonconstant holomorphic function on a one-variable complex domain is an open map (Open mapping theorem for holomorphic functions).

[L4]

Balls in Cm are the Euclidean balls of Balls, polydiscs and the distinguished boundary in Cm, and convex subsets are those containing the segment between any two of their points (A convex subset of Rm contains every line segment between two of its points).

Proof

technique · direct
1.1

Let aO. Choose an open ball BO centred at a. If f were constant on B, then ff(a) would vanish on the nonempty open set B, and [L1] would force f to be constant on all of U, contrary to the hypothesis. So there is bB with f(b)f(a).

givenL1L4
2.1

Define W:={ξC:a+ξ(ba)B}. Because B is convex, W is a nonempty open disc about 0 containing 1. The affine map (ξ):=a+ξ(ba) is holomorphic, so [L2] makes g:=f holomorphic on W; and g(0)=f(a)f(b)=g(1), so g is nonconstant.

step 1.1L2L4
3.1

By [L3], the image g(W) is open in C and contains g(0)=f(a). Since g(W)f(B)f(O), the point f(a) is interior to f(O). As aO was arbitrary, every point of f(O) is interior, so f(O) is open by [L5]. Therefore f is an open map.

step 2.1L3L5

Remarks

  • Why the theorem is scalar-valued. The proof restricts to a complex line and then invokes the one-variable open mapping theorem. That argument produces an open image only in C, not for maps into higher-dimensional targets.
RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Conventions on this page, and what the several-variable identity theorem does not say

Remark

Coordinates and multi-indices on this page are indexed from 0, exactly as on the library's Euclidean pages. Thus a point of Cm is z=(z0,,zm1), a multi-index is α=(α0,,αm1) (Ck maps and multi-index derivative notation in Euclidean space), and the polydisc notation is coordinatewise (Balls, polydiscs and the distinguished boundary in Cm).

Holomorphic means complex differentiable, and nothing more. Holomorphic functions on an open subset of Cm does not build in continuity, local boundedness or power-series representability. That is why Osgood's lemma: continuous and separately holomorphic implies holomorphic and Locally bounded and separately holomorphic implies holomorphic are theorems rather than tautologies, and why Multi-indexed power series in Cm and their absolute convergence is a separate object rather than the definition of holomorphy.

The distinguished boundary is the one that carries the Cauchy theory here. The polydisc Cauchy formula is an iterated sequence of one-variable contour integrals over the coordinate circles whose product is the distinguished boundary; it does not define an integral over that boundary as a set. The power-series coefficients and Cauchy estimates use those same circles, rather than the whole topological boundary of a closed polydisc. The companion example A function whose modulus attains its maximum only on the distinguished boundary of a bidisc shows exactly why: the topological boundary contains points where the modulus of a holomorphic function can be far from maximal.

The several-variable identity theorem is weaker than the one-variable one. A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically assumes a nonempty open set of zeros. It does not say that an accumulation point of the zero set is enough, and that stronger statement is false in several variables: the companion false statement A holomorphic function on a domain in C2 vanishing on a set with an accumulation point vanishes identically records the witness. The gap is structural, not cosmetic. In one variable, a nonzero holomorphic function has isolated zeros; in several variables a zero set may contain whole positive-dimensional complex pieces.

5 · Examples, counterexamples and false statements

None yet.

Sources