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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26
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A bounded separately holomorphic function on a polydisc is Lipschitz on every smaller polydisc

Statement

Let m≥1, let a∈Cm, let ρ be a polyradius, let f:Δρ(a)→C be separately holomorphic (Separately holomorphic functions) with ∣f∣≤M throughout Δρ(a), and let 0<θ<1. Then for all z,w∈Δ‾θρ(a)

∣f(w)−f(z)∣ ≤ M1−θ∑k<m∣wk−zk∣ρk ≤ M1−θ(∑k<m1ρk)∥w−z∥.

In particular f is Lipschitz, hence continuous, on Δ‾θρ(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 ∣f∣≤M 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), Δ‾r(a) and Γr(a) are defined coordinatewise by ∣zk−ak∣<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 ∣g∣≤K on the circle ∣ζ−a′∣=r′, one has ∣g(n)(a′)∣≤n!K/r′n (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L4]

For an open convex V⊆C, a holomorphic g on V and z′,w′∈V, g(w′)−g(z′)=(w′−z′)∫01g′(z′+t(w′−z′)) 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 a′≤b′, ∥∫a′b′F∥2≤∫a′b′∥F∥2 (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 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 ∫a′b′h≤∫a′b′H when h≤H pointwise (If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[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∣=∣z∣∣w∣ and ∣z+w∣≤∣z∣+∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, 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
1.1givenL2L9

Fix z,w∈Δ‾θρ(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)−aj∣≤θρj for every j<m and each v(k) lies in Δ‾θρ(a)⊆Δρ(a) by [L2].

2.1step 1.1L7

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

2.2step 1.1L1L2

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 ∣ξ−ak∣<ρk, so [L1] makes gk holomorphic on the disc D(ak,ρk), and ∣gk∣≤M there.

3.1step 2.2L3L8L10

Let ∣ξ−ak∣≤θρk and let 0<s<(1−θ)ρk. For ∣ζ−ξ∣<(1−θ)ρk one has ∣ζ−ak∣<ρk by [L8], so gk is holomorphic on D(ξ,(1−θ)ρk) and bounded by M on the circle ∣ζ−ξ∣=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,(1−θ)ρk) by [L10] gives ∣gk′(ξ)∣≤M/((1−θ)ρk).

4.1step 1.1step 3.1L2L4L5L6L8

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 ∣ξ−ak∣≤θρk by [L8]. By [L4], [L6] and [L5], ∣gk(wk)−gk(zk)∣≤∣wk−zk∣∫01∣gk′(zk+t(wk−zk))∣ dt≤∣wk−zk∣⋅M(1−θ)ρk, using step 3.1 on the segment.

5.1step 2.1step 2.2step 4.1L7L8∎

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)∣≤M1−θ∑k<m∣wk−zk∣/ρk; each ∣wk−zk∣≤∥w−z∥ by the dictionary, which yields the second displayed bound and makes f Lipschitz on Δ‾θρ(a). Only the slices of f and the uniform bound M were used, never continuity of f in the remaining variables.

Depends on

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