proof of Hartogs’ theorem


Suppose that g is a smooth differentialMathworldPlanetmath (0,1)-form with compact support in Cn. Then there exists a smooth functionMathworldPlanetmath ψ such that

¯ψ=g, (1)

and ψ has compact support if n2.

Let z=(z1,,zn)n be our coordinates. We note that (0,1)-form means a differential formMathworldPlanetmath given by


The operator ¯ is the so-called d-bar operator and we are looking for a smooth function ϕ solving the equation inhomogeneous ¯ equation. It is important that g has compact support, otherwise solutions to (1) are much harder to obtain.


Written out in detail we can think of g as n different functions g1,,gn, where are date gk satisfy the compatibility condition

gkz¯l=glz¯k for all kl.

Then we write equation (1) as

ψz¯k=gk for all k.

We assume also that gk have compact support.

This system of equations has a solution (many equations in fact). We can obtain an explicit solution as follows.


ψ is smooth by differentiating under the integral. When n2, this solution will also have compact support since g1 has compact support and as z tends to infinity (ζ,z2,,zn) also tends to infinity no matter what ζ is. The reader should notice that there is one direction which does not work. But if ψ has boundedPlanetmathPlanetmathPlanetmath supportPlanetmathPlanetmath except for the line defined by z2=z3==zn=0, then the support must be compactPlanetmathPlanetmath by continuity of ψ. It should also be clear why ψ does not have compact support if n=1.

One might be wondering why we picked z1 and g1 in the construction of ψ. It does not matter, we will get different solutions we we use zk and gk, but it will still have compact support. Further one might wonder why we only use one part of the data, and still get an actual solution. The answer here is that the compatibility condition relates all the data, so we only need to look at one.

We still must check that this really is a solution. We apply the compatibility condition. Let k2.


Note that the integral can be taken over a large ball B that contains the support of gk. We apply the generalized Cauchy formula, where the boundary part of the integral is obviously zero since it is over a set where gk is zero.


Hence ψz¯k=gk.

When k=1, change coordinates to see that


Next differentiate in z¯k and change coordinates back and apply the generalized Cauchy formula as before to get that ψz¯1=g1.

Proof of Theorem.

Let Un, K a compact subset of U and f be a holomorphic functionMathworldPlanetmath defined on UK and UK to be connectedPlanetmathPlanetmath. By the smooth version of Urysohn’s lemma we can find a smooth function φ which is 1 in a neighbourhood of K and is compactly supported in U. Let f0:=(1-φ)f, which is identically zero on K and holomorphic near the boundary of U (since there φ is 0). We let g=¯f0, that is gk=f0z¯k. Let us see why gk is compactly supported. The only place to check is on UK as elsewhere we have 0 automatically,


By Lemma Lemma. we find a compactly supported solution ψ to ¯ψ=g.

Set f~:=f0-ψ. Let us check that this is the desired extention. Firstly let us check it is holomorphic,


It is not hard to see that ψ is compactly supported in U. This follows by the fact that UK is connected and the fact that ψ is holomorphic on the set where g is identically zero. By unique continuation of holomorphic functions, support of ψ is no larger than that of g.

Title proof of Hartogs’ theorem
Canonical name ProofOfHartogsTheorem
Date of creation 2013-03-22 17:46:48
Last modified on 2013-03-22 17:46:48
Owner jirka (4157)
Last modified by jirka (4157)
Numerical id 5
Author jirka (4157)
Entry type Proof
Classification msc 32H02