proof of Hartogs’ theorem

proof of Hartogs’ theorem

Lemma

.

Suppose that

g

is a smooth

differential

(

0

,

1

)

-form with compact support
in

C

n

.

Then there exists a

smooth function

ψ

such that

∂

¯

⁢

ψ

=

g

,

(1)

and

ψ

has compact support if

n

≥

2

.

Let

z

=

(

z

1

,

…

,

z

n

)

∈

ℂ

n

be our coordinates.
We note that

(

0

,

1

)

-form means a

differential form

given by

∑

k

=

1

n

g

k

⁢

(

z

)

⁢

d

⁢

z

¯

k

.

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.

Proof.

Written out in detail we can think of

g

as

n

different

functions

g

1

,

…

,

g

n

,

where are date

g

k

satisfy the compatibility condition

∂

⁡

g

k

∂

⁡

z

¯

l

=

∂

⁡

g

l

∂

⁡

z

¯

k

⁢

for all

k

,

l

.

Then we write equation (

1

) as

∂

⁡

ψ

∂

⁡

z

¯

k

=

g

k

⁢

for all

k

.

We assume also that

g

k

have compact support.

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

ψ

⁢

(

z

)

=

1

2

⁢

π

⁢

i

⁢

∫

ℂ

g

1

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

.

ψ

is smooth by differentiating under the

integral

.
When

n

≥

2

,

this solution will also have compact support since

g

1

has compact support and as

z

tends to infinity

(

ζ

,

z

2

,

…

,

z

n

)

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

bounded

support

except for the line defined by

z

2

=

z

3

=

⋯

=

z

n

=

0

,

then
the support must be

compact

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

z

1

and

g

1

in the construction of

ψ

.

It does not matter,
we will get different solutions we we use

z

k

and

g

k

,

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

k

≥

2

.

∂

⁡

ψ

∂

⁡

z

¯

k

⁢

(

z

1

,

z

2

,

…

,

z

n

)

=

1

2

⁢

π

⁢

i

⁢

∫

ℂ

∂

⁡

g

1

∂

⁡

z

¯

k

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

=

1

2

⁢

π

⁢

i

⁢

∫

ℂ

∂

⁡

g

k

∂

⁡

z

¯

1

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

.

Note that the integral can be taken over a large ball

B

that contains
the support of

g

k

.
We apply the

generalized Cauchy formula

, where the

boundary

part of the integral is obviously zero since it is over
a set where

g

k

is zero.

1

2

⁢

π

⁢

i

⁢

∫

ℂ

∂

⁡

g

k

∂

⁡

z

¯

1

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

=

1

2

⁢

π

⁢

i

⁢

∫

B

∂

⁡

g

k

∂

⁡

z

¯

1

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

=

g

k

⁢

(

z

1

,

…

,

z

n

)

.

Hence

∂

⁡

ψ

∂

⁡

z

¯

k

=

g

k

.

When

k

=

1

, change coordinates to see that

1

2

⁢

π

⁢

i

⁢

∫

ℂ

g

1

⁢

(

ζ

,

z

2

,

…

,

z

n

)

ζ

-

z

1

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

=

1

2

⁢

π

⁢

i

⁢

∫

ℂ

g

1

⁢

(

ζ

+

z

1

,

z

2

,

…

,

z

n

)

ζ

⁢

𝑑

ζ

∧

d

⁢

ζ

¯

.

Next

differentiate

in

z

¯

k

and change coordinates back and apply
the generalized Cauchy formula as before to get that

∂

⁡

ψ

∂

⁡

z

¯

1

=

g

1

.

∎

Proof of Theorem.

Let

U

⊂

ℂ

n

,

K

a compact subset of

U

and

f

be a

holomorphic function

defined on

U

∖

K

and

U

∖

K

to
be

connected

.
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

f

0

:=

(

1

-

φ

)

⁢

f

,

which is identically zero on

K

and holomorphic
near the boundary of

U

(since there

φ

is 0).
We let

g

=

∂

¯

⁢

f

0

, that is

g

k

=

∂

⁡

f

0

∂

⁡

z

¯

k

. Let us see why

g

k

is compactly supported. The
only place to check is on

U

∖

K

as elsewhere we have 0
automatically,

∂

⁡

f

0

∂

⁡

z

¯

k

=

∂

∂

⁡

z

¯

k

⁢

(

(

1

-

φ

)

⁢

f

)

=

-

f

⁢

∂

⁡

φ

∂

⁡

z

¯

k

.

By Lemma

Lemma

.

we find a compactly supported solution

ψ

to

∂

¯

⁢

ψ

=

g

.

Set

f

~

:=

f

0

-

ψ

. Let us check that this is the desired
extention. Firstly let us check it is holomorphic,

∂

⁡

f

~

∂

⁡

z

¯

k

=

∂

⁡

f

0

∂

⁡

z

¯

k

-

∂

⁡

ψ

∂

⁡

z

¯

k

=

g

k

-

g

k

=

0

.

It is not hard to see that

ψ

is compactly supported in

U

.

This follows
by the fact that

U

∖

K

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

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