proof of Hartogs’ theorem
Let be our coordinates. We note that -form means a differential form given by
The operator is the so-called d-bar operator and we are looking for a smooth function solving the equation inhomogeneous equation. It is important that has compact support, otherwise solutions to (1) are much harder to obtain.
Written out in detail we can think of as different functions where are date satisfy the compatibility condition
Then we write equation (1) as
We assume also that have compact support.
This system of equations has a solution (many equations in fact). We can obtain an explicit solution as follows.
is smooth by differentiating under the integral. When this solution will also have compact support since has compact support and as tends to infinity also tends to infinity no matter what is. The reader should notice that there is one direction which does not work. But if has bounded support except for the line defined by then the support must be compact by continuity of . It should also be clear why does not have compact support if
One might be wondering why we picked and in the construction of It does not matter, we will get different solutions we we use and 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 .
Note that the integral can be taken over a large ball that contains the support of . We apply the generalized Cauchy formula, where the boundary part of the integral is obviously zero since it is over a set where is zero.
When , change coordinates to see that
Next differentiate in and change coordinates back and apply the generalized Cauchy formula as before to get that ∎
Proof of Theorem.
Let , a compact subset of and be a holomorphic function defined on and to be connected. By the smooth version of Urysohn’s lemma we can find a smooth function which is 1 in a neighbourhood of and is compactly supported in Let which is identically zero on and holomorphic near the boundary of (since there is 0). We let , that is . Let us see why is compactly supported. The only place to check is on as elsewhere we have 0 automatically,
By Lemma Lemma. we find a compactly supported solution to .
Set . Let us check that this is the desired extention. Firstly let us check it is holomorphic,
It is not hard to see that is compactly supported in This follows by the fact that is connected and the fact that is holomorphic on the set where is identically zero. By unique continuation of holomorphic functions, support of is no larger than that of ∎
|Title||proof of Hartogs’ theorem|
|Date of creation||2013-03-22 17:46:48|
|Last modified on||2013-03-22 17:46:48|
|Last modified by||jirka (4157)|