proof of Hartogs’ theorem

Lemma.

Suppose that $g$ is a smooth differential $\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$-form with compact supportin ${\mathrm{C}}^n\mathrm{.}$ Then there exists a smooth function $\psi$ such that

$$\overline{\partial }\psi =g,$$(1)

and $\psi$ has compact support if $n\mathrm{\ge }\mathrm{2}\mathrm{.}$

Let $z=(z_1,\mathrm{\dots },z_n)\in {ℂ}^n$ be our coordinates.We note that $(0,1)$-form means a differential form given by

$$\sum _{k=1}^n{g}_k(z)d{\overline{z}}_k.$$

The operator $\overline{\partial }$ is theso-calledd-bar operator and we are looking for a smooth function $\varphi$ solving theequation inhomogeneous $\overline{\partial }$ equation.It is important that $g$ has compact support, otherwise solutionsto (1) are much harder to obtain.

Proof.

Written out in detail we can think of $g$ as $n$ different functions$g_1,\mathrm{\dots },g_n,$where are date $g_k$ satisfy the compatibility condition

$$\frac{\partial g_k}{\partial {\overline{z}}_l}=\frac{\partial g_l}{\partial {\overline{z}}_k}\text{ for all }k\text{, }l.$$

Then we write equation (1) as

$$\frac{\partial \psi }{\partial {\overline{z}}_k}=g_k\text{ for all }k\text{.}$$

We assume also that $g_k$ have compact support.

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

$$\psi (z)=\frac{1}{2\pi i}{\int }_{ℂ}\frac{g_1(\zeta ,z_2,\mathrm{\dots },z_n)}{\zeta -z_1}𝑑\zeta \wedge d\overline{\zeta }.$$

$\psi$ is smooth by differentiating under the integral.When $n\ge 2,$this solution will also have compact support since $g_1$has compact support and as $z$tends to infinity $(\zeta ,z_2,\mathrm{\dots },z_n)$ also tends to infinityno matter what $\zeta$ is. The reader should notice that there is onedirection which does not work. But if $\psi$ has bounded supportexcept for the line defined by $z_2=z_3=\mathrm{\cdots }=z_n=0,$ thenthe support must be compact by continuity of $\psi$.Itshould also be clear why $\psi$ does not have compact supportif $n=1.$

One might be wondering whywe picked $z_1$ and $g_1$ in the construction of $\psi .$ It does not matter,we will get different solutions we we use $z_k$ and $g_k,$ but it willstill have compact support.Further one might wonder why we onlyuse 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 mustcheck that this really is a solution.We apply the compatibility condition. Let $k\ge 2$.

Note that the integral can be taken over a large ball $B$that containsthe support of $g_k$.We apply the generalized Cauchy formula, where theboundary part of the integral is obviously zero since it is overa set where $g_k$ is zero.

Hence $\frac{\partial \psi }{\partial {\overline{z}}_k}=g_k$.

When $k=1$, change coordinates to see that

$$\frac{1}{2\pi i}{\int }_{ℂ}\frac{g_1(\zeta ,z_2,\mathrm{\dots },z_n)}{\zeta -z_1}𝑑\zeta \wedge d\overline{\zeta }=\frac{1}{2\pi i}{\int }_{ℂ}\frac{g_1(\zeta +z_1,z_2,\mathrm{\dots },z_n)}{\zeta }𝑑\zeta \wedge d\overline{\zeta }.$$

Next differentiate in ${\overline{z}}_k$ and change coordinates back and applythe generalized Cauchy formula as before to get that$\frac{\partial \psi }{\partial {\overline{z}}_1}=g_1.$∎

Proof of Theorem.

Let $U\subset {ℂ}^n$, $K$ a compact subset of $U$ and $f$ be aholomorphic function defined on $U\setminus K$ and $U\setminus K$ tobe connected.By the smooth version of Urysohn’s lemma we canfind a smooth function $\phi$ which is 1 in a neighbourhood of$K$ and is compactly supported in $U.$ Let$f_0:=(1-\phi )f,$ which is identically zero on $K$ and holomorphicnear the boundary of $U$ (since there $\phi$ is 0).We let $g=\overline{\partial }{f}_0$, that is $g_k=\frac{\partial f_0}{\partial {\overline{z}}_k}$. Let us see why $g_k$ is compactly supported. Theonly place to check is on $U\setminus K$ as elsewhere we have 0automatically,

$$\frac{\partial f_0}{\partial {\overline{z}}_k}=\frac{\partial }{\partial {\overline{z}}_k}((1-\phi )f)=-f\frac{\partial \phi }{\partial {\overline{z}}_k}.$$

By Lemma Lemma. we find a compactly supported solution $\psi$to $\overline{\partial }\psi =g$.

Set $\tilde{f}:=f_0-\psi$. Let us check that this is the desiredextention. Firstly let us check it is holomorphic,

$$\frac{\partial \tilde{f}}{\partial {\overline{z}}_k}=\frac{\partial f_0}{\partial {\overline{z}}_k}-\frac{\partial \psi }{\partial {\overline{z}}_k}=g_k-g_k=0.$$

It is not hard to see that $\psi$ is compactly supported in $U.$ This followsby the fact that $U\setminus K$ is connected and the factthat $\psi$is holomorphic on the set where $g$ is identically zero. By unique continuationof holomorphic functions, support of $\psi$ is no larger than that of $g.$∎