Extending a function from the boundary of an open set

[Graven Water]

see http://camoo.freeshell.org/quest1.pdf

Laura

If you like reading the source better it is below:

I'm reading Lars Hormander's book on multivariable complex
analysis and I'm puzzled by something.

In Theorem 2.3.2' the suppositions are:

Let $\Omega$ be a bounded open set in $\mathbf C^n$, $n>1$, such that the
complement of $\bar\Omega$ is connected and $\partial\Omega\in C^4$.
Let $\rho$ be a real valued function in $C^4$ such that $\rho=0$
precisely on $\partial\Omega$ and grad($\rho) \neq 0$ on
$\partial\Omega$. Let $u\in C^4(\bar\Omega)$ and
$\bar\partial u\wedge\bar\partial\rho=0$ on $\partial\Omega$.
Then, one can find an analytic function $U\in C^1(\bar\Omega)$ such
that $U=u$ on $\partial\Omega$.

I don't know what he means by $\partial\Omega\in C^4$, but that
isn't the question. I figure maybe if it's a typo maybe it'll become
clear what it means later.

Otherwise: $\bar\Omega$ is the closure of $\Omega$

$\partial\Omega$ is the boundary of $\Omega$

$C^k(S)$ is $k$ times continuously differentiable complex-valued
functions on $S$.

$\displaystyle\bar\partial u=\sum_{i=1}^n
\frac{\partial u}{\partial\bar{z_i}}
d\bar{z_i}$

Now here's the question. He says in the proof that by assumption,
$\bar\partial u=h_0\bar\partial\rho+\rho h_1$, \newline$h_0\in C^3(\bar\Omega)$.

I was trying to figure out how you determine $h_0$. You could do a
least
squares fit of $\bar\partial u$ to a multiple of $\bar\partial\rho$.
But that has a problem: the least squares fit would blow up when
grad$(\rho)=0$, which it would be somewhere in $\Omega$.

So how do you determine $h_0$ in $\Omega$? It's determined on
$\partial\Omega$, can that be extended to a $C^3$ function on
$\bar\Omega$?

[W^3]

In article <200901082333...@grex.org>,
pb...@grex.org (Graven Water) wrote:

You really should post in plain text and not TEX.

[Graven Water]

I seem to have figured this one out. See
http://camoo.freeshell.org/quest1.pdf if you're curious, I put my answer
at the end.
Laura