Conformal Extension from a closed set to open

Let $Q = \{(x,y): x,y\geq 0\}$ be the 1st quadrant of $\mathbb R^2$, and $f$ is a function defined on it such that all the partial derivative(any order) of $f$ exists and continuous. By Whitney extension theorem (1934 Proceedings of AMS) we know there exists a functions $\tilde{f}$ and open set $\tilde{U}$, such that $\tilde{f}$ is defined on $\tilde{U}$ and $\tilde{f}$ is smooth, $Q\subset \tilde{U}$ and $\tilde{f}$ restricted to $Q$ is $f$. Also $\tilde{f}$ is not unique but for each $p\in Q$, $d\tilde{f}_p$ is same. Now define $df_p:= d\tilde{f}_p$

Now for each $p\in Q$ , let $R(p)$ be a constant time rotation matrix at $p$, This mean $R(p)= c(p) \left[ {\begin{array}{cc} \cos \theta(p) & \sin \theta(p),\\ -\sin \theta(p) & \cos \theta(p) \end{array} } \right]$

$R(p)$ is $2\times 2$ matrix (pls someone fix the tex), and $c(p)$ is differentiable function on $Q$ to $\mathbb R$. and we have for each $p\in Q$, $df_p= R(p)$, does there exits any $\tilde{f}$ and $\tilde{U}$ as above such that for each $q\in \tilde{U}$ we have $d\tilde{f}_q= S(q)$. Where $S(q)$ is constant time rotational matrix and $S(p)= R(p)$ for each $p\in Q$.

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I fixed the LaTeX. – David Roberts Sep 18 '11 at 0:05
Thanks for fixing latex – zapkm Sep 18 '11 at 1:36
What is a 'constant time rotation(al) matrix'? You didn't mention 'time' before that phrase. Without this definition, I have no idea how to help you. I know what a rotation matrix is, so you could start there. – Robert Bryant Sep 18 '11 at 21:56
@Robert Bryant, I edited the question.. does it make sense now??? waiting for suggestion or answer – zapkm Sep 19 '11 at 5:38

Ah, so you just mean that, when you regard $\mathbb{R}^2$ as $\mathbb{C}$ and you have a complex-valued function $f$ on $Q$, the closed first quadrant of $\mathbb{C}$, that satisfies the Cauchy-Riemann equations up to and including the boundary of $Q$, then does it extend holomorphically across the boundary. (I've never heard of your matrices $R(p)$ being called 'constant time rotation matrices. Where did you get that name?)
Well, the answer is 'no'. Using the Riemann Mapping Theorem, one can construct examples of such $f$ that are not real-analytic on the boundary of $Q$, which implies that $f$ does not extend holomorphically across the boundary, which is what you are asking for.
The corner is not really relevant. You can even assume that $f$ is defined and smooth on the closed half-plane $x\ge0$ and construct examples such that $f$ does not extend across the line $x=0$ at any point. It suffices to consider a domain such as $x\ge g(y)$ where $g:\mathbb{R}\to\mathbb{R}$ is, say, smooth and bounded, but nowhere real-analytic. Then let $f$ biholomorphically map the domain $x>0$ onto the domain $x>g(y)$ and carry $\infty$ to $\infty$. Then $f$ will extend smoothly to a mapping that carries the line $x=0$ to the curve $x=g(y)$, but $f$ won't be real-analytic anywhere along that line, so it can't extend holomorphically to any domain that contains it.
For "a constant time rotation matrix" read "a constant times a rotation matrix", i.e., a scalar $c(p)$ multiplied by a rotation matrix, i.e., a complex number considered as a $2\times2$ real matrix. – Andreas Blass Sep 19 '11 at 14:05
Ah, that makes sense! However, I still don't understand why one would call $c(p)$ a `constant'. – Robert Bryant Sep 19 '11 at 21:00