MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p\in Q$.

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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
Thanks a lot for the answer... – zapkm Sep 20 '11 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.