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Does there exist a conformal smooth extension of a smooth function? Smooth extension is guaranteed by Whitney extension theorem. does that theorem also says for conformality. Precisely the question is the following:

Let $ D= \{z: |z|\leq 1,I[z]\geq 0 \}$. Define $f\colon D \to\mathbb R^2$ such that all the derivative of $f$ exists and continuous and for all $x\in D$ and $u,v\in\mathbb R^2$, we have $\langle f'(x)(u) ,f'(x)(v)\rangle=\alpha (x)\langle u,v\rangle$, where $\alpha $ is some positive function on D.

Now does there exists an open set $U$ in $\mathbb R^2$ such that $D\subset U$, and a smooth function $g$ on $U$ which restricted to $D$ is $f$ and $\langle g'(x)(u),g'(x)(v)\rangle=\beta(x)\langle u,v\rangle$ for all $x \in U$ for $\beta $ some positive function on $U$.

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I presume you really want $\langle f'(x)(u),f'(x)(v)\rangle=\alpha(x) \langle u,v\rangle$ (which would be the usual definition of conformal) rather than $\langle f(x),f(y)\rangle=\alpha\langle x,y\rangle$? – Neil Strickland Aug 30 '11 at 14:18
@neil yes you are right.. i mean $< f'(x)(u), f'(x) (v)>=\alpha(x)< u,v>$ – zapkm Aug 30 '11 at 15:08
up vote 3 down vote accepted

Firstly, conformal is the same as complex-analytic, and it is much more convenient to think in those terms.

Secondly, there are well-known examples of functions such as $f(z)=\sum_{n=0}^\infty z^{n!}$ that are analytic on the open unit disc but diverge at a dense set of points on the unit circle so cannot be continued analytically. This is not quite what you asked for, but is suggestive. You can search for 'analytic natural boundary' to find out more.

I think it is also true that there are functions that are continuous on the closed disc and holomorphic on the open disc but cannot be extended analytically any further. My feeling is that there should be elementary examples, but I cannot seen any at the moment. However, you can also think about the Riemann Mapping Theorem, which says that for any simply connected proper open subset $U\subset\mathbb{C}$, there is an analytic bijection $g$ from the open unit disc to $U$. You can choose $U$ so that the boundary is quite wild, and then it will be unlikely that $g$ has an analytic extension. For example, we can take $U$ to be the complement of the Mandelbrot set in the Riemann sphere. I don't have the relevant book to hand, but I think it is known that in this case the map $g$ can be extended continuously over the closed unit disc. I would bet that it cannot be extended analytically any further than the closed disc, but I do not know a proof of that.

However, it is possible that I have still not addressed the question that you really need. Perhaps you want to assume that $f$ is conformal on the boundary in some sense, rather than just assuming that it is continuous on the closed disc and conformal in the interior. If so, you should spell things out in more detail, because there seem to be some rather subtle issues about exactly what the right definition should be, and I have never understood this properly myself.

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There are some examples of functions continuous on the closed disk and holomorphic on the open disk but which cannot be extended in this answer… – j.c. Aug 30 '11 at 16:10
@jc: thanks for the pointer, if I were you I would promote that to an answer, and it would be much better than mine. – Neil Strickland Aug 30 '11 at 16:29
@neil @jc, Thanks a lot.. atleast i will get some idea.. @neil.. yeah you are right.. i am unable to explain the thing.. actually i want one map which is somehow not only conformal in interior but it is conformal on the boundary.. but what does mean by conformal on the boundary.. I will try to be more precise.. then i will post the question.. by the way thanks a lot.. – zapkm Aug 30 '11 at 17:39

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