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Hi, my question is :

Let $D$ be the open unit disk in $\mathbb{C}=R^2$ and $f:D\to D$ be a real-analytic diffeomorphism. Let us think of the canonical embedding : $\mathbb{C}=R^2\subset \mathbb{C^2}. $ Does there exist a complex analytic diffeomorphism $F$ ( analytic in two complex variables ) whose domain is either $D^2\subset \mathbb{C^2}$ or the complex 2-dimensional unit open polydisk $\Omega ={{(z,w): |z|^2+|w|^2 < 1}}$ in $\mathbb{C^2}$ such that its restriction to $D\subset D^2$ or $D\subset \Omega $ is $f$ ? By restriction , I mean $ F(z,0) = f(z) $ in the case of $ D \subset D^2 \subset \mathbb {C}^2 $ .

The range of $F$ does not necessarily have to be $D^2$ or $\Omega$, but it would be even better if they are !

If this is a very well-known result, you can cite a reference.

Is the same result true in 1-dimension as well , i.e. replacing $D$ by $I\subset R$ and changing the complex-analytic/conformal diffeomorphism $F$ accordingly , i.e. asking that domain of $F$ is $I^2$ or $D$ with restriction $ f $ ?

Thank you .

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    $\begingroup$ Your $\Omega$ isn't a polydisk, it's a sphere! $\endgroup$
    – Thomas K
    Commented Jul 5, 2011 at 12:23
  • $\begingroup$ Sorry,my mistake, it should be an $\leq$ sign there in the definition of $\Omega$ $\endgroup$
    – Analysis Now
    Commented Jul 5, 2011 at 13:33
  • $\begingroup$ The fact that the extension exist is elementary: Real analytic map will be locally equal to the sum of their Taylor series. If the series converges in a small neighborhood of $x$ in $\mathbb{R}^2$, it converges in a corresponding neighborhood in $\mathbb{C}^2$. The problem is that the convergence may be on a very small polydisk at every point, as Gavrilov's excellent example shows. $\endgroup$
    – Thierry Zell
    Commented Jul 5, 2011 at 13:56
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    $\begingroup$ Minor note: your $\Omega$ is a closed ball. The unit open polydisk is defined by the inequalities $|z| < 1$ and $|w| < 1$. $\endgroup$
    – S. Carnahan ♦
    Commented Jul 6, 2011 at 4:14

2 Answers 2

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Indeed, a real-analytic map can always be extended to some complex neighbourhood. The problem is, the neighbourhood may be very small. Consider, for example, the map $$f:I\to I,\, I=[-1,1],$$ defined by $$f:x\mapsto x+\frac{a^3x(x-1)^2}{x^2+a}.$$ For small $a>0$ this is a diffeomorphism of $I$, but it cannot be extended very much due to the poles near $x=0$. The similar map (though a bit more contrived) can be designed for a disc. So, the answer to the question is no.

Of course, all this is well known but what is a proper reference I cannot say.

P.S. This is an answer to the question as I understand it. There are some points I do not understand. $\Omega$, as it is defined, is a sphere. And, I hope, the restriction is not defined by $F(z,0)=(f(z),0)$: if it is, you can always take $F(z,w)=(f(z),w)$.

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  • $\begingroup$ I put $=$ sign wrongly before to define $\Omega$ , which I corrected to $\leq $ sign now. Thanks for the answer and pointing that out. $\endgroup$
    – Analysis Now
    Commented Jul 5, 2011 at 13:38
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I'm not 100% sure I understand your question, so pardon me if I'm saying something unrelated.

It seems to me that you are asking whether a given real-analytic function $f \colon \mathbb{R}^2 \to \mathbb{R}^2$ (perhaps only defined on suitable open subsets), given by

$$f(x,y) = [ f_1(x,y), f_2(x,y) ]$$

is in fact complex-analytic, i.e. there is $F \colon \mathbb{C} \to \mathbb{C}$ such that

$$f(x,y) = F(x+iy) = u(x,y) + iv(x,y)\text{ .}$$

This is patently the case if and only if $u$ and $v$, hence $f_1$ and $f_2$, satisfy the Cauchy-Riemann equations, and thus any pair of real-analytic functions $f_i \colon \mathbb{R}^2 \to \mathbb{R}$, $i=1,2$, which do not satisfy CR furnish a counter-example.

On the other hand, what you may have wanted to ask, is whether a real-analytic function extends to a complex-analytic function, e.g. if for a real $f(x)$ there is a complex $F(z)$ such that $F(\Re{z}) = f(\Re{z})$. This can indeed always be done on some Stein neighbourhood, but you don't in general have control over the size of the neighbourhood.

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  • $\begingroup$ Thomas: I believe you misread the question. The OP is asking about extending $f$ from $\mathbb{R}^2$ to $\mathbb{C}^2$, not to $\mathbb{C}$. $\endgroup$
    – Thierry Zell
    Commented Jul 5, 2011 at 13:51
  • $\begingroup$ @Thierry: The OP says she wants $F(z,0) = f(z)$, which lead me to believe that she may have somehow wanted to go from some function on reals $g(x,y)$ to a function of a complex variable $G(x+iy)$. I agree that the question should be about complexifying a function $f(x,y)$ by allowing complex variables $x,y$ instead of real variables. (Which is always possible on a small neighbourhood if $f$ is real-analytic.) $\endgroup$
    – Thomas K
    Commented Jul 5, 2011 at 15:01
  • $\begingroup$ @ Thomas K : I am rerally talking about complex analytic extension in two complex varioables, treating the real variables $ x,y\in R $ as complex numbers. That is why asked :Does there exist a complex analytic diffeomorphism $F$ ( analytic in two complex variables ) whose domain is either $D^2\subset \mathbb{C^2}$ ? However , I am still interested in finding or looking up a sufficient condition to ensure that such an extension exists, for which I would not have difficulty like the example provided by Alex Gavrilov. $\endgroup$
    – Analysis Now
    Commented Jul 5, 2011 at 19:54
  • $\begingroup$ P.S. : By $D^2\subset \mathbb{C}^2 $, I really meant the cartesian product of two unit open disks in the copmplex plane. $\endgroup$
    – Analysis Now
    Commented Jul 5, 2011 at 19:55
  • $\begingroup$ @Lvriemsurf: Hm, I'm probably just confused... let's drop open subsets for convenience. You want a holomorphic map $F \colon \mathbb{C}^2 \to \mathbb{C}^2$. Let's write $\phi(z) = F(z,0)$ for the restriction to the first factor. Now you write that you want $\phi(z) = f(z)$, where $f$ is a real function $f \colon \mathbb{R}^2 \to \mathbb{R}^2$. Since $\phi$ is holomorphic, isn't that asking for two real analytic function of $(x,y)$ to be a holomorphic function in $x+iy$? I must be missing something... $\endgroup$
    – Thomas K
    Commented Jul 5, 2011 at 23:06

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