What are the criterions for holomorphic functions except $\frac{\partial f}{\partial \overline z}=0$ and $f$ has a power series extension?

I was considering the problem, which is the extension of a bounded non-vanishing holomorphic function $f$ such that $|f(z)|=1$ when $|z|=1$ from the closure of the unit disk to the whole plane. After guessing the function $g(z)=1/ \overline{f(1/\overline{z})}$ for $|z|>1$, I can't prove that g is holomorphic.

However, I know if $f$ is holomorphic, then we also have $\frac{\partial \overline{f}}{\partial z}=\frac{\partial f}{\partial \overline z}=0$, but we don't have $\frac{\partial f(\overline z)}{\partial z}=\frac{\partial f}{\partial \overline z}$ so the first criterion can't be applied. Also I can't check whether g is holomorphic by the power series criterion since if $f(1/\overline z)=\Sigma a_n (1/\overline z-z_0)^n$ when $|z|>1$, I don't know whether $\frac{1}{\Sigma a_n (1/\overline z-z_0)^n}$ has a power series.

I would really appreciate any help. Thanks


You need the chain rule in $\partial/\partial z$ and $\partial/\partial \bar z$ form. If you have $h(z,\bar z) = f(g(z,\bar z))$, then we have $$\frac{\partial h}{\partial \bar z}=\frac{\partial f}{\partial w}\frac{\partial g}{\partial \bar z}+\frac{\partial f}{\partial \bar w}\frac{\partial \bar g}{\partial \bar z}.$$ You can deduce this from the usual chain rule.

By the way, there are identities like $$\overline{\left(\frac{\partial h}{\partial z}\right)}=\frac{\partial \bar h}{\partial \bar z} \quad\text{and} \quad\overline{\left(\frac{\partial h}{\partial \bar z}\right)}=\frac{\partial \bar h}{\partial z}.$$

  • $\begingroup$ And may I ask how does this prove the fact that $g$ is holomorphic? $\endgroup$ – Alex Aug 18 '13 at 17:56
  • $\begingroup$ Apply this with $h(z)=\bar f(1/\bar z)$. $\endgroup$ – Ted Shifrin Aug 18 '13 at 18:07
  • $\begingroup$ Oh, it did work, thank you very much!I didn't expect the chain rule would be so helpful. $\endgroup$ – Alex Aug 18 '13 at 18:09

The extended function

$$g(z) = \begin{cases} f(z) &, \lvert z\rvert \leqslant 1\\ 1/\overline{f(1/\overline{z})} &, \lvert z \geqslant 1\end{cases},$$

where $f \colon \overline{\mathbb{D}} \to \mathbb{C}$ is continuous and holomorphic in $\mathbb{D}$ with $\lvert f(z) \rvert = 1$ for $\lvert z\rvert = 1$, is always a meromorphic function on the entire plane $\mathbb{C}$, with poles in the points obtained from reflection of the zeros of $f$ in the unit circle. That is a special case of the Schwarz reflection principle.

In the particular case that $f$ has no zeros in the unit disk, $f$ is constant by the minimum modulus principle, and the holomorphy (constantness) of $g$ is readily verified.

  • $\begingroup$ From what I have learnt from Chap 2 of Stein's book, the Schwarz reflection principle is about the real axis, and the proof is straight with Morera's Theorem and power series extension. But here I don't get how we can get the condition for the Schwarz reflection principle to hold(namely, why the extension g outside $\mathbb{\overline D}$ is holomorphic) $\endgroup$ – Alex Aug 18 '13 at 17:55
  • $\begingroup$ Ah, it's way more general than that. For circular arcs, you can get it by composing with Möbius transformations mapping the arcs into the real line. $\endgroup$ – Daniel Fischer Aug 18 '13 at 17:59
  • $\begingroup$ Okay, I see what you mean. But is there any way to prove that g is holomorphic without the idea of Mobius transformation or meromorphic functions? Because it appearred in Chap 2 of the book and certainly the author believes that I should be able to prove it without such knowledge. thx $\endgroup$ – Alex Aug 18 '13 at 18:01
  • $\begingroup$ @ Daniel Fisher: Aha, now I do understand what you mean, the Mobius transformation really is amazing. And the minimum modulus principle is really quick. Thanks $\endgroup$ – Alex Aug 26 '13 at 22:15

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