Let $\mathcal{O}(D)$ denote the set of holomorphic functions for $D \subseteq \mathbb{C}$ open.

Let $f \in \mathcal{O}(\mathbb{C})$ and define $g: \mathbb{C} \to \mathbb{C}$ by $g(z) := f(\overline{z})$. In which points is $g$ holomorphic?

I used the Wirtinger calculus. We have $$\frac{\partial g}{\partial \overline{z}} = \frac{\partial f(\overline{z})}{\partial \overline{z}} = \frac{\partial f}{\partial w}\frac{\partial \overline{z}}{\partial \overline{z}} + \frac{\partial g}{\partial \overline{w}}\frac{\partial z}{\partial \overline{z}} = \frac{\partial f}{\partial w}$$ Now that $g$ is holomorphic in a point $z_0$, we must have that $$\frac{\partial f}{\partial w}(z_0) = 0$$ How do I proceed further? Can I explicitely describe the set? Is there another way? I also tried it with the other characterization, i.e. since $f \in \mathcal{O}(\mathbb{C})$, we have for all $z_0 \in \mathbb{C}$ a function $\varphi: \mathbb{C} \to \mathbb{C}$ which is continuous at $z_0$ such that $$f(z) = f(z_0) + \varphi(z)(z - z_0)$$ holds for any $z \in \mathbb{C}$.

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    $\begingroup$ If $f$ is non-constant, then $g$ is nowhere holomorphic. $\endgroup$ – J.R. Jun 18 '17 at 10:45

Thanks to J.R. I have figured it out. If $f$ is constant, then $g$ is clearly entire. So assume that $f$ is not constant and that $g$ is holomorphic in a point $z_0$. Hence $$\frac{\partial g}{\partial \overline{z}} = 0$$ in some neighbourhood $U$ of $z_0$. But by the above deduction we have that $$\frac{\partial f}{\partial w} = f' = 0$$ in $\{ \overline{w} : w \in U\}$. Since $U$ is a neighbourhood of $z_0$ we may find $B_r(z_0) \subseteq U$ and thus $f'$ vanishes in $\{\overline{w} : w \in B_r(z_0)\}$. Since this is a domain, we have that $f$ is constant there and thus by the identity principle, $f$ is constant on $\mathbb{C}$, contradicting our assumption. Hence $g$ is not holomorphic in any point of $\mathbb{C}$.

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  • $\begingroup$ In fact, $g(\overline z)$ is holomorphic if and only if it admit a power serie in $\overline z$. $\endgroup$ – user171326 Jun 18 '17 at 16:30
  • $\begingroup$ @N.H. Thanks, I will check this. How does this help in this particular example? $\endgroup$ – TheGeekGreek Jun 18 '17 at 18:31
  • $\begingroup$ If $f(z)$ is holomorphic and is a power series in $\overline z$ it follows that $f$ is constant. $\endgroup$ – user171326 Jun 18 '17 at 19:28

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