If composition of one function and the other holomorphic function is holomorphic, then the other should be holomorphic?

Actually, this is an exercise on Rudin's Real and Complex Analysis:

Suppose $\Omega_1, \Omega_2$ are plane regions, $f$ and $g$ are nonconstant complex functions in $\Omega_1$, $\Omega_2$ resp. and $f(\Omega_1) \subset \Omega_2$, so that $h=g \circ f$ can be defined. If we know that $f$ and $h$ are holomorphic, is $g$ holomorphic as well? What if we know that $g$ and $h$ are holomorphic?

The easiest example will be a constant function, but the problem does not allow this. I found it difficult to find counter examples. Can anyone help me?

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Consider the chain rule for complex functions... –  Niccolò Oct 3 '12 at 16:58
@Niccolò: I can't make sense of what you say. What is the chain rule for complex functions, one of which is not assumed holomorphic? –  Georges Elencwajg Oct 3 '12 at 17:38
@GeorgesElencwajg: Maybe Niccolò means something like $\overline{\partial}(g\circ f) = (\partial g\circ f)\overline{\partial} f + (\overline{\partial} g\circ f)\overline{(\partial f)}$ (whenever reasonable). Therefrom one can indeed deduce the vanishing of $\overline{\partial} g$ on $f(\Omega_1)$, except the points where $\partial f=0$. –  Ben Oct 3 '12 at 18:13
Dear @Ben, maybe he means that and maybe not: it's for him to say (I actually doubt that many students outside Germany master the Wirtinger calculus as well as you do). Also, the formula may not be used because $g$ (resp. $f$) is not assumed real-differentiable. And anyway this will not help since both questions have a negative answer and can only be answered by giving two counterexamples. –  Georges Elencwajg Oct 3 '12 at 18:51
Lieber @Ben, um Missverständnisse vorzubeugen: ich selber bin ein großer Anhänger des Wirtinger Kalküls. Er scheint leider außerhalb des Kreises der Spezialisten der komplexen Analysis in mehreren Veränderlichen nicht so breit bekannt zu sein, wie er es verdient.Ich freue mich deshalb um so mehr, dass Sie diesen Kalkül erwähnen und beherrschen! –  Georges Elencwajg Oct 3 '12 at 21:18

2 Answers

I suspect Rudin's problem was intended to be a little broader in scope.

Let me deal with the second part first by giving an counterexample where $g,h$ are homomorphic, but $f$ is not.

Define the function $\lambda: \mathbb{C}\setminus \{0\} \to \mathbb{C}$ by $\lambda(z) = \log |z| + i \arg z$, where I am taking $\arg: \mathbb{C}\setminus \{0\} \to [-\pi, \pi)$. This is the principal branch of $\log$ extended to include the ray $\{(x,0)| x < 0\}$. $\lambda$ is not continuous on this ray, hence not analytic on $\mathbb{C}\setminus \{0\}$. The point being that $e^{\lambda(z)}= z$ (on the appropriate region).

Then take $\Omega_1 = \mathbb{C}\setminus \{0\}$, $\Omega_2 = \mathbb{C}$. Let $g:\Omega_2 \to \Omega_2$ be given by $g(z) = e^z$, and let $h:\Omega_1 \to \Omega_2$ be given by $h(z) = z$. Again, it is easy to see that $f = \lambda$ satisfies the equation $h(z) = g(f(z))$, for $z \in \Omega_1$. Thus $f$ need not be holomorphic.

Now for the first part. It turns out that if $f$ and $h$ are holomorphic, then $g$ is too. We must show that $g$ is differentiable on $f(\Omega_1)$. To simplify life, take $\Omega_2 = f(\Omega_1)$. Since $f$ is not constant, $\Omega_2$ is a region (Rudin, The Open Mapping Theorem).

First we show that $g$ is continuous. Let $\hat{w} \in \Omega_2$, then $\hat{w} = f(\hat{z})$ for some $\hat{z} \in \Omega_1$. By Rudin, Theorem 10.32, we can write $f(z) = \hat{w}+(\phi(z))^m$ in some neighborhood $V$ of $\hat{z}$, for some integer $m\geq 1$, and an invertible, analytic map $\phi$. Note that $\phi(\hat{z}) = 0$, and $|\phi(z)| = |f(z)-\hat{w}|^\frac{1}{m}$. Now let $w_k \to \hat{w}$, with $w_k \in f(V)$, and choose $z_k \in V$ such that $f(z_k) = w_k$. Then we have $|\phi(z_k)| \to 0$, and since $\phi$ is invertible and analytic, we have $z_k \to \hat{z}$. Then we have $g(w_k) = h(z_k)$, and since $h(\hat{z}) = g(\hat{w})$, we have that $g$ is continuous at $\hat{w}$.

Now suppose $f'(\hat{z}) \neq 0$. Then by Rudin, Theorem 10.30, $f$ has a local analytic inverse $\psi$ such that $\psi(f(z)) = z$ in a neighborhood $U$ of $\hat{z}$. This in turn gives $f(\psi(f(z))) = f(z)$, and since $f$ is non constant, $f(U)$ is a region. Then for $w \in f(U)$, we have $f(\psi(w)) = w$. Thus we have $g(f(\psi(w)) = g(w) = h(\psi(w))$ in $f(U)$. Hence $g$ is analytic at $f(\hat{z})$.

By Rudin, Theorem 10.18, we see that the zeros of a non-constant analytic function are isolated. In particular, the zeros of $f'$ are isolated. So, suppose $f'(\hat{z}) = 0$. Then $f'(z) \neq 0$ for $z \neq \hat{z}$ in a neighborhood of $\hat{z}$. Hence $g$ is analytic for $z \neq \hat{z}$ in this neighborhood. Then Morera's theorem, along with Rudin, Theorem 10.13 and continuity of $g$ shows that $g$ is also analytic at $f(\hat{z})$.

Hence $g$ is analytic on $\Omega_2$.

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I learned a lot from you. Thank you! –  Detectives Oct 4 '12 at 4:21
Oh.. I think I find an error. –  Detectives Oct 5 '12 at 11:19
lambda(e^z) != z isn't it? –  Detectives Oct 5 '12 at 11:20
since lambda just maps a point to horizontal strip.. Not all of C –  Detectives Oct 5 '12 at 11:21
Good catch, sorry about that. I have modified the domain, I think it still works. –  copper.hat Oct 5 '12 at 14:46

if $\Omega_2$ is strictly larger than $f(\Omega_1)$, then $g \circ f$ doesn't depend of what $g$ does outside $f(\Omega_1)$, so you can pick $g$ non holomorphic there and still have $f$ and $h$ holomorphic.

As for the other question, if $g$ is many-to-one, such as $g(z) = z^2$, you can switch the arguments given to $g$ as you want as long as they get sent to the same image : here if $f$ is any function such that for any $z$, $f(z) \in \{-z;z\}$, then $g = g \circ f$, where $g$ and $g$ are holomorphic, but $f$ may not be.

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I think $g$ is defined on the region $f(\Omega_1)$ by $h = g \circ f$. Defining $g$ to be non analytic elsewhere avoids the point of the question, I think. –  copper.hat Oct 3 '12 at 16:57
definitely. It's better to assume f(omega1) = omega2. Thank you very much. –  Detectives Oct 4 '12 at 4:20