A question from a past qualifying exam at my university reads:
Let $f$ be a nonconstant analytic function on the unit disk $D$ and
let $U = f(D)$. Show that if $\phi$ is a function on $U$ (not necessarily even
continuous) and $\phi \circ f$ is analytic on $D$, then $\phi$ is analytic on $U$.
My approach so far is as follows. Because differentiable functions are continuous, then $\phi\circ f$ is continuous. One can go on to show that $\phi$ must be continuous on $U$. I now want to use Morea's theorem to show that $\phi$ is analytic, i.e. show that $\int_{\partial T}\phi(z)dz=0$ for any triangle $T\subset U$. For $z\in U$, $\phi(z)=\phi(f(w))$ where $f(w)=z$. So I want to rewrite the previous integral in terms of $\phi(f(w))$ and apply Cauchy's theorem to conclude the integral vanishes. However, I don't know how to control the preimage of $T$ under $f$. Will it even be connected? If I can show that $f^{-1}(T)$ is a domain with peicewise-smooth boundary, then I should be down. How should I proceed?
Best Answer
Hint: If $u_0=f(z_0)$ and $f'(z_0)\neq 0$ ($u_0$ is a regular value) then $f$ has an analytic local inverse $\psi$ between respective neighborhoods. Now, write $\phi(u)=(\phi\circ f) \circ \psi(u)$ for $u$ close to $u_0$ and conclude in this case. You will have to treat the critical values of $f$ aside.
Hint for the critical case $f'(z_0)=0$: Note that as $f$ is non-constant and analytic it is an open mapping. Use this to see that $\phi$ must be continuous at $u_0=f(z_0)$. Finally deduce that $\phi$ is analytic at $u_0$.