One of my friend asked me the following problem:
Problem. Suppose that $f$ is a holomorphic function on a convex open set $U$ which satisfies the following property: For all distinct $z, w \in U$, there exists $\zeta$ on the line segment joining $z$ and $w$ such that $$ \frac{f(w) - f(z)}{w - z} = f'(\zeta). $$ Then show that $f$ is a polynomial of degree at most 2.
This problem was given in the lecture note right after some basic properties of complex differentiability are introduced.
I was able to prove this problem, though my argument relies on heavy machinery (such as inverse function theorem) which is beyond the chapter containing it.
Q. My question is, is there any elementary solution?
For a reference, here is my heavy solution:
Step 1. If $f'' \equiv 0$ on $U$, there is nothing to prove. So we may assume that $f''(z_0) \neq 0$ for some $z_0 \in U$. Our goal is to show that
Claim. There exists a neighborhood $V \subseteq U$ of $z_0$ such that for any $z \in V$, $$ \frac{f(z) - f(z_0)}{z - z_0} = f'(z_0 + \tfrac{1}{2}(z - z_0)).$$
Indeed, assuming that this claim is true, we obtain
\begin{align*} \sum_{n=0}^{\infty} \frac{f^{(n+1)}(z_0)}{(n+1)!} (z - z_0)^n &= \frac{f(z) - f(z_0)}{z - z_0} \\ &= f'(z_0 + \tfrac{1}{2}(z - z_0)) = \sum_{n=0}^{\infty} \frac{f^{(n+1)}(z_0)}{n! 2^n} (z - z_0)^n. \end{align*}
Comparing coefficients of both sides, we have
$$f^{(3)}(z_0) = f^{(4)}(z_0) = \cdots = 0$$
and hence $f$ is a polynomial of degree at most 2 on $V$. Then by the principle of analytic continuation, this continues to hold on all of $U$ as desired.
Step 2. So it remains to prove the claim. To this end, notice that
- $g(z) := (f(z) - f(z_0))/(z - z_0)$ has removable singularity at $z = z_0$, hence extends to a holomorphic function on $U$.
- By the inverse function theorem, there exists a neighborhood $V \subseteq U$ of $z_0$ on which $f'$ is invertible. By shrinking $V$ further, we may assume that $V$ is convex.
From the assumption, we know that $g(V) \subseteq f'(V)$. Thus
$$ \zeta(z) := (f')^{-1}(g(z)) $$
is a well-defined holomorphic function on $V$ such that
$$ \frac{\zeta(z) - z_0}{z - z_0} \in [0, 1] \quad \text{for all } z \in V \setminus\{z_0\}. $$
This implies that $(\zeta(z) - z_0)/(z - z_0)$ is constant, and this value can be determined by
$$ \frac{\zeta(z) - z_0}{z - z_0} = \lim_{w \to z_0} \frac{\zeta(w) - z_0}{w - z_0} = \zeta'(z_0) = \frac{g'(z_0)}{f''(z_0)} = \frac{1}{2}. $$
This completes the proof. ////