Suppose that $f$ is an entire function satisfying $f(2z)=\frac{f(z)+f(z+1)}{2}$. Show that $f$ is constant.

Hint 1. Let $R$ be a large positive real number, and consider how $f$ behaves on the closed disc $D$ of radius $R$ centred at $0\in\mathbb{C}.$

Hint 2.

Use the Maximum Modulus Principle, the given identity and the fact that $R$ is sufficiently large to derive a contradiction concerning the maximum value $\lvert f\rvert$ takes on $D$.

Hint 3.

The maximum value $\lvert f\rvert$ takes on $D$ is of the form $\lvert f(2w)\rvert$ for some $w$ strictly inside the disc $D$. Now use the triangle inequality.

Solution.

Let $R\geq2,$ and let $D$ be the closed disc of radius $R$ centred at $0$. Assume, for a contradiction, that $f$ is not constant. The maximum value of $\lvert f \rvert$ on $D$ must be on the boundary, so is of the form $\lvert f(2w) \rvert$ for some $w$ satisfying $\lvert w \rvert = R/2$. Since $R\geq2$, it follows that $w+1$ is in $D$ also. Therefore $\lvert f(2w) \rvert > \lvert f(w) \rvert$ and $\lvert f(2w) \rvert \geq \lvert f(w+1) \rvert$. By the triangle inequality, $ \lvert f(2w) \rvert \leq \frac{ \lvert f(w) \rvert + \lvert f(w+1) \rvert}{2} < \lvert f(2w) \rvert.$ This contradiction proves the result: the non-constancy of $f$ contradicts the Maximum Modulus Principle, so $f$ must be constant.