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

Question

I've been working old qualifying exam problems from my university. I've been struggling with the following:

Suppose $f$ is an entire function with the property that $f(2z)=\frac{f(z)+f(z+1)}{2}$ for all $z$. Prove that $f$ must be a constant function.

Here are my thoughts

Write $f(z)=\sum_{n=0}^{\infty}c_nz^n$. Then \begin{align*} 2\sum_{n=0}^{\infty}c_n2^nz^n&=\sum_{n=0}^{\infty}c_nz^n+\sum_{n=0}^{\infty}c_n(z+1)^n\\ &=\sum_{n=0}^{\infty}c_nz^n+\sum_{n=0}^{\infty}\sum_{k=0}^{n}c_n\binom{n}{k}z^k. \end{align*} By equating the coeffients of the power series, we have $$ c_n2^{n+1}=c_n+\sum_{k=0}^{\infty}c_k\binom{k}{n}, $$ or $$ c_n2^{n+1}=2c_n+\sum_{k=n+1}^{\infty}c_k\binom{k}{n}. $$ Thus $$ (2^{n+1}-2)c_n=\sum_{k=n+1}^{\infty}c_k\binom{k}{n}. $$ Furthermore, \begin{align*} (2^{n+1}-2)c_n-(2^{n+2}-2)c_{n+1}&=\sum_{k=n+1}^{\infty}c_k\binom{k}{n}-\sum_{k=n+2}^{\infty}c_k\binom{k}{n+1}\\ &=c_{n+1}(n+1)+\sum_{k=n+2}^{\infty}c_k\left(\binom{k}{n}-\binom{k}{n+1}\right) \end{align*}

However here I am stuck. I realize that these equations should contain everything needed to solve this problem, but I don't see how to use them. Are there any easier ways to approach this problem? Any help is greatly appreciated. Thank you.

Answer 1

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.

Answer 2

Suppose that $f(z)$ was unbounded. Since $f(z)$ is entire and unbounded, there exists $z_0\in\mathbb{C}$ such that

$$|f(z_0)|\geq 2|f(z)|$$

for all $\{z\in\mathbb{C}:|z|\leq 2\}$. Now, consider the disk of radius $z_0$ centered at zero. Define

$$S=\{z\in\mathbb{C}:|z|\leq|z_0|\text{ and }|f(z)|\text{ is maximized}\}$$

Since $f(z)$ is entire and the disk is a compact set, we are assured that $S$ is non-empty. Now, define $z_1$ to be any element of $S$ such that

$$|z_1|\leq |z|\text{ for }z\in S$$

Then

$$|f(z_1)|\geq|f(z_0)|\geq 2|f(z)|$$

for all $\{z\in\mathbb{C}:|z|\leq 2\}$. This also implies that $|z_1|>2$. More importantly, this implies

$$\left|\frac{z_1}{2}\right|<z_0\text{ and }\left|\frac{z_1}{2}+1\right|<z_0$$

Then taking $z=\frac{z_1}{2}$ gives us

$$f(z_1)=\frac{f\left(\frac{z_1}{2}\right)+f\left(\frac{z_1}{2}+1\right)}{2}$$

$$|f(z_1)|\leq \frac{\left|f\left(\frac{z_1}{2}\right)\right|+\left|f\left(\frac{z_1}{2}+1\right)\right|}{2}$$

However, since $|f(z_1)|$ is the maximum modulus inside the radius $|z_0|$, we know

$$|f(z_1)|\leq \frac{\left|f\left(\frac{z_1}{2}\right)\right|+\left|f\left(\frac{z_1}{2}+1\right)\right|}{2}\leq \frac{|f(z_1)|+|f(z_1)|}{2}=|f(z_1)|$$

Equality can only be achieved when

$$\left|f\left(\frac{z_1}{2}\right)\right|=\left|f\left(\frac{z_1}{2}+1\right)\right|=|f(z_1)|$$

But this implies that

$$\frac{z_1}{2}\in S$$

This is a contradiction as

$$\left|\frac{z_1}{2}\right|<|z_1|$$

We conclude that $f(z)$ is bounded. Since it is a bounded, entire function, by Liouville's Theorem $f(z)$ is constant.