# How big can $U\subset\mathbb{C}$ be if there exists a non-constant holomorphic $f\colon U\to\mathbb{C}$ with $2f(2z)=f(z)+f(z+1)?$

This question was inspired by this other question.

Let $$U\subset\mathbb{C}$$ be open, and let $$f\colon U\to \mathbb{C}$$ be holomorphic and non-constant. Suppose $$f$$ also satisfies the following identity: for all $$z\in U$$ with $$z+1,2z\in U,$$ we have $$f(2z) = \frac{f(z)+f(z+1)}{2}.$$ The question linked above is basically about proving that such an $$f$$ cannot be entire, i.e., $$U\neq\mathbb{C}.$$ In fact, my solution proves a very slightly stronger result, see the following.

Let $$D\subset U$$ be the closed disc of radius $$R$$ centred at $$0.$$ Assume, for a contradiction, that $$R\geq2.$$ 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 $$w$$ with $$\lvert w \rvert = R.$$ 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, $$\lvert f(2w) \rvert \leq \frac{\lvert f(w) \rvert + \lvert f(w+1) \rvert}{2} < \lvert f(2w) \rvert,$$ a contradiction.

This proves that $$\{z:\lvert z \rvert \leq 2\} \not\subset U.$$ With regard to the above linked question, it therefore follows that $$U\neq\mathbb{C}$$, but I'm curious about what else can be said about $$U$$.

Question. How big can $$U$$ be?

Let's be specific, for the sake of defining the parameters of an answer.

In the first place, does there exist a non-constant holomorphic $$f\colon \{z:\lvert z\rvert >2, \text{Re}(z)>0\}\to\mathbb{C}$$ such that $$2f(2z)=f(z)+f(z+1)$$ for all $$z?$$

Does there exist a $$U$$ admitting such an $$f$$ such that $$U$$ only misses out a discrete subset of $$\mathbb{C}?$$

• I presume you want the equation to hold whenever $z, z+1, 2z \in U$ (so, in other words, $U$ may contain $z_0, z_0+1$ but not $2z_0$ and then the equation for $z_0$ holds by default so to speak Jul 9, 2020 at 21:54
• @Conrad: that's a very good point, and yes, I think that should be the assumption. Editing as appropriate. Jul 9, 2020 at 21:58

Although this answer does not fully address OP's question, I hope it provides some useful information on it.

Here, we will assume:

• $$X \subseteq \mathbb{C}$$ is a set containing an open neighborhood $$U$$ of $$[0, 2]$$.

• For each $$z \in X$$, both $$\frac{z}{2}$$ and $$\frac{z}{2}+1$$ are elements of $$X$$. (For instance, this holds when $$X$$ is a convex set containing $$U$$.)

• $$f : X \to \mathbb{C}$$ is a function that satisfies $$f(z) = \frac{1}{2}\left( f\left(\frac{z}{2}\right) + f\left(\frac{z}{2}+1\right) \right) \tag{1}$$ for any $$z \in X$$.

• $$f$$ is continuous on $$U$$.

Then we claim that $$f$$ is constant. Indeed, it is straightforward to verify that

$$f(z) = \sum_{k=0}^{2^n - 1} f\left(\frac{z}{2^n} + \frac{2k}{2^{n}} \right) \frac{1}{2^n} \tag{2}$$

holds for all $$n \geq 1$$ and $$z \in U$$. So by the continuity assumption, as $$n\to\infty$$ we have

$$f(z) = \int_{0}^{1} f(2x) \, \mathrm{d}x,$$

which is independent of $$z$$. Therefore any such $$f$$ must be constant.

Here are some quick follow-up questions:

1. What can we say about $$f$$ when the equation $$\text{(1)}$$ is required to hold only when all of $$z$$, $$\frac{z}{2}$$, and $$\frac{z}{2}+1$$ are simultaneously in $$X$$ (as in OP's original formulation)?

2. The identity $$\text{(2)}$$ seems to suggest that there might exist a holomorphic function on $$\mathbb{C}\setminus[0,2]$$. Is it indeed possible to pursue in this direction?

Note that $$\cot\frac{\pi (t+1)}{2}=-\tan(\frac{\pi t}{2})$$ hence $$\cot(\frac{\pi t}{2})+\cot\frac{\pi (t+1)}{2}=2\frac{\cos (\pi t)}{\sin (\pi t)}=2 \cot (\pi t)$$

hence $$f(t)=\cot(\frac{\pi t}{2})$$ satisfies $$f(t)+f(t+1)=2f(2t)$$ and is analytic on $$\mathbb C-2\mathbb Z$$

• Well spotted! This is probably the inspiration of whoever designed the original question (the one I linked to). This is the top contender for an "accepted" tick so far. I'll leave the question open for a day longer, in case anyone does find other interesting examples or has other interesting thoughts to suggest. Jul 10, 2020 at 18:48
• Considering @Sangchul analysis above, it seemed that an $f$ with some simple positive poles may work and after a bit of fiddling, I realized that the cotangent properly normalized works Jul 10, 2020 at 21:08