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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 $\mathbb{C}?$

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  • $\begingroup$ 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 $\endgroup$
    – Conrad
    Jul 9, 2020 at 21:54
  • $\begingroup$ @Conrad: that's a very good point, and yes, I think that should be the assumption. Editing as appropriate. $\endgroup$
    – Will R
    Jul 9, 2020 at 21:58

2 Answers 2

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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?

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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 $\mathbb C-2\mathbb Z$

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  • $\begingroup$ 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. $\endgroup$
    – Will R
    Jul 10, 2020 at 18:48
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Jul 10, 2020 at 21:08

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