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Let $\overline{D(z_0,r)}\subset \Omega $ where $\Omega $ open and connected and $f$ is holomorphic $f:\Omega \to \mathbb{C}$, suppose $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.

Let $g(z)=\frac{f(z)-z_0}{f(z^2)-z_0}$ then $|g(z)|\le 1$ in $D(0,r')$ (taking small $r'$ such that $f(z)\neq z_0$) and $|g(0)|=1$ so from maximum module theorem $g$ is constant also if $f(0)=z_0$ I can take $g=\frac{f(z)-z_0-r'/2}{f(z^2)-z_0-r'/2}$. But I don't see how $f$ is constant if $g$.

I am sure that's not the right approach (and maybe has mistakes what I am writing) to this problem, but I can't think of anything better I tried all the usual "tricks" ($e^f$, identity theorem )

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    Check this: https://math.stackexchange.com/q/1480433/42969, or this: https://math.stackexchange.com/q/3566798/42969 – Martin R Apr 16 '22 at 16:57
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    You probably want $z_0 = 0$ and $r < 1$, so that $z\in D$ implies $z^2 \in D$. – Martin R Apr 16 '22 at 16:59
  • Let $\Omega=\mathbb{C}\setminus {z:|z|\le 1}$ and $f(z)=z$. Then $|f(z)|=|z|\le |z|^2=|f(z^2)|$ –  Apr 16 '22 at 18:34

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