f is an entire function, suppose $|f(z^{2})| \leq 2|f(z)|$ for all C, then f is a constant.
I 'm trying to use Liouville's theorem, but it seems that it isn't helpful.
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Let $M(r)=\max_{|z| \le r} |f(z)|$. The hypothesis yields that $M(e^{2^k}) \le 2^k M(e)$ for all integer $k \ge 0$, so $f$ has sublinear growth. Since analytic functions are harmonic, we conclude that $f$ is constant by [1].
[] A harmonic function with sublinear growth at infinity is constant
Yuval Peres
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one can do it directly by decomposing $|z| \ge 2$ in unions of annuli $A_1,A_2..A_n,..$ given by $2 \le |z| \le 4, 4 \le |z| \le 16, 16 \le |z| \le 16^2, ..$ etc for which $z^2$ maps one into the next; if $C$ is the maximum of $f$ on the annulus $2 \le |z| \le 4$, we get $|f(z)| \le 2^nC$ for $z$ in the annulus $A_{n+1}$ where $\log |z| \ge C_12^n$ so we get $|f(z)| \le C_2 \log |z|$ for all $|z| \ge 2$ and from here we conclude by cauchy – Conrad Jan 16 '21 at 02:23