I am currently working on some review problems in complex analysis and came upon the following conundrum of a problem.

"If $f(z)$ is an entire function, and satisfies $|f(z^2)|\le|f(z)|^2$, prove that f(z) is a polynomial."

My intuition tells me to show that f(z) has a pole at infinity by showing that infinity is not an essential or removable singularity. However, I am getting stuck after this.

Thanks for the help,


1 Answer 1


Let $$ M=\sup_{|z|=2}|f(z)| $$ Then, with the condition given, it can be proven inductively that $$ \sup_{|z|=2^{2^n}}|f(z)|\le M^{2^n} $$ which implies $$ |f(z)|\le|z|^{2\log_2(M)} $$ We can use Cauchy's Theorem to give $$ f^{(n)}(z)=n!\int_{\gamma_R}\frac{f(w)\,\mathrm{d}w}{(w-z)^{n+1}} $$ where $\gamma_R$ is the circle of radius $R$. If we choose $n\gt2\log_2(M)$, then if we let $R\to\infty$, $$ \begin{align} |f^{(n)}(z)| &\le\int_{\gamma_R}\frac{|w|^{2\log_2(M)}\,\mathrm{d}w}{(z-w)^{n+1}}\\ &\sim2\pi R^{2\log_2(M)-n}\\[12pt] &\to0 \end{align} $$ Since the $n^\text{th}$ derivative is identically $0$, $f$ must be a polynomial.


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