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in Conway's Complex Analysis textbook he writes that

if $f$ is entire and $|f|\leq 1+|z|^{1/2}$ then f is constant.

Obviosuly from Liouville's theorem I only need to show that $f$ is always bounded.

For $|z|\leq 1$ it's obvious that $|f|\leq 2$, but how about $|z|>1$?

Thanks in advance.

The statement seems stronger than Liouville since $1+\sqrt{|z|}\rightarrow\infty$ as $|z|\rightarrow \infty$. I'm not sure if trying to reduce to Liouville is the right approach. –  EuYu Oct 5 '12 at 14:12

1 Answer 1

up vote 5 down vote accepted

We have $\left|\frac{f(z)-f(0)}z\right|\leq\frac{2+\sqrt{z}}{|z|}$ for $|z|\geq 1$, so $\sup_{|z|\geq 1}\left|\frac{f(z)-f(0)}z\right|=:C<\infty$. As $\lim_{z\to 0}\frac{f(z)-f(0)}z=f'(0)$, $g(z):=\frac{f(z)-f(0)}z$ is entire and bounded, hence constant: $f(z)=Cz+f(0)$. We get $|C|\cdot |z|\leq |f(z)|+|f(0)|\leq 2+\sqrt{|z|}$, so necessarily $C=0$.


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