if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant? How can we prove that if $f:\mathbb{C}\rightarrow\mathbb{C}$ is holomorphic (analytic) and $|f(z)| \leq 1+|z|^{1/2} \forall z$, then $f$ is constant?
Liouville's theorem springs to mind, but I can't see how to use it since $1+|z|^{1/2}$ is not holomorphic. The maximum modulus principle doesn't seem easily usable either. And the principle of isolated zeroes can't really be applied since all we know is an inequality, not an equation.
Many thanks for any help with this!
 A: A slightly different way: $|z f(1/z)| \leq |z| + |z|^{1/2}$ for $z \neq 0$ so $z f(1/z)$ extends to an entire function $\sum_{k \geq 1} a_k z^k$ by Riemann's extension theorem.  Then $f(z) = \sum_{k \geq 1} a_kz^{1-k}$.  This implies that all coefficients $a_k$ vanish except possibly $a_1$.
A: Since $f\in\mathcal{O}(\mathbb{C})$, then
$$
f(z)=\sum\limits_{n=0}^\infty c_n z^n
$$
for all $z\in \mathbb{C}$.
Moreover, for all $R>0$ we have integral representation for coefficients
$$
c_n=\frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz
$$
Then, we get an estiamtion
$$
|c_n|\leq\frac{1}{2\pi}
\oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq
\frac{1}{2\pi}\oint\limits_{\partial B(0,R)}\frac{1+|z|^{1/2}}{|z|^{n+1}}|dz|=
\frac{1}{2\pi}\frac{2\pi R(1+R^{1/2})}{R^{n+1}}
$$
Hence for $n\in\mathbb{N}$ we obtain
$$
|c_n|\leq\frac{1}{2\pi}\lim\limits_{R\to+\infty}\frac{2\pi R(1+R^{1/2})}{R^{n+1}}=0
$$
which implies $c_n=0$. Finally we get
$$
f(z)=c_0+\sum\limits_{n=1}^\infty c_n z^n=c_0=\mathrm{const}
$$
Here you can find generalized version of this answer 
