Let $f$ be an entire function. If there exists a positive real constant $K$ such that $|f(z)| \le K(|z|^2 + |z|)$ for all $z \in \Bbb C$, then find a formula for $f(z)$.
Attempt:
Define a function $g(z) = \frac{f(z)}{z^2+z}$ for all complex numbers $z \notin \{0,1\}$. Then, $g$ is bounded (by $1$), it is holomorphic in $\Bbb C \setminus \{0,1\}$, and have two removable singularities at $z=-1$ and $z=0$. Hence, $g$ can be extended to a function $G$ which is holomorphic in all of $\Bbb C$, and coincides with $g$ on $\Bbb C \setminus \{0,1\}$. But, I stuck in determine the function $G$. If such $G$ can be determined, by Liouville's theorem, it follows that $g$ is constant, and hence, $f(z) = C(z^2 + z)$ for all $z \in \Bbb C$.
How to approach it?
Thanks in advanced.
Best Answer
We must have $f(0)=0$, so $f(z)=zg(z)$ for some entire function $g$. For this $g$ we have $|g(z)|\leq K(|z|+1)$. Consider the function $$h(z)=\frac{g(z)-g(0)}{z}.$$ $h$ is entire and moreover $$|h(z)|\leq \left|\frac{g(z)-g(0)}{z}\right|\leq\frac{|g(z)|+|g(0)|}{|z|}\leq \frac{K(|z|+1)+K}{|z|}=K+\frac{2K}{|z|}.$$ This implies that $\lim_{z\to\infty}|h(z)|\leq K$. Now since $|h(z)|\leq K+1$ on $|z|>R$ for some $R$ and $h(z)$ is bounded on $|z|\leq R$ since it is compact, we have that $h$ is bounded on all of $\mathbb{C}$ and it is therefore constant.
Then $$\frac{g(z)-g(0)}{z}=C$$ and $g(z)=Cz+g(0)$. This means that $g(z)$ is of the form $az+b$. Now $|b|=|g(0)|\leq K$ and $|a|=\lim_{z\to\infty}|h(z)|\leq K.$ We now get that $f$ must be of the form $az^2+bz$ where $|a|\leq K$ and $ |b|\leq K$.
Moreover if $f$ is any function of the form $az^2+bz$ where $a$ and $b$ are as above then it satisfies $|f(z)|=|az^2+bz|\leq |a||z|^2+|b||z|\leq K(|z|^2+|z|)$.