to prove a Holomorphic function to be constant

Question

Suppose that $f$ is holomorphic on $\mathbb{C}$, and suppose that the function $g(z) = f(z)/z$, defined for $z \neq 0$, satisfies $g(z) → 0$ as $|z| \to \infty$. Prove that $f$ is constant.

I have come across this question which I have not managed to find a way to prove it. I though maybe if I use the fact that $f$ will be continuous and differentiable on any region in $\mathbb{C}$ I would be able to find a bound for $f$ and then apply Liouville's Theorem but couldn't find a way. any help is appreciated.

Answer 1

Show that $$h(z):=\begin{cases}g(1/z)&z\ne 0\\0&z=0\end{cases}$$ is entire. Show that you can write it as $h(z)=zk(z)$. Show that $f(z)=k(1/z)$ for $z\ne 0$ and conclude that $f$ is bounded.

Answer 2

Consider $\gamma(z)=\frac{f(z)-f(0)}{z}$. Based on your hypothesis on $g$, it follows that $\gamma(z)\to 0$ as $|z|\to \infty$. Now, $\gamma$ is clearly holomorphic on $\Bbb{C}\setminus\{0\}$. Also, we have $\gamma(z)\to f'(0)$ as $z\to 0$, since $f$ is holomorphic at the origin as well. In particular we have $\gamma$ is bounded near the origin, so $\gamma$ has a holomorphic extension to $\Bbb{C}$, which we shall still denote as $\gamma$ (this is Riemann's removable singularity theorem), i.e $\gamma$ is an entire function.

Since $\gamma(z)\to 0$ as $|z|\to \infty$, in particular it follows that $\gamma$ is bounded. So, by Louiville's theorem $\gamma$ is constant (what is this constant equal to?) Therefore...

Answer 3

The hypothesis implies that $g(z)$ has a removable singularity at $\infty$, i.e. $g(1/z)$ has a removable singularity at $0$. Hence for $z \in \mathbb{C} \setminus {0}$ $$\frac{f(1/z)}{1/z} = \sum_{n = 0}^{\infty}a_nz^n.$$ Hence $$f(1/z) = \sum_{n = -1}^{\infty}b_nz^n.$$ Hence for $z \in \mathbb{C} \setminus {0}$, $$f(z) = \sum_{n = -\infty}^{1}c_nz^n.$$ Since $f$ is holomorphic on $\mathbb{C}$, $$f(z) = \sum_{n = 0}^{1}c_nz^n = c_0 + c_1z.$$ Then the hypothesis $f(z)/z \to 0$ as $|z| \to \infty$ implies that $c_1 = 0$.