Show, a holomorphic function is constant I am given that $\left| \frac{g'(z)}{g(z)}\right|\leq \frac{1}{\left|z\right|^2} \hspace{0.3cm}(*)$. I want to show that if $g$ is holomorphic in $\mathbb{C}$, it is constant. I am not sure if additional assumptions are needed. Maybe it is needed, that $g$ has only a finite number of roots. Does the result change if $(*)$ is only valid outside a compact set around zero?
 A: Since a zero of $g$ leads to a simple pole of $\dfrac{g'}{g}$, the bound immediately yields that $g$ can have no other zeros than $0$. Write $g(z) = z^k\cdot e^{h(z)}$, where $k$ is the order of the zero of $g$ in $0$ (if $g(0) \neq 0$, then $k = 0$). Then
$$\frac{g'(z)}{g(z)} = \frac{k}{z} + h'(z).\tag{1}$$
Now we have
$$\lvert h'(z)\rvert = \left\lvert \frac{g'(z)}{g(z)} - \frac{k}{z}\right\rvert \leqslant \left\lvert \frac{g'(z)}{g(z)}\right\rvert + \frac{k}{\lvert z\rvert} \leqslant \frac{1}{\lvert z\rvert^2} + \frac{k}{\lvert z\rvert}.\tag{2}$$
Since $h'$ is entire, it is bounded on the unit disk, hence $(2)$ shows it is bounded on all of $\mathbb{C}$, so it is constant. But $(2)$ also shows $\lim\limits_{\lvert z\rvert\to\infty} h'(z) = 0$, so $h'\equiv 0$, and $g(z) = C\cdot z^k$, so $(1)$ becomes
$$\frac{g'(z)}{g(z)} = \frac{k}{z}.$$
That is only compatible with
$$\left\lvert \frac{g'(z)}{g(z)}\right\rvert \leqslant \frac{1}{\lvert z\rvert^2}\tag{$\ast$}$$
if $k = 0$.
If $(\ast)$ is only required outside some disk, say it should hold for $\lvert z\rvert > R$, we still see that $g$ can have only finitely many zeros, and we have a representation
$$g(z) = e^{h(z)}\prod_{k=1}^m (z-z_k)^{n_k},$$
and correspondingly
$$\frac{g'(z)}{g(z)} = h'(z) + \sum_{k=1}^m \frac{n_k}{z-z_k}.$$
By the same reasoning as after $(2)$, we conclude $h'\equiv 0$, and since all $n_k \geqslant 0$, $(\ast)$ implies that $n_k = 0$ for $1\leqslant k \leqslant m$, i.e. $g$ must be constant. If $g$ were not supposed an entire holomorphic function, but only meromorphic in the entire plane, then $g$ could be non-constant.
