TITLE: An entire function is identically zero?
QUESTION [14 upvotes]: I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim:  
Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$.  
Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated!

REPLY [4 votes]: We already know that $f(0) = 0$. Now from $|f(z)| \leq \log(1 + |z|)$ and we get that on every circle of radius $R$ about the origin,
$$|f^{(n)}(0)| \leq  \frac{\log(1  + |R|) \times n!}{R^n} \hspace{5mm} \forall n \geq 1$$
by the Cauchy estimate. Since $R^n$ grows faster than $\log (1+ |R|)$ when $n \geq 1$, we see that $f^{(n)}(0) = 0$ for all $n \geq 1$. But we also know $f(0) = 0$ so that $f^{(n)}(0) = 0$ for all $n \geq 0$. Since $f$ is entire the identity principle implies $f \equiv 0$.